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Александр Григорьевич Озеров (3 августа 1849 года, Манглиси — 1922) — русский . Работал в Грузии.
Биография
Родился в семье Григория Ивановича Озерова, штаб-лекаря Эриванского карабинерного полка, и Любови Карловны Ган, из семьи немецких колонистов.
Окончил Ставропольскую губернскую гимназию с золотой медалью. В 1867 году поступил в Петербургское строительное училище, которое окончил в 1872 году с занесением его имени на мраморную доску в актовом зале училища. По окончании обучения получил чин X класса и был причислен в Техническо-строительный комитет Министерства внутренних дел.
Вернулся в Грузию и получил должность младшего архитектора строительного отделения тифлисского губернского правления. В 1878 году с ознакомительной целью был командирован в Париж на Всемирную выставка (1878). За предоставленный отчёт о командировке в 1881 году удостоен благодарности министра внутренних дел.
С 1879 по 1885 год — городской архитектор Тифлисской городской управы. В 1888 году назначен губернским инженером строительного отделения кутаисского губернского правления. В 1894 году занял должность архитектора при тифлисской городской больнице.
Жил на улице Боржомской, 14 (1891—1922).
Похоронен на старом Кукийском кладбище.
Потомки А. Г. Озерова жили в Тбилиси до начала 1990-х годов.
Известные работы
Дом А. Чавчавадзе в Цинандали (1886—1887);
Тифлисское Общество взаимного страхования от огня.
Дом Сараджева. Улица Ираклия II.
Абанотубани и Мирзоевские бани (1880, реконструкция);
Красный зал Тифлисской Городской Думы (1886, реконструкция);
Жилые дома. Улица Коте Абхази, 22-26 (1890-е — 1900-е);
Жилой дом. Угол улиц Мераба Коставы и Николадзе (1905);
3-я женская гимназия. Улица Ладо Асатиани, 28 (1905);
Жилые дома. Улица Геронтия Кикодзе, 3 и 6 (1906);
Летние помещения Тифлисского кружка в саду Йогана Майера (Сад Роз). проспект Давида Агмашенебели, 73 (1909);
Здание поликлиники. Угол улицы Пушкина, 10 и Вачнадзе (1910);
2 жилых дома. Улица Тодадзе, 3 (1910);
Дом братьев Форер (Союз «Катарсис»). проспект Давида Агмашенебели, 121 (1910);
Завод Сараджева. Ольгинская улица, 18 (1910, не сохранился);
Жилой дом 23 у Верийского базара (1910—1911);
Тифлисский Почтамт (вначале проектировался, как доходный дом). проспект Давида Агмашенебели, 44 (1912);
Доходный дом рядом с Почтамтом. Улица Чорохская (1913);
«Палас-отель» (здание Министерства культуры). Проспект Руставели, 19 (1914);
Жилой дом. Улица Човелидзе, 7;
Земельное училище. Улица Петриашвили, 3;
Жилые дома на улице Боржомской, 14, 16 и 18.
Примечания
Литература
Тбилиси: Энциклопедия - Тб., 2002. - с. 716
Ссылки
ალექსანდრე ოზეროვი
Похороненные на Кукийском кладбище | {
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{"url":"https:\/\/webwork.maa.org\/moodle\/mod\/forum\/discuss.php?d=2486&parent=5139","text":"## WeBWorK Problems\n\n### Re: getting a numerical MathObject to display its formula in text\n\nby Dick Lane -\nNumber of replies: 0\nI want to recant about half-way for the first paragraph (and a bit more for the last).\n\nOne list\/array of MathObjects can be used, an item selected, and then it alone is put thru the ->{original_formula} filter to get the typeset-able version. \u00a0The benefit to programming effort can vary with the structure of the problem.\n\nIn a problem about converting rectangular coordinates to spherical, I began with an array of [x,y,z,rho,theta,phi] 6-tuples which were chosen to be \"nice\", selected one 6-tuple, used its values for answer-checking and put just its items thru the ->{original_formula} filter for use in the problem statement and in a solution. \u00a0To offer a bit more variety to students, I used a scale factor for [x,y,z,rho] and discovered that \"original formula\" is not inherited. \u00a0This was easy to fix: just put the scale factor into the computations which produce the array of 6-tuples, e.g., rather than having [sqrt(3),0,1,2,0,pi\/4] be extracted from the array and then scale [x,y,z,rho], have [a sqrt(3),0,a,2 a, 0, pi\/4] be in the original array (Perl note: of array references).\n\nI would still prefer to have TexString be the default behavior for a Computed object in a texString context, but ->{original_formula} did simplify my example greatly.\n\ntechnical note about data structure for my example:\n@DTA = ( [6-tuple of Computed items] , [another 6-tuple] , ... ) ;\n$pick = random( 0 , scalar(@DTA) , 1 ) ; ($x, $y,$z, $rho,$theta, $phi) = @{$DTA[ \\$pick ] } ;","date":"2021-10-21 13:54:04","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.6668480038642883, \"perplexity\": 4174.967566082679}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-43\/segments\/1634323585424.97\/warc\/CC-MAIN-20211021133500-20211021163500-00658.warc.gz\"}"} | null | null |
Q: PDO cant connect, but mysql_connect can (without PW) I'm really having a strange Issue here, and it's really starting to annoy me. It's about the different behaviour of connections. I'm just trying to set up CakePHP, but the PDO cant conenct to the mysql Server.
Okey, step by step: It's a fresh computer, I just installed XAMPP (on Win7) and downloaded CakePHP. Nothing else was done. On phpMyAdmin, I created a user 'test' with PW 'test' and he owns the Database 'test'. Simple, right?
Here, the row of the User / Rights table in phpMysqlAdmin:
User Host Password Global Rights GRANT
test % Yes USAGE No
Now, to the real Issue:
This works:
$link = mysql_connect('localhost', 'test'); //<- not using the 3. parameter, 'password'
But, what should work, doesnt:
$link = mysql_connect('localhost', 'test', 'test'); //<- using the 3. parameter, 'password'
And since I cant seem to 'remove' the 'using Password: YES' for PDO's, I cant connect with PDO's either (and cake Uses PDO's):
$dbh = new PDO('mysql:host=localhost;dbname=test', 'test', 'test');
Error Messages:
Error!: SQLSTATE[28000] [1045] Access denied for user
'test'@'localhost' (using password: YES)
It cant be so complicated, I just want to connect to a freshly installed DB, with the correct credentials. What am I doing wrong? I read trough many similar Questions, but didnt find a solution for my own Problem.
And yes, the password really is 'test' - I have no idea why the mysql_connect() whitout PW, can connect - is it using the username as PW by default?
This really cant be that hard
Thanks for the help,
Wish you a nice day.
EDIT (answer to question)
Here the entry for the DB rights (in user: test): - the user really seems to have all rights for DB 'test'
Database Rights GRANT
test ALL PRIVILEGES Nein
A: GRANT ALL ON test.* TO test@localhost IDENTIFIED BY "test";
Run the above query at the mysql shell and everything should work.
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CONTENTS
1. HALLUCINATIONS, EVEN IN THE SANE
2. FIND THEM AMONG THE DEAD
3. HIGHLY EDUCATED, MODERN PROFESSIONALS
4. IF I SHOULD DIE BEFORE I WAKE
5. HURT, YOU WERE A CHILD AGAIN
6. UNCLEAN SPIRITS ENTERED THE SWINE
7. SOUL IS A UNIVERSAL FEATURE
8. BONES THAT FED OUT THEIR COLD
9. TO THE WHITE CASTLE
10. I WASN'T MYSELF, BUT THE IMAGE OF ME
HALLUCINATIONS, EVEN IN THE SANE
WHEN I INSISTED ON KEEPING THE BABY, NED THREW HIS HANDS into the air palms-forward. He looked like a mime climbing a wall—one of the few times I've ever seen him look clumsy.
Then he dropped his hands and turned away, shaking his head. It was a terminal shake. Afterward his schedule got fuller, his long work hours longer, his attention more completely diverted.
And I have to admit it wasn't just him who turned away. After we differed on that point, the point concerning the baby, I began to give up on Ned too.
So I was alone preparing. It had been an accident, technically more his fault than mine, but who's haggling? And once it happened I felt I needed to accept it—I wanted to. I drove by myself to buy the various infant containers. I chose the doll-sized pieces of newborn clothing, set up a nursery and glued stars on the ceiling; I crept in at night sometimes to see how they glowed. I went alone to doctor's appointments to listen for the heartbeat and see the first pictures, and when the time came I went through labor with mostly just medical staff keeping me company.
Ned did stop by the hospital, apparently, and spent some time talking on his cell phone in the lounge, but he stepped out again for a work lunch, later for work cocktails, and finally for a late work dinner. After dinner he drove home and went to sleep.
None of this was too far beyond the pale, I guess, when it comes to unfortunate marriages. After about twenty hours I lay against the pillows holding her slippery body. Her eyes, against my expectation, were wide open and there was a perplexing chaos of sound in my ears, too many voices in the room for the number of people—soundtracks that overlapped. A kindly nurse was telling me about the other babies he'd seen born with their eyes open when a stream of words intruded, covering his. I heard it most distinctly when the nurse paused.
Later I would hear volumes and forget almost all of it, but the first phrase I picked out stayed with me despite my exhaustion. It started out as a string of foreign words, only one of which resolved, to my ear, into anything recognizable—something like "power," powa or poa. And then it was English: The living spring from the dead.
Delirium, was what I thought, and I dispensed with it by falling fast asleep. It was only when I woke up later, and the baby was brought back to me, also awake, that the stream of chatter started up again and was impossible to ignore.
AT FIRST I was mostly irritated, and went to get my ears looked at. Once, when I was a kid, I'd had an infected ear and heard a wavy music when I pressed my head against the pillow. Maybe this had a physical explanation, maybe some ear-brain interface was being disrupted. But my ears checked out fine. The baby didn't enjoy the doctor's visit, and the voice talked on—only for me, of course—throughout her noisy crying.
Next I made an appointment with a neurologist and insisted on an expensive scan: nothing.
For weeks I combed through psychology case studies, ready to discover the evidence against my sanity. I read up on post-partum depression, though I didn't feel depressed. Of course I might be in denial, I knew; I had a newborn baby, after all, and a husband who had no time for either of us.
But I didn't feel sad. I suffered from no flatness of affect. I was tired and confused—I felt besieged by the noise—but it was frustration, not despair.
I also gave schizoid conditions due consideration. No mother wants a woman with psychotic features bringing up her child, even if that woman is her. So reading accounts of patients who heard voices became my avocation for a while, since, as it turns out, mental illness isn't required to hallucinate. Hallucinations, even in the sane, are quite common. They accompany certain drugs and medicines and an impressive list of diseases; they can be caused by blindness or sensory deprivation or even seem to come out of nowhere.
A stream of advice is often heard by people in extremis, fighting injury or the elements. Voices are heard by the sane in wartime or under other forms of duress, prison or isolation or grief. Sometimes the voices have no obvious cause, their origins buried in the electric labyrinth of the brain.
I was prepared to accept the hallucination hypothesis—the baby's presence, her rapt attention caused me to hallucinate voices speaking to me—but I was curious beyond that and needed to cover my bases. I also went to worst-case scenarios, to the bizarre and outlandish. I studied the occult, including demonology, for instance—spent hours on the Internet reading myths and legends of demonic possession. I made trips to the library, the baby snug in her carrier, and moved from articles about people with auditory hallucinations to those who identified their visitors very specifically, brooking no disagreement.
Demons, they said.
They saw demons with claws, horns and pointed teeth, of course, but often demons appeared in the shape of seductive women and yet others were amorphous shapes that shifted beneath the faces of loved ones. Briefly those faces would distort, then swiftly resume their devious guise, pull over themselves the skin of normalcy.
Or people heard demons that had no physical form but only spoke, mostly in biblical tongues like Aramaic or Hebrew. Experts were consulted and that was often their verdict: what the demon-visited persons were hearing was Aramaic or Hebrew or Greek. The demons tended to speak in dead scripts, as though frozen in the time of early Christianity—the demons clung to the old, reluctant to embrace the new.
I was glad Lena's mouth didn't move when the words issued, as in some possession stories. Because it was only sound and words, invisible, the experience also conjured TV shows involving ESP. I looked into spoon-bending hoaxes and watched shows that featured ghost-finding teams that crept through haunted houses trying to capture stray ectoplasm.
I was worn down by the elements of my routine—the stream of words and my bewilderment during the days, the nights half-sleepless, a mesh of hours spent fitfully dozing or nursing my daughter when she woke up. Ned had moved out of our bedroom while I was pregnant and never moved back, claiming his restless sleep would bother me. Often he didn't come home at all, in those first months when Lena's crying disturbed the nightly peace, but stayed over at the office. It wasn't long before I began to understand that at the office was a euphemism.
And when the baby was sleeping but I couldn't sleep, I wallowed in pulp fiction. I read thick paperbacks set in old houses, where the devil took the form of flies and buzzed on windowpanes, or in upscale prewar apartment buildings in Manhattan, where babies were fed evil baby food and raised by Satan cults. Plus there were the movies about antichrists and child possessors, the one with the black-haired boy named Damien, the one with the blank-faced girl who floated over her bed, rasping obscenities. When I was too tired to read, with the baby mostly sleeping and the speaker fallen silent, I'd curl up in front of the screen with cheese popcorn.
But in the end the B-movie fiends were too showy for me to take seriously, almost self-parodies. Besides, the stream of words wasn't malicious and my daughter committed no alarming actions. She ate and slept, lay bundled in my arms. Time passed and she rolled over, sat up, crawled; also gurgled and drooled.
She never fixed upon me a bold, sinister eye.
So by and by I let the demons go, telepathy I dismissed out of hand, schizoaffective disorders I further renounced.
I went with the hallucination theory.
Hallucination has the qualities of real perception: vivid, substantial, and located in external space. It is distinct from a delusional perception, in which correctly sensed stimuli are given additional, often bizarre, significance. —Wikipedia 5.10.2009
PEOPLE WITH MIGRAINES see colors and shapes fading and forming anew on the wall. Others, with visual hallucinations, believe strangers are sitting beside them dressed in old-fashioned garb. Next to these people's apparitions my own affliction didn't seem so grave.
It was true that the disturbance was constant, and I didn't find an identical case in the articles I read, but this struck me as more or less a technical detail. At first I called it the voice, as others like me did. Because I wasn't alone: there were whole support groups given over to non-psychotics who heard things, including a so-called Hearing Voices Movement (its mission: to empower chronic voice-hearers). There were affirming Listservs.
I avoided them studiously. I began to write in this Word file instead, a diary whose sporadic, rambling texts I'd tinker with for years. Over time I redacted, adding and subtracting until the entries formed a narrative that clarified my own story—at least to me.
I spoke to no one about what I believed I heard. I sought out no company in my infirmity.
WHERE WE LIVE now is a seaside motel in the off-season. We're on the edge of rocky bluffs, so I can see a car coming when it's a speck on the long gravel road.
There are few guests this time of year; in summertime they get the kind of tourists who, says Don the motel manager, bicker sharply over the bright-orange sandwich crackers in the vending machine re: advisability of purchasing.
But in the wintertime it's quiet here and there are weekly rates. The carpets aren't much to write home about, having an ashy cast. The tables in the rooms are brown Formica with black cigarette burns; our shower curtains are mildewed. I like their pale-blue imprint of daisies. I also like the cliffs, the rocks, the trees and the gray water stretching to the east. I like the sharp nearness of pine needles against a blurry sheen of sea.
And my little girl loves it. She loves the people and the place; small events make her giddy with pleasure. She spins, cartwheels, races and laughs easily. She doesn't have much, but she doesn't need much. She has her books and toys and art supplies. Some of the toys are old and bedraggled, since she doesn't want to throw out anything—the second I suggest a disused toy might be taken to the charity bin in town she feels a rush of protectiveness and clings pathetically, lavishing praise upon the object that had been utterly forgotten until then.
Watching her protect a ratty mouse, a dog-eared, broken-spined, finger-smeared picture book, it's almost possible to believe that everything in the world is precious, that each humble item that exists has a delicate and singular value.
It's possible to believe that all matter should be treated tenderly.
LENA WAS BORN in a hospital in Alaska. Up to that time I taught as an adjunct at the university and her father was in business: and he's still in business today, though he's expanded his purview.
I was fond of Anchorage. It's a sprawling city of mostly ugly buildings, but no other city I know has bears roaming downtown. I'd be picnicking with the baby near the central business district, watching the sunset from the Cook Inlet shore, and black bears would come rustling through the undergrowth a few feet away. Feeling a tug of panic, I grabbed Lena and retreated to the car, but still I treasured having them so close. The moose roamed Anchorage too, and you could encounter them on a casual run through city parks—more dangerous than the bears, if you believed the statistics.
Of all the actions I've taken, leaving Alaska was the hardest. Not because I enjoyed living there, though I did, but because it's a bold move to take a child so far away from the man who's her father. Even when he doesn't accept the position.
I did have his approval at first for our departure. The part of the split he resented was financial: he didn't like that I took half the value of our savings account and our CDs with me. (I left the stock, I left the mutual funds, but still.) Aside from money quibbles he was glad we'd left, at first; for more than a year he didn't mind at all. He'd been indifferent to me for a long time, as he's indifferent to most people who aren't of use to him.
As for Lena, he hadn't wanted her in the first place and he never warmed to her. Our leave-taking gave him the same liberty it gave us—namely the open-ended chance to be who we were, instead of trapped.
I'd send him the occasional email telling him what she'd learned, what she was doing, an anecdote here or there to keep her real. I clung to the belief that any father would want that, and more than that I felt I owed it to her, to try to keep him existent as a father, however marginal. He rarely responded to these, and his occasional replies were brief and rife with hasty misspellings.
But over the past few months he's decided to make himself a candidate, and candidates want family since family looks reassuring on them. So now we're useful again and he's searching for us. I think he wants a moving snapshot for the campaign trail, two female faces behind him as he stands on the podium.
When I first met Ned he claimed not to have any politics. I should have known enough to be wary of that, but instead I made excuses to myself. Politics were for crooks, he said. But later politics grew in him like metastasis, branching into a network threaded throughout his veins and nerves and bones. It's not that he's left the business world behind, it's just that he now believes politics are a sector of his enterprise.
His platform includes a prolife agenda, for instance, which "values the sanctity of every human soul," and also "believes in the greatness of the American family." The word family, on his glossy-but-down-home webpage in its hues of red, white and blue, is a code for you, where you also means right, deserving, genuine and better than those others, you know, the ones who aren't you. Ned believes in "the American family" the same way processed food companies do, companies that make products for cleaning floors or unclogging toilets—the kind of easy code that makes public speech moronic.
But even if he'd been a genuine family man, I wouldn't have wanted to be a part of his platform.
Once he nearly caught up with us, before I understood that emails can be traced. It was stupid of me and caused a close call and as a result I'm wiser now—or craftier, in that I don't send emails anymore. We move, we don't use credit cards, I don't write my own name when I sign things. I bought a fake driver's license from a computer-savvy teen in Poughkeepsie. If a cop pulled me over I'd have to use the real one, which matches my registration, but I drive cautiously and keep the car in good repair and so far that hasn't happened.
I'm not in any system, that I know of, I'm not a fugitive. Ned wouldn't report me. It would make him look bad, defeat his whole purpose in reclaiming us.
The only authority I'm running from is him.
EVEN THOUGH it's cold out, we spend a lot of time on the beach, the rocks and pebbles and sand. At dawn we take the first walk, following a narrow path down the face of the cliff. I carry a thermos of coffee and she carries a basket divided into one section for treasures, another for litter. Not every form of litter is welcome: she can't pick up medical waste, newly broken glass, rotting food, or old, yellow-white balloons.
I'd like for us to settle down and live a steady life, so she can go to school and have friends. Lena begs not to go to school and claims she wants our life to stay the same forever.
She's six years old. She doesn't know better.
It seems to me that if we can escape his grasp till after the election, we may have a fair shot at an undisturbed existence. If he wins he won't need us.
On the other hand, if he loses and decides to take another shot in another cycle, he may search harder. He may get more determined.
When we discuss her father, who's only a vestigial memory for her, I rely on platitudes like "Our lives took different paths," or "Sometimes people decide to stop living in the same place." The matter of the separation, unlike the matter of the voice I used to hear—on which I hope always to keep my own counsel—will one day require unpleasant conversation, but so far she's satisfied with generalities. She's not overly interested, since she never saw much of him. Much as she never caught his interest, he never seemed to capture hers either. When we did share an address he seldom came home: he traveled, he worked late, he cultivated his casual friends and many acquaintances. He never read bedtime stories or sat down with us for meals.
He was a sasquatch in a photograph, a fuzzy obscure figure moving in far-off silhouette.
DON, WHO'S BEEN so good to us, is a pear-shaped man. This feature endears him to Lena, whose favorite stuffed animal is a plush, duck-like bird with a small head and giant baggy ass. Don has a shuffling gait, seems erudite by hospitality-industry standards, and like us appears to be hiding here—not hiding from one person but from crowds of people, possibly, or from a faster pace. He has a job that involves people, true, but seldom too many at one time, and when people do show up they're in his territory, his cavernous and dimly lit domain.
I imagine he keeps the motel ramshackle so as not to attract too much traffic—so as to keep the trickle of company thin. His family owns the business and seems to accept the small returns.
When a stray overnight guest comes through, Don's civil but hardly overjoyed. Lena, by contrast, is always excited. She acts as though she, not he, is the owner: she's the mistress of all she surveys, with the hosting duties this brings. To her the motel is first-rate; she sees no mildew or cigarette burns. Because I can't leave her with strangers, this means I meet many guests too, tagging along in the background as she gives them the tour.
Most are highly tolerant of her—eager children receive a plenary indulgence, especially dimple-cheeked girls—and her exuberance is contagious. She explains the rules about clean towels with gusto, as though the rules, if not the towels, are sacrosanct; she showcases the antique ice machine with pride of ownership.
"This ice is only for people's drinks," she says sternly. "So don't pick it up and put it back, OK? And don't stand with your hands stuck in the ice, even if you like the shiver."
WHEN NED CAUGHT up with us we were staying at a cabin in New Hampshire near the summit of a low mountain. It was a large, wooden cabin with a dozen bunk beds for hikers and three caretaker-cooks. Only a few dozen feet from the porch was a waterfall with a flat-topped boulder at its edge, where Lena liked to sit trailing her hand in the water and basking in the sun. The water wasn't deep.
We only got away that time because Ned made a mistake; he did a flyover. Maybe he wanted to preside from the air while his employees cornered us; maybe not. I still don't know if he was personally there.
But helicopters were rare along that part of the Appalachian Trail, coming in only with major equipment or for medical emergencies. I was on the porch with one of the cooks when that one chop-chop-chopped overhead and she looked up and said, "Huh, a private helicopter. It's not the local guy."
That was all I needed to pull Lena off her sunny rock and leave our sleeping bags behind. I did it only because my stomach twisted when the cook said what she said: I followed my instincts and we bushwhacked down the mountainside—I said it was a game, going off-trail, and the one who made it to the bottom with no scratches on her legs or arms would win a double-scoop cone. When we reached the road I had some light scratches on my forearms while Lena had none; new mosquito bites itched and swelled around my ankles, and our shoes were soaked from slogging through a stagnant creek.
Still, Lena was gleeful at the prospect of her ice cream reward.
The car wasn't parked in the trailhead lot most of the hikers used but in a shaded pullout I'd found. After a short walk on the shoulder of the road we got in and drove off.
And I knew we'd been right to run when the cook, who had become a friend, called me. She said four men had come, two from each direction since the trail stretched out on either side of the cabin. They converged on it fifteen minutes after we'd left. They weren't dressed for hiking: their shoes were shiny leather ruined by mud. So she told them only that we'd left the day before, and after some unhappy muttering and some prowling around the grounds and questioning of other guests, the four men went away.
NED MARRIED ME for my family's money, because he had none of his own and wanted some; I married him because I thought it was love. I was wrong too, it wasn't love—I don't mean to pin it all on him. I had a crush, if I'm being honest, and I didn't know the difference.
Ned's a very attractive man, a man many people use the word handsome or magnetic to describe. Even straight men have said this of him, the same way they'll concede it, often grudgingly, of famous actors or athletes. Both before and after we were married, men and women alike would confide in me about their attraction to Ned. He makes people covet him, inspires a desperate greed. And he knows this all too well—it's key to his strategy for gathering investors. Ned is his own asset, his own front man, a property that sells itself. Both men and women want to own him or sleep with him, but failing that they're just grateful to be part of his enterprise.
It goes far beyond standard-issue good looks.
He always had a talent for captivating an audience. From the first moment he meets you he establishes eye contact, and he doesn't relinquish it easily. But he's not only a mesmerist. He can embody audience convincingly as well, when listening is called for. When he receives a personal disclosure he seems to listen intently, even adoringly.
In fact he isn't listening but intently, tactically appearing to listen—no mean feat in itself.
He's humorless, though, which for me proved slowly deadening. Ned always laughs when others laugh, taking the social cues, but laughter doesn't come naturally to him. And while he could occasionally say a funny thing, back in our early days together, it wasn't intentional.
There were other, more minor details of Ned that should have been red flags for me too—his allegiance, for example, to a certain brand of cologne. Before Ned I'd never been with any man who wore cologne. The smell of it didn't bother me: this particular cologne was inoffensive, even subtle. But once, when a bottle of it was knocked off a bathroom counter and broke on the tile floor, I saw a strange edge of rage in him.
In general I had no eyes for red at all in the infatuated months before we got married. Any flags of bright color were lost in the hills and dales of a hazy, indulgent country.
And my feelings were irrelevant, in the end, since he had close to none for me. I was surprisingly late to this realization. We tend to believe what we wish to, and I was no exception. I hoped that Ned loved me, and hope shaded into assumption without me recognizing it.
Before I got pregnant he found me attractive enough too, I guess, but this disappeared with the pregnancy, which he found repulsive. He pursued other women with unqualified success. He had no lasting feelings for any of them either, as far as I could tell, but each was new in her turn, and Ned prizes novelty. Novelty and momentum are his two passions.
In saying he married me for money, I don't mean to imply I was an heiress—my family had the complacent, middling inherited wealth that passes without much notice unless you happen to be Ned, brought up in poverty, entrepreneurial, and with an incentive to research. He could have held out for someone with far more money and far, far better connections, for I had none.
Now, looking back, I'm surprised he didn't. I was a small fish, very small. I had barely enough. But he was impatient to get his enterprises off the ground. And his disinterest in the marriage probably reflected his own awareness of that hasty choice—the fact that he'd settled for much less than he was capable of getting.
WITH A HANDFUL of exceptions I found that when I tried to write down what the voice said, I couldn't. A fog would descend. Phrases that seemed sharply etched to me when I heard them, sense and structure cut like a skyscraper against a crisp sky, would crumble and fade as soon as I tried to record them.
I heard the words in the stream as English or French or Spanish, or sometimes it would be modern English in an accent or dialect, say Australian English or an English with Welsh accents. Other times it was English that sounded like Shakespeare or Middle English, like Chaucer maybe, which I'd read in college. But whenever the format changed I half-forgot what had come before, as though the switch between lexicons and grammars occurred imperceptibly. Since I couldn't identify the languages that weren't English or Spanish or French I figured my imagination was making up a stream of nonsense, sounds that resembled other tongues but were only a sham.
That was a game I'd liked to play when I was a kid. I even played it a few times with Lena, speaking in rapid-fire gibberish, pretending it was an unknown exotic language, say Urdu or Tahitian.
And the voice never went silent, except when Lena was sleeping. It changed from low tones to high, speech to singing, singing to humming to clicking sounds that had a rhythmic quality, on occasion devolving into grumbling or even yelling. I drew the line at yelling—at those times I'd call a babysitter and go out.
I'd shut the door behind me and step into the street, and right away I didn't hear a thing.
WHEN I LEFT Ned, I took enough money to live on for a while. It was only a fraction of the legacy from my family that he'd funneled into his businesses, but I didn't want to fight over money. Ned wanted it more than I ever had and taking too much would bring out the edge in him.
So I took only what I felt I needed. I made a budget carefully, knowing I wasn't going to work again until Lena started school. I'd worked steadily all my adult life and I thought I could use a break; I was well pleased to be only her mother and teacher for those years. I didn't plan to have a second child.
The money keeps us afloat, Lena and me, and in that respect we're fortunate.
I PUZZLED OVER the link between the baby's presence and my hallucination. There wasn't generally supposed to be such a clear connection, in the hallucinations of the sane, between what was heard or seen and the fixations of the hallucinating person—not in the descriptions that I read, anyway. This made my case seem more psychological than purely neurological, and I worried about it periodically. Because the presence of my infant carried with it a voice that had the appearance of fluency in all tongues and gave an impression of encyclopedic knowledge—some kind of frightened projection of my overpowering responsibility as a mother, possibly, was one of my interpretations.
Sometimes the stream of sound wasn't a voice but music, welcome relief: old standards, dramatic epics by well-known composers, folk tunes, pop riffs. It liked Woody Guthrie, whose music I didn't remember encountering before except for the song "This Land Is Your Land," which I knew from summer camp. Research on the snatches of lyrics I could recall yielded his name, and I thought I must have been exposed as a child, and quashed the recollection.
But most often the content was words—what sounded like recitations of texts of all kinds, poems, fictions both literary and mass-market, movie scripts and stage plays, histories, dictionaries, textbooks, biographies, news stories. The subjects were as diverse as the genres: single-celled organisms, hockey scores, feathers on dinosaurs, celebrity suicides, the pattern of Pleistocene extinctions, the fate of the tribe known as the Nez Perce; relativity, particle accelerators, Greek myths, the troubled term Anthropocene, the chemistry of a callus on the hand of Heidelberg man.
I was impressed by the knowledge base from which my mind appeared to be drawing. I marveled at it, even. Buried in my unconscious must be some capacity for photographic memory, I thought.
That surprised me.
Nothing salacious ever came from the voice—that is, there were curses, there was profanity, there were even vague references to sex and reproduction, but there was never a suggestion of lechery directed toward me personally. Still, I felt perversion was implicit in the combination of a baby nursing while a stream of elevated diction flowed up from somewhere beyond the O of her mouth. I had to distance myself from the voice when I was nursing her: it might be my hallucination, but, much in the way I might detest my head lice or my chicken pox, should those happen to manifest, I was forced, at those times, to treat it as a pest.
On occasion I'd try hard to write down what I heard despite my confusion, with doggedness but a lack of clarity, determined to record the substance of the hallucinated event. I still carry with me some scraps of paper—deep in the trunk, where I stuck the file after the last time I picked through it. I'd had to write the words down fast to get any of them, seldom had time to get to the keyboard, so the notes are scribbled on the backs of envelopes, grocery and housewares receipts, once along the edge of a worn dollar bill. Many seemed nonsensical: Windlessness = illusion planet is static in space ∴ windlessness entropic. Or "social animals + writing: ERRATUM."
Neighbors and friends came over fairly often in the first year of Lena's life and (of course) they never heard the voice, not even the faintest hint of it—I made sure. I'd ask, in a roundabout, casual way, if anyone was hearing anything unusual as we sat there, but my questions always met with offhand dismissals.
Joan of Arc had heard a voice advising her to help raise the siege of Orleans, but as far as I could tell the voice had no specific instructions for the likes of me.
I PASSED THROUGH stages with my hallucination. Sometimes I wished I could hide from it, other times I was determined to study it steadfastly until I could pick out the details and know it more perfectly. After almost a year I fit myself into a certain orbit, adjusting my routine to its disruptions. I shrank and disappeared in the brightness of its perpetual day but at night, when it was silent and so was Lena, I tracked across the dark relief in solitary flight.
I relied heavily on the fact that babies sleep for longer than adults and I also depended on her midday nap, an hour and a half like clockwork. The babysitters gave me some time off, and for the rest I'd found ways to fit myself into the spaces between words, to distract myself sometimes, at other times to tolerate nearness and even, when well-rested, to listen.
In general I felt besieged, my defenses walled up around me, but every now and then something in the fall of words would strike. I'd feel my throat clench in grief or recognition, be on the brink of tears and then not be.
At those times—it's hard to describe and I feel like a fool even trying—I didn't understand why emotion was overwhelming me but I also didn't waste time belaboring the question. I had distinct sensations and I stilled everything to feel them: sometimes I thought I was being cut bloodlessly, cut so that a clear, frigid air entered me and the rest of the outside followed; or possibly I spilled out, it may have been the other way around. I'd feel as though I had the long view, past the end of my life, past the horizon, dispersing into ether.
I loved that feeling the way a drug might be loved, I think, quick as it was, freeing—but also with an icy burn, a searing touch I imagined as the cold of space and couldn't stand for long. There was the euphoria of ascent, the vertigo of height.
Then the feeling would vanish abruptly. I'd just be there, in my house or on the street or in a store, wherever, with Lena. And I'd be desperate to see her clear eyes gazing at me with no interference—to be alone with her instead of in the company of slime molds, cyanobacteria genomics, cuneiform or the dancing of bees.
And finally it wasn't the substance or character of the voice I resented but its proximity—the fact that it was so close, and that it never ceased. I urgently wanted to be rid of the torrent of sound and image, the stream of convolved murmurings that often evoked either oppressive problems or, at the very least, the broad dramatic canvas of a universe that went on forever beyond our cozy walls. What I wished for was my child by herself, the child I'd counted on only with me—the two of us in peace and privacy.
I wanted the normal pleasures of babies, the smell of her soft cheek against my face, to hold her in my lap at bedtime and be able to read picture books to her without hearing, as I read, the constant burble of a parallel story.
But I adjusted, for the most part. I felt I knew the voice for the invention that it was, unconscious, a product of haywire neurology; albeit with some resistance, with some anxiety, I'd learned to live around it.
And then that changed.
WE WERE HAVING a rare family moment. One of Ned's affairs had just ended in a mildly humiliating way (I figured out later) and at the same time he'd had a major setback at work—failed at a takeover of a small company that made some minor machine part for shrimp trawlers. He'd flown in that afternoon from Dutch Harbor and was home for dinner, albeit with the crabby attitude of someone who's racking his brain but just can't think of somewhere else to be. I stood at the stove cooking as the baby sat in her high chair eating spinach puree and cheese; as always, in those days, the voice was droning on in the background.
"Turn off that racket, for Chrissake," said Ned irritably, before he'd finished his first drink.
At first I didn't know what he was talking about. I was accustomed to talking over the noise in the background when I had company.
"Turn what off?" I asked, and looked around me as if to see the source.
"That AM radio, that shock-jock shit you're listening to," he said.
I cocked my head and caught a few obscenities. The voice didn't shy away from coarse invective: this piece must have been some standup routine, a foulmouthed rant. It liked to take a run through those, from time to time. I was pretty sure the FCC wouldn't have let those words onto the airwaves and got distracted for a second thinking Ned should've realized that too.
Then I realized the implications of what he had said—the sheer impossibility—and after a double take I walked away from the stove and sat down, stunned.
He was hearing it.
"Well, shit, OK. I'll turn it off myself," he said, and went to the radio on the stereo, where he overlooked the darkness of the control panel and spun the volume knob to zero.
The voice didn't miss a beat and Ned said fuck, it must be coming from the neighbors' and he wasn't in the mood to walk over there and yell at them. There followed a tirade about said neighbors, who were hippies, a category Ned reviled. He ranted about their refusal to wear deodorant and their seaweed-harvesting business; he shoveled his dinner down, took an aspirin and went to bed with earplugs in.
Earplugs had never worked for me.
I'd lifted Lena from her high chair and she was sitting on a mat with arches over it, soft toys that dangled from the arches. When Ned disappeared down the hallway I heard the voice, rising again and switching into a milder patter. For once I was able to record what it said—a couple of quotations. On my laptop I found attributions to famous writers, and I wrote the quotes down. "It requires wisdom to understand wisdom: the music is nothing if the audience is deaf." "None so deaf as those who will not hear."
While Ned and Lena slept I went into a panic. I stayed up all night; I tried to fall asleep again and again, but I couldn't, and so by 3 a.m. I gave up and put sneakers on and went walking—at times even running—in the dark, in the cold, through the silent neighborhood.
The houses all seemed like statues, the cars, the trees all seemed deliberately placed to me. Of course, most of them had been deliberately placed, deliberately built or planted there, and yet their placement suddenly possessed a different character. It was as though they watched me, as though their positions had been decided by some unified and motive force . . . I was getting paranoid, I thought: first a delusion of hallucination, and now paranoia had come for me.
Ned had heard it. Ned, indifferent, superficial, and seemingly sane as the next guy, had heard the voice. Someone else had heard it, therefore it couldn't be purely hallucination. I had been wrong.
Starting at that moment when Ned cursed, and on and on forevermore, in my mind, it could not be and was not a hallucination.
It was something else.
MY PARENTS' RELIGION had always seemed like a curious habit to me. While I was growing up I drove to services with them on Sundays, I said grace before evening meals, I went through the motions agreeably. But as soon as I was old enough to have my own opinion their churchgoing fell into a category like the next-door neighbor's golf hobby, the macramé wall hangings accomplished by a wall-eyed teacher I had for fifth grade. I saw the neighbor bundle his clubs into the back of the car on days with pleasant weather; I watched the teacher sorting wooden beads to string into an orange owl. I wondered what shaped the particular details of their interests, where their strange avidity came from.
I thought about mortality, sure, and I felt the pull of soulful music, but I never met with elevated feeling sitting beside my parents and listening to their minister. For me it couldn't be found in the cramped and unlovely building of their church, the boring sermons, the congregants next to us (mostly aged, with skin tags and wadded sleeve-tissues). It would have been as out of place there as it was, for me, in the plaid of the neighbor's golf bag, the yarn of the owl.
What seemed as though it might partake of the awesome or sublime was away from these close-up elements, away from the grainy texture of everyday. It was in cloud passage, in the galactic sweep; it was the stars beyond count, footage of herds of beasts thundering over grasslands or flocks darkening the sky in migration. I saw it in the play of light over rivers, the rush of multitudes, large beauty: a utopian sunset, the black cloudbank of a looming storm.
Meaning can be attached to it or not, I thought when I was younger, but either way the sacred has to live apart.
Later I saw that the sacred was the apart, the untouchable and the untouched. Divinity is only visible from afar.
THE NEXT MORNING I watched Ned like a hawk as soon as he woke up. I stared at him when he came into the kitchen and poured his coffee (with the voice nattering on to me the whole time as usual). But he said nothing. He didn't seem flustered or confused in the least, only impatient as he always was to get away—impatient to begin the real life of his day, out of our house, with people who mattered.
He never seemed to hear the voice again, or if he did, he never mentioned it.
Had I believed I was psychotic, no doubt I would have been relieved by what had happened—would have construed his hearing the voice as evidence of my sanity.
But I hadn't gone with the psychosis explanation in the first place, so I hadn't been seriously worried for my sanity. I'd comfortably believed in the power of a faulty and deeply complex neurology, and now that had been taken from me.
FIND THEM AMONG THE DEAD
I WAS GRATEFUL THAT I NEVER RECEIVED THE VOICE'S ASSESSMENT of Lena or me, that I was neither mentioned nor addressed directly. There were comments on what we encountered, though, the content of the patter overlapping with an image that flashed across a TV screen, a person driving the car beside us, a squirrel on a branch, a fresh berm at a building site. I'd see Lena's eyes alight on something and seconds later the voice would rush out a series of connected phrases, usually too swift and polysyllabic to be memorable to me, even when they were in English.
I got used to watching Lena's attention fasten onto a scene as only a baby's attention will, without seeming to focus—that round-eyed, often unblinking gaze of passive-seeming intake. But unlike with other babies this would be followed by commentary as the voice bounced over the object or landscape like a sound wave, a light wave, a stream of particles. I didn't get the feeling it was moving her, only that it was following her eyes, her fingers, her tongue. The model was accompaniment, not possession.
And what words came did appear, sometimes, to pass a kind of judgment. Their position seemed to be guided by aesthetics rather than morals—or no, that wasn't it either. More like, the morals were the aesthetics. What was ugly was wrong, but what was ugly was not the same as, for instance, what was brutal: ugliness was less the jarring or crude than the false or dishonest. Based on some standard I could never measure, the voice would be dismissive of systems or events, individuals or ideas, products of human ingenuity. It would rebuke the odd politician or captain of industry, engineer, or physicist; it would take even artists or musicians to task for crimes against humanity. And yet somehow the impressions I took from it were both less and more than opinions. They glittered like sun on water and glanced off again before I could fix my eyes on them.
Only a small number of the voice's observations were given over to the conditions of my life and Lena's, the rooms and scenes we moved through, but periodically there were upticks in interest. For damaged persons we encountered on the street, when we crossed paths with someone sick or in pain or disabled, often the voice would let loose a benediction, recite a snatch of poetry or hum a piece of music. To a shakily walking grandmother: "Bright star, would I were steadfast as thou art." To a kid with Down syndrome, "The Carriage held—but just Ourselves—and Immortality." Of all the lines of poetry, those were the only two I wrote down right away and looked up.
For an emaciated man we passed in the halls of a cancer ward, where we were visiting someone else, the voice had the famous lines from Chief Joseph after the battle that finally defeated him, which I searched via key words.
I want to have time to look for my children; maybe I shall find them among the dead. Hear me, my chiefs! From where the sun now stands, I will fight no more forever.
Upon Ned's entry into our space there was always the same phrase, a faintly aggressive chant. In fact the chant was a tipoff that Ned was arriving. Typically it started up on cue a few seconds early, before I even recognized his presence.
You can keep your Army khaki, you can keep your Navy blue, I have the world's best fighting man to introduce to you.
Google revealed this to be a Marine Corps cadence, one of the verses cadets call out when they're marching.
But Ned was never in the military.
A NEW GUEST came to us today. She's maybe a decade younger than I am, probably in her mid-twenties, and according to Don may stay a while.
She has an air of recovery, or so I thought as my daughter took her through the tour. She was nice to Lena in the cautious way of people who aren't used to the company of children but react graciously when it's imposed on them: patience, no talking down, a genuine interest.
Lena says the woman is a princess—probably because she's slim, tall and pretty, with long hair—and has spun a tale about her already. The princess fell from her throne through the deeds of an evil troll. She awaits an act of magic, here beside the sea. Lena says a team of seahorses will arrive pulling a giant white shell, and in the shell the princess will be borne away to her own kingdom.
At this point the story gets convoluted, because the princess can't be taken away; that would mean her leaving us. Instead she will sleep in a shimmering palace on the waves, a palace hidden from us now that hovers invisibly beyond the whitecaps. A bridge of waves will stretch from the beach outside the motel to the princess's ancestral home, a white castle of pearl, and we will walk over this bridge to banquets held in our honor, for we may live there too. Inside the castle keep, a special room will belong to us, connected to the princess's royal chamber by a spiral staircase. The chamber is full of sparkling fountains and cushions of cloud. It features a four-poster canopy bed and live-in midget ponies.
The ponies are velvety to the touch and curl up on the bed like dogs, their legs tucked beneath them.
But Lena reassures me that we won't have to sacrifice our lodgings at the motel for this resplendence. No, we'll still treasure our motel home. We'll still frequent these faithful lodgings with their yellowing shower curtain and moldy grout between the tiles. We'll have two houses, she says, that's all—"one for regular and one for special occasions."
I'd go with her. I'd take the miniature dog-ponies and the pillows of cloud.
PEOPLE WHO SAY they feel the presence of the Almighty hovering close to them, their personal savior, or tell how faith dwells in their hearts—the advantage they have is that if God overwhelms them, they're free to retreat. Or if the knowledge is so overwhelming it can't be contained, sometimes they let it out with shaking and strange articulations, crying and falling, ecstasy. I admire the idea of this, though I've never shaken in ecstasy myself.
I like to imagine I could, under the right conditions.
My point is, abandon to the spirit has an appointed time and place: the spirit can't be on you all the time. I never thought of the voice as God, while it was with Lena and me; such a thought would have been an outrage. When I write about God right now, that three-letter word—so loaded, so presumptuous—it's a word that I use in hindsight, as close a description as I can get of that stray cascade of ambient knowledge that distinguished itself from the static of everything else and filtered down to me.
So the voice wasn't God to me then, but in the months after Ned heard it, when I couldn't think of it as hallucination anymore, I was confused and stowed my questions in a locked compartment. Some things were unexplained; well, some things had always been. But I listened to it differently once I couldn't believe it was my own confabulation anymore. I gave it more credibility.
My brain's a little above average, according to standard aptitude tests, but not far above: I was always bad at calculus, I had no patience for high school chemistry. Whatever intelligence I have isn't rated for the ornate subtlety of the divine. Most of the time the voice was still wallpaper or elevator music as it streamed past and over me, citing, listing, cajoling, eulogizing, heckling. If I stopped what I was doing and concentrated on it, it quickly dazzled the faculties.
But there was no aspect of feeling chosen, no conviction of being purposefully anointed. We might have been sitting in a lounge chair on the green grass of my lawn, reading, when suddenly a bank of cumulus moved in and rain began pattering onto the pages of my book and the skin of my arms and we had to go in. I never believed the nimbus had chosen her or me or us on the basis of special qualities. I have other failings but I'm not subject to visions of personal grandiosity.
When I looked at holiday crèches or paintings of the infant Jesus I recognized the parallels—that Jesus as an infant had been believed to contain divinity, at least in retrospect—but there the similarity ended for me. I didn't think Lena was a prophet or a messiah.
More or less, in the time after Ned heard, I put off the question of causation, deferring inquiry.
The question of origin was too much for me.
LENA'S SECRET PRINCESS is named Kay and hails from the fair land of Boston. She's a med student there, or possibly a resident or a nurse. She has a hospital job holding babies, according to Lena, so maybe she's assigned to a maternity ward. She seems reluctant to discuss her work so I haven't pressed her.
I let Lena eat lunch on the bluffs with her and they went out wrapped in scarves and wearing puffer coats, though it was mild, for Maine in fall, and the big jackets were overkill. They spread a blanket on the dry grass. I could see them from the back window in our room—the room's best feature, a picture window that offers a view of the cliff edge and the sea. Lena chattered constantly—I watched her small head bobbing and her hands moving—and Kay smiled indulgently as she followed Lena's gestures. And yet somehow Lena seemed to be looking after Kay, not the reverse; the young woman's face was shuttered, and only when Lena spoke did she become animated.
It's one of the bargains I've made with myself, to let Lena have the company of relative strangers as long as I'm nearby and can keep an eye on them. I try to compensate for the lack of other children in her life and the rarity with which she sees her extended family. Of course, it doesn't compensate for that; she's an extroverted little girl, always has been, and likes to caper and perform. People are Lena's game.
For her a trip to the post office in town is a trip to see Mrs. Farber, the gum-popping straight talker who presides over the counter; a trip to buy groceries to stock our kitchenette is a visit to Roberto, the skinny cashier with the soul patch and exuberance about cartoons. She knows all the cashiers' favorite colors, pet names, and birthdays. A trip to the big-box store a couple of towns inland is a carnival of anecdotes during which Lena recounts our previous trips at great and exhausting length. She has perfect recall of people she's met even once. "Julio, he's a Pisces that means fish, cars are his hobby, like racing cars that go fast. He has a niece named Avery, the tooth fairy brought her a charm bracelet with clovers on it. Faneesha likes those yucky cookies with figs in them, she learned to tap-dance in Michigan but once she ran over a worm that came out flat."
I COULD OCCASIONALLY discern what I thought were shadings of emotion in the voice, shadings of will. Maybe those shadings were my interpretation, but thinking about it now I'm not surprised, because after all the voice was words, sometimes converted to music or other sound, and I don't see how words can follow each other without implying emotion. Even the effort to control emotion is an act of words, while every effort to control words is an act of emotion.
I didn't catch much at a time but there were recurrent themes in the patter that I learned to recognize. The voice made light of what it held to be false ideas—for example, the yearning for an all-powerful father who grants wishes and absolves. On that subject it seemed to evince something like condescension, rattling off mocking wordplay when we passed a church marquee or once, another time, while I stood at the front door trying to get rid of a Witness. Omnimpotence, the voice said more than once. Omnimpotent being, omnimpotent force. A great and ancient omnimpotence.
Sometimes it sang an eerie lullaby. Oh little man, tie your own shoes, it would sing, on the heels of a passage about the all-powerful father. There was a fire-and-brimstone sermon it liked to recite by an old-time preacher; it interspersed this text with laugh tracks and sang the cradlesong afterward. Oh little man, dry your own tears. Oh little man, there is no knee. There is no knee to dandle on. Bury your dead, oh little man. Let darkness fall over the land.
Property was an object of mockery too—the ownership of land, of pets, and even of inanimate objects seemed held to be an elaborate charade, maybe a shared psychotic disorder. The voice inflected words like owner or rich with irony—as though these should be bracketed, in perpetuity, in quotation marks. Once it said Fool, you are owned by the sun.
I couldn't find an attribution anywhere. No results.
But in general such great swaths of what it said were borrowed or adapted that they were already familiar—part of the background of culture somehow, part of the landscape of the commonplace. I sometimes wondered if all of it was borrowed, if it was all pure appropriation, a colorful textile made only of copies.
I'd started reading in philosophy, every so often, and that was when I came upon its first word to me, the sound I'd heard in the hospital before it spoke English. That word was Phowa, or poa, meaning "mindstream" in Sanskrit—the transference of consciousness at the moment of death, was one meaning.
Phowa (Wylie: 'pho ba; also spelled Powa or Poa phonetically; Sanskrit: sa krānti) is a Vajrayāna Buddhist meditation practice describable as a "transference of consciousness" or "mindstream." —Wikipedia 6.20.2009
Sometimes there were brief flickers of foreboding, brief intimations of the voice's departure, but I tried not to invest too much in those. I didn't want to be disappointed so I didn't hope too hard. When I caught a glimpse of a future leave-taking, a tiny slip of possibility, I didn't trot out the streamers or confetti or whistles, the bejeweled gowns and conical party hats, the jeroboams of champagne.
I waited quietly, holding my cards close to the vest.
CURIOUSLY TWO MORE guests have arrived at the motel right on the heels of Kay. By the standards of this place, it's a madding crowd.
They're two middle-aged men, a couple, and I can't help but feel that they, like Kay, are in some state of dismay. Maybe it's conjugal, a conjugal problem, but I feel like it's something else. One of them seems to be consoling the other half the time, he has a steadying hand on the other guy's shoulder practically whenever I see them.
They checked in at the cocktail hour—I have a glass of wine before dinner most days, while Lena and I play "Go Fish" or "War"—and shortly after that we heard a knock on our room door. When I opened it there was Don, the two men standing behind him, politely waiting, and Don peered past me and asked Lena if she wanted to conduct a tour. Typically she has to pester him for that; she'll run along the row of room doors to the lobby as soon as she sees a car pull in and beg to be the tour guide, and Don will check with the new guests to see if they're sufficiently captive to her charms. But this time Don sought her out, and it thrilled her, of course.
So we set out, the four of us—Don peeled off toward the lobby again—and I talked to the balder of the two men while Lena kept up her monologue with the other, a gaunt, handsome blond called Burke who seems to need consolation. The balding one, Gabe, said they wanted to take advantage of the off-season rates, they don't go in for tanning anyway, the cancerous harm of the sun's rays; winter beaches are just fine. Nor do they like to swim, he said, except in pools that are very clean. They also do not fish, surf, parasail, or favor any other ocean-related activities.
It became clear to me—as we stood near the ice machine and I listened to Gabe rattle on about bikini- and Speedo-clad crowds lying on beaches, the rude spectacle of this—that the two men knew Don, that Don was a personal friend of theirs, and that was why he'd felt all right bringing them back to our room.
At that moment I saw Don coming out of the lobby again, this time with Kay; they walked with their heads inclined toward each other, talking low. And it struck me with certainty that Don knew Kay, too. In fact it could well be that everyone else staying here already knew Don; that Lena and I were now the only guests who had not known Don before we came to stay at his motel.
I felt a little jarred.
And now I couldn't remember how I'd found the place, when we first came to stay. Had I driven past a billboard? Had I sorted through online reviews of budget motels? But I couldn't remember a billboard or a review. All I recalled was driving up the long gravel road in an exhausted reverie, hardly thinking, and turning into the small parking lot, shaded with pine trees. I'd liked the peeling wooden sign.
Welcome to THE WIND AND PINES.
I had a feeling of unease, flashing back to the movies I'd watched when the voice was first with me, a vision of black-clad people leaning over a baby carriage. I thought of a sedate old apartment building that was in truth a hive of sinister insects, where behind the ornately carved doors, in sleepy luxury, the neighbors quietly worshiped some dark beast.
I wondered, if I asked Don how he knew them all, whether he would tell me a simple story about how he'd gotten to meet them or would avoid answering my question. I felt a temptation to try this, to confront Don shockingly, demanding information.
But my misgivings are absurd, I realize that. The motel is Don's home, and motel managers can have friends to stay like anyone else.
WHEN THE VOICE fell silent relief washed through me like bliss. I know everyone has reliefs as the days run their course: the feeling of relief is as familiar as a hiccup or jolt of fear. But this relief was the swiftest joy of my life.
Lena said her first word early in a day, so indistinctly that at first I took it for a murmur. She crawled across the rug and began idly banging on my shoe with a red sippy cup. I was skimming the news on my computer, a mug of coffee at my elbow, when she repeated the word, Ma-ma, Ma-ma, until I pulled out of my reverie and looked down.
Then she stopped saying it, her mouth falling open as she gazed at me. And in the wake of her utterance a new silence fell around us like a sheath.
I sat in startlement for a few seconds—it seemed to me that the silence had its own soft, rising hum.
This was it, this was how it happened: this was its departure. Her first word had supplanted the voice. And suddenly I knew, in a rush, what had been suggested to me, what had been hinted at opaquely in the preceding weeks—the voice had a life cycle. It passed through those who were newly born, in the time before they spoke, and when they spoke it moved on, displaced by the beginning of speech. It lived in the innocence before that speech, the time that was free of words.
The end, the end, I thought: the beginning.
I picked her up and laughed, bouncing us both around.
For a while, after she said that first word and the voice fell silent, I was worried it would return. This reflexive, ritual worry recurred whenever I found myself in an anxious frame of mind.
But the voice didn't return, and by and by I persuaded myself to stop fearing.
And during the new silence I spent weeks, even months in an altered state—the euphoric state of a lottery winner, as I imagine it, or maybe a newly minted Nobel laureate, a state of incredulous rapture. I've never won a lottery, I've never been given a prize, but I had this. I floated wherever I went, my baby in the stroller ahead of me or on my back—my tiny girl toddling contentedly beside me, holding my hand. I smiled a lot, people said, shone like a bride.
Ignorance is bliss, few sayings are so demonstrable, and I was blissful without the voice, I drifted on thermals. I loved the freshness of the new quiet and sometimes sat deliberately in a hushed room, picking out faint noises from the street. And the opposite too—I played favorite songs loudly, held Lena and danced with her. Excitedly I prompted her to speak, I asked for repetitions of the word Mama, for other words, whatever. I would lean down over her little face with such joy in the movement!—lean close to her, lean eagerly—no one between us, nothing but sparkling air.
Since the voice fell silent I've often been able to put the whole episode behind me. There've been many days, many nights, whole weeks when I've been able to forget the untenable aspects of that time, the first year of my daughter's life.
I've frequently been successful in my denial strategy, and it's probably this success that has allowed me to live a life that, aside from my domestic problems and our flight, could almost be called normal.
SINCE GABE AND BURKE arrived, the routine has changed. Actual maids come now, since the linen laundry is more than Don can handle by himself. They're a couple of teenagers from town who do their work with earbuds in and haven't introduced themselves to us.
Plus Don has opened up a spare room off the lobby and begun to cook. The food he offers is simple and good—special dishes for Lena, a children's menu with pancakes or cinnamon rolls in the morning, macaroni and once bite-sized hamburgers at night. These didn't tempt Lena since she doesn't eat meat and never has; she feels too sorry for killed animals.
The motel guests have been gathering in the café for breakfast and dinner, and since Don keeps limited hours—as befits a chef with a base clientele of five—we're usually all there at the same time. And it's not just the guests anymore; stragglers from town have also been appearing here. First there were two or three old people wanting a break from microwave dinners, then a portly state trooper; Faneesha, the UPS driver, came at the end of her rounds and was instantly commandeered by Lena. Every night there are a couple more customers.
The first evening it felt strange to dine in the room off the lobby. I hadn't realized how much of a restaurant's mood comes from an illusion of permanence. The place seemed like an oversize supply closet, despite the flowers and candles and checkered tablecloths. But already by the second dinnertime it didn't seem preposterous to call it a café—even the lighting seemed altered, though the lamps and candles were in the same places. It had gotten more welcoming overnight.
Lena was intent on the patrons, and on the fourth night she hit the jackpot: a kid came in who was only a year older. He was with his father, whose attention was captured by a cell phone, and the boy too had an electronic toy, a glossy plastic robot that emitted tinny music and recorded the children's voices to play back. The two of them traded it to and fro, giggling at the senseless insults they made the robot pronounce. Lena got so enraptured she forgot to eat.
I was absorbed in the question of Thanksgiving, whether Lena and I should visit my parents. They're not too far from here, but on the other hand Ned knows the house. I was weighing the risks while the guests talked and laughed and Don carried food back and forth with the help of a teenage girl from down the beach. A song was playing in the background, a sad folk song about a love-struck, gunshot bandito dying alone in the hills, and I looked out over the ocean, reached to rest my fingertips on the cold window. I thought of other Thanksgivings, suffused in an amber glow.
When I turned back to the room again, my fingers still tingling, the guests all seemed familiar. It was one of those soft sinkholes of time when separate elements coalesce—we were a blur of sympathy, the air between us pockets of space in one great body, one saltwater being, unplumbed depths where the ancestors came from, primeval well of genes . . . the feeling stretched like a generosity, the gift of oneness. Who cared about those differences we had, those minor distinctions that kept us apart?
But then that lofty idea turned trivial, from second to second its shine faded. It's your commonality that's frivolous, I scolded myself, you want to think we're so many eggs under the down of a nesting bird—you want to be held there forever, sheltered in the warmth of a body that watches over you. You want it as almost everyone wants it, to pretend that we're one. To let the burden of our separation be lifted at long last.
That was all it was, I told myself: desire. Was it the case that every hopeful sentiment, each stir of communion and vision of eternity, is nothing but a projection of desire?
It's what we want that we see, not what is, I thought. Scraped bare, we're nothing but machines for wanting.
I felt a maudlin pity for us. Together now for the blink of an eye, I thought drunkenly, before we tread off into separate futures and one fine day, though motes of our bodies still persist, the last traces of our inner selves vanish. The private selves evanesce, the secret worlds that only we knew. The nameless company of ourself, that warm sleeve of being—goodbye, old friend.
With the voice, very rarely, I'd also felt these moments of loss, as though I was looking back at myself from somewhere past my death. At times I'd felt a cold freedom then, when my irritation faded and tears caught in my throat. The long view, the far distance of the stratosphere, clean and thin as high air. The axis where distance and closeness met, the axis on which the world spun.
But back then, at those rare times of elevation, the common ground had felt like truth. Now it was only a wish.
Drunkenness, I thought, could pass for a connection to God.
At my elbow Lena was making the plastic robot dance and laughing at it. There were plenty of people around. I should have felt content but I was distant, like an elder sitting apart, watching others that spun and shrieked, so busy in the midst of life.
IT WAS AT the tail end of that golden summer when the voice went quiet, coming down off the high, that I realized we had to leave Ned. I had no more patience for his complete detachment, his reluctance to come home and rudeness when he did—a rudeness that positioned us as his unpleasant burden. In my own home I had to feel like someone else's dead weight, and I couldn't keep carrying it. We had to leave Ned and the string of young women in whose name he missed our weekly family dinners, who left their sunglasses in his spotless BMW, and after the BMW was gone, in his more electable Ford truck.
Neither the car nor truck ever contained a baby seat. Later I racked my brain trying to recall a single instance when Ned had driven the baby anywhere, but I came up with nothing.
I do remember, though, that one of his girlfriends wore lacy pink boyshort underwear, which found its way into the pocket of a jacket I took to be dry-cleaned—I hadn't checked the pockets beforehand and the panties were handed across sheepishly afterward by a drycleaner. They hung in their own plastic bag, a doll-sized scrap of fabric dwarfed by the hanger.
The drycleaner had cleaned them for free, he said to me shyly.
By then I'd known for a while that I didn't love Ned. But now, rather than existing in an amiable neutrality toward him which I'd tried, even before Lena was born, to cultivate and fit into the space where love should be, I'd come to actively dislike him. I turned the corner one day with nothing else to preoccupy me and caught sight of my own dislike, plain as day.
It couldn't be talked away, couldn't be handled in therapy (which Ned, in any case, would never have gone in for). It was as solid as a dining room table. His coldness toward me I might have tolerated for Lena's sake, had he been any vague semblance of a father, but his dismissal of her got more and more unbearable. I had the devotional urgency of new mothers and couldn't help feeling that a baby was a standing debt, a debt to a forming soul.
His lack of paternal feeling was unsurprising, in the end, since he'd never promised anything else. And it was true I'd forced him into parenthood by having Lena instead of getting a D&C as he wanted me to. I'd told myself that when the baby was born he'd come around a bit. I never expected him to be a candidate for Father of the Year, but maybe, I hoped, part of a circle would be described, a slow curve into warmth. Surely a real, living child would thaw his chill. It was what happened, I believed.
Now I'm not sure where I got that belief—maybe from a TV movie. I committed a cardinal error of women, by which I mean an error to which women in particular seem prone: the error of expecting someone else to change toward them, to grow into alignment. I expected love, change, and alignment from Ned, and all these expectations were baseless. The category of children was as alien to him as if he himself had sprung fully formed from the forehead of Zeus. His own dim trailer park childhood had ceased to exist after he emerged from it—in his mind, despite the odds, nearly a perfect man.
It didn't help that around that time he was nurturing his budding interest in politics. He wasn't a candidate yet, he wouldn't be for a while, but he was angling, forging careful alliances. Though he'd never professed religious faith, he started attending church "for the connections." He gathered new opinions around him like sacks he was hefting—sacks that bulged ominously, misshapen sacks full of hidden, gross things. Tired catchphrases would spring from his conversation in passing: "No handouts for welfare mothers," say, but also, a fetus was sacred.
It was hard not to take his remarks personally when they concerned, as they often did, categories such as motherhood or women. But at the same time the remarks felt like objects to me—prefabricated items he had purchased quickly in a store, items he was busily stuffing into his shopping cart without close scrutiny.
"BURKE'S BUYING DRINKS for everyone," said Kay, twisting in her chair to talk to me from the next table. "It's his birthday. We only have beer or wine, but Don's serving a pretty good Shiraz."
I accepted the pour of wine into my glass and raised it; we toasted Burke, Gabe saying something I didn't catch about rare hothouse flowers (Burke is a horticulturist). There was a rowdy crowd from town that night, some large-bodied, friendly-looking women out celebrating a remission; one of them had a tumor that had responded well to treatment. Everyone drank on Burke's dime and I embraced once again the sentimental illusions offered by wine—what was wrong with them, after all? I'd clearly been hasty.
"You know what they say about horticulture, right?" Gabe was saying, still on his long-winded toast. "Dorothy put it best: 'Well, you can lead a horticulture, but you can't make her think.' "
I watched Burke laugh and raise his glass; I recalled a half-joke the voice had told. Have you heard the one about the Buddhist fly? It was a lovely iridescent fly, ran the riff, that flew through a room buzzing I am one with the universe, I am one with the universe. The fly felt the descending peace of its enlightenment, the liberating lift of air beneath its gossamer body. How beautiful it was! How beautiful the very air! How blessed was its flight!
The swatter fell.
You were not one with the universe, my friend, said the voice. But now you are.
But Don was serving a good Shiraz. In its flow I decided Lena and I should go see my family, we should sit at the table with them and be thankful for what we had. I recalled our dusty old centerpiece of orange-and-red silk leaves and decrepit Indian corn, which my mother always trots out with an enthusiasm that borders on the poignant.
I HAVEN'T FILED for divorce and custody yet, though I could and probably should—partly because I know it would hurt Ned's career and therefore anger him, partly because it also presents complications for me, since I removed our child from him without a written agreement.
It took me years to leave, years of deciding and planning—far longer than it had taken me to get married in the first place—and by the time I was ready it was past Lena's fourth birthday. I should have divorced him before we left, when he had no legal leverage over me. I don't know why I didn't—ineptitude. I must still have been spellbound, and I didn't know how serious his politics would get, I didn't anticipate a fight. I expected a quiet, long-distance divorce about which he would be indifferent, as he was about me, as long as he got to keep a lot of money.
Or maybe I was afraid, just afraid to take a direct and final action. Maybe it was common cowardice.
When I told him we were leaving he never once objected: there was no tension around our departure. And I only decided to evade him later, when he started stalking us instead of asking for a visit. It was only in the White Mountains that I knew his motives were strong and impersonal, as, with Ned, any motives must be. It was then that unease crept into me.
But there's no proof I didn't spirit her away against his will, only a few emails after the fact that wouldn't bind anyone legally.
WHEN I DECIDED to make the trip I hadn't told Don the details of our domestic situation. He only knew I wanted to keep a low profile at the motel—that much was obvious. So I finally took him into my confidence about Ned, I told him the story. I included Ned not wanting a child, his proven disinterest, until his Alaskan PR campaign, in a family reunion; I left out, needless to say, our visitation by the possibly divine.
To my relief Don didn't see me as a kidnapper. Rather he was alarmed for us, he tried to convince me to skip the dangers of a Thanksgiving in my parents' house and spend the day with him and the other guests. He promised to cook a prize turkey, with something vegetarian for Lena; he would bake pies, pumpkin, fake mincemeat, and pecan.
But I felt bad for keeping her from her grandparents so long, and from my brother Solomon, Solly for short, and others in her family she'd spent too little time with—only a rare Christmas, a few weeks' summer vacation she'd been too young to remember well. Alaska is far from Rhode Island. On Ned's side she'd never known relatives; even if he hadn't been estranged from his parents, he wouldn't have taken her to meet them since he never took her anywhere.
We had to go, I said. I was betting Ned wouldn't dare approach me in my family's presence—my family with whom he'd always played the part of a thoughtful, upright man, my family without whose financial gifts to us he never could have started his first business, from which all else had sprung.
I was more afraid, I told Don, that he would corner us afterward, because it was when we were away from my family that he could coerce me effectively. An in-person encounter between Ned and me is my main anxiety. The prospect fills me with the fatalistic certainty that I wouldn't be able to pull away from him right off, not with Lena's eyes on us. Somehow I'm certain of this despite its weakness, its irrationality, despite the fact that I know it would be wrong, dead wrong for me and for her too.
If Ned gets to us physically I fear he'll outmaneuver me. From the day I left him and felt the welcome release of distance the prospect of his presence has terrified me. Always since then, whenever I think of seeing him again, I'm a deer in the headlights.
If he was watching my parents' house for the holidays, some men in suits and leather shoes might follow us when we left.
"If you have to go, have someone in your family drive back with you," suggested Don.
"But he could still follow us, and then he'd know we were here," I said. "From then on. And we'd just have to move out. I don't want to go yet, and Lena doesn't either."
Don nodded.
"If you want, I can meet you somewhere in my car. We can do a switch—you go into a store, you go through the back, we leave in my car. Whoever was driving your car could bring it back here once he'd given up and stopped following them."
I was startled that he'd go to such lengths to help us.
"There are different ways to do it," he said. "But the key is, you have to be careful. Don't think of complex dodges as ridiculous. It's worth it."
He said he'd known a woman who was abused and had helped sneak her in and out of shelters. But always, sooner or later, she would lose patience and decide to make a generous gesture, she would throw caution to the winds and be caught and beaten again.
SOMETIMES I CONSIDER wishfully whether, when she's grown up, it might be possible to tell Lena about the voice and stop being alone with it. I keep this record for that reason also: not to feel so alone.
Before I had Lena, when something upset me I talked to my friends about it in the standard way. But after she was born, when that ragged, uninvited disruption entered my life, I found I couldn't talk about it to my friends. Maybe we weren't close enough or maybe I was averse to risk. It can't be taken lightly, the rumor of mental confusion.
So this hybrid document is what I have instead, my journal entries mixed with thoughts that came to me later. I don't mean for Lena to read it—it's password-protected—because I understand that even if I fantasize about telling her, it would be the kind of unburdening adulterers sometimes do, a kind of selfishness dressed up as truth. The rules of sound parenting weigh against it. No, I write for myself or for no one. I have no stake in convincing an audience of my trustworthiness; my welfare isn't of general interest. I'm someone who was rained on for a period of months, rained on by word instead of water.
When it comes to my daughter, trustworthiness is the first thing I offer. I value it above all else.
BEING WITHOUT a car made me nervous, but on the other hand I'm nostalgic for trains, and Lena, it turns out, loves them. For her a train is a social bonanza: a long container of possible friends with the added bonus of scenery out the windows. It's far superior to our sedan, where she's limited to my company.
Skipping down the aisle of the café car—where a drooping, whey-faced man looked at us glumly as he wiped down the counter in front of a near-empty display of potato chips—Lena said she wanted to live in the train forever. That's how she expresses approval, sometimes adding a touch of the morbid: "I want to eat ice cream forever and ever, till I'm even older than you are," she's said to me before. "I want to stay in the motel that long, I want to walk on the beach." At her age even a day has an eternal quality, so that forever and ever is less a linear stretch of time than a form of reassurance. "I want to live on this exact train forever and ever till I die. Until I die, Mommy! Until I die!"
I told her about sleeping cars and she decided we needed that kind of train instead, where we would have curtains to draw across our bunks for privacy, supplies of chocolate bars and chips, warm sweaters for fall and tank tops for summer; we would ride in our train over green hill and dale, mountain and plain, bedazzled by the sights, enraptured by our fellow travelers.
We finally stepped out onto the platform of the old station near my parents' town. The sun was low in the dull-gray sky and a wind whipped our hair around. Lena was perfectly happy to forget about trains in favor of the reward of seeing her grandparents again. She clutched my hand and scanned the station wide-eyed, though she can only have half-remembered what my parents look like.
But she knew them right off, probably by their tremulous smiles. I was looking along a row of lockers, past restroom doors and soda vending machines, trying to cultivate the vigilance Don had urged. But all I saw was a couple of teenagers slumped on a bench beside their old-school boombox, belligerent sounds issuing.
"Nana! Grumbo!" cried Lena, and ran forward.
Her pet name for her grandfather, invented I'm not sure how, has always been redolent of a booze-soaked clown—ill-suited to the personage of my father, whose bearing afforded him, in the past, a quiet dignity. These days he doesn't know his name, he draws a blank equally on his history and the identities of his family, but still the mantle of that dignity hasn't entirely dropped from him. He holds fast to my mother when he walks, a dreamy look on his face suggesting a dim and lovely scene back in the recesses of his mind, a hidden spring from which he alone may drink.
Lena hugged them excitedly, ambassador of affection. Young children are the standard-bearers of visible love, I thought, watching. After we grow up and get sparing with our physical affection, children are sorely needed to bridge the gap. I love my parents but the urge to touch them seems to have mostly faded. Without Lena we'd be stranded in the lonely triangle of adulthood, the lovable child I ceased to be hovering sadly between us.
"Do you still have the kittens?" squealed Lena, who remembered kittens from a visit when she was three.
One day she'll separate herself with an adult coldness she'll be unable to control, uninterested in controlling; one day she'll probably touch me as rarely as I touch my parents now. She'll come and go, returning only for visits.
The thought is so acute, the outcome so near-certain I cringe, thinking: This is why parents want grandchildren. Really they want their own children back again, they long to feel that vanished and complete love.
I watched my parents' beaming faces as they bent to encircle her with their arms—my father doing so in a spirit of general camaraderie, not specific attachment. He doesn't recognize Lena across time but since his memory went he has learned to obey my mother; he simply believes her when she tells him that he knows or loves someone. He has agreed to go along with it. In a way this trust is the crowning glory of their lives, a final achievement. He knows my mother and through her he accepts the rest.
I'm often teary when I first see them again—my mother a little bit grayer but still solid and known, my father a meek shade diminished almost to nothing.
IN MY PARENTS' house, where I grew up, it's hard to convince myself to stay alert for watchers in the shadows. Their neighborhood's staid, the houses upright and boxy and spacious, the trees sheltering. It's well-mapped terrain for me and its textures make for a sense that nothing surprising can happen here.
So I'd relaxed my guard by the time Ned called.
Lena was playing with my brother in the backyard, where a rusty swing set and jungle gym remain from our childhood. Solly's good with children, though he has none of his own—he's younger than I am and prefers the bachelor life, long work hours punctuated by trips to Atlantic City to play poker and weekends drinking and watching sports with college friends—and Lena's smitten with him.
I sat on a stool beside the kitchen island, cutting pie dough, and watched out the window as he lifted her up to the monkey bars. My mother was unhurried in her preparations and the house was quiet, though the next day people I barely knew would come teeming in—old colleagues of my father's, a group from the homeless shelter my mother invites every year, a couple of church friends. When the outdated rotary phone on the kitchen wall rang, she answered it with a voice that faltered at first, though it was perfectly pleasant.
She mouthed Ned to me across the room and I slid off the stool, helpless.
"It's so good of you to remember us," she said politely. "Are you having a nice Thanksgiving?"
I went out of the kitchen and lifted the receiver of the hallway phone.
". . . missing my two girls, of course," I heard Ned say.
I recognized his angle instantly: loving husband, abandoned callously.
"You know, Lindsay . . . it's pretty tough to be alone. It's tough, over the holidays."
Ned's automaton nature is well hidden from guileless observers. My mother has never fathomed my leaving him, which was alien to her and which I can't hope to explain fully—especially as I've chosen not to mention, for example, his many affairs or the fact that he pressured me to end the pregnancy. That would upset her too much. I've said only vaguely that there were infractions and that Ned and I don't love each other. But that's an obstacle whose scale, she seems to feel, falls short of the requirements for divorce.
My mother's loyal and chooses to respect my wishes—most recently not to let Ned learn that Lena and I were coming. But she dislikes subterfuge of any kind, which goes against both her instincts and her ethical code; she can't shake off her early positive impressions of Ned, probably shaped by his good looks and the refined manners he affects in certain company (initially acquired from books on etiquette so he could pass among the rich as one of their own; then honed by practical experience). I believe she's always thought of Ned as that nice, handsome boy.
And of course my father is now effectively neutral.
My brother, on the other hand, has never trusted Ned; when I first introduced them he said to me privately, "Well, he's sure as shit white! That is some Crest toothpaste, bright-white shit you got yourself there!" He said it in a joking fashion, grinning at me affectionately and cuffing my arm to take the sting off. I knew what he meant: Ned's whiteness, unlike Solly's or mine, has a fifties Boy Scout aspect. It seems to extend deeper than his skin, which is as unblemished as his straight and beautifully formed teeth. And as the months and years passed Solly never warmed to Ned—for which I was eventually glad.
"I can imagine," said my mother weakly.
"I miss them. I really do. I fully understand, Lindsay, you don't like to get into a difficult position, an intermediary position, and I respect that and I would never ask it of you. I just—I miss them"—here a quaver came into his voice—"and I thought I'd feel a little better if I touched base with you. It's a family time. That's all."
"It is, yes," said my mother carefully, after a pause. "Well, Ned. I'm so sorry to hear you're feeling lonely."
"I've got to admit," said Ned, sighing, "part of me just can't believe she doesn't mean to come back. Part of me still holds out hope. I've recommitted myself to the church, Lindsay, and to my faith. And marriage—as a sacrament . . ."
"Yes," said my mother hastily. "Of course. I'm glad for your faith, Ned."
"Faith is what pulls you through," said Ned, "when nothing else will, just . . . nothing. I've had to face that, Lindsay. So in that way, this has been strengthening for me. For my relationship with God. At first I didn't want to see how much I needed this to bring me back to Him . . ."
I felt a wave of nausea at Ned's string of clichés and my mother's vulnerability to them so I put down the receiver. I stood there in the hallway, the faint squawk of the conversation still audible, arrested by an image. Ned's God was a life coach—the kind for whom you had to be at least a mid-six-figure earner. Ned's God was a superstar, a braggart and a motivational speaker, presiding from an office whose walls were lined with awards, diplomas and framed pictures taken with celebrities. Ned's God would have to take an interest in the workings of his personal ego.
Even Ned's Lagerfeld cologne, I thought, would be a matter of no small interest to the God he conceived of.
I smiled at that and the movement of smiling let me lift the receiver again.
Thankfully the talk of religiosity had passed and Ned had moved on to a discussion of his electoral goals, his new mandate to serve the people, and his humble wish for Lena and me to be with him in what was, it turned out, chiefly a humanitarian crusade for public office. He deployed some pieces of text from his website, evoking the twin needs to restore values and build communities (wisely passing over those pieces of his rhetoric that would not jibe with my parents' political leanings, moderate Democrat). He said the word humbled several times: he was humbled by the growing "grass-roots" support for his candidacy and also humbled by the "tireless dedication" of the campaign's volunteers.
Finally, it seemed to me, he was quite humbled by his own humility.
Later I'd try to explain his cynicism to my mother, the connection between his recent discovery of the joys of piety and his career. It was painstaking because she doesn't want to impute evil motives to anyone, much less a son-in-law and in spiritual matters—a generous but inconvenient aspect of her personality. I'd step lightly, not arguing my end too hard, but still she wouldn't be entirely persuaded.
"I'd just ask, if you do talk to her over this holiday weekend," said Ned mournfully, getting ready to wrap up, "I'd just ask that you give her my love. She doesn't take my calls anymore, and so I can't . . . say it to her myself. But I want to, Lindsay. You know? I may not have given her the . . . well, the full and complete attention that she obviously needed. I know that now. If I had it to do again, I would. There were pressures, of course . . . but I shouldn't have let my passion for my work come between us. I'd try as hard as I could to give her the attention that she really needs."
There were subtle stresses on certain words. And I knew. I knew he knew not only that I was in the house, but also that I was listening.
"WELL, DEAR," said my mother, coming in after she hung up. "Ned tells me he misses you. I must say he didn't say much about his little girl. I think that's very strange."
Probably the harshest thing she's ever said about Ned.
I mulled it over in my bed that night, what advantage he hoped to gain by calling. If he was letting me know he was watching, why? The element of stealth had been sacrificed, which must mean he wouldn't be showing up in person. So there was that.
I thought of his false regretful tone, saying the full and complete attention that she obviously needed. The implication, not too deeply buried, that I was secretly demanding, that I was a woman with hidden and deep reserves of need, was intended less for my mother than for me—for me to get a taste of poison, to see how sly he could be.
Maybe coming here physically was too much of a risk, though it was hard to believe his contest for the Alaska state senate was going to expose him to the media in far-off Rhode Island. He's egotistical, but not unrealistic.
But he knows Solly sees through him, and likely he didn't want to have to deal with my family—whose money was still in play for him—on their own ground.
I lay restless on the bed I shared with Lena, who was snoring lightly. I listened to the radiator knock. In the end I decided that, along with laying the groundwork with my mother for our eventual "reconciliation," Ned must want me to feel a threat. To know that he can still touch me.
HIGHLY EDUCATED, MODERN PROFESSIONALS
AFTER TEN DAYS AT MY PARENTS' HOUSE WE'VE COME BACK TO THE motel. Snow has fallen and lies in the evergreen branches in perfect white tufts. Meanwhile another three rooms have been filled.
Since it's a small place, rooms numbering one to ten, this means we're close to capacity.
"Business is booming!" I said to Don with forced cheer.
I felt put-upon, since the motel is supposed to be my personal refuge.
He nodded and smiled warmly.
"We're glad to have you back," he said.
Don's elderly father is among the new tenants, so not every new guest is a paying one, I guess. He totters around in faded plaid shirts, leaning on a cane, and smiles apologetically. When his arthritis is bad he lives here so Don can take care of him. There's also a pair of mannish, gangly sisters from Vermont, whom I haven't seen up close but who give an impression of short hair and protruding teeth.
The fourth new resident is a guy not too many years out of college who seems an unlikely person to land alone at an obscure motel on the coast of Maine in early December. He's handsome, with a five o'clock shadow, and unlike Kay—not far from him in age—has an arrogant manner. Maybe he's a drug dealer seeking shelter or a day laborer whose work has disappeared with the cold; maybe he has a trust fund but is aimless and deranged.
But I haven't met the new guests yet, save for sightings of Don Sr., because as soon as we got back from Thanksgiving Lena came down with the flu. Since we went to see the doctor in town she's been confined to her bed. She sleeps for most of the day; I stay with her, I read to her and I write this account. Occasionally, feeling stir-crazy, I emerge for a few minutes, locking the door behind me, and stroll to the lobby or amble to the edge of the bluffs and stare out over the ocean. I leave the picture-window drapes open so I can check on her.
WHEN WE FIRST got here, months ago now, I went over the clutch of notes I'd made during Lena's first year—some from the time of belief in hallucination and some from afterward, the uncertain time.
After the fact it was easy to find a thread that ran throughout them, a thread that reinforced my idea that the voice had said "Phowa," that it might have referred, on Lena's first day, to the transmigration of souls. I patched together pieces of text and saw a story there, I thought—did I imagine it, or was it real? I read the pieces as a story of consciousness, believing the voice had always known it would fade when its host began to speak on her own.
I uncovered references to the human brain, to "Broca's area" and "Wernicke's area," which at the time I'd assumed were geographic but which an online search told me had to do with the capacity for speech. There were terms like remote insult and neural plasticity. Yet there was also a lexicon of religious terms, of Hindu words like jiva, mentions of the Sikh brotherhood, the passing along of the soul from one body to another until its liberation.
There were allusions to Jainism and to African faiths—Àtúnwá, I was even able to decipher: a Yoruba belief in the rebirth of the ancestors.
KAY OFFERED to babysit Lena tonight while I went to the café to grab some dinner, to give me time out of the room. Lena was already sleeping when I got back after the meal and Kay and I talked in hushed tones, standing near the doorway.
She told me she was a med student and volunteered in a neonatal intensive care unit. There, she said, one of her tasks was "cuddling"—their name for holding babies, just as Lena had said. These were sick babies, some born without a chance of lasting, and they liked the touch of skin. Incubators and other machines weren't enough.
"My shift was late-night," she said. "You know, when the mothers were sleeping. Or some of the NICU babies didn't have mothers who could hold them, they'd be addicts or occasionally they'd have died in childbirth."
She'd hold one of these fragile infants and her next shift, if the infant had gone, she never followed up—it was the policy and she tried to observe it. But this part of her work had proved too much for her. Eventually it had driven her to antidepressants that didn't work and she'd spun out and taken a leave of absence.
She was unfit to be a doctor, she said, shaking her head, but she'd wanted to be a doctor all her life.
"I don't have much longer to decide, the program will give my place to someone else," she said, and looked down at the floor to hide the fact that her eyes were filling.
She's just a girl, I thought with a pang, grown up thin and sad. I wondered where her parents were, if they knew how miserable she was. Since I had Lena I see my own child in any young woman; before that they were only adults, but now they're former children.
How did a young girl come to be alone in a cold motel, I thought, a row of rooms, because she was deemed mature enough? Not long ago she'd lived safely, I imagined, in her parents' home, and now here she was, wretched. Alone.
Not everyone, really hardly anyone, is suited to the job of constant dying babies, I said to her as gently as I could. Most doctors wouldn't be equal to that particular task . . . she nodded but I could tell she'd heard this before and it was useless to her, though she was too polite to say so.
I felt low after she went away and curled up next to my daughter in her bed.
DURING LENA'S BOUT with the flu I was more solitary with my thoughts than I am usually, and I don't think it was healthy. I started to wonder if Ned did know where we were, if he'd known for ages, if I'd been wrong to think we were on our own the whole time. I felt more and more paranoid and I made up theories—he was watching us using satellites and GPS, he'd turned my laptop camera into a spy device.
In the movies it was easy.
The paranoia's still with me, exaggerated and ridiculous as paranoia has to be. I live alongside it the way I would an unpredictable roommate. A suspicion rises that we're not as far away from him as I assumed we were, that Ned hovers unseen. Then I reassure myself, which works mildly: the nervousness subsides, until it rises again.
He's always known my parents' telephone number—it's the same number they've had since I was a child, I say to myself. So what if he called while Lena and I were there? It was Thanksgiving and I knew he might call, or worse. Our material circumstances haven't changed, I tell myself, I have no real evidence of his proximity here at the motel.
It's only that his voice—a warm South Carolina drawl that's alluring until you detect the insincere overtone—and his manipulative conversation with my mother have infected me, exactly as he intended. It's me realizing, hearing that voice for the first time in two years, that I've gone from what I thought was love to neutrality to dislike to open hostility. I'm contaminated by the discord between loathing Ned now and once having adored him: I remember my adoration acutely and wince. I don't know how much is shame and how much is confusion. My former, deluded self was a loose construction of poorly angled mirrors and blind spots, I can see that now.
But Lena's better. She woke up smiling and full of energy yesterday morning with no fever, and we've started lessons again. I'm relieved but out of sorts anyway, because besides my paranoia about Ned I'm also grappling to understand the staying of the guests.
In Lena's and my case I know why we're lying low. We have two scarce commodities: disposable income and my willingness to spend it on a dingy motel in Maine in December. I hold my willingness to pay for this cold privilege to be an idiosyncratic feature. But here are the other guests, also apparently willing and able to pay and stay.
They can't all be in hiding from estranged husbands; they can't all be, say, drug dealers on the lam. And even if they are all friends or relations of Don's, that fails to fully explain their presence, short of a simultaneous eviction from their homes. It's disorienting and is preoccupying me. Technically it's none of my business, though, and I'm reluctant to broach the subject with Don.
And the college drug dealer with the five o'clock shadow has been making overtures to Kay. He approached her in the café this morning and offered small talk about genres of orange juice.
"Who likes the kind with orange pulp?" he asked. "Where are these orange pulp drinkers? I don't want to drink the pulp. Do you want to drink the pulp?"
There was a certain expectant force to his approach that I recognized with curiosity. Pick-up lines have changed since the advent of Seinfeld; now they often take the form of one person asking another about a mundane detail, a baffling social or consumer habit. Maybe the idea is to forge an alliance in the face of seemingly senseless choices made by others. Anyway Kay shrugged at the orange-juice pulp opener, but she smiled at him.
Later she told me he isn't a college drug dealer but a guy who makes and spends fortunes selling Hollywood movies to foreign markets. His youth combined with his skill in this realm makes him a prodigy at profit, a producer or studio executive or other dealmaker, I can't recall the title she gave me. So he is rich, but not aimless or deranged, and his wealth, combined with the youth and good looks, makes it even more unlikely that The Wind and Pines would find itself by chance at the top of his list of winter vacation spots.
"What's he doing here?" I asked her. "I mean, why here?"
I wanted to ask, Why are any of us here? Why here? But it was too pointed.
"Not sure," she said, as though it was all the same where he was.
"Well, how about you?" I asked. "I don't mean why aren't you in Boston, I understand that. I mean how did you end up at this motel?"
Again she looked indifferent to the question but passingly curious about why it had been asked, the way a person might look if you asked them, with intense and focused interest, where they bought their toothpaste.
"I was here last summer," she said, flipping through a magazine about trout. "I came back for the rest. It's restful. You know. And Don's such a nice guy. Isn't he?"
"Don's great. But last summer," I persisted—because it was gnawing at me, the casual presence of everyone, their unlikely presence, their stubborn persistence—"how'd you find it in the first place?"
"Just the website," she said, and put down the trout magazine in favor of a yellowing copy of Cat Fancy.
As she reached for it one of her long sleeves rode up, and I saw a red scar along the wrist.
BURKE CAME TO HELP with Lena's lessons; he's her tutor in botany. They planted seeds in a doll-sized greenhouse we put together from a kit, Burke bent over beside her, avuncular and kindly. The greenhouse has rows of light-green pots maybe two inches in diameter, a line of small lightbulbs and transparent plastic walls. It sits on our windowsill.
Lena had said she wanted to grow a beanstalk, so Burke brought her several kinds of beans to plant. He cautioned her the stalk might not be large enough to climb on; it might not reach the sky. She nodded and told him that was just as well, because she didn't want to meet a giant or a giantess, she didn't want to hear a cannibal giant say "Fee, fi, fo, fum. I smell the blood of an Englishman."
She isn't an Englishman, she said to Burke, but she still thought the giant might want her, even if she's a girl and an American. She didn't want to hear that giant talk about smelling blood.
Burke patted her head.
"I promise, sweetie," he said, "there won't be any giants speaking to you from this beanstalk."
As soon as he said it his face went pale. He stood there for a few seconds and sat down heavily on my bed, leaned over and stuck his head between his knees.
I was taken aback—Burke had seemed more solid and self-assured lately, seemed to require less comforting.
"Are you OK?" I asked, leaning over him, laying my hand against his back and taking it off self-consciously.
He looked up and nodded.
"Sorry," he said. "Panic-attack type . . . sorry. I'm fine. Heading back to my room."
Lena cocked her head, confused; I watched the door close behind him.
"Here," I said, picking up a library picture book on plants, "let's read this part about how seeds germinate. Most seeds contain an embryo and food package . . ."
IT OCCURRED TO ME, reading about the transmigration of souls, that my early assumption of some kind of nonhuman power or supernatural omniscience had been impressively unfounded. It might have been just a person's thoughts that had got loose, the memories or knowledge base of, say, some overeducated, possibly unhinged individual whose stream of consciousness flowed along carrying the debris of a lifetime. Could be that Lena caught the ruminations of a scientist or scholar.
Maybe this is a ghost story after all.
Or maybe the information that's now carried by so many frequencies just caught in her as it passed, lodged in her body—the live feed of a humble taxpayer somewhere, erudite but alive. Maybe some unseen field around my infant simply filtered particles from the immense cloud of content carried by those millions of waves that pass through us all the time.
THE SISTERS FROM Vermont, it turns out, aren't sisters from Vermont: I'm bad at pegging guests' identities. Their teeth aren't even protruding, just large and blocky, and they're cousins from somewhere on the mid-Atlantic coast near Baltimore. Both of them are named Linda, a name that's common in their extended family; they're in their early fifties, friendly, good-natured and hearty. One is an administrator at a university while the other is retired from her career at a famous aquarium in Florida where marine animals do tricks for crowds.
When the Lindas went to town for groceries today we hitched a ride with them. They dropped us off at the library so Lena could exchange her picture books—one of which is too young for her, about a bear who's a splendid friend, the other of which turned out to feature cows rising in armed revolt. (They hold roughly drawn Uzis in their hooves; this puzzled Lena's literal mind due to the cows' lack of opposable thumbs.) To answer the question of the guests who don't leave I have to be more outgoing than I have been until now, so I'm trying.
The Lindas, being friendly, are helpful in this chore. Big Linda, as Lena calls her, told us about someone she knew who was bitten by a bull sea lion. "Right on the keester, kiddo. And let me tell you it made a mighty broad target," she chortled. She told Lena that performing seals at zoos and aquariums are not seals at all but sea lions; that some sea lions work for the U.S. Navy, finding things in the ocean; and that male sea lions can be four times the size of the females—weighing, put in the other Linda, up to one thousand pounds. That's half a ton.
Lena calls the other one Main Linda because she met her first. Main Linda goes swimming in very cold water, Lena said to me, once every year to help raise money for the Special Olympics. Lena's resolved to join her in one of these polar bear plunges, as she calls them. I have to restrain her from practicing.
The Lindas have embraced their nicknames.
When the two of us finished at the library we walked over to the local diner to have lunch. A beefy middle-aged man sat down beside us at the counter—beside Lena, I should say, with me on her other side. He ordered a Reuben, introduced himself as John and proceeded to engage her in a conversation about her gold and silver metallic markers. He was inoffensive, on the face of it, a neighborly fellow patron, yet I thought I detected something off-color in his expression as he glanced over the top of her head at me, a hint of a leer, some glint of beady self-interest.
So I hurried Lena at her lunch a bit. We shared a piece of sickly-sweet cherry pie for dessert, leaving bright jelly smears on the plate. Then we left, with the beefy man smiling after us as the door swung to.
Big Linda was waiting for us in her bulky car; Main Linda, who was buying birdseed in the hardware store down the block, remained to be picked up.
"Big Linda?" said Lena hesitantly, as we pulled away from the curb. "Do sea lions have really sharp teeth?"
While we waited in the car again, this time outside the hardware store, the two of them discussed sea lion dentition, a subject that was, to me, of limited interest. I sank into the warm seat in a half-dream, full of the sickly-sweet pie, grown even more sickly in retrospect, and mused on my attraction to the town's librarian, who seems out of place here. He's good-looking; his skin is a coffee shade but the geometry of his face seems less African than Eastern, maybe Malaysian or Indian, I don't know. It's noteworthy mostly because there aren't too many colorful immigrants in this part of Maine—in some parts there are Somalis and Asians but around here most everyone I've seen is plain old white.
When Mainers rise up against immigration it's often been Canadians they accuse of stealing jobs; once Maine loggers blockaded the Canadian border.
I stared out the window, which was fogged up and yielded no defined shapes, only hazy panels of white and gray. I realized I was thinking of sex, of the idea of sex or rather, to be precise, the idea of no sex—no sex at all. I mulled over my asexual existence as a mother, gazing at the foggy window, mulled over the asexual existence of many mothers, whose bodies, formerly toasted politely as sex objects when not worshiped outright, had been diverted from the sexual to the post-sexual. In the natural plumpness of motherhood they were summarily dropped by male society like so much fast-food detritus in a mall food court.
I wondered if it was impossible that I would ever be a sex object again, if I should embrace that impossibility or try to reclaim my status as a sex object—by, say, enrolling in pole-dancing classes as one of my old college friends had done after her divorce, enacting a middle-aged crypto-feminist stripper fantasy that seemed to keep her entertained.
I decided I wouldn't enroll in pole-dancing classes.
After a minute along those lines I gave up thinking. As I swiped at the condensation on the car window I caught sight of the beefy man, John, walking toward us down the sidewalk from the diner. A light snow had just begun to fall and his slab of pink face was a blur, so I couldn't tell if the small blue eyes were pointed in our direction; before the blur resolved he turned and disappeared through a door.
The falling snow made me want to shore us up snugly for the winter and brought a pang of homesickness for our house in Alaska, which had always been more mine and Lena's than Ned's, for all the time he never spent there. Ned should have gone, I thought, Ned should have left.
But I myself had chosen otherwise, no one had chosen my course of action for me, and so Ned had not left the house—rather I was the one who fled. I forsook my existence, my local friends, the belongings I'd slowly and carefully amassed over the years of my life up till then, most of which would mean nothing to him . . . I left it all, except for some file cabinets of photos and documents, a few boxes of books and a handful of childhood keepsakes I'd stowed in a small storage unit. I'd given up everything to keep Lena close to me and get us clear: nothing else had mattered.
This was still true, I reflected, still perfectly true, my intent remained the same, it was my decisions that were questionable.
In our old house in suburban Anchorage—a city where every street was suburban except three or four downtown blocks—I'd kept us warm through several Alaska winters, Lena and myself, I'd cooked soup and stews and lasagna and other hearty foods in a kitchen shining with copper pots and brimming with heat . . . I'd loved it there, I'd arranged all the spaces exactly the way I wanted them. It had been a golden burrow.
All we had now was a small microwave, its walls cool and thin. People we hardly knew, though they were nice enough. The motel's walls were thin too. I had no solid walls, I thought. Would a wind rise around us this winter?
A wind would rise, I could feel it already, rise off the gray ocean and howl at the thin motel walls.
And then there was my parents' home, far nearer than the house in Anchorage, with its solid brick, wooden floors, and soft throw rugs, vines dormant on their trellises till spring. My mother would welcome us, I thought, if only I could shake this phobia of Ned, if only I could just face Ned and stand up to Ned, if I was willing to call Ned's bluff.
Instead we were living in a room like a cardboard box, with no source of warmth except the wall heater. We were socked in, I thought, perched on the rim of the frigid Atlantic—unknown in a group of other itinerants passing through, their lives as opaque to us as ours must be to them.
This cold, flimsy box was where my irrational impulses had brought us, I thought, my formless certainties.
I HEARD THE YOUNG mogul pacing along the walkway outside the rooms this evening, pacing and talking on his cell phone. Lena was sitting in the bathtub blowing bubbles—she luxuriates in long baths, though without nagging she doesn't bother to ply a washcloth—and I was reading a magazine in one of our two armchairs, the bathroom door cracked open between us. We'd had dinner early because of her bedtime, but most of the other guests were at the café.
The windows of our room weren't open, since the temperature was below freezing; the heater thrummed, so at first I didn't hear his words. But his voice got louder; he grew agitated as the call went on.
"That's not fucking relevant," he snapped. "Can we not do this analytical bullshit? If I wanted an analyst I'd lie on a couch and jerk off for two hundred bucks an hour. Hell, put me in a Skinner box. Fix me! . . . I couldn't give a fuck."
I glanced at Lena to see if she was paying attention. But the bathroom was farther away from the door than I was, in my homely armchair backed up to the heater, and bobbing in front of her was a waterproof MP3 speaker shaped like a yellow duck and playing sea shanties. She was impervious to the young mogul's call, dipping a rubber whale toy in and out of the water as it consorted with the duck.
"It has nothing to do with that crap. I'm telling you. My mother was fine. My father was fine. They were both fucking fine. They're still fucking fine. Everyone should have such fucking decent and doting parents . . . no pervert uncles! Jesus Christ."
I swept a drape aside to look out, making a sour mother-face that went unseen. If he moved off before Lena caught on I'd be relieved—and it wasn't so much the swearing that annoyed me as the force of his anger. He stalked by in his elegant leather coat and kicked one of the square wooden posts that was holding up the overhang.
"Well yeah. I told you that already. Not so much now. Before. Doing coke raises your chances of that shit. Plus oxy . . . what? Harvard. Aren't there brain scans? Some other radioactive shit?"
I decided to join Lena in the bathroom, where I shut the door behind us and ran in some fresh water to rinse the shampoo from her hair. I was thinking the young mogul would be bad news for Kay, if she submitted to the sitcom pickup tactics.
I don't know if Kay needs an angry young mogul.
ON TV THERE were numerous "exposés" of small children remembering past lives. One two-year-old boy was born with the memories of a fighter pilot shot down in World War II, they said, and repeatedly enacted scenes of the pilot's fiery cockpit death. He showed a high level of competence at identifying bombers used on the Western Front. A girl of four painted watercolors apparently based on her great-aunt's early life as an orphan in Minneapolis, although the two had never met before the great-aunt perished of cirrhosis.
Their parents had been skeptical at first, the voice-overs told viewers, but over time had clearly seen no other explanation fit the bill.
"Young Alex's parents are highly educated, modern professionals," intoned one narrator. "They did not wish to accept the evidence that past lives are real."
A FEW DAYS before Christmas my car stalled out so I left Lena with the Lindas and got into the cab of a tow truck, where I sat beside a driver who pulled my car into the only car-repair place in town. Imagine my displeased surprise—although I shouldn't have been surprised, since after all the town is small—when I was greeted by the beefy man from the diner.
He was the owner, apparently, since he wore a button-down collar shirt while the other, thinner man behind the counter wore polyester-mesh with the name of the franchise appliquéd. The beefy man—John—reclined with his arms crossed in a posture of managerial ease; I stood across the counter and smiled wanly. I felt the discomfort I always feel in car-repair places, the low-level dread of condescension followed by cost inflation, and wished to call upon my considerable expertise on the workings of internal combustion. Unfortunately I had none.
Waiting for the man to finish typing and Beefy John to finish watching him, I looked around at the walls, at ugly posters for automotive service packages, tires, motor oils.
One poster was markedly different: it was for something called American Family Radio. I peered closely at it, an airbrushed-looking photo of a plump, pink-faced man in headphones, shining smugly. Inscribed beneath his face and what I guessed was the name of his radio show were the smaller words The AFA Works to: (1) Restrain Evil by Exposing the Works of Darkness . . .
"Ma'am?" said Beefy John, finally.
I tore myself away from the fine print.
"Hey there, and how's that pretty little girl of yours? What can we do you for today?"
"My car keeps stalling out," I said. "A Honda. It's a Civic hybrid—getting a little old, maybe. But it's always been pretty reliable. I can leave it overnight."
"A Honda, huh? Well sure, we can take a look at that rice burner for you," said Beefy John, and his smile said he was bestowing a favor. "B.Q. here will help you with the paperwork." He smiled again before he clapped the underling on the shoulder, tapped his forehead in my direction in a mock salute and disappeared into the back office.
"B.Q.?" I asked.
" 'At's me," said the underling, typing.
"What does the Q stand for? If you don't mind the question."
"Quiet," he said.
"Quiet?"
"Be Quiet. Always saying that to me when I was a kid."
B.Q. looked up from the keyboard and grimaced. His teeth were a rotting brown from the gums up, old-bone yellow and tobacco brown.
Handing over my keys I realized Don wasn't due to pick me up for almost half an hour; it was bitterly cold outside and I needed to be warm while I waited. But Beefy John in his satisfied recline, his crossed arms, the words that pretty little girl of yours, the jagged mossy teeth of B.Q.—they made me uncomfortable. The words land shark came to me as I signed the work order, B.Q. leaning forward unnecessarily from the other side of the counter, so close that I could smell the residue of cigarettes. B.Q. wasn't a shark, surely, he seemed more ruined than fierce, but the teeth . . . I considered whether his meth use was current or past, whether teeth ravaged by meth could be reclaimed. Then I pivoted and walked out into the winter, pretending to have a goal.
Once I was on the sidewalk I slowed down and ambled, watching my breath fog and feeling the cold on my cheeks until I fetched up in front of the library and went in. I hadn't had that goal in mind, of all the thousands of possibilities offered by libraries no single one presented itself to me, but there, right away, was the librarian I was attracted to. I had nothing to say as he looked up from the front desk, nothing at all. And yet I felt better already.
"Sorry, just coming in from the cold," I blurted.
"What we're here for," he said.
I couldn't think of any more small talk so I wandered along the shelves looking at titles, plucking out books at random. I seemed to be in a section either for children or for adults who were childlike: true-life accounts of balloonists, explorers. Pictures of famous caves. Prehistoric animals turned to fossil—trilobites that looked like beetles, ammonites that looked like snails. Real-life Monsters. Haunted Houses of New Orleans. The more I looked at the variety of subjects, the more hopeful I felt.
Maybe we could travel, I thought. Not just in my small car—across the world. To the Himalayas, say, jungles, dormant volcanoes with crater lakes, those acid lakes that shimmer turquoise in the sun . . . we stood on the decks of ships, rode camels over Saharan dunes toward the pyramids, wandered the Prado, the Great Wall of China, treaded the paths of picturesque ruins. What, in the end, would keep us from the world? I'd planned to give her a solid, settled childhood, where she could have the same friends for years and run through the same backyards, a childhood much like my own. But maybe she didn't need that. Maybe we could sail away, out of this chill into a summer country.
I hadn't thought of the voice in a while, I thought (suddenly thinking of it). These days a memory of it will flash through me and what I notice is myself forgetting, the rarity of that flash. It's like sickness—the whole world when you're in its grips, but once gone, quickly dismissed. Within days you take good health for granted once again.
"We only have a fake log," said the librarian, behind me. "It's not as warm as the real thing."
Privately joyful that he'd spoken to me, feeling as though I'd performed a small but neat trick, I followed him to a reading room. In the hearth an electric log glowed orange behind its fiberglass bark. The chairs were overstuffed, the high ceilings dark, but still I noticed, trailing after him, peering with difficulty at the fingers of his left hand, that he wore no ring, and I was pleased. I felt like a cliché noticing, a woman who read glossy, man-pleasing magazines, a member of some predatory horde . . . he had broad shoulders, an elegant posture.
"I'm so glad there is a library," I said. "In a town this small. With only one gas station and no fast-food chain."
"The building was a gift from a wealthy benefactor," he said. "He made his fortune in lumber. His wife died young and he never remarried. He died without anyone to inherit his fortune. Brokenhearted, they say."
"Oh."
"So he left his house to the town for a library. In short, his tragedy was our gain," said the librarian.
"Oh," I said again. "Yes!" I couldn't think of anything else to say.
Luckily he smiled at me.
When he went back to his desk I sat gazing into the glowing seams of the artificial wood and wondered whether to ask him out. I wasn't sure I could. It'd be a cold call; I had nothing.
And yet I might be restless enough to do it, I thought, I was bored and agitated at once these days. I was constantly aggravated by the open question of the gathering of motel guests, frustrated by the problem of their continuing presence—and then, bookended with that problem, there were the limitations of my existence and the tedious routine of our schedule. I felt drawn to the librarian but at the same time ambivalent about the prospect of not being alone, that is, not being alone with my daughter, the two of us a capsule . . . the two of us close together after the leave-taking of the voice and our running away from Ned.
Of course it was premature to speculate, I knew nothing about him, but still, I thought, why actually try to know someone if you don't wish to know anyone at all?
Still, in the end you seek out company again. After the noise has passed, after the great clamor's hushed and the crowds have thinned—then a silence descends upon your room.
And though at first the silence is perfect, the silence is thought and peace, after a while the silence passes too.
IT WAS EMBARRASSING to ask him out and I had to buoy myself up with bravado: it didn't matter if he said no. I had nothing to lose. The worst that could happen was that my life would remain the same.
In the few moments after, waiting for him to decline the invitation as I rested my fingertips on the edge of his desk, I thought of a girl from high school: she'd been average-looking and not particularly good-natured—in fact she was manipulative, crude, and often picked on easy scapegoats, the poor kids with hygiene problems, the loners. Despite this she always had a boyfriend, and her boyfriends were kinder and far better-looking than she. Waiting for rejection, I remembered her clearly.
Years after high school was over, when I was home from college on vacation, I ran into her on the street. We stepped into a nearby bar for a drink. I had an awareness of being only half there, as though the other half of me had continued along the sidewalk without acknowledging her presence. But we had caught each other's eyes, we hadn't flinched and glanced away in time—so there we were, perched on adjoining barstools with little in common.
We quickly ran out of old friends to mention and teetered on the brink of leaving, but we eventually succumbed to inertia and ordered more drinks. On the third she told me the key to men was that they always wanted sex but rarely had the luxury of expecting propositions. And they were tired of always having to be the ones to ask, she said. From the day they hit puberty they wanted to lay that burden down, so all you had to do, she said, was suggest sex and they would take you up on it. This applied equally with most married men, she said—to be honest, with any of them. Failure was rare, she said, and tipped back her glass all the way.
It was admirable, the ease with which she approached the question. It didn't change my own behavior, however, which in that arena was passive; possibly this was part of why I found myself married to Ned.
In fact, looking back, you could say my passivity in that arena was the start of my greatest failure.
But seeing her unremarkable face in the bar mirror, I felt awed by her attitude, part aggression and part simple confidence. I believed someone should shake her hand or pin a medal on her lapel, but that someone would not be me: for I, even as I was impressed, felt a lucid dislike.
Then the librarian said yes, and I was grateful to the girl from high school.
STILL, THOUGH, EVEN if the bogus exposés and hair-sprayed New Age gurus hawking their bestselling books about past lives had a point, there was no explanation I could find for my having heard the voice. There was no reason I should have had to hear anything at all, if little Lena had contained a reborn soul.
It wasn't as though she herself had spoken, like the little boy with his encyclopedic knowledge of Mosquitos and Messerschmitts. She'd painted no old-fashioned watercolors depicting orphanage memories from 1934.
"I'VE BEEN WONDERING," I said to Don as he drove me back to the motel, his backseat a neat row of paper grocery bags. "I was thinking this place would be quiet over the winter. I don't get the draw for all these people in the off-season. I thought you only ever had a full house in the summer, but now it's almost Christmas. Did you—I mean, just out of—were you planning on all of them arriving?"
Don was silent for a few moments as we ascended the long, slow road that leads up to the bluffs, changing from pavement to gravel as it goes. He reached out a gloved finger and scratched the side of his nose, shrugging lightly as he spun the steering wheel with the other hand.
"I'm trying to help them out," he said.
On the expanse of ground beside the parking lot Lena was playing, wearing her hot-pink earmuffs. She appeared to be piling the previous day's graying snow onto a grim effigy vaguely suggestive of a snowman. Around the dumpy figure was a large impact crater where she'd scraped snow off the dead grass.
Main Linda watched her from the doorway to her room, her hands around a steaming mug, ensconced in a parka with a fur-lined hood like she was Peary at the North Pole.
I realized the light was leaving: a long, knife-thin shadow was falling toward the sea from the dirty pillar of the snowman, which had frayed sticks for hands, pieces of trash stuck on its torso for decoration and what appeared to be a rusty zipper for a mouth.
I didn't like the look of it.
She ran to greet me when I stepped out of Don's car, her nose red and running profusely above her scarf, bundled-up arms flung wide. She's always excited to see me again—though if I'm being honest, as long as she has someone else to talk to, she's almost equally excited to watch me go.
Though the U.S. is an overwhelmingly Christian country . . . 24% of the public overall and 22% of Christians say they believe in reincarnation—that people will be reborn in this world again and again. —Pew Research/www.pewforum.org
AT DINNER in the motel café I took a census of the guests. Lena was making the rounds; having lost interest in her food quickly—for her, food is never the point of a meal—she was stopping at every table, talking to each guest, leaving me alone to watch her progress and consider the obliqueness of Don's answer.
There were Burke and Gabe; there were the Lindas, Main and Big. There was Kay, eating at a table with the angry young mogul who, less shaven every day, was leaning across the table to talk to her confidingly. Before long he'd be sporting a full mountain-man beard. There was Don's father, sharing a large table with Faneesha while Don cooked and the waitress served, and there were the newest guests of all, an arty couple from New York, maybe in their early forties, who had the room right next to Lena's and mine at the far end of the row. They dressed tastefully and didn't seem to talk to anyone.
And then there were the regulars from town, including a woman who dressed in multiple shades of blue and always ordered the chicken pot pie and an old man who, before Don opened the café, had eaten only frozen meals since his wife died, Lena said. But I was interested in the motel guests, the motel guests only and why they were here.
Don couldn't have meant to imply his help consisted of letting friends stay for free—the young mogul needed no such help and the chic couple had arrived in two separate gleaming cars, each of which had to have cost six figures. So that couldn't have been what he meant.
On the other hand Kay was distressed, Burke was distressed, the young mogul was distressed too.
Maybe Don offered some other form of assistance.
IT TOOK ME till this morning to ask the Lindas. I asked while Main Linda was driving me to the auto shop; I asked her with no subterfuge.
"So why are you guys here?"
"My cousin took early retirement after some work-related stress," she said briskly. "Down in Orlando, where she lives. She's on her own, mostly, her ex-husband lives in Vancouver, the sons have grown up and left the nest. I get a long winter break. The two of us have been close since we were ten. I brought her up to make her take a breather."
"But why here?" I asked. "Specifically?"
Main Linda cocked her head.
"Our family used to have a house in the area. Not on the beach, inland. Came up every summer. We shared the place with the cousins. There was a candy store, we walked there every Saturday. Jawbreakers. Gobstoppers. You remember those? Giant round hard candies you could barely fit in your mouth, started out black and you went through all the colors as they shrank? Disgusting actually, kids taking the things out of their mouths all the time to look at the different rainbow hues, then sticking them in again. Filthy. Dyed tongues. Saliva. Yeah, we loved it though. Also, there were those Atomic Fireballs."
"We had those."
"Naming a candy after a nuclear mushroom cloud. Only in America, right?"
"Yeah. What I meant, though, was how did you choose this particular motel?"
"Liked Don from the beginning. And heck, the price was right," said Main Linda. "I'm cheap as crap. Always have been, always will be."
"Good to know," I said, but I was disappointed in my weak powers of detection. People revealed little to me, and I couldn't even tell whether they meant to be evasive or were just uninterested in detail.
Maybe Don opened up his motel to those in need. But why disguise it?
I was at a dead end, I realized, falling silent as I sat in Main Linda's heated passenger seat, and how did you get out of a dead end? You had no choice. All you could do was give up, turn around and drive the other way, drive back where you'd come from.
I mean I don't want to leave the motel or the town, I want to keep my date with the librarian, for instance, a prospect that pleases me out of keeping with its likely outcome. But the sense I have of failing to understand the motel's gathering has started to disrupt my sleep: I lie awake nights distracted by my ongoing failure to grasp why these people are here. Maybe there's nothing to fathom in the first place but maybe there is, and the uncertainty doesn't sit well with me.
And I'm not so sure anymore I need to be hiding us. Increasingly my past interpretations strike me as arbitrary and I pick through them, second-guessing.
There's a chance I could stand up to Ned, I thought, sitting in the car, a chance he couldn't make Lena and me do anything we didn't want to do. Maybe I'm just a coward, I thought, hunkering here, as I was a coward about divorcing him. The line between cowardice and caution was blurred to me.
For a moment, Ned started to look less like a threat than an inconvenience and the future seemed almost simple.
Sitting in Main Linda's car I lapsed into a daydream of peaceful retreat—retreat to my parents' house, their quiet street where snow fell in pristine layers over the lawns. Only the few residents of the block drove down that street in winter, only the neighbors' footsteps marred the sidewalk; the snow lay pure and gently curved on the bushes and old trees of the neat gardens. There would be no cold cement catwalk stretching between the bedroom and dining room, as there was here—no questions to speak of, either, beyond the mundane questions of the design and order of days.
I didn't relish the part where I, fully grown, would be choosing to live in my parents' house again, but they would be good to me and I could help my mother with my father, when she needed me. In that way I could do my part. We would stay there and Lena would go to school; I could get a new job, though I'd long since fallen off the tenure track—a community college might have me, or maybe a private high school. I could almost believe in a return to routine, an end to stealth.
I felt the wings of the normal touch my shoulders, ready to settle on me with a bland, insulating protection. I felt hopeful.
"Here you go, dear," said Main Linda, and I saw we were already at the auto shop. There weren't many cars in the lot: Saturday. "You want me to wait here till you make sure your car's ready?"
"No, that's fine," I said.
"You sure? It's no problem."
"That's OK. I've wasted enough of your time already. He said it was all done. You go ahead, Linda, and thanks so much. I'll see you back at the motel."
I regret those words.
IF I SHOULD DIE BEFORE I WAKE
B.Q. WASN'T IN THE OFFICE; BEEFY JOHN WAS ALONE. HE HUNG UP the phone as I went in stamping slush off my boots, shuffling them back and forth on the black rubber mat and making the electronic doorbell chime.
"Enjoying your weekend?" he asked.
I leaned over and scanned the bill on the counter, trying to pay attention to the line items as he explained what had been wrong with my car's workings.
As usual when a mechanic talks to me I put considerable effort into looking interested, even respectful. I was intent on that effort, though it warred against my instinctive dislike of John, when I detected someone behind me, felt or heard the brush of thick, expensive fabric against itself. I registered that the doorbell hadn't chimed this time and there was a scent, subtle but clear, which I had to identify—much as I wished not to—as a familiar cologne.
Beefy John, still talking about the car, looked steadily over my shoulder; I turned.
"Hey there, honey," said Ned.
THERE WERE THREE thinly padded, black folding chairs along the wall, beside a fake potted plant with dusty leaves. I sat down on one. The fake plant was two times a stand-in, I thought, as a fake plant it stood in for a real one, and then the dust on it, the full neglect, made it seem so purely symbolic that it became an imitation not only of a plant but of an imitation plant.
I wished I could stare at that homely fake plant forever, and never, ever look upon Ned's face.
I was ignoring Beefy John too, or ignoring the blank space left by him, because he must have retreated into the private recesses of the establishment. I felt a vacancy in the space over the counter. Had he given me back my car keys? It was as though I'd lost time, I'd skipped some minutes and found things changed. Instead of looking up I was staring at the fake plant and at myself—but from a great lunar or stellar distance, across a reach of airless space. I might have been a pushpin on a map, a piece on a board game, any tiny, manufactured item on a wide background.
I couldn't choose a direction for my attention. I failed to assimilate.
"Relax, sweetheart, it's all good," said Ned.
His presence and the vapid words were separate—the words, I thought as I gazed at a streak in the plastic leaves' dust, an impressively hollow comfort. In the instant when I turned from the counter I'd caught a flash of his handsome face, enough to register his features; but now I was insanely reluctant to raise my gaze to him again.
It was insane, I realized that—some kind of rapid breakdown. But I couldn't change the angle of my head. I sat heavy in the chair, sack-like. After a minute he lowered himself into a squat in front of me.
And even squatting he stayed graceful, not subordinate the way a squat can make you. I kept my head bowed as long as I could, avoiding the solid offense of his beauty. Before me rose an immaculate camelhair coat, unbuttoned; a well-cut dark-blue suit beneath it, complete with downy-white shirt and silver tie; crisp, businesslike wrinkles on each side of his knees where the cloth was stretched taut. Yes: even the wrinkles in his slacks possessed a symbolic efficiency. They bracketed his sculpted knees concisely, minutely telegraphing competence, even mastery.
I remembered being in bed with him, in bed where he'd always been so perfect that it disguised his lack of emotion. It didn't occur to me to wonder about what wasn't given.
Ned was still exactly the man he intended to be.
Inevitably I found myself looking into his face. He had a light and pleasant tan that must have looked as out of place in the Alaskan winter as it did in Maine. I tried to calm myself by picturing him in a sunbed at Planet Beach, slathering lotion onto his body, arranging the little goggles onto his face. I remembered how the fatless musculature of his torso was maintained with daily bouts of grunting resistance training. But it was no use, no matter how hard I tried to belittle him I couldn't reduce the feeling of beauty and threat he imparted.
Except for the anxiety of his nearness, though, I found I was less susceptible to his looks than I remembered being. I could see him impersonally by placing the barrier of my dislike between us. As I did this, his looks became less the features of a living person and more a formal structure—less animal than mineral, transmuted into a polymer that encased him in its petrochemical sheen.
Had he already sent his guys to the motel? Henchmen, I repeated silently, henchmen, a comical word I'd never thought I'd have a use for. Was Lena already with them? Had her babysitters been pushed aside or persuaded?
I felt a twinge of panic. What should I do? What was the right course of action? Call the Lindas? Don? 911? My cell phone was in my bag, on the counter; there were my car keys beside it. I could grab them and run.
I couldn't decide. I was useless. I tried to stall.
"A suit and tie? On Saturday?"
He smiled at me indulgently, as though what was coming from my mouth was empty breath. There was no need for him to acknowledge my speech.
"Look, honey, you and me just need a little face time. We need to put our two heads together and be reasonable here, figure out what's best for everyone."
"I don't know what you mean."
In fact I did not know what I meant: he was terrifying me. I shook my head. I wasn't in charge of myself, just flustered and stuck. It was exactly what I'd been afraid of since the day he started pursuing us. He'd never laid a finger on me in anger, Ned had never been violent physically. He'd only been false and cold.
Despite this nonviolent history he chilled me to the bone.
"I know you want to come home," he said.
The arrogance of it flummoxed me—as though he was speaking to a third party, a cameraman, maybe, who was watching and evaluating our performances and knew nothing whatsoever about us.
"I don't want to at all," I rushed. "I don't have a home with you and I don't want a home with you. You know what I want, don't you, Ned? I just want a divorce."
"Oh now. Listen. You're getting yourself all in a bunch, aren't you? Relax! We'll go down the street and get a bite. John here tells me y'all have a diner in this town that serves Mexican Coke. All the way up here in the pine-tree state. Go figure. You like that Mexican Co-cola, don't you? Cane sugar, not corn syrup? We need to bring that old-style Coke back to the U. S. of A. I'll put a bill in Congress, on down the road when I get there."
He'd ramped up his Southern accent several notches, the Southern manners of speech he'd partly suppressed in his first flush of adulthood. Maybe he'd raised the good ol' boy quotient for electability—Alaska has a certain kinship with the South, a redneck commonality without the heat or black people. Southern accents may be a bankable asset, I thought. Ned had always considered Alaska a frontier, the main reason he'd asked me to move there in the first place—not that he cared about the wild and scenic aspects, not that he was attracted to the state's unpopulated beauty. It was the mythology of fortune-seeking that he liked, the small but abundant niches in various markets in the state that called to him.
Because while it was true that Alaska had glaciers and polar bears, albeit melting and starving/drowning, it was a frontier in other ways too—a colony still in development, into which, therefore, generous moneys pour from oil companies and Washington. Ned had been right, I guessed, to see his future in a place where men loved both their guns and their government and corporate handouts. He liked the cojones of Alaskans, was what he always said, the way they swaggered like lone cowboys and professed to hate all vestiges of government but at the same time clung fiercely to the coattails of that government—both to their own small government and its big, rich uncle in D.C.
Anyway he'd rediscovered his Southernness. And he was on a first-name basis with Beefy John.
"How'd you know I was here?" I asked.
All of it hung at the margins, all was fuzzy irrelevance except for Lena—where was she, who had her right at this moment? I struggled to think of anything else, stalling until I saw clearly what I should do. I expected a decision to come: presently I would render a decision, a decision would descend and land on me.
I waited for it.
"I make friends easy, honey," Ned said smoothly. "You know me."
" 'Fraid I got to close up, folks," interrupted Beefy John, emerging from the back office, grinning broadly. The pink skin on his nose and cheeks shone under the fluorescents. "Don't keep Saturday hours, normally."
That was how I came to scrape my keys toward me on the counter and follow Ned out into the parking lot. Trudging through the slush I considered the fact that Beefy John had opened the shop on Saturday and then Ned had been there. Conspiracy, I thought, conspiracy, I'd been stalked, I'd been tracked, I hadn't been paranoid at all.
Could Don help me?
I got into my car and of course couldn't stop Ned from following in his own—a rented SUV driven by someone else, some kind of bodyguard or other employee—in a dutiful procession to the diner a block down, a procession that made me feel like a condemned person. The diner served beer and wine, at least . . . and what could Ned do to me there, in broad daylight? I didn't care how early it seemed to be; it was a zero hour for me, the time of reckoning. I had to stay clearheaded for Lena, but also I desperately needed to calm down.
I ordered a beer.
STRANGE THINGS EXIST, astonishing oddities—transparent butterflies, three-foot-wide parasites that look like orange flowers, babies born pregnant with their own twins. There are fish like sea serpents, fifty-five feet long, lizards whose species are all female; there's the mysterious roar from outer space, the contagiousness of yawns, the origin of continental drift.
What I want to know is whether the unknowns in nature are only unexplained phenomena or whether there are genuine anomalies—whether a true anomaly exists. I doubt that it's possible for an event to occur only once, to one person, and as I look and look for an answer the more it seems to me that what are called anomalies aren't unique but only symptoms of gaps in understanding. Some of them are just exceptions to the systems people have invented, showing the limits and biases of those invented systems. Or, in physics or astronomy, anomalies are names for states or forces that haven't been figured out.
It was always improbable that whatever happened, way back then, happened only to Lena and me.
a·nom·a·ly [uh-nom-uh-lee] noun, plural a·nom·a·lies.
1. deviation from the common rule, type, arrangement, or form.
Synonyms: abnormality, exception, peculiarity.
I CAN'T RECALL the pattern of our conversation at the diner. Ned, when he wants to, can have a way of saying nothing specific, conveying only a broad intent. And that intent was exactly what I'd been afraid of: he wanted Lena and me back with him, he wanted us to be his TV family.
His position, as far as I could tell—or his pollster's—was that he was much too good-looking to run as a bachelor or divorced man. And the fact that he was married was already public, so now he had to produce the wife.
No emotions were summoned to build a case for this, no passionate declarations or rhetorical flourishes; Ned simply projected his plan. He has the knack of power, I thought as I drank my beer and picked at the corner of a limp grilled-cheese sandwich, intermittently wiping my fingertips on a napkin. It was undeniable. No wonder he's running for state senate. This first race may be small-time but at some point it'll be a governorship or a senate seat, he's in that forum now, and then probably Congress, just as he'd said.
I wondered how I'd ever become connected with such a man, much less married to him—a person who's mechanistic in his view of others, an individual streamlined to exploit them.
We'd met through a woman I'd only half-liked who had a history at prep schools like Choate and a new, expensive silver-blue car, a brand of car I felt should never be owned by an undergraduate. She'd been a student of mine while I was working toward my PhD, a student in a class I taught as part of my grant package.
This woman had thrown a dinner party the summer I finished grad school, while I was still living in Providence and working as a cashier in a gourmet food store, after the assistant teaching gig had ended. (My family money wasn't given to Solly or me to spend as we saw fit; our parents expected us to work like everyone else. Much of the money Ned took came to us at our wedding, by which time my parents were apparently convinced I wouldn't become a wastrel. Now, of course, I wish they'd never handed it over.) I'd gone to the party because I was lonely and needed to feel like a guest for once instead of a cashier, needed to say something to someone else other than Did you find everything you were looking for today?
Ned was at the party too, Ned the frat-boy Boy Scout, and somehow not a year later I married him. It must have been partly the setting that carried the evening: a rambling green garden with flowers on trellises and weeping willows and ponds arranged around a house with a colonial aspect, columns, wraparound porches, shining wood floors and chandeliers. I'd had no one to talk to there while my hostess was busy; I hovered awkwardly on the porch, looking out at the garden with my wine in hand, till Ned approached.
He'd been washed in those August colors, a borrowed glow that took a long time to fade since, unlike him, I harbored romantic delusions—that pre-nostalgic filmmaking of the self that separates events into vignettes and montages, curates time into a gallery of sepia-toned images. What were the chances of meeting someone like Ned, a man with movie-star charisma at large among the civilians?
Even as inexperienced as I was then, I was foolish to overlook the indicators of his mercenary bent, blind not to notice his edge of narcissism—an edge that was leading. I must have been quite stupid, I reflected, sitting across from him over grilled cheese. The selfish stupidity of youth had been upon me.
For a minute I sat listless, not even attempting to remove myself from his slick enchantment. In one corner of the diner was the man who had to be his driver or bodyguard, with nothing in front of him on his own table but a cell phone and a glass of what looked like iced tea. He wore a wire in one ear like a Secret Service officer.
It was laughable, I thought, to have a man like that working for you when all you were doing was running for a Podunk state senate.
Lena. I knew the Lindas would still be looking after her, as long as they could. For an hour or two they wouldn't even wonder where I was.
"Have you sent someone to find her?" I interrupted Ned after a while.
He was talking about television or radio, a program he'd been on or was going to be on, some anecdote to which I was incapable of listening. I wasn't mentioning the motel, of course, in case he didn't know where she was after all—in case the car-repair place had been his only touchstone.
"Did you send some of your guys over to where we're living? Is that what you did?"
Ned raised one arm for the waitress, who had already fawned over him. She smiled hopefully, her lipstick bright as she rushed over to our booth, and this eager subservience allowed me to see her as he would: a worker bee possessing only the slightest shading of utility.
Still, no being with any utility, however slight, was undeserving of Ned's charm when he was on active duty. He made small talk with the waitress while ignoring my question about our daughter. As he did so I weighed the advantages and disadvantages of running outside and jumping in my car and I decided that, on balance, I had little to lose. I had to get back to Lena anyway, sooner or later I would have to go to her and inevitably, if he hadn't already found out where we were living, lead him there. I was impatient to be with her again, to see her and be near.
And so, abruptly—while Ned was holding the middle-aged waitress in thrall to his shining attention and I was hearing her say she'd been married to three different members of the same MC—I rose and hurried out the door, not looking behind me.
There was no flurry of activity back there, Ned didn't ever tend to exhibit undue haste, but still it hadn't been two minutes before I could see his rented car in my rearview mirror.
I let out a breath I hadn't known I was holding: a childish part of me had hoped to lose him by bolting, though realistically I knew better.
I DROVE TO the motel with mounting panic, knowing it wasn't the best move. But I had to be with her. I talked to myself as I drove, tapping the steering wheel restlessly at the lone stoplight between the town and the motel. Of course he can't take her from me, with all his concern about public relations. Calm down. Calm down, calm down, calm down.
Glass half-full, I said to myself, now you have to face up to the situation, iron it out. Maybe Ned's not dead wrong after all, there's no need to hyperventilate—be practical. Next steps. He said it himself, we just need to sit down and figure out what's best for all of us. I agree for the most part, I told myself, nodding as I pressed down on the gas pedal again. For the most part I agree, right? We need to figure out what's best for all of us.
Except him.
Ned's election to a position of state power was what he wanted, but it wasn't what I wanted—I felt it was against the interests of many, indeed most. It's actually my obligation, I thought, not only to think of Lena and myself but also of how not to get Ned elected. He relies on an implicit system of beliefs I think are cold as ice, a system of assumptions more than beliefs that has nothing to do with either reason or kindness. Ned's beliefs are like the programmed responses of a computer, I thought, they require no justification, in his view, beyond the fact that he has chosen to embrace them.
Maybe I could accomplish all these goals at once, protect my daughter and myself, try to weigh in against my husband's election: File for divorce on grounds of adultery, as a spurned wife would on TV.
But now the motel sign was up ahead of me, here came the parking lot, and I felt despairing. I'd never gathered evidence while we still lived together because once I knew the marriage was lost I assumed Ned wanted out of it too. So I had no proof of his many affairs. Most likely he was certain of this. I'd known some of their names and faces, but he would have covered his bases and I couldn't believe those women would help me. Of the two of us, Ned is by far the more persuasive. And—except for the one instance I knew of when a woman broke it off with him—he tended to let them down easy. He wasn't a bridge-burner; on the contrary. Even the one who'd been disgruntled had to be in his pocket now.
Don, I thought again. Could Don help me?
I didn't want Lena to see her father yet; I wanted to prepare her. I didn't know what to do. He'd park as soon as I did, he'd be right behind me.
She wasn't in front, anyway, wasn't playing in the snow this time, though her snow effigy remained, lumpish, melting. She might be in the Lindas' room, I figured. Maybe they would help me. Although—what could they even do? Ned wasn't a wife-beater. Ned wasn't a clear and present danger. Ned wore a camelhair coat and shone like the noonday sun.
I couldn't sit in the car thinking, I had to press forward. I'd call and tell them to keep her in the room—so I ran to the lobby, Ned's car somewhere behind me, headed up that long gravel road. I ran to the lobby, but Don wasn't there: the front desk was unattended. I looked behind me, out the glass door, then ducked into the café room and closed its door. It was empty. I took out my phone and dialed Main Linda's cell, butterfingers. I got her outgoing message and left a voicemail. I asked her to keep Lena in her room, not to come out until I called again, could she please do me this favor? Please?
I hung up, still trembling.
Lena could be anywhere, exposed . . . I'd go around the back, look in the picture windows . . . what if the Lindas didn't have their cells with them? I snuck back into the kitchen looking for a back door: EMERGENCY EXIT ONLY, with a metal bar. I pressed the bar and it ka-thunked, no alarm. Then I was outside, crunching along the dead grass and snow behind the building, along the rear windows of the rooms.
But all the curtains were closed.
When I turned the corner of the building I saw Lena walking beside Big Linda, wearing her pink puffer coat—they'd just come up from the beach, because Lena was carrying her basket. And a few feet away from her, leaning relaxedly back against the hood of his parked, black SUV—there stood Ned.
He held a large, gift-wrapped box topped with an explosion of professional ribbons. The wrapping was covered in silver glitter and festooned with candy canes.
"Baby girl," he said, and the teeth had never been whiter in his head.
I TRIED TO appear gracious after that, to the extent I could—that was my tactic, for lack of better. I pretended calm as I reintroduced Lena to her father, then introduced him to Big Linda and Main Linda when she, too, appeared huffing and puffing at the top of the staircase down the cliff. Lena did remember him from two years before, though she'd been four when we left, but she didn't greet him with the exuberance she'd shown her grandparents. She gave him a restrained embrace, clearly struggling to understand his sudden presence in our midst.
"A surprise visit," I said, trying to deliver a cheerful smile.
"You're so big," Ned said to her, and to the Lindas, his Southern drawl in full effect: "She's like a little doll version of her beautiful mama! Isn't she?"
An off-base gambit, since Lena's skin is lighter than mine, her hair gold instead of brown; in fact she looks more like Ned. She didn't preen under this particular praise either, just waited patiently.
"She does have those high cheekbones," said Main Linda politely.
I could tell the Lindas were wary of Ned and felt a rush of gratitude for that.
"Why don't we go inside?" said Ned, looking from me to Lena. "Chilly out here, idn't it? And you can open your present, honeypie! I bet you'll like it a whole lot."
I didn't see a choice: it was cold, and getting colder all the time. The damage was done: he already knew where we lived.
"You take this?" he asked, and handed me the unwieldy gift before I could answer. He reached down and grabbed one of Lena's hands, forcing her to struggle with the basket and have to kneel down to pick up fallen shells. The three of us began walking, me lagging beside them, hesitant, Ned moving slowly because, I guess, he didn't know where our room was. After a moment I turned around. The Lindas hadn't moved much; they were watching. I couldn't read their expressions.
"Linda, could you mention to Don that my husband is here?"
It was all I could do. Don was the only one who'd know what it meant to me that Ned had found us.
As we made our way along the walkway to the room Lena began to chatter, as she would with any new guest, telling Ned how the motel worked: how towels and clean linens were organized, that she knew how to slide the keycard into the slot herself. I was following them by then, looking at Ned's back, Lena's face turned sideways to him, and trying to figure out what it meant to her to be holding her father's hand.
"I'll show you, see?" she said, and slid her hand out of Ned's to turn and hold it out to me. "Can I have the key, Mommy?"
Duly I handed it over, circling the gaudy gift with one arm while I rummaged in a pocket. I was too aware of Ned looming, his pheromones, or whatever the fuck, casting over me a vibrant net.
Lena clicked the door open, proud of her competence.
"Whoa," said Ned, when he stepped inside. "Not exactly the Ritz, is it. You can do better, can't you, Anna?"
"Ritz like crackers?" asked Lena.
"It's grown on us," I said lightly.
Ned tried to shut the door after I brushed in past him, but I propped it a few inches open with the rubber wedge.
"Let's let in some fresh air," I said.
"Fresh freezing air," said Ned.
"How come it's a Christmas present?" asked Lena, as I set the box down on our small table. "It isn't Christmas yet."
"You know that song, baby?" said Ned, sitting himself in the armchair with magisterial ease, crossing his legs. "'The Twelve Days of Christmas'? On the first day of Christmas, my true love gave to me—you know that one?"
"A partridge in a pear tree," said Lena.
"That's right! Smart girl. But don't worry, this isn't a partridge."
Lena approached the box shyly at first, then began to rip it open chaotically as I started to pour water into the carafe for the in-room coffee. Coffee didn't appeal to me in the least, especially not from that little plastic-wrapped packet. But ours was a small room with not many options for looking busy.
"Look, a new friend for Lucky Duck," said Lena, pulling out a fluffy white sheep. "Is it . . . a goat, Mommy?"
She'd turned to me to ask, instead of asking Ned.
"It's a lamb, baby," he said. "And it's made from real sheepskin."
Lena was instantly upset. Ned couldn't have known it was a misstep since he knew nothing of what she ate and didn't eat, of her softheartedness.
She blinked away tears and said nothing, holding the sheep at arm's length.
"Go on, give it a squeeze," said Ned.
Reluctantly she did so, first one way and then another, until the lamb began reciting, in a high-pitched, childish voice, "Now I lay me down to sleep / I pray the Lord my soul to keep / If I should die before I wake / I pray the Lord my soul to take."
"Hey look, Ducky," she said, gamely trying to make the best of a sheepskin tragedy. She picked up her ratty, baggy duck from the bed and pressed the two stuffed animals together. The duck was a dingy gray compared to the snow-white, fleecy lamb. "Be nice to her, Ducky. Her skin got cut off her."
Ned raised an eyebrow.
"I'm not going to make her talk that much, OK?" she asked Ned. "It's babyish. And I don't like what they made her say."
"Bit morbid, isn't it?" I said to Ned. "I've always thought that prayer was cloying."
"Well, that's OK, sweetie," said Ned to Lena, ignoring me. He didn't look pleased, though, which made my spirits lift briefly, then just as soon worried me. "It's yours. You do whatever you want with it. But listen, you didn't read my card yet. I wrote it for you special."
"I can read it. I can read a whole book," said Lena.
Ned plucked a card from beneath the efflorescence of ribbon.
"Will you read it out loud to me, baby doll?" he asked.
He'd already achieved a proprietorial air with Lena, an air of ownership.
She took the card out of its envelope, revealing an airbrushed-looking kitten with eyes the size of saucers.
"Dear Lena, I—missed you—very much," read Lena. "To my best girl ever, love x's and o's Daddy."
I was nervous that Ned was right on the edge of saying something to her I didn't wish him to say.
"That's really nice," I said. "Lena, your father's here for a quick visit." I kept my voice easy as the coffeemaker started to burble. "I'm sure he won't be able to stay for long. It's so nice he brought you the lamb, isn't it?"
"OK. Want me to read to you?" Lena asked him. She'd had a breakthrough in her reading and liked to perform her favorite book. "Want me to read Ferdinand?"
Ned arranged her atop his lap for the purpose, flicking on the lamp beside him. They made a Norman Rockwell picture sitting there, their hair burnished the same shade of gold—you'd think the man cherished the little girl deeply, looking over the top of her small head at the open book with his eyes down, his handsome, almost noble features and form arranged in a cast of paternal protection.
You'd think that unless you were me—or unless, maybe, you caught sight of one of his elegant feet jiggling minutely but rapidly under the armchair as he pretended to listen. He hadn't seen his daughter for years, but there was his foot, shod tastefully in black leather, already impatient.
Lena read slowly and haltingly about the peaceful bull who only wanted to sit and smell the flowers, not travel to the big city and fight the toreador. But the men from the city came and took him away, forcing him into the bullring.
Ned took a leisurely glance at his watch and smiled when he saw me seeing him do it.
Would he go away soon? Please? I couldn't even make a trip to the bathroom while he was here with her, I'd never leave them alone. What was his plan?
When she was done she jumped off his lap and scurried to the bathroom herself, announcing she was going to pee. Ned picked the storybook off his knees as though it was soiled, with two fingers, and deposited it on the table. Then he brushed off his slacks where she'd been sitting.
"That bull was light in the loafers," he said.
"Ned. What are you doing right now?" I kept my voice low. "We're not going to come back to you."
"How 'bout a compromise?"
He pointed at the coffeemaker, meaning Give me a cup. I turned, feeling cold, and started to pour one. It was better than looking at him.
"I propose this, darlin'. Some photo shoots, interviews. Couple appearances. Then y'all can take a vacation. I'll only need you now and then. It doesn't have to be 24-7, if we manage it right. Anchorage is a big enough city."
"But I don't want to support your campaign, Ned." I handed him the cup. "I don't like what you stand for."
"We may have policy differences here or there," he said, shrugging, and sipped. "Now, that's just foul."
"It's really not good," I agreed.
He set down the bad coffee on the storybook. It slopped out and made a ring; I grabbed the book and wiped it.
"Bottom line is, we're family."
"As it turns out, that's not my bottom line at all."
Then Lena was out of the bathroom again, looking at us expectantly.
"Why don't you go play, sweetie?" he said, barely glancing her way. "Let the grownups talk."
"She doesn't play outside by herself," I said. "The motel's on the edge of a cliff."
He slipped his phone from a coat pocket.
"My driver can babysit."
"No thanks," I said firmly. "We don't know him."
"Hello?"
It was Don, knocking at the cracked-open door with perfect timing.
"Come in!" I said, relieved.
He stepped inside, nodded curtly at Ned without smiles or introductions, and held his hand out to Lena.
"I've got a job for you," he said. "You want to help me?"
"I'm the assistant!" crowed Lena.
And Don towed her efficiently out of the room.
I was so grateful to see her go that I felt my shoulders unclench.
"Look, I'm not asking you to give any stump speeches, honeypie," said Ned, stretching out a hand and pushing the door closed behind them. "You don't have to say a word. You can be deaf, dumb and blind. Hell, I like you better that way. Just smile and hold my hand sometimes. And get the girl to do the same. You soldier through till the election, smiling all the time, I'll give you a friendly, neat divorce as your very own victory gift. Plus full custody. With visitation rights, of course. Couldn't be looking like a deadbeat dad."
"And you'd actually put that in writing. Before the fact."
"All official. With confidentiality agreements on the timing and conditions there, of course."
"Even if you lose? You'd sign off beforehand on it, no matter how the election goes?"
"I won't lose. Not with the friends I have and your two pretty faces beside me. But sure, I'll sign."
"Because I know you want more than the state senate. Won't you want a wife and kid when you run for something bigger, too?"
"I'll cross that bridge. Let me worry."
I was asking questions, but I wasn't seriously considering the request.
"Don't you think I could get sole custody now?" I said. "I mean Ned. You've come to one of her birthday parties. Ever. And that was by accident, if I remember correctly."
"You might could get custody," said Ned, and smiled again. "But maybe not. Running off like you did."
"You wouldn't want that fight," I said. "Publicly. You'd never want it. Especially not now."
"You'd be amazed how I can spin things, when I need to. I might decide to play the victim. People do love their victims, in America."
We gazed at each other across the room. That is, I looked past Ned, not wanting to look at him, so I don't know if he really looked at me either. I tried to remember another time he'd been so direct, and all I could come up with was when he asked me to get married. It had been at a restaurant with white tablecloths and obsequious waiters—he likes being served by such waiters and I hate it. When waiters are too fawning I hear the falseness they've brought to it, possible snide remarks in the kitchen.
Now he was relaxed in the chair, facing me, while I was in a defensive posture, backed up against the counter of our kitchenette as far from him as I could be. My hands were braced against the edge.
"I need time to think," I said. "And while I think, I need you to not be here. And not spend time with Lena, either."
He shrugged. "The clock's ticking."
"Why? Isn't the election a whole year away?"
"Primary's in August. My party controls the governor's office and the House; the Senate's a 10–10 split, but with redistricting we could take over there too, come November. We've been low-key till now, but it's time for a higher gear."
"You're not going to start campaigning before Christmas, are you?"
He picked up Lena's Lucky Duck from where it lay, studied it for a few moments, and then dropped it.
"Getting my ducks in a nice little row."
There was a knock on the door, so I crossed the room and opened it. The Lindas stood there, smiling pleasantly, waiting. Ned rose from his chair and smiled too, at them first, then at me.
"Well, got to be getting back," he said. "You mull it over, honey. So great to see my girls again. Ladies? A pleasure."
The Lindas moved aside for him, and just like that he was gone.
I DON'T HAVE confidence we can run away again. For one thing it would clearly look illegal, now that he's sought us out. And for another he's obviously better at stealth than I am, and he does have friends. Whether Beefy John tipped him off or was only a witness, he has sources of information and I'm clearly not equipped to detect them.
The Lindas told me Lena was helping Don in the café; they sat and listened while I explained. I told them what my position was; they were sympathetic. And I didn't have to persuade them Ned wasn't the charmer people always think he is—maybe, as post-reproductive women, they were outside the field of his pheromones.
Almost as soon as Ned was gone the guests seemed to come out of the woodwork: the motel returned to life, with movement and light in the rooms, people talking and walking between them, breath visible in the cold. Don brought Lena back, and Kay and Burke were with them and made remarks about Ned's shining car, his bodyguard/driver, his tailored coat and even the lamb, which lay abandoned in a corner of the room atop its pile of bright wrapping.
Laughter and conversation echoed from the walkway into our room. The day had passed quickly; before I knew it late afternoon was casting its long shadows.
Burke stayed a while after Don and Kay left, helping Lena tend to her bean plants in the miniature greenhouse. Some of them had sprouted; one was growing fast, already too tall for the container, and this they moved into a small pot he'd brought with him.
Eventually he got up to go and I thanked him for coming by, for all he did for Lena. As he was going out the door he turned and looked at me.
"You know, we have to look after each other," he said quietly. "The people who've heard it."
HURT, YOU WERE A CHILD AGAIN
I DIDN'T STOP BURKE FROM LEAVING, DIDN'T DO ANYTHING BUT watch as he headed off down the walkway. When he stepped into his own room I closed the door without noise and sat down on the bed.
Lena had her sheep on her lap and had found a buttoned opening in its stomach. Out of the opening, while I sat looking at her in a daze, she pulled a white-plastic box.
"That's how she talks," she said, and pushed a large, flat button on the box, which obligingly bleated out its eerie, falsetto prayer. "See? When you press the tummy she talks. It's for babies. Mommy. I'm six. Can I throw away the talking part?"
"Of course," I said feebly.
The strength had been pressed out of me; I was breathless and flat.
She turned a small screw neatly with her fingernail, impressing me, and extracted two batteries, which she placed neatly on her bedside table. She marched over to the trash can and dumped the box without ceremony.
"It's not the lamb's fault," she said. "When she talks it makes me think how they took off her skin."
"Oh, honey," I said, reaching. "Don't worry about that. OK? It's sheepskin. No reason to think it's from a baby. Maybe that sheep lived a long and happy life. Maybe it died of old age."
"Maybe," said Lena doubtfully.
"Can I see her for a minute?" I asked. It was occurring to me that the lamb could be a nanny cam, hold some kind of tracker. I'd been paranoid, this was paranoid, but then again in broad strokes I'd also been correct.
I held it and stared into its glass eyes, squeezed the face, inspected the nose and mouth.
With Lena in front of the TV I poured myself the glass of wine I'd been wanting. The people who've heard it, I thought. It had to mean what I thought it meant. So this wasn't a random selection of winter travelers in Maine.
It was an enclave.
But I'd never told anyone about the voice—no one. That was what made my hands shake as I drank my wine.
"I'm going to take a bath, honey," I told her, and carried my glass into the bathroom with me, leaving the door open. I thought the soak might calm me.
I'd have to ask Don, I thought as the water ran, it was the only course of action, I'd ask him now, and this time he'd have to tell me. Or I'd ask Burke how we came to be here, how it was that someone had known and how they'd summoned me, if that was what had happened.
Probably the voice wasn't anything supernatural, you credulous primitive, I thought. I sat there in the hot water and finally leaned out to set the empty goblet on the floor, heard the slight scratch of its circular base on the tile.
Probably it was sound waves, radio waves, technology: that was the best idea I'd had. I'd been so childish to think of magic when it was likely the product of science—some manipulative brainchild of one of these peripatetic characters.
Maybe it had been one of them all along.
I ALMOST FORGOT Ned that evening, preoccupied by what Burke had said. I debated whether to go to dinner and face that crowd. We could always make food in our kitchenette or even drive to town.
But Lena wanted to go because another child was coming, the boy with the robot. She knew this and planned to sit with him. I was worried about the emotional effects of Ned's sudden appearance, although she seemed to have taken it in stride. I wanted to watch her closely and give her the small assurances she asked for, so I said yes.
And when we entered the café it felt homey. We sat down with the little boy's family, at their invitation, and as I exchanged small talk with his parents I studied my fellow guests, wondering who among them was in Burke's club and who was not. The Lindas? The chic couple? Kay? The angry young mogul?
The mogul, yes. I'd heard him on the telephone that night, yelling; and now I thought, That's what it was about. He'd told someone what he'd heard, the person on the other end. I watched him and Kay at their table alongside the wall, leaning close as they confided in each other. Maybe they were discussing it right now, I thought.
The mogul's name was Navid, Kay had told me. It meant good news.
And Kay: Kay with the babies at the NICU. Had she heard it from one of them?
I'd accepted the voice, then gratefully dismissed it when it ceased. Once it had loosed me from my moorings so that I had to tread water in a fluid world; finally, when it fell silent, I'd stepped onto solid ground again. But now there was a new unknown, of how and why I'd got to the motel and how the others had, and the earth was shifting beneath my feet again. How much I hated that jarring movement, the rush of fear! I'd tucked it all behind me and moved on; I'd adapted to it as best I could and concentrated on bringing up my girl.
Surely there was nothing else I could have done.
IT HAPPENED THAT I didn't have to buttonhole Don. With his customary placidness he stopped by our table. The family from town had left and Lena was picking at a berry cobbler. He had a tray of cobbler dishes in his hands, which he set down on the table next to us before he placed his hand on the back of my chair; I studied the waves of whipped cream on top of the pie.
Don's friendly, familiar slump suggested nothing too significant was happening; and yet he knew.
"The others found us through a website," he said. "Call it a support group."
"But I didn't," I said.
Lena wasn't listening but waving her spoon and making faces at Faneesha, who sat across the room making them back at her. I thought of the Hearing Voices Movement; I thought of support groups in general, and how I'd never been drawn to them.
"Well, you needed something else," said Don. "You recovered and they're still struggling. You needed a different kind of assistance."
"That's true," I said. "Thank you so much for today. Your timing was perfect."
"No trouble," said Don. "But we're still worried about you."
"What you just said, though, it doesn't explain how I knew where to come."
"You could think of it like salmon," he said, cocking his head. "Or migrating birds. They know where to go, but no one really knows how they know."
"Ducks fly south in winter," said Lena, who'd put down her fork. She had no idea what we were discussing, but lack of context has never stopped her.
"That's right, Lena," said Don solemnly. "That they do."
"Except Lucky Duck," said Lena. She patted him on the chair next to her. "This guy's lazy."
"But ducks and geese and salmon migrate in groups," I said to Don. "They have other ducks and salmon."
"Mostly. But not always," said Don lightly. "Individuals of many species engage in solitary migrations. Humpback whales, for instance. Young songbirds often make their first trips alone. Scientists say direction and distance are written into their genes."
"They travel for food or breeding, don't they?" I said. "But I didn't travel for those reasons." Because Lena was there, I couldn't be more specific and I wanted to keep it casual.
"Well, I don't know about that," said Don, and took a bowl off his tray before he picked it up to move on. "Have some cobbler. It's on the house."
BACK IN the room I went online briefly.
In some butterfly species, for example the monarch, no single individual completes a migratory journey, which is spread over a number of generations. Instead the animals reproduce and die while underway, and it is left to the next generation to complete the next leg of the journey. —Wikipedia 2015
"Are you mad at Don, Mama?" asked Lena when I was putting her to bed.
She clutched both the duck and the sheep.
"What? No, I'm not mad at him," I said.
"Don's too nice to be mad at."
"Don's definitely nice," I said. "And we're getting to know him better, aren't we."
"People ask questions to know each other better," she said.
"Exactly."
"Are you mad at my father?"
"Hmm. Well, that's a good question."
"You don't like him."
"I wouldn't put it that way."
"How would you put it, then."
At that moment she sounded over forty.
"I'd say . . . well, I'd say we turned out not to have as much in common as I first thought we did."
"I don't know if I like him either. I love him, because everyone loves their father."
"Right. Of course you do."
"We used to live with him."
"Yes. We certainly did."
"He never gave me a present before. Even though it's not Christmas. Did he give me a present ever before this?"
"Hmm. He must have, mustn't he?"
"I like my sheep."
"That's good. It's a nice sheep."
"I like living here. With you and me."
"I know you do. I do too."
"We live at Don's motel."
"For now. But not forever, sweetie. You know that."
"I know. One day we have to go. That's why they call it a motel. It's not a house or apartment."
"No."
"One day we have to live in one of those."
"I expect so. We'll have neighbors, I bet. You'll like that, too."
"OK. I'm going to go to sleep now."
"I'm glad to hear it."
I COULDN'T SLEEP, so I wrote down that exchange, figuring it might give me needed insight, further on, into my failings as a parent.
Then I consoled myself by thinking that at least I was a good enough parent to try to keep account of those failings.
I lay in the other bed, letting the TV play muted in front of me, laptop on my knees. Don had to be some kind of counselor, some kind of advisor to those who'd heard . . . but now that I wasn't the only one who spoke of "hearing," the word seemed cultish to me and I didn't like it, not at all. The word hearing had an unpleasant ring suddenly—now it was a matter for shame, almost, rather than one of the senses—and "the voice" wasn't the plain and straightforward moniker I'd taken it for but a worshipful honorific.
Now it was the Voice.
I wondered if what the other guests had heard was different from what I had—assuming it wasn't just Burke, of course, assuming he spoke for more of them. Not all of the guests had babies, in fact none of them did. As far as I knew, only Kay had necessarily had regular contact with infants. So maybe they'd encountered it, as she had, in the infants of others.
I went over the guest roster, as on TV a pretty woman was murdered with a knife. I knew the voice's life cycle, or I thought I had. But I knew nothing. You don't even remember how this supposed knowledge came to you, I told myself—it was never spelled out. If the voice had brought me here, how? What had driven us from old friends' welcoming houses to these Maine bluffs, with this peculiar group?
Maybe Don was onto something, maybe the migration was encoded in my genes.
Many mechanisms have been proposed for animal navigation: there is evidence for a number of them, including orientation by the sun, orientation by the stars and by polarized light, magnetoception, and other senses such as echolocation and hydrodynamic reception . . . investigators have often been forced to discard the simplest hypotheses. —Wikipedia 2015
How could the other guests have heard the voice? I tried to recall exactly when they had seemed upset. Burke was the only one who'd showed emotion to me, aside from the angry young man on his cell phone and Kay talking about the NICU.
I picked up my computer and scrolled back in this document to what I'd written about Burke. Talking to Lena about giants and beanstalks: that was when he'd lost it. And now I saw it, and it was obvious. His dismay had been brought on by something he himself had said, that Lena didn't have to worry about giants saying "Fee, fi, fo, fum" from beanstalks—a voice, talking down from the clouds.
There was my evidence, right there.
I heard a text alert on my cell phone and rose from the bed to fumble in my bag. I missed you tonight, it read. And then another: Did I get the date wrong?
I'd entered his name and number into my contacts list, and there it was: Will Garza. I'd forgotten my first date in almost a decade.
I apologized in a low voice, with the door to the bathroom closed so that I wouldn't wake Lena, and found myself relaxing as I listened to his deep and pleasant tone. I talked a bit about Lena, for whom he'd once suggested a book about a donkey named Sylvester who found a wishing pebble and got turned into a boulder. She liked it almost as much as Ferdinand. I told him my husband had followed us here. I told him almost every material fact about our situation, leaving out the part where I used to hear a voice.
He said he had never been married, that he had most often lived alone, that he preferred books to people. His parents had been from Argentina but he had grown up in New York before he moved to Maine and had relocated here when the rest of the family had returned to Argentina. They ran a small bakery there, and his father cultivated oak trees.
But he'd stayed here because this, he said, for better or worse, felt more like his country.
His given name was Guillermo but he'd always gone by the shortest Anglicized version, Will, not liking the initials G.G. as a boy and living among Anglos. He used to be a feral librarian, he said, before he went back to school.
That was what they called them, he said, librarians without a master's degree.
Olfactory cues may be important for salmon, which return from the ocean to spawn and die in the very streams where they hatched. Some scholars believe they use their magnetic sense to navigate within reach of the stream, then their sense of smell to identify the river at close range. —Wikipedia 2015
THIS MORNING I woke up simple-minded, as though a dream had narrowed my focus. I had to ask Don the question, the large question was all I was interested in, and now I would take him by the shoulders and shake him and ask it. Don! Don! Don! Who was it? Who was speaking to us?
But the urge passed. I guess I couldn't handle an answer, an answer would be too unsettling. I don't want to be part of some enclave of believers, some marginal sect. I've always avoided joining. I don't even have one of those plastic grocery-store cards that make the food cheaper. I haven't enrolled in any frequent flyer programs; though I can't fix a flat tire I've never paid dues to Triple A; even my friend's book club in East Anchorage, which mostly involved eating and drinking, was of little interest to me.
When I was alone I could accept, with difficulty, having heard what I heard, but to find myself among others who might confess and describe it, impute their own meanings—it makes me claustrophobic. And who is Don, even, to hand down high knowledge to me? I like him, I do, but when it comes to the greatest mystery of my life I have no reason to privilege a motel owner's beliefs over my own.
I do want to ask why, if several of them are in on this, they hid it from me until now. Why didn't they let me in before, if this is why we're here? When I asked how they came to be at this motel, why didn't anyone answer me?
Ned called and I let it go to voicemail, to which I listened promptly. He said he needed a decision, and followed this with an amicably phrased threat to show up again if he didn't hear from me right away. He's always been restless; after all, it's why he married me.
I struggled under the pressure of his impatience, trying to shrug it off as I made toast and spooned out yogurt for Lena. I wondered if I could put him off. I wasn't ready to see him again so soon, much less decide my course of action. This might be a subject, I decided, I could safely broach with Don, possibly Don would have some solid counsel for me, with his background defending wives under duress.
I'd table the other conversation for now, I'd focus my energies on fending off Ned.
So I called the Lindas, who like any excuse to go for a walk, and asked if one of them had time to take Lena down to the beach. I called the front desk and asked Don if he could meet to discuss my situation. I still needed help, I said.
I met him in his back office.
"You have a few options, as I see it," he said. "One, you can leave the country. But that wouldn't be wise, legally. Two, you can hide somewhere, the way you've been doing, but better. On that choice I could help with logistics. But that's complicated legally too, since you're not alleging abuse. He could use it against you, certainly. Three, you simply file for divorce now. Maybe he makes good on his threat, maybe he doesn't. He could be bluffing."
He stopped.
"That's all?" I said.
"Or four, you can do what he says. Sign the papers first, with your lawyer, and then do what he wants you to."
"Isn't there a five?"
"I don't trust him," said Don. "Four's a more dangerous option than it may seem."
"But so is three," I said. "He could try to get custody. Having her with me trumps everything."
"I know."
Don studied me, waiting.
I CALLED A COUPLE of friends, pacing my room while Lena and the Lindas wandered up and down the beach. You shouldn't be rushed in this decision, they said, tell him you need a week. They were kind, but their support didn't help me, beyond the warmth of reassurance.
It seemed to me I had weak information about my choices, so I made more calls. I asked Don for a family lawyer's number, he had a personal friend who would take my call even today, he said, so back in my room used his name to get her on the phone. But she didn't tell me much more than I already knew, and while I was half-listening another call came in—Ned's voicemail had said he'd love to have lunch with "his girls," it'd be no problem for him to "swing by."
When I called him back my call went to his voicemail, which pleased me. I said I'd need till Tuesday, but don't come for lunch. Ease up.
I was wary of calling a lawyer in Anchorage. Ned knew so many people in the city that I couldn't be sure of steering clear of his contacts or friends. When I thought of lawyers there I saw two faces of lawyers he'd slept with, a young blonde and a middle-aged hardbody who ran marathons. A few other lawyers were investors of his. But Juneau, at least, wasn't his territory yet—maybe I could find a lawyer there, one who wasn't beholden to him. So when Lena came back from her walk I assigned her some reading and scanned search results.
Then I remembered Will Garza. He was intelligent, I thought, and kind and easy to talk to. I let Lena watch television, since it generates more background noise than reading, and stepped outside to make my call. We barely knew each other, of course, so I hadn't asked anything of him. But now it struck me that maybe I could ask his advice, and he, unlike my distant friends, was here.
We decided to meet; it needed to be someplace warm, someplace Lena would play hard and ignore us. Will remembered an outlet mall in the hinterlands, a mall with an indoor playground you paid for. It sounded to me like the worst place in the world for a first date, but I needed someone to talk to more than I needed to set a scene, at that moment, and I said yes.
The place was full of inflated slides and bouncy houses, with tinny pop music playing and bright lights shining and the red, blue and yellow decor of fast-food restaurants and clowns; it smelled like sweat and dirty socks and off-gassing vinyl. For me there was nothing to like, for Lena there was everything. She'd put her shoes in a cubby before I finished paying and was off climbing, running between the machines, making friends: not two minutes had passed before she was holding hands with an older girl as they tumbled down a wide blue slide.
I sat self-consciously under the fluorescents on a sticky chair and waited, following Lena with my eyes as she pulled the older girl from one puffy structure to the next. I wondered if my face was clean but was too self-conscious to check it in the cell phone's camera. I'd seen that a few times: people trying to look as though they were doing something else on their phones when it was clear from the angle of their head, sometimes a set of pursed lips or a hair toss, that they were studying their faces.
Then he got there, carrying a cardboard tray of drinks—a hot chocolate for Lena and a coffee for me, which he handed over without saying anything. There was a little milk in the fourth cup, he said, did I take my coffee with milk?
He brought with him a microclimate of calm. I was drawn to it, his warm calmness that set the stage for trust.
African ball-rolling dung beetles exploit the sun, the moon, and the celestial polarization pattern to move along straight paths, away from the intense competition at the dung pile . . . this finding represents the first convincing demonstration for the use of the starry sky for orientation in insects and provides the first documented use of the Milky Way for orientation in the animal kingdom. —Abstract, "Dung Beetles Use the Milky Way for Orientation," Dack, Baird et al., Current Biology, Volume 23, Issue 4
I'VE DECIDED TO call Ned's bluff, though I have no idea how he'll react. I'm afraid, but I took a couple of Valiums, dredged up from the bottom of my cosmetics bag, and thought of how he's never had a genuine wish to be in the same room with Lena. He's never wanted to be near her, listen to her, keep her safe—never.
It makes me angry to think of this, makes me feel a burning anger. Remembering his disinterest I can't believe a court will ever side with him when it comes to my little girl, I can't believe it's a realistic possibility. Even if his constituency were to believe him, I think, even if he did successfully paint himself as a victim in the eyes of electors, surely a court would not, I tell myself.
So while Don's family lawyer made up the papers—I could file from Maine, as it turned out—I stalled, putting off Ned's voicemails and texts with short texts of my own. I'd tell him on Tuesday, not a minute before I'd said I would, and meanwhile Lena and I spent time with the others at the motel and with Will; we weren't alone much. We kept busy, went to a movie in the afternoon, to dinner in the motel café.
I envisioned a hard, bad conversation with Ned when the deadline came. Because of that I was constantly nervous, I almost trembled with a brimming anxiety. I picked at my food, I tried to keep busy so that I didn't have time to succumb to fear, and on Sunday night I could barely sleep.
I had dreams of small, furry dogs being mauled by something they couldn't see.
Don suggested Lena and I could switch rooms and stay right off the lobby. We could trade with Burke and Gabe, there was no difference between our room and theirs except for location, and that way we'd be near Don—near help, in other words, in the event that Ned started banging on our door Tuesday night. Lena was jubilant, when we told her about the change, at the thought of trying a new room—it might as well have been a trip to Disneyland. She fantasized about trying all the rooms, one at a time. "Then Don's room, then Kay's room, then the Lindas' . . ."
We would move Tuesday morning, before I called Ned and told him I was filing for divorce; by Monday night, on the momentum of Lena's excitement, we had our small bags neatly packed and waiting just inside the door.
But we didn't move to Gabe and Burke's room the next day, because I woke up Tuesday morning and Lena was gone.
IT WOULD BE futile to try to evoke the desperation I felt when I saw she wasn't there.
My head was pounding—I'd been drugged—sharp pains like nails or tacks in my temples. Still that was nothing to what I felt, nothing, and I picked up the phone as soon as I saw her empty bed, the wrinkled sheets, as soon as I called her name and got silence, and then I sat up and saw her suitcase was gone too, Lucky Duck, her puffer coat. The chain on the door hung in two pieces.
All this took five seconds—less.
And then I was standing and running to the door, I was throwing it open and running up and down on the cement walkway in nothing but underwear and the long T-shirt I slept in, calling her name. Bare feet on ice, on the ridges in the pavement. I tore the pads on my toes, fell in a panic and scraped the skin off my knees, flailing.
I found Don in the lobby and I called Ned, hysterical, but of course he didn't pick up. Don sat me down on a brown-and-orange couch with coarse upholstery, whose pattern I still remember well, how I picked at the threads as I sobbed . . . I'll spare myself writing more about this. The point is she was gone, and the worst time in my life started.
I didn't keep a written record during the days after she was taken, but it's not those days anymore and it helps me to write now.
So Will came, Don was there, Kay and the Lindas, Burke and Gabe, even the well-dressed couple with two expensive cars. Everyone was around me after that, though I only half-noticed them. They were a blur of people who weren't my little girl, the blur of irrelevance.
They said things, they called the police and the police were coming, they said, hovering—we'd stay right here and wait for them. A blanket? A heating pad for my feet? I was in shock, said one of them.
I registered goodwill but I hadn't known what desolation was, before Ned took Lena, I'd never known what it felt like to be destroyed.
THERE CAME A TEXT on my phone while I was still almost catatonic. It was a text from Ned, I understood when Will held the phone up for me, though it didn't have Ned's name beside it, only a string of unfamiliar digits. Don said it was probably a prepaid.
The text bubble read Call off the lawyer.
"So he already knew she was filing," I heard Don murmur to Will.
"Surveillance," nodded Will.
"And sedation to make them both sleep through it," said Don. "How? The bottled water in the room? Something I cooked?"
He was on edge: everyone was.
There were security cameras, of course, the motel had a camera aimed at the parking lot, one in the lobby, a couple more. But when Don tried to view the footage his software told him the files had been damaged and couldn't be retrieved.
Beside me was an egg-salad sandwich on a paper plate. I remember it distinctly: the pores and craters of the beige whole-wheat bread, the fact that it looked nothing like food. I realized, seeing it, that there was no food for me—no food existed, in this world, nothing would ever be eaten.
The sandwich sat beside me, aging. I didn't touch it, and though I did relent about food in general—evidently—to this day the sight of an egg-salad sandwich makes me queasy.
Someone got my laptop and at their request I managed to click through a number of frames, I clicked here and there, tears running down my face, until I was able to bring up a photograph of Ned. Don emailed it to himself, then went back into the motel office and printed it out, though everyone present remembered what Ned looked like.
A new text: No police.
"He's got to be kidding," said Will.
We were still in the lobby. I think guests must have been coming and going by then, no longer crowding near. Will and Don and I sat on the couches while Main Linda kept busy making tea in the café. The yellow-beige sandwich had gone away—good riddance to it, unappetizing forever. Instead a coffee cup sat next to me on an end table, the surface of its cold, weak coffee as still as stone.
". . . are there people outside action movies who'd actually agree to that condition?" Will was asking.
"Where are the cops? I'm going to call them again," said Don.
Another text came in.
If you call the cops again [end of text bubble] I'll call my FBI friends [end of text bubble] and make a counterclaim of kidnapping.
"He can hear us," said Don, and stood up hastily. "Still listening, aren't you?"
Big ears to hear you with.
We gazed at each other, Will and Don and I. They looked round-eyed. I don't think I did. I wasn't surprised. I was on a plateau, the final plane of hell, I thought, a flat, dry place.
"He does know someone in the FBI," I said.
There'd been this asshole from the Anchorage field office. A couple of times he and Ned had driven to a rifle range called Rabbit Creek—I remembered because I thought of small rabbits running scared as the two men fired their weapons. They went for drinks afterward at some sports bar, where Ned stayed sober and the FBI guy got sloppy drunk. I hadn't understood what Ned wanted with him, some kind of "ASAC," Ned had said, assistant special agent—a sullen man with pitted cheeks, a spare tire and a comb-over.
I'd expected him to look like Mulder from The X-Files, I realized when they stopped by the house once, Mulder had been my main teenage exposure to an FBI idea and it lingered.
But he didn't look like Mulder at all. Sadly unlike Mulder.
And surely they'd had precisely nothing in common, I thought now, nothing but the FBI guy's future utility. Ned was a bet hedger, a fortifier and consolidator, effective at building networks and circuits. They met at a boxing gym and the FBI guy had apparently been drawn to Ned, as so many people were—as I had been.
Considering this I started to feel a spur of practicality again, my ruined center cauterized for a time so that it stopped infecting the rest of me. I could keep it together as long as I didn't think of Lena being alone or afraid. It was her emotions I feared for when I let myself fear, her trust of the world being damaged, eroded bitterly as I sat there with my hands tied, unable to reach her.
I didn't even consider physical harm. I couldn't stand to: that possibility was walled off in me.
Quickly all of us stood up and started searching for the microphone. There sat the laptop and my cell phone, which seemed the most likely, so Don called in the angry young mogul to inspect my devices. Apparently Navid knew about electronics. He came in, scruffy in his mountain-man beard and plaid shirt, and took my computer apart piece by piece. He seemed attentive, not angry at all, and I felt grateful and guilty for not liking him before; I would like him from now on, I would like anyone who helped me get Lena back, more than that I would love them abjectly, I'd be abject, I thought.
At some point I noticed I was digging my fingernails too deep into the heel of one hand. They were too short to draw blood, but the bruises would be there for weeks.
Navid took apart my phone, making me agitated—it was my only link to Lena, and what if it got broken?—but he put it back together again without finding anything.
Was the mike on my person? I didn't wear jewelry and I had no buttons, even, except for the one on my jeans. Don and Navid inspected my shoes—by this time a pair had been brought to me, along with a pile of clothes, and I'd dragged myself to the bathroom beside the café and put them on, the jeans and woolen socks and a pair of worn sneakers—but they found nothing there either. I didn't have my purse in the lobby so the bug, we figured, had to be elsewhere. We switched to inspecting the furniture.
It was confusing, since Ned wasn't likely to have heard about my plan to file for divorce through a microphone in the lobby.
After a fruitless search we trailed out of there, Don and Will and I, and into Don's office, but I was still nervous, I couldn't know when Ned was listening since we hadn't found the bug. I felt conflicted about calling the police again, we couldn't figure out why they hadn't arrived yet, so I insisted we go analog for a while, talking to each other by writing things down on a pad of lined paper and passing it among us.
Don and Will thought Ned must have got to the local police somehow, they suggested he wouldn't be able to do more than delay them and we should call again, get someone different on the phone. If that failed we should try another jurisdiction—the feds, probably, since none of the Mainers believed Ned's threats about the FBI could possibly amount to much.
He was bluffing, Don said, it was highly unlikely his contacts in Anchorage could strong-arm agents in Boston.
But I still felt overheard. I couldn't even trust my clothes, despite the fact that we'd inspected them: everything was suspect. Back in my room I stripped them all off; I stepped into the bathroom and made another 911 call—they transferred me to the sheriff's office and I reported the kidnapping again—they said they were dispatching a car, they promised two officers would arrive within the half hour.
After I pressed the END button I stepped into the shower and let hot water beat down on my face.
What about those chips people implanted in pets, I thought—what about them? Could I have been implanted with a chip? Could I pick it out from under the skin, as I'd once seen in some otherwise forgettable movie?
Scratch, scratch, blood, and a loosened nub of metal dug out of the flesh.
IF I HAD been guided to the motel by some sense beyond the usual five, some navigational instinct having to do with magnetism or light, I wanted to know what for.
THE STATE POLICE finally got to us hours after we'd first called. It was two officers, polite and attentive in their note taking. We made them sit with us in the back office, where we felt Ned might not be able to hear, and I told them everything I could think of—about Beefy John, B.Q., Ned's driver, his rented SUV. Black and American, was all I could say, and of course he might easily have switched it out. A couple of times I had to stop, and the cops waited patiently, their faces presenting sympathy.
I wrote down the address of our house in Anchorage, where as far as I knew Ned still lived. I had no idea where he'd been staying locally—there weren't other motels nearby, said Don, you had to drive at least half an hour for the closest lodgings open this time of year.
"Or he could be staying with local contacts," said Will. "That mechanic, maybe? John something . . . Pruell, maybe," he told the police.
"Ned—my husband isn't the type to sleep in his car," I mumbled. "He never stays in hotels under four stars."
The policemen looked at each other.
"That narrows it down," said one. "He ain't in Maine."
I had a tin ear. My sense of humor had left with Lena.
We were surprised at how soon the cops went away again. I'd thought they would stay near, I thought there'd be a task force, something—in movies policemen walked around the house or apartment of the kidnapped child's family, tapping phones, watching at windows. But in fact the two policemen left after their brief interview of me and an even briefer search for the concealed microphone (they found nothing). Their expressions were mild.
"We'll do our best to find your daughter, ma'am," said one. But I didn't like how he said it—casually, as though it wasn't life or death.
In the silence after the lobby doors swung shut Don said Ned had to have got to them, that their placid demeanor was unnatural. He said we should assume they weren't going to move quickly and I had to just call the FBI. But I wasn't so sure, I was more afraid of Ned's capabilities than they were, so instead I went online and then I borrowed Will's phone, distrusting my own. I hired a private investigation company based in Portland.
They'd assign a team right away, they said.
I called my parents next. My mother seemed shell-shocked, as though Lena's abduction was a sheer unreality, and offered to help with money. Her voice was so faint that I could barely hear her.
I COULDN'T SIT in the motel, I found, waiting for someone else to look for my daughter. I couldn't stand it. I didn't want to talk to anyone who didn't already know what had happened.
So Will and I got into his truck, a beater with worn Mexican blankets over the seats, and at my request he drove slowly up and down the icy streets, up and down, back and forth, prowling. The streets were fairly empty of traffic, only the silence of blinking Christmas lights on house fronts and in yards. There were teams of reindeer pulling sleighs, yellow outlines of bells.
Now and then someone would honk behind us or angrily pass, swerving to make a point. It felt as bleak as it looked, the houses spread out, the odd signal flashing the white walking figure to an empty corner. But I had it in my head that I needed to drive every street, and Will was willing to humor me, likely knowing I was on the edge of hysteria.
There was a worn map in the glove box, there was a half-dried-out pink highlighter in the armrest compartment, and Will pulled over and showed me how to mark the route we'd already driven. Even though it signified nothing, since we weren't knocking on doors or looking in windows, I colored furiously. As he drove I stared out the window, checking driveways for black SUVs, trying to imagine the potential of each business or house to be harboring her. I tried to intuit Lena's presence. Would I feel it? Would the other animals' senses come to my aid now—detection of the Earth's magnetic field, navigation by smell?
When we stopped at a stop sign or rare traffic light I'd trail the highlighter down the map, along the road we'd driven, which gave me a brief, businesslike feeling. Then I'd raise my face from the map. The next moment, I thought, the next moment it will be . . . I willed myself to see a face at the window, to see her small figure in the puffer coat.
"We need to stop now and go home," said Will after a while. He said it bluntly but kindly.
I was a child myself now: as soon as you were a victim, as soon as you were deeply hurt, you were a child again.
Helplessness was the one true fountain of youth.
IT WASN'T CLEAR what Ned wanted to accomplish. He'd ordered me to cancel the divorce filing, sure, but that could easily be restarted once Lena was returned. And any contract would have been signed under duress, and not binding.
After his first texts I heard nothing for days. Christmas passed without anyone seeming to celebrate it. Or if they did, I didn't see. It passed and faded and never was.
I went over and over how my girl must be feeling, alone with someone she barely knew, whether her father loomed as a threatening figure or had made himself charming and likable to reassure her. I worked to craft this kind of picture for myself, Ned as a babysitter, performing an imitation of affection—I sculpted this image painstakingly, smoothing my fingers along the edges, pushing it into a shape I could live with. But it collapsed whenever I wasn't vigilant and I wondered what he was telling her, what particular architecture of lies she was living in and what part of them she believed.
I couldn't help recalling Ned's phone conversation with my mother, his sly undermining of me, whether he was doing the same with Lena. But it was her relation to the whole world I feared for most, the way she might be changed. I got a prescription for tranquilizers the day after she was taken and tried, with Will's help, to make a routine for myself around the investigators' progress reports, which they gave to me twice a day.
Not even the voice had affected me like this, made my whole body weak with terror or my knees buckle whenever the knowledge of it struck me. It was my abject state that took me to Don's meetings. Between the kidnapping and the first meeting I attended there was only one exchange with Don about the hearing of voices—one moment when he bowed his head to me and apologized for having kept me out.
"You've been in recovery longer than most of them," he said. "You've done far better with it. For them it's still new. They didn't bring you in before because they weren't ready."
I said nothing to Will about the meetings, didn't even intend to go myself—I only started to attend them because I'd been by myself in my room and, without Lena, was hit by the lightning bolt that had been striking me constantly since she was taken. It was a stupefaction that refused to diminish: as soon as I had a loose, idle moment I was scorched down the center by remorse, burnt black by the feeling of guilt. My fault. My fault.
At those moments I'd do anything for distraction, and so it happened that one time I left my room headed for anywhere—looking for the moving figures of people, the sounds they made, the industry of normal lives—and as I passed through the lobby I saw the café door cracked open.
The tables had been pushed back to the walls, chairs set out in a circle. I'd been to an Al-Anon meeting once keeping a friend company, and this had the same encounter-group feeling. There were baked goods and coffee arrayed on one of the tables, a hot-water container and a basket of tea bags. I settled myself on a chair a bit back from the rest—an outlier, satellite chair—and as the fog of panic receded, I took hold of myself and worked not to think of Lena. One minute, I said to myself, one minute first, then two; one minute at a time, one day was an eternity.
Navid wasn't there, but the rest of the guests were accounted for.
"It's been four months since I retired," said Big Linda.
It didn't grow clear to me then where they'd heard their voices or how, only that the content of their perceptions varied. They'd heard different sounds, drawn different conclusions and had different responses. Linda had heard a voice at work, somehow, and told no one until much later; Burke had told Gabe about hearing a voice immediately, and Gabe had believed him schizophrenic . . . but in fact, that first day, I barely heard what was said. I drifted on the back of my Valium, lulled by the drone of voices.
And my fear of a cult, at least, was assuaged by the drabness of the plastic chair edge in front of me and the matter-of-fact trudge of Main Linda over to the snacks table. There was no grim power to be felt amid that mundane scene of guests selecting baked goods beneath the tube fluorescents. Main Linda piled sandwich cookies onto a paper plate printed with rainbows, then returned to her chair licking the powdered sugar off a finger.
Don didn't address a single word to me at that initial meeting, just let me sit there behind the ranks, saying nothing.
Not for the first time I thought how groups of people had a habit of making even the exceptional banal. Was it a national characteristic or a trait of all humanity? Crowds could be grandiose, that was true, but small groups in small rooms . . . it took me back to my parents' church, where I'd sat bored and staring around, looking high and low for any object of interest. More often than not I'd failed to find such an object and ended up gazing at the dirty Kleenex wadded into someone's sleeve. I remembered the backs of my legs sweating on the smooth wood of the pew, heard wet coughs off to one side, saw dandruff on shoulders and, in sandal weather, hoary toenails.
Still: there'd been hymns, and some of them were dull but many were beautiful and sad. Although I hadn't felt that sadness till after, long after we had left the church.
It was remembered music that was beautiful.
UNCLEAN SPIRITS ENTERED THE SWINE
THE INVESTIGATORS IMPRESSED ME WITH A SENSE OF COMPETENCE as I looked at their faces on the screen or scanned the neat pdf records of their efforts and expenditures, the rows of line items. I thought how easy I must be to fool—experience had shown this with sparkling transparence.
My questions were lame and I was often sedated. So I made Don and Will, and also the Lindas, ask questions for me. They huddled around and gazed into the laptop's camera. The investigators' clean, concerned faces stared back at us from a gray office only a couple hours' drive away. Were they really present, I wondered, in an office building in Portland? Or were they a shallow illusion of service?
Absurd how all transactions had become talking heads, the whole culture a mass of flat images of heads with mouths moving: we barely needed our bodies. There were hardly even dialogues anymore, rather there were a million monologues a day, each head with its mouth, each mouth with its talk. Still I listened with obsessive attention as the investigators fielded the questions, tried to show us they were pursuing all possible avenues.
Whether or not they were skilled or diligent, they hadn't found Lena by the next time Ned texted me.
He wanted to talk, he wrote.
Four days had gone by, the longest days I had lived.
WILL HAD TO DIAL the unfamiliar, prepaid number for me, my hands were shaking so hard, and when we finally got Ned on the line he wouldn't talk long—maybe in case someone was trying to trace the call. I don't know.
"I'munna need a photo op at the announcement, at least one TV show in Anchorage, down the road. Ads, maybe. Events availability. Magazine profiles, what have you. Like I said. And if I don't get 'em, this is just what happens, honey. Kid's just not with you anymore. She's gone. There's no cops out there gonna help you. It's my call what happens. If you want to fix it, I need your full onboarding."
Onboarding, I saw Will mouth silently, gazing down.
"Anything," I said. I could barely breathe—I was taking shallow breaths, quickly, afraid I might hyperventilate. "Give her the phone. Please. Ned. Please."
"She's having a good time with her toys," said Ned. His tone was indifferent.
"I need to hear her voice, Ned, and I need her back, please. I'll do whatever. Today, Ned, please, I need her back today. You win. Completely, Ned, you won, you win. Please?"
There was a long silence. With my free hand I grabbed the fabric of my skirt and scrunched my fingernails into it, into the tops of my thighs.
"Some other time, darlin'," said Ned. "I want you to recall exactly how this feels."
"It's killing me," I said.
But he'd hung up.
YOU DIDN'T NEED a picture ID to take a six-year-old kid onto a plane, I said to Will, perched on a stool at his kitchen island, a bottle of wine open in front of me. The shaking had stopped and I was self-medicating. There had been a small, odd reassurance in Ned's saying she was playing with her toys, maybe just that I was able to picture her. You didn't even need a birth certificate—nothing. No piece of paper attesting to the child's identity, the child's relationship to you. Unless you were trying to leave the country, they didn't ask for anything. You could walk onto a plane with any kid in the world, as long as that kid didn't open her mouth and give you away.
And the country was endless.
Children have no identity here, I said, no one cares who they are. Although the same could be said of adults, I added. More or less, the only interest our country takes in our identities is as taxpayers, consumers or criminals, I said. They could be anywhere, the investigators had reminded us, anywhere in the country, they could be in Vegas or Boca—they could be back in Anchorage.
I couldn't easily picture Lena standing quietly while Ned checked her in at a flight gate, but it was possible. He might have made threats. He might have threatened her. Or drugged her again.
Or she might be somewhere offshore, I thought. Ned might have a boat. She might be on the ocean.
"Don't think along those lines," said Will, and put his hand over mine. "You have to stop yourself going down that road. There's nothing helpful there."
I looked at him and felt flattened and paralyzed: depression weakened my limbs. My whole body felt inert with the exception of a core of fear that burned with its own perpetual energy like a star being born, born, and reborn.
"Come," said Will.
I stood with difficulty, with lassitude, barely moving until he took my arm. He made me lie down on the couch across from his fireplace, covered me in a blanket.
"But I have to be at the motel," I said. "In case he shows up with her."
Will said nothing, because he didn't need to: I could hear the words he won't without anyone saying them. He only lifted the back of my head and set a pillow beneath it, smoothing a lock of hair from my eyes. He turned off the overhead lights, leaving only a table lamp or two, and sat down in an armchair somewhere behind me, where he began reading. I gazed at the fire, absorbed in its abstraction, and listened to the crisp cut of a page turning.
Most women probably wanted a man who acted more like a woman, I considered—more like a mother, even. You wanted to be taken care of. As long as he wasn't womanish, I thought, as long as he had central masculine characteristics such as strength and confidence, in most other respects an ideal man was more like a woman.
Later I fell asleep.
MORE THAN BEFORE, with Lena gone I lost myself in research. Whatever was said in the meetings was a catalyst for my searches. There was something necessary in the order that research gave me, in the finding of lists, the recording of definitions. This is what x is. This is what y is.
Soothing.
A recent area of development is the discovery that . . . the ability to produce "sentences" is not limited to humans. The first good evidence of syntax in nonhumans, reported in 2006, is from the greater spot-nosed monkey (Cercopithecus nictitans) of Nigeria, showing that some animals can take discrete units of communication and build them up into a sequence that then carries a different meaning from the individual "words." —The Times of London 12.2013
AT THE SECOND meeting I'd taken twice the usual dosage of my tranquilizers but I'd also been drinking coffee steadily.
I still sat back from the others, mug in hand, but this time I leaned forward on my chair, almost perched. I succeeded in sealing off my anxiety over Lena only by pretending that my life with her, my devoted focus on her, did not exist at all. Fortified in this way, holding an image in my mind of a wall placed between emotion and me, between my life and myself—by blocking out my life outside the room—I was able to listen with a manufactured singularity of purpose.
Regina spoke first. I'd barely heard her talk before but now she was painfully eager. She has what I guess is a Dutch accent, and what she said corrected me: it wasn't just preverbal infants. There too my assumptions had been unfounded.
I listened to what she said and it never struck me to disbelieve. She'd been exposed through someone named Terence, and though she didn't describe him he clearly wasn't a baby. She was an ad exec who began hearing the voice when Terence was with her in her corner office. At first it spoke to her only in ditties and slogans; whenever she was with Terence, these ditties and slogans were audible, though he didn't seem to hear them. Almost right away it began happening when they were at home, too, she said, so now I assumed Terence was her husband—that they had worked together and gone home together too.
The man she'd come with sat across from her, nodding. But he couldn't be Terence; the way she talked about the absent person was almost patronizing. She'd cycled through various fixed ideas, she said, one of which had to do with wires in the walls, the audio of her TVs, computers, and many other interlinked devices. In service to that idea she'd hired contractors to tear into the walls, looking for speakers, receivers, anything that could be transmitting—she watched the workers like a hawk to determine whether wires existed where they should not. She'd pretended to be opening the walls for other reasons, she'd actually pretended to want to renovate, she said, had her company pay through the nose to renovate her corner office. Then she renovated her home, where, as a pretext for opening the walls and having the electricians carefully inspect all wiring, she paid to install complex systems that controlled the house's appliances, temperature, and lights.
Nothing had been found, the contractors dismissed her as a neurotic rich woman—which she was, she admitted in her tight, well-bred, Dutch voice. She was a neurotic rich woman, but so what?
Finally she went online and she found Don, she said: "I found all of you. And it was such a relief."
"She didn't tell me any of it," said her companion, who also had a Dutch accent. "She never told me what she was hearing, why she had taken on these construction projects, until we were on our way here."
"You know," said Regina. "I feared that Reiner would dismiss me. For being mentally ill, you know? People just get dismissed. It's how we get rid of people these days, we throw their opinions in the garbage can by calling them crazy. Whenever a man talks about his ex-wife, he says she's crazy! You notice? Because she must be crazy, right? To want to get a divorce from him."
"Ik wil geen scheiden, schat," said Reiner fondly.
"He says, 'I don't want to get a divorce,' " translated Regina.
Quaking aspen trees make clones of themselves to build colonies, becoming one large organism connected by its root system. They are able to survive forest fires because, although individual trees may burn, the roots underground remain intact. One colony in Utah is 80,000 years old.
Not to have to have children, I thought as I read about the aspens in Wikipedia, or at least not to have children that were separate from you—and yet to live throughout history, your family not only close around you for all that time but part of your own body.
Not to have to be alone.
I envied those aspens.
NAVID TOOK A TURN at the next meeting, shuffling his feet on the linoleum and clearing his throat nervously. I felt the attention of the group fasten on him: he must not have talked much before.
He'd been on set, he said, he loved being on set, and even though his job almost never required it, he did it as often as he could. He had one assistant, he said, just out of film school whose job it was to hang around a movie set all day and then, when finally a scene was ready to shoot, to text Navid so he could drive over. Best money he ever spent, he said, best money . . . he trailed off. I saw Kay catch his eye and smile at him, encouraging.
When he had started hearing, he said, it was a period of hard work and, he admitted, chronic drug use, and so his assumption was that what he heard was a cocaine artifact. Well, also crack, he said, because sex on crack, you know, was really excellent, he added awkwardly. "Or maybe you don't know, ha ha," and he looked around at the room of non-crack-users and emitted a nervous laugh.
No one else laughed.
His problem was, he said, he didn't know where the voice was coming from—there were so many people on the soundstage that he couldn't isolate it. He'd gone home to his house in the Hills when the shoot was over and hadn't heard it there; the house was empty except for his housekeeper cook and one other staff and the big rooms hung heavy with quiet. But as soon as he was on set again—the same movie, but there was a large cast, there was a massive crew, it was a big movie—the voice started up.
It drove him crazy, he said, because visiting the set was the only real perk of his job. He'd never cared much about the money, he liked to be there seeing movies get made, it was the whole reason for his career, and this movie, this movie in particular was his baby, he'd nurtured it from the cradle, it was his project. He even tried not doing drugs, but that didn't help (he smiled, self-mocking) so he went back off the wagon a couple of days later.
The voice performed speeches, he said, as far as he could tell it was speeches from hundreds of different films, scripted monologues and dialogues—not all of which he recognized. It might as well have been making its way through the AFI Catalog.
"I mean, I guess—" he interrupted himself, and looked at Don. "I guess I'm wanting to know what we're all doing, like, what is this? Why me? I just wanted to do my thing, make the pictures and sell them, you know, stay on trajectory. I was making, like, this almost perfect arc. And then there was this—it was pretty much noise, like static, like really fucking annoying, I mean I'd punch walls, man, I put a hole in drywall once—which—and cracked a Lexus window—anyway. So this is what I'm saying—assuming it is some kind of higher power or whatever, then what's the goddamn point of it? It fucks us over, and for fucking what?"
Instead of answering, though, when all of us turned and looked at him, Don just nodded.
"Go on," he said gently, as though Navid hadn't asked him a question.
"We come here, we talk, we tell our stories and whatever, say what we heard or felt, what our perception was. We have this—with the cookies and the donuts and that shit. Group therapy. But it's, like, circles. Around and around. Are there answers? Will anyone ever fucking tell me why and how this shit happened to me?"
Don kept nodding solemnly.
We sat there in an uncomfortable silence. But Don was waiting too, clearly, as though he didn't get that he was being directly asked—as though he didn't feel the pregnancy of the pause. Still no one wanted to say anything. There was a force field around Don, it seemed.
"He means, Don, do you have an explanation for us," said Kay softly.
The group seemed embarrassed, people fiddling with coffee cups or adjusting their positions on the hard chairs.
"Basically I'm one of you," said Don, after a few seconds. "They don't offer degrees in this, I'm afraid. I need you to understand that I try to be here for you, I want to do everything I can to help, but I'm not a credentialed expert."
There were a couple of nods, but faces went slack and shoulders sank with a disappointment so tangible I could feel it even from the cheap seats, sitting behind a row of backs. They'd wanted him to explain in simple terms what had happened to them, they'd thought he might really have the key.
I had too. I was no different.
Someone's cell phone rang from a bag under a chair and around the circle the guests shuffled their feet, started to pull on gloves and wrap scarves around their necks. I noticed they'd come bundled in full winter gear, even though most had only twenty feet to walk from their rooms.
"One thing," said Don. "Navid. When you say why us, it's not that we're the only ones. We're a subset—we heard more clearly than most. But we're not the only ones by a long shot. None of you are alone."
"'My name is legion, for we are many,'" said Gabe.
There was a glazed look in his eyes.
Listless, wanting something to occupy me when I got back to my room, I searched for the quote online.
It was from Mark 5, when Jesus cast demons out of a man and into a herd of pigs.
He said to him, "Come out of the man, unclean spirit!"
Then He asked him, "What is your name?"
And he answered, saying, "My name is Legion; for we are many."
The unclean spirits entered the swine; and the herd rushed down the steep bank into the sea, about two thousand of them; and they drowned.
IN A SUDDEN acceleration they started holding the meetings twice a day. I built the meetings into my routine, though I always had my cell phone ringer on high waiting for Ned to call. Part of me lived only for the second when he'd call again, or even better—a perfect ending—the investigators would call and tell me everything was solved, Lena was there with them, safe and sound and beyond excited to see me.
My limbic brain waited for that, the call that would effect reanimation, while the rest of the neural circuits were dedicated to not feeling alone while Will worked, marking time as I listened and watched at the meetings. I abandoned this journal. I had no wish to think, I had no wish to record. Until I found her I would distract myself with whatever this was, some talk-therapy hunt for God or even more ominous possibilities—none of it frightened me anymore. That was the difference: the second-worst thing (not the worst: I blocked the worst) had already happened. Whatever phenomenon they were painstakingly trying to uncover, there in the cafeteria beside the folding table of cookies, it was easier to consider than Lena.
Once I would have paid through the nose for a cogent explanation of the voice; now I sought that understanding mostly to stop agonizing over what I couldn't do or was not doing to find her. Part of me stubbornly refused to believe I couldn't just walk until I found her—treading through snow, knocking at doors—and felt a rotten guilt. Part of me couldn't believe she wasn't still neatly indicated, as I was, by a small blue dot on the map on my phone, moving as I did, going where I went.
I grilled myself over my incompetence, how I had come to let myself be roofied. Nights when I wasn't with Will were the worst, but I couldn't ask him to take care of me every minute so I pretended to need "time alone" some nights, whenever I could stand to. I often passed the time by retracing my steps in the hours before she was taken, seeing a simple blueprint of our room from above. In bird's-eye view I moved around performing mundane actions, the oval of my head between the knobs of my shoulders, the tips of my shoes beneath. There was Lena, a smaller oval, the same shapes in miniature.
I tried to reframe each movement to determine how the drug was introduced, think of myself brushing my teeth—was it in the toothpaste?—or brushing my hair. Maybe it hadn't been a pill at all, maybe it was some kind of narcotic that was absorbed through the skin. I played back that hour before I went to bed, when Lena was already sleeping. It couldn't have been the toothpaste because she uses a different one—a children's flavor called Silly Strawberry—and she must have been sedated, as I was, otherwise she would have woken up as they carried her out, she would have kicked and screamed.
Sedated or not, I told myself, I would have been woken by a scream. Since she was a year old I've jerked awake at the slightest sound, a murmur or one-word whisper of sleep talk.
The cops had taken away the half-empty wine bottle and the plastic motel cup I'd drunk from, claiming they were going to test them; the wine was all I remembered eating or drinking, after our restaurant dinner one town over.
But they didn't report any results. They were useless, Don said, they've been bought off or distracted or co-opted, he had no idea how but it seemed to be the case.
There was also the possibility of a needle, that I was injected while I slept and never found the pinprick hole. I couldn't figure it out no matter how many times I set up that blueprint in my mind's eye. No matter how often I took us through the paces, I could never narrow it down.
We never found Ned's recording device, and together the two unknowns obsessed me.
WHAT IF ONE of the aspen trees was cut down, while the rest of the organism remained? Did the remainder grieve?
TRYING TO AVOID images of how Lena was living in that moment I lay on top of the neatly made motel bed and stared at the ceiling. I thought how, in our normal, middle-class circumstance, we almost relish the idea of dark forces that lurk in the shadows. We watch movies, read books made glamorous by black-and-red palettes of horror, the hint of an otherworldly malice running like quicksilver through the marrow of our bones. We like to call the dark rumors demonic, like to have monsters to fear instead of time, aging, the falling away of companions.
Even people who scoff at the supernatural can embrace the demonic with a gothic fervor, hold in themselves an abiding fascination with that beauty of darkness and blood.
BIG LINDA HAD been working, she said—her work for decades had been training orcas like Shamu. She's pursued that vocation for most of her adult life.
She hadn't been doing the shows for a while, though, she'd gotten middle-aged and taken on more of a supervisory role, because to get in the pool with the animals you had to be in peak physical form. There'd been human deaths, of course, she said, maybe you read about them, saw them in the news, and trainers knew the real story, that it wasn't trainer error that caused those deaths but rather psychosis, because the great, predatory whales lived captive lives of aching, maddening frustration, shut up in their small cement tanks.
Some were more aggressive than others. Tilikum, she said. Blackfish.
Of course killer whales aren't whales in the sense of baleen whales, the kind of whales that cruise gently through the deep, slowly straining millions of krill and copepod through large maws full of white comb-like structures (she told us). The orcas were toothed whales, big dolphins really, though also apex predators, if we were familiar with the term. They were so highly intelligent that parts of their brain appeared a good deal more complex than our own—the part that processes emotion, she said, was so highly developed that some neurologists believe orcas' emotional lives are more complex than those of humans.
We know so little about them, she said, even the scientists, but they have language, even different dialects. They have culture. There are three kinds of orcas in the wild, all with their different cultures.
"They are astonishing creatures," she said, her voice trembling. "Some peoples hold them to be sacred."
I think I wasn't the only one to feel how much she cared, in the moment when she said that—how palpable her passion was—and how also, on this large, horse-faced older woman, passion like that looked almost pitiable.
Anyway, her favorite whale was a youngster who'd been bred and born in captivity, which is still fairly rare, she said, they die off more quickly than they can reproduce, the captive ones. His mother and father were popular with the crowds who visited the aquarium-amusement park where she worked (swiftly I shut down the mental link children, blocked an image of children laughing, splashed by the orca's leap).
Big Linda was alone one morning at the pool—the pools they live in, she added, only have to be twice the length of an orca's body. Main Linda cleared her throat, jerking Big Linda out of her sad reverie.
There was a silence, a pleasant tranquillity, said Big Linda. This was Florida in summer; there were palm trees overhead, the smell of heating pavement.
"I can't say what it was like, exactly," she went on, shaking her head and staring at the floor in front of her. The others also looked at the floor, as though listening to the shameful confessions of an addict. "I don't know how to describe it."
I saw Burke nodding slowly, pensive, also not lifting his eyes from the linoleum. I had no idea what Linda was getting at, couldn't make sense of it in the least, and was gazing distractedly at the side table, thinking about eating a cookie—they had some that were an unnatural shade of pink, those long rectangular wafers stamped with a waffle pattern that seem like play food. Lena had play food—she had fruit and vegetables made of wood that you could slice and put back together with Velcro. She had berry pie slices made of plastic. No! Stop.
"First I thought I was making it up," said Linda, "truth is I'd been real unhappy there lately, I don't like how we keep the animals—you have to understand, we only stay, most of the trainers stay because we're sorry for them, deeply sorry. We stay to do what we can for these creatures. For years I couldn't leave because of that, I'm so attached to them, you know, the little guy especially. Not that little, of course, since he's fourteen feet long." She laughed nervously.
I got up, telling myself to block out the lingering image of Lena at play, and gingerly approached the snack table; I put one of the waffle cookies on the tip of my tongue. Like balsa wood with sugar, I thought, and sawdust between the layers—sawdust with sugar. Still I chewed it, studiously not letting my thoughts stray back to Lena with her toys.
"Point is I was stressed out. Still. I finally had to admit to myself that something was there. I mean not the clicks and whistles and chirps, the usual elements of calls that we occasionally hear, you know, the vocalizing . . . it wasn't that."
I stopped mechanically chewing the balsa wood/sawdust wafer and turned toward the circle, where others were also gazing at her, their faces unreadable to me. She meant she'd heard the killer whale, I thought, and had an abrupt urge to laugh.
Instead I swallowed the mouthful and sat down on my chair again, careful to make no noise. I wanted to be very polite. It was Big Linda, I thought, who'd always been so kind to us—to think of ridiculing her made me wince. I would be unfailingly polite, I would be more attentive than I had been before, and I would suppress the instinct to laugh. It'd be hysterical laughter anyway, I told myself: again I had signs of incipient hysteria, as I had after Ned heard the voice. Both euphoria and hysteria had risen in me as I jogged along our street in the dark. Now they threatened to rise in me again.
But I was still a wretch. My misery came crashing back. I felt no lightheartedness at all; I was as heavy as lead.
"I always heard it, whenever I was at the tank, and I couldn't tell you how I got anything from it, but I knew—something about the way it was, somehow the rhythms were linked, how he'd be moving around and I'd be hearing it. I knew it was connected to him. He'd just been separated from his mother, you know, he'd just been weaned, but in the wild the male orcas stay at their mothers' sides for their whole lives. He'd been taken away from her, you could tell he was lost, basically, and then there was this—it was a kind of wall of sound, I guess, a wall of sound that also felt like a wall of feeling."
In the end—to me at least—a baby, a whale, there was nothing more nonsensical there than anywhere else.
Male humpback whales have been described by biologists as "inveterate composers" of songs that are "strikingly similar" to the products of human musical tradition. —Wikipedia 2015
I TRIED TEXTING Ned's various numbers, the temporary cell phones he'd used recently as well as his old number, the one he'd had for years. I repeatedly typed messages such as I'll do anything you want me to, I accept your terms, Give her back and I'll do whatever you say. For several nights there was no amount of abjection I wouldn't stoop to.
Finally I pulled up short and pretended to be made of granite, went from spineless to fossilized. There wasn't a middle ground. I knew it wouldn't last, either, the rock-like immobility, the erasure of my real life.
It was unbearable to submit to my profound weakness and so the only choice was to shore up surface strength.
Plants might be able to eavesdrop on their neighbors and use the sounds they "hear" to guide their own growth, according to a new study that suggests plants use acoustic signaling to communicate with one another. Findings published in the journal BMC Ecology suggest that plants can not only "smell" the chemicals and "see" the reflected light of their neighbors, they may also "listen" to the plants around them. —National Geographic News
ONE EVENING AROUND dusk there was a call from a new number, and when I picked it up after one ring, as I picked up all calls—instantly, slavishly—I heard her.
"Mommy?" said Lena, on the brink of tears.
"I'm here! I'm here!" is all I remember saying.
The phone was passed from Lena to someone else, an adult voice I didn't recognize. A contract was being faxed, it said, and I would have to sign it in front of a notary. We both understood, technically, that it wasn't binding, wouldn't hold up in court since it was being signed under duress, etc., but Ned also knew I knew that if I didn't stick to its terms this would simply happen again.
"But worse," said the person, inflectionless.
After I signed the contracts and they were delivered, Lena would be brought back to me.
These events unrolled quickly. The contracts were received and signed, Will and Don read them, as well as Reiner, who turned out to be a corporate lawyer. Will drove me to a notary at the fire station that stayed open all night, and after that a messenger took the packet from me. Then we went back to the motel and waited.
I took no pill and drank no wine, determined to be sober as a judge. Instead of drinking I walked around and around the outside of the motel, my heart beating fast, my cheeks hot, until my calves burned and the soles of my feet were sore. Freezing, I walked for hours. Every brief headlight near the end of the road made me breathless.
It was after midnight when the car pulled up and two men got out, two men I didn't know, though I wondered in passing if I recognized one of them as a cop.
Then Lena was here, I had her with me again, and the motel guests were close, and Don and Will, Don's father smiling widely as he leaned on his wavering cane. Everyone was hugging Lena or patting her, congratulating me, whatever. We were in the warm lobby without having walked there—we'd floated, I think now, and when I finally looked up there were no men and there was no car. Vanished.
SO NED HAS BECOME a condition again, a feature of life. Our end date is still the election, contractually, after which Lena and I should be released—but for now we're indentured. We're flying to Alaska next week for the official candidacy announcement, to do our duty as mannequins.
Ned's staff booked the tickets; Ned's staff booked the rental car. We're staying in our old house for almost a week. Without speaking to me at all, only sending me emails containing flight confirmation numbers and the rental car details, Ned's staff took charge of the arrangements.
Lena's still saying little about her time in kidnapping—I can't tell how deep the injury may go, though Don found us a counselor forty-five minutes away and we drive to see her three days a week. It doesn't seem to be the case that anything of substance occurred while she was in Ned's hands. That is, as long as she hasn't blocked a trauma. All that happened, apparently—once the initial violation had occurred when she was drugged and taken from me—was that she stayed in a hotel suite with a babysitter. And of course she was frightened because they told her I was sick.
It sounds like it was one of those big chain hotels, more like apartments in an office park, possibly in Massachusetts somewhere, the PIs say, with generic but pleasant enough bedrooms off a central living room and kitchen. The babysitter had her own room, and so did Lena, between which the doors were left open.
Apparently she only saw Ned once. The first morning he stayed away and had the babysitter tell her that she was safe, I was safe, the illness wasn't life-threatening. Everyone was safe, but she was staying there for her own protection in case the sickness was contagious. He made his single in-person appearance that evening, bearing ice cream and an expensive, wholesome-looking doll wearing a red-velvet ice-skating outfit. After that he sent her toys daily through the caregiver: animated movies, books, doll clothes.
She kept the doll for longest, toward which she felt a parental responsibility, but finally she asked me to take it to the same donation bin in the grocery-store parking lot where we'd taken the other items he'd sent. The gifts must have left a sour taste in her mouth.
The babysitter, a kindly, bland-sounding woman, prepared their meals: whatever Lena wanted, up to and including large ice-cream sundaes, chocolate layer cake, and piles of frosted cookies. For exercise she was taken to the indoor hotel pool, which, to hear Lena tell it, was always deserted, except for the babysitter and her. She liked the hot tub, which kids weren't allowed to go in: she had received the babysitter's special permission.
She watched a lot of TV.
Now that she's back I can stand to hear about it, I want to know every detail she imparts. Her experience has taken her sense of security and consistency from her—her exuberance has been curtailed. She doesn't sob or clutch at me, but she moves more cautiously than she used to, she's more measured.
One afternoon a guest checked in—a tired man from Quebec who didn't appear to hear any voices; he was so tired he barely even heard ours—and Don asked if she wanted to offer him a tour. She was polite and dutiful, mainly, I think, to protect Don's feelings. She didn't want to seem ungrateful. Yet the tour was subdued. She skipped the ice machine entirely.
I'm so angry at Ned for taking it from her, that free, unreasonable joy that was her greatest possession.
SO MY FEAR has turned mostly to anger, which is much easier to live with—I see now why it's popular.
But I continue to need distraction so to expend my nervous energy, maybe dispel the rage, I scroll and scroll and click and click once she's tucked in at night.
I've been going to the meetings faithfully, knowing we're leaving, trying to absorb as much as I can before I say goodbye to this strange circle. I can't take Lena with me to the meetings and there's no one I trust to watch her when I'm occupied except Will, so I've been vague about the meetings, implying only that they're about "recovery"—my own therapy, as she has hers. Fifteen minutes before they start I drop Lena at the library.
I've been trying to learn if anything unites the motel guests beyond the fact of having heard—whether, for instance, a message was conveyed to anyone. For me there hadn't seemed to be a message, as I've written, for me the voice had been like weather, but I shared Navid's questions, we all did: they were basic. I wanted to know if the voice had carried portents for others—if they'd felt like the Maid of Orleans, if any had believed they were receiving instructions or prophecies. It was a whale that spoke to Big Linda; well, whales have often figured in myths and stories. It seems well within the standard imaginative canon.
And just yesterday Burke spoke to the group at length.
"Chinese native," he mumbled, looking down at his feet. Burke has the bearing of an absentminded professor. "Acer griseum. Paperbark maple. Beautiful, peeling red bark, this great, faded red I've never seen anywhere else. I remember having the impression that it was melodies made by the flow of cellular division, the phloem and xylem. The movement of sugar in the trunk."
For him the voice—something like humming or singing, he said, a pure music sometimes like a chorale, sometimes like a Glass symphony—seemed to issue from a certain tree in the arboretum where he worked. The tree sang and its music was holy.
"But you know. Maybe it wasn't really coming from the maple tree or Shamu," said Navid. "Maybe they were both sort of like one of those ventriloquist's dummies—like the sound or the song were being thrown onto them."
I spoke for the first time. I said I'd been quite sure, when I was hearing the voice, that it was closely associated with Lena. It was either part of her or attached to her, but she was no ventriloquist's dummy. I said how its monologues would follow the movements of her eyes, at times, commenting on what those eyes beheld.
"Assuming it's not technology or communications from extraterrestrials," said Big Linda, "maybe it can have many kinds of living hosts."
"ET, really?" said Navid. "Hadn't gone there. But now that you mention it."
It seemed we were almost considering levity, or at least some of us were, and others were resisting and disapproving, at least that was how I interpreted the silence.
Kay spoke, softly as always.
"I know something," she said.
Heads turned.
"I mean—I don't have all the answers, I don't mean that," she went on carefully. "But I know part of it. I thought everyone did, until this meeting, hearing what Linda said, what all of you have, I thought we all knew that part of it, but now I think maybe that, with us hearing things, maybe I have this particular piece, and others have other pieces. I guess?"
Kay has that insecure person's mannerism of ending her statements with question marks.
"What piece?" asked Navid.
"It—so what we heard is, how can I put it," she said nervously. She was looking down at her hands in her lap, as though embarrassed by her claim to knowledge. "It exists in most things that live. It's language, or the innate capacity for language, is a better way to put it. You could say it's the language of sentience."
"Trees don't have language. Trees don't have opinions," objected Navid, kicking the floor with his toes.
Kay looked up at him. It was a different look from those she usually gave him, I realized. It was sympathy.
"It's not that we're the only ones who have it, or hear it, or are it," she went on, so quiet that I had to strain to hear. "What's different about us, different from how it is with the other animals and even the plants—what happened with Lena and Anna and in my case with Infant Vasquez? What's different is that we're the only ones it leaves."
Communication is observed within the plant organism, i.e. within plant cells and between plant cells, between plants of the same or related species, and between plants and non-plant organisms, especially in the root zone . . . plant roots communicate with rhizome bacteria, fungi and insects in the soil. These interactions . . . are possible because of the decentralized "nervous system" of plants. —Wikipedia 2016
IT WAS A LONG meeting, a meeting that went on for three hours instead of one, and by the time we dispersed afterward it seemed that Kay had always had a clearer understanding than any of the rest of us—Kay's hospital infant, an infant with a hole in its heart that lived for only three days, had somehow imparted more to her than the voice had told the rest of us in months. Even years.
Kay had heard more. Or Kay had listened with a greater aptitude for hearing.
I hadn't thought I was special, just equal. Equal, at least, I always assumed. But by the time I left the meeting I was unsure, unsure and diminished.
After the meeting I suspected I wasn't equal, and more, that there was no equality. Our idea of equality is a fiction useful mostly for the purposes of fairness, for law and economics. Elsewhere it's an empty husk, a costume we put on when we get up in the morning. In the length of our legs and arms, the breadth of our shoulders, the tendons that give us strength or weakness, our beauty or lack of it, sharp or dull intelligence—we aren't equal at all, and we never have been.
SOUL IS A UNIVERSAL FEATURE
NOT MANY TOURISTS FLY INTO ANCHORAGE IN WINTER. IN SUMMER there are backpackers galore: the small airport is full of tower-like packs with attachments dangling from them and duffel bags lumped on the floor in archipelagos of nylon and canvas. Among them you see hippies milling, hikers, hunters, fishermen, naturalists and wilderness fans of all stripes, talking excitedly about their planned itineraries as they wait for their car rides or small-plane connections. They crowd beneath the terminal's fluorescents in a fug of B.O. and patchouli and bug spray, headed for Denali and other points west or north.
But January is quiet in Alaska. When we flew in, the airport was almost deserted. It had that peculiar desolation of an empty public space, and in the silence our roller-bags squeaked and our footsteps rang out. Lena squealed at the sight of a rearing grizzly in a glass cage, which a placard claims is the largest bear ever shot. Paws raised, it looms over the polished expanse of floor in a perfect embodiment of overkill. She stood beside me and gripped my hand as she read aloud the sign at the bottom of the case: WORLD RECORD KODIAK BROWN BEAR. The bear's reared-up stance was upright, almost gentlemanly.
Ned wasn't there to meet us, happily, only a driver at the curb. Everything had been choreographed by his staff; there was a schedule with places, times, and tasks listed: 4:30 p.m. Consultant Appt. 1: Wardrobe. He's as disinclined to be in my company as I am to be in his. No good words will ever pass between us now.
We had an appointment with his lead media person right off, in his campaign office; we were instructed today, before the first press conference tomorrow. There are even clothes I have to wear, looks custom-designed for me as though I'm Sarah Palin. Clothes have been picked for Lena, too, apparently. Really? I thought. Even for the small time?
Ned has to do everything with corporate shine, he needs to be at the top of his game from the start. And he requires similar performances from his associates.
So we met with them and tried on the clothes. It was tedious standing around as they recorded our sizes and made adjustments, trying to keep Lena in one place. A hair and makeup person came and practiced painting our faces, taking pictures of us colored in different palettes. Lena was turned out like Shirley Temple at first and looked like a beauty pageant contestant, so I said no. The media consultant trotted out a second outfit, slightly less frilly, and agreed not to curl her hair into ringlets.
I know I won't be able to stand Ned's platforms and opinions, much less concur with them, so I'm doing my best to learn nothing more than I have to about what I'm shilling for. This is a farce I'm acting in. Except for one dinner with some women's church group, I don't have any conversations on my to-do list. I hold Lena's hand whenever I feel doubt, press her to my side when I find I'm quizzing myself on how I could have been so easily brought to heel.
But I'm not willing to take risks: I stay close to her all the time. I was given a second chance, I was rescued after a shipwreck, and my goal isn't ambitious. It's just to keep our heads above water.
After the meeting with the wardrobe consultant we were driven to the house, once our home. I felt anxious walking in, not sad or nostalgic; the abduction had erased even the vestigial possibility of that. But I did feel off-kilter entering the place. Lena was merely intrigued and ran around trying to identify what she remembered.
Ned has a housekeeper so everything is neat, and he's replaced the furniture I chose with items that are new and more generic. There's beige upholstery and beige drapes, a bland beige background everywhere; there are cut flowers on mantels and tables, as though the premises are being kept at the ready for a meet-and-greet. Behind shining cabinet doors there's a huge flat-screen TV, and photographs of snow-covered mountains have been placed on the white walls, no doubt by a decorator connected to his media team. They're Alaskan mountains, of course—discreetly labeled at the bottom lest anyone doubt Ned's loyalties. Chugach Range. 2008. Wrangell-St. Elias National Park.
Wrangell-St. Elias, I remembered telling Ned once, was larger than Switzerland. He'd shrugged: to him national parks were a waste of rich mineral and timberland.
But now he has pictures of them.
"Where'd my room used to be?" asked Lena. "Did I have my own room?"
"You did," I said. "But mostly you slept in the bed with me."
We stood at the door of the very small room that had been the nursery, which now contains an exercise bike and free weights.
"It doesn't look like my room," objected Lena.
"Your daddy likes to stay fit," I said.
THE NEXT CONSULTANT made her practice standing beside me in front of a video camera. She showed us the footage on her laptop, showed Lena how she was fidgeting and playing with her hair. Lena should stand still and smile and keep her hands clasped together, she said, or at least let them hang by her sides. She shouldn't move around, said the consultant, because it would distract from Ned.
"Your daddy's going to make a little speech, and then he'll answer questions."
"What if I have an itch?" asked Lena.
The consultant smiled and said the whole thing would be over before she even knew it.
The initial response to an anomaly is typically to ignore it; this is how the scientific community has responded to the seeming anomaly of consciousness.
Then, when the anomaly ceases to be ignored, the common reaction is to try to explain it within the current paradigm . . . to date, no such effort in any discipline—be it chemistry, quantum physics, chaos theory, or computing—has proved fruitful.
No matter what theory is put forward, the central question remains: How can immaterial consciousness ever arise from matter?
When it comes to consciousness itself, science falls curiously silent. There is nothing in physics, chemistry, biology, or any other science that can account for our having an interior world. —Peter Russell, huffingtonpost.com 12.2013
I DON'T WANT to see my Anchorage friends, because to see them again now would bring them into this queasy distortion of my life, the fake alliance with Ned. It makes me ashamed, even though I'm looking down the barrel of his gun.
Some know about the kidnapping, some don't; others know about how it resolved, others don't. I can't stand to do the mental accounting of who knows what, can't bear to revisit the ordeal—it was hard enough writing it down for myself. I don't need to listen to sympathy or indignation on my behalf.
And from the few calls I made while I was panicking, I have the lingering feeling that most of them don't believe I was trapped into making this deal with Ned. None of my friends here seem to understand the urgency of my fear. They live in a personal world where rules are followed and fairness reigns; they're mostly white and mostly middle-class, meaning they feel entitled to justice for themselves and expect it for all the other people in their lives. Corruption belongs elsewhere, other countries, Wall Street or Congress, lobbyists.
They tried to be sympathetic when I talked to them, as people have to in the face of a missing child, but I felt, behind their commitment to sympathy, a steady seep of disbelief as though they suspected I was exaggerating or dramatizing. I was failing to stay normal, so either my perceptions were biased or I'd mistaken the facts of the case.
Because their take is that Ned's a good guy, basically. Too handsome and too charming, one of my friends wrote me, and sometimes you resent him for that. But as soon as you see him again you forget the resentment—you like him again the moment he speaks to you. He's maybe a bit of a playa, she wrote. There've always been rumors, but there are always going to be rumors when a man's that HOT-HOT-HOT [sic]. Men aren't monogamous anyway, they're just not built that way, and I'm sure it was hard to live in the shadow of the light he sheds . . .
That was the kind of email I got from my Anchorage friends about Ned. He's not a credible kidnapper to them. They figure he probably just missed his kid. Maybe he missed her desperately.
The first time we saw him was an hour before the press conference to announce he was running. We sat in his campaign office, waiting to go into the room with the small stage and podium where the reporters were going to be; Lena was in modified pageant gear, only half as gaudy as the outfit they'd first put her in, and no ringlets. I was in a suit that made me look like a first lady, and they shellacked my hair with spray so that it was big on top and swooped up at the bottom. The makeup artist gave me pink lips.
Ned came in while they were working on us, making his usual pretense of jocular fatherhood—bending to hug Lena, then grab her face and say "Got your nose!" (She jerked back at this, banging into the hairdresser standing behind her.) He acted as though he'd already greeted me, as though we'd spent hours together earlier that day—for the benefit of the staff, possibly, he squeezed my arm as he passed—preparing himself, maybe, for the public embraces we'd been asked to perform.
We hadn't seen each other since the day he showed up in Maine. Since before he took my daughter.
"My girls ready?" he asked.
My girls triggered my gag reflex, surprising me, and I fled to the bathroom. I didn't throw up, in that closet-sized half bath full of rolled-up campaign posters and yard signs, but it was close.
Once we were up in front of the flashbulbs and digital recorders, my nausea turned to a stunned thoughtlessness. When Ned spoke I barely registered the content of what he was saying. Everything but Lena, who held my hand, was scenery, and when I embraced that I felt less nervous.
When people say "scenery" they can mean either a stage set or the beauty of the natural world—the two are interchangeable, in the word scenery. In that strange word the entire landscape, up to and including mountains and the moon, is only a background, probably two-dimensional, for the human figures in front of it. But it helped me, in those minutes, to think we were just playacting.
The press didn't ask many political questions; mostly the reporters there were interested in giving Ned opportunities to talk about his success at business, to brag about his companies, of which the room seemed to be full of boosters. There was one timid question from someone at the back about a drillship that had almost run aground in Unalaska Bay, but the other reporters moved on quickly when Ned waved that one away. The room was stacked with his allies.
Just when I thought we'd got off scot-free and things were winding down, a reporter waved at Lena.
"What do you have to say about your daddy running for office, honey?"
Lena blinked and said nothing, and then, as the silence lingered: "He's my daddy."
Her tone was confused, almost questioning, but because she's a kid and her voice is high and thin, this bland remark gave the room an excuse for aw-shucks laughter. People shuffled out, grinning and shooting the breeze.
WE NEEDED TO be seen out on the town together, so Ned made reservations at upscale restaurants for all our dinners on this trip, except for the very first night when he took us to a pizza place that's a local favorite.
The "narrative," as he calls it, meaning the group of fabrications we give out for public consumption, is that I have a dying parent back East, and Lena and I are staying there to help my mother suffer through the time of decline and hospice. My father gets to be the one who's dying.
"Lymphoma on top of the ol' dementia," Ned said.
I hope my mother or Solly don't see any of the coverage of Ned's campaign, that none of it makes its way onto YouTube. I imagine how their faces would crumple, seeing my father used that way.
At dinner I had to talk directly to him at close quarters. I had to look closely at his smooth features, his deep-blue eyes that glance off me now, never resting for long, straying around whatever space we're in as though even a table leg is more compelling than my face. I welcome it in practice, but it hits me how he used to work those eyes so hard to make me believe he was earnest.
The Moose's Tooth was crowded as usual—there are always lines there—and our booth was sandwiched close between two others. Ned's fake-Secret Service bodyguards took the nearest two-top, but still we were back to back with other diners and I could tell Ned felt everyone must be watching him, so the fake cheer of our conversation had to pass muster. It was surprisingly difficult to smile and nod and be a wifely mainstay.
I found I couldn't eat. The restaurant's pizza, which I used to love, reminded me of egg salad. So I drank my one glass of white wine, picked at a salad and listened to Ned rattle off his campaign reports. My single glass of white wine was mandated by his staff, as it didn't look feminine to drink beer, it didn't look Christian to have a second glass, and red might stain my teeth. I drank my quota slowly, savoring it as I watched Lena doodle on a child's menu and Ned reeled off a list of coming events, repeated sound bites about his exchanges with campaign donors, why they believed in him and his values "in their own words." There were the usual anecdotes about small-town Americans, a farmer named Milt, a grandma named Pearl. He seemed to be running lines, rehearsing his material with a very small focus group.
After a while I looked up from Lena's artwork and found myself staring at elements of his face and carefully detesting each. You'd think a facial feature in itself would tend to be inoffensive, particularly a well-formed one, but I discovered that if I concentrated even an earlobe could be invested with spite.
Lena spoke quietly, softly about the plot of a Disney movie while I stared at the earlobe and savored my distaste. There were a couple of moments when I felt deranged looking at him, considering my loathing, but mostly I relished it.
I couldn't believe we'd make it out of the restaurant without running into someone I knew. Ned had instructed me to prepare my Anchorage friends on the specifics of the narrative even if I didn't plan to see them; he'd sent me a list of talking points as an attachment to one of his blank emails, including a timeline: when my father became terminally ill, when we were notified of the diagnosis, when we left Anchorage to help my mother take care of him.
The timeline projected forward, even stipulating when my father would enter hospice. These would all occur, of course, in the months before the election, explaining our absences from Alaska.
So I'd emailed my friends and bcc'd Ned as he instructed, putting the talking points into a "personalized letter." Partly because of this, the prospect of actual in-person encounters dismayed me. As we were rising to go—Ned had, to my relief, spent half the meal talking into his phone's headset—we were intercepted by a group of people from city government, civil servant types who were mainly Ned's contacts but whom I'd spoken to a few times at parties. Their faces betrayed a certain hesitation at my presence, which made me wonder who Ned was sleeping with these days, whether these people knew the marriage was a sham. I wondered how it was possible that everyone didn't know, since Lena and I had been away two full years. Yet they acted as though nothing was out of the ordinary and I reminded myself that Ned took care of business, Ned kept his ducks in a row. For the past few months we'd been staying with my terminally ill father . . . the narrative, unbeknownst to me, has been in place for some time.
I made my excuses and led Lena away, Ned grabbing his jacket and glad-handing behind us.
WHILE LENA AND I sleep in the house that used to be ours, Ned's supposedly staying at a B&B tucked away in the foothills of east Anchorage. He thought we'd be noticed coming and going from a hotel, whereas he can move around discreetly. I'm not sure why, since he's the public figure with the striking face and still lives full-time in the city. On the other hand, so far no one has found out that we're sleeping separately, so maybe he's correct in his calculations.
He has a "house," these days, not a house, much as he has a "family." His car, driven by the chauffeur, had dropped us off and pulled away quietly in the dark: entering the building I felt stealthy, though it's hard to feel stealthy in puffer coats and mukluks.
Lena and I have been sharing the master bedroom, which feels like a hotel room—as though no one familiar has slept there before, certainly not me. Along with the rest of the place, its redecoration was drastic. There's the skin of a polar bear on the wall—Ned must have bought it from a native, I thought, or possibly on the black market—a bold choice, given the politics. Maybe it signals his radicalism; in the bedroom, maybe he reveals his radical anti-government core. But it doesn't quite ring true, since the king bed's piled high with satiny showroom cushions that only his interior decorator could have chosen. They do feature masculine colors.
Lena fell right to sleep despite the bearskin, curled up with Lucky Duck, and I went back to the living room, where I flicked on the gas fire in the fireplace. I took a bottle of wine out of Ned's new wine refrigerator, poured myself a glass, and sat on the sofa with a blanket, feet tucked under me, to call Main Linda.
She said the mood among the motel guests has changed, it's gone from a support group to the scene of a dispute. Navid and Kay were a couple, and now they're estranged. Navid says Kay kept her understanding of the voice from him—"intentionally, privately kept her knowledge to herself," as he apparently put it, like a "hoarder of information." Kay's hurt by this and says she never hid anything.
Meanwhile Burke and Gabe argue that Kay's assertion that the voice is language, the language of sentience, is unimportant. Of course it's language, that's a truism, Burke wrote in an email to me. Words. Yeah. We know. The question is where that language is coming from.
"Do you realize how Regina heard?" said Main Linda in her gruff voice. "The whole time I thought she was talking about a kid, when she talked about Terence, I honestly thought it was a retarded kid, sorry, developmentally disabled. Turns out that Terence was one of those little, yappy dogs. Probably wore ribbons. And miniature vests. She heard the voice of God from a Pomeranian! Or maybe a shih tzu. She showed us a picture on her phone. She used to carry him around in a Fendi handbag."
I couldn't help it, I laughed. I thought of a curly dog trotting around at Regina's heels, speaking the way the voice had spoken to me.
"It died," added Main Linda.
A linguistics scholar had been called in, she said, an expert who'd been talking to Kay. He seemed, said Main Linda, to be somewhat outside the mainstream of linguistic studies, though still (she'd looked it up) fairly well published in peer-reviewed journals. He had theories about grammar genes, about animal communication systems.
"The FOXP2 gene," said Linda. "This English family, I guess, has this speech defect down through the generations? And it ends up they have a defective copy of one gene. Or maybe it's a protein, but anyway, I guess the idea is language is maybe genetic. I only half-listened. Don reached out to this linguistics guy because Kay, I guess, does a speaking-in-tongues thing. Like, she can spew out a bunch of languages she isn't supposed to know. Stuff she supposedly heard from Vasquez. Plus she can do insanely complicated chemistry diagrams. Idiot savant shit. All Greek to me. Hey. Can we talk about normal crap?"
"We have consultants who pick out our clothes for us," I told her lightly. "And there's a family photo shoot for some glossy local rag, basically a real-estate brochure. Tomorrow. Ned's using someone else's dog. Can you believe it? A dog-for-hire!"
"That's low," said Main Linda. "A trophy dog? Is that even legal?"
"A golden retriever."
"Hope God doesn't talk through it."
"Do you believe Don knows more than he says?" I asked, pouring my second glass of wine.
I'd gotten restless sitting and was cruising through the rooms, taking a closer tour of Ned's model home. There was a picture of him fishing, the standard fishing photo with a giant salmon dangling from one hand. Kenai Peninsula, read the caption. Ned never fished. He hated the smell of fish and never ate it. A guide must have taken him and he must have learned some lingo to be able to shoot the shit with other fishers and hunters. Everyone fished in Alaska, practically, in season salmon falls from the skies here like rain and everyone has a smoker in their backyard, but Ned hadn't allowed fish in our kitchen.
"Don wants to keep things friendly, that seems to be his role, you know?" said Main Linda. "Moderator."
"I don't see how any of this can be proved or not proved," I said. "It was a phenomenon. But it's not as though any of us were given instructions. It's not like we have a task to do. Is it?"
I stopped in the hallway. Beyond the standard fishing photo, the standard hunting photo (deer on truck), the photo of Ned in crampons hiking up a glacier (looking down from the heights, smiling), there were numerous family photos. Some of them looked like upscale versions of mall shots while some were "candid" action shots: Ned, me, and Lena. All of us together, in different variations. Lena was a baby on a rug, Lena was a toddler in Ned's arms, all three of us stood beside a Christmas tree; there we were cross-country skiing, with Lena standing on a pair of junior skis, poles held in snowflake-mittened hands.
Except that none of the scenes, with the exception of Lena sitting on the rug all by herself, were real.
Ned had never done any of those things with us.
"Oh my God," I said.
I set my wineglass down on a table and flicked on the overhead light, leaned in to look closely. The pictures looked authentic. They were carefully framed and artfully staggered on the wall. Some seemed recent; they featured Lena's face pretty much the way she looked now. Ned must have taken the photos from my phone and used those images.
While I was sleeping a drugged sleep, when he was taking Lena.
Or he had open access to my phone.
"There's a whole wall of family pictures," I said. "They never happened at all. Family vacations, skiing—there's us on matching snowmobiles and us fishing. There's Ned with a dead buck and a truck and rifle. Redneck wholesome. They've been messed with to put us together when we never were. I don't believe it."
"Brazen," said Main Linda. "That guy's got some big ones on him, you gotta admit."
After we hung up I took pictures of the pictures, sat on the couch and scrolled through looking at them, comparing the faces in them to the faces already on my phone's camera roll: Lena with her snowman, Lena on the beach, Lena with Faneesha the UPS driver. I texted a couple of matches to Will, nearing the bottom of the wine bottle, and then called him.
He communicated his reserve with few words. He wasn't happy that we'd gone up to Alaska, wasn't happy with anything concerning contact between Ned and us. Ned is probably sociopathic, he has suggested. He feels no empathy.
And I have to admit, when I find a list on some website of the behavioral characteristics of a sociopath, there's only one box I wouldn't check for my husband.
Superficial charm and good intelligence
Absence of delusions
Absence of nervousness or neurotic manifestations
Unreliability, untruthfulness, and insincerity
Lack of remorse and shame
Inadequately motivated antisocial behavior
Pathologic egocentricity and incapacity for love
General poverty in major affective reactions
Unresponsiveness in general interpersonal relations
Sex life impersonal and poorly integrated
Failure to follow any life plan
We have no control over his actions, Will reminds me, no one does, possibly not even him. Much of a sociopath's game is aimed at controlling people and outcomes, Will says. All you can do about a sociopath is steer clear of him. Ned's a time bomb, Will has insisted since the abduction, and we don't know that it's finished exploding.
Still, neither of us was able to come up with another course of action for me—not one that wouldn't risk Lena being taken again or hinge on police cooperation.
So here I am.
Now almost every piece of information I give Will about Ned seems to escalate his anxiety, so I find myself trying to avoid mentions—from thousands of miles away there's no use alarming him. He's done too much to help already: I'm confused about why he has time for all this for us, for me. I wonder what I've ever done for him other than need his help.
There's an imbalance of generosity.
Panpsychism is one of the oldest philosophical theories, the view that mind or soul (Greek: ψυχή) is a universal feature of all things, and the primordial feature from which all others are derived. —Wikipedia 2016
ON THE WAY to our potluck dinner with the church group Lena sat bolt upright in the back of Ned's Town Car holding Lucky Duck. She doesn't relax around her father since the kidnapping—her rigid stance stops just short of afraid, bespeaks reserve and attentiveness.
In what I felt was an egregious lapse in taste on the part of the consultants, we were made to wear matching dresses. Sitting there in the Town Car in my dress that was the same as a six-year-old's, I felt beyond foolish but hadn't bothered to protest. Also it was too cold for dresses by far; there was slushy snow on the ground; dresses don't look too good with puffer coats atop them.
But of course Ned couldn't have cared less about my discomfort or opinion. And Lena was pleased, saying the twin dresses reminded her of dolls you can order from a catalog in "look-alike" form, with features custom-selected to mimic your own hair and eyes and skin. It was one of those dolls that Ned had offered her during the kidnapped period.
"You didn't bring the lamb I gave you?" he asked from the passenger seat, texting rapidly, not bothering to turn.
"Lamb got sick," said Lena gravely, a doctor delivering the bad news. "She had to go in koranteen."
"Quarantine," I said.
"Quarantine," said Lena. "She got a cancer in her tail."
"Sounds serious," said Ned.
"Uh-huh. She's almost dead," said Lena.
Ned did turn and look at her. I was surprised too.
"I see," he said.
It piqued his interest for a second, but then he went back to pushing buttons. He was holding the phone at a different angle now, and I could see he wasn't texting about business or the campaign; no, he was playing Angry Birds.
Once we pulled up at the church, though—it was a potluck in the basement—he snapped into his public mode, his face suddenly animated. The light of Ned's personality has an ON/OFF button, which when he's alone with us now is typically set on OFF. It's fine with me, in fact I prefer it since he's nearly a robot when the switch is off, far easier to tolerate shut down. The ON switch makes me anxious with its vibrant, fizzing current.
When he's switched off I can almost ignore him.
"Hey Mom. Lamb's not dying," she whispered to me, as Ned was getting out of the front seat. "Lamb's fine."
My instructions for this more fluid assignment were to avoid all topics of conversation except the shortlist Ned had specifically allowed: food, weather, his qualities as a good husband, and, if additional content was absolutely needed, I could reminisce about the times when Lena "did cute things. IE u can take out phone, show Haloween bunny fotos" [sic].
Only Lena made any waves, as it turned out, and even those were small ones.
"Do you like my dress?" she asked, as she and I stood awkwardly near a food line, trying to be nice to some middle-aged ladies in the congregation after Ned pronounced a blessing that was also a stump speech.
"Why yes! I do!" said the woman.
"My daddy made me wear it."
"I see!"
"My mommy doesn't like matching dresses," she said.
"Oh?"
"I do. They have them in a catalog. You can order your own doll to look like you and even order the same dress. Like not in doll size but for a real person. My mom said matching outfits might be OK for dolls but not for real people."
The women eyeballed each other, smiles faltering.
"Oh, now, I like the dress fine," I hurried. "I just think it looks better on you, honey."
I set a hand on her shoulder as I turned to the ladies. The penalty for poor performance will be, Ned had written in an email, and left the sentence unfinished.
"She's very fond of those dolls," I made myself say, trying to pass. "She studies the catalogs as though they're the greatest story ever told."
"My daughter had one of those dolls, too," said the first lady. "I still have it in her bedroom! In a little bitty chair."
"Girls just love them," agreed a second.
"I don't know," said a third, shaking her head. "I think that company's liberal. Don't they sell Jew dolls too?"
Lena gazed at her.
"You must be starving, sweetie. Let's go and scoop you up a plate of food," I said, as smoothly as I could.
"What's Jewdolls?" said Lena as I steered her away.
"Honey, these people aren't your daddy's friends," I said in an undertone as I plunked potato salad she'd never eat onto a paper plate. Technically it was true, after all. "They're more like people he needs to impress. And it's our job to help him because we're his family. It's not for long. For now we have to just smile, OK?"
"I think that lady might be mean," said Lena.
"We can talk about all of it later," I said. "We'll talk it through. For now, though, would you do me a favor? Just try to smile and be friendly?"
"If you're nice to mean people, Faneesha says you're mean too."
All in all I was surprised at how down-homey the church event was, with its paper tablecloths and deviled eggs whose yoke-ridges had gone crusty. There must have been someone in the congregation to whom Ned owed a personal favor. We got away finally with Lena sulking, face screwed up into a mask of resentment, but no open conflict.
Her father talked about sports to his driver as we headed over to the magazine shoot, where, in high-tech outdoor gear, he would run and throw a Frisbee across a field of snow to be caught by the dog he had rented.
LENA LOVES VIDEO chats and I'd promised her she could do one with our Maine friends, so we opened my laptop in Ned's living room and hooked up to my cell phone's hotspot. She talked first to Kay and then the Lindas and Don.
When she got tired of talking and settled down with a TV show I carried the open laptop into the bedroom, panning around at the dead polar bear and the pictures of snow-covered mountains.
"Why don't you take that outside?" said Don.
"It's freezing," I said. "Are you kidding?"
"You don't have privacy in the house. Which you'd do well to keep in mind—I hear you had a sensitive conversation with Linda recently. And possibly Will?"
I'd registered when we first walked in that the house was probably set up for surveillance, I had no reason to think otherwise, but then I'd conveniently forgotten. I still have the habit from my old life of not feeling watched, somehow, a habit that's been hard to cast off even after I was roofied and had my child stolen—I can be paranoid one minute and the next relapse into my lifelong, previous routine of feeling unwatched.
But my conversation with Linda hadn't been too revealing, I told myself: the part about the voice would have been of no interest to Ned, at least, though he or his proxy would have heard me exclaiming over the faked pictures.
"OK," I said.
I stepped onto the small back patio, gloves on, a blanket over my shoulders. It was getting dark, the sky indigo already and not overcast at all. A few stars were out. If I turned to my left I could see through a large picture window into the living room, where Lena sat on the couch watching her TV show, her face small and expressionless in profile. The scenes of the television screen flashed their varying colors over the room.
I grabbed the laptop and strolled away from the building, into the expanse of dried grass.
"So it turns out your husband's bankrolled by a major PAC," Don said. "This isn't going to be a local or state career, if he succeeds. It'll be the governor's race next or a Senate seat. He's going national."
"I'm not surprised at all," I said. "That's what he has to want. He's always been ambitious."
"I have a friend with D.C. connections. He said big plans may be in the works for your man Ned."
"I'm not surprised at all," I repeated.
The angle of the picture changed, with Don's head sliding beneath the bottom frame and jumping back into view. Behind him I could see Will.
"So are you thinking that after the election he'll smile and let you walk away into the sunset?" said Will. "Is that what you're hoping?"
"I mean, there's a contract. Don, your lawyer read it."
"The contract lays out the terms of the divorce, custody and so forth," said Don. "But it's only a piece of paper. It's not a guarantee of a happy ending."
"What do you mean?"
"We're not feeling so great about your safety."
It was hard to see their faces, both of their features in shadow. The tops of their heads were blurred in front of a sconce that haloed white light.
"There are plenty of ways to make a contract irrelevant," went on Don. "Say after the election you had an accident. Then Ned could be a grieving widower and loving father rolled into one. He'd have Lena as a permanent prop. It would look very nice on him, in terms of electability."
"But I'm not going to die after the election. That's . . ."
"It's really easy to die."
"Don. I was married to this guy."
"Look," said Will urgently. "You don't think, once he's elected, that he'll want to be a divorced guy, do you? That title won't be his first choice."
"Well—"
"And he likes to have his first choice. He really likes it. Right? We know that about him."
"But you're—but he's not physically violent. He never even hit—"
"He drugged you. And Lena. No reason to assume he's not capable. He wouldn't have to do it personally."
"You don't have any—I mean, there's no proof of any of this, though, right?"
"Clearly we don't have Ned bugged," said Don. "He has you bugged. All we have for evidence is our familiarity with him. His record."
"Life's not a TV procedural," said Will. He sounded stiff and almost condescending—unusual for him. "We don't live in a place with instant forensic identification of every killer. It's common for murders to go unsolved."
I didn't know them that well, I thought, I barely knew them. Don seemed more than ever to have entered my life under a guise, leaked into it through a minor opening I hadn't known was there. This slumping man with his womanly hips, I thought. I still didn't understand him. Was I even supposed to know him, was it even right that we were familiar, or was it part of some dimly occluded design that might hurt Lena or me? Indeed, had it already? And Will—there I felt soft-centered, the pull of attraction and fondness and gratitude, but he was new, and I hadn't shown good judgment in the past.
Pointedly I should be the last person to trust someone because I wanted to sleep with him.
But maybe they weren't the sketchy ones after all—I was the one who'd married a man devoid of emotion. I might be the one who couldn't be trusted. I'd caved to Ned, and in my weakness I'd brought them in too—into something that shouldn't involve them at all.
"You need to get away from him," said Don.
After the blobby icon replaced their faces on the screen I walked back to my former patio and stood there shivering, imagining the dark shapes of bears in the woods behind the house. Many times in the past I'd spotted them there, humped figures barely distinct under the interwoven shadows of branches—except for once when a mother and cub lumbered into the backyard looking for garbage scraps. They must be hibernating now.
Around me on the patio were some plants that used to be mine, shriveled brown threads I couldn't identify anymore, though I remembered picking out their pots in a big box store. I remembered patting down the soil around the green seedlings. I should have taken them inside or given them away . . . they'd lived for years while I was in this house, growing, flowering, then suddenly been abandoned out here on the flagstones when I left. They would have died in the first frost.
I thought of all the green surrounding the house in summer, the green in the woods, long trailing banks of green, great oval storms of leaves, how despite that huge green outside I'd pored over and tended these small green outcroppings. But then I'd walked away from them.
What could I take care of?
I went inside the house, annoyed.
But despite my annoyance—he's never been physically violent, I repeated to myself several times, walking around the house in my sock feet—I found myself hesitating as I took a fresh bottle out of the brushed-stainless wine cellar, letting the heavy glass-and-stainless door close with its small suck.
I don't know that much about Ned's life before me, actually. I know he started working at age twelve, I know the story of that: he ran errands for petty criminals, then not so petty. At last he scammed his way into a prestigious university, but dropped out after two years, switching to a business school with a degree he finished online. All that was the tip of the iceberg, the part he pretty much had to tell me, but the rest of it was a blank.
He'd always been closemouthed. No matter how gently I asked, he wasn't interested in rehashing ancient history.
It occurred to me, looking at the bottle, that he'd never been a wine drinker. He'd only ever accepted a glass of wine when there was no liquor or beer available. And wine wasn't likely to be part of his image makeover; it was too bourgeois for the image he was cultivating, bearing rumors of Europe or at least California. This was Alaska, where Europeans were fags and Californians were too. He might as well drink espresso and drive a Volvo instead of his hulking Ford truck.
Maybe the wine had been selected for me.
I'd drunk one before, so far with no ill effects, but still.
I put the bottle back.
Some cultural and religious traditions see mind as a property exclusive to humans, whereas others ascribe properties of mind to animals and deities. —Wikipedia 2016
OUR LAST COMMITMENT was today, a dinner with some of the donors and staff. We leave tomorrow and don't have to come back till spring.
I've been torn since the call with Will and Don. Their theory of Ned as a murderer has set me half-against them. It's irrational but I can't help it—I'm set off at a new distance. Their conviction seems to skew them to outlier status. On one hand there's Ned, for whom I feel fear and loathing, and on the other there are these men who've been kind enough to help me, given me time and care. But their murder thesis is an awkward weight on my shoulders I have to shrug off.
I float in isolation between Ned and them, not touching either of the shores.
In the morning I pulled my old belongings from storage, lugged them to the post office and sent them to my parents' house, Lena tagging along with her face in a picture book. In the afternoon I visited my closest friend here, the only one who didn't seem to think the reported kidnapping, or its poor resolution, was the result of my own weakness. Charley, who taught with me at the university and is soon to retire, has disliked Ned from the start, much as Solly did, and it relaxed me to be with someone I didn't have to convince of anything—Charley has a serene bearing and little surprises her. From the trees in her garden hang bamboo wind chimes and homemade birdfeeders.
We sat in her sprawling house full of natural light and drank tea, watching out the big bay window as Lena made snow angels in the backyard.
It was during the snow angels that Ned showed up: his driver had dropped us at Charley's, our whereabouts weren't secret, but I hadn't expected him to take an interest. He'd always dismissed Charley with her hand-knit sweaters, her chunky necklaces made of shell and rock; yet now he rang the doorbell and when she let him in he was with a beautiful girl, doe-eyed and long-limbed, draped in furs and wearing giant, shaggy boots that gave an impression of an adorable yeti.
She might have been twenty-two, she might have been nineteen. She would have been more usual in SoHo or Milan.
Trying to be polite, I think, she pointed at a sculpture on Charley's mantel and asked if it was "done by an Eskimo." When Charley said no, it was a Chinese Buddha, she went on to say Oh with a round, pretty mouth, frozen in wonder. The words were blank as paper: that lovely child was so slow to make connections that it almost hurt to listen to her talk. Maybe she was sixteen, not nineteen or twenty-two, I thought, and it was simple childishness.
Ned bringing her was of course, given his PR focus, his obsessive commitment to the slick campaign, startling. It seemed needlessly risky and certainly meant to be needling. He may have thought I was still capable of jealousy. But I felt only pity for her as she sat, nestled into his side on Charley's deep sofa, long legs drawn up.
Charley, who cared as little for what Ned thought as he cared for her, asked her outright how old the girl was at one point, but Ned intercepted the question and asked Charley how old she thought the girl was.
When Charley said "Too young for you," he smiled and trailed his fingers along the gazelle's spaghetti-thin upper arm. With her furs off she wore only a tight dress sparkling with gold flecks, and the arms were full of holes made to look like knife slashes.
They didn't stay long, only long enough to accept Charley's offer of coffee with disinterested shrugs and then leave before it was finished brewing. The two of them stood briefly at the big bay window, from which Ned—one arm strung over the young gazelle's shoulders—watched Lena run across the snow for a few seconds while his girlfriend looked down at her phone, texting with lightning speed. When it came to texting she wasn't slow at all.
"Place hasn't changed one little bit," he said to Charley as they were leaving, in a clearly insulting tone. He turned, smirked and pointed at me. "We'll pick you up at six. Cocktail dress in the master bed."
Charley looked at me for a long time after the door closed, shaking her head. Meanwhile Lena was still playing in the back by herself; she'd never noticed they were there.
BONES THAT FED OUT THEIR COLD
MY BROTHER'S APARTMENT IS SMALL. HE MAKES DECENT MONEY for a young guy working at a start-up, but this is Manhattan—where he was lucky to get five hundred square feet in a building with roof access.
So he sleeps on the couch and Lena and I take the bedroom. He wakes us up by coming in to open his closet; Solly's a sluggish awakener and every morning he stands there tousled and half-asleep, swaying faintly and staring at his row of shirts on their hangers. The shirt indecision paralyzes him.
I promised we wouldn't stay for long, this is a quick visit, but he waved away that promise when we arrived and said we could stay forever, if we wanted to. Lena nodded solemnly.
"Forever, Uncle Solly," she agreed.
Forever means two weeks. I feel safe in this prewar ziggurat with its thick walls and overheated air. I don't love the city at this time of year—the way white snow turns to gray slush, how the freeze of the sidewalk reaches right through your boot soles. But it's good to see Solly, and I need a break before we go back to Maine.
Whenever I call Will he brings up his worry about Ned, his fear that Ned's going to have me hurt or killed. It makes the conversations strained. I was so pleased by his quiet bearing when I first met him, his calmness that had an almost mystical quality. But now that quality is gone, its glassy surface has been broken and doesn't seem to be smoothing out again. He's still soft-spoken and kind, but there's wariness when he talks to me. I know he feels he should be here—whether he wants to or not, he believes he should be near enough to guard me, that it's somehow his responsibility, which is preposterous.
Conspiracy theories are a mostly male hysteria, it seems to me. That style of paranoia isn't my own—it has a self-importance I don't relate to. Even now, when I know for a fact I've been conspired against, it's hard for me to believe in conspiracies.
Ned acted against me not because of who I am but because of who he is—I'm just the one he happened to marry. And the kidnapping was only a conspiracy in that he hired some people and used others.
Without Will in front of me, though, the attraction is more abstract. Was it only a wishful idea? It was my idea, I know that, I asked him out and brought him in—but the newness of knowing him and Don makes them feel less like fixtures in my life and more like bystanders. Only the Lindas, with their earthiness, seem concrete and reliable.
Lena and I need relief from the closeness of the small apartment, so we do her lessons in a coffee shop. After the morning rush has subsided the place is colonized by mothers and their goggle-eyed toddlers, who stagger around banging plastic toys on the backs of chairs and gumming them; the women chatter to each other, brooding on nests of scarves and coats. Lena takes the roaming toddlers under her wing, holding their hands and showing them colorful objects. She's popular with the mothers for this.
Most days when Solly gets home from work the two of them go out to a nearby playground; she doesn't mind the creaking freeze of the swings, the burn of the icy slide. Sometimes I walk out of the lobby with them, wave goodbye as they cross eastward to the park and then veer west myself. I walk to the Hudson River, past a bagel shop, bodegas, some kind of pretentious cigar lounge, and an opaque window whose neon sign reads HYPNOSIS. QUIT SMOKING / LOSE WEIGHT / MANAGE GRIEF.
YESTERDAY IT WAS the Lindas first on Skype, then Kay. When Lena and Kay had finished singing together, a tuneless song about a mermaid, she ran off to build a LEGO castle and I slid into Solly's desk chair in her place.
I was dismayed at how Kay looked. She had the same hollow-eyed face she'd had when she first arrived at the motel—ghostly pale. She and Navid hadn't reconciled; after her meeting with the linguistics scholar Navid had spun off, his behavior erratic. He said he couldn't trust her again because she had concealed too much.
But we don't know how much we know, she said unsteadily, or we don't know how little other people know. None of us ever possess this knowledge. We can't know what others are thinking.
"It's like a kind of instinct we go on, right? After we get reassured we're not crazy. You know what Don told me?" she asked.
It was hard to hear her so I raised the volume on the laptop's speakers.
"He told me there are crowds of people who never get to that point, they never cross that barrier. People who hear and never stop thinking they're just insane, spend their whole lives on Thorazine or getting ECT. Living their lives all alone. And sad. We're just this small fraction of people who, basically, refused to believe in our insanity."
She hadn't meant to keep secrets, she just hadn't talked enough, she guessed. And now Navid was gone, flown back to Los Angeles. If all this was, he'd said, was some kind of off-brand encounter group, he might as well bite the bullet and do the real twelve steps. And when it came to AA, he had said, or NA or GA or CA, L.A. was the nation's capital.
"I'm sorry," I said, watching her cock her head to one side in the jittery connection. I had the fleeting illusion that she was preparing to keel over sideways in slow motion.
But she didn't say anything, just gazed at me, so I kept on talking.
"I don't think you were holding out on us, but I still want to know everything you know."
"There are so many words for it," she said.
I felt alarmed as I gazed at the fuzzy image of her face, the brown half-moons beneath her eyes. She always looks pretty, with the waifish delicacy of a ballet dancer, but there was a distraction to her expression.
She's not paying attention to her own welfare and no one else is, either. She has no one to take care of her yet I suspect she needs help. I want to call her mother; I wish I had her mother's telephone number.
It can't be my job, though, to look after Kay as well as Lena—not now, especially, when I've failed so dismally with my own daughter. I'm not equipped.
"It is language," she said. "The same kind that makes your body work without you telling it to. You know how the brain runs your kidneys, say, or tells an embryo how to grow in a pregnant woman? What's the difference between that kind of implicit, like, limbic OS for our biology—and for the biology of all animals—and just a miracle?"
"I don't know," I said.
"It's part of deep language that runs these operating systems for us. You see? It's not the language we speak. I mean our language comes from it, like all language, but our own specific language is like the surface of the ocean, the very top line of the water. Just the line. Deep language—I mean I happen to call it that, but there are other names—it's the rest of the ocean beneath, see, Anna? It's the rest of the water below, and it's everything the rest of the ocean holds, that makes that thin line of surface possible."
She was doing something with her hands behind her head—scooping her hair into a ponytail as wings of the hair fell forward around her face. She kept talking faster and faster and shook her head as she did this, making it hard for me to hear; the volume was already at maximum. I wondered if she was manic.
That has to be it.
"See Western medicine doesn't come close to understanding the body, that's part of what I learned in med school and my residency, for doctors, we have to act like we know things, 'project an air of competence,' is what they said to me"—here she used air quotes—"but let's be serious, it's a crapshoot, with anything in the least rare, whether you can get to a diagnosis that works and maybe jury-rig a cure for it. Medicine's more guesswork than the AMA wants patients to even think about, if they knew how much of a gray area there is they wouldn't believe a thing we said—"
"Mama," said Lena, behind me. "I can't find it."
"Shh, honey. Just for a minute. I'm trying to hear Kay."
"We'd never be able to tell our brains how to manage the body's systems, so much more sophisticated than our self-awareness," went on Kay, and now she was fiddling with an earring and in the process turning her face away from the computer's microphone. ". . . colonies of microbes—billions! Not to save our lives! What I got from Infant Vasquez, what I didn't have time to tell Navid, is that system . . . one aspect of deep language . . . the other—"
"Mama," repeated Lena, apparently deciding Kay's desperate monologue was background noise. "I can't find the bottom LEGO piece, you know the one you make into the floor? I can't find that big flat green piece to even build them on, Mom. I swear, I looked everywhere!"
"In a minute, honey, just a minute, OK?" I said, flapping a hand at her impatiently, but I'd already missed what Kay was saying.
Then Solly and his new girlfriend burst in the door stamping snow off their feet, his girlfriend whom I'd never met before was smiling at me expectantly, so I made my excuses to Kay and got up from the computer.
Language extinction has occurred quite slowly throughout human history, but is now happening at a breakneck pace due to globalization and neocolonialism—so rapidly that, by 2100, 50 to 90 percent of languages spoken in today's world are expected to be extinct. —Wikipedia 2016
LUISA WAS SITTING with Solly and me in his kitchen/dining room/living room (Lena had gone to bed) when we got the call from my mother.
Solly put her on speakerphone.
Our father had been losing weight and sweating at night, she said—so much that he soaked the sheets. They'd gone in to see the family doctor and the doctor had sent them to a specialist, where he'd been biopsied.
"Why didn't she tell us this before?" asked Solly, after punching the mute button. "A biopsy?"
"I didn't want to bother you, in case it wasn't anything," she said.
I guess the mute button doesn't work.
"Sorry," muttered Solly, but he was already distracted by the import of that.
"I'm afraid it did come back positive," she said. "A fairly common cancer of the blood. 'Hematological malignancy,' they said. We don't have the staging on it yet, but we should know soon and I don't want you to get too worried just yet. OK? It's not necessarily a dire prognosis, depending on the staging, of course, whether it's metastasized—it doesn't have too low a five-year survival rate. More than half of all patients pull through. Maybe even three-quarters, we'll see. So your father's chances aren't so bad."
Luisa squeezed Solly's hand, her dark eyes glittering. Solly and I looked at each other steadily.
"Do they have a treatment plan yet?" asked Solly.
"There will probably be chemo," said my mother. "Possibly radiation, possibly surgery. I'll share all of that with you as soon as I know more, dear."
"Blood cancer," I said, after a silence. I'd begun to feel uneasy—beyond even the facts of the case I felt a creeping apprehension. "That's where . . . isn't that . . ."
"It's where the white blood cells divide faster than normal cells, or live longer than they're supposed to," said my mother. "He has at least a couple of primary tumors, which they tell me is a common presentation. With this kind of a lymphoma."
AFTER WE HUNG UP I told Solly what Ned had said to me before: lymphoma. I described it to him before he left for Luisa's place for the night, right before I took out my laptop and began typing this.
But he shrugged it off as though the detail either wasn't accurate or wasn't relevant. Our father has a disease, our father has a potentially terminal illness of the kind we all fear for the insidious poison of its medicine, the emaciation of bodies, shedding of hair, desiccating of bones and aging of skin. That was all Solly had room for, and I can't blame him.
And our father will have to endure all that without ever understanding his illness. He'll be like a child throughout the suffering, confused and blinking as my mother herds him gently on.
I think of those scenes to come and I also think of my father when we were young and he was middle-aged instead of old—how he read us stories using different voices, some deep, some squeaky, here a quaking mouse, here a growling lion. I think of how he carried us on his shoulders—"so you can pretend to be giants."
He had so much dignity back then, but he was willing to cast it off to entertain his children. He tickled us until we grew out of being tickled, he made corny jokes until we grew out of those too.
Now I feel an ache of remorse when I think how we stopped laughing at his jokes. I would laugh so hard, if I could have a do-over. I can see that to Solly we're only losing my father now, where to me we lost him some time ago—or maybe it's fairer to say that Solly seems to be able to lose him twice, while for me once was all I could do.
Still Ned's casual assertion a few weeks ago, his matter-of-fact statement that my father would get sick with lymphoma—which at that time I assumed was just a fictional element of the so-called narrative—vibrates so hard I almost get a headache. I've actually been taking painkillers when the thought of it starts to make my temples send out their thin flashes of pain.
But Ned's foreknowledge vies with the diagnosis for my attention and I can't let it go. It may be coincidence—or maybe it's information gleaned from surveillance. Could he be surveilling them as well as me, tracking my father's diagnosis? Observed by Ned or his consultants, did my mother find out weeks ago and only tell us now? And what use would it be for Ned to spy on my parents anymore, when he already has my cooperation, when I've already done what he wanted me to do?
I'm going to ask my mother tomorrow when she heard the diagnosis. I'll reassure her that it's not a problem if she decided to delay telling us—we understand completely. But I need to know when she heard.
YESTERDAY, she said.
And the whole earth was of one language, and of one speech . . . And the Lord said, Behold: The people is one, and they have all one language . . . and now nothing will be restrained from them . . . Let us go down, and there confound their language, that they may not understand one another's speech. —Genesis 11:1–7
I HAVE IT—I have it here on my desktop, a written record.
It's in the "templates," as he and his staff call them: the schedule for the narrative, with our travel dates; the list of his positions on issues, which I'm supposed to know even though I won't parrot them, and a partial list of planned public appearances, both with Lena and me and without us; the breakdown of campaign employees by job description, plus key volunteers. All this is supposed to be memorized before our next stint in Alaska.
It's so repellent that I hadn't looked at it after a cursory glance, but here it is. The templates are connected to my laptop's calendar, which I don't use for anything else, with events assigned to months or weeks or days. The events pop up, color-coded, and I can't take them off again—I tried once and it gave me a message about contacting the administrator.
Apparently I don't have permission.
The developments connected to my father, and therefore my extended absences from Anchorage and Ned's campaign, are lime-green bars extending across several different blocks of days on the calendar.
They're labeled like this, on various dates:
LYMPHOMA STAGE 3. DIAGNOSIS, PROCESSING
TREATMENT MODULE 1: SURGERY, CHEMOTHERAPY
TREATMENT MODULE 2: RADIATION
METASTASIS: BONE MARROW, CEREBROSPINAL FLUID
And there's one I didn't notice before, a little further on.
PALLIATIVE CARE/MEMORIAL SERVICE
"Lymphoma Stage 3" is assigned to this month, the month we're in right now: February.
I called Ned, I left a voicemail for him asking how he knew, but I strongly doubt he'll tell me anything at all.
He typically has his staff email me when information needs to be exchanged; he and I don't communicate.
"STAGE 3," said my mother, on the phone again.
I'M PASTING IN an email I got from Kay, strange and dense. I think she may be bipolar.
You said you wanted to hear everything I know. So OK. So I have trouble explaining how I know it & what it is—writing isn't my thing. I mean I was more the organic chem type!!! I used to get visions of like resonance structures & chair conformations & stuff, when I was holdig Infant V. But so. You know how I told you we r the only ones it leaves, what I meant was, it doesn't leave the whales or the crocodiles, it doesn't leave the plants & the trees, & that's not because, like, theyre dumb. Theyre not. Deep language is in all living things but all the others, it stays with. Only not humans. Its because the other things, apes, cats, even the grasses in a field, don't live just for themselves. They live for the group. They live for all, this whole of being. We used to be like that to, once a long time ago, once in our evolution, I don't know when but once. But slowly it chaged & now we live for ourselves. So the deep language does'nt stay with us when we get our own, our surface language, you coud call it. We split off from it then & are forever alone. God leaves us Anna.
God leaves us.
I can't tell how much is rumination or fabrication, whether some is intuition, how much she was given to know. In short I'm not sure if she has much authority.
But I'm keeping her message. I read it over in quiet times.
MORE GUESTS ARE leaving the motel, Big Linda reports, all vowing to keep in touch—I've started to check in on the Listserv, where so far Navid's the only one absent. Regina and Reiner have gone back to their professions in the city, and Gabe has decamped too. He cited the needs of a lonely Bedlington terrier, pining away under the care of a neighbor back at the condo he shares with Burke, who's soon to follow him home.
And what did they accomplish with the meetings? I get Navid's impatience, though I wish he'd been nicer to Kay. Unlike me, the rest of the guests knew about each other before they came—they had an earlier version of the Listserv. They'd already exchanged messages containing much of what they'd say later, alongside the table of watery coffee and stale cookies. So I was the only new element. And they can't have got much from me.
I never illuminated anything.
I account, on my fingers, for all the elements of these events I keep failing to understand. I wish I had an abacus—confusion like this calls for a deliberate, manual counting, a ritual of organization. Digits or beads, bones or a rosary. Even assuming there does exist an ambient language that underlies life, what some people call God, others possibly photosynthesis or humpback song or the opinions of a dog, I have the same questions that I always did. I want to know why I heard it, and why through Lena; why it fell silent when she slept; why it departed when she said her first word. I want to know not only its rules but its purpose, but all of that remains opaque to me.
There are the practical questions, too: How did I know to go to the motel? How was Ned able to find me? How did John know to contact him, when I took my car in to his shop?
And how did Ned know my father's diagnosis?
On the face of it my questions about Ned are in a different category. And yet there's the lymphoma diagnosis. This is new, this introduces a fresh mystery, and it counts just as fluidly on my fingers as the questions that came before.
It was recorded digitally, "Lymphoma Stage 3," a number of weeks, not days but weeks, before the doctors even biopsied my father. It was set in stone then, it has a path, a history that can be verified—the fact that he had that information, or at least that he acted as though he had it.
Lymphoma Stage 3.
I WAS ALONE in the subway today, coming back from getting my hair cut, during which appointment Lena had stayed at home with Solly and Luisa. I was on a crowded platform at Columbus Circle with my bag over my shoulder and a book in my other hand—I must have been standing distractedly at the front hem of the crowd, my paperback curled back on itself in my hand as I read.
Then there were the lights and roar of the train. I felt a push behind me like a head butting against the small of my back and suddenly I was teetering, one of my legs over the edge, before someone grabbed my arm and my book flew out of my hand and I was jerked back, the tendons in my neck strained and one shoulder wrenched.
With a rush the train was screeching to a stop, people surging past me as the doors opened, jostling me and turning me around. I felt a weird heat prickle where my scalp meets my face, was breathless and seeing spots of light. Somehow I found my way to a bench, newly vacated and still ass-warm.
I never knew who pushed me or if it had been an accident, or who caught my arm and saved me either—maybe the push was just the movement of the crowd, that's the likeliest explanation. Right? But as the train pulled away I noticed a child staring at me through a train door, a dark-hooded child with a white face, and the child's head turned as the train moved, the child was staring at me fixedly . . . it had been a forceful push, so forceful it seemed it must have been purposeful.
Or so I felt as I sat there.
As soon as I got over the shock a wave of gratitude washed over me, a pure beam of gratitude struck out toward my unknown rescuer—how impossible it always is, I thought intently, to remember how lucky we are each second we remain alive.
When the train was long gone and the platform bare, I got up shakily and walked back to the edge. On the tracks was my book, ripped up and streaked with gray, its pages spread over the black. I gazed down at it for a while and then sat down again to wait for the next train. I wanted to call someone, maybe Will, maybe Solly, but of course there was no signal in the tunnel. And what had happened, anyway?
When the next train finally came rushing in I found I was trembling. I had to press my back against the cool, grimy tile of the wall. Presently I left and hailed a taxi.
Since then my day has been cast in a fractured light. I go back and forth between telling myself it was pure accident and wondering if Don and Will's fears deserve more serious consideration.
I SENT AN EMAIL to Navid. Can I find out online, I asked him, who's financing Ned's campaign? I wouldn't mind knowing who Ned's backers are, what interests they represent and how deeply embedded their money is in institutions. Maybe one of them has connections to hospital records, who knows, some shadowy X-rays that were interpreted before the biopsy without my parents' knowledge—some link that would provide an explanation for that premature diagnosis.
Ned left his family of origin when he was in his teens, left and never looked back. His father had disappeared when he was an infant, his mother was strung out or drunk all the time, and there were no others. He lived outside the house anyway, from when he was twelve or thirteen, only returning to sleep. This was what I had gleaned, anyway, from the couple of times he'd talked about it to me.
But somewhere, now, he has another family. I want to know who his new family is.
WALKING BY MYSELF to get a carton of milk, I suddenly spun on my heel and entered the business with the HYPNOSIS sign. I hadn't planned it.
There was no receptionist, only a counter with a fiber-optic lamp sitting on it, an abstract medley of colored lights pulsing. I wondered how a hypnosis business made the street-level storefront rent in this neighborhood; I rang a push-button bell on the counter and heard an electronic chime. After a minute a woman came in from the back, a woman with a soft, homely face and wavy hair. She was about to close up for the evening, she said, could she help me?
There was a voice, an auditory hallucination I used to have, when my child was a baby, I told her. I wanted to remember it now—wanted to hear it again to see if I could figure out what it had said. Could that kind of memory retrieval occur through hypnosis?
She asked me if the voice had issued instructions, had told me to do anything I didn't want to do.
I said no. No instructions.
She asked me a couple more questions I guessed were supposed to screen for mental illness, then hemmed and hawed briefly. She said there were no guarantees, that it was up to me, in a sense, what was accessed, but sure, she was willing to give it a shot. She could implant a suggestion that this "voice" return, she said; she could invite my mind to generate the "voice" again.
She had me sign a waiver and I made an appointment.
NAVID WROTE BACK saying he'd research Ned's funding. He was good at following money trails, he said. Somehow he doesn't seem to blame me as he blames Kay and Don; with me he doesn't seem to have a bone to pick. Or maybe, because of what happened to Lena, he's just sorry for me.
USING VACATION TIME, Solly's going to visit my parents and taking Luisa with him. He wants to be there to help out, as he puts it, but has urged Lena and me to stay here in the apartment without him. All four of us descending on my parents would be a burden and not what the doctor ordered.
It is possible that all languages spoken today are related through direct or indirect descent from a single ancestral tongue.
—Wikipedia 2016
THE PRACTICE OF HYPNOTISM seems to hover in the alt-medicine gray area, near chiropractors and acupuncturists. It's viewed as sporadically effective in treating certain bad habits and disorders, but tarnished by its history of showmanship.
The hypnotherapist had me lie down on a huge, brown recliner with wide arms—my arms had to be stretched out, hands laid flat, feet raised. She dimmed the lights, put on music, and asked me if the room's temperature was comfortable.
And I had to admit the temperature was comfortable. The air felt like a soft extension of my skin, without too much moisture or too much aridity. I could stay here, I thought.
A person could remain.
"Remember, I can't make you do anything you don't want to do," she said. "This is a completely safe space."
She had me close my eyes and listen to her voice describing a wooden rowboat over a deep blue lake. Out we went into the lake, rowing, rowing. Maybe I dove off the side of the boat or sank into the water, deeper, deeper, deeper; or maybe I was just looking down, looking into the water from the dry bench of the boat. I recall the color blue, the clarity of the lake water.
During this tranquil immersion a jellyfish floated up from the depths. I don't know whether it was associated with the therapist's words or only with my thoughts, but I gazed at it—a pink-white bulb with tendrils rippling. Although I wasn't asleep or dreaming I knew in the way of dreams, the passing of information that happens there where one thing is simultaneously another, that the jelly, having no place in fresh water, was an emissary from the ocean Kay had spoken of.
There was something to know here, something to discover. So when I left I made another appointment.
AROUND LUNCHTIME YESTERDAY we got a call from downstairs: Will was in the lobby.
We went down in the elevator to meet him and there he stood, talking to the doorman.
Lena ran to him and hugged him and then turned her attention to the doorman, her friend. Will stepped away from them and turned to me, a woolen cap in his hands, the shoulders of his coat sparkling with melting snowflakes, and I was so happy to feel my stomach flip, to know how much I still liked him. More, even. His eyes, skin, mouth.
"I brought your car," he said.
I'd been selfish. I'd given him nothing, and I'd added insult to injury by doubting him. Yet here he was.
He didn't ask to stay with us, in fact he had a friend's place lined up, but then he did stay.
It was good but curious, after so long a time—like walking through a forgotten wood. Like wandering beneath old trees, whose faint smell reminds you of a person you may once have been.
Not only does Will know now about the motel's Hearing Voices Movement—as I've come to call it privately—but he's known about it all along.
He knows the backstory of the motel guests; he's familiar with our group pathology. And he has known about it all, he says, since a couple of years after he got to know Don, when he first moved to Maine. Don has always lived there, as far as Will knows, like his father before him. He's a feature of the landscape and has never seemed to do anything but what he does now.
"But that's the thing. What does he do?" I asked.
Will shrugged.
"He's a host."
We were in bed. I was so glad to be there, though at first it took me a while to relax about Lena, who was fast asleep in the bedroom and still too near for my sense of propriety.
"So confused people who hear voices have been coming there for years," I said. "All of us with that same complaint."
"You don't all seem the same to me," he said.
The only unity I'd found in the guests was economic: none of them were poor. There were men and women, young and old, white, Asian and Iranian and Dutch Americans, straight and gay. We had no profession or other clear trait in common save money—everyone was at least middle-class, no one was on food stamps. I'm a former academic, Kay's a med student, Navid a producer; Burke is a botanist and between them the Lindas have three master's and a PhD.
"That's true," said Will. "Because the poor don't weigh in on the channels Don uses to bring his guests together. He can't find them because he can't separate them in the social-service world from the population with schizoid conditions. They may be institutionalized or on the streets or just toiling, but they don't tend to be online so much. He doesn't have a way to get to them."
But Don never found me online, or if he did I didn't know of it. I wondered if Will knew that too.
"Why does he want to bring them in?" I asked.
Maybe it was just group therapy, as Navid had alleged, I was thinking.
"He says it's just his role," said Will.
After breakfast we sat on Solly's cheap, caving-in couch, which pushed us together comfortably in the middle, as Lena played with a magic coloring book Will had brought for her. Depending on how you flip through the book, its pages are blank or black-and-white or startling full color. Lena had wanted to do the trick in our coffee shop, but only a one-year-old had been present, on whom the trick was wasted. Babies think magic is normal, she said.
She flipped through the book as Will and I sat against each other, my laptop on my knees, his arm around my shoulders. Then I brought up the schedule and stared at it. Where before it had annoyed me, now the bristling field of white seemed ominous. Onscreen it didn't seem inert, as any other file would, but almost radioactive: it bore the weight of grim prediction.
By Ned's reckoning, it appears—or the reckoning of his aide or campaign runner or secretary, whoever created this schedule—my father will begin hospice in June and die before Independence Day.
NAVID CALLED ME on the phone Will just bought me, his face popping up on the screen. I'd never bothered to use my cell that way before. He was wearing a headset and seemed to be sitting in a car: I saw the curves of a headrest behind him.
"Are you alone?"
I was trying to figure out how to hold the phone so that he didn't see the inside of my nose or ear. "Lena and Will are here. Can we just talk normally?"
"Yeah. I wanted to see it was you," he said. "Now I've seen."
"Did you find out anything?"
"So his donors fall into two categories. Industry kingmakers, the ones that run the politicians, first. Then there are others—also rich but not as rich, one or two have as much as half a billion in revenue, sometimes their wealth is shared among smaller entities or they're hidden behind so-called educational groups, these 501(c)4s—a big corporate entity of biblical literalists that owns hundreds of radio channels, for example. Those guys are his other backers."
"It's not so surprising," I said. "He's been talking the talk."
"It's how he found you," said Navid. "Turns out these guys have citizen networks. I wouldn't call it grassroots, there's too much money moving around for that. Or let me put it another way: there's money at the top and blue-collars at the bottom. Far's I can tell, the money at the top talks about ending the separation of church and state, making biblical law the law of the country. Like sharia, right? But Christian. End-times bringers. They use this shit to get the blue-collars to do their dirty work. It's cynical. So your husband's friends put out their version of an APB, you go down as a threat on the list they give out to their little-guy helpers across the country, and you're a target. I'm guessing it was your VIN that tipped them off. You took your car into the shop, right?"
"My VIN," I repeated slowly.
I thought of Beefy John and the radio poster on his wall.
"Ned knew my father's cancer diagnosis before the doctors told my mother. Before there was even a biopsy. So I'm thinking maybe there was a scan or something, maybe the doctors knew earlier but just needed the biopsy to confirm to the family. Maybe someone connected to him had access and knew the probability."
"I couldn't tell you for sure," said Navid. "I can't get into hospital records. There are people who can, but it's not my bailiwick."
It struck me after I hung up with him that he'd spoken faster than he used to when he was staying at the motel, he was energetic and focused but without the anger. High.
WHEN I PASSED along Navid's discoveries Will gave me a look like I told you so—why I'm not sure, since there was nothing in what Navid had said that would link Ned to murder.
I keep trying to see clearly. There are clear signs out there, I feel sure, but all I can make out are the blurred edges. I feel a ghost of pressure on my lower back, the push that felt like the crown of a head. I remember a hooded child with a white face. Male or female, I don't know, but whatever it was stared out at me from a window of the subway train.
Either it was a simple accident or it was Ned's agency, no matter who was acting on his behalf. If it wasn't connected to Ned it shouldn't matter, since in that case it must have been pure accident. And it's a characteristic of accidents that they don't often identically repeat.
A programming language is an artificial language used to communicate instructions to a machine . . . thousands have been created rapidly in the computer field, and still many more are being created every year. —Wikipedia 2016
ALONE IN HER SMALL walk-up on Beacon Street, Kay took an overdose of sleeping pills last night. She lived but they had to resuscitate her, and now she's in a coma.
There was a bipolar diagnosis, as it turns out.
I can't bring myself to tell Lena. I should have been there to look out for Kay, should have done what I could: something. I seem to plod along in my own tracks, following footsteps I made before; this is always how I proceed, I don't look sideways, I'm not willing to stop. I was inside my own concern, blindered like that—worried about abstractions, worried about the future when for so many people, Kay for instance, the present is already a state of emergency.
I overlooked my duty for the sake of my convenience.
Will tries to tell me it's not my fault. I know I didn't cause it, but I didn't stop it either. I see what he's doing and I know it comes from affection, but listen: This is what we do for the people we're close to, all in the name of comforting. We ease the path for them to excuse their own failings.
We let them off the hook and call it love.
The truth is bare—I abandoned her, that tall, sad girl.
WHEN I WENT back to the hypnotist I was like an addict. I rushed out of the apartment with the usual weight of guilt clutched to myself. The sessions are the only times I've left Lena with Will. And I do trust him, but he's not family.
I saw a city, mile upon mile of buildings, a cluster of tall commercial ones at the center and then, moving outward, the residential blocks, the tree-lined streets. The buildings were dilapidated but elegant, there was a detail of ornament to them like the tiny lines on an engraving, the careful, hair-thin lines of pictures on paper money.
The cave-in began in the distance, with the smallest, farthest buildings disappearing first, only visible as yellow clouds of dust billowed up and curled in. Like puffy hands clenching, I thought: beneath the furls of dust buildings were collapsing. Above them something dark raked down from a cloudbank, fouling the air.
I was standing in sand, sand that used to be an ocean and would be ocean again. I stood on the edge of the city as dust rose from the falling buildings. But these buildings were made out of words, locked into each other like bricks and beams—small words, minuscule words, inscrutable as seams.
Ned was coming, flying in from the west. His advent turned the distant sky black. Before him he whipped up a slave army, a crowd of gruesome flying things that drove billions of insects before them, clearing a path. The cloud was made of words too but the words were deformed, they meant confusion or blankness or insidious poison. What light filtered through them was cadmium yellow and leaked a slow disease.
I HAVEN'T SLEPT well lately; I often sit staring at the screen of my laptop while Lena and Will are sleeping. I sit there and stare as the screen resolves into dull letters or right angles of light and fades into disinterest again. It was open to my inbox and at some point I noticed, on the left panel of the page, that my spam folder said 172. I clicked on it and was about to Delete all spam messages now when I saw, buried between an Enlarge Your Manhood and Hot Women in Your Neighborhood, another email from Kay.
I felt sick for a second, scared it was a suicide note or a goodbye letter—so sure I sat there for a long time gazing out Solly's window at the yellow and white squares in the buildings, tall rows of windows rising into the night sky. It's a sight I've always loved, assumed everyone loves: columns of lights in tall buildings at night in the city. Beneath them was the irregular black solid of the park's treetops.
Finally I looked back at the screen and it wasn't as dire as I'd thought. The date and timestamp were there as always, on the right: Kay's message hadn't been sent the day she took the pills; it had gone into my spam folder two days before.
Still, five more minutes passed before I was willing to click on it. I sat on and on at Solly's desk, counting the rows of yellow squares hovering midair, wondering what forms of life moved in the darkness of the park below.
The problem is, now, were going to be nothing BUT surface language. & no safeties, no backups, no checks & balances. The future is nothing but language, see, not languageS but language. Monolithic. The little ones are dying off @lightning speed. Programming Language, ad talk, 1 speech for all, a juggernot, that's where we're going Anna. All the native languages dead, all we'l have left is shells & false things & tongues spoken for profit &/or by machines. Don't u c Anna this is the tru end of God. When everything that lives the deep language dies. This is the end of God and not the fake god made up to look like us, not that fake god anna, the real god, the god tht IS evolution & speciation & Life, a god that did make the world, u see?—b/c this god is the beautiful unconscious, it is billion processes & intuitions under all of biology & personality & art, the thousands and millions of cultures of both Man and Beast. We're killing that deep god ana, the speakers of false language are suffocating the deep, they are the oil on the water beneath which all suffocates & dies
Satan is God weaponized
God weaponized by man
Now is the point of danger b/c true language is the Soul Anna, tru deep language the soul & the soul can be ruined. God needs us Anna, as much as we need god
I WONDER IF Ned's allies are mostly true believers or, like him, mere opportunists.
Will believes, like Don with his geese and songbird migrations, that I found my way to them via some kind of homing instinct, since a couple of others over the years have showed up without prior contact. He thinks it's part of the background orchestration of the deeper language, an urge that underlies our patterns of survival.
It isn't that I learned nothing at the motel, only that as soon as I learned it I seemed to always have known it, yet still feel I know nothing at all. Burke with his speaking tree, Linda with her theme-park whale, Kay with Infant Vasquez—I picture Burke's maple in its arboretum, planted halfway across the world from where it evolved, a lone specimen with a plaque in front of it bearing its names, both Latin and common. So unlike the aspen that grew not far away from that arboretum—those cloned aspen, connected underneath the earth, that lived as one for what could be millennia . . . I watch a pigeon strut around on Solly's windowsill, dirty but free, and wonder about the orca in its pool, its home only twice the length of its body.
They did have something in common, all those the voice spoke through: they were captives. Even Infant Vasquez, who quickly died, or Lena, who lived on and spoke. All infants are kept creatures, after all. I remember how snatches of poetry were given out to unfortunates when we passed them; I think of prisoners and victims and martyrs, the persistent notion of their closeness to God. I think of how a tinge of the divine rests on the hurt or unfortunate, how so many of them wear a kind of halo of gilded pity.
But if the injured and wretched are closer, what does it point to? Likely we give the poor and weak and sick their halos reflexively, I think, to make it easier to detach from them and not have to do fuck all. We give them sympathy in the place of help. We say they're not like us, they're sanctified and only half-human. They might as well be on a cross.
I recall acutely how abjection makes you a part of a herd. The kidnapping left me feeling robbed, not just of my assumptions about freedom but of my personality—no one has personality when their leg's being amputated, no one has personality when their eye's being poked out. You don't have any selfhood when you're suffering extremely: in suffering you could be anyone. Whether that makes you everyone, though, is a different question.
And I don't like the proposition that suffering puts us closer to each other. That suffering isolates the sufferer—this is equally valid.
So Will has comforted me over Kay. He's trying to be kind, of course, and I'd do the same if our positions were reversed, you don't question the rightness of trying to comfort someone. As behaviors go, it's universally acclaimed. Yet he told me there wasn't anything I could have done, when in fact there was: I could have done more than nothing.
I think of the duress that can be brought to bear on a soul, how selfhood, which we depend on so completely, is a luxury good.
I turn my palms up reflexively, thinking of those who suffer their whole lives. As though the gesture would make me one of them.
WE LEAVE SOON, after one last hypnosis session. Kay has been moved to a hospital in Boston, near where her parents live. We will visit her there on the way to see my parents.
LYING IN THE RECLINER I found myself walking along an institutional hallway, following green footsteps on the white floor—the footsteps were color-coded to the different wings and there were colored lines along the ceilings, too. I walked with deliberate steps until I came to a room.
An older woman sat in a chair, knitting with blue-gray yarn. The nightstand was crowded with propped-open cards. But instead of lying inert in her coma, Kay hovered above the bed. Her levitation had a Buddhist quality—though her posture was comfortable, not a straight-backed, cross-legged stance as in meditation or yoga. She slumped a bit, relaxed, and remained in the air smiling down at me, with a serene quality that's rare inside the confines of real life.
I wanted to rise to where she was, but I couldn't, so at an angle from each other, she high and me low, we gazed out the window. Out there was the crumbling city of words, much as I'd seen it before, though farther in the distance, dust rising from its slow-motion collapse. Kay nodded and stared. Her face had a kind of shining, imperturbable sadness like a bronze statue in a park, somehow civic.
I followed her gaze back to the window again and saw it wasn't a window after all but a computer screen.
She wouldn't explain at first, though her face kept on gleaming with a smooth and oddly official grief: yes, her grief seemed ceremonial. It was a stately mourning, like a dignitary presiding over a state funeral.
Expository words scrolled quickly along the windowsill.
IF Our symbols are corrupt. IF Our tools are made of symbols. IF We are made of our tools. ∴ We are made of our symbols. ∴ WE ARE CORRUPT WE ARE CORRUPT WE ARE CORRUPT WE ARE CORRUPT WE
The last sentence ran on repeating forever, scrolling across the bottom of the screen like a stock-market ticker tape.
"Think of social-media websites," said Kay.
For some reason she insisted on speaking silently, using a comic-book speech bubble.
"Are you kidding?" I asked.
"Think of all those sites, all those apps, the billions of selfies. Now we filter ourselves through them. Sometimes it's our whole presentation of ourselves to the world. That's all that enters the social sphere—that imprint of our ego is all that ever meets up with the collective."
"Seriously?"
I was sorely disappointed that here, under hypnosis, an oracle appeared and spoke to me, and the subject turned out to be social media.
The oracle had actually said the word apps.
"Lena will be all symbols, by the time she's grown up," said Kay. "I'm sorry to inform you. It's a fact. Nothing but symbols, your little girl."
The lights dimmed in her room, and in the corners dark beings flitted. I couldn't see them but I knew they were only half-alive, hybrids of flesh and machine, and they moved through the pipes in the walls, among the wires and conduits. Those too, the long tubes and threads that were supposed to be inanimate, moved sluggishly behind the drywall. Between the girders they pulled themselves in. Closer and closer they approached.
"Why do you pretend to know everything?" I asked her. "Are you right about it all? Or are you just sick?"
Kay's face kept on shining, turned away from me, but the knitting mother looked up from her bedside chair. Now the hands in her lap, holding a panel of blue-gray yarn that might have been a scarf, were made of metal: robot hands, with clicking needles. Her face was contorted with rage.
"This isn't a dream, Kay. It's more like a horror movie," I said.
She was supposed to be trustworthy—she'd watched over my daughter's sleep, cried to me and told me about her life. But telling a feeling isn't the same as knowing someone, I thought regretfully. We think it is. A piece of the Freudian inheritance. People tell their emotions, tell their emotional story, and think that equates to knowing each other.
The pipes in the walls turned from ducts or sacks to the old bones of patients, bones that fed out their cold onto me so that the hairs rose on the back of my neck and my forearms. Yet when I tilted my head back the ceiling hadn't gone brittle at all but was warm and rotten, like pink foam breathing.
Kay turned her head slowly and looked at me, and when she smiled I saw her teeth were gray, not regular teeth but some kind of ugly digital code that shifted and moved in her mouth.
It looked a bit like hieroglyphs, a bit like 1's and 0's.
I thought: What have they done to her?
Suddenly her mouth opened wide, wider and wider, far too wide. And something ugly streamed out.
"Your little girl won't even need her face," she said.
TO THE WHITE CASTLE
FOR A WHILE LENA AND I ARE GOING TO STAY WITH WILL. I DON'T want to move back into the motel—memories of the kidnapping give the place an edge of chilled hardness for me, replacing the clean sea air, the pine needles I loved for their scent and sharpness, with an atmosphere of dread.
Will wants to be my bodyguard, and if he had his way I'd never be out of his sight. This has a cloying aspect, but more and more, during our last days in New York, I found myself hugging the sides of the buildings as I hurried down the sidewalk. I'd catch myself glancing around to make sure that no one was following me, no one was looking at me too purposefully.
I may not be any safer in Maine, but I want to see trees again that weren't planted by city planners. I'd like to take Lena sledding. I remember Will's house as neat and tasteful, floor lamps instead of fluorescents, old rugs and a lot of bookshelves. And next to Solly's apartment it's the Taj Mahal.
I'VE FOUND a replacement for the hypnosis sessions and this afternoon, our first of three days with my parents in Providence, tried it for the first time. Lena was sitting at my father's feet putting on a show with puppets she'd made out of paper bags; Will was fixing a broken step on the porch. So I retreated to my childhood bedroom, which still bore the dusty traces of my teenage self—the pocked bulletin board that had held printouts of pop-star faces, snapshots of me with my arms around friends, a stray ribbon or two.
One ribbon that's been pinned to my corkboard for twenty years says just PARTICIPANT.
I lay down carefully on the bed on my back, stuck in my earbuds, and cued up a twenty-minute hypnosis track downloaded from a website: "Goodbye to Stress."
All it did was put me to sleep, but I'll try again tonight.
Later Will and my mother cornered me in the kitchen; she plied me with peppermint tea and announced she wanted to have a serious talk about "personal security."
Somehow Will had convinced her I need protection. At least, she said, I could agree that there was a risk and humor her by letting Will install a home security system. Then she could rest easy, she said (and here she looked careworn and shaky—more elderly, I realized, than she ever had before). She already had my father to worry about; she didn't want to have to worry about Lena and me too.
I pictured a couple of sluggish rent-a-cops pulling up fifteen minutes too late, shooting the breeze about their personal lives as they casually dismounted from a company car whose doors were emblazoned with a bogus-looking shield. I don't like the idea of being guarded by electronics, of being sealed off from the world outside. More surveillance, I was thinking—all it's done in the past is harm us. It was surveillance that allowed my daughter to be taken from me.
But my mother looked drained. Resistance was futile.
"It's already being set up," said Will gravely.
Panic welled up: I'd done everything Ned asked, everything I could possibly do to meet his demands, and still maybe it wasn't enough.
My mother advised me to carry mace whenever I go out.
"Or maybe pepper spray, dear," she amended. "I think it's better. For their health. The criminals', I mean."
Hypnosis is ". . . a special psychological state with certain physiological attributes, resembling sleep only superficially and marked by a functioning of the individual at a level of awareness other than the ordinary conscious state." —Encyclopædia Britannica, 2004
It was a quiet and uneventful visit to Kay, who lay, much as you'd expect, motionless on a stainless-steel bed hooked up to machines. We had her to ourselves, as her parents had just gone to get lunch, a nurse told me. Kay's face was a ghostly shell, but Lena sat beside her bravely and held her hand. She only cried later, as we were walking out. I'd told her Kay took too much medicine by mistake.
The private room didn't bear much resemblance to the one I'd envisioned under hypnosis—no surprise there—but one thing struck me as we were leaving: a pile of knitting, two needles sticking out of it, on a low shelf on her beside table.
The yarn was blue-gray.
WE HAD a car accident today.
Or almost had an accident, I should say. We avoided an accident, but it was close.
We were maybe half an hour northeast of Boston on the freeway. It was my turn to drive and I was fiddling a bit with the radio when abruptly the car started weaving back and forth across the lanes, fishtailing. My right hand flew back to the wheel as I felt the loss of control in the pit of my stomach and tried to keep the car straight. I almost hit someone on my left but veered away just in time, and then the car almost crashed into a guardrail on our right.
In the end we veered away from that too, luckily, and somehow I steered us onto the first off-ramp, pulling over onto a wide shoulder without any more near-collisions.
It happened too fast for Lena—startled out of a nap by the car's fishtailing motion—even to get frightened. When I'd pulled up the emergency brake I turned to look at her; she smiled at me uncertainly and rubbed her eyes.
Will and I got out and walked around the car: all four of the tires were flat.
The three of us rode in the tow truck to the car-repair place, where we hung around in a brown-tiled lounge area that smelled of disinfectant while they sprayed foam on the tires, performed some other tests. We were sure I'd driven over a spilled cargo of nails or other sharp objects—what else could have caused four same-time flats?—but finally they seemed to have exhausted their diagnostic tools.
Never seen anything like it, they said. There were no holes or slits, no punctures at all: the tires were perfectly good except for the fact that the treads on the rear ones were a bit too worn for comfort.
They wanted to sell us two new tires.
"Maybe these mechanics are in league with Ned too," I said nervously. Lena was feeding coins into a vending machine, out of earshot, and I watched her as I spoke.
"I thought I was the paranoid one," said Will. "Still. Maybe we should replace all four tires, huh?"
"I don't want to be chickenshit," I said. "But OK."
Will drove after that while I tried to play a word game with Lena, thinking of animals whose names started with the last letter of the animal before. But she soon tired of it and asked to use my tablet for a game, making hairstyles on cartoon people whose faces looked like square potatoes.
When we got to his house I was relieved. I'd sat in the passenger seat with the muscles in my stomach clenched—sat forward the whole way, strained, unable to relax enough to lean back in the seat. The guy who'd installed the alarm system was waiting for us, his van idling in the driveway behind Will's truck, now covered in drifts of hardened snow. Will warned me as we were driving into town, so I wouldn't take fright, I guess—that's what I've come to, apparently. I have to be warned about the presence of men in vans.
We all went up to the door, rubbing our gloved hands together in the cold, the installer chugging along beside us, a drunk-nosed man with a beard. He let us in and walked us through the system, whose electronic display looked out of place amid the weathered wood trim and old furniture. Lena was puzzled by the setup, asking why we needed to touch a display to come in. We hadn't needed to before.
"It's like Doug," said Will. "Solly's apartment building has a doorman to watch over it, right? But we don't have Doug in my house so we're using this little guy right here." He rested his hand on the console.
She'd loved Doorman Doug, of course, who brought her puzzle books that featured the Mario Brothers, with a few of their yellowing pages scrawled over long ago by his now-teenage sons. Lena did not prefer the Mario Brothers. They were strangers to her.
But she liked Will's explanation and named the alarm console New Doug.
ON OUR FOURTH afternoon back in Maine, while Will was off at the library, Lena wanted a nap; I was tired too so I lay down beside her on the double bed in Will's guest room, which he'd given her for our time here. The walls, covered in antique wallpaper of faded but regal-looking lions, were festooned with her taped-up decorations, drawings she'd done of fairies and princesses, photos of Kay, Faneesha, Solly, herself standing with both Lindas beside her snow effigy, its head already half-melted.
I dozed off not long after she did and was only woken by a wrong smell. It was familiar, but still I took a minute to put a name to it: smoke. And it was too warm in the room, I realized—sweat had beaded on my forehead and under my arms.
Had I left something on the stove, maybe a kettle? I left Lena sleeping and started down the stairs.
But there was smoke at the bottom, enough of it to hide the view below, and a block of hot air hit me. I turned around again to get Lena—and where was my phone? Downstairs, damn it, somewhere past the smoke, I'd left it charging down there. Will's landline was on the first floor too.
I shook her awake and bundled her into a thick sweater and we ran to the bedroom where Will and I slept, which had French doors that opened onto a balcony. I wrenched the doors open and stepped out onto the rickety wooden platform, which hung over the back of the house. The view was of the large and unkempt yard, brown grass mostly covered in thin patches of ice and crusts of snow. At the back of it were trees, over which rooftops were faintly visible, but not close enough to yell at.
Most of the neighbors were probably at work, I thought, since it was the middle of a weekday.
"Honey, I think we have to climb down," I said.
"It's too slippery!" cried Lena, her voice squeaking. She touched the ice along the wooden rail.
But Lena's a much better climber than I am, a climber who shimmies up to the canopy of trees and freely climbs rock faces I'd never try, and we got out safely, she first, me after, though I fell the last couple of feet. I twisted my ankle, scraped my elbows a bit. We went around to the front yard and still saw no fire, just smoke leaking out the crack at the bottom of the front door. We ran next door, knocking and waiting, and just as the neighbor's door opened we watched the roof cave in.
THE HOUSE ISN'T a total loss. A fire engine pulled up not long after the neighbor called 911, siren shrieking, and we stood by shivering as the firemen plied the hoses, stood with our eyes smarting as smoke billowed out of a broken front window.
I picked up Lena and held her on my hip the whole time—she's old for holding like that, but still light at forty-some pounds. She didn't cry. She was openmouthed but not outwardly frightened.
Other than the section of roof that collapsed, only the kitchen and living room are badly damaged. Mostly they're waterlogged. Will's homeowner's insurance will cover the repairs, but those repairs will take a while. It was an electrical fire, the cops told us when we met with them at the station. There's no evidence of arson, they said.
I assumed it was Ned, somehow this too was Ned's doing. But the firemen shrugged and said the house is old, its wiring is pre-code. One of them brought me an informational brochure, nodding helpfully as though the handout would fully explain everything.
On the front it has a picture of a fifties-style couple in their kitchen—she beside the stove, he sitting straight-backed at the table, wearing a suit and tie, with a cup of coffee and a plate of eggs in front of him. The man and woman are both slim and attractive, and smile at each other in a satisfied fashion. But sticking through the open door behind their backs, as though to peer in and wave, are plump, decorative tongues of flame, apparently unseen.
Each year, household wiring and lighting cause an estimated average of 32,000 home fires in the United States. On average, these fires result in 950 injuries and 220 deaths. They cause more than $670 million in property damage.
Even the insurance forensics guys who came to inspect the house shook their heads as though the fire had been inevitable—we'd been asking for it by being so brash as to live in the house at all.
So it's back to The Wind and Pines, where Don has set us up with two adjoining rooms close to the lobby. He keeps the security system updated since the kidnapping: he gave himself a crash course in the software after it happened. So we're still surveilled, and the homeowners' insurance is paying for our rooms until the repairs are done.
There are other motels in driving distance, of course, but Don is Will's friend and Lena's so fond of him, and besides the Lindas are still here, the sole holdouts of the group, still setting out on their beachcombing walks every morning, still not ready to part ways from each other and go home.
In the end, coming back here, it seems we didn't have much choice.
WE ALL ATE in the motel café tonight, Will and Don and the Lindas and Lena and I. Somehow it felt like we were trying too hard to have a regular meal. No one from town was there; the café's first emptiness had returned.
Don and I were left alone together after dinner, when the Lindas went to show Lena some video clips of kittens who were friends with tortoises; Will headed back to our rooms to unpack. We'd maybe had a couple too many glasses of wine, Don and I. Or at least I had. Don was drinking whiskey.
"At first, when it began," he said, "I did worry. I knew there were antagonists who might also be attracted, antagonists like your husband. We're a magnet for them."
"You mean—a magnet? How?" I asked.
"Some people, historically, have heard the voice when—let's say when danger is already near. But after a while, this year, I relaxed my vigilance because no one showed up. No one to worry about. And then they did. I'm sorry I wasn't better prepared, Anna."
"You did your best," I said.
We sat in silence, likely both wondering if that was true.
"Kay sent you some emails," said Don after a minute. "Didn't she?"
"She was so upset. And with her diagnosis—I didn't know what to make of them," I said.
"You can credit them. She knew," said Don softly.
I met his gaze for a moment, but there was something too plain or too frank there and I had to look away.
"She knows," I corrected, a little halfhearted.
"If you pay attention to the culture," said Don, "you can see these threads of recognition. There are interferences and smokescreens all over, but the threads are perceptible if you know where to find them. Kay was right. And she's sick, yes. She suffers from an illness of long standing. She's struggled very hard against it. But she also has rare insight. These years are decisive, Anna. We're in the midst of a great acceleration and a great implosion. These years are our last chance."
I sat there sipping my wine and wondering if Don was, finally, a crank. I think like that when bold pronouncements are made; I wonder if both sides are nothing but cranks, with one simply more powerful than the other. Ned's Bible-thumping friends think they're right and all others are wrong—their powerful fear of other groups that turns to hatred and plays into the hands of the profiteers. But the profiteers themselves, with their millions of tentacles sunk deep into every crack in the earth, don't give a shit about being right. They're powerful. When you have enough power, right or wrong is for kids. Then there's Don, with just us, this small crowd of overeducated, confused liberals who also believe the other side is dead wrong, his small stable of adherents to the Hearing Voices Movement.
"No," Don said into the silence.
I guess I'd spoken out loud, though I could have sworn I hadn't. But I was drunk enough not to worry about it.
I probably still am.
And I did know what he was talking about, I knew what he meant by last chance. He meant what Kay had written to me in her rambling and half-coherent email. He meant the world that had evolved over millions of years, the mass of living things through which all forms of intelligence cycle, through which a billion variations move and express themselves, the ark of creation over eras and eons. He meant the spirit and expression of all creatures and all people, their cultures and tongues and arts and musics, from the vaunted to the unknown; he meant what was organic and alive, the broad, branching tree of evolution that was history and biology and all kinds of astonishing bodies full of ancient knowledge.
He meant that it was on its way out.
THE PUSH IN FRONT of the subway train, all four tires going out on a fast road, the house fire while we were fast asleep—they seem too multiple for sheer coincidence, but they don't add up to an understandable pattern. Also, after the subway push someone had grabbed me and pulled me back. That was the first attempt, if I want to see it that way. The second: our tires went out on the Interstate, but in the end we hit no other cars—not the car so close on our left, not the dinged-up, rusting gray guardrail on the right. And the third: Will's house burning. But I woke up and I smelled the smoke, and ten minutes later Lena and I were standing safely outside in the snow, watching an empty building burn.
Will barely believed in the fire when I called him at work. He's seemed to be in a mild state of shock ever since, a man who's been pushed too far: many of his dear old books were destroyed, all the books on his living room shelves.
I want to tell him: Really, Will. You don't have to be in this with me. I'm grateful. And I don't know the difference anymore between gratitude and love. But I'm willing to cut you loose.
I know he wouldn't go.
I wonder what's more important, the fact that all these events occurred in the first place or the fact that they were only close calls, that in each case none of us have succumbed.
So far.
Since the fire I'm obsessed with when the next "accident" will occur, when the new onslaught will begin.
The subway episode was ten days ago. The car accident was less than a week. The fire was the day before yesterday. They fall closer together now.
I lie awake thinking of Lena, of what will become of her if something happens to me, or if she is also a target. She was there two out of three times, after all. I harbor wild thoughts, such as: Maybe I should have fallen in front of the subway train, because at least then I was alone. At least she might be safe right now. But I fear what would become of her if I die, so there's cold comfort there.
I lie awake worrying about Ned having custody. It's Solly I'd want to raise her, I guess, but since Ned and I aren't even divorced I'm pretty sure there's no way to legally exclude him. If he wanted guardianship, regardless of his craven reasons, he would get it. And I lie awake berating myself for my lack of leverage. I've brought this down on our heads, but I cast bitterness in Ned's direction too. I blame myself but I also know hatred.
I never knew it before him.
I TOLD WILL I was going to turn in with Lena last night, that I was exhausted—because I was—and then I lay in bed wearing Lena's earphones, which are large and shiny plastic discs in the shape of monkey faces. I thought of what Don had said to me, what Kay had written, of how I'd seen a city crumble beneath a cloud of dust.
Lena rolled away from me as I prepared to say Goodbye to Stress, and before long fell asleep clutching her duck.
The images didn't feel like a dream. I was aware of the room as I lay there, the shape of the TV cabinet, the bathroom door slightly ajar, the mirror on the dresser showing glints in the dark. I lay in an indoor twilight holding those dim motel-room shapes in front of me as I began to sink under. Did I keep my eyes open?
Into the dark room came a thin, stooped man. My impulse was to fling my body over Lena, shielding her and keeping her safe with me forever. But I couldn't move.
The thin man turned to look at me, and I recognized him. With his bloodshot eyes and tobacco-stained mouth, his gray, grubby mechanic's workshirt with the franchise logo on the pocket, I recognized him instantly: B.Q.
I felt repulsion, then fear; I knew I couldn't turn onto my side or cover her with my arms, I knew I had to lie just as I was, belly up and exposed. And she was exposed beside me. That was the worst of it.
But next I understood he was a weak and broken person. He had never been a threat to us. He worked for Beefy John, that was all—he drew a paycheck.
"She told you herself," he said sadly. "But you didn't listen, Mrs. Mrs., she sent me with a message because she can't bring it. She can't say anything anymore. So here it is. True language is the deep magic. As old as time. God of the hills and water. God of the sun and trees."
He stood at the foot of the bed looking down at Lena, and as he reached out toward her I felt I had to stop him—but instead of touching her he swooped farther down and grabbed something else: Hurt Sheep, which had fallen off the bed and onto the floor.
He picked up the stuffed animal and kept on walking across the motel room, headed toward the window now, where he stood and drew the drapes open.
In the night sky there was a deep-blue light, a kind of royal blue out over the ocean, and stars twinkled in it, the four-pointed stars you might see in paintings. They made me think of the three kings, of the Nativity.
I turned my head and watched him leave by the window. After a couple of seconds I could see quite well, almost as though I was standing at the window myself. He walked out through the glass and into the air and kept going, the sheep tucked under one arm, to where Kay waited, standing on the furling crest of a wave.
"HEY. MAMA. WHERE'S Hurt Sheep?" asked Lena in the morning. "Hurt Sheep was right exactly here!"
"Maybe under the bed. There's lots of space down there. Remember to check beneath things, when you're looking," I said, brushing my teeth.
Later I helped her and we looked everywhere.
No, I thought, no no no. Come on now.
"Maybe she's gone. Oh! Yeah. I guess she went with Kay," said Lena, and shrugged, cocking her head.
"What do you mean, love?"
"It's a good place for Hurt Sheep. That's OK, Mom. She went with Kay. I told you before. Remember? In the boat, to the white castle."
We are sending this message to our daughter Kay's friends, her fellow medical professionals and students, and others who knew her. This is to let you know with our deep sadness, that in the evening of this past Friday, we authorized the medical staff of Brigham and Women's Hospital, to remove, Kay from her ventilator and other support equipment. This was the most difficult decision, a parent can ever make, but as she left a "Living Will" document on her Computer, we know for certain, that it is what she wished.
Please do not reply to this Email, because neither Kay's father, nor I, will continue to use Kay's Email address, which we would view as a violation, of her personal privacy. We used it only to access her many Contacts, which we could not find, in another way. Neither of us uses an Email, and this is the only time, we will send a message with Kay's Email Account. However, regular mail can be sent to us at the address below.
Also below, is listed a charity that was close to Kay's heart, for any gifts made in her memory.
Our deepest thanks to all of you for your visits, cards, flowers, and for the love, you also held for our beloved daughter.
I WASN'T MYSELF, BUT THE IMAGE OF ME
IT'S LATER NOW—MUCH, MUCH LATER.
I was in the shower one evening before Lena's bedtime, just after Kay's death. One of the two rooms we were renting off the lobby—the room that used to be Burke and Gabe's—had a shower curtain in its small bathroom that Lena had pointed out right away. Where our old curtain had borne a pattern of blue flowers, this one had golden sheaves of wheat repeating on a background of creamy white.
I remember noticing, as I stood there letting the water drum down onto my shoulders, the cleanness and freshness of this new shower curtain with its sheaves of wheat. I noticed the sparkling-white quality of the small tiles on the shower walls, how they contrasted with the worn and grimy tiles of our previous motel-room shower stall, frankly a sorry bathroom feature. We were living the high life now, I recall saying to myself.
I washed my hair with plenty of shampoo. I saw no need to rush, since Lena was safe in the room next door with Will, reading to him from her bedtime books. I'd just rinsed out the lather and was looking around for my razor—had I left it on the sink counter?—when I felt a scratch at my ankle and glanced down to see a thin trickle of blood. What had cut me? I must have rubbed my other foot across the ankle—my big toe, on the other foot, had a freakishly long toenail.
Unattractive. I didn't like it. How had it gotten so long without me noticing? I felt embarrassed, despite being alone. I'd clip it right now, as soon as I shaved my legs and stepped out and toweled off.
But wait, the other toenails were long too—they all were, on both feet. They were almost obscene; they looked like a bird's talons, like bird claws stuck onto a mammal. How could Will not have noticed, either? Maybe he'd been too polite to say anything. The front edges of the nails had to be nearly a centimeter long. Beyond disgusting.
I'll get out right away and grab the clippers from the bag next to the sink, I thought. It was both strange and vile: my toenails had never been so long in my life. Must be because it's winter, I told myself, you wear thick socks all the time, even to bed usually, hating to have cold feet—that must be how you missed it. I was about to turn off the water when I caught sight of my ankles, my calves. The hairs on them were as long as the toenails, practically. Jesus, I thought. How could that have happened?
My gaze hit the wall tiles. I'd thought they were so clean, but now I saw some of the caulked cracks between them contained lines of mildew. I'd get the maids in here first thing tomorrow, I'd get down on my own hands and knees . . . wait. My fingernails were almost as long as the toes. Hard to believe I hadn't cut up my scalp with them while I was lathering. My gaze flicked back to the wall tiles and I saw a line of mildew was creeping up the grout.
It was visibly extending itself before my eyes, indeed all over the white surface of miniature tiles on the shower wall mildew was creeping along the lines of caulking. In a grid of right angles a black mold was spiking out farther and farther along the network of tiles, straight angles in every direction.
"What is this," I said, "what is this," and tore the curtain back without even turning the water off. Wait—the water had flooded, the floor was soaked, and everything was damp. A lightbulb flickered above the vanity. In passing I noticed the tub was full, backed up, the water a sludgy gray, and a rim of scum ran around the tub over the waterline. I panicked, throwing a towel around my middle, tying it over my chest—it too smelled stale, possibly moldy. I pulled the door open and ran out into the room: there were Will and Lena reading on the bed, pillows propped behind them, with a picture book open across their laps.
Relief: she was there. She was safe.
But all around us the room seemed to be changing, though I couldn't put my finger on it at first.
"Goodnight, little house. Goodnight, mouse," read Lena. Her voice was muffled.
"Goodnight, comb. Goodnight, brush," read Will. His voice, too, sounded like it was coming through a barrier.
They looked relaxed, as I'd left them, but around the bed they lay on other features shifted and altered. The desk lamp turned off and on rapidly, at irregular intervals; dust piled on surfaces and then seemed to go away, as though either blown or wiped; an object vanished and reappeared somewhere else, a toy on the round table, a glass. They didn't take notice. Through a chink in the drapes I saw flashes of light outside. But it was night, and there shouldn't have been light on that ocean side—so I ran past the foot of the bed to pull the drapes open where the big picture window looked over the cliffs and sea.
And I saw it was day. But then it was night, again, night in the sky and rapidly back to day. Boats appeared on the surface of the water, both far and nearer, then disappeared in an eye-blink, only to reappear elsewhere; the sky switched from morning to midday to evening to night within the space of seconds, and then did it again—this time with different cloud formations, other ships.
"Will, Will! What's happening?" I shrieked, turning to look at him and Lena where they sat with their backs against the headboard, their legs stretched out on the bedspread.
But they seemed to be walled off. When I leaned over the bed to reach out to them something in the air resisted me. I couldn't punch through the space around them, though I tried, increasingly desperate. Lena and Will looked the same as ever but I could see my hair growing in front of my eyes, my hair was getting longer and longer on my shoulders, inch by inch it moved down the front of my shirt, my hairs were visibly lengthening.
My little girl was looking calmly at her picture book, touching the drawings. She looked so normal, just here, just the way she should be. But I—I looked up at myself in the mirror. There was an ominous element to the growth of my hair, the choppy, almost digital-looking growth of the ends, so fast it was visible to the naked eye. There was something badly wrong. I wasn't myself, but the image of me.
Lena's fingernails were normal where they lay on the bottom edge of the pages of her book, bitten off a bit but normal: Goodnight, nobody, said the text on the page.
Beneath my own lengthening fingernails a line of dirt crept, growing along with the keratin.
I'd seen this somewhere, I thought, seen this somewhere before.
"OK," I said, and made myself take deep breaths, count slowly. One of the hypnotic visions or a vivid nightmare—in any case nothing physically real, that was clear from the nails, from the hair—impossibility. I had to figure out the rules of the nightmare; possibly I could control it and wake myself up. I turned my back on Will and Lena and walked to the window again, where birds appeared on the cliff edge and then flicked away. The grass was greener, yes, the ice melted and springtime was here, even the color of the ocean changed from gray to a bluer hue, even the color of the sky.
I heard a voice in the other bedroom and went back through the interior door, reluctant to let Lena and Will out of my sight but pulled there somehow—still, all this was an effect, wasn't it? An effect, I remember telling myself as the light kept changing up around me, lights shifted and went from dark to dim to bright. It was disorienting. But part of me also worried that I'd been drugged again and this would turn out to be another kidnapping, so I made sure the chain was on the room door. Dream or not, lock the door, I said as I went. Dream or not, lock the door.
The voice was coming from my laptop, open on the bed where I'd left it during my shower. I came up beside it and I could see the screen: Ned's face. It was a video call, his head in a window on the screen—talking to someone else as I came up, his face in profile, but he turned and looked at me.
"A little fast-forwarding," he said.
"What? What do you mean?"
"I hit the fast-forward button," he repeated. "Didn't you see? The kid. Your boy in there. They're not going so fast, are they? You're all alone."
They were at regular speed, I realized. But I was sped up.
"You're growing old," said Ned, and smiled again. "See?"
I looked down: new wrinkles on my hands. Old hands. Somehow I'd moved through time alone—and yet still I spoke at normal speed, or else I couldn't have talked to Ned; I still thought normally. Didn't I?
"It's impossible," I said, more to myself than him. "It's just a bad dream."
"That's what you do with losers, right? Isolate them. You're one of the losers, wifey."
"But how—why are you doing this? I was cooperating, Ned. I did what you asked, didn't I? I don't get it."
"I've got the primaries in a few weeks and I need my pretty wife where I want her. A mental case, alone and needy. Makes them do what they're told. Obedient. And a nice little bereavement in the family. Sympathy vote's the icing on the cake. I look good in black. Well. I look good in everything."
"A bereavement?"
"I took your time from you. You've missed a whole lot. Just take a look."
Outside the picture window the sun was bright. Gnats and flies hung in the air. There were bunches of grass near the edge of the cliff and they were full green, bowing and dancing in the breeze.
"Ain't we got fun?" said Ned.
Doris Day was singing it in the background. Not much money, oh but honey, ain't we got fun . . . There's nothing surer: the rich get rich and the poor get children . . .
I had a cold feeling. I was brittle as bone.
Had he made me a ghost?
I'd disappeared—I'd gone, slipped out of being like water down a drain. Was my girl alone now? Was Will looking after her?
"Like I said, we're going out today," he said. "We have a public appearance. Believe me, darlin', it's easier if you don't fight it. Don't get yourself all bothered. You won't get anywhere, I promise. You're confused, sure. You're a sick woman. You're weak. But it won't be forever. You don't have to go on that much longer like this. Just do what I say. OK? Put on the gown."
I looked behind me and saw a black dress laid out on the second bed.
"I'll see you outside," he said. "Be on your good behavior, now. You see what I can do."
His face went gray and for some reason I reached out and touched the screen softly. But it wasn't warm, and fine dust came off on my fingertips. The laptop wasn't even on. I raised my face: Lena and Will were standing in the doorway. Will wore a suit and Lena's eyes were puffy.
We weren't in the motel at all but in my parents' house; I stood in my old bedroom. There was a rush of confusion that was almost a thrill, almost velocity. Then it stilled. Here was my corkboard with its colored pushpins and ribbons. PARTICIPANT. The air was humid and close; my parents had never had central air. I heard my father's voice: they never "held with it." I was wearing the black dress now, I saw, glancing down—no memory of changing into it—and toe-pinching black shoes with heels so high I could barely walk on them. I'd never have picked out those shoes.
I wouldn't struggle. Don't fight it, Ned had said sleazily. But it did hurt more if you struggled.
Prey animals had the sense to play dead.
So I leaned down and picked Lena up, though her weight made me stagger on the too-thin heels. But she was real and solid. I knew from her red eyes that she'd been crying and I squeezed her hard, maybe too urgently. Had all of us been frozen there? Had we all been suspended on Ned's whim, or only me? I tried to see if Lena looked older . . . I was flailing. It was possible, faintly possible that her face was more angular suddenly, but whatever slight change I might imagine wasn't obvious like my long talons. I tried to keep them from scraping her back as I held her; I'd rip them off. They were like parasites on me.
"Mommy, I'm hot," complained Lena.
I put her down and as I turned away bit at the longest nail, ripped the white edge of a thumbnail off with my teeth. But then—they weren't long anymore.
And the hairs on my legs? I leaned down to look beneath my tights. They were black tights, semi-sheer, and I could see no hairs through them. The skin on my calves was smooth. I straightened up again and was holding out my hands, looking at them dazedly, when Ned appeared behind Lena in the hall. He wore a black suit, true to his word, and a silver-gray tie, and looked like he'd stepped off the pages of a magazine.
"My father," I said, and it hit me whose death this was—I wasn't the ghost after all.
It had happened without me. He was all gone, and I'd missed him. I'd been absent. There was a picture in one of my mother's photo albums: my father as a tiny boy in a white suit, sitting on the back of a horse. Or maybe it was a donkey. It was a blurry, black-and-white picture.
That little boy, I thought.
How would my mother ever forgive me for missing it? How would my brother?
Had my father lain in bed, had he grown thinner, the way the dying do? He might not have missed me. I hoped he hadn't but I would never know.
"You were always a daddy's girl," said Ned.
"You were a rotten son-in-law," I said, as though it was news.
He kept smiling, as always. His smile never wavered now. It was a rictus.
"You took his money and you even took his dying," I said.
"Mommy?" said Lena. It was as though she hadn't heard me; I was glad and ashamed, ashamed for speaking that way in front of her. "Can we go now? Nana says they're going to play a pretty song for Grumbo at the funeral. She said they're going to play 'The Skye Boat Song.'"
"Take my arm, kiddo," said Ned, bowing his head in Lena's direction.
She clung to Will for a second, she would much rather have walked with Will, it was awkwardly obvious, but finally she lifted her hand up to Ned's.
I walked right behind them, fearfully close; as I stepped into place at their heels, I clutched Will's arm for a moment where she'd let go of it.
"Let go of that thing right this fucking second," said Ned through gritted teeth. But he was facing away from us. As though he had eyes in the back of his head. "You're my wife. You remember it."
"How did you know how sick my father was?" I asked weakly. "How did you know before we did?"
"Whatever you need to know, I'll fucking tell you," ground out Ned. Then he turned and whispered over his shoulder, almost tenderly, "Bitch."
My stomach flipped but Lena was looking elsewhere and waving at someone: she hadn't heard the tone or the words. Again she seemed to be immune. She was usually so observant—it was as though Ned had a wand.
We stepped out onto the front porch, where I saw my parents' grass was yellow and dry. There were flags flapping from porches down the street: it was Independence Day. Out past the awning, where the shade stopped, reigned a bright blank July heat, cicadas whining in the trees. A small group of photographers stood on the lawn. Had Ned hired them? Would a real news outlet spend money on pictures of a candidate's in-law's funeral?
Ned wore a solemn expression, making the occasion momentous—such was the power of his bearing—and curved a graceful arm around me in a supportive gesture. He was between Lena and me, seeming to shelter us both, there on the porch: the father of the family, presiding over a sad wife and innocent little girl.
Will had fallen behind somewhere—that he had even been allowed to come was surprising. Ned couldn't have liked it; maybe my mother had pushed. There were limousines at the curb, and my mother was getting into the first. We joined her there, Ned and Lena and I (I looked back and saw Will headed across the dry grass for limo number two). My mother slid in beside Solly and Luisa, already seated.
In the cool car with the air-conditioning blowing into our faces Lena sat between Ned and me and sang in a high little voice.
Speed, bonnie boat, like a bird on the wing,
Onward! the sailors cry;
Carry the lad that's born to be King
Over the sea to Skye.
Across from us my mother wore an expression both peaceful and relieved, maybe. Alone now, without my father, but probably also relieved. She avoided looking at Ned as though there was a blank space where he sat.
Lena, who only knew the chorus, sang it again.
I tried to discern from my mother's face, then from Solly's whether they were angry at me for being trapped by Ned this whole time as my father was dying.
But Solly wasn't looking at me at all: he was looking at Ned with open contempt, with raw hostility. Luisa nestled into his side, her eyes cast down. Miserable, I thought, and polite. My mother patted Luisa's knee and they smiled at each other sympathetically.
I turned my head toward Ned, slowly and slightly so that he wouldn't notice. He'd dropped his falsely protective arm off me when we got into the limo and also dropped Lena's hand; now he was looking down at his phone, as usual.
There was his neck, its even tan, the sweep of one lock of hair over his forehead, his perfectly clean ear. There was the faint scent of his cologne. I kept looking, I kept gazing at the graceful tendon of his neck, the clean shave along his jawline. And just when I was about to turn away—feeling my eyeballs throb dully from being rolled to one side too long—I saw a movement on the skin. Just for a second, just for an instant, I saw an L-shape made up of pink-and-white squares flash onto the skin before they disappeared.
I swear I saw him pixelate.
I didn't say anything, my tongue was stuck in my throat, but as we got out of the car I found myself scrabbling at his sleeve. Lena was walking ahead holding my mother's hand; I had Solly's and Luisa's backs in front of me. We were on display again as we stepped onto the cemetery's gravel footpath—I didn't see the photographers yet but there were mourners around us, others were parking and walking over to the gravesite—and so, again in the open air, Ned turned to me smiling. The smile was perfect, too: restrained, as though in grief, and yet compassionate.
"How are you doing it?" I asked, a bit pathetic. "What are you doing?"
"I'm playing with you, honey, that's all," he said softly, and tapped one temple. "You let me in when you started 'clearing your mind.' That New Age horseshit is good for one thing: access. Safer when you had the therapist in the room, but then you started to do it all by your lonesome, didn't you. With the little earbuds in, all walled off from other people and with your mind wide open."
"The hypnosis tapes?" I squeaked.
"You threw open the doors and I walked in. So now I'm tinkering. I'm just tinkering around a bit with the little wife's thalamic nerve projections. I can do that now. I can make you see what I want you to see."
He'd effected some kind of amnesia. If not a dream he'd given me, it wasn't far from it, I guess, a thought, an idea, a mental frame. Drugs, maybe? Could this be pharmacological, and his mind-control brags just a component of his intricate manipulation?
"But I don't know what you mean," I said. "How can you—
anyone—?"
"I have the skills," he said. "Ever since I took the kid. Added bonus. You just take what you want. You know that, sweet thing? The more you take, the more you get. It just starts to pour in. Talk about miracles."
"I don't get it," I said. "I don't . . ."
"The same way money gives you everything, so does power. It's like one of them math curves, rising steeper and steeper. That's how power grows once you grab it. How'd you get through thirty-some years without even knowing that much? Stupid. I can make things happen without even being there. I kept you on your toes. The subway, right? The freeway. And the house. It's nice for me, watching."
"But not—that isn't possible."
"Not only possible. Easy. With neurons so much is easy. Didn't your little Hearing Voices club tell you that? Haven't you learned anything?"
"So you're saying you can get into my—"
"I have the keys to the kingdom."
"What kingdom?"
"I can slide my fingers," and he leaned over and whispered close, "right into the holes in your head."
His breath was moist and stale on my ear and a sight flashed before me, a black pit. Out of it climbed naked people in stuttering, stiff movements, herky-jerky. I'd seen that movie, I thought, a Japanese horror movie, I'd seen it and it scared me. They were like puppets pulled and released on unseen strings, and their thin limbs were hairy and banded as the tails of rats.
"Like I did with the little doctor girl," he said. "You can't let people like that just keep going. She saw way too much. And then she opened her little bitch mouth. So she had to go. Didn't she."
I turned and stared at his smile. Then I bolted ahead, my stiletto heels biting into the turf, until I was near enough to grab Lena's hand and use the contact to steady myself. I walked forward holding that little hand tightly, my mother on her other side, and looked down at her face that I love so much, trusting and bright.
I gazed at her face that banished fear and thought of not looking back—no matter what, I said to myself, no one can make me look back now.
AT THE RECEPTION (carefully steering clear of Ned, who was at the far side of my parents' house glad-handing the mourners) I took Will's arm and pulled him into the kitchen with me, where we could talk. I watched Lena through the open door, carrying a tray of food with my mother at her side. I felt cracked and hollow.
Drinking wine didn't make me less parched but at least it loosened the tendons on my neck. I was living in a half-life, I thought, a life of distorted lenses where I couldn't trust anymore that a man's skin wouldn't pixelate beside me. Even my thoughts weren't my own, and without them I wasn't myself. Alone had been free, I saw that now—alone had frightened me but the air was clear there. Now I was in prison, without the privacy of my mind. With those claws in my thoughts I wasn't myself—I wasn't anyone.
Will and I stood and gulped from our goblets beside the trays of brought food, the donated lasagnas and plates of brownies crowded onto the island. I made myself focus on the practical and asked him what had happened over the past weeks.
I didn't say months. I was trying to test the waters.
"You mean—in the news?" he asked.
"I mean with us," I said. "What have we been doing?"
"Besides your father—helping take care of him? Besides the illness?"
"We've been here at the house for a long time," I repeated, tentative. "Just here with my parents."
I saw in his face: Of course. Yes.
So it had been a nightmare, I'd been here, where I needed to be, with them. That motel room and fluttering fast-forward of days and weeks and months had been a memory Ned implanted when he took away the rest.
"Is he threatening you again?" asked Will urgently. "Did he say something threatening?"
"It's not what he threatens," I said. "He said—he said he did it to Kay. He said she saw t-too clearly. Somehow he did it, Will. She d-didn't do it to herself."
I was starting to stammer, a habit I thought I'd gotten rid of as a child. Will reached out and held my shoulders.
"And now I don't have the right memories. This—it's like I wasn't here till today. I don't remember anything since March. Right after the fire, after Kay. And he says it was him. In my head. He did it to her and he can do it to me."
"You don't have the right memories," repeated Will.
I mumbled what Ned had said—his fingers, the holes in my head. My hands had started shaking. "You were reading Goodnight Moon. I had this—I thought my hair was growing just, just fast—"
"Anna," said Will, and moved his hands onto my own to hold them still.
"That's all I have since then, all I have since March, since we moved back to the motel—it was the week after your house burned, remember? Listen. He robbed me of this time. These months. His face talked to me from the computer . . . ."
I looked down at my nails, my nails on the fingers held by his fingers, wanting some evidence to show for all of it, but the evidence was gone. It was my interior life Ned commandeered, that was all. Not time. He couldn't do that. I was half-comforted.
"My fingernails were never long," I said dimly.
Will was looking into my eyes intently, but I couldn't describe it any better. As he stared at me, waiting worriedly for me to explain myself, I thought of checking this journal—I'd open this document, see what I'd recorded. Maybe, in the real months that had been taken from me, I'd written real entries. I'd check tonight, I decided.
"I have to tell Don," said Will. He was patting his jacket pockets, searching. "This has to be it. This is what we expected. He wants you to look fucked up. Depressed and grief-stricken. After your father's death, you're going to . . . he's going to do it. Maybe just pills, like Kay, but he's going to make you—he's down in the polls. He could actually lose this. He and his people are desperate. He needs the sympathy vote."
He found his phone and dialed it.
"Sympathy?" I asked. I noticed I was holding the stem of my wineglass too hard. I set the goblet down on the countertop, then picked it up and drained it. "I don't understand it, Will. Do you?"
But then he was saying he couldn't get cell service in here and headed out the kitchen door, slipping his phone against his cheek. "Stay here, stay right here," he called back. "OK? Don't get near him."
Lena, I thought: Where was she? Still offering her tray of food to guests? I'd forgotten to watch her for maybe five minutes by then and my mother might not be vigilant enough. She knew Ned had taken my daughter before, but she didn't know this new Ned, this Ned phase-shifted into pixels and a grin that was a rictus. This one who said bitch instead of honey, whose skin had pulled back from his face to reveal bone and metal . . . I looked down the hallway at the milling people, pushing away the fact that Will had asked me to stay here: it didn't include panic over having to look for my girl. Then I was out of the kitchen, rushing to get to the living room.
There was my mother, talking to an old woman with a walker, and there was Solly, there was Luisa.
I couldn't see Lena. I didn't see her.
I pushed my way through the people, made it to the front door, and hesitated. There was a ringing in my ears and my hands felt too numb to turn the knob.
But then I was outside, and I must have left my heels behind because I was standing on the front porch in nylons, feeling the rubber nubbins of their welcome mat against my soles. Closest to me were Main and Big Linda, right there on the path from the street, and Lena was holding Main Linda's hand and picking with a stick at the sole of her shoe—it seemed to have a piece of gum stuck to it. She waved the stick when she saw me, grinning.
They were watching our suburban Fourth of July parade, whose route comes down our street every third year. There were floats and bunting that glittered red, white, and blue; there were some kids in an off-key marching band, a girl turning cartwheels. Up came a horse-drawn buggy decorated with stars-'n'-stripes and the name of a car dealership, and then, in the bed of a pickup truck, a human-sized blow-up statue of liberty with a big head. Its torch flames were made of yellow plastic streamers, blown upward from a small fan below in the truck bed. They snapped and fluttered in the breeze.
The Lindas were looking out at the street and didn't turn. Nearby Don and his aged father stood near a waxy rhododendron bush, the father leaning on his cane; up toward the sidewalk, on the burnt July grass, were the other motel guests, Navid and the Dutch couple. There were Burke and Gabe, just getting out of a car parked at the curb. I thought maybe I should talk to them, thank them for coming, but they were watching the parade, all of their faces turned toward it. I would wait, I decided.
I walked down the sagging wood steps and went over to Lena, feeling the grass poke between my toes; I took the gum-stick from her gently and tossed it into a bush. With her hand in mine I turned to look at the parade.
But as I gazed at it—the high school marching band passing—the marchers changed. Their uniforms faded to drab brown and gray; some of them were wearing hoodies or hats, some dragged bags after them, scraping the street—their instruments were gone, and instead of the instruments they carried sacks full of trash, sacks leaking fluids I couldn't make out, leaving brown-red streaks on the road. Their heads hung down. Their passage had a dreadful weight.
As I stared at them, all at once, they raised their faces to me. Hideous. Some seemed to be wearing gluey, primitive masks; some looked like burn victims and others were pimply teenagers, some were middle-aged with bad teeth and glasses. Some were diseased, their eyes red-rimmed, lesions that looked like eczema or leprosy splitting the skin of their faces. The worst were crones with thinning hair, clumps of ragged gray sticking to yellow scalps.
But they were all Lena.
"No," I said out loud.
They were Lena old, young, wretched, in a hundred distortions. That's why we have to die first, I thought, panicked: before they get so old. I shook off the urge to throw up.
Around me the motel guests were watching the parade and smiling. They didn't see what I saw.
I had to be defiant. It wasn't the time to play dead.
"Is this show all for me, Ned?"
The parade shifted so that, for a moment, I saw normality—a second of cheerleaders with pompoms. Then the ruined Lenas were back, deformed and crooked, shambling. They made noises low in their throats. I saw a toddler so thin she was almost a skeleton.
"This is ridiculous," I said, summoning a desperate bluster. "Give it up, Ned." I moved my eyes off the parade and fixed them on the solid, actual Lena beside me. No one seemed to be hearing what I said.
I looked over my shoulder and saw the front windows of the house and sure enough there was Ned, his grin a death's-head rictus through the glass.
"They got here fast, didn't they," came a voice. Will's.
He was on the porch. I pulled Lena with me, stepping back onto the lawn to meet him as he walked down the steps and grazed my cheek gently with the backs of his curled fingers. Once he was near us the yard felt more physical, the house—and when I turned back to the street the parade was normal, just a small-town parade befitting my parent's sleepy suburb.
My body slumped in relief.
"I thought we were supposed to be the ones that didn't go crazy," I said, and leaned against Will, my whole body sagging against his side.
"You're not crazy," he said. "He just wants you to feel that way. And look like it. So your suicide's credible. Do you believe me?"
I cocked my head at him and nodded slowly.
"The others are here," I said.
"They came when they heard about your father."
"But Will. Ned's calling the shots. He's still—he's in my head, messing around with me. The parade? To me it looked different."
"So," said Main Linda, approaching. "Hey. I'm sorry for your loss."
"We're all so sorry," added Big Linda.
Navid hugged me lightly. He wore a dark suit almost as expensive as one of Ned's, I noticed. And he was clean-shaven again.
In the street several jeeps passed by with banners supporting the armed forces.
"Soldiers," said Lena helpfully.
"Brave young Americans," came Ned's voice from behind us. "How do you like the parade, Anna?"
I opted not to turn around. The others barely acknowledged his presence either, but I felt them tense and stiffen, I felt their mood turn gray.
"Can we see fireworks?" asked Lena.
"It has to be dark for fireworks," said Will.
"That's later on tonight," said Big Linda.
"But can I stay up late?"
"Of course you can," I said.
"I'll see you then," said Ned, and he strode down to the sidewalk, two suited bodyguards converging on him as he went, the engine of a parked car revving.
Watching him get into the backseat of the car, hearing the curt slams of three car doors in a row as the bodyguards got into the front, was when it hit me: one job remains to me. However bad it is now, I saw—his cartoon-thug tactics, the way he used my love for my daughter against me—it will be far worse if he wins. And not just for Lena and me, not only for us, not at all.
I've been blindered for months—maybe the whole length of my life. These visions and pixels make it obvious. Around me is the desperation of others, the arms of supplicants growing out of the dirt, and I've walked through those fields as though there's nothing there but tall grass. I should have played dirty long ago.
The living spring from the dead, was the first thing I had heard.
I smile thinking of it. Maybe the dead had been me.
I won't have Lena if I don't even have myself. And Ned has a sociopath's overconfidence, that's his weakness. Maybe he's made mistakes that can be used against him, one or more of his obvious, arrogant, flagrantly taken risks.
"Why are they really here?" I asked Will as we headed back into the house with Lena. The motel guests were drawing closer together on my parents' lawn.
"To be of service," he said.
And there was war in heaven: Michael and his angels fought against the dragon. —King James Bible, Revelation 12:7
62 percent of Americans . . . think recent natural disasters are evidence of global climate change while 49 percent say such disasters are evidence of biblical end times. —Washington Post, 11.21.2014
I WRITE THIS in my old room with the bulletin board, where among the dust bunnies on a closet shelf I found a fortune-teller made of pink construction paper. It's numbered with blunted pencil on the finger flaps, and inside each flap is an outcome scrawled in miniature writing. My friends and I made them up, giggling hysterically, during a sleepover when I was in sixth grade. You will be Famous (for Burping the national Anthem) / You will be Rich but Really Dumb / Our Love will never Die.
In the corner is a crate of my old records, on top of which an LP lies flat. It bears a once-famous logo, a black-and-white dog staring into the cone of a gramophone beneath the words His Master's Voice.
When I looked into this Word file to see what I might have written during my lost spring, all I found after the cut and pasted-in email from Kay's parents were two fragments.
I assume I wrote them, but have no memory of it.
Say God is a complex grammar that doesn't coexist with our own language, its ego-driven structures. Say Kay is right and dolphins or whales can be its hosts for their whole lives, instead of funneling it briefly as Lena did, because the form of language that emerges in those animals doesn't displace the deep grammar the way ours does. Say that deep language, whose name may also be God, stays with them because their communication systems, though capable of individuation, are not devoted to the self. Say we're left on our own, as Kay had it, when we pronounce our first words and God deserts us, and it's in that respect that we're different from the other beasts and different from the aspen trees. Then it has to be said also that instead of being raised above the other kinds of life—instead of being special as we have always claimed—we're only more alone.
That one was peppered with errors of rapid typing that I've fixed. The other was this:
Some people hear more, some less, some nothing at all. What we hear is what we can hear, its content minutely tailored to our character and biases. That means, if I believe her, that even we, who should be outside the range of any dogmatic faith, even we only ever know the God our personality describes.
Lena, living out a fixation on the cute, has made the screen on our tablet into a picture of a fawn in a snowy forest glade, looking over its shoulder with big dark eyes as flakes fall and soften the world around it. The only moving elements are its eyes, which every so often blink, and the snow. I look at it now, while she runs through the sprinkler in the backyard with Will.
I can see them out my window if I scooch my desk chair sideways—there. Better.
The fawn with its dark, slowly blinking eyes takes me back to Ned's beautiful girlfriend, the young model or model lookalike. Where is she now? I could do so much good mischief if I had just a little time-stamped footage of her and Ned together.
Of course any action I take is a risk to Lena and there's no way to attack him from anonymity. Everything's obvious now that I know he actively wishes me harm. It's transparent that nothing binds him to the norms of decency—no guarantee exists, none ever did. There was never a contract to rely on, some solid agreement that could be wielded in a court, only my naive belief that such abstractions have any weight at all.
All my credulity is out the window now, that frail screen of written-on paper I let myself believe would keep the world predictable. Ned's handshakes add up to less than nothing—or nothing but the flag my gullibility flew on. Don and Will and their fears, well, those fears were only a slice of the malice that Ned is.
Maybe the knowledge is chilling—it is, when I don't block it out stubbornly—but it also means I have no reason to jump at his command anymore, there's nothing left to make me do what he wants me to. So he's now lost his postcard family. He shouldn't have shown himself, and I wonder why he did, because he could have strung me along forever, practically, while I still believed he could be bargained with.
Now I know he doesn't bargain. He only pretends to.
And he has nothing left to get from me. Nothing but the last thing.
It must be his narcissism that's to blame, maybe he couldn't help showing me how powerful he is. Maybe he had to flaunt it.
And all I have left is this: my girl and her uncertain future.
She used to look forward to every new day.
THE OTHER MOTEL GUESTS insisted on staying close, crowding around Lena and me as we walked across the parking lot. We'd left Solly and Luisa and my mother at home, my mother so she could help the caterers clean up, Solly and Luisa so they could pack to leave for Manhattan the next day. But they urged Will and me to take Lena out. Let the child see the bombs bursting in air, at least, on this death-textured day.
The show was being put on at a Minor League baseball stadium about twenty minutes from my parents' that's always been the perfect place for pyrotechnics; we used to go there for the Fourth when I was a kid. Until I went to college I came here every Independence Day, first with my parents and Solly, then with a swarm of classmates, and by my last years in high school with guys in their hand-me-down family cars—not watching the show, just using the dark and crowds and noise as camouflage. That was when the stadium was small and dilapidated, with wooden seats, before it was renovated into the slick behemoth it is now.
I led the others into the elevator and we went all the way to the last row, where the view was worst for sports but best for fireworks. It had been a hot day but now a breeze swept up and chilled me; I'd remembered to bring a jacket for Lena, but not one for myself. As we filed along into our seats my arms came up in goosebumps.
Ned would have to work hard to find us, I figured, and of course there was no good reason he should try: no photo op, no obvious prospect of gain. Then again, hounding me seemed to be its own reward. He was always able to trace my movements. And he'd said he was coming.
Tumbling acrobats erupted onto the field, frolicked and ran off again; lumpy costumed creatures ran out next, waggling their top-heavy bodies, possibly cartoon animals connected to a TV show. Tinny pop music blared loudly from a bad sound system to accompany their antics, then mercifully cut off. Vendors of popcorn and glow-in-the-dark novelties moved among the seats; I bought Lena a whistle in the colors of the rainbow. Finally the main show started, a local orchestra that tuned up and launched into a jumbled rendition of the 1812 Overture.
Maybe it wasn't jumbled, I thought, maybe the flaw was in my ear, maybe Ned's long fingers had twisted even the music that I heard.
Lena bounced off my lap in delight when the first firecrackers shot into the sky. She'd moved along to sit with Main Linda by the time, a few minutes later, her father appeared at the end of our row.
He wasn't wearing a suit and tie for once; in a jacket and a polo shirt he was strictly business casual. Where was his bodyguard, I thought, his driver, the fake Secret Service guy? There was always one of them at least, but I saw no one near. Maybe they were seated somewhere, hidden in the crowd.
I moved out toward him instantly, exactly as though I wanted to be in his company and sought it out—and I did want to, I wanted nothing more than to reach him at that moment. Instead of a rush of adrenaline or heavy dread it was a stolid calm that guided my progress; I barely noticed the guests as I inched myself along between their jutting-out knees and the seatbacks in front of us. It was almost romantic, as though, beneath the falling pink stars and showers of green, there was no one anymore but Ned and me.
Stepping onto the catwalk I remembered parking with a boyfriend senior year, just a few steps from the stadium wall. I knew the moldy smell of the seats in his car, the lacy brown rust along the bottom of the doors, and how we'd thought of the fireworks as our soundtrack—the world was about us. We were sure of that as we made out, moving in the darkness of the car with our long, lean arms and legs bound up in each other, that soft skin tingling over the curves, thrilled by the conviction that this here, this was the only and the all. There was no question, then, that the world had been created as our scenery.
That was the bliss of being young, the pure egoist joy. But if you get old and don't grow out of it, I thought, looking up at my husband, you are ruined.
Maybe he'd never had a chance for that. Maybe he never had that kind of youth. Maybe he could only feel it now.
I leaned in as though I wanted to kiss him, and though I don't think he believed or wished for that anymore he must have been surprised for a moment. He's always assumed I'm harmless, pathetically harmless, and that gave me a couple of seconds' grace to slam my hands against his chest. I was feeling nothing for him then but a pity that stretched all the way back to his childhood, all the way back to before he was him.
At the second of contact I saw how the guests had been drawn together, dots gathering around a node or birds flocking to a flyway. I saw Ned and his ominous host converging on us like a machine army—even the child in the subway train, even the air in the tires of my car, even the fire that had burned the house, all these were his armaments. I saw in every granule and wave how my husband's power had seemed impossible, how it had borne the sheen of dark magic for me but was constituted of energy, energy subverted.
And when the heels of my hands came off him again, the images faded.
But it wasn't easy to send him over the rail. I didn't have enough weight behind me or enough leverage; maybe the angle of my approach was weak. I felt the bulk of his chest against my hands, the shock of his unyielding body as he leaned back. The chest was the wrong place to hit, a mistake that almost cost me my life: he was well-balanced with the rail against the backs of his thighs. Instead of toppling backwards he grabbed me and steadied himself—a strong man as well as a beautiful one. With his disciplined allegiance to fitness he'd always had strength. Discipline equals strength, though the coldness of the equation is depressing—unfair, it seemed to me as I felt instantly shocked and made foolish by the feebleness of my attack.
I'd felt its prospect tingle on my skin and seconds later that prospect had ebbed. My chance had passed. Why does strength hold itself so stubbornly away, why can't it be that we can summon it out of feeling or impulse, out of just wanting to? Fear made my legs weak. I couldn't move.
One hand grabbed my right shoulder and the other dug into my left wrist like a claw, and then it was twisting me there, by the wrist, and I don't know if I gasped or shrieked.
But all the time he was smiling.
Then he raised his hand from my shoulder and, still smiling, punched my face with it, sideways and hard. I felt my nose crunch and the pain was blinding; my eyes squeezed shut and now he was punching me again. And again. My tongue felt a loose molar and my mouth was full of blood.
I was willing to fall with him if I had to. I feared being crippled, but dying I could stand, as long as I could hold a picture of Lena in safety as I fell, Lena in Solly's care, Solly and Luisa keeping her safe from harm. Before I could push myself forward and topple us both there was a rush of others around me, a cluster of people, and it's hard to say what the geometry was. Ned must have known I wasn't alone, but only then did his smile flicker. Or so I believe. I couldn't see much by then, was blinded by the blood in my eyes.
I know there were others all around me but I couldn't say if we made noise, I couldn't say how our hands moved or our feet, couldn't say much about who did what, whose bodies pushed or pulled, all I can say is that at a certain point I swiped at my eyes and saw we were by ourselves.
Ned was gone.
And when we looked around us—after we leaned over the rail and stared down into a pool of black that didn't tell us anything—we found the crowd seemed to have ignored our scuffle. But I wasn't paying attention, I was preoccupied by the pain in my face, the blood dripping down my chin. My nose made a high wheezing noise when I breathed. Will took my hand, Navid was squeezing my shoulder, and then we turned and in a rush we pounded down the stairs—other than me it was all men, Will and Gabe and Burke and Navid and Don; Lena was away from all of it, back in the seats, deep into the row surrounded by the Lindas.
We pounded down and out and around, running hard until we got to the right stretch of pavement. In the dark I breathed my fast, wheezing breaths, tried not to faint from the acuteness of the pain. There wasn't a floodlight anywhere near us and I couldn't see his face. I wasn't sure I even wanted to.
Finally someone found a penlight—I think it was hanging off a keychain—and its weak light was dancing over Ned's head and shoulders, a small spot unequal to the task, lighting the planes of his face in a piecemeal way. I clenched my hands into fists so the pressure would anchor me against unreality.
But he looked as real as anything lying there, real and even alive, his magnetism intact despite the white polo shirt that should have left him looking like an out-of-place golfer. His jacket was spread open at his sides like wings, his arms were flung out, eyes nearly closed, well-shaped mouth just a bit open. The skin of his face was stainless, almost without a pore, its same delicate hue of salon gold.
Only the pebbly asphalt around his head was stained.
NEWS OF HIS death ran in Alaskan media outlets: heroically trying to save a fellow climber, he'd lost his footing in the mountains and plummeted. On the main street in Anchorage there were altars of flowers and photographs. People held candlelight vigils, although (said Charley) they were notably more modest than for fallen celebrities.
There are cameras at the stadium but maybe it was too dark for them to capture what had happened: in any case none of us were ever contacted, none of us were questioned. I have to conclude this was intentional—that it wouldn't have jibed with the narrative.
We stayed in my parents' house for two weeks after the Fourth. I had to have my nose reset and the bruises around my eyes are still fading; the tooth I lost was at the back so the gap doesn't show.
Lena asked about the nose and the bruises, of course. I thought about not telling her, but then I thought again and I did. "I pushed your daddy," I said. "Listen. I'm not proud of it. It's not the way to solve problems. But then he hit me back. Harder. Men shouldn't ever hit women."
"He should have only pushed you back," said Lena, pragmatic. "It's not fair. I'm glad I don't like him."
"Lena," I said, holding her hands, "your daddy's not coming back. We won't see him again."
"That's good," she said.
I'VE BEEN HELPING my mother with the funeral aftermath. Solly had to go back to work, so we said goodbye to him and Luisa and waved to them from the front porch as they drove off.
After they left we moved at our leisure through tidy rooms, curating the many vases of cut flowers as their rotting stems sloughed off into the clouding water. We sorted clothes and shoes into boxes for donation, read and acknowledged condolence cards; we cleaned out my father's desk, his chest of drawers, the file cabinets and high-up shelves at the back of his closet. I drove my mother to the bank to fill out forms, went online to switch her utilities and other services out of his name, made sure she filed a claim with the life-insurance carrier.
While we were going about these dull tasks, Will walked with Lena to a nearby park, a nearby pool. He took her to the movies, to a beach in Connecticut, and once to a state fair, where they went on a Tilt-A-Whirl, ate funnel cakes with powdered sugar and, by shooting a water gun, won her an orange stuffed giraffe.
Those public places, open to the world, the two of them were able to wander through in liberty.
For me it was a melancholy, dreary time with a curious softness. I kept waiting—I wait even now—but so far I've found no moral torment in being a murderer.
None at all.
IF WHAT SLIPS through to us from the deeper language is filtered and textured by our own interests and affections—our ties to babies or animals or trees—maybe I heard only what I could.
Maybe our gods are as small as we are or as large, varying with the size of our empathy. Maybe when a man's mind is small his God shrinks to fit.
Because if you're the kind of person who wants to know what's at the end of the universe, what's at the edge of being, and you grow older and older and comprehension settles on you that you'll never know, despair can well up. The question of what we don't see, what's beyond our capacity—in the space where the answer should be, in the knowledge that nothing will ever give us that answer—we have to pass through all the dark nights we live until we die. Never to see what's at the end of infinity, never to see the future of what we love, even the hidden lives of our children—
the knowledge breaks our hearts. It nearly cracks us open as we walk.
It's enough of a burden, that futile desire to know more than we ever can. But worse than the mind's natural limits, far worse is the invasion of its privacy. Ned's desecration of my thoughts, that was a distortion I could never have kept living with, that conversion of the world's airy expansiveness and wild unknowns into gray squares. Compared to that violence the presence of divinity was gentle.
With language, with the splendid idea of an intelligence that lasted forever, at least I still had my own perceptions, my own moods. I had room for doubt, plenty of space for movement. That room and space could be inhabited. But Ned's monotony of empty assertions in the service of self-promotion, self-replication and mastery for its own sake, his reach that extended past the boundaries of even the body—that was a weapon without end.
DON CALLED ME tonight, just called me on the cell phone. Slowly I'm learning to live with his pronouncements. It wasn't over, he said, as I had to know: in fact we were still at the start. My husband happened to be the first we met, he said, the first we encountered personally, but another had already risen to take his place. There are many like him.
They are legion, said Don. They speak in false tongues and want to own the world.
No, scratch that, he said. We both know they own the world already, but now they want even more.
Now they want to make it over in their own image.
"Are you ready?" he asked me.
I THINK OF what Kay wrote in her mania.
Deep language is in all living things but all the others, it stays with. Only not humans . . . God leaves us, Anna, God leaves us.
Yet we're the children of that language—not the only children, that boast was always a rookie mistake, but among their multitudes. We still swim in the shallows of that vast and ancient sea, the water that runs through us, a coding of genes and flesh that lives on in beings and cultures. We are those bonds that make our nervous systems, our circulation, our lungs exert their miraculous intelligence without our direction—the beneath and always, the insane, preposterous motion of life.
Let God leave us, Kay, if what you mean is constant company. Let God leave us! Let us grow up. Let us walk forward on our own. Because we need the silence of the holy: we need the sacred and equally we need its maddening silence. And in the curious privacy and relief of that silence we can go out into the chaos and commit a thousand acts of minor and gleeful splendor all our own. If it's our tragedy to be left by God, then let it also be our luck.
Our loneliness is our strength. It's not the same as being alone—almost the opposite. Loneliness is the sense of others, present but beyond our reach.
We feel a terrible tenderness, a terrible gratitude, and at the end we see that face and know the moment is here. The beast has come for us at last.
ALSO BY LYDIA MILLET
Mermaids in Paradise
Magnificence
Ghost Lights
How the Dead Dream
Love in Infant Monkeys
Oh Pure and Radiant Heart
Everyone's Pretty
My Happy Life
George Bush, Dark Prince of Love
Omnivores
FOR YOUNG READERS
Pills and Starships
The Shimmers in the Night
The Fires Beneath the Sea
Deepest thanks to Maria Massie and Tom Mayer, beloved agent and beloved editor, and to all at Norton who worked on this book, including Elizabeth Riley, Ryan Harrington, Nancy Palmquist, Don Rifkin, Ingsu Liu, David High, Bill Rusin, Deirdre Dolan, Dan Christiaens, Golda Rademacher, Karen Rice, Meredith McGinnis, Steve Colca, and Julia Druskin.
Copyright © 2016 by Lydia Millet
All rights reserved
First Edition
For information about permission to reproduce selections from this book,
write to Permissions, W. W. Norton & Company, Inc.,
500 Fifth Avenue, New York, NY 10110
For information about special discounts for bulk purchases, please contact
W. W. Norton Special Sales at specialsales@wwnorton.com or 800-233-4830
Book design by Chris Welch Design
Production manager: Julia Druskin
ISBN 978-0-393-28554-3
ISBN 978-0-393-28555-0 (e-book)
W. W. Norton & Company, Inc.
500 Fifth Avenue, New York, N.Y. 10110
www.wwnorton.com
W. W. Norton & Company Ltd.
Castle House, 75/76 Wells Street, London W1T 3QT
| {
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Nikolaï Karlovitch Reitzenstein (en ), né le , décédé le à Pétrograd, est un amiral russe, membre du Conseil de l'Amirauté de 1909 à 1916.
Biographie
Originaire d'une famille de la noblesse de Bohême installée en Russie au . Diplômé du corps naval des Cadets en 1874, Nikolaï von Reitzenstein navigua sur l'Océan Pacifique de 1886 à 1887. Il exerça le commandement à bord du Tchaïka et du Raketa (1882-1883). En 1889 il fut nommé commandant du destroyer Narva. Il servit en qualité d'officer supérieur sur le clipper Djiguit de 1891 à 1894. En 1895, il fut nommé commandant de la canonnière Erch. Il est transféré sur le navire de transport Europe dont il exerça le commandement de 1895 à 1898) puis sur le croiseur Askold.
Reitzenstein débuta la Guerre russo-japonaise (1904-1905) au grade de capitaine de premier rang. Il fut nommé contre-amiral le et chef d'une escadre de croiseurs basée à Vladivostok. Il occupa les fonctions d'adjoint du chef d'escadre de la Flotte du Pacifique à partir du . Le 14 mars de la même année, il reçut le commandement d'un escadron à Port-Arthur. Au cours de la bataille de la mer Jaune, le croiseur Askold subit de graves dommages, le . Il fut capturé par les Japonais et immobilisé dans le port de Shanghai jusqu'à la fin du conflit opposant les Russes aux Japonais.
De 1907 à 1908, Reitzenstein occupa les fonctions de commandant d'un détachement d'artillerie de marine appartenant à la Flotte de la mer Baltique. En 1909, il fut admis au Conseil de l'Amirauté. Il présida les Commissions spéciales de la flotte de l'Amour de 1910 à 1916 et, de 1912 à 1916, celle de la défense du littoral.
Il fut mis à la retraite, le .
Décès et inhumation
Nikolaï Karlovitch Reitzenstein décéda le à Petrograd et fut inhumé au cimetière de Novodiévitchi de Saint-Pétersbourg.
Distinctions
: Ordre hawaïen de Kalakaua (Chevalier)
: Ordre de l'Aigle rouge (deuxième classe)
1902 : Ordre de la Légion d'Honneur (officier)
1907 : Saint-Stanislas (première classe)
1910 : Ordre de Sainte-Anne (première classe)
1913 : Ordre de Saint-Vladimir (deuxième classe)
: Ordre de l'Aigle blanc
Notes et références
Sources
В.Я. Крестьянинов, С.В. Молодцов, Крейсер Аскольд, Велень, СПБ, 1993.
Liens internes
Guerre russo-japonaise (1904-1905)
Bataille de Port-Arthur
Liens externes
eps.dvo.ru
www.rustrana.ru
Amiral de l'Empire russe
Militaire de l'Empire russe
Récipiendaire de l'ordre de Saint-Vladimir de 2e classe
Récipiendaire de l'ordre de Saint-Vladimir de 3e classe
Récipiendaire de l'ordre de Saint-Vladimir de 4e classe
Récipiendaire de l'ordre de l'Aigle blanc (russe)
Récipiendaire de l'ordre de Sainte-Anne
Récipiendaire de l'ordre de Saint-Stanislas (russe)
Naissance en août 1854
Naissance dans le gouvernement de Saint-Pétersbourg
Naissance à Saint-Pétersbourg
Décès en novembre 1916
Décès à Pétrograd
Décès à 62 ans
Personnalité inhumée au cimetière de Novodevitchi (Saint-Pétersbourg)
Militaire russe de la guerre russo-japonaise | {
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TURN UP THE VOLUME
Dope Blog for Music Junkies and Gig Addicts
How Was 2021 For THE CHRONICLES OF MANIMAL AND SAMARA
Posted on 18 Days Ago by TUTV!
Imagine the late, legendary American poet/writer Sylvia Plath fronting a noise-exploring band, creating a striking combination of spoken-word poetry and metallic soundscapes. That's what THE CHRONICLES OF MANIMAL AND SAMARA accomplish with their sonic art. The way they embed their acute and outspoken views on political and social issues in sound and vision is pretty special.
No wonder the duo's – Daphne Ang and Andrea Papi – LP called
FULL SPECTRUM, is Turn Up The Volume's the debut album of 2021.
Want to learn more about this DIY tandem? Check out this interview. As usual,
we start a Q & A with music. Here comes one of their two singles 'Count The Dead'…
Your stunning debut album FULL SPECTRUM came out last February.
What was the experience like, then and now, of having made a first longplayer?
"Thank you! It seems so long ago that we wrote and recorded the songs in Full Spectrum.
It has truly been a journey to remember. It was tough of course, being independent artists, we only have each other to count on to get the job done.
Whatever you hear, read, see, or watch, has been made entirely by the both of us, everything from designing graphics and cover artwork, making our own videos, copywriting, of course, the music and lyrics.
Of course, there have been ups and downs in this journey, but at least we didn't
have to do it alone. Also very happy to have made so many friends along the way."
Already working on a follow-up?
"Yes, in fact we started working on the songs in our second album, Trust No Leaders, about a month after we released Full Spectrum. We wrote and recorded the album between April and August this year. It is now mixed and mastered and due for release sometime in 2022.
It's definitely much heavier, and darker than our first album, both musically and lyrically. Our first album was about the metaphysical and the existential. But the new album looks inwards into the human condition, often going into dark places.
The current crisis and the events of our time have made it necessary for us to analyse, investigate, and expose the issues in urgent need of systemic change and look inwards into the human condition. But despite its dark tone, overall it pushes for making a future that prioritises people and the environment. It is about the awakening of the collective human spirit. It is about what it means to be human.
They are similar in the way that the lyrics are both influenced by theatre and literature,
the first album drew from Ancient Greek theatre to Shakespearean drama, and the new
album tends to be more precise and structured, as it draws elements from Carl Jung's works, Classical Persian poetry, Gnostic texts, theology, and modern theatre."
You released also two new singles after the LP. What's COUNT THE DEAD about?
"Count the Dead actually continues the themes addressed in Love in the Time of Pestilence, and is the second installment of songs which address the ongoing pandemic.
We composed this song as a protest against world leaders, whose negligence and recklessness have resulted in one of the largest avoidable losses of lives in generations.
We wanted to draw attention to the social and economic injustices that plague society, and the violent health inequalities that have been further exacerbated by the global pandemic, which has now claimed more than 5 million lives worldwide.
At a time when the world appears to be more divided than ever before,
discourse has never been more important, and urgent."
The video for the other single THE PROPHET is very impressive.
What's the process of the construction of the clip like?
Andrea: "We usually have a clear script before the music composition.
Samara is the best to make it real in visual form."
Daphne: "Thank you! I think it's our best video so far. Andrea and I will first discuss the storyline and the scenes we want to present in the video. I then go away and construct them scene by scene to go with the music and lyrics. This usually involves several layers per scene. With every new video we make, I learn new techniques and tricks. But all you really need is imagination, patience…and several Adobe photo and video editing software programs!"
Your spoken word performances are a fundamental part of the whole TCOMAS sound, Daphne. Any intention to sing on a track(s) in the future?
Daphne: "There's going to be surprises in the new album but I can't say it right now. But definitely no plans to abandon this. It's our form of expression. I feel that it allows for the use of phrases, metaphors and imagery in a way that standard lyrical songwriting does not.
It is powerful, cathartic, and moving, and when combined within music, has the power to connect with the personal and collective unconscious on a really deeper level and singing would."
What are the consequences of BREXIT for musicians?
"BREXIT and COVID killed and destroyed the economy
in the UK and guess who's gonna pay for it?"
Any future plans to play live?
"Yes, of course, we are dying to. We would really love to, but unfortunately, this pandemic has made us seriously consider the implications of what it will be to go back 'to normal' when obviously the time is still not ripe and until we can ensure the safety of all our fans, and everyone in attendance, we will have to hold it off until better days come for all of us."
Which movie would you pick to visualize the TCOMAS sound?
Andrea: "It's difficult to pinpoint as each song sounds like a movie on its own. But let's try, shall we? For Full Spectrum, I would say From Dusk Till Dawn (1996) directed by Robert Rodriguez and written by Quentin Tarantino as well Fear and Loathing in Las Vegas (1998) directed by Terry Gilliamould.
For the new album, Trust No Leaders, probably somewhere along the lines of Apocalypto (2006) meets American History X (1998) with a touch of Once Upon a Time in America (1984), with Sir Alfred Joseph Hitchcock as the director."
Daphne: "Stanley Kubrick's 2001: A Space Odyssey (1968), Peter Greenaway's The Cook, the Thief, His Wife & Her Lover (1989), and Darren Aronofsky's The Fountain (2006). Film and the moving image play a big role in our music, not just in influencing the videos we make but also in the writing process as well. Being visual artists as well, image and sound always go hand in hand for us."
Suppose you were asked to rewrite and put new music to the British Royal Anthem 'God Save The Queen'. No restrictions whatsoever. What would be the outcome, in sound and vision?
"The Sex Pistols have already done it! We'd stick with their version, because it is epic and irreplaceable. Especially this event where they played God Save The Queen ('we mean it man') on The River Thames in London in 1977 on the Silver Jubilee of Elizabeth II 'beneath
the bridges of London'…before they were stopped by the police."
The best track you heard all year?
'Us Against December Skies' by Harakiri for the Sky
THE Event – good or bad – of 2021?
Andrea:"The good one is that we are still here rocking more than ever before!"
Daphne: "We are really proud that we released our debut album and also
managed to write and record our second album in one year…and also start
a side project business:TCOMAS Studio."
Andrea: "The worst thing happened in 2021: The Holy See now cannot be sued for cases
of sexual abuse committed by priests of various countries — as decided by the European Court of Human Rights."
Daphne: "Delta and Omicron."
Are you fans of traditional Xmas carols? If so, which one is your favorite?
"No, not at all, sorry! But we love to share our culinary traditions with each other, so we will whip up a Christmas feast and probably play some non-Xmas Jazz. Okay, okay, if we were to choose, it would have to be sung by Frank Sinatra, and only Frank Sinatra."
Suppose you were asked to a DJ set on 31st December.
Name 3 songs you would certainly play?
"The new year is always a time to reflect upon history through great music.
For that, can we get 4 each, please? Our playlist will be the following songs, in this order.
House Of The Rising Sun by The Animals (1964)
Blowin' In The Wind by Bob Dylan
Volare by Domenico Modugno (1958)
The Revolution Will Not Be Televised by Gil Scott-Heron (1971)
You Want It Darker by Leonard Cohen (2016)
Johnny Cash's cover of Hurt (2002) by Trent Reznor (1995)
Lazarus by David Bowie (2015/2016)
Disturbed's 2015 cover of Sound Of Silence by Simon & Garfunkel (1964)
Three things you really love to happen
in 2022 for Samara and Manimal.
"We'd really like to go on a holiday!
A successful album launch
A Covid-free world (might be wishful thinking)."
Thank you for this interview, Daphne and Andrea.
May the road rise with The Chronicles of Manimal and Samara in 2022!
Back to the music.
Stream FULL SPECTRUM here…
TCOMAS: Facebook
Full Spectrum How Was 2021 For THE CHRONICLES OF MANIMAL AND SAMARA Interview
LIAM GALLAGHER KNEBWORTH 2022
Quote Of The Day – SLEAFORD MODS' Half JASON WILLIAMSON About The Working Class Area | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 2,286 |
package server
import (
"fmt"
"net/http"
"os"
"github.com/gorilla/mux"
"github.com/servicebroker/servicebroker/k8s/service_controller/model"
)
type Server struct {
controller *Controller
}
func CreateServer(serviceStorage model.ServiceStorage) (*Server, error) {
return &Server{
controller: CreateController(serviceStorage),
}, nil
}
func (s *Server) Start() {
router := mux.NewRouter()
router.HandleFunc("/v2/services", s.controller.Services).Methods("GET")
router.HandleFunc("/v2/service_plans", s.controller.Inventory).Methods("GET")
// Broker related stuff
router.HandleFunc("/v2/service_brokers", s.controller.ListServiceBrokers).Methods("GET")
router.HandleFunc("/v2/service_brokers", s.controller.CreateServiceBroker).Methods("POST")
router.HandleFunc("/v2/service_brokers/{broker_id}", s.controller.GetServiceBroker).Methods("GET")
router.HandleFunc("/v2/service_brokers/{broker_id}", s.controller.DeleteServiceBroker).Methods("DELETE")
// TODO: implement updating a service broker.
// router.HandleFunc("/v2/service_brokers/{broker_id}", s.Controller.UpdateServiceBroker).Methods.("PUT")
router.HandleFunc("/v2/service_instances", s.controller.ListServiceInstances).Methods("GET")
router.HandleFunc("/v2/service_instances", s.controller.CreateServiceInstance).Methods("POST")
router.HandleFunc("/v2/service_instances/{service_id}", s.controller.GetServiceInstance).Methods("GET")
router.HandleFunc("/v2/service_instances/{service_id}", s.controller.DeleteServiceInstance).Methods("DELETE")
// TODO: implement list service bindings for this service instance.
// router.HandleFunc("/v2/service_instances/{service_id}/service_bindings", s.controller.ListServiceInstanceBindings).Methods("GET")
router.HandleFunc("/v2/service_bindings", s.controller.ListServiceBindings).Methods("GET")
router.HandleFunc("/v2/service_bindings", s.controller.CreateServiceBinding).Methods("POST")
router.HandleFunc("/v2/service_bindings/{binding_id}", s.controller.GetServiceBinding).Methods("GET")
router.HandleFunc("/v2/service_bindings/{binding_id}", s.controller.DeleteServiceBinding).Methods("DELETE")
http.Handle("/", router)
cfPort := os.Getenv("PORT")
if cfPort == "" {
cfPort = "10000"
}
fmt.Println("Service Controller started on port " + cfPort)
err := http.ListenAndServe(":"+cfPort, nil)
fmt.Println(err.Error())
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 4,099 |
Acier damassé peut désigner :
acier de Damas de cristallisation (voir wootz) ;
acier de Damas de corroyage (voir corroyage). | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 7,203 |
The Panel will review and provide independent expert advice on EPA's draft Hydraulic Fracturing Study Plan that will investigate the potential public health and environmental protection research issues that may be associated with hydraulic fracturing.
It will be led by David A. Dzombak, professor of environmental engineering at Carnegie Mellon University.
Legere, Laura. "Peer-review panel for EPA fracking study includes six Pa. scientists." The Scranton Times-Tribune. Jan. 18, 2011.
A panel of geologists, toxicologists, engineers and doctors that will peer-review a high-profile Environmental Protection Agency study of hydraulic fracturing will include six scientists from Pennsylvania, more than any other state.
The panel will review the techniques and analysis the EPA uses to draft a study of the potential environmental and health impacts of hydraulic fracturing - the process used in natural gas exploration of injecting a high-pressure mix of chemically treated water and sand underground to break apart a rock formation and release the gas.
...In a memo announcing the new panel, the EPA found "no conflicts of interest or appearances of a lack of impartiality for the members of this panel." | {
"redpajama_set_name": "RedPajamaC4"
} | 6,104 |
Belief in the Prophets
By Javed Ahmad Ghamidi on July 01, 2012
Al-Islam
(Translated. by Dr Shehzad Saleem)
People through whom the Almighty completed providing His guidance to mankind are called prophets. They were, in fact, human beings; however, the Almighty selected them for this purpose on the basis of His all embracing knowledge and wisdom. As such, prophethood is God-given and cannot be acquired through self-effort or training. It is evident from the Qur'ān that only those people were chosen as prophets who were able to shield themselves from the lures of their inner-self as well as those of Satan, guard themselves against sin and were the righteous and pious among their people.
The Almighty sent these prophets to every community. He had promised Adam (sws) that He would guide his progeny through guidance revealed by Him. This guidance was given to mankind through these very prophets. After receiving revelation from God, they told what the truth was to people, gave glad tidings to those who believed in Him and warned those who did not believe in Him of a dreadful fate.
Prophets were not needed to make people recognize their Lord or give them a means to distinguish between good and evil. These facts are ingrained in their nature and innately found in them. Thus the need for prophets did not arise to inform man of these things; it arose because of two other reasons.
Firstly, for completion of guidance. This means that man be reminded of whatever is ingrained in his nature in concise form and of whatever he has known eternally and all its details be specified for him.
Secondly, for itmām al-hujjah. This means to awaken man from his slumber of unawareness and after providing the testimony of his intellect and knowledge provide another testimony through these prophets to such an extent that no one is left with any excuse to deny the truth.
Through the prophethood of Muhammad (sws), both these objectives had been achieved at the global level and accomplished to the utmost. Hence the institution of prophethood was terminated. The Qur'ān declared him to be the last prophet. After him, no prophet or messenger shall come.
Every right-minded person has no difficulty in recognizing a prophet. If a person has a discerning mind and a vibrant heart, then the very person of a prophet is a miracle. However, besides this, the Almighty blesses a prophet with potent signs which, though may not induce his adversaries to openly acknowledge him, are enough to leave them with no excuse to deny his veracity. It is evident from the Qur'ān that these signs are given to every prophet and their nature depends upon his times and circumstances. We shall mention a few of these in the following paragraphs:
i. A prophet generally comes in accordance with the prediction of the prophet who precedes him and comes as a fulfillment of this prediction. Viewed thus, he is not an un-introduced personality. People are familiar with him and also await him. It is known from the Bible that the Prophet John (sws) foretold the coming of Jesus (sws) all over Jerusalem. The predictions of the advent of Muhammad (sws) are mentioned both in the Old and the New Testaments. One of the primary objectives of the advent of Jesus (sws) was to prophesy about the coming of an unlettered prophet in Arabia. The Qur'ān has presented as a conclusive proof of its own authenticity the fact that the scholars of the Israelites recognize the Qur'ān the way an estranged father recognizes his much awaited and promised son. This means that they also fully recognized Muhammad (sws).
ii. Whatever a prophet presents as the word and message of God is without any contradiction and inconsistency. Even the most ultimate of geniuses of this world like Socrates and Aristotle, Kant and Einstein, Ghālib and Iqbāl, Rāzī and Zamakhsharī cannot make such a claim about the works they have produced. However, the Qur'ān has vehemently asserted about itself that there is not a semblance of contradiction in the philosophy and ideology it presents and there is not the slightest evolution or development in its style that a person can detect. Is it possible that for years a person should give speeches on a variety of topics in different situations and circumstances and when these speeches are compiled from the beginning to the end they form such a harmonious and congruous discourse that it has no contradiction of views and does not reflect any effects of mood change of the speaker and also does not depict any revision or change of views? Only the Qur'ān has this characteristic.
iii. A prophet is blessed with miracles from God. The Qur'ān has specified that one of the reasons for which extra-ordinary miracles were given to Jesus (sws) and Moses (sws) was to authenticate them as messengers of God. No one can reject these miracles by pronouncing them as magic or trickery. The reason for this is that the reality behind such acts is very much known to those adept in such disciplines and they too are compelled to acknowledge such acts as miracles.
The miracle given to the Prophet (sws) to validate his prophethood is the Qur'ān. When those who are aware of the stylistic features of Arabic language and of its literary tradition and also have a literary appreciation of the language read the Qur'ān, they clearly feel that this could not have been produced by human beings. Thus, at more than one instance, it has challenged its addressees who do not regard it to be of divine origin and regard it to be a fabrication of Muhammad (sws) to produce just a single sūrah that can match the majestic style of the Qur'ān. If a person from amongst them in their own opinion can produce such a discourse without any literary and academic background, then they should also have no problem in doing so.
This Book of God is still with us. Fourteen centuries have passed ever since it was revealed. During this period, our world has undergone tremendous changes. Many ideologies and thoughts were presented by man only to be rejected by him later. Theories regarding man's being and those regarding the universe around him were put forth every now and then. Each underwent a process of acceptance or rejection in various periods of time. His intellectual journey took him through various paths and destinations; however, this Book of God is the only book which is as unassailable and sound today as it was fourteen centuries ago regarding the various facts it presents and which have remained under academic discussion in the past few centuries. Knowledge and intellect helplessly acknowledged its superiority at that time the way they do today. Every statement it has given has stood the test of time. The world, in spite of its astounding scientific and academic discoveries, has failed to present a better alternative to the views it presents.
iv. The Almighty informs a prophet of certain unknown things which are impossible to come into the knowledge of any other human being. One example of this is the predictions made by divine revelations which were fulfilled to the utmost. Some of these predictions are mentioned in the Qur'ān and some in the Hadīth narratives. Every student of the Qur'ān is aware of the predictions regarding the supremacy of the Prophet Muhammad (sws) in Arabia, the conquest of Makkah and the people entering the folds of Islam in multitudes. Another great prediction made by the Qur'ān was the victory of the Romans after being defeated by the Persians.
When this prediction was made, then in the words of Edward Gibbon: "No prophecy could be more distant from its accomplishment, since the first twelve years of Heraclius announced the approaching dissolution of the empire."[1]However, it was fulfilled at the very time it was meant to and in March 628 AD the Roman emperor returned to Constantinople with such splendour that four elephants were pulling his chariot and numerous people stood outside the city with lanterns and branches of olive to welcome back their hero.
v. Prophets who are also designated as Messengers (rusul) are a symbol of divine justice in this world and decide the fate of their people in this very world. The way this is brought about is that if these Messengers abide by their covenant with God they are rewarded in this world and if they deviate from it they are punished in this world. The result is that the very existence of these Messengers becomes a sign of God and it is as if they witness God walking on earth with these Messengers and administering justice. It is this situation which becomes the basis of passing judgement by the Almighty both in this world and in that to come. Consequently, the Almighty grants supremacy to these Messengers and punishes those who reject their message.
The Almighty has directed people to obey a prophet. He has made it very clear in His Book that a prophet is not merely to be revered, he is to be obeyed also. He is not sent that people merely regard him to be a prophet and then leave him aside; he is not merely one who counsels and preaches; he is a guide who must be obeyed. The very objective of his coming is that whatever guidance he provides in all affairs of life must be followed without any hesitation. Moreover, obeying a prophet is not merely a ritual. The Qur'ān requires from a believer to obey him with the spirit of following him and with full sincerity, full reverence and full devotion.
Ghamidi on Farahi
Surah Luqman
Surah Ahzab
Sūrah al-Rūm
Surah al-'Ankabut
Surah al-Naml (2/2)
Surah al-Shu'ara' (2/2)
Surah al-Furqan
People of Paradise
Surah al-Ra'd
Indications of Islam, Faith, Virtue and the Day of Judgement
Surah al-Nur (2/2)
Surah Mu'minun (2/2)
Surah al-Hajj (2/2)
Surah al-Anbiya (2/2)
Surah Taha (3/3)
Surah Maryam (2/2)
Surah al-Kahf (2/2)
Surah Bani Isra'il (1/3)
Surah al-Nahl (2/2)
Surah Ibrahim – Surah al-Hijr (2/2)
Surah Yusuf (3/3)
Surah Hud (3/3)
Surah Yunus (2/2)
Surah al-Tawbah (61-129) (2/2)
Surah al-Tawbah (38-60)
Surah al-Tawbah (1-37) (2/2)
Is Democracy Compatible with Islam?
Surah al-Anfal (41-75) (2/2)
Surah al-Anfal (31-40)
Surah Nisa' (153-176)
Surahs Muzzammil and Muddaththir
Surah al-Anfal (1-30)
Surah al-A'raf (184-205)
Islamic Punishments
Surah al-A'raf (123-153)
Surah al-A'raf (80-122)
Surah al-A'raf (57-79)
Surah al-A'raf (26-56) (2/2)
Surah al-A'raf (1-25)
Surah An'am (128-165) (2/2)
Surah An'am (100-127)
Surah An'am (74-99)
The Noble Wives of the Prophet (sws)
Surah An'am (1-24)
Surahs Falaq-Nas
Surahs Lahab-Ikhlas
Surahs Kafirun-Nasr
Surahs Ma'un-Kawthar
Surah Fil – Surah Quraysh
Surah 'Asr – Surah Humazah
Surah Qari'ah – Surah Takathur
Surah Zilzal – Surah 'Aadiyat
Surahs Qadr-Bayyinah
Surahs Tin – 'Alaq
State and Government
Surahs Duha-Alam Nashrah
Surahs Shams-Layl
Islam and the State: A Counter Narrative
Surahs Fajr-Balad
The Basis of Legislation
The Shari'ah of Preaching
Surahs A'la - Ghashiyah
Variant Readings
Surahs Mutaffifin - Inshiqaq
Surahs Buruj – Tariq
Itmam al-Hujjah [1] of God's Messengers
Theory of Evolution (2)
Surahs 'Asr-Humazah
Surahs Nazi'at-'Abas
Surahs Mursalat-Naba
Dealings and Practices of God
Wudū and Nail Polish
Surahs Qiyamah-Dahr
Surahs Takwir - Infitar
Sūrahs Muzzammil and Muddaththir
Downfall of the Muslims
Javed Ahmad Ghamidi on Hadith
Surahs Nuh and Jinn
An Interview with the Indian Media
Surahs Haqqah and Ma'arij
Roles and Responsibilities of Muslims in the West
Surahs Mulk - Qalam
The Punishment of Intentional Murder
Hajj and 'Umrah
Sūrahs Hashr – Mumtahinah
Sūrahs Hadīd – Mujādalah
The Right to make a Will
Talks of the Prophet Muhammad (sws)
Surah Waqi'ah
The General and the Specific
Sūrah Qamar and Sūrah Rahmān
Surah Tur and Surah Najm
Inheritance of an Orphaned Grandchild
The Sharī'ah of Preaching
The Source of Religion
Sūrah Mā'idah (90-120)
Sūrahs Qāf and Dhāriyāt (Part 2/2)
Sūrah Mā'idah (32-63) part (1/2)
Age of 'Ā'ishah (rta) at her Marriage
Sūrah Mā'idah (1-31) part (1/2)
Sūrah Nisā' (153-176)
Belief in the Hereafter
Sūrah Nisā' (58-100)
The Consensus of Muslims
Sūrah Nisā' (36-57)
Islam and the State
Our Call to Humanity | {
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{"url":"http:\/\/nlab-pages.s3.us-east-2.amazonaws.com\/nlab\/show\/Joseph+Varilly","text":"# nLab Joseph Varilly\n\n## Selected writings\n\nOn the Moyal product as the polarized groupoid convolution algebra of the corresponding symplectic groupoid, hence as an example of geometric quantization of symplectic groupoids:\n\nDiscussion of the spectral Riemannian geometry of the fuzzy 2-sphere:\n\ncategory: people","date":"2022-01-16 19:39:04","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 1, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.8691176772117615, \"perplexity\": 2876.5721712845066}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2022-05\/segments\/1642320300010.26\/warc\/CC-MAIN-20220116180715-20220116210715-00692.warc.gz\"}"} | null | null |
Provincial heritage sites in South Africa are places that are of historic or cultural importance within the context of the province concerned and which are for this reason declared in terms of Section 28 of the National Heritage Resources Act (NHRA) or legislation of the applicable province. The designation was a new one that came into effect with the introduction of the Act on 1 April 2000 when all former national monuments declared by the former National Monuments Council and its predecessors became provincial heritage sites as provided for in Section 58 of the Act.
Both provincial and national heritage sites are protected under the terms of Section 27 of the NHRA or legislation of the relevant province and a permit is required to work on them. Provincial heritage sites are declared and administered by the relevant provincial heritage resources authority whilst national heritage sites are the responsibility of SAHRA.
KwaZulu-Natal is the only province to have its own heritage legislation and provincial heritage sites are known as either 'heritage landmarks' or 'provincial landmarks' depending upon whether they are privately or government owned.
Most provincial heritage sites are still marked with an old national monuments badge, but provincial heritage resources authorities in KwaZulu-Natal, the Northern Cape and Western Cape have developed their own badges.
List of Heritage sites by province
The lists have been split up by province. Some districts have been split off from their province for site performance reasons.
List of heritage sites in Eastern Cape
List of heritage sites in Albany
List of heritage sites in Graaff-Reinet
List of heritage sites in Port Elizabeth
List of heritage sites in Free State
List of heritage sites in Gauteng
List of heritage sites in KwaZulu-Natal
List of heritage sites in Pietermaritzburg
List of heritage sites in Limpopo
List of heritage sites in Mpumalanga
List of heritage sites in North West
List of heritage sites in Northern Cape
List of heritage sites in Colesberg
List of heritage sites in Kimberley
List of heritage sites in Richmond
List of heritage sites in Victoria West
List of heritage sites in Western Cape
List of heritage sites in Paarl
List of heritage sites in Simonstown
List of heritage sites in Stellenbosch
List of heritage sites in Swellendam
List of heritage sites in Table Mountain
List of heritage sites in the Cape
List of heritage sites in Tulbagh
List of heritage sites in Worcester
List of heritage sites in Wynberg
See also
List of heritage sites in South Africa
Heritage objects (South Africa)
South African Heritage Resources Agency
Amafa aKwaZulu-Natali
Heritage Western Cape
National Monuments Council (South Africa and Namibia)
References
External links
South African Heritage Resource Agency
Provincial Heritage Resources Authorities:
KwaZulu Natal - Amafa/Heritage KwaZulu Natal
Western Cape - Heritage Western Cape
Free State - Heritage Free State
Eastern Cape - Eastern Cape Provincial Heritage Resources Authority
Mpumalanga - Mpumalanga Provincial Heritage Resources Authority
Limpopo - Limpopo Heritage Resources Authority
North West - North West Provincial Heritage Resources Authority
Northern Cape - Ngwao-Boswa Jwa Kapa Bokone
Gauteng - Provincial Heritage Resources Authority Gauteng
Searchable database of protected sites, objects and shipwrecks
Heritage registers in South Africa
Historic sites in South Africa
South African heritage resources
South African heritage sites | {
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Gordoservon or Gordoserbon or Gordoserba (; ) was an early medieval Byzantine city, and a bishopric, suffragan of the Metropolis of Nicaea, in the region of Bithynia, Asia Minor. It is mentioned in several ecclesiastical sources from the period between the 7th and the 9th century. Most notably, the city is mentioned in the acts of the Council of Trullo (691-692), as a seat of bishop Isidore, who attended the council.
The exact location of this city, and etymology of its name, have been a subject of interest for scholars, who proposed several solutions for both questions.
History
In the 7th century, the Byzantine Emperors Constans II (in 657–658) and Justinian II (in 688–689) led expeditions against the Balkan Slavs as far as rivers Struma and Vardar in the region of Macedonia. Many of the conquered tribes were transferred to the Opsikion district of northwestern Asia Minor. Part of those Asia Minor Slavs deserted to the Arabs in 665 and again in 692. As the name of the city could suggest that among its founders were Serbs, some modern scholars consider that the colony was founded by these Slavs, and variously date it to 649, 667, 680, or 688–689. According to Sima Ćirković it is possible that some Serbs which populated Gordoservon were brought from an area near Thessaloniki.
Similarly, in 1129–1130 some Serbs were likely settled in Bithynia by John II Komnenos, due to the mention of a settlement called Servochōria () near Nicomedia, mentioned in the 13th century source Partitio regni Graeci (1204). Some identified Gordoserba with this Servochōria, but the connection is uncertain.
Up to the 20th century, Gordo-Servorum or Gordoservae was commonly equated with nova Juliopolis, which in turn was equated with Gordium (capital of Phrygia) or another place with the same name Gordion, Gordenorum, Gordiu-come(nis), Gordiū-tīchos which became known as Juliopolis (Iuliogordus) according to several 1st-century BCE up to 2nd century CE sources. William Mitchell Ramsay (1890) connected Justinianopolis-Mela, called Nova Justinianopolis Gordi (680), with the bishoprics of Gordoserboi or Gordoserba in Bithynia, Gordorounia or Gordorinia in Phrygia Salutaris, and Gordou-Kome, the former name of Juliopolis in Galatia, and that an ancient country or district along the Sangarios River was called Gordos. Additionally, he argued that Gordoserba was formed into bishopric by Justinian I in the 6th century. Siméon Vailhé, writing for the Catholic Encyclopedia (1913) considered, like Michel Le Quien, that Juliopolis of Nicaea of Bithynia was identical to Gordoserboi, because otherwise the exact location, titulars, and bishops are unknown; and that it should not be confused with Juliopolis of former Gordium.
However, Peter Charanis, analyzing the sources on the early Slavs of Asia Minor, noted that the sources are ambiguous on the exact date of migration, especially concerning Constans II, and that the first certain mention of the place is in 692, during the Quinisext Council, where was mentioned Isidore "ἀνάξιος ἐπίσκοπος Γορδοσέρβων τῆς Βιθυνῶν ἐπαρχίας" ("unworthy bishop of Gordoserba of the province of the Bithynians"). If the settlement is related to the Serbs then it contradicts the date of the Ecthesis of pseudo-Epiphanius (640), a list of cities and bishoprics which mentions Gordoservorum or Gordoserboi in the Metropolis of Nicaea in the province of Bithynia. Charanis and other scholars doubt the Slavic-Serbian origin of the city because among the known bishops (Isidoros, Neophytos, Stephanos) there are none with Slavic names, and due to the uncertainty around the etymology of the Serbian ethnonym.
Etymology
Ladislav Zgusta considered that "-serba" has nothing to do with Slavs and pointed to toponyms such as Άνάζαρβος and Ανάζαρβα Καμουή σαρβον (Anazarbus), while argued that if Gordoserba and Servochōria are identical then both cannot have a connection to John II Komnenos's activity in the 12th century, and contrary to Zgusta, Servochōria most probably means "Serbian land". Predrag Komatina also argued Serbian connection, but denied that "gordo-" derives from Proto-Slavic *gordъ (fortification, city) because Gordos was a name for a district where the settlement was situated and hence the meaning would have been "the place of the Gordos Serbs" rather than "the city of the Serbs".
References
Sources
Populated places of the Byzantine Empire
History of the Serbs
Populated places in Bithynia
Byzantine Bithynia
Defunct dioceses of the Ecumenical Patriarchate of Constantinople | {
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Q: Let {$x_n$} be such that $\lim \inf_{n \to \infty }|x_n|=0$. Prove that there exists a {$x_{n_{k}}$} where $\sum_{k=1}^{\infty}x_{n_{k}}$ converges.
Problem: Let $\{x_n\}$ be such that $\displaystyle\lim \inf_{n \to \infty }|x_n|=0$. Prove that $\exists$ a subsequence $\{x_{n_{k}}\}$ where the subseries $\displaystyle\sum_{k=1}^{\infty}x_{n_{k}}$ converges.
I know that $\displaystyle\lim\inf_{n \to \infty } |x_n| = \sup_{n \geqslant 1} \inf_{k\geqslant n} |x_k|=0$.
However, I am confused as to how to take this to the limit of a series, and conclude that infinite series converges. Any help would be appreciated!
A: Let $\displaystyle a_n=\inf_{j\geq n} |x_j|$. By hypothesis, $a_n \to 0$ as $n\to\infty$. Let $\{b_k\}$ be a positive sequence such that $\sum_{k=1}^\infty b_k <\infty.$ Obviously, $b_k \to 0$. Since $a_n \to 0$, for each $k$, there exists $n_k$ such that $a_{n_k}<b_k$. By definition, $b_k$ is not a lower bound for the set $A_{n_k}=\{|x_m| : m\geq n_k\}$. Hence, there exists $|x_{m(n_k)}|\in A_{n_k}$ such that $a_{n_k}\leq |x_{m(n_k)}|<b_k$.
Finally, by comparison, $\displaystyle \sum_{k=1}^\infty |x_{m(n_k)}|<\sum_{k=1}^\infty b_k<\infty,$ so the convergence follows (absolutely).
| {
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{"url":"https:\/\/www.doubtnut.com\/question-answer\/which-of-the-following-equations-have-same-graphs-i-y-x-2-ii-y-x2-4-x-2-iii-x-2y-x2-4-644524710","text":"Home\n\n>\n\nEnglish\n\n>\n\nClass 12\n\n>\n\nMaths\n\n>\n\nChapter\n\n>\n\nTest Papers\n\n>\n\nWhich of the following equatio...\n\nUpdated On: 27-06-2022\n\nGet Answer to any question, just click a photo and upload the photo and get the answer completely free,\n\nText Solution\n\nI and IIonlyI and III onlyII and III onlyAll have different graphs\n\nSolution\u00a0:\u00a0All have different graphs.\n\nStep by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.\n\nTranscript\n\nit is that which of the following equations have same graph first to second equation Levis and third equation X + 2 Y is equal to x square - 4 so we have given questions and we have to find that which graph has same so purpose equation Y is equal to 6 months to so that its graph will be straight line so it is straight line which is bicycle 6.2 and it will be defined on X belongs to real number it will depend on whole are whole number line and checkout second equation X square - 4 x + 2 is equal to\n\nX + 2 \/ 1 + 2 and now he can we find that X is not equal to minus 2 and then we can cancel X + 2 Y we can tell that is wise to the expense to now this is a straight line graph of state and but it will be need it will be not defined at x is equal to minus to show its domain will be explored you are excluded because it is not defined at -2 and now here to check out her equation X + 20 y x + 2 into Y is equal to x square - 4 aur candles that X + 2 into Y is equal to X + 2 into x minus 2 and now we can take X + 2 into x minus 2 on the left hand side through antique\n\nExpressway is a common for a healthy can that it is that X + 2 into why minus of x minus 2 it is equal to zero now from here we can tell that is we have two equations how to graph that X is equals to minus 2 and this whole equation be defined at X is equals to and Y is equals to expand super here you can read does that it contains two graph that X is equals to minus y is equals to show both the curves have different because that women Sadiq so we can touch that is all graphs have different curves all equations have different graph","date":"2022-06-29 06:41:50","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.3849738538265228, \"perplexity\": 410.3698231106672}, \"config\": {\"markdown_headings\": true, \"markdown_code\": false, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2022-27\/segments\/1656103624904.34\/warc\/CC-MAIN-20220629054527-20220629084527-00355.warc.gz\"}"} | null | null |
{"url":"http:\/\/openstudy.com\/updates\/4f26e062e4b0a2a9c267893b","text":"## A community for students. Sign up today\n\nHere's the question you clicked on:\n\n## anonymous 4 years ago What is the area of a sector with radius 6\" and measure of arc equal to 120\u00b0?\n\n\u2022 This Question is Closed\n1. Hero\n\n$\\frac{120}{360} = \\frac{x}{ \\pi (6)^2}$\n\n2. Hero\n\nSolve for x\n\n3. Hero\n\nIn general use this formula: $\\frac{\\theta}{360} = \\frac{x}{2 \\pi r}$ where x = area of sector\n\n4. anonymous\n\nArea would be $(\\pi*r^{2}) * 120\/360$ = $(\\pi*6^2)\/3 \\approx 37.7 units$\n\n#### Ask your own question\n\nSign Up\nFind more explanations on OpenStudy\nPrivacy Policy","date":"2017-01-23 21:34:16","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.5776748061180115, \"perplexity\": 4470.545747226323}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": false}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2017-04\/segments\/1484560283008.19\/warc\/CC-MAIN-20170116095123-00078-ip-10-171-10-70.ec2.internal.warc.gz\"}"} | null | null |
A couple in a small village in the Scottish Borders is reportedly seeking a live-in nanny for their two small children. On the surface, it sounds like a helluva deal. Get paid about £50,000 per year (with the legally mandated 28 holiday days) to help the kids with meals and homework... plus you get a bedroom with a private bathroom and kitchen. Sounds sweet, right? Except there's one caveat: Their house is haunted.
According to the post on Childcare.co.uk, the couple has lived in their home for about a decade, having been told when they bought it that it could very well be filled with the souls of the departed. While they said they haven't personally seen any evil spirits mucking up the place, they've lost several nannies over the past few years— all of whom quite possibly left the house like this.
Five nannies have left the role in the last year, each citing supernatural incidents as the reason, including strange noises, broken glass and furniture moving. This has obviously been a period of great upheaval for our children. | {
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\section{Introduction}\label{sec:intro}
The two-component Camassa--Holm (2CH) system, which was first derived in
\cite[Eq.~(43)]{OlverRosenau}, is given by
\begin{subequations}
\label{eq:chsys}
\begin{align}
\label{eq:chsys11}
u_t-u_{txx}+3uu_x-2u_xu_{xx}-uu_{xxx}+\rho\rho_x&=0,\\
\label{eq:chsys12}
\rho_t+(u\rho)_x&=0,
\end{align}
\end{subequations}
or equivalently
\begin{subequations}
\label{eq:rewchsys10}
\begin{align}
\label{eq:rewchsys11}
u_t+uu_x+P_x&=0,\\
\label{eq:rewchsys12}
\rho_t+(u\rho)_x&=0,
\end{align}
\end{subequations}
where $P$ is implicitly defined by
\begin{equation}
\label{eq:rewchsys13}
P-P_{xx}=u^2+\frac12u_x^2+\frac12\rho^2.
\end{equation}
The Camassa--Holm equation \cite{CH,CHH} is obtained by considering the case when $\rho$ vanishes
identically. The aim of this article is to present the construction of a Lipschitz metric for this
system on the real line with vanishing asymptotics, that is, $u\in H^1$ and $\rho\in L^2$. The
conservative solutions to \eqref{eq:rewchsys10} are constructed in \cite{GHR:12} for nonvanishing
asymptotics. A Lipschitz metric for the system with periodic boundary conditions is given in
\cite{GHR:12b}. We here combine the two approaches by constructing a Lipschitz metric for
conservative, decaying solutions. The preservation of the energy is needed in the proofs so that the
constuction of the metric only applies to vanishing asymptotics. Here we rather describe and
motivate the general ideas behind the construction, which we hope can be of interest in the study of
other related equations. For more background on the two-component Camassa--Holm system, we refer to
\cite{GHR:12} and the references therein. For related papers, see
\cite{BC,BHR,HolRay:06a,HolRay:06b}.
\section{Relaxation of the equations by the introduction of Lagrangian coordinates}
\label{sec:semilag}
The change of coordinates from Eulerian to Lagrangian coordinates has relaxation properties which
are well-known for the Burgers equation, viz.
\begin{equation}
\label{eq:burger}
u_t+uu_x=0.
\end{equation}
Lagrangian coordinates are defined by characteristics
\begin{equation*}
y_t(t,\xi)=u(t,y(t,\xi)),
\end{equation*}
which give the position of a particle which moves in the velocity field $u$ and its velocity, known
as the Lagrangian velocity, is given by
\begin{equation*}
U(t,\xi)=u(t,x), \quad x=y(t,\xi).
\end{equation*}
The method of characteristics consists of rewriting \eqref{eq:burger} in terms of the Lagrangian
variables and yields
\begin{equation}
\begin{aligned}
y_t&=U,\\
U_t&=0.
\end{aligned}\label{eq:linburger}
\end{equation}
Comparing \eqref{eq:burger} to \eqref{eq:linburger}, we observe that we start with a
\textit{nonlinear} and \textit{partial} (derivatives with respect to $t$ and $x$) differential
equation and end up with a \textit{linear} and \textit{ordinary} (derivative only with respect to
$t$) differential equation. We get rid of the nonlinear convection term, and \eqref{eq:linburger} is
nothing but Newton's law, which states that the acceleration is constant in the absence of forces. A
well-known drawback of the change of coordinates from Eulerian to Lagrangian coordinates is that it
doubles the dimension of the problem: We start with a scalar equation and end up with a system of
dimension two. This is an important issue and we will deal with it in Section \ref{sec:relab}.
However, in return, we gain the possibility to represent a larger class of objects or, more
precisely in our case, to increase the regularity of the unknown functions. Let us make this
imprecise statement clearer by an example and, to do so, we drop the dependence in $t$ in the
notation, as we look at singularities in the space variable. The function $u(x)$ can be represented
by its graph $(x,u(x))$ but this graph can itself be represented as a parametric curve, namely,
$(y(\xi), U(\xi))$ and, as we know, the set of graphs is smaller than the set of parametric
curves. As far as regularity is concerned, the Heaviside function
\begin{equation*}
h(x)=
\begin{cases}
0&\text{ if }x<0,\\
1&\text{ if }x\geq 0,
\end{cases}
\end{equation*}
is only of bounded variation but it can be represented in Lagrangian coordinates by the following
pair of more regular (in this case Lipschitz) functions
\begin{align}
\label{eq:initcollpeakon}
y(\xi)&=
\begin{cases}
\xi&\text{ if }\xi<0,\\
0&\text{ if }\xi\in[0,1),\\
\xi - 1&\text{ if }\xi\geq1,\\
\end{cases}
&H(\xi)&=
\begin{cases}
0&\text{ if }\xi<0,\\
\xi&\text{ if }\xi\in[0,1),\\
1&\text{ if }\xi\geq1
\end{cases}
\end{align}
Indeed, $(x,h(x))$ and $(y(\xi), H(\xi))$ represent one and the same curve, except for the vertical line
joining the origin to the point $(0,1)$. We will return to this example later. The solution of the
Camassa--Holm equation (i.e., where $\rho$ vanishes identically) experiences in general wave
breaking (i.e., loss of of regularity in the sense that the spatial derivative becomes unbounded
while keeping the $H^1$ norm finite) in finite time (\cite{cons:98,cons:98b,cons:00}) and the
antisymmetric peakon-antipeakon solution, which is described in \cite{HolRay:06b} and depicted in
Figure \ref{fig:coll}, helps us to understand how the solutions can be prolonged in a way which
preserves the energy.
\begin{figure}
\begin{center}
\includegraphics[width=6cm]{coll1}
\includegraphics[width=6cm]{coll2}
\caption{Anti symmetric peakon-antipeakon collision, before (on the left) and after (on the
right) collision.}
\label{fig:coll}
\end{center}
\end{figure}
At collision time $t_c$, we have
\begin{align*}
\lim_{t\to t_c}u(t, x)&=0\text{ in }L^\infty,&\lim_{t\to t_c} u_x(t,0) = -\infty,
\end{align*}
while the $H^1$ norm is constant so that $\lim_{t\to
t_c}\norm{u(t,\,\cdot\,)}_{H^1}=\norm{u(0,\,\cdot\,)}_{H^1}$. To obtain the conservative solution, we need
to track the amount and the location of the concentrated energy. The function $u$ alone cannot
provide this information as $u(t_c,\,\cdot\,)$ is identically zero. Thus, we have to introduce an extra
variable to describe the solutions. In Lagrangian variables, it takes the form of the
\textit{cumulative energy} $H(t,\xi)$, which is given by
\begin{equation}
H(t,\xi)=\int_{-\infty}^{y(t,\xi)} (u^2+u_x^2+\rho^2)(x) dx.
\end{equation}
We will introduce later its counter-part in Eulerian variables. Equation \eqref{eq:chsys12}
transports the density $\rho$. Formally, after changing variables, we have $\rho(x)\,dx =
\rho(y)\,dy = \rho(y)y_\xi\,d\xi$, so that the Lagrangian variable corresponding to $\rho$ is given
by
\begin{equation}\label{eq:rhotor}
r(t,\xi)=\rho(t,y(t,\xi))y_\xi(t,\xi).
\end{equation}
Next, we rewrite \eqref{eq:rewchsys10} in the Lagrangian variables $(y,U,H,r)$. We obtain the
following system
\begin{equation}
\label{eq:equivsys}
\begin{aligned} \zeta_t &= U,\\ U_t &= -Q,\\ H_t &= U^3-2PU,\\r_t&=0,
\end{aligned}
\end{equation}
where $\zeta (t,\xi)=y(t,\xi)-\xi$,
\begin{equation}\label{repP} P(t,\xi)=\frac{1}{4} \int_{\mathbb{R}} \exp(-\vert y(t,\xi)-y(t,\eta)\vert )
(U^2y_\xi +H_\xi)(t,\eta)d\eta,
\end{equation}
and
\begin{equation}\label{repQ} Q(t,\xi)=-\frac{1}{4} \int_\mathbb{R} \mathop{\rm sign}(y(t,\xi)-y(t,\eta)) \exp(-\vert
y(t,\xi)-y(t,\eta)\vert ) (U^2y_\xi +H_\xi)(t,\eta)d\eta.
\end{equation}
See \cite{GHR:12} for more details on this derivation. After differentiation, we obtain
\begin{subequations}
\label{eq:govsysder}
\begin{align}
\label{eq:govsysder1}
y_{\xi t}& =U_\xi,\\
\label{eq:govsysder3}
U_{\xi t}&=\frac{1}{2}H_\xi+(\frac12 U^2-P)y_\xi, \\
\label{eq:govsysder4}
H_{\xi t}&=(3U^2-2P)U_\xi - 2QUy_\xi,\\
\label{eq:govsysder5}
r_t&=0.
\end{align}
\end{subequations}
This system is semilinear and we recognize some features observed earlier for the Burgers equation:
We start from a nonlinear partial differential equation and we end up with a system of ordinary
differential equations which is \textit{semilinear}. We consider the system as an ordinary
differential equation because the order of the spatial derivative is the same on both sides of the
equation, so that the existence and uniqueness of solutions can be established by a contraction
argument. Finally, it is important to recall in this section the geometric nature of the
Camassa--Holm equation. The equation is a geodesic in the group of diffeomorphism for the $H^1$
norm, see, e.g., \cite{cons:01b}, as the Burgers equation for the $L^2$ norm. Using the connection
between geometry and fluid mechanics, as presented in \cite{arnold}, the function $t\mapsto
y(t,\xi)$ can then be understood as a path in the group of diffeomorphisms. Thus besides the
relaxation properties we have just described, this interpretation adds a direct geometrical
relevance to use of Lagrangian coordinates, see also \cite{EKL:11} for the system.
\section{Semigroup in Lagrangian coordinates}\label{sec:lag}
In \cite[Theorem 3.2]{GHR:12}, we prove by a contraction argument that short-time solutions to
\eqref{eq:equivsys} exist in a Banach space, which we will here denote $E$ and define as
follows. Let $V$ be the Banach space defined by
\begin{equation*} V=\{f\in L^\infty \ |\ f_\xi\in L^2\}
\end{equation*}
and the norm of $V$ is given by $\norm{f}_V=\norm{f}_{L^\infty}+\norm{f_\xi}_{L^2}$. We set $E$
\begin{equation*} E=V\times H^1\times V\times L^2
\end{equation*} with the following norm
$\norm{X}=\norm{\zeta}_{V}+\norm{U}_{H^1}+\norm{H}_{V}+\norm{r}_{L^2}$ for any $X=(\zeta, U, H, r)\in E$. Given a constant $M>0$, we denote by $B_M$ the ball
\begin{equation}
\label{eq:defBM} B_M=\{X\in E\ |\ \norm{X}\leq M\}.
\end{equation}
Short-time solutions of \eqref{eq:govsysder} cannot in general be extended to global solutions. The
challenge is to identify an appropriate set of initial data for which one can construct global
solutions that at the same time preserve the structure of the equations, allowing us to return to
the Eulerian variables. There are intrinsic relations between the variables in \eqref{eq:govsysder}
that need to be conserved by the solution. This is handled by the set $\ensuremath{\mathcal{G}}$ defined below. In
particular, the set $\ensuremath{\mathcal{G}}$ is preserved by the flow.
\begin{definition}
\label{def:F} The set $\ensuremath{\mathcal{G}}$ is composed of all $(\zeta, U, H, r)\in E$ such that
\begin{subequations}
\label{eq:lagcoord}
\begin{align}
\label{eq:lagcoord1} &(\zeta,U,H,r)\in \left[W^{1,\infty}\right]^3\times L^\infty,\\
\label{eq:lagcoord2} &y_\xi\geq0, H_\xi\geq0, y_\xi+H_\xi>0 \text{ almost everywhere, and
}\lim_{\xi\rightarrow-\infty}H(\xi)=0,\\
\label{eq:lagcoord3} &y_\xi H_\xi=y_\xi^2U^2+U_\xi^2+r^2\text{ almost everywhere},
\end{align}
\end{subequations} where we denote $y(\xi)=\zeta(\xi)+\xi$.
\end{definition}
The condition $y_\xi\geq 0$ implies that the mapping $\xi\mapsto y(\xi)$ is \textit{almost} a
diffeomorphism. The solution develop singularities exactly when this mapping ceases to be a
diffeomorphism, that is, when $y_\xi=0$ in some regions. The condition \eqref{eq:lagcoord3} shows
that the variables $(y,U,H,r)$ are strongly coupled. In fact, when $y_\xi\neq0$, we can recover $H$
from \eqref{eq:lagcoord3}. It reflects the fact that $H_\xi$ represents, in Lagrangian coordinates,
the energy density of $u$ and $\rho$ (that is, $(u^2+u_x^2+\rho^2)dx$ in Eulerian coordinates) and
therefore, when the solution is smooth, it can be computed from the variables $y$, $U$, and
$r$. Note that the coupling between $H$ and $(y,U,r)$ disappears when $y_\xi=0$, which is precisely
the moment when collisions occur and when we need the information $H$ provides on the energy to
prolong the solution. The identity makes also clear the smoothing property of the Camassa--Holm
system. If $r_0\geq c>0$ for some constant $c$, this property is preserved and then $y_\xi$ never
vanishes. The solution keeps the same degree of regularity it has initially, see \cite{GHR:12}.
As in \cite[Theorem 3.6]{GHR:12}, we obtain the Lipschitz continuity of the semigroup
\begin{theorem}
\label{th:global} For any $\bar X=(\bar y,\bar U,\bar H, \bar r)\in\ensuremath{\mathcal{G}}$, the system
\eqref{eq:equivsys} admits a unique global solution $X(t)=(y(t),U(t),H(t), r(t))$ in
$C^1(\mathbb{R}_+,E)$ with initial data $\bar X=(\bar y,\bar U,\bar H,\bar r)$. We have $X(t)\in\ensuremath{\mathcal{G}}$ for
all times. If we equip $\ensuremath{\mathcal{G}}$ with the topology induced by the $E$-norm, then the mapping
$S\colon\ensuremath{\mathcal{G}}\times\mathbb{R}_+\to\ensuremath{\mathcal{G}}$ defined by
\begin{equation*} S_t(\bar X)=X(t)
\end{equation*} is a Lipschitz continuous semigroup. More precisely, given $M>0$ and $T>0$, there exists a
constant $C_M$ which depends only on $M$ and $T$ such that, for any two elements
$X_\alpha,X_\beta\in\ensuremath{\mathcal{G}}\cap B_M$, we have
\begin{equation}
\label{eq:stabSt} \norm{S_tX_\alpha-S_tX_\beta}\leq C_M\norm{X_\alpha-X_\beta}
\end{equation} for any $t\in[0,T]$.
\end{theorem}
\section{Relabeling symmetry}\label{sec:relab}
The equations are well-posed in Lagrangian coordinates. We want to transport this result back to
Eulerian coordinates. If the two sets of coordinates were in bijection, then it would be
straightforward but, as mentioned earlier, Lagrangian coordinates increase the number of unknowns
from two ($u$ and $\rho$) to four (the components of $X$), which indicates that such a bijection
does not exist. There exists a redundancy in Lagrangian coordinates and the goal of this section is
precisely to identify this redundancy, in order to be able to define the correct equivalence
classes. This redundancy is also present in the case of the Burgers equation when we define the
Cauchy problem for both \eqref{eq:burger} and \eqref{eq:linburger}. To the initial condition
$u(0,x)=u_0(x)$ for \eqref{eq:burger}, there corresponds infinitely many parametrizations of the
initial conditions for \eqref{eq:linburger} given by
\begin{align*}
y(0,\xi)&=f(\xi),&U(0,\xi)=u_0(f(\xi)),
\end{align*}
for an arbitrary diffeomorphism $f$. As also mentioned earlier, the representation of a graph is
uniquely defined by a single function while there are infinitely many different parametrizations of
any given curve. We will use the term \textit{relabeling} for this lack of uniqueness in the
characterization of one and the same curve.
We now define the relabeling functions as follows.
\begin{definition}
\label{def:Gr} We denote by $G$ the subgroup of the group of homeomorphisms from $\mathbb{R}$ to
$\mathbb{R}$ such that
\begin{subequations}
\label{eq:Hcond}
\begin{align}
\label{eq:Hcond1} f-\id\text{ and }f^{-1}-\id &\text{ both belong to }W^{1,\infty},\\
\label{eq:Hcond2} f_\xi-1& \text{ belongs to }L^2,
\end{align}
\end{subequations} where $\id$ denotes the identity function. Given $\kappa>0$, we denote by
$G_\kappa$ the subset $G_\kappa$ of $G$ defined by
\begin{equation*} G_\kappa=\{f\inG\ |\
\norm{f-\id}_{W^{1,\infty}}+\norm{f^{-1}-\id}_{W^{1,\infty}}\leq\kappa\}.
\end{equation*}
\end{definition}
We refine the definition of $\ensuremath{\mathcal{G}}$ in Definition \ref{def:F} by introducing the subsets $\ensuremath{\mathcal{F}}_\kappa$
and $\ensuremath{\mathcal{F}}$ as
\begin{equation*}
\ensuremath{\mathcal{F}}_\kappa=\{X=(y,U,H,r)\in\ensuremath{\mathcal{G}}\ |\ y+H\in G_\kappa\},
\end{equation*} and
\begin{equation}
\label{eq:defF}
\ensuremath{\mathcal{F}}=\{X=(y,U,H,r)\in\ensuremath{\mathcal{G}}\ |\ y+H\in G\}.
\end{equation}
The regularity requirement on the relabeling functions given in Definition \ref{def:Gr} and the
definition of $\ensuremath{\mathcal{F}}$ are introduced in order to be able to define the action of $G$ on $\ensuremath{\mathcal{F}}$, that is,
for any $X=(y,U,H,r)\in\ensuremath{\mathcal{F}}$ and any function $f\inG$, the function $(y\circ f,U\circ f,H\circ f,
r\circ f f_\xi)$ belongs to $\ensuremath{\mathcal{F}}$ and we will denote it by $X\circ f$. This corresponds to the
relabeling action. Note that relabeling acts differently on \textit{primary} functions, as $y$, $U$
and $H$ (in this case, we have $(U, f)\mapsto U\circ f$) and on \textit{derivatives} or
\textit{densities}, as $y_\xi$, $U_\xi$, $H_\xi$ and $r$ (in that case we have $(r, f)\mapsto r\circ
f f_\xi$). The space $\ensuremath{\mathcal{F}}$ is preserved by the governing equation \eqref{eq:equivsys} and, as
expected,
the semigroup of solutions in Lagrangian coordinates preserves relabeling, i.e., we have the
following result.
\begin{lemma}[{\cite[Theorem 4.8]{GHR:12}}]
\label{lem:equivPi} The mapping $S_t$ is equivariant, that is,
\begin{equation*} S_t(X\circ f)=S_t(X)\circ f
\end{equation*}
for any $X\in\ensuremath{\mathcal{F}}$ and $f\in G$.
\end{lemma}
Now that we have identified the redundancy of Lagrangian coordinates as the action of relabeling, we
want to handle it by considering equivalence classes. However, equivalence classes are rather
abstract objects which will be hard to work with from an analytical point of view. We consider
instead the section defined by $\ensuremath{\mathcal{F}}_0$, which contains one and only one representative for each
equivalence class, so that the quotient $\ensuremath{\mathcal{F}}/G$ is in bijection with $\ensuremath{\mathcal{F}}_0$. Let us denote by $\Pi$
the projection of $\ensuremath{\mathcal{F}}$ into $\ensuremath{\mathcal{F}}_0$ defined as
\begin{equation*}
\Pi(X)=X\circ(y+H)^{-1}
\end{equation*}
for any $X=(y,U,H,r)\in\ensuremath{\mathcal{F}}$. By definition, we have that $X$ and $\Pi(X)$ belong to the same
equivalence class. We can check that the mapping $\Pi$ is a projection, i.e., $\Pi\circ\Pi = \Pi$,
and that it is also invariant, i.e., $\Pi(X\circ f)=\Pi(X)$. It follows that the mapping $[X]\mapsto
\Pi(X)$ is a bijection from $\ensuremath{\mathcal{F}}/G$ to $\ensuremath{\mathcal{F}}_0$.
\section{Eulerian coordinates}\label{sec:euler}
In the method of characteristics, once the equation is solved in Lagrangian coordinates, we recover
the solution in Eulerian coordinates by setting $u(t,x) = U(t,y^{-1}(t,x))$, where $y^{-1}(t,x)$
denotes---assuming it exists---the inverse of $\xi\mapsto y(t,\xi)$. The Burgers equation and the
Camassa--Holm equation develop singularity because $y$ does not remain invertible. In the case of
the Burgers equation, $u$ becomes discontinuous but the Camassa--Holm equation enjoys more
regularity and $u$ remains continuous. This is a consequence of the preservation of the $H^1$ norm,
but it can also be seen from the Lagrangian point of view. Indeed, even if $y$ is not invertible, we
can define $u(t,x)$ as
\begin{equation*}
u(t,x) = U(t,\xi)\text{ for any }\xi\text{ such that }x=y(t,\xi).
\end{equation*}
This is well-defined because if there exist $\xi_1$ and $\xi_2$ such that $x=y(t,\xi_1)=y(t,\xi_2)$,
then $y_\xi(t,\xi)=0$ for all $\xi\in[\xi_1,\xi_2]$ because $y$ is non-decreasing, see
\eqref{eq:lagcoord2}. Then, by \eqref{eq:lagcoord3}, we get $U_\xi(t,\xi)=0$ so that
$U(t,\xi_1)=U(t,\xi_2)$. Furthermore, as we explained earlier in the case of a peakon-antipeakon
collision, some information is needed about the energy to prolong the solution after
collision. If $y$ is invertible, we recover the energy density in Eulerian coordinates as
\begin{equation}
\label{eq:euldefenerg1}
(u^2+u_x^2+\rho^2)\,dx= \frac{H_\xi}{y_\xi}\circ y^{-1}\,d\xi,
\end{equation}
which corresponds to the push-forward of the measure $H_\xi\,d\xi$ with respect to $y$, i.e.,
\begin{equation}
\label{eq:euldefenerg2}
(u^2+u_x^2+\rho^2)\,dx = y_\#(H_\xi\,d\xi).
\end{equation}
However, when $y$ is not invertible \eqref{eq:euldefenerg1} cannot be used and $y_\#(H_\xi\,d\xi)$
may not be absolutely continuous so that \eqref{eq:euldefenerg2} will not hold either. It motivates
the introduction of the energy $\mu$ defined here as $y_\#(H_\xi\,d\xi)$, which represents the
energy of the system. The set $\ensuremath{\mathcal{D}}$ of Eulerian coordinates is defined as follows.
\begin{definition}
\label{def:D} The set $\ensuremath{\mathcal{D}}$ consists of all triples $(u,\rho,\mu)$ such that
\begin{enumerate}
\item $u\in H^1$, $\rho\in L^2$, and
\item $\mu$ is a positive Radon measure whose absolutely continuous part, $\mu_{ac}$, satisfies
\begin{equation}
\label{eq:relmumuac}
\mu_{ac}=(u^2+u_x^2+\rho^2)dx.
\end{equation}
\end{enumerate}
\end{definition}
It can be shown (see \cite[Section 4]{GHR:12}) that the identity \eqref{eq:lagcoord3} is somehow
equivalent to \eqref{eq:relmumuac} but it is clear that, from an analytical point of view, it easier
to deal with an algebraic identity like \eqref{eq:lagcoord3} than with a property like
\eqref{eq:relmumuac} which immediately requires tools from measure theory. We can show that $\ensuremath{\mathcal{D}}$ and
$\ensuremath{\mathcal{F}}_0$ are in bijection, and the mappings between the two are given in the following definition. The
first one has been already explained.
\begin{definition} Given any element $X $ in $\ensuremath{\mathcal{F}}_0$, then $(u,\rho,\mu)$ defined as follows
\begin{equation*}
u(x)=U(\xi) \text{ for any } \xi \text{ such that } x=y(\xi),
\end{equation*}
\begin{equation*}
\rho(x)=y_\#(r d\xi),\quad \mu=y_\#(H_\xi d\xi),
\end{equation*} belongs to $\ensuremath{\mathcal{D}}$. We denote by $M:\ensuremath{\mathcal{F}}_0\to \ensuremath{\mathcal{D}}$ the map which to any $X$ in $\ensuremath{\mathcal{F}}_0$
associates $(u,\rho,\mu)$.
\end{definition}
The mapping, which we denoted by $L$, from $\ensuremath{\mathcal{D}}$ to $\ensuremath{\mathcal{F}}_0$ is defined as follows.
\begin{definition}
\label{def:L}
For any $(u,\rho,\mu)$ in $\ensuremath{\mathcal{D}}$ let
\begin{equation}
\label{eq:defL}
\left\{
\begin{aligned} y(\xi)&=\sup\{y \mid \mu((-\infty,y))+y<\xi\},\\
H(\xi)& =\xi-y(\xi),\\
U(\xi)&=u\circ y(\xi),\\
r(\xi)&=\rho\circ y(\xi)y_\xi(\xi).
\end{aligned} \right.
\end{equation}
\end{definition}
We can see that the lack of regularity of $u$, which will occur when $\mu$ is singular or very
large, is transformed into regions where the function $y$ is constant or almost constant. Using the
relabeling degree of freedom, we manage to rewrite functions in $L^2$ and measures as bounded
functions (in $L^\infty$). For example, for the peakon-antipeakon collision depicted in Figure
\ref{fig:coll}, the initial data given by $u_0(x)=\rho_0(x)=0$ and $\mu=\delta(x)\,dx$, which
corresponds to the collision time, $t_c$, when the total energy is equal to one, yields
$r(\xi)=U(\xi)=0$ with $y(\xi)$ and $H(\xi)$ as defined in \eqref{eq:initcollpeakon}. We can check that,
in this case $\delta(x)\,dx=y_\#(H_\xi\,d\xi)$. Finally, we define the semigroup $T_t$ of
conservative solutions in the original Eulerian variables $\ensuremath{\mathcal{D}}$ as
\begin{equation*}
T_t:=M\Pi S_tL.
\end{equation*}
\section{Lipschitz metric for the semigroup}\label{sec:lip}
We apply the construction of the semigroup $T_t$ in Section \ref{sec:euler}, and we can check, as
done in \cite[Theorem 5.2]{GHR:12}, that, for given initial data $(u_0,\rho_0,\mu_0)$, if we denote
$(u(t), \rho(t), \mu_t)=T_t(u_0,\rho_0,\mu_0)$, then $(u, \rho)$ are weak solutions to
\eqref{eq:rewchsys10}. Moreover,
\begin{equation*}
\mu_t(\mathbb{R})=\mu_0(\mathbb{R})
\end{equation*}
so that the solutions are conservative. Our goal is to define a metric on $\ensuremath{\mathcal{D}}$ which makes the
semigroup Lipschitz continuous. The Lipschitz continuity is a property of a semigroup which can be
used to establish its uniqueness, see \cite{Bressan} and \cite[Theorem 2.9]{Bressan:book}. By our construction, a metric for the semigroup $T_t$ is
readily available. We can simply transport the topology of the Banach space $E$ from $\ensuremath{\mathcal{F}}_0$ to $\ensuremath{\mathcal{D}}$
and obtain, for two elements $(u,\rho,\mu)$ and $(\tilde u,\tilde \rho,\tilde \mu)$,
\begin{equation}
\label{eq:defdD1}
d_\ensuremath{\mathcal{D}}\big((u,\rho,\mu),(\tilde u,\tilde \rho,\tilde \mu)\big) = \norm{L(u,\rho,\mu)-L(\tilde u,\tilde \rho,\tilde \mu)}_E.
\end{equation}
We have
\begin{equation*}
d_\ensuremath{\mathcal{D}}\big(T_t(u,\rho,\mu),T_t(\tilde u,\tilde \rho,\tilde \mu)\big) = \norm{\Pi S_tL(u,\rho,\mu)-\Pi S_t L(\tilde u,\tilde \rho,\tilde \mu)}_E.
\end{equation*}
It can be proven that the projection $\Pi$ is continuous (see \cite[Lemma 4.6]{GHR:12}), but it is
not Lipschitz (at least, we have been unable to prove it). Thus, even if $S_t$ is Lipschitz
continuous, the semigroup $T_t$ is only continuous with respect to the metric $d_\ensuremath{\mathcal{D}}$ defined by
\eqref{eq:defdD1}. In the definition \eqref{eq:defdD1} of the metric, we let the section $\ensuremath{\mathcal{F}}_0$ play
a special role, but this section is arbitrarily chosen. The set $\ensuremath{\mathcal{F}}_0$ is by construction nonlinear
(because of \eqref{eq:lagcoord3}) and to use a linear norm to measure distances does not respect
that. In fact, we want to measure the distance between equivalence classes. A natural starting point
is to define, for $X_\alpha,X_\beta\in\ensuremath{\mathcal{F}}$, $\bar J(X_\alpha,X_\beta)$ as
\begin{equation}
\label{eq:defJXX}
\bar J(X_\alpha,X_\beta)=\inf_{f, g\inG}\norm{X_\alpha\circ f-X_\beta\circ
g}.
\end{equation}
The function $\bar J$ is relabeling invariant, that is, $\bar J(X_\alpha\circ f,X_\beta\circ g)=\bar
J(X_\alpha,X_\beta)$ and measures precisely the distance between two equivalence classes. However, we
have to deal with the fact that the linear norm of $E$ does not play well with relabeling: It is not
invariant with respect to relabeling, i.e., we do not have
\begin{equation}
\label{eq:invnorm}
\norm{X\circ f} = \norm{X}.
\end{equation}
However, such a norm exists. Let
\begin{equation*}
B=\{X\in L^\infty\ |\ X_\xi \in L^1\}.
\end{equation*}
Then,
\begin{equation*}
\norm{X\circ f}_{B} =\norm{X\circ f}_{L^\infty} + \norm{X_\xi\circ f f_\xi}_{L^1} = \norm{X}_{L^\infty} + \norm{X_\xi}_{L^1} = \norm{X}_{B}.
\end{equation*}
To cope with the lack of relabeling invariance of $\bar J$, we introduce $J$ defined as follows.
\begin{definition}
\label{def:J}
Let $X_\alpha,X_\beta\in\ensuremath{\mathcal{F}}$, we define $J(X_\alpha,X_\beta)$ as
\begin{equation}
\label{eq:defJ} J(X_\alpha,X_\beta)=\inf_{f_1,f_2\inG}\big(\norm{X_\alpha\circ
f_1-X_\beta}+\norm{X_\alpha-X_\beta\circ f_2}\big).
\end{equation}
\end{definition}
The function $J$ is not relabeling invariant, but we have $J(X_\alpha,X_\beta)=0$ if $X_\alpha$ and
$X_\beta$ both belong to the same equivalence class. Moreover, the relabeling invariance is not strictly
needed for our purpose and the following weaker property is enough. Given $X_\alpha,X_\beta\in \ensuremath{\mathcal{F}}$
and $f\inG_\kappa$, we have
\begin{equation}
\label{eq:Jrelabbound} J(X_\alpha\circ f,X_\beta\circ f)\leq CJ(X_\alpha,X_\beta)
\end{equation}
for some constant $C$ which depends only on $\kappa$, see \cite{GHR:13}. Note that, if the norm $E$
were invariant, that is, \eqref{eq:invnorm} were fulfilled, then the function $J$ and $\bar J$ would be
equivalent, because we would have $\bar J\leq J\leq 2\bar J$.
\begin{remark}
We will make use of the following notation. The variable $X$ is used as a standard notation for
$(y,U,H,r)$. By the $L^\infty$ norm of $X$, we mean
\begin{equation}
\label{eq:notation1}
\norm{X}_{L^\infty}=\norm{y-\id}_{L^\infty}+\norm{U}_{L^\infty}+\norm{H}_{L^\infty},
\end{equation}
and, by the $L^2$ norm of the derivative $X_\xi$, we mean
\begin{equation}
\label{eq:notation2}
\norm{X_\xi}_{L^2}=\norm{y_\xi-1}_{L^2}+\norm{U_\xi}_{L^2}+\norm{H_\xi}_{L^2}+\norm{r}_{L^2},
\end{equation}
and, similarly,
\begin{equation}
\label{eq:notation3}
\norm{X_\xi}_{L^\infty}=\norm{y_\xi-1}_{L^\infty}+\norm{U_\xi}_{L^\infty}+\norm{H_\xi}_{L^\infty}+\norm{r}_{L^\infty}.
\end{equation}
\end{remark}
From $J$, we obtain a metric $d$ by the following construction.
\begin{definition} Let $X_\alpha,X_\beta\in\ensuremath{\mathcal{F}}_0$, we define $d(X_\alpha,X_\beta)$ as
\begin{equation}
\label{eq:defdist} d(X_\alpha,X_\beta)=\inf \sum_{i=1}^NJ(X_{n-1},X_n)
\end{equation} where the infimum is taken over all finite sequences $\{X_n\}_{n=0}^N\subset\ensuremath{\mathcal{F}}_0$ which
satisfy $X_0=X_\alpha$ and $X_N=X_\beta$.
\end{definition}
\begin{lemma}
\label{lem:distance}
The mapping $d:\ensuremath{\mathcal{F}}_0\times\ensuremath{\mathcal{F}}_0\to\mathbb{R}_+$ is a distance on $\ensuremath{\mathcal{F}}_0$, which is bounded as follows
\begin{equation}
\label{eq:dequiv} \frac12\norm{X_\alpha-X_\beta}_{L^\infty}\leq
d(X_\alpha,X_\beta)\leq2\norm{X_\alpha-X_\beta}.
\end{equation}
\end{lemma}
\begin{proof}
The first part of the proof is identical to \cite{GHR:13} and we reproduce it here for
convenience. For any $X_\alpha,X_\beta\in\ensuremath{\mathcal{F}}_0$, we have
\begin{equation}
\label{eq:linfcompj} \norm{X_\alpha-X_\beta}_{L^\infty}\leq 2 J(X_\alpha,X_\beta).
\end{equation} We have
\begin{align} \notag \norm{X_\alpha-X_\beta}_{L^\infty}&\leq\norm{X_\alpha-X_\alpha\circ
f}_{L^\infty}+\norm{X_\alpha\circ f-X_\beta}_{L^{\infty}}\\
\label{eq:xamxbediffg} &\leq
\norm{X_{\alpha,\xi}}_{L^\infty}\norm{f-\id}_{L^\infty}+\norm{X_\alpha\circ
f-X_\beta}_{L^{\infty}}.
\end{align}
It follows from the definition of $\ensuremath{\mathcal{F}}_0$ that $0\leq y_\xi\leq1$, $0\leq H_\xi\leq1$
and $\abs{U_\xi}\leq1$ so that $\norm{X_{\alpha,\xi}}_{L^\infty}\leq 3$. We also have
\begin{equation}
\label{eq:bdfminid}
\norm{f-\id}_{L^\infty}=\norm{(y_\alpha+H_\alpha)\circ f-(y_\beta+H_\beta)}_{L^\infty}\leq\norm{X_\alpha\circ f-X_\beta}_{L^\infty}.
\end{equation} Hence, from \eqref{eq:xamxbediffg}, we get
\begin{equation}
\label{eq:bdlinfour}
\norm{X_\alpha-X_\beta}_{L^\infty}\leq4\norm{X_\alpha\circ
f-X_\beta}_{L^\infty}.
\end{equation} In the same way, we obtain
$\norm{X_\alpha-X_\beta}_{L^\infty}\leq4\norm{X_\alpha-X_\beta\circ f}_{L^\infty}$ for any
$f\inG$. After adding these two last inequalities and taking the infimum, we get
\eqref{eq:linfcompj}. For any $\varepsilon>0$, we consider a finite sequence $\{X_n\}_{n=0}^N\subset\ensuremath{\mathcal{F}}_0$ such that
$X_0=X_\alpha$ and $X_N=X_\beta$ and $\sum_{i=1}^NJ(X_{n-1},X_n)\leq d(X_\alpha,X_\beta)+\varepsilon$. We
have
\begin{align*} \norm{X_\alpha-X_\beta}_{L^\infty}&\leq
\sum_{n=1}^{N}\norm{X_{n-1}-X_n}_{L^\infty}\\ &\leq2\sum_{n=1}^{N}J(X_{n-1},X_n)\\
&\leq2(d(X_\alpha,X_\beta)+\varepsilon).
\end{align*} After letting $\varepsilon$ tend to zero, we get
\begin{equation}
\label{eq:LinfbdJ}
\norm{X_\alpha-X_\beta}_{L^\infty}\leq 2 d(X_\alpha,X_\beta).
\end{equation}
The second inequality in \eqref{eq:dequiv} follows from the definitions of $J$ and $d$. Indeed, we
have
\begin{equation*}
d(X_\alpha,X_\beta)\leq J(X_\alpha,X_\beta)\leq2\norm{X_\alpha-X_\beta}.
\end{equation*}
It is left to prove that $d$ defines a metric. The symmetry is intrinsic in the definition of $J$
while the construction of $d$ from $J$ takes care of the triangle inequality. From
\eqref{eq:dequiv}, we get that $d(X_\alpha,X_\beta)=0$ implies $(y_\alpha, U_\alpha,
H_\alpha)=(y_\beta, U_\beta, H_\beta)$. By \eqref{eq:lagcoord3}, we get that
$r_\alpha^2=r_\beta^2$, but we cannot yet conclude that $r_\alpha=r_\beta$. Let us define
$R_\alpha(\xi)=\int_{-\infty}^\xi r_\alpha(\eta)e^{-\abs{\eta}}\,d\eta$ and
$R_\beta(\xi)=\int_{-\infty}^\xi r_\beta(\eta)e^{-\abs{\eta}}\,d\eta$. Then, we have, for any
$f\inG$,
\begin{multline}
R_\alpha(\xi) - R_\beta(\xi) = -\int_{\xi}^{f(\xi)}r_\alpha(\eta) e^{-\abs{\eta}}\,d\eta +
\int_{-\infty}^\xi r_\alpha\circ f f_\xi (e^{-\abs{f(\eta)}}-e^{-\abs{\eta}})\,d\eta\\ +
\int_{-\infty}^\xi(r_\alpha\circ f f_\xi -r_\beta) e^{-\abs{\eta}}\,d\eta,
\end{multline}
which implies
\begin{align*}
\norm{R_{\alpha}-R_{\beta}}_{L^\infty}& \leq \norm{f-\id}_{L^\infty}+
\norm{\int_{-\infty}^\xi r_\alpha\circ f f_\xi
(e^{-\abs{f(\eta)}}-e^{-\abs{\eta}})\,d\eta}_{L^\infty}\\
&\quad+\norm{r_\alpha\circ f f_\xi -r_\beta}_{L^2}.
\end{align*}
We have that
\begin{align*}
\int_{-\infty}^\xi r_\alpha\circ f f_\xi (e^{-\abs{f(\eta)}}-e^{-\abs{\eta}})\,d\eta &=
\int_{-\infty}^\xi r_\alpha\circ f f_\xi
e^{-\abs{f(\eta)}}(1-e^{\abs{f(\eta)}-\abs{\eta}})\,d\eta
\end{align*}
implies
\begin{align*}
\norm{\int_{-\infty}^\xi r_\alpha\circ f f_\xi (e^{-\abs{f(\eta)}}-e^{-\abs{\eta}})\,d\eta}_{L^\infty}&\leq \norm{e^{\abs{f(\xi)}-\abs{\xi}} - 1}_{L^\infty}\norm{r_\alpha}_{L^2}\norm{e^{-\abs{\xi}}}_{L^2}\\
&\leq C\norm{r_\alpha}_{L^2} \norm{f-\id}_{L^\infty},
\end{align*}
for $C=e$ if we assume that $\norm{f-\id}_{L^\infty}\leq 1$. Since $X_\alpha\in \ensuremath{\mathcal{F}}_0$ so that
$y_\xi\leq 1$, we get from \eqref{eq:lagcoord3} that $\norm{r_\alpha}_{L^2}\leq
\norm{H_\alpha}_{L^\infty}^{1/2}$. Collecting the results obtained so far, we find that
\begin{equation}
\label{eq:bdralphabeta}
\norm{R_{\alpha}-R_{\beta}}_{L^\infty} \leq (2 + C \norm{H_\alpha}_{L^\infty}^{1/2})\norm{X_\alpha\circ f - X_\beta}
\end{equation}
for any $\norm{f-\id}_{L^\infty}\leq 1$. Let us now assume that $d(X_\alpha, X_\beta) = 0$. For
any $\varepsilon>0$, we can find a sequence such that
\begin{equation*}
\sum_{n=1}^N\norm{X_n\circ f_n - X_{n-1}}\leq \varepsilon.
\end{equation*}
Using \eqref{eq:bdfminid} and \eqref{eq:bdlinfour}, we get $\norm{f_n-\id}_{L^\infty}\leq\varepsilon$
and prove by induction that
\begin{equation}
\label{eq:indstat}
\norm{H_n}_{L^\infty}\leq\sum_{i=1}^{n}\norm{X_i\circ f_i - X_{i-1}}_{L^\infty} + \norm{H_\alpha}_{L^{\infty}},
\end{equation}
for all $n\leq N$. Indeed, we have
\begin{align*}
\norm{H_{n+1}}_{L^\infty} & = \norm{H_{n+1}\circ f_{n+1}}_{L^\infty}\\
&\leq \norm{H_{n+1}\circ f_{n+1} - H_n}_{L^\infty} + \norm{H_n}_{L^\infty}\\
&\leq \sum_{i=1}^{n + 1}\norm{X_i\circ f_i - X_{i-1}}_{L^\infty} + \norm{H_\alpha}_{L^{\infty}},
\end{align*}
after using the induction hypothesis. From \eqref{eq:indstat}, we get
\begin{equation*}
\norm{H_n}_{L^\infty}\leq \varepsilon + \norm{H_\alpha}.
\end{equation*}
Hence, by choosing $\varepsilon\leq 1$, and using repeatedly \eqref{eq:bdralphabeta}, we obtain
\begin{align*}
\norm{R_{\alpha}-R_{\beta}}_{L^\infty} &\leq \sum_{n=1}^N \norm{R_{n} - R_{n-1}}_{L^\infty}\\
&\leq (2 + C(\varepsilon + \norm{H_\alpha}_{L^\infty})^{1/2})\sum_{n=1}^N\norm{X_\alpha\circ f - X_\beta} \\
&\leq (2 + C(\varepsilon + \norm{H_\alpha}_{L^\infty})^{1/2})\varepsilon.
\end{align*}
After letting $\varepsilon$ tend to zero, this last inequality implies that $R_\alpha=R_\beta$ so that
$r_\alpha=r_\beta$, which concludes the proof that $d$ is a metric.
\end{proof}
The Lipschitz estimate for the semigroup $S_t$ given in \eqref{eq:stabSt} is valid for initial data
in $B_M$. Hence, as we want to use the same Lipschitz estimate for any of the $X_n$ in the sequence
defining the metric in \eqref{eq:defdist}, we have to redefine this metric and require that all
$X_n$ belong to $\ensuremath{\mathcal{F}}_0\cap B_M$. The problem is that $B_M$ is not preserved by the semigroup
$S_t$, and we will not be able to use the same distance at later times. This is why we introduce the
set
\begin{equation*}
\ensuremath{\mathcal{F}}^M=\{X=(y,U,H,r)\in \ensuremath{\mathcal{F}}\ |\ \norm{H}_{L^\infty}\leq M\},
\end{equation*}
which is preserved by \textit{both} relabeling and the semigroup. Note that $\ensuremath{\mathcal{F}}^M$ has a simple
physical interpretation as it corresponds to the set of all solutions which have total energy
bounded by $M$. Moreover, following closely the proof of \cite[Lemma 3.4]{GHR:13}, we obtain that
for $X\in \ensuremath{\mathcal{F}}_0$, the sets $B_M$ and $\ensuremath{\mathcal{F}}^M$ are in fact equivalent, i.e., there exists $\bar M$
depending only on $M$ such that
\begin{equation}
\label{eq:eqFBM}
\ensuremath{\mathcal{F}}_0\cap\ensuremath{\mathcal{F}}^M\subset B_{\bar M}.
\end{equation}
We set $\ensuremath{\mathcal{F}}_0^M=\ensuremath{\mathcal{F}}_0\cap\ensuremath{\mathcal{F}}^M$ and define the metric $d^M$ as follows.
\begin{definition}\label{def:metric} Let $d^M$ be the distance on $\ensuremath{\mathcal{F}}_0^M$ which is defined, for any
$X_\alpha,X_\beta\in\ensuremath{\mathcal{F}}_0^M$, as
\begin{equation}
\label{eq:defdM}
d^M(X_\alpha,X_\beta)=\inf \sum_{n=1}^NJ(X_{n-1},X_n)
\end{equation}
where the infimum is taken over all finite sequences $\{X_n\}_{n=0}^N\subset\ensuremath{\mathcal{F}}_0^M$ such that
$X_0=X_\alpha$ and $X_N=X_\beta$.
\end{definition}
\begin{figure}[h]
\centering
\includegraphics[width=10cm]{distance}
\caption{Illustration for the construction of the metric. The \textit{horizontal} curves
represent points which belong to the same equivalence class.}
\label{fig:distance}
\end{figure}
We can now state our main stability theorem
\begin{theorem}
\label{th:stab} Given $T>0$ and $M>0$, there exists a constant $C_{M,T}$ which depends only on $M$
and $T$ such that, for any $X_\alpha,X_\beta\in\ensuremath{\mathcal{F}}_0^M$ and $t\in[0,T]$, we have
\begin{equation}
\label{eq:stab} d^M(\Pi S_tX_\alpha, \Pi S_tX_\beta)\leq C_{M,T} d^M(X_\alpha,X_\beta).
\end{equation}
\end{theorem}
In fact due to the use of equivalent notations, the proof of the theorem is identical to
\cite[Theorem 3.6]{GHR:13}. Here, we propose to present a simplified proof where we assume that the
norm of $E$ is invariant with respect to relabeling, that is, \eqref{eq:invnorm} holds. By doing so,
we hope that some general ideas behind the construction of the metric becomes clearer. Much of the
construction can be understood from the illustration in Figure~\ref{fig:distance}. In this figure,
we denote $X_\alpha^t=\Pi S_t(X_\alpha\circ f_0)$, $X_\beta^t=\Pi S_t(X_\beta\circ g_1)$ and
$X_1^t=\Pi S_t(X_1\circ g_0)=\Pi S_t(X_1\circ f_1)$. Let us imagine the (very improbable) case where
the infimum in \eqref{eq:defdM} and the infimum in \eqref{eq:defJ} both are reached, so that
$d^M(X_\alpha, X_\beta) = \norm{X_\alpha\circ f_0-X_1\circ g_0} + \norm{X_1\circ f_1-X_\beta\circ
g_1}$. Then, we have
\begin{align*}
d^M(X_\alpha^t,X_\beta^t)&\leq J(X_\alpha^t,X_1^t) + J(X_1^t,X_\beta^t)\\
&= J(S_t(X_\alpha\circ f_0), S_t(X_1\circ g_0)) + J(S_t(X_1\circ f_1),S_t(X_\beta\circ g_1))\\
&\leq \norm{S_t(X_\alpha\circ f_0)-S_t(X_1\circ g_0)} + \norm{S_t(X_1\circ f_1)-S_t(X_\beta\circ
g_1)}\\
&\leq C_{M,T} \big(\norm{X_\alpha\circ f_0-X_1\circ g_0} + \norm{X_1\circ f_1-X_\beta\circ g_1}\big)\\
&= C_{M,T} d^M(X_\alpha, X_\beta),
\end{align*}
which corresponds to the Lipschitz estimate of Theorem \ref{th:stab}.
\begin{proof}[Simplified proof of Theorem \ref{th:stab}] As we mentioned earlier, when the norm is
invariant, then $J$ and $\bar J$ are equivalent. Here, it is simpler to consider $\bar J$. For any
$\varepsilon>0$, there exist a finite sequence $\{X_n\}_{n=0}^N$ in $\ensuremath{\mathcal{F}}_0^M$ and functions
$\{f_n\}_{n=0}^{N-1}$, $\{g_n\}_{n=0}^{N-1}$ in $G$ such that $X_0=X_\alpha$, $X_N=X_\beta$ and
\begin{equation}
\label{eq:sumXnm1}
\sum_{i=1}^N\norm{X_{n-1}\circ f_{n-1}-X_{n}\circ g_{n-1}}\leq
d_M(X_\alpha,X_\beta)+\varepsilon.
\end{equation}
Since $B_{\bar M}$, where $\bar M$ is defined so that \eqref{eq:eqFBM} holds, is preserved by
relabeling, we have that $X_{n}\circ f_n$ and $X_{n}\circ g_{n-1}$ belong to $B_{\bar M}$. From
the Lipschitz stability result given in \eqref{eq:stabSt}, we obtain that
\begin{equation}
\label{eq:normSxnm1}
\norm{S_t(X_{n-1}\circ f_{n-1})-S_t(X_{n}\circ g_{n-1})}\leq C_{M,T} \norm{X_{n-1}\circ f_{n-1}-X_{n}\circ g_{n-1}},
\end{equation}
where the constant $C_{M,T}$ depends only on $M$ and $T$. Introduce
\begin{equation*}
\bar X_n=X_n\circ f_n,\ \bar X_n^t=S_t(\bar X_n), \text{ for }n=0,\ldots,N-1,
\end{equation*}
and
\begin{equation*}
\tilde X_n=X_{n}\circ g_{n-1},\ \tilde X_n^t=S_t(\tilde X_n), \text{ for }n=1,\ldots,N.
\end{equation*}
Then \eqref{eq:sumXnm1} rewrites as
\begin{equation}
\label{eq:sumXnm1b}
\sum_{i=1}^N\norm{\bar X_{n-1}-\tilde X_{n}}\leq
d_M(X_\alpha,X_\beta)+\varepsilon
\end{equation}
while \eqref{eq:normSxnm1} rewrites as
\begin{equation}
\label{eq:normSnnot}
\norm{\bar X_{n-1}^t-\tilde X_{n}^t}\leq C_{M,T} \norm{\bar X_{n-1}-\tilde X_{n}}.
\end{equation}
We have
\begin{equation*}
\Pi(\bar X_0^t)=\Pi\circ S_t(X_0\circ f_0)=\Pi\circ (S_t(X_0)\circ f_0)=\Pi\circ S_t(X_0)=\bar S_t(X_\alpha)
\end{equation*}
and similarly $\Pi(\tilde X_N^t)=\Pi S_t(X_\beta)$. We consider the sequence which consists of
$\{\Pi \bar X_n^t\}_{n=0}^{N-1}$ and $\bar S_t(X_\beta)$. Using the property that $\ensuremath{\mathcal{F}}^M$ is
preserved both by relabeling and by the semigroup, we obtain that $\{\Pi \bar X_n^t\}_{n=0}^{N-1}$
and $\bar S_t(X_\beta)$ belong to $\ensuremath{\mathcal{F}}^M$ and therefore also to $\ensuremath{\mathcal{F}}_0^M$. The endpoints are $\Pi
S_t(X_\alpha)$ and $\Pi S_t(X_\beta)$. From the definition of the metric $d_M$, we get
\begin{align}
\notag d_M(\bar S_t(X_\alpha),\bar S_t(X_\beta))&\leq\sum_{n=1}^{N-1} \bar J(\Pi \bar
X_{n-1}^t,\Pi\bar X_n^t)+\bar J(\Pi \bar
X_{N-1}^t,\bar S_t(X_\beta))\\
\label{eq:dtilS}
&=\sum_{n=1}^{N-1} \bar J(\bar X_{n-1}^t,\bar X_n^t)+\bar J(\bar X_{N-1}^t,\tilde
X_N^t),
\end{align}
due to the invariance of $\bar J$ with respect to relabeling. By using the
equivariance of $S_t$, we obtain that
\begin{equation}
\label{eq:tilXrelbarX}
\begin{aligned}
\tilde X_n^t=S_t(\tilde X_n)&=S_t((\bar X_n\circ f_n^{-1})\circ g_{n-1})\\
&=S_t(\bar X_n)\circ (f_n^{-1}\circ g_{n-1})=\bar X_n^t\circ (f_n^{-1}\circ g_{n-1}).
\end{aligned}
\end{equation}
Hence we get from \eqref{eq:dtilS} that
\begin{align*}
d_M(\bar S_t(X_\alpha),\bar S_t(X_\beta))&\leq\sum_{n=1}^{N-1} \bar J(\bar X_{n-1}^t,\tilde
X_n^t)+\bar J(\bar
X_{N-1}^t,\tilde X_N^t)\\
&\leq \sum_{n=1}^N\norm{\bar X_{n-1}^t-\tilde
X_n^t} &\text{ by \eqref{eq:dequiv} }\\
&\leq C_{M,T} \sum_{n=1}^N\norm{\bar X_{n-1}-\tilde
X_n} &\text{ by \eqref{eq:normSnnot}}\\
&\leq C_{M,T} (d_M(X_\alpha,X_\beta)+\varepsilon).
\end{align*}
After letting $\varepsilon$ tend to zero, we obtain \eqref{eq:stab}.
\end{proof}
The Lipschitz stability of the semigroup $T_t$ follows then naturally from Theorem~\ref{th:stab}. It
holds on sets of bounded energy. Let $\ensuremath{\mathcal{D}}^M$ be the subsets of $\ensuremath{\mathcal{D}}$ defined as
\begin{equation}
\ensuremath{\mathcal{D}}^M=\{ (u,\rho,\mu)\in\ensuremath{\mathcal{D}} \mid \mu(\mathbb{R})\leq M\}.
\end{equation}
On the set $\ensuremath{\mathcal{D}}^M$ we define the metric $d_{\ensuremath{\mathcal{D}}^M}$ as
\begin{equation}\label{eq:defDM}
d_{\ensuremath{\mathcal{D}}^M}((u,\rho,\mu),(\tilde u, \tilde\rho, \tilde\mu))
=d^M (L(u,\rho,\mu), L(\tilde u,\tilde\rho,\tilde \mu)),
\end{equation}
where the metric $d^M$ is defined as in Definition~\ref{def:metric}. This definition is well-posed
as, from the definition of $L$, we have that if $(u,\rho,\mu)\in \ensuremath{\mathcal{D}}^M$, then $L(u,\rho,\mu)\in
\ensuremath{\mathcal{F}}_0^M$.
\begin{theorem}
The semigroup $(T_t, d_\ensuremath{\mathcal{D}})$ is a continuous semigroup on $\ensuremath{\mathcal{D}}$ with respect to the metric
$d_D$. The semigroup is Lipschitz continuous on sets of bounded energy, that is: Given $M>0$ and a
time interval $[0,T]$, there exists a constant $C_{M,T}$, which only depends on $M$ and $T$ such
that for any $(u,\rho,\mu)$ and $(\tilde u,\tilde\rho,\tilde \mu)$ in $\ensuremath{\mathcal{D}}^M$, we have
\begin{equation}
d_{\ensuremath{\mathcal{D}}^M}(T_t(u,\rho,\mu),T_t(\tilde u, \tilde\rho, \tilde\mu))\leq C_{M,T}d_{\ensuremath{\mathcal{D}}^M}((u,\rho,\mu),(\tilde u,\tilde\rho,
\tilde \mu))
\end{equation} for all $t\in [0,T]$. Let $(u,\rho,\mu)(t)=T_t(u_0,\rho_0,\mu_0)$, then $(u(t,x),\rho(t,x))$ is weak solution
of the Camassa--Holm equation \eqref{eq:rewchsys10}.
\end{theorem}
We conclude the section about this metric by mentioning that, even if the construction of the metric
is abstract, it can be compared with standard norms, cf. \cite[Section 5]{GHR:13}, so that it can be
used in practice, for example in the study of numerical schemes \cite{CR:12,HR:08}.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 3,455 |
Jennie van Ackeren Dieterle
* August 02, 1909 – April 15, 1999 †
botanist, university teacher, taxonomist, explorer, curator
botanist (1909-1999)
1,397 specimens used in 55 works
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Roberts, J., & Florentine, S. (2021). Biology, distribution and control of the invasive species Ulex europaeus (Gorse): A global synthesis of current and future management challenges and research gaps. Weed Research. doi:10.1111/wre.12491 https://doi.org/10.1111/wre.12491
Ulex europaeus (Gorse) is one of the most invasive shrubs in the world, being now found in more than 50 countries where it economically and environmentally degrades the land. This highly versatile shrub can live more than 30 years and produce over 18,000 fertile seeds annually that can remain viable…
Xu, J., Chai, N., Zhang, T., Zhu, T., Cheng, Y., Sui, S., … Liu, D. (2021). Prediction of temperature tolerance in Lilium based on distribution and climate data. iScience, 24(7), 102794. doi:10.1016/j.isci.2021.102794 https://doi.org/10.1016/j.isci.2021.102794
There are plenty publications providing guidance for resistant taxa selection by experimental researches while the number of experimental taxa is often restricted. In this study, we presented a concise method to predict the temperature tolerance of wild Lilium in China based on open access botanical…
Klisz, M., Puchałka, R., Netsvetov, M., Prokopuk, Y., Vítková, M., Sádlo, J., … Koprowski, M. (2021). Variability in climate-growth reaction of Robinia pseudoacacia in Eastern Europe indicates potential for acclimatisation to future climate. Forest Ecology and Management, 492, 119194. doi:10.1016/j.foreco.2021.119194 https://doi.org/10.1016/j.foreco.2021.119194
As a consequence of native tree species decline and distribution range contraction in Europe, acclimation of the non-native tree species at the edge of their distribution is gaining importance. Although non-native tree species may provide sustainable ecosystem services, as a potentially invasive spe… | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 5,060 |
Last week, Pune-based product startup Innovize Tech announced that it has received funding of $350k from the Indian Angel Network. (Note: Indian Angel Network had also invested in another Pune-based startup Druva.) Read More.. | {
"redpajama_set_name": "RedPajamaC4"
} | 1,527 |
{"url":"https:\/\/adventures.michaelfbryan.com\/posts\/rust-best-practices\/bad-habits\/","text":"When you are coming to Rust from another language you bring all your previous experiences with you.\n\nOften this is awesome because it means you aren\u2019t learning programming from scratch! However, you can also bring along bad habits which can lead you down the wrong rabbit hole or make you write bad code.\n\nThis one is a pet peeve of mine.\n\nIn most C-based languages (C, C#, Java, etc.), the way you indicate whether something failed or couldn\u2019t be found is by returning a \u201cspecial\u201d value. For example, C#\u2019s String.IndexOf() method will scan an array for a particular element and return its index. Returning -1 if nothing is found.\n\nThat leads to code like this:\n\nstring sentence = \"The fox jumps over the dog\";\n\nint index = sentence.IndexOf(\"fox\");\n\nif (index != -1)\n{\nstring wordsAfterFox = sentence.SubString(index);\nConsole.WriteLine(wordsAfterFox);\n}\n\n\nYou see this sort of \u201cuse a sentinel value to indicate something special\u201d practice all the time. Other sentinel values you might find in the wild are \"\", or null (someone once referred to this as their \u201cbillion-dollar mistake\u201d).\n\nThe general reason why this is a bad idea is that there is absolutely nothing to stop you from forgetting that check. That means you can accidentally crash your application with one misplaced assumption or when the code generating the sentinel is far away from the code using it.\n\nWe can do a lot better in Rust, though. Just use Option!\n\nBy design, there is no way to get the underlying value without dealing with the possibility that your Option may be None. This is enforced by the compiler at compile time, meaning code that forgets to check won\u2019t even compile.\n\nlet sentence = \"The fox jumps over the dog\";\nlet index = sentence.find(\"fox\");\n\n\/\/ let words_after_fox = &sentence[index..]; \/\/ Error: Can't index str with Option<usize>\n\nif let Some(fox) = index {\nlet words_after_fox = &sentence[fox..];\nprintln!(\"{}\", words_after_fox);\n}\n\n\n(playground)\n\nBack in the 70\u2019s, a naming convention called Hungarian Notation was developed by programmers writing in languages where variables are untyped or dynamically typed. It works by adding a mnemonic to the start of a name to indicate what it represents, for example the boolean visited variable might be called bVisited or the string name might be called strName.\n\nYou can still see this naming convention in languages Delphi where classes (types) start with T, fields start with F, arguments start with A, and so on.\n\ntype\nTKeyValue = class\nprivate\nFKey: integer;\nFValue: TObject;\npublic\nproperty Key: integer read FKey write FKey;\nproperty Value: TObject read FValue write FValue;\nfunction Frobnicate(ASomeArg: string): string;\nend;\n\n\nC# also has a convention that all interfaces should start with I, meaning programmers coming to Rust from C# will sometimes prefix their traits with I as well.\n\ntrait IClone {\nfn clone(&self) -> Self;\n}\n\n\nIn this case, just drop the leading I. Rust\u2019s syntax guarantees that it just isn\u2019t possible to confuse a trait for a normal type, so it isn\u2019t helping anyone. This is in contrast with C# where interfaces and classes are largely interchangeable.\n\nThis is also seen inside functions where people will conjure up new names for something as they convert it from one form to another. Often these names are silly or contrived, providing negligible additional information to the reader.\n\nlet account_bytes: Vec<u8> = read_some_input();\nlet account_str = String::from_utf8(account_bytes)?;\nlet account: Account = account_str.parse()?;\n\n\nI mean, if we\u2019re calling String::from_utf8() we already know account_str will be a String so why add the _str suffix?\n\nUnlike a lot of other languages, Rust encourages shadowing variables when you are transforming them from one form to another, especially when the previous variable is no longer accessible (e.g. because it\u2019s been moved).\n\nlet account: Vec<u8> = read_some_input();\nlet account = String::from_utf8(account)?;\nlet account: Account = account.parse()?;\n\n\nThis is arguably superior because we can use the same name for the same concept.\n\nOther languages frown on shadowing because it can be easy to lose track of what type a variable contains (e.g. in a dynamically typed language like JavaScript) or you can introduce bugs where the programmer thinks a variable has one type but it actually contains something separate.\n\nNeither of these is particularly relevant to a strongly typed language with move semantics like Rust, so you can use shadowing freely without worrying about shooting yourself in the foot.\n\n## An Abundance of Rc<RefCell<T>> Link to heading\n\nA common pattern in Object Oriented languages is to accept a reference to some object so you can call its methods later on.\n\nOn its own there is nothing wrong with this, Dependency Injection is a very good thing to do, but unlike most OO languages Rust doesn\u2019t have a garbage collector and has strong feelings on shared mutability.\n\nPerhaps this will be easier to understand with an example.\n\nSay we are implementing a game where the player needs to beat up a bunch of monsters until they have inflicted a certain amount of damage (I dunno, maybe it\u2019s for a quest or something).\n\nWe create a Monster class which has a health property and a takeDamage() method, and so we can keep track of how much damage has been inflicted we\u2019ll let people provide callbacks that get called whenever the monster receives damage.\n\ntype OnReceivedDamage = (damageReceived: number) => void;\n\nclass Monster {\nhealth: number = 50;\n\ntakeDamage(amount: number) {\namount = Math.min(this.health, amount);\nthis.health -= amount;\n}\n\non(event: \"damaged\", callback: OnReceivedDamage): void {\n}\n}\n\n\nLet\u2019s also create a DamageCounter class which tracks how much damage we\u2019ve inflicted and lets us know when that goal is reached.\n\nclass DamageCounter {\ndamageInflicted: number = 0;\n\nreachedTargetDamage(): boolean {\nreturn this.damageInflicted > 100;\n}\n\nonDamageInflicted(amount: number) {\nthis.damageInflicted += amount;\n}\n}\n\n\nNow we\u2019ll create some monsters and keep inflicting a random amount of damage until the DamageCounter is happy.\n\nconst counter = new DamageCounter();\n\nconst monsters = [new Monster(), new Monster(), new Monster(), new Monster(), new Monster()];\nmonsters.forEach(m => m.on(\"damaged\", amount => counter.onDamageInflicted(amount)));\n\nwhile (!counter.reachedTargetDamage()) {\n\/\/ pick a random monster\nconst index = Math.floor(Math.random()*monsters.length);\nconst target = monsters[index];\n\/\/ then damage it a bit\nconst damage = Math.round(Math.random() * 50);\ntarget.takeDamage(damage);\n\nconsole.log(Monster ${index} received${damage} damage);\n}\n\n\n(TypeScript Playground)\n\nNow let\u2019s port this code to Rust. Our Monster struct is fairly similar, although we need to use Box<dyn Fn(u32)> for a closure which accepts a single u32 argument (all closures in JavaScript are heap allocated by default).\n\ntype OnReceivedDamage = Box<dyn Fn(u32)>;\n\nstruct Monster {\nhealth: u32,\n}\n\nimpl Monster {\nfn take_damage(&mut self, amount: u32) {\nfor callback in &mut self.received_damage {\n}\n}\n\n}\n}\n\nimpl Default for Monster {\nfn default() -> Self {\nMonster { health: 100, received_damage: Vec::new() }\n}\n}\n\n\nNext comes our DamageCounter, nothing interesting here.\n\n#[derive(Default)]\nstruct DamageCounter {\ndamage_inflicted: u32,\n}\n\nimpl DamageCounter {\nfn reached_target_damage(&self) -> bool {\nself.damage_inflicted > 100\n}\n\nfn on_damage_received(&mut self, damage: u32) {\nself.damage_inflicted += damage;\n}\n}\n\n\nAnd finally our code that inflicts damage.\n\nfn main() {\nlet mut counter = DamageCounter::default();\nlet mut monsters: Vec<_> = (0..5).map(|_| Monster::default()).collect();\n\nfor monster in &mut monsters {\n}\n\nwhile !counter.reached_target_damage() {\nlet index = rng.gen_range(0..monsters.len());\nlet target = &mut monsters[index];\n\nlet damage = rng.gen_range(0..50);\ntarget.take_damage(damage);\n\nprintln!(\"Monster {} received {} damage\", index, damage);\n}\n}\n\n\n(playground)\n\nBut herein lies our first problem, when we try to compile the code rustc gives us not one, but four compile errors for the monster.add_listener() line \ud83e\udd23\n\nerror[E0596]: cannot borrow counter as mutable, as it is a captured variable in a Fn closure\n--> src\/main.rs:47:48\n|\n| ^^^^^^^ cannot borrow as mutable\n\nerror[E0499]: cannot borrow counter as mutable more than once at a time\n--> src\/main.rs:47:39\n|\n| ---------^^^^^^^^------------------------------------\n| | | |\n| | | borrows occur due to use of counter in closure\n| | counter was mutably borrowed here in the previous iteration of the loop\n| cast requires that counter is borrowed for 'static\n\nerror[E0597]: counter does not live long enough\n--> src\/main.rs:47:48\n|\n| ------------------^^^^^^^----------------------------\n| | | |\n| | | borrowed value does not live long enough\n| | value captured here\n| cast requires that counter is borrowed for 'static\n...\n60 | }\n| - counter dropped here while still borrowed\n\nerror[E0502]: cannot borrow counter as immutable because it is also borrowed as mutable\n--> src\/main.rs:50:12\n|\n| -----------------------------------------------------\n| | | |\n| | | first borrow occurs due to use of counter in closure\n| | mutable borrow occurs here\n| cast requires that counter is borrowed for 'static\n...\n50 | while !counter.reached_target_damage() {\n| ^^^^^^^ immutable borrow occurs here\n\n\nThere are a number of things wrong with this line, but it can be boiled down to:\n\n\u2022 The closure captures a reference to counter\n\u2022 The counter.on_damage_received() method takes &mut self so our closure needs a &mut reference. We add the closures in a loop so we end up taking multiple &mut references to the same object at the same time\n\u2022 Our listener is a boxed closure without any lifetime annotations, meaning it needs to own any variables it closes over. We would need to move the counter into the closure, but because we do this in a loop we\u2019ll have a \u201cuse of moved value\u201d error\n\u2022 After passing the counter to add_listener() we try to use it in our loop condition\n\nOverall it\u2019s just a bad situation.\n\nThe canonical answer to this is to wrap the DamageCounter in a reference-counted pointer so we can have multiple handles to it at the same time, then because we need to call a &mut self method we also need a RefCell to \u201cmove\u201d the borrow checking from compile time to run time.\n\n fn main() {\n- let mut counter = DamageCounter::default();\n+ let mut counter = Rc::new(RefCell::new(DamageCounter::default()));\nlet mut monsters: Vec<_> = (0..5).map(|_| Monster::default()).collect();\n\nfor monster in &mut monsters {\n+ let counter = Rc::clone(&counter);\n+ }));\n}\n\n- while !counter.reached_target_damage() {\n+ while !counter.borrow().reached_target_damage() {\nlet index = rng.gen_range(0..monsters.len());\nlet target = &mut monsters[index];\n...\n}\n}\n\n\n(playground)\n\nWell\u2026 it works. But this approach tends to get messy, especially when you are storing non-trivial things like a Rc<RefCell<Vec<Foo>>>> (or its multi-threaded cousin Arc<Mutex<Vec<Foo>>>>) inside structs 1.\n\nIt also opens you up to situations where the RefCell might be borrowed mutably multiple times because your code is complex and something higher up in the call stack is already using the RefCell. With a Mutex this will cause a deadlock while the RefCell will panic, neither of which is conducive to a reliable program.\n\nA much better approach is to change your API to not hold long-lived references to other objects. Depending on the situation, it might make sense to take a callback argument in the Monster::take_damage() method.\n\nstruct Monster {\nhealth: u32,\n}\n\nimpl Monster {\nfn take_damage(&mut self, amount: u32, on_damage_received: impl FnOnce(u32)) {\n}\n}\n\nimpl Default for Monster {\nfn default() -> Self { Monster { health: 100 } }\n}\n\n...\n\nfn main() {\nlet mut counter = DamageCounter::default();\nlet mut monsters: Vec<_> = (0..5).map(|_| Monster::default()).collect();\n\nwhile !counter.reached_target_damage() {\nlet index = rng.gen_range(0..monsters.len());\nlet target = &mut monsters[index];\n\nlet damage = rng.gen_range(0..50);\n\nprintln!(\"Monster {} received {} damage\", index, damage);\n}\n}\n\n\n(playground)\n\nA nice side-effect of this is that we get rid of all the callback management boilerplate, meaning this version is only 47 lines long instead of the Rc<RefCell<_>> version\u2019s 62.\n\nOther times it may not be acceptable to give take_damage() a callback parameter, in which case you could return a \u201csummary\u201d of what happened so the caller can decide what to do next.\n\nimpl Monster {\nfn take_damage(&mut self, amount: u32) -> AttackSummary {\n}\n}\n\nstruct AttackSummary {\n}\n\n...\n\nfn main() {\nlet mut counter = DamageCounter::default();\nlet mut monsters: Vec<_> = (0..5).map(|_| Monster::default()).collect();\n\nwhile !counter.reached_target_damage() {\nlet index = rng.gen_range(0..monsters.len());\nlet target = &mut monsters[index];\n\nlet damage = rng.gen_range(0..50);\nlet AttackSummary { damage_received } = target.take_damage(damage);\n\nprintln!(\"Monster {} received {} damage\", index, damage);\n}\n}\n\n\n(playground)\n\nThis is my preferred solution; from experience, it tends to work well for larger codebases or when the code is more complex.\n\nAnother hang-over from writing a lot of C is using the wrong integer type and getting frustrated because you need to cast to\/from usize all the time.\n\nI\u2019ve seen people run into this so many times in the wild, especially when indexing.\n\nThe underlying problem is that C programmers are all taught to use int for indexing and for-loops, so when they come to Rust and they need to store a list of indices, the programmer will immediately reach for a Vec<i32>. They then get frustrated because Rust is quite strict when it comes to indexing and standard types like arrays, slices, and Vec can only be indexed using usize (the equivalent of size_t), meaning their code is cluttered with casts from i32 to usize and back again.\n\nThere are a number of perfectly legitimate reasons for why Rust only allows indexing by usize:\n\n\u2022 It doesn\u2019t make sense to have a negative index (accessing items before the start of a slice is UB), so we can avoid an entire class of bugs by indexing with an unsigned integer\n\u2022 A usize is defined to be an integer with the same size as a normal pointer, meaning the pointer arithmetic won\u2019t have any hidden casts\n\u2022 The std::mem::size_of() and std::mem::align_of() functions return usize\n\nOf course, when stated this way the solution is clear. Choose the right integer type for your application but when you are doing things that eventually be used for indexing, that \u201cright integer type\u201d is probably usize.\n\n<rant>\n\nThere\u2019s an old Rust koan on the User Forums by Daniel Keep that comes to mind every time I see a grizzled C programmer reach for raw pointers or std::mem::transmute() because the borrow checker keeps rejecting their code: Obstacles.\n\nYou should go read it. It\u2019s okay, I\u2019ll wait.\n\nToo often you see people wanting to hack around privacy, create self-referencing structs, or create global mutable variables using unsafe. Frequently this will be accompanied by comments like \u201cbut I know this program will only use a single thread so accessing the static mut is fine\u201d or \u201cbut this works perfectly fine in C\u201d.\n\nThe reality is that unsafe code is nuanced and you need to have a good intuition for Rust\u2019s borrow checking rules and memory model. I hate to be a gate keeper and say \u201cyou must be this tall to write multi-threaded unsafe code\u201d 2, but there\u2019s a good chance that if you are new to the language you won\u2019t have this intuition and are opening yourself and your colleagues up to a lot of pain.\n\nIt\u2019s fine to play around with unsafe if you are trying to learn more about Rust or you know what you are doing and are using it legitimately, but unsafe is not a magical escape hatch which will make the compiler stop complaining and let you write C with Rust syntax.\n\n<\/rant>\n\nA common practice in C is to prefix functions with the name of the library or module to help readers understand where it comes from and avoid duplicate symbol errors (e.g. rune_wasmer_runtime_load()).\n\nHowever, Rust has real namespaces and lets you attach methods to types (e.g. rune::wasmer::Runtime::load()). Just use them - it\u2019s what they are there for.\n\nThe for-loop and indexing is the bread and butter for most C-based languages.\n\nlet points: Vec<Coordinate> = ...;\nlet differences = Vec::new();\n\nfor i in 1..points.len() [\nlet current = points[i];\nlet previous = points[i-1];\ndifferences.push(current - previous);\n]\n\n\n(playground)\n\nHowever, it\u2019s easy to accidentally introduce an off-by-one error when using indexing (e.g. I needed to remember to start looping from 1 and subtract 1 to get the previous point) and even seasoned programmers aren\u2019t immune from crashing due to an index-out-of-bounds error.\n\nIn situations like these, Rust encourages you to reach for iterators instead. The slice type even comes with high-level tools like the windows() and array_windows() methods to let you iterate over adjacent pairs of elements.\n\nlet points: Vec<Coordinate> = ...;\nlet mut differences = Vec::new();\n\nfor [previous, current] in points.array_windows().copied() {\ndifferences.push(current - previous);\n}\n\n\n(playground)\n\nYou could even remove the for-loop and mutation of differences altogether.\n\nlet differences: Vec<_> = points\n.array_windows()\n.copied()\n.map(|[previous, current]| current - previous)\n.collect();\n\n\n(playground)\n\nSome would argue the version with map() and collect() is cleaner or more \u201cfunctional\u201d, but I\u2019ll let you be the judge there.\n\nAs a bonus, iterators can often allow better performance because checks can be done as part of the looping condition instead of being separate3 (Alice has a good explanation here).\n\nOnce you start drinking the Kool-Aid that is Rust\u2019s iterators you can run into the opposite problem - when all you have is a hammer everything looks like a nail.\n\nLong chains of map(), filter(), and and_then() calls can get quite hard to read and keep track of what is actually going on, especially when type inference lets you omit a closure argument\u2019s type.\n\nOther times your iterator-based solution is just unnecessarily complicated.\n\nAs an example, have a look at this snippet of code and see if you can figure out what it is trying to do.\n\npub fn functional_blur(input: &Matrix) -> Matrix {\nassert!(input.width >= 3);\nassert!(input.height >= 3);\n\n\/\/ Stash away the top and bottom rows so they can be\n\/\/ directly copied across later\nlet mut rows = input.rows();\nlet first_row = rows.next().unwrap();\nlet last_row = rows.next_back().unwrap();\n\nlet top_row = input.rows();\nlet middle_row = input.rows().skip(1);\nlet bottom_row = input.rows().skip(2);\n\nlet blurred_elements = top_row\n.zip(middle_row)\n.zip(bottom_row)\n.flat_map(|((top, middle), bottom)| blur_rows(top, middle, bottom));\n\nlet elements: Vec<f32> = first_row\n.iter()\n.copied()\n.chain(blurred_elements)\n.chain(last_row.iter().copied())\n.collect();\n\nMatrix::new_row_major(elements, input.width, input.height)\n}\n\nfn blur_rows<'a>(\ntop_row: &'a [f32],\nmiddle_row: &'a [f32],\nbottom_row: &'a [f32],\n) -> impl Iterator<Item = f32> + 'a {\n\/\/ stash away the left-most and right-most elements so they can be copied across directly.\nlet &first = middle_row.first().unwrap();\nlet &last = middle_row.last().unwrap();\n\n\/\/ Get the top, middle, and bottom row of our 3x3 sub-matrix so they can be\n\/\/ averaged.\nlet top_window = top_row.windows(3);\nlet middle_window = middle_row.windows(3);\nlet bottom_window = bottom_row.windows(3);\n\n\/\/ slide the 3x3 window across our middle row so we can get the average\n\/\/ of everything except the left-most and right-most elements.\nlet averages = top_window\n.zip(middle_window)\n.zip(bottom_window)\n.map(|((top, middle), bottom)| top.iter().chain(middle).chain(bottom).sum::<f32>() \/ 9.0);\n\nstd::iter::once(first)\n.chain(averages)\n.chain(std::iter::once(last))\n}\n\n\n(playground)\n\nBelieve it or not, but that\u2019s one of the more readable versions I\u2019ve seen\u2026 Now let\u2019s look at the imperative implementation.\n\npub fn imperative_blur(input: &Matrix) -> Matrix {\nassert!(input.width >= 3);\nassert!(input.height >= 3);\n\n\/\/ allocate our output matrix, copying from the input so\n\/\/ we don't need to worry about the edge cases.\nlet mut output = input.clone();\n\nfor y in 1..(input.height - 1) {\nfor x in 1..(input.width - 1) {\nlet mut pixel_value = 0.0;\n\npixel_value += input[[x - 1, y - 1]];\npixel_value += input[[x, y - 1]];\npixel_value += input[[x + 1, y - 1]];\n\npixel_value += input[[x - 1, y]];\npixel_value += input[[x, y]];\npixel_value += input[[x + 1, y]];\n\npixel_value += input[[x - 1, y + 1]];\npixel_value += input[[x, y + 1]];\npixel_value += input[[x + 1, y + 1]];\n\noutput[[x, y]] = pixel_value \/ 9.0;\n}\n}\n\noutput\n}\n\n\n(playground)\n\nI know which version I prefer.\n\nIn most other mainstream languages it is quite common to see the programmer write a check before they do an operation which may throw an exception. Our C# IndexOf() snippet from earlier is a good example of this:\n\nint index = sentence.IndexOf(\"fox\");\n\nif (index != -1)\n{\nstring wordsAfterFox = sentence.SubString(index);\nConsole.WriteLine(wordsAfterFox);\n}\n\n\nCloser to home, you might see code like this:\n\nlet opt: Option<_> = ...;\n\nif opt.is_some() {\nlet value = opt.unwrap();\n...\n}\n\n\nor this:\n\nlet list: &[f32] = ...;\n\nif !list.is_empty() {\nlet first = list[0];\n...\n}\n\n\nNow both snippets are perfectly valid pieces of code and will never fail, but similar to sentinel values you are making it easy for future refactoring to introduce a bug.\n\nUsing things like pattern matching and Option help you avoid this situation by making sure the only way you can access a value is if it is valid.\n\nif let Some(value) = opt {\n...\n}\n\nif let [first, ..] = list {\n...\n}\n\n\nDepending on where it is used and how smart LLVM or your CPU\u2019s branch predictor are, this may also generate slower code because the fallible operation (opt.unwrap() or list[index] in that example) needs to do unnecessary checks 3.\n\nIn many languages, it is normal to call an object\u2019s constructor and initialize its fields afterward (either manually or by calling some init() method). However, this goes against Rust\u2019s general convention of \u201cmake invalid states unrepresentable\u201d.\n\nSay you are writing an NLP application and have a dictionary containing all the possible words you can handle.\n\nThis is one way you could create the dictionary:\n\nlet mut dict = Dictionary::new();\n\/\/ read the file and populate some internal HashMap or Vec\n\n\nHowever, writing Dictionary this way means it now has two (hidden) states - empty and populated.\n\nAll downstream code that uses the Dictionary will assume it\u2019s been populated already and write code accordingly. This may include doing things like indexing into the dictionary with dict[\"word\"] which may panic if \"word\" isn\u2019t there.\n\nNow you\u2019ve opened yourself up to a situation where passing an empty dictionary to code that expects a populated dictionary may trigger a panic.\n\nBut that\u2019s completely unnecessary.\n\nJust make sure the Dictionary is usable immediately after constructing it instead of populating it after the fact.\n\nlet dict = Dictionary::from_file(\".\/words.txt\")?;\n\nimpl Dictionary {\nfn from_file(filename: impl AsRef<Path>) -> Result<Self, Error> {\nlet mut words = Vec::new();\nfor line in text.lines() {\nwords.push(line);\n}\nOk(Dictionary { words })\n}\n}\n\n\nInternally the Dictionary::from_file() might create an empty Vec and populate it incrementally, but it won\u2019t be stored in the Dictionary\u2019s words field yet so there is no assumption that it is populated and useful.\n\nHow frequently you fall into this anti-pattern depends a lot on your background and coding style.\n\nFunctional languages are often completely immutable so you\u2019ll fall into the idiomatic pattern naturally. After all, it\u2019s kinda hard to create a half-initialized thing and populate it later when you aren\u2019t allowed to mutate anything.\n\nOn the other hand, OO languages are much happier to let you initialize an object after it has been constructed, especially because object references can be null by default and they have no qualms about mutability\u2026 You could argue this contributes to why OO languages have a propensity for crashing due to an unexpected NullPointerException.\n\nTo point out the obvious, a really nice property of immutable objects is that you can rely on them to never change. However, in languages like Python and Java, immutability isn\u2019t transitive - i.e. if x is an immutable object, x.y isn\u2019t guaranteed to be immutable unless it was explicitly defined that way.\n\nThis means it\u2019s possible to write code like this\u2026\n\nclass ImmutablePerson:\ndef __init__(self, name: str, age: int, addresses: List[str]):\nself._name = name\nself._age = age\n\n@property\ndef name(self): return self._name\n@property\ndef age(self): return self._age\n@property\n\n\nThen someone else comes along and accidentally messes up the address list as part of their normal code.\n\ndef send_letters(message: str, addresses: List[str]):\n# Note: the post office's API only works with with uppercase letters so we\n# need to pre-process the address list\n\nclient = PostOfficeClient()\n\nperson = ImmutablePerson(\"Joe Bloggs\", 42, [\"123 Fake Street\"])\n\nsend_letters(\n)\n\n\n\nWhile I admit the example is a bit contrived, it\u2019s not uncommon for functions to modify the arguments they are given. Normally this is fine, but when your ImmutablePerson assumes its addresses field will never change, it\u2019s annoying for some random piece of code on the other side of the project to modify it without you knowing.\n\nThe typical solution to this is to preemptively copy the list so even if the caller tries to mutate its contents, they\u2019ll be mutating a copy and not the original addresses field.\n\nclass ImmutablePerson:\n...\n\n@property\n\n\nIn general, you\u2019ll see defensive copies being used anywhere code wants to be sure that another piece of code won\u2019t modify some shared object at an inopportune time.\n\nConsidering this is an article about Rust, you\u2019ve probably guessed what the root cause of this is - a combination of aliasing and mutation.\n\nYou\u2019ve also probably guessed why defensive copies aren\u2019t really necessary when writing Rust code - lifetimes and the \u201cshared immutable XOR single mutable\u201d rule for references means it just isn\u2019t possible for code to modify something without first asking its original owner for mutable access or explicitly opting into shared mutation by using a type like std::sync::Mutex<T>.\n\n1. Out of curiosity, how many people noticed there are 4 >\u2019s in Rc<RefCell<Vec<Foo>>>> but only 3 <\u2019s?\u00a0\u21a9\ufe0e","date":"2023-02-08 20:00:42","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.2040858268737793, \"perplexity\": 4970.241109102977}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2023-06\/segments\/1674764500904.44\/warc\/CC-MAIN-20230208191211-20230208221211-00293.warc.gz\"}"} | null | null |
Q: How to make headers glossy in IE7,8,9 using CSS I am using below code to get a glossy look for a header label. It works fine in FF, Chrome and IE10. But when loaded in lower versions of IE like IE7,8,9. Glossy finish is lost. Can any body help me to get glossy look on the header?? Please refer to the below images for glossy look and normal look.
Glossy Header
Normal Header
.docTitle
{
font-weight: bold;
width: 99.3%;
font: bold 12px/100% Arial,Verdana,helvetica;
padding: 8px 5px;
text-decoration: none;
color: rgb(255, 255, 255);
text-shadow: 0 -1px 0 rgba(0,0,0,.9);
background: #255182;
background: -webkit-linear-gradient(top, #2f4f88, #385993 49%, #21427d 50%, #255182);
background: -moz-linear-gradient(top, #2f4f88, #385993 49%, #21427d 50%, #255182);
background: -ms-linear-gradient(top, #2f4f88, #385993 49%, #21427d 50%, #255182);
background: -o-linear-gradient(top, #2f4f88, #385993 49%, #21427d 50%, #255182);
background: linear-gradient(to bottom, #2f4f88, #385993 49%, #21427d 50%, #255182);
}
A: see the "notes" and "resources" on this page: http://caniuse.com/#feat=css-gradients. Older IE does not support the css3 gradients, but there are workarounds if its important enough to you. I personally would be fine with a "non-glossy" appearance in IE.
http://css3pie.com/ is one option for emulating support in IE.
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{"url":"https:\/\/homework.study.com\/explanation\/if-f-x-integral-25-x-1-t-dt-then-find-f-x.html","text":"# If {eq}\\displaystyle F(x) = \\int_{25}^{x} \\frac{1}{t} \\ dt {\/eq}, then find {eq}\\displaystyle F'(x) {\/eq}\n\n## Question:\n\nIf {eq}\\displaystyle F(x) = \\int_{25}^{x} \\frac{1}{t} \\ dt {\/eq}, then find {eq}\\displaystyle F'(x) {\/eq}\n\n## Leibniz's Rule:\n\nIt is a special method used to find the derivative of a definite integral without integration. If the definite integral is {eq}\\displaystyle \\int\\limits_{A\\left( x \\right)}^{B\\left( x \\right)} {f\\left( x \\right)} dx {\/eq}, then the following formula can be used to find the derivative of the definite integral.\n\n{eq}\\displaystyle \\dfrac{d}{{dx}}\\int\\limits_{A\\left( x \\right)}^{B\\left( x \\right)} {f\\left( x \\right)} dx = f\\left[ {B\\left( x \\right)} \\right]\\dfrac{d}{{dx}}B\\left( x \\right) - f\\left[ {A\\left( x \\right)} \\right]\\dfrac{d}{{dx}}A\\left( x \\right) {\/eq}\n\nThe above formula is known as Leibniz's formula.\n\nBecome a Study.com member to unlock this answer!\n\nGiven Data\n\n\u2022 The given definite integral is {eq}\\displaystyle F(x) = \\int\\limits_{25}^x {\\dfrac{1}{t}} \\;dt {\/eq}.\n\nUsing the Leibniz's rule to...","date":"2023-03-24 00:15:34","metadata":"{\"extraction_info\": {\"found_math\": false, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.9855991005897522, \"perplexity\": 7845.958967320762}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2023-14\/segments\/1679296945218.30\/warc\/CC-MAIN-20230323225049-20230324015049-00161.warc.gz\"}"} | null | null |
{"url":"http:\/\/simple.wikipedia.org\/wiki\/Drake_equation","text":"# Drake equation\n\nIn 1961, Frank Drake wrote down an equation for the chance of a contactable alien civilization from another planet in the Milky Way Galaxy. This is known as the Drake Equation (sometimes called the Green Bank equation or the Green Bank Formula). Carl Sagan mentioned the Drake equation often so it that has been mistaken for the 'Sagan equation'.\n\n## The equation\n\nThe Drake equation states that:\n\n$N = R^{\\ast} \\cdot f_p \\cdot n_e \\cdot f_{\\ell} \\cdot f_i \\cdot f_c \\cdot L \\!$\n\nwhere:\n\nN = the number of civilizations in our galaxy with which communication might be possible;\n\nand\n\nR* = the average rate of star formation per year in our galaxy\nfp = the fraction of those stars that have planets\nne = the average number of planets that can potentially support life per star that has planets\nf = the fraction of the above that actually go on to develop life at some point\nfi = the fraction of the above that actually go on to develop intelligent life\nfc = the fraction of civilizations that develop a technology that releases detectable signs of their existence into space\nL = the length of time for which such civilizations release detectable signals into space.[1][2]\n\n## References\n\n1. \"PBS NOVA: Origins - The Drake Equation\". Pbs.org. Retrieved 2010-03-07.\n2. \"Space: SETI: Frank Drake Biography\". Retrieved November 17, 2011.","date":"2014-07-29 11:01:06","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 1, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.8058319687843323, \"perplexity\": 1703.864037180874}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2014-23\/segments\/1406510267075.55\/warc\/CC-MAIN-20140728011747-00020-ip-10-146-231-18.ec2.internal.warc.gz\"}"} | null | null |
06/09/14--15:47: _Texas GOP endorses ...
06/10/14--10:39: _California judge st...
06/10/14--11:16: _Access to students'...
06/10/14--11:48: _CIA cracks down on ...
06/10/14--13:32: _Twitter expects Wor...
06/10/14--13:54: _House approves bill...
06/10/14--14:45: _Gun control failure...
06/10/14--14:48: _Surge of children a...
06/10/14--15:02: _News Wrap: Investig...
06/10/14--15:10: _In major blow to Ma...
06/10/14--15:12: _How did Sunni insur...
06/10/14--15:20: _How 'Hard Choices' ...
06/10/14--15:33: _Unexpected legislat...
06/10/14--15:39: _Debating tenure pro...
06/10/14--15:49: _Young people learn ...
06/10/14--17:46: _House Majority Lead...
06/11/14--11:58: _In Brazil, soccer n...
06/11/14--12:23: _Cantor to resign Ho...
06/11/14--12:42: _Bill and Melinda Ga...
06/11/14--12:44: _Suffocating cells f...
06/09/14--15:47: Texas GOP endorses 'treatment' for homosexuality
FORT WORTH, Texas — The Texas Republican Party now endorses so-called "reparative therapy" for gays, under a new platform given final approval at its annual convention Saturday.
The new anti-gay language never came up for debate before roughly 7,000 delegates ratified a Texas GOP platform that tea party groups succeeded in pushing further to the right, including winning a harder line on immigration.
One influential tea party group called Texas Eagle Forum had urged the party to support psychological treatments that seek to turn gay people straight. It comes after Republican New Jersey Gov. Chris Christie last fall signed a law banning such therapies on minors, and California has a similar law.
The Fort Worth Convention Hall cheered when party leaders announced that Christie finished a distant 11th in a 2016 presidential straw poll.
"There's a very, very small group of people who want to keep the party in the past. We were here today to try to pull the party into the future," said Rudy Oeftering, vice president of the gay conservative group Metroplex Republicans. "The only way the party can go into the future is to start listening to young people, to start listening to people who have gay family members."
Oeftering and allies had lined up to speak against the therapy language that had been added earlier this week. But they never got a chance to address delegates, because a parliamentary motion to approve the full platform was called first.
Under the new plank, the Texas GOP recognizes "the legitimacy and efficacy of counseling, which offers reparative therapy and treatment for those patients seeking healing and wholeness from their homosexual lifestyle."
The American Psychological Association and other major health organizations have condemned such counseling, which generally try to change a person's sexual orientation or to lessen their interest in engaging in same-sex sexual activity. The groups say the practice should not be used on minors because of the danger of serious psychological harm.
"The platform reflects what the people in the Republican Party have asked for, and that should be no surprise: family values, protection of marriage between one man and one woman and everything that goes along with that," said Jonathan Saenz, president of the conservative group Texas Values and a convention delegate.
Gay conservatives did come away with a rare victory at the convention: Winning the removal of decades-old language in the state party platform that states, "homosexuality tears at the fabric of society."
Follow Paul J. Weber on Twitter @PaulJWeber
The post Texas GOP endorses 'treatment' for homosexuality appeared first on PBS NewsHour.
06/10/14--10:39: California judge strikes down teacher protections in landmark case
Los Angeles Superior Court Judge Rolf Treu struck down multiple California laws providing job protections for the state's public school teachers, calling them unconstitutional, reports KPCC in Los Angeles.
The lawsuit was filed in 2012 by Student Matters, a group representing nine California students. They argued the state's laws — granting teachers tenure after two years on the job, governing teacher dismissals and requiring teacher layoffs be based on seniority — codify inequality in the state's public schools. The plaintiffs called witnesses who testified these laws mean predominantly low-income, black and Latino students are most roiled by budget cuts and most likely to be staffed with ineffective teachers.
California and the state's two teachers unions argued that factors outside of a classroom teacher's control, including years of dramatic state funding cuts, concentrated poverty, racial segregation, inadequate housing and limited access to healthcare, are to blame for most of the so-called achievement gap between high- and low-performing schools. Witnesses also argued that administrators have sufficient leeway to dismiss ineffective teachers.
NewsHour Weekend examined the issues at play in the two month-long trial in March. Each side called students, teachers, school district administrators and education researchers to testify.
California and the teachers unions are expected to appeal the ruling. National education advocacy groups like the National Center for Teacher Quality, StudentsFirst and the Education Trust backed the Students Matter case.
Laws to modify or end teacher tenure have passed in at least 18 states in recent years. Students Matter is considering similar lawsuits in more than half a dozen other states, according to Politico .
The post California judge strikes down teacher protections in landmark case appeared first on PBS NewsHour.
06/10/14--11:16: Access to students' online health information a boon to school nurses
Although the school nurse is a familiar figure, school-based health care is unfamiliar territory to many medical professionals, operating in a largely separate health care universe from other community-based medical services.
Now, as both schools and health care systems seek to ensure that children coping with chronic conditions such as diabetes and asthma get the comprehensive, coordinated care the students need, the schools and health systems are forming partnerships to better integrate their services. In these projects, some funded by the health law, school health professionals gain access to students' electronic health records and/or specialists and other health system resources. Such initiatives currently exist or are on the drawing board in Delaware, Miami and Beaverton, Ore., among other locations.
School nurses today do a lot more than bandage skinned knees. They administer vaccines and medications, help diabetic students monitor their blood sugar, and prepare teachers to handle a student's seizure or asthma attack, among many other things.
A 2007 study found that 45 percent of public schools have a full-time nurse on site, while 30 percent have one who works part time. In addition to school nurses, 12.5 percent of school districts have at least one school-based health center that offers both health services and mental health or social services, according to the federal Centers for Disease Control and Prevention's 2012 Schools Health Policies and Practices Study. School nurses often work closely with school-based health centers, referring students there as needed.
"Chronic disease management is what school nurses spend most of their time doing," says Carolyn Duff, president of the National Association of School Nurses. "We do care for students in emergencies, but we spend more time planning to avoid emergencies."
School-based health care providers may bill Medicaid for some services, but rarely bill private insurers.
"The juice isn't worth the squeeze," says John Schlitt, interim president for the School-Based Health Alliance, an advocacy organization for school-based health centers. "It takes so much energy to track these bills from the commercial insurers, many just stop trying."
Although school nurses see many students regularly, they don't always have the most up-to-date information about the students' health. School nurses must get permission from parents to communicate with a child's doctor. Once the doctor gives them a care plan for the child, they generally rely on the doctor and/or parents for updates and changes.
"When things change, we don't always get told in a timely manner," says Nina Fekaris, a school nurse in the Beaverton, Ore., school district. "It works, but it takes a lot of coordination."
School nurses in Delaware voiced similar concerns a few years ago to administrators at Nemours Children's Health System that serves residents around the state.
"Lots of nurses expressed that they had difficulty communicating with providers" at Nemours, says Claudia Kane, program manager of the Student Health Collaboration at Nemours. In 2011, the health system got together with the Delaware School Nurses Association and the state Department of Education to develop a program that, with parental approval, now gives school nurses read-only access to the electronic health records of more than 1,500 students who have complex medical conditions or special needs such as diabetes, asthma, attention deficit hyperactivity disorder, seizure disorders or gastrointestinal problems.
Now that she has access to the Nemours system, Beth Mattey can check the recent lab test results of a student who has diabetes.
"It's helpful for me to monitor his [blood sugar levels] and work with him to make sure he's in better control," says the Wilmington school nurse who is president-elect of the National Association of School Nurses.
When a student put a staple through his finger, she was able to check to make sure he went to the doctor and got treatment.
"Checking with him directly involves calling him out of class," she says.
Eventually school nurses will be able to put information into the Nemours electronic records system as well, says Kane. In the meantime, Nemours doctors, some of whom were initially skeptical about allowing school nurses access to health system medical records, are warming up to the arrangement. It encourages communication between Nemours physicians and school nurses, and eases the burden of routine tasks because Nemours doctors no longer have to fax over care plans or instructions to the school nurse every couple of months for students who are part of the program, says Kane.
The Nemours Student Health Collaboration project is operating in all Delaware public school districts as well as half of charter schools and about a third of private schools. Nemours plans to extend the program to school-based health centers next, says Kane.
"Our primary care practices are going through the process to become certified as medical homes," says Kane. "School nurses have a big role in care coordination, and this program is integrated as a piece of that."
Kaiser Health News is an editorially independent program of the Henry J. Kaiser Family Foundation, a nonprofit, nonpartisan health policy research and communication organization not affiliated with Kaiser Permanente.
The post Access to students' online health information a boon to school nurses appeared first on PBS NewsHour.
06/10/14--11:48: CIA cracks down on employee harassment; internal postings call actions insufficient
Photo by Wikimedia user BetacommandBot
WASHINGTON — Fifteen CIA employees were found to have committed sexual, racial or other types of harassment last year, including a supervisor who was removed from the job after engaging in "bullying, hostile behavior," and an operative who was sent home from an overseas post for inappropriately touching female colleagues, according to an internal CIA document obtained by The Associated Press.
The examples, sent several weeks ago in an email to the CIA's workforce by the director of the agency's Office of Equal Employment Opportunity, were meant to show how the agency is enforcing a zero-tolerance policy toward harassment. But the announcement sparked heated commentary in postings on the CIA's internal networks, officials acknowledged, with some employees arguing the agency does not sufficiently ferret out and punish misconduct.
The CIA's personnel systems seem to be fundamentally broken, and harassment frequently goes unreported, one officer said in an excerpt of an employee posting obtained by the AP. The authenticity of the posting was not disputed by the agency.
CIA officials took issue with that assertion after agreeing to discuss the workforce message on the condition that they not be quoted by name.
The agency officials made available CIA Director John Brennan's March workforce message reaffirming the zero-tolerance policy, saying, "Words or actions that harm a colleague and undermine his or her career are more than just unprofessional, painful and wrong — they are illegal and hurt us all." Brennan assured employees that he would not tolerate acts of reprisal against those who complained of harassment.
The agency won't release its employee workplace surveys or details about complaints, on the grounds that such numbers are classified. The CIA takes that position even though the size of its workforce — 21,459 employees in 2013, not counting thousands of contractors — was disclosed in the "black budget" leaked last year by former National Security Agency contractor Edward Snowden.
The message to employees on harassment, which CIA officials said was the first of its kind, said 15 out of 69 complaints in the 12 months ending Sept. 30, 2013, were found to be true.
In the interest of "transparency," the message said, officials shared summaries of four examples involving three unidentified CIA employees and a contractor:
A supervisor who engaged in bullying, hostile behavior and retaliatory management techniques was removed from the job, given a letter of reprimand, and ordered to undergo leadership and harassment training.
A male officer who sexually harassed female colleagues at an overseas post was sent back to the U.S. and given a letter of counseling and mandatory harassment training.
An employee who used a racial slur and threatened a contractor was given a letter of reprimand.
A contractor who groped a woman was removed from his tour and "reviewed for possible termination."
In response to the memo, CIA officials acknowledged, many employees complained that none of the government employees involved were fired or demoted.
The CIA officials said the idea was to deter the behavior, not punish the offenders.
The officials declined to name the disciplined employees or describe their jobs. One recent disciplinary action was not included in the examples, officials said: Jonathan Bank, the CIA's director of Iran operations, who was removed from his post at headquarters in March after it was found he created a hostile work environment that caused morale to plummet. He is now assigned to the Pentagon.
Many large organizations grapple with workplace harassment, but the CIA faces some unique challenges. For example, the agency, which trains its case officers to manipulate people and lead secret lives, had for years been a place where trysts between managers and subordinates were common, former CIA officials say. And since most of the agency's business is conducted in secret, there has been almost no public accountability for misconduct by senior officials, as there has been in the military.
In 2010, a senior clandestine service manager was forced to quietly retire after he had an affair with a female subordinate. But that was because her husband complained to Leon E. Panetta, then the CIA director, said two former officials who refused to be named because they could lose their security clearances for discussing internal CIA matters. Other similar workplace relationships resulted in no action, they said.
In 2012, then-CIA director David Petraeus sent a message to agency staff members outlining a new effort to curb sexual harassment in war zones, where CIA men and women often live in close quarters under stressful conditions. Petraeus himself later admitted he was having an affair with his biographer and resigned his post.
The agency has faced complaints of gender bias in the past. In 2007, a group of female officers filed a class-action complaint with the federal Equal Employment Opportunity Commission, alleging that women who had affairs with foreigners were treated more harshly than their male counterparts. An EEOC judge dismissed the case, however, on the grounds that there were not enough women in the class. The women pursued their cases separately, and some were paid settlements, said former CIA officer Janine Brookner, the lawyer who brought the case.
In 1995, the agency paid $990,000 to settle a class-action lawsuit by 450 women. The settlement included promotions, raises and better assignments for about 100 female officers.
Neither the CIA nor its National Clandestine Service has ever been headed by a woman. CIA officials point out that the agency now has a female No. 2, deputy director Avril D. Haines. Another woman, Fran P. Moore, is director of intelligence, the agency's analytical arm. Female analysts also played a key role in the effort to find Osama bin Laden.
The post CIA cracks down on employee harassment; internal postings call actions insufficient appeared first on PBS NewsHour.
06/10/14--13:32: Twitter expects World Cup 2014 to be most tweeted event ever
Photo by Wikimedia user YarikUkraine
When the 2014 World Cup kicks off June 12, it may be scoring a new goal for Twitter.
Guilherme Ribenboim, Brazil's general director for Twitter, said Tuesday that the microblogging company expects that the month-long international soccer competition will easily be the most tweeted event in the website's history, eclipsing the 150 million tweets sent during the 2012 London Olympic Games.
"The World Cup will take all that to a whole different level," Ribenboim said, in an interview with Reuters.
In addition to viewer tweets, the athletes themselves will also be dribbling into the social media field. Twitter claims to be working with soccer players and organizations, telling them how to use the website and engage with the fans.
According to FIFA, the governing body that organizes and oversees international soccer, 3.2 billion people watched at least part of the 2010 World Cup. Twitter's user base currently sits at 255 users worldwide.
The post Twitter expects World Cup 2014 to be most tweeted event ever appeared first on PBS NewsHour.
06/10/14--13:54: House approves bill that allows veterans facing medical delays to seek outside care
Pedestrians walk past the U.S. Department of Veterans Affairs (VA) headquarters in Washington, D.C. Photo by Andrew Harrer/Bloomberg via Getty Images
WASHINGTON — United and eager to respond to a national uproar, the House overwhelmingly approved legislation Tuesday to make it easier for patients enduring long waits for care at Veterans Affairs facilities to get VA-paid treatment from local doctors. Lawmakers were so keen to vote for the bill, they did it twice.
The 426-0 final vote was Congress' strongest response yet to the outcry over backlogs and falsified data at the beleaguered agency. Senate leaders plan debate soon on a similar, broader package that has also drawn bipartisan support, underscoring how politically toxic it could be to be seen as ignoring the problem.
House members didn't want to be left out of their roll call. An unusual second vote, superseding the chamber's 421-0 passage of the bill barely an hour earlier, was taken after a handful of lawmakers missed the first one. They included Veterans Affairs Committee Chairman Jeff Miller, R-Fla., the bill's author, who said he had been in his office.
The VA, which serves almost 9 million veterans, has been reeling from mounting evidence that workers fabricated statistics on patients' waits for medical appointments in an effort to mask frequent, long delays. A VA audit this week showed that more than 57,000 new applicants for care have had to wait at least three months for initial appointments.
Reverse incentives of VA health care made fixing the numbers easier than fixing the system
More than one in 10 VA schedulers report they were told to falsify records
"I cannot state it strongly enough – this is a national disgrace," Miller said during the debate.
"We often hear that the care that veterans receive at the VA facilities is second to none – that is, if you can get in," said Rep. Mike Michaud of Maine, top Democrat on the committee. "As we have recently learned, tens of thousands of veterans are not getting in."
The controversy led Eric Shinseki to resign as head of the VA on May 30, but the situation remains a continuing embarrassment for President Barack Obama and a potential political liability for congressional Democrats seeking re-election in November.
Monday night, a top VA official told the veterans committee that there is "an integrity issue here among some of our leaders."
Philip Matkovsky, who helps oversee the VA's administrative operations, said of patients' long waits and efforts to hide them, "It is irresponsible, it is indefensible, and it is unacceptable. I apologize to our veterans, their families and their loved ones."
Matkovsky did not specify which VA officials had questionable integrity. The agency has started removing top officials at its medical facility in Phoenix, a focal point of the department's problems, and investigators have found indications of long waits and falsified records of patients' appointments at many other facilities.
Richard Griffin, acting VA inspector general, told lawmakers his investigators were probing for wrongdoing at 69 agency medical facilities, up from 42 two weeks ago. He said he has discussed evidence of manipulated data with the Justice Department, which he said was still considering whether crimes occurred.
"Once somebody loses his job or gets criminally charged, it will no longer be a game and that will be the shot heard around the system," Griffin said.
The VA drew intensified public attention two months ago with reports of patients dying while awaiting agency care and of cover-ups at the Phoenix center. The VA, the country's largest health care provider, serves almost 9 million veterans.
The House bill would let veterans facing long delays for appointments or living more than 40 miles from a VA facility choose to get care from non-agency providers for the next two years. Some veterans already receive outside care, but the bill would require the VA to provide it for veterans enduring delays or who live far away.
In Chicago, the American Medical Association added its voice, urging President Barack Obama to take immediate action to enable veterans to get timely access to care from outside the VA system. The nation's largest doctors group also recommended that state medical societies create and make available registries of outside physicians willing to treat vets.
VA performance bonuses have also been an issue in recent disclosures. And the House bill would ban bonuses for all VA employees through 2016 and require an independent audit of agency health care. An earlier House-passed bill would make it easier to fire top VA officials.
Miller said VA would save $400 million annually by eliminating bonuses, money the agency could use for expanded care.
Senators have written a similar bill, which Senate Majority Leader Harry Reid, D-Nev., said his chamber would consider "as soon as it is ready."
Senate Minority Leader Mitch McConnell, R-Ky., said the chamber should debate the bill immediately, instead of first considering a Democratic measure letting borrowers refinance student loans at lower rates.
"Veterans have been made to wait long enough at these hospitals," McConnell said.
On Monday, the VA released an internal audit showing more than 57,000 new patients had to wait at least three months for initial appointments. It also found that over the past decade, nearly 64,000 newly enrolled veterans requesting appointments never got them, though it was unclear how many still wanted VA care.
The audit covered 731 VA medical facilities. It said 13 percent of scheduling employees said they'd been instructed to enter falsified appointment dates, and 8 percent used unofficial appointment lists, both practices aimed at improving agency statistics on patient wait times.
As a result, the agency said it was ordering further investigations at 112 locations where interviews revealed indications of fabricated scheduling data or of supervisors ordering falsified lists.
Sloan Gibson, the acting VA secretary, directed several steps to address Monday's audit, including a short-term boost in medical services at overburdened facilities, including using mobile units.
The agency has contacted 50,000 veterans awaiting appointments and plans to reach 40,000 others to accelerate care, letting them choose VA treatment or local non-VA health care providers.
Associated Press writer Matthew Daly contributed to this report from Washington, Medical Writer Lindsey Tanner from Chicago.
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06/10/14--14:45: Gun control failure has been biggest frustration as president, Obama says
Photo by George Frey/Bloomberg via Getty Images
WASHINGTON — President Barack Obama says his biggest frustration so far as president is that American society hasn't been willing to take steps to strengthen gun control.
Obama is reflecting on frequent mass shootings during a question-and-answer session on social media site Tumblr. He says the U.S. should be ashamed.
Obama says there's no place else in the world where mass shootings are a once-a-week occurrence. He says the country must do some soul-searching.
Obama says he respects gun rights. But he says he was stunned after the Newtown school shooting that Washington couldn't even pass universal background checks. He's citing the political strength of the National Rifle Association and gun manufacturers.
Obama says until there's public opinion a fundamental shifts, the problem won't change. He says as a parent, that's terrifying.
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06/10/14--14:48: Surge of children apprehended at border overwhelms ill-equipped facilities
The U.S.-Mexico border fence in Nogales, Ariz. on June 27, 2009. Photo by Flickr user Ryan Bavetta
PHOENIX — The federal government is scrambling to house a surge of unaccompanied Central American children and teenagers apprehended crossing the border illegally, many in Texas' Rio Grande Valley.
Obama administration officials said Monday federal agencies are requesting more than $2 billion from Congress to pay for more shelters.
Unaccompanied migrants under the age of 18 are only supposed to be held in Department of Homeland Security facilities for fewer than 72 hours before they're transferred to the custody of the Department of Health and Human Services. They are supposed to be housed in shelters run by that agency's Office of Refugee Resettlement.
But a shortage of space in those shelters means the children have been languishing in facilities that are not equipped for them.
In the last eight months, border agents made more than 47,000 child apprehensions. That is more than a 90 percent increase from the same time period last year.
Many of these young migrants are fleeing violence and gangs in Central America. Some have heard rumors that U.S. immigration policy is lenient for children who cross alone.
Obama administration addresses influx of child migrants
Emergency facility for child migrants opens on Arizona border
El Paso shelters overflow of immigrant families
The Obama administration has called this an "urgent humanitarian situation requiring a unified and coordinated federal response."
Back in late May, before this issue became known publicly as a presidential priority, a 24-year-old Honduran migrant named Marleny Bueso Ponce was detained in a Border Patrol station in Arizona. There she met a boy who had been caught at the border without his parents.
"He was crying that he wanted to phone his mother, to tell her that he loved her, that he missed her," Bueso Ponce said.
Bueso Ponce and her own child were paroled after a couple days, as has been typical for families apprehended at the border. But she says the boy stayed behind at the holding cell when she left.
She says the boy told her he'd already been held there for 11 days.
That would be a violation of a federal statute that says unaccompanied children must be transferred out of such DHS facilities within 72 hours.
Bueso Ponce's account of the boy's story can't be verified. But Obama administration officials say cases like this one are happening. On Monday they acknowledged on a press call that migrant children have been held in short-term DHS facilities for longer than three days.
The officials spoke to reporters on the condition of anonymity.
"Border Patrol stations were not designed for any kind of long-term custody," said Michelle Brané of the Women's Refugee Commission in Washington, D.C. "They are completely ill-equipped to deal with anybody long-term, and they are particularly inappropriate for children to be in for any length of time."
Brané said such facilities have no showers, beds or recreation areas.
"And not having a shower, for example, we have been hearing kids have been in facilities for up to two weeks," Brané said. "That is a very long time to be in the same clothes you have traveled in and crossed rivers in."
As the number of unaccompanied children crossing the border has surged, there hasn't been enough space for them in HHS shelters. The federal government has been trying to set up more, but Brané says that doesn't happen overnight.
"You need to find beds quickly but you also want them to safe," Brané said. "So that is where I think this bottleneck is coming. Places need to be licensed. They need to have the appropriate staff caring for them. They need to have protection mechanisms for the children."
Obama administration officials said, HHS is requesting an additional $2 million from Congress for this effort. DHS is also requesting an additional $160 million.
In the meantime, the federal government has been setting up emergency housing for child migrants on military bases in California, Texas and Oklahoma.
Over the weekend, federal officials started adding showers and other amenities at a processing center in Nogales, Arizona. It will serve as a way-station for up to 1,500 children at a time, before they're transferred to more permanent sites.
Tony Banegas, the honorary Honduran consul in Arizona, visited the Nogales site over the weekend. Many of the children there had been flown in from South Texas, where they were apprehended.
Banegas said the conditions there were still a work in progress.
"They need mattresses, they need toothpaste," Banegas said. "Better food, warm food."
Banegas said he is grateful for the effort federal agents are making. But he says it is still very difficult for the children who are housed there.
"Some are young and they miss their family, they don't know what is going to happen," he said. "It's scary."
Eventually these children will be transferred to a shelter, and then may be reunited with family members or placed in foster care.
They're still in deportation proceedings, though. So they'll either be ordered to return to their home countries, or win the right to stay in the U.S.
Obama administration officials said they had planned for an increase in unaccompanied migrant children this year, but were caught by surprise by the size of the increase.
Central American women caught crossing the border with children are also overwhelming federal facilities in South Texas. Community shelters and churches in El Paso are helping to house these families after they are released from border facilities.
This story was reported by the Fronteras: Changing Americas Desk, a multimedia collaboration among seven public radio stations. It is led by KJZZ in Phoenix and KPBS in San Diego and funded by the Corporation for Public Broadcasting as part of its Local Journalism Center initiative.
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06/10/14--15:02: News Wrap: Investigation launched into friendly fire deaths of five U.S. soldiers in Afghanistan
JUDY WOODRUFF: The international coalition in Afghanistan launched an investigation today after five U.S. special operations soldiers were killed, apparently by friendly-fire.
It happened in Zabul province in the south, and may be one of the worst such incidents in almost 14 years of the Afghan war.
For more, I spoke earlier with NPR reporter Sean Carberry, who's in Kabul.
Sean Carberry, thank you very much for talking with us.
What is known exactly about what happened?
SEAN CARBERRY, NPR: Well, U.S. special operations forces and Afghan forces were carrying out a clearing operation in Zabul province, which is one of the more unsecured provinces in the country.
And according to Afghan officials, when the troops were on their way back to the base from this operation, it came under attack by Taliban militants. At that point, they called in for air support, and the airstrike apparently hit the friendly forces, killing five U.S. troops and one Afghan force.
NATO and U.S. officials have not officially confirmed that it was a friendly-fire incident. They have indicated they're investigating that. However, Afghan officials said the airstrike hit the friendly forces, again, killing five U.S. and one Afghan troops.
JUDY WOODRUFF: So it's not confirmed, but it sounds like the evidence points to friendly-fire. Is it known what type aircraft?
SEAN CARBERRY: Reports are that it was a B-1 bomber, so this is not an Apache helicopter or a gunship or something like that. So, this was most likely heavy munitions, a heavy bomb that would have been dropped in this instance.
JUDY WOODRUFF: Sean, we're just a few days away from the presidential election runoff there in Afghanistan. What is the security situation there overall? You're in Kabul.
SEAN CARBERRY: Well, security has tightened.
In the last few days, we have seen an increase in the number of checkpoints around the city. Security officials have said that they're now on essentially high alert going into the elections. In terms of the security incidents, there haven't been as many as people have been expecting. There have been fewer attacks than there were in the run-up to the first round of voting in April.
However, last weekend, militants did carry out a suicide attack against presidential candidate Abdullah Abdullah. He and his entourage survived. But that's the most high-profile attack we have seen, but officials are expecting more violence for election day this year because it's coming in peak fighting season, whereas the first round was at the tail end of spring and, at that point, the thaw hadn't happened. Fighters weren't as active at that point.
JUDY WOODRUFF: Sean Carberry, we thank you very much, talking to us from Kabul.
SEAN CARBERRY: You're welcome, Judy.
JUDY WOODRUFF: U.S. officials in Washington declined comment on the incident, saying they are waiting for the investigation.
GWEN IFILL: Gunmen in Pakistan attacked a police training facility today near the Karachi Airport. It followed Sunday night's Taliban assault that killed 26 people at the airport itself.
Today's incident forced a temporary suspension of flights and triggered a brief shoot-out with security forces. But airport officials downplayed its severity.
AZAM KHAN, Director General, Airport Security Force (through interpreter): The wrong word was used, that there was an attack. There was no such situation. There was a firing incident, which was within our capability to manage. However, the word attack was used by the media, which created panic. You saw the response, police, rangers, army. Everyone was immediately here.
GWEN IFILL: The Taliban said it was also behind today's attack and warned of more violence to come.
JUDY WOODRUFF: In Syria, infighting between an al-Qaida-linked group and other rebel factions has taken a heavy toll in recent weeks. The Syrian Observatory for Human Rights reported today more than 630 people have been killed in the east, near the Iraqi border, since the end of April. At least 130,000 others have fled the region.
GWEN IFILL: There was more fallout in Congress today over that prisoner swap that freed Army Sergeant Bowe Bergdahl in Afghanistan. The Senate's number two Democrat, Dick Durbin of Illinois, said there was no time to notify lawmakers in advance because the deal to free five Guantanamo detainees in exchange for Bergdahl was finalized just one day before it happened.
But Republican Jeff Sessions rejected that reasoning after a closed briefing by defense officials.
SEN. JEFF SESSIONS, R, Ala.: I'm not satisfied in any way that Congress shouldn't have been consulted in this matter. Just as a matter of courtesy, whether it was in law or not, a matter of this importance should have been discussed with at least key leaders in the Congress.
GWEN IFILL: Democratic Senator Carl Levin of Michigan came out of that same briefing, saying military leaders supported the prisoner swap.
SEN. CARL LEVIN, D, Mich.: Armed Services Committee Chairman: What the media has not carried, I think, is the critical question: Do our top military uniformed leaders support this agreement? Did they — were they involved in it? They assured me they were. And did they support it? They assure me they strongly support it because of the ethos of getting our people back.
GWEN IFILL: The House begins hearings on the Bergdahl release tomorrow. Defense Secretary Chuck Hagel will testify before the Armed Services Committee.
JUDY WOODRUFF: The House has voted unanimously to make it easier for veterans to get in to see doctors. The bill that passed today would pay for vets to get care outside the Veterans Affairs system, if they have had long waits or if they live more than 40 miles from a VA hospital. A similar bill is pending in the Senate.
GWEN IFILL: Crew members of the South Korea ferry that sank in April appeared in court today at an emotional hearing. Relatives of the more than 300 who died in the sinking wrestled with officials and packed the courtroom. Some shouted "murderer" when the captain entered. The 15 crew members face charges ranging from negligence to homicide for abandoning the ship.
JUDY WOODRUFF: And this is primary election day for six more states. Republican Senator Lindsey Graham of South Carolina and House Majority Leader Eric Cantor of Virginia are both expected to hold off Tea Party challengers. Primaries are also scheduled in Maine, Nevada, and North Dakota; and Arkansas is holding a series of runoffs.
GWEN IFILL: Wall Street had a relatively quiet day. The Dow Jones industrial average gained not quite three points to close near 16,946. The Nasdaq rose more than one point to close at 4,338. And the S&P 500 slipped half-a-point to end at 1,950.
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06/10/14--15:10: In major blow to Maliki government, Sunni militants force Iraqi army out of Mosul
JUDY WOODRUFF: In a major blow to the government of Iraqi Prime Minister Nouri al-Maliki, and to that country's stability, Sunni militants have taken over Iraq's second largest city.
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Fighters from the Islamic State of Iraq and the Levant pushed Iraqi army units from parts of Mosul overnight after days of fighting. The largely Sunni-Muslim city in the north is a strategic hub for Iraq's oil industry, as well as a gateway to Syria.
The militants, also known as ISIS, or ISIL, captured military depots, equipment and weapons in Mosul. They also seized provincial government headquarters and freed more than 1,000 prisoners. Thousands of residents fled north toward the Kurdish autonomous region, jamming roads. Some were Iraqi soldiers, who left their uniforms in the streets.
In Baghdad, newly-reelected Prime Minister Nouri al-Maliki addressed the most serious challenge yet in his eight-year tenure.
NOURI AL-MALIKI, Prime Minister, Iraq (through interpreter): I call upon the parliament to live up to its responsibility to declare a state of emergency and general mobilization. We have to declare a comprehensive mobilization and the highest alert in political, financial and popular capabilities to defeat terrorism and bring life to normal in all areas occupied by terrorists, either in Mosul or any other city.
JUDY WOODRUFF: But Maliki's Shiite-led government has largely failed in reconciling with Iraq's Sunni population. The Islamic State has taken advantage of the breach. The Sunni extremist group previously took over Fallujah and parts of Ramadi in western and central Iraq.
It's also a principal combatant in Syria's civil war, but has fought against other rebel groups as fiercely as many of its units have fought against the army of President Assad. The group's ambitions there have led to a rupture with al-Qaida's core organization, which sides with the Syrian rebellion.
The attack on Mosul now threatens to draw nearby Kurdish forces into the fighting as well.
That drew this reaction from the U.S. State Department today.
JEN PSAKI, State Department Press Secretary: The threat ISIL is presenting is not just threat to Iraq or the stability of Iraq, but it is a threat to the region. And this growing menace exemplifies the importance of Iraqis from all communities working together to confront this common enemy.
JUDY WOODRUFF: Iraq's parliament has announced it will meet Thursday to decide on a state of emergency.
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06/10/14--15:12: How did Sunni insurgents gain momentum in Iraq?
JUDY WOODRUFF: For more on what this takeover means for Iraq, for the region and beyond, I'm joined by Laith Kubba. He's senior director for the Middle East and North Africa at the National Endowment for Democracy. And Kimberly Kagan, she's president of the Institute for the Study of War.
And we welcome you both.
So, Kimberly Kagan, let me start with you.
We know these insurgents have been creating havoc for some time, launching attacks. How important is this particular attack, taking over the city of Mosul?
KIMBERLY KAGAN, Institute for the Study of War: This attack that the Islamic State of Iraq and Sham, or the Levant, has launched on Muslims is incredibly important, because it is the beginning of a campaign and a push beyond Mosul into the areas toward Baghdad that the Islamic State of Iraq wants to govern.
It seeks to establish an emirate or a state and govern terrain inside Iraq, as well as governing the terrain inside of Syria, in Raqqa, where it has announced the beginning of its emirate. I believe Mosul will be its new capital.
JUDY WOODRUFF: Laith Kubba, how do you see it? And why is this happening now? How is it that this insurgent group has gotten to this level of where they can wreak this kind of chaos in the country?
LAITH KUBBA, National Endowment for Democracy: Well, over the last two years, I think they have been growing steadily.
All the signs were there. Nobody wanted to read them. They moved from being an offshoot, a terrorist group — and there, people might think we can live with terrorist groups and the skirmishes they create, but this has become an army of 10,000 to 15,000, very well-equipped with rocket launchers, some air missile — missiles, and they are so coordinated
I think they have become a magnet for more soldiers to join them. This latest attack, not only one attack — it's coordinated attacks on five cities in Iraq — has given them immense momentum and credibility, and they're going to become a serious threat to the region.
I'm not really sure if they can hold territory for long, but certainly they have achieved their objective in saying, we're a force, and I think hundreds will join them, and this is going to become a regional problem.
JUDY WOODRUFF: Kimberly Kagan, how did they get to this to have this capability? Because it wasn't the case when the U.S. pulled out of Iraq.
KIMBERLY KAGAN: The Islamic State of Iraq is no longer a terrorist group. It is an army.
It is an insurgency, and it has grown in its capabilities inside of Iraq by really fielding an army, by freeing prisoners that were held in Iraqi prisons, by testing the Iraqi security forces, by taking control piecemeal.
And if we try to confront it as a terrorist organization, we will misunderstand its nature. It is an insurgency. It is fighting for terrain, and it has really come of importance as Prime Minister Maliki started to exclude Sunnis from his government and create resentment among the Sunni population in Iraq.
JUDY WOODRUFF: And that's what I want to ask you about, Laith Kubba. What is the role of the Maliki government, in that this is happening under their noses?
LAITH KUBBA: I think there are two elements here.
There is one element — the emergence of ISIS is very much a product of what is going on in Syria. But I think…
LAITH KUBBA: But I think the failure of politics in Baghdad and the failure certainly of the Iraqi army is a direct result of what's going on in Baghdad.
Bear in mind, the prime minister is the commander in chief. He's been prime minister for eight years. Iraq is an oil-producing country. And its army could not stand their ground in front of hundreds of attackers? Just think of a country that managed to keep Iran at check for eight years, where the same country now is not capable of keeping an insurgency at check?
I think, for Iraqis, this is an evidence that something is fundamentally wrong in the way their country is governed. Even the electoral process that is repeated over and over again is not producing a good government.
JUDY WOODRUFF: Kimberly Kagan, what's to stop them, ISIS, or ISIL, depending on what you call them, from just going — doing what they want and taking over any territory they want?
KIMBERLY KAGAN: I think that the organization has momentum.
The organization has aimed for a year to dissolve the Iraqi security forces and has named a campaign, the soldiers' hardest campaign, that it's been conducting for a year with that goal. I think that the Iraqi security forces are breaking and will continue to break, even though Prime Minister Maliki has declared a state of emergency.
And the only question now is, are they able really to defend Baghdad and its environs? And will they be able to mobilize, as they already have, Shia militants grouped backed by Iran and trained by Hezbollah that have fought in Syria in order to defend the Shia areas of Baghdad and south of Baghdad that Maliki relies on?
JUDY WOODRUFF: Is that what we are looking for, some kind of showdown around Baghdad?
LAITH KUBBA: Well, just to emphasize the point, Iraq, Iraqis expect a state with an army, not militias that are sponsored or sent to Syria to fight or called upon to defend a city.
Iraq should have an army, and I think it's a sign of how far things have gone to see militias roaming trying to defend cities.
JUDY WOODRUFF: Less than a minute, but both of you — Laith Kubba, to you first.
Effect on the region? Why does this matter in the broader region? You have got the civil war going on right next door in Syria. The effect on countries in the area?
LAITH KUBBA: The fact that this army now controls territory, so well-equipped and so capable and with momentum, I would worry about Jordan, because that's a very soft front. They can push there any time. And this group now is increasing in numbers and sophistication.
They're going to be there for a long time.
JUDY WOODRUFF: And, again, the same question I asked a minute ago, what's to stop them from just marching and going where they want?
KIMBERLY KAGAN: Well, I certainly hope that the Iraqi security forces can stop them, but, realistically, what we see now is a safe haven that has developed in Iraq and Syria from which Islamist militants can both launch attacks and train foreign fighters and send them out, and also govern terrain and oppress people.
That is precisely the kind of safe haven that this administration and its predecessor have stated that the United States will not tolerate in the region.
JUDY WOODRUFF: And that of course raises questions that we will be continuing to look at and want to ask about in the days to come.
Kimberly Kagan, Laith Kubba, we thank you both.
LAITH KUBBA: Thank you.
KIMBERLY KAGAN: Thank you.
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06/10/14--15:20: How 'Hard Choices' may help Clinton define herself ahead of possible presidential run
GWEN IFILL: Former Secretary of State Hillary Clinton headed back out on the trail today, not for a political campaign, but to support her new memoir.
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With a hopscotch schedule of media appearances, political observers are scanning the book, and her interviews, for 2016 tea leaves.
The book tour formally began this morning, with the author arriving at a Barnes & Noble in New York to applause and an army of cameras. But the buildup started weeks ago.
HILLARY RODHAM CLINTON, Former Secretary of State: It's really about the hard choices everybody has to make in life.
GWEN IFILL: After days of leaks and secretly purchased advance copies, Hillary Clinton's second memoir, "Hard Choices," is now officially on the bookshelves.
In New York today, book buyers lined up to meet the former first lady, senator and secretary of state. The carefully-orchestrated rollout only stokes speculation that she is launching a 2016 presidential practice run. She's on the cover of "People" magazine. She's made high-profile appearances at a number of recent forums.
And she's sitting down for a slew of television interviews, starting last night with ABC News' Diane Sawyer.
DIANE SAWYER: When are you going to decide whether you are running for president?
HILLARY CLINTON: You know, I'm going to decide when it feels right for me to decide, because…
DIANE SAWYER: Still by the end of this year?
HILLARY CLINTON: Well, certainly not before then.
I just kind of want to get through this year, travel around the country, sign books, help in the midterm elections in the fall, and then take a deep breath, and kind of go through my pluses and minuses about what I will and will not be thinking about as I make the decision.
GWEN IFILL: In that same interview, Clinton said she made millions in paid speeches because she and former President Bill Clinton were dead broke when they left the White House in 2001.
After Republicans said this proves Clinton is now out of touch with average Americans, she returned to ABC this morning to defend herself.
HILLARY CLINTON: I fully appreciate how hard life is for so many Americans today. It's an issue that I have worked on and cared about my entire adult life.
Bill and I were obviously blessed. We worked hard for everything we got in our lives, and we have continued to work hard and we have been blessed in the last 14 years. But I want to use the talents and resources I have to make sure other people get the same chances.
GWEN IFILL: Early reviews have been decidedly mixed.
The New York Times called it a "subtle, finely calibrated work." The Washington Post described it as "a careful book," and Slate said Clinton goes on at great length, but not great depth. Part travelogue and party policy treatise, "Hard Choices" clocks in at 600 pages, focusing largely on Clinton's time as secretary of state.
It includes photos from Clinton's time campaigning for President Obama, working with Vice President Biden, who is also considering a 2016 run himself, as well as her meetings with world leaders from Africa to Asia.
For more on the rollout of Mrs. Clinton's memoir and what it might tell us about her future in politics, we are joined by Ann Lewis, a longtime adviser to both Hillary and President Bill Clinton, Chris Lehane, a Democratic strategist who worked in the Clinton administration, and Amy Chozick, national political reporter for The New York Times who is covering Mrs. Clinton.
Amy Chozick, words that come to mind are orchestrated, calibrated, structured, this rollout. How organized is it?
AMY CHOZICK, The New York Times: Absolutely.
I think it was incredibly well-thought-out, right up until the Mother's Day excerpt about Hillary Clinton's mother, Dorothy Rodham, that was kind of our introduction to the book, and also showed a very softer side of her, I would say, and it just continued from then, the "People" magazine cover, right up to today, the event at Barnes & Noble in New York.
GWEN IFILL: What was that like? Was it crazy?
AMY CHOZICK: It was insane. There were 1,000 people outside. A lot of people had slept out front ready for Hillary. The outside group had a giant bus that looked like a campaign bus signing people up.
I talked to a woman who pulled her 11-year-old daughter out of school to attend. So it definitely had the sort of feel and enthusiasm of a campaign. That said, of course, this was the middle of downtown Manhattan, so not exactly a proxy for a nationwide presidential campaign.
GWEN IFILL: Ann Lewis, you have been involved in Ready for Hillary since it kicked off. Ready for what, exactly?
ANN LEWIS, Former Clinton Adviser: Whatever Hillary Clinton decides to do.
I think Ready for Hillary, which now has about two million members or more, growing every day, said, here's this wonderful public servant. And anybody who reads the book is going to see how much Hillary Clinton cares and policy, about doing the right thing, how proud she is of our country.
We want her to go as far as she would like to go. And we want her to know that, if she wants to go further, we will be there.
GWEN IFILL: Is it just me, or does this seem to be much more carefully rolled out, structured, not — if not in collaboration with Ready for Hillary, but certainly much more — done with more of an eye of the political future than the last book rollout in 2008?
ANN LEWIS: Well, the last book, after all, was in a very different climate.
And if you go back and look — and, by the way, a different media landscape. Look at all the ways we're talking now about the book that didn't exist for the last time. There was no Twitter at the time, right, that says Hillary's book.
So you got — there's a very high-tech, very sophisticated kind of conversations that are going on throughout I guess the blogosphere, and then you have also got the low-tech like the bus. Doesn't get much more low-tech than that.
GWEN IFILL: Chris Lehane, you have a candidate here, presumably, or potentially a candidate, with 100 percent name recognition.
How do you handle — if this is indeed a political rollout as well as a literary one, how do you handle that differently from any other candidate?
CHRIS LEHANE, Democratic Strategist: Yes, great question. I'm not sure whether the book will or will not be the political equivalent of "War and Peace," but it's certainly been handled about as well as you could, at a "War and Peace" level.
And by that, I mean, what I think is really smart, and I think probably informed a little bit by Hillary Clinton's experience in the 2008 presidential campaign, if this is really an effort to control the narrative, define herself on her own terms.
And even the timing of this, right — typically, presidential aspirant books, if that is indeed what this may or may not be, typically those happen a year out. This is happening several years out, goes to the fact — I think Ann was alluding to this — that we live in a perpetual campaign world.
And it's really imperative to control, own, drive your narrative, your profile, your character definition. And even down to the title of this book and the content, she's doing that, and I think it's a smart strategy and the execution has been extremely effective.
GWEN IFILL: Amy Chozick, but the execution also includes getting in front of interviewers and answering questions in ways that might kind of take you off topic.
Like, she was asked last night about Monica Lewinsky. She's been asked Benghazi and she made the comment to Diane Sawyer that she and Bill Clinton needed to earn that money in order to pay for their houses and their mortgages, which Republicans jumped on. So there's also a potential for some slip-ups here.
AMY CHOZICK: Yes, absolutely.
And I was actually — I actually welcomed the slip-ups, because I thought that signaled that if she's a candidate in 2016, maybe she won't be so scripted and everything that she says is polled, to figure out how Americans feel about it before she utters every line.
I kind of found the gaffe sort of refreshing in a way, at least as a political reporter.
GWEN IFILL: So, Ann Lewis, you have been through a lot of this with the Clintons. And I wonder if part of this isn't also trying to define yourself before everybody races in and defines you on just the kinds of issues that we were just talking — that have now been described as her gaffes.
ANN LEWIS: Yes, good point.
Let me say, I think the first part of it is Hillary Clinton talking about the last four years. As secretary of state, she was really America's ambassador to the war. She wants people to know why that is so important. We need to be leaders. Here's what's at stake. Here are all these countries and comments.
And she wants to do it in her own voice. This is something she cares about very deeply. And people who read the book are going to get that sense from her, both the policy, the substance, but also the personality she brings to it. She can do that better for herself than anybody else can talking about her.
GWEN IFILL: Chris Lehane, you talked about the title of the book, "Hard Choices."
One of the things she said today, yesterday — there have been so many interviews, it's hard to keep track, but she said hard choices are what presidents do. That's the closest she's come, it seems to me, in saying that "Hard Choices" is about considering the presidency.
CHRIS LEHANE: Yes, well, as we were talking about, right, that sometimes you're not supposed to judge a book by its title, but I think in these types of books, there's an awful lot to the title, because ultimately what you're trying to do is to give folks a sense of who you are as a person.
And particularly if you end up being a presidential candidate — and she obviously has been one before and has a lot of experience — that sort of breaks into two categories. One is your character, your personality, and the other is your vision for the country, and I think the title "Hard Choices" sort of does a good job of encapsulating both.
By the way, it would fit very nicely on a bumper sticker. And I think, again, it is a smart way to package all of this, and at the end of the day, she is talking about the fact that when you are a president or you are secretary of state or you are in the U.S. Senate, that you do have to make hard choices.
That is part of being a leader. And again I think this goes to both elements, the character, personality, who she is, but also that vision for the country. And, you know, based on what I have seen from the book, she is talking retrospectively about her four years, but extrapolating from that in terms of what she cares about and how potentially she would look at issues prospectively. So I think it does a very good job of both.
GWEN IFILL: Chris, I want to ask you this, and Amy as well.
You have both talked to enough people inside Hillary-land to know, to answer this. Is there anybody who doesn't think that she's running?
First, Chris, then Amy.
CHRIS LEHANE: Well, I think that, every single signal and sign out there would certainly suggest that she's doing everything possible to make sure that she's in a position to run.
Obviously, I defer to her. It's her choice, and I think we all recognize it and respect that. But this book — I mean, this book, to me, is a prism by which you can evaluate the process and the approaches that she's taking. And it's certainly indicative of someone who's doing everything possible to put themselves in the strongest position to be able to say, yes, indeed, I am a candidate for president in 2016.
GWEN IFILL: Amy, as you do your interviews and you talk to people, and you devote so much time to tracking her footprints, anybody say, no, she's not going to do it?
AMY CHOZICK: A source yesterday told me that it now looks like 99 percent sure — that this person was sure that she was running.
Another source told me that, the duller the book, the more chance that she's running. So read into that what you will. But I do think that there is thinking that the book was cautious, so that she could leave her options open.
Of course, there are friends of hers, Cheryl Mills being a key one, who have said that they don't want her to run, that they worry about the scrutiny of the campaign, and they just want her to kind of enjoy her life.
GWEN IFILL: Does it freeze the field for other Democrats who might be considering it, Amy?
AMY CHOZICK: She — it was interesting, because Clinton addressed this in one of the interviews, and she said that she wasn't worried that it was freezing out the field.
But I definitely do think that the party is anxious for her to make up her mind, and so that they can get behind someone else if for any reason it is not her.
GWEN IFILL: Ann Lewis, what's the hardest choice Hillary Clinton has to make between now and, say, the end of the year?
ANN LEWIS: Well, I think by the end of the year — and she has referred to this — she will probably have to decide whether, in fact, she's going forward.
Can I stop for a moment and disagree? I almost never disagree with Amy. And she's been a great observer on this. But I think people who read this — and a lot of people are going to read this book — will not think it's dull. You know, it is — could more interesting…
AMY CHOZICK: I don't think it's dull. I'm just saying this person said that.
ANN LEWIS: Oh, good. All right.
ANN LEWIS: I just want to get that on the record.
GWEN IFILL: She's leaving open the possibility it may not be exciting.
ANN LEWIS: Oh, I think it is going to be thrilling.
But going beyond that, the hardest choice — and Hillary Clinton has said this — she is right now in, for her, a very unusual place. That is, she can take her time when she gets up in the morning. She can go out and walk. She can spend that time with her husband.
She really, at the same time, has this opportunity to step forward, to be a leader again for the United States. She has to decide that. And, again, nobody should underestimate. That is a very tough choice.
GWEN IFILL: Well, and a lot of people are weighing pro and con to see what she does with that choice.
Ann Lewis, Amy Chozick, Chris Lehane, thank you all very much.
CHRIS LEHANE: Thank you.
AMY CHOZICK: Thanks for having me.
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06/10/14--15:33: Unexpected legislature resignation tips Virginia fight over Medicaid expansion
JUDY WOODRUFF: The battles over whether to expand Medicaid coverage under the federal health care law are still playing out in a number of states this summer, the latest, Virginia, where Republicans in the state Senate grabbed control of that chamber yesterday, and prevented Governor Terry McAuliffe, a Democrat, from opting into the expansion.
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It happened after another Democrat, a state senator, unexpectedly resigned amid claims that he was trading his seat for another position and an appointment for his daughter.
Meanwhile, there are a handful of states, including Indiana, Pennsylvania and Utah, where Republican governors are considering participating in Medicaid expansion, but only with more requirements and restrictions from those who would be covered.
So far, 26 states, plus the District of Columbia, have agreed to expand. The Obama administration says six million people have gained Medicaid coverage. But there may be nearly two million waiting for their applications to be processed.
Julie Rovner, who is now with Kaiser Health News, joins us now.
And welcome back to the program, Julie.
JULIE ROVNER, Kaiser Health News: Nice to be here.
JUDY WOODRUFF: So, why have some states still not made a decision about whether to expand Medicaid?
JULIE ROVNER: Well, of course, they don't have to. In 2012, the Supreme Court said that this was optional. This, of course, was not going to be optional as the health law was passed.
The health law expected and anticipated that every state would expand Medicaid. Congress of course knew when it passed the law in 2010 that states didn't have a lot of extra money lying around to help pay for Medicaid, which is of course a shared program between federal and state governments. Basically, the federal government pays a little more than half of the cost for it.
But the federal government said that for this expansion population, they would pay 100 percent of the cost for the first three years and it would phase down to 90 percent, so it would still be most of the cost going into perpetuity, but when the Supreme Court made it optional, half the states, as you pointed out, jumped in, and the other half, most of them Republican states, have said, we don't even know if we can afford that 10 percent that we would have to kick in after the three years, when it's going to be, you know, fully federally funded.
So, they're still arguing about in the states.
JUDY WOODRUFF: And as we were just pointing out, in one of these states where it's on the cusp, they're trying to figure out what to do, Virginia, with a new Republican governor, having had…
JULIE ROVNER: A new Democratic governor.
JUDY WOODRUFF: I'm sorry, a new Democratic governor, having had a Republican governor — the state legislature, state Senate, now has this switch, a very unusual situation there.
JULIE ROVNER: Yes.
Well, of course, the new Democratic governor ran with this as his number one priority to expand Medicaid, and, of course, it's a Republican House, and it was a split. The Senate, it was 50/50, with, of course, the lieutenant governor being the deciding vote. So this one resignation by this state senator has slipped the state Senate to the Republicans.
And basically there's this budget standoff. And it could have — it was thinking that it was going to close — shut down the state over the budget. Now, of course, it looks like the budget will not have the Medicaid expansion it in, much to the dismay of the Democratic governor.
JUDY WOODRUFF: Terry McAuliffe.
JULIE ROVNER: There is still a possibility — that's right. There is still a possibility that there could be a special session, that Medicaid could come back.
There are a couple of moderate Republicans in the Senate who do support the Medicaid expansion, but certainly it puts it in much more doubt, just with this one expected and unknown-why resignation.
JUDY WOODRUFF: But when — but then you have, Julie, these other states that are trying to figure out, Republican governors, Utah, Pennsylvania, Indiana, just to name a few of them, politics peculiar to each state in play in every one.
JULIE ROVNER: Yes, and we're starting to see kind of a theme with these Republican governors who would like the federal money.
It's a lot of money for a large uninsured population in many of the states. Pennsylvania has perhaps 400,000 people who could be covered. And what many of these Republican governors are doing — we have seen this in Iowa already — is they're saying, OK, we want to expand Medicaid, but we want to do it our way. So we want to have perhaps these people go into private plans. We have already seen that in Arkansas.
We want to have these people perhaps pay a little bit more in cost sharing, so have this low-income population pay something for their coverage. That's not traditionally been allowed in Medicaid. And we want to have perhaps incentives for them to have — do healthy behaviors, maybe stop smoking or lose weight.
So those tend to be the kinds of things they're asking for, and they haven't necessarily got that yet from the federal government.
JUDY WOODRUFF: And what is the pushback that you're hearing from advocates for the poor, who say even a small amount that some of these folks are required to pay can prevent them from having access?
JULIE ROVNER: Well, there's a large body of research that suggests that cost sharing does deter people from getting care and it particularly deters poor people from getting care. And there's a concern that if — that even if you put small amounts of cost sharing on, that people who need care, particularly people with chronic ailments, won't get that care, so it can be counterproductive
And they're very concerned about the idea of having — quote, unquote — "skin in the game" for the Medicaid population in particular.
JUDY WOODRUFF: So while those dramas play out and with very real consequences, just quickly, Julie, this notion that there are several million people who have applied for Medicaid don't have their applications responded to yet.
JULIE ROVNER: That's right. And this is partly because in putting in the Affordable Care Act, every state had to change the way it calculated eligibility for the Medicaid population.
This was to basically make it standardized across all of the states. And so basically, as of January 1, every state had to change the way it calculated eligibility. Also, there was some difficulty in healthcare.gov, the Web site, reporting — basically sending the information to the states and getting it back and forth.
So there is this backlog. Most of it turns out to be in California and Illinois, two big states. This was some work done by my Kaiser Health News colleague Phil Galewitz. But there's also a backlog, it turns out, in some states that didn't expand Medicaid, states like North Carolina, and Georgia. Hopefully, just over the next few weeks, as these states get their I.T. problems straightened out, the backlog will dissipate, if not go away.
JUDY WOODRUFF: Julie Rovner, always on top of it, thank you very much.
JULIE ROVNER: Thank you.
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06/10/14--15:39: Debating tenure protections for public school teachers
GWEN IFILL: Teachers unions lost a major court case in California today, which could make it easier to fire ineffective teachers. A California judge ruled that the state's tenure protections for public school teachers are unconstitutional.
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The case was brought on behalf of and by nine students, who said those policies effectively denied their right to a quality education.
We begin with some background.
Los Angeles Superior Court Judge Rolf Treu struck down three laws, saying they offered teachers job security at the expense of many students.
Elizabeth Vergara and her sister Beatriz are two of the students who brought the complaint. A year apart in age, they attend the same Los Angeles area high school. The sisters told the NewsHour this spring that when they were in middle school, they both sat through a history class taught by the same ineffective teacher:
ELIZABETH VERGARA, Plaintiff: He would just be at his desk, like, just using his computer or sleeping. I didn't even learn anything. Like, I was getting behind.
GWEN IFILL: In 2012, they joined with seven other California students to file suit against the state, saying their education suffered because teacher and tenure laws prevented schools from acting in their best interest.
Teachers are eligible for tenure in California after 18 months. The students sought to get rid of that and other laws, including seniority protections and dismissal procedures they say allow poorly performing teachers to stay in the classroom.
But attorneys for the state and teachers unions told the judge the laws are key to recruiting and retaining a skilled teaching force, and were meant to ensure educators aren't dismissed unfairly.
Fourth grade teacher Kelly Iwamoto considers that due process protection crucial:
KELLY IWAMOTO, Teacher: Because I speak out very frequently about resources being brought to our district, for lowering class sizes. And if I'm vocal, and then someone doesn't like what I'm saying, then I can be let go for that. And I don't think that's fair.
GWEN IFILL: Several other states have either eliminated or limited tenure protection through law or union contracts. Lawyers for the state said they plan to appeal today's ruling.
Now to the decision's potential impact.
Russlynn Ali serves as an advisory board member of Students Matter, which represented the plaintiffs. She is a former assistant secretary for civil rights at the Department of Education. And Joshua Pechthalt is the president of the California Federation of Teachers, one of the key parties who lost in today's ruling.
Joshua Pechthalt, there are six million students in California, 325,000 students. Is this the beginning of the end of teacher tenure?
JOSHUA PECHTHALT, President, California Federation of Teachers: No, I don't think so.
We're going to appeal this. And I think even more importantly, we're going to continue to work with parents and community around a vision for a quality public education that both raises up kids and also values the men and women who work with them, which is the complete opposite of what this ruling attempts to do.
GWEN IFILL: Russlynn Ali, how do we know that it's the teachers who are failing the students and not something else?
RUSSLYNN ALI, Former Assistant Secretary, U.S. Department of Education: Well, what the judge decided today, made it quite clear that, taken together, the statutes that guarantee lifetime employment after 18 months that don't look at factors like whether teachers are actually teaching kids and whether they're learning, statutes that make termination nearly impossible, costing hundreds of thousands of dollars, and statutes that ensure that those most new to the profession are first fired, and any time there needs to be a reduction of force for budget reasons, that, taken together, those statutes violate the quality right to an education guaranteed by California's Constitution.
GWEN IFILL: Joshua Pechthalt, there are those who say that this is about funding. Is that part of the problem?
JOSHUA PECHTHALT: Well, I think funding is — yes, I think funding is a big part of this.
It's funny that we're going to — at least if the Vergara plaintiffs have their way, we are going to shape education policy based on layoffs. The California Federation of Teachers and the California Teachers Association worked very hard to pass a progressive tax measure which stopped massive years of budget cuts.
And more obviously has to be done. But we know the solutions for public education. We know what works. We have districts that are successful. We have schools that are successful. And in those districts and in those schools, it's where the community works collaboratively.
Education is a collaborative endeavor. The slogan it takes a village to educate a child, that happens in a cooperative way, not by attacking teachers and teacher rights.
GWEN IFILL: Can I ask you something from something from the judge's ruling today, in which he talked about the last hired, first fired part of the statute? And he said it was clear to him that keeping a senior and incompetent teacher was preferable to keeping a junior — or to letting go a junior effective teacher. What do you say to that argument?
JOSHUA PECHTHALT: Well, so, number one, we don't want to see any teachers let go during layoffs.
So, that's the most important point. But the reason to have seniority is because seniority creates an objective criteria for determining who's in the classroom. If it's about teacher effectiveness, that's very subjective. And, really, what these folks ultimately want to get at is evaluating teachers based on test scores.
That's the only measure that they are going to be able to come up with when you're assessing thousands and thousands of teachers in a particular district. And we have seen years now of having test scores drive an educational curriculum. It distorts what happens in the classroom. It narrows the curriculum, and it's not good education policy. And these folks are — have the money and resources to follow that agenda.
GWEN IFILL: Let me allow — pardon me, but let me allow Russlynn Ali to respond to that.
Is there another agenda at work here?
RUSSLYNN ALI: The only agenda that the plaintiffs in this case, that the lawyers, that all of us, it's about what is right for students. This is not about pitting students against teachers.
That — Josh, that it would be framed that way, it does a disservice to what we all need to, as California citizens, want to accomplish. And that is to ensure that our students have access to the best and most motivated teachers, that they are inspired to learn, that we transform the way our schools work, so that everyone learns better.
GWEN IFILL: And do you judge that by test scores?
RUSSLYNN ALI: Well, we ought not use quality-blind determinations, right?
This notion that seniority and how long you have been in the profession is all that guarantees whether you stay in the profession or get promoted in the profession is quality-blind. We are not talking about widgets. We are talking about children and their lives. We need to know whether they are learning. That is not to suggest that we ought to use test scores alone, but as part of an evaluation system that gauges true learning in the classroom, we ought to look at the assessments that we give kids.
GWEN IFILL: Joshua Pechthalt…
RUSSLYNN ALI: Joshua also intimated that…
GWEN IFILL: I'm sorry.
RUSSLYNN ALI: Joshua also intimated that we were deciding policy based on layoffs, right?
What happens today is policy decided based on layoffs. The churn factor, that is, that in schools that serve mostly students of color, are some of our lowest-performing students and our highest-poverty schools. They suffer vacancies at a rate that is extraordinary. Those students in Los Angeles, for example, are at 68 percent more likely to have a grossly ineffective teacher if they are English-language learners.
They're almost as likely to have an ineffective teacher if they are African-American students. We are talking about equity. We are talking about justice. Those principles ought not to fly in the face of what is in the best interest of teachers. What is in the best interest of students is in the best interest of teachers. These laws are not.
GWEN IFILL: We have less than 30 seconds, Joshua Pechthalt. What are the prospects for appeal?
JOSHUA PECHTHALT: Well, we are hopeful.
And we think the evidence is compelling that blowing up the education code in California doesn't help with equity. If it — if doing away with seniority and due process rights was so effective, then why is it that in the states where these rights don't exist for teachers, education is also suffering, and it's suffering for poor and working-class kids?
There are obviously other, more compelling reasons that shape public education, not simply the teacher in front of the classroom. We want the best teachers in front of the classroom, and we want to put the resources in to raise everybody up, not create mechanisms for getting rid of people.
GWEN IFILL: We will see what happens with this in the next — now the judge has put a stay on its enforcement.
Joshua Pechthalt and Russlynn Ali, thank you both very much.
JOSHUA PECHTHALT: Thank you very much.
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06/10/14--15:49: Young people learn trade of preservation with hands-on work at Shenandoah
JUDY WOODRUFF: Finally tonight: Across the country, there's a new effort under way to preserve America's historical sites, while at the same time teach a new generation the art and the importance of that work.
Jeffrey Brown has that story, part of his ongoing series Culture at Risk.
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JEFFREY BROWN: High atop Central Virginia's Shenandoah Mountains, these students are continuing work begun 75 years ago.
They have been renovating the Skyland Stables, which for generations have provided national park visitors with horses to ride along 200 miles of equestrian trails. This is the pilot project of the Hands-On Preservation Experience, or HOPE Crew, a new nationwide initiative from the National Trust for Historic Preservation.
Monica Rhodes is its volunteer coordinator.
MONICA RHODES, National Trust for Historic Preservation: What the National Trust hopes to bring to this — through this new initiative is an opportunity for a younger, more diverse audience to get involved with these buildings, to really interact with their environment and contribute to their country.
JEFFREY BROWN: The Trust is partnering with the National Park Service, the Corps Network and other groups to bring a new generation into the preservation fold across the country.
In this case, water damage and wear from the horses had damaged some of the wood on this rustic U-shaped stable. It also needed a new fence.
MONICA RHODES: There are number of historic buildings in the nation. The Park Service alone has about a $4.5 billion backlog of deferred maintenance, so these are historic buildings that are in need of repair and rehabilitation right now.
JEFFREY BROWN: Working with an expert craftsman, a team of students carried out the restoration work. Many are from cities far from the mountains, now studying at the Harpers Ferry Job Corps Center in West Virginia.
JARMAINE BUDD: I'm not used to being around horses. So, yes, their loud noises are pretty startling.
JEFFREY BROWN: While staying nearby, students like Jarmaine Budd earn $10 an hour, far less than the $40 to $60 an hour a contractor might charge. In exchange, they get hands-on experience in a new kind of work, which Budd says requires greater attention to detail.
JARMAINE BUDD: The challenging part about it is the matching of the wood. And the cuts are a little rougher. And some of the cuts have to be a little more cleaner and more precise than inside a house, where a little half-inch, you won't really tell that difference, but, out here, it's a big margin.
JEFFREY BROWN: Elijah Smith is from Washington, D.C.
ELIJAH SMITH: I think it's important to save old buildings, because when you go back, you can see what you did right, what you did wrong, how you want to add ideas to it. And the older something is, the more value it is to it. It brings more people to it.
JEFFREY BROWN: In the 1930s it was the Civilian Conservation Corps, under President Franklin Roosevelt, that originally built the national park.
During that time, the stables and the nearby Skyland resort were also purchased by the National Park Service. To preserve them for new generations to come, craft expert David Logan, who owns a vintage restoration company in Virginia, says the HOPE Crew has done the heavy lifting.
DAVID LOGAN, Vintage, Inc.: What I have done is guided the team just on some approaches for replacing siding, ways of cutting out the old, and then how to handle the oak to let it move, and just little tips and advice.
JEFFREY BROWN: But he sees a real dearth of young people learning his trade.
DAVID LOGAN: In our area, the people that are really skilled are from 40 to 65. We rarely get young people coming into this field, and I think that's very unfortunate.
JEFFREY BROWN: From a conservation standpoint, Logan argues that older structures like these deserve to be saved.
And student Nicholas Edwards agrees.
NICHOLAS EDWARDS: You are using more lumber to make a new building. And then, if you can use something's that been standing up there for a while, it could be there for a little bit longer.
JEFFREY BROWN: Monica Rhodes says, in addition to conservation, this work is about preserving history and the memories people have for a place.
MONICA RHODES: It's continuity. You know, it continues a historical conversation that started in the late '30s into 2014. So it really contributes to a sense of what Shenandoah National Park is.
JEFFREY BROWN: She hopes upcoming projects like this one, in Missouri, Texas, New Mexico, and Montana, will help instill those values and inspire more people to go into this field.
For his part, Jarmaine Budd says he wants to do more preservation work. And after this project is complete, he hopes to return to the stables to see how his own work has held up over time.
JARMAINE BUDD: Come back to a place that we helped do, that we built, to show everybody else, this is what we did.
JEFFREY BROWN: Creating another connection to the park and to history.
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06/10/14--17:46: House Majority Leader Eric Cantor defeated in Virginia primary
Photo by Andrew Harrer/Bloomberg via Getty Images
RICHMOND, Va. — House Majority Leader Eric Cantor was defeated Tuesday by a little-known economics professor in Virginia's Republican primary, a stunning upset and major victory for the tea party.
Cantor is the second-most powerful member of the U.S. House and was seen by some as a possible successor to the House speaker.
His loss to Dave Brat, a political novice with little money marks a huge victory for the tea party movement, which supported Cantor just a few years ago.
Brat had been a thorn in Cantor's side on the campaign, casting the congressman as a Washington insider who isn't conservative enough. Last month, a feisty crowd of Brat supporters booed Cantor in front of his family at a local party convention.
His message apparently scored well with voters in the 7th District.
"There needs to be a change," said Joe Mullins, who voted in Chesterfield County Tuesday. The engineering company employee said he has friends who tried to arrange town hall meetings with Cantor, who declined their invitations.
Tiffs between the GOP's establishment and tea party factions have flared in Virginia since tea party favorite Ken Cuccinelli lost last year's gubernatorial race. Cantor supporters have met with stiff resistance in trying to wrest control of the state party away from tea party enthusiasts, including in the Cantor's home district.
Brat teaches at Randolph-Macon College, a small liberal arts school north of Richmond. He raised just more than $200,000 for his campaign, according to the most recent campaign finance reports.
Beltway-based groups also spent heavily in the race. The American Chemistry Council, whose members include many blue chip companies, spent more than $300,000 on TV ads promoting Cantor. It's the group's only independent expenditure so far this election year. Political arms of the American College of Radiology, the National Rifle Association and the National Association of Realtors also spent money on ads to promote Cantor.
Brat offset the cash disadvantage with endorsements from conservative activists like radio host Laura Ingraham, and with help from local tea party activists angry at Cantor.
Much of the campaign centered on immigration, where critics on both sides have recently taken aim at Cantor.
Brat has accused the House majority leader of being a top cheerleader for "amnesty" for immigrants in the U.S. illegally. Cantor has responded forcefully by boasting in mailers of blocking Senate plans "to give illegal aliens amnesty."
It was a change in tone for Cantor, who has repeatedly voiced support for giving citizenship to certain immigrants brought illegally to the country as children. Cantor and House GOP leaders have advocated a step-by-step approach rather than the comprehensive bill backed by the Senate. They've made no move to bring legislation to a vote and appear increasingly unlikely to act this year.
Cantor, a former state legislator, was elected to Congress in 2000. He became majority leader in 2011.
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06/11/14--11:58: In Brazil, soccer no longer 'opiate' of the masses
Brazilian street artist Paulo Ito drew a picture of a crying child, who is served a soccer ball to appease his hunger, on the wall of a public school in Sao Paulo, as Brazil readies for the 2014 World Cup. Photo by Miguel Tovar/LatinContent/Getty Images
Much has changed in Brazil since the last time it hosted the FIFA World Cup in 1950, including a more skeptical, informed and middle class audience.
Roberto DaMatta, a professor at Catholic University of Rio de Janeiro, recently recalled the day in July 1950 when he went with his father to watch the Brazil-Yugoslavia match in the cavernous Maracana stadium — a game the Brazilians won 2-0.
"I was 15 years old and I realized we had built the greatest stadium in the world," said DeMatta at a Wilson Center event. He said he and his father marveled at the team, the stadium and how they had gotten there by public transportation.
Now, public transportation workers are striking for higher wages, threatening to further tangle to Sao Paulo's infamous traffic. It's one of the areas people point to when they gauge whether or not Brazil is ready for World Cup, which starts Thursday.
In addition to the striking workers, Brazilians have staged protests over the $11 billion cost of hosting the tournament, saying the money would be better spent on hospitals, schools and airports.
Compared to 1950, the Brazilian population — equipped with computers and mobile phones — is not just more aware of the country's problems but in general feels entitled to get better services, said DaMatta. "Football is no longer the opiate of the people."
In 1950, Brazil was still in its early stages of industrial and urban development and had a relatively new democratic regime, said Joao Augusto de Castro Neves, Latin America director of Eurasia Group. At the time, Brazilians felt optimistic about hosting the World Cup, not least of which because the tournament had been on hiatus during the 1940s due to World War II, he said.
"There was a great sense of a new moment. Brazil was emerging from a dictatorship; it was this young country hosting this game and building the stadiums. It was a sign of a country coming of age."
There still were problems related to stadium construction in 1950, but the bar was much lower then, in terms of the infrastructure needed to host such a huge commercial event, said Neves. In some ways, "Brazil in 1950 hosting the World Cup was similar to South Africa hosting it four years ago, as the first World Cup on the African continent," he said.
But attitudes and demographics have changed. With industrialization and urbanization just getting started in 1950, there was a smaller middle class and more people living in rural areas. Now, about 85 percent of Brazil is considered urban. That shift increased infrastructure needs, traffic and air pollution, and launched a much larger middle class with its own demands, said Neves.
"They're not satisfied only by having these huge state-of-the-art stadiums, they want everything that's supposed to surround these stadiums like subway systems, airports and good roads, not to mention hospitals and schools," he said.
The World Cup itself also has changed. "It's an industry now — it wasn't like that 60 years ago — demanding more investments across the board from the hosting countries," said Neves. "It's a profitable venture, but necessarily for the country itself."
In Brazil, 12 cities are hosting the games, but half don't have meaningful soccer teams, so the grand stadiums might not get much use later, Neves explained. "The stadiums will become while elephants in some cities."
Large-scale sporting events, such as the World Cup and the Olympics, are supposed to benefit the host country but they can end up losing money, he said. "More and more, that vision is becoming clear" to the general public, prompting a backlash including the demonstrations in Brazil.
Mauricio Moura, a researcher for Harvard University, said by telephone from Sao Paulo that a survey of 5,000 Brazilians recently conducted for the university showed only 51 percent are in favor of the World Cup — the lowest rating for any previous World Cup where data is available.
"For Brazilians, having soccer as the main sport, it's very strange to have about 45 percent of people against the World Cup," said Moura. On the other hand, when Brazilians were asked if they would support the national team, 88 percent of respondents said "yes."
There's a separation between what people are feeling for the team and the soccer games, and frustration toward the Brazilian government and World Cup organization itself, he said.
So the sour taste in Brazil over the World Cup and the fewer flying flags might all change once the golden shirts take the field. "This is the first time we are seeing this dissonance" in the streets, but when World Cup begins, there will be at least some temporary peace, said DaMatta.
Why isn't the U.S. more competitive on the world men's soccer stage? Weigh in during our live Twitter chat at 1 p.m. EDT on Thursday using hashtag #NewsHourChats. We'll also explore Brazil's challenges in hosting the tournament on Wednesday's broadcast. View all of our World Cup coverage.
Follow @NewsHourWorld
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06/11/14--12:23: Cantor to resign House leadership after primary loss
Photo by Nicholas Kamm/AFP/Getty Images.
Updated June 11, 3:23 p.m. EDT | WASHINGTON — Repudiated at the polls, House Majority Leader Eric Cantor intends to resign his leadership post at the end of next month, officials said Wednesday, clearing the way for a potentially disruptive Republican shake-up just before midterm elections with control of Congress at stake.
Cantor was expected to announce his plans at a late-afternoon meeting of the party's rank and file, less than 24 hours after the Virginia Republican lost a primary election to David Brat, a little-known and underfunded rival backed by tea party groups.
Before the announcement, jockeying already had broken out among fellow Republicans eager to move up the House leadership ladder — or establish a foothold on it.
Rep. Kevin McCarthy of California, the party whip and third-ranking leader, informed fellow Republicans he intended to run to succeed Cantor, and Rep. Pete Sessions of Texas also made clear his interest.
Rep. Peter Roskam of Illinois, the chief deputy whip, and Rep. Steve Scalise of Louisiana quickly jumped into the expected race to succeed McCarthy.
Cantor's office declined to confirm his decision, which was reported by numerous Republican aides as well as lobbyists who said they had been informed of the plans. His intention was to declare his decision to step down from the leadership on July 31.
One Republican said he feared the effects of Cantor's defeat could be debilitating for the party and the government.
Interviewed on MSNBC, Rep. Peter King, R-N.Y., said he was worried that Cantor's stunning loss may lead to even more congressional gridlock. Asked if he thought immigration legislation was dead, he replied, "I'm concerned that Ted Cruz supporters, Rand Paul supporters, are going to use this as an excuse" to shut down the government.
"This is not conservatism to me," King said. "Shutting down the government is not being conservative."
The resignation would mark a swift end to a quick rise to power for Cantor, 51, who was elected to Congress in 2000, was appointed to the leadership two years later, and then rose steadily to become the second-most powerful Republican in the House. In that post, he was the most powerful Jewish Republican in Congress, and occasionally was seen as a potential rival to Speaker John Boehner but more often as a likely successor.
He was defeated Tuesday by primary rival David Brat, an economics professor making his first run for office in an underfunded campaign that benefited from the support of tea party groups.
Brat campaigned as a foe of immigration legislation, and said Cantor was likely to help immigrants living in the United States illegally gain amnesty if given a new term in the House.
Interviewed on MSNBC, Brat declined to spell out any policy specifics.
"I'm a Ph.D. in economics, and so you analyze every situation uniquely," he said.
Brat begins the fall campaign as a decided favorite in the race against Democratic rival Jack Trammell in a solidly Republican Richmond-area district.
His primary triumph was by far the biggest of the 2014 campaign season for tea party forces, although last week they forced veteran Mississippi Sen. Thad Cochran into a June 24 runoff and they hope state Sen. Chris McDaniel will achieve victory then.
Tuesday's outcome may well mark the end of Cantor's political career, although at his age he has plenty of time to attempt a comeback.
The impact of Cantor's surprise loss on the fate of immigration legislation in the current Congress seemed clear. Conservatives will now be emboldened in their opposition to legislation to create a path to citizenship for immigrants living in the country illegally, and party leaders who are sympathetic to such legislation will likely be less willing to try.
Cantor has compiled a solidly conservative voting record in his tenure, but he was sometimes viewed with suspicion by tea party activists who said he had been in Congress too long and was insufficiently committed to blocking immigration legislation. Many party officials argue that Republicans must temper their hard line on immigration if they are to compete effectively in future presidential elections.
Already on Wednesday, Hillary Rodham Clinton, a potential Democratic contender, said Cantor "was defeated by a candidate who basically ran against immigrants."
Democrats, underdogs in the struggle for control of the House this fall, sought to cast Cantor's defeat as evidence that the Republican Party and tea party groups were one.
"The Republican Party has been completely swallowed by the tea party. I mean, any debate over whether the tea party controls the Republican Party has ended," Rep. Debbie Wasserman Schultz of Florida, the Democratic national chair, said on MSNBC.
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06/11/14--12:42: Bill and Melinda Gates Foundation urges delay for Common Core actions
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One of the most ardent supporters of the Common Core education guidelines, the Bill and Melinda Gates Foundation, is urging participating states to delay major accountability decisions based on assessment tests aligned to the standards.
The foundation's director of education, Vicki Phillips, suggested in an open letter released Tuesday that significant actions tied to the testing, such as teacher evaluation and student promotion, be pushed back by two years.
"…no evaluation system will work unless teachers believe it is fair and reliable, and it's very hard to be fair in a time of transition," Philips wrote. "The standards need time to work. Teachers need time to develop lessons, receive more training, get used to the new tests and offer their feedback."
Since 2010, 46 states and the District of Columbia adopted the Common Core State Standards, which are benchmarks in mathematics and English language arts, outlining what skills a student should have at the end of each grade.
But some teachers' unions and parent groups have complained that the standards have been rolled out too quickly. In fact, implementation issues and other concerns have prompted three states — Indiana, Oklahoma, and South Carolina — to drop Common Core this year. Lawmakers in several other states, including Florida, Illinois and South Dakota, have introduced legislation reassessing their involvement with the both the standards and related tests.
According to a recent article published in the Washington Post, The Bill and Melinda Gates Foundation was not only the primary funder behind the development of the Common Core, but also "built political support across the country, persuading state governments to make systemic and costly changes."
The post Bill and Melinda Gates Foundation urges delay for Common Core actions appeared first on PBS NewsHour.
06/11/14--12:44: Suffocating cells for science
Moh El-Naggar, an assistant professor of physics at the University of Southern California, holds a chip containing bacteria cells that are the basis for much of his team's research on how microbes move electrons. Photos by Kent Treptow.
If you think that all living things need oxygen to breathe, you're not only wrong, but hopelessly human-centric. But don't be too hard on yourself. Most mammals are biased toward multicellular organisms.
It's true that humans, along with mammals, birds, even insects and fish, require oxygen for survival. But not bacteria. What bacteria lack in intellect, they make up for in the extraordinary adaptability of their metabolism.
On Earth some 3 to 4 billion years ago, primitive microbes thrived in the absence of oxygen — anaerobically — by "breathing" iron-based minerals. Many species of bacteria still have the ability to do this.
"Chemists study chemicals. Biologists study what's alive. Physics is more mercurial than that."
That's right. Cells can breathe rock. And to do so, they must send their electrons outside of their cell bodies, sometimes moving them great distances.
Chemists understand respiration differently than the rest of us. At the cellular level, it is simply this: taking electrons from food and giving them to the thing they're breathing, whether that's oxygen or rock. When oxygen is involved, this is a fairly simple process. Since oxygen is a soluble molecule, it gets easily diffused from the outside to the inside of the cell through the cell membrane.
"But what do you do if the thing you're trying to pass your electrons to, a.k.a. the thing you're breathing, is outside the cell?" asked Moh El-Naggar, an assistant professor at University of Southern California, who runs a lab dedicated to basic research. "Somehow you have to build a bridge to get your electrons to the outside."
At his lab, El-Naggar and his team obsess about this process: how the electron gets from Point A to Point B.
"It is a beautiful problem at the interface of physics and biology," he said.
Unlocking the secrets of how these processes work could lead to breakthroughs in semiconductors, fuel cells, solar power and understanding how life can thrive in the harshest environments.
Moh El-Naggar, an assistant professor of physics at the University of Southern California, sits at the probe station where he studies how bacteria shuttle electrons great distances. Photo by Kent Treptow
In May 2009, El-Naggar made a discovery, from which all of the experiments in his lab have since sprung: A few years earlier, a pair of scientists discovered that microbes grow long, hairy filaments or fibers that are electrically conductive. El-Naggar had a hypothesis. These fibers, he suspected, serve as a conductive bridge between the cell and the rock that they're breathing. In other words, the path the electrons take to move from the cell body to material outside the cell.
"What do you do if the thing you're trying to pass your electrons to, a.k.a. the thing you're breathing, is outside the cell?"
Testing this theory was a two-year process, which required growing cells and building fantastically tiny rods to study the conductive properties of the fibers, or as they're now called, "bacterial nanowires."
It was late one night at Lawrence Berkeley National Laboratory that El-Naggar connected all the pieces, quite literally. He had flown out that morning with a silicon chip in a carry-on Canon camera case. The chip contained thousands of dead, but well-preserved cells, and their newly sprouted nanowires.
Moh El-Naggar holds a chip containing bacteria cells that are the basis for his research on how microbes move electrons. Photo by Kent Treptow
In a clean-room facility there, he used a powerful electron microscope to build dozens of tiny platinum electrodes – leads, he calls them — each measuring a micron long. That's about 100th the width of a human hair. Then, with an electron beam, he deposited electrodes onto his chip, placing them at each end of several nanowires.
Once the nanowires and electrodes were connected, he moved the chip to a probe station containing an optical microscope and a voltmeter, fastened the chip to a plate and turned a knob to pump in electricity. He wanted to know if electrical current could move through the nanowires. And over the course of several minutes, he studied a figure on the screen.
"People think we get more excited about seeing stuff. But every good experiment needs a good control. And sometimes nothing is as exciting as something."
"What you'd expect to see is that you get more current flowing through the nanowires, the more voltage you apply to them," he said. "And that's exactly what we saw."
El-Naggar believed that current was generated by electrons surging through the nanowires to the platinum leads. But to be sure, he had to cut the nanowires and repeat the experiment. If he was right, breaking the fibers would kill the current. Using an ion beam — a tiny scalpel made out of pure energy — he sliced the wires. And then he performed the experiment again, hoping that he would see… nothing.
"It's kind of weird, right?" he said. "People think we get more excited about seeing stuff. But every good experiment needs a good control. And sometimes nothing is as exciting as something."
As he'd hoped, the numbers stayed frozen at zero. Alone in the lab, he pumped his fist in the air and then sent an email to his collaborators. One of his postdocs had vowed to grow his beard until the experiment was complete. In his email, El-Naggar recommended a shave. He had a bushy beard, he recalls now. "And it was beginning to look pretty bad."
The finding was published in the journal PNAS in 2010.
El-Naggar now oversees a lab of five graduate students, two postdoctoral researchers and a small army of undergraduates. Earlier this year, he was awarded the Presidential Early Career Award for Scientists and Engineers. And his Twitter posts look like this:
Spread the word: I have an opening for a bioelectrochemistry postdoc to work on single microbe and single mitochondrion electron transport.
— Moh El-Naggar (@BioPhysicalMoh) March 31, 2014
Intimidating, right? But he is anything but.
"He's the best advisor the physics department has had," said Benjamin Gross, a postdoctoral candidate in his lab, who studies the basic biophysics of how electrons move. "Some advisors ignore you; some are anxious for you to publish good work," but El-Naggar, he said, gets in the trenches with the young scientists. "People love working here."
One big discovery, many applications
The whole story of El-Naggar's lab since that night in Berkeley is about trying to move the ball forward on the electron transport discovery. And his team has taken that basic concept and launched it into wildly different directions of research.
One member of his team, Yamini Jangir, traveled to Death Valley, where, among the sand dunes and jagged salt flats and hellish 115-degree temperatures, she collected water samples from a giant pressurized well. In that water, bacteria thrives deep underground in low oxygen conditions.
Graduate student Jamini Jangir, 27, points out Shewanella bacteria cells on a computer screen at the University of Southern California. Photo by Kent Treptow
Back in the lab, Jangir built her own reactor out of a polypropylene jar, plastic piping, titanium wires and aluminum foil. Her hope is to recreate in the lab the anaerobic conditions from the Death Valley subsurface that allow this bacteria to thrive. Statistics for these bacteria are grim now — a mere 1 percent survive in the lab setting. The ultimate goal: to develop a technique that would allow scientists to cultivate underground bacteria in a lab setting, and thus study them better.
Yamini Jangir has built her own reactor using a polyproylene jar, plastic piping, titanium wires, a nitrogen bubbling system and aluminum foil. Photo by Kent Treptow
Another graduate student, Ian McFarlane is trying to make semiconductors in the greenest possible way – using only cells. And Benjamin Gross, a postdoctoral candidate, is studying the basic mechanism of how cells shuttle their electrons. (McFarlane and Gross both rely on a bacteria called Shewanella that was discovered in Lake Oneida in upstate New York.)
Against the wall in the wet lab is a fridge with glass bottles filled with arsenic and sulfur — nutrients for the bacteria — and a sign that says "No Food or Drink." A "Poison" label is taped onto many of the bottles, which are sealed with black rubber caps to keep the oxygen out. A handful have a skull and crossbones scrawled across their midsection in black Sharpie.
Hand-drawn skull-and-crossbones mark a hazardous waste container used by scientists at a lab at the University of Southern California. Scientists at USC are doing basic research to determine how certain bacteria are able to transfer electrons great distances from themselves to other cells or inorganic minerals. Photograph by Kent Treptow for PBSNewsHour
On McFarlane's computer screen are dozens of nanofibers, each about 15 microns long and ranging from 10 to 300 nanometers wide — roughly 1,000 times thinner than a sheet of paper. He hopes these will someday have a practical use as semiconductor devices, possibly as cheap solar panel material. The key, he explains, is having the bacteria do the work.
"They're my little factories," he said. "Rather than having a person make insulin, you have a bacteria make insulin. Rather than having a person make nanoscale fibers, you have bacteria do it."
McFarlane moves to the far end of the room, where he's done just that. He pulls a glass jar from a refrigerator. The jar contains a thick yellow substance resembling foam insulation. But when he shakes it, the consistency changes, becoming more liquid, like orange juice. The resemblance is so strong that he can no longer drink orange juice from a clear glass, he says.
Graduate student Ian McFarlane holds a bottle containing billions of semiconductor nanofibers. Photo by Kent Treptow
Inside the jar are billions of nanofibers created by combining arsenic and sulfur, adding bacteria and removing oxygen.
"I love the smell of rotten eggs," McFarlane said. "It means it's working."
Removing the oxygen forces the bacteria to use a backup power source. In other words, to breathe the arsenic and sulfur. And the nanofibers that grow during that process act as primitive semiconductors.
Magnified on his computer, MacFarlane studies the fibers. Some look like smooth rods – others sport a crystalline pattern, like a spinal cord.
Ian McFarlane points to a set of bacterial nanofibers on his computer. Photo by Kent Treptow
He's creating fully-functioning transistors out of these semiconductors, El-Naggar explains. Understanding their shape and structure could help engineers put the incredible metabolism of these microbes to practical use. Engineers, they hope, could one day harness the bacteria's breathing system to power fuel cells, for example.
So many dead cells
A few feet from McFarlane, undergraduate James Lu painstakingly prepares slides for Gross, who studies Shewanella bacteria cells one at a time as they pass their electrons onto electrodes. They are clumsy-looking slides with black and white wires sticking out, cover glass embedded with electrodes, and a plastic chamber, mounted onto a plate with double sticky tape.
Postdoctoral candidate Benjamin Gross injects millions of bacteria cells into a chamber. Photo by Kent Treptow
On his computer, Gross pulls up a graphic of a cell. The cell takes in food, he explains — acetate or lactate, for example. It converts the food into something called ATP – Adenosine triphosphate. And then it spits out the electron.
"It needs to get rid of the electron, or it will gum up the works," he says.
Gross moves to the lab counter, and using a syringe, draws half a milliliter of solution from a glass bottle and injects it into the chamber on the plate. Then he caps the two ends so they're airtight. He has just injected millions of cells into the chamber.
"We're suffocating the children," Gross says. He impersonates the bacteria. "Fine. I'll breathe your electrode."
You can't see them now, but the glass on the plate is patterned with electrodes made of indium tin oxide, a material that's widely used in smartphone touch screens, solar cells and as a defroster for Boeing 747 aircraft windshields.
Earlier, he had bubbled nitrogen into the solution and added the amino acid, cysteine. Both work jointly to starve out the oxygen, so the cells will be forced to breathe the ITO, just like MacFarlane's cells were forced to breathe sulfur and arsenic.
"We're suffocating the children," Gross says, and impersonates the bacteria. "Fine. I'll breathe your electrode."
Benjamin Gross manipulates a laser as he tries to trap and isolate a cell on a computer screen. Photo by Kent Treptow
In a dry lab across the hall is a device called an optical trap that Gross spent six months building. It relies on an infrared laser to trap and isolate cells, a laser that could blind you, he warns, causing excruciating pain if your eyes were to wander directly into the light without safety glasses.
"Uh oh." His face becomes serious. "It looks like a little atomic bomb went off on my electrode."
He slides the plate under the microscope, and by turning a knob, peers at its contents, which appear murky green on a computer screen. Hard lines appear in the image.
"There. I found an electrode," he says. "Now we kind of go fishing."
He is fishing for cells, which ideally will search out the ITO electrodes and stick to them. Then, he will use highly sensitive electronics to detect the amount of electrical current. Lines spiking on the computer indicate that the cell is squirting electrons into the electrode. And his research has shown that about a million electrons a second cross from the bacteria and land on the electrode.
El-Naggar explains that Gross has taken his initial experiment a step further.
"The kinds of measurements I did myself in Berkeley in 2009 were non-biological measurements," El-Naggar said. "Crude tools to assess the conductivity of the wires. Now we're at a stage where we're grabbing individual cells live and measuring respiration."
Benjamin Gross spent six months building this optical trap used to isolate cells and measure their conductivity. Photo by Kent Treptow
Suddenly, one of those live cells appears – a tiny blob darting through across the screen.
"Oh, there's a little guy," Gross says. "Where'd you go, buddy?"
And just as suddenly, the image turns black.
The electrode, he explained, had gotten fried by the machine's laser. It's two full days of work — gone.
"So many dead cells," he said. "It happened so fast."
Put simply, he explained, the laser was too strong and the electrodes too thick. It's the kind of troubleshooting that happens all the time.
"Every Ph.D. student is essentially trying to do something original, so you can't just buy a kit at the store. You engineer from spare parts whatever you need to get it done. We end up being our own artisans to a degree."
"A large number of questions we're answering every day are like, 'What's the best way to glue these things together?' or whatever mundane thing we're doing, he says. "Once you have that figured out, you use it to answer a deeper question.
Among those deeper questions: What kind of electrode material is best? Should the electrode be rough or smooth? What are the best ways to grow the bacteria, so they're most likely to stick to the electrode? How strong should the laser be to avoid these tiny massacres?
Failure, El-Naggar said, comes with the territory: "I have never myself designed and built an experiment that worked the second or third time. The key thing is to have the right temperament and attitude about science. A lot of people would be incredibly frustrated to waste two days of work."
A proud physics tradition
El-Naggar's goal is to be a cheerleader and protector of his students. He seeks funding, speaks at conferences and plans projects so that they have more time in the lab.
"I try to protect my students by keeping these things away from them, so they can focus on the exciting science, the research," he said.
Moh El-Naggar, an assistant professor of physics at the University of Southern California, stands at the white board where he and his team brainstorm ideas about their work. Photo by Kent Treptow
This is especially true considering the do-it-yourself philosophy of his lab, where students build most components of the experiments themselves.
"Every Ph.D. student is essentially trying to do something original, so you can't just buy a kit at the store," Gross said. "You engineer from spare parts whatever you need to get it done. We end up being our own artisans to a degree."
That homegrown quality is a proud physics tradition, El-Naggar added.
"Chemists study chemicals. Biologists study what's alive. Physics is more mercurial than that," he says. "It's about the way we do things. Physics is one field that isn't defined by the subject of the study as much as by the approach of the study. And the way we do things doesn't matter, as long as we have a very quantitative approach."
In this case, each experiment, each question answered, brings the scientists closer to the basic physics of how organisms move electrons very long distances. But it goes beyond that.
On one hand, El-Naggar says he hopes the research will ultimately lend itself to new technologies: using microorganisms to clean wastewater, power fuel cells, or build better solar power materials, for example. But more importantly, he says, it's about curiosity.
"We do this because we're curious about the world," El-Naggar says. "Here is one particular example of something microorganisms have been doing for billions of years, and we're only learning how they do it now."
And there's a lot of power in that.
The post Suffocating cells for science appeared first on PBS NewsHour. | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 4,371 |
\section{Introduction}
Since the pioneer work of Aldous in \cite{Aldous1}, the study of continuum random trees (CRT) is considered as a powerful tool to study properties of large random discrete trees. In particular, it has been conjectured in \cite{Aldous1}, that the Brownian CRT is a universal limit for numerous models of trees with large height.
This has been verified over and over. Furthermore the Brownian CRT model has been extended, for discrete trees with smaller height, toward two main distinct directions. On the one hand, L\'evy trees are introduced, in Le Gall Duquesne \cite{Duquesne1,Duquesne2}, as limits of Galton-Watson trees. On the other hand, inhomogeneous continuum random trees (ICRT) are introduced by Aldous, Camarri and Pitman, in \cite{IntroICRT1,IntroICRT2}, as limits of $\mathcal{P}$-trees. Those two distinct but similar models leave the following main problem: Finding a universal model for limits of random discrete trees (with no restriction on the height).
To solve this problem, we prove in a forthcoming paper \cite{Uniform}, that ICRT appears as limits of uniform random trees with fixed degree sequence. Since many models of interest can be studied under the spectrum of those trees, this proves that ICRT are universal. In particular L\'evy trees are ICRT with random parameters. The aim of the present paper is twofold: obtain refined information about the ICRT, the universal limit object in particular concerning compactness and fractal dimensions, and introduce some tools for convergence that will be used in \cite{Uniform}.
Our main results are derived from a new version of the stick-breaking construction of the ICRT from Aldous, Pitman \cite{IntroICRT1}. Stick-breaking constructions generate a $\mathbb{R}$-tree (a loopless geodesic space see Le Gall \cite{Legall} for an extensive treatment) and are separated in two steps:
\begin{compactitem}
\item the line $\mathbb{R}^+$ is first cut into the segments ("sticks") $[0, Y_1],\, (Y_1,Y_2], (Y_2,Y_3] \dots$
\item the segments are then re-arranged sequentially in a tree-like fashion by gluing $(Y_i,Y_{i+1}]$ at a point $Z_i\leq Y_i$. (see Figure \ref{SB})
\end{compactitem}
Such a construction has been introduced by Aldous \cite{Aldous1} for the Brownian CRT. Recently Amini, Devroye, Griffiths, Olver in \cite{Amini} studied a case where cuts are fixed with $(Y_{i+1}-Y_i)_{i\in \mathbb{N}}$ decreasing. The condition of monotonicity has been removed by Curien and Haas in \cite{Curien} where they construct a probability measure on $\mathcal{T}$, give a sufficient criterion for compactness of $\mathcal{T}$ and compute the Hausdorff dimension of $\mathcal{T}$. We use similar methods in a setting where cuts and glue points are generated according to a random measure $\mu$ on $\mathbb{R}^+$.
\paragraph{Plan of the paper} In the next section we present the new construction for ICRT. Our main results are then stated in Section \ref{Section 3}. In Section \ref{Section 4} we study the measure $\mu$ and cuts. A probability measure is constructed from $\mu$ in Section \ref{Section 5}. The compactness and fractal dimensions are the topics of Sections \ref{Section 6} and \ref{Section 7} respectively.
\section{Model and definition of the fractal dimensions} \label{Section 2}
\subsection{The ICRT and its construction} \label{def}
Let us first present a generic deterministic stick-breaking construction. It takes for input two sequences in $\mathbb{R}^+$ called cuts ${\textbf y}=(y_i)_{i\in \mathbb{N}}$ and glue points ${\textbf z}=(z_i)_{i\in \mathbb{N}}$, which satisfy
\begin{equation} \forall i<j,\ \ y_i<y_j \qquad ; \qquad y_i\limit \infty \qquad ; \qquad \forall i\in \mathbb{N},\ \ z_i\leq y_i, \label{2609} \end{equation}
and creates an $\mathbb{R}$-tree by recursively "gluing" segment $(y_i,y_{i+1}]$ at position $z_i$ (see Figure \ref{SB}), or rigorously, by constructing recursively a consistent sequence of distances $(d_n)_{n\in \mathbb{N}}$ on $([0,y_n])_{n\in \mathbb{N}}$.
\begin{figure}[!h] \label{SB}
\centering
\includegraphics[scale=0.6]{ICRT_glue5.eps}
\caption{A typical step of the stick-breaking construction: the "gluing" of $(y_i,y_{i+1}]$ at $z_i$. }
\label{SB}
\end{figure}
\begin{algorithm} \label{Alg1} \emph{Generic stick-breaking construction.}
\begin{compactitem}
\item[--] Let $d_0$ be the trivial distance on $\{0\}$.
\item[--] For each $n\geq 1$ define $d_n$ on $[0, y_n]$ such that for each $x\leq y$:
\[ d_n(x,y):=
\begin{cases}
d_{n-1}(x,y) & \text{if } x,y\in [0, y_{n-1}] \\
d_{n-1}(x,z_{n-1})+|y-y_{n-1}| & \text{if } x \in [0, y_{n-1}], \, y \in (y_{n-1}, y_n] \\
|x-y| & \text{if } x,y\in (y_{n-1}, y_n]
\end{cases} \]
where by convention $y_0:=0$ and $z_0:=0$.
\item[--] Let $d$ be the unique metric on $\mathbb{R}^+$ which agrees with $d_n$ on $[0, y_n]$ for each $n\in \mathbb{N}$.
\item[--] Let $\SB({\textbf y},{\textbf z})$ be the completion of $(\mathbb{R}^+,d)$.
\end{compactitem}
\begin{remark}
There is a more general way of gluing metric space. (see \cite{Glue} for definition or \cite{Seni} for similar work in this context). We prefer to work directly on $\mathbb{R}^+$ for practical reasons. %
\end{remark}
\end{algorithm}
We now introduce the probability space that will be used in the paper. Note that the space $\Upsilon:=\bigcup_{{\textbf y},{\textbf z}} \SB({\textbf y},{\textbf z})$ is in bijection with the space of couples of sequences $({\textbf y},{\textbf z})$ that satisfy \eqref{2609}, hence one can naturally define the weak topology on $\Upsilon$. Then we work on a complete probability space such that every random variable defined below are measurable for the weak topology
Now, let $\Omega$ be the space of sequences $\{\theta_i\}_{i\in \mathbb{N}}$ in $\mathbb{R}^+$ such that:
\[ \sum_{i=0}^\infty \theta_i^2=1 \quad ; \quad \theta_1\geq \theta_2 \geq \dots \quad ; \quad \theta_0\neq 0 \text{ or } \sum_{i=1}^\infty \theta_i=\infty. \] The ICRT of parameter $\Theta\in \Omega$ is the random $\mathbb{R}$-tree constructed via the following algorithm.
\begin{algorithm} \label{Alg2} \emph{Classical construction of the $\Theta$-ICRT from \cite{IntroICRT1,IntroICRT2}} \begin{compactitem}
\item[--] Let $(A_i, B_i)_{i\in \mathbb{N}}$ be a Poisson point process of intensity $\theta_0^2$ on $\{(a,b)\in \mathbb{R}^{+2}: b\leq a \}$.
\item[--] Let $((A_{i,j} )_{j\in\mathbb{N}})_{i\in \mathbb{N}}$ be a family of independent Poisson point processes of intensity $(\theta_i)_{i\in \mathbb{N}}$ on $\mathbb{R}^+$ and independent of $(A_i, B_i)_{i\in \mathbb{N}}$.
\item[--] Sort the elements of the (almost surely) locally finite set $\bigcup_{i=0}^\infty \{A_i\} \cup\bigcup_{i=1}^\infty \bigcup_{j=1}^\infty \{A_{i,j} \}$ as $U=(U_i)_{i\geq 1}$ with $U_1<U_2<\dots$
\item[--] For $i\geq 1$, let $V_i= \begin{cases} Y_j \, \, \, \, \text{if } U_i \text{ is of the form} & A_j \\ A_{i,0} \text{ -----------------------} & A_{i,j} \end{cases}$ and let $V=(V_i)_{i\geq 1}$.
\item[--] The (old) $\Theta$-ICRT is defined as $(\mathcal{T}^*,d^*)=\SB(U,V)$.
\end{compactitem}
\end{algorithm}
For technical reasons, it is convenient to deal with the following alternative construction.
\begin{algorithm} \label{Alg3} \emph{New construction of the $\Theta$-ICRT}
\begin{compactitem}
\item[--] Let $(X_i)_{i\in \mathbb{N}}$ be a family of independent exponential random variables of parameter $(\theta_i)_{i\in \mathbb{N}}$.
\item[--] Let $\mu$ be the measure on $\mathbb{R}^+$ defined by $\mu=\theta_0^2 dx+\sum_{i=1}^{\infty} \delta_{X_i} \theta_i$.
\item[--] For each $l\in\mathbb{R}^+$ let $\mu_l$ be the restriction of $\mu$ to $[0,l]$.
\item[--] Let $(Y_i)_{i\in \mathbb{N}}$ be a Poisson point process on $\mathbb{R}^+$ of rate $\mu[0,l]dl$. \item[--] Let $(Z_i)_{i\in \mathbb{N}}$ be a family of independent random variables with respective laws $\frac{\mu_{Y_i}}{\mu[0,Y_i]}$, $i\in \mathbb{N}$.
\item[--] The (new) $\Theta$-ICRT is defined as $(\mathcal{T},d)=\SB(Y,Z)$.
\end{compactitem}
\end{algorithm}
\begin{remark}
The construction may fail because $\mu[0,l]$ may be infinite for some $l$. However, since $\mu[0,l]$ is of finite expectation, this almost surely never happens. (See Lemma \ref{2})
\end{remark}
The constructions in Algorithms \ref{Alg2} and \ref{Alg3} are equivalent that is:
\begin{lemma} \label{equiv}
$(\mathcal{T}^*,d^*)$ and $(\mathcal{T},d)$ have the same distribution.
\end{lemma}
\begin{proof} First conditionally on $\{A_{i,0}\}_{i\in \mathbb{N}}$, $\{U_i,V_i\}_{i\in \mathbb{N}}$ is a Poisson point process on $\Delta:=\{(a,b)\in \mathbb{R}^{+2}: b\leq a \}$ of intensity
\[ \theta_0^2dx dy+\sum_{i=1}^\infty \theta_i \mathbf{1}_{A_{i,0}\leq x} dx \times \delta_{A_{i,0}}. \]
Also, conditionally on $(X_i)_{i\in \mathbb{N}}$, $(Y_i,Z_i)_{i\in\mathbb{N}}$ is a Poisson point process on $\Delta$ of intensity
\[ \theta_0^2dx dy+\sum_{i=1}^\infty \theta_i \mathbf{1}_{X_i\leq x} dx \times \delta_{X_i}. \]
So since $(X_i)_{i\in \mathbb{N}}$ and $\{A_{i,0}\}_{i\in \mathbb{N}}$ have the same distribution, $(U,V)$ and $(Y,Z)$ also have the same distribution. Finally $(\mathcal{T}^*,d^*)=\SB(U,V)$ and $(\mathcal{T},d)=\SB(Y,Z)$ have the same distribution.
\end{proof}
Finally let us introduce some notation that will simplify many expressions later.
\begin{definition*}
For $n\in \mathbb{N}$ let $l_n:=Y_{n}-Y_{n-1}$ denotes the length of the $n$th segment, and let $m_{n}:=\mu(Y_{n-1},Y_{n}]$ denote its weight. Then let $M_n:=\mu[0,Y_n]=m_1+\dots+m_n$.
\end{definition*}
\subsection{Fractal dimension} \label{2.5}
In the entire section $X$ is a metric space and for every $x\in X$, $\varepsilon>0$, $B(x,\varepsilon)$ denotes the closed ball centered at $x$ with radius $\varepsilon$. We recall the definitions of the fractal dimensions we compute in this paper.
\begin{definition*} (Minkowski dimensions) For every $\varepsilon>0$ let $N_\varepsilon$ be the minimal number of closed balls of radius $\varepsilon$ to cover $X$. Define the Minkowski lower box and upper box dimensions respectively by
\[ \underline{\dim}(X):=\liminf_{l\to \infty} \frac{ \log N_{1/l}}{\log l} \quad \text{and} \quad \overline{\dim}(X):= \limsup_{l\to \infty} \frac{ \log N_{1/l}}{\log l}. \]
\end{definition*}
\begin{definition*} (Packing dimension) For every $s\geq 0$ and $A\subset X$ let
\[ P^s_0(A):= \limsup_{\delta\to \infty} \left \{ \sum_{i\in I} \diam(B_i)^s \Bigg | \, \{B_i\}_{i\in I} \text{ are disjoint balls $B(x,r)$ with $x\in A$ and $r\leq \delta$}\right \}. \]
and
\[ P^s(X):=\inf \left \{\sum_{i=1}^{\infty} P^s_0(A_i) \Bigg | X\subset \bigcup_{i=1}^{\infty} A_i \right \}. \]
Then $P^s$ is a decreasing function of $s$, and we define the packing dimension of $X$ as
\[ \dim_P(X):= \sup \{s, P^s(X)<\infty\}. \]
\end{definition*}
\begin{definition*} (Hausdorff dimension) For every $s,r\geq 0$ write
\[ H^s_r(X):= \inf_{\diam(A_i)\leq r} \left \{ \sum_{i=1}^{\infty} \diam(A_i)^s \Bigg | X \subseteq \bigcup_{i=1}^{\infty} A_i \right \}. \]
The Hausdorff dimension of $X$ is defined by
\[ \dim_H(X):=\sup \left \{s, \sup_{r\in \mathbb{R}^+}H_r^s(X)<\infty \right \}. \]
\end{definition*}
To compute the Packing dimension and Hausdorff dimension of the ICRT we will use the following extension of Theorem 6.9, and Theorem 6.11 from \cite{fractal}. (\cite{fractal} deals with subsets of Euclidian space, but the same arguments hold for every metric space.)
\begin{lemma}\label{Hausdorff} Let $p$ be a Borel probability measure on $X$ and $s\in \mathbb{R}^+$.
\begin{compactitem}
\item[a)]If $p$-almost everywhere $\liminf p(B(x,\varepsilon))\varepsilon^{-s}<+\infty$ as $\varepsilon\to 0$, then $\dim_P(X)\geq s$.
\item[b)]If $p$-almost everywhere $p(B(x,\varepsilon))=O(\varepsilon^{s})$ as $\varepsilon\to 0$, then $\dim_H(X)\geq s$.
\end{compactitem}
\end{lemma}
We have the well-known inequalities (see e.g. Chapter 3 of Falconer \cite{FalconPunch}):
\begin{lemma} \label{FalconPunch} For every metric space $X$ we have
\[ \dim_H(X)\leq \underline{\dim}(X) \leq \overline{\dim}(X) \quad \text{and} \quad \dim_H(X)\leq \dim_P(X) \leq \overline{\dim}(X). \]
\end{lemma}
\section{Main results} \label{Section 3}
The first theorem defines a probability measure on ICRT.
\begin{theorem} \label{THM1}
Almost surely there is a probability measure $p$ on the tree $\mathcal{T}$ such that
\[ p_l:=\frac{\mu_l}{\mu[0,l]} \limit^{\text{weakly}}_{l\to \infty} p. \]
Furthermore $p$ has support $\mathcal{T}$, has no atoms and gives measure $1$ to the set of leaves (the set of $x\in \mathcal{T}$ such that $\mathcal{T}\backslash\{x\}$ is connected).
\end{theorem}
This probability is also the limit of other natural empirical measures on $\mathcal{T}$:
\begin{proposition} \label{other}
Let $\mu^{\leadsto}$ be the Lebesgue measure on $\mathbb{R}^+$ and $\mu^{\bullet}=\sum_{i=1}^{\infty} \delta_{Y_i}$. For every $l\in \mathbb{R}^+$ let $\mu^{\leadsto}_l$ (resp. $\mu^{\bullet}_l$) be the restriction of $\mu^{\leadsto}$ $(resp. \mu^{\bullet})$ to $\mathcal{T}_l=([0,l],d)$. Also let for every $l\in \mathbb{R}^+$, $p_l^{\leadsto}=\frac{\mu_l^{\leadsto}}{\mu^{\leadsto}_l[0,l]}$ and $p_l^{\bullet}=\frac{\mu_l^{\bullet}}{\mu^{\leadsto}_l[0,l]}$. Then
\[ p^{\bullet}_l\limit^{weakly}_{l\to \infty} p \quad \text{ and }\quad p^{\leadsto}_l\limit^{weakly}_{l\to \infty} p. \]
\end{proposition}
Intuitively speaking this comes from the fact that $\mu$ "dictates" how segments are glued together so the convergence of $p_l$ implies the convergence of many others quantities.
\begin{remark} Proposition \ref{other} shows that $p$ corresponds to the probability measure introduced in Aldous Pitman \cite{IntroICRT1}. In particular independent leafs sampled by $p$ "behaves" like $(Y_i)_{i\in \mathbb{N}}$. (\cite{IntroICRT1} Corollary 8) \end{remark}
Then we prove the conjecture of Aldous, Miermont, Pitman in \cite{ExcICRT} about compactness.
\begin{theorem} \label{THM2}
The ICRT is almost surely compact if and only if
\begin{equation} \int^{\infty} \frac{1}{l\mathbb{E}[\mu[0,l]]}<\infty. \label{1404} \end{equation}
\end{theorem}
\begin{remark}
The conjecture in \cite{ExcICRT} is based on a comparison between the ICRT and Levy trees introduced by Le Gall Le Jan \cite{IntroLevy1}. Levy trees are characterized by their Laplace exponent $\psi$ and are compact if and only if $\int^{\infty} \frac{1}{\psi(l)} <\infty$ (see \cite{Duquesne2}). The formulation of the conjecture in \cite{ExcICRT} is based on an analog of the Laplace exponent in the setting of ICRT, which behaves like $l\mathbb{E}[\mu[0,l]]$ (see Lemma \ref{condition}) which turns out to be equivalent to \eqref{1404}.
\end{remark}
For the proof of Theorem \ref{THM2}, we first translate the condition in \eqref{1404} into a more convenient one: it turns out (Lemma \ref{condition}) that
\[ \int^{\infty} \frac{1}{l\mathbb{E}[\mu[0,l]]}<\infty \quad \text{if and only if} \quad \sum_{n=1}^{\infty} \frac{\log \mathcal{X}_{2^n} }{\mathcal{X}_{2^n}} <\infty,\] where for every $l\in \mathbb{R}^+$, $\mathcal{X}_l$ is the real number such that $\mathbb{E}[\mu[0,\mathcal{X}_l]]=l$ (see Lemma \ref{2} for existence and uniqueness).
To prove that the condition is sufficient, we will upper bound the law of the distance between a random point in $\mathcal{T}_{\mathcal{X}_{2^n}}$ and its projection on $\mathcal{T}_{\mathcal{X}_{2^{n-1}}}$. We then use this bound to prove that \[ d_H(\mathcal{T}_{\mathcal{X}_{2^n}},\mathcal{T}_{\mathcal{X}_{2^{n-1}}})\leq C\frac{\log \mathcal{X}_{2^n} }{\mathcal{X}_{2^n}}, \]
where $d_H$ denotes the Hausdorff distance on subsets of $\mathcal{T}$. For the Hausdorff topology, Cauchy sequences of compact sets converge toward a compact set so this proves that $\sum_{n=1}^{\infty} \frac{\log \mathcal{X}_{2^n} }{\mathcal{X}_{2^n}} <\infty$ implies that $\mathcal{T}$ is compact.
The fact that the condition is necessary follows from an adaptation of an argument of Amini, Devroye, Griffiths, Olver in \cite{Amini}. We show that, for some fixed constants $c,C\in (0,\infty)$ and for all $k$ large enough:
\[ c \sum_{n=k+1}^{\infty} \frac{\log \mathcal{X}_{2^n} }{\mathcal{X}_{2^n}} \leq d_H \left (\mathcal{T},\mathcal{T}_{\mathcal{X}_{2^k}} \right ) \leq C \sum_{n=k+1}^{\infty} \frac{\log \mathcal{X}_{2^n} } {\mathcal{X}_{2^n}}. \]
We then proceed to the computation of some fractal dimensions.
\begin{theorem} \label{THM3}
Almost surely
\[\dim_P(\mathcal{T}) =\overline{\dim}(\mathcal{T}) =1+\limsup_{l\to \infty} \frac{\log l}{\log \mathbb{E}[\mu[0,l]]}.\]
Furthermore if $\log l=\mathbb{E}[\mu[0,l]]^{o(1)}$ then
\[\dim_H(\mathcal{T})=\underline{\dim}(\mathcal{T})=1+\liminf_{l\to \infty} \frac{\log l}{\log \mathbb{E}[\mu[0,l]]}.\]
\end{theorem}
\begin{remark} If one replaces $l\mathbb{E}[\mu[0,l]]$ by the Laplace exponent $\psi$ then one recovers the formulas for the fractal dimensions of Levy trees obtained by Duquesne and Le Gall \cite{Duquesne1}.
\end{remark}
To prove Theorem \ref{THM3}, it suffices by Lemma \ref{FalconPunch} to upper bound the Minkowski dimensions and to lower bound the Packing and Hausdorff dimension. To upper bound $\overline{\dim}(\mathcal{T})$ and $\underline{\dim}(\mathcal{T})$ we use some cover of $\mathcal{T}$ which relies on $\log l=\mathbb{E}[\mu[0,l]]^{o(1)}$. Then we derive the lower bound on $\dim_P(\mathcal{T})$ and $\dim_H(\mathcal{T})$ from Lemma \ref{Hausdorff}.
\section{Preliminaries} \label{Section 4}
This section should be seen as a tool box: we gather here a collection of lemmas that will be used repeatidly throughout the paper. Most of them are straightforward.
\subsection{Fundamental properties of $\mu$}
\begin{lemma} \label{2}
The map $l\to\mathbb{E}[\mu[0,l]]$ is differentiable and its derivative decreases to $\theta_0^2$ as $l\to \infty$ we thus have as $l\to \infty$:
\[ \mathbb{E}[\mu[0,l]]=\theta_0^2l+o(l). \]
\end{lemma}
\begin{proof} By Fubini's theorem,
\begin{equation} \mathbb{E}\left [\mu[0,l]-\theta_0^2l \right ]=\mathbb{E} \left [\sum_{i=1}^{\infty}\theta_i\mathbf{1}_{X_i\leq l} \right ]=\sum_{i=1}^{\infty}\theta_i\mathbb{P} \left (X_i\leq l \right )=\sum_{i=1}^{\infty}\theta_i(1-e^{-\theta_il}).\label{107} \end{equation}
Each term of the sum is positive and increasing so we can differentiate term by term:
\[ \frac{d}{dl}\mathbb{E}\left [\mu[0,l]-\theta_0^2l \right ]=\sum_{i=1}^\infty \theta_i^2e^{-\theta_il}.\]
Since $\sum_{i=1}^\infty \theta_i^2<\infty$, by bounded convergence the last term decreases to $0$ as $l\to \infty$.
\end{proof}
Lemma \ref{2} implies that the map $l\mapsto \mathbb{E}[\mu[0,l]]$ is strictly increasing, continuous, and diverges, so is invertible. Thus for every $l\in \mathbb{R}^+$, there is a well-defined real number $\mathcal{X}_l$ with $\mathbb{E}[\mu[0,\mathcal{X}_l]]=l$.
\begin{lemma} \label{5}
We have almost surely
\[\mu[0,l]\asym_{l\to\infty} \mathbb{E} \left [\mu[0,l] \right ].\]
\end{lemma}
\begin{proof} For every $l\in \mathbb{R}^+$ the variance of $\mu_l$ is given by:
\[\Varr [\mu[0,l]]= \Varr \left [\theta_0^2l+\sum_{i=1}^{\infty}\theta_i\mathbf{1}_{X_i\leq l} \right ]=\sum_{i=1}^{\infty} \Varr \left [\theta_i\mathbf{1}_{X_i\leq l} \right ] \leq \sum_{i=1}^{\infty}\theta_i^2 \leq 1.\]
Therefore for every $n\in \mathbb{N}$,
\[ \mathbb{P} \left ( \left | \mu[0,\mathcal{X}_{n^2}] -\mathbb{E} \left [\mu[0,\mathcal{X}_{n^2}] \right ] \right | > n \right ) \leq \frac{1}{n^2}. \]
By definition of $\mathcal{X}_n$ we deduce by the Borel--Cantelli lemma that for every $n$ large enough
\[ n^2-n\leq \mu[0,\mathcal{X}_{n^2}] \leq n^2+n. \]
We thus have almost surely $\mu[0,\mathcal{X}_{n^2}] \sim \mathbb{E}[\mu[0,\mathcal{X}_{n^2}]]=n^2$. This result is then extended to every $l\in\mathbb{R}^+$ by monotonicity of $l\mapsto \mu[0,l]$.
\end{proof}
Note that Lemmas \ref{2} and \ref{5} implies that for every $l$ large enough $\mu[0,l]\leq l$.
The following lemma should be seen as an estimate for the "density" and "jump" of $l\mapsto \mu[0,l]$.
\begin{lemma} \label{6}
Almost surely there exists $L_0\in \mathbb{R}$ such that for every $l\geq L_0$ and $0\leq \delta \leq l$,
\[ \mu[l,l+\delta] \leq 2 \delta \frac{\mathbb{E} \left [ \mu \left [0,l\right ] \right ]}{l}+\frac{13\log (l)}{l}. \]
\end{lemma}
\begin{proof} First let us prove a concentration inequality for $\mu[l,l+\delta]$. We have by Fubini's Theorem,
\begin{equation} \mathbb{E} \left [e^{\frac{l}{2}\mu[l,l+\delta] }\right ]= \mathbb{E} \left [e^{\frac{l}{2}\theta_0^2\delta}\prod_{i=1}^{\infty}e^{\frac{l}{2} \theta_i\mathbf{1}_{l\leq X_i\leq l+\delta}} \right ]=e^{\frac{l}{2}\theta_0^2\delta} \prod_{i=1}^{\infty} \left (1+ (e^{\frac{l}{2}\theta_i}-1)\mathbb{P} \left (l\leq X_i\leq l +\delta \right )\right ). \label{WAFWAF} \end{equation}
Furthermore we have for every $i\in \mathbb{N}$, since $X_i$ is an exponential random variable of parameter $\theta_i$
\[ (e^{\frac{l}{2}\theta_i}-1)\mathbb{P} \left (l\leq X_i\leq l +\delta \right )=(1-e^{-\frac{l}{2}\theta_i})\mathbb{P} \left ( \frac{l}{2} \leq X_i\leq \frac{l}{2} +\delta \right )\leq \frac{l}{2}\theta_i\mathbb{P} \left ( \frac{l}{2} \leq X_i\leq \frac{l}{2} +\delta \right ), \]
Therefore by \eqref{WAFWAF} and \eqref{107},
\begin{align} \mathbb{E} \left [e^{\frac{l}{2} \mu[l,l+\delta]}\right ]
& \leq \exp \left (\frac{l}{2}\theta_0^2\delta+ \sum_{i=1}^{\infty} \frac{l}{2} \theta_i \mathbb{P} \left ( \frac{l}{2} \leq X_i\leq \frac{l}{2} +\delta\right ) \right ) \notag
\\ & = \exp \left ( \frac{l}{2} \mathbb{E} \left [ \mu \left [\frac{l}{2},\frac{l}{2}+\delta\right ] \right ] \right ).\label{15101}
\end{align}
Moreover by Lemma \ref{2}, $t\mapsto \mathbb{E}[\mu[0,t]]$ is concave and increasing, hence,
\begin{equation} \mathbb{E} \left [ \mu \left [\frac{l}{2},\frac{l}{2}+\delta\right ] \right ] \leq \frac{2\delta}{l}\mathbb{E} \left [ \mu \left [0,\frac{l}{2}\right ] \right ] \leq \frac{2\delta}{l} \mathbb{E} \left [ \mu \left [0,l\right ] \right ].\label{15102} \end{equation}
Finally it follows from Markov's inequality, \eqref{15101}, and \eqref{15102} that for every $l,l',t\in \mathbb{R}^+$,
\begin{equation} \mathbb{P} \left ( \mu[l,l+\delta ] \geq\frac{2\delta}{l}\mathbb{E} \left [ \mu \left [0,l\right ] \right ] +\frac{2t}{l}\right )\leq e^{-t}. \label{1310} \end{equation}
We now derive the desired result from \eqref{1310}. First by the Borel--Cantelli Lemma, there exists almost surely an $N\in \mathbb{N}$, such that for every $n\geq N$ and $n\leq m \leq 8n$,
\[ \mu[\sqrt{n},\sqrt{m}] \leq 2\frac{(\sqrt{m}-\sqrt{n})}{\sqrt{n}}\mathbb{E} \left [ \mu \left [0,\sqrt{n}\right ] \right ]+\frac{6\log (n)}{\sqrt{n}}. \]
Now fix $l\geq N+10$, $0\leq \delta \leq l$ then let $n:=\max\{i, \sqrt{i}\leq l\}$ and let $m:= \min\{i, l+\delta\leq \sqrt{i}\}$. Since $l\mapsto \mu[0,l]$ is non decreasing, we have,
\begin{align} \mu[l,l+\delta]\leq \mu[\sqrt{n},\sqrt{m}] & \leq 2\frac{\sqrt{m}-\sqrt{n}}{\sqrt{n}}\mathbb{E} \left [ \mu \left [0,\sqrt{n}\right ] \right ]+12\frac{\log (\sqrt{n})}{\sqrt{n}} \notag
\\ & \leq 2 \frac{\delta+2/l}{l-1/l} \mathbb{E} \left [ \mu \left [0,l\right ] \right ]+12\frac{\log (l)}{l} . \label{1510manger}\end{align}
Finally by Lemma \ref{2}, $\mathbb{E} \left [ \mu \left [0,l\right ] \right ]=O(l)$ as $l\to \infty$ and the desired result follows from \eqref{1510manger}.
\end{proof}
\subsection{Key results on cuts and sticks}
\begin{lemma} \label{8}
Almost surely there exists $L_0\in \mathbb{R}^+$ such that for every $l\geq L_0$ there are at most $2l\mu[0,l]\leq 2l^2$ cuts on $[0,l]$.
\end{lemma}
\begin{proof}
Conditionally on $\mu$, $\{Y_i\}_{i\in \mathbb{N}}$ is a Poisson point process with rate $\mu[0,l]dl$ so the number of cuts in $[0,l]$ is stochastically dominated by a Poisson random variable $\alpha$ with mean $l\mu[0,l]$ and for $l$ large enough
\[ \mathbb{P} \left (\alpha\geq \frac 3 2 l\mu[0,l] \right)\leq \frac{1}{l^2}. \]
Thus by the Borel--Cantelli lemma almost surely for every $l\in \mathbb{N}$ large enough, there are at most $\frac{3}{2}l\mu[0,l]$ cuts on $[0,l]$. This can be easily extended to all $l\in \mathbb{R}^+$ large enough using Lemmas \ref{2} and \ref{5}. We omit the straightforward details.
\end{proof}
\begin{lemma} \label{9}
Almost surely there exists $i_0\in \mathbb{N}$ such that for every $i\geq i_0$:
\[ l_{i+1} \leq \frac{5\log(Y_i)}{M_i}. \]
\end{lemma}
\begin{proof}
Because the cuts are made at rate $\mu[0,l]dl$, for every $i\in \mathbb{N}$, $(Y_{i+1}-Y_i)\mu[0,Y_i]$ is stochastically dominated an exponential random variable with mean one. Therefore
\[ \mathbb{P} \left ((Y_{i+1}-Y_{i})\mu[0,Y_i] \geq 2\log(i)\right )\leq 1/i^2. \]
So by the Borel--Cantelli lemma and Lemma \ref{8}, for every $i$ large enough,
\[ Y_{i+1}-Y_i \leq \frac{2\log(i)}{\mu[0,Y_i]} \leq \frac{2\log(2Y_i^2)}{\mu[0,Y_i]} \leq \frac{5\log(Y_i)}{\mu[0,Y_i]}. \qedhere \]
\end{proof}
\begin{lemma} \label{10}
Almost surely there exists $L_0\in \mathbb{R}^+$ such that for all $l\geq L_0$ and $i\in \mathbb{N}$ with $Y_i\geq l$,
\[ m_{i+1}\leq \frac{\log^2{l}}{l}.\]
\end{lemma}
\begin{proof}
We have by Lemmas \ref{9}, \ref{5}, and \ref{6}, as $i\to \infty$,
\[ l_{i+1}\leq \mu\left [Y_i, Y_i+\frac{5 \log Y_i}{\mu[0,Y_i]} \right ] \leq O\left ( \frac{ \log Y_i}{\mu[0,Y_i]} \frac{\mu[0,Y_i]}{Y_i} +\frac{\log Y_i}{Y_i} \right )=o\left ( \frac{\log^2 Y_i}{Y_i}\right ).\qedhere \]
\end{proof}
\subsection{An estimate of distances in $\mathcal{T}$}
\begin{definition*} For every random variables $A$, $B$ on $\mathbb{R}$ we recall that $A$ is stochastically dominated by $B$ if and only if for every $t\in \mathbb{R}^+$, $\mathbb{P}(A\geq t)\leq \mathbb{P}(B\geq t)$. In this case we write $A\leq_{\st} B$. Also for every $l\in \mathbb{R}^+$, let $\Exp(l)$ denotes an exponential random variable of mean $l$.
\end{definition*}
\begin{lemma} \label{pizza} For every $x,y \in \mathbb{R}^+$, conditionally on $\mu$, $d(\mathcal{T}_x,y)\leq_{\st} \Exp (\frac{4}{\mu[0,x]} )$.
\end{lemma}
\begin{remark}
Proving an equivalent of Lemma \ref{pizza} is crucial for each studies on stick-breaking constructions, notably for compactness \cite{Curien,Seni} and convergence \cite{Aldous1,Uniform}. Although the proof presented below use strong property on $\mu$, more general methods can be found in \cite{Aldous1,Uniform,Curien,Seni}. Finally, we believe that such methods can be useful for the study of several other classes of algorithms.
\end{remark}
\begin{proof} To simplify the notation let for every $l\in \mathbb{R}^+$, $\mathcal{F}_l:=\sigma \big (\mu, \big \{(Y_i,Z_i)\big \}_{i\in \mathbb{N}}\cap [l,+\infty]\times \mathbb{R}^+ \big )$. We first prove that if $\mu[0,x]\geq 2\mu[0,y)$ then conditionally on $\mathcal{F}_y$, $d(\mathcal{T}_x,y)\leq_{\st}\Exp (\frac{2}{\mu[0,x]})$. If $y\leq x$ then $d(\mathcal{T}_x,y)=0$. We assume henceforth that it is not the case. Let us "follow" the geodesic path from $y$ to $\mathcal{T}_x$. More precisely we define the following sequence by induction (see Figure \ref{ImageProof}). Let $z_0:= y$, then for every $i\geq 0$, let $k_i:=\max\{k\in \mathbb{N}: Y_{k}<z_i\}$ and let $y_i:=Y_{k_i}$, and $z_{i+1}:=Z_{k_i}$. Additionaly let $T$ denotes the smallest integer such that $z_{T+1}\leq x$. Note that
\begin{equation} d(y,\mathcal{T}_x)=\sum_{i=0}^{T} \big (z_i-\max(y_i,x)\big). \label{07100} \end{equation}
\begin{figure}[!h] \label{ImageProof}
\centering
\includegraphics[scale=0.6]{Proof2.eps}
\caption{A typical construction of $(y_i,z_i)_{i\in \mathbb{N}}$. Note that in general we do not know if $y_T\in \mathcal{T}_x$.} \label{ImageProof}
\end{figure}
Now recall that conditionally on $\mu$, $\{(Y_i,Z_i),i\in \mathbb{N}\}$ is a Poisson point process, so $\{(y_i,z_i),\mathcal{F}_{y_i}\}_{i\geq 0}$ is a Markov chain. Also note that $T+1=\inf\{n: z_n<x\}$ is a stopping time for $\{(y_i,z_i),\mathcal{F}_{y_i}\}_{i\geq 0}$.
Moreover, for every $i\in \mathbb{N}$ conditionally on $(\mu,y_i,z_i)$, $z_{i+1}$ has law $p_{y_i}$. Hence if $y_i\geq x$,
\[ \mathbb{P} (z_{i+1}\leq x | \mu,y_i,z_i ) =p_{y_i}[0,x] \geq \frac{\mu[0,x]}{\mu[0,y)}\geq \frac{1}{2}. \]
So $T$ is stochastically dominated by a geometric random variable of parameter $1/2$. Furthermore, if $i\leq T$, conditionally on $(\mu,y_i)$, $\{Y_i\}_{i\in \mathbb{N}}\cap[x,y_i)$ is a Poisson point process of rate $\mu[0,l]\geq \mu[0,x]$ so $z_i-\max(y_i,x)\leq_{\st}\Exp(\frac{1}{\mu[0,x]})$. Finally it follows from \eqref{07100} that $d(\mathcal{T}_x,y)\leq_{\st} \Exp(\frac{2}{\mu[0,x]} ).$
Let us now treat the general case. As previously, we bound $d(\mathcal{T}_x,y)$ by following the geodesic path between $\mathcal{T}_x$ and $y$. More precisely, let for every $i\geq 0$, $x_i:=\inf\{a\in \mathbb{R}^+,\mu[0,a]\geq 2^i\mu[0,x]\}$ and let $y_i$ be the nearest point from $y$ on $[0,x_i]$. Note that
\begin{equation} d(x,y)=\sum_{i=0}^{+\infty} d(y_{i},y_{i+1}). \label{24102} \end{equation}
Then for every $i\geq 0$, since $2\mu[0,x_i]\geq \mu[0,x_{i+1})$, the first case yields, conditionally on $\mathcal{F}_{y_{i+1}}$,
\begin{equation} d(y_i,y_{i+1})=d(\mathcal{T}_{x_i},y_{i+1})\leq_{\st} \Exp \left (\frac{2}{\mu[0,x_i]} \right )\leq_{\st} \Exp \left (\frac{2^{1-i}}{\mu[0,x]} \right ). \label{24103} \end{equation}
Finally since for every $j>i$, $d(y_j,y_{j+1})$ is $\mathcal{F}_{y_{i+1}}$ measurable, it follows from \eqref{24102} and \eqref{24103} that
$ d(x,y)\leq_{\st} \Exp (\frac{4}{\mu[0,x]})$.
\end{proof}%
\section{The mass measure} \label{Section 5}
First we prove Lemma \ref{E=MC2} that describes precisely the evolution of the mass $\mu$ as we add branches to the tree. Then we prove that $(p_l)_{l\geq 0}$ is tight and use Lemma \ref{E=MC2} to prove that for every bounded Lipschitz function $(p_l(f))_{l\geq 0}$ converges. It proves, by the Portmanteau Theorem, that $(p_l)_{l\geq 0}$ converges weakly toward a probability measure $p$ (Theorem \ref{THM1}). Then we adapt the argument to prove Proposition \ref{other}.
\subsection{The mass conservasion lemma} \label{5..2}
\begin{definition*} For every $l\in \mathbb{R}^+$ let the projection of $x$ in $\mathcal{T}_l$ be the nearest point from $x$ in $\mathcal{T}_l$. Also for every $S\subset\mathcal{T}$, let $S^{\uparrow l}$ be the set of $x\in\mathcal{T}$ such that the projection of $x$ in $\mathcal{T}_l$ is in $S$.
\end{definition*}
\begin{lemma} \label{E=MC2}
Almost surely $(\mu,(Y_i)_{i\in\mathbb{N}})$ satisfy the following property. For every $a$ large enough, conditionally on $\mathcal{T}_{Y_a}$, for every measurable set $S\subset \mathcal{T}_{Y_a}$, the following assertions hold.
\begin{compactitem}
\item[(i)] Almost surely $\{ p_l(S^{\uparrow Y_a} ) \}_{l\in \mathbb{R}^+}$ converges toward a real number $p(S^{\uparrow Y_a} )$.
\item[(ii)] If $\mu(S)\geq \frac{\log^6 Y_a}{Y_a}$ with probability at least $1-\frac{1}{Y_a^5}$, for every $l\geq Y_a$
\[ \left (1-\frac 1 {\log Y_a} \right )p_{Y_a} (S) \leq p_{l} \left (S^{\uparrow Y_a} \right ) \leq \left (1+\frac 1 {\log Y_a} \right) p_{Y_a} (S).\]
\item[(iii)] If $\mu(S)\leq \frac{\log^6 Y_a}{Y_a}$ with probability at least $1-\frac{1}{Y_a^5}$, for every $l\geq Y_a$
\[ p_{l} \left (S^{\uparrow Y_a} \right ) \leq \frac{\left (\log Y_a\right )^{6}}{Y_aM_a}.\]
\end{compactitem}
\end{lemma}
\begin{proof} First for every $i\geq a$, let $A_i:= \mu_{Y_i} \left (S^{\uparrow Y_a} \right )$ and $\mathbb{F}_i:= \sigma \left (\mu, \{Y_n\}_{n\in \mathbb{N}}, \{Z_n\}_{1 \leq n < i} \right )$ . Note that for every $i\geq a$, since $Z_{i}$ has law $\frac{\mu_{Y_i}}{M_i}$, we have $(Y_i,Y_{i+1}]\subset S^{\uparrow Y_a}$ with probability $\frac{A_i}{M_i}$ so
\[ \mathbb{P} \left (A_{i+1}=A_i+m_{i+1} \right)=\frac{A_i}{M_i} \quad ; \quad \mathbb{P} \left (A_{i+1}=A_i \right )=\frac{M_i-A_i}{M_i} .\]
Thus $(A_i,\mathbb{F}_i)_{i\geq a}$ can be seen as a P\'olya urn in the sense of Lemma \ref{P\'olya}. Furthermore by Lemma \ref{10}, we have almost surely for every $a$ large enough, $\max_{n>a} m_n\leq \frac{\log^2 Y_a}{Y_a}$, hence by Lemma \ref{P\'olya} (b), for every $t\in [0,1]$,
\begin{align} \mathbb{P} \left ( \left . \sup_{i\geq a} \left | \frac{A_i}{M_i} - \frac{A_a}{M_a} \right | > t\frac{A_a}{M_a} \right | A_a \right )
\leq 2 \exp \left (-\frac{t^2}{8}\frac{A_a Y_a}{\log^2 Y_a} \right ) .
\label{2610} \end{align}
Also still by Lemma \ref{10} we have for every $a\in \mathbb{N}$ large enough, $i\geq a$, and $Y_i\leq l \leq Y_{i+1}$,
\[p_l(S^{\uparrow Y_a})=\frac{\mu_l(S^{\uparrow Y_a})}{\mu[0,l]}\leq \frac{\mu_{Y_i}(S^{\uparrow Y_a})+m_{i+1}}{\mu[0,Y_i]} = \frac{A_i}{M_i}+\frac{m_{i+1}}{M_i}\leq \frac{A_i}{M_i}+\frac{\log^2 Y_a}{Y_aM_a}, \]
and similarly
\[p_l(S^{\uparrow Y_a})=\frac{\mu_l(S^{\uparrow Y_a})}{\mu[0,l]}\geq \frac{\mu_{Y_{i+1}}(S^{\uparrow Y_a})-m_{i+1}}{\mu[0,Y_{i+1}]} = \frac{A_{i+1}}{M_{i+1}}-\frac{m_{i+1}}{M_{i+1}}\geq \frac{A_{i+1}}{M_{i+1}}-\frac{\log^2 Y_a}{Y_aM_a}. \]
Therefore,
\begin{equation} \sup_{l\geq Y_a} \left | p_l(S^{\uparrow Y_a}) - \frac{A_a}{M_a} \right | \leq \sup_{i\geq a} \left | \frac{A_i}{M_i} - \frac{A_a}{M_a} \right | + \frac{\log^2 Y_a}{Y_aM_a}. \label{0610} \end{equation}
The claims in (i) (ii) (iii) are applications of the inequalities in \eqref{2610} and \eqref{0610}.
Consider first (i). Note that \eqref{2610} implies that $\{\frac{A_i}{M_i}\}_{i\in \mathbb{N}}$ is almost surely Cauchy, and hence converges. Furthermore $\{\frac{\log^2 Y_a}{Y_aM_a}\}_{a\in\mathbb{N}}$ almost surely converges to 0. (i) then follows from \eqref{0610}.
Towards (ii), we have by assumption $A_a\geq \frac{\log^6 Y_a}{Y_a}$ so if $a\geq 10$, $\frac{\log^2 (Y_a)}{Y_aM_a}\leq \frac{1}{2\log Y_a}\frac{A_a}{M_a}$.
Therefore by \eqref{0610} it suffices to estimate the right-hand side of \eqref{2610} with $t=\frac{1}{2\log Y_a}$:
\[2 \exp \left (-\frac{t^2}{8}\frac{A_a Y_a}{\log^2 Y_a} \right ) \leq 2 \exp \left (-\frac{1}{32\log^2 Y_a}\frac{\log^{6} Y_a}{\log^2 Y_a} \right )=o\left (\frac{1}{Y_a^5} \right ) \]
and $(ii)$ follows. $(iii)$ can be treated similarly using \eqref{2610} with $t= \frac{\log^6 Y_a}{2Y_aA_a}$. We leave the details to the reader. This concludes the proof.
\end{proof}
\subsection{Weak convergence of $\mu_n$ : proof of Theorem \ref{THM1}}
In this section we prove Theorem \ref{THM1}. Let us start with the tightness of $(p_l)_{l\in \mathbb{R}^+}$ which follows from the following lemma.
\begin{lemma} \label{tight} For every $n\in \mathbb{N}$ let $A_n$ be the set of $x\in \mathcal{T}$ such that, $d(x,[0,\mathcal{X}_{2^n}]) \leq 8n/2^{n}$ and $B_n:= \bigcap_{m \geq n} A_m$.
The following assertions hold:
\begin{enumerate}
\item[(i)] Almost surely for every n large enough, for every $l\geq 0$, $p_l \left (B_n \right )\geq 1-2^{-2n}$.
\item[(ii)] For every $n$ large enough $B_n$ is compact.
\end{enumerate}
\end{lemma}
\begin{proof} First for every $n,m\in \mathbb{N}$, such that $m\geq n$, conditionally on $\mu$ we have by Fubini's theorem, Lemma \ref{pizza}, and Lemma \ref{5},
\begin{equation*} \mathbb{E} \left [ \left . p_{\mathcal{X}_{2^m}}\left (\mathcal{T}\backslash A_n \right ) \right | \mu \right ] = \int_{0}^{\mathcal{X}_{2^m}} \mathbb{P}\left( x\notin A_n \right ) \frac{d\mu(x)}{\mu[0,\mathcal{X}_{2^m}]} \leq e^{-\frac{8n}{2^n} \frac{\mu[0,\mathcal{X}_{2^{n}}]}{4}}=e^{-2n(1+o(1))}.\end{equation*}
It directly follows by Markov's inequality and the Borel--Cantelli lemma that almost surely for every $n$ large enough and $m\geq n$, \[p_{\mathcal{X}_{2^m}}\left (\mathcal{T} \backslash A_n \right ) \leq 2^{-2n-3}.\]
Therefore for every $n\in \mathbb{N}$ and $l\geq \mathcal{X}_{2^n}$, writing $k$ for the smallest integer such that $l\leq \mathcal{X}_{2^{k}}$ we have by Lemma \ref{5},
\[ p_{l}\left (\mathcal{T} \backslash A_n \right ) \leq \frac{\mu[0,\mathcal{X}_{2^k}] }{\mu[0,l]} p_{\mathcal{X}_{2^{k}}}\left (\mathcal{T} \backslash A_n \right ) \leq \frac{\mu[0,\mathcal{X}_{2^k}] }{\mu[0,\mathcal{X}_{2^{k-1}}]} p_{\mathcal{X}_{2^{k}}}\left (\mathcal{T} \backslash A_n \right ) \leq 2^{-2n-1}. \]
Note that the latter is also true for $l\leq \mathcal{X}_{2^n}$ since in this case $\mathcal{T}_l \subset \mathcal{T}_{\mathcal{X}_{2^n}}\subset A_n$. (i) then follows from a union bound on $n$.
Toward $(ii)$, note that $A_m$ is a closed set for $m\geq n$, so $B_n$ is a closed set as well. Therefore it suffices to show that any sequence $(x_i)_{i\in \mathbb{N}}$ in $B_n$ has an accumulation point. Fix $(x_i)_{i\in \mathbb{N}}$ then for every $m\in \mathbb{N}$ let $x_{i}^m$ be the projection of $x_i$ on $[0,\mathcal{X}_{2^m}]$. Since for every $m\in \mathbb{N}$, $\mathcal{T}_{\mathcal{X}_{2^m}}$ is compact, by a diagonal extraction procedure there exists an increasing function $\phi: \mathbb{N} \mapsto \mathbb{N}$ such that for every $m\in \mathbb{N}$, $(x_{\phi(i)}^m)_{i\in \mathbb{N}}$ converges. Hence, for every $m\geq n$ there exists $N\in \mathbb{N}$ such that for every $a,b\geq N$, $d(x_{\phi(a)}^m,x_{\phi(b)}^m )\leq 1/m $ and so
\begin{align*} d(x_{\phi(a)},x_{\phi(b)}) \leq d(x_{\phi(a)}, x_{\phi(a)}^m)+ d(x_{\phi(a)}^m, x_{\phi(b)}^m)+d(x_{\phi(b)}^m, x_{\phi(b)})\leq \frac{8m}{2^m}+\frac{1}{m}+\frac{8m}{2^m} .\end{align*}
Therefore $(x_{\phi(i)})_{i\in \mathbb{N}}$ is Cauchy and thus converges since $\mathcal{T}$ is complete by definition. Since $(x_i)_{i\in \mathbb{N}}$ is arbitrary, $B_n$ is compact.
\end{proof}
\begin{definition*}Let $\mathbb{F}$ be the set of positive, 1-Lipschitz functions that are bounded by 1 on $\mathcal{T}$.
For every finite measure $\nu$ on $\mathcal{T}$ and measurable function $f:\mathcal{T}\to \mathbb{R}$ let $\nu(f):=\int_{\mathcal{T}} f(x) d\nu(x)$. \end{definition*}
\begin{lemma} \label{manteau} Almost surely, for every $f\in \mathbb{F}$, $p_l(f)\limit p(f)$ as $l\to \infty$.
\end{lemma}
\begin{proof} First for every $a\in \mathbb{N}$ let $\{I_i^a\}_{1\leq i \leq N_a}$ be a partition of $\mathcal{T}_{Y_a}=([0,Y_a],d)$ into intervals of diameter at most $1/a$. Then for every $a\in \mathbb{N}$ and $1\leq i \leq N_a$ let $J_i^a:= (I^a_i)^{\uparrow Y_a}$ and let $x^a_i\in I^a_i$. Note that for every $a\in \mathbb{N}$, $\{J_i^a\}_{1\leq i \leq N_a}$ is a partition of $\mathcal{T}$. So for every $l\geq Y_a$ and $f\in \mathbb{F}$,
\begin{equation} p_l(f) = \sum_{i=1}^{N_a} p_l \left (\mathbf{1}_{J_i^a} f \right) = \sum_{i=1}^{N_a} p_l\left ( J_i^a \right )f(x_i^a)+ \sum_{i=1}^{N_a} p_l \left (\mathbf{1}_{J_i^a} \left (f-f(x_i^a)\right ) \right). \label{11101} \end{equation}
By Lemma \ref{E=MC2} (i), almost surely for every $f\in \mathbb{F}$ the first sum converges toward $\sum_{i=1}^{N_a} p\left ( J_i^a \right )f(x_i^a)$ as $l$ goes to infinity. Let us bound the second sum in order to prove that $(p_l(f))_{l\in \mathbb{R}^+}$ is Cauchy. For every $a\in \mathbb{N}$ let $k_a$ be the largest integer such that $\mathcal{X}_{2^{k_a}}\leq Y_a$. We have for every $f\in \mathbb{F}$:
\[ \sum_{i=1}^{N_a} p_l \left (\mathbf{1}_{J_i^a} \left (f-f(x_i^a)\right ) \right) = \sum_{i=1}^{N_a} p_l \left (\mathbf{1}_{J_i^a\cap B_{k_a}} \left (f-f(x_i^a)\right ) \right) + p_l \left (\mathbf{1}_{\mathcal{T}\backslash B_{k_a}} \left (f-f(x_i^a)\right ) \right) \]
and \[ \left | \sum_{i=1}^{N_a} p_l \left (\mathbf{1}_{J_i^a} \left (f-f(x_i^a)\right ) \right) \right | \leq \sum_{i=1}^{N_a} p_l \left (\mathbf{1}_{J_i^a\cap B_{k_a}} \left | f-f(x_i^a) \right | \right) + p_l(\mathcal{T}\backslash B_{k_a}). \]
Furthermore for every $a\in \mathbb{N}$ and $1\leq i \leq N_a$, recall that by definition $I^a_i$ has diameter at most $\frac{1}{a}$ and that $d_H([0,\mathcal{X}_{2^{k_a}}],B_{k_a})\leq 8k_a2^{-k_a}$. Therefore $J_i^a\cap B_{k_a}=(I^a_i)^{\uparrow Y_a} \cap B_{k_a}$ has diameter at most $\delta_a:= \frac{1}{a}+\frac{16k_a}{2^{k_a}}$. Hence for every $f\in \mathbb{F}$,
\begin{equation*} \left | \sum_{i=1}^{N_a} p_l \left (\mathbf{1}_{J_i^a} \left (f-f(x_i^a)\right ) \right) \right |
\leq \sum_{i=1}^{N_a} p_l \left (J_i^a \cap B_{k_a} \right ) \delta_a + p_l(\mathcal{T}\backslash B_{k_a}) \leq \delta_a+p_l(\mathcal{T}\backslash B_{k_a}).\notag
\end{equation*}
Moreover by Lemma \ref{tight} for every $a$ large enough $p_l(\mathcal{T}\backslash B_{k_a})\leq 2^{-2k_a}$. Finally for every $f\in \mathbb{F}$,
\begin{equation} \limsup_{a\to \infty} \limsup_{l\to \infty} \left | \sum_{i=1}^{N_a} p_l \left (\mathbf{1}_{J_i^a} \left (f-f(x_i^a)\right ) \right) \right | =0, \label{0710} \end{equation}
which implies together with \eqref{11101} that $(p_l(f))_{l\in \mathbb{R}^+}$ is Cauchy and thus converges.
\end{proof}
\begin{proof}[Proof of Theorem \ref{THM1}]
First by lemma \ref{tight}, $(p_l)_{l\in \mathbb{R}^+}$ is tight. The convergence of $(p_l)_{l\in \mathbb{R}^+}$ then directly follows from Lemma \ref{manteau} and the Portmanteau theorem.
Towards proving that $p$ has full support, we first prove that $\mu$ has almost surely full support. Note that it suffices to prove that for every $a<b\in \mathbb{R}^+$, almost surely $\mu[a,b]>0$. If $\theta_0>0$ then $\mu[a,b]\geq (b-a)\theta_0^2>0$. So we assume henceforth that $\theta_0=0$. Note that in this case, $\sum_{i=1}^\infty \theta_i=\infty$. Moreover, recall that $\{X_i\}_{i\in \mathbb{N}}$ is a family of independent exponential random variables of parameter $\{\theta_i\}_{i\in\mathbb{N}}$ so that,
\[ \sum_{i=1}^\infty \mathbb{P}(X_i\in[a,b])=\sum_{i=1}^\infty e^{-\theta_i a}\left (1-e^{-\theta_i(b-a)} \right ) =\infty. \]
Therefore by the Borel--Cantelli lemma, for every $a,b\in \mathbb{R}^+$ almost surely there exists an $i\in \mathbb{N}$ such that $X_i\in [a,b]$ and so $\mu[a,b]\geq \theta_i>0$. Thus, $\mu$ has almost surely full support.
Next we prove that $p$ also has full support. Fix $x\in \mathbb{R}^+$ and $\varepsilon>0$. Additionally for every $a\in \mathbb{N}$ let $k_a$ be the largest integer such that $\mathcal{X}_{2^{k_a}}\leq Y_a$. Note that for every $a\in \mathbb{N}$ large enough, by definition of $B_{k_a}$, $B(x,\varepsilon)^{\uparrow Y_a}\cap B_{k_a}$ has diameter at most $\varepsilon+16k_a2^{-k_a}\leq 2\varepsilon$. It follows that,
\begin{equation} p \left (B(x,2\varepsilon)\right )\geq p \left (B(x,\varepsilon)^{\uparrow Y_a}\cap B_{k_a} \right ) \geq p \left (B(x,\varepsilon)^{\uparrow Y_a} \right ) - p(\mathcal{T}\backslash B_{k_a}). \label{07102}\end{equation}
On the one hand, recall that almost surely $\mu(B(x,\varepsilon))>0$. Thus by Lemma \ref{E=MC2} (ii), for every $a$ large enough, with probability at least $1-1/Y_a^5$,
\[ p \left (B(x,\varepsilon)^{\uparrow Y_a} \right ) \geq \frac{1}{2}p_{Y_a} (B(x,\varepsilon)) =\frac{\mu_{Y_a}(B(x,\varepsilon))}{2M_a}. \]
On the other hand, by Lemmas \ref{tight} (i), \ref{5}, and the definition of $k_a$, for every $a$ large enough,
\[ p(\mathcal{T}\backslash B_{k_a})\leq 2^{-2k_a} \leq 2\mu[0,\mathcal{X}_{2^{k_a}}]^{-2}\leq 2\mu[0,Y_a]^{-2} =o\left (1/M_a\right ).\]
Therefore by \eqref{07102}, almost surely $p(B(x,2\varepsilon))>0$. Since $x$, $\varepsilon$ were arbitrary and since rational numbers are dense on $\mathcal{T}$, it follows that $p$ has full support.
Finally, we prove that almost surely $p$ gives measure $1$ to the set of leaves and is non-atomic.
For every $\varepsilon>0$ and $S\subset \mathcal{T}$, let $B(S,\varepsilon)=\{x\in \mathcal{T}: d(x,S)<\varepsilon\}$. Then let $(\varepsilon_a)_{a\in\mathbb{N}}$ be a sequence of positive real numbers decreasing sufficiently fast so that for every $a>0$ and $0\leq i <a$ we have $\mu_{Y_a}(B((Y_i,Y_{i+1}],\varepsilon_a))\leq 2 \mu_{Y_a}(Y_a,Y_{a+1}]$. By Lemma \ref{E=MC2} (ii) (iii), for every $a$ large enough and $0\leq i<a$, with probability at least $1-1/Y_a^{5}$, for every $l\geq Y_a$,
\begin{equation} p_{l} \left (B((Y_i,Y_{i+1}],\varepsilon_a)^{\uparrow Y_a} \right) \leq \max \left \{ 2 p_{Y_a} \Big (B((Y_i,Y_{i+1}],\varepsilon_a) \Big); \frac{\left (\log Y_a\right )^{6}}{Y_aM_a}\right \}. \label{0810} \end{equation}
Since by Lemma \ref{8} $a=O(Y_a^2)$, the Borel--Cantelli lemma implies that almost surely \eqref{0810} is true for every $a$ large enough, $0\leq i<a$ and $l\geq Y_a$. Furthermore by Lemma \ref{10} $M:=\max_{i\in \mathbb{N}} \mu(Y_i,Y_{i+1}]<\infty$. Also note that $\frac{\left (\log Y_a\right )^{6}}{Y_a}\limit 0$ as $a\to +\infty$. Therefore for every $a$ large enough, $0\leq i<a$, and $l\geq a$:
\begin{equation} p_{l} \left (B((Y_i,Y_{i+1}],\varepsilon_a)^{\uparrow Y_a} \right) \leq \frac{4M}{M_a}. \label{08102}\end{equation}
Moreover since for every $a\in \mathbb{N}$ the projection on $\mathcal{T}_{Y_a}$ (see \ref{5..2} for definition) is a continuous fonction, for every $0\leq i <a$, $B((Y_i,Y_{i+1}],\varepsilon_a)^{\uparrow Y_a}$ is open. Thus by letting $l\to \infty$ in \eqref{08102}, the Portmanteau theorem yields for $a$ large enough:
\begin{equation} p\left (B((Y_i,Y_{i+1}],\varepsilon_a)^{\uparrow Y_a} \right) \leq \frac{4M}{M_a}, \label{2102} \end{equation}
which tends to $0$ as $a\to\infty$. So for every $i\in \mathbb{N}$, $p[Y_i,Y_{i+1}]=0$. Summing over all $i\in \mathbb{N}$ we get $p(\mathbb{R}^+)=0$ and so $p$ gives measure $1$ to the set of leaves. Note that \eqref{2102} also yield for every $a\in \mathbb{N}$,
\[ \sup_{x\in \mathcal{T}} p\{x\} = \max_{0\leq i <a} \sup_{x\in [Y_i,Y_{i+1}]^{\uparrow Y_a}} p\{x\} \leq \frac{4M}{M_a},\]
which implies, taking $a\to \infty$, that $p$ is non-atomic.
\end{proof}
\subsection{Other convergences toward $p$ : proof of Proposition \ref{other}}
In this section we prove Proposition \ref{other}. We will in fact prove the following stronger result.
\begin{lemma} \label{other2}Let $\mu^{\alpha}$ be a positive random Borel measure on $\mathbb{R}^+$ which is $\sigma(\mu,\{Y_i\}_{i\in \mathbb{N}})$ measurable. Let for every $l\in \mathbb{R}^+$, $\mu^{\alpha}_l$ be the restriction of $\mu^{\alpha}$ to $\mathcal{T}_l=([0,l],d)$ and $p^{\alpha}_l:=\frac{\mu^{\alpha}_l}{\mu^{\alpha}[0,l]}$. Suppose that almost surely the following assertions hold:
\begin{compactitem}
\item[(i)] For every $l>0$ $\mu^{\alpha}[0,l]<\infty$, and $\mu^{\alpha}(\mathbb{R}^+)=+\infty$.
\item[(ii)] There exists $\varepsilon>0$ such that $\mu^{\alpha}(Y_{i-1},Y_{i}]=o(\mu^{\alpha}[0,Y_i]^{1-\varepsilon})$.
\item[(iii)] For all $\varepsilon>0$, $\sum_{i=1}^{n} \mu^{\alpha}(Y_{i-1},Y_i]\mathbf{1}_{Y_i-Y_{i-1}>\varepsilon} =o(\mu^{\alpha}[0,Y_n])$. \end{compactitem}
Then almost surely $\{p^{\alpha}_l\}_{l\in \mathbb{R}^+}$ converges weakly toward $p$.
\end{lemma}
In order to prove Lemma \ref{other2}, we first show the following strong law of large number.
\begin{lemma} \label{strong2} Let $\mu^{\alpha}$ be such as in Lemma \ref{other} and $S\subset \mathcal{T}$ be a random measurable set such that for every $n$ large enough, $p_{Y_n}(S)$ is $\sigma(\mu,\{Y_i\}_{i\in \mathbb{N}}, \{Z_i\}_{1\leq i <n})$ measurable. We have almost surely,
\[ \limsup_{n\to \infty} \sum_{i=1}^n p^{\alpha}(Y_{i-1},Y_{i}] \mathbf{1}_{Z_{i-1}\in S} \leq \limsup_{l\to \infty} p_l(S).\]
\end{lemma}
\begin{proof}Let $\{U_n\}_{n\in \mathbb{N}}$ be a family of independent uniform random variables on $[0,1]$. Since for every $n\in \mathbb{N}$, conditionally on $(\mu,\{Y_j\}_{j\in \mathbb{N}}, \{Z_j\}_{1\leq j <n} )$, $Z_n$ has law $p_{Y_n}$, we may couple $\{U_n\}_{n\in \mathbb{N}}$ and $\{Z_n\}_{n\in \mathbb{N}}$ in such a way that for every $n$ large enough, $Z_n\in S$ if and only if $U_n\leq p_{Y_n}(S)$. Therefore, by Lemma \ref{strong} and assumptions $(i)$ and $(ii)$, almost surely for every $t>\limsup_{l\to \infty} p_l(S) $,
\begin{equation}\limsup_{n\to \infty} \frac{\sum_{i=1}^n \mu^{\alpha}(Y_{i-1},Y_{i}]\mathbf{1}_{Z_{i-1}\in S}}{\mu^{\alpha}[0,Y_n]} \leq \limsup_{n\to \infty}\frac{\sum_{i=1}^n \mu^{\alpha}(Y_{i-1},Y_{i}]\mathbf{1}_{U_{i-1}\leq t}}{\mu^{\alpha}[0,Y_n]}=t. \label{cle} \end{equation}
Taking $t\to \limsup_{l\to \infty} p_l(S)$ in \eqref{cle} yields the desired inequality.
\end{proof}
\begin{proof}[Proof of Lemma \ref{other2}] First by the Portmanteau's theorem it suffices to prove that for every $f\in \mathbb{F}$, $p_l^\alpha(f)\to p(f)$ where $\mathbb{F}$ is the set of positive, 1-Lipschitz functions that are bounded by 1 on $\mathcal{T}$. Moreover since we work with probability measures and since for every $f\in \mathbb{F}$, $(1-f)\in \mathbb{F}$, it suffices to prove instead that for every $f\in \mathbb{F}$, $\limsup p_l^\alpha(f)\leq p(f)$. To this end, we proceed as in the proof of Lemma \ref{manteau} and will hence use the same notations. In addition, for $\varepsilon>0$ let $\Lambda_{\varepsilon}:=\bigcup_{i, Y_{i+1}-Y_i\leq \varepsilon} (Y_{i},Y_{i+1})$ and for every $x\in \mathcal{T}$, let $\zeta(x):=Z_{\max\{i, Y_i\leq x\}}$.
Now fix $\varepsilon>0$ and recall from the proof of Lemma \ref{manteau} that for every $a\in \mathbb{N}$, $\{J_i^a\}_{1\leq i \leq N_a}$, is a partition of $\mathcal{T}$, so for every $f\in\mathbb{F}$ and $l\geq Y_a$,
\begin{equation} p_l^{\alpha}(f) \leq \sum_{i=1}^{N_a} p_l^{\alpha}(f\mathbf{1}_{z(\cdot)\in J_i^a\cap B_{k_a}}\mathbf{1}_{\Gamma_e})+p_l^{\alpha}(f\mathbf{1}_{z(\cdot)\notin B_{k_a}}) + p_l^{\alpha}(f\mathbf{1}_{\mathcal{T}\backslash \Gamma_\varepsilon}).\label{0910} \end{equation}
We now upper bound each term of \eqref{0910} separately. First, recall that $J_i^a\cap B_{k_a}$ have diameter at most $\delta_a$, thus for every $1\leq i \leq N_a$ and $s\in S_i^a:=\{x,z(x)\in J_i^a\cap B_{k_a}\}\cap\Gamma_\varepsilon$ we have $d(s,x_i^a)\leq \varepsilon+\delta_a$. Therefore for every $f\in \mathbb{F}$,
\begin{equation}
\sum_{i=1}^{N_a} p_l^{\alpha}(f\mathbf{1}_{S_i^a}) \leq \sum_{i=1}^{N_a} p_l^{\alpha}((f(x_i^a)+\varepsilon+\delta_a)\mathbf{1}_{S_i^a}) \leq \sum_{i=1}^{N_a} f(x_i^a)p_l^{\alpha}\left (\mathbf{1}_{z(\cdot)\in J_i^a}\right )+\varepsilon +\delta_a. \label{2010}
\end{equation}
Furthermore by Lemma \ref{E=MC2} (i) almost surely for every $a\in \mathbb{N}$ and $1\leq i \leq N_a$, $p_l\left (J_i^a\right )\limit p(J_i^a)$ as $l\to \infty$, hence by Lemma \ref{strong2} almost surely
\[ \limsup_{l\to \infty} p_l^{\alpha}(\mathbf{1}_{z(\cdot)\in J_i^a})\leq p(J_i^a).\]
Therefore since $\delta_a\to 0$ as $a\to \infty$, we have by \eqref{2010} and \eqref{0710} for every $f\in \mathbb{F}$,
\[ \limsup_{a\to \infty} \limsup_{l\to \infty} \sum_{i=1}^{N_a} p_l^{\alpha}(f\mathbf{1}_{S_i^a})\leq \limsup_{a\to \infty} \sum_{i=1}^{N_a} f(x_i^a)p(J_i^a) +\varepsilon \leq p(f)+\varepsilon.\]
Next we have by Lemma \ref{tight} (i) and Lemma \ref{strong2}, almost surely for every $a$ large enough,
\[ \limsup_{l\to \infty} p_l^{\alpha}(\mathbf{1}_{z(\cdot)\notin B_{k_a}}) \leq 2^{-2k_a}. \] Futhermore by assumption (iii), $p_l^{\alpha}(\mathbf{1}_{\mathcal{T}\backslash \Gamma_\varepsilon})\to 0$ as $l\to 0$. Finally \eqref{0910} yields, for every $f\in\mathbb{F}$,
\[ \limsup_{l\to \infty} p_l^{\alpha}(f)\leq \limsup_{a\to \infty} p(f)+ \varepsilon+2^{-2k_a} = p(f)+\varepsilon. \]
Taking $\varepsilon\to 0$ in the previous inequality concludes the proof.
\end{proof}
\begin{proof}[Proof of Proposition \ref{other}]
We now justify that $\mu^{\leadsto}$ and $\mu^{\bullet}$ satisfy the assumptions of Lemma \ref{other2}. First (i) and the $\sigma(\mu,\{Y_i\}_{i\in \mathbb{N}})$ measurability for $\mu^{\leadsto}$ and $\mu^{\bullet}$ are straightforward from their definitions. (ii) for $\mu^{\leadsto}$ is an immediate consequence of Lemma \ref{9}. (ii) for $\mu^{\bullet}$ comes from $\mu^{\bullet}(Y_n,Y_{n+1}]=1$. (iii) is a little tedious to prove and follows directly from the fact that conditionally on $\mu$, $\{Y_i\}_{i\in \mathbb{N}}$ is a Poisson point process with rate $\mu[0,l]dl$ and that by Lemma \ref{5} almost surely $\mu[0,l]\limit \infty$ as $l\to \infty$. We omit the details. This concludes the proof of Proposition \ref{other}.
\end{proof}
\section{Compactness} \label{Compactness} \label{Section 6}
\subsection{Equivalent condition}
In this section, we obtain a condition equivalent to that of Theorem \ref{THM2} which is more convenient to study the compactness of the ICRT from the bounds provided by Lemmas \ref{BIG LEMMA} and \ref{manger}. Additionally we also prove that the condition conjectured in \cite{ExcICRT} is also equivalent to that of Theorem \ref{THM2}. For $l\geq 0$, recall that $\mathcal{X}_l$ is defined by $\mathbb{E}[\mu[0,\mathcal{X}_l]]=l$ and let
\[\psi(l):= \frac{\theta_0^2}{2} l^2+ \sum_{i=1}^{\infty} (e^{-l\theta_i}-1+l\theta_i). \]
\begin{lemma} \label{condition} The following conditions are equivalent:
\[ (i) \quad \int^{+\infty} \frac{dl}{l\mathbb{E}[\mu[0,l]]} <+\infty \quad , \quad (ii) \quad \int^{+\infty} \frac{dl}{\psi(l)} <+\infty \quad , \quad(iii) \quad \sum^{+\infty} \frac{\log \mathcal{X}_{2^n}}{2^n}<\infty. \]
\end{lemma}
\begin{proof}
Since for every $x\in \mathbb{R}^+$, $e^{-x}-1+x\leq x(1-e^{-x}) \leq 2 \left ( e^{-x}-1+x \right )$, for every $l\geq 0$:
\[ \frac{\theta_0^2}{2} l^2+ \sum_{i=1}^{\infty} \left (e^{-l\theta_i}-1+l\theta_i \right ) \leq \theta_0^2 l^2+\sum_{i=1}^{\infty}l\theta_i \left (1-e^{-\theta_il} \right ) \leq \theta_0^2 l^2+ \sum_{i=1}^{\infty} 2 \left(e^{-l\theta_i}-1+l\theta_i \right). \]
So by \eqref{107} for every $l\geq 0$, $\psi(l)\leq l\mathbb{E}[\mu[0,l]]\leq 2\psi(l)$. It follows readily that (i) and (ii) are equivalent. Furthermore
\[ \int_{\mathcal{X}_1}^{\infty} \frac{dl}{l\mathbb{E}[\mu[0,l]]} = \sum_{k=0}^{\infty} \int_{\mathcal{X}_{2^k}}^{\mathcal{X}_{2^{k+1}}} \frac{dl}{l\mathbb{E}[\mu[0,l]]} \leq \sum_{k=0}^{\infty} \int_{\mathcal{X}_{2^k}}^{\mathcal{X}_{2^{k+1}}} \frac{dl}{l2^k}
= \sum_{k=1}^{\infty} \frac{\log \mathcal{X}_{2^k}}{2^k}-\log \mathcal{X}_1, \]
and similarly
\[ \int_{\mathcal{X}_1}^{\infty} \frac{dl}{l\mathbb{E}[\mu[0,l]]} = \sum_{k=0}^{\infty} \int_{\mathcal{X}_{2^k}}^{\mathcal{X}_{2^{k+1}}} \frac{dl}{l\mathbb{E}[\mu[0,l]]} \geq \sum_{k=0}^{\infty} \int_{\mathcal{X}_{2^k}}^{\mathcal{X}_{2^{k+1}}} \frac{dl}{l2^{k+1}}= \sum_{k=1}^{\infty} \frac{\log \mathcal{X}_{2^k}}{2^{k+1}}-\frac{\log \mathcal{X}_1}{2}. \]
So (i) and (iii) are equivalent.
\end{proof}
\subsection{The condition of Theorem \ref{THM2} is sufficient for compactness}
The aim of this section is to prove Lemma \ref{BIG LEMMA} below. This Lemma implies that under condition (iii) of Lemma \ref{condition}, $(\mathcal{T}_{\mathcal{X}_{2^k}})_{k\in \mathbb{N}}$ is a Cauchy sequence of compact sets for the Hausdorff topology and thus converges toward a compact set. Since $(\mathcal{T}_{\mathcal{X}_{2^k}})_{k\in\mathbb{N}}$ is increasing (for $\subset$) toward $\mathcal{T}$, $\mathcal{T}$ is the only possible limit, and hence is compact.
\begin{lemma} \label{BIG LEMMA}
Almost surely, for every $k$ large enough:
\[d_{H}(\mathcal{T}_{\mathcal{X}_{2^{k-1}}}, \mathcal{T}_{\mathcal{X}_{2^{k}}})\leq 21 \frac{\log \mathcal{X}_{2^k}}{2^k}. \]
\end{lemma}
\begin{proof} For every $k\in \mathbb{N}$ and $x\in \mathcal{T}$, let $E_k(x)$ denotes the event $d(x,[0,\mathcal{X}_{2^{k-1}}])>20 \log \mathcal{X}_{2^k} 2^{-k}$.
First by Fubini's theorem and Lemma \ref{pizza}, we have conditionally on $\mu$:
\[ \mathbb{E} \left [ \left . \int_{0}^{\mathcal{X}_{2^k}} \mathbf{1}_{E_k(x)} dx \right | \mu \right ] = \int_{0}^{\mathcal{X}_{2^k}} \mathbb{P} \left ( \left . E_k(x) \right | \mu \right ) dx \leq \mathcal{X}_{2^k} \exp \left ( -5 \frac{\log \mathcal{X}_{2^k}}{2^k} \mu[0,\mathcal{X}_{2^{k-1}}] \right ).
\]
Then by Lemma \ref{5} as $k$ goes to infinity $\mu[0,\mathcal{X}_{2^{k}}]\sim 2^k$. So for every $k$ large enough:
\[ \mathbb{E} \left [ \left . \int_{0}^{\mathcal{X}_{2^k}} \mathbf{1}_{E_k(x)} dx \right | \mu \right ] \leq \mathcal{X}_{2^k}^{-4/3}. \]
Furthermore by Lemma \ref{2}, $2^k=O(\mathcal{X}_{2^k})$ so $\sum \mathcal{X}_{2^k}^{-1/3}<\infty$. Hence by Markov's inequality and the Borel--Cantelli lemma, for every $k$ large enough:
\[ \int_{0}^{\mathcal{X}_{2^k}} \mathbf{1}_{E_k(x)}dx < \mathcal{X}_{2^k}^{-1}. \]
Note that it implies that, for every $k$ large enough and $x\in [0,\mathcal{X}_{2^k}]$,
\[ d(x,[0,\mathcal{X}_{2^{k-1}}])\leq 20 \frac{\log \mathcal{X}_{2^k}}{2^k} +\mathcal{X}_{2^k}^{-1}, \]
since otherwise the geodesic path from $x$ to $[0,\mathcal{X}_{2^{k-1}}]$ would contain a segment $S$ of length at least $\frac{1}{\mathcal{X}_{2^k}}$ such that $d(S,[0,\mathcal{X}_{2^{k-1}}])> 20\log \mathcal{X}_{2^k} 2^{-k}$. Finally by Lemma \ref{2}, for every $k$ large enough $\mathcal{X}_{2^k}\geq 2^k$, hence $\mathcal{X}_{2^k}^{-1} \leq \log \mathcal{X}_{2^k} 2^{-k}$. This concludes the proof.
\end{proof}
\subsection{The condition of Theorem \ref{THM2} is necessary for compactness}
The following section is organized as follow: Lemma \ref{Long} defines and proves the existence of "long" segments, Lemma \ref{glue} proves that they tend to "aggregate". Lemma \ref{manger} deduces a lower bound on $d_H(\mathcal{T}_{\mathcal{X}_{2^k}}, \mathcal{T})$ from the two previous lemmas, thus proving that the condition is necessary. Finally Lemma \ref{NonCompact} gives a more precise view of the geometry of the tree in the non-compact case: "the tree is infinite in every direction".
\begin{lemma} \label{Long} For every $n\in \mathbb{N}$ let $L_n:=\frac{\log \mathcal{X}_{2^n}}{2^{n+2}}$ and let $\mathcal{I}_n$ be the set of segments $[Y_a+L_n, Y_{a+1}]$ with
\[ Y_a\in [\mathcal{X}_{2^n},\mathcal{X}_{2^{n+1}}) \quad ; \quad Y_a+L_n\leq Y_{a+1} \quad ; \quad \mu \left [Y_a+L_n, Y_{a+1} \right ] \geq \frac{1}{\mathcal{X}_{2^{n+1}}^2}. \]
Almost surely for every $n$ large enough we have $ \# \mathcal{I}_n\geq 2^{n+2}\mathcal{X}_{2^n}^{1/3}$.
\end{lemma}
\begin{proof}
Write $ \mathcal{I}'_n$ for the set of segments $[Y_a+L_n, Y_{a+1}]$ with
\[ Y_a\in [\mathcal{X}_{2^n},\mathcal{X}_{2^{n+1}}) \quad ; \quad Y_a+L_n\leq Y_{a+1} \quad ; \quad \mu \left [Y_a+L_n, Y_{a+1} \right ] < \frac{1}{\mathcal{X}_{2^{n+1}}^2}. \]
First by Lemmas \ref{8} and \ref{5}, for every $n$ large enough, there are at most $2^{n+2}\mathcal{X}_{2^{n+1}}$ cuts on $\left [0,\mathcal{X}_{2^{n+1}} \right ]$, hence $\# \mathcal{I}'_n \leq 2^{n+2}\mathcal{X}_{2^{n+1}}$. Furthermore, by Lemma \ref{10}, for every $n$ large enough and $I\in\mathcal{I}_n$, we have $\mu( I) \leq \log^2 \mathcal{X}_{2^n}/ \mathcal{X}_{2^n}$ so
\[
\sum_{I\in \mathcal{I}_n\cup \mathcal{I}'_n} \mu \left ( I \right )
=\sum_{I\in \mathcal{I}_n} \mu \left ( I \right ) +\sum_{I\in \mathcal{I}'_n} \mu \left ( I \right )
\leq \# \mathcal{I}_n\frac{\log^2 \mathcal{X}_{2^n}}{\mathcal{X}_{2^n}}+ \frac{2^{n+2}}{\mathcal{X}_{2^{n+1}}} .
\]
Therefore, since $\mathcal{X}_{2^{n+1}}\geq \mathcal{X}_{2^n}$, it suffices to prove that, writing $S_n:= \bigcup_{I\in \mathcal{I}_n\cup \mathcal{I}'_n} I$,
\begin{equation} \mu \left ( S_n\right ) > 2^{n+2}{\mathcal{X}_{2^n}^{-2/3}}{\log^2 \mathcal{X}_{2^{n}}}+\frac{2^{n+2}}{\mathcal{X}_{2^n}}\label{0210}. \end{equation}
Note that for every $x\in [\mathcal{X}_{2^n},\mathcal{X}_{2^{n+1}}]$, $x\in S_n$ if and only if there is a cut in $[\mathcal{X}_{2^n},x]$ and no cut in $[x-L_n,x]$. So if there is a cut in $[\mathcal{X}_{2^n},\mathcal{X}_{2^n}+1]$,
\[ \mu \left ( S_n \right )
\geq
\int_{\mathcal{X}_{2^n}+L_n+1}^{\mathcal{X}_{2^{n+1}}} \emptyset_{x-L_n,x}d\mu(x), \]
where for every $x\leq y$, $\emptyset_{x,y}:=\mathbf{1}_{\forall i\in \mathbb{N},\,Y_i\notin [x,y]}$.
Let $A_n$ denotes the right-hand side above. Since, conditionally on $\mu$, $(Y_i)_{i\in \mathbb{N}}$ is a Poisson point process of rate $\mu[0,l]dl$, for every $n$ large enough
\[ \mathbb{P} \left ( \left . A_n >\mu \left ( S_n \right ) \right | \mu\right ) \leq \mathbb{P} \left ( \left . \emptyset_{\mathcal{X}_{2^n},\mathcal{X}_{2^n}+1}=0 \right | \mu\right )
\leq e^{-\mu \left[0,\mathcal{X}_{2^n} \right ]}
\leq e^{-2^{n-1}}. \]
Therefore, by the Borel--Cantelli lemma, almost surely for every $n$ large enough $\mu \left ( S_n \right )\geq A_n$.
We now lower bound $A_n$ via a second moment method. We have, still by the properties of $(Y_i)_{i\in \mathbb{N}}$,
\begin{equation} \mathbb{E}[ A_n|\mu] \geq \int_{\mathcal{X}_{2^n}+L_n+1}^{\mathcal{X}_{2^{n+1}}} e^{- \mu \left[0,\mathcal{X}_{2^{n+1}} \right ]L_n}d\mu(x) = e^{-\mu \left[0,\mathcal{X}_{2^{n+1}} \right ]L_n }\mu \left [\mathcal{X}_{2^n}+L_n+1,\mathcal{X}_{2^{n+1}}\right ].\label{18002}
\end{equation}
Furthermore note that $\frac{1+L_n}{\mathcal{X}_{2^n}}\to0$ as $n\to \infty$, hence by Lemmas \ref{5} and \ref{2} almost surely as $n\to \infty$,
\begin{equation*}\mu[\mathcal{X}_{2^n}+L_n+1,\mathcal{X}_{2^{n+1}}] = \mathbb{E}[\mu[0,\mathcal{X}_{2^{n+1}}]](1+o(1))-\mathbb{E}[\mu[0,\mathcal{X}_{2^n}+L_n+1]](1+o(1))
\sim 2^n.
\end{equation*}
It follows from \eqref{18002}, Lemma \ref{5}, and the definition of $L_n$ that, as $n\to \infty$,
\begin{equation} \mathbb{E}[A_n| \mu] \geq \mathcal{X}_{2^n}^{-1/2+o(1)} 2^n. \label{1800} \end{equation}
Moreover we have by Fubini's theorem,
\[ \Varr[ A_n|\mu]=\int_{\mathcal{X}_{2^n}+L_n+1}^{\mathcal{X}_{2^{n+1}}}\int_{\mathcal{X}_{2^{n}}+L_n+1}^{\mathcal{X}_{2^{n+1}}} \Covv \left [ \left . \emptyset_{x-L_n,x}\emptyset_{y-L_n,y} \right | \mu \right ]d\mu(x)d\mu(y). \]
Note that for every $x,y\in \mathbb{R}^+$, $\Covv \left [ \left . \emptyset_{x-L_n,x} \emptyset_{y-L_n,y} \right | \mu \right ]\leq \mathbb{E} \left [ \emptyset_{y-L_n,y} | \mu \right ]$, and that conditionally on $\mu$, $\emptyset_{x-L_n,x}$ and $\emptyset_{y-L_n,y}$ are independent when $|y-x|>L_n$. It follows that,
\begin{align} \Varr[ A_n|\mu] & \leq \int_{\mathcal{X}_{2^n}+L_n+1}^{\mathcal{X}_{2^{n+1}}} \mathbb{E} \left [ \emptyset_{y-L_n,y} | \mu \right ] \int_{y-L_n}^{y+L_n} d\mu(x)d\mu(y) \notag
\\ & \leq \mathbb{E}[A_n|\mu] \max_{\mathcal{X}_{2^n}+L_n+1\leq y \leq \mathcal{X}_{2^{n+1}}} \mu[y-L_n,y+L_n]. \label{1801}\end{align}
Furthermore by Lemma \ref{6}, for every $n$ large enough and $y\in [\mathcal{X}_{2^n}+1,\mathcal{X}_{2^{n+1}}-L_n]$,
\begin{equation} \mu[y,y+2L_n] \leq 4L_n \frac{\mathbb{E}[\mu[0,y]]}{y} + 13\frac{\log y}{y}
\leq 4 \frac{\log \mathcal{X}_{2^n}}{2^{n+2}} \frac{\mathbb{E}[\mu[0,\mathcal{X}_{2^{n+1}}]]}{\mathcal{X}_{2^n}}+13\frac{\log \mathcal{X}_{2^n}}{\mathcal{X}_{2^n}}. \label{1802} \end{equation}
Put together \eqref{1801} and \eqref{1802} yield as $n\to \infty$,
\begin{equation} \Varr[ A_n|\mu] \leq \mathbb{E}[A_n|\mu] \frac{17\log \mathcal{X}_{2^n}}{\mathcal{X}_{2^n}}. \label{1610} \end{equation}
Therefore, by Chebyshev's inequality, \eqref{1610}, and \eqref{1800}, we have as $n\to \infty$,
\begin{equation*} \mathbb{P}\left ( \left .A_n \leq \frac{\mathbb{E}[A_n|\mu]}{2} \right | \mu \right )
\leq 4\frac{\Varr[ A_n|\mu]}{\mathbb{E}[ A_n|\mu]^2} \leq \frac{O(1)\log \mathcal{X}_{2^n}}{\mathcal{X}_{2^n}\mathbb{E}[ A_n|\mu]}\leq 2^{-n}\mathcal{X}_{2^n}^{-1/2+o(1)}.
\end{equation*}
So by the Borel--Cantelli lemma, almost surely for every $n$ large enough $A_n \geq \frac{\mathbb{E}[A_n|\mu]}{2}$.
Finally the inequality in \eqref{0210} follows from \eqref{1800} and the fact that for every $n$ large enough $\mu(S_n)\geq A_n$. This concludes the proof.
\end{proof}
Formally we call the segments in $\bigcup_{n\in \mathbb{N}} \mathcal{I}_n$ "long". The following lemma proves that those long segments tend to "glue" to one another.
\begin{lemma} \label{glue} For every $I\in \bigcup_{n\in \mathbb{N}} \mathcal{I}_n$ let $a_I$ denotes the only integer such that $I\subset (Y_{a_I},Y_{a_I+1}]$. Almost surely for every $n,m\in \mathbb{N}$ large enough with $n<m$ and $\mathcal{X}_{2^m}\geq \mathcal{X}_{2^{n+1}}^8$, for every $I\in \mathcal{I}_{n}$ there exists $I' \in \mathcal{I}_{m}$ such that $Z_{a_{I'}}\in I$. In this case we say that $I'$ is glued on $I$.
\end{lemma}
\begin{proof}
Conditionally on $\mathcal{F}:=\sigma(\mu,(Y_i)_{i\geq 1})$, $(Z_i)_{i\geq 1}$ are independent random variables with law $(p_{Y_i})_{i\in \mathbb{N}}$ so for every $i\in \mathbb{N}$ and $I\in \mathcal{I}_{n}$
\begin{align*} \mathbb{P} \left ( \left. \forall I' \in \mathcal{I}_{m}, Z_{a_{I'}}\notin I \right | \mathcal{F} \right )
= \prod_{I'\in \mathcal{I}_{m}} \left (1-\frac{\mu(I)}{\mu[0,Y_{a_{I'}}]} \right )
\leq \exp \left (-\#\mathcal{I}_{m} \frac{\mu(I)}{\mu\left [0,\mathcal{X}_{2^{m+1}} \right ]} \right ).
\end{align*}
Furthermore we have by definition of $I_{n}$, $\mu(I)\leq \mathcal{X}_{2^{n+1}}^{-2}\leq \mathcal{X}_{2^m}^{-1/4}$. It follows from Lemmas \ref{Long} and \ref{5} that for every $m$ large enough,
\[ \mathbb{P} \left ( \left. \forall I' \in \mathcal{I}_{m}, Z_{a_{I'}}\notin I \right | \mathcal{F} \right ) \leq \exp \left (-\mathcal{X}_{2^m}^{-1/4}\frac{ 2^{m+2}\mathcal{X}_{2^m}^{1/3} }{2^{m+2}} \right )= \exp \left( -\mathcal{X}_{2^m}^{1/12} \right ). \]
Moreover, by Lemma \ref{8}, for every $i$ large enough $\#\mathcal{I}_{n}\leq 2\mathcal{X}_{2^{n+1}}^2$, and by Lemma \ref{2} for every $m\in \mathbb{N}$, $\mathcal{X}_{2^{m}}\geq 2^{m}$. So for every $m$ large enough,
\[\mathbb{P} \left ( \left. \exists I\in \mathcal{I}_{n}, \, \forall I' \in \mathcal{I}_{m}, Z_{a_{I'}}\notin I \right | \mathcal{F} \right ) \leq 2\mathcal{X}_{2^{n}}^2 e^{-\mathcal{X}_{2^{m}}^{1/12}} \leq f(\mathcal{X}_{2^m})\leq f(2^m), \]
where $f:x\mapsto 2x^2e^{-x^{1/12}}$. Since $\sum_{n=0}^{\infty}\sum_{m=n}^{\infty}f(2^m)<\infty$ the Borel--Cantelli lemma yields the desired result.
\end{proof}
\begin{lemma} \label{manger}
Almost surely for every $k$ large enough:
\[d_{H}(\mathcal{T}_{\mathcal{X}_{2^k}}, \mathcal{T})\geq \frac{1}{128} \sum_{n=k}^{\infty} \frac{\log \mathcal{X}_{2^{n}}}{2^n}. \]
\end{lemma}
\begin{proof} First define by induction $(n_i)_{i\in \mathbb{N}}$ such that $n_0=k$ and such that for every $i\geq 0$, $n_{i+1}=\inf\{n\in \mathbb{N}: n>n_i, \mathcal{X}_{2^{n}}\geq \mathcal{X}_{2^{n_i}}^8\}$. Note that for $i\in \mathbb{N}$, $\mathcal{X}_{2^{n_{2i+2}}}\geq \mathcal{X}^8_{2^{n_{2i+1}}} \geq \mathcal{X}_{2^{n_i+1}}^8,$ so by Lemma \ref{glue}, there exists a sequence $\{I_i\}_{i\in \mathbb{N}}$ such that for every $i\in \mathbb{N}$, $I_i\in \mathcal{I}_{n_{2i}}$ and $I_{i+1}$ is glued on $I_i$. On this event, note that for every $j\in \mathbb{N}$ and $x\in I_j $, $x$ is at distance at least $\sum_{i=0}^{j-1} L_{n_{2i}}$ of $\mathcal{X}_{2^k}$ so $d_{H}(\mathcal{T}_{\mathcal{X}_{2^k}}, \mathcal{T}) \geq \sum_{i=0}^{\infty} L_{n_{2i}}$. Similarly we have $d_{H}(\mathcal{T}_{\mathcal{X}_{2^k}}, \mathcal{T}) \geq \sum_{i=0}^{\infty} L_{n_{2i+1}}$, hence
\[ d_{H}(\mathcal{T}_{\mathcal{X}_{2^k}}, \mathcal{T})\geq \frac{1}{2}\sum_{i=0}^{\infty} L_{n_{i}}.\]
Finally we compare $\sum_{i=0}^{\infty} L_{n_{i}}$ with $\sum_{n=k+1}^{\infty} \frac{\log \mathcal{X}_{2^{n}}}{2^n}$. By definition of $\{n_i\}_{i\in \mathbb{N}}$ we have:
\begin{equation*} \sum_{n=k}^{\infty} \frac{\log \mathcal{X}_{2^{n}}}{2^n}
= \sum_{i=0}^{\infty} \sum_{n=n_i}^{n_{i+1}-1}\frac{\log \mathcal{X}_{2^{n}}}{2^n} \leq \sum_{i=0}^{\infty} \sum_{n=n_i}^{n_{i+1}-1}\frac{8\log \mathcal{X}_{2^{n_i}}}{2^n} \leq 64 \sum_{i=0}^{\infty} \frac{\log \mathcal{X}_{2^{n_i}}}{2^{n_i+2}} = 64 \sum_{i=0}^{\infty} L_{n_{i}}. \notag
\end{equation*}
This concludes the proof.
\end{proof}
The previous lemma proves that when $\sum \frac{\log \mathcal{X}_{2^{n}}}{2^n}=\infty$ the tree is not compact, thus finishing the proof of Theorem \ref{THM1}. The next lemma gives a more precise description of the geometry of the tree in the non-compact case: "the tree is infinite in every direction".
\begin{lemma} \label{NonCompact}
Suppose that $\sum \frac{\log \mathcal{X}_{2^{n}}}{2^n}=\infty$ then almost surely for every $a<b<c$, $[a,b]^{\uparrow c}$ has infinite diameter.
\end{lemma}
\begin{remark} An equivalent result is proved in Le Gall and Le Jan \cite{IntroICRT1} for non-compact L\'evy trees: the set of values taken by the height process on any non-trivial open interval contain a half line $[a,\infty)$.
\end{remark}
\begin{proof}
First one may adapt the argument of the proof of Lemma \ref{manger} to prove that for every $k\in \mathbb{N}$ large enough and $I\in \mathcal{I}_k$,
\begin{equation} \diam((Y_{a_I},Y_{a_I+1}]^{\uparrow Y_{a_{I}+1}}) \geq L_{k}+\frac{1}{2} \left (\sum_{i=1}^{\infty} L_{n_{2i+1}}+\sum_{i=1}^{\infty} L_{n_{2i}} \right ) \geq \frac{1}{64} \sum_{n=n_2}^{\infty} \frac{\log \mathcal{X}_{2^{n}}}{2^n}= \infty, \label{18102}\end{equation}
where $\{n_i\}_{i\in \mathbb{N}}$ is defined in the proof of Lemma \ref{manger} and $a_I$ in Lemma \ref{glue}. We leave the details to the reader.
We now fix $a<b<c\in \mathbb{R}^+$. Since conditionally on $\mathcal{F}:=\sigma(\mu,(Y_i)_{i\geq 1})$, $(Z_i)_{i\geq 1}$ are independent random variables with law $(p_{Y_i})_{i\in \mathbb{N}}$, we have for every $m\in \mathbb{N}$,
\[ \mathbb{P} \left ( \left . \forall I \in \mathcal{I}_m, Z_{a_I}\notin [a,b] \right | \mathcal{F} \right )=\prod_{I'\in \mathcal{I}_{m}} \left (1-\frac{\mu[a,b]}{\mu[0,Y_{a_{I'}}]} \right ) \leq \exp \left (-\frac{\mu[a,b]\# \mathcal I_m}{\mu[0,\mathcal{X}_{2^{m+1}}]} \right). \]
Since $\mu$ has full support it follows from Lemmas \ref{5} and \ref{Long} that the right-hand side above converges to $0$ as $m\to \infty$. Therefore for every $n\in \mathbb{N}$, there exists almost surely $m\geq n$ and $I\in \mathcal I_{m}$ such that $Z_{a_I}\in [a,b]$. It follows from \eqref{18102} that if $m$ is large enough $[a,b]^{\uparrow Y_{a_I}}$ has infinite diameter, hence $[a,b]^{\uparrow c}$ also has infinite diameter. Since $a<b<c$ are arbitrary and since rational numbers are dense on $\mathbb{R}^+$, the desired claim follows.
\end{proof}
\section{Fractal dimensions : proof of theorem \ref{THM3}} \label{Section 7}
In this section we prove Theorem \ref{THM3}. By Lemma \ref{FalconPunch}, it suffices to upper bound the Minkowski dimensions and to lower bound the Packing and Hausdorff dimension. We obtain the upper bounds from some simple cover of $\mathcal{T}$ and we derive the lower bounds from Lemma \ref{Hausdorff}. \subsection{Upper bound for the Minkowski dimensions}
First from the change of variables $u=\mathcal{X}_l$, note that the upper bound for the Minkowski dimensions given by Theorem \ref{THM3} are equivalent to
\[ \text{(a)} \quad \overline{\dim}(\mathcal{T})\leq \limsup_{l\to \infty} \frac{ \log l\mathcal{X}_l}{\log l} \quad \text{and} \quad \text{(b)} \quad \underline{\dim}(\mathcal{T})\leq \liminf_{l\to \infty} \frac{ \log l\mathcal{X}_l}{\log l} \ \ \text{when} \ \ \log \mathcal{X}_l=l^{o(1)} . \]
Then for every $l\in \mathbb{R}$, $\mathcal{T}_{\mathcal{X}_{l}}$ has total length $\mathcal{X}_{l}$, hence one can construct a cover of $\mathcal{T}_{\mathcal{X}_l}$ using $l\mathcal{X}_l$ balls of radius $2/l$. By increasing the radius of those balls by $d_H(\mathcal{T}_{\mathcal{X}_l},\mathcal{T})$ one obtains a cover of $\mathcal{T}$. So for every $l\in \mathbb{R}^+$,
\begin{equation} N_{2/l+d_H(\mathcal{X}_l,\mathcal{X})} \leq l\mathcal{X}_l. \label{1102} \end{equation}
The claims (a) and (b) are applications of the inequality in \eqref{1102}.
Toward proving (a), we may assume that $\log \mathcal{X}_l =O(\log l)$ since otherwise the bound is trivial. It follows from Lemma \ref{BIG LEMMA} that $d_{H}(\mathcal{T}_{\mathcal{X}_{2^{k-1}}}, \mathcal{T}_{\mathcal{X}_{2^{k}}})=O(k/2^k)$ and summing over all $k\geq \log_2(l)$, we obtain $d_H(\mathcal{X}_{l},\mathcal{T})=O (\log(l)/l )$. Therefore by \eqref{1102},
\[ \overline{\dim}(\mathcal{T})=\limsup_{l\to \infty} \frac{ \log N_{1/l}}{\log l} = \limsup_{l\to \infty} \frac{ \log N_{2/l+d_H(\mathcal{T}_{\mathcal{X}_l},\mathcal{T})}}{-\log \left (2/l+d_H(\mathcal{T}_{\mathcal{X}_l},\mathcal{T}) \right )} \leq \limsup_{l\to \infty} \frac{\log (l\mathcal{X}_l) }{ \log l}. \]
and (a) follows. (b) can be treated similarly by observing that Lemma \ref{BIG LEMMA} and $\log \mathcal{X}_l=l^{o(1)}$ implies that $d_H(\mathcal{X}_{l},\mathcal{T})= l^{-1+o(1)}$. We leave the details to the reader. This concludes the proof.
\subsection{Lower bound for the Packing dimension and the Hausdorff dimension}
In this section we show that almost surely,
\[ \dim_P(\mathcal{T})\geq \alpha :=1+\limsup_{l\to \infty} \frac{\log l}{\log \mathbb{E}[\mu[0,l]]} \quad \text{and} \quad \dim_H(\mathcal{T})\geq \beta :=1+\liminf_{l\to \infty} \frac{\log l}{\log \mathbb{E}[\mu[0,l]]}. \]
To this end, by lemma \ref{Hausdorff} it suffices to prove that if $A$ is a random variable with law $p$ then almost surely for every $\delta>0$, $\liminf p(B(A,\varepsilon)) \varepsilon^{-\alpha-\delta}<\infty$ and $p(B(A,\varepsilon))=O(\varepsilon^{\beta+\delta})$ as $\varepsilon\to 0$. The two previous inequalities can be proved via an elementary computation using $\mu[0,l]\sim \mathbb{E}[\mu[0,l]]$ (Lemma \ref{5}),
\[ \text{(a)} \quad p(B(A,d(A,\mathcal{T}_{Y_i})))\leq \frac{1}{Y_i^{1+o(1)} \mu[0,Y_i]} \quad \text{ and }\quad \text{(b)} \quad d(A,\mathcal{T}_{Y_{i+1}})\geq \mu[0,Y_i]^{-1+o(1)}. \]
We omit the details and focus on the proof of (a) and (b).
Toward (a), let $\gamma$ be the geodesic path from $0$ to $A$ and let for every $i\in \mathbb{N}$, $j_i:=\min\{j\geq i : (Y_j,Y_{j+1}]\cap \gamma \neq \emptyset\}$. Note that since by Theorem \ref{THM1} almost surely $A\notin \mathbb{R}^+$,
\[ B(A,d(A,\mathcal{T}_{Y_i})) \subset \{Z_{j_i}\}\cup (Y_{j_i},Y_{j_{i}+1}]^{\uparrow Y_{j_{i}+1}}. \]
Furthermore we have by Lemma \ref{10}, $\mu(Y_{j_i},Y_{j_{i+1}}]\leq \frac{\log^2 Y_i}{Y_i}$. Therefore by Lemma \ref{E=MC2} (iii), conditionally on $(\mu, \{Y_j\}_{j\in \mathbb{N}})$, for every $i$ large enough with probability at least $1-Y_{j_i}^{-5}\geq 1-Y_{i}^{-5}$,
\begin{equation} p\left ( B(A,d(A,\mathcal{T}_{Y_i})) \right ) \leq p\left (]Y_{j_i},Y_{j_{i+1}}]^{\uparrow Y_{j_{i+1}}} \right )\leq 2 \frac{\log^6 Y_{j_i}}{Y_{j_i} \mu[0,Y_{j_i}]} \leq 2 \frac{\log^6 Y_i}{Y_i\mu[0,Y_i]}. \label{NAVIGOA} \end{equation}
Moreover by Lemma \ref{8}, we have $i=O(Y_i^2)$, hence $\sum_{i=1}^\infty Y_i^{-5}<\infty$. The Borel--Cantelli lemma then yields that almost surely \eqref{NAVIGOA} holds for every $i$ large enough, hence (a) holds.
Toward (b), let us first upper bound $\{p(S_n)\}_{n\in \mathbb{N}}$ where for $n\in \mathbb{N}$, $S_n$ denotes the set of $x\in \mathcal{T}$ such that $d(x,[0,\mathcal{X}_{2^n}])\leq \delta_n:=\frac{1}{2^n n^6}$. Let
for every $n\in \mathbb{N}$,
\[a_n:=\max\{a : Y_a\leq \mathcal{X}_{n^22^n}\} \quad \text{ and } \quad S'_n:=\left \{x\in \left (\mathcal{X}_{2^n}+\delta_n, Y_{a_n} \right ] : \left [x-\delta_n,x \right ]\cap \{Y_i\}_{i\in \mathbb{N}}\neq \emptyset \right \}. \]
Note that for every $n\in \mathbb{N}$, $S_n\subset ([0,\mathcal{X}_{2^n}+\delta_n]\cup S'_n)^{\uparrow Y_{a_n}}.$
Therefore by Lemma \ref{E=MC2} (ii), \ref{5} and \ref{10}, almost surely for every $n$ large enough:
\begin{equation} p(S_n)\leq 2p_{Y_{a_n}} \left ([0,\mathcal{X}_{2^n}+\delta_n]\cup S'_n \right ) = 2\frac{\mu[0,\mathcal{X}_{2^n}+\delta_n]+\mu(S'_n)} {\mu[0,\mathcal{X}_{n^22^n}]-\mu(Y_{a_n},\mathcal{X}_{n^22^n}]}\leq 4\frac{2^n+\mu(S'_n)} {n^22^n}. \label{1210} \end{equation}
Furthermore since conditionally on $\mu$, $(Y_i)_{i\in \mathbb{N}}$ is a Poisson point process of rate $\mu[0,l]dl$, we have by Fubini's theorem, for every $n\in \mathbb{N}$:
\begin{align*}
\mathbb{E}[ \mu(S'_n) |\mu]
& \leq \int_{0}^{\mathcal{X}_{n^22^n}} \mathbb{P} \left [ \left . \left [x-\delta_n,x \right ]\cap\{Y_i\}_{i\in \mathbb{N}} \neq \emptyset \right |\mu \right ] d\mu(x) \\
& \leq \int_{0}^{\mathcal{X}_{n^22^n}} 1-e^{-\delta_n\mu \left[0, \mathcal{X}_{n^22^n} \right ]}d\mu(x) \\
& \leq \delta_n\mu \left[0, \mathcal{X}_{n^22^n} \right ]^2.
\end{align*}
It directly follows from Lemma \ref{5} that almost surely $\mathbb{E}[ \mu(S'_n) |\mu]=O(2^n/n^2)$ as $n\to \infty$. Thus by Markov's inequality and the Borel--Cantelli lemma almost surely $\mu(S'_n)=O(2^n)$. Therefore by \eqref{1210}, $p(S_n)=O\left (1/n^2 \right )$, hence by the Borel--Cantelli lemma almost surely for every $n$ large enough, $A\notin S_n$.
Finally let for every $i\in \mathbb{N}$, $n_i:=\inf\{n\in \mathbb{N}, Y_{i+1}\leq \mathcal{X}_{2^{n}}\}$. We have by Lemma \ref{5} and \ref{10} almost surely $\mu[0,Y_i] \geq 2^{n_i+O(1)}$. Hence, since for every $i$ large enough $A\notin S_{n_i}$, we have,
\begin{equation*} d(A,Y_{i+1})\geq d(A,\mathcal{X}_{2^{n_i}}) \geq \frac{1}{2^{n_i} {n_i}^6} \geq \mu[0,Y_i]^{-1+o(1)}. \label{NAVIGOB} \end{equation*}
This concludes the proof of $(b)$ and therefore of Theorem \ref{THM3}.
\paragraph{Acknowledgment}
Thanks are due to Nicolas Broutin for interesting conversations and numerous advice on earlier versions of this paper.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 2,189 |
Le elezioni parlamentari in Grecia del 1946 si tennero il 31 marzo.
Risultati
Collegamenti esterni
1946
Grecia | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 2,288 |
3 NJ institutions make American Banker's best banks list (updated)
By: Dawn Furnas
Three New Jersey financial institutions made American Banker's 10th annual Best Banks to Work For 2022.
A total of 90 banks across the nation made the magazine's recently released list, which was based on two surveys managed by fellow BridgeTower Media firm Best Companies Group.
The first survey gauged employee satisfaction in eight categories; the second looked at the institutions' policies and employee benefits.
American Banker noted that the asset size for each bank is as of June 30, 2022, the most recent information available from the FDIC (Federal Deposit Insurance Corp.), and that the banks supplied their employee headcounts. Click here to read the full report.
Making the cut from New Jersey are:
No. 33: Peapack-Gladstone Bank, Bedminster
American Banker said Brydget Falk-Drigan, chief human resources officer at Peapack-Gladstone, credited top-down transparent communication from leaders such as President and CEO Douglas Kennedy as one reason for satisfied employees. The magazine said Peapack-Gladstone reported $6.1 billion in assets and nearly 500 employees.
In an email to NJBIZ, Denise Pace-Sanders, senior vice president, managing principal, brand and marketing director for Peapack-Gladstone, said, "This recognition is incredibly important to the Bank because it is powered by our employees. They believe Peapack-Gladstone Bank is a great place to work. And, the fact that they've felt that way for five years in a row makes all of us collectively very proud of the work we do, and the support we give each other, our clients and our communities."
No. 42: Manasquan Bank, Wall Township
James Vaccaro, chairman, president and CEO of Manasquan Bank, told American Banker that the "best way to motivate employees to do their best is by creating an environment where everyone must contribute and everyone understands the strategic goals."
In a separate statement about the recognition, Vaccaro said the bank was "incredibly proud to be honored."
"I am continually inspired by our team of high-performing professionals, who work together each day to deliver upon our mission of improving the lives of our clients and enhancing the communities we live in," Vaccaro said.
The bank has $2.8 billion in assets and more than 220 employees.
No. 76: Lakeland Bank, Oak Ridge
Lakeland President and CEO Thomas Shara told American Banker that the institution prioritizes employees' work-life balance and has created wellness programs based on staff feedback. Shara said Lakeland considers employees' "workload, mental and financial health and socialization."
"It is an honor to be named one of the Best Banks to Work For in the financial industry," Shara said in a statement emailed to NJBIZ. "I am so proud of this distinguished recognition and thrilled to share it with our amazing associates who make Lakeland a premier bank and continue to go above and beyond to serve our customers every day."
The statement noted this is the second time that Lakeland has been recognized by the annual program.
Lakeland, which is in merger talks with Provident Bank for $1.3 billion, reported $10.4 billion in assets and more than 900 employees.
Editor's note: This story was updated at 9:57 a.m. ET Nov. 22 to include a statement from Lakeland Bank.
banking best companies group 1:30 pm Mon, November 21, 2022 NJBIZ
Dawn Furnas
NJBIZ Business Events
2023 NJBIZ Small Business Challenges Panel Discussion
2023 NJBIZ Leaders in Digital Technology
NJBIZ Leaders in Finance 2023
NJBIZ Leaders in Law 2023 | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 2,147 |
A well-established client, with a high-profile client base that includes an array of FTSE 100 companies is looking for a SQL Developer to join their growing Database team.
The SQL Developer will join an established team and be heavily involved in an array of client-based projects, working with large amounts of data for web-based reporting, including data migration, database design and database optimisation.
If you are available and interested, please apply in the first instance and you will be contacted to discuss the position further. | {
"redpajama_set_name": "RedPajamaC4"
} | 4,029 |
import config
import json
import itertools
# Migrates json file versions using a version number
# File version prior to adding a version number are 0
# Version 1 adds the version to all json files
# Version 2+ change the structure of the files
# to add migrations add version_{n}_to_{n+1} methods in the subclasses of JSONMigrator
# currently [IntrumentMigrator, ChannelMigrator,SweepMigrator, MeasurementMigrator]
# to make the changes in the json dict
class JSONMigrator(object):
""" Base class for the JSON Migration
Includes the base method for migrating
Assumes migration methods are of the form version_{n}_to_{n+1}
Includes version_0_to_1
"""
def __init__(self, fileName, libraryClass, primaryKey, max_version = 1):
self.fileName = fileName
self.jsonDict = None
self.min_version = 0
self.max_version = max_version
self.primaryKey = primaryKey
self.primaryDict = None
self.libraryClass = libraryClass
def version(self):
try:
return self.jsonDict['version']
except:
return 0
def load(self):
try:
with open(self.fileName, 'r') as FID:
self.jsonDict = json.load(FID)
except IOError:
print('json file {0} not found'.format(self.fileName))
self.jsonDict = None
if not self.json_input_validate():
self.jsonDict = None
return
# cache primary dictionary
self.primaryDict = self.jsonDict[self.primaryKey]
def json_input_validate(self):
if not self.is_class(self.jsonDict, self.libraryClass) or self.primaryKey not in self.jsonDict:
print "Error json file is not a " + self.libraryClass
return False
return True
def is_class(self, dict, searchClasses):
if type(searchClasses) == type(''):
searchClasses = [searchClasses]
if type(dict) != type({}):
return False
if 'x__class__' not in dict.keys():
return False
for className in searchClasses:
if dict['x__class__'] == className:
return True
return False
def save(self):
# store primary dictionary back into json dictionary
self.jsonDict[self.primaryKey] = self.primaryDict
# write out file
with open(self.fileName,'w') as FID:
json.dump(self.jsonDict, FID, indent=2, sort_keys=True)
def migrate(self):
messages = []
self.load()
if self.jsonDict:
while self.version() < self.max_version:
migrate_function = "version_{0}_to_{1}".format(self.version(), self.version() + 1)
msg = "Migrating: {0}.{1}()".format(self.__class__.__name__, migrate_function)
messages.append(msg)
function = getattr(self,migrate_function)
function()
self.jsonDict['version'] = self.version() + 1
self.save()
return messages
def get_items_matching_class(self, classes):
return [a for a in self.primaryDict if self.is_class(self.primaryDict[a],classes)]
def version_0_to_1(self):
# does nothing but bump version number
pass
class IntrumentMigrator(JSONMigrator):
""" Migrator for the Intrument Manager JSON File """
def __init__(self):
super(IntrumentMigrator, self).__init__(
config.instrumentLibFile,
"InstrumentLibrary",
"instrDict",
3)
def version_1_to_2(self):
# Migration step 1
# Change Labbrick64 class to Labbrick
chClasses = ['Labbrick64']
lb64 = self.get_items_matching_class(chClasses)
for lb in lb64:
self.primaryDict[lb]['x__class__'] = "Labbrick"
def version_2_to_3(self):
# Migration step 2
# Follow X6 channel schema change
scopes = self.get_items_matching_class(['X6'])
for x6 in scopes:
for channel in self.primaryDict[x6]['channels'].values():
channel['enableDemodResultStream'] = channel['enableResultStream']
channel['enableRawResultStream'] = channel['enableResultStream']
channel['demodKernel'] = channel['kernel']
channel['rawKernel'] = ''
del channel['enableResultStream']
del channel['kernel']
class ChannelMigrator(JSONMigrator):
""" Migrator for the Channel Manager JSON File """
def __init__(self):
super(ChannelMigrator, self).__init__(
config.channelLibFile,
'ChannelLibrary',
'channelDict',
3)
def version_1_to_2(self):
# Migration step 1
# Move SSBFreq from Physical Channel for Qubits to the Logical Qubit Channel
# two phases
# 1) copy all of the data from physical channel
# 2) delete data after copying is done in cases there are many-to-one mappings
lcClasses = ['Qubit', 'Measurement']
logicalChannels = self.get_items_matching_class(lcClasses)
for lc in logicalChannels:
pc = self.primaryDict[lc]['physChan']
if pc not in self.primaryDict:
print 'Error: Physical Channel {0} not found.'.format(pc)
continue
if 'SSBFreq' not in self.primaryDict[pc]:
print "Warning: did not find SSBFreq for PhysicalChannel: ", pc
continue
frequency = self.primaryDict[pc]['SSBFreq']
self.primaryDict[lc]['frequency'] = frequency
lcClasses = ['PhysicalQuadratureChannel']
iqChannels = self.get_items_matching_class(lcClasses)
for iq in iqChannels:
if 'SSBFreq' not in self.primaryDict[iq]:
continue
del self.primaryDict[iq]['SSBFreq']
def version_2_to_3(self):
#Migration step 2
#Set default digitizer trigger to digitizerTrig
measChannels = self.get_items_matching_class('Measurement')
for mc in measChannels:
if 'trigChan' in self.primaryDict[mc]:
if self.primaryDict[mc]['trigChan'] != '':
continue
self.primaryDict[mc]['trigChan'] = 'digitizerTrig'
class SweepMigrator(JSONMigrator):
""" Migrator for the Sweeps JSON File """
def __init__(self):
super(SweepMigrator, self).__init__(
config.sweepLibFile,
"SweepLibrary",
"sweepDict",
1)
class MeasurementMigrator(JSONMigrator):
""" Migrator for the Sweeps JSON File """
def __init__(self):
super(MeasurementMigrator, self).__init__(
config.measurementLibFile,
"MeasFilterLibrary",
"filterDict",
1)
def migrate_all():
migrators = [IntrumentMigrator,
ChannelMigrator,
SweepMigrator,
MeasurementMigrator]
messages = []
for migrator in migrators:
m = migrator()
msg = m.migrate()
messages.append(msg)
return list(itertools.chain(*messages))
if __name__ == '__main__':
migrate_all()
| {
"redpajama_set_name": "RedPajamaGithub"
} | 2,344 |
Q: namespace usage I'm trying to start using namespaces the correct (or at least best) way.
The first thing I tried to do was to avoid putting using namespace xxx; at the beginning of my files. Instead, I want to using xxx::yyy as locally as possible.
Here is a small program illustrating this :
#include <iostream>
#include <cstdlib>
#include <ctime>
int main() {
using std::cout;
using std::endl;
srand(time(0));
for(int i=0; i<10;++i)
cout << rand() % 100 << endl;
return 0;
}
If I omit the lines using std::cout; or using std::endl, the compiler will complain when I'm trying to use cout or endl.
But why is this not needed for srand, rand and time ? I'm pretty sure they are in std, because if I try to specifically pour std:: in front of them, my code is working fine.
A: If you use cstdlib et al. the names in them are placed in both the global and the std:: namespaces, so you can choose to prefix them with std:: or not. This is seen as a feature by some, and as a misfeature by others.
A: If you really want to know, take a close look at the ctime and cstdlib headers. They were built backwards-compatible.
Note: all this using vs. using namespace business is about readability. If your IDE would allow to just not show the namespaces when you don't want to see them, you wouldn't need these constructs...
A: I prefer to omit using and just have the std::cout every time just to maintain readability. although this is probably only useful in larger projects
A: As long as we on the subject, there's also a thing called Koenig Lookup which allows you to omit a namespace identifier before a function name if the arguments it take come from the same namespace.
For example
#include <iostream>
#include <algorithm>
#include <vector>
void f(int i){std::cout << i << " ";}
int main(int argc, char** argv)
{
std::vector<int> t;
// for_each is in the std namespace but there's no *std::* before *for_each*
for_each(t.begin(), t.end(), f);
return 0;
}
Well, it's not related directly but I though it may be useful.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 3,160 |
{"url":"http:\/\/raphaelkoh.me\/Course-Notes\/courses\/phys234\/wave.html","text":"\nWavefunctions\n\nThe Hamiltonian\n\nSimilar to how position and momentum play a key role in expressing the total energy of a particle in classical physics, position and momentum are the primary observables in quantum mechanics. The quantum mechanical Hamiltonian operator for a particle moving in one dimension is: $$\\hat{H}=\\frac{\\hat{p}_x^2}{2m}+V(\\hat{x})$$\n\nEnergy Eigenvalue Equation\n\nThere are few quantum mechanical problems that can be solved using abstract kets. It is generally more convenient to express quantum states as spatial functions known as wavefunctions, denoted as $\\psi(x)$. This is a representation of the abstract quantum state, known as the position representation: $$\\ket{\\psi}\\doteq\\psi(x)$$ This representation uses the position eigenstates as the preferred basis. The energy eigenstates, written as wave functions, are denoted as $\\ket{E_i}\\doteq\\phi_{E_i}(x)$. Thus, the energy eigenvalue equation becomes: $\\hat{H}\\phi_{E_i}(x)=E_i\\phi_{E_i}(x)$. To solve this, we represent the operators in the Hamiltonian in the position representation. $$\\hat{x}\\doteq x,\\quad\\hat{p}\\doteq -i\\hbar\\frac{d}{dx}$$ Thus, the eigenvalue equation becomes: $$\\bigg(-\\frac{\\hbar^2}{2m}\\frac{d^2}{dx^2}+V(x)\\bigg)\\phi_E(x)=E\\phi_E(x)$$\n\nThe Wavefunction\n\nUnlike ket vectors, wavefunctions represent a continuous distribution of probabilities. The wavefunction is defined as the probability amplitude for the quantum state to be measured in the position eigenstate: $$\\psi(x)=\\braket{x}{\\psi}$$ The probability distribution can be calculated by taking the absolute square of the probability amplitude. $$P(x)=|\\braket{x}{\\psi}|^2=|\\psi(x)|^2$$ Probability of finding a particle in $a< x < b$: $$P_{a < x < b}=\\int_a^b|\\psi(x)|^2dx$$ Probability for finding a state with energy $E_n$: $$P_{E_n}=|\\braket{E_n}{\\psi}|^2=|\\int_{-\\infty}^\\infty\\phi^*_n(x)\\psi(x)dx|^2=|c_n|^2$$ $$\\phi_n(x)=\\braket{x}{E_n}$$ where $\\phi_n(x)$ is the energy eigenstate with energy $E_n$.\n\nTo find the expansion coefficient $c_n$, $$c_n=\\int_{-\\infty}^\\infty\\phi^*_n(x)\\psi(x)dx$$ Note this is identical to the ket notation of $c_n=\\braket{a_n}{\\psi}$\n\nNormalization condition: $$\\int_{-\\infty}^\\infty P(x)dx=\\int_{-\\infty}^\\infty|\\psi(x)|^2dx=\\int_{-\\infty}^\\infty\\psi^*(x)\\psi(x)dx=1$$\n\nBra-ket $\\rightarrow$ Wavefunction\n\n1. $\\ket{\\psi}\\rightarrow\\psi(x)$\n2. $\\bra{\\psi}\\rightarrow\\psi^*(x)$\n3. $\\braket{}{}\\rightarrow\\int_{-\\infty}^\\infty dx$\n4. $\\hat{A}\\rightarrow A(x)$\n\nExpectation Value\n\n$$\\angle{\\hat{A}}=\\braketo{\\psi}{\\hat{A}}{\\psi}=\\int_{-\\infty}^\\infty\\psi^*(x)A\\psi(x)dx$$ $$\\angle{\\hat{x}}=\\braketo{\\psi}{\\hat{x}}{\\psi}=\\int_{-\\infty}^\\infty\\psi^*(x)x\\psi(x)dx$$ $$\\angle{\\hat{p}}=\\braketo{\\psi}{\\hat{p}}{\\psi}=\\int_{-\\infty}^\\infty\\psi^*(x)\\big(-i\\hbar\\frac{d}{dx}\\big)\\psi(x)dx$$ $$\\angle{H}=\\sum_n |c_n|^2E_n$$\n\nProperties of Energy Eigenstates\n\n1. Normalization: $$\\braket{E_n}{E_n}=1\\longleftrightarrow\\int_{-\\infty}^\\infty|\\phi_n(x)|^2dx=1$$\n2. Orthogonality: $$\\braket{E_n}{E_m}=\\delta_{nm}\\longleftrightarrow\\int_{-\\infty}^\\infty\\phi_n^*(x)\\phi_m(x)dx=\\delta_{nm}$$\n3. Completenes: $$\\ket{\\psi}=\\sum_n c_n\\ket{E_n}\\longleftrightarrow\\psi(x)=\\sum_n c_n\\phi_n(x)$$\n\nThe SE Recipe for Wavefunctions\n\n1. Diagonalize $H$ to find energy eigenvalues and energy eigenstates. $$H\\phi_n(x)=E_n\\phi_n(x)$$\n2. Write $\\psi(0)$ as a superposition of energy eigenstates $$\\psi(x,0)=\\sum_n c_n\\phi_n(x),\\quad c_n=\\int_{-\\infty}^\\infty\\phi_n^*(x)\\psi(x,0)dx$$\n3. Multiply each term in the superposition (in the energy\/momentum basis) by a phase, $e^{-iE_nt\/\\hbar}$, to get $\\psi(x,t)$. $$\\psi(x,t)=\\sum_n c_n e^{-iE_nt\/\\hbar}\\phi_n(x)$$\n4. Calculate observable $\\hat{A}$. $$\\angle{A}=\\int_{-\\infty}^\\infty\\psi^*(x,t)A\\psi(x,t)dx$$","date":"2018-10-23 04:04:24","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.9536664485931396, \"perplexity\": 366.8275639538618}, \"config\": {\"markdown_headings\": false, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2018-43\/segments\/1539583516003.73\/warc\/CC-MAIN-20181023023542-20181023045042-00184.warc.gz\"}"} | null | null |
Home / Historical Biographies / Charles Henderson Stinson
Charles Henderson Stinson
Stinson: Prominent and Progressive Pennsylvanians of the 19th Century, vol 3.
1867 5 Republican
1869 Speaker 5 Republican
Chester, Delaware, Montgomery
Senator Charles H. Stinson was the son of the Honorable Robert M. and Elizabeth Porter Stinson, born at Norriton Township, Montgomery County June 28, 1825. A respected Montgomery County agriculturist, his father served as a county justice and an 1836 Anti-Masonic state representative. Charles received an impressive education, attending local schools before entering John McNair's "select school" in Abbington, in 1840. Stinson ultimately graduated with honors from Dickinson College in 1845, and for two years thereafter, explored the various mountain ranges of Pennsylvania, returning to Norristown as a private tutor. He joined his brother's (George) law practice in 1847 as an understudy, continuing legal study under Addison May after his brother died in 1848. He joined the Montgomery bar in 1849, established a private practice in Montgomery (1849), and married Emily Freedly of Norristown in 1855. Stinson, an early Whig, aligned with the new Republican Party in 1856. During the Civil War, he enlisted as a private in an emergency company of the Montgomery Volunteers during the Gettysburg campaign. Charles resumed his Montgomery County legal practice in 1865, entering the state Senate in October 1867, serving as Speaker in 1870. Senator Stinson enjoyed a collegial relationship with Senator Wilmer Worthington, emerging as the cornerstone of the Judiciary General Committee.
Stinson's major legislative accomplishments included framing the 1870 Treasury Reform Act, providing the means to monitor state deposits and expenditures through detailed monthly reports and reporting systems; and the same session's Mine Safety Act, responding to the September 1869 Avondale, Pennsylvania coalmine disaster – a calamity that claimed 108 lives.
Charles voted for adoption of the U.S. Constitution's Fifteenth (suffrage) Amendment, however, supported the controversial Liberian Colonization program. In 1868, he was a Grant "stalwart" and prohibitive-tariff Republican. The senator chaired Accounts (1868), Estates and Escheats (1869), and served second chair on Judiciary (1869) under Chairman Harry White. After a brief political career, Charles returned to private practice, refusing appeals for various judgeships and senatorial seats. In 1882, however, Governor Hoyt convinced Charles to accept an appointment as president judge of the Thirty-eighth Judicial District, replacing the late Judge Henry Ross. Among Senator Stinson's many civic contributions, he served as a trustee for the Pennsylvania Hospital of the Insane. He also reserves the distinction as the first hospital administrator to appoint a woman as the head of a department: the move led to that institution's further hiring of women to key managerial positions. Stinson organized the First National Bank of Norristown (1864) and co-founded the Norristown Music Hall Association. He worked as a solicitor for the Pennsylvania Railroad and became a key figure in clearing the legal path for construction of the Schuylkill Valley Railroad's Montgomery County branches. His sister, Dr. Mary H. Stinson, is one of nation's first matriculated women physicians and a founder of the Stinson Home for Aged Women in Norristown. The Honorable Charles H. Stinson died in Norristown on March 13, 1899. | {
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Public schools in Larue County, KY. Find schools, get student / teacher ratios & counts, demographics and other facts.
Get detailed information on elementary schools, middle schools & high schools in the Larue County area.
Demographics for Larue County, KY. Data includes population, income, unemployment rate & other facts. | {
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Delta Virginis (δ Virginis / δ Vir) è una stella della costellazione della Vergine. Ha una magnitudine apparente di +3,38 e dista 202 anni luce dal sistema solare.
È conosciuta anche con i nomi tradizionali di Minelava e Auva, entrambi provenienti dall'arabo. Sulle rive dell'Eufrate veniva chiamata Lu Lim, la gazzella o la capra, presso gli Indù era conosciuta come Apa o Apas, le acque, mentre per i cinesi era Tsze Seang, il secondo ministero dello stato.
Osservazione
Si tratta di una stella situata nell'emisfero celeste boreale, ma molto in prossimità dell'equatore celeste; ciò comporta che possa essere osservata da tutte le regioni abitate della Terra senza alcuna difficoltà e che sia invisibile soltanto nelle aree più interne del continente antartico. Nell'emisfero nord invece appare circumpolare solo molto oltre il circolo polare artico. Essendo di magnitudine 3,3, la si può osservare anche dai piccoli centri urbani senza difficoltà, sebbene un cielo non eccessivamente inquinato sia maggiormente indicato per la sua individuazione.
Il periodo migliore per la sua osservazione nel cielo serale ricade durante i mesi della primavera boreale, che corrispondono alla stagione autunnale nell'emisfero australe. Il periodo di visibilità rimane indicativamente lo stesso, grazie alla posizione della stella non lontana dall'equatore celeste.
Caratteristiche fisiche
Delta Virginis è una gigante rossa di tipo spettrale M3III con una temperatura superficiale di 3600 K circa. La massa stimata è 1,4 volte quella del Sole, mentre il raggio è enormemente più grande (circa 60 volte maggiore). La stella è certamente nello stadio finale della sua esistenza, anche se non si conosce per certo l'esatta fase in cui si trova; potrebbe infatti avere un nucleo inerte di elio e star aumentando la sua brillantezza, oppure potrebbe aver già iniziato la fusione dell'elio in carbonio nel suo nucleo tramite il processo tre alfa, o ancora potrebbe avere un nucleo inerte di carbonio al suo interno ed essere ormai giunta alla fine della sua normale vita stellare.
Come molte stelle in questo stadio è una variabile: classificata come variabile semiregolare, la sua magnitudine fluttua tra la +3,32 a +3,40.
Delta Virginis ha una compagna a 80 secondi d'arco di distanza che potrebbe essere legata gravitazionalmente ad essa. Si tratta di una nana arancione di classe K che impiegherebbe almeno 200 000 anni ad orbitare attorno alla principale, ad una distanza di oltre 5000 UA.
Note
Voci correlate
Stelle principali della costellazione della Vergine
Gigante rossa
Giganti rosse
Stelle di classe spettrale M
Variabili semiregolari | {
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Fomalhaut (α PsA / α Piscis Austrini / Alfa Piscis Austrini) è una stella situata nella costellazione del Pesce Australe. Avendo magnitudine +1,16 essa è la stella più luminosa della costellazione nonché la diciottesima più brillante del cielo visto dalla Terra. È una stella dell'emisfero australe ma le sue possibilità di osservazione dall'emisfero boreale sono ampie. Si tratta di una stella bianca di sequenza principale, simile a Sirio e Vega, distante 25 anni luce. Ha un raggio e una massa all'incirca doppi rispetto a quelli del Sole e una temperatura superficiale di circa 8.500 K.
La sua caratteristica più nota e studiata è quella di possedere un esteso disco circumstellare di gas e polveri. Si sospetta inoltre che la stella possa ospitare un sistema planetario e nel 2008 è stato annunciato che un pianeta orbitante intorno alla stella era stato osservato nella banda del visibile. Si sarebbe trattato del primo pianeta osservato direttamente nel visibile. Tuttavia osservazioni e studi successivi hanno messo fortemente in dubbio l'esistenza del pianeta.
Il nome della stella è , che significa "la bocca della balena".
Osservazione
Fomalhaut appare come una stella dal colore bianco-azzurro; la sua individuazione in cielo è facilitata, oltre che dalla sua luminosità (si tratta della diciottesima stella più brillante del cielo ad occhio nudo) dalla sua singolare posizione, lontana da altre stelle luminose e quindi in risalto in un campo celeste povero di stelle appariscenti. Ha una declinazione di 30°S, il che favorisce gli osservatori posti a latitudini meridionali. Tuttavia, la sua posizione non troppo discosta dall'equatore celeste fa in modo che sia visibile dalla gran parte delle aree abitate della Terra. In particolare essa diventa invisibile solo a partire dal 60°N, escludendo in tal modo l'Alaska, il Canada settentrionale, la Groenlandia, l'Islanda, le regioni scandinave settentrionali, e buona parte della Russia. Si mostra relativamente alta in cielo a partire dalle regioni mediterranee, mentre è circumpolare a sud del 60°S, cioè solo nelle regioni antartiche.
Il periodo più propizio per la sua osservazione è quello che va dal mese di agosto a quello di dicembre, nei cieli serali dell'emisfero nord, mentre a sud dell'equatore si mostra da giugno a gennaio; la sua presenza nei cieli dopo il tramonto del Sole nell'emisfero nord indica l'approssimarsi della stagione autunnale.
Ambiente galattico
La nuova riduzione dei dati astrometrici del telescopio spaziale Hipparcos risalente al 2007 ha portato a un nuovo calcolo della parallasse di Fomalhaut, che è risultata essere 129,81 ± 0,47 mas. Pertanto la distanza di Fomalhaut dalla Terra è pari a 1/0,12981 pc, ossia 7,70 pc, equivalenti a 25,12 anni luce. Fomalhaut è quindi una stella relativamente a noi vicina, che condivide lo stesso ambiente galattico del Sole. In particolare, si trova come il Sole all'interno della Bolla Locale, una "cavità" del mezzo interstellare presente nel Braccio di Orione, uno dei bracci galattici della Via Lattea. Le coordinate galattiche di Fomalhaut sono 20,48° e -64,90°. Una longitudine galattica di circa 20° significa che la linea ideale che congiunge il Sole e Fomalhaut, se proiettata sul piano galattico, forma con la linea ideale che congiunge il Sole con il centro galattico un angolo di circa 20°. Ciò significa che Fomalhaut è leggermente più vicina al centro galattico di quanto non lo sia il Sole. Una latitudine galattica di quasi -65° significa tuttavia che la distanza che separa il Sole da Fomalhaut è per la maggior parte dovuta al fatto che le due stelle non sono allineate sullo stesso piano e che Fomalhaut si trova parecchio a sud rispetto al piano su cui sono posti il Sole e il centro galattico.
Fomalhaut fa parte dell'Associazione di Castore, un'associazione stellare composta da stelle relativamente vicine al Sole, che condividono lo stesso movimento rispetto al sistema di riposo locale. Questa associazione, scoperta nel 1990, comprende almeno 16 membri fra cui, oltre a Fomalhaut e Castore (da cui prende il nome), Vega, Alderamin () e Zubenelgenubi (). È probabile che le stelle abbiano una origine comune e che quindi siano nate tutte più o meno nello stesso periodo di tempo. Basandosi sulle tracce evolutive delle varie stelle appartenenti all'associazione e su altri dati, come l'abbondanza di litio, l'età dell'associazione è stata stimata in 200 ± 100 milioni di anni.
La stella più vicina a Fomalhaut è TW Piscis Austrini, una nana arancione di classe spettrale K4 V e di magnitudine apparente 6,48, distante solo 0,9 anni luce da Fomalhaut. Data questa vicinanza e il fatto che le due stelle condividono lo stesso moto proprio nel cielo e lo stesso movimento rispetto al sistema di riposo locale, è probabile che esse siano fisicamente legate. TW Piscis Austrini dovrebbe avere una temperatura superficiale di 4.594 K, una massa di 0,73 M☉, un raggio di 0,688 R☉ e una luminosità di 0,189 L☉. È una variabile BY Draconis.
La seconda stella più vicina a Fomalhaut è LHS 3885, una nana arancione di classe spettrale K7 V e magnitudine apparente 7,86, distante 3,5 anni luce. A 5,6 anni luce si trova invece FK Aquarii, una nana rossa di classe spettrale M2 V e magnitudine apparente 9,0.
Caratteristiche
Fomalhaut è una stella bianca di sequenza principale di classe spettrale A4 V o A3 V. Lo studio della stella è stato facilitato dal fatto che è stato possibile misurare direttamente il suo diametro angolare mediante tecniche interferometriche. Utilizzando il Narrabri Stellar Intensity Interferometer, situato a Narrabri, nel Nuovo Galles del Sud, Hanbury Brown et al. (1974) hanno ottenuto un valore di 2,10 ± 0,14 mas; questo valore è stato corretto e reso più preciso delle osservazioni condotte da Di Folco et al. (2004), che hanno utilizzato lo strumento VINCI del Very Large Telescope: essi hanno ottenuto un valore di 2,228 ± 0,031 mas. Alla distanza calcolata da Hipparcos, ciò equivale a un raggio di 1,840 ± 0,023 R☉. Poiché dalla magnitudine apparente della stella e dalla sua distanza è possibile ricavare la sua luminosità assoluta, da questa e dal raggio è possibile dedurre la temperatura superficiale, utilizzando la legge di Stefan-Boltzmann: essa risulta essere 8.760 ± 100 K. La luminosità della stella è invece calcolata in 17,8 ± 0,8 L☉. La nuova riduzione dei dati di Hipparcos, risalente al 2007, ha permesso di correggere questi valori in 8.590 ± 73 K e 16,63 ± 0,48 L☉.
Per valutare la massa della stella è necessario conoscere, oltre alla sua posizione sul diagramma H-R, data dalla temperatura e dalla luminosità, anche la sua metallicità, ossia l'abbondanza di elementi più pesanti dell'elio. Tuttavia gli studi che sono stati fatti riguardo alla metallicità di Fomalhaut non concordano circa il suo valore, ma solo sul fatto che essa sia più bassa rispetto a quella del Sole. La metallicità è determinata misurando il rapporto fra l'abbondanza di ferro e l'abbondanza di idrogeno nella fotosfera della stella. Uno studio spettroscopico del 1997 riporta una abbondanza di metalli pari a 93% di quella del Sole. Un altro studio, pubblicato lo stesso anno, deduce la metallicità di Fomalhaut da quella di TW Piscis Austrini, supponendo che le due stelle abbiano una comune origine. Ne risulta un valore 78% di quello solare. Un modello evolutivo proposto da Di Folco et al. (2004) ha condotto a un valore di 79%, mentre una misurazione spettroscopica del 2008 ha dato un valore molto più basso di 46%. Di Folco et al. (2004) hanno stimato la massa di Fomalhaut seguendo le tracce evolutive di stelle di diversa massa sul diagramma H-R e assumendo una metallicità pari al 93% di quella del Sole; ne è risultata una massa di 2,00 ± 0,20 M☉. Mamajek (2012) ha invece utilizzato la nuova riduzione dei dati di Hipparcos e una metallicità simile a quella del Sole per ricavare un valore di 1,95 ± 0,02 M☉.
C'è invece maggiore accordo fra le misurazioni della velocità di rotazione della stella. Il sito SIMBAD riporta tre diverse misurazioni della velocità di rotazione per il seno dell'inclinazione dell'asse di rotazione rispetto alla nostra visuale (v × sin i): esse variano da 85 km/s a 93 km/s. Il prof. Kaler nella voce dedicata a Fomalhaut nel suo database Stars riporta invece un valore di 102 km/s.
L'età stimata dell'Associazione di Castore, lo stato evolutivo dell'associata TW Piscis Austrini e le tracce evolutive di stelle simili a Fomalhaut hanno fatto a lungo ritenere che essa sia una stella giovane, con un'età compresa tra 100 e 300 milioni di anni ed un tempo di vita residuo stimato in un miliardo di anni, prima che l'astro evolva in gigante rossa. Tuttavia Mamajek (2012) ha messo parzialmente in dubbio queste ipotesi: basandosi su nuove stime più precise della luminosità e della temperatura superficiale della stella, su tracce evolutive più aggiornate, egli ha dedotto una età di 450 ± 40 milioni di anni. Dopo avere confermato che Fomalhaut e TW PsA formano un sistema fisico, Mamajek (2012) ha nuovamente cercato di determinare l'età di quest'ultima, basandosi sulla velocità di rotazione della stella, sulla quantità di raggi X emessi e sull'abbondanza di litio. Dopo avere pesato tutti questi fattori, lo studioso è arrivato alla conclusione che il sistema ha una età di 440 ± 40 milioni di anni.
Disco circumstellare
Nel 1983 il telescopio spaziale IRAS rilevò che Fomalhaut, Vega, e emettevano un eccesso di radiazione infrarossa. Tale radiazione fu interpretata essere emessa da grani di polvere orbitanti intorno a queste stelle. Si suppose che tali grani formassero un disco circumstellare, all'interno del quale erano in formazione nuovi pianeti.
Nel 1998 un team di scienziati statunitensi e britannici riuscì ad ottenere la prima immagine del disco, fotografandolo a lunghezze d'onda inferiori al millimetro. Dall'immagine appariva una vasta cavità centrale, sgombra da gas e altro materiale, approssimativamente delle dimensioni dell'orbita di Nettuno. Ciò portava gli scienziati del team a paragonare il disco circumstellare alla Fascia di Kuiper che circonda il Sole e a stimarne la massa in poche masse lunari. L'attività di formazione di pianeti, se aveva avuto luogo, era quindi già terminata e aveva forse prodotto la cavità centrale, ripulendola di materiale. I ricercatori ipotizzavano che la fascia contenesse, oltre che grani di piccole dimensioni, anche comete e forse corpi di dimensioni maggiori, probabilmente frutto della frammentazione di un pianeta preesistente.
Nel 2005 fu possibile ottenere delle immagini della cintura di Fomalhaut nella banda del visibile utilizzando il coronografo della camera ad alta risoluzione del telescopio spaziale Hubble. Le immagini avevano una risoluzione di 0,5 UA, 100 volte maggiore di quelle precedenti. Ciò permise di comprendere con relativa precisione quale fosse la forma del disco: il suo semiasse maggiore è lungo 140,7 ± 1,8 UA, quello minore 57,5 ± 0,7 UA ed è inclinato di 65,9° rispetto al piano della volta celeste, mentre la longitudine del nodo ascendente è pari a 156,0° ± 0,3°. Fomalhaut non si trova esattamente al centro del disco ma spostata rispetto ad esso di 15,3 UA. La cintura ha una larghezza di 25 UA, sicché il suo raggio minore è di 133 UA e quello maggiore 158 UA: la cavità centrale è quindi molto più grande di quanto precedentemente si fosse supposto. L'eccentricità del disco è stimata essere 0,11 ± 0,01. A partire dalla sua magnitudine apparente di 16,2 e dalla sua albedo di 0,05-0,1, i ricercatori hanno stimato una massa totale di 50-100 M⊕, molto più alta rispetto alle stime precedenti. Il disco è molto schiacciato con uno spessore di sole 3,5 UA. La temperatura dei grani che formano il disco varia da 40 a 75 K. La sua composizione chimica presunta consiste nel 43% di acqua ghiacciata, nel 32% di silicati, nel 13% di carbonio amorfo e nel 10% di solfuro ferroso.
Gli scienziati si interrogavano intanto circa l'origine del disco e circa l'interpretazione di addensamenti di materiale che le osservazioni stavano rilevando al suo interno. Nel modello di Wyatt e Dent (2002) il disco è il risultato della frantumazione di planetesimi della dimensione di qualche chilometro, in seguito a scontri e collisioni. I due scienziati partono dall'assunzione che il disco non può essere presente dalla formazione della stella in quanto la radiazione proveniente da essa avrebbe dovuto già dissolverlo. Ciò implica che deve esistere un meccanismo che rimpiazza continuamente il materiale perduto ed esso è individuato nella frantumazione dei planetesimi in grani di dimensioni inferiori al millimetro. Gli addensamenti di materiale sono interpretati come il risultato o della collisione di due planetesimi, che ha originato del materiale che non si è ancora disperso, oppure alla risonanza con un ipotetico pianeta interno al disco. I due studiosi ipotizzano che a loro volta i planetesimi si siano formati dalla collisione di corpi di dimensioni maggiori.
Il modello della frantumazione dei planetesimi in corpi via via più piccoli è stato poi confermato e affinato da studi successivi. In particolare, nel modello proposto da Acke et al. (2012) la massa totale dei grani si aggira intorno alle 10 M⊕, il resto della massa del disco (circa 110 M⊕) è composto da planetesimi, che scontrandosi forniscono continuamente al disco materiale, senza il quale si dissolverebbe in tempi relativamente brevi. Nel modello infatti il tasso di evaporazione dei grani del disco è molto elevato, circa 0,03 masse lunari all'anno. Ne segue che l'intero materiale del disco di polveri viene rimpiazzato totalmente ogni 1700 anni. Per alimentarlo è necessario vengano frantumati due planetesimi del diametro di 10 km l'uno ogni giorno, oppure 2000 planetesimi del diametro di 1 km al giorno. Nell'intero disco devono essere presenti circa 100 miliardi di planetesimi della dimensione di 10 km, oppure 10 000 miliardi di planetesimi della dimensione di 1 km. Il fatto che si stimi che la nube di Oort, che circonda il Sole, contenga un numero comparabile di comete e la somiglianza dei grani con quelli prodotti dal disgregamento delle comete portano gli studiosi a paragonare i planetesimi che si trovano nel disco di Fomalhaut alle comete che popolano la nube di Oort.
Osservazioni compiute dal telescopio spaziale Spitzer nel 2003 suggerirono l'esistenza di un altro disco di polvere, interno a quello già osservato. Si tratta di un disco più difficile da osservare rispetto al precedente in quanto più vicino alla brillante stella centrale. In particolare, apparve difficile determinare se esso fosse un disco che si estendeva in modo continuo a partire dal disco più esterno o se esso fosse un disco separato il cui margine esterno si trova a una distanza minore alle 20 UA dalla stella. La presenza di un disco interno è stata confermata dalle osservazioni compiute da terra nel 2004 tramite l'interferometro del Very Large Telescope, ma ancora una volta si rivelò difficile determinarne le dimensioni e la forma. Gli scienziati che compirono le osservazioni ipotizzarono comunque che il disco debba trovarsi entro un raggio di 6 UA dalla stella.
Approfondimenti effettuati con il telescopio ALMA nel 2017 hanno consentito di osservare il bagliore dell'apocentro e determinare che la composizione dell'anello del disco polveroso intorno alla stella è simile all'ambiente cometario del nostro sistema solare
Fomalhaut b
Ipotesi sull'esistenza
Fin dalle prime scoperte riguardo al disco circumstellare di Fomalhaut, alcuni studiosi hanno avanzato l'ipotesi che alcune delle sue caratteristiche potevano essere spiegate tramite l'esistenza di uno o più pianeti. Holland et al. (1998) avanzavano l'ipotesi che la presenza della cavità interna al disco circumstellare, creduta allora essere della dimensione dell'orbita di Nettuno, fosse indizio della presenza di almeno un pianeta. Come si è detto, Wyatt e Dent (2002) supponevano che una delle possibili spiegazioni della presenza di addensamenti all'interno del disco fosse l'esistenza di un pianeta. Stapelfeldt et al. (2004) si richiamavano alla medesima ipotesi per spiegare alcune asimmetrie presenti nel disco. Uno dei lavori più accurati in questo filone è quello di Quillen (2006), la quale riconduceva alcune caratteristiche del disco, quali la sua eccentricità e il fatto che il suo confine interno sia molto netto, alla presenza di un pianeta collocato appena all'interno del disco stesso. La studiosa supponeva che il pianeta avesse una massa compresa fra quella di Nettuno e quella di Saturno e che compisse un'orbita avente una eccentricità simile a quella del disco stesso (e ≈ 0,1)
Osservazione
Il 13 novembre 2008 un team di scienziati diretti dall'astrofisico Paul Kalas presentò le immagini del disco circumstellare ottenute nel 2004 e nel 2006 tramite lAdvanced Camera for Surveys del telescopio spaziale Hubble. Nelle vicinanze del bordo interno del disco era apprezzabile un puntino luminoso che si era spostato di 184 ± 22 mas, corrispondenti a 1,41 ± 0,17 UA, in 1,7 anni. Il punto fu interpretato dal team come la prima immagine di un pianeta extrasolare mai ottenuta; il moto del punto fu interpretato come il moto di rivoluzione del pianeta intorno alla stella centrale. Il pianeta fu chiamato Fomalhaut b. Le ragioni che spinsero il team a interpretare il punto come l'immagine di un pianeta furono essenzialmente due: in primo luogo esso è troppo poco luminoso (ha una luminosità di 3,4 × 10−7 L⊙) per essere una stella di piccola massa o una nana bruna; in secondo luogo, la presenza di un corpo delle dimensioni di una nana bruna o di una stella di piccola massa nelle vicinanze del disco avrebbe provocato il suo disgregamento.
Caratteristiche
Fomalhaut b dista 115 UA dalla sua stella, che equivale a circa dieci volte la distanza di Saturno dal Sole. Lo spostamento del pianeta, rilevato dal telescopio Hubble, ha permesso di stimare il suo periodo di rivoluzione, pari a circa 872 anni terrestri. La velocità orbitale è 3,9 km/s, mentre l'eccentricità dell'orbita è 0,13. La massa del pianeta dovrebbe essere non superiore a 3 volte quella del pianeta Giove e non inferiore a quella di Nettuno (che equivale a 0,054 masse gioviane); si tratterebbe dunque di un gigante gassoso. Chiang et al. (2009) hanno sviluppato un modello in cui un singolo pianeta della massa inferiore a 3 MJ è responsabile delle caratteristiche osservabili del disco, ossia della sua eccentricità e del suo bordo interno molto netto. Tale modello è compatibile con le osservazioni di Fomalhaut b compiute da Kalas e colleghi. Di per sé il modello di Chiang et al. (2009) è compatibile con l'esistenza di altri pianeti più interni rispetto a Fomalhaut b, a patto che la massa di quest'ultimo sia di molto inferiore a 3 MJ. Tuttavia, le immagini catturate nella banda M dall'MMT Observatory escludono l'esistenza di altri giganti gassosi con una massa superiore a 2 MJ a una distanza compresa fra 10 e 40 UA dalla stella. Segue un prospetto sulle principali caratteristiche del sistema planetario.
* Non confermato
Dubbi sull'esistenza di Fomalhaut b
Due caratteristiche del punto luminoso osservato nel 2008 sono apparse subito problematiche per la sua identificazione con un pianeta. La prima è che esso non ha alcuna corrispondenza nella banda dell'infrarosso, mentre un pianeta dell'età di almeno 200 milioni di anni, alla distanza in cui Fomalhaut b è stato osservato, dovrebbe essere abbastanza freddo (circa 400 K) per emettere un notevole quantitativo della sua radiazione nelle frequenze dell'infrarosso. Tuttavia, già Marengo et al. (2009), sulla base di una serie di osservazioni condotte dal telescopio spaziale Spitzer, sottolineavano che, data la sensibilità di Spitzer, la mancata osservazione del pianeta nelle frequenze dell'infrarosso poneva severi limiti circa la possibilità di esistenza di un pianeta nelle regioni in cui Fomalhaut b era stato osservato. In particolare, per sostenere la tesi dell'identificazione del punto luminoso con un pianeta, era necessario rivedere il modello dell'atmosfera di Fomalhaut b, in modo da rendere conto della mancata osservazione nell'infrarosso. In attesa di tale revisione, continuavano questi studiosi, l'ipotesi più probabile circa il punto luminoso osservato era che si trattasse di una regione del disco che avesse riflettuto la luce della stella.
Questi dubbi sono stati rilanciati con maggiore forza da Janson et al. (2012) sulla base di nuove più precise osservazioni compiute da Spitzer, che pongono limiti ancora più severi alla luminosità di Fomalhaut b nell'infrarosso. Gli autori respingono anche l'ipotesi avanzata nel 2008 da Kalas e colleghi, secondo la quale l'alta luminosità di Fomalhaut b nel visibile sia determinata da un disco di accrescimento intorno al pianeta: tale ipotesi viene ritenuta improbabile, dato l'imponente tasso di accrescimento che sarebbe richiesto per spiegare la luminosità osservata, di dimensioni simili a quello di una stella T Tauri. Janson et al. (2012) ritengono molto più probabili altri due scenari rispetto a quello dell'esistenza di un gigante gassoso: il primo riconduce il punto luminoso al recente scontro di due planetesimi; il secondo a un pianeta molto più piccolo di quello ipotizzato da Kalas e colleghi, della massa inferiore a 10 M⊕, e quindi roccioso o ghiacciato, intorno al quale orbitano uno sciame di planetesimi che collidono fra loro, producendo il punto luminoso osservato.
L'altra caratteristica del punto luminoso che suscita perplessità è la sua variabilità. Fra il 2004 e il 2006 il punto ha diminuito la sua luminosità di circa mezza magnitudine. La teoria che identifica il punto con un pianeta gassoso deve formulare ipotesi aggiuntive per spiegare questa variabilità. Kalas e colleghi ne avanzano due: oltre alla già citata presenza di un disco di accrescimento intorno al pianeta, essi avanzano l'ipotesi che il pianeta sia circondato da un sistema di anelli, simile a quello di Saturno: gli anelli rifletterebbero variamente la luce della stella centrale, mentre il pianeta si sposta, e questo spiegherebbe la variazione di luminosità. Tuttavia Janson et al. (2012) respingono anche questa seconda ipotesi in ragione del fatto che appare probabile che attualmente l'origine del punto luminoso osservato si trova fra la stella e la Terra e quindi appare improbabile che un sistema di anelli possa riflettere grossi quantitativi di luce da quella posizione.
Negli anni 2010, il punto continuò a affievolirsi e a ingrandirsi, fino a sparire nelle immagini riprese nel 2014 dal telescopio spaziale Hubble. Con uno studio pubblicato nel 2020, Andras Gaspar e George H. Rieke hanno avallato l'ipotesi di Jansson et al. che in realtà il presunto pianeta non fosse altro che una nube di polveri derivata da una collisione tra due asteroidi, e che il pianeta non sia mai esistito. Gaspar e Rieke ritengono che l'impatto sia avvenuto non molto tempo prima delle immagini riprese da Hubble nel 2004, e che a quel punto la nube di polveri si fosse espansa a oltre 300 milioni di chilometri, divenendo invisibile agli occhi dell'Hubble.
Le caratteristiche del disco implicano l'esistenza di pianeti?
Al di là dell'identificazione del punto luminoso osservato da Hubble con un pianeta gassoso, resta la questione se le caratteristiche del disco implichino o meno l'esistenza di un pianeta. Come si è detto, il disco è eccentrico e Fomalhaut è collocato in uno dei fuochi dell'ellisse. Tale forma ellittica è stata supposta essere generata da un pianeta avente un'orbita eccentrica che ha "scolpito" il disco in modo da indurre una simile eccentricità nel disco stesso. Il fatto che il disco sembri avere confini molto netti ha portato a simili conclusioni circa la presenza di un pianeta che ne "scolpisca" la forma. La necessità dell'esistenza di pianeti per spiegare la morfologia del disco è stata ribadita da Boley et al. (2012), i quali ipotizzano l'esistenza di due pianeti pastori che scolpiscono rispettivamente i confini interni ed esterni del disco, che appaiono essere entrambi molto netti quando osservati mediante il radiointerferometro ALMA.
Tuttavia non tutti gli studiosi sono d'accordo con questa conclusione. Lyra e Kuchner (2012) hanno modellato l'idrodinamica dei dischi asteroidali. Il disco è supposto contenere, oltre a polvere e planetesimi, grandi quantità di gas. La polvere trasferirebbe il calore ricevuto dalla stella centrale al gas, il quale si espanderebbe: la pressione del gas favorirebbe l'accumularsi della polvere in particolari regioni del disco, mentre il gas, raffreddandosi, tenderebbe a raddensarsi in regioni ripulite dalla polvere stessa. Ne seguirebbe una anti-correlazione fra la densità della polvere e quella del gas: ciò avrebbe il risultato di confinare la polvere in regioni dai confini molto netti. Le onde di polvere create dal gas ad altra pressione sarebbero anche responsabili della eccentricità del disco. Nessun pianeta quindi sarebbe necessario per spiegare le caratteristiche fisiche del disco.
Etimologia e significato culturale
Il nome Fomalhaut deriva dall'arabo classico فم الحوت fam al-ḥūt (al-janūbī), che significa "la bocca del pesce/della balena (australe)", con evidente riferimento alla posizione della stella all'interno della costellazione del Pesce Australe. Tale nome ha una origine antica, tanto che si trova già in un almanacco del 1340, con la sua traduzione latina Os Piscis Merīdiāni. Altre traduzioni latine dello stesso nome arabo sono Os Piscis Merīdionālis ed Os Piscis Notii. Probabilmente nessun'altra stella possiede un nome proprio con più varianti ortografiche di Fomalhaut. Alcune di esse sono: Fumahant, Fumahaut, Fomahand, Fontabant, Phomaut, Phomault, Phomant, Phomaant, Phomhaut, Fomalcuti.
Un altro nome della stella è Difda al Auwel, derivante dall'arabo colloquiale الضفدع الأول al-ḍifdiʿ al-awwal, "la prima rana". La seconda rana è Deneb Kaitos, ; tale nome deriva probabilmente dal fatto che Fomalhaut precede Deneb Kaitos nel suo moto apparente nel cielo.
Nel Settecento, Fomalhaut venne erroneamente identificata con la persiana Hastorang (che nell'Avestā è l'Orsa Maggiore).
È considerata una delle quattro stelle regali guardiane del cielo, che sovraintendono alle altre stelle, le altre tre sono Aldebaran, Regolo e Antares. Fomalhaut sarebbe la sentinella delle stelle del sud(3°52' Costellazione Pesce Australe), Aldebaran di quelle dell'est (9° Costellazione Gemini), Regolo di quelle del nord (29° Costellazione Leo) e Antares di quelle dell'ovest (9° Costellazione Sagittario).
Queste quattro stelle marcano i due solstizi e i due equinozi e quindi dividono il cielo in quattro parti. Le quattro stelle furono poi identificate con i quattro arcangeli: Fomalhaut con Gabriele, Aldebaran con Michele, Regolo con Raffaele e Antares con Uriel. Altre associazioni sono quelle con i quattro cavalieri dell'Apocalisse o con i cavalli che trainavano i carri citati nel libro di Zaccaria.
Fomalhaut è citata nella Divina Commedia di Dante Alighieri in Purgatorio, VIII, vv. 89-90 come una delle "tre facelle" che illuminano l'emisfero meridionale, essendo le altre due Canopo e Achernar.
La stella appare più volte negli scritti dello scrittore statunitense Howard Phillips Lovecraft, come nel romanzo La ricerca onirica dello sconosciuto Kadath e, nel XIV sonetto della raccolta Fungi from Yuggoth, in cui descrive brevemente anche il pianeta Nithon che vi orbiterebbe intorno.
Inoltre la stella è l'oggetto della spedizione spaziale alla base del romanzo Ritorno dall'Universo di Stanisław Lem.
Note
Bibliografia
Testi generici
Sulle stelle
Carte celesti
- Atlante celeste liberamente scaricabile in formato PDF.
Cultura di massa
Voci correlate
Cintura asteroidale
Fomalhaut b
HR 8799, una stella attorno alla quale sono stati fotografati tre pianeti (b, c e d)
Pianeti extrasolari confermati
Stelle con pianeti extrasolari confermati
Vega, un'altra stella brillante con un disco di polveri
Altri progetti
Collegamenti esterni
Stelle di classe spettrale A
Stelle bianche di sequenza principale
Stelle con pianeti extrasolari
Dischi circumstellari | {
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Why do you have to be so diligent with the Word? Because the devil is diligent with his junk. He is constantly throwing fiery darts at you. He works diligently to make sure the world is surrounding you with fear, sickness, poverty and every other kind of garbage he can use to destroy you. But you can protect your spirit from those things by conditioning yourself to respond to every one of them with the Word of God. Change your routine, and make a decision right now to begin that spiritual conditioning.
Speak these words as a part of your daily routine. As you read your Bible each day, locate and highlight more to add to your collection. Then, watch your words, guard your heart and get results! | {
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angular
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Mahdi Army 'obeys' truce order
Militia fighters say they could resume violence if provoked by US troops.
Al-Sadr reportedly suspended activities to
"rehabilitate" the Mahdi Army [AFP]
Ahmed al-Shaibani, a senior Sadr aide, warned US forces not to take advantage of the order, while another said the suspension might last only a week if US and Iraqi forces did not stop detaining the cleric's followers.
"We say to the Americans, don't be happy. The resistance does not end," Shaibani said.
'Wait and see'
The Pentagon cautiously welcomed the six-month truce.
"The proof of Muqtada al-Sadr's sincerity is in the pudding," Brigadier General Richard Sherlock, deputy director for operational planning at the US department of defence, said.
Who is Muqtada al-Sadr?
"We have to wait and see. We need to see actions that correspond to that as opposed to just calls for calm."
The US military has welcomed an order given by Muqtada al-Sadr, the Iraqi Shia leader, to freeze his militia's activities following deadly clashes in southern Iraq.
Lieutenant Colonel Chris Garver, the US military spokesman in Iraq said: "Any time someone in Iraq, especially a leader, wants to use non-violent methods to solve problems ... we encourage this."
An aide said al-Sadr had suspended armed action by his 60,000 member militia to remove rogue elements.
Hazim al-Araji, an aide to al-Sadr, read a statement from the Shia leader saying "we declare the freezing of Mahdi Army without exception in order to rehabilitate it in a way that will safeguard its ideological image within a maximum period of six months starting from the day this statement is issued."
Asked if Wednesday's unexpected order meant no attacks on US troops, another senior aide who declined to be identified said: "All kinds of armed actions are to be frozen, without exception."
Corrupt leaders
Abu Ali, another Mahdi Army commander in Sadr City, said the suspension order would expose corrupt leaders.
"There are some bad leaders in the Mahdi army who should have been replaced a long time ago"
Abu Ali,
Mahdi Army commander
"There are some bad leaders in the Mahdi Army who should have been replaced a long time ago ... Through this order the corrupt elements will show their faces, because they will not comply," he said.
US forces have previously said that many attacks against them and Iraqi security forces are the work of "rogue" Mahdi Army groups, who may not operate under al-Sadr's full control.
The decision had been taken after 52 people were killed in the southern Iraqi city of Karbala in fighting between the Mahdi Army and another Shia bloc - the Supreme Iraqi Islamic Council (SIIC), whose armed wing controls much of the south.
On Thursday, Iraqi security forces arrested 72 fighters following the Karbala clashes, the defence ministry said.
A ministry statement said a number of weapons had also been confiscated during a search of homes across the southern city.
Test of authority
Analysts said al-Sadr's six-month order would be a test of his authority over the group, which is believed to have fragmented and, according to the US, has received funding, training and weapons from Iran.
Also on Wednesday, Nuri al-Maliki, the Iraqi prime minister, said that fighters from the Mahdi Army and the SIIC who were involved in the violence in Karbala had wanted to blow up the Imam Hussein shrine, one of the holiest to Shia Muslims.
"From our initial investigation, we found some evidence of who did this act ... the intention of this act was to storm into the shrine of Imam Hussein and blow it up," al-Maliki said from inside the shrine during a visit to Kerbala, 110km south of Baghdad.
The Mahdi Army denied the allegation.
While his ministers have left the cabinet, al-Sadr's political bloc holds 30 seats in parliament and is still part of the ruling Shia alliance. | {
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{"url":"https:\/\/meta.stackoverflow.com\/questions\/318202\/how-can-i-know-if-im-banned-from-answering-without-trying-to-answer","text":"How can I know if I'm banned from answering without trying to answer?\n\nI gave some bad answers on Stack Overflow and as a result I'm banned from answering again.\n\nI know that to lift up the ban, I need to gain reputation by asking questions and eventually the ban will be removed.\n\nBut how do I know my ban was removed without trying to actually answer questions?\n\nWhat happened several times already, that I spent some time on writing an answer to later discover I'm banned.\n\n\u2022 I think hiding that info until you try to post something helps them prevent people from guessing what the algorithms are for avoiding bans. \u2013\u00a0BSMP Mar 3 '16 at 16:04\n\u2022 Have you considered seeing if you could fix your existing answers? \u2013\u00a0Jon Clements Mar 3 '16 at 16:11\n\u2022 I deleted the bad answers I gave. \u2013\u00a0Alex Weitz Mar 3 '16 at 16:14\n\u2022 @AlexWeitz IIRC, deleted answers count against you. The recommendation is to edit them into good answers and undelete \u2013\u00a0Justin Mar 3 '16 at 17:03\n\u2022 You can use deleted:1 in the search query to find your own deleted answers. \u2013\u00a0Jester Mar 3 '16 at 17:33\n\u2022 @Jester - Don't you still need enough rep to see deleted posts? \u2013\u00a0BSMP Mar 3 '16 at 18:50\n\u2022 I think you can see your own without problem, it's not the question that has been deleted. \u2013\u00a0Jester Mar 3 '16 at 19:02\n\u2022 @Jester Using user:660921 deleted:1 gives me all my answers, and just deleted:1 gives me all answers on the site. It looks like you need 10k rep for this :-\/ \u2013\u00a0Martin Tournoij Mar 4 '16 at 0:08\n\u2022 Yeah, looks like it. See relevant meta question. \u2013\u00a0Jester Mar 4 '16 at 0:27\n\u2022 Good luck digging your answers out of a ditch... I'm told that works really, really well. \u2013\u00a0canon Mar 4 '16 at 0:32\n\u2022 When I first came to Stack Overflow I didn't know how to ask and answer questions well so my account got banned. Rather than waiting however long it takes to get unbanned, I just started over with a new account. \u2013\u00a0Suragch Mar 4 '16 at 0:43\n\u2022 @Suragch I too like to live on the edge, but being as badass as you is my dream :D \u2013\u00a0Alexander Derck Mar 4 '16 at 10:46\n\u2022 @Suragch nowadays system limits those who deleted account and restarted to one question per week - \"f you're blocked at the time you remove your account and return, you'll be limited to one question per week until you can establish yourself as a contributor to the site...\" \u2013\u00a0gnat Mar 4 '16 at 11:09\n\u2022 It's in the URL of your userpage: http:\/\/stackoverflow.com\/users\/3501205\/alex-weitz So it's 3501205. \u2013\u00a0Martin Tournoij Mar 4 '16 at 18:10\n\u2022 Hmm, this seems counter productive. It is hardly going to encourage the crafting of quality answers if it isn't clear that the system is going to allow the user to post it. \u2013\u00a0Martin Smith Mar 5 '16 at 20:29\n\nThere seems to be no way to know until you try.\n\nThere is an excellent explanation of this at MSE by Shog9. It was written by a former Stack Exchange Community Manager to explain question bans, but the general reasoning appears to apply to answers as well:\n\n...ban is applied when a user who qualifies for it tries to ask a question. Until that point, they're like that cat in a box, both banned and unbanned.\n\n[\u2026]\n\nAs usual, I'm not going to publish the actual formula, but I'll make a few notes as to the nature of these bans:\n\n\u2022 There are multiple, cooperating algorithms at work. Some of them result in permanent bans, others just apply stricter rate-limits [\u2026]\n\u2022 How do you know the formula? I heard that only moderators and the people working on SO know the formulas? \u2013\u00a0Ashish Ahuja Mar 4 '16 at 11:54\n\u2022 @AshishAhuja moderators don't know the formula either. If you happen to click the link to the post I quote (\"excellent explanation...\") you'll find out that its author is Stack Exchange employee, Community Manager - he knows the formula \u2013\u00a0gnat Mar 4 '16 at 11:57","date":"2021-04-22 15:21:28","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.23152218759059906, \"perplexity\": 1384.453996364764}, \"config\": {\"markdown_headings\": false, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-17\/segments\/1618039610090.97\/warc\/CC-MAIN-20210422130245-20210422160245-00193.warc.gz\"}"} | null | null |
La seconde bataille de Ras Lanouf est une bataille de la guerre civile libyenne. Elle est la première victoire de Mouammar Kadhafi sur le front de l'est.
Déroulement
Le les rebelles gagnent du terrain vers Ben Jawad après la victoire lors de la première bataille de Ras Lanouf mais ce mouvement de repli des loyalistes n'est qu'un leurre. Les loyalistes ont amené les rebelles dans une nasse ou ils enferment 600 rebelles. Le , les rebelles d'Abdul Fatah Younis tentent de contre-attaquer, mais sont repoussés, avec de lourdes pertes. Le , les derniers rebelles de Ben Jawad capitulent. Les rebelles se replient sur Ras Lanouf.
Du 8 au , Ras Lanouf est violemment bombardé par les loyalistes. Le , les rebelles regagnent du terrain. Le les chars de Kadhafi contournent Ras Lanouf, pour éviter l'encerclement et la destruction totale du gros des forces rebelles Abdul Fatah Younis envoie 800 hommes au sud pour contenir les tanks de Kadhafi. Le , les rebelles se retirent de Ras Lanouf et grâce au sacrifice des 800 rebelles envoyés par Abdul Fatah Younis, les rebelles évitent l'encerclement et se replient sur El Agheila. Le , 500 rebelles tentent une contre-offensive qui se solde par un échec complet.
Notes et références
Vidéographie
Libye, la bataille de Brega (version longue partie 2), France 24, .
Articles connexes
Guerre civile libyenne
Protestations et révolutions dans le monde arabe en 2010-2011
Ras Lanouf 02 | {
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Please join me in a celebration of picture books in 2014. After tracking our book reading habits for (almost) every week in 2013, we're inviting you to join in too.
If you're a parent or carer of young children, you probably read more books than you think. The challenge for the year is to read 300 different picture books. This comes to 5 or 6 different books a week on average.
There will be a weekly post every Monday where I'll list our different books read for the previous Monday through Sunday. The linky will be open until the following Sunday. You don't have to link up every week, or even any week (unless you want to enter a giveaway).
I will try to have a giveaway once a month, on the first Monday of the month. There will be a variety of prizes across all age ranges, depending on what review books we receive! Open to UK addresses only I'm afraid but if you're based elsewhere and have someone in the UK I can send to, that's not a problem.
Because it's challenging but achievable whilst still reading that favourite book over and over again. There are 365 days in the year so it's less than one different book each day. Young children will easily ask for at least three picture books every day, that's over one thousand stories read in the year, so it's just making sure that around a third of the stories are new to you that year. It's also a nice round number.
I can't afford new books.
You don't need to. It's keeping track of what you're already reading. The count starts on 1st January and every book you read is 'new' the first time you read it in 2014. Even if you have no books in the house already, if you can get to a library once a fortnight and borrow 11 books each time then you can do it.
I work full time and can't read with my children every day.
5 or 6 picture books can be read in two or three nights per week, but it will be more of a challenge to read 300 different books in the year. Can you challenge yourself to find the time to read 300 books in total if you can get two sessions of reading time each week?
Can I count the books children read to me?
Yes. I usually don't, but if the only reading time you get each day is listening to a child read their school reading scheme, then use the time to enjoy the books as much as possible.
Do I have to review all the books we read?
No. You don't have to list them all either as long as you're keeping track. You can review favourites, or just write the titles down for future reference, if you want.
Do I have to join in the linky?
No. But you might want to do a monthly post looking at how you're approaching the challenge of reading and pick out some favourite reads to share.
Does it have to be 300 different picture books?
That's my challenge, but I understand that it won't be possible for everyone. Challenging yourself to read 300 books to your child(ren) during the year, even the same book 300 times, is still a challenge.
Does it have to be picture books?
Well, that's what the challenge is primarily aimed at (although I also included highly illustrated early readers if they were short enough), but if your children prefer having chapter books read to them, maybe make it 300 chapters instead?
I missed the start date!
You can join #300PBs at any point, but may take on a smaller challenge to compensate. For example, 275 books if you join in February; 150 books if you join in July. The title and tag will remain the same for simplicity.
Nice idea. I'm in. What challenge are you setting yourself? I'll be rerunning my read52 🙂 even though my linky was very much not loved!
I'm doing my own Good Reads 50 books in 2014 challenge after doing it this year, and don't think we'd manage 300 different picture books (takes my son ages to get into new books) but we might track our books along with you. Working full time definitely means short on time as often he's too tired to go to bed with a book which is our traditional time, but we're toilet training so books are being read in volume during toilet breaks!
Do the 300 reads challenge instead of different, and I bet you make it in no time 🙂 I'm doing 52 books too for me, but need to make myself read more out loud for the children even though I love books!
Keris Stainton recently posted..Happy New Year!
I'm in! I've joined the Classics Club for myself and these will definitely help my blogging! Happy New Year!
Love books so going to do my best to keep up with this, although I think it might scare me to realise just how many books Monkey has!
We'll definitely be doing our best to read at least 300. Looking at the stats for last year we managed to review 521 and I'm pretty sure we read far more. Utterly rewarding and definitely part of why C did so well with her first literacy steps in Year 1.
readItDaddy recently posted..ReadItMummiesandDaddies2013 – Our Weekly Theme Campaign. Was it a success?
What a great idea. I'm in. My 8 month old has over 50 books already (I've just counted) and a library card (which I've not used yet so this challenge will change that). I love reading them books to her.
I'm in! In the five years I've been reading to my girls I've never tracked the books we read so this is a great idea!
We did it and loved it – thank you.
Laura (@chickywiggle) recently posted..300 Picture Book Challenge …… the year is over. | {
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} | 5,114 |
\section{Introduction}\label{sec:introduction}
Question answering (QA) has been a major area of research in Artificial Intelligence and Machine Learning
since the early days of computer science~\citep{voorhees1999trec, moldovan2000structure, brill2002analysis, ferrucci2010building}.
The need for a performant open-book QA solution was exacerbated by rapid growth in available information in niche
domains, the growing number of users accessing this information, and the expanding need for more efficient operations.
QA systems are especially useful when a user searches for specific information and does not have the time - or simply
does not want - to peruse all available documentation related to their search to solve the problem at hand.\par
In this article, open-book QA is defined as the task whereby a system (such as a computer software) answers natural
language questions from a set of available documents (open-book).
These questions can have yes-no-none answers, short answers, long answers, or any combination of the above.
In this work, we did not train the system on our domain-specific documents or questions and answers, a technique
called zero-shot learning~\citep{brown2020language}.
The system should be able to perform with a variety of document types and question and answers without training.
We defined this approach as "zero-shot open-book QA".
The proposed solution is tested on AWS Documentation dataset.
However, as the models within this solution are not trained on the dataset, the solution can be used in other similar
domains such as finance and law. \par
Software technical documentation is a critical piece of the software development life cycle process.
Finding the correct answers for one's questions can be a tedious and time-consuming process.
Currently, software developers, technical writers, and marketers are required to spend substantial
time writing documents such as technology briefs, web content, white papers, blogs, and reference guides.
Meanwhile, software developers and solution architects have to spend time searching for specific information
they need.
Our approach to QA aims to help them find it faster.\par
Our work's key contributions are:
\begin{enumerate}
\item introduce a new dataset in open-book QA,
\item propose a two-module architecture to find answers without context,
\item experiment on ready-to-use information retrieval systems,
\item infer text and yes-no-none answers in a single forward pass once we find the right document.
\end{enumerate}
\par
The rest of the paper is structured as follows: First, related previous work is summarized.
Then the dataset is described.
Next, details on implementing the zero-shot open-book QA pipeline are provided.
In addition, the experiments are explained, and finally the results along with limitations and next steps are presented.
\section{Related work}\label{sec:related-work}
There are a number of datasets in the literature for natural language QA
~\citep{rajpurkar2016squad, joshi2017triviaqa, khashabi2018looking, richardson2013mctest, lai2017race,
reddy2019coqa, choi2018quac, tafjord2019quarel,mitra2019declarative}, as well several solutions to tackle
these challenges~\citep{seo2016bidirectional, vaswani2017attention, devlin2018bert, he2011summarization, kumar2016ask,
xiong2016dynamic, raffel2019exploring}.
The natural language QA solutions take a question along with a block of text as context and attempts to find the
correct answer to the original question within the context.
Open-book QA solutions take a question along with a set of documents that may contain the answer to the question,
then the solution attempts to find the answer to the original question within the available set of documents.
Open-book QA solutions have been explored by several research teams including Banerjee et al.~\citep{banerjee2019careful},
which performs QA using fine-tuned extractive language models, and the work of Yasunaga et al.~\citep{yasunaga2021qa},
which performs QA using GNNs.
\par
In this paper, we propose an approach that differs from the previous body of work as we do not receive the context but
assume that the answer lies in a set of readily available documents (open-book);
In addition, we are not allowed to train our models on the given questions or set of documents (zero-shot).
Our proposed solution attempts to answers questions from a set of documents with no prior training or
fine-tuning (zero-shot open-book question answering).
\section{Data}\label{sec:data}
Real world open-book QA use cases require significant amounts of time, human effort, and cost to access or
generate domain-specific labeled data.
For our solution, we intentionally did not use any domain-specific labeled data and ran experiments on popular
QA datasets and pre-trained models.
We used feedback from customers to generate a set of 100 questions as the test dataset and used QA datasets, explained in
section~\ref{subsec:squad} and~\ref{subsec:natural-questions}, for training.
\subsection{AWS Documentation Dataset}
\label{subsec:aws-documentation-dataset}
Herein, we present the AWS documentation corpus \footnote{https://github.com/siagholami/aws-documentation} ,
an open-book QA dataset, which contains 25,175 documents along with 100 matched questions and answers.
These questions are based on real customer questions on AWS services.
There are two types of answers: text and yes-no-none answers.
Text answers range from a few words to a full paragraph sourced from a continuous series of words in a document
or from different locations within the same document.
Yes-no-none(YNN) answers can be yes, no, or none for cases where the returned result is empty and does not lead to a
binary answer (i.e., yes or no).
All questions in the dataset have a valid answer within the accompanying documents.
Table~\ref{tab:examples} shows a few examples from the dataset.
\begin{table}[h]
\caption{\footnotesize{Three sample questions from the test dataset }}
\label{tab:examples}
\centering
\begin{tabular}{p{0.4\linewidth} p{0.45\linewidth} p{0.1\linewidth}}
\toprule
Question & Text Answer & YNN Answer \\
\midrule
What is the maximum number of rows in a dataset in Amazon Forecast? & 1 billion & None \\\\
Can I stop a DB instance that has a read replica? & You can't stop a DB instance that has a read replica. & No \\\\
Is AWS IoT Greengrass HIPAA compliant? & Third-party auditors assess the security and compliance of AWS IoT Greengrass as part of multiple AWS compliance programs. These include SOC PCI FedRAMP HIPAA and others. & Yes \\
\bottomrule
\end{tabular}
\end{table}
\subsection{SQuAD Datasets}
\label{subsec:squad}
The Stanford Question Answering Dataset (SQuAD)\footnote{https://rajpurkar.github.io/SQuAD-explorer/}
is a reading comprehension dataset~\citep{rajpurkar2016squad}, including questions created by crowdworkers on Wikipedia articles.
The answers to these questions is a segment of text from reading passages, or the question might be unanswerable.
SQuAD1.1 comprises 100,000 question-answer pairs on more than 500 articles.
SQuAD2.0 adds 50,000 unanswerable questions written adversarially by crowdworkers to look similar to answerable ones.
\subsection{Natural Questions Dataset}
\label{subsec:natural-questions}
The Natural Questions (NQ) dataset \footnote{https://ai.google.com/research/NaturalQuestions} includes 400,000
questions and answers created on Wikipedia articles~\citep{kwiatkowski2019natural}.
Questions consist of real queries which answers can be long (a few sentences), short (a few words) if present on
the page, or null if no long or short answer is present.
\section{Approach}\label{sec:approach}
Our approach consists of two high-level modules: retriever and extractor.
Given a question, the retriever tries to find a set of documents that contain the answer;
Then, from these documents, the extractor tries to find the answer.
Figure~\ref{fig:architecture} illustrates a high level workflow of the solution,
and Table~\ref{tab:solution-example} shows an example of the question,retrieved documents, and extracted
answers using the solution.
\begin{figure}[h]
\label{fig:architecture}
\centering
\includegraphics[width=\linewidth]{figures/architecture.png}
\caption{\footnotesize{High level workflow of the solution in one pass}}
\end{figure}
\begin{table}[h]
\caption{\footnotesize{Our solution with an example}}
\label{tab:solution-example}
\begin{tabular}{p{.4\linewidth} p{0.55\linewidth}}
\toprule
Question: & What are the Amazon RDS storage types? \\
\midrule
Retriever set of documents: & CHAP\_Storage.txt, CHAP\_Limits.txt, CHAP\_BestPractices.txt \\\\
Extractor Text Answer: & General Purpose, SSD, Provisioned IOPS, Magnetic \\\\
Extractor Yes/No Answer: & None \\
\bottomrule
\end{tabular}
\end{table}
\subsection{Retrievers}\label{subsec:retrievers}
Given a question with no context, our approach relies on the retriever to find the right documents
that contains the answer.
The need for a retriever stems from the fact that our extractors are fairly large models and it is time and cost
prohibitive for the extractor to go through all available documents.
For example in our AWS Documentation dataset from Section~\ref{subsec:aws-documentation-dataset}, it will take hours for a
single instance to run an extractor through all available documents.
We ran experiments with simple information retrieval systems with a keyword search along with deep
semantic search models to list relevant documents for a question.
We used precision at K ($P@K$) metric to evaluate our retrievers.
Precision at K is the proportion of retrieved items in the top-k set that are relevant:
\[ P@K = \frac{\text {number of retrieved documents that are relevant}}{\text {total number of retrieved documents}}\]
\subsubsection{Whoosh}\label{subsubsec:whoosh}
Whoosh\footnote{https://whoosh.readthedocs.io/} is a fast, pure Python search engine library.
The primary design impetus of Whoosh is that it is pure Python and can be used anywhere Python is running,
as no compiler or Java is required.
\subsubsection{Amazon Kendra}\label{subsubsec:amazon-kendra}
Amazon Kendra\footnote{https://aws.amazon.com/kendra/} is a semantic search and question answering service
provided by AWS for enterprise customers.
Kendra allows customers to power natural language-based searches on their own AWS data by using a deep learning-based
semantic search model to return a ranked list of relevant documents.
Amazon Kendra's ability to understand natural language questions enables it to return the most relevant passage and
related documents.
\subsection{Extractors}
\label{subsec:extractors}
Given a question with no context, the retriever finds a set of documents.
Then the output of the retriever will pass on to the extractor to find the right answer for a question.
We created our extractors from a base model which consists of different variations of
BERT~\citep{devlin2018bert} language models and added two sets of layers to extract yes-no-none answers and text answers.
Our approach attempts to find yes-no-none answers and text answers in the same pass.
Our model takes the pooled output from the base BERT model and classifies it in three categories: yes, no, and none.
Furthermore, our model takes the sequence output from the base BERT model and adds two sets of dense layers with
sigmoid as activation.
The first layer tries to find the start of the answer sequences, and the second layer tries to find the end of the
answer sequences.
There can be multiple starts and ends for a single text answer.
Figure~\ref{fig:extractor-model} illustrates the extractor model architecture.
\begin{figure}[h]
\label{fig:extractor-model}
\centering
\includegraphics[width=\linewidth]{figures/extractor-model}
\caption{\footnotesize{Extractor model architecture }}
\end{figure}
\par
For our base model, we compared BERT (tiny, base, large)~\citep{devlin2018bert} along with RoBERTa~\citep{liu2019roberta},
AlBERT~\citep{lan2019albert}, and distillBERT~\citep{sanh2019distilbert}.
We implemented the same strategy as the original papers to fine-tune these models.
We also used the same hyperparameters as the original papers: L is the number of transformer blocks (layers),
H is the hidden size, and A is the number of self-attention heads.
\subsubsection{Extractor Model}
\label{subsubsec:extractor-model}
We define a training set datapoint as a four-tupele $(d,s,e,yn)$ , where $d$ is a document containing the answers,
$s, e$ are indices to the start and end of the text answer, and $yn$ defines the yes-no-none answer.
The loss of our model is:
\[ L = \log p (s, e, yn | d) \]
\[ L = \frac{1}{3} (\log ps (s | d) + \log pe (e | d) + \log pyn (yn | d) ) \]
where each probability $p$ is defined as:
\[ ps (s | d) = \frac {1} {1 + \exp(-f_{start} (s,d; \theta))} \]
\[ pe (e | d) = \frac {1} {1 + \exp(-f_{end} (e,d; \theta))} \]
\[ pyn (yn | d) = \frac {\exp (-f_{yn} (yn,d; \theta))} {\sum{yn'} \exp(-f_{yn} (yn',d;\theta))' } \]
where $\theta$ is the base model parameters and $f_{start}$, $f_{end}$, $f_{yn}$ represent three outputs from the
last layer of the model.
\par
At inference, we pass through all text from each document and return all start and end indices with scores higher
than a threshold. We used F1 and Exact Match (EM) metrics to evaluate our extractor models.
\section{Experiments}\label{sec:experiments}
In our experiments, we used pre-trained or ready-to-use information retrieval systems because these systems are
readily available and building a custom retriever with better performance is not economical.
Our experiments show that Amazon Kendra's semantic search is far superior to a simple keyword search and that
the bigger the base model (BERT-based), the better the performance.
The retriever results are shown in Table~\ref{tab:retrievers}. \par
\begin{table}[h]
\caption{\footnotesize{Retrievers performance}}
\label{tab:retrievers}
\centering
\begin{tabular}{p{0.15\linewidth} p{0.05\linewidth} p{0.05\linewidth} p{0.05\linewidth} p{0.05\linewidth} p{0.05\linewidth} p{0.05\linewidth} p{0.05\linewidth} p{0.05\linewidth} p{0.05\linewidth} p{0.05\linewidth}}
\toprule
Retriever & $P@1$ & $P@3$ & $P@5$ & $P@7$ & $P@9$ & $P@13$ & $P@22$ & $P@30$ & $P@40$ & $P@60$ \\
\midrule
Whoosh & .05 & .06 & .06 & .06 & .06 & .06 & .06 & .06 & .06 & .06 \\
Kendra & .66 & .79 & .86 & .87 & .9 & .91 & .92 & .93 & .94 & .95 \\
\bottomrule
\end{tabular}
\end{table}
Regarding our extractors, we initialized our base models with popular pretrained BERT-based models as described in
Section~\ref{subsec:extractors} and fine-tuned models on SQuAD1.1 and SQuAD2.0~\citep{rajpurkar2016squad} along with
natural questions datasets~\citep{kwiatkowski2019natural}.
We trained the models by minimizing loss L from Section~\ref{subsubsec:extractor-model} with the AdamW
optimizer~\citep{devlin2018bert} with a batch size of 8.
Then, we tested our models against the AWS documentation dataset (Section~\ref{subsec:aws-documentation-dataset})
while using Amazon Kendra as the retriever. Our final results are shown in Table~\ref{tab:extractors}.
\begin{table}[h]
\caption{\footnotesize{Extractor performance}}
\label{tab:extractors}
\centering
\begin{tabular}{p{.3\linewidth} p{0.4\linewidth} p{.1\linewidth} p{.1\linewidth} }
\toprule
Extractor Base Model & Extractor Hyperparameters & F1 & EM \\
\midrule
BERT Tiny & L=2, H=128 & 0.128 & 0.09 \\
RoBERTa Base & L=12, H=768 & 0.154 & 0.09 \\
DistilBERT & L=12, H=768 & 0.158 & 0.08 \\
AlBERT Base & L=12, H=768 & 0.199 & 0.11 \\
BERT Base & L=12, H=768 & 0.245 & 0.16 \\
BERT Large & L=24, H=1024 & 0.247 & 0.16 \\
AlBERT XXL & L=12, H=4096 & 0.422 & 0.39 \\
\bottomrule
\end{tabular}
\end{table}
\section{Limitations and Future Work}
\label{sec:limitations-and-future-work}
Our solution has a number of limitations.
Below we describe some of these and suggest directions for future work.
We were able to achieve 49\% F1 and 39\% EM for our test dataset due to the challenging nature of
zero-shot open-book problems.
The performance of the solution proposed in this article is fair if tested against technical software documentation.
However, it needs to be improved before finding use in real-world software products.
Additionally, more testing is needed if we want to further expand the applicability of this solution in other domains
(e.g., medical corpus, laws and regulations).
Furthermore, the solution performs better if the answer can be extracted from a continuous
block of text from the document.
The performance drops if the answer is extracted from several different locations in a document.
Moreover, all questions had a clear answer in the AWS documentation dataset,
which is not always the case in the real-world.
As our proposed solution always returns an answer to any question, '
it fails to recognize if a question cannot be answered.\par
For future work, we plan to experiment with generative models such as GPT-2~\citep{radford2019language} and
GPT-3~\citep{brown2020language} with a wider variety of text in pre-training to improve the F1 and EM score
presented in this article.
\section{Conclusion}\label{sec:conclusion}
In this paper, we presented a new solution for zero-shot open-book QA with a two-step architecture to
answer natural language questions from an available set of documents.
With this novel solution, we were able to achieve 49\% F1 and 39\% EM with no domain-specific labeled data.
We hope this new dataset and solution helps researchers create better solutions for zero-shot open-book use cases
in similar real-world environments.
\section*{Declaration of competing interest}\label{sec:Declaration of competing interest}
The authors declare that they have no known competing financial interests or personal relationships
that could have appeared to influence the work reported in this paper.
\section*{CRediT authorship contribution statement}\label{sec:CRediT authorship contribution statement}
\textbf{Sia Gholami}: Conceptualization, Methodology, Software, Investigation, Writing - original draft.
\textbf{Mehdi Noori}: Writing - review \& editing.
\section*{Acknowledgements}\label{sec:Acknowledgements}
The contributions of Sia Gholami and Mehdi Noori were funded by Amazon Web Services.
\vskip 0.2in
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 1,851 |
\section{Introduction
Deciding whether a polynomial is nonnegative on an (often compact) semialgebraic set and the closely related problem of computing the (approximate) minimum value of a polynomial are fundamental problems of computational algebraic geometry and theoretical computer science, with many applications from discrete geometry and algorithmic theorem proving to the design and analysis of dynamical systems such as power networks, to name a few. This problem is well-known to be decidable \cite{Tarski1951, Renegar1992abc} but strongly NP-hard. The perhaps most studied, and arguably practically most successful, computational approach to it has been to certify the nonnegativity of the polynomial by writing it as a (weighted) sum of squared polynomials---a technique known as \emph{sum-of-squares decomposition}. A variety of results from real algebraic geometry such as Putinar's \emph{Positivstellensatz} \cite{Putinar1993} guarantee that every polynomial that is strictly positive over a compact semialgebraic set has such a representation.
Lower bounds on the global minima of polynomials and weighted sum-of-squares (WSOS) decompositions are usually computed numerically, using semidefinite programming (e.g., \cite{sostools, gloptipoly, Lasserre2001}) or non-symmetric cone optimization \cite{PappYildiz2019}, which is sufficient in many of the practical applications mentioned above. However, in many theoretical contexts, such as in computational algebraic geometry and automated theorem proving, it is required that the computed bounds be certified rigorously, in exact arithmetic.
Computing rational WSOS decompositions for polynomials with rational coefficients is a challenging problem even in the univariate case \cite{MagronSafeyElDinSchweighofer2019}. Symbolic methods such as those that rely on quantifier elimination or root isolation are exponential in the degree of the input polynomial. The optimal value of the semidefinite program is an algebraic number, but the study of the algebraic degree of the positive semidefinite cone \cite{NieRanestadSturmfels2010} suggests that one cannot hope for easily computable and verifiable certificates from taking a purely symbolic computing approach to the semidefinite programming problems that come from sums-of-squares. Therefore, a number of authors have proposed hybrid methods that ``round'' or ``project'' efficiently computable but inexact numerical sum-of-squares certificates to rigorous rational ones \cite{PeyrlParrilo2008, MagronSafeyElDin2018, DostertDeLaatMoustrou2020}; see also \cite{KaltofenLiYangZhi2008,KaltofenLiYangZhi2012,BrakeHauensteinLiddell2016}.
Our contribution is twofold. In \mbox{Section \ref{sec:dualcertificates}}, we propose a new framework for certifying that a polynomial is WSOS. The approach relies on convex programming duality and allows the efficient construction of rational WSOS decompositions from suitable rational vectors from the dual cone. In contrast to conventional WSOS certificates, which can be viewed as different representations of the polynomial whose nonnegativity they certify, dual certificates are distinct from the certified polynomials themselves---in particular, every polynomial in the interior of the WSOS cone has a full-dimensional cone of dual certificates, which makes it particularly easy to identify one with an efficient numerical method and to find a rational dual certificate with small numerators.
In \mbox{Section \ref{sec:algorithm}}, we discuss various algorithmic applications of dual certificates. We propose an efficient hybrid algorithm, \mbox{Algorithm \ref{alg:Newton}}, for computing and certifying rational WSOS lower bounds for polynomials over a compact semialgebraic set using dual certificates. The algorithm is almost entirely numerical, and has lower computational complexity than off-the-shelf semidefinite programming software applied to the same problem.
The algorithm provides, in each iteration, a certifiable WSOS bound with a dual certificate that can be converted (in polynomial time) to a rational WSOS certificate without any additional rounding or projection of the numerical solutions. The sequence of WSOS bounds converges to the optimal WSOS bound at a linear rate. In \mbox{Section \ref{sec:univariate}}, we deduce explicit bounds on the number of iterations of Algorithm \ref{alg:Newton} in the univariate case.
\subsection{Preliminaries}
In the rest of this section we introduce some notation and briefly review some convex optimization and interior-point theory that we rely on throughout the paper.
\subsubsection{Weighted SOS polynomials and positive semidefinite matrices} Recall that a convex set $K\subseteq\mathbb{R}^n$ is called a \emph{convex cone} if for every $\mathbf{x}\in K$ and $\lambda \geq 0$ scalar, the vector $\lambda\mathbf{x}$ also belongs to $K$. A convex cone is \emph{proper} if it is closed, \emph{full-dimensional} (meaning $\operatorname{span}(K)=\mathbb{R}^n$), and \emph{pointed} (that is, it does not contain a line). We shall denote the interior of a proper cone $K$ by $K^\circ$.
\paragraph{Sum-of-squares (SOS) polynomials} Let $\mathcal{V}_{n,2d}$ denote the cone of $n$-variate polynomials of degree $2d$. We say that a polynomial $p \in \mathcal{V}_{n,2d}$ is \emph{sum-of-squares} (SOS) if there exist polynomials $q_1,\dots,q_k \in \mathcal{V}_{n,d}$ such that $p = \sum_{i=1}^kq_i^2$. Define $\Sigma_{n,2d}$ to be the cone of $n$-variate SOS polynomials of degree $2d$. The cone $\Sigma_{n,2d}\subset \mathcal{V}_{n,2d} \equiv \mathbb{R}^{\binom{n+2d}{n}}$ is a proper cone for every $n$ and $d$.
\paragraph{Weighted sum-of-squares} More generally, let $\mathbf{g} = (g_1,\dots,g_m)$ be some given nonzero polynomials and let $\mathbf{d} = (d_1,\dots,d_m)$ be a nonnegative integer vector. We denote by $\mathcal{V}_{n,2\mathbf{d}}^\mathbf{g}$ the space of polynomials $p$ for which there exist $r_1 \in \mathcal{V}_{n,2d_1}, \dots, r_m \in \mathcal{V}_{n,2d_m}$ such that $p = \sum_{i=1}^m g_ir_i$. A polynomial $p \in\mathcal{V}_{n,2\mathbf{d}}^\mathbf{g}$ is said to be \emph{weighted sum-of-squares} (WSOS) if there exist $\sigma_1 \in \Sigma_{n,2d_1}, \dots, \sigma_m \in \Sigma_{n,2d_m}$ such that $p = \sum_{i=1}^mg_i\sigma_i$. It is customary to assume that $g_1=1$, that is, the ordinary ``unweighted'' sum-of-squares polynomials are also included in the WSOS cones. Let $\Sigma_{n,2\mathbf{d}}^\mathbf{g}$ denote the set of WSOS polynomials in $\mathcal{V}_{n,2\mathbf{d}}^\mathbf{g}$. Under mild conditions, the cone $\Sigma_{n,2\mathbf{d}}^\mathbf{g} \subset \mathcal{V}_{n,2\mathbf{d}}^\mathbf{g}$ is a proper cone; for example, it is sufficient that the set \[\{\mathbf{x}\in\mathbb{R}^n\,|\,g_i(\mathbf{x})>0,\,i=1,\dots,m\}\] is a unisolvent point set for the space $\mathcal{V}_{n,2\mathbf{d}}^\mathbf{g}$ \cite[Prop.~6.1]{PappYildiz2019}. In particular, this implies that both $\Sigma_{n,2\mathbf{d}}^\mathbf{g}$ and its dual cone have a non-empty interior.
\deletethis{
\david{The next paragraph is new, and it ended up being far too long, but I thought we need to say something about this. Perhaps we could assume that the audience knows all of this and our quick ref to Putinar on p.1 is sufficient.}
WSOS cones are central in polynomial optimization, as they are tractable inner approximations of the cone of nonnegative polynomials, owing to their connection to semidefinite optimization; see below. Consider the set
\[S_\mathbf{g} \defeq \{\mathbf{x}\in\mathbb{R}^n\,|\,g_i(\mathbf{x})\geq0,\,i=1,\dots,m\}.\]
It is immediate that every polynomial in $\Sigma_{n,2\mathbf{d}}^\mathbf{g}$ is nonnegative on $S_\mathbf{g}$. As a partial converse, Putinar's celebrated \emph{Positivstellensatz} \cite{Putinar1993} guarantees that if the associated \emph{quadratic module}
\[M_\mathbf{g} \defeq \left\{\sum_{i=0}^m g_i\sigma_i\,\middle|\,\sigma_i \text{ is SOS}, i=1,\dots,m; g_0 = 1\right\}\]
satisfies the property that $R-\|\cdot\|^2 \in M_\mathbf{g}$ for some $R$, then every polynomial that is strictly positive on $S_\mathbf{g}$ also belongs to $M_\mathbf{g}$; that is, it belongs to $\Sigma_{n,2\mathbf{d}}^\mathbf{g}$ for every sufficiently large degree vector $\mathbf{d}$.
}
\paragraph{WSOS polynomials and positive semidefinite matrices}
We will denote the set of $n\times n$ real symmetric matrices by $\mathbb{S}^n$, and the cone of positive semidefinite $n\times n$ real symmetric matrices by $\mathbb{S}^n_+$. When the dimension is clear from the context, we use the common shorthands $\mathbf{A}\succcurlyeq 0$ to denote that the matrix $\mathbf{A}$ is positive semidefinite and $\mathbf{A}\succ 0$ to denote that the matrix $\mathbf{A}$ is positive definite. We will routinely identify polynomials with their coefficient vectors in a fixed basis of $\mathcal{V}_{n,2\mathbf{d}}^\mathbf{g}$. Thus, $\mathcal{V}_{n,2\mathbf{d}}^\mathbf{g}$ and $\left(\mathcal{V}_{n,2\mathbf{d}}^\mathbf{g}\right)^*$ are identified with $\mathbb{R}^U$, where $U = \dim\left(\mathcal{V}_{n,2\mathbf{d}}^\mathbf{g}\right)$.
The following well-known theorem (rooted in the works of Shor, Lasserre, Parrilo, and Nesterov; here reproduced in the notation of the latter) illustrates the connection between $\Sigma_{n,2\mathbf{d}}^\mathbf{g}$ and the cone of positive semidefinite matrices.
\begin{proposition}[\protect{\cite[Thm.~17.6]{Nesterov2000}}]\label{thm:Nesterov} Fix an ordered basis $\mathbf{q} = (q_1,\dots,q_U)$ of $\mathcal{V}^\mathbf{g}_{n,2\mathbf{d}}$ and an ordered basis $\mathbf{p}_{i} = (p_{i,1},\dots,p_{i,L_i})$ of $\mathcal{V}_{n,d_i}$ for $i = 1,\dots,m$. Let $\Lambda_i: \mathcal{V}_{n,2\mathbf{d}}^\mathbf{g} \left(\equiv \mathbb{R}^U\right) \to \mathbb{S}^{L_i}$ be the unique linear mapping satisfying $\Lambda_{i}(\mathbf{q}) = g_i\mathbf{p}_i\mathbf{p}_i^T$, and let $\Lambda_i^*$ denote its adjoint. Then $\mathbf{s} \in \Sigma_{n,2\mathbf{d}}^\mathbf{g}$ if and only if there exist matrices \mbox{$\mathbf{S}_1\succcurlyeq \mathbf{0}, \dots, \mathbf{S}_m \succcurlyeq \mathbf{0}$} satisfying
\begin{equation}\label{eq:Nesterov-Lambdastar}
\mathbf{s} = \sum_{i=1}^m\Lambda_i^*(\mathbf{S}_i).
\end{equation}
Additionally, the dual cone of $\Sigma_{n,2\mathbf{d}}^\mathbf{g}$ admits the characterization
\begin{equation}\label{eq:Nesterov-Lambda}
\left(\Sigma^\mathbf{g}_{n,2\mathbf{d}}\right)^* = \left\{\mathbf{x} \in \mathcal{V}_{n,2\mathbf{d}}^\mathbf{g} \left(\equiv \mathbb{R}^U\right) \ | \ \Lambda_i(\mathbf{x}) \succcurlyeq \mathbf{0} ~~ \forall i=1,\dots,m\right\}.
\end{equation}
\end{proposition}
The proof of Proposition~\ref{thm:Nesterov} is constructive: given matrices $\mathbf{S}_i\in\mathbb{S}^{L_i}_+\,(i=1,\dots,m)$, one may explicitly construct a (weighted) sum-of-squares decomposition of the polynomial $\mathbf{s}$. Thus, the collection of matrices $(\mathbf{S}_1,\dots,\mathbf{S}_m)$ itself can be interpreted as a WSOS certificate of the polynomial $\mathbf{s}$.
To lighten the notation, throughout the rest of the paper we assume that the weight polynomials $\mathbf{g}=(g_1,\dots,g_m)$ and the degrees $\mathbf{d}=(d_1,\dots,d_m)$ are fixed, and denote the cone $\Sigma_{n,2\mathbf{d}}^\mathbf{g}$ by $\Sigma$ and the space of polynomials $\mathcal{V}_{n,2\mathbf{d}}^\mathbf{g}$ by $\mathcal{V}$. Additionally, we denote by $\Lambda$ the $\mathbb{R}^U\to \mathbb{S}^{L_1} \oplus \cdots \oplus \mathbb{S}^{L_m}$ linear map $\Lambda_1(\cdot)\oplus \cdots \oplus\Lambda_m(\cdot)$ from Proposition~\ref{thm:Nesterov}. With this notation, the condition \eqref{eq:Nesterov-Lambdastar} can be written as $\mathbf{s}=\Lambda^*(\mathbf{S})$ for some positive semidefinite (block diagonal) matrix $\mathbf{S}\in\mathbb{S}^{L_1} \oplus \cdots \oplus \mathbb{S}^{L_m}$. Similarly, Eq.~\eqref{eq:Nesterov-Lambda} simplifies to
\begin{equation}\label{eq:SigmaStar}
\Sigma^* = \{\mathbf{x}\in\mathbb{R}^U\,|\,\Lambda(\mathbf{x})\succcurlyeq \mathbf{0}\}.
\end{equation}
The interior of this cone is simply
\begin{equation}\label{eq:intSigmaStar}
\Ssc = \{\mathbf{x}\in\mathbb{R}^U\,|\,\Lambda(\mathbf{x})\succ \mathbf{0}\}.
\end{equation}
\subsubsection{Barrier functions and local norms in convex cones}
The analysis of the dual certificates introduced in Section~\ref{sec:dualcertificates} relies heavily on the theory of barrier functions for convex cones. In this section, we give a brief overview of the parts of this theory and some additional notation that will be needed throughout the rest of the paper.
It is convenient to identify the spaces $\mathcal{V}$ and $\mathcal{V}^*$ with $\mathbb{R}^U$ ($U=\dim(\mathcal{V})$), equipped with the standard inner product, $\langle \mathbf{x}, \mathbf{y} \rangle = \mathbf{x}^\mathrm{T}\mathbf{y}$ and the induced Euclidean norm $\|\cdot\|$.
Let $\Lambda: \mathbb{R}^U \to \mathbb{S}^L$ be the unique linear mapping specified in Proposition~\ref{thm:Nesterov} above, and let $\Lambda^*$ denote its adjoint. Central to our theory is the \emph{barrier function} $f: \Ssc \to \mathbb{R}$ defined by
\begin{equation}\label{eq:def:f}
f(\mathbf{x}) \defeq -\ln(\det(\Lambda(\mathbf{x})).
\end{equation}
Note that by Eq.~\eqref{eq:intSigmaStar}, $f$ is indeed defined on its domain. The function $f$ is twice continuously differentiable; we denote by $g(\mathbf{x})$ its gradient at $\mathbf{x}$ and by $H(\mathbf{x})$ its Hessian at $\mathbf{x}$. Since $f$ is strictly convex on its domain, $H(\mathbf{x})\succ \mathbf{0}$ for all $\mathbf{x} \in \Ssc$. Consequently, we can also associate with each $\mathbf{x}\in\Ssc$ the \emph{local inner product} $\langle\cdot,\cdot\rangle_\mathbf{x} : \mathcal{V}^* \times \mathcal{V}^* \to \mathbb{R}$ defined as $\langle\mathbf{y},\mathbf{z}\rangle_\mathbf{x} \defeq \mathbf{y}^\mathrm{T} H(\mathbf{x}) \mathbf{z}$ and the \emph{local norm} $\|\cdot\|_\mathbf{x}$ induced by this local inner product. Thus, $\|\mathbf{y}\|_\mathbf{x} = \|H(\mathbf{x})^{1/2}\mathbf{y}\|$. We define the local (open) ball centered at $\mathbf{x}$ with radius $r$ by $B_{\mathbf{x}}(\mathbf{x},r) \defeq \{\mathbf{y}\in\mathcal{V}^*\,|\, \|\mathbf{y} - \mathbf{x}\|_{\mathbf{x}} < r\}$.
Analogously, we define the \emph{dual local inner product} $\langle \cdot,\cdot \rangle_\mathbf{x}^*: \mathcal{V}\times\mathcal{V}\to\mathbb{R}$ by $\langle \mathbf{s}, \mathbf{t} \rangle_\mathbf{x}^* \defeq \mathbf{s}^\mathrm{T} H(\mathbf{x})^{-1}\mathbf{t}$; the induced \emph{dual local norm} $\|\cdot\|_{\mathbf{x}}^*$ satisfies the identity $\|\mathbf{t}\|_{\mathbf{x}}^* = \|H(\mathbf{x})^{-1/2}\mathbf{t}\|$.
We remark that the function in \eqref{eq:def:f} falls into the broader category of \emph{logarithmically homogeneous self-concordant barriers} (or LHSCBs for short), which are expounded upon in the classic texts \cite{NesterovNemirovskii1994} and \cite{Renegar2001}. Throughout, we will invoke several useful results concerning LHSCBs for the function \eqref{eq:def:f}; these are enumerated in the following lemma:
\begin{lemma}\label{thm:f-properties}
Using the notation introduced in this section, the following hold for every $\mathbf{x}\in\Ssc$:
\begin{enumerate}
\item We have $B_{\mathbf{x}}(\mathbf{x},1) \subset \left(\Sigma^*\right)^\circ$, and for all $\mathbf{u}\in B_{\mathbf{x}}(\mathbf{x},1)$ and $\mathbf{v} \neq 0$, one has
\begin{equation}\label{eq:self-concordance}
1 - \|\mathbf{u} - \mathbf{x}\|_\mathbf{x} \leq \frac{\|\mathbf{v}\|_\mathbf{u}}{\|\mathbf{v}\|_\mathbf{x}}\leq (1 - \|\mathbf{u} - \mathbf{x}\|_\mathbf{x})^{-1}.
\end{equation}
\item The gradient $g$ of $f$ can be computed as
\begin{equation}\label{eq:g}
g(\mathbf{x}) = -\Lambda^*(\Lambda(\mathbf{x})^{-1}),
\end{equation}
and the Hessian $H(\mathbf{x})$ is the linear operator satisfying
\begin{equation}\label{eq:H}
H(\mathbf{x}) \mathbf{w}= \Lambda^*\!\left(\Lambda(\mathbf{x})^{-1}\Lambda(\mathbf{w})\Lambda(\mathbf{x})^{-1}\right) \;\; \text{ for every } \mathbf{w} \in \mathbb{R}^U.
\end{equation}
\item The function $f$ is \emph{logarithmically homogeneous}; that is, it has the following properties:
\begin{equation}\label{eq:log-homogeneity}
g(\alpha \mathbf{x}) = {\alpha}^{-1} g(\mathbf{x}) \;\text{ and }\; H(\alpha \mathbf{x}) = {\alpha}^{-2}H(\mathbf{x}) \;\; \text{ for every } \alpha > 0,
\end{equation}
furthermore
\begin{equation}\label{eq:gH-identities}
H(\mathbf{x})\mathbf{x} = -g(\mathbf{x})\quad\text{and}\quad \|g(\mathbf{x})\|_{\mathbf{x}}^* = \|\mathbf{x}\|_{\mathbf{x}} = \sqrt{\langle -g(\mathbf{x}),\mathbf{x}\rangle} = \sqrt{\nu},
\end{equation}
where $\nu = \sum_{i=1}^m L_i$ is the \emph{barrier parameter} of $f$.
\item \label{item:bijection} The gradient map $g:\Ssc\to \mathbb{R}^U$ defines a bijection between $\left(\Sigma^*\right)^\circ$ and ${\Sigma}^\circ$, In particular, for every $\mathbf{s} \in {\Sigma}^\circ$ there exists a unique $\mathbf{x} \in \left(\Sigma^*\right)^\circ$ satisfying $\mathbf{s} = -g(\mathbf{x})$.
\item If $\|\mathbf{u}-\mathbf{x}\|_\mathbf{x} < 1$, then
\begin{equation}\label{eq:lemma5}
\|g(\mathbf{u}) - g(\mathbf{x})\|_{\mathbf{x}}^* \leq \frac{\|\mathbf{u} - \mathbf{x}\|_\mathbf{x}}{1 - \|\mathbf{u} - \mathbf{x}\|_\mathbf{x}}.
\end{equation}
\item If $\|g(\mathbf{u}) - g(\mathbf{x})\|_{\mathbf{x}}^* < 1$, then
\begin{equation}\label{eq:revlemma5}
\|\mathbf{u} - \mathbf{x}\|_{\mathbf{x}}\leq \frac{\|g(\mathbf{u}) - g(\mathbf{x})\|_{\mathbf{x}}^*}{1 - \|g(\mathbf{u}) - g(\mathbf{x})\|_\mathbf{x}^*}.
\end{equation}
\end{enumerate}
\end{lemma}
\begin{proof}\phantom{ }\\
\begin{enumerate}
\item This is Renegar's definition of self-concordance applied to the function $f$, which is a composition of an affine function and a well-known self-concordant function, and is thus self-concordant; see \cite[Sec.~2.2.1 and Thm.~2.2.7]{Renegar2001}.
\item Straightforward calculation.
\item Straightforward calculation using the identities \eqref{eq:g} and \eqref{eq:H}. We remark that these identities hold for all LHSCBs \cite[Thm.~2.3.9]{Renegar2001}.
\item See \cite[Sec.~3.3]{Renegar2001}.
\item See \cite[Lemma~5]{PappYildiz2017corrigendum}.
\item This is an application of the previous claim to the conjugate barrier function of $f$. {}
\end{enumerate}
\end{proof}
\section{Dual certificates}\label{sec:dualcertificates}
We begin this section by introducing our central object, the cone of dual certificates corresponding to a WSOS polynomial (\mbox{Definition \ref{def:dual-certificate}}) and showing in \mbox{Theorem \ref{thm:main}} how we can use dual certificates to construct an explicit (weighted) sum-of-squares decomposition of WSOS polynomials in closed form. We continue using the notation introduced in the previous section, and let $\Sigma$ denote a general WSOS cone $\Sigma_{n,2\mathbf{d}}^\mathbf{g}$ with non-empty interior and $H$ denote the Hessian of the barrier function $f$ defined in \eqref{eq:def:f}.
\begin{definition}\label{def:dual-certificate}
Let $\mathbf{s}\in\Sigma$. We say that the vector $\mathbf{x}\in\Ssc$ is a \emph{dual certificate of $\mathbf{s}$}, or simply that \emph{$\mathbf{x}$ certifies $\mathbf{s}$}, if $H(\mathbf{x})^{-1}\mathbf{s} \in \Sigma^*$. We denote by
\[\mathcal{C}(\mathbf{s}) \defeq \{\mathbf{x}\in\Ssc\,|\,H(\mathbf{x})^{-1}\mathbf{s} \in \Sigma^*\}\]
the set of dual certificates of $\mathbf{s}$.
Conversely, for every $\mathbf{x}\in\Sc$, we denote by
\[\mathcal{P}(\mathbf{x}) \defeq \{\mathbf{s}\in\Sigma\,|\,H(\mathbf{x})^{-1}\mathbf{s} \in \Sigma^*\}\]
the set of polynomials certified by the dual vector $\mathbf{x}$.
\end{definition}
The following theorem justifies the terminology introduced above. Through \mbox{Eq. \eqref{eq:Sdef}} below, we can construct a WSOS certificate $\mathbf{S}$ for the polynomial $\mathbf{s}$ in the spirit of \mbox{Proposition \ref{thm:Nesterov}} by an efficiently-computable closed-form formula, and thus we may interpret the dual vector $\mathbf{x}\in\mathcal{C}(\mathbf{s})$ itself as a certificate of the polynomial $\mathbf{s}$.
\begin{theorem}\label{thm:main}
Let $\mathbf{x}\in\Ssc$ be arbitrary. Then the matrix $\mathbf{S}=\mathbf{S}(\mathbf{x},\mathbf{s})$ defined by
\begin{equation}\label{eq:Sdef}
\mathbf{S}(\mathbf{x},\mathbf{s}) \defeq \Lambda(\mathbf{x})^{-1}\Lambda\!\left(H(\mathbf{x})^{-1}\mathbf{s}\right)\Lambda(\mathbf{x})^{-1}
\end{equation}
satisfies $\Lambda^*(\mathbf{S}) = \mathbf{s}$. Moreover, $\mathbf{x}$ is a dual certificate for $\mathbf{s}\in\Sigma$ if and only if $\mathbf{S}\succcurlyeq 0$.
\end{theorem}
\begin{proof}
The first statement can be shown by applying the Hessian formula from \mbox{Lemma \ref{thm:f-properties}}:
\[
\Lambda^*(\mathbf{S}) \overset{\eqref{eq:Sdef}}{=} \Lambda^*\left(\Lambda(\mathbf{x})^{-1}\Lambda\!\left(H(\mathbf{x})^{-1}\mathbf{s}\right)\Lambda(\mathbf{x})^{-1}\right) \overset{\eqref{eq:H}}{=} H(\mathbf{x})H(\mathbf{x})^{-1}\mathbf{s} = \mathbf{s},
\]
For the second statement, note that $\mathbf{S} \succcurlyeq \mathbf{0}$ if and only if $\Lambda\!\left(H(\mathbf{x})^{-1}\mathbf{s}\right) \succcurlyeq \mathbf{0}$, which is equivalent to $\mathbf{x}\in\mathcal{C}(\mathbf{s})$ by the definition of $\mathcal{C}(\mathbf{s})$ and the characterization \eqref{eq:SigmaStar} of $\Sigma^*$.
\end{proof}
\deletethis{
\begin{example}
Consider the univariate polynomial given by $s(t)=3 - 4 t - 6 t^2 + 6 t^3 + 6 t^4$. If we represent polynomials in the monomial basis, $\Lambda(\mathbf{x})$ is the Hankel matrix corresponding to $\mathbf{x}$. This is certified by every vector of the form $\mathbf{x} = (\frac{3}{2}, 0, 1, -\frac{1}{2}, x_4)$, $x_4>\frac{11}{12}$, because $\mathbf{x}\in (\Sigma^*)^\circ$, and
\[\mathbf{S}(\mathbf{x},\mathbf{s}) = \Lambda(\mathbf{x})^{-1}\Lambda\!\left(H(\mathbf{x})^{-1}\mathbf{s}\right)\Lambda(\mathbf{x})^{-1} =
\begin{pmatrix}
3 & -2 & -4 \\
-2 & 2 & 3 \\
-4 & 3 & 6 \\
\end{pmatrix} \succcurlyeq \mathbf{0}.\]
Different certificates will yield different decompositions, e.g., $\mathbf{x} = (\frac{3}{2}, 0, 1, 0, 5) \in (\Sigma^*)^\circ$ produces
\[\mathbf{S}(\mathbf{x},\mathbf{s}) =
\begin{pmatrix}
3 & -2 & -\frac{27}{7} \\
-2 & \frac{12}{7} & 3 \\
-\frac{27}{7} & 3 & 6 \\
\end{pmatrix} \succcurlyeq \mathbf{0}.\]
\vfill \qed
\end{example}
}
From a high-level perspective, the matrix $\mathbf{S}(\mathbf{x},\mathbf{s})$ is defined in \eqref{eq:Sdef} by a ``closed-form formula''. We will make some more precise statements on the complexity of this formula in Section \ref{sec:complexity} below.
Recall from Lemma~\ref{thm:f-properties} (claim \ref{item:bijection}) that for every $\mathbf{s}\in\Sigma^\circ$ there exists a unique $\mathbf{x} \in \Ssc$ satisfying $\mathbf{s}=-g(\mathbf{x})$. This vector is a dual certificate of $\mathbf{s}$, since
\[H(\mathbf{x})^{-1}\mathbf{s} = -H(\mathbf{x})^{-1}g(\mathbf{x}) \overset{\eqref{eq:gH-identities}}{=} \mathbf{x} \in \Ssc.\]
Thus, every polynomial in the interior of the WSOS cone $\Sigma$ has a dual certificate.
\begin{definition}
When $-g(\mathbf{x}) = \mathbf{s}\, (\in\!\Sigma^\circ)$, we say that $\mathbf{x}$ is the \emph{gradient certificate of $\mathbf{s}$}.
\end{definition}
It is immediate from the definition that if $\mathbf{x}$ is a dual certificate of $\mathbf{s}$, then so is every positive multiple of $\mathbf{x}$. (One may also confirm directly that the matrix $\mathbf{S}$ constructed in \eqref{eq:Sdef} is invariant to a positive scaling of $\mathbf{x}$.)
Also note that when $\mathbf{x}$ is the gradient certificate of $\mathbf{s}=-g(\mathbf{x})$, then $\mathbf{S}(\mathbf{x},\mathbf{s})$ is positive definite. Since $\mathbf{S}$ is continuous on $\Ssc\times\Sc$, it is immediate that all vectors in some ($\mathbf{s}$-dependent) neighborhood of $\mathbf{x}$ are dual certificates of $\mathbf{s}$. Conversely, the gradient certificate of $\mathbf{s}$ is also a dual certificate of every polynomial in some ($\mathbf{x}$-dependent) neighborhood of $\mathbf{s}$. Our next lemma is a quantitative version of this observation.
\begin{lemma}\label{thm:sufficient-cone} Suppose $\mathbf{t}\in\Sigma^\circ$ and let $\mathbf{x}\in\Ssc$ be any vector that satisfies the inequality
\begin{equation}\label{eq:sufficient-cone}
\mathbf{t}^\mathrm{T}\left(\mathbf{x}\vx^\mathrm{T} - (\nu - 1)H(\mathbf{x})^{-1}\right)\mathbf{t} \geq 0.
\end{equation}
Then $\mathbf{x}\in\mathcal{C}(\mathbf{t})$, equivalently, $\mathbf{t}\in\mathcal{P}(\mathbf{x})$. In particular, if $\mathbf{s}=-g(\mathbf{x})$ for some $\mathbf{x}\in\Ssc$, then $\mathbf{x}$ is a dual certificate for every polynomial $\mathbf{t}$ satisfying $\|\mathbf{t}-\mathbf{s}\|_\mathbf{x}^*\leq 1$.
\end{lemma}
\begin{proof} We start with the second claim. From the definitions of the local norm and the dual local norm, we have
\begin{equation}\label{eq:distance-identities}
\|\mathbf{t} - \mathbf{s}\|_{\mathbf{x}}^* =
\|H(\mathbf{x})^{-1/2}(\mathbf{t} - \mathbf{s})\| =
\|H(\mathbf{x})^{1/2}(\mathbf{x} - H(\mathbf{x})^{-1}\mathbf{t})\| =
\|\mathbf{x} - H(\mathbf{x})^{-1}\mathbf{t}\|_\mathbf{x} .
\end{equation}
Thus, $\|\mathbf{t} - \mathbf{s}\|_{\mathbf{x}}^* \leq 1$ is equivalent to $H(\mathbf{x})^{-1}\mathbf{t} \in \overline{B_\mathbf{x}(\mathbf{x},1)}$. Since $B_\mathbf{x}(\mathbf{x},1) \subseteq \left(\Sigma^*\right)^\circ$ from the first claim of Lemma \ref{thm:f-properties}, $\overline{B_\mathbf{x}(\mathbf{x},1)} \subseteq \Sigma^*$, and $\mathbf{x} \in \mathcal{C}(\mathbf{t})$ by definition
The first claim of the Lemma is the ``conic version'' of the second claim. To prove it, suppose that the inequality in \eqref{eq:sufficient-cone} holds. Then the univariate quadratic polynomial
\[z \mapsto (1 - \nu)z^2 + \left( 2 \langle \mathbf{t},\mathbf{x}\rangle \right)z - \langle \mathbf{t}, H(\mathbf{x})^{-1}\mathbf{t}\rangle\]
has a nonnegative discriminant, therefore it has a root $\delta$. Moreover, since $(1-\nu) < 0$ and $\langle \mathbf{t}, H(\mathbf{x})^{-1}\mathbf{t} \rangle > 0$, it follows that
$\delta > 0$.
Using the identities in Eq.~\eqref{eq:gH-identities}, we have
\begin{align*}
0 &\leq (1 - \nu)\delta^2 + \left( 2 \langle \mathbf{t},\mathbf{x}\rangle \right)\delta - \langle \mathbf{t}, H(\mathbf{x})^{-1}\mathbf{t}\rangle\\
&= \delta^2 \left(1 - \langle g(\mathbf{x}),H(\mathbf{x})^{-1} g(\mathbf{x})\rangle\right) - \delta\left( 2 \langle \mathbf{t},H(\mathbf{x})^{-1}g(\mathbf{x})\rangle \right) - \langle \mathbf{t}, H(\mathbf{x})^{-1}\mathbf{t}\rangle\\
&=\delta^2 - \langle\mathbf{t} + \delta g(\mathbf{x}), H(\mathbf{x})^{-1}(\mathbf{t} + \delta g(\mathbf{x})) \rangle \\
&=\delta^2 - \|H(\mathbf{x})^{-1/2}(\mathbf{t} + \delta g(\mathbf{x}))\|^2.
\end{align*}
We conclude that $\|H(\mathbf{x})^{-1/2}(\mathbf{t} + \delta g(\mathbf{x}))\| < \delta$ for some $\delta > 0.$ Then using Lemma~\ref{thm:f-properties} again, we have
\begin{align*}
1 &\geq \frac{1}{\delta} \|H(\mathbf{x})^{-1/2}(\mathbf{t} + \delta g(\mathbf{x}))\| \\
&\overset{\eqref{eq:gH-identities}}{=} \left\|\delta H(\mathbf{x})^{1/2} \left(\delta^{-2}H(\mathbf{x})^{-1}\mathbf{t} - \delta^{-1}\mathbf{x}\right)\right\| \\
&\overset{\eqref{eq:log-homogeneity}}{=}\left\|H\left(\delta^{-1}\mathbf{x}\right)^{1/2}\left(H\left(\delta^{-1}\mathbf{x}\right)^{-1}\mathbf{t} - \delta^{-1}\mathbf{x}\right)\right\|,
\end{align*}
so by the identities \eqref{eq:distance-identities} and the first part of our proof, $\mathbf{t}$ is certified WSOS by $\frac{1}{\delta}\mathbf{x}$. Since all positive multiples of $\mathbf{x}$ certify $\mathbf{t}$, and $\delta$ is positive, it follows that $\mathbf{x}$ certifies $\mathbf{t}$. \end{proof}
\begin{corollary}\label{thm:x-ycert} Let $\mathbf{x}, \mathbf{y} \in \Sigma^*$ and $\mathbf{s}, \mathbf{t} \in \Sigma$, with $-g(\mathbf{x}) = \mathbf{s}$ and $-g(\mathbf{y}) = \mathbf{t}$. If $\|\mathbf{x} - \mathbf{y}\|_\mathbf{x} < \frac{1}{2}$, then $\mathbf{x}$ certifies $\mathbf{t}$. \end{corollary}
\begin{proof}If $\|\mathbf{x} - \mathbf{y}\|_\mathbf{x} < \frac{1}{2}$, then by Lemma \ref{thm:f-properties},
\[
\|\mathbf{s} - \mathbf{t}\|_{\mathbf{x}}^* =\|g(\mathbf{x}) - g(\mathbf{y})\|_{\mathbf{x}}^* \overset{\eqref{eq:lemma5}}{\leq} \frac{\|\mathbf{x} - \mathbf{y}\|_{\mathbf{x}}}{1 - \|\mathbf{x}-\mathbf{y}\|_\mathbf{x}} < 1.
\]
Then by Lemma 2, $\mathbf{x}$ certifies $\mathbf{t}$. \end{proof}
\begin{example} \label{ex:simple} Consider the univariate polynomial $t$ given by $t(z) =1 - z + z^2 + z^3 - z^4$. To show that $t$ is nonnegative on the interval $[-1,1]$, it suffices to show that the coefficient vector $\mathbf{t} = (1 , -1, 1, 1, -1)$ is a member of $\Sigma^{\mathbf{g}}_{1,2\mathbf{d}}$, with weights $\mathbf{g}(z) = (1,1-z^2)$ and degree vector $\mathbf{d} = (2,1)$. For this example, we represent all polynomials in the monomial basis. In this setting, the blocks of $\Lambda(\mathbf{x})$ have a Hankel structure; precisely, the $\Lambda : \mathbb{R}^5 \to \mathbb{S}^3 \oplus \mathbb{S}^2$ operator is given by
\[
\Lambda(x_0,x_1,x_2,x_3,x_4) =
\begin{pmatrix}
x_0 & x_1 & x_2 \\
x_1 & x_2 & x_3 \\
x_2 & x_3 & x_4
\end{pmatrix} \oplus
\begin{pmatrix}
x_0 - x_2 & x_1 - x_3 \\
x_1 - x_3 & x_2 - x_4
\end{pmatrix};
\]
see, for example, \cite[Sec.~II.2]{KarlinStudden1966}.
The adjoint operator is given by
\[\Lambda^*(\mathbf{S}^1\oplus\mathbf{S}^2) = \left(S^1_{00}+S^2_{00},
2 S^1_{01}+2 S^2_{01},
2 S^1_{02}+S^1_{11}-S^2_{00}+S^2_{11},
2 S^1_{12}-2 S^2_{01},S^1_{22}-S^2_{11}\right).\]
Consider the vector $\mathbf{x}=(5, 0, 5/2, 0, 15/8)$. This vector is the gradient certificate of the constant one polynomial, since simple arithmetic yields that $-g(\mathbf{x})=\Lambda^*(\Lambda(\mathbf{x})^{-1})=(1,0,0,0,0)$. The same certificate also certifies the nonnegativity of the polynomial $t$ above. To confirm this, we compute $H(\mathbf{x})^{-1}$:
\[
H(\mathbf{x})^{-1} =\frac{5}{384} \begin{pmatrix}
384 & 0 & 192 & 0 & 144 \\
0 & 240 & 0 & 180 & 0 \\
192 & 0 & 176 & 0 & 152 \\
0 & 180 & 0 & 165 & 0 \\
144 & 0 & 152 & 0 & 149
\end{pmatrix},
\] and by Theorem \ref{thm:main}, it is sufficient to verify that
\[ \frac{128}{5}\Lambda\left(H(\mathbf{x})^{-1}\mathbf{t}\right) = \begin{pmatrix}
144 & -20 & 72\\
-20 & 72 & -5 \\
72 & -5 & 49
\end{pmatrix} \oplus
\begin{pmatrix}
72 & -15 \\
-15 & 23 \\
\end{pmatrix}\succcurlyeq \mathbf{0}. \]
With some additional work, we can also compute from $\mathbf{x}$ rational matrices $\mathbf{S}_1$ and $\mathbf{S}_2$ to certify that the polynomial using Proposition \ref{thm:Nesterov}: plugging our certificate into the formula \eqref{eq:Sdef}, we obtain
\[
\mathbf{S}_1 = \frac{1}{40}\begin{pmatrix}
22 & -5 & -26 \\
-5 & 18 & 5 \\
-26 & 5 & 52
\end{pmatrix}
\quad\text{and}\quad
\mathbf{S}_2 = \frac{1}{40}\begin{pmatrix}
18 & -15 \\
-15 & 92 \\
\end{pmatrix}.
\]
These matrices, in turn, can be factored using the $LDL^\mathrm{T}$ form of Cholesky decomposition to compute an explicit rational sum-of-squares representation of $t$:
\begin{align*}
t(z) =&
\frac{11}{20} \left(-\frac{13 z^2}{11}-\frac{5 z}{22}+1\right)^2 +
\frac{371}{880} \left(z-\frac{20 z^2}{371}\right)^2 +
\frac{3937 z^4}{7420} + \\
& + \left(1-z^2\right) \left(
\frac{9}{20} \left(1-\frac{5 z}{6}\right)^2 +
\frac{159 z^2}{80}
\right).
\end{align*}
\end{example}
\subsection{Algorithmic considerations}\label{sec:complexity}
Depending on the choice of the $\Lambda$ operator (that is, in essence, the choice of bases $\mathbf{p}$ and $\mathbf{q}$ in the construction of the semidefinite representation of $\Sigma$ following \mbox{Proposition \ref{thm:Nesterov}}), the computation of $\mathbf{S}(\mathbf{x},\mathbf{s})$ can be made efficient, even polynomial-time in the bit model. Suppose that for a given rational $\mathbf{x}\in\Ssc$, the matrices $\Lambda(\mathbf{x})$ and $H(\mathbf{x})$ are rational and can be computed efficiently. Then for any $\mathbf{s}\in\mathbb{R}^U$, the computation of $\mathbf{S}(\mathbf{x},\mathbf{s})$ amounts to (1) computing a rational Cholesky ($LDL^\mathrm{T}$) factorization of $\Lambda(\mathbf{x})$ and $H(\mathbf{x})$ (which are positive definite by definition); (2) computing the vector $\mathbf{w}=H(\mathbf{x})^{-1}\mathbf{s}$ using the Cholesky factors of $H(\mathbf{x})$ computed in the previous step; and (3) computing $\Lambda(\mathbf{w})$ and then $\mathbf{S}(\mathbf{x},\mathbf{s})$ using the Cholesky factors of $\Lambda(\mathbf{x})$.
Therefore, computing $\mathbf{S}(\mathbf{x},\mathbf{s})$ is efficient as long as $\Lambda(\cdot)$ and $H(\cdot)$ can be computed efficiently.
For any reasonable choice and representation of $\Lambda$, the computation of $\Lambda(\cdot)$ and $\Lambda^*(\cdot)$ are efficient, as they are linear operators, typically explicitly represented in matrix form with rational entries. Studying the same question in the context of numerical methods for SOS optimization, the authors in \cite[Sec.~6]{PappYildiz2019} showed that when polynomials are represented as Lagrange interpolants, the Hessian $H(\mathbf{x})$ can be computed with $\mathcal{O}(m\max_i\{L_i\}U^2)$ arithmetic operations. One can also argue directly from the identity \eqref{eq:H}, that (since $\Lambda$ and $\Lambda^*$ are efficiently computable) the Hessian can be computed efficiently; the bottleneck once again is the inversion or factorization of $\Lambda(\mathbf{x})$. We note the monomial and Chebyshev polynomial bases as two additional interesting special cases (both in the univariate and multivariate setting): in these cases, $\Lambda(\mathbf{x})$ is a low displacement-rank matrix. For example, when the polynomials are univariate, each block of $\Lambda$ is a Hankel (or Hankel+Toeplitz) matrix if using the monomial (or Chebyshev) basis. Therefore the inversion of $\Lambda$ and the computation of $H$ can be handled using discrete Fourier transforms or the superfast (nearly-linear-time) algorithms of Pan and others \cite{Pan2001}.
\section{Computing rigorously certified lower bounds with dual certificates}\label{sec:algorithm}
With our theoretical infrastructure and notation in place, we now turn to the question of computing certified lower bounds and dual certificates for these bounds. In \mbox{Section~\ref{sec:EveryPolyHasaBound}} we show that under the condition that the constant one polynomial is in the interior of our WSOS cone, every polynomial has a dual certifiable lower bound. (We argue that this is a mild, essentially without loss of generality, condition in \mbox{Section \ref{sec:Discussion}}.) We also show that after a suitable preprocessing (required only once for every WSOS cone), such a certified bound can be computed by a closed form formula for any polynomial.
In \mbox{Section \ref{sec:c-update}} we discuss efficient algorithms to compute the best lower bound that a given certificate certifies for a given polynomial and show that using dual certificates, inexact numerical certificates (that come, for example, from numerical sum-of-squares optimization approaches) can be turned into rigorous rational certificates with minimal additional effort.
In \mbox{Section \ref{sec:rational-certificates}}, we discuss how rational certificates can be constructed efficiently from the dual certificates obtained with a numerical method. We exploit this result in \mbox{Section \ref{sec:Newton}}, in which we present a new algorithm (Algorithm \ref{alg:Newton}) for approximating the best WSOS lower bound for a given polynomial with arbitrary accuracy. The algorithm returns both an approximate lower bound and a rational certificate certifying the bound. We also show that Algorithm \ref{alg:Newton} is linearly convergent to the optimal bound. In \mbox{Section \ref{sec:C-bounds}}, we detail how to compute a bound on the linear rate of convergence of \mbox{Algorithm \ref{alg:Newton}}. This in turn makes it possible to compute WSOS lower bounds that are certifiably within a prescribed $\varepsilon$ from the optimal bound. Finally, in \mbox{Section \ref{sec:smaller-rational-certs}}, we investigate how the certificate returned by \mbox{Algorithm \ref{alg:Newton}} can be rounded to a nearby rational certificate with smaller denominators.
Throughout this section, and the rest of the paper, the boldface vector $\mathbf{1}$ represents the constant one polynomial (or, precisely, its coefficient vector) in the WSOS cone $\Sigma\, (=\Sigma_{n,2\mathbf{d}}^\mathbf{g})$, in the space of polynomials $\mathcal{V} (=\mathcal{V}_{n,2\mathbf{d}}^\mathbf{g})$.
\subsection{Universal dual certificates} \label{sec:EveryPolyHasaBound}
Suppose that $\mathbf{1}\in\Sc$. Then $\mathbf{1}$ has a gradient certificate ${\vx_{1}}$, and as we have seen above, $\mathbf{1}\in\mathcal{P}({\vx_{1}})^\circ$, that is, ${\vx_{1}}$ certifies an entire full-dimensional cone of polynomials with $\mathbf{1}$ in its interior. Conversely, an entire cone of certificates, with ${\vx_{1}}$ in its interior, certifies $\mathbf{1}$. Our next theorem shows that each of these certificates also certifies \emph{some} WSOS lower bound for every polynomial:
\begin{lemma}\label{thm:EveryPolyHasaBound}
Let $\mathbf{x}\in\Ssc$ be any certificate for which $\mathbf{1}\in\mathcal{P}(\mathbf{x})^\circ$ and $r\in(0,1/2]$. Then for every polynomial $\mathbf{t} \in \mathcal{V}$, the inclusion $\mathbf{x}\in\mathcal{C}(\mathbf{t}+c\mathbf{1})$ holds for every sufficiently large scalar $c$. Specifically, if ${\vx_{1}}$ is the gradient certificate of $\mathbf{1}$ and $\mathbf{y}_c$ is the gradient certificate of $\mathbf{t}+c\mathbf{1}$, then the inclusion ${\vx_{1}} \in \mathcal{C}(\mathbf{t}+c\mathbf{1})$ and the inequality
\begin{equation*
\|c^{-1}\mathbf{x}_1 - \mathbf{y}_c\|_{c^{-1}{\vx_{1}}} \leq r,
\end{equation*}
hold for every
\begin{equation}\label{eq:c0-bound}
c \geq \frac{1+r}{r}\|\mathbf{t}\|^*_{\mathbf{x}_1}.
\end{equation}
\end{lemma}
\begin{proof}The first statement is immediate from the fact that $\mathcal{P}(\mathbf{x})$ is a cone and the assumption that $\mathbf{1}\in\mathcal{P}(\mathbf{x})^\circ$: the dual vector $\mathbf{x}$ certifies all small perturbations of $\mathbf{1}$, including every polynomial of the form $(c^{-1}\mathbf{t} + \mathbf{1})$, and thus also $\mathbf{t}+c\mathbf{1}$, for every sufficiently large $c$. We prove the second statement in detail.
Using the definitions of the local dual norm and logarithmic homogeneity \eqref{eq:log-homogeneity} from Lemma~\ref{thm:f-properties}, we have
\begin{equation*
\|(\mathbf{t} + c \mathbf{1}) - c \mathbf{1}\|_{c^{-1}\mathbf{x}_1}^* \overset{(\text{by def.})}{=} \|H(c^{-1}\mathbf{x}_1)^{-1/2}\mathbf{t}\| \overset{(\ref{thm:f-properties})}{=} c^{-1}\|H(\mathbf{x}_1)^{-1/2}\mathbf{t}\| \overset{(\text{by def.})}{=} c^{-1}\|\mathbf{t}\|_{\mathbf{x}_1}^*
\end{equation*}
Our assumed inequality \eqref{eq:c0-bound} thus yields
\begin{equation}\label{eq:initial-3-eq}
\|(\mathbf{t} + c \mathbf{1}) - c \mathbf{1}\|_{c^{-1}x_1}^* = c^{-1}\|\mathbf{t}\|_{\vx_{1}}^* \leq \frac{r}{r+1}.
\end{equation}
Using logarithmic homogeneity again, we see that $c^{-1}\mathbf{x}_1$ is the gradient certificate for $c \mathbf{1}$. Therefore, invoking Lemma~\ref{thm:sufficient-cone}, we deduce from the inequality \eqref{eq:initial-3-eq} that $c^{-1}\mathbf{x}_1$ is a dual certificate for $\mathbf{t} + c\mathbf{1}$.
Moreover, via the inequality \eqref{eq:revlemma5} in Lemma~\ref{thm:f-properties}, we conclude that
\[
\|c^{-1}\mathbf{x}_1 - \mathbf{y}_c\|_{c^{-1}\mathbf{x}_1} \overset{\eqref{eq:revlemma5}}{\leq} \frac{\|\mathbf{t}\|^*_{c^{-1}\mathbf{x}_1}}{1-\|\mathbf{t}\|^*_{c^{-1}\mathbf{x}_1}} \overset{\eqref{eq:initial-3-eq}}{\leq} r,
\]
as claimed.
\end{proof}
We emphasize that the certificate ${\vx_{1}}$ (or any $\mathbf{x}$ with $\mathbf{1}\in\mathcal{P}(\mathbf{x})^\circ$) in Lemma \ref{thm:EveryPolyHasaBound} only needs to be computed once for any particular WSOS cone $\Sigma_{n,2\mathbf{d}}^\mathbf{g}$. Once ${\vx_{1}}$ (and the corresponding $H({\vx_{1}})^{-1}$) are computed, a certifiable lower bound and a corresponding certificate can be computed in closed form for any polynomial $\mathbf{t} \in \mathcal{V}$, with minimal effort.
When the weight polynomials $\mathbf{g}$ are sufficiently simple, the gradient certificate of $\mathbf{1}$ may even be easily expressible in closed form, as in the following example.
\begin{example}\label{ex:Chebyshev}
Consider the cone of nonnegative univariate polynomials of degree $2d$ over the interval $[-1,1]$, which is well known to be the same as the WSOS cone $\Sigma_{n,2\mathbf{d}}^\mathbf{g}$ with $n=1$, $m=2$, degree vector $\mathbf{d} = (d,d-1)$, and weight polynomials $\mathbf{g}(z) = (1,1-z^2)$ \cite{BrickmanSteinberg1962}. Furthermore, suppose that all polynomials are represented in the basis of Chebyshev polynomials of the first kind, that is, both of the ordered bases $\mathbf{p}$ and $\mathbf{q}$ in \mbox{Proposition \ref{thm:Nesterov}} that determine the operator $\Lambda$ are Chebyshev basis polynomials. Then both diagonal blocks of $\Lambda$ are Hankel+Toeplitz matrices (we omit the rather tedious details), and the gradient certificate of $\mathbf{1}=(1,0,\dots,0) \in \mathbb{R}^{2d+2}$ is simply the vector
\[
{\vx_{1}} = (2d+1,0,\dots,0).
\]
This can be proven by direct calculation verifying the equality $-g({\vx_{1}}) = \Lambda^*(\Lambda({\vx_{1}})^{-1}) = \mathbf{1}$.
The Hessian at this certificate is the diagonal matrix
\begin{equation}\label{eq:Hx1}
H({\vx_{1}}) = \frac{1}{2d+1}\operatorname{diag}\left(1, \frac{4d}{2d+1}, \frac{4d-2}{2d+1}, \dots, \frac{2}{2d+1}\right).
\end{equation}
Analogous results can be derived for polynomials of odd degree using $\mathbf{d} = (d,d)$, and weight polynomials $\mathbf{g}(z) = (1-z,1+z)$.
\end{example}
\subsection{Optimal and near-optimal lower bounds from a given dual certificate}\label{sec:c-update}
Suppose we have found a dual certificate $\mathbf{x}$ that certifies the nonnegativity of the polynomial $\mathbf{t}-c\mathbf{1}$. What is the \emph{best} lower bound certified by the same certificate? By definition, the answer is the solution of the one-dimensional optimization problem
\[
c_\text{max} \defeq \max \left\{\gamma\in\mathbb{R}\,|\,\mathbf{t}-\gamma\mathbf{1} \in \mathcal{P}(\mathbf{x})\right\}.
\]
As discussed in Section~\ref{sec:dualcertificates}, if the inverse Hessian $H(\mathbf{x})^{-1}$ (or the Cholesky or $LDL^\mathrm{T}$ factorization of $H(\mathbf{x})$) is already computed, then membership in $\mathcal{P}(\mathbf{x})$ is easy to test by verifying the positive semidefiniteness of $\Lambda(H(\mathbf{x})^{-1}(\mathbf{t}-\gamma\mathbf{1}))$. Therefore, an arbitrarily close lower approximation of $c_\text{max}$ can be found efficiently, in time proportional to the logarithm of the approximation error, by binary search on the optimal $\gamma$. (An initial lower bound on $c_\text{max}$ is the currently certified lower bound $c$ assumed to be part of the input; an upper bound on $c_\text{max}$ can be computed, e.g., by evaluating the polynomial $\mathbf{t}$ at any point in its domain.)
The repeated matrix factorization makes the algorithm outlined above too expensive to use as a subroutine. A weaker bound can be computed \emph{in closed form} using \mbox{Lemma \ref{thm:sufficient-cone}}: if
\[
c_\text{max}^\prime \defeq \max \left\{\gamma \in\mathbb{R}\,\middle|\, (\mathbf{t}-\gamma\mathbf{1})^\mathrm{T}\left(\mathbf{x}\vx^\mathrm{T} - (\nu - 1)H(\mathbf{x})^{-1}\right)(\mathbf{t}-\gamma\mathbf{1}) \geq 0 \right\},
\]
then $\mathbf{t}-c_\text{max}^\prime \in \mathcal{P}(\mathbf{x})$. For a given certificate $\mathbf{x}$, if the inverse Hessian $H(\mathbf{x})^{-1}$ (or the Cholesky or $LDL^\mathrm{T}$ factorization of $H(\mathbf{x})$) is already computed, then solving this optimization problem amounts to finding the roots of a univariate quadratic function.
\begin{example}\label{ex:simple-cmax}
Continuing with Example \ref{ex:simple} (with $\mathbf{t} = (1, -1, 1, 1, -1)$, weights $\mathbf{g}(z) = (1, 1 - z^2)$), we compute $c_\text{max}$ and $c_\text{max}^{\prime}$ for $t$ using the certificate $\mathbf{x} = (5, 0, 5/2, 0, 15/8)$. For comparison, the minimum of the polynomial is $\frac{1}{512} \left(619-51 \sqrt{17}\right)\approx 0.798$.
To compute $c_{\text{max}}$, we compute the largest $\gamma$ such that $\Lambda(H(\mathbf{x})^{-1}(\mathbf{t} - \gamma\mathbf{1})) = \Lambda_1(H(\mathbf{x})^{-1}(\mathbf{t} - \gamma\mathbf{1})) \oplus \Lambda_2(H(\mathbf{x})^{-1}(\mathbf{t} - \gamma\mathbf{1}))$ is positive semidefinite but not positive definite. We compute the characteristic polynomials of the $\gamma$-parametrized matrices, as $\Lambda(H(\mathbf{x})^{-1}(\mathbf{t} - \gamma\mathbf{1}))$ is on the boundary of the PSD cone when the constant term of the characteristic polynomial vanishes.
The constant term of the characteristic polynomial of $\Lambda_1(H(\mathbf{x})^{-1}(t - \gamma\mathbf{1}))$, itself a polynomial in $\gamma$, has smallest real root at $\gamma = \frac{1}{64}\left(67 - 5\sqrt{17}\right)$.
Meanwhile, the constant term of the characteristic polynomial of $\Lambda_2(H(\mathbf{x})^{-1}(t - \gamma\mathbf{1}))$ has smallest real root at $\gamma = \frac{1}{32} \left(41 - 5\sqrt{10}\right)$. Therefore, as $\frac{1}{64}\left(67 - 5\sqrt{17}\right) < \frac{1}{32} \left(41 - 5\sqrt{10}\right)$, we conclude that $c_{\text{max}} = \frac{1}{64}\left(67 - 5\sqrt{17}\right) \approx 0.724$.
To compute $c_{\text{max}}^{\prime}$, we expand and reduce
\[
(\mathbf{t}-\gamma\mathbf{1})^\mathrm{T}\left(\mathbf{x}\vx^\mathrm{T} - (\nu - 1)H(\mathbf{x})^{-1}\right)(\mathbf{t}-\gamma\mathbf{1}) = \frac{205}{64} - \frac{45\gamma}{4} + 5\gamma^2.
\]
Note that $\frac{205}{64} - \frac{45\gamma}{4} + 5\gamma^2 \geq 0$ when $\gamma \leq \frac{1}{8}\left(9 - 2\sqrt{10}\right)$ and when $\gamma \geq \frac{1}{8}\left(9 + 2\sqrt{10}\right)$. We conclude $c_{\text{max}}^{\prime} = \frac{1}{8}\left(9 - 2\sqrt{10}\right) \approx 0.334$.
\end{example}
\subsection{Rational certificates} \label{sec:rational-certificates}
Lemma \ref{thm:sufficient-cone} and Corollary \ref{thm:x-ycert} have important consequences for both numerical (finite-precision), exact-arithmetic, and hybrid algorithms for computing sum-of-squares certificates.
\paragraph{Rigorous certificates from numerical methods} The fact that every polynomial $\mathbf{s}\in\Sigma^\circ$ has a full-dimensional cone of certificates means that exact dual certificates can in principle be computed by purely numerical, inexact algorithms. Consider, for example, a hypothetical algorithm that aims to compute the gradient certificate $\mathbf{y}$ of some polynomial $\mathbf{t}-\gamma\mathbf{1}\in\Sigma^\circ$ to certify $t-\gamma\geq0$, but computes instead only an approximation $\mathbf{x}\approx \mathbf{y}$ in finite-precision arithmetic. As long as the inherent errors of the finite-precision computation are small enough to ensure $\|\mathbf{x}-\mathbf{y}\|_\mathbf{x} \leq 1/2$, \mbox{Corollary \ref{thm:x-ycert}} guarantees that $\mathbf{x}$ is a certificate of nonnegativity for $t-\gamma$. Since floating-point numbers are, by definition, rational, every sufficiently accurate numerical solution of $-g(\mathbf{x}) = \mathbf{t}-\gamma\mathbf{1}$ is automatically a rational dual certificate of nonnegativity. Additionally, as long as the coefficient vector $\mathbf{t}$ and the lower bound $\gamma$ are also rational, any such numerical dual certificate $\mathbf{x}$ can be directly converted to an exact rational primal certificate $\mathbf{S} \succcurlyeq \mathbf{0}$ satisfying $\Lambda^*(\mathbf{S}) = \mathbf{t}-\gamma\mathbf{1}$ via the formula of Eq.~\eqref{eq:Sdef}.
This property sets dual certificates apart from conventional certificates: a numerical solution to the semidefinite programming (feasibility) problem
\[ \text{find an } \mathbf{S}\succcurlyeq \mathbf{0} \text{ satisfying } \Lambda^*(\mathbf{S}) = \mathbf{t}-\gamma\mathbf{1} \]
will generally satisfy the equality constraints $\Lambda^*(\mathbf{S}) = \mathbf{t}-\gamma\mathbf{1}$ only within some numerical tolerance, thus $\mathbf{S}$ will not be a rigorous certificate, even if we can guarantee (by the appropriate choice of optimization algorithm) that at least the cone constraint $\mathbf{S}\succcurlyeq\mathbf{0}$ is always satisfied. Hence, additionally post-processing (rounding or projection steps, such as those in the hybrid methods of \cite{PeyrlParrilo2008} and \cite{MagronWang2020}) are needed. In contrast, any dual certificate $\mathbf{x}$ from the full-dimensional cone $\mathcal{C}(\mathbf{t}-\gamma\mathbf{1})$ is a rigorous certificate that can be turned into a rational WSOS decomposition.
In Section \ref{sec:Newton}, we present an efficient algorithm (Algorithm \ref{alg:Newton}) to compute certifiable rational lower bounds with matching dual certificates that can be implemented as an entirely numerical method using these ideas.
\paragraph{Rounding rational certificates to simpler certificates}
If desired, one may also ``round'' a rational certificate $\mathbf{x}$ to a ``nearby'' rational certificate with smaller components using Lemma \ref{thm:sufficient-cone} and Corollary \ref{thm:x-ycert}, in order to obtain a simpler WSOS decomposition. As with the numerical certificates above, any certificate $\mathbf{x}$ can be replaced by another one with smaller denominators (e.g., by applying Diophantine approximation component-wise, or using the LLL algorithm for simultaneous approximation of $\mathbf{x}$ with a ``smaller'' rational vector) as long as the rounded certificate is still within $\mathcal{C}(\mathbf{t}-\gamma\mathbf{1})$. The sizes of the denominators can be bounded using Lemma \ref{thm:sufficient-cone} and Corollary \ref{thm:x-ycert}. We will revisit this idea in the context of Algorithm \ref{alg:Newton} in Lemma \ref{thm:roundx}.
\subsection{Computing optimal WSOS bounds}\label{sec:Newton}
We now present an iterative method to compute the best WSOS lower bound for a given polynomial $\mathbf{t}$. The pseudocode of the algorithm is shown in \mbox{Algorithm \ref{alg:Newton}}. After a high-level description of the method, we show that it converges linearly to the optimal WSOS bound below (\mbox{Theorem \ref{thm:linear-convergence}}).
\afterpage{
\begin{algorithm}[t]
\DontPrintSemicolon
\SetKwInOut{Input}{input}
\SetKwInOut{Output}{outputs}
\SetKwInOut{Params}{parameters}
\Input{A polynomial $\mathbf{t}$; a tolerance $\varepsilon>0$.}
\Params{An oracle for computing the barrier Hessian $H$ for $\Sigma$; the gradient certificate ${\vx_{1}}$ for the constant one polynomial; a radius $r\in(0,1/4]$.}
\Output{A lower bound $c$ on the optimal WSOS lower bound $c^*$ satisfying $c^*-c\leq \varepsilon$; a dual vector $\mathbf{x}\in\Ssc$ certifying the nonnegativity of $\mathbf{t}-c\mathbf{1}$.}
\medskip
Compute $c_0 = -\frac{1+r}{r}\left({\mathbf{t}^\mathrm{T} H(\mathbf{x}_1)^{-1}}\mathbf{t})\right)^{1/2}$. Set $c = c_0$ and $\mathbf{x} = -\frac{1}{c_0}\mathbf{x}_1$.
\label{line:init}
\Repeat{$\Delta c \leq \rho_rC\varepsilon$ \do}{
Set $\mathbf{x} := 2\mathbf{x} - H(\mathbf{x})^{-1}(\mathbf{t} -c\mathbf{1} )$. \label{line:x-update}
Find the largest real number $c_+$ such that
\[
\|\mathbf{x} - H(\mathbf{x})^{-1}(\mathbf{t} - c_+\mathbf{1})\|_\mathbf{x} \leq \frac{r}{r+1}.
\]\label{line:c-update}
Set $\Delta c := c_+ - c$.
Set $c := c_+$.
}\label{line:stopping}
\Return{$c$ and $\mathbf{x}$.}
\caption{Compute the best WSOS lower bound and a dual certificate}\label{alg:Newton}
\end{algorithm}
}
Previously, in Lemma~\ref{thm:EveryPolyHasaBound}, we showed that for a sufficiently large $c$, $\mathbf{t} + c\mathbf{1}$ can be certified by $c^{-1} {\vx_{1}}$; this result justifies the initialization of the algorithm in Line \ref{line:init}. In order to increase the lower bound, the algorithm iterates two steps: certificate updates (Line \ref{line:x-update}) and bound updates (Line \ref{line:c-update}). The bound updates are similar to the $c_\text{max}^{\prime}$ bound in \mbox{Section \ref{sec:c-update}}; we will precisely justify this step in Lemma~\ref{thm:constant-update}. The certificate updates are motivated as follows: since each bound update attempts to push $c$ towards the best bound certifiable by $\mathbf{x}$, the certificate $\mathbf{x}$ sits near the boundary of $\mathcal{C}(\mathbf{t}-c\mathbf{1})$ after each bound update. To allow for a sufficient additional increase of the bound in the subsequent iteration, the certificate $\mathbf{x}$ is updated to be closer to the gradient certificate $\mathbf{y}$ of the current $\mathbf{t}-c\mathbf{1}$. This certificate $\mathbf{y}$ would be prohibitively expensive to compute in each iteration; instead, the update step in Line \ref{line:x-update} can be interpreted as a single Newton step from $\mathbf{x}$ towards the solution of the nonlinear system $-g(\mathbf{y}) = \mathbf{t}-c\mathbf{1}$.
\begin{example}\label{ex:simple-alg}
We continue with the setup of Examples \ref{ex:simple} and \ref{ex:simple-cmax}: we consider the univariate polynomial whose coefficient vector in the monomial basis is $\mathbf{t} = (1, -1, 1, 1, -1)$, over the interval $[-1,1]$, represented by the weights $\mathbf{g}(z) = (1,1-z^2)$. Algorithm \ref{alg:Newton} with $r=1/4$, with inputs $\mathbf{t}$ and tolerance $\varepsilon = 10^{-7}$ in double-precision floating point arithmetic outputs the bound $c\approx 0.798284319$
and a certificate vector $\mathbf{x}$. Note that the exact minimum of $t$ is $\frac{1}{512}(619-51\sqrt{17})\approx 0.798284401$.
A plot of the difference between the current certified lower bound $c$ and the minimum $c^*$ in each iteration is shown in Figure \ref{fig:simple-alg}, illustrating the linear convergence of \mbox{Algorithm \ref{alg:Newton}} for this polynomial.
\afterpage{
\begin{figure}
\centering
\includegraphics[scale=.30]{ListLogPlotPointSize01.eps}\\
\caption{The convergence of the sequence of certified lower bounds computed by Algorithm \ref{alg:Newton} to the minimum of the polynomial studied in Examples \ref{ex:simple} and \ref{ex:simple-alg}, illustrating the linear convergence shown in Theorem \ref{thm:linear-convergence} below.}
\label{fig:simple-alg}
\end{figure}
}
The exact rational representation of the floating point bound is
\[
c = 2^{-53} \cdot 7190305926654593,
\]
and the rational vector certifying the nonnegativity of $\mathbf{t}-c\mathbf{1}$ is
\[
\mathbf{x} = 2^{-33} \begin{pmatrix}173493184462864992\\ 67729650226350000\\ -120611300436615200\\ -161900156381728960\\ -5796381308580693\end{pmatrix}.
\]
Note that no rounding or projection steps are needed to compute a rigorous certificate. In spite of the inherent errors of floating point computation, as long as the magnitude of the errors is sufficiently small to ensure $\|\mathbf{x}-\mathbf{y}\|_\mathbf{x} \leq 1/2$ (recall that with exact arithmetic we would have $\|\mathbf{x}-\mathbf{y}\|_\mathbf{x} \leq 1/4$ in each iteration), the computed numerical certificate $\mathbf{x}$ is automatically a rational certificate for the computed SOS lower bound $c$.
\end{example}
The computationally most expensive part in each iteration is having to compute (after each certificate update) a Cholesky factorization of the Hessian $H(\mathbf{x})$ (or the inverse Hessian $H(\mathbf{x})^{-1}$). With that available, the bound update and the next certificate update are very efficient: by an argument analogous to the discussion on $c_{\max}^\prime$ in the previous section, the bound update amounts to solving a univariate quadratic polynomial, and the certificate update is essentially a matrix-vector multiplication. As discussed in Section~\ref{sec:complexity}, the computation and factorization of the Hessian is efficient for popular choices of polynomial bases.
We now turn to the analysis of the algorithm, deferring the discussion on the stopping criterion until later. To simplify the statements of the results, we will use the following notation throughout the rest of the section. We define $\mathbf{x}_+ \defeq 2\mathbf{x} - H(\mathbf{x})^{-1}(\mathbf{t} -c\mathbf{1} )$ to be the updated certificate in Line \ref{line:x-update} to help distinguish the certificates before and after the update.
Finally, we let $\mathbf{y}$ be the vector satisfying $-g(\mathbf{y}) = \mathbf{t} - c\mathbf{1}$ and $\mathbf{y}_+$ be the vector satisfying $-g(\mathbf{y}_+) = \mathbf{t} - c_+ \mathbf{1}$.
In the next series of Lemmas we show that the bound update from $c$ to $c_+$ is well-defined, and is always an increase, by bounding the distance between $\mathbf{x}$ and $\mathbf{y}$ in each step of the iteration. We also establish that throughout the algorithm, the iterates satisfy $\|\mathbf{x}-\mathbf{y}\|_\mathbf{x} \leq r$. (At the beginning of the first iteration this holds by Lemma \ref{thm:EveryPolyHasaBound}.) The first result, Lemma~\ref{thm:xtoxplus}, shows that $\mathbf{x}_+$ is closer than $\mathbf{x}$ to the gradient certificate of $\mathbf{t} - c\mathbf{1}$ in their respective local norms
\begin{lemma}\label{thm:xtoxplus} Let $\mathbf{x}_+$ and $\mathbf{y}$ be defined as above, and assume that $\|\mathbf{x}-\mathbf{y}\|_\mathbf{x} \leq r$ for some $r<\frac{1}{3}$. Then $\|\mathbf{x}_+ - \mathbf{y}\|_{\mathbf{x}_+} \leq \frac{r^2}{1 - 2r}$.
\end{lemma}
\begin{proof} Recall that the update in Line \ref{line:x-update} of Algorithm \ref{alg:Newton} is a single (full) Newton step towards the solution of the linear system $-g(\mathbf{y}) = \mathbf{t} - c\mathbf{1}$. Equivalently, the update $\mathbf{x}_+ - \mathbf{x}$ is a Newton step toward the minimizer of the convex self-concordant function
\[
f_c(\mathbf{x}) \defeq (\mathbf{t} - c\mathbf{1})^T\mathbf{x} + f(\mathbf{x}).
\]
Applying \cite[Thm.~2.2.3]{Renegar2001}) to $f_c$, we have
\[
\|\mathbf{x}_+ - \mathbf{y}\|_{\mathbf{x}} \leq \frac{\|\mathbf{x} - \mathbf{y}\|_\mathbf{x}^2}{1 - \|\mathbf{x} - \mathbf{y}\|_{\mathbf{x}}} = \frac{r^2}{1 - r}.
\] Coupling this result with the definition of self-concordance (Eq.~\eqref{eq:self-concordance}), we have
\[
\|\mathbf{x}_+ - \mathbf{y}\|_{\mathbf{y}} \leq \frac{\|\mathbf{x}_+ - \mathbf{y}\|_\mathbf{x}}{1 - \|\mathbf{x} - \mathbf{y}\|_\mathbf{x}} \leq \frac{\|\mathbf{x} - \mathbf{y}\|_\mathbf{x}^2}{(1 - \|\mathbf{x} - \mathbf{y}\|_\mathbf{x})^2} \leq \frac{r^2}{(1-r)^2} < 1.
\]
We conclude that $\mathbf{x}_+ \in B_{\mathbf{y}}(\mathbf{y},1)$, and we can thus invoke the inequality \eqref{eq:self-concordance} for another change of norms to conclude that
\[\|\mathbf{x}_+ - \mathbf{y}\|_{\mathbf{x}_+} \leq \frac{\|\mathbf{x}_+- \mathbf{y}\|_{\mathbf{y}}}{1 - \|\mathbf{x}_+ -\mathbf{y}\|_{\mathbf{y}}} \leq \frac{\frac{r^2}{(1-r)^2}}{1-\frac{r^2}{(1-r)^2}} = \frac{r^2}{(1-r)^2 - r^2} = \frac{r^2}{1 - 2r}.
{}\]
\end{proof}
We remark that, while $\mathbf{x}$ certifies $\mathbf{t} - c\mathbf{1}$ whenever $\|\mathbf{x} - \mathbf{y}\|_{\mathbf{x}} <\frac{1}{2}$, and each step of our proof is valid for all $0<r<\frac{1}{2}$, we can only have $\frac{r^2}{1-2r} \leq r$ whenever $0 < r < \frac{1}{3}$. Therefore, using Lemma~\ref{thm:xtoxplus}, we can guarantee that $\|\mathbf{x}_+ - \mathbf{y}\|_{\mathbf{x}_+} \leq \|\mathbf{x} - \mathbf{y}\|_{\mathbf{x}}$ when $\|\mathbf{x} - \mathbf{y}\|_{\mathbf{x}} < \frac{1}{3}$. Below, we need to further limit $r$ to ensure that the bound update is an improvement.
\begin{lemma}\label{thm:constant-update}
Suppose that $\|\mathbf{x}_+ - \mathbf{y}\|_{\mathbf{x}_+} \leq \frac{r^2}{1-2r}$ for some $0<r\leq\frac{1}{4}$.
Then $c_+>c$ and $\|\mathbf{x}_+ - \mathbf{y}_+\|_{\mathbf{x}_+} \leq r$.
\end{lemma}
\begin{proof}
We begin by showing that
\[\|\mathbf{x}_+ - H(\mathbf{x})^{-1}(\mathbf{t} - c\mathbf{1})\|_{\mathbf{x}_+} < \frac{r}{r+1},\]
which implies that Step \ref{line:c-update} of the algorithm indeed increases the lower bound to $c_+>c$.
Suppose $\|\mathbf{x}_+ - \mathbf{y}\|_{\mathbf{x}_+} \leq \frac{r^2}{1-2r}$.
Recall from Eq.~\eqref{eq:gH-identities} that $H(\mathbf{x}_+)\mathbf{x}_+ = -g(\mathbf{x}_+)$. Using this identity and the definition of the local norm, we deduce that
\begin{equation}
\begin{split}\label{eq:gdiff-identities}
\|-g(\mathbf{x}_+) + g(\mathbf{y})\|_{\mathbf{x}_+}^* &= \|H(\mathbf{x}_+)^{-1/2}\left(H(\mathbf{x}_+)\mathbf{x}_+ - (\mathbf{t}-c\mathbf{1})\right)\| \\
&=\|H(\mathbf{x}_+)^{1/2}\mathbf{x}_+ - H(\mathbf{x}_+)^{-1/2}(\mathbf{t} -c\mathbf{1})\| \\
&=\|\mathbf{x}_+ - H(\mathbf{x}_+)^{-1}(\mathbf{t}-c\mathbf{1})\|_{\mathbf{x}_+}.
\end{split}
\end{equation}
Using this in tandem with inequality \eqref{eq:lemma5} from Lemma \ref{thm:f-properties}, we have
\begin{equation*}
\begin{split}
\|\mathbf{x}_+ - H(\mathbf{x})^{-1}(\mathbf{t} - c\mathbf{1})\|_{\mathbf{x}_+} &\overset{\eqref{eq:gdiff-identities}}{=} \|-g(\mathbf{x}_+) + g(\mathbf{y})\|_{\mathbf{x}_+}\\
&\overset{\eqref{eq:lemma5}}{\leq} \frac{\|\mathbf{x}_+ - \mathbf{y}\|_{\mathbf{x}_+}}{1 - \|\mathbf{x}_+ - \mathbf{y}\|_{\mathbf{x}_+}} \leq \frac{\frac{r^2}{1-2r}}{1 - \frac{r^2}{1-2r}} < \frac{r}{r+1}
\end{split}
\end{equation*}
for every $r\leq\frac{1}{4}$, proving our first claim.
To see the second statement, we observe that
\begin{equation}\label{eq:gdiff-identities2}
\|-g(\mathbf{x}_+) + g(\mathbf{y}_+)\|_{\mathbf{x}_+}^* = \|\mathbf{x}_+ - H(\mathbf{x}_+)^{-1}(\mathbf{t}-c_+\mathbf{1})\|_{\mathbf{x}_+} = \frac{r}{r+1} < 1\end{equation}
by the definition of the bound update step in Line~\ref{line:c-update} and our discussion above. Now inequality~\eqref{eq:revlemma5} from Lemma~\ref{thm:f-properties} yields
\begin{equation*
\|\mathbf{x}_+ - \mathbf{y}_+\|_{\mathbf{x}_+} \leq \frac{\|-g(\mathbf{x}_+) + g(\mathbf{y}_+)\|_{\mathbf{x}_+}^*}{1 - \|-g(\mathbf{x}_+) + g(\mathbf{y}_+)\|_{\mathbf{x}_+}^*} \leq \frac{ r/( r+1)}{1 - ( r/( r+1))} = r.{}
\end{equation*}
\end{proof}
\deletethis{
\begin{lemma}\label{thm:constant-update} Suppose that $\|\mathbf{x}_+ - \mathbf{y}\|_{\mathbf{x}_+} \leq \frac{r^2}{1-2r}$ for some $0<r\leq\frac{1}{4}$
Then \[\|\mathbf{x}_+ - H(\mathbf{x})^{-1}(\mathbf{t} - c\mathbf{1})\|_{\mathbf{x}_+} < \frac{r}{r+1}.\]
\end{lemma}
\begin{proof} Suppose $\|\mathbf{x}_+ - \mathbf{y}\|_{\mathbf{x}_+} \leq \frac{r^2}{1-2r}$.
Recall from Eq.~\eqref{eq:gH-identities} that $H(\mathbf{x}_+)\mathbf{x}_+ = -g(\mathbf{x}_+)$. Using this identity and the definition of the local norm, we deduce that
\begin{equation}
\begin{split}\label{eq:gdiff-identities}
\|-g(\mathbf{x}_+) + g(\mathbf{y})\|_{\mathbf{x}_+}^* &= \|H(\mathbf{x}_+)^{-1/2}\left(H(\mathbf{x}_+)\mathbf{x}_+ - (\mathbf{t}-c\mathbf{1})\right)\| \\
&=\|H(\mathbf{x}_+)^{1/2}\mathbf{x}_+ - H(\mathbf{x}_+)^{-1/2}(\mathbf{t} -c\mathbf{1})\| \\
&=\|\mathbf{x}_+ - H(\mathbf{x}_+)^{-1}(\mathbf{t}-c\mathbf{1})\|_{\mathbf{x}_+}.
\end{split}
\end{equation}
Then using this in tandem with inequality \eqref{eq:lemma5} from Lemma \ref{thm:f-properties}, we have
\[
\|\mathbf{x}_+ - H(\mathbf{x})^{-1}(\mathbf{t} - c\mathbf{1})\|_{\mathbf{x}_+} \overset{\eqref{eq:gdiff-identities}}{=} \|-g(\mathbf{x}_+) + g(\mathbf{y})\|_{\mathbf{x}_+} \overset{\eqref{eq:lemma5}}{\leq} \frac{\|\mathbf{x}_+ - \mathbf{y}\|_{\mathbf{x}_+}}{1 - \|\mathbf{x}_+ - \mathbf{y}\|_{\mathbf{x}_+}} \leq \frac{\frac{r^2}{1-2r}}{1 - \frac{r^2}{1-2r}} < \frac{r}{r+1}
\]
for every $r\leq\frac{1}{4}$.
\end{proof}
As a consequence, since $\|\mathbf{x}_+ - H(\mathbf{x})^{-1}(\mathbf{t} - c\mathbf{1})\|_{\mathbf{x}_+} < \frac{r}{r+1}$, we can guarantee that there exists a constant $c_+ > c$ such that $\|\mathbf{x}_+ - H(\mathbf{x})^{-1}(\mathbf{t} - c_+\mathbf{1})\|_{\mathbf{x}_+} = \frac{r}{r+1}$.
Now, we show that the certificate $\mathbf{x}$ and the gradient certificate $\mathbf{y}$ satisfy $\|\mathbf{x} - \mathbf{y}\|_{\mathbf{x}} \leq r$ at the beginning of each iteration. This is clearly true for the first iteration, as shown by the inequality \eqref{eq:initial-r} in \mbox{Lemma \ref{thm:EveryPolyHasaBound}}.
\begin{lemma}\label{thm:r-invariant}
Suppose that $\|\mathbf{x} - \mathbf{y}\|_{\mathbf{x}} \leq r \leq 1/4$. Then $\|\mathbf{x}_+ - \mathbf{y}_+\|_{\mathbf{x}_+} \leq r$.
\end{lemma}
\begin{proof}
Analogously to Eq.~\eqref{eq:gdiff-identities}, $\|-g(\mathbf{x}_+) + g(\mathbf{y}_+)\|_{\mathbf{x}_+}^* = \|\mathbf{x}_+ - H(\mathbf{x}_+)^{-1}(\mathbf{t}-c_+\mathbf{1})\|_{\mathbf{x}_+}. $
Therefore
\begin{equation}\label{eq:gdiff-identities2}
\|-g(\mathbf{x}_+) + g(\mathbf{y}_+)\|_{\mathbf{x}_+}^* = \|\mathbf{x}_+ - H(\mathbf{x}_+)^{-1}(\mathbf{t}-c_+\mathbf{1})\|_{\mathbf{x}_+} = \frac{r}{r+1} < 1\end{equation}
by the definition of the bound update step in Line~\ref{line:c-update} and our discussion above. Now inequality~\eqref{eq:revlemma5} from Lemma~\ref{thm:f-properties} yields
\begin{equation*
\|\mathbf{x}_+ - \mathbf{y}_+\|_{\mathbf{x}_+} \leq \frac{\|-g(\mathbf{x}_+) + g(\mathbf{y}_+)\|_{\mathbf{x}_+}^*}{1 - \|-g(\mathbf{x}_+) + g(\mathbf{y}_+)\|_{\mathbf{x}_+}^*} \leq \frac{ r/( r+1)}{1 - ( r/( r+1))} = r.{}
\end{equation*}
\end{proof}
Coupling \mbox{Lemma \ref{thm:r-invariant}} with Corollary~\ref{thm:x-ycert}, this reasserts that $\mathbf{x}$ is indeed a certificate of $\mathbf{t} - c\mathbf{1}$ at the end of each iteration of the algorithm.
}
The next lemma uses Lemma~\ref{thm:xtoxplus} in showing that the improvement in the lower bound can be bounded from below by a constant times the local norm of $\mathbf{1}$.
\begin{lemma}\label{thm:m-m*} Define
$\rho_r \defeq \frac{r(1-3r-2r^2)}{1-r-2r^2}$
Then at the end of each iteration of \mbox{Algorithm \ref{alg:Newton}}, $c_+ - c \geq \frac{\rho_r}{\|\mathbf{1}\|_{\mathbf{y}}^*}$, where $\mathbf{y}$ is the gradient certificate of $\mathbf{t} - c\mathbf{1}$.
\end{lemma}
\begin{proof}
From the identities \eqref{eq:gdiff-identities2} and the definition of $c_+$ in Line \ref{line:c-update} of the algorithm, we have
\[
\frac{r}{r+1} = \|\mathbf{x}_+ - H(\mathbf{x}_+)^{-1}(\mathbf{t} - c_+\mathbf{1})\|_{\mathbf{x}_+} = \|-g(\mathbf{x}_+) + g(\mathbf{y}_+)\|_{\mathbf{x}_+}^*.
\]
Upper bounding the right-hand side by the triangle inequality gives
\begin{equation}\label{eq:thm:m-m*1}
\frac{r}{r+1} -\|-g(\mathbf{x}_+) + g(\mathbf{y})\|_{\mathbf{x}_+}^* \leq \|-g(\mathbf{y}_+) +g(\mathbf{y})\|_{\mathbf{x}_+}^* = \|(c_+ - c)\mathbf{1}\|_{\mathbf{x}_+}^*.
\end{equation}
Thus, to lower bound $(c_+ - c)$, it suffices to upper bound $\|-g(\mathbf{x}_+) + g(\mathbf{y})\|_{\mathbf{x}_+}^*$.
From Lemma~\ref{thm:xtoxplus}, we know that $\|\mathbf{x}_+ - \mathbf{y}\|_{\mathbf{x}_+} \leq \frac{r^2}{1-2r}$. Using the inequality \eqref{eq:lemma5} in Lemma~\ref{thm:f-properties}, we have
\begin{equation}\label{eq:thm:m-m*2}
\|-g(\mathbf{x}_+) + g(\mathbf{y})\|^*_{\mathbf{x}_+} \leq\frac{\|\mathbf{x}_+ - \mathbf{y}\|_{\mathbf{x}_+}}{1 - \|\mathbf{x}_+ - \mathbf{y}\|_{\mathbf{x}_+}} \leq \frac{\frac{r^2}{1-2r}}{1 - \frac{r^2}{1-2r}} = \frac{r^2}{1 - 2r - r^2}.
\end{equation}
Combining the inequalities in \eqref{eq:thm:m-m*1} and \eqref{eq:thm:m-m*2}, we have
\begin{equation*
(c_+ - c)\|\mathbf{1}\|_{\mathbf{x}_+}^* \geq \frac{r}{r+1} - \frac{r^2}{1 - 2r - r^2}.
\end{equation*}
Finally, changing norms again with inequality \eqref{eq:self-concordance},
\begin{align*}
(c_+ - c)\|\mathbf{1}\|_{\mathbf{y}_+}^* &\geq (c_+ - c)\|\mathbf{1}\|_{\mathbf{x}_+}^* (1 - \|\mathbf{y}-\mathbf{x}_+\|_{\mathbf{x}_+}^*)\\
& \geq \left(\frac{r}{r+1} - \frac{r^2}{1 - 2r - r^2}\right)\left(1-\frac{r^2}{1-2r}\right) = \rho_r.
\end{align*}
\end{proof}
We remark that if $r$ is chosen so that $0 < r \leq \frac{1}{4}$, then $\rho_r > 0$, and, for example, $\rho_r > 2/21$ for $r=1/6$. Therefore in each iteration of the algorithm, the improvement of the bound can be bounded from below by a quantity proportional to $(\|\mathbf{1}\|_{\mathbf{y}}^*)^{-1}$, where $\mathbf{y}$ is the current gradient certificate.
Now, we turn our attention to the convergence of Algorithm \ref{alg:Newton}. When $\mathbf{1}\in\Sc$, the optimal WSOS lower bound $c^*$ for a polynomial $\mathbf{t}$ is the unique scalar $\gamma$ for which $\mathbf{t}-\gamma\mathbf{1}$ is on the boundary of $\Sigma$.
In Theorem \ref{thm:lincon}, we show that the norm $\|\mathbf{1}\|_{\mathbf{y}}^*$ can be related to the distance $(c^*-c)$ between the current bound and the optimal WSOS lower bound. We will then combine this result with Lemma~\ref{thm:m-m*} above to show that the algorithm converges linearly to the optimal WSOS lower bound of $\mathbf{t}$. The analysis also motivates the stopping criterion for the algorithm.
In what follows, we let $\lambda_{\max}(\mathbf{M})$ denote the largest eigenvalue of the matrix $\mathbf{M}$ and $\lambda_{\min}(\mathbf{M})$ denote the smallest eigenvalue. We also remark that $\|\cdot\|_1$, $\|\cdot\|$ and $\|\cdot\|_\infty$ refer to the standard 1-norm, 2-norm, and infinity norm of vectors, respectively (not to be confused with the local norms used above).
\begin{theorem}\label{thm:lincon} Suppose that $\mathbf{t} - c^*\mathbf{1}$ is on the boundary of $\Sigma$. Let $\mathbf{y}$ denote the gradient certificate of some $\mathbf{t}-c\mathbf{1}$ with $c<c^*$. Then there exists a constant $C$ (depending only on the operator $\Lambda$) such that $c^*-c \leq (C\|\mathbf{1}\|_\mathbf{y}^*)^{-1}$.
\end{theorem}
\begin{proof}
Recall that $-g(\mathbf{y}) = \mathbf{t} - c\mathbf{1}$. Define the constant \[k_1 \defeq \min\{\mathbf{1}^\mathrm{T}\mathbf{v} \ | \ \mathbf{v} \in \Sigma^*, \|\mathbf{v}\|_\infty = 1\}.\] Observe that the minimum exists (as $\Sigma^*$ is a closed and non-trivial cone) and $k_1 > 0$, because $\mathbf{1} \in \Sigma^\circ$. Using the shorthand $\alpha \defeq c^* - c > 0$, we now have
\begin{align*}
\nu &\overset{\eqref{eq:gH-identities}}{=}\left \langle -g\left(\frac{\mathbf{y}}{\|\mathbf{y}\|_\infty}\right), \frac{\mathbf{y}}{\|\mathbf{y}\|_\infty}\right \rangle \\
&\overset{\eqref{eq:log-homogeneity}}{=} \|\mathbf{y}\|_\infty \left\langle \mathbf{t} - c\mathbf{1}, \frac{\mathbf{y}}{\|\mathbf{y}\|_\infty} \right\rangle \\
&= \|\mathbf{y}\|_\infty\left(\left\langle \mathbf{t} - c^*\mathbf{1}, \frac{\mathbf{y}}{\|\mathbf{y}\|_\infty}, \right\rangle + (c^*-c) \left\langle \mathbf{1}, \frac{\mathbf{y}}{\|\mathbf{y}\|_\infty}\right\rangle \right) \\
&\geq0 + \|\mathbf{y}\|_\infty\alpha k_1 = \|\mathbf{y}\|_\infty\alpha k_1,
\end{align*}
from which we conclude that
\begin{equation}\label{eq:linconk1}
\|\mathbf{y}\|_\infty \leq \frac{\nu}{\alpha k_1}.
\end{equation}
Recall from Eq.~\eqref{eq:H} that $H(\mathbf{y})\mathbf{w} = \Lambda^*(\Lambda(\mathbf{y})^{-1}\Lambda(\mathbf{w})\Lambda(\mathbf{y})^{-1})$. Therefore, $\mathbf{w}^\mathrm{T} H(\mathbf{y})\mathbf{w} = \langle \mathbf{w}, \Lambda^*(\Lambda(\mathbf{y})^{-1}\Lambda(\mathbf{w})\Lambda(\mathbf{y})^{-1}) \rangle = \text{tr}(\Lambda(\mathbf{w})\Lambda(\mathbf{y})^{-1}\Lambda(\mathbf{w})\Lambda(\mathbf{y})^{-1})$.
Moreover, observe that for every $\mathbf{A}\succcurlyeq \mathbf{0}$ and real symmetric matrix $\mathbf{B}$ of the same size, we have
\[
\text{tr}(\mathbf{A})\lambda_{\min}(\mathbf{B}) \leq \text{tr}(\mathbf{A}\mathbf{B}) \leq \text{tr}(\mathbf{A})\lambda_{\max}(\mathbf{B}).
\]
Using this fact, we have that for every $\mathbf{w}\in\mathbb{R}^U$,
\begin{align*}
\mathbf{w}^TH(\mathbf{y})\mathbf{w} &= \text{tr}\left(\Lambda(\mathbf{w})\Lambda(\mathbf{y})^{-1}\Lambda(\mathbf{w})\Lambda(\mathbf{y})^{-1}\right) \\
&\geq\lambda_{\min}(\Lambda(\mathbf{y})^{-1})~\text{tr}\left(\Lambda(\mathbf{w})\Lambda(\mathbf{y})^{-1}\Lambda(\mathbf{w})\right)\\
&=\lambda_{\min}(\Lambda(\mathbf{y})^{-1})~\text{tr}\left(\Lambda(\mathbf{w})^2\Lambda(\mathbf{y})^{-1}\right) \\
&\geq\lambda_{\min}(\Lambda(\mathbf{y})^{-1})^2~\text{tr}(\Lambda(\mathbf{w})^2)\\
&=\lambda_{\max}(\Lambda(\mathbf{y}))^{-2}~\text{tr}(\Lambda(\mathbf{w})^2).
\end{align*}
We conclude that
\begin{equation}\label{eq:c64.2}
\lambda_{\min}(H(\mathbf{y})^{1/2}) \geq \frac{k_2}{\lambda_{\max}(\Lambda(\mathbf{y}))},
\end{equation}
wherein we define
\[k_2 \defeq \min\{\sqrt{\text{tr}(\Lambda(\mathbf{w})^2)} \ | \ \|\mathbf{w}\| = 1\}.\]
We remark that $k_2 = \sigma_{\min}(\Lambda) > 0$ (since $\Lambda(\mathbf{w}) \neq \mathbf{0}$ whenever $\mathbf{w} \neq \mathbf{0}$).
Next, recall that $\|\mathbf{1}\|_{\mathbf{y}}^* = \|H(\mathbf{y})^{-1/2}\mathbf{1}\|$ and note $\|H(\mathbf{y})^{-1/2} \| = \frac{1}{\lambda_{\min}\left(H(\mathbf{y})^{1/2}\right)}$. Define
\[
k_3 \defeq \max\left\{\lambda_{\max}(\Lambda(\mathbf{y}))\ \middle|\ \mathbf{y}\in\Sigma^*, \|\mathbf{y}\|_\infty = 1\right\}.
\]
These identities and our previous inequalities give
\begin{equation*
\|\mathbf{1}\|_{\mathbf{y}}^* =
\|H(\mathbf{y})^{-1/2}\mathbf{1}\| \leq
\frac{\|\mathbf{1}\|}{\lambda_{\min}\left(H(\mathbf{y})^{1/2}\right)} \overset{\eqref{eq:c64.2}}{\leq}
\frac{\lambda_{\max}(\Lambda(\mathbf{y}))\|\mathbf{1}\|}{k_2} \leq
\frac{k_3\|\mathbf{y}\|\|\mathbf{1}\|}{k_2} \overset{\eqref{eq:linconk1}}{\leq} \frac{k_3\nu \|\mathbf{1}\|}{k_1k_2\alpha}.
\end{equation*}
Defining $C \defeq \frac{k_1k_2}{k_3\nu \|\mathbf{1}\|}$, we conclude that
\[
\alpha = c^*-c \leq (C\|\mathbf{1}\|_\mathbf{y}^*)^{-1}. {}
\]
\end{proof}
We remark that the parameter $\nu=\sum_{i=1}^m L_i$ is a parameter of the WSOS cone $\Sigma$ entirely independent of the representation of the polynomials. The parameter $k_1$ depends on the basis in which the WSOS polynomials are represented (but otherwise does not depend on $\Lambda$), while $k_2$ and $k_3$ are properties of the $\Lambda$ operator representing $\Sigma$
Coupling \mbox{Lemma \ref{thm:m-m*}} with \mbox{Theorem \ref{thm:lincon}}, we have also proven our main result about the convergence of our algorithm:
\begin{theorem}\label{thm:linear-convergence} Algorithm \ref{alg:Newton} is globally linearly convergent to
$c^* = \max\{c \ | \ \mathbf{t} - c\mathbf{1} \in \Sigma\}$, the optimal WSOS lower bound for the polynomial $\mathbf{t}$. More precisely, in each iteration of Algorithm \ref{alg:Newton}, the improvement of the lower bound $\Delta c = c_+ - c$ satisfies
\begin{equation}\label{eq:distance-bound}
\frac{\Delta c}{c^* - c} \geq \rho_r C,
\end{equation}
with the absolute constant $\rho_r>0$ defined in Lemma \ref{thm:m-m*} and the $\Lambda$-dependent constant $C>0$ defined in Theorem \ref{thm:lincon}.
\end{theorem}
Theorem \ref{thm:linear-convergence} motivates the stopping criterion (\mbox{Line \ref{line:stopping}}) of \mbox{Algorithm \ref{alg:Newton}}. The current bound $c$ is guaranteed to satisfy $c\leq c^* \leq c+\varepsilon$ as soon as $\Delta c \leq \rho_r C\varepsilon$.
Alternatively, we can rearrange the same inequality to provide an explicit upper bound on the number of iterations of the algorithm. After $k$ iterations of \mbox{Algorithm \ref{alg:Newton}} we have
\[
c^* - c_k \leq (1 - \rho_rC)^k(c^* - c_0),
\]
therefore, for a fixed cone (and parameter $C$), the algorithm terminates after
$ \mathcal{O}\left(\log\frac{c^*-c_0}{\varepsilon}\right) $
iterations. Additionally, it is typically easy to bound from above the global minimum of the input polynomial $\mathbf{t}$ (e.g., by evaluating it at any point in its domain), and thus bound $c^*$ from above, and when an explicit bound on the magnitude of the elements in $\{\mathbf{x}\in\mathbb{R}^n\,|\,g_i(\mathbf{x})\geq0,\,i=1,\dots,m\}$
is known, it is also straightforward to upper bound $c^*$ by $\kappa_\mathbf{g}\|\mathbf{t}\|$ with some constant $\kappa_\mathbf{g}$ dependent only the weight functions $\mathbf{g}$. Similarly, from the first step of Algorithm~\ref{alg:Newton}, $c_0 \geq -\frac{1+r}{r}\lambda_{\max}(H({\vx_{1}})^{-1})\|\mathbf{t}\|$, bounding the initial bound $c_0$ from below by a $\Lambda$-dependent constant multiple of $\|\mathbf{t}\|$. Thus, for a fixed cone (and representation $\Lambda$), the algorithm terminates after $\mathcal{O}(\log\frac{\|\mathbf{t}\|}{\varepsilon})$
iterations.
We also remark that although our primary goal is to obtain certified rational \emph{lower} bounds on the polynomial, dual certificates also provide \emph{upper bounds} on the optimal WSOS bound via Theorem~\ref{thm:linear-convergence}, whenever the $\Lambda$-dependent constant $C$ in is known (or can be bounded from below) for a particular cone $\Sigma$. In particular, although in the analysis heavily relies on the quantity $\|\mathbf{1}\|_\mathbf{y}^*$, which is not efficiently computable (we do not have access to the gradient certificate $\mathbf{y}$), the inequality \eqref{eq:distance-bound} provides a computable upper bound on $c^*$.
\subsection{Bounding constants in Theorem \ref{thm:lincon}} \label{sec:C-bounds}
In general, we cannot hope to find sharp closed-form bounds for the constant $C$ in Theorems \ref{thm:lincon} and \ref{thm:linear-convergence}, but we can compute cone-specific bounds on each of the constants $k_1, k_2,$ and $k_3$ in the formula for $C$ by convex optimization.
Recall that $k_1 = \text{min}\{\mathbf{1}^\mathrm{T}\mathbf{v} \ | \ \mathbf{v} \in \Sigma^*, \|\mathbf{v}\|_\infty = 1\}$. Although the norm constraint is not convex, we have
\begin{equation*
k_1 = \min_{1 \leq i \leq U} \{\min\{k_{1,i}^-, k_{1,i}^+\}\},
\end{equation*}
with
\begin{equation}\label{eq:k1+}
k_{1,i}^+ = \min\{\mathbf{1}^\mathrm{T}\mathbf{v} \ | \ \mathbf{v} \in \Sigma^*, \|\mathbf{v}\|_\infty \leq 1\text{ and } v_i = 1\} \quad (i=1,\dots,m)
\end{equation}
and
\begin{equation}\label{eq:k1-}
k_{1,i}^- = \min\{\mathbf{1}^\mathrm{T}\mathbf{v} \ | \ \mathbf{v} \in \Sigma^*, \|\mathbf{v}\|_\infty \leq 1\text{ and } v_i = -1 \}. \quad (i=1,\dots,m)
\end{equation}
Therefore, $k_1$ can be computed (numerically) by solving $2U$ convex optimization problems. (For a rigorous lower bound, we can use dual methods that determine approximately optimal but feasible solutions of the dual optimization problems of \eqref{eq:k1+} and \eqref{eq:k1-}.)
Recall that $k_2 = \min\{\text{tr}(\Lambda(\mathbf{w})^2) \ | \ \|\mathbf{w}\| = 1\}$. Hence, the constant $k_2$ is the smallest singular value of the linear operator $\Lambda$ and can be computed to high accuracy using singular value decomposition. Alternatively, we have
\begin{equation*
\text{tr}(\Lambda(\mathbf{w})^2) = \sum_{i=1}^L \Lambda_i(\mathbf{w})^2 = \mathbf{w}^\mathrm{T} \mathbf{M} \mathbf{w},
\end{equation*} for a positive semidefinite rational matrix $\mathbf{M}$ that is easily computable from $\Lambda$; lower bounding $k_2$ amounts to lower bounding the smallest eigenvalue of the matrix $\mathbf{M}$.
Recall that the constant $k_3 = \max\left\{\lambda_{\max}(\Lambda(\mathbf{y}))\ \middle|\ \mathbf{y}\in\Sigma^*, \|\mathbf{y}\|_\infty = 1\right\}$.
Using the Gershgorin circle theorem, we know that
\begin{equation}\label{eq:k3.1}
\lambda_{\max}(\Lambda(\mathbf{y}))
= \max_{1 \leq k \leq m} \lambda_{\max}(\Lambda_k(\mathbf{y}))
\leq \max_{1 \leq k\leq m} \|\Lambda_k(\mathbf{y})\|_\infty.
\end{equation}
So $k_3$ can be bounded from above by the largest absolute row sum of all of the $\Lambda_k$ operators.
Since the values of $\|\mathbf{1}\|$ and $\nu$ are known, having bounded $k_1$ and $k_2$ from below by positive quantities and $k_3$ from above, $C$ can be bounded from below by a positive, efficiently computable constant. In Section \ref{sec:univariate} we revisit this question and find closed-form bounds for the case of univariate nonnegative polynomials over an interval.
\subsection{Smaller rational certificates}\label{sec:smaller-rational-certs}
As discussed in Section \ref{sec:rational-certificates}, if desired, we may round the certificate returned by Algorithm \ref{alg:Newton} quite freely to any nearby certificate with smaller denominators, using \mbox{Corollary \ref{thm:x-ycert}}. Recall that in infinite precision, Algorithm \ref{alg:Newton} returns a certificate with $\|\mathbf{x}-\mathbf{y}\|_\mathbf{x}\leq r \leq 1/4$. As long as the rounded certificate $\mathbf{x}_N$ satisfies $\|\mathbf{x}_N-\mathbf{y}\|_{\mathbf{x}_N}<1/2$, it also certifies the bound returned by the algorithm. The following lemma quantifies how small the denominators of such a rounded certificate can be.
\begin{lemma}\label{thm:roundx}
Suppose that $\|\mathbf{x}-\mathbf{y}\|_\mathbf{x}\leq r_1 < 1/2$ and choose any large enough integer denominator $N$ to satisfy
\begin{equation}\label{eq:roundx2}
\|H(\mathbf{x})^{1/2}\| \leq \frac{2N}{\sqrt{U}}\left( \frac{r_2-r_1}{1 + r_2}\right)
\end{equation}
Then every $\mathbf{x}_N \in \frac{1}{N}\mathbb{Z}^U$ with $\|\mathbf{x}_N-\mathbf{x}\|\leq \frac{\sqrt{U}}{2N}$ satisfies $\|\mathbf{x}_N - \mathbf{y}\|_{\mathbf{x}_N} \leq r_2$.
In particular, taking $r_1 = r$ and $r_2 = 1/2$, we may round the certificate $\mathbf{x}$ returned by Algorithm \ref{alg:Newton} componentwise to the nearest rational vector with denominator $N$, and the resulting vector $\mathbf{x}_N$ is a rational certificate for the lower bound $c$ returned by the algorithm.
\end{lemma}
\allowdisplaybreaks
\begin{proof}
By self-concordance (inequality \eqref{eq:self-concordance} in Lemma \ref{thm:f-properties}), we have
\begin{align*}
\|\mathbf{x}_N - \mathbf{y}\|_{\mathbf{x}_N} &\overset{\eqref{eq:self-concordance}}{\leq} \frac{\|\mathbf{x}_N - \mathbf{y}\|_{\mathbf{x}}}{1-\|\mathbf{x}_N - \mathbf{x}\|_\mathbf{x}} \\
&\leq \frac{\|\mathbf{x}_N - \mathbf{y}\|_\mathbf{x}}{1 - \frac{\sqrt{U}}{2N}\|H(\mathbf{x})^{1/2}\|}\\
&\overset{\eqref{eq:roundx2}}{\leq} \frac{1+r_2}{1+r_1}\|\mathbf{x}_N - \mathbf{y}\|_{\mathbf{x}} \\
&\leq \frac{1+r_2}{1+r_1}\left( \|\mathbf{x}_N - \mathbf{x}\|_\mathbf{x} + \|\mathbf{x} - \mathbf{y}\|_{\mathbf{x}} \right)\\
&\leq \frac{1+r_2}{1+r_1}\left( \frac{\sqrt{U}}{2N}\|H(\mathbf{x})^{1/2}\| + r_1 \right) \\
&\overset{\eqref{eq:roundx2}}{\leq} \frac{1+r_2}{1+r_1}\left( \frac{\sqrt{U}}{2N}\frac{2N}{\sqrt{U}}\left( \frac{r_2 - r_1}{1+r_1}\right) + r_1 \right) \\
&= r_2.{}
\end{align*}
\end{proof}
\section{Univariate polynomials}\label{sec:univariate}
In the univariate case, we can bound the number of iterations of Algorithm \ref{alg:Newton} by providing explicit bounds on the constant $C$, adapting the arguments from those in Section \ref{sec:C-bounds}. For brevity, we only treat the even-degree case in detail.
\begin{theorem
Suppose that $n=1$ and $\deg t = 2d$. Using the Chebyshev basis to to represent all polynomials and weights $\mathbf{g}(z) \defeq (1, 1 - z^2)$ (as in Example \ref{ex:Chebyshev}), \mbox{Algorithm \ref{alg:Newton}} terminates after at most $\mathcal{O}(d^2\log\frac{\|\mathbf{t}\|d}{\varepsilon})$ iterations and requires $\mathcal{O}(d^5\log\frac{\|\mathbf{t}\|d}{\varepsilon})$ floating point operations overall.
\end{theorem}
\begin{proof}
We start by bounding the constant $C$ from Theorem \ref{thm:lincon} as a function of all relevant parameters by bounding each of $k_1,k_2$ and $k_3$ in the formula for $C$.
\begin{enumerate}
\item $k_1 \geq 1$. Recall that $k_1 = \min \{ \mathbf{1}^\mathrm{T} \mathbf{v} \ | \ \mathbf{v} \in \Sigma^*, \|\mathbf{v}\|_\infty = 1\}$.
Since nonnegative polynomials and weighted sum-of-squares polynomials coincide in the univariate case \cite{BrickmanSteinberg1962}, every vector $\mathbf{v} \in \left(\Sigma_{1,2d}^\mathbf{g} \right)^*$ can be written as a conic combination of moment vectors; precisely, we write $\mathbf{v} = \sum_{i=1}^n \alpha_i \mathbf{q}(z_i)$, wherein $z_i \in [-1,1]$ and $\alpha_i \geq 0$ for each $i$ \cite[Sec.~II.2]{KarlinStudden1966}. Then, we have \[
\mathbf{1}^\mathrm{T}\mathbf{v} = \mathbf{1}^\mathrm{T}\left(\sum_{i=1}^n \alpha_i \mathbf{q}(z_i) \right) = \sum_{i=1}^n \alpha_i \left(\mathbf{1}^\mathrm{T} \mathbf{q}(z_i)\right) = \sum_{i=1}^n\alpha_i.
\]
If $\|\mathbf{v}\|_\infty = 1$, there exists some $j$ such that $|\sum_{i=1}^n \alpha_i \mathbf{q}_j(z_i)| = 1$. Since for the Chebyshev basis each $\mathbf{q}(z_i) \in [-1,1]^{2d+1}$, it follows that
\[
\mathbf{v}_j = 1= \left|\sum_{i=1}^n \alpha_i \mathbf{q}_j(z_i)\right| \leq \sum_{i=1}^n \left| \alpha_i \mathbf{q}_j(z_i) \right| \leq \sum_{i=1}^n |\alpha_i| = \sum_{i=1}^n \alpha_i,
\]
since $\alpha_i \geq 0$. Thus, $\sum_{i=1}^n \alpha_i \geq 1$. It follows that
\[
\mathbf{1}^\mathrm{T}\mathbf{v} = \sum_{i=1}^n \alpha_i \geq 1,
\]
therefore $k_1 \geq 1$.
\item $k_2 \geq \frac{1}{2}\sqrt{3 - \sqrt{5}} \approx 0.437$. Recall that $k_2 = \min \{ \text{tr}(\Lambda(\mathbf{w})^2) \ | \ \|\mathbf{w}\| = 1\}$.
We have
\[
\text{tr}(\Lambda(\mathbf{w})^2) = \text{tr}(\Lambda_1(\mathbf{w})^2) + \text{tr}(\Lambda_2(\mathbf{w})^2) \geq \text{tr}(\Lambda_1(\mathbf{w})^2).
\]
Note that
\begin{equation}\label{eq:TiTj}
2p_i(x)p_j(x) = p_{i+j}(x) + p_{|i - j|}(x) \quad \text{ for every }i,j=0,1,\dots
\end{equation}
Coupling this identity with the fact that $\Lambda_1(\mathbf{q}) = g_1\mathbf{p}\vp^\mathrm{T}= \mathbf{p}\vp^\mathrm{T}$ (recall that the first weight is $g_1=1$), we deduce that
\begin{equation}\label{eq:TiTjLambda}
\Lambda_1(\mathbf{w})_{i,j} = \frac{1}{2}w_{i + j} + \frac{1}{2}w_{|i - j|}.
\end{equation}
Therefore, the zeroth row (and the zeroth column) of $\Lambda_1(\mathbf{w})$ is $(w_0,w_1,\dots,w_d)$ and the last row (and the last column) is $(\frac{1}{2}(w_d + w_d), \frac{1}{2}(w_{d-1} + w_{d+1}), \dots, \frac{1}{2}(w_0 + w_{2d}))$, and so we have
\[
\text{tr}(\Lambda_1(\mathbf{w}))^2 \geq \sum_{i=0}^d w_i^2 + \sum_{j=0}^d \frac{1}{4}(w_{j} + w_{2d - j})^2 = \mathbf{w}^\mathrm{T}\mathbf{M}\mathbf{w},
\]
where $\mathbf{M}$ is the $2d + 1 \times 2d + 1$ matrix (indexed from zero) given by
\[ M_{i,j} = \begin{cases}
\frac{5}{4} & \text{ if } i=j < d \\
2 & \text{ if } i=j=d \\
\frac{1}{4} & \text{ if } i + j = 2d, i\neq j,\text{ or if }i=j > d\\
0 &\text{ otherwise }
\end{cases} \]
Therefore
\[
\mathbf{w}^\mathrm{T} \mathbf{M} \mathbf{w} \geq \|\mathbf{w}\|^2 \lambda_{\min}(\mathbf{M}) = \|\mathbf{w}\|^2 \left(\frac{1}{4}(3 - \sqrt{5})\right).
\]
We conclude that
\[
k_2 \geq \frac{1}{2}\sqrt{3 - \sqrt{5}} \approx 0.437.
\]
\item $k_3 \leq d+1$. Recall that
\[k_3 = \max\{\lambda_{\max}(\Lambda(\mathbf{y})) \ |\ \mathbf{y} \in \Sigma, \|\mathbf{y}\|_\infty = 1\},\] and that based on the inequality \eqref{eq:k3.1}, we need only bound the largest absolute row sum of $\Lambda_1(\mathbf{y})$ and $\Lambda_2(\mathbf{y})$.
For $\Lambda_1(\mathbf{y})$, the identity \eqref{eq:TiTjLambda} and $\|\mathbf{y}\|_\infty = 1$ yield the bound
\[
\sum_{j=0}^d |\Lambda_1(\mathbf{y})_{i,j}| = \sum_{j=0}^d\left|\frac{1}{2}\mathbf{y}_{i+j} + \frac{1}{2}\mathbf{y}_{|i - j|}\right| \leq d +1.
\]
For $\Lambda_2(\mathbf{y})$, observe that $1 - t^2 = \frac{1}{2}(p_0(t) - p_2(t))$. Coupling this with the identity \eqref{eq:TiTj}, we deduce that
\begin{align*}
\frac{1}{2}(p_0(t) - p_2(t))p_i(t)p_j(t)
=& \frac{1}{8} \big((2 p_{i + j}(t) + 2 p_{|i - j|} (t) - p_{i + j + 2}(t) - \\
& p_{|i + j - 2|}(t) - p_{|i - j| + 2}(t)-p_{||i - j| - 2|}(t)\big),
\end{align*}
so
\[
\Lambda_2(\mathbf{y})_{i,j} = \frac{1}{8} \left(2 \mathbf{y}_{i + j} + 2\mathbf{y}_{|i - j|} - \mathbf{y} _{i + j + 2} -
\mathbf{y}_{|i + j- 2|} - \mathbf{y}_{|i - j| + 2} - \mathbf{y}_{||i -j| - 2|}\right).
\]
Then, assuming $\|\mathbf{y}\|_\infty = 1$, we obtain the bound
\[
\sum_{j=0}^d |\Lambda_2(\mathbf{y})_{i,j}| \leq \sum_{j=0}^d \left|\frac{1}{8} \left(2+ 2 +1 +
1 + 1+ 1\right)\right| \leq d+1.
\]
Thus, $k_3 \leq \max\{\max\{\|\Lambda_1(\mathbf{y})\|_\infty, \|\Lambda_2(\mathbf{y})\|_\infty\} \ |\ \|\mathbf{y}\|_\infty = 1\} \leq d+1$.
\end{enumerate}
Lastly, since $\nu = 2d + 1$ and $\|\mathbf{1}\| = 1$, combining the above bounds on $k_1$, $k_2$ and $k_3$ we get
\[
C = \frac{k_1k_2}{k_3\nu\|\mathbf{1}\|} \geq \frac{\sqrt{3 - \sqrt{5}}}{2(d+1)(2d+1)}.
\]
From \eqref{eq:distance-bound} and the following discussion, the number of iterations is proportional to
\begin{equation}\label{eq:number-of-iterations}
\mathcal{O}\left(\frac{1}{\log\frac{1}{1-\rho_r C}}\log\frac{c^*-c_0}{\varepsilon}\right).
\end{equation}
From the series expansion $\frac{1}{\log\frac{1}{1-z}} = z^{-1}-\frac{1}{2}-\dots$ we see that the first term in \eqref{eq:number-of-iterations}, which only depends on the input through the degree $d$, is $\mathcal{O}(C^{-1}) = \mathcal{O}(d^2)$. To bound the numerator of second term, recall that for a coefficient vector $\mathbf{t}$ in the Chebyshev basis, $|c^*|\leq \|\mathbf{t}\|_1 \leq (2d+1)^{1/2}\|\mathbf{t}\|_2$, and from the intialization of \mbox{Algorithm \ref{alg:Newton}} we also have
\[|c_0| \leq \frac{1+r}{r}\|H(\mathbf{x}_1)^{-1/2}\|_2\|\mathbf{t}\|_2 \overset{\eqref{eq:Hx1}}{\leq} \frac{1+r}{r}\|\mathbf{t}\|_2\frac{2d+1}{\sqrt{2}}. \]
Thus, $|c^*-c_0|$ is of order $\mathcal{O}(\|\mathbf{t}\|_2 d)$, and the claim about the number of iterations follows.
The bottleneck of each iteration is the computation and factorization of the Hessian, which require $\mathcal{O}(d^3)$ floating point operations. Therefore, the total number of floating point operations is $\mathcal{O}(d^5\log\frac{\|\mathbf{t}\|d}{\varepsilon})$. The $\mathcal{O}(\cdot)$ notation hides only absolute constants and the user-defined constant parameter $\rho_r$ from Lemma \ref{thm:m-m*}.
\end{proof}
\section{Discussion}\label{sec:Discussion}
\paragraph{Primal versus dual certificates}
Conventional nonnegativity certificates are representations of the certified polynomials that make their nonnegativity apparent. This is a fundamental issue for numerical methods for computing nonnegativity certificates, as the certificate they compute is typically a rigorous WSOS certificate for a slightly different polynomial from the one we seek to certify.
Dual certificates address this issue: through the formula \eqref{eq:Sdef}, not only can we interpret any rational dual vector from $\mathcal{C}(\mathbf{s})$ as a certificate, but we can also compute, via a closed-form formula, a rational certificate for the polynomial $\mathbf{s}$ with rational certificates. Since every polynomial (in the interior of the SOS cone) has a full-dimensional cone of dual certificates, even an inexact numerical method computing low-accuracy solutions to an SOS optimization problem can return dual certificates that can be turned into a rational certificate this way. For example, Algorithm \ref{alg:Newton} can be implemented as a purely numerical method, followed by an application of the formula \eqref{eq:Sdef} to compute a rational certificate for the computed bound. Although the certificate $\mathbf{x}$ only loosely tracks the gradient certificate of $\mathbf{t}-c\mathbf{1}$, we can guarantee that $\mathbf{x}$ certifies the current bound. This also means that, unlike most numerical or hybrid methods that require high-accuracy solutions from the numerical component of the algorithm, Algorithm \ref{alg:Newton} provides a certified bound even if terminated early; only the quality of the bound suffers.
Recent work in numerical methods for non-symmetric cones has resulted in a few additional algorithms that can directly optimize over the cone of WSOS certificates circumventing semidefinite programming, including \cite{KarimiTuncel2019} and \cite{PappYildiz2020}; in principle, these can also be coupled with the methods presented in Section \ref{sec:dualcertificates}.
\paragraph{Efficiency} In general,
it is difficult to make general statements about the asymptotic running time of \mbox{Algorithm \ref{alg:Newton}} as a function of every interesting parameter (the degree and the number of unknowns of the input polynomial, etc.) as these also depend on the specific weight polynomials and the chosen representation ($\Lambda$ operator). As noted, the computational cost per iteration is a low-degree polynomial for $\Lambda$ operators corresponding to popular bases in numerical methods (e.g., Chebyshev and interpolant bases), and the method is linearly convergent, that is, for a given polynomial it requires a number of iterations proportional to $\log(1/\varepsilon)$ to compute a certified rational bound within $\varepsilon$ of the optimal bound $c^*$.
While we only derive an explicit bound on the exact linear rate and the initial gap $c^* - c_0$ in the univariate case in Section \ref{sec:univariate}, it may also be possible to derive such bounds in other interesting special cases, such as the cases of multivariate polynomials over simple semialgebraic sets such as the unit sphere or the unit cube. It is encouraging that the complexity of the algorithm in the univariate case is lower than the complexity of applying off-the-shelf semidefinite programming algorithms to the conventional semidefinite reformulations of the univariate SOS optimization problem.
\paragraph{Assumptions}
Throughout, we have made the fundamental assumption that the constant one polynomial is in the interior of the WSOS cone $\Sigma = \Sigma_{n,2\mathbf{d}}^\mathbf{g}$. (Naturally, in any remotely interesting situation, positive constant polynomials must belong to $\Sigma$, but not necessarily to the interior.) This is a mild assumption both from a theoretical and practical perspective. In many cases, it can be verified directly and ensured to hold a priori. Computationally, it can be verified via convex optimization, and if it does not hold, $\Sigma$ can be extended, with the inclusion of a single additional weight that is nonnegative on the nonnegativity set of the existing weights, to satisfy this condition without changing $\operatorname{span}(\Sigma)$.
\bibliographystyle{siamplain}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 1,803 |
Q: Spark SQL - Get Column Names of a Hive Table in a String I'm trying to get the column names of a Hive table in a comma separated String. This is what I'm doing
val colNameDF = spark.sql("show columns in hive_table")
val colNameStr = colNameDF.select("col_name").collect.mkString(", ")
And the output I get is
res0: String = [col_1], [col_2], [col_3]
But what I want is col_1, col_2, col_3. I can remove [ and ] from the String, but I'm curious as to whether we can get the column names without the brackets in the first place.
Edit: The column names in the Hive table don't contain [ ]
A: Instead of show columns, Try below approach as it is faster than yours.
val colNameDF = spark.sql("select * from hive_table").limit(0)
Or
val colNameDF = spark.table("hive_table").limit(0)
val colNameStr = colNameDF.columns.mkString(", ")
A: The collect returns to you an array of Row which is particularly represented internally as array of values, so you need to trick it like this:
val colNameDF = spark.sql("show columns in hive_table")
val colNameStr = colNameDF.select("col_name").collect.map(r=>r.getString(0)).mkString(", ")
A: Building on @Srinivas' answer above, here is the equivalent Python code. It is very fast:
colNameStr = ",".join(spark.table(hive_table).limit(0).columns)
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 5,960 |
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var is = {
obj: function obj(value) {
return (typeof value === 'undefined' ? 'undefined' : _typeof(value)) === 'object' && !!value;
},
all: function all(value) {
return is.obj(value) && value.type === _keys2.default.all;
},
error: function error(value) {
return is.obj(value) && value.type === _keys2.default.error;
},
array: Array.isArray,
func: function func(value) {
return typeof value === 'function';
},
promise: function promise(value) {
return value && is.func(value.then);
},
iterator: function iterator(value) {
return value && is.func(value.next) && is.func(value.throw);
},
fork: function fork(value) {
return is.obj(value) && value.type === _keys2.default.fork;
},
join: function join(value) {
return is.obj(value) && value.type === _keys2.default.join;
},
race: function race(value) {
return is.obj(value) && value.type === _keys2.default.race;
},
call: function call(value) {
return is.obj(value) && value.type === _keys2.default.call;
},
cps: function cps(value) {
return is.obj(value) && value.type === _keys2.default.cps;
},
subscribe: function subscribe(value) {
return is.obj(value) && value.type === _keys2.default.subscribe;
},
channel: function channel(value) {
return is.obj(value) && is.func(value.subscribe);
}
};
exports.default = is;
/***/ }),
/***/ 2:
/***/ (function(module, exports) {
(function() { module.exports = this["lodash"]; }());
/***/ }),
/***/ 254:
/***/ (function(module, exports, __webpack_require__) {
"use strict";
Object.defineProperty(exports, "__esModule", {
value: true
});
exports.createChannel = exports.subscribe = exports.cps = exports.apply = exports.call = exports.invoke = exports.delay = exports.race = exports.join = exports.fork = exports.error = exports.all = undefined;
var _keys = __webpack_require__(255);
var _keys2 = _interopRequireDefault(_keys);
function _interopRequireDefault(obj) { return obj && obj.__esModule ? obj : { default: obj }; }
var all = exports.all = function all(value) {
return {
type: _keys2.default.all,
value: value
};
};
var error = exports.error = function error(err) {
return {
type: _keys2.default.error,
error: err
};
};
var fork = exports.fork = function fork(iterator) {
for (var _len = arguments.length, args = Array(_len > 1 ? _len - 1 : 0), _key = 1; _key < _len; _key++) {
args[_key - 1] = arguments[_key];
}
return {
type: _keys2.default.fork,
iterator: iterator,
args: args
};
};
var join = exports.join = function join(task) {
return {
type: _keys2.default.join,
task: task
};
};
var race = exports.race = function race(competitors) {
return {
type: _keys2.default.race,
competitors: competitors
};
};
var delay = exports.delay = function delay(timeout) {
return new Promise(function (resolve) {
setTimeout(function () {
return resolve(true);
}, timeout);
});
};
var invoke = exports.invoke = function invoke(func) {
for (var _len2 = arguments.length, args = Array(_len2 > 1 ? _len2 - 1 : 0), _key2 = 1; _key2 < _len2; _key2++) {
args[_key2 - 1] = arguments[_key2];
}
return {
type: _keys2.default.call,
func: func,
context: null,
args: args
};
};
var call = exports.call = function call(func, context) {
for (var _len3 = arguments.length, args = Array(_len3 > 2 ? _len3 - 2 : 0), _key3 = 2; _key3 < _len3; _key3++) {
args[_key3 - 2] = arguments[_key3];
}
return {
type: _keys2.default.call,
func: func,
context: context,
args: args
};
};
var apply = exports.apply = function apply(func, context, args) {
return {
type: _keys2.default.call,
func: func,
context: context,
args: args
};
};
var cps = exports.cps = function cps(func) {
for (var _len4 = arguments.length, args = Array(_len4 > 1 ? _len4 - 1 : 0), _key4 = 1; _key4 < _len4; _key4++) {
args[_key4 - 1] = arguments[_key4];
}
return {
type: _keys2.default.cps,
func: func,
args: args
};
};
var subscribe = exports.subscribe = function subscribe(channel) {
return {
type: _keys2.default.subscribe,
channel: channel
};
};
var createChannel = exports.createChannel = function createChannel(callback) {
var listeners = [];
var subscribe = function subscribe(l) {
listeners.push(l);
return function () {
return listeners.splice(listeners.indexOf(l), 1);
};
};
var next = function next(val) {
return listeners.forEach(function (l) {
return l(val);
});
};
callback(next);
return {
subscribe: subscribe
};
};
/***/ }),
/***/ 255:
/***/ (function(module, exports, __webpack_require__) {
"use strict";
Object.defineProperty(exports, "__esModule", {
value: true
});
var keys = {
all: Symbol('all'),
error: Symbol('error'),
fork: Symbol('fork'),
join: Symbol('join'),
race: Symbol('race'),
call: Symbol('call'),
cps: Symbol('cps'),
subscribe: Symbol('subscribe')
};
exports.default = keys;
/***/ }),
/***/ 276:
/***/ (function(module, exports, __webpack_require__) {
"use strict";
Object.defineProperty(exports, "__esModule", {
value: true
});
exports.wrapControls = exports.asyncControls = exports.create = undefined;
var _helpers = __webpack_require__(254);
Object.keys(_helpers).forEach(function (key) {
if (key === "default") return;
Object.defineProperty(exports, key, {
enumerable: true,
get: function get() {
return _helpers[key];
}
});
});
var _create = __webpack_require__(410);
var _create2 = _interopRequireDefault(_create);
var _async = __webpack_require__(412);
var _async2 = _interopRequireDefault(_async);
var _wrap = __webpack_require__(414);
var _wrap2 = _interopRequireDefault(_wrap);
function _interopRequireDefault(obj) { return obj && obj.__esModule ? obj : { default: obj }; }
exports.create = _create2.default;
exports.asyncControls = _async2.default;
exports.wrapControls = _wrap2.default;
/***/ }),
/***/ 410:
/***/ (function(module, exports, __webpack_require__) {
"use strict";
Object.defineProperty(exports, "__esModule", {
value: true
});
var _builtin = __webpack_require__(411);
var _builtin2 = _interopRequireDefault(_builtin);
var _is = __webpack_require__(139);
var _is2 = _interopRequireDefault(_is);
function _interopRequireDefault(obj) { return obj && obj.__esModule ? obj : { default: obj }; }
function _toConsumableArray(arr) { if (Array.isArray(arr)) { for (var i = 0, arr2 = Array(arr.length); i < arr.length; i++) { arr2[i] = arr[i]; } return arr2; } else { return Array.from(arr); } }
var create = function create() {
var userControls = arguments.length <= 0 || arguments[0] === undefined ? [] : arguments[0];
var controls = [].concat(_toConsumableArray(userControls), _toConsumableArray(_builtin2.default));
var runtime = function runtime(input) {
var success = arguments.length <= 1 || arguments[1] === undefined ? function () {} : arguments[1];
var error = arguments.length <= 2 || arguments[2] === undefined ? function () {} : arguments[2];
var iterate = function iterate(gen) {
var yieldValue = function yieldValue(isError) {
return function (ret) {
try {
var _ref = isError ? gen.throw(ret) : gen.next(ret);
var value = _ref.value;
var done = _ref.done;
if (done) return success(value);
next(value);
} catch (e) {
return error(e);
}
};
};
var next = function next(ret) {
controls.some(function (control) {
return control(ret, next, runtime, yieldValue(false), yieldValue(true));
});
};
yieldValue(false)();
};
var iterator = _is2.default.iterator(input) ? input : regeneratorRuntime.mark(function _callee() {
return regeneratorRuntime.wrap(function _callee$(_context) {
while (1) {
switch (_context.prev = _context.next) {
case 0:
_context.next = 2;
return input;
case 2:
return _context.abrupt('return', _context.sent);
case 3:
case 'end':
return _context.stop();
}
}
}, _callee, this);
})();
iterate(iterator, success, error);
};
return runtime;
};
exports.default = create;
/***/ }),
/***/ 411:
/***/ (function(module, exports, __webpack_require__) {
"use strict";
Object.defineProperty(exports, "__esModule", {
value: true
});
exports.iterator = exports.array = exports.object = exports.error = exports.any = undefined;
var _is = __webpack_require__(139);
var _is2 = _interopRequireDefault(_is);
function _interopRequireDefault(obj) { return obj && obj.__esModule ? obj : { default: obj }; }
var any = exports.any = function any(value, next, rungen, yieldNext) {
yieldNext(value);
return true;
};
var error = exports.error = function error(value, next, rungen, yieldNext, raiseNext) {
if (!_is2.default.error(value)) return false;
raiseNext(value.error);
return true;
};
var object = exports.object = function object(value, next, rungen, yieldNext, raiseNext) {
if (!_is2.default.all(value) || !_is2.default.obj(value.value)) return false;
var result = {};
var keys = Object.keys(value.value);
var count = 0;
var hasError = false;
var gotResultSuccess = function gotResultSuccess(key, ret) {
if (hasError) return;
result[key] = ret;
count++;
if (count === keys.length) {
yieldNext(result);
}
};
var gotResultError = function gotResultError(key, error) {
if (hasError) return;
hasError = true;
raiseNext(error);
};
keys.map(function (key) {
rungen(value.value[key], function (ret) {
return gotResultSuccess(key, ret);
}, function (err) {
return gotResultError(key, err);
});
});
return true;
};
var array = exports.array = function array(value, next, rungen, yieldNext, raiseNext) {
if (!_is2.default.all(value) || !_is2.default.array(value.value)) return false;
var result = [];
var count = 0;
var hasError = false;
var gotResultSuccess = function gotResultSuccess(key, ret) {
if (hasError) return;
result[key] = ret;
count++;
if (count === value.value.length) {
yieldNext(result);
}
};
var gotResultError = function gotResultError(key, error) {
if (hasError) return;
hasError = true;
raiseNext(error);
};
value.value.map(function (v, key) {
rungen(v, function (ret) {
return gotResultSuccess(key, ret);
}, function (err) {
return gotResultError(key, err);
});
});
return true;
};
var iterator = exports.iterator = function iterator(value, next, rungen, yieldNext, raiseNext) {
if (!_is2.default.iterator(value)) return false;
rungen(value, next, raiseNext);
return true;
};
exports.default = [error, iterator, array, object, any];
/***/ }),
/***/ 412:
/***/ (function(module, exports, __webpack_require__) {
"use strict";
Object.defineProperty(exports, "__esModule", {
value: true
});
exports.race = exports.join = exports.fork = exports.promise = undefined;
var _is = __webpack_require__(139);
var _is2 = _interopRequireDefault(_is);
var _helpers = __webpack_require__(254);
var _dispatcher = __webpack_require__(413);
var _dispatcher2 = _interopRequireDefault(_dispatcher);
function _interopRequireDefault(obj) { return obj && obj.__esModule ? obj : { default: obj }; }
var promise = exports.promise = function promise(value, next, rungen, yieldNext, raiseNext) {
if (!_is2.default.promise(value)) return false;
value.then(next, raiseNext);
return true;
};
var forkedTasks = new Map();
var fork = exports.fork = function fork(value, next, rungen) {
if (!_is2.default.fork(value)) return false;
var task = Symbol('fork');
var dispatcher = (0, _dispatcher2.default)();
forkedTasks.set(task, dispatcher);
rungen(value.iterator.apply(null, value.args), function (result) {
return dispatcher.dispatch(result);
}, function (err) {
return dispatcher.dispatch((0, _helpers.error)(err));
});
var unsubscribe = dispatcher.subscribe(function () {
unsubscribe();
forkedTasks.delete(task);
});
next(task);
return true;
};
var join = exports.join = function join(value, next, rungen, yieldNext, raiseNext) {
if (!_is2.default.join(value)) return false;
var dispatcher = forkedTasks.get(value.task);
if (!dispatcher) {
raiseNext('join error : task not found');
} else {
(function () {
var unsubscribe = dispatcher.subscribe(function (result) {
unsubscribe();
next(result);
});
})();
}
return true;
};
var race = exports.race = function race(value, next, rungen, yieldNext, raiseNext) {
if (!_is2.default.race(value)) return false;
var finished = false;
var success = function success(result, k, v) {
if (finished) return;
finished = true;
result[k] = v;
next(result);
};
var fail = function fail(err) {
if (finished) return;
raiseNext(err);
};
if (_is2.default.array(value.competitors)) {
(function () {
var result = value.competitors.map(function () {
return false;
});
value.competitors.forEach(function (competitor, index) {
rungen(competitor, function (output) {
return success(result, index, output);
}, fail);
});
})();
} else {
(function () {
var result = Object.keys(value.competitors).reduce(function (p, c) {
p[c] = false;
return p;
}, {});
Object.keys(value.competitors).forEach(function (index) {
rungen(value.competitors[index], function (output) {
return success(result, index, output);
}, fail);
});
})();
}
return true;
};
var subscribe = function subscribe(value, next) {
if (!_is2.default.subscribe(value)) return false;
if (!_is2.default.channel(value.channel)) {
throw new Error('the first argument of "subscribe" must be a valid channel');
}
var unsubscribe = value.channel.subscribe(function (ret) {
unsubscribe && unsubscribe();
next(ret);
});
return true;
};
exports.default = [promise, fork, join, race, subscribe];
/***/ }),
/***/ 413:
/***/ (function(module, exports, __webpack_require__) {
"use strict";
Object.defineProperty(exports, "__esModule", {
value: true
});
var createDispatcher = function createDispatcher() {
var listeners = [];
return {
subscribe: function subscribe(listener) {
listeners.push(listener);
return function () {
listeners = listeners.filter(function (l) {
return l !== listener;
});
};
},
dispatch: function dispatch(action) {
listeners.slice().forEach(function (listener) {
return listener(action);
});
}
};
};
exports.default = createDispatcher;
/***/ }),
/***/ 414:
/***/ (function(module, exports, __webpack_require__) {
"use strict";
Object.defineProperty(exports, "__esModule", {
value: true
});
exports.cps = exports.call = undefined;
var _is = __webpack_require__(139);
var _is2 = _interopRequireDefault(_is);
function _interopRequireDefault(obj) { return obj && obj.__esModule ? obj : { default: obj }; }
function _toConsumableArray(arr) { if (Array.isArray(arr)) { for (var i = 0, arr2 = Array(arr.length); i < arr.length; i++) { arr2[i] = arr[i]; } return arr2; } else { return Array.from(arr); } }
var call = exports.call = function call(value, next, rungen, yieldNext, raiseNext) {
if (!_is2.default.call(value)) return false;
try {
next(value.func.apply(value.context, value.args));
} catch (err) {
raiseNext(err);
}
return true;
};
var cps = exports.cps = function cps(value, next, rungen, yieldNext, raiseNext) {
var _value$func;
if (!_is2.default.cps(value)) return false;
(_value$func = value.func).call.apply(_value$func, [null].concat(_toConsumableArray(value.args), [function (err, result) {
if (err) raiseNext(err);else next(result);
}]));
return true;
};
exports.default = [call, cps];
/***/ }),
/***/ 448:
/***/ (function(module, __webpack_exports__, __webpack_require__) {
"use strict";
__webpack_require__.r(__webpack_exports__);
// CONCATENATED MODULE: ./node_modules/@wordpress/redux-routine/build-module/is-generator.js
/**
* Returns true if the given object is a generator, or false otherwise.
*
* @see https://www.ecma-international.org/ecma-262/6.0/#sec-generator-objects
*
* @param {*} object Object to test.
*
* @return {boolean} Whether object is a generator.
*/
function isGenerator(object) {
// Check that iterator (next) and iterable (Symbol.iterator) interfaces are satisfied.
// These checks seem to be compatible with several generator helpers as well as the native implementation.
return !!object && typeof object[Symbol.iterator] === 'function' && typeof object.next === 'function';
}
// EXTERNAL MODULE: ./node_modules/rungen/dist/index.js
var dist = __webpack_require__(276);
// EXTERNAL MODULE: external {"this":"lodash"}
var external_this_lodash_ = __webpack_require__(2);
// EXTERNAL MODULE: ./node_modules/is-promise/index.js
var is_promise = __webpack_require__(129);
var is_promise_default = /*#__PURE__*/__webpack_require__.n(is_promise);
// CONCATENATED MODULE: ./node_modules/@wordpress/redux-routine/build-module/is-action.js
/**
* External dependencies
*/
/**
* Returns true if the given object quacks like an action.
*
* @param {*} object Object to test
*
* @return {boolean} Whether object is an action.
*/
function isAction(object) {
return Object(external_this_lodash_["isPlainObject"])(object) && Object(external_this_lodash_["isString"])(object.type);
}
/**
* Returns true if the given object quacks like an action and has a specific
* action type
*
* @param {*} object Object to test
* @param {string} expectedType The expected type for the action.
*
* @return {boolean} Whether object is an action and is of specific type.
*/
function isActionOfType(object, expectedType) {
return isAction(object) && object.type === expectedType;
}
// CONCATENATED MODULE: ./node_modules/@wordpress/redux-routine/build-module/runtime.js
/**
* External dependencies
*/
/**
* Internal dependencies
*/
/**
* Create a co-routine runtime.
*
* @param {Object} controls Object of control handlers.
* @param {Function} dispatch Unhandled action dispatch.
*
* @return {Function} co-routine runtime
*/
function createRuntime() {
var controls = arguments.length > 0 && arguments[0] !== undefined ? arguments[0] : {};
var dispatch = arguments.length > 1 ? arguments[1] : undefined;
var rungenControls = Object(external_this_lodash_["map"])(controls, function (control, actionType) {
return function (value, next, iterate, yieldNext, yieldError) {
if (!isActionOfType(value, actionType)) {
return false;
}
var routine = control(value);
if (is_promise_default()(routine)) {
// Async control routine awaits resolution.
routine.then(yieldNext, yieldError);
} else {
yieldNext(routine);
}
return true;
};
});
var unhandledActionControl = function unhandledActionControl(value, next) {
if (!isAction(value)) {
return false;
}
dispatch(value);
next();
return true;
};
rungenControls.push(unhandledActionControl);
var rungenRuntime = Object(dist["create"])(rungenControls);
return function (action) {
return new Promise(function (resolve, reject) {
return rungenRuntime(action, function (result) {
if (isAction(result)) {
dispatch(result);
}
resolve(result);
}, reject);
});
};
}
// CONCATENATED MODULE: ./node_modules/@wordpress/redux-routine/build-module/index.js
/* harmony export (binding) */ __webpack_require__.d(__webpack_exports__, "default", function() { return createMiddleware; });
/**
* Internal dependencies
*/
/**
* Creates a Redux middleware, given an object of controls where each key is an
* action type for which to act upon, the value a function which returns either
* a promise which is to resolve when evaluation of the action should continue,
* or a value. The value or resolved promise value is assigned on the return
* value of the yield assignment. If the control handler returns undefined, the
* execution is not continued.
*
* @param {Object} controls Object of control handlers.
*
* @return {Function} Co-routine runtime
*/
function createMiddleware() {
var controls = arguments.length > 0 && arguments[0] !== undefined ? arguments[0] : {};
return function (store) {
var runtime = createRuntime(controls, store.dispatch);
return function (next) {
return function (action) {
if (!isGenerator(action)) {
return next(action);
}
return runtime(action);
};
};
};
}
/***/ })
/******/ })["default"]; | {
"redpajama_set_name": "RedPajamaGithub"
} | 2,754 |
{"url":"https:\/\/socratic.org\/questions\/we-know-that-99-985-of-all-hydrogen-atoms-are-hydrogen-1-only-0-015-are-hydrogen","text":"# We know that 99.985% of all hydrogen atoms are hydrogen-1. Only 0.015% are hydrogen-2 atom What is the approximate atomic mass of hydrogen?\n\nJan 20, 2017\n\nThe approximate atomic mass of hydrogen is 1 u.\n\n#### Explanation:\n\nTo get the atomic mass of an element, we must take a weighted average of the atomic masses of its isotopes.\n\nThat is, we multiply the atomic mass of each isotope by its percentage expressed as a decimal fraction. Then we add them together.\n\nThe atomic mass of hydrogen-1 is 1 u and the atomic mass of hydrogen-2 is 2 u.\n\nThe atomic mass of hydrogen is then\n\n$\\text{0.999 85 \u00d7 1 u + 0.000 15 \u00d7 2 u\" = \"0.999 85 u + 0.000 30 u\" = \"1 u}$ (1 significant figure)","date":"2020-01-29 07:54:55","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 1, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.8466470837593079, \"perplexity\": 1047.8474936706302}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2020-05\/segments\/1579251789055.93\/warc\/CC-MAIN-20200129071944-20200129101944-00282.warc.gz\"}"} | null | null |
in: Boy Meets World
The Boy Meets World logo.
Boy Meets World is a 90's sitcom about a boy, Cory Matthews, and his journey to adulthood. With him are his best friend Shawn Hunter, his girlfriend Topanga Lawrence, his charming yet silly older brother Eric, and his teacher and next-door-neighbor George Feeny.
1 Timeslot
2 Syndication
3 Humor
Timeslot
Boy Meets World ran from September 24, 1993 to May 5, 2000. Initially, the show aired at 8:30 p.m. in ABC's TGIF lineup, but was moved to 9:30 p.m. in the middle of season 4 due to increasingly adult subject matter. (This was internally noted in the episode Shallow Boy, in which a character complains about his favorite show moving to 9:30.).
Currently, Disney-ABC Domestic Television (sister company to Touchstone Television, now ABC Studios) handles the syndication rights to the series.
Boy Meets World reruns began airing in off-network syndication on September 8, 1997 and continued until September 2000. After the show officially ended its run on ABC in 2000, Disney Channel acquired the rights to air the series, lasting from 2000–2007. The series began airing on the network again in 2014 (albeit sporadically) in honor of the spin-off series Girl Meets World; however to the consternation of some longtime fans of the show, many episodes aired on Disney Channel were edited for suggestive content deemed inappropriate for the channel's 7-14 year old intended target audience and at least three episodes were banned due to subject matter within the episode's plotline also deemed inappropriate. They each only aired once on the network and were never seen again.
ABC Family also aired the show from 2004 until August 2007, in a way inheriting the rights to the show from Disney Channel, but ABC Family did not air the Disney Channel versions of the episodes, instead it ran different syndicated prints which restored portions of scenes Disney Channel did not show during its run on that network (though small portions of certain scenes from episodes during the earlier seasons were cut due to time constraints), incorporated each season's corresponding opening title sequence and even used the original versions of the season one teaser scenes that featured the opening titles after the teaser, bringing them back in line with the original ABC telecasts (prior syndicated versions of season one episodes had the cast and creators' names shown during the teaser scene). It was announced on April 1, 2010 that ABC Family had re-obtained the rights to Boy Meets World, and began airing it at 7 a.m. (ET/PT) on weekdays beginning April 12, 2010; the series replaced Sister, Sister.
MTV2 also aired the show since 2011. Every episode is rated TV-PG (even episodes with a sub-rating). This show is also aired on TeenNick since 2016.
The humor of Boy Meets World developed with its main star. Initially, it was a bit childish, although entertained. Cory was a little sarcastic kid. As the show aged, the humor became more outlandish and farcical. Especially in season 7, physical humor is emphasized. Also, the show is not above cross-dressing for a laugh, which was the focus of two separate episodes (Chick Like Me and What a Drag!).
The humor is also notable due to a somewhat constant stream of parodies, including of itself, especially in Eric Hollywood. The movie Scream is parodied in And Then There Was Shawn, and the movie The Truman Show in And In Case I Don't See Ya.
Boy Meets World Season 1
Boy Meets World Wikipedia entry
Popular Fan Sites | {
"redpajama_set_name": "RedPajamaCommonCrawl"
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{"url":"https:\/\/www.suepke.net\/category\/linux-programming\/","text":"### Archive\n\nArchive for the \u2018Linux & Programming\u2019 Category\n\n## rsslibnotify \u2013 My very first python script\n\nUPDATE: I included tante\u2019s comments and extended the program. The config-file now allows you to define the checking interval and will be automatically created if nonexisting.\n\nI noticed that gwibber removed their rss support for some reason. However, it would be really nice to have the option for libnotify to \u2026well\u2026 notify you on gnome when there is something new on rss\/atom feeds you find important. Since I always wanted to learn python anyway, I hacked together a small script, which does exactly that. Just create a config file in ~\/.rsslibnotify with a number of rss\/atom feeds separated by newlines and you will be informed every ten minutes via libnotify if there\u2019s something new. (see update above)\n\nfeedparser and pynotify need to be installed prior. Btw. feedparser is giving me errors while installing, however seems to be working without a problem.\n\nOf course this script is far from perfect. Keep in mind it\u2019s my first python script and I worked at most one to two hours on it ;).\n\n#!\/usr\/bin\/python # coding=utf-8 \u00a0 # #\u00a0\u00a0\u00a0 Copyright 2010 by Daniel S\u00fcpke # #\u00a0\u00a0\u00a0 This program is free software: you can redistribute it and\/or modify #\u00a0\u00a0\u00a0 it under the terms of the GNU General Public License as published by #\u00a0\u00a0\u00a0 the Free Software Foundation, either version 3 of the License, or #\u00a0\u00a0\u00a0 (at your option) any later version. # #\u00a0\u00a0\u00a0 This program is distributed in the hope that it will be useful, #\u00a0\u00a0\u00a0 but WITHOUT ANY WARRANTY; without even the implied warranty of #\u00a0\u00a0\u00a0 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.\u00a0 See the #\u00a0\u00a0\u00a0 GNU General Public License for more details. # #\u00a0\u00a0\u00a0 You should have received a copy of the GNU General Public License #\u00a0\u00a0\u00a0 along with this program.\u00a0 If not, see <http:\/\/www.gnu.org\/licenses\/>. \u00a0 # # rsslibnotify v0.3 # \u00a0 import pynotify import feedparser import time import os.path import sys \u00a0 config_file = os.path.expanduser('~\/.rsslibnotify') feeds = [] titles = [] \u00a0 try: file = open(config_file, 'r') except: file = open(config_file, 'w') file.write('# amount of seconds to wait before checking for new items\\n') file.write('interval 600\\n\\n') file.write('# feeds to check\\n') file.write('feed https:\/\/suepke.net\/feed\/\\n') file.write('feed http:\/\/rss.golem.de\/rss.php?feed\\x3dRSS1.0\\x26ms\\x3drss\\n') \u00a0 print('Config file not found! Created example file at ' + config_file + '. Please edit.') sys.exit(0) \u00a0 for line in file.readlines(): config_line = line.rstrip().partition('#')[0] # Remove \\n and comments \u00a0 if config_line.startswith('feed '): feeds.append(config_line.partition('feed ')[2]) if config_line.startswith('interval '): interval = int(config_line.partition('interval ')[2]) \u00a0 titles.append('') \u00a0 file.close() \u00a0 if not feeds: print('There were no feeds in your config file at ' + config_file + '. Exiting...') sys.exit(0) \u00a0 if not interval: print('No check interval has been defined. Assuming ten minutes.') interval = 600 \u00a0 # Check for new feeds every interval seconds while True: i = 0 \u00a0 for feed in feeds: d = feedparser.parse(feed) \u00a0 # Assume new content if there is a new title if d.entries[0].title != titles[i]: titles[i] = d.entries[0].title \u00a0 if pynotify.init('rssnotify'): n = pynotify.Notification(d.feed.title, d.entries[0].title) n.show() else: print('there was a problem initializing the pynotify module') i += 1 \u00a0 time.sleep(interval)\nCategories: Tags:\n\n## Uni Oldenburg VPN with ubuntu 10.04\n\nFinally got the VPN of my university working. I will share how I got it working in case you study\/work here and want to use it, too.\n\nI found the VPN to be more stable when started from a shell:\nsudo apt-get install vpnc sudo nano \/etc\/vpnc\/default.conf\n\nEdit the config as following:\nIPSec gateway 10.9.8.7 IPSec ID student IPSec secret student #IKE Authmode hybrid Xauth username YOUR Xauth password YOUR\n\nThen just connect to the wlan and start the vpn with sudo vpnc.\n\nIf you want to use the network-manager, you need to install another package. However, I always get an error message about invalid secrets. If you have more luck, please let me know.\nsudo apt-get install network-manager-vpnc sudo service network-manager restart\n\nCategories: Tags:\n\n## Solving Sound Problems with Dosbox in Ubuntu 8.10 and 9.04\n\nHey guys, since I get a lot of hits to this article, I\u2019d really appreciate if you leave a short note when this helped you, so we can see if this is a general configuration problem. Thanks!\n\nSince ubuntu 8.10 I was having stuttering sound with dosbox. I just made a fresh install of 9.04 and still was experiencing such problems (pulseaudio ftw again\u2026). I searched the web, alas was unable to find a solution that was actually working. After trying a couple of things, I finally seem to have found a solution which is working (at least for me): I set the mixing rate within dosbox to match that of pulseaudio. To make it short, change your conf in ~\/.dosbox\/ to\n\n[mixer]\n# nosound: Enable silent mode, sound is still emulated though.\n# rate: Mixer sample rate, setting any device's rate higher than this will probably lower their sound quality.\n# Possible values: 22050, 44100, 48000, 32000, 16000, 11025, 8000, 49716.\n# blocksize: Mixer block size, larger blocks might help sound stuttering but sound will also be more lagged.\n# Possible values: 2048, 4096, 8192, 1024, 512, 256.\n# prebuffer: How many milliseconds of data to keep on top of the blocksize.\nnosound=false\n#rate=22050\nrate=44100 # Change this line\nblocksize=2048\nprebuffer=10\n\nAlso, to get rid of the initial midi warning I changed the following, though it may not be necessary:\n\n[midi]\n# mpu401: Type of MPU-401 to emulate.\n# Possible values: intelligent, uart, none.\n# mididevice: Device that will receive the MIDI data from MPU-401.\n# Possible values: default, win32, alsa, oss, coreaudio, coremidi, none.\n# midiconfig: Special configuration options for the device driver. This is usually the id of the device you want to use. See README for details.\nmpu401=intelligent\nmididevice=default\nmidiconfig=128:0 # Change this line\n\nYet another pulseaudio problem solved.\n\nUpdate: Also, setting output to opengl helps a lot, at least if you have compiz enabled.\n\nCategories: Tags:\n\n## Enabling Automatic Upgrades for Ubuntu\n\nUbuntu allows you to install security updates automatically in the background (under Synaptic -> Repositories -> Updates), but there was only the option Install Security updates without notification, no option for just installing all updates. I searched for a way to automate it and found the solution in launchpad:\n\nJust open \/etc\/apt\/apt.conf.d\/50unattended-upgrades and enable all updates, then you can have daily automatic upgrades without notification. No more need to confirm all the same upgrades on different computers again and again, neat!\n\nCategories: Tags:\n\n## Conditional Commenting in LaTeX\n\nIn my Ph.D. thesis are quite a lot of comments\/reminders for myself. I defined two commands for this, one for a red footnote and one for red bold text. But if someone else is to read parts of the document, he shouldn\u2019t necessarily see all my comments, so I was searching for a method to suppress all comments with a single flag. I found the package ifthen and it\u2019s exactly what I was looking for. If you need to use comments for yourself in a document you\u2019re writing, consider using the following:\n\nCategories: Tags:\n\n## Embedding Windows in Linux with VirtualBox\n\nI am obviously a big fan of Linux and use it at home as well as in my office. But in both places I still require Windows for two reason:\n\n\u2022 At home I require it for gaming. Wine is ok, but can still be a major pain in the ass with configuration and lesser known games.\n\u2022 In the office I am obliged to use Microsoft Office. I prefer LaTeX, but compared to OpenOffice the 2007 version of MsOffice is the big winner in my opinion.\u00a0 But since working under Windows otherwise is out of question for me (I am way more productive with Linux\/Gnome), a dual-boot like I use for gaming is no option.\n\nSo, in my office, I use VirtualBox to embed Windows inside my Linux desktop. Formerly I used VMware, which admittedly is more powerful, but provides a less appealing user interface and, at that time, did not provide the seamless mode, which I will make use of. To give an impression, here is a screenshot of my desktop:\n\nUsing VirtualBox to embed Windows inside Gnome\n\nI will not go through the installation process, as it\u2019s pretty self-explanatory. But what VirtualBox additionally offers is the aforementioned seamless mode (default combination: Right-CTRL + L), which gives you the taskbar over your normal gnome panel. This is really nice, since it allows easy access and switching between Windows and Linux applications. While Alt-Tab is not used optimally (it only lets you access applications of one OS at a time), you can switch between Windows and Linux applications by pressing Right-CTRL before using Alt-Tab.\n\nAnother great feature is embedding any host folder (e.g. your home folder) directly inside the Windows explorer as seen on the screenshot. This is achieved by setting a shared folder in Devices->Shared Folders and then adding a network drive. This is as simple as right-clicking on My Computer (unsure about the name, I only have German Windows \ud83d\ude09 ) and adding the drive. From now it\u2019s possible to modify and share files between host and virtual machine without any hassle.\n\nAlso, I configured cups to allow web access to the printers configured in Linux, so the Windows VM could easily print by using these. But I haven\u2019t configured this yet since my last installation, so I cannot give a howto right now \ud83d\ude09 .\n\nI hope you enjoy the progress of free VMs like VirtualBox as much as I do, as it allows Windows apps, that are sadly still required, to be used inside your Linux without a hassle and with nice integration. I hope that Alt-Tab is improved, then the whole thing would feel like a natural part of your Linux desktop. Additionally, VirtualBox now has Direct3D support (which I haven\u2019t tested yet), so maybe one day we will be able to even do gaming inside our Linux distributions without any more need for a dual boot.\n\n### Update:\n\nFor all this, you need to install the Guest additions. You can find these in the menu devices.\n\nCategories: Tags:\n\n## A Java Wizard Toolkit\n\nWell, here goes my first post. This is a toolkit I started developing back in 2006. I am sure there are other, better implementations and I did not develop this small project any further, but I think it\u2019s also a good example on how to program in OO (though I am sure there is much room for improvement, any flames appreciated \ud83d\ude09 ). So, here it goes:\n\n### Introduction\n\nThe Wizard Toolkit is a java framework \/ toolkit for the easy use and creation of wizards. This toolkit lets you focus on defining and validating input fields while the control flow and data storage is automated by the surrounding toolkit.\n\nWizards are a useful way to gather information for different purposes. They interact in form of dialogs, allowing step-by-step data input. Wizards can be used to gather information, for creating specific output or for setting preferences.\n\n### Tutorial\n\nThe following example serves as a step-by-step tutorial demonstrating the possibilities provided by the toolkit and how to implement a custom wizard based on it. It will consist of three parts, which will be found in most wizards:\n\n\u2022 A welcome dialog\n\u2022 Some input forms\n\u2022 A final dialog for checking \/ presenting all input\n\nThe toolkit provides two classes for implementing a wizard:\n\n\u2022 Wizard: The main class of any wizard, extending JDialog. Manages the control flow and displays the panels provided. After finishing the input process, it contains the data entered.\n\u2022 WizardPanel: An extension of JPanel. Every WizardPanel stands for one step in the dialog process. It may contain everything a JPanel can contain, usually text or input fields.\n\n#### The welcome dialog\n\nThe first step in a dialog system should always be a welcoming screen, describing what needs to be done in the next steps. This way the user can get an overview of the data required or the process he is about to start. So we create a first class extending WizardPanel:\n\nNote: All following sources are excerpts from the example files. For copyright issues look in the according java files.\n\npublic class Panel1 extends WizardPanel { public Panel1() { super(\u201cWelcome\u201d); add(new JLabel(\u201cWelcome to the Wizard toolkit demonstration!\u201d)); } }\n\nThe title Welcome is set in the constructor. It is used for the header and for the menu. You might add further welcome information or for example add licence information here.\n\nTheoretically one WizardPanel is sufficient to create the Wizard class, but in the next step we will add the skeleton of the other two panels and then create the Wizard.\n\n#### Starting the Wizard\n\npublic class Panel2 extends WizardPanel { public Panel2() { super(\"Options\"); } }\npublic class Panel3 extends WizardPanel { public Panel3() { super(\u201cFinish\u201d); } }\npublic class WizardExample { public static void main(String[] args) { WizardPanel[] panels = new WizardPanel[] { new Panel1(), new Panel2(), new Panel3() }; Wizard wizard = new Wizard((JFrame) null, \u201cWizard Example\u201d, panels); wizard.setVisible (true); System.out.println(wizard.getProperties()); } }\n\nThe Wizard class needs three arguments: A parent dialog or frame (may be null), a title and an array with WizardPanels to display. The constructor automatically initializes everything needed for the process and starts after being set visible. When the user finished using the wizard, all data may be gathered via getProperties. With the WizardPanels implemented so far, calling the constructor will open the following dialog:\n\nWelcoming Screen of the Wizard\n\nMenu, header and buttons are created and managed by the toolkit. Only the main panel \/ content panel needs to be implemented. Back and Finish cannot be selected, as there is no previous WizardPanel and this is not the last WizardPanel. The buttons behaviour may be modified. Selecting Next will display the next dialog, which will be further implemented now.\n\n#### The options dialog\n\nAfter the first dialog, the data input forms begin. In this tutorial a WizardPanel with some sample input fields will be implemented, including verification and storage of the data entered. At first, some input components will be added in the panel\u2019s constructor:\n\npublic class Panel2 extends WizardPanel { private JCheckBox box1, box2; private JRadioButton rbutton1, rbutton2; private JTextField textField; \u00a0 public Panel2() { super(\"Options\"); \u00a0 box1 = new JCheckBox(\"Some option\"); box2 = new JCheckBox( \"If this radio button is selected, you cannot select \"next\"\"); rbutton1 = new JRadioButton(\"Select me\"); rbutton2 = new JRadioButton( \"If this button is selected, clicking \"next\" will open a Dialog Box.\"); textField = new JTextField(30); \u00a0 rbutton1.setSelected(true); \u00a0 setLayout(new GridBagLayout()); \u00a0 GridBagConstraints gc = new GridBagConstraints(); ButtonGroup rbuttons = new ButtonGroup(); \u00a0 rbuttons.add(rbutton1); rbuttons.add(rbutton2); \u00a0 gc.gridx = gc.gridy = 0; gc.anchor = GridBagConstraints.NORTHWEST; add(new JLabel(\"Checkbox options:\"), gc); \u00a0 gc.gridy++; add(box1, gc); \u00a0 \/\/ Add other Components \/\/ ... }\n\nThe toolkit enables the Back-Button as there is a previous dialog. But we do not want the user to go back, as there is only a welcome screen. On the other hand, we want to disable the forward button if box2 is selected (as described in its text). The toolkit\u2019s control over the buttons may be overridden to implement panel-specific behaviour:\n\n protected boolean backButtonEnabled() { return false; } \u00a0 protected boolean nextButtonEnabled() { return !box2.isSelected(); }\n\nThe Wizard toolkit must be specifically told when to refresh the buttons. Therefore it is necessary to add an ActionListener to the component and tell it to refresh the buttons upon selection:\n\npublic class Panel2 extends WizardPanel implements ActionListener { \/\/ ... \u00a0 public Panel2() { \/\/ ... \u00a0 \/\/ Refresh buttons upon selecting \/ deselecting this checkbox box2.addActionListener(this); box2.setActionCommand(\"box1\"); \u00a0 \/\/ ... } \u00a0 public void actionPerformed(ActionEvent e) { refreshButtons(); } }\n\nIf the second radio button is selected, a confirmation box will be opened upon clicking Next. This can be done via verification of the data input. To use this method, all data must be made available to the toolkit. Every information that needs to be stored must be added to a Property object. This data is gathered by the toolkit.\n\n protected Properties getProperties() { Properties properties = new Properties(); \u00a0 properties.setProperty(\"box1\", String.valueOf(box1.isSelected())); properties.setProperty(\"box2\", String.valueOf(box2.isSelected())); properties.setProperty(\"rbutton1\", String .valueOf(rbutton1.isSelected())); properties.setProperty(\"rbutton2\", String .valueOf(rbutton2.isSelected())); properties.setProperty(\"textfield\", textField.getText()); \u00a0 return properties; } \u00a0 protected boolean verifyChanges(Properties properties) { if (Boolean.parseBoolean(properties.getProperty(\"rbutton2\"))) { return JOptionPane .showConfirmDialog( this, \"Are you sure you want to continue with radio button 2 selected?\", \"Please confirm\", JOptionPane.YES_NO_OPTION) == JOptionPane.YES_OPTION; } \u00a0 return ALLOW; }\n\nEvery WizardPanel which gathers information must provide the getProperties method to store its data. Every WizardPanel that needs verification must provide the verifyChanges method. This method either returns ALLOW (true) or DENY (false). If DENY is returned, the toolkit will do nothing upon selecting Next or Finish. Note: The properties given as parameter here are not only the properties of this WizardPanel but the properties of all Panels. This way, data can be verified over multiple steps.\n\nSecond Dialog of the Wizard\n\nConfirmation before Advancing to the next Step\n\n#### The finish dialog\n\nOn the last dialog we want to display all information entered by the user so far. Therefore we cannot use the constructor for initializing the components, as we do not know what the user has entered previously. Even worse, the user may use the back button and change the data. So we need a way to update all components upon displaying the WizardPanel. This is done via setComponents, which is called upon every display of the WizardPanel and is provided all properties so far:\n\npublic class Panel3 extends WizardPanel { public Panel3() { super(\"Finish\"); \u00a0 setLayout(new GridBagLayout()); } \u00a0 protected void setComponents(Properties properties) { removeAll(); \u00a0 GridBagConstraints gc = new GridBagConstraints(); \u00a0 gc.gridx = gc.gridy = 0; add(new JLabel(\"All required information have been entered.\"), gc); \u00a0 gc.gridy++; add(new JLabel(\" \"), gc); \u00a0 gc.anchor = GridBagConstraints.NORTHWEST; gc.gridy++; add(new JLabel(\"Boxes selected:\"), gc); \u00a0 if (Boolean.valueOf(properties.getProperty(\"box1\"))) { gc.gridy++; add(new JLabel(\"Checkbox 1\"), gc); } \u00a0 if (Boolean.valueOf(properties.getProperty(\"box2\"))) { gc.gridy++; add(new JLabel(\"Checkbox 2\"), gc); } \u00a0 if (Boolean.valueOf(properties.getProperty(\"rbutton1\"))) { gc.gridy++; add(new JLabel(\"Radio button 1\"), gc); } \u00a0 if (Boolean.valueOf(properties.getProperty(\"rbutton2\"))) { gc.gridy++; add(new JLabel(\"Radio button 2\"), gc); } \u00a0 gc.gridy++; add(new JLabel(\" \"), gc); \u00a0 gc.gridy++; add(new JLabel(\"Text entered: \"), gc); \u00a0 gc.gridy++; add(new JLabel(properties.getProperty(\"textfield\")), gc); } }\n\nFinal Dialog of the Wizard, showing Summary\n\n### Conclusion\n\nWith this step, all implementation for the example is finished, showing all possibilities this framework provides. You may verify data on the fly or after selecting buttons and override the button behaviour. The toolkit lets you specify custom behaviour when necessary and provides standard control flow when applicable, supporting a very easy and structured way to implement Wizards.\n\n### Source code (SVN)\n\nSorry, this outdated information, if you need the source code, download it from the section below.\n\nThis software is developed as open source under the general public licence. You may freely use or modify this software in terms specified by the GPL. The source code is accessible via Subversion with the repository lying at:\n\nfeanorscurse.dyndns.org\/home\/svn\/toolkit\n\nPlease inform me before publishing any code changes.\n\nThe current stable version is 0.2. You may download the toolkit as jar-files or access the source. If you are planning to use this toolkit, it would be nice to leave a comment \ud83d\ude42 .\n\nWizard Toolkit v0.2\n\nWizard Toolkit v0.2 with example application\n\nCategories: Tags:","date":"2019-01-23 11:09:03","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.2534245252609253, \"perplexity\": 6374.144623396252}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2019-04\/segments\/1547584331733.89\/warc\/CC-MAIN-20190123105843-20190123131843-00430.warc.gz\"}"} | null | null |
\section{Introduction}
The structure of neutron-rich nuclei is a central theme of study in the field of nuclear structure. Of particular interest is the quest to understand the evolution of shell-structure and collectivity with isospin.
The emergence of the Islands of Inversion at $N$=8, 20, and 40 are prime examples of such evolution and have provided strong evidence
regarding the important role played by the neutron-proton force~\cite{Otsuka20, Sor08, Poves, Ots01, Heyde1, Warburton90}
Another intriguing and dramatic impact of the action of the neutron-proton force is seen in the so-called oxygen neutron-dripline anomaly, at $N=16$, which is extended to $N=22$ in for the F isotopes
with just the addition of one $ d_{5/2}$ proton. In a recent work~\cite{tang20}, the structure of $^{25}$F has been investigated
via $(p,2p)$ quasi-free knockout experiments with exclusive
measurements using a $^{25}$F beam at RIBF/RIKEN. %
The analysis of measured cross-sections
and derived spectroscopic factors may imply that the core of $^{25}$F
consists of $\sim$ 35\% $^{24}$O$_{gs}$ and $\sim$ 65\%
excited $^{24}$O. As discussed by the authors, their results suggest that the addition of the $0d_{5/2}$ proton considerably changes
the neutron structure in $^{25}$F from that in $^{24}$O, and calls for a revision to the $np$ tensor interaction in the
widely used effective interactions, which appears to be too weak to
reproduce the observations. In contrast, studies of neutron decay from unbound excited states in $^{24}$O~\cite{Calem09} and one-neutron removal from $^{24}$O~\cite{Ritu09} were indicative of a $N$=16 shell closure and the doubly-magic nature of this nucleus. The relatively high excitation energy $E_x = 4.7 \pm 0.1$ and the small $B(E2) \approx 1/2$ WU (Weisskopf units) of the $2^+_1$ state~\cite{Tshoo12} has further supported this interpretation.
In this article we follow up on our earlier work~\cite{aom} and interpret the above results in terms of
a collective picture, within the framework of the Particle-Rotor Model (PRM)~\cite{Larsson, Rag} to provide further insight into the nature of the effective $^{24}$O core in $^{25}$F.
\section{The Structure of $^{25}$F}
The structure of odd-A nuclei usually
offers fingerprints that can disentangle the competition of
single-particle and collective degrees of freedom if they
can be regarded, at least a priori, as one
nucleon coupled to a core. Considering $^{24}$O as our core, an inspection of the Nilsson diagram~\cite{Sven} in Fig.~\ref{fig-1} suggests that the odd proton will occupy the single-$j$ multiplet
originating from the $d_{5/2}$ orbit, namely the levels $[220]\frac{1}{2}$, $[211]\frac{3}{2}$, and $[202]\frac{5}{2}$, with its Fermi energy at the $\Omega=\frac{1}{2}$, as indicated by the wavy line in Fig.~\ref{fig-1}\\
\begin{figure}
\centering
\includegraphics[width=9cm,angle=0]{Fig1rev.pdf}
\caption{Nilsson levels relevant for the structure of positive parity proton states in $^{25}$F, with the red dashed-lines
representing the single-$j$ approximation of the $d_{5/2}$ multiplet. The shaded area indicates the anticipated $\epsilon_2$ deformation and the wavy line the Fermi level of the odd proton. Energies are in units of the harmonic oscillator frequency, $\hbar\omega_0$. }
\label{fig-1}
\end{figure}
The effects of rotation on the single-particle motion are well understood, and the Particle Rotor Model (PRM) has been very successful in explaining the observed near-{\sl yrast} structures in deformed nuclei ~\cite{Frank2}.
The PRM Hamiltonian can be written as~\cite{Larsson,Rag}:
\begin{equation}
H= H_p + \frac{\hbar^{2}}{2\mathscr{I}}{\vec{R}^2}
\label{eq:eq1}
\end{equation}
where $H_p$ is the Nilsson Hamiltonian~\cite{Sven} for the particle in the absence
of rotation, $ \mathscr{I}$ and $\vec{R} $ the moment of inertia and the angular momentum of the core respectively.
Replacing $\vec{R} = \vec{I} -\vec{j}$ in Eq.~\ref{eq:eq1} gives the usual expression:
\begin{equation}
H= E_\Omega + \frac{\hbar^{2}}{2\mathscr{I}}I(I+1) + H_c
\label{eq:eq2}
\end{equation}
\noindent
where $E_\Omega$ are the intrinsic level energies and $H_c$ is the Coriolis coupling term
\begin{equation}
H_c= - \frac{\hbar^{2}}{2\mathscr{I} }(I_+j_-+I_-j_+)
\label{eq:eq3}
\end{equation}
where $I_\pm$ and $j_\pm$ are the ladder operators for the total and single particle angular momenta, respectively. This coupling is particularly important for small deformations and large $j$, and increases with the rotational frequency, $\omega_{rot}$. The Coriolis $K$-mixing gives rise to a wave-function of the general form
\begin{equation}
\psi_I = \sum_{K} \mathcal{A}_{K} | I K \rangle
\label{eq:eq4}
\end{equation}
\begin{figure}
\centering
\includegraphics[width=8cm]{Fig2b.pdf}
\caption{Schematic representation of the strongly coupled and decoupled limits of the PRM. The latter is used here in our description of the $^{25}$F.
The symmetry axis is labeled $\hat{3}$. Collective rotation takes place around a perpendicular axis ($\hat{1}$, $\hat{2)}$. Figure adapted from Ref.~\cite{Frank2}.}
\label{fig-2}
\end{figure}
\noindent
The ratio of the Coriolis matrix elements in Eq.~\ref{eq:eq3} ($H_c \sim\hbar^2Ij/\mathscr{I} \sim j \hbar\omega_{rot} $) to the intrinsic level spacings
($\Delta E \sim \epsilon_2 \hbar\omega_0$) serves as a control parameter defining the characteristics of the coupling between collective and intrinsic angular momenta.
For $H_c/\Delta E \ll 1$, the particle remains strongly coupled to the core maintaining the projection of its angular momentum on the symmetry axis, $\Omega$, as a good quantum number.
When $H_c/\Delta E \gg 1$, a rotation-aligned coupling limit is anticipated~\cite{Frank1,Frank2}. In this case, the {\sl yrast} band has spins $I=j,~j+2,~j+4,~...$, and the energy spacings equal that of the core; this type of band is referred to as a decoupled band. The two limiting cases are illustrated in Fig.~\ref{fig-2}.
\section{Results}
\subsection{Level energies}
\begin{figure}
\centering
\includegraphics[width=6cm]{LevelSchemePRM.pdf}
\caption{ Left: the experimental level scheme of $^{25}$F from Ref.~\cite{Vajta14}. Right: Results of the PRM calculations. Energies are in keV.}
\label{fig-3}
\end{figure}
The experimental level scheme of $^{25}$F~\cite{Vajta14}, shown in Fig.~\ref{fig-3} (left side), exhibits an interesting pattern having the first two $yrast$ states with spins $5/2_1^+$ and $9/2_1^+$ and a conspicuous
$1/2_1^+$ state in between. In analogy with our interpretation in Ref.~\cite{aom} of the structure of $^{29}$F~\cite{Pieter}, the $yrast$ states can be associated with members of the decoupled band based on the $d_{5/2}$ multiplet
and for which we have $(j \omega_{rot})/( \epsilon_2\omega_0) > 1$. The $1/2_1^+$ must have anti-parallel coupling of $\vec{j}$ with the core rotation, $\vec{R}$. It follows that in the decoupled limit ($\epsilon_2 \rightarrow 0$) the energy of the $1/2^+_1$ state with respect to the
ground state is proportional to the rotational energy of the core, $E_{2^+}(core)$.
Together with the $9/2_1^+$ state they provide a proxy for the $2^+$ energy of the effective $^{24}$O core in $^{25}$F. Adjusting to the energies of the $1/2_1^+$ and $9/2_1^+$ states gives $E_{2^+}(core)\approx $ 3.2 MeV, in line with a modest quadruple deformation, $\epsilon_2 \approx 0.15$, and consistent with the conditions required for the appearance of a decoupled band.
The results obtained of the PRM calculations, shown also in Fig.~\ref{fig-3} (right side) are in good agreement with the experimental data and give support to the rotational model description. Furthermore, in the rotation-aligned coupling limit the amplitudes $\mathcal{A}_{K}$ entering in Eq.~\ref{eq:eq4} are given by the Wigner $\mathbb{d}$-function evaluated at $\pi/2$, the angle between the symmetry ($\hat{3}$) and rotation axes ($\hat{1},\hat{2}$,)~\cite{Frank2}:
\begin{equation}
\mathcal{A}_{K} \approx \mathbb{d}^{5/2}_{5/2,K}(\pi/2)
\label{eq:eq5}
\end{equation}
\noindent
In $^{26}$F~\cite{Lepa13}, the $1^+$ ground and $4^+$ isomeric states
can be associated with the anti-parallel and parallel couplings of the odd-neutron, in the $d_{3/2}$ Nilsson multiplet, to the structure of $^{25}$F. The former, favored by the Gallagher-Moszkowski rule~\cite{GM58}, gives a $1^+$ as the lowest state and the latter a $4^+$ as the bandhead of a doubly-decoupled band.
\subsection{Spectroscopic factors}
We now proceed with the calculation of spectroscopic factors for the $(-1p)$ knockout reaction and compare them to those reported in Ref.~\cite{tang20}.
Following the formalism discussed in Ref.~\cite{Elbek}, which we recently applied to a similar case in $^{18,19}$F~\cite{aom2}, we obtain the expression:
\begin{equation}
S_{i, f} (j\ell) = \big( \sum_{K} \mathcal{A}_{K} \theta_{i, f}(j\ell,K) \big)^2
\label{eq:eq6}
\end{equation}
\begin{equation}
\begin{split}
\theta_{i, f }(j\ell,K)& = \sqrt2 \langle I_{i}j K \Omega_\pi | I_{f} 0\rangle C_{j,\ell} \langle\phi_f|\phi_i\rangle
\end{split}
\label{eq:eq7}
\end{equation}
where $ \mathcal{A}_{K}$ are given in Eq.~\ref{eq:eq5}, $\langle | \rangle$ is a Clebsch-Gordan coefficient and $\langle\phi_f|\phi_i\rangle$ represents the core overlap between the initial and final states, typically assumed to be 1. Since we are considering a single-$j$ approximation for
$d_{5/2}$ Nilsson multiplet, the amplitudes $C_{j,\ell}$ are equal to 1. Special care should be taken to assure consistency between the relative phases of the $\mathcal{A}_{K}$ Coriolis-mixed amplitudes and the Clebsch-Gordan coefficients entering in the sum.
In Table~\ref{table1}, the spectroscopic factors are compared to the measurement reported in Ref.~\cite{tang20}. The PRM is able to explain the level scheme of $^{25}$F but predicts a small fragmentation of the $d_{5/2}$ proton strength, with $\approx$ 15\% going to the $2^+$ and $4^+$ of $^{24}$O (PRM1). However, the premise of a substantial difference between the initial and final cores requires that the overlap in Eq.~\ref{eq:eq7} should be considered explicitly.
\begin{table}[ht]
\centering
\caption{ Comparison of the measured spectroscopic factors to the PRM results, with (PRM1) and without core overlap (PRM2). Also shown are shell-model calculations using the SDPF-MU interaction.}
\bigskip
\begin{tabular}{c|c|cccc}
\hline\hline
Final State & $S_{exp}$ & && $S_{th}$ & \\
in $^{24}$O & Ref.~\cite{tang20} & &PRM1& PRM2 & SDPF-MU \\
\hline
Ground &~0.36(13)~ & & 0.85 & 0.56& 0.95 \\
Excited &~0.65(25)~ & & 0.15 & 0.44& 0.05\\
\hline\hline
\end{tabular}
\label{table1}
\end{table}
We use the method described in Refs.~\cite{TNA1, TNA2} and obtain\footnote[1]{ A simple volume overlap gives $ \langle\phi_f|\phi_i\rangle \approx \frac{1}{1+ \epsilon_2+ \frac{2}{3}\epsilon_2^2 +~ ...} \sim0.85$} $\langle\phi_f|\phi_i\rangle \approx $ 0.81 bringing the PRM result closer to the observations (PRM2).
For reference we also include the shell-model results using the SDPF-MU interaction given in~\cite{tang20}. Note, if the authors of Ref.~\cite{tang20} had corrected the gs to gs spectroscopic factor by a quenching factor $\sim$ 0.6, usually observed in $(p,2p)$ reactions~\cite{leila}, the agreement would have been excellent.
Obviously, additional studies of proton addition and removal reactions will be of interest, specifically proton knockout or $(d,^3He)$ from $^{26}$Ne come to mind. Here, anticipating that $\langle\phi_f|\phi_i\rangle \sim 1$, we predict an spectroscopic factor for the $0^+$ to $5/2^+$ transition $S_{if} \approx 1.25$.
\section{Conclusion}
The rotational model is able to describe the structure of $^{25}$F as arising from the coupling of a proton $d_{5/2}$ Nilsson multiplet
to an effective core of modest deformation, $\epsilon_2 \sim 0.15$. These conditions anticipate that the development of a decoupled band
should be favorable and indeed, PRM calculations show that the rotation aligned coupling scheme is in agreement
with the observed levels. Using the formalism developed for studies of single-nucleon transfer reactions in deformed nuclei,
we calculated the proton spectroscopic factors for the $^{25}$F($5/2^+) (-1p) ^{24}$O reaction. Agreement with the experimental
data~\cite{tang20} is obtained by the fragmentation of the $d_{5/2}$ strength due to both deformation and a core overlap.
The Nilsson plus PRM picture suggests that the extra proton, with a dominant component in the down-sloping $\frac{1}{2}[220]$ level, polarizes $^{24}$O and stabilizes its dynamic deformation.
Thus, the effective
core in $^{25}$F can be interpreted as a slightly deformed rotor with $E_{2^+}(core)\approx$ 3.2 MeV and $\epsilon_2 \approx0.15$,
compared to the real doubly-magic $^{24}$O with $E_{2^+}\approx$ 4.7 MeV and weak vibrational quadrupole collectivity.
Furthermore, electromagnetic observables for the three lowest experimental levels obtained in the PRM (Table~\ref{Table 2}),
suggest that measurements of the magnetic and quadrupole moments of the $5/2^+$ state as well as a Coulomb excitation measurement of the transition probabilities, will definitely shed further light on the validity of our interpretation.
\begin{table}[ht]
\centering
\caption{ Electromagnetic properties of the low-lying levels of $^{25}$F in the PRM. Magnetic moments have been calculated using $g_R=Z/A$ and $g_s=0.7 (g_s)_{free}$.}
\bigskip
\begin{tabular}{c|c|c|c|c}
\hline\hline
$I^\pi$ & $E_{x}$ & $\mu$ & $Q$ & $B(E2;I \rightarrow \frac{5}{2}^+$)\\
& ~ [MeV] ~ &~ [$\mu_N$] ~ & $ ~[efm^2]$ ~ & [WU]\\
\hline
$\frac{5}{2}^+$& 0 & 3.9 & -4.5 & ---\\
$\frac{1}{2}^+$& 1.4 & 1.9 & 0 &3.9\\
$\frac{9}{2}^+$& 3.1 & 4.6 & -7.8 &1.9\\
\hline\hline
\end{tabular}
\label{Table 2}
\end{table}
\begin{acknowledgments}
This material is based upon work supported by the U.S. Department of Energy, Office of Science, Office of Nuclear Physics under Contract No. DE-AC02-05CH11231. We would like to thank K.~Wimmer and R.~Kanungo for their comments on the manuscript.
\bigskip
\bigskip
\end{acknowledgments}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 2,388 |
package org.apache.giraph.graph.partition;
import java.util.Collection;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;
import org.apache.giraph.graph.VertexEdgeCount;
import org.apache.giraph.graph.WorkerInfo;
import org.apache.log4j.Logger;
import com.google.common.collect.Lists;
import com.google.common.collect.Maps;
/**
* Helper class for {@link Partition} related operations.
*/
public class PartitionUtils {
/** Class logger */
private static Logger LOG = Logger.getLogger(PartitionUtils.class);
public static int workerCount;
private static class EdgeCountComparator implements
Comparator<Entry<WorkerInfo, VertexEdgeCount>> {
@Override
public int compare(Entry<WorkerInfo, VertexEdgeCount> worker1,
Entry<WorkerInfo, VertexEdgeCount> worker2) {
return (int) (worker1.getValue().getEdgeCount() -
worker2.getValue().getEdgeCount());
}
}
private static class VertexCountComparator implements
Comparator<Entry<WorkerInfo, VertexEdgeCount>> {
@Override
public int compare(Entry<WorkerInfo, VertexEdgeCount> worker1,
Entry<WorkerInfo, VertexEdgeCount> worker2) {
return (int) (worker1.getValue().getEdgeCount() -
worker2.getValue().getEdgeCount());
}
}
/**
* Check for imbalances on a per worker basis, by calculating the
* mean, high and low workers by edges and vertices.
*/
public static void analyzePartitionStats(
Collection<PartitionOwner> partitionOwnerList,
List<PartitionStats> allPartitionStats) {
Map<Integer, PartitionOwner> idOwnerMap =
new HashMap<Integer, PartitionOwner>();
for (PartitionOwner partitionOwner : partitionOwnerList) {
if (idOwnerMap.put(partitionOwner.getPartitionId(),
partitionOwner) != null) {
throw new IllegalStateException(
"analyzePartitionStats: Duplicate partition " +
partitionOwner);
}
}
Map<WorkerInfo, VertexEdgeCount> workerStatsMap = Maps.newHashMap();
VertexEdgeCount totalVertexEdgeCount = new VertexEdgeCount();
for (PartitionStats partitionStats : allPartitionStats) {
WorkerInfo workerInfo =
idOwnerMap.get(partitionStats.getPartitionId()).getWorkerInfo();
VertexEdgeCount vertexEdgeCount =
workerStatsMap.get(workerInfo);
if (vertexEdgeCount == null) {
workerStatsMap.put(
workerInfo,
new VertexEdgeCount(partitionStats.getVertexCount(),
partitionStats.getEdgeCount()));
} else {
workerStatsMap.put(
workerInfo,
vertexEdgeCount.incrVertexEdgeCount(
partitionStats.getVertexCount(),
partitionStats.getEdgeCount()));
}
totalVertexEdgeCount =
totalVertexEdgeCount.incrVertexEdgeCount(
partitionStats.getVertexCount(),
partitionStats.getEdgeCount());
}
List<Entry<WorkerInfo, VertexEdgeCount>> workerEntryList =
Lists.newArrayList(workerStatsMap.entrySet());
if (LOG.isInfoEnabled()) {
Collections.sort(workerEntryList, new VertexCountComparator());
LOG.info("analyzePartitionStats: Vertices - Mean: " +
(totalVertexEdgeCount.getVertexCount() /
workerStatsMap.size()) +
", Min: " +
workerEntryList.get(0).getKey() + " - " +
workerEntryList.get(0).getValue().getVertexCount() +
", Max: "+
workerEntryList.get(workerEntryList.size() - 1).getKey() +
" - " +
workerEntryList.get(workerEntryList.size() - 1).
getValue().getVertexCount());
Collections.sort(workerEntryList, new EdgeCountComparator());
LOG.info("analyzePartitionStats: Edges - Mean: " +
(totalVertexEdgeCount.getEdgeCount() /
workerStatsMap.size()) +
", Min: " +
workerEntryList.get(0).getKey() + " - " +
workerEntryList.get(0).getValue().getEdgeCount() +
", Max: "+
workerEntryList.get(workerEntryList.size() - 1).getKey() +
" - " +
workerEntryList.get(workerEntryList.size() - 1).
getValue().getEdgeCount());
}
}
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 8,297 |
import {DataType as proto} from '../../third-party/tensorflow/tensorflow/core/framework/types.proto';
class DataType {
_alias: string | undefined;
_value: object;
static create(dtype: string | DataType) {
if (dtype instanceof DataType)
return dtype;
return new DataType(dtype);
}
constructor(alias?: string) {
switch (alias && alias.toLowerCase()) {
case 'float':
this._value = proto.DT_FLOAT.value;
break;
case 'double':
this._value = proto.DT_DOUBLE.value;
break;
case 'int8':
this._value = proto.DT_INT8.value;
break;
case 'int16':
this._value = proto.DT_INT16.value;
break;
case 'int32':
this._value = proto.DT_INT32.value;
break;
case 'int64':
console.warn('Current filetype is not supported');
this._value = proto.DT_INT64.value;
break;
case 'uint8':
this._value = proto.DT_UINT8.value;
break;
case 'uint16':
this._value = proto.DT_UINT16.value;
break;
default:
this._value = proto.DT_INVALID.value;
}
this._alias = alias;
}
get value() {
return this._value;
}
// TODO: needs implementation?
convert(value: any) {
return value;
}
getArrayBufferView() {
switch (this._alias) {
case 'float':
return Float32Array;
case 'double':
return Float64Array;
case 'int8':
return Int8Array;
case 'int16':
return Int16Array;
case 'int32':
return Int32Array;
case 'int64':
//TODO
console.warn('Current filetype is not supported');
return Int32Array;
case 'uint8':
return Uint8Array;
case 'uint16':
return Uint16Array;
default:
//TODO DT_INVALID
return Int8Array;
}
}
}
export const FLOAT = DataType.create('float');
export const DOUBLE = DataType.create('double');
export const INT8 = DataType.create('int8');
export const INT16 = DataType.create('int16');
export const INT32 = DataType.create('int32');
export const INT64 = DataType.create('int64');
export const UINT8 = DataType.create('uint8');
export const UINT16 = DataType.create('uint16');
export const STRING = DataType.create('string');
export default DataType
| {
"redpajama_set_name": "RedPajamaGithub"
} | 2,638 |
This was the first book by Sy Montgomery I had ever read, and it was good, but I don't think I'll be rushing out to read any others. I've been interested in octopuses for a few years now, and I liked reading about Montgomery's observations of them, and was especially moved by the relationships she formed with them, and how they affected her. However, I felt the book was a little bogged down by things I personally wasn't that interested it, although they would probably be fascinating to other readers. I cannot stand to read about scuba diving with burst eardrums. I know someone who burst her eardrum while scuba diving, and the vertigo that haunted her afterwards meant she had to stop working, socializing, pretty much do anything for the next four years. I also wasn't terribly interested in reading about the other fish that hang out at the aquarium. However, I did like reading about octopuses, and how amazing they are, although the subtitle of the book about the "wonder of consciousness" was a little misleading. I was hoping to read a little more about philosophy, and how we could apply it to creatures we have always thought of as being without feeling, or thought, motivated purely by instinct. The point was touched on briefly a few times, but this was mostly a memoir about falling in love with various octopuses. | {
"redpajama_set_name": "RedPajamaC4"
} | 7,372 |
{"url":"https:\/\/www.electro-tech-online.com\/threads\/eagle-drc-stop-mask-error-layer-29-tstop.149405\/","text":"# Eagle DRC \"Stop MAsk error\", layer 29 (tStop)\n\nStatus\nNot open for further replies.\n\n#### earckens\n\n##### Member\nOut of the blue after multiple checks, and with always .sch and .brd open at the same time I get these \"Stop Mask\" errors where funny little marks appear in the stop mask (top and bottom) of several components leading to these errors.\nThese marks are not present in the part in the library and only appear after a DRC is done on the newly opened file.\nWould anyone please want to have a look at this? I am baffled.\n\nIn the .rar attachment you find the library file, the schematic and the board file.\n\nErik\n\n#### Attachments\n\n\u2022 70.9 KB Views: 66\n\u2022 38.4 KB Views: 51\n\n#### JonSea\n\n##### Well-Known Member\nI've never seen that error before. Some quick looking shows the error occurs any place the silk screen layer intersects holes (i.e., pads) in the soldermask. This isn't a problem for making a successful board, as issues like this get taken care of in the process - you'll never see silkscreen across a pad, but it will affect how the board looks - Read an important comment below.\n\nThis illustration shows one of the errors.\n\nOne this to note in this illustration - the blue cross-hatching is the Tstop layer. \"Fixing\" these errors is pretty simple. Don't display the Tstop and Bstop layers. This is why I've never seen it before. The illustration below shows the same area with the Tstop layer turned off - no more errors.\n\nOne real problem I did notice - all of your fonts are \"proportional\" - proportional fonts do not reproduce properly in Gerber files. Typically, they are too long and don't show up as expected. Change any labels that will be on the silkscreen to Vector - then what-you-see-is-what-you-get. You can do this easily by grouping everything, and changing font properties all at once.\n\nHmmm... maybe this has been fixed in version 7. The picture below is the board as it looks in Eagle. The second is the Gerber silkscreen layer.\n\n#### earckens\n\n##### Member\nHi JonSea, I use Eagle 7.6.0\nGood to learn I need to convert to vector font; I will try and find the reason why it is not in vector, and how to cenvert it to vector (possibly wwith a script?).\nI see the silkscreen does not include names or values; I am planning to include those in the silkscreen.\nIndeed with the tStop and bStop turned of no errors: does this in any way affect the pcb manufacturer if tStop and bStop are included in the gerber?\nAnd if the tNames and tValues are included with the 21 layer for the sikscreen?\n\nAnd most important: why do these errors show up? They also do in places where the silkscreen does not interfere with holes or pads.\n\nI will be off a few days, we were on vacation in Southern France and my missus wants us on our return tomorrow to have a stopover in the Champagne region, to get some boxes for the end of year festivities. Hopefully sober back by Monday.\n\nErik\n\n#### JonSea\n\n##### Well-Known Member\nWhen you turn a layer \"off\", you're only controlling whether you see it on the screen. When I opened your bed file, every layer was turned on...maybe you toggled all layers on to cause this \"error\" to show up. I used the default DRC and I saw only about 5 errors, which all went away when the tStop layer was off.\n\nTo include values in the silkscreen layer, you need to enable the tValues layer in the silkscreen layer in the Gerber CAM job. Whatever layers are ticked will be included in the Gerber layer. I'll post a picture shortly.\n\n#### JonSea\n\n##### Well-Known Member\nAdd tValues to the CAM job.\n\n#### spec\n\n##### Well-Known Member\nI will be off a few days, we were on vacation in Southern France and my missus wants us on our return tomorrow to have a stopover in the Champagne region, to get some boxes for the end of year festivities.\nLife is hell for some people.\n\nspec\n\n#### earckens\n\n##### Member\nLife is hell for some people.\n\nspec\nRight, back from hell. Last night we had a champagne tasting, we got \"somewhat\" drunk; good thing about this stuff is that you do not get a hangover, headache or bad taste the next morning. So this morning we did set off right away to another two houses and had another six or seven tastings, luckily we had had breakfast. You have to take it while you can . (PS, just for the record: we normally drink at most one glass a day ).\n\nSo, sober again, and I just converted all text to vector, problem solved except for 3 obvious stop mask errors: the ones JonSea did show in post 2.\n\nJonSea, I owe you a bottle of champagne, if you get around to fetch it here; thank you very much!\n\nErik\n\n#### JonSea\n\n##### Well-Known Member\nHmmm....a long way to go for champagne...is it a good bottle?","date":"2020-12-05 18:41:47","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.3872566819190979, \"perplexity\": 3089.6685398597147}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2020-50\/segments\/1606141748276.94\/warc\/CC-MAIN-20201205165649-20201205195649-00630.warc.gz\"}"} | null | null |
{"url":"http:\/\/en.wikiversity.org\/wiki\/User_talk:Hillgentleman\/20070519","text":"User talk:Hillgentleman\/20070519\n\nLast edit of page User talk:Hillgentleman\/20070519: 2008-04-28, 09:57, by Crochet.david.bot\n\nHello Hillgentleman! Welcome to Wikiversity! If you decide that you need help, check out Wikiversity:Help desk, ask the support staff, or ask me on my talk page. Please remember to sign your name on talk pages using four tildes (~~~~); this will automatically produce your name and the date. Below are some recommended guidelines to facilitate your involvement. Happy Editing! -- Trevor MacInnis 15:00, 27 August 2006 (UTC)\nGetting Started\nGetting more Wikiversity rules\nGetting Help\nGetting along\nGetting technical\n\nWelcome to Wikiversity; translation\n\n\"Thus, the only relevant term here is 'research' and, as such, this term implies a set of critical methodologies. that transcend the merely literary.\"\n\n\u2022 Does \"critical\" here mean \"crucial\" or \"criticism-related\"?\n\u2022 Does \"literary\" here mean \"on the surface of the words\" or \"literature-related\"?\nMy best guess is that \"critical methodologies\" means \"crucial methodologies\". The original author seemed to be arguing that \"real\" research involves methods that go beyond what can be achieved by any evaluation of the existing published literature. I think \"literary\" here means \"literature-related\", as in what could arise from study and evaluation of previously published results. --JWSchmidt 01:58, 11 October 2006 (UTC)\n\nThanks!\n\nThanks for editing the Strategic Studies pages. I read over those selections of text so often that after a point, I'm incapable of spotting stylistic errors. Thanks, your work is much appreciated.--Dnjkirk 05:36, 22 October 2006 (UTC)\n\nThe End of Eternity\n\nI just noticed some comments you left at Talk:The End of Eternity. I have been thinking about starting a Wikiversity learning project called Exploring science through fiction. Would you be interested in such a project? --JWSchmidt 04:05, 28 October 2006 (UTC)\n\n\"What is your plan?\" <-- I'm mostly interested in biology, but the project could cover all sciences. The basic idea would be to have a reading list of science fiction then discuss the science issues raised by the fiction. \"Most of the good science fictions do not include the new advances in science\" <-- I'm not really concerned with finding and using \"good science fiction\". The main point is to just use fiction as a starting point that will lead into learning real science. --JWSchmidt 16:14, 28 October 2006 (UTC)\n\npossible science fiction\/science topics\n\n\u2022 Books such as Paths to Otherwhere by James Hogan could be used to introduce the Many-worlds interpretation and quantum theory. Some of Stephen Baxter's books (such as the Manifold Trilogy) could introduce topics that cross from physics to biology such as the idea that there are physical processes like stellar gamma-ray generating events that seriously impact the evolution of biological organisms. Books such as Marvin Minsky's The Turing Option could be used to introduce artificial intelligence as compared to biological intelligence. Books such as Assignment Nor'Dyren by Sydney J. Van Scyoc could be used as introductions to genetic engineering. --JWSchmidt 19:19, 28 October 2006 (UTC)\n\nMotto\n\nI think learning by doing would be an acceptable motto- it motivates me, and it describes wikiversity. It is mentioned a lot; was it ever proposed?--Hillgentleman 12:36, 26 October 2006 (UTC)\n\nThanks for the suggestion. This option could be suggested as an option in the learn and teach motto discussion section. -- Reswik 02:41, 31 October 2006 (UTC)\n\nHonesty and Civility\n\nHi Hillgentleman, and let me first take a moment to thank you for the comments and work you've been putting into discussions about Wikiversity - both here and on Meta. I've been meaning to stop by and say that for some time now.\u00a0:-)\n\nOn honesty, I agree that it should be an over-riding principle (or \"code\"), but I'm interested in why you see Wikiversity as being so different from Wikipedia in this regard. Why do you see it as more difficult to deal with dishonesty here? Should we have something more definitive in place perhaps? What, for example do you think of Wikiversity:Disclosures as a related policy? Would honesty fit in with that, or does it still need to be separate, in your opinion?\n\nIn any case, I think this kind of discussion would be most appropriate for the Colloquium. This kind of thinking is fundamental to shaping the kind of project that we can all be a part of and inspire others to be a part of it too. Thanks for your help. Cormaggio 13:23, 10 November 2006 (UTC)\n\nRe:The name\n\nHello again\u00a0:-), the name \"Wikiversity\" was coined as a potential name for the Wikibooks project - see this mail (and its responses), m:Talk:Wikiversity\/Old, and m:Talk:Wikiversity\/Archive_2. However, the project became \"Wikibooks\", and Wikiversity then became a subproject within Wikibooks - until this August, when it became a separate project. Cormaggio 19:41, 11 November 2006 (UTC)\n\nSysop?\n\nWell, i was wondering, is Hillgentleman a Sysop yet? Just asking\n\nWould you accept if i nominated you for sysop, as you are one of the few emmbers who actually adds eduational matirials.\n\nTreasure hunt\n\n\"An idea of an idea\" - I like it\u00a0:-). Yes, it's an interesting idea - to test how usable Wikiversity is. The only thing I'd say is that surely we are trying to educate people to be able to use\/navigate Wikiversity more efficiently - so why would we want to prevent someone from using Google or Recent Changes? (And how would we find that out? Through their accounts of their paths? But if we said it wasn't allowed in the rules, wouldn't people want to cover that up?\u00a0:-) Just a thought.) However, if you wanted to try this out, you could do it in your userspace for now, and I'll happily play the guinea pig to try it out\u00a0:-). Keep the ideas coming! Cormaggio 11:38, 23 November 2006 (UTC)\n\nConfused\n\nCould you explain, what you meant to say with this edit? It seems somewhat out of place at Wikiversity or do I misunderstand something? sebmol ? 16:15, 24 November 2006 (UTC)\n\nHow did you get that? I'm seeing this: [1], coming from the IP before Hillgentleman.--Rayc 19:07, 24 November 2006 (UTC)\nMaybe it was a mistake in my recent changes view. I think that section should just be removed without further discussion. sebmol ? 20:55, 24 November 2006 (UTC)\n\nHi there\n\nHi I'm Student Galaxy, a new user on Wikiversity, and I can certainly say. There's a lot needing to be done, I have made 100 contributions already, and I'm looking for stuff to do. Student Galaxy 12:38, 26 November 2006 (UTC)\n\nI do football research\n\nI do football research. Please do not comment on that. I' am only helping Wikiversity. I have seen a lot of matches, practiced a lot, and played a few matches. I listen to the words of coaches, and if I can teach it I can. I' am just trying to help. Student Galaxy 12:55, 26 November 2006 (UTC)\n\nOh by the way, I play defence and in goal. Quite good at it.\n\nTopic \"software testing\"\n\nHello my friend, as I see you made a change here. Was this because you wanted me to help because of this? Unfortunately my problem still persist. Do you know a hint? --Erkan Yilmaz 20:58, 27 November 2006 (UTC)\n\nThank you very much Hillgentleman. --Erkan Yilmaz 07:56, 28 November 2006 (UTC)\n\nHello again, I answered, perhaps you could also give it a try? I saw your \"see also\" on the page - in the end, perhaps this is the only solution.\u00a0:-( --Erkan Yilmaz 21:38, 29 November 2006 (UTC)\n\nthank you for the link, will give it a try soon. --Erkan Yilmaz 12:12, 30 November 2006 (UTC)\n\nBasic atomic physics\n\nThanks for your attention to Basic atomic physics! I take it Mathematics of Theoretical Physics ought to be Mathematics for Theoretical Physics. I can move it there if appropriate.\n\nAs for the removed text:\n\n $\\frac{\\partial^2 f}{\\partial\\varphi^2}$ This example, though intimidating to the freshman, is simply a formula used to describe spherical coordinates. See Jacobolus and Dysprosia\n\n...I was attempting to direct the reader to a \"real world\" dialog that, to me, illustrates how the language (math for physics) is shared within a relevant context, resourcefully, between those who, I feel, know the topic well.\n\nI tried to set the tone of the article to what I hope is a friendly one, alerting the reader to a common \"gothcha\" when thinking about deep science \u2014 one that gets me often, in fact. Feel free to expand (or contract) the article as you see fit.\n\nI hope the tougher article Mathematics of Theoretical Physics (or Mathematics for Theoretical Physics) will capture your attention, also. Thanks again. CQ 21:48, 27 November 2006 (UTC)\n\nIBC - Interlangual Beta Club\n\nHi, I replied here. Could you have a look please? thx. --Erkan Yilmaz 13:31, 3 December 2006 (UTC)\n\nFold\n\nSure, but I am not entirely sure what you mean by \"folding\" as our termingology may vary. Do you mean archiving a talk page? Despite what you mean, I am sure that I can help you or direct you to someone who can. Happy New Year! -- J.Steinbock 06:15, 1 January 2007 (UTC)\n\nOkay, there is just one more thing that needs to be established for clarity. You want to have the contents from w:zh-classical:user:hillgentleman to go on the Beta Wikiversity under your user namespace? -- J.Steinbock 22:26, 1 January 2007 (UTC)\nOkay, I clearly understand the issue now\u00a0:). I will work on this and should get back to you sometime soon. -- J.Steinbock 00:32, 2 January 2007 (UTC)\\\nI have been reviewing this, but I can't seem to find a solution. There is always one problem or another that arises. I would suggest that you ask Rayc and contact him on his talk page. I'm sorry that I couldn't find the solution. Best wishes. -- J.Steinbock 21:55, 6 January 2007 (UTC)\n\nthe name \"wikiversity\"\n\nOriginally there was a Nupedia project. I suppose that the name meant \"new encyclopedia\". Then it was decided to add a wiki to the Nupedia project. The original Nupedia participants objected and a new project was started called Wikipedia. A few years later people started thinking about a new wiki-based project that would go beyond the limited restrictions of Wikipedia and deal with general aspects of education and wiki-based learning. I do not think that \"wikiuniversity\" was ever a serious option for the name. It became traditional to use names such as \"wikibooks\" and \"wikiquote\". A few people objected to the name \"wikiversity\" but there was never any alternative name that had any significnt community support. I'm not sure if \"wikiversity\" is actually a registered trademark of the Wikimedia Foundation. If you want to find out more, I suggest contacting Brad Patrick of the Trademarks committee (BradPatrick user page on the meta-wiki).--JWSchmidt 20:15, 5 January 2007 (UTC)\n\nHello.\n\nHello.","date":"2014-04-23 12:18:00","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 1, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.5860676169395447, \"perplexity\": 1992.3054378723766}, \"config\": {\"markdown_headings\": false, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2014-15\/segments\/1398223202548.14\/warc\/CC-MAIN-20140423032002-00599-ip-10-147-4-33.ec2.internal.warc.gz\"}"} | null | null |
Q: VBA complie error when trying to pass range as argument I am passing two range arguments in a function. One is working fine but second is giving compile error.
Here I am calling the function -
Set SourceRange = Sheets("QueryResult").Range("QueryResult")
Set DestinationRange = Sheets("TradeUnderlyingCptyWWRTemplate").Range("CounterpartyName")
Call PopulateDetails(23, SourceRange, DestinationRange, 2, 8)
And here is the function:
Function PopulateDetails(SourceColumns As Integer, Srce As Range, Destination As Range, DestinationColums As Integer, DestinationRows As Integer)
Dim CellName As String
Dim a, b, i As Integer
For b = 0 To DestinationRows - 1
For a = 0 To DestinationColums - 1
On Error GoTo Err:
Sheets("TradeUnderlyingCptyWWRTemplate").Select
Destination.Offset(b, a).Select
CellName = Destination.Offset(b, a).Name.Name
For i = 0 To SourceColumns - 1
If (Destination.Offset(b, a).Name.Name = Srce.Offset(0, i).Value And Destination.Offset(b, a).Value = "") Then
Destination.Offset(b, a).Value = Srce.Offset(1, i).Value
Exit For
End If
Next
Err:
On Error Resume Next
Err.Clear
CellName = ""
Next
Next
End Function
However, it is giving compile error:
byref argument type mismatch
with SourceRange highlighted in the code.
When I change the function to only have four arguments (deleting Source argument), it works fine.
A: You need to change
Dim SourceRange, DestinationRange As Range
To
Dim SourceRange As Range
Dim DestinationRange As Range
Otherwise it is equivalent to
Dim SourceRange As Variant
Dim DestinationRange As Range
| {
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<Alloy>
<Window class="container">
<View>
<Label class="label">Titanium IOC Example</Label>
<Label class="label">Oscar Brito - @aetheon</Label>
</View>
</Window>
</Alloy> | {
"redpajama_set_name": "RedPajamaGithub"
} | 2,233 |
Q: JavaScript forEach statement differences when argument is before or in the brackets? I came across this method in JavaScript but is wondering why the 2 blocks of code makes a difference. Why when I put A before the forEach, it works and when I pass it as an argument it doesn't work anymore?
function test(A) {
var str = '';
/* block 1 */
A.forEach(function(element){
str += element;
});
/* block 2
forEach(A,function(element){
str += element;
});
*/
return str; // returns 123 from block 1, but error for block 2
}
console.log(test([1,2,3]));
A: There is no foreach statement in JavaScript. There is a .forEach method on arrays, but there is no global forEach function.
A: Because that's just how JS works. forEach is a property (method) of an array. It's not a global function.
| {
"redpajama_set_name": "RedPajamaStackExchange"
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Arts | Culture, Interviews, Music, Music | Theatre October 4, 2013
Sitting Down With UB40
The story of UB40, and how this group of young friends from Birmingham transcended their working-class origins to become the world's most successful reggae band is not the stuff of fairytales as might be imagined. The group's led a charmed life in many respects it's true, but it's been a long haul since the days they'd meet up in the bars and clubs around Moseley, and some of them had to scrape by on less than £8 a week unemployment benefit. The choice was simple if you'd left school early. You could either work in one of the local factories, like Robin Campbell did, or scuffle along aimlessly whilst waiting for something else to happen.
By the summer of 1978, something else did happen, and the nucleus of UB40 began rehearsing in a local basement. Robin's younger brother Ali, Earl Falconer, Brian Travers and James Brown all knew each other from Moseley School of Art, whilst Norman Hassan had been a friend of Ali's since school. Initially, they thought of themselves as a "jazz-dub-reggae" band, but by the time Robin was persuaded to join and they'd recruited Michael Virtue and Astro – who'd learnt his craft with Birmingham sound-system Duke Alloy – the group had already aligned themselves to left-wing political ideals and forged their own identity, separate from the many punk and Two Tone outfits around at that time. The group had nailed their colours to the mast by naming themselves after an unemployment benefit form. Their political convictions hadn't been gleaned secondhand either, but cemented in place whilst attending marches protesting against the National Front, or rallies organised by Rock Against Racism.
California, Dawn Garcia, DG's Passion, From the Editor, Letters to the Editor, Los Angeles, Newsworthy June 21, 2013
What is HATE.
HATE. A word that just makes you feel dirty. Ashamed. A society living in a veil of confusion and fear using their own prejudice to alienate and isolate. The only thing we need in this world is one thing. Just one: LOVE. There is a song that is now playing loudly all over the airwaves by Macklemore and Ryan Lewis featuring Mary Lambert. The lyrics are impossible to tune out because the truth that pierces through every spoken word is the most honest profoundly necessary truth this world is hungry for.
dawngarcia
IG 2.0
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🎥📺🌱🗞
Open to wherever life leads.
Rep: @wme
Copyright 2012-2016 ATOD Magazine™ - Powered by Austin Creatives | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 3,723 |
module Softlayer
module Billing
class Item
module Account
module Media
module Data
module Transfer
autoload :Request, 'softlayer/billing/item/account/media/data/transfer/request'
end
end
end
end
end
end
end
| {
"redpajama_set_name": "RedPajamaGithub"
} | 3,299 |
Косье Вейзенбек (; ) — нидерландская скрипачка и музыкальный педагог.
Окончила Амстердамскую консерваторию у Давины ван Вели. Преподавала в Гаагской консерватории, затем в Амстердамской консерватории. Воспитала целый ряд значительных музыкантов, среди которых выделяется Янин Янсен. Основала (1984) и бессменно возглавляла детский ансамбль скрипачей «De Fancy Fiddlers», в котором играют юные музыканты из разных стран. В 2007 году удостоена внеочередной премии Фонда имени Антона Кершеса за исключительный вклад в музыкальную педагогику Нидерландов.
Источники
Ссылки
De Fancy Fiddlers
Фотография Косье Вейзенбек
Выпускники Амстердамской консерватории
Академические музыканты Нидерландов
Преподаватели Гаагской консерватории
Преподаватели Амстердамской консерватории | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 621 |
El Club Natación Maristas es un club acuático con sede en el colegio Maristas "El Salvador" en Bilbao, y fundado en 1994.
En el club se practican dos disciplinas: natación y waterpolo.
Historia
El club se funda en 1994, tras la desaparición en 1993 de la Asociación Deportiva Maristas, que era el antiguo club Maristas fundado a principios de 1970. La Asociación Natación Maristas El Salvador es una asociación privada que intenta fomentar la práctica de la natación y el waterpolo, principalmente entre los alumnos del colegio El Salvador, de los Hermanos Maristas, de Bilbao. Aunque también está abierto a otros socios que no son del colegio. En el 2000 empieza la sección de waterpolo. A finales de la década del 2000 llegó a tener 160 socios entre nadadores y jugadores de waterpolo.
Para desarrollar estos objetivos el club cuenta con las instalaciones deportivas del colegio El Salvador, situadas en la calle Iturribide 78 de Bilbao.
En 2009, su sección de waterpolo se une a la de la Deportiva Náutica Portugalete. En 2012, el equipo de waterpolo masculino senior retoma su camino en solitario (ya sin Deportiva Náutica Portugalete) para competir en el Campeonato de Ascenso de la Liga de Euskal Herria de Waterpolo.
Presidentes
Rafael Herreros
José Luis Valle
Santiago Gil
Palmarés de waterpolo
1 Copa Euskal Herria de waterpolo femenino (2007)
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Subscribe To Where Every Game of Thrones Valyrian Steel Sword and Weapon Is Right Now Updates
Where Every Game of Thrones Valyrian Steel Sword and Weapon Is Right Now
Corey Chichizola
After a hiatus that felt just about as endless as the Long Night, Game of Thrones is finally returning. Only six episodes are left before Westeros' story is concluded (at least, until the prequel premieres), and the Season 7 cliffhanger saw the Night King and a zombified Viserion destroy a section of the wall at Eastwatch, finally allowing the army of the dead to march on Westeros.
The Great War between the living and the dead has never been nearer, and the characters are going to need as many dragonglass and Valyrian Steel weapons as they can get their hands on. The latter is extremely rare, so here are where all of the Valyrian steel blades still in the mix are currently located on Game of Thrones.
Longclaw is perhaps Game of Thrones' most iconic Valyrian Steel weapon. It has been wielded throughout most of the series by none other than Jon Snow (or should we call him Aegon now?). The sword is a family heirloom of House Mormont, and was gifted to Jon in the Night's Watch by Lord Commander Mormont before his death. Jon attempted to return it to Jorah in "Beyond the Wall" last season, but Dany's chief advisor maintained that it belonged to Jon.
While we're glad to see our favorite former bastard with his trusty weapon, I have to wonder what little Lyanna Mormont would think about Jorah's decision. She's currently the head of House Mormont, and she may want her family's blade back eventually. Jon has already used Longclaw to kill two White Walkers, and smart money says he's got more to chop down in the future.
Widow's Wail
After Tywin Lannister got his hands Ned Stark's former greatsword Ice, he melted down the Valyrian Steel blade into two swords for House Lannister. This scene was the cold open of Season 4, which hints at how important the swords will likely be in the coming battle against the dead. One sword was made for King Joffrey and gifted to him before the Purple Wedding. Obviously he didn't get to use it much (except on a large book given to him by Tyrion), and the sword has since been claimed by Jaime Lannister.
With Jaime finally departing King's Landing to join the battle against the dead at the end of Season 7, it seems like Jaime may end up slaying some White Walkers of wrights in the future. Although it will probably take some convincing for him to be welcomed into the Targaryen/Stark alliance.
Oathkeeper
The second sword forged from Ice's Valyrian Steel is Oathkeeper. This sword was originally made by Tywin for Jaime, although he had just lost his hand when that happened. Because Jaime wasn't yet trained to battle with his left hand, he gifted the sword to Brienne of Tarth. Brienne and Jaime formed a strong relationship during their journey back to King's Landing, and saved each other's life more than once.
There are also hints of some romantic feelings between the two (with no love for Tormund), so Jaime's gift to Brienne is a significant one. She named it Oathkeeper after her vow to Catelyn Stark, and she's used it in a bevy of important battles throughout the seasons, including her fight with The Hound and her slaying of Stannis Baratheon. Now that Brienne has seen a wight in person, I'm sure she'll be killing plenty of dead soldiers before Season 8 ends.
Heartsbane
Until recently, Heartsbane was one of the lesser known Valyrian steel swords in Game of Thrones, but it may end up being quite important now that the dead are marching on Winterfell. Heartsbane is the Valyrian steel sword of House Tarly (a.k.a. Sam's family). When Sam briefly returned to Horn Hill in Season 6 (and before he pissed off fans), he was promptly embarrassed and shamed by his father Randyll. Right before he left on his way to The Citadel he stole Heartsbane, to his family's chagrin. Sam himself is unable to wield Heartsbane, but -- spoilers ahead for Episode 2 of Season 8 -- he handed it over to Ser Jorah Mormont, who definitely knows his way around a sword.
Now that both Randyll and Dickon have been killed by Dany, no one is going to come asking for the sword. While Sam likely won't take it up in battle, perhaps he'll find a way to harness the power of Valyrian steel, or at least figure out what makes it able to kill White Walkers. If he (and maybe Bran) are able to do this in Season 7 and maybe even create more Valyrian steel, then the living may have a chance after all. They do have a new blacksmith handy!
The Dagger
This little dagger has been a major plot point in Game of Thrones. Back in Season 1, a sellsword infiltrated Winterfell and attempted to assassinate Bran Stark with the blade. Catelyn and Summer were able to fight him off, but the dagger itself was used by the not-so-dearly departed Littlefinger to begin the conflict between House Stark and House Lannister. In Season 7, it made a major reappearance as Littlefinger handed it to Bran, who then gave it to Arya.
With Arya now armed with a Valyrian steel weapon of her own, I can't wait to see her kill some undead enemies when the final season of Game of Thrones premiers. While she's still got Cersei on her name of people to kill, I wouldn't be surprised if Arya ends up using her dagger to protect Winterfell, especially her sister.
Bonus Sword: Dawn
If Longclaw is Game of Thrones' most iconic Valyrian steel weapon, then Dawn has to be the least-known. The ancestral sword of House Dayne has barely even been mentioned, and only eagle-eyed fans spotted it in the Tower of Joy flashback sequences. After Ser Arthur Dayne was slain, Ned picked up Dawn and took it with him into the tower in Dorne, where he found Lyanna dying and was entrusted with Jon/Aegon.
Game of Thrones never definitively revealed what happened to Dawn, although if the events of the show follow the books, then Ned returned Dawn to House Dayne after the fighting was over. Still, the camera lingered on Dawn just long enough for fans to identify it. Was that a sign that Dawn has a part to play in the Great War? Or was it just a nod to book fans? It's not one of the many things we already know about Season 8, so we'll have to wait and see.
Season 8 of Game of Thrones premieres Sunday, April 14 at 9 p.m. ET on HBO.
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\section{Introduction}
In this paper, we study in detail a $2$-categorical generalization of the construction of the category of elements, that we call the \predfn{$2$-$\catfont{Set}$-enriched Grothendieck construction}. Being this also a natural extension of the usual Grothendieck construction (to admit $2$-copresheaves $\catfont{B}\to\catfont{CAT}$ with $\catfont{B}$ a $2$-category), the results presented here apply to the latter as well. An explicit definition of the \predfn{$2$-$\catfont{Set}$-enriched Grothendieck construction} firstly appeared in Street's~\cite{street_limitsindexedbycatvalued}, but we present it in a more abstract way, that is particularly relevant to, but not only, (higher dimensional) elementary topos theory. Furthermore, after showing a connection between the \predfn{$2$-$\catfont{Set}$-enriched Grothendieck construction} and what we call the \predfn{\pteor{op}lax normal $2$-\pteor{co}limits}, we use the latter to state and prove a pointwise Kan extension result for the former.
In dimension $1$, the construction of the {category of elements}, that we shall think of as the $\catfont{Set}$-enriched Grothendieck construction, is a well known process (for which we take as main reference Section 1.5 of Mac Lane and Moerdijk's~\cite{maclanemoerdijk_sheavesingeoandlogic}) of reorganization of the data of a (co)presheaf in terms of a {discrete (op)fibration}, that forms an equivalence of categories. Such construction brings many advantages, as it allows to consider a family of sets indexed over a category without mentioning morphisms that land into the universe of all sets, and it also makes the change of base operation (written in the assignment on morphisms of the (co)presheaf) more evident. But also the process itself gives important consequences, among which the reduction of the weighted $\catfont{Set}$-enriched limits to conical ones (that also brings to the explicit formula for the $\catfont{Set}$-enriched Kan extensions). Surprisingly, more is true. Indeed the wish of conicalizing the $\catfont{Set}$-enriched limits actually guides and almost determines the definition of the construction of the category of elements. And we will show how the wish of giving an essential solution to the problem of conicalizing weighted $2$-limits guides the definition of the \predfn{$2$-$\catfont{Set}$-enriched Grothendieck construction} (\nameit{Construction}\ref{2SetGrothconstrtoconicalize}), justifying the explicit definition given by Street in~\cite{street_limitsindexedbycatvalued} (and the definition of the usual Grothendieck construction as well).
This will also bring to the \BF{first main result} of this paper (\nameit{Theorem}\ref{redlaxnormalconical}), which is a new, more intuitive and elementary proof of the fact that every weighted $2$-limit can be essentially conicalized, up to admitting $2$-cells inside the cones and using what we call the \predfn{lax normal conical $2$-limits} (firstly proved by Street in~\cite{street_limitsindexedbycatvalued}). The \predfn{lax normal conical $2$-limits}, that are actually a particular case of a $2$-dimensional notion of limit introduced by Gray in~\cite{gray_formalcattheory}, can be very useful, giving a theory that is equivalent to but in many situations simpler than the one of weighted $2$-limits. Indeed one can sometimes handle $2$-cells inside the cones quite easily, but really needs to have one selected universal cone rather than a bunch of cones forming a cylinder. Despite their potential, these limits have been almost forgotten, at least until Descotte, Dubuc and Szyld's paper~\cite{descottedubucszyld_sigmalimandflatpseudofun} (where they use their pseudo version, called by them \predfn{sigma-limits}).
\begin{defne}\label{defexplicgrothconstr}
Explicitly, in dimension $1$, the {construction of the category of elements} is based on the idea of taking the disjoint union of a family of sets in order to reorganize such sets as fibres of a unique function into the set of indexes. But this idea is extended to take into account also families of sets that are indexed over a category (rather than just a set), building a notion of morphism between objects of the disjoint union as a property on the morphisms between indexes. Indeed, a functor $F\:\catfont{B}\to\catfont{Set}$ is reorganized as a functor $\groth{F}\:\Grothop{F}\to \catfont{B}$, that we call the \dfn{construction of the category of elements, or the $\catfont{Set}$-enriched Grothendieck construction, of $F$}, with $\Grothop{F}$ such that:
\begin{description}
\item[an object of $\Grothop{F}$] is a pair $(B,X)$ with $B\in \catfont{B}$ and $X\in F(B)$;
\item[a morphism $(B,X)\to (C,X')$] is a morphism $f\:B\to C$ in $\catfont{B}$ such that $F(f)(X)=X'$.
\end{description}
Analogously, in dimension $2$, the idea of taking some disjoint union of a family of categories indexed over a category is realized by the usual Grothendieck construction (see Section 1.1 of Jacobs's book~\cite{jacobs_catlogicandtypetheory}). But we can then extend it to take into account also families of categories that are indexed over a $2$-category (rather than just a category), building a notion of $2$-cell in the corresponding $\Grothop{F}$ as a property on the $2$-cells in the category of indexes. A $2$-functor $F\:\catfont{B}\to\catfont{CAT}$ with $\catfont{B}$ a $2$-category is then reorganized as a $2$-functor $\groth{F}\:\Grothop{F}\to \catfont{B}$, that we call the \dfn{$2$-$\catfont{Set}$-enriched Grothendieck construction of $F$}, with $\Grothop{F}$ such that:
\begin{description}
\item[an object of $\Grothop{F}$] is a pair $(B,X)$ with $B\in \catfont{B}$ and $X\in F(B)$;
\item[a morphism $(B,X)\to (C,X')$ in $\Grothop{F}$] is a pair $(f,\alpha)$ with $f\:B\to C$ a morphism in $\catfont{B}$ and $\alpha\:F(f)(X)\to X'$ a morphism in $F(B)$;
\item[a $2$-cell $(f,\alpha)\aR{}(g,\beta)\:(B,X)\to (C,X')$] is a $2$-cell $\delta\:f\aR{}g$ in $\catfont{B}$ such that $\alpha=\beta\circ F(\delta)_X$.
\end{description}
\end{defne}
In dimension $1$, it is then known that the construction of the category of elements can be captured in a more abstract way, which is particularly interesting for the ($2$-dimensional) elementary topos theory.
\begin{prop}\label{catofelementsiscomma}
Let $F\:\catfont{B}\to \catfont{Set}$ be a copresheaf. The construction of the category of elements of $F$ is equivalently given by the comma object
\begin{eqD}{catofelemcomma}
\sq*[l][7][7][\opn{comma}][2.7][2.2][0.6]{\Grothop{F}}{\catfont{1}}{\catfont{B}}{\catfont{Set}}{}{\groth{F}}{1}{F}
\end{eqD}
As a direct consequence, it is also given by the pullback between $F$ and the lax limit of the arrow $1\:\catfont{1}\to \catfont{Set}$, where the latter coincides with the forgetful functor $\tau\:\catfont{Set}_{\bullet}\to \catfont{Set}$ from pointed sets to sets.
\begin{eqD*}
\begin{cd}*
\Grothop{F}\PB{rd} \arrow[d,"\groth{F}"'] \arrow[r] \& \catfont{Set}_{\bullet} \arrow[d,"{\tau}"'] \arrow[r]\& \catfont{1} \arrow[d,"{1}"]\arrow[ld,Rightarrow,shorten <=2.7ex,shorten >=2.2ex,"\opn{comma}"{pos=0.6}] \\
\catfont{B}\arrow[r,"F"'] \& \catfont{Set} \arrow[r,equal]\& \catfont{Set}
\end{cd}
\end{eqD*}
\end{prop}
According to Weber's~\cite{weber_yonfromtwotop}, \nameit{Proposition}\ref{catofelementsiscomma} shows the construction of the category of elements as the archetypal $2$-dimensional classification process, exhibiting $\catfont{Set}$ as the canonical ($2$-dimensional) universe for the $2$-category $\catfont{CAT}$. What gets classified is precisely the discrete opfibrations with small fibres. Weber actually only considered the point of view of the second part of \nameit{Proposition}\ref{catofelementsiscomma}, but we prefer the first, that of equation~\refs{catofelemcomma}, as it shows the singleton as the verum inside the category $\catfont{Set}$ of generalized truth values. Equation~\refs{catofelemcomma} also makes it clear that the passage from the $1$-dimensional elementary topos theory to the $2$-dimensional one is obtained by upgrading the classification by pullbacks to one by comma objects. As a consequence, we end up classifying discrete opfibrations, since they are (defined by representability from) what is classified in the archetypal elementary $2$-topos $\catfont{CAT}$.
We are interested in understanding a general pattern, for example from an enriched point of view. Of course it is natural to consider $1\:\catfont{1}\to \catfont{Set}$ as the verum truth value, and this has an immediate generalization to the general enriched setting, but we would like to understand the reason why we should take the comma object to regulate the classification process in dimension $2$. What we believe is the deep reason behind this is that the comma objects are the archetypal example of \predfn{exact square in \catfont{CAT}} (\nameit{Definition}\ref{defexactsquare}), with the notable consequence that comma objects manage to express every copresheaf as a pointwise left Kan extension of the constant copresheaf at $1$ (\nameit{Theorem}\ref{copresheavesarekanextensions}).
\begin{defne}\label{defexactsquare}
An \dfn{exact square in \catfont{CAT}} is a diagram
\sq[l][7][7][\lambda][2.7][2.2]{\catfont{L}}{\catfont{A}}{\catfont{B}}{\catfont{C}}{g}{p}{q}{f}
in \catfont{CAT} such that, for every category $\catfont{M}$, the associated {Beck-Chevalley natural transformation} given by the pasting
\begin{cd}
\& \m{\catfont{L}}{\catfont{M}} \arrow[d,Rightarrow,shift right = 6.5ex,shorten <= 1ex, shorten >= 1ex,"{\epsilon^p}"'] \arrow[ld,bend right,"{\Lan{p}}"'] \arrow[r,leftarrow,"{-\circ g}"] \arrow[d,leftarrow,"{-\circ p}"'] \& \m{\catfont{A}}{\catfont{M}} \arrow[d,Rightarrow,shift left = 7ex,shorten <= 1ex, shorten >= 1ex,"{\eta^q}"] \arrow[d,leftarrow,"{-\circ q}"] \arrow[ld,Rightarrow,shorten <=2.7ex,shorten >=2.2ex,"{-\ast \lambda}"{pos=0.5}] \arrow[r,equal,"{}"] \& \m{\catfont{A}}{\catfont{M}} \arrow[ld,bend left,"{\Lan{q}}"]\\
\m{\catfont{B}}{\catfont{M}} \arrow[r,equal,"{}"] \& \m{\catfont{B}}{\catfont{M}} \arrow[r,leftarrow,"{-\circ f}"'] \& \m{\catfont{C}}{\catfont{M}}
\end{cd}
is an isomorphism, where $\Lan{p}$ and $\Lan{q}$ denote the pointwise left Kan extensions respectively along $p$ and along $q$, and $\epsilon^p$ and $\eta^q$ are respectively the counit of the adjunction formed by the first one and the unit of the adjunction formed by the second one.
\end{defne}
\begin{teor}\label{copresheavesarekanextensions}
Let $F\:\catfont{B} \to \catfont{Set}$ be a copresheaf. Then
$$F=\Lan{\groth{F}}{\Delta 1}$$
where $\Delta 1\:\Grothop{F}\to \catfont{Set}$ is the functor constant at $1$.
\end{teor}
\begin{proof}
This proof does not seem to appear in the literature, but might be folklore. By \nameit{Proposition}\ref{catofelementsiscomma}, the construction of the category of elements can be expressed as a comma object, forming thus an exact square. Considering $\catfont{M}=\catfont{Set}$ in the definition of exact square, the component of the associated Beck-Chevalley transformation on the copresheaf $1\:\catfont{1}\to \catfont{Set}$ exhibits
$$\Lan{\groth{F}}{\Delta 1}\cong \Lan{1}{1} \circ F.$$
And since $1\:\catfont{1}\to \catfont{Set}$ is dense, we have that $\Lan{1}{1}=\Id{\catfont{Set}}$ (a reference for this basic result on density is Kelly's~\cite{kelly_basicconceptsofenriched}).
\end{proof}
\begin{rem}\label{remtheorkanextisuseful}
\nameit{Theorem}\ref{copresheavesarekanextensions} (together with \nameit{Proposition}\ref{catofelementsiscomma}) implies a huge portion of the theory around the construction of the category of elements, including the canonical extension of the construction to a functor $\groth{-}\:\m{\catfont{B}}{\catfont{Set}}\to \slice{\catfont{Cat}}{\catfont{B}}$, the fully faithfulness of $\groth{-}$, the stability under pullback of the discrete (op)fibrations with small fibres (that form the essential image of $\groth{-}$) and the conicalization of the weighted $\catfont{Set}$-enriched limits. This does not seem to appear in the literature, but might be folklore.
For example, the extension of the construction of the category of elements to a fully faithful functor $\groth{-}\:\m{\catfont{B}}{\catfont{Set}}\to \slice{\catfont{Cat}}{\catfont{B}}$ can be obtained by the chain of isomorphisms
$$\HomC{\m{\catfont{B}}{\catfont{Set}}}{F}{G}\cong \HomC{\m{\Grothopdiag{F}}{\catfont{Set}}}{\Delta 1}{G\circ \groth{F}}\cong \HomC{\slice{\catfont{Cat}}{\catfont{B}}}{\Grothopdiag{F}}{\Grothopdiag{G}}$$
(for $F$ and $G$ arbitrary), where the first isomorphism is given by $F=\Lan{\groth{F}}{\Delta 1}$ and the second one is given by the universal property of the comma object $\Grothop{G}$.
\end{rem}
The \BF{second} (\nameit{Theorem}\ref{grothconstrislaxcomma}) and \BF{third} (\nameit{Theorem}\ref{teorpointkanextforgroth}) \BF{main results} of this paper can be condensed in the following theorem, that gives a categorification of both \nameit{Proposition}\ref{catofelementsiscomma} and \nameit{Theorem}\ref{copresheavesarekanextensions}. Exactly as its $1$-dimensional analogue yields a lot of useful applications (see \nameit{Remark}\ref{remtheorkanextisuseful}), so can this result.
\begin{teor}\label{theorcondensed}
Let $F\:\catfont{B}\to \catfont{CAT}$ be a $2$-functor. The \predfn{$2$-$\catfont{Set}$-enriched Grothendieck construction} is equivalently given by the \predfn{lax comma object} \pteor{with a new universal property, described in \nameit{Definition}\ref{completeunivproplaxcomma}, that refines both the ones given by Gray in~\cite{gray_formalcattheory} and by Lambert in~\cite{lambert_discretetwofib}}
\begin{eqD*}
\sq*[l][7][7][\lax \opn{comma}][2.7][2.2][0.65]{\Grothop{F}}{\catfont{1}}{\catfont{B}}{\catfont{CAT}}{}{\groth{F}}{\catfont{1}}{F}
\end{eqD*}
Furthermore, this square \pteor{filled with a \predfn{lax normal natural transformation}} exhibits $F$ as the \predfn{pointwise left Kan extension of $\Delta 1$ along $\groth{F}$ in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$} \pteor{it lives in a tridimensional world that admits the lax natural transformations, see \nameit{Remark}\ref{needof2catlax}}, where this concept of pointwise Kan extension is defined originally in this paper in \nameit{Definition}\ref{defpointkanextensionlaxthreecat}, using the concept of \pteor{non necessarily conical anymore} \predfn{oplax normal $2$-colimit} that we introduce in Section~\ref{sectionlaxnormallimits}.
$$F=\Lan{\groth{F}}{\Delta 1}.$$
\end{teor}
\nameit{Theorem}\ref{theorcondensed} shows in which sense the \predfn{$2$-$\catfont{Set}$-enriched Grothendieck construction} can be thought of as the archetypal $3$-dimensional classification process, in the sense of a would-be tridimensional elementary topos theory. What gets classified are the \predfn{split $2$-$\catfont{Set}$-opfibrations} with small fibres (introduced in Lambert's~\cite{lambert_discretetwofib}) as \predfn{discrete $2$-fibrations}). Analogously to how the classification by pullbacks (in dimension $1$) is upgraded to one by comma objects to obtain the notion of $2$-dimensional elementary topos (see above), we believe that it should then be upgraded to one by lax comma objects (defined as in \nameit{Definition}\ref{completeunivproplaxcomma}), in order to reach a definition of \predfn{tridimensional elementary topos}. It is interesting to notice that we have to move out of $2\h[2]\mbox{-}\CAT$ (see \nameit{Remark}\ref{needof2catlax}) in order to capture the laxness that permeates the Grothendieck construction. So the archetypal elementary $3$-topos seems to be $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$, inscribed in a sequence
$$\catfont{Set}\quad\leadsto\quad \catfont{CAT} \quad\leadsto\quad 2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$$
that we believe is best explained by what we call a \predfn{$2$-$\catfont{V}$-enrichment} (\nameit{Definition}\ref{deftwovenrichment}):
$$\catfont{V}\quad\leadsto\quad \VCAT{\catfont{V}} \quad\leadsto\quad \twoVCAT{\catfont{V}}$$
(whence comes the name we give to the \predfn{$2$-$\catfont{Set}$-enriched Grothendieck construction}).
The \BF{fourth main result} of this paper (\nameit{Proposition}\ref{pointkanisalsoweak}) is that a \predfn{pointwise left Kan extension in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$} (\nameit{Definition}\ref{defpointkanextensionlaxthreecat}) along a \predfn{$2$-$\catfont{Set}$-opfibration} (\nameit{Definition}\ref{deftwosetopf}) is always also a \predfn{weak left Kan extension} (\nameit{Definition}\ref{defweakkanextensionlaxthreecat}). This justifies and explains more the concept of pointwise Kan extension in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$ that we propose here. Such result is based on a new $\oplaxn\mbox{-}\lax$ generalization of the parametrized Yoneda lemma, presented in \nameit{Theorem}\ref{parametrizedyonedaoplaxnlax}, that shows how the oplax normal naturality is the minimum amount of strictness required to expand the data on the identities to a lax natural transformation.
\subsection*{Outline of the paper}
In Section~\ref{sectionlaxnormallimits}, we produce, simultaneously (\nameit{Construction}\ref{2SetGrothconstrtoconicalize}), both the \predfn{$2$-$\catfont{Set}$-enriched Grothendieck construction} and the \predfn{lax normal conical $2$-limits}. We show with new proofs that the lax normal conical $2$-limits and the weighted $2$-limits give equivalent theories (\nameit{Theorem}\ref{redlaxnormalconical}, \nameit{Theorem}\ref{laxnormalareweighted}) and consider a colimit version as well. In Section~\ref{sectiontwosetgrothconstr}, we refine the universal property of the lax comma object to suit a lax $3$-categorical ambient (\nameit{Definition}\ref{completeunivproplaxcomma}) and we conceive the $2$-$\catfont{Set}$-enriched Grothendieck construction as the archetypal $3$-dimensional classifier (\nameit{Theorem}\ref{grothconstrislaxcomma}). We then inscribe this in a weak enrichment idea (\nameit{Remark}\ref{remarchetypalthreedimclassifproc}). In Section~\ref{sectionkanextensions}, we present and apply a pointwise Kan extension result for the $2$-$\catfont{Set}$-enriched Grothendieck construction (\nameit{Theorem}\ref{teorpointkanextforgroth}), after giving an original definition of \predfn{pointwise Kan extension in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$} (\nameit{Definition}\ref{defpointkanextensionlaxthreecat}). We conclude by showing that a pointwise Kan extension in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$ is always a weak one as well (\nameit{Proposition}\ref{pointkanisalsoweak}), using a $\oplaxn\mbox{-}\lax$ generalization of the parametrized Yoneda lemma (\nameit{Theorem}\ref{parametrizedyonedaoplaxnlax}).
\section{(Op)lax normal conical 2-(co)limits}
\label{sectionlaxnormallimits}
In this section, we show how the wish of giving an essential solution to the problem of conicalization of the weighted $2$-limits produces, simultaneously, both the \predfn{$2$-$\catfont{Set}$-enriched Grothendieck construction}, thus justifying the explicit definition of \nameit{Definition}\ref{defexplicgrothconstr} that Street gave in~\cite{street_limitsindexedbycatvalued}, and the \predfn{lax normal conical $2$-limits}, that are a particular case of a $2$-dimensional limit introduced by Gray in~\cite{gray_formalcattheory}. This brings to the first main result of this paper (\nameit{Theorem}\ref{redlaxnormalconical}), which is a new, more intuitive and elementary proof of the fact (firstly proved by Street in~\cite{street_limitsindexedbycatvalued}) that every weighted $2$-limit can be reduced to a \predfn{lax normal conical} one - that is, one of conical form but with coherent $2$-cells inside - using the \predfn{$2$-$\catfont{Set}$-enriched Grothendieck construction}.
These two kinds of $2$-limit give equivalent theories, as it is also true that a \predfn{lax normal conical $2$-limit} is a weighted one - proved here with a new more explicit proof or originally by Street in~\cite{street_limitsindexedbycatvalued} - but \predfn{lax normal conical} ones are in many situations simpler to use. Indeed one can sometimes handle $2$-cells inside the cones quite easily, but really needs to have one selected universal cone rather than a bunch of cones forming a cylinder.
The following idea may be kept in mind to understand the differences between these two kinds of $2$-limit. When one moves to dimension $2$, the conical limits do not suffice anymore, as the functors from the terminal $\catfont{1}$ to a category $\catfont{C}$ cannot capture the whole of $\catfont{C}$, but just its objects. A solution to this problem is to consider at least also the functors from $\catfont{2}$ to $\catfont{C}$ (or all the functors from anything to $\catfont{C}$), in order to capture the morphisms of $\catfont{C}$ as well; this brings to the concept of weighted $2$-limit. But there is another solution, that is considering the functors from the terminal $\catfont{1}$ to $\catfont{C}$ and now also the natural transformations between them; this brings towards the concept of \predfn{lax normal conical $2$-limit}.
We then notice that, in order to essentially conicalize the weighted $2$-colimits, it is more natural to use the \predfn{oplax normal conical $2$-colimits}. It is the non necessarily conical anymore \predfn{oplax normal $2$-colimits} that we will use in Section~\ref{sectionkanextensions} to propose a notion of colimit in a \predfn{$2$-$\catfont{Set}$-category} and with that one of \predfn{pointwise left Kan extension in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$}.
\begin{rec}
Let $\catfont{V}$ to be a complete and cocomplete symmetric closed monoidal category. Let $F\:\catfont{A} \to \catfont{C}$ be a $\catfont{V}$-functor with $\catfont{A}$ a small $\catfont{V}$-category, and let $W\:\catfont{A}\to \catfont{V}$ be another $\catfont{V}$-functor (called the \dfn{weight}, in place of the classical constant at $1$ functor $\Delta 1$, that now does no longer suffice). The \dfn{$\catfont{V}$-limit of $F$ weighted by $W$}, denoted as $\wlim{W}{F}$, is (if it exists) an object $L\in \catfont{C}$ together with an isomorphism in \catfont{V}
\begin{equation}\label{isoweightlim}
\HomC{\catfont{C}}{U}{L}\cong \HomC{\m{\catfont{A}}{\catfont{V}}}{W}{\HomC{\catfont{C}}{U}{F(-)}}
\end{equation}
$\catfont{V}$-natural in $U\in \catfont{C}\ensuremath{^{\operatorname{op}}}$, where $\m{\catfont{A}}{\catfont{V}}$ is the $V$-category of $\catfont{V}$-copresheaves on $\catfont{A}$ valued in $\catfont{V}$ enriched over itself. (For example, if $\catfont{V}=\catfont{Cat}$, then $\m{\catfont{A}}{\catfont{V}}$ is the $2$-category of $2$-functors, $2$-natural transformations and modifications from $\catfont{A}$ to \catfont{CAT}.) Notice that, when $\wlim{W}{F}$ exists, the isomorphism of equation~\refs{isoweightlim} with $U=L$ gives us in particular a $\catfont{V}$-natural transformation
$$\lambda\:W\aR{}\HomC{\catfont{C}}{L}{F(-)}$$
by considering the identity on $L$; this $\lambda$ is called the \dfn{universal cylinder}.
Given instead $F\:\catfont{A}\to\catfont{C}$ and $W\:\catfont{A}\ensuremath{^{\operatorname{op}}}\to\catfont{V}$ two $\catfont{V}$-functors with $\catfont{A}$ small, the \dfn{$\catfont{V}$-colimit of $F$ weighted by $W$}, denoted as $\wcolim{W}{F}$, is defined by the universal property
\begin{equation*}
\HomC{\catfont{C}}{\wcolim{W}{F}}{U}\cong \HomC{\m{\catfont{A}\ensuremath{^{\operatorname{op}}}}{\catfont{V}}}{W}{\HomC{\catfont{C}}{F(-)}{U}}
\end{equation*}
$\catfont{V}$-natural in $U\in \catfont{C}$. And in place of $\lambda$ we find the \dfn{universal cocylinder}
$$\mu\:W\aR{}\HomC{\catfont{C}}{F(-)}{C}.$$
\end{rec}
\begin{rec}\label{recstrategyconicalize}
Although the classical constant at $1$ weight $\Delta 1$, called the \dfn{conical weight}, does no longer suffice in the general enriched setting, we pay attention to when a weighted limit can be \dfn{reduced to a conical one} (also said \dfn{conicalized}), i.e.\ reduced to one weighted by $\Delta 1$.
It is well known (see Kelly's~\cite{kelly_basicconceptsofenriched}) that in the $\catfont{Set}$-enriched setting every weighted limit can be conicalized, using the construction of the category of elements (that we shall think of as the $\catfont{Set}$-enriched Grothendieck construction). A known strategy to show this (used by Kelly in~\cite{kelly_basicconceptsofenriched}) is to show that the particular weighted $\catfont{V}$-colimits
$$W\cong \wcolim{W}{\yy},$$
for $W\:\catfont{A}\to \catfont{V}$ an enriched presheaf with $\catfont{A}$ small and $\yy\:\catfont{A}\ensuremath{^{\operatorname{op}}}\to\m{\catfont{A}}{\catfont{V}}$ the Yoneda embedding (that are immediatly seen to hold by the Yoneda lemma), are conicalizable in a nice way for $\catfont{V}=\catfont{Set}$, and deduce that then every weighted $\catfont{Set}$-limit is conicalizable using the lemma of continuity of a limit in its weight, that we recall here below. The idea is that if we manage to write any weighted limit in the left hand side of equation~\refs{isolemmacont} with $W=\Delta 1$, we obtain in the right hand side its conicalization, provided that $H$ is nice enough (that is, similar enough to the Yoneda embedding) to make the limit $\wlim{H(-)}{F}$ simplify.
\end{rec}
\begin{lemm}[Continuity of a limit in its weight]\label{continuityinweight}
Let $W\:\catfont{B}\ensuremath{^{\operatorname{op}}}\to\catfont{V}$, $H\:\catfont{B}\to\m{\catfont{A}}{\catfont{V}}$ and $F\:\catfont{A}\to\catfont{C}$ be $\catfont{V}$-functors, and suppose that $\wcolim{W}{H}$ and each $\wlim{H(B)}{F}$ with $B\in \catfont{B}$ exist. Then
\begin{equation}\label{isolemmacont}
\wlim{\wcolim{W}{H}}{F} \cong \wlim{W}{\left(\wlim{H(-)}{F}\right)}
\end{equation}
either side existing if the other does.
The colimit version, with $W$ and $F$ as above but now with $H\:\catfont{B}\to\m{\catfont{A}\ensuremath{^{\operatorname{op}}}}{\catfont{V}}$, is
$$\wcolim{\wcolim{W}{H}}{F} \cong \wcolim{W}{\left(\wcolim{H(-)}{F}\right)}.$$
\end{lemm}
\begin{proof}
See Kelly's~\cite{kelly_basicconceptsofenriched}.
\end{proof}
\begin{rem}\label{needoflaxcones}
We are interested in the problem of conicalization of the weighted $2$-limits. This is, strictly speaking, not possible, but we search for an essential solution to it nevertheless, motivated by the advantages that having a universal essential cone selected from the universal cylinder could bring. We will exploit the strategy described in \nameit{Recall}\ref{recstrategyconicalize} to give a new, more intuitive and elementary proof of the fact that every weighted $2$-limit can be reduced to a \predfn{lax normal conical} one.
So we focus on conicalizing first the $2$-colimits
$$W\cong \wcolim{W}{\yy},$$
for $W\:\catfont{A}\to \catfont{CAT}$ an enriched presheaf with $\catfont{A}$ small and $\yy\:\catfont{A}\ensuremath{^{\operatorname{op}}}\to\m{\catfont{A}}{\catfont{CAT}}$ the $2$-Yoneda embedding. In particular, we want to encode the universal cocylinder
$$\mu\:W\aR{}\HomC{\m{\catfont{A}}{\catfont{V}}}{\y{-}}{W}$$
(given by the Yoneda lemma) in terms of a universal cocone
$$\t{\mu}\:\Delta 1\aR{}\HomC{\m{\catfont{A}}{\catfont{V}}}{H(-)}{W}$$
with $H$ some $\catfont{V}$-functor $\catfont{B}\to \m{\catfont{A}}{\catfont{V}}$. The difficulty we encounter is that, in dimension $2$, the components of $\mu$ are functors rather than mere functions, and a cocone is too limited to encode the extra data given by the assignments on morphisms. The idea is to then admit $2$-cells inside the cocone $\t{\mu}$ in order to encode the images under the components $\mu_A$ of $\mu$ (with $A\in \catfont{A}$) of the morphisms in $W(A)$. After all, such images are morphisms in $\HomC{\m{\catfont{A}}{\catfont{CAT}}}{\y{A}}{W}$, and thus $2$-cells between $\y{A}$ and $W$.
This is explored in the following construction, that originally shows how both the \predfn{$2$-$\catfont{Set}$-enriched Grothendieck construction} and the \predfn{lax normal conical $2$-limits} arise simultaneously from the wish to essentially solve the problem of conicalizing the weighted $2$-limits.
\end{rem}
\begin{cons}[The $2$-$\catfont{Set}$-enriched Grothendieck constr.]\label{2SetGrothconstrtoconicalize}
Following \nameit{Remark}\ref{needoflaxcones}, we search for a more relaxed notion of $2$-natural transformation and for a $2$-functor $H\:\catfont{B}\to \m{\catfont{A}}{\catfont{CAT}}$ such that any cocylinder
$$\varphi\:W\aR{}\HomC{\m{\catfont{A}}{\catfont{CAT}}}{\y{-}}{U}$$
with $U\:\catfont{A}\to \catfont{CAT}$ a $2$-functor can be encoded in terms of a relaxed $2$-natural transformation
$$\t{\varphi}\:\Delta 1 \aR[\opn{relaxed}]{} \HomC{\m{\catfont{A}}{\catfont{CAT}}}{H(-)}{U}\:\catfont{B}\ensuremath{^{\operatorname{op}}}\to\catfont{CAT},$$
that is a relaxed version of a cocone.
As we will want to apply \nameit{Lemma}\ref{continuityinweight} (continuity of a limit in its weight) to deduce that every weighted $2$-limit can then be analogously conicalized, we search for $H$ of the form
$${\left(\Grothopdiag{W}\right)}\ensuremath{^{\operatorname{op}}}\arr{{\groth{W}}\ensuremath{^{\operatorname{op}}}} \catfont{A}\ensuremath{^{\operatorname{op}}}\arr{\yy} \m{\catfont{A}}{\catfont{CAT}},$$
where $\Grothop{W}$ and $\groth{A}$ are, up to now, just symbols, but will be found to be the \predfn{$2$-$\catfont{Set}$-enriched Grothendieck construction} (as defined explictly by Street in~\cite{street_limitsindexedbycatvalued}, see \nameit{Definition}\ref{defexplicgrothconstr}). Indeed, we will need the limit $\wlim{H(-)}{F}$ of the right hand side of equation~\refs{isolemmacont} to simplify, and we can achieve this using the Yoneda lemma if $H$ factorizes through the Yoneda embedding.
For every $A\in \catfont{A}$ and $X\in W(A)$, we have a morphism
$$\varphi_A(X)\:\y{A}\to U,$$
and we want to form the cocone $\t{\varphi}$ exactly with these morphisms. So, for every $A\in \catfont{A}$ and $X\in W(A)$ we need an object in $\Grothop{W}$ whose image with respect to $\groth{W}$ is $A$. We call such object $(A,X)$. Since these are the only objects we need for our conicalization process, we take the objects of $\Grothop{W}$ to be precisely all the pairs $(A,X)$ with $A\in \catfont{A}$ and $X\in W(A)$ and define $\groth{W}$ on objects to be the projection on the first component. We then take
$$\t{\varphi}_{(A,X)}\mathrel{\mathrel{\mathop\ordinarycolon}\mkern-1.2mu=} \varphi_A(X).$$
But we also need to encode inside $\t{\varphi}$ the assignment of every $\varphi_A$ with $A\in \catfont{A}$ on morphisms $\alpha\:X\to X'$ in $W(A)$. And, after \nameit{Remark}\ref{needoflaxcones}, the idea is to use a relaxed version of a cocone. We would like $\t{\varphi}$ to be at least a lax natural transformation, so that it is not ill-behaved. So, for every morphism $\xi\:(A,X) \to (A',X')$ in $\Grothop{W}$, we can have a $2$-cell
\begin{cd}[6][7]
1 \arrow[r,"{\t{\varphi}_{(A,X)}}"] \arrow[d,equal]\& \HomC{\m{\catfont{A}}{\catfont{CAT}}}{\y{A}}{U}\arrow[d,"{-\circ \y{\h\groth{W}\ensuremath{^{\operatorname{op}}}(\xi)}}"]\arrow[dl,Rightarrow,"{\t{\varphi}_\xi}"{pos=0.445}, shorten <= 2.8ex, shorten >= 4.9ex]\\
1 \arrow[r,"{\t{\varphi}_{(A',X')}}"'] \& \HomC{\m{\catfont{A}}{\catfont{CAT}}}{\y{A'}}{U}
\end{cd}
For every $A\in \catfont{A}$ and every morphism $\alpha\:X\to X'$ in $W(A)$, we need a morphism $(A,X)\to (A,X')$ in $\Grothop{W}$ whose image with respect to $\groth{W}$ is $\id{A}$, so that we can take the component of $\t{\varphi}$ on it to be $\varphi_A(\alpha)$. Wishing to write the action of $\groth{W}$ again as a projection on the first component, we call such morphism $(A,X)\to (A,X')$ as $(\id{A},\alpha)$.
Now, we want to encode the $2$-naturality of $\varphi$ into the relaxed naturality of $\t{\varphi}$. For every morphism $f\:A\to A'$ in $\catfont{A}$ and every $X\in W(A)$, the naturality of $\varphi$ expresses the equality
$$\varphi_{A'}(W(f)(X))=\varphi_A(X)\circ \y{f}.$$
So, for every $f\:A\to A'$ in $\catfont{A}$ and $X\in W(A)$, we need a morphism
$$\underline{f}^X\:(A,X)\to (A',W(f)(X))$$
in $\Grothop{W}$ such that $\groth{W}\left(\underline{f}^X\right)=f$ and $\t{\varphi}_{\underline{f}^X}=\id{}$.
It is now natural to take $\underline{\id{A}}^X=(\id{A},\id{X})$ for every $A\in \catfont{A}$ and $X\in W(A)$ and ask any of such equal morphisms to be the identity on $(A,X)$. We will see below that the two kinds of morphisms $(\id{A},\alpha)$ and $\underline{f}^X$ are enough for our needs, but we need to close the union of these morphisms under composition. Of course, it is clear how to compose morphisms of the same kind with each other in a natural way. For the composition of different kinds of morphisms, we notice that, given $f\:A\to A'$ in $\catfont{A}$ and $\alpha\:X\to X'$ in $W(A)$, the two morphisms $\underline{f}^{X'}\circ (\id{A},\alpha)$ and $(\id{A'},W(f)(\alpha))\circ \underline{f}^X$ in $\Grothop{W}$ will have the same associated structure $2$-cell of $\t{\varphi}$, by lax naturality of $\t{\varphi}$, since by naturality of $\varphi_A$
$$\varphi_A(\alpha)\circ \y{f}=\varphi_{A'}(W(f)(\alpha)).$$
We then take such two morphisms in $\Grothop{W}$ to be equal, so that we will be able to recover the naturality of $\varphi$ (on morphisms) starting from $\t{\varphi}$. At this point, every finite composition of morphisms in $\Grothop{W}$ can be reduced to a composite
$$(A,X)\ar{\underline{f}^X} (A',W(f)(X)) \ar{(\id{A'},\alpha)} (A',X')$$
for some $f\:A\to A'$ in $\catfont{A}$ and $\alpha\:W(f)(X)\to X'$ in $W(A')$. So we define the morphisms in $\Grothop{W}$ to be precisely all the formal composites $(\id{A'},\alpha)\circ \underline{f}^X$, that we call $(f,\alpha)$, with $f\:A\to A'$ in $\catfont{A}$ and $\alpha\:W(f)(X)\to X'$ in $W(A')$. And we take the identities and the composition as described above. In particular, we see that $\underline{f}^X=(f,\id{W(f)(X)})$, whence we shall use the right hand side notation for such morphism from now on. And we can also get an explicit\v formula for the composition of an arbitrary diagram $(A,X)\ar{(f,\alpha)}(A',X')\ar{(f',\alpha')}(A'',X'')$ in $\Grothop{W}$, that is\v[-1]
\begin{center}
\linesep{1.5}
\begin{tabular}{LL}
(f',\alpha')\circ (f,\alpha)=(\id{A''},\alpha')\circ (f',\id{})\circ (\id{A'},\alpha)\circ (f,\id{})=\\
=(\id{A''},\alpha')\circ (\id{A''},W(f')(\alpha))\circ (f',\id{})\circ (f,\id{})=(f'\circ f,\h \alpha'\circ W(f')(\alpha)).
\end{tabular}
\end{center}
It is readily seen that we have just given the data for a category $\Grothop{W}$. Since we want $\groth{W}$ to be a functor, for an arbitrary morphism $(f,\alpha)$ in $\Grothop{W}$ we need to have
$$\groth{W}(f,\alpha)=\groth{W}(\id{A'},\alpha)\circ \groth{W}(f,\id{})=f$$
So, with the notation we have chosen, $\groth{W}$ is still acting as a projection on the first component. As we want $\t{\varphi}$ to be at least a lax natural transformation, the structure $2$-cell of $\t{\varphi}$ associated to an arbitrary morphism $(f,\alpha)$ in $\Grothop{W}$ needs to be
$$\t{\varphi}_{(f,\alpha)}=\t{\varphi}_{(\id{A'},\alpha)\circ (f,\id{})}=\varphi_{A'}(\alpha)\circ \id{}=\varphi_{A'}(\alpha).$$
We now want to encode the $2$-dimensional part of the $2$-naturality of $\varphi$ into the $2$-dimensional part of the relaxed naturality of $\t{\varphi}$. As we want $\t{\varphi}$ to be at least lax natural, we will have, for every $2$-cell $\Xi\:(f,\alpha)\aR{}(g,\beta)\:(A,X)\to (A',X')$ in $\Grothop{W}$,
\begin{eqD}{twodimlaxphitilde}
\begin{cd}*
1 \arrow[r,"{\t{\varphi}_{(A,X)}}"] \arrow[d,equal]\& \HomC{\m{\catfont{A}}{\catfont{CAT}}}{\y{A}}{U}\arrow[d,"{-\circ \y{f}}"]\arrow[dl,Rightarrow,"{\t{\varphi}_{(f,\alpha)}}"{pos=0.445}, shorten <= 2.8ex, shorten >= 4.9ex]\\
1 \arrow[r,"{\t{\varphi}_{(A',X')}}"'] \& \HomC{\m{\catfont{A}}{\catfont{CAT}}}{\y{A'}}{U}
\end{cd} = \quad
\begin{cd}*
1 \arrow[r,"{\t{\varphi}_{(A,X)}}"] \arrow[d,equal]\& \HomC{\m{\catfont{A}}{\catfont{CAT}}}{\y{A}}{U}\arrow[d,bend right,shift right=0ex,"{-\circ \y{g}}"'{name=B}]\arrow[d,bend left,shift left = 2ex,"{-\circ \y{f}}"{name=A}]\arrow[dl,Rightarrow,shift right=1ex,"{\t{\varphi}_{(g,\beta)}}"{pos=0.55}, shorten <= 4.65ex, shorten >= 3.2ex]\\
1 \arrow[r,"{\t{\varphi}_{(A',X')}}"'] \& \HomC{\m{\catfont{A}}{\catfont{CAT}}}{\y{A'}}{U}
\arrow[Rightarrow,from=A,to=B,"{-\ast \y{\h\groth{W}\ensuremath{^{\operatorname{op}}}(\Xi)}}"'{inner sep=1ex},shorten <=1.55ex, shorten >=1.5ex]
\end{cd}
\end{eqD}
The $2$-naturality of $\varphi$ expresses the following equality, for every $2$-cell $\delta\:f\aR{}g\:A\to A'$ in $\catfont{A}$ and for every $X\in W(A)$:
$$\varphi_{A'}(W(\delta)_X)=\varphi_A(X)\h\y{\delta}\:\varphi_{A'}(W(f)(X))\aR{}\varphi_{A'}(W(g)(X)).$$
So, for every $2$-cell $\delta\:f\aR{}g\:A\to A'$ in $\catfont{A}$ and every $X\in W(A)$, we need a $2$-cell in $\Grothop{W}$, that we call $\underline{\delta}^X$ or just $\delta$, such that
$$\underline{\delta}^X\:(f,W(\delta)_X)\aR{}(g,\id{})\:(A,X)\to (A',W(g)(X))$$
and $\groth{W}(\underline{\delta}^X)=\delta$. These are the only $2$-cells in $\Grothop{W}$ that we need in order to encode the $2$-naturality of $\varphi$ and they are closed under both the vertical composition and the horizontal composition inherited from $\catfont{A}$, but we have to close them under whiskering with each of the two kinds of $1$-cells in $\Grothop{W}$. The horizontal composition of those $2$-cells inherited from $\catfont{A}$ tells us in particular how to whisker with morphisms of type $(h,\id{})$. Precisely, given a $2$-cell $\delta\:f\aR{}g\:A\to A'$ and morphisms $h\:B\to A$ and $k\:A'\to C$ in $\catfont{A}$, and given $X\in W(A)$ and $Y\in W(B)$,
$$(k,\id{})\h\underline{\delta}^X= \underline{(k\h\delta)}^X \quad\text{and}\quad\underline{\delta}^{W(h)(Y)}\h (h,\id{})= \underline{(\delta \h h)}^Y.$$
Now, since for every $\delta\:f\aR{}g\:A\to A' $ in $\catfont{A}$ and every $\alpha\:X\to X'$ in $W(A)$, we have that the axiom of equation~\refs{twodimlaxphitilde} of $\t{\varphi}$ on the two whiskerings $\underline{\delta}^{X'}\h(\id{A},\alpha)$ and $(\id{A'},W(g)(\alpha))\h \underline{\delta}^{X}$ in $\Grothop{W}$ is exactly the same, by the analogous swapping property we asked the composition of $1$-cells in $\Grothop{W}$, we ask such two whiskerings in $\Grothop{W}$ to be equal. So, at this point, every whiskering and then every horizontal composition of $2$-cells in $\Grothop{W}$ can be reduced to a whiskering of the form $(\id{},\beta)\h \underline{\delta}^X$ for some $2$-cell $\delta\:f\aR{}g\:A\to A'$ in $\catfont{A}$, $X\in W(A)$ and $\beta\:W(g)(X)\to X'$ in $W(A')$.
So we define the $2$-cells in $\Grothop{W}$ to be precisely the formal whiskerings $(\id{},\beta)\h \underline{\delta}^X$ for some $2$-cell $\delta\:f\aR{}g\:A\to A'$ in $\catfont{A}$, $X\in W(A)$ and $\beta\:W(g)(X)\to X'$ in $W(A')$. Equivalently, a $2$-cell $(f,\alpha)\aR{}(g,\beta)\:(A,X)\to (A',X')$ in $\Grothop{W}$ is a $2$-cell $\delta\:f\aR{}g$ in $\catfont{A}$ such that
$$\alpha=\beta\circ W(\delta)_X.$$
For this, we will call such $2$-cell just $\delta\:(f,\alpha)\aR{}(g,\beta)$. The identities are given by taking $\delta$ to be the appropriate identity $2$-cell. The horizontal composition is the one obtained by the description above; namely, given $\delta\:(f,\alpha)\aR{}(g,\beta)\:(A,X)\to(A',X')$ and $\varepsilon\:(f',\alpha')\aR{}(g',\beta')\:(A',X')\to(A'',X'')$ in $\Grothop{W}$,\v[-1]
\begin{center}
\linesep{1.5}
\begin{tabular}{LL}
\varepsilon\ast \delta = \left((\id{},\beta')\h \underline{\varepsilon}^{X'}\right)\ast \left((\id{},\beta)\h \underline{\delta}^X\right)=\left((\id{},\beta')\circ (\id{},W(g')(\beta))\right)\h \left(\underline{\varepsilon}^{W(g)(X)}\ast \underline{\delta}^X\right)=\\
=(\id{},\beta'\circ W(g')(\beta))\h \underline{(\varepsilon\ast \delta)}^X
\end{tabular}
\end{center}
and so it corresponds to the $2$-cell $\varepsilon\ast \delta$ in $\catfont{A}$. It is natural to define the vertical composition $(f,\alpha)\aR{\delta}(g,\beta)\aR{\delta'}(h,\gamma)$ in $\Grothop{W}$ to be the $2$-cell in $\Grothop{W}$ that corresponds to the $2$-cell $\delta'\circ\delta$ in $\catfont{A}$. And since we want $\groth{W}$ to be a $2$-functor, we need $\groth{W}$ to send a $2$-cell $(\id{},\beta)\h\underline{\delta}^X$ to the $2$-cell $\delta$ in $\catfont{A}$.
It is then straightforward to show that we have given $\Grothop{W}$ the structure of a $2$-category and that $\groth{W}\:\Grothop{W}\to \catfont{A}$ is a $2$-functor. Moreover, we see that $\Grothop{W}$ is small if $\catfont{A}$ is small. We call $\groth{W}$ (or sometimes also just $\Grothop{W}$) the \dfn{$2$-$\catfont{Set}$-enriched Grothendieck construction} of $W\:\catfont{A}\to \catfont{CAT}$. Notice that we have also described the right notion of relaxed $2$-natural transformation that we need $\t{\varphi}$ to satisfy in order to encode the $2$-naturality of $\varphi$. It is a form of marked lax natural transformation, that is, a lax natural transformation such that certain structure $2$-cells are asked to be identities. And such slight strictness is necessary to encode the strict axioms of $2$-naturality of $\varphi$. This brings to \nameit{Definition}\ref{deflaxnormal} and to the new proof of the result that every weighted $2$-limit can be reduced to a \predfn{lax normal conical} one (for which we have given the ideas in this construction).
\end{cons}
In short, we read from \nameit{Construction}\ref{2SetGrothconstrtoconicalize} the following explicit definition of the $2$-$\catfont{Set}$-enriched Grothendieck construction, that coincides with one of Street's~\cite{street_limitsindexedbycatvalued}.
\begin{defne}\label{defexpltwosetgroth}
Let $F\:\catfont{B}\to\catfont{CAT}$ be a $2$-functor with $\catfont{B}$ a $2$-category. The \dfn{$2$-$\catfont{Set}$-enriched Grothendieck construction of $F$} is the $2$-functor $\groth{F}\:\Grothop{F}\to \catfont{B}$, given by the projection on the first component, with $\Grothop{F}$ such that:
\begin{description}
\item[an object of $\Grothop{F}$] is a pair $(B,X)$ with $B\in \catfont{B}$ and $X\in F(B)$;
\item[a morphism $(B,X)\to (C,X')$ in $\Grothop{F}$] is a pair $(f,\alpha)$ with $f\:B\to C$ a morphism in $\catfont{B}$ and $\alpha\:F(f)(X)\to X'$ a morphism in $F(B)$;
\item[a $2$-cell $(f,\alpha)\aR{}(g,\beta)\:(B,X)\to (C,X')$ in $\Grothop{W}$] is a $2$-cell $\delta\:f\aR{}g$ in $\catfont{B}$ such that $\alpha=\beta\circ F(\delta)_X$;
\item[the compositions and identities] are as described in \nameit{Construction}\ref{2SetGrothconstrtoconicalize}.
\end{description}
\end{defne}
\begin{rem}\label{twosetgrothistrueusual}
The $2$-$\catfont{Set}$-enriched Grothendieck construction is a natural extension of the usual Grothendieck construction. We believe that the latter should actually be conceived as the restriction of the former to $2$-presheaves into $\catfont{CAT}$ on a locally discrete $2$-category. After all, in dimension $1$ (see \nameit{Definition}\ref{defexplicgrothconstr}), the idea of reorganizing a family of sets as the fibres of a function with domain the disjoint union of the family was fully realized by the construction of the category of elements, admitting families of sets indexed by a category rather than by just a set. So we conceive the $2$-$\catfont{Set}$-enriched Grothendieck construction as the complete Grothendieck construction that lives in dimension $2$, rather than a particular case of a construction that lives in dimension $3$, and we will do the same with the notion of fibration that it produces (see Section~\ref{sectiontwosetgrothconstr}).
It is interesting to notice that the $2$-$\catfont{Set}$-enriched Grothendieck construction and then also the usual Grothendieck construction are bound up with some form of laxness, being produced together with the relaxed notion of $2$-natural transformation that is necessary to encode the weighted $\catfont{Cat}$-(co)cylinders (\nameit{Construction}\ref{2SetGrothconstrtoconicalize}). We will explore this in more detail in Section~\ref{sectiontwosetgrothconstr}.
\end{rem}
\begin{defne}\label{deflaxnormal}
Let $W\:\catfont{A}\to \catfont{CAT}$ be a $2$-functor with $\catfont{A}$ small, and consider $2$-functors $M,N\:\Grothop{W}\to \catfont{D}$. A \dfn{lax normal natural transformation $\alpha$ from $M$ to $N$}, denoted $\alpha\:M\alaxn{}N$,\v is a lax natural transformation $\alpha$ from $M$ to $N$ such that the structure $2$-cell on every morphism
$$\p{f,\id{W(f)(X)}}\:(A,X)\to (B,W(f)(X))$$
in $\Grothop{W}$ is the identity.
\end{defne}
\begin{rem}\label{motivationlaxnormal}
The lax normal natural transformations are a particular case of a more general notion of marked lax natural transformation introduced by Gray in~\cite{gray_formalcattheory}. Street showed in~\cite{street_limitsindexedbycatvalued} that this less general notion is totally sufficient to build all the general limits considered by Gray (showing that it is sufficient to build all the weighted limits and that all the limits considered by Gray are particular weighted ones).
Notice that, in some sense, the laxness of the lax normal natural transformations only belongs to the vertical part of $\Grothop{W}$. This idea will be understood even better after \nameit{Theorem}\ref{laxnormalareweighted} (the lax normal conical $2$-limits are weighted ones).
\end{rem}
\begin{defne}\label{laxnormalconicallimits}
Let $W\:\catfont{A}\to \catfont{CAT}$ be a $2$-functor with $\catfont{A}$ small, and let $F\:\Grothop{W} \to \catfont{C}$ be a $2$-functor. Notice that $\Grothop{W}$ is small, since $\catfont{A}$ is small. The \dfn{lax normal conical $2$-limit of $F$}, denoted as $\laxnlim{\Delta 1}{F}$, is (if it exists) an object $L\in \catfont{C}$ together with an isomorphism of categories
$$\HomC{\catfont{C}}{U}{L}\cong \HomC{\mlaxn{\Grothopdiag{W}}{\catfont{CAT}}}{\Delta 1}{\HomC{\catfont{C}}{U}{F(-)}}$$
$2$-natural in $U\in \catfont{C}\ensuremath{^{\operatorname{op}}}$,\v where $\mlaxn{\Grothop{W}}{\catfont{CAT}}$ is the $2$-category of $2$-functors, lax normal natural transformations and modifications from $\Grothop{W}$ to \catfont{CAT}. (Notice that indeed lax normal natural transformations compose well vertically.)
\noindent When $\laxnlim{\Delta 1}{F}$ exists, taking $U=L$ and considering the identity on $L$ gives us in particular a lax normal natural transformation
$$\lambda\:\Delta 1 \alaxn{}\HomC{\catfont{C}}{L}{F(-)},$$
called the \dfn{universal lax normal cone}.
For the definition of \predfn{lax normal conical $2$-colimit} in $\catfont{C}$, we just apply the definition of lax normal conical $2$-limit to a $2$-functor $F\:\Grothop{W}\to\catfont{C}\ensuremath{^{\operatorname{op}}}$. As usual, we prefer to consider instead $F\ensuremath{^{\operatorname{op}}}\:{\left(\Grothop{W}\right)}\ensuremath{^{\operatorname{op}}}\to\catfont{C}$, but this time we cannot rename ${\left(\Grothop{W}\right)}\ensuremath{^{\operatorname{op}}}$ as some $\Grothop{Z}$. So the definition we propose is the following.
Let $W\:\catfont{A}\to\catfont{CAT}$ be a $2$-functor with $\catfont{A}$ small, and let $F\:{\left(\Grothop{W}\right)}\ensuremath{^{\operatorname{op}}}\to \catfont{C}$ be a $2$-functor. The \dfn{lax normal conical $2$-colimit of $F$}, denoted as $\laxncolim{\Delta 1}{F}$, is (if it exists) an object $C\in \catfont{C}$ together with an isomorphism of categories
$$\HomC{\catfont{C}}{C}{U}\cong \HomC{\mlaxn{\Grothopdiag{W}}{\catfont{CAT}}}{\Delta 1}{\HomC{\catfont{C}}{F(-)}{U}}$$
$2$-natural in $U\in \catfont{C}$.
\noindent When $\laxncolim{\Delta 1}{F}$ exists we have, in place of $\lambda$, the \dfn{universal lax normal cocone}.
\end{defne}
\begin{rem}\label{fromgrothisnotrestrictive}
Notice that considering just $2$-functors $F$ of the form $F\:\Grothop{W}\to\catfont{C}$ in \nameit{Definition}\ref{laxnormalconicallimits} is not restrictive at all. Indeed any $2$-category $\catfont{B}$ can be seen as the $2$-$\catfont{Set}$-enriched Grothendieck construction of the $2$-functor $\Delta 1\:\catfont{B}\to\catfont{CAT}$ constant at $\catfont{1}$. We have in fact that $\Grothop{\Delta 1}\cong \catfont{B}$ and that $\groth{\Delta 1}$ is the identity $2$-functor up to this isomorphism.
It is important, though, to have a specified expression of the domain of the diagram $F$ in terms of a $2$-$\catfont{Set}$-enriched Grothendieck construction of some $2$-functor $W\:\catfont{A}\to\catfont{CAT}$ with $\catfont{A}$ small, as we need it in order to be able to say ``normal". The idea is that we want the laxness of the cones to only belong to the categories that form the family $W$ themselves rather than also to the connections between them (see also the proof of \nameit{Theorem}\ref{laxnormalareweighted}, that shows how the lax normal conical $2$-limits are weighted).
\end{rem}
\begin{exampl}\label{conicalarelaxnormalconical}
Every conical $2$-limit is a lax normal conical one. Indeed, let $F\:\catfont{A}\to\catfont{C}$ be a $2$-functor. By \nameit{Remark}\ref{fromgrothisnotrestrictive}, we can view $\catfont{A}$ as the $2$-$\catfont{Set}$-enriched Grothendieck construction of the $2$-functor $\Delta 1\:\catfont{A}\to\catfont{CAT}$. And the lax normal natural transformations between $\catfont{A}=\Grothop{\Delta 1}$ and $\catfont{CAT}$ are just strict $2$-natural, as every morphism of $\Grothop{\Delta 1}$ is of the form $(f,\id{})$.
\end{exampl}
\begin{rem}
We would now like to see that the lax normal conical $2$-limits are particular weighted $2$-limits. The proof we present (\nameit{Theorem}\ref{laxnormalareweighted}) does not seem to appear in the literature. There is a solution to this problem (without proof) in Street's paper~\cite{street_limitsindexedbycatvalued}, but that solution is actually about all the general $2$-limits introduced by Gray and is more complicated than the one presented here (the weight is given as the coidentifier of a certain $2$-cell with horizontal codomain the weight of lax conical limits). And since Street proved in~\cite{street_limitsindexedbycatvalued} that all the general $2$-limits introduced by Gray are built from the lax normal conical ones, we think a more explicit solution of the weight for lax normal conical $2$-limits could be valuable. It is also interesting to notice that the weight for lax normal conical $2$-limits (that is in the proof of \nameit{Theorem}\ref{laxnormalareweighted}) is much simpler than the weight for lax conical $2$-limits (that can be found in Street's~\cite{street_limitsindexedbycatvalued}). Indeed, the latter involves quotients of lax $2$-dimensional slices, while ordinary $1$-dimensional slices are enough for the former.
\end{rem}
\begin{teor}\label{laxnormalareweighted}
Lax normal conical $2$-limits are particular weighted $2$-limits.
\end{teor}
\begin{proof}
Let $Z\:\catfont{A}\to \catfont{CAT}$ be a $2$-functor with $\catfont{A}$ a small $2$-category, and let $F\:\Grothop{Z} \to \catfont{C}$ be a $2$-functor. We want to write the universal property of the lax normal conical $2$-limit of $F$ as the universal property of the $2$-limit of $F$ weighted by some $2$-functor $\operatorname{W}^{\laxn}\:\Grothop{Z}\to\catfont{CAT}$. It suffices to show that for every $2$-functor $N\:\catfont{C}\ensuremath{^{\operatorname{op}}}\to\m{\Grothop{Z}}{\catfont{CAT}}$, calling $N^U\mathrel{\mathrel{\mathop\ordinarycolon}\mkern-1.2mu=} N(U)$, there is an isomorphism of categories
\begin{equation}\label{laxnormalclas}
\HomC{\mlaxn{\Grothopdiag{Z}}{\catfont{CAT}}}{\Delta 1}{N^U}\cong \HomC{\m{\Grothopdiag{Z}}{\catfont{CAT}}}{\operatorname{W}^{\laxn}}{N^U}
\end{equation}
$2$-natural in $U\in \catfont{C}\ensuremath{^{\operatorname{op}}}$. Let then $\varphi\:\Delta 1 \alaxn{} N^U$ be a lax normal natural transformation; we want to convert it into a $2$-natural transformation $[\varphi]\:\operatorname{W}^{\laxn}\aR{}N^U$.
Given $(B,X')\in \Grothop{Z}$, we have a map $\varphi_{(B,X')}\:\catfont{1}\to N^U(B,X')$, i.e.\ an object $\varphi_{(B,X')}\in N^U(B,X')$. But we also have, for every morphism $(f,\alpha)\:(A,X)\to (B,X')$ in $\Grothop{Z}$, a $2$-cell
\sq[l][6][7][\varphi_{(f,\alpha)}][2][2][0.5][l]{\catfont{1}}{{N^U}({A,X})}{\catfont{1}}{{N^U}({B,X'})}{\varphi_{(A,X)}}{}{{N^U}({f,\alpha})}{{\varphi}_{(B,X')}}
i.e.\ a morphism $\varphi_{(f,\alpha)}$ in $N^U(B,X')$. So we want $\operatorname{W}^{\laxn}(B,X')$ to be some sort of slice that parametrizes all these data. But the idea behind lax normal natural transformations is that the laxness is all concentrated in the vertical part of $\Grothop{W}$, that is, on the morphisms of the form $(\id{A},\alpha)$ in $\Grothop{W}$ for some $A\in \catfont{A}$ and some morphism $\alpha\:X\to X'$ in $W(A)$. We can then parametrize just the vertical part of the data, defining
\begin{fun}
\operatorname{W}^{\laxn} & \: & \Grothop{Z} \hphantom{..}& \longrightarrow & \catfont{CAT} \\[1ex]
&& \fib[n][3]{(B,X')}{(g,\beta)}{(C,X'')} &\mapsto & \fib[n][3]{\slice{Z(B)}{X'}}{\beta\circ Z(g)(-)}{\slice{Z(C)}{X''}}
\end{fun}
where the action of $\beta\circ Z(g)(-)$ on morphisms\v is given by $Z(g)(\operatorname{dom}(-))$. (This can be compared with the weight for lax conical $2$-limits, that considers quotients of the lax slices of the entire $\Grothop{Z}$ on $(B,X')$.)
Given a $2$-cell $\delta\:(g,\beta)\aR{}(h,\gamma)\:{(B,X')}\to{(C,X'')}$, we define $\operatorname{W}^{\laxn}(\delta)$ to be the natural transformation\v[-2]
\tc+[8]{\slice{Z(B)}{X'}}{\slice{Z(C)}{X''}}{\beta\circ Z(g)(-)}{\gamma \circ Z(h)(-)}{Z(\delta)_{\operatorname{dom}(-)}}
Indeed, given $(X\ar{\alpha} X')\in \slice{Z(B)}{X'}$, we do have that $Z(\delta)_X$ is a morphism\v in $\slice{Z(C)}{X''}$ from $\beta\circ Z(g)(\alpha)$ to $\gamma\circ Z(h)(\alpha)$, as $Z(\delta)$ is a natural transformation and $\delta$ is a $2$-cell in $\Grothop{Z}$.\v And the naturality of $Z(\delta)_{\operatorname{dom}(-)}$ holds by naturality of $Z(\delta)$. It is then easy to show that $\operatorname{W}^{\laxn}$ is a $2$-functor.
Now, we construct $[\varphi]\:\operatorname{W}^{\laxn}\aR{}N^U$ in the following way. Given $(B,X')\in \Grothop{Z}$, we set
$$[\varphi]_{(B,X')}(\id{X'})\mathrel{\mathrel{\mathop\ordinarycolon}\mkern-1.2mu=} \varphi_{(B,X')}$$
whence we have to choose by naturality that $$[\varphi]_{(B,X')}(X\ar{\alpha}X')=N^U(\id{B},\alpha)\left(\varphi_{(B,X)}\right).$$
And we can then set
$$[\varphi]_{(B,X')}\left(\begin{cd}*[1.1][0.55]
X\arrow[rd,"{\alpha}"']\arrow[rr,"{\alpha}"] \&\& X'\arrow[ld,equal] \\
\& X'
\end{cd}\right)\mathrel{\mathrel{\mathop\ordinarycolon}\mkern-1.2mu=} \varphi_{(\id{B},\alpha)},$$
whence we have to choose (looking again at the naturality of $[\varphi]$ that we want to obtain)
$$[\varphi]_{(B,X')}\left(\begin{cd}*[1.2][0.6]
S\arrow[rd,"{\theta\circ\alpha}"']\arrow[rr,"{\alpha}"] \&\& T\arrow[ld,"{\theta}"] \\
\& X'
\end{cd}\right)=N^U(\id{B},\theta)\left(\varphi_{(\id{B},\alpha)}\right).$$
The component $[\varphi]_{(B,X')}$ is then a functor, thanks to the lax naturality of $\varphi$. And it is easy to show that $[\varphi]$ is a natural transformation, using that $\varphi$ is a lax normal natural transformation and how composition is defined in $\Grothop{Z}$. We show that $[\varphi]$ is $2$-natural. So take a $2$-cell $\delta\:{(g,\beta)}\aR{}{(h,\gamma)}\:{(B,X')}\to {(C,X'')}$ in $\Grothop{Z}$ and $(X\ar{\alpha}X')\in \slice{Z(B)}{X'}$; we prove that
$$N^U(\id{C},\gamma\circ Z(h)(\alpha))\left(\varphi_{(\id{C},Z(\delta)_X)}\right)=N^U(\delta)_{N^U(\id{B},\alpha)\left(\varphi_{(B,X)}\right)}.$$
But this is obtained combining the following two equations:
$$\varphi_{(\id{C},Z(\delta)_X)}=\varphi_{(g,Z(\delta)_X)}=N^U\left(\underline{\delta}^X\right)_{\varphi_{(B,X)}}$$
$$\delta\h(\id{B},\alpha)=(\id{C},\gamma\circ Z(h)(\alpha))\h\underline{\delta}^X.$$
Consider now a modification $\Theta\:\varphi\aM{}\psi$ between two lax normal natural transformations $\varphi,\psi\:\Delta 1 \alaxn{} N^U$.\v We want to convert it into a modification $[\Theta]\:[\varphi]\aM{}[\psi]\:\operatorname{W}^{\laxn}\aR{}N^U$. Given $(B,X')\in \Grothop{Z}$, we define $[\Theta]_{(B,X')}$ to be the natural transformation with general component on $(\alpha\:X\to X')\in\slice{Z(B)}{X'}$ the morphism in $N^U(B,X')$
$$[\Theta]_{(B,X'),\alpha}\mathrel{\mathrel{\mathop\ordinarycolon}\mkern-1.2mu=} N^U(\id{B},\alpha)\left(\h\Theta_{(B,X)}\right)\:N^U(\id{B},\alpha)\left(\varphi_{(B,X)}\right)\to N^U(\id{B},\alpha)\left(\psi_{(B,X)}\right).$$
Then $[\Theta]_{(B,X')}$ is natural since $\Theta$ is a modification, and $[\Theta]$ is a modification since $\Theta$ is a modification between lax normal natural transformations. Moreover this construction is certainly functorial.
Conversely, starting from a $2$-natural transformation $\sigma\:\operatorname{W}^{\laxn}\aR{} N^U$, we convert it into a lax normal natural transformation $\ov{\sigma}\:\Delta 1 \alaxn{} N^U$. Given $(B,X')\in \Grothop{Z}$, we define $$\ov{\sigma}_{(B,X')}\mathrel{\mathrel{\mathop\ordinarycolon}\mkern-1.2mu=} \sigma_{(B,X')}(\id{X'}).$$
And given a morphism $(f,\alpha)\:(A,X)\to (B,X')$ in $\Grothop{Z}$, we define
$$\ov{\sigma}_{(f,\alpha)}\mathrel{\mathrel{\mathop\ordinarycolon}\mkern-1.2mu=} \sigma_{(B,X')}\left(\begin{cd}*[0.6][0.3]
Z(f)(X)\arrow[rd,"{\alpha}"']\arrow[rr,"{\alpha}"] \&\&[0.8ex] X'\arrow[ld,equal] \\
\& X'
\end{cd}\right)$$
noticing that $$\sigma_{(B,X')}(Z(f)(X)\ar{\alpha}X')=N^U(f,\alpha)\left(\sigma_{(A,X)}(\id{X})\right)=N^U(f,\alpha)\left(\ov{\sigma}_{(A,X)}\right)$$
by naturality of $\sigma$. We then have that $\ov{\sigma}$ satisfies the $1$-dimensional part of being a lax normal natural transformation, by functoriality of the components of $\sigma$ and naturality of $\sigma$ (on morphisms). It remains to prove the $2$-dimensional part of $\ov{\sigma}$ being lax natural. So take a $2$-cell $\delta\:{(g,\beta)}\aR{}{(h,\gamma)}\:{(B,X')}\to {(C,X'')}$ in $\Grothop{Z}$; we want to show that\v[-3]
\begin{equation*}
\ov{\sigma}_{(h,\gamma)}\circ N^U(\delta)_{\h\ov{\sigma}_{(B,X')}}=\ov{\sigma}_{(g,\beta)}.
\end{equation*}
By the $2$-naturality of $\sigma$, we have that
$$N^U(\delta)_{\h\ov{\sigma}_{(B,X')}}=N^U(\delta)_{\h{\sigma}_{(B,X')}(\id{X'})}=\sigma_{(C,X'')}\left(\begin{cd}*[0.6][0.3]
Z(g)(X')\arrow[rd,"{\beta}"']\arrow[rr,"{Z(\delta)_{X'}}"] \&\& Z(h)(X')\arrow[ld,"{\gamma}"] \\
\& X''
\end{cd}\right)$$
and we conclude by functoriality of $\sigma_{(C,X'')}$.
Considering now\v a modification $\Xi\:\sigma\aM{}\rho\:\operatorname{W}^{\laxn} \aR{} N^U$, we convert it into a modification $\ov{\Xi}\:\ov{\sigma}\aM{}\ov{\rho}\:\Delta 1 \alaxn{} N^U$. Given $(B,X')\in \Grothop{Z}$, we define $$\ov{\Xi}_{(B,X')}\mathrel{\mathrel{\mathop\ordinarycolon}\mkern-1.2mu=} \Xi_{(B,X'),\id{X'}}.$$
Then $\ov{\Xi}$ is a modification since $\Xi$ is a modification, and this construction is certainly functorial.
It is now easy to see that the two constructions we have produced are inverses of each other, giving us an isomorphism of categories as in equation~\refs{laxnormalclas} for every $U\in \catfont{C}\ensuremath{^{\operatorname{op}}}$. And the $2$-naturality of such isomorphism holds trivially just by the fact that $N$ lands in $\m{\Grothop{Z}}{\catfont{CAT}}$.
\end{proof}
\begin{rem}
The proof of \nameit{Theorem}\ref{laxnormalareweighted} works the same also for the lax normal conical $2$-colimit of a $2$-functor $F\:{\left(\Grothop{Z}\right)}\ensuremath{^{\operatorname{op}}}\to \catfont{C}$. Indeed the corresponding natural isomorphism of categories that we then want to prove is exactly the same we had in equation~\refs{laxnormalclas}. The only thing that changes is that $N$ becomes $N\:\catfont{C}\to\m{\Grothop{Z}}{\catfont{CAT}}$, but the naturality in $U$ of the isomorphism of categories in equation~\refs{laxnormalclas} continues to hold trivially just by the fact that $N$ lands in $\m{\Grothop{Z}}{\catfont{CAT}}$. We thus obtain the following corollary (of the proof of \nameit{Theorem}\ref{laxnormalareweighted}).
\end{rem}
\begin{coroll}\label{laxnormalcolimareweighted}
Lax normal conical $2$-colimits are particular weighted $2$-colimits, and the weight that expresses them is $\operatorname{W}^{\laxn}$.
\end{coroll}
\begin{rem}
We now present our new proof of the fact (firstly proved by Street in~\cite{street_limitsindexedbycatvalued}) that every weighted $2$-limit can be reduced to a lax normal conical one (and thus essentially conicalized). It will be based on the intuitive and elementary \nameit{Construction}\ref{2SetGrothconstrtoconicalize}.
\end{rem}
\begin{teor}\label{redlaxnormalconical}
All the weighted $2$-limits can be reduced to lax normal conical ones. More precisely, given $2$-functors $F\:\catfont{A}\to \catfont{C}$ and $W\:\catfont{A}\to\catfont{CAT}$ with $\catfont{A}$ small, we have that
$$\wlim{W}{F}\cong \laxnlim{\Delta 1}{\left(F\circ \groth{W}\right)}$$
either side existing if the other does, where $\groth{W}$ is the $2$-$\catfont{Set}$-enriched Grothendieck construction of $W$ (see \nameit{Definition}\ref{defexpltwosetgroth}).
\end{teor}
\begin{proof}
Looking at \nameit{Remark}\ref{needoflaxcones}, we first focus on essentially conicalizing the weighted $2$-colimits
$$W\cong \wcolim{W}{\yy}$$
with $W\:\catfont{A}\to \catfont{CAT}$ a $2$-presheaf with $\catfont{A}$ small and $\yy\:\catfont{A}\ensuremath{^{\operatorname{op}}}\to\m{\catfont{A}}{\catfont{CAT}}$ the $2$-Yoneda embedding, and then apply the lemma of continuity of a limit in its weight (\nameit{Lemma}\ref{continuityinweight}) to analogously essentially conicalize any weighted $2$-limit. Looking at \nameit{Construction}\ref{2SetGrothconstrtoconicalize}, given $W$ as above, we prove that there is an isomorphism of categories
\begin{equation}\label{isoredlaxnormal}
\scalebox{0.95}{$\HomC{\m{\catfont{A}}{\catfont{CAT}}}{W}{\HomC{\m{\catfont{A}}{\catfont{CAT}}}{\y{-}}{U}}\cong \HomC{\mlaxn{\Grothopdiag{W}}{\catfont{CAT}}}{\Delta 1}{\HomC{\m{\catfont{A}}{\catfont{CAT}}}{{\left(\yy\circ\h{\groth{W}}\ensuremath{^{\operatorname{op}}}\right)}(-)}{U}}$}
\end{equation}
$2$-natural in $U\in\m{\catfont{A}}{\catfont{CAT}}$, so that we can conclude that
$$W\cong \wcolim{W}{\yy}\cong \laxncolim{\Delta 1}{\left(\yy\circ\h{\groth{W}}\ensuremath{^{\operatorname{op}}}\right)}.$$
We have already described in \nameit{Construction}\ref{2SetGrothconstrtoconicalize} how to obtain from a $2$-natural transformation
$$\varphi\:W\aR{}\HomC{\m{\catfont{A}}{\catfont{CAT}}}{\y{-}}{U};$$
with $U\:\catfont{A}\to \catfont{CAT}$ a $2$-presheaf, a lax normal natural transformation
$$\t{\varphi}\:\Delta 1 \alaxn{} \HomC{\m{\catfont{A}}{\catfont{CAT}}}{{\left(\yy\circ\h{\groth{W}}\ensuremath{^{\operatorname{op}}}\right)}(-)}{U}\:\Grothopdiag{W}\to\catfont{CAT}.$$
Indeed, defining
$$\t{\varphi}_{(A,X)}\mathrel{\mathrel{\mathop\ordinarycolon}\mkern-1.2mu=} \varphi_A(X)$$
$$\t{\varphi}_{(f,\alpha)}=\varphi_B(\alpha)$$
for every morphism $(f,\alpha)\:(A,X)\to (B,X')$ in $\Grothop{W}$, we have that $\t{\varphi}$ satisfies the $1$-dimensional part of being a lax normal natural transformation, since $\varphi$ is natural and the components of $\varphi$ are functors, and it also satisfies the $2$-dimensional part of being lax natural by the definition of the $2$-cells in $\Grothop{W}$ and the $2$-naturality of $\varphi$.
Take now a modification
$$\Theta\:\varphi\aM{}\psi\:W\aR{} \HomC{\m{\catfont{A}}{\catfont{CAT}}}{\y{-}}{U};$$
we want to convert it into a modification between lax normal natural transformations
$$\t{\Theta}\:\t{\varphi}\aM{}\t{\psi}\:\Delta 1 \alaxn{} \HomC{\m{\catfont{A}}{\catfont{CAT}}}{{\left(\yy\circ\h{\groth{W}}\ensuremath{^{\operatorname{op}}}\right)}(-)}{U}.$$
Given $(A,X)\in \Grothop{W}$, we define
$$\t{\Theta}_{(A,X)}\mathrel{\mathrel{\mathop\ordinarycolon}\mkern-1.2mu=} \Theta_{A,X}.$$
$\t{\Theta}$ is then a modification since $\Theta$ is so and such assignment is certainly functorial. We now construct its inverse.
Take a lax normal natural transformation
$$\sigma\:\Delta 1 \alaxn{} \HomC{\m{\catfont{A}}{\catfont{CAT}}}{{\left(\yy\circ\h{\groth{W}}\ensuremath{^{\operatorname{op}}}\right)}(-)}{U}\:\Grothopdiag{W}\to\catfont{CAT};$$
we want to convert it into a $2$-natural transformation
$$\widehat{\sigma}\:W\aR{}\HomC{\m{\catfont{A}}{\catfont{CAT}}}{\y{-}}{U}.$$
We define the component of $\widehat{\sigma}$ on $A\in \catfont{A}$ as
\begin{fun}
\widehat{\sigma}_{A} & \: & W(A) & \longrightarrow & \HomC{\m{\catfont{A}}{\catfont{CAT}}}{\y{A}}{U} \\[1ex]
&& \fib[n][2.85]{X}{\alpha}{X'} & \mapsto & \fib[n][2.85]{\sigma_{(A,X)}}{\sigma_{(\id{A},\alpha)}}{\sigma_{(A,X')}}
\end{fun}
Then $\widehat{\sigma}_{A}$ is a functor, since $\sigma$ is lax natural. We prove that $\widehat{\sigma}$ is natural. So take $f\:A\to B$ a morphism in $\catfont{A}$. We need to show that the following square is commutative:
\sq[5][7]{W(A)}{\HomC{\m{\catfont{A}}{\catfont{CAT}}}{\y{A}}{U}}{W(B)}{\HomC{\m{\catfont{A}}{\catfont{CAT}}}{\y{B}}{U}}{\widehat{\sigma}_A}{W(f)}{-\circ \y{f}}{\widehat{\sigma}_B}
This holds by the fact that $\sigma$ is lax normal and that, given $\alpha\:X\to X'$ in $W(A)$, the following two morphisms are equal in $\Grothop{W}$:
$$(A,X)\ar{(f,\id{W(f)(X)})}(B,W(f)(X))\ar{(\id{B},W(f)(\alpha))}(B,W(f)(X'))$$
$$(A,X)\ar{(\id{A},\alpha)}(A,X')\ar{(f,\id{W(f)(X')})}(B,W(f)(X')).$$
We now prove that $\widehat{\sigma}$ is $2$-natural. Given a $2$-cell $\delta\:f\aR{}g\:A\to B$, we need to show that for every $X\in W(A)$ we have
$$\sigma_{(\id{B},W(\delta)_{X})}=\sigma_{(A,X)}\h\y{\delta}.\v$$
But $\delta$ gives a $2$-cell $\underline{\delta}^X\:{(f,W(\delta)_X)}\aR{}{(g,\id{})}\:{(A,X)}\to{(B,W(g)(X))}$, whence by $\sigma$ being lax normal
$$\sigma_{(A,X)}\h\y{\delta}=\sigma_{(f,W(\delta)_X)}=\sigma_{(\id{B},W(\delta)_X)}.$$
Take now a modification between lax normal natural transformations
$$\Xi\:\sigma\aM{}\rho\:\Delta 1 \alaxn{} \HomC{\m{\catfont{A}}{\catfont{CAT}}}{{\left(\yy\circ\h{\groth{W}}\ensuremath{^{\operatorname{op}}}\right)}(-)}{U};$$
we want to convert it into a modification
$$\widehat{\Xi}\:\widehat{\sigma}\aM{}\widehat{\rho}\:W\aR{} \HomC{\m{\catfont{A}}{\catfont{CAT}}}{\y{-}}{U}.$$
We define the component of $\widehat{\Xi}$ on $A\in\catfont{A}$ to be the natural transformation with component on $X\in W(A)$
$$\widehat{\Xi}_{A,X}\mathrel{\mathrel{\mathop\ordinarycolon}\mkern-1.2mu=} \Xi_{(A,X)}.$$
We have that $\widehat{\Xi}$ is a modification since $\Xi$ is a modification between lax normal natural transformations. And this construction is certainly functorial.
It is now easy to see that the two constructions we have produced are inverses of each other, giving us an isomorphism of categories as in equation~\refs{isoredlaxnormal}. And the $2$-naturality of such isomorphism of categories holds trivially. We have thus proved that every $2$-presheaf $W\:\catfont{A}\to\catfont{CAT}$ with $\catfont{A}$ small can be expressed as
$$W\cong \laxncolim{\Delta 1}{\left(\yy\circ\h{\groth{W}}\ensuremath{^{\operatorname{op}}}\right)}.$$
Consider now $2$-functors $F\:\catfont{A}\to\catfont{C}$ and $W\:\catfont{A}\to\catfont{CAT}$ with $\catfont{A}$ small. Then by the argument above and \nameit{Corollary}\ref{laxnormalcolimareweighted} we have
$$W\cong \laxncolim{\Delta 1}{\left(\yy\circ\h{\groth{W}}\ensuremath{^{\operatorname{op}}}\right)}\cong \wcolim{\operatorname{W}^{\laxn}}{\left(\yy\circ\h{\groth{W}}\ensuremath{^{\operatorname{op}}}\right)}.$$
By \nameit{Lemma}\ref{continuityinweight} and \nameit{Theorem}\ref{laxnormalareweighted}, we obtain
$$\wlim{W}{F}\cong\wlim{\operatorname{W}^{\laxn}}{\left(\wlim{\left(\yy\circ\h {\groth{W}}\ensuremath{^{\operatorname{op}}}\right)(-)}{F}\right)}\cong\wlim{\operatorname{W}^{\laxn}}{\left(F\circ \groth{W}\right)}\cong\laxnlim{\Delta 1}{\left(F\circ \groth{W}\right)}$$
either side existing if the other does, where the isomorphism in the middle is easy to prove.
\end{proof}
\begin{rem}\label{univlaxnormalcone}
The proof of \nameit{Theorem}\ref{redlaxnormalconical}, together with \nameit{Lemma}\ref{continuityinweight} and the proofs of \nameit{Theorem}\ref{laxnormalareweighted} and \nameit{Corollary}\ref{laxnormalcolimareweighted}, also shows how to obtain the correspondence between the universal cylinder of a weighted $2$-limit and the universal lax normal cocone of the associated lax normal conical $2$-limit. Calling the two, respectively,
$$\lambda\:W\aR{} \HomC{\catfont{C}}{L}{F(-)}$$
$$\hats{\lambda}\:\Delta 1 \alaxn{} \HomC{\catfont{C}}{L}{\left(F\circ \groth{W}\right)(=)}$$
for $F\:\catfont{A}\to\catfont{C}$ and $W\:\catfont{A}\to \catfont{CAT}$ two $2$-functors with $\catfont{A}$ small, the correspondence is given by the following equations, for every $(f,\alpha)\:(A,X)\to(B,X')$ in $\Grothop{W}$:
\begin{equation}\label{lambdahats}
\hats{\lambda}_{(A,X)}=\lambda_A(X)\quad\text{ and }\quad\hats{\lambda}_{(f,\alpha)}=\lambda_B(\alpha).
\end{equation}
\end{rem}
\begin{prop}\label{preservereflect}
A weighted $2$-limit is preserved or reflected precisely when its associated lax normal conical $2$-limit is so.
\end{prop}
\begin{proof}
Clear after \nameit{Remark}\ref{univlaxnormalcone}.
\end{proof}
\begin{rem}\label{remoplaxnormalvslaxnormal}
As weighted $2$-colimits in $\catfont{C}$ are just weighted $2$-limits in $\catfont{C}\ensuremath{^{\operatorname{op}}}$, we automatically obtain from \nameit{Theorem}\ref{redlaxnormalconical} the reduction of weighted $2$-colimits in $\catfont{C}$ to lax normal conical ones. More precisely, given $2$-functors $F\:\catfont{A}\to \catfont{C}$ and $W\:\catfont{A}\ensuremath{^{\operatorname{op}}}\to\catfont{CAT}$ with $\catfont{A}$ small, we obtain that
$$\wcolim{W}{F}\cong \laxncolim{\Delta 1}{\left(F\circ {\groth{W}}\ensuremath{^{\operatorname{op}}}\right)},$$
where $\groth{W}\:\Grothop{W}\to\catfont{A}\ensuremath{^{\operatorname{op}}}$.
But notice that, when $W\:\catfont{A}\ensuremath{^{\operatorname{op}}}\to\catfont{CAT}$, there is a more natural Grothendieck construction we can do on $W$, i.e.\ the one which produces the projection on the first component $\h\groth{W}\:\Groth{W}\to\catfont{A}$, with $\Groth{W}$ defined as follows:
\begin{description}
\item[an object of $\Groth{W}$] is a pair $(A,X)$ with $A\in \catfont{A}$ and $X\in F(A)$;
\item[a morphism $(A,X)\to (B,X')$ in $\Groth{W}$] is a pair $(f,\alpha)$ with $f\:A\to B$ a morphism in $\catfont{A}$ and $\alpha\:X\to W(f)(X')$ a morphism in $W(A)$;
\item[a $2$-cell $(f,\alpha)\aR{}(g,\beta)\:(A,X)\to (B,X')$ in $\Groth{W}$] is a $2$-cell $\delta\:f\aR{}g$ in $\catfont{A}$ such that $W(\delta)_{X'}\circ \alpha=\beta$;
\item[the compositions and identities] are analogous to the ones described in \nameit{Construction}\ref{2SetGrothconstrtoconicalize}.
\end{description}
In dimension $1$, given $Z\:\catfont{A}\ensuremath{^{\operatorname{op}}}\to\catfont{Set}$, we have that ${\left(\Grothop{Z}\right)}\ensuremath{^{\operatorname{op}}}$ and $\Groth{Z}$ coincide, but this is not true in dimension $2$.
We can obtain a formula of reduction of the weighted $2$-colimits to some kind of conical ones that uses this more natural Grothendieck construction, changing ``lax" into ``oplax". Our idea, that does not seem to appear in the literature, is that it is more natural to reduce weighted $2$-limits to lax normal conical ones, but weighted $2$-colimits to \predfn{oplax normal conical} ones. And this brings to \nameit{Theorem}\ref{oplaxnormalareweighted} and \nameit{Theorem}\ref{redoplaxnormalconical}, which are original.
\end{rem}
\begin{defne}
Let $W\:\catfont{A}\ensuremath{^{\operatorname{op}}}\to \catfont{CAT}$ be a $2$-functor with $\catfont{A}$ small, and consider $2$-functors $M,N\:{\left(\Groth{W}\right)}\ensuremath{^{\operatorname{op}}}\to \catfont{D}$. An \dfn{oplax normal natural transformation $\alpha$ from $M$ to $N$}, denoted $\alpha\:M\aoplaxn{}N$,\v is an oplax natural transformation $\alpha$ from $M$ to $N$ such that the structure $2$-cell on every morphism
$$\p{f,\id{W(f)(X)}}\:(A,W(f)(X))\al{} (B,X)$$
in ${\left(\Groth{W}\right)}\ensuremath{^{\operatorname{op}}}$ is the identity.
\end{defne}
\begin{defne}\label{defoplaxnormalconical}
Let $W\:\catfont{A}\ensuremath{^{\operatorname{op}}}\to \catfont{CAT}$ be a $2$-functor with $\catfont{A}$ small, and let $F\:\Groth{W}\to \catfont{C}$ be a $2$-functor. Notice that $\Groth{W}$ is small, since $\catfont{A}$ is small. The \dfn{oplax normal conical $2$-colimit of $F$}, denoted as $\oplaxncolim{F}$, is (if it exists) an object $C\in \catfont{C}$ together with an isomorphism of categories
$$\HomC{\catfont{C}}{C}{U}\cong \HomC{\moplaxn{{\left(\Grothdiag{W}\right)}\ensuremath{^{\operatorname{op}}}}{\catfont{CAT}}}{\Delta 1}{\HomC{\catfont{C}}{F(-)}{U}}$$
$2$-natural in $U\in \catfont{C}$,\v where $\moplaxn{{\left(\Grothdiag{W}\right)}\ensuremath{^{\operatorname{op}}}}{\catfont{CAT}}$ is the $2$-category of $2$-functors, oplax normal natural transformations and modifications from ${\left(\Groth{W}\right)}\ensuremath{^{\operatorname{op}}}$ to \catfont{CAT}. (Notice that indeed oplax normal natural transformations compose well vertically.)
\noindent When $\oplaxncolim{F}$ exists, taking $U=C$ and considering the identity on $C$ gives us in particular an oplax normal natural transformation
$$\mu\:\Delta 1 \aoplaxn{}\HomC{\catfont{C}}{F(-)}{C},$$
called the \dfn{universal oplax normal cocone}.
\end{defne}
\begin{teor}\label{oplaxnormalareweighted}
Oplax normal conical $2$-colimits are particular weighted $2$-colimits. More precisely, given $2$-functors $Z\:\catfont{A}\ensuremath{^{\operatorname{op}}}\to\catfont{CAT}$ and $F\:\Groth{Z}\to\catfont{C}$ with $\catfont{A}$ small, the weight that realizes $\oplaxncolim{F}$ is
\begin{fun}
\operatorname{W}^{\oplaxn} & \: & {\left(\Groth{Z}\right)}\ensuremath{^{\operatorname{op}}} \hphantom{.}& \longrightarrow & \catfont{CAT} \\[1ex]
&& \fib[o][2.85]{(B,X')}{(g,\beta)}{(C,X'')} &\mapsto & \fib[n][2.85]{\coslice{X'}{Z(B)}}{Z(g)(-)\circ\beta}{\coslice{X''}{Z(C)}}\\[6ex]
&&\tcv*'[4.5]{(B,X')}{(C,X'')}{(g,\beta)}{(h,\gamma)}{\delta} & \mapsto &\h[-3] \tc*+[7]{\coslice{X'}{Z(B)}}{\coslice{X''}{Z(C)}}{\beta\circ Z(g)(-)}{\gamma \circ Z(h)(-)}{Z(\delta)_{\operatorname{cod}(-)}}\\
\end{fun}
where the action of $Z(g)(-)\circ\beta$ on morphisms\v is given by $Z(g)(\operatorname{cod}(-))$.
\end{teor}
\begin{proof}
The proof is analogous to the one of \nameit{Theorem}\ref{laxnormalareweighted}.
\end{proof}
\begin{teor}\label{redoplaxnormalconical}
All the weighted $2$-colimits can be reduced to oplax normal conical ones. More precisely, given $2$-functors $F\:\catfont{A}\to \catfont{C}$ and $W\:\catfont{A}\ensuremath{^{\operatorname{op}}}\to\catfont{CAT}$ with $\catfont{A}$ small, we have that
$$\wcolim{W}{F}\cong \oplaxncolim{\left(F\circ \groth{W}\right)}$$
either side existing if the other does, where $\groth{W}\:\Groth{W}\to\catfont{A}$ is the $2$-$\catfont{Set}$-enriched Grothendieck construction of $W$.
\end{teor}
\begin{proof}
Exactly as in the proof of \nameit{Theorem}\ref{redlaxnormalconical}, we can prove that every $2$-presheaf $W\:\catfont{A}\ensuremath{^{\operatorname{op}}}\to\catfont{CAT}$ with $\catfont{A}$ small can be expressed as
$$W\cong \wcolim{W}{\yy}\cong \oplaxncolim{\left(\yy\circ\h{\groth{W}}\right)}\cong \wcolim{\operatorname{W}^{\oplaxn}}{\left(\yy\circ\h{\groth{W}}\right)},$$
where $\yy\:\catfont{A}\to\m{\catfont{A}\ensuremath{^{\operatorname{op}}}}{\catfont{CAT}}$ is the $2$-Yoneda embedding, and the last isomorphism is given by \nameit{Theorem}\ref{oplaxnormalareweighted}.
Using \nameit{Lemma}\ref{continuityinweight}, we obtain
$$\wcolim{W}{F}\cong \wcolim{\operatorname{W}^{\oplaxn}}{\left(\wcolim{\left(\yy\circ\h {\groth{W}}\right)(-)}{F}\right)}\cong \wcolim{\operatorname{W}^{\oplaxn}}{\left(F\circ \groth{W}\right)},$$
either side existing if the other does, and we conclude by \nameit{Theorem}\ref{oplaxnormalareweighted}.
\end{proof}
\begin{rem}\label{univoplaxnormalcocone}
Exactly as in \nameit{Remark}\ref{univlaxnormalcone}, in the notation of \nameit{Theorem}\ref{redoplaxnormalconical}, we can calculate the correspondence between the universal cocylinder of $\wcolim{W}{F}$ and the universal oplax normal cocone of $\oplaxncolim{\left(F\circ \groth{W}\right)}$. We find that it is the same as the one in equation~\refs{lambdahats}.
\end{rem}
\begin{prop}
A weighted $2$-colimit is preserved or reflected precisely when its associated oplax normal conical $2$-colimit is so.
\end{prop}
\begin{proof}
Clear after \nameit{Remark}\ref{univoplaxnormalcocone}.
\end{proof}
\begin{exampl}\label{univoplaxnormalcoconepresheaves}
By the proof of \nameit{Theorem}\ref{redoplaxnormalconical}, every $2$-presheaf $W\:\catfont{A}\ensuremath{^{\operatorname{op}}}\to\catfont{CAT}$ with $\catfont{A}$ small can be expressed as
$$W\cong \oplaxncolim{\left(\yy\circ\h{\groth{W}}\right)}.$$
The universal oplax normal cocone is given by
$$\forall \fib[o][2.4]{(B,X')}{(f,\alpha)}{(A,X)} \text{ in } {\Grothdiag{W}} \quad\quad\quad{
\begin{cd}*[4.5][2.25]
\y{A} \arrow[rr,"{\ceil{X}}",""'{name=Q}] \arrow[d,"{\y{f}}"'] \&\& W\\
\y{B} \arrow[rru,bend right,"{\ceil{X'}}"']\& \phantom{.} \arrow[Rightarrow,from=Q,"{\ceil{\alpha}}"{pos=0.36},shift right=1.5ex,shorten <=1ex,shorten >= 3.1ex]
\end{cd}}$$
In particular, taking $\catfont{A}=\catfont{1}$, we have that a $2$-presheaf on $\catfont{A}$ is just a small category $\catfont{D}$, and its $2$-$\catfont{Set}$-enriched Grothendieck construction just gives the functor to the terminal ${\catfont{D}}\to{\catfont{1}}$. We then obtain that every small category $\catfont{D}$ is the oplax normal conical $2$-colimit of $\Delta 1\:\catfont{D}\to \catfont{CAT}$, with oplax normal cocone given by
$$\forall \fib[o]{D}{f}{C} \text{ in } \catfont{D} \quad\quad\quad {\begin{cd}*[4][2]
\catfont{1} \arrow[rr,"C",""'{name=Q}] \arrow[d,equal] \&\& \catfont{D}\\
\catfont{1} \arrow[rru,bend right,"D"']\& \phantom{.} \arrow[Rightarrow,from=Q,"{f}"{pos=0.355},shift right=1.5ex,shorten <=0.8ex,shorten >= 2.45ex]
\end{cd}}$$
\end{exampl}
\begin{rem}\label{remoneisoplaxnormalconicaldense}
\nameit{Example}\ref{univoplaxnormalcoconepresheaves} shows that $\catfont{1}$ is ``oplax normal conical dense" in \catfont{CAT}. And this may also serve as another motivation for the (op)lax normal conical $2$-(co)limits. Indeed, this is sufficient to express all the (co)powers as (op)lax normal conical $2$-(co)limits. And when $\catfont{C}$ is a (co)complete $2$-category, every $2$-(co)limit in $\catfont{C}$ can be written in terms of (co)powers and conical $2$-(co)limits. Remember also that by \nameit{Example}\ref{conicalarelaxnormalconical} every conical $2$-(co)limits is an (op)lax normal conical one. We can now understand how the (op)lax normal conical (co)limits are the result of the idea (anticipated at the beginning of this section) of considering the $2$-cells between $\catfont{1}$ and a category $\catfont{D}$ in order to capture the whole of $\catfont{D}$, rather than considering the functors from $\catfont{2}$ (that brings instead to the weighted $2$-limits).
\end{rem}
\section{The 2-$\catfont{Set}$-enriched Grothendieck construction}\label{sectiontwosetgrothconstr}
In this section, we explore in detail the $2$-$\catfont{Set}$-enriched Grothendieck construction (\nameit{Definition}\ref{defexpltwosetgroth}), that firstly appeared in Street's~\cite{street_limitsindexedbycatvalued}, from a more abstract point of view, that is particularly relevant to, but not only, the (higher dimensional) elementary topos theory. This brings to the second main result of this paper (\nameit{Theorem}\ref{grothconstrislaxcomma}), which is expressing the $2$-$\catfont{Set}$-enriched Grothendieck construction as a \predfn{lax comma object in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$}, as defined here in \nameit{Definition}\ref{completeunivproplaxcomma} with a universal property that refines both the existing ones of Gray's~\cite{gray_formalcattheory} (unraveled by Kelly in~\cite{kelly_clubsanddoctrines}) and Lambert's~\cite{lambert_discretetwofib}. We explain how such laxness is needed to capture the $2$-$\catfont{Set}$-enriched Grothendieck construction (and the usual Grothendieck construction as well), and inscribe it in an original idea of $2$-$\catfont{V}$-enrichment, whence comes the name we gave to the studied construction.
The lax $3$-category $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$ of $2$-categories, $2$-functors, lax natural transformations and modifications is thus conceived as the archetypal $3$-dimensional elementary topos, suggesting to use the \predfn{lax comma objects in a lax $3$-category} (\nameit{Definition}\ref{completeunivproplaxcomma}) to regulate the classification process in a would-be \predfn{elementary $3$-topos}. Compare this with \nameit{Proposition}\ref{catofelementsiscomma}, that according to Weber's~\cite{weber_yonfromtwotop} presents $\catfont{CAT}$ as the archetypal elementary $2$-topos, with the construction of the category of elements as classification process, and suggests to regulate the classification by comma objects in dimension $2$. What gets classified in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$ is the \predfn{discrete $2$-fibrations} with small fibres (a generalized version of the usual Grothendieck fibrations) introduced by Lambert in~\cite{lambert_discretetwofib}, that are a locally discrete version of Hermida's $2$-fibrations (defined in~\cite{hermida_somepropoffibasafibredtwocat}). Insisting on a $2$-$\catfont{Set}$-enrichment point of view, we will call them \predfn{$2$-$\catfont{Set}$-opfibrations}.
We think that the $2$-$\catfont{Set}$-enriched Grothendieck construction should be taken into consideration towards the development of a general enriched Grothendieck construction (that considers general enriched fibrations). This will be explored in future work.
\begin{rem}\label{nextstepislaxcomma}
The guiding idea of this section is to categorify \nameit{Proposition}\ref{catofelementsiscomma}, presenting then the $2$-$\catfont{Set}$-enriched Grothendieck construction as the archetypal $3$-dimensional classification process (in the sense of a would-be tridimensional elementary topos theory). Analogously to the passage from pullbacks, that regulate the classification in dimension $1$, to comma objects, in dimension $2$, we expect to find now a further ``comma version" of the comma object that is suitable to a tridimensional ambient. The first instance of such process is given by \nameit{Proposition}\ref{laxnatinsidegrothconstr}, that shows the strictest filled square analogous to the one of \nameit{Proposition}\ref{catofelementsiscomma} we can have in the setting of the $2$-$\catfont{Set}$-enriched (but also just the usual) Grothendieck construction.
\end{rem}
\begin{prop}\label{laxnatinsidegrothconstr}
Let $F\:\catfont{B}\to \catfont{CAT}$ be a $2$-functor and consider its $2$-$\catfont{Set}$-enriched Grothendieck construction. There is a lax normal natural transformation $\lambda$ of the form
\begin{eqD*}
\sq*[l][7][7][\laxn][2.7][2.2][0.52]{\Grothop{F}}{\catfont{1}}{\catfont{B}}{\catfont{CAT}}{}{\groth{F}}{\catfont{1}}{F}
\end{eqD*}
\end{prop}
\begin{proof}
Given $(A,X)\in \Grothop{F}$, we define the component of $\lambda$ on $(A,X)$ to be the functor $\catfont{1}\to F(A)$ corresponding to $X\in F(A)$. Given now a morphism $(f,\alpha)\:(A,X)\to (B,X')$ in $\Grothop{F}$, we define the structure $2$-cell of $\lambda$ on $(f,\alpha)$ to be the natural transformation corresponding to $\alpha$. It is straightforward to see that this is indeed a lax normal natural transformation, by construction of $\Grothop{F}$.
\end{proof}
\begin{rem}\label{needof2catlax}
\nameit{Proposition}\ref{laxnatinsidegrothconstr} forces us to move out of $2\h[2]\mbox{-}\CAT$ in order to capture the $2$-$\catfont{Set}$-enriched (but also just the usual) Grothdendieck construction from an abstract point of view. Indeed, we need to at least admit the lax natural transformations as $2$-cells. And if we wish to recover the Grothendieck construction of pseudofunctors or of general lax functors into $\catfont{CAT}$, we also need to admit the lax functors as $1$-cells of our ambient. We will just consider strict $2$-functors for simplicity, to avoid at least the problems with the whiskering, but we actually expect everything to hold for lax functors as well (despite not having investigated much of it yet).
We call $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$ the lax $3$-category of $2$-categories, $2$-functors, lax natural transformations and modifications. In his paper~\cite{lambert_discretetwofib}, Lambert has indeed proved that this forms a lax $3$-category, that is, a category enriched over the $1$-category of $2$-categories and normal lax functors.
Be careful that $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$ has no underlying $2$-category, since the interchange rule now only holds in a lax version, in the sense that we now have a modification between the two possible lax natural transformations. Indeed, consider two lax natural transformations
\begin{cd}
\catfont{A}\arrow[r,bend left,"{F}",""'{name=A}]\arrow[r,bend right,"{G}"',""{name=B}]\&\catfont{B} \arrow[r,bend left,"{H}",""'{name=C}]\arrow[r,bend right,"{K}"',""{name=D}]\&\catfont{C};
\arrow[from=A,to=B,Rightarrow,"{\alpha}"]
\arrow[from=C,to=D,Rightarrow,"{\beta}"]
\end{cd}
Then for every $A\in \catfont{A}$, the component $\alpha_A$ is a morphism $F(A)\ar{}G(A)$ in $\catfont{B}$ and we can consider the structure $2$-cell of $\beta$ on such morphism. We obtain that $\beta_{\alpha_A}$ is a $2$-cell in $\catfont{C}$ of the form
\begin{cd}
H(F(A))\arrow[r,"{\beta_{F(A)}}"]\arrow[d,"{H(\alpha_A)}"']\& K(F(A))\arrow[d,"{K(\alpha_A)}"]\arrow[ld,Rightarrow,"{\beta_{\alpha_{A}}}",shorten <=2.7ex,shorten >=2.2ex]\\
H(G(A))\arrow[r,"{\beta_{G(A)}}"']\& K(G(A))
\end{cd}
And the $\beta_{\alpha_A}$'s collect into a modification $\beta_\alpha$, since the axiom we should check on a morphism $f\:A\to A'$ in $\catfont{A}$ is given by the $2$-dimensional property of $\beta$ being a lax natural transformation, applied to the $2$-cell $\alpha_f$ in $\catfont{B}$.
\end{rem}
\begin{rem}
\nameit{Remark}\ref{nextstepislaxcomma} and \nameit{Proposition}\ref{laxnatinsidegrothconstr} lead us to use the concept of \predfn{lax comma object}. This concept appears and is heavily used in Gray's book~\cite{gray_formalcattheory}, but without giving a complete universal property suitable to the lax $3$-categorical ambient $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$. Kelly just unraveled in~\cite{kelly_clubsanddoctrines} the partial universal property presented by Gray, while Lambert attempted in~\cite{lambert_discretetwofib} to give a better one but missing the uniqueness result in the $2$-dimensional part and giving only a partial $3$-dimensional part. We give in \nameit{Definition}\ref{completeunivproplaxcomma} (see also \nameit{Proposition}\ref{laxcommaisobject}) the complete universal property of the lax comma object, that refines both the ones of Gray (and Kelly) and Lambert.
In order to distinguish the explicit definition (given in Gray's~\cite{gray_formalcattheory}) from the complete universal property of the lax comma object, we will call the former ``\predfn{lax comma}" and the latter ``\predfn{lax comma object in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$}". However, we will use the same symbol for both, justified by \nameit{Proposition}\ref{laxcommaisobject}.
\end{rem}
\begin{defne}
Let $F\:\catfont{A}\to \catfont{C}$ and $G\:\catfont{B}\to \catfont{C}$ be $2$-functors. The \dfn{lax comma from $F$ to $G$} is the $2$-category $\laxcomma{F}{G}$ that is given by the following data:
\begin{description}
\item[an object of $\laxcomma{F}{G}$] is a triple $(A,B,h)$ with $A\in \catfont{A}$, $B\in \catfont{B}$ and $h\:F(A)\to G(B)$ a morphism in $\catfont{C}$;
\item[a $1$-cell $(A,B,h)\to (A',B',h')$ in $\laxcomma{F}{G}$] is a triple $(f,g,\varphi)$ with $f\:A\to A'$ in $\catfont{A}$, $g\:B\to B'$ in $\catfont{B}$ and
\sq[l][7][7][\varphi][2.3][2.3]{F(A)}{G(B)}{F(A')}{G(B')}{h}{F(f)}{G(g)}{h'}
a $2$-cell in $\catfont{C}$;
\item[a $2$-cell $(f,g,\varphi)\aR{}(f',g',\varphi')\:(A,B,h)\to (A',B',h')$] is a pair $(\alpha,\beta)$ with $\alpha\:f\aR{}f'$ in $\catfont{A}$ and $\beta\:g\aR{}g'$ in $\catfont{B}$ such that
\twonats[1.95][8][7][1.05][3.45][2]{F(A)}{G(B)}{F(A')}{G(B')}{F(f')}{F(f)}{G(g')}{G(g)}{h}{h'}{F(\alpha)}{G(\beta)}{\varphi}{\varphi'}
\item[the composition] of $1$-cells is given by pasting and that of $2$-cells is inherited by the ones in $\catfont{A}$ and $\catfont{B}$.
\end{description}
The \dfn{oplax comma from $F$ to $G$} is the co of the lax comma from $F\ensuremath{^{\operatorname{co}}}$ to $G\ensuremath{^{\operatorname{co}}}$.
\end{defne}
The following proposition shows the partial universal property of the lax comma object presented by Gray in~\cite{gray_formalcattheory}.
\begin{prop}\label{univproplaxcommagray}
Let $F\:\catfont{A}\to \catfont{C}$ and $G\:\catfont{B}\to \catfont{C}$ be $2$-functors. The lax comma from $F$ to $G$ is equivalently given by the enriched conical limit in $2\h[2]\mbox{-}\CAT$ of the diagram
\begin{cd}[4][4]
\catfont{A} \arrow[rd,"{F}"']\&\& \ARoplax{\catfont{C}} \arrow[rd,"{\operatorname{cod}}"] \arrow[ld,"{\operatorname{dom}}"']\&\& \catfont{B}\arrow[ld,"{G}"]\\
\& \catfont{C} \&\& \catfont{C}
\end{cd}
where $\ARoplax{\catfont{C}}$ is the lax comma from $\Id{\catfont{C}}$ to $\Id{\catfont{C}}$ \pteor{in some sense, the fundamental one}.
\end{prop}
\begin{rem}
But there is a better universal property that the lax comma satisfies. Indeed, it is a \predfn{lax comma object in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$}, as defined here below.
\end{rem}
\begin{defne}\label{completeunivproplaxcomma}
Let $\catfont{Q}$ be a lax $3$-category and consider $1$-cells $F\:\catfont{A}\to \catfont{C}$ and $G\:\catfont{B}\to \catfont{C}$ in $\catfont{Q}$. The \dfn{lax comma object in $\catfont{Q}$ from $F$ to $G$} is, if it exists, an object $\laxcomma{F}{G}\in \catfont{Q}$ together with a $2$-cell
\begin{cd}
\laxcomma{F}{G} \arrow[r,"{\partial_0}"]\arrow[d,"{\partial_1}"']\& \catfont{A}\arrow[d,"{F}"]\arrow[ld,Rightarrow,"{\lambda}",shorten <= 2.7ex, shorten >=2.2ex]\\
\catfont{B}\arrow[r,"{G}"']\& \catfont{C}
\end{cd}
in $\catfont{Q}$ that is universal in the following lax $3$-categorical sense:
\begin{enum}
\item for every $2$-cell $\gamma\:F\circ P\aR{}G\circ Q\:\catfont{M}\to \catfont{C}$, there exists a unique $1$-cell $V\:\catfont{M}\to \laxcomma{F}{G}$ such that
\begin{eqD*}
\begin{cd}*[6.5][6.5]
\catfont{M}\arrow[rrd,bend left,"{P}"]\arrow[rdd,bend right,"{Q}"']\\[-5ex]
\&[-5ex]\& \catfont{A}\arrow[d,"{F}"]\arrow[ld,Rightarrow,shift right=3.4ex,"{\gamma}"{pos=0.53},shorten <= 2.2ex, shorten >=1.4ex]\\
\&\catfont{B}\arrow[r,"{G}"']\& \catfont{C}
\end{cd}\quad = \quad
\begin{cd}*[6.5][6.5]
\catfont{M}\arrow[rrd,bend left,"{P}"]\arrow[rdd,bend right,"{Q}"']\arrow[rd,"{V}",dashed,shorten <=-0.2ex,shorten >=-0.2ex]\\[-5ex]
\&[-5ex]\laxcomma{F}{G} \arrow[r,"{\partial_0}"]\arrow[d,"{\partial_1}"']\& \catfont{A}\arrow[d,"{F}"]\arrow[ld,Rightarrow,"{\lambda}",shorten <= 2.7ex, shorten >=2.2ex]\\
\&\catfont{B}\arrow[r,"{G}"']\& \catfont{C}
\end{cd}
\end{eqD*}
\item for every $1$-cells $V,W\:\catfont{M}\to \laxcomma{F}{G}$ and every $3$-cell
\begin{eqD*}
\begin{cd}*
\catfont{M}\arrow[dd,"{W}"']\arrow[rd,"{V}"]\\[-5ex]
\&[-4ex]\laxcomma{F}{G}\arrow[ld,Rightarrow,"{\Delta}",shorten <=0ex, shorten >=0ex] \arrow[r,"{\partial_0}"]\arrow[dd,"{\partial_1}"']\& \catfont{A}\arrow[dd,"{F}"]\arrow[ldd,Rightarrow,"{\lambda}",shorten <= 2.7ex, shorten >=2.2ex]\\[-5ex]
\laxcomma{F}{G}\arrow[rd,"{\partial_1}"']\\[-5ex]
\&\catfont{B}\arrow[r,"{G}"']\& \catfont{C}
\end{cd} \quad\aMM{\Xi}\quad
\begin{cd}*
\catfont{M}\arrow[rr,"{V}"]\arrow[rd,"{W}"']\&[-5ex]\&[-7ex] \laxcomma{F}{G}\arrow[rd,"{\partial_0}"]\arrow[ld,Rightarrow,"{\Gamma}"]\\[-4ex]
\&\laxcomma{F}{G} \arrow[rr,"{\partial_0}"]\arrow[d,"{\partial_1}"']\&\&[-5ex]\catfont{A}\arrow[d,"{F}"]\arrow[lld,Rightarrow,"{\lambda}",shorten <= 2.7ex, shorten >=2.2ex]\\
\&\catfont{B}\arrow[rr,"{G}"']\&\& \catfont{C}
\end{cd}
\end{eqD*}
for $2$-cells $\Gamma$ and $\Delta$, there exists a unique $2$-cell $\nu\:V\aR{}W$ such that
$$\partial_0\h \nu=\Gamma, \quad \partial_1\h \nu=\Delta, \quad \lambda_{\nu}=\Xi;$$
notice that we are precisely asking that $\Xi$ corresponds to the $3$-cell given by the lax interchange rule in $\catfont{Q}$ of
\begin{cd}
\catfont{M}\arrow[r,bend left,"{V}"{pos=0.57},""'{name=A}]\arrow[r,bend right,"{W}"'{pos=0.57},""{name=B}]\&\laxcomma{F}{G} \arrow[r,bend left,"{F\circ \partial_0}"{pos=0.43},""'{name=C}]\arrow[r,bend right,"{G\circ \partial_1}"'{pos=0.43},""{name=D}]\&\catfont{C};
\arrow[from=A,to=B,Rightarrow,"{\nu}"]
\arrow[from=C,to=D,Rightarrow,"{\lambda}"]
\end{cd}
\item for every $2$-cells $\nu,\omega\:V\aR{}W\:\catfont{M}\to \laxcomma{F}{G}$ and every pair of $3$-cells $\Phi\:\partial_0\h \nu\aM{}\partial_0\h \omega$ and $\Psi\:\partial_1\h \nu\aM{}\partial_1\h \omega$ such that
\begin{eqD}{tridimpartcompleteunivproplaxcomma}
\lambda_{\omega}\circ
\begin{cd}*
\catfont{M}\arrow[dd,"{W}"']\arrow[rd,"{V}"]\\[-5ex]
\&[-4ex]\laxcomma{F}{G}\arrow[ld,Rightarrow,bend right,"{\partial_1\h \nu}"'{pos=0.42},""'{name=A},shorten <=0.4ex, shorten >=0.4ex]\arrow[ld,Rightarrow,bend left,"{\partial_1\h \omega}"{pos=0.58},""{name=B},shorten <=0.4ex, shorten >=0.4ex] \arrow[r,"{\partial_0}"]\arrow[dd,"{\partial_1}"]\& \catfont{A}\arrow[dd,"{F}"]\arrow[ldd,Rightarrow,"{\lambda}",shorten <= 2.7ex, shorten >=2.2ex]\\[-5ex]
\laxcomma{F}{G}\arrow[rd,"{\partial_1}"']\\[-5ex]
\&\catfont{B}\arrow[r,"{G}"']\& \catfont{C}
\arrow[from=A,to=B,triple,"{\Psi}",shorten <=1.1ex,shorten >=0.9ex]
\end{cd} \quad=\quad
\begin{cd}*
\catfont{M}\arrow[rr,"{V}"]\arrow[rd,"{W}"']\&[-5ex]\&[-7ex] \laxcomma{F}{G}\arrow[rd,"{\partial_0}"]\arrow[ld,Rightarrow,bend right,"{\partial_0\h \nu}"'{pos=0.58},""'{name=C},shorten <=0.4ex,shorten >=0.4ex]\arrow[ld,Rightarrow,bend left,"{\partial_0\h\omega}"{pos=0.42},""{name=D},shorten <=0.4ex,shorten >=0.4ex]\\[-4ex]
\&\laxcomma{F}{G} \arrow[rr,"{\partial_0}"']\arrow[d,"{\partial_1}"']\&\&[-5ex]\catfont{A}\arrow[d,"{F}"]\arrow[lld,Rightarrow,"{\lambda}",shorten <= 2.7ex, shorten >=2.2ex]\\
\&\catfont{B}\arrow[rr,"{G}"']\&\& \catfont{C}
\arrow[from=C,to=D,triple,"{\Phi}",shorten <=1.1ex,shorten >=0.9ex]
\end{cd}\circ \lambda_{\nu}
\end{eqD}
there exists a unique $3$-cell $\Theta\:\nu\aM{}\omega$ such that $\partial_0\h \Theta=\Phi$ and $\partial_1\h \Theta=\Psi$.
\end{enum}
\end{defne}
\begin{prop}\label{laxcommaisobject}
Let $F\:\catfont{A}\to \catfont{C}$ and $G\:\catfont{B}\to \catfont{C}$ be $2$-functors. Then there is a lax natural transformation
\begin{cd}
\laxcomma{F}{G} \arrow[r,"{\partial_0}"]\arrow[d,"{\partial_1}"']\& \catfont{A}\arrow[d,"{F}"]\arrow[ld,Rightarrow,"{\lambda}",shorten <= 2.7ex, shorten >=2.2ex]\\
\catfont{B}\arrow[r,"{G}"']\& \catfont{C}
\end{cd}
that exhibits the lax comma $\laxcomma{F}{G}$ as the lax comma object in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$ from $F$ to $G$. Moreover, in condition $(ii)$ of lax comma object in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$, if $\Gamma$ and $\Delta$ \pteor{that can be general lax natural transformations} are both strict $2$-natural \pteor{resp. pseudo-natural} then also $\nu$ is so.
\end{prop}
\begin{proof}
Firstly, we construct $\lambda$. Given $(A,B,h)\in \laxcomma{F}{G}$, we define the component of $\lambda$ on it to be $h$. Given a morphism $(f,g,\varphi)\:(A,B,h)\to (A',B',h')$ in $\laxcomma{F}{G}$, we define the structure $2$-cell of $\lambda$ on it to be $\varphi$. It is then straightforward to show that $\lambda$ is a lax natural transformation.
For condition $(i)$ of lax comma object in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$, since $\lambda$ picks the third component of objects and morphisms in $\laxcomma{F}{G}$ and the other two components are determined by the projections through $\partial_0$ and $\partial_1$, we have to define $V$ as
$$V(M)\mathrel{\mathrel{\mathop\ordinarycolon}\mkern-1.2mu=} (P(M),Q(M),\gamma_M)$$
$$V(m)\mathrel{\mathrel{\mathop\ordinarycolon}\mkern-1.2mu=} (P(m),Q(m),\gamma_m)$$
for every $M\in \catfont{M}$ and every morphism $m$ in $\catfont{M}$. This is easily checked to be a $2$-functor, and it works by construction.
For $(ii)$, take an arbitrary $\Xi$ as above. We see that the three requests
$$\partial_0\h \nu=\Gamma, \quad \partial_1\h \nu=\Delta, \quad \lambda_{\nu}=\Xi$$
force us to construct the component of $\nu$ on an arbitrary $M\in \catfont{M}$ to be the morphism $(\Gamma_M,\Delta_M,\Xi_M)$ in $\laxcomma{F}{G}$. Then the structure $2$-cell of $\nu$ on a morphism $m\:M\to M'$ in $\catfont{M}$, being a $2$-cell in $\laxcomma{F}{G}$, is determined by its projections through $\partial_0$ and $\partial_1$. So we are forced to define $\nu_m$ to be the $2$-cell $(\Gamma_m,\Delta_m)$ in $\laxcomma{F}{G}$. This is indeed a $2$-cell since $\Xi$ is a modification. It is straightforward to check that $\nu$ is a lax natural transformation, since $\Gamma$ and $\Delta$ are so. And we immediatly see that if both $\Gamma$ and $\Delta$ are strict $2$-natural (resp. pseudo-natural) then also $\nu$ is so. The observation that $\nu$ is then the unique lax natural transformation $V\alax{}W$ such that the modification corresponding to the lax interchange rule in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$ of $\nu$ and $\lambda$ coincides with $\Xi$ follows from \nameit{Remark}\ref{needof2catlax}.
For $(iii)$, let $M\in \catfont{M}$. Since the component $\Theta_M$ will be a $2$-cell in $\laxcomma{F}{G}$, it is determined by its projections through $\partial_0$ and $\partial_1$. So we need to define
$$\Theta_M\mathrel{\mathrel{\mathop\ordinarycolon}\mkern-1.2mu=} (\Phi_M,\Psi_M).$$
That this is indeed a $2$-cell $\nu_M\aR{}\omega_M$ in $\laxcomma{F}{G}$ is guaranteed by equation~\refs{tridimpartcompleteunivproplaxcomma}, taking components on $M$. The condition that the $\Theta_M$'s need to satisfy in order for them to collect into a modification $\Theta$ is then an equality between $2$-cells in $\laxcomma{F}{G}$, and thus it suffices to check its projections through $\partial_0$ and $\partial_1$. But those two resulting conditions are given by the fact that both $\Phi$ and $\Psi$ are modifications.
\end{proof}
\begin{rem}
Notice from \nameit{Definition}\ref{completeunivproplaxcomma} that the lax comma object in a lax $3$-category really is an upgrade of the comma object to a lax $3$-dimensional ambient. Indeed, a lax comma object in a $2$-category is precisely a comma object, since any $\Xi$ of \nameit{Definition}\ref{completeunivproplaxcomma} is then forced to be the identity, and the tridimensional part becomes trivial. Interestingly, the uniqueness in the $2$-dimensional part of the universal property of the lax comma object in a lax $3$-category is obtained by considering the lax interchange rule.
The universal property of \nameit{Proposition}\ref{univproplaxcommagray} is obtained precisely by restricting ourselves to consider as $\Gamma$ and $\Delta$ only strict $2$-natural transformations.
\end{rem}
\begin{teor}\label{grothconstrislaxcomma}
Let $F\:\catfont{B}\to \catfont{CAT}$ be a $2$-functor. The $2$-$\catfont{Set}$-enriched Grothendieck construction, defined explicitly in \nameit{Construction}\ref{2SetGrothconstrtoconicalize}, is equivalently given by the lax comma object
\begin{eqD}{grothconstrislaxcommadiagram}
\sq*[l][7][7][\lax \opn{comma}][2.7][2.2][0.65]{\Grothop{F}}{\catfont{1}}{\catfont{B}}{\catfont{CAT}}{}{\groth{F}}{\catfont{1}}{F}
\end{eqD}
in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$ and such lax comma object is presented by the lax normal natural transformation $\lambda$ of \nameit{Proposition}\ref{laxnatinsidegrothconstr}.
As a consequence, it is then also given by the strict $3$-pullback in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$ between $F$ and the replacement $\tau$ of $\catfont{1}\:\catfont{1}\to \catfont{CAT}$ obtained by taking the lax comma object of $\catfont{1}\:\catfont{1}\to \catfont{CAT}$ along the identity of $\catfont{CAT}$ \pteor{that is a lax $3$-dimensional version of the lax limit of the arrow $\catfont{1}\:\catfont{1}\to \catfont{CAT}$}:
\begin{eqD*}
\begin{cd}*
\Grothop{F}\PB{rd} \arrow[d,"\groth{F}"'] \arrow[r] \& \catfont{CAT}_{\bullet,\operatorname{lax}} \arrow[d,"{\tau}"'] \arrow[r]\& \catfont{1} \arrow[d,"{\catfont{1}}"]\arrow[ld,Rightarrow,shorten <=2.7ex,shorten >=2.2ex,"\lax \opn{comma}"{pos=0.65}] \\
\catfont{B}\arrow[r,"F"'] \& \catfont{CAT} \arrow[r,equal]\& \catfont{CAT}
\end{cd}
\end{eqD*}
The domain of $\tau$ is a lax pointed version of $\catfont{CAT}$, whence the notation $\catfont{CAT}_{\bullet,\operatorname{lax}}$.
\end{teor}
\begin{proof}
The proof is a straightforward calculation. The fact that $\groth{F}$ is then also the strict $3$-pullback of $\tau$ is readily checked by showing that such strict $3$-pullback satisfies the universal property of the lax comma object $\laxcomma{\catfont{1}}{F}$ in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$ that we have presented in \nameit{Definition}\ref{completeunivproplaxcomma}, using the universal properties of $\laxcomma{\catfont{1}}{\Id{\catfont{CAT}}}$ and of the strict $3$-pullback together with some basics of the calculus of pasting.
\end{proof}
\begin{rem}
Before commenting on \nameit{Theorem}\ref{grothconstrislaxcomma} (we will in \nameit{Remark}\ref{remarchetypalthreedimclassifproc}), we use it to canonically extend the $2$-$\catfont{Set}$-enriched Grothendieck construction by functoriality.
\end{rem}
\begin{prop}\label{twosetgrothcanonicallyextendstwofunctor}
By \nameit{Theorem}\ref{grothconstrislaxcomma}, the $2$-$\catfont{Set}$-enriched Grothendieck construction canonically extends to a $2$-functor
$$\groth{-}\:\m{\catfont{B}}{\catfont{CAT}}\to \slice{2\h[2]\mbox{-}\CAT}{\catfont{B}}$$
and to one with domain $\mlax{\catfont{B}}{\catfont{CAT}}$ as well, for every $2$-category $\catfont{B}$.
\end{prop}
\begin{proof}
Given a $2$-natural transformation $\varphi\:F\aR{}G\:\catfont{B}\to \catfont{CAT}$ - also a lax natural transformation would equally work - we define $\groth{\varphi}$ as the unique morphism $\groth{\varphi}\:\Grothop{F}\to \Grothop{G}$ induced by the universal property of the lax comma object $\Grothop{G}$ in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$ applied to the lax natural transformation
\begin{cd}[7][8]
{\Grothop{F}}\arrow[d,"{\groth{F}}"'] \arrow[r,"{}"]\& {\catfont{1}}\arrow[ld,Rightarrow,shift right=1ex,"{\lambda^F}"'{pos=0.475, inner sep=0.25ex},shorten <=3.2ex, shorten >=3.3ex] \arrow[d,"{\catfont{1}}"]\\
{\catfont{B}} \arrow[r,bend left,"{F}"{pos=0.64},""'{name=A}]\arrow[r,bend right,"{G}"',""{name=B}] \& {\catfont{CAT}}
\arrow[Rightarrow,from=A,to=B,shorten <=0.4ex,shorten >=0.4ex,"\varphi"]
\end{cd}
where $\lambda^F$ is the lax natural transformation that presents $\Grothop{F}$ as a lax comma object in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$. Explicitly, for every $2$-cell $\delta\:(f,\alpha)\aR{}(g,\beta)\:(B,X)\to (C,X')$ in $\Grothop{F}$
$$\groth{\varphi}(B,X)=(B,\varphi_B(X)) \quad \text{and} \quad \groth{\varphi}(f,\alpha)=(f,\varphi_C(\alpha))\quad \text{and} \quad \groth{\varphi}(\delta)=\delta.$$
Given a modification $\Theta\:\varphi\aM{} \psi\:F\aR{}G\:\catfont{B}\to \catfont{CAT}$, we define $\groth{\Theta}$ as the unique $2$-natural transformation induced by the universal property of the lax comma object $\Grothop{G}$ in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$ applied, in the notation of \nameit{Definition}\ref{completeunivproplaxcomma} to $V=\groth{\varphi}$, $W=\groth{\psi}$, $\Gamma=\id{}$, $\Delta=\id{}$ and $\Xi$ given by
\begin{cd}[7][8]
{\Grothop{F}}\arrow[d,"{\groth{F}}"'] \arrow[r,"{}"]\& {\catfont{1}}\arrow[ld,Rightarrow,shift right=1ex,"{\lambda^F}"'{pos=0.475},shorten <=3.2ex, shorten >=3.3ex] \arrow[d,"{\catfont{1}}"]\\
{\catfont{B}} \arrow[r,bend left=35,"{F}"{pos=0.64},""'{name=A}]\arrow[r,bend right=35,"{G}"',""{name=B}] \& {\catfont{CAT}}
\arrow[Rightarrow,from=A,to=B,bend right=40,shift right=1ex,shorten <=0.4ex,shorten >=0.4ex,"\varphi"',""{name=L}]
\arrow[Rightarrow,from=A,to=B,bend left=40,shift left=1ex,shorten <=0.4ex,shorten >=0.4ex,"\psi",""'{name=R}]
\arrow[triple,from=L,to=R,shorten <=0ex,shorten >=-0.2ex,"\Theta"{inner sep=1ex}]
\end{cd}
Explicitly, the component of $\groth{\Theta}$ on an object $(B,X)\in \Grothop{F}$ is
$$\groth{\Theta}_{(B,X)}=\left(\id{B},\Theta_{B,X}\right).$$
It is straightforward to show that $\groth{-}$ is indeed a $2$-functor.
\end{proof}
\begin{rem}\label{remarchetypalthreedimclassifproc}
\nameit{Theorem}\ref{grothconstrislaxcomma} shows in which sense the $2$-$\catfont{Set}$-enriched Grothendieck construction can be thought of as the archetypal $3$-dimensional classifier, in the sense of a would-be elementary $3$-topos theory. Weber proposed in~\cite{weber_yonfromtwotop} to convert, in the passage from dimension $1$ to dimension $2$, all the monomorphisms into discrete opfibrations. As we said in the introduction, also keeping the classifier to have the terminal as domain and thus be the inclusion of a verum inside of generalized truth values but changing the classification process to be regulated by comma objects is equally good, and actually preferable. Now, in order to reach the dimension $3$, we propose to either upgrade the classification process into one regulated by lax comma objects (as defined here in \nameit{Definition}\ref{completeunivproplaxcomma}) or to consider pullbacks of the notion of fibration that the $2$-$\catfont{Set}$-enriched Grothendieck construction produces. We will describe such notion of fibration in detail below, see \nameit{Definition}\ref{deftwosetopf}.
It is interesting to notice that we had to move out of $2\h[2]\mbox{-}\CAT$ (see also \nameit{Remark}\ref{needof2catlax}), in order to capture the laxness that permeates the Grothendieck construction (seen already from \nameit{Construction}\ref{2SetGrothconstrtoconicalize}, where we simultaneously constructed the lax normal conical $2$-limits and the $2$-$\catfont{Set}$-enriched Grothendieck construction). So the archetypal elementary $3$-topos seems to be $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$. This idea is also reinforced by the fact that Buckley, in his paper~\cite{buckley_fibredtwocatandbicat}, continuing the work of Bakovi\'c's~\cite{bakovic_fibrofbicat}, found the need of considering trihomomorphisms $F\:\catfont{B}\ensuremath{^{\operatorname{coop}}}\to 2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$ in order to capture non-split Hermida's $2$-fibrations (presented in Hermida's~\cite{hermida_somepropoffibasafibredtwocat}) via a suitable Grothendieck construction. So what seems to work is the sequence
$$\catfont{Set}\quad\leadsto\quad \catfont{CAT} \quad\leadsto\quad 2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$$
We believe that such sequence is best explained by what we call a \predfn{$2$-$\catfont{V}$-enrichment}, given $\catfont{V}$ a nice enough monoidal category, so that we view the sequence as
$$\catfont{V}\quad\leadsto\quad \VCAT{\catfont{V}} \quad\leadsto\quad \twoVCAT{\catfont{V}}$$
with $\catfont{V}=\catfont{Set}$. Our idea is that enriching again over $\VCAT{\catfont{V}}$ should take into account the fact that $\VCAT{\catfont{V}}$ is a $2$-category, and then be a \predfn{weak enrichment} rather than an ordinary enrichment (that would only consider the underlying category of $\VCAT{\catfont{V}}$). We would like to thank Francesco Dagnino for the interesting discussions that brought to this idea of $2$-$\catfont{V}$-enrichment.
As described with more detail below, weakly enriching over $\VCAT{\catfont{Set}}=\catfont{CAT}$ actually produces bicategories, lax functors, lax natural transformations and modifications, but we will restrict to consider $2$-categories and $2$-functors, for simplicity. Nevertheless, we will be able to capture the lax natural transformations in this way, and the $2$-$\catfont{V}$-enrichment idea will guide us to propose a notion of colimit and of pointwise Kan extension in the still poorly studied context of $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$, in Section~\ref{sectionkanextensions}. We also notice that the further step in the sequence above could bring towards a version of the Gray-categories, but we have not investigated this yet.
\end{rem}
We recall the following remark on the ordinary enrichment.
\begin{rec}\label{remenrichment}
If $(\catfont{V},\otimes,I)$ is a monoidal category with coproducts such that $-\otimes -$ preserves coproducts in each variable, we can define a $\catfont{V}$-enriched category as pair $(S,\catfont{A})$ with $S$ a set, that will be the set of objects, and $\catfont{A}$ a monoid in the monoidal category $\m{S\times S}{\catfont{V}}$ of functors (actually given by mere functions) from $S\times S$ to $\catfont{V}$ and natural transformations (actually given by the mere components), that we think of as the monoidal category of square matrices indexed by $S$ with entries in $\catfont{V}$, with the tensor product given by matrix multiplication and tensor unit given by the identity matrix (with $I$ all over the main diagonal and the initial object elsewhere). The multiplication of the monoid $\catfont{A}$ gives indeed the composition of the enriched category, the unit gives the identities and the axioms of monoid precisely ask the composition to be associative and unital.
We can then also define $\catfont{V}$-enriched functors on this line, using the following construction. Given a $\catfont{V}$-enriched category $(T,\catfont{B})$ and a function $F\:S\to T$, we can define a monoid $F^{\ast} \catfont{B}$ in $\m{S\times S}{\catfont{V}}$, taking
$$F^{\ast} \catfont{B} =\left(S\times S\ar{F\times F} T\times T \ar{\catfont{B}}\catfont{V}\right)$$
and defining the multiplication and the unit by whiskering those of $\catfont{B}$ with $F\times F$ on the left. Indeed $F^{\ast}\catfont{B}$ is a monoid, by pasting calculations, since all the required axioms just involve cells of strictly higher levels than $F\times F$.
Given $(S,\catfont{A})$ and $(T,\catfont{B})$ two $\catfont{V}$-enriched categories, a $\catfont{V}$-enriched functor $(S,\catfont{A})\to (T,\catfont{B})$ can be defined as a pair $(F,\overline{F})$ with $F\:S\to T$ a function and $\overline{F}\:\catfont{A}\to F^{\ast}\catfont{B}$ a morphism between monoids in $\m{S\times S}{\catfont{V}}$.
\end{rec}
\begin{rem}
We now give a weak $2$-categorical generalization of \nameit{Recall}\ref{remenrichment}. The concept of \predfn{weak enrichment} is explored in Garner and Shulman's~\cite{garnershulman_enrichedcatasafreecocompl}, but in the terms presented below (on the line of \nameit{Recall}\ref{remenrichment}) it does not seem to appear in the literature.
We will use the concepts of pseudomonoid in a $2$-category and of lax morphism between them. A definition of them can be found in Vasilakopoulou's~\cite{vasilakopoulou_monoidalgrothconstr}.
\end{rem}
\begin{cons}[Weak enrichment]\label{consweakenrichment}
Let $(\catfont{K},\otimes,I,\alpha,\lambda,\rho)$ be a monoidal $2$-category, i.e.\ a $2$-category $\catfont{K}$ that is monoidal in the $1$-dimensional sense but such that the tensor product is a $2$-functor $\catfont{K}\times \catfont{K}\to \catfont{K}$. And assume that $\catfont{K}$ has coproducts and that $-\otimes -$ preserves them in each variable. Then, for every set $S$, the $2$-category $\m{S\times S}{\catfont{K}}$ is $2$-monoidal as well, with tensor product given by matrix multiplication and tensor unit given by the identity matrix. Indeed, the matrix multiplication can be extended to a $2$-functor using the $2$-dimensional property of the (now enriched) coproducts, with the $2$-functoriality given by the fact that everything can be discharged on components and that $-\otimes -\:\catfont{K}\times \catfont{K}\to \catfont{K}$ is a $2$-functor.
We define a \dfn{$\catfont{K}$-weakly enriched category} as a pair $(S,\catfont{A})$ with $S$ a set, thought as the set of objects, and $\catfont{A}$ a pseudomonoid in the monoidal $2$-category $\m{S\times S}{\catfont{K}}$ of square $S$-indexed matrices with entries in $\catfont{K}$ (whose $1$-cells are the $2$-natural transformations and whose $2$-cells are the modifications). Notice that a strict $2$-monoid in $\m{S\times S}{\catfont{K}}$ is the same thing as a monoid in the monoidal category $\m{S\times S}{{\catfont{K}}_{\h[3]0}}$, and thus precisely an enriched category over ${\catfont{K}}_{\h[3]0}$ with object set $S$.
Now, notice that if $(T,\catfont{B})$ is a $\catfont{K}$-weakly enriched category and $F\:S\to T$ is a function, then
$$F^{\ast} \catfont{B} =\left(S\times S\ar{F\times F} T\times T \ar{\catfont{B}}\catfont{V}\right)$$
is a pseudomonoid in $\m{S\times S}{\catfont{K}}$, defining all the needed structure cells as those of $\catfont{B}$ whiskered with $F\times F$ on the left. That $F^{\ast} \catfont{B}$ is a pseudomonoid holds by pasting calculations, since all the required axioms just involve cells of strictly higher levels than $F\times F$.
Given two $\catfont{K}$-weakly enriched categories $(S,\catfont{A})$ and $(T,\catfont{B})$, we define a \dfn{$\catfont{K}$-weakly enriched functor $(S,\catfont{A})\to (T,\catfont{B})$} as a pair $(F,\overline{F})$ with $F\:S\to T$ a function and $\overline{F}\:\catfont{A}\to F^{\ast}\catfont{B}$ a lax morphism between lax monoids in $\m{S\times S}{\catfont{K}}$.
Given now $(F,\overline{F}), (G,\overline{G})\:(S,\catfont{A})\to (T,\catfont{B})$ two $\catfont{K}$-weakly enriched functors, we define a \dfn{$\catfont{K}$-weakly enriched natural transformation $\varphi\:(F,\overline{F})\alax{}(G,\overline{G})$} as a collection of $1$-cells
$$\varphi_A\:I\to \HomC{\catfont{B}}{F(A)}{G(A)}$$
in $\catfont{K}$ for every $A\in S$ and $2$-cells
\begin{cd}[2.2][4]
\& \catfont{A}(A,B) \otimes I \arrow[r, "G\otimes \varphi_{A}",""'{name=A}] \&[1.9ex] \catfont{B}(GA,GB) \otimes \catfont{B}(FA,GA) \arrow[rd, "\opn{comp}"{inner sep=0.3ex}] \\
\catfont{A}(A,B) \arrow[rd, "\lambda_{\catfont{A}(A,B)}^{-1}"'{inner sep=0.3ex}] \arrow[ru, "\rho_{\catfont{A}(A,B)}^{-1}"{inner sep=0.3ex}] \&\&\&[-3ex] \catfont{B}(FA,GB) \\
\& I \otimes \catfont{A}(A,B) \arrow[r, "\varphi_{B}\otimes F"',""{name=B}] \& \catfont{B}(FB,GB) \otimes \catfont{B}(FA,FB) \arrow[ru, "\opn{comp}"'{inner sep=0.3ex}]
\arrow[Rightarrow,from=A,to=B,shift left=2.85ex,"{\h[1]\varphi_{A,B}}",shorten <=2.3ex,shorten >=2ex]
\end{cd}
in $\catfont{K}$ for every pair $(A,B)\in S\times S$ such that, for every $A,B,C\in S$, with notations like $\catfont{A}_{A,B}\mathrel{\mathrel{\mathop\ordinarycolon}\mkern-1.2mu=}\HomC{\catfont{A}}{A}{B}$ and $\catfont{B}^{F,G}_{A,B}\mathrel{\mathrel{\mathop\ordinarycolon}\mkern-1.2mu=}\HomC{\catfont{B}}{F(A)}{G(B)}$ and omitting the tensor product of objects, the pasting in $\catfont{K}$\v[1]
\newcommand{\z}[0]{}
\begin{cd}[3][4.75]
\&\& \catfont{B}^{FG}_{AA}
\arrow[rrrdd,bend left=20,equal,"{}"]\arrow[rd,"{\lambda^{-1}}"{inner sep=0.2ex}]
\arrow[d,equal,shift left=5.15ex,shorten <= 3.8ex,shorten >=-2.3ex]
\\
I
\arrow[rru,bend left=18,"{\varphi_A}"{inner sep=0.2ex}]\arrow[rr,"{\id{G(A)}}"]\arrow[rd,"{\id{A}}"'{inner sep=0.2ex}]\&
\arrow[d,Rightarrow,shift left=0.5ex,"{\overline{G}\h[1]}"'{pos=0.4},shorten <= 0.35ex,shorten >=0.7ex]\&
\catfont{B}^{GG}_{AA}
\arrow[rd,"{\rho^{-1}}"{inner sep=0.2ex}]
\arrow[d,equal,shift left=2.6ex,shorten <= 1.35ex,shorten >=0.35ex]\&
I\z \catfont{B}^{FG}_{AA}
\arrow[rd,"{\id{G(A)}\otimes 1}"{inner sep=0.2ex}]
\arrow[r,iso,shift left=0.2ex,"{\opn{unit}_\catfont{B}^{-1}}"{inner sep=1ex}]\&\vphantom{.}
\\
\& \catfont{A}_{AA}
\arrow[r,"{\rho^{-1}}"]\arrow[ru,"{G}"{inner sep=0.2ex}]\arrow[rd,"{\lambda^{-1}}"'{inner sep=0.2ex}]\&
\catfont{A}_{AA}\z I
\arrow[r,"{G\otimes 1}"]\&
\catfont{B}^{GG}_{AA}\z I
\arrow[r,"{1\otimes \varphi_A}"]
\arrow[d,Rightarrow,shift right=4ex,"{\varphi_{A,A}}"{pos=0.495},shorten <= 0.4ex,shorten >=0.17ex]\&
\catfont{B}^{GG}_{AA}\z \catfont{B}^{FG}_{AA}
\arrow[r,"{\opn{comp}}"]\&
\catfont{B}^{FG}_{AA}
\\
\&\& I\z \catfont{A}_{AA}
\arrow[r,"{1\otimes F}"']\&
I\z \catfont{B}^{FF}_{AA}
\arrow[r,"{\varphi_A\otimes 1}"']\&
\catfont{B}^{FG}_{AA}\z \catfont{B}^{FF}_{AA}\v[2]
\arrow[ru,"{\opn{comp}}"'{inner sep=0.2ex}]
\end{cd}
is equal to the pasting\v[1]
\begin{cd}[3][4.75]
\&\& \catfont{B}^{FG}_{AA}
\arrow[rrrd,equal,bend left=16]\arrow[rd,"{\rho^{-1}}"{inner sep=0.2ex}]
\arrow[d,equal,shift right=5ex,shorten <= 1ex,shorten >=0.7ex]
\\
I
\arrow[rru,bend left=17,"{\varphi_A}"{inner sep=0.2ex}] \arrow[rrd,"{\id{F(A)}}"'{inner sep=0.1ex}]\arrow[rdd,"{\id{A}}"'{inner sep=0.2ex}]\arrow[rr,"{\rho^{-1}}"]\&
\arrow[dd,Rightarrow,shift left=0.5ex,"{\overline{F}\h[1]}"'{pos=0.575},shorten <= 4.9ex,shorten >=2.4ex]\&
I\z I
\arrow[rd,"{1\otimes \id{F(A)}}"'{pos=0.4,inner sep=0.1ex}]\arrow[r,"{\varphi_A\otimes 1}"]
\arrow[d,equal,shift right=4ex,shorten <= 0.67ex,shorten >=1.3ex]\&
\catfont{B}^{FG}_{AA}\z I
\arrow[r,"{1\otimes \id{F(A)}}"]
\arrow[r,iso,shift left=3.9ex,"{\opn{unit}_\catfont{B}^{-1}}"{inner sep=1ex},start anchor={[xshift=-3ex]}]
\arrow[d,equal,shift left=3.5ex,shorten <=-0.2ex,shorten >=1.65ex]\&
\catfont{B}^{FG}_{AA}\z \catfont{B}^{FF}_{AA}
\arrow[r,"{\opn{comp}}"']\&
\catfont{B}^{FG}_{AA}
\\
\&\&
\catfont{B}^{FF}_{AA}
\arrow[r,"{\lambda^{-1}}"']\&
I\z \catfont{B}^{FF}_{AA}
\arrow[ru,"{\varphi_A\otimes 1}"'{inner sep=0.2ex}]
\\
\&\catfont{A}_{AA}\v[2]
\arrow[ru,"{F}"']
\end{cd}
and the pasting in $\catfont{K}$\v[1]
\begin{cd}[3.5][-4.9]
\&[0.2ex] \catfont{A}_{BC}\z\left(\catfont{A}_{AB}\z I\right)
\arrow[rr,"{1\otimes G\otimes \varphi_A}"]\&[0.6ex]\&[-0.4ex]
\catfont{A}_{BC}\z\left(\catfont{B}^{GG}_{AB}\z \catfont{B}^{FG}_{AA}\right)
\arrow[rr,"{1\otimes\opn{comp}}"]
\arrow[d,Rightarrow,shift right=7.7ex,"{1\otimes \varphi_{A,B}}"{pos=0.4},shorten <= 0.25ex,shorten >=1.68ex]\&[-0.85ex]\&[-0.65ex]
\catfont{A}_{BC}\z \catfont{B}^{FG}_{AB}
\arrow[rr,"{G\otimes 1}"]
\arrow[d,equal,shift left=6ex,shorten <= 0.87ex,shorten >=2.9ex]\&[-0.65ex]\&[0.35ex]
\catfont{B}^{GG}_{BC}\z\catfont{B}^{FG}_{AB}
\arrow[rdd,"{\opn{comp}}"{inner sep=0.2ex}]
\\
\catfont{A}_{BC}\z\catfont{A}_{AB}
\arrow[ru,"{1\otimes \rho^{-1}}"{pos=0.47,inner sep=0.14ex}]\arrow[rr,"{1\otimes \lambda^{-1}}"] \arrow[rrd,"{1\otimes F}"{inner sep=0.2ex}]\arrow[rdd,"{F\otimes F}"{inner sep=0.2ex}]\arrow[ddd,"{\opn{comp}}"'{inner sep=0.2ex}]\&\&
\catfont{A}_{BC}\z \left(I\z \catfont{A}_{AB}\right)
\arrow[rr,"{1\otimes \varphi_B\otimes F}"] \&\vphantom{.}\&
\catfont{A}_{BC}\z\left(\catfont{B}^{FG}_{BB}\z\catfont{B}^{FF}_{AB}\right)
\arrow[rr,"{G\otimes 1\otimes 1}"]\arrow[ru,"{1\otimes \opn{comp}}"{inner sep=0.2ex}]
\arrow[d,equal,shift right=8ex,shorten <= 0.9ex,shorten >=0.95ex]\&\vphantom{.}\&
\catfont{B}^{GG}_{BC}\z\left(\catfont{B}^{FG}_{BB}\z\catfont{B}^{FF}_{AB}\right)
\arrow[ru,"{1\otimes \opn{comp}}"{inner sep=0.2ex}]
\\
\&\& \catfont{A}_{B,C}\z\catfont{B}^{FF}_{AB}
\arrow[rr,"{\rho^{-1}\otimes 1}"]\arrow[rd,"{\lambda^{-1}\otimes 1}"'{inner sep=0.2ex}]\&\&
\left(\catfont{A}_{BC}\z I\right) \z \catfont{B}^{FF}_{AB}
\arrow[rr,"{G\otimes \varphi_B\otimes 1}"]
\arrow[d,Rightarrow,shift right=0ex,"{\varphi_{B,C}\otimes 1}"{pos=0.43},shorten <= 0.77ex,shorten >=1.33ex]\&\&
\left(\catfont{B}^{GG}_{BC}\z \catfont{B}^{FG}_{BB}\right) \z \catfont{B}^{FF}_{AB}
\arrow[u,"{\alpha}"]\arrow[rd,"{\opn{comp}\otimes 1}"'{pos=0.4,inner sep=0.2ex}]
\arrow[rr,iso,shift left=3.24ex,"{\opn{assoc}_\catfont{B}^{-1}}"{inner sep=1ex},start anchor={[xshift=-4ex]}]\&\&
\catfont{B}^{FG}_{AC}
\\
\& \catfont{B}^{FF}_{BC}\z \catfont{B}^{FF}_{AB}
\arrow[rd,"{\opn{comp}}"{inner sep=0.2ex}]
\arrow[d,Rightarrow,shift right=1.6ex,"{\overline{F}\h[1]}"'{pos=0.42},shorten <= 0.81ex,shorten >=1.29ex]
\arrow[rr,iso,shift right=4.36ex,"{\opn{assoc}_\catfont{B}}"{inner sep=1ex},start anchor={[xshift=7.2ex]}]\&\&
\left(I \z \catfont{A}_{BC}\right)\z \catfont{B}^{FF}_{AB}
\arrow[rr,"{\varphi_C\otimes F\otimes 1}"']\&\vphantom{.}\&
\left(\catfont{B}^{FG}_{CC}\z \catfont{B}^{FF}_{BC}\right)\z \catfont{B}^{FF}_{AB}
\arrow[rr,"{\opn{comp}\otimes 1}"']\&\&
\catfont{B}^{FG}_{BC}\z \catfont{B}^{FF}_{AB}
\arrow[ru,"{\opn{comp}}"{inner sep=0.2ex}]
\\[-0.1ex]
\catfont{A}_{AC}
\arrow[rr,"{F}"']\&\vphantom{.}\&
\catfont{B}^{FF}_{AC}
\arrow[rr,"{\lambda^{-1}}"']\&\&
I\z \catfont{B}^{FF}_{AC}
\arrow[rr,"{\varphi_C\otimes 1}"']\&\&
\catfont{B}^{FG}_{CC}\z \catfont{B}^{FF}_{AC}\v[2]\arrow[rruu,bend right=55,"{\opn{comp}}"{pos=0.37,inner sep=0.2ex}]
\end{cd}
is equal to the pasting\v[1]
\begin{cd}[3.5][-0.9]
\& \catfont{A}_{BC}\z\left(\catfont{A}_{AB}\z I\right)
\arrow[rr,"{1\otimes G\otimes \varphi_A}"]\&[0.2ex]\&[0.2ex]
\catfont{A}_{BC}\z\left(\catfont{B}^{GG}_{AB}\z \catfont{B}^{FG}_{AA}\right)
\arrow[rr,"{1\otimes\opn{comp}}"]\&[-0.2ex]\&[-0.2ex]
\catfont{A}_{BC}\z \catfont{B}^{FG}_{AB}
\arrow[rrd,bend left=17,"{G\otimes 1}"{inner sep=0.2ex}]\&[-0.2ex]\&[-0.2ex]
\\
\& \catfont{B}^{GG}_{BC}\z\catfont{B}^{GG}_{AB}
\arrow[rr,"{\opn{comp}}"]
\arrow[d,Rightarrow,shift left=4.5ex,"{\overline{G}\h[1]}"'{pos=0.37},shorten <= 0.43ex,shorten >=1.42ex]\&\&
\catfont{B}^{GG}_{AC}
\arrow[rr,"{\rho^{-1}}"]
\arrow[d,equal,shift left=9.5ex,shorten <= 0.79ex,shorten >=2.76ex]\&\&
\catfont{B}^{GG}_{AC}\z I
\arrow[rd,"{1\otimes \varphi_A}"{inner sep=0.2ex}]
\arrow[rr,iso,shift left=0.5ex,"{\opn{assoc}_\catfont{B}^{-1}}"{inner sep=1ex},start anchor={[xshift=-0.58ex]}]\&\&
\catfont{B}^{GG}_{BC}\z\catfont{B}^{FG}_{AB}
\arrow[dd,"{\opn{comp}}"]
\\
\catfont{A}_{BC}\z\catfont{A}_{AB}
\arrow[ruu,bend left=25,"{1\otimes \rho^{-1}}"{inner sep=0.12ex}] \arrow[ru,"{G\otimes G}"{inner sep=0.2ex}]
\arrow[rr,"{\opn{comp}}"']\&\vphantom{.}\&
\catfont{A}_{AC}
\arrow[rr,"{\rho^{-1}}"]\arrow[ru,"{G}"{inner sep=0.2ex}]\arrow[rd,"{\lambda^{-1}}"'{inner sep=0.2ex}]\&\vphantom{.}\&
\catfont{A}_{AC}\z I
\arrow[rr,"{G\otimes \varphi_A}"]
\arrow[d,Rightarrow,shift left=0.5ex,"{\varphi_{A,C}}"{pos=0.43},shorten <= 0.75ex,shorten >=1.235ex]\&\&
\catfont{B}^{GG}_{AC}\z \catfont{B}^{FG}_{AA}
\arrow[rd,"{\opn{comp}}"{inner sep=0.2ex}]
\\
\&\&\& I\z \catfont{A}_{AC}
\arrow[rr,"{\varphi_C\otimes F}"']\&\vphantom{.}\&
\catfont{B}^{FG}_{CC}\z \catfont{B}^{FF}_{AC}
\arrow[rr,"{\opn{comp}}"']\&\&
\catfont{B}^{FG}_{AC}\v[1.5]
\end{cd}
\end{cons}
\begin{rem}
We observe that the definition of weakly enriched natural transformation we have given in \nameit{Construction}\ref{consweakenrichment} cannot be easily generalized to a fully lax enriched context in which we consider lax monoids rather than pseudomonoids in $\m{S\times S}{\catfont{K}}$. Indeed the pastings we need to consider have to use the associator of $\catfont{B}$ in both the directions.
We also notice that \nameit{Construction}\ref{consweakenrichment} is particularly useful when $\catfont{K}$ is $\VCAT{\catfont{V}}$, since we then obtain a construction that we can apply to a starting ordinary $\catfont{V}$ (creating further dimension levels that did not initially belong to $\catfont{V}$).
\end{rem}
\begin{defne}[The $2$-$\catfont{V}$-enrichment]\label{deftwovenrichment}
Let $\catfont{V}$ be a nice enough monoidal category such that $\VCAT{\catfont{V}}$ with the tensor product of $\catfont{V}$-categories (so we need at least a braiding in $\catfont{V}$ to have such tensor product) becomes a monoidal $2$-category with coproducts such that its tensor product preserves them in each variable.
We call \dfn{$2$-$\catfont{V}$-enriched category (resp. functor, natural transformation)} a $\left(\VCAT{\catfont{V}}\right)$-weakly enriched category (resp. functor, natural transformation). We call $\twoVCAT{\catfont{V}}$ the tridimensional structure they form.
\end{defne}
\begin{exampl}\label{exampletwosetenrichment}
Consider $\catfont{V}=\catfont{Set}$. Then $2$-$\catfont{Set}$-enriched categories, functors and natural transformations are, respectively, bicategories, lax functors and lax natural transformations. A notion of $2$-$\catfont{V}$-enriched modification could be given as well, such that the $2$-$\catfont{Set}$-enriched modifications are precisely the modifications between lax natural transformations.
So we can now think of $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$ as the lax $3$-category of $2$-$\catfont{Set}$-enriched categories, restricted to strict weakly enriched categories and functors for simplicity.
\end{exampl}
\begin{rem}\label{remnametwosetenrichedgroth}
We can now clarify the name ``$2$-$\catfont{Set}$-enriched Grothendieck construction", after \nameit{Example}\ref{exampletwosetenrichment}. We have given this name from the observation that it is the Grothendieck construction that lives in the $2$-$\catfont{Set}$-enriched world (it could also be extended to consider bicategories and lax functors into $\catfont{CAT}$). It takes enriched presheaves with values in $\VCAT{\catfont{Set}}=\catfont{CAT}$, that is the base of the weak enrichment we consider, and converts them into an extended notion of Grothendieck fibration, that is worth to be called \predfn{$2$-$\catfont{Set}$-fibration}.
We think that the $2$-$\catfont{Set}$-enriched Grothendieck construction should be taken into consideration towards the development of a general enriched Grothendieck construction. This will be explored in future work. In particular, we believe that for this the $2$-$\catfont{Set}$-enrichment should be distinguished from the ordinary $\catfont{Cat}$-enrichement, and we expect the latter to be much less useful than the former.
\end{rem}
\begin{rem}
We now proceed to describe the \predfn{$2$-$\catfont{Set}$-opfibrations} over a $2$-category $\catfont{B}$, that are the notion of opfibration produced by the $2$-$\catfont{Set}$-enriched Grothendieck construction, in the sense that they will precisely be (up to restricting to small fibres) what forms the essential image of the $2$-functor $\groth{-}\:\m{\catfont{B}}{\catfont{CAT}}\to \slice{2\h[2]\mbox{-}\CAT}{\catfont{B}}$ described in \nameit{Proposition}\ref{twosetgrothcanonicallyextendstwofunctor}. They have been firstly introduced by Lambert in~\cite{lambert_discretetwofib} with the name ``discrete $2$-fibrations" (up to co-op changes), justified by the fact that they are equivalently the locally discrete Hermida's $2$-fibrations (for Hermida's $2$-fibrations, see~\cite{hermida_somepropoffibasafibredtwocat}). We believe, however, that the \predfn{$2$-$\catfont{Set}$-fibrations} are what fully realizes the concept of fibration in dimension $2$, while Hermida's $2$-fibrations belong more to dimension $3$.
We shall look at \nameit{Construction}\ref{2SetGrothconstrtoconicalize} (where we intuitively produced the $2$-$\catfont{Set}$-enriched Grothendieck construction) to investigate which lifting properties hold for $\groth{F}$ for some $2$-functor $F\:\catfont{B}\to \catfont{CAT}$ and could characterize the functors that are isomorphic to some $\groth{F}$. Since, calling $H_0$ the underlying functor of a $2$-functor $H$, we have that
$$\groth{F}_0=\groth{F_0},$$
we obtain that $\groth{F}_0$ is a usual Grothendieck opfibration. In the notation of \nameit{Construction}\ref{2SetGrothconstrtoconicalize}, the cartesian liftings are given by the $\underline{f}^X$. But we now also have liftings of the $2$-cells in $\catfont{B}$. Indeed, for every $2$-cell $\delta\:f\aR{}g\:B\to B'$ in $\catfont{B}$ and every $X\in F(B)$, we have a $2$-cell in $\Grothop{F}$
$$\underline{\delta}^X\:(f,F(\delta)_X)\aR{}(g,\id{})\:(B,X)\to (B',F(g)(X))$$
such that $\groth{F}(\underline{\delta}^X)=\delta$. As a consequence, for every morphism $(g,\beta)$ in $\Grothop{F}$, the $2$-cell
$$\delta=(\id{B},\beta)\h \underline{\delta}^X$$
in $\Grothop{F}$ is a $2$-cell that lifts $\delta$ to the morphism $(g,\beta)$ that lives above the codomain $g$ of $\delta$. Recall from \nameit{Construction}\ref{2SetGrothconstrtoconicalize} that all the $2$-cells in $\Grothop{F}$ emerge in this way. As we noticed there, having a $2$-cell $\delta$ in $\Grothop{F}$ as above is actually a property for $\delta$. Thus, fixed $\delta\:f\aR{}g$ in $\catfont{B}$ and $(g,\beta)$ in $\Grothop{F}$ that lives above the codomain $g$ of $\delta$, the liftings of $\delta$ to $(g,\beta)$ are unique. And we conclude that $\groth{F}$, further than being an ordinary Grothendieck opfibration, is also locally a discrete fibration (see \nameit{Definition}\ref{deftwosetopf}).
It is interesting to notice the change of direction from opfibration on objects to fibration locally. This is given by the fact that we are classifying from $\catfont{1}\:\catfont{1}\to \catfont{CAT}$ using lax comma objects. We will see in \nameit{Remark}\ref{coopflavourstwosetenrichedgroth} that we have other classifiers as well, with all the co-op flavours, but we believe that this is the most natural one for what we want to call \predfn{$2$-$\catfont{Set}$-opfibration}.
\end{rem}
\begin{defne}\label{deftwosetopf}
Let $\catfont{B}$ be a $2$-category. A \dfn{$2$-$\catfont{Set}$-opfibration over $\catfont{B}$} (or \dfn{discrete $2$-fibration} in the language of Lambert's~\cite{lambert_discretetwofib}) is a $2$-functor $P\:\catfont{E}\to \catfont{B}$ such that
\begin{enum}
\item the underlying functor $P_0$ of $P$ is an ordinary Grothendieck opfibration;
\item for every pair $X,Y\in \catfont{E}$ the functor
$$P_{X,Y}\:\HomC{\catfont{E}}{X}{Y}\to \HomC{\catfont{B}}{P(X)}{P(Y)}$$
is a discrete fibration.
\end{enum}
We say that $P$ is \dfn{split} if $P_0$ is so.
\end{defne}
\begin{rem}
The following theorem is proved in Lambert's~\cite{lambert_discretetwofib}. One can find there some examples of $2$-$\catfont{Set}$-opfibrations as well.
We will extend such theorem to a complete $2$-equivalence betwen $\m{\catfont{B}}{\catfont{CAT}}$ with various laxness flavours on morphisms and corresponding $2$-categories of $2$-$\catfont{Set}$-opfibrations in Section~\ref{sectionkanextensions}, after showing that the lax comma object square of \nameit{Theorem}\ref{grothconstrislaxcomma} exhibits a \predfn{weak Kan extension} in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$.
\end{rem}
\begin{teor}\label{essentialimagegivenbytwosetopf}
Let $\catfont{B}$ be a $2$-category. The essential image of the $2$-functor
$$\groth{-}\:\m{\catfont{B}}{\catfont{CAT}}\to \slice{2\h[2]\mbox{-}\CAT}{\catfont{B}}$$
is given by the split $2$-$\catfont{Set}$-opfibrations with small fibres.
\end{teor}
\begin{rem}\label{coopflavourstwosetenrichedgroth}
The following table shows the four co-op versions of the $2$-$\catfont{Set}$-enriched Grothendieck construction, with the corresponding notions of fibration. We have named the various notions of fibration from an enriched point of view, where the op always makes sense while the co does not.
\begin{center}\h[-18]
\begin{minipage}{3.3cm}
\begin{eqD*}
\h[-7]\sq*[l][5.5][4.5][\lax \opn{comma}][2.7][2.2][0.56]{\Grothop{F}}{\catfont{1}}{\catfont{B}}{\catfont{CAT}}{}{\groth{F}}{\catfont{1}}{F}
\end{eqD*}
\begin{center}{\footnotesize $2$-$\catfont{Set}$-opfibrations:}
\noindent {\footnotesize opfibrations, locally discrete fibrations}
\end{center}
\end{minipage}\quad
\begin{minipage}{3.38cm}
\begin{eqD*}
\h[-7]\sq*[o][5.5][4.5][\lax \opn{comma}][2.2][2.7][0.6]{\Groth{F}}{\catfont{1}}{\catfont{B}}{\catfont{CAT}\ensuremath{^{\operatorname{op}}}}{}{\groth{F}}{\catfont{1}}{F}
\end{eqD*}
\begin{center}{\footnotesize $2$-$\catfont{Set}$-fibrations:}
\noindent {\footnotesize fibrations, locally discrete opfibrations}
\end{center}
\end{minipage}\quad
\begin{minipage}{3.47cm}
\begin{eqD*}
\h[-7]\sq*[l][5.5][3.4][\oplax \opn{comma}][2.7][2.2][0.56]{\Grothcoop{F}}{\catfont{1}}{\catfont{B}}{\catfont{CAT}\ensuremath{^{\operatorname{co}}}}{}{\groth{F}}{\catfont{1}}{F}
\end{eqD*}
\begin{center}{\footnotesize $2$-$\catfont{Set}$-coopfibrations:}
\noindent {\footnotesize opfibrations, locally discrete opfibrations}
\end{center}
\end{minipage}\quad
\begin{minipage}{3.53cm}
\begin{eqD*}
\h[-7]\sq*[o][5.5][3.3][\oplax \opn{comma}][2.2][2.7][0.6]{\Grothco{F}}{\catfont{1}}{\catfont{B}}{\catfont{CAT}\ensuremath{^{\operatorname{coop}}}}{}{\groth{F}}{\catfont{1}}{F}
\end{eqD*}
\begin{center}{\footnotesize $2$-$\catfont{Set}$-cofibrations:}
\noindent {\footnotesize fibrations, locally discrete fibrations}
\end{center}
\end{minipage}
\end{center}
\end{rem}
\section{A pointwise Kan extension result}\label{sectionkanextensions}
In this section, we investigate a pointwise Kan extension result for the $2$-$\catfont{Set}$-enriched Grothendieck construction. The difficulty we encounter is that no notion of (pointwise) Kan extension in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$ seems to appear in the literature. The pointwise version in particular is hard to establish, as it requires a concept of colimit ``internal to $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$", i.e.\ in a $2$-$\catfont{Set}$-category, that does not seem to have been introduced yet.
We firstly give a definition of weak Kan extension in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$ (actually, in any lax $3$-category), and prove the weak version of the pointwise Kan extension result we aim at establishing. This weak result will be enough to prove $2$-fully faithfulness results for the $2$-$\catfont{Set}$-enriched Grothendieck construction (in three laxness flavours) and complete \nameit{Theorem}\ref{essentialimagegivenbytwosetopf} to $2$-equivalences between $2$-copresheaves (on a $2$-category) and $2$-categories of $2$-$\catfont{Set}$-opfibrations.
Then, we propose a notion of ``internal colimit in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$", i.e.\ of colimit in a $2$-$\catfont{Set}$-enriched category, and originally define the concept of \predfn{pointwise left Kan extension in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$ along a $2$-$\catfont{Set}$-opfibration} using it. We prove that the lax comma object square produced by the $2$-$\catfont{Set}$-enriched Grothendieck construction (\nameit{Theorem}\ref{grothconstrislaxcomma}) exhibits a \predfn{pointwise left Kan extension in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$}. This is the third main result of this paper, presented in \nameit{Theorem}\ref{teorpointkanextforgroth}.
In the last part of this section, we show the fourth main result of this paper, which is that a pointwise left Kan extension in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$ (as defined here) is always a weak one as well. This (presented in \nameit{Proposition}\ref{pointkanisalsoweak}) will be based on an $\oplaxn\mbox{-}\lax$ generalization of the parametrized Yoneda lemma that does not seem to appear in the literature. Such lemma, proved in \nameit{Theorem}\ref{parametrizedyonedaoplaxnlax}, also sheds more light on the concept of lax normal natural transformation (\nameit{Definition}\ref{deflaxnormal}). Indeed it shows how the oplax normal naturality is the minimum amount of strictness that is required to expand the data on the identities to a lax natural transformation.
\begin{defne}\label{defweakkanextensionlaxthreecat}
A diagram
\begin{cd}[7][5]
\catfont{B} \arrow[rr,"{F}"]\arrow[d,"{K}"']\&\arrow[dl,Rightarrow,shift right=0.75ex,"{\lambda}"{pos=0.355, inner sep= 0.35ex},shorten <=1.65ex,shorten >= 3.35ex]\& \catfont{C} \\
\catfont{A}\arrow[rru,"L"']
\end{cd}
in a lax $3$-category $\catfont{Q}$ (that is a category enriched over the $1$-category of $2$-categories and lax functors), exhibits $L$ as the \dfn{weak left Kan extension of $F$ along $K$}, written $L={\lan{K}{F}}$, if pasting with $\lambda$ gives an isomorphism of categories
\begin{equation}\label{eqweakkanextensionlaxthreecat}
\HomC{\HomC{\catfont{Q}}{\catfont{A}}{\catfont{C}}}{L}{U}\cong \HomC{\HomC{\catfont{Q}}{\catfont{B}}{\catfont{C}}}{F}{U\circ K}
\end{equation}
for every $U\in \HomC{\catfont{Q}}{\catfont{A}}{\catfont{C}}$ (the $2$-naturality in $U$ is granted automatically).
\end{defne}
\begin{rem}\label{remhomtwocategoriestwocatlax}
Lambert showed in~\cite{lambert_discretetwofib} that $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$ is a lax $3$-category with hom-$2$-categories $$\HomC{2\h[2]\mbox{-}\CAT_{\h[-3]\lax}}{\catfont{A}}{\catfont{C}}\mathrel{\mathrel{\mathop\ordinarycolon}\mkern-1.2mu=} \mlax{\catfont{A}}{\catfont{C}}$$
where $\mlax{\catfont{A}}{\catfont{C}}$ is the $2$-category of $2$-functors from $\catfont{A}$ to $\catfont{C}$, lax natural transformations and modifications.
So, in the notation of \nameit{Definition}\ref{defweakkanextensionlaxthreecat}, the isomorphism of categories of equation~\refs{eqweakkanextensionlaxthreecat} becomes, for every $U\in \mlax{\catfont{A}}{\catfont{C}}$,
$$\HomC{\mlax{\catfont{A}}{\catfont{C}}}{L}{U}\cong \HomC{\mlax{\catfont{B}}{\catfont{C}}}{F}{U\circ K}.$$
\end{rem}
\begin{prop}\label{isomorphismsweakkanextension}
Let $F\:\catfont{A}\to \catfont{CAT}$ be a $2$-functor and consider its $2$-$\catfont{Set}$-enriched Grothendieck construction
\sq[l][7][7][\lax \opn{comma}][2.7][2.2][0.65]{\Grothop{F}}{\catfont{1}}{\catfont{A}}{\catfont{CAT}}{}{\groth{F}}{\catfont{1}}{F}
Call $\lambda^F$ the lax normal natural transformation that presents such lax comma object in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$. Then, for every $U\:\catfont{A}\to \catfont{CAT}$, pasting with $\lambda^F$ gives an isomorphism of categories
$$\HomC{\mlax{\catfont{A}}{\catfont{CAT}}}{F}{U}\cong \HomC{\mlax{\Grothopdiag{F}}{\catfont{CAT}}}{\Delta 1}{U\circ \groth{F}},$$
where $\Delta 1\:\Grothop{F}\to \catfont{CAT}$ is the $2$-functor constant at $1$. Moreover, this restricts to isomorphisms
$$\HomC{\mps{\catfont{A}}{\catfont{CAT}}}{F}{U}\cong \HomC{\msigma{\Grothopdiag{F}}{\catfont{CAT}}}{\Delta 1}{U\circ \groth{F}}$$
$$\HomC{\m{\catfont{A}}{\catfont{CAT}}}{F}{U}\cong \HomC{\mlaxn{\Grothopdiag{F}}{\catfont{CAT}}}{\Delta 1}{U\circ \groth{F}},$$
where $\opn{ps}$ means to restrict to pseudonatural transformations and $\opn{sigma}$ means to restrict to \predfn{sigma natural transformations}, that have been defined in Descotte, Dubuc and Szyld's paper~\cite{descottedubucszyld_sigmalimandflatpseudofun} and are a pseudo version of the lax normal natural transformations \pteor{asking the special structure $2$-cells to be isomorphisms rather than identities}.
\end{prop}
\begin{proof}
The first isomorphism is proved in Bird's PhD thesis~\cite{bird_limitsintwocatoflocpresented}. Looking at the explicit construction, it is straightforward to see that it restricts to the other two. In this, what works very well is that $\lambda^F$ is a lax normal natural transformation, and thus always contributes with an identity in the special structure $2$-cells.
\end{proof}
\begin{rem}
The third isomorphism of \nameit{Proposition}\ref{redlaxnormalconical} offers a shorter but less intuitive and elementary proof to \nameit{Theorem}\ref{redlaxnormalconical} (reduction of the weighted $2$-limits to lax normal conical ones). Indeed this is what Street did in~\cite{street_limitsindexedbycatvalued}.
\end{rem}
\begin{teor}\label{corollweakkanextension}
Let $F\:\catfont{B}\to \catfont{CAT}$ be a $2$-functor. Then the lax normal natural transformation $\lambda^F$ of \nameit{Proposition}\ref{isomorphismsweakkanextension} exhibits
$$F=\lan{\groth{F}}{\Delta 1}.$$
\end{teor}
\begin{proof}
The proof is a combination of \nameit{Remark}\ref{remhomtwocategoriestwocatlax} and \nameit{Proposition}\ref{isomorphismsweakkanextension}.
\end{proof}
\begin{rem}
Exactly as can be done in dimension $1$ (see \nameit{Remark}\ref{remtheorkanextisuseful}), we can use \nameit{Theorem}\ref{corollweakkanextension} (or actually \nameit{Proposition}\ref{isomorphismsweakkanextension} to prove the $2$-fully faithfulness of the $2$-$\catfont{Set}$-enriched Grothendieck construction (in three laxness flavours) and complete \nameit{Theorem}\ref{essentialimagegivenbytwosetopf} to $2$-equivalences between $2$-copresheaves and $2$-$\catfont{Set}$-opfibrations. The fact that the first $2$-functor $\groth{-}$ of \nameit{Theorem}\ref{twofullyfaithfulnessgroth} is $2$-fully faithful is proved also in Bird's~\cite{bird_limitsintwocatoflocpresented}, but without mentioning that it comes from a weak Kan extension result. None of the three $2$-equivalence results of \nameit{Theorem}\ref{twofullyfaithfulnessgroth} seems to appear in the literature.
\end{rem}
\begin{teor}\label{twofullyfaithfulnessgroth}
Let $\catfont{A}$ be a $2$-category. The $2$-$\catfont{Set}$-enriched Grothendieck construction \pteor{extended to consider lax natural transformations as in the proof of \nameit{Proposition}\ref{twosetgrothcanonicallyextendstwofunctor}} produces a $2$-equivalence
$$\groth{-}\:\mlax{\catfont{A}}{\catfont{CAT}}\ensuremath{\stackrel{\raisebox{-1ex}{\kern-.3ex$\scriptstyle\sim$}}{\rightarrow}} \Fib[t][n][\catfont{A}]$$
where $\Fib[t][n][\catfont{A}]$ is the full sub-$2$-category of $\h[1]\slice{2\h[2]\mbox{-}\CAT}{\catfont{A}}$ given by the split $2$-$\catfont{Set}$-opfibrations with small fibres. Moreover this restricts to $2$-equivalences
$$\groth{-}\:\mps{\catfont{A}}{\catfont{CAT}}\ensuremath{\stackrel{\raisebox{-1ex}{\kern-.3ex$\scriptstyle\sim$}}{\rightarrow}} \Fib'[t][n][\catfont{A}]$$
$$\groth{-}\:\m{\catfont{A}}{\catfont{CAT}}\ensuremath{\stackrel{\raisebox{-1ex}{\kern-.3ex$\scriptstyle\sim$}}{\rightarrow}} \Fib"[t][n][\catfont{A}]$$
where $\Fib'[t][n][\catfont{A}]$ and $\Fib"[t][n][\catfont{A}]$ restrict $\Fib[t][n][\catfont{A}]$ respectively\v[0.4] to cartesian functors \pteor{as underlying functors of the $1$-cells} and to cleavage preserving functors.
\end{teor}
\begin{proof}
We already know by \nameit{Theorem}\ref{essentialimagegivenbytwosetopf} that the essential image of $\groth{-}$ (in each of the three versions) is given by the split $2$-$\catfont{Set}$-opfibrations. So we are missing the $2$-fully faithfulness of the three $2$-functors.
Let $F,G\:\catfont{A}\to \catfont{CAT}$ be $2$-functors. Then combining \nameit{Proposition}\ref{isomorphismsweakkanextension} (or the weak Kan extension result \nameit{Theorem}\ref{corollweakkanextension}) and \nameit{Theorem}\ref{grothconstrislaxcomma} (the $2$-$\catfont{Set}$-enriched Grothendieck construction is a lax comma object in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$) we obtain the composite isomorphism of categories
$$\HomC{\mlax{\catfont{A}}{\catfont{CAT}}}{F}{G}\cong \HomC{\mlax{\Grothopdiag{F}}{\catfont{CAT}}}{\Delta 1}{G\circ \groth{F}}\cong \HomC{\slice{2\h[2]\mbox{-}\CAT}{\catfont{A}}}{\Grothopdiag{F}}{\Grothopdiag{G}}$$
where the first functor is given by pasting with the $2$-cell $\lambda^F$ that presents the lax comma object $\Grothop{F}$ and the second functor is the inverse of pasting with $\lambda^G$ on objects and producing the modification associated to the lax interchange rule (see \nameit{Remark}\ref{needof2catlax}) on morphisms. Indeed the second functor is surely a bijection on objects, and it is also a bijection on morphisms by part $(ii)$ of \nameit{Definition}\ref{completeunivproplaxcomma} (the $2$-dimensional part of the universal property of the lax comma obejct) with $\Gamma=\id{}$ and $\Delta=\id{}$ (and so in particular strict $2$-natural transformations). Since the composite functor precisely coincides with the functor on morphisms associated to $\groth{-}$ between $\catfont{A}$ and $\catfont{C}$ (see the proof of \nameit{Proposition}\ref{twosetgrothcanonicallyextendstwofunctor}), this completes the proof of the first $2$-equivalence.
The composite isomorphism above then restricts to the following two:
$$\scalebox{0.95}{$\HomC{\mps{\catfont{A}}{\catfont{CAT}}}{F}{G}\cong \HomC{\msigma{\Grothopdiag{F}}{\catfont{CAT}}}{\Delta 1}{G\circ \groth{F}}\cong \HomC{\Fib'[t][n][\catfont{A}]}{\Grothopdiag{F}}{\Grothopdiag{G}}$}$$
$$\scalebox{0.975}{$\HomC{\m{\catfont{A}}{\catfont{CAT}}}{F}{G}\cong \HomC{\mlaxn{\Grothopdiag{F}}{\catfont{CAT}}}{\Delta 1}{G\circ \groth{F}}\cong \HomC{\Fib"[t][n][\catfont{A}]}{\Grothopdiag{F}}{\Grothopdiag{G}}$}$$
by part $(i)$ of \nameit{Definition}\ref{completeunivproplaxcomma}, since whiskering $\lambda^G$ on the left with a $2$-functor $\Grothop{F}\to \Grothop{G}$ looks at the second component of the morphisms in $\Grothop{G}$. Indeed, if we start for example from $\gamma\:\Delta 1 \alaxn{} G\circ \groth{F}$, the associated functor $V^\gamma\:\Grothop{F}\to \Grothop{G}$ is such that, for every morphism $(f,\id{})$ in $\Grothop{F}$, $$\pr{1}(V^\gamma (f,\id{}))=f \quad\text{and}\quad \pr{2}(V^\gamma (f,\id{}))=(\lambda^G\h V^\gamma)_{(f,\id{})}=\gamma_{(f,\id{})}=\id{}.$$
And then the first $2$-equivalence restricts to the other two.
\end{proof}
\begin{rem}
It is interesting to notice the bond between cartesian or cleavage preserving functors and sigma or lax normal natural transformations. \nameit{Theorem}\ref{twofullyfaithfulnessgroth} also suggests how to define the various flavours of $1$-cells between $2$-$\catfont{Set}$-opfibrations, showing their preasheaf counterparts as well. In particular there are no requests on the morphism part of such $1$-cells, but this perfectly fits the fact that the $2$-$\catfont{Set}$-opfibrations are discrete fibrations on morphisms.
\end{rem}
\begin{rem}\label{remaimatpointwise}
We now aim at proving that, for some notion of \predfn{pointwise Kan extension in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$}, the weak left Kan extension of \nameit{Theorem}\ref{corollweakkanextension} is \predfn{pointwise} as well. This would mean to have a complete $2$-categorical analogue of \nameit{Theorem}\ref{copresheavesarekanextensions}.
Firstly, in order to define a \predfn{pointwise Kan extension in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$}, we need to understand better the notion of (co)limit in the setting of $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$. There surely is a natural notion of ``external limit" in this context, i.e.\ a limit in the whole $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$, which is that of limit enriched over the $1$-category of $2$-categories and lax functors, since $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$ is enriched over such $1$-category. Indeed, for $\catfont{C}$ a $2$-category, this recovers the natural isomorphism
$$\HomC{2\h[2]\mbox{-}\CAT_{\h[-3]\lax}}{\catfont{U}}{\laxcomma{\Id{\catfont{C}}}{\Id{\catfont{C}}}}\cong \HomC{2\h[2]\mbox{-}\CAT_{\h[-3]\lax}}{\catfont{2}}{\HomC{2\h[2]\mbox{-}\CAT_{\h[-3]\lax}}{\catfont{U}}{\catfont{C}}}$$
presented by Lambert in~\cite{lambert_discretetwofib} as the universal property of the lax comma object $\laxcomma{\Id{\catfont{C}}}{\Id{\catfont{C}}}$ being the power of $\catfont{C}$ by $\catfont{2}$ in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$.
But we are more interested in an ``internal" notion of colimit, i.e.\ a notion of colimit in a $2$-$\catfont{Set}$-enriched category (after \nameit{Example}\ref{exampletwosetenrichment}). This should include the notion of colimit in a $2$-category, but be able to express the laxness as well. And there is a notion of colimit that fulfils both these requests and naturally emerges in the world of $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$ together with the $2$-$\catfont{Set}$-enriched Grothendieck construction: that is, the oplax normal conical $2$-colimit (see \nameit{Remark}\ref{remoplaxnormalvslaxnormal}, \nameit{Definition}\ref{defoplaxnormalconical} and \nameit{Theorem}\ref{redoplaxnormalconical}). Indeed the $2$-$\catfont{Set}$-enriched Grothendieck construction for $2$-functors $\catfont{A}\ensuremath{^{\operatorname{op}}}\to \catfont{CAT}$ still lives in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$, as showed in \nameit{Remark}\ref{coopflavourstwosetenrichedgroth}. But here we will need a slightly more general notion. We propose to use now non-necessarily conical oplax normal $2$-colimits as colimits in a $2$-$\catfont{Set}$-category (see \nameit{Definition}\ref{defoplaxnormaltwocolimit}).
Notice that there is a structure of marking included in the definition of such colimit, in the sense that further than specifying the diagram and the weight we also need to specify the way in which we view the domain of the diagram as a Grothendieck construction. And actually, we will need to consider both the Grothendieck constructions that live in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$, for $2$-functors of both the forms $\catfont{A}\to\catfont{CAT}$ and $\catfont{A}\ensuremath{^{\operatorname{op}}}\to \catfont{CAT}$ (this will bring to the definition of \predfn{opmarked oplax normal $2$-colimit}).
Recall that in Section~\ref{sectionlaxnormallimits} we showed that at least the oplax normal conical $2$-colimits are as expressive as the weighted $2$-colimits. But there is a change of perspective here. We prefer specifying a diagram, a weight and a marking rather than condensing the three to some modified diagram and weight.
\end{rem}
\begin{defne}\label{defoplaxnormaltwocolimit}
Let $M\:\catfont{A}\ensuremath{^{\operatorname{op}}}\to \catfont{CAT}$ (the marking), $F\:\Groth{M}\to \catfont{C}$ (the diagram) and $W\:{\left(\Groth{M}\right)}\ensuremath{^{\operatorname{op}}}\to \catfont{CAT}$ (the weight) be $2$-functors with $\catfont{A}$ small.
The \dfn{oplax normal $2$-colimit of $F$ marked by $M$ and weighted by $W$}, denoted as $\oplaxnmcolim{M}{W}{F}$, is (if it exists) an object $C\in \catfont{C}$ together with an isomorphism of categories
$$\HomC{\catfont{C}}{C}{U}\cong \HomC{\moplaxn{{\left(\Grothdiag{M}\right)}\ensuremath{^{\operatorname{op}}}}{\catfont{CAT}}}{W}{\HomC{\catfont{C}}{F(-)}{U}}$$
$2$-natural in $U\in \catfont{C}$.
\noindent When $\oplaxnmcolim{M}{W}{F}$ exists, taking $U=C$ and considering the identity on $C$ gives us in particular an oplax normal natural transformation
$$\mu\:W \aoplaxn{}\HomC{\catfont{C}}{F(-)}{C},$$
called the \dfn{universal oplax normal cocylinder}.
We will need to consider also the case in which the domain of $F$ is expressed as $\Grothop{M}$ for some $2$-functor $M\:\catfont{A}\to \catfont{CAT}$, and $W\:{\left(\Grothop{M}\right)}\ensuremath{^{\operatorname{op}}}\to \catfont{CAT}$. The \dfn{oplax normal $2$-colimit of $F$ opmarked by $M$ and weighted by $W$}, denoted as $\oplaxnopmcolim{M}{W}{F}$, is (if it exists) an object $C\in \catfont{C}$ together with an isomorphism of categories
$$\HomC{\catfont{C}}{C}{U}\cong \HomC{\moplaxn{{\left(\Grothopdiag{M}\right)}\ensuremath{^{\operatorname{op}}}}{\catfont{CAT}}}{W}{\HomC{\catfont{C}}{F(-)}{U}}$$
$2$-natural in $U\in \catfont{C}$.
\end{defne}
\begin{defne}
Recall from \nameit{Remark}\ref{fromgrothisnotrestrictive} what we can now call \dfn{the trivial marking} and denote as $\opn{triv}$. That is, given $\catfont{A}$ a $2$-category, we can view $\catfont{A}$ as the $2$-$\catfont{Set}$-enriched Grothendieck construction of $\Delta 1\:\catfont{A}\to \catfont{CAT}$ (that produces the identity on $\catfont{A}$ as $2$-$\catfont{Set}$-opfibration). And $\oplaxn$ with respect to the trivial marking coincides with the strict $2$-naturality (see also \nameit{Example}\ref{conicalarelaxnormalconical}).
\end{defne}
\begin{rem}
We can now rephrase \nameit{Theorem}\ref{redoplaxnormalconical} as follows:
\noindent \textit{In $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$ every trivially-marked weighted $2$-colimit can be equivalently expressed as a marked trivially-weighted $2$-colimit.}
\noindent \textit{More precisely, given $2$-functors $F\:\catfont{A}\to \catfont{C}$ and $W\:\catfont{A}\ensuremath{^{\operatorname{op}}}\to \catfont{CAT}$ with $\catfont{A}$ small,}
$$\oplaxnmcolim{\opn{triv}}{W}{F}\h[2]\cong\h[2] \oplaxnmcolim{W}{\Delta 1}{\left(F\circ \groth{W}\right)}$$
So we could say that in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$ all the (trivially-marked) weighted $2$-colimits are (now strictly speaking) conicalizable, up to changing the marking.
\end{rem}
\begin{exampl}
Let $F\:\catfont{A}\ensuremath{^{\operatorname{op}}}\to \catfont{CAT}$ be a $2$-functor with $\catfont{A}$ small. Then by \nameit{Example}\ref{univoplaxnormalcoconepresheaves}
$$F\cong \oplaxnmcolim{F}{\Delta 1}{\left(\yy\circ\h \groth{F}\right)}$$
In particular, taking $\catfont{A}=1$, we obtain that for every small category $\catfont{D}$
$$\catfont{D}\cong \oplaxnmcolim{\catfont{D}}{\Delta 1}{\Delta 1}.$$
Notice that the marking given by $\catfont{D}$ is ``chaotic", in the sense that $\oplaxn$ with respect to it simply becomes $\oplax$.
So in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$ the $2$-functor $\catfont{1}\:\catfont{1}\to \catfont{CAT}$, that we can think of as the $3$-dimensional classifier of $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$, is conically dense (with respect to \nameit{Definition}\ref{defoplaxnormaltwocolimit}), considering the chaotic marking. This completes the idea of \nameit{Remark}\ref{remoneisoplaxnormalconicaldense}.
Recall that the $1$-dimensional analogue of this, which is that $1\:\catfont{1}\to \catfont{Set}$ (that is the $2$-dimensional classifier of $\catfont{CAT}$) is (conically) dense, was very useful in proving \nameit{Theorem}\ref{copresheavesarekanextensions} (the pointwise Kan extension result for the $\catfont{Set}$-enriched Grothendieck construction).
\end{exampl}
\begin{rem}
We are now ready to propose an original notion of pointwise left Kan extension in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$ along a $2$-$\catfont{Set}$-opfibration, using our definition of colimit in such setting (\nameit{Definition}\ref{defoplaxnormaltwocolimit}). Our idea is to keep the corresponding diagram and weight considered in the (ordinary) enriched setting (see Kelly's~\cite{kelly_basicconceptsofenriched} for the classical definition), but adding the marking that we naturally have when we extend along a $2$-$\catfont{Set}$-opfibration.
\end{rem}
\begin{defne}\label{defpointkanextensionlaxthreecat}
Consider a diagram
\begin{cd}[7][5]
\catfont{B} \arrow[rr,"{F}"]\arrow[d,"{K}"']\&\arrow[dl,Rightarrow,shift right=0.75ex,"{\lambda}"{pos=0.355, inner sep= 0.35ex},shorten <=1.65ex,shorten >= 3.35ex]\& \catfont{C} \\
\catfont{A}\arrow[rru,"L"']
\end{cd}
in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$ with $\catfont{B}$ small and $K$ a $2$-$\catfont{Set}$-opfibration. Then by \nameit{Theorem}\ref{essentialimagegivenbytwosetopf}, $K$ is isomorphic in the slice $\slice{2\h[2]\mbox{-}\CAT}{\catfont{A}}$ to $\groth{M}$ for some $2$-functor $M\:\catfont{A}\to \catfont{CAT}$. We can assume $K$ is in the form $\groth{M}$, up to whiskering the diagram with the isomorphism in the slice. Assume then that $\lambda$ is a lax normal natural transformation with respect to $M$.
We say that $\lambda$ exhibits $L$ as the \dfn{pointwise left Kan extension of $F$ along $K$}, written $L=\Lan{K}{F}$, if for every $A\in \catfont{A}$
$$L(A)\cong \oplaxnopmcolim{M}{\HomC{\catfont{A}}{K(-)}{A}}{F}$$
with universal oplax normal cocylinder
\begin{equation}\label{eqpointkanextensiontwocatlax}
\HomC{\catfont{A}}{K(-)}{A}\aRR{L}\HomC{\catfont{C}}{(L\circ K)(-)}{L(A)}\aRR[\oplaxn]{\HomC{\catfont{C}}{\lambda_{-}}{\id{}}}\HomC{\catfont{C}}{F(-)}{L(A)};
\end{equation}
or equivalently if for every $A\in \catfont{A}$ and every $C\in \catfont{C}$ the functor
$$\HomC{\catfont{C}}{L(A)}{C}\arr{}\HomC{\moplaxn{\catfont{B}\ensuremath{^{\operatorname{op}}}}{\catfont{CAT}}}{\HomC{\catfont{A}}{K(-)}{A}}{\HomC{\catfont{C}}{F(-)}{C}}$$
given by the oplax normal natural transformation of equation~\refs{eqpointkanextensiontwocatlax} is an isomorphism of categories (notice that the $2$-naturality in $C$ and $A$ is granted, where the latter is using that $L$ is a $2$-functor).
\end{defne}
\begin{rem}
We now prove the pointwise Kan extension result for the $2$-$\catfont{Set}$-enriched Grothendieck construction we were aiming at (from \nameit{Remark}\ref{remaimatpointwise}). Such result is original.
\end{rem}
\begin{teor}\label{teorpointkanextforgroth}
Let $F\:\catfont{A}\to \catfont{CAT}$ be a $2$-functor with $\catfont{A}$ a small $2$-category. Then the $2$-$\catfont{Set}$-enriched Grothendieck construction lax comma object square
\sq[l][7][7][\lax \opn{comma}][2.7][2.2][0.65]{\Grothop{F}}{\catfont{1}}{\catfont{A}}{\catfont{CAT}}{}{\groth{F}}{\catfont{1}}{F}
exhibits
$$F=\Lan{\groth{F}}{\Delta 1}.$$
\end{teor}
\begin{proof}
By \nameit{Theorem}\ref{grothconstrislaxcomma} (and \nameit{Proposition}\ref{laxnatinsidegrothconstr}) we know that the lax natural transformation $\lambda$ that presents the lax comma object is lax normal.
Given $A\in \catfont{A}$ and $C\in \catfont{CAT}$, we prove that the oplax normal natural transformation
\begin{equation*}
\HomC{\catfont{A}}{\groth{F}(-)}{A}\aRR{F}\HomC{\catfont{CAT}}{(F\circ \groth{F})(-)}{F(A)}\aRR[\oplaxn]{\HomC{\catfont{CAT}}{\lambda_{-}}{\id{}}}\HomC{\catfont{CAT}}{\Delta 1(-)}{F(A)},
\end{equation*}
that we call $\mu$, is $2$-universal. Explicitly, $\mu$ has components
\begin{fun}
\mu_{(B,X)} & \: & \HomC{\catfont{B}}{B}{A} & \longrightarrow & \HomC{\catfont{CAT}}{1}{F(A)} \\[1ex]
&& \tc*[6]{B}{A}{u}{v}{\theta} &\longmapsto & \tc*[9][30][pos=0.59][pos=0.59][,shift right=0.4ex]{1}{F(A)}{F(u)(X)}{F(v)(X)}{F(\theta)_X}
\end{fun}
for every $(B,X)\in \Grothop{F}$ and structure $2$-cells
$${\left(\mu_{(g,\gamma)}\right)}_u=F(u)(\gamma)\:F(u\circ g)(X')\to F(u)(X)$$
on every $(g,\gamma)\:(B,X)\al{} (B',X')$ in $\Grothop{F}$, for every $u\:B\to A$ in $\catfont{A}$. Given
$$\sigma\:\HomC{\catfont{A}}{\groth{F}(-)}{A}\aoplaxn{}\HomC{\catfont{CAT}}{\Delta 1(-)}{C},$$
we prove that there exists a unique functor $s\:F(A)\to C$ such that
$$(s\circ -)\circ \mu =\sigma.$$
We see that there is at most one such $s$, as we need, for every morphism $\alpha\:X\to X'$ in $F(A)$,
$$s(X)=s\left(\mu_{(A,X)}(\id{A})\right)=\sigma_{(A,X)}(\id{A})$$
$$s(\alpha)=s\left({\left(\mu_{(\id{A},\alpha)}\right)}_{\id{A}}\right)=\left(\sigma_{(\id{A},\alpha)}\right)_{\id{A}}.$$
And this $s$ works thanks to the fact that $\sigma$ is oplax normal, since it is readily shown to be functorial (using that $\sigma$ is oplax) and for every $(g,\gamma)\:(B,X)\to (B',X')$ in $\Grothop{F}$ and every $\theta\:u\aR{}v\:B\to A$ in $\catfont{A}$, considering $\underline{u}^X=(u,\id{})$ and $\underline{\theta}^X\:(u,F(\theta)_X)\aR{}(v,\id{})$,
$$\sigma_{(B,X)}(u)=\sigma_{(A,F(u)(X))}(\id{A})=s(F(u)(X))=s\left(\mu_{(B,X)}(u)\right)$$
$$\sigma_{(B,X)}(\theta)={\left(\sigma_{(u,F(\theta)_X)}\right)}_{\id{}}={\left(\sigma_{(\id{},F(\theta)_X)}\right)}_{\id{}}=s(F(\theta)_X)=s\left(\mu_{(B,X)}(\theta)\right)$$
$${\left(\sigma_{(g,\gamma)}\right)}_{u}={\left(\sigma_{(\id{},\gamma)}\right)}_{u}={\left(\sigma_{(u,F(u)(\gamma))}\right)}_{\id{}}={\left(\sigma_{(\id{},F(u)(\gamma))}\right)}_{\id{}}=s(F(u)(\gamma))=s\left({\left(\mu_{(g,\gamma)}\right)}_u\right)$$
We now prove the $2$-dimensional universality of $\mu$. Given $$\Xi\:\sigma\aM{}\sigma'\:\HomC{\catfont{A}}{\groth{F}(-)}{A}\aoplaxn{}\HomC{\catfont{CAT}}{\Delta 1(-)}{C},$$
we prove that there exists a unique natural transformation $\xi\:s\aR{} s'\:F(A)\to C$ such that
$$(\xi\ast -)\h \mu =\Xi.$$
We see that there is at most one such $\xi$, as we need, for every $X\in F(A)$,
$$\xi_X=\xi_{\mu_{(A,X)}(\id{A})}=\Xi_{(A,X),\id{A}}$$
And this $\xi$ works since it is readily shown to be natural (using that $\Xi$ is a modification) and for every $(B,X)\in \Grothop{F}$ and $u\:B\to A$ in $\catfont{A}$
$$\Xi_{(B,X),u}=\Xi_{(A,F(u)(X)),\id{A}}=\xi_{F(u)(X)}=\xi_{\mu_{(B,X)}(u)}.$$
We have thus shown that $\mu$ is $2$-universal and this concludes the proof.
\end{proof}
\begin{rem}
The rest of this section is dedicated to the proof that every pointwise left Kan extension in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$ along a $2$-$\catfont{Set}$-opfibration (as defined in \nameit{Definition}\ref{defpointkanextensionlaxthreecat}) is a weak left Kan extension in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$ as well.
For this, we need a $\oplaxn\mbox{-}\lax$ generalization of the parametrized Yoneda lemma (\nameit{Theorem}\ref{parametrizedyonedaoplaxnlax}) that does not seem to appear in the literature. This lemma will also shed further light on the concept of oplax normal natural transformation. Indeed the idea is the following: a fully lax parametrized Yoneda lemma is not possible, since it is thanks to the strict naturality that we can classically expand the datum on the identity to a complete natural transformation, but our version shows the minimal strictness that we need in order to do so.
Interestingly, such expansion in the fully strict $2$-natural case classically depends on the naturality of what will be our parameter $A$. Instead, we will need to expand through the slight strictness of the oplax normal naturality in $B$. And an expansion through $B$ is harder to achieve than one through $A$.
\end{rem}
\begin{defne}\label{defoplaxnlax}
Let $G,H\:\catfont{B}\ensuremath{^{\operatorname{op}}}\times \catfont{C}\to \catfont{E}$ be $2$-functors. An \dfn{$\text{\textbf{oplax}}^\text{\textbf{n}}\mbox{-}\text{\textbf{lax}}$ natural transformation} $\alpha$ from $G$ to $H$ is a collection of morphisms
$$\alpha_{B,C}\:G(B,C)\to H(B,C)$$
in $\catfont{E}$ for every $(B,C)\in \catfont{B}\ensuremath{^{\operatorname{op}}}\times\catfont{C}$ and, for every $f\:B'\to B$ in $\catfont{B}\ensuremath{^{\operatorname{op}}}$ and $g\:C\to C'$ in $\catfont{C}$, structure $2$-cells
\begin{eqD*}
\sq*[o][6][6][\alpha_{f,C}][3.1][3]{G(B',C)}{H(B',C)}{G(B,C)}{H(B,C)}{\alpha_{B',C}}{G(f,\id{})}{H(f,\id{})}{\alpha_{B,C}}\qquad
\sq*[l][6][6][\alpha_{B,g}][3.1][3]{G(B,C)}{H(B,C)}{G(B,C')}{H(B,C')}{\alpha_{B,C}}{G(\id{},g)}{H(\id{},g)}{\alpha_{B,C'}}
\end{eqD*}
such that $\alpha_{-,C}$ is oplax normal natural in $B\in \catfont{B}\ensuremath{^{\operatorname{op}}}$, $\alpha_{B,-}$ is lax natural in $C\in \catfont{C}$ and\v[0.5]
\begin{eqD*}
\begin{cd}*[5][-3]
\&[-1ex] G(B,C)\arrow[rr,"{\alpha_{B,C}}"]\arrow[rd,"{G(\id{},g)}"{inner sep=0.2ex},""'{name=A}]\&[-1ex]\& H(B,C)\arrow[rd,"{H(\id{},g)}"{inner sep=0.2ex}]\arrow[d,Rightarrow,shift right=2.85ex,"{\alpha_{B,g}}"{pos=0.44},shorten <=0.8ex,shorten >=2.2ex]\\
G(B',C)\arrow[rd,"{G(\id{},g)}"'{inner sep=0.2ex},""{name=B}]\arrow[ru,"{G(f,\id{})}"{inner sep=0.2ex}]\&\&G(B,C')\arrow[rr,"{\alpha_{B,C'}}"]\&\vphantom{.}\arrow[d,Rightarrow,shift right=2.85ex,"{\alpha_{f,C'}}",shorten <=1.05ex,shorten >=1ex]\& H(B,C')\\
\& G(B',C')\arrow[ru,"{G(f,\id{})}"'{inner sep=0.2ex}]\arrow[rr,"{\alpha_{B',C'}}"']\&\&H(B',C')\arrow[ru,"{H(f,\id{})}"'{inner sep=0.2ex}]
\arrow[equal,from=A,to=B,shorten <=3.64ex,shorten >=3.64ex]
\end{cd}\h[2] = \h[2]
\begin{cd}*[5][-3]
\& G(B,C)\arrow[rr,"{\alpha_{B,C}}"]\arrow[d,Rightarrow,shift left=2.85ex,"{\alpha_{f,C}}"'{pos=0.44},shorten <=0.8ex,shorten >=2.2ex]\&\&[-1ex] H(B,C)\arrow[rd,"{H(\id{},g)}"{inner sep=0.2ex},""'{name=A}]\\
G(B',C)\arrow[rr,"{\alpha_{B',C}}"]\arrow[rd,"{G(\id{},g)}"'{inner sep=0.2ex}]\arrow[ru,"{G(f,\id{})}"{inner sep=0.2ex}]\&\vphantom{.}\arrow[d,Rightarrow,shift left=2.85ex,"{\alpha_{B',g}}"',shorten <=1.05ex,shorten >=1ex]\&H(B',C)\arrow[ru,"{H(f,\id{})}"{inner sep=0.2ex}]\arrow[rd,"{H(\id{},g)}"'{inner sep=0.2ex},""{name=B}]\&\&[-1ex] H(B,C')\\
\& G(B',C')\arrow[rr,"{\alpha_{B',C'}}"']\&\&H(B',C')\v[2]\arrow[ru,"{H(f,\id{})}"'{inner sep=0.2ex}]
\arrow[equal,from=A,to=B,shorten <=3.64ex,shorten >=3.64ex]
\end{cd}
\end{eqD*}
We call the last axiom the compatibility axiom.
A \dfn{modification} $\Theta\:\alpha\aM{}\beta\:G\aR[\opln\mbox{-}\lax]{}H$ between $\oplaxn\mbox{-}\lax$ natural transformations is a collection of $2$-cells
\tc+[7]{G(B,C)}{H(B,C)}{\alpha_{B,C}}{\beta_{B,C}}{\Theta_{B,C}}
in $\catfont{E}$ that forms both, fixing $C$, a modification $\alpha_{-,C}\aM{}\beta_{-,C}$, and, fixing $B$, a modification $\alpha_{B,-}\aM{}\beta_{B,-}$.
\end{defne}
\begin{teor}[The $\oplaxn\mbox{-}\lax$ parametrized Yoneda lemma]\label{parametrizedyonedaoplaxnlax}
Let $K\:\catfont{B}\to \catfont{A}$ be a $2$-$\catfont{Set}$-opfibration (with small fibres) and $F\:\catfont{B}\ensuremath{^{\operatorname{op}}}\times \catfont{A}\to \catfont{CAT}$ be a $2$-functor. There is a bijection between
$$\alpha_{B,A}\:\HomC{\catfont{A}}{K(B)}{A}\to F(B,A)$$
$\oplaxn\mbox{-}\lax$ natural in $(B,A)\in \catfont{B}\ensuremath{^{\operatorname{op}}}\times \catfont{A}$ and
$$\eta_B\:1\to F(B,K(B))$$
extraordinary lax natural in $B\in \catfont{B}$ \pteor{see Hirata's~\cite{hirata_notesonlaxends} for a definition of extraordinary lax natural transformations and modifications between them}.
Moreover this bijection extends to an isomorphism of categories, considering as morphisms of the two categories respectively the modifications between $\oplaxn\mbox{-}\lax$ natural transformations and the modifications between extraordinary lax natural transformations.
\end{teor}
\begin{proof}
By \nameit{Theorem}\ref{essentialimagegivenbytwosetopf}, we can assume that $K$ is in the form $\groth{M}\:\Grothop{M}\to \catfont{A}$ for a $2$-functor $M\:\catfont{A}\ensuremath{^{\operatorname{op}}} \to \catfont{CAT}$. Given
$$\alpha_{(B,Y),A}\:\HomC{\catfont{A}}{K(B,Y)}{A}\to F((B,Y),A)$$
$\oplaxn\mbox{-}\lax$ natural in $((B,Y),A)\in {\left(\Grothop{M}\right)}\ensuremath{^{\operatorname{op}}}\times \catfont{A}$ (see \nameit{Definition}\ref{defoplaxnlax}), we construct $\eta_{(B,Y)}$ as the composite
$$1\ar{\id{B}} \HomC{\catfont{A}}{B}{B}\ar{\alpha_{(B,Y),B}}F((B,Y),B)$$
Then $\eta_{(B,Y)}$ is extraordinary lax natural in $(B,Y)\in \Grothop{M}$, with structure $2$-cell on $(g,\gamma)\:(B,Y)\to (B',Y')$ in $\Grothop{M}$ given by the pasting
\begin{eqD}{diagramfromalphatoeta}
\begin{cd}*[5][5]
1 \arrow[r,"{\id{B}}"]\arrow[d,"{\id{B'}}"']\& \HomC{\catfont{A}}{B}{B} \arrow[ld,equal,shorten <=5.05ex,shorten >=5.05ex]\arrow[d,"{g\circ -}"] \arrow[r,"{\alpha_{(B,Y),B}}"]\&F((B,Y),B) \arrow[dd,"{F(\id{},g)}"] \arrow[ld,Rightarrow,"{\alpha_{(B,Y),g}}"{pos=0.61},shorten <=4.45ex,shorten >=2.25ex]
\\
\HomC{\catfont{A}}{B'}{B'}\arrow[d,"{\alpha_{(B',Y'),B'}}"'] \arrow[r,"{-\circ g}"']\& \HomC{\catfont{A}}{B}{B'}\arrow[rd,"{\alpha_{(B,Y),B'}}"{inner sep =0.2ex}]\arrow[ld,Rightarrow,"{\alpha_{(g,\gamma),B'}}"{pos=0.56},shorten <=3.9ex,shorten >=3.35ex]
\\
F((B',Y'),B') \arrow[rr,"{F((g,\gamma),\id{})}"']\&\&F((B,Y),B')
\end{cd}
\end{eqD}
Indeed it is true in general that, given $2$-functors $T,S\:\catfont{C}\ensuremath{^{\operatorname{op}}}\times \catfont{C}\to \catfont{E}$ the composite
$$J\ar{\iota_C}T(C,C)\ar{\beta_{C,C}}S(C,C)$$
is extraordinary lax in $C\in \catfont{C}$ if $\iota_C$ is extraordinary lax in $C$ and $\beta_{C,D}$ is $\oplaxn\mbox{-}\lax$ in $(C,D)\in \catfont{C}\ensuremath{^{\operatorname{op}}}\times \catfont{C}$, with structure $2$-cells given by a pasting like that of equation~\refs{diagramfromalphatoeta} (with now a possibly non-identity $2$-cell also in the upper left square). Such structure $2$-cells surely respect the identities, and they also respect the composition by the compability axiom of $\oplaxn\mbox{-}\lax$ (and oplax and lax naturality). The two dimensional axiom is satisfied as well by moving the external $2$-cell through the diagram using the three $2$-dimensional properties that we have. We can then apply this result to $\eta_{(B,Y)}$ since $\id{K(B,Y)}$ is extraordinary natural in $(B,Y)$ and $\alpha_{(B,Y),K(B',Y')}$ is $\oplaxn\mbox{-}\lax$ natural in $((B,Y),(B',Y'))$ (as $\alpha{(B,Y),A}$ is $\oplaxn\mbox{-}\lax$ in $((B,Y),A)$).\v[0.3]
Now, given $\eta_B\:1\to F((B,Y),B)$ extraordinary lax natural in $(B,Y)\in \Grothop{M}$, we expand it to functors
$$\alpha_{(B,Y),A}\:\HomC{\catfont{A}}{K(B,Y)}{A}\to F((B,Y),A)$$
$\oplaxn\mbox{-}\lax$ natural in $((B,Y),A)\in {\left(\Grothop{M}\right)}\ensuremath{^{\operatorname{op}}}\times \catfont{A}$ as follows, using the oplax normal naturality in $(B,Y)$ (that is the only strictness we have). Given $u\:B\to A$ in $\catfont{A}$, considering $\underline{u}^{Y}=(u,\id{})$, the structure $2$-cell $\alpha_{(u,\id{}),A}=\id{}$ will give us a commutative square
\sq[n][5.5][10]{\HomC{\catfont{A}}{A}{A}}{F((A,M(u)(Y)),A)}{\HomC{\catfont{A}}{B}{A}}{F((B,Y),A)}{\alpha_{(A,M(u)(Y)),A}}{-\circ u}{F((u,\id{}),A)}{\alpha_{(B,Y),A}}
So, looking at how we constructed $\eta$ from $\alpha$, in order to reach the bijection we want, we define
$$\alpha_{(B,Y),A}(u)\mathrel{\mathrel{\mathop\ordinarycolon}\mkern-1.2mu=} F((u,\id{}),A)\left(\eta_{(A,M(u)(Y))}\right).$$
Given $\theta\:u\aR{} v\:B\to A$ in $\catfont{A}$, considering
$$\underline{\theta}^{Y}\:(u,M(\theta)_Y)\aR{}(v,\id{})\:(B,Y)\to (A,M(v)(Y))$$
and using that $\alpha_{(v,\id{}),A}=\id{}$, we will have by the $2$-dimensional axiom of oplax normal naturality that
$$\alpha_{(B,Y),A}(\theta)=F(\underline{\theta}^Y,A)_{\h\alpha_{(A,M(v)(Y)),A}(\id{A})}\circ {\left(\alpha_{(u,M(\theta)_Y),A}\right)}_{\id{A}}$$
So we firstly define the components of the structure $2$-cells that express the oplax normal naturality of $\alpha_{(B,Y),A}$ in $(B,Y)$ and then we will read how to define the action of $\alpha_{(B,Y),A}$ on morphisms $\theta$.
Looking at the diagram of equation~\refs{diagramfromalphatoeta} applied to $(\id{B},\gamma)\:(B,Y)\to (B,Y')$ in $\Grothop{M}$, we see that, in order to have a bijection between the $\alpha$'s and the $\eta$'s, we need to define
$${\left(\alpha_{(\id{B},\gamma),B}\right)}_{\id{B}}\mathrel{\mathrel{\mathop\ordinarycolon}\mkern-1.2mu=} \eta_{(\id{B},\gamma)}.$$
Whence, given arbitrary $(g,\gamma)\:(B,Y)\to (B',Y')$ in $\Grothop{M}$ and $u\:B\to A$ in $\catfont{A}$, since
$$(u,\id{})\circ (g,\gamma)=(\id{A},M(u)(\gamma))\circ (u\circ g,\id{}),$$ we need to define
$${\left(\alpha_{(g,\gamma),A}\right)}_u\mathrel{\mathrel{\mathop\ordinarycolon}\mkern-1.2mu=} F((u\circ g,\id{}),A)\left(\eta_{(\id{A},M(u)(\gamma))}\right).$$
And at this point we define, by the argument above,
$$\alpha_{(B,Y),A}(\theta)\mathrel{\mathrel{\mathop\ordinarycolon}\mkern-1.2mu=} F(\underline{\theta}^Y,A)_{\eta_{(A,M(v)(Y))}}\circ F((u,\id{}),A)\left(\eta_{(\id{A},M(\theta)_Y)}\right)$$
for every $\theta\:u\aR{}v\:B\to A$ in $\catfont{A}$.
Looking at the diagram of equation~\refs{diagramfromalphatoeta} applied to $(g,\id{})\:(B,Y)\to (B',M(g)(Y))$ in $\Grothop{M}$, we see that, in order to have a bijection between the $\alpha$'s and the $\eta$'s, we need to define
$${\left(\alpha_{(B,Y),g}\right)}_{\id{B}}\mathrel{\mathrel{\mathop\ordinarycolon}\mkern-1.2mu=} \eta_{(g,\id{})}.$$
Whence, given an arbitrary $f\:A\to A'$ in $\catfont{A}$ and $u\:K(B,Y)\to A$ in $\catfont{A}$, by the compatibility axiom of $\oplaxn\mbox{-}\lax$ applied to $(u,\id{})\:(B,Y)\to (A,M(u)(Y))$ in $\Grothop{M}$ and $f\:A\to A'$ in $\catfont{A}$, we need to define
$${\left(\alpha_{(B,Y),f}\right)}_u\mathrel{\mathrel{\mathop\ordinarycolon}\mkern-1.2mu=} F((u,\id{}),A')\left(\eta_{(f,\id{})}\right).$$
Now, we verify that such assignments work. To show that $\alpha_{(B,Y),A}$ is a functor, consider
$$u\aR{\theta} v\aR{\rho}w\:B\to A$$
in $\catfont{A}$. Then
$$\alpha_{(B,Y),A}(\rho\circ \theta)\mathrel{\mathrel{\mathop\ordinarycolon}\mkern-1.2mu=} F(\underline{(\rho\circ \theta)}^Y,A)_{\eta_{(A,M(w)(Y))}}\circ F((u,\id{}),A)\left(\eta_{(\id{A},M(\rho\circ\theta)_Y)}\right)$$
while $\alpha_{(B,Y),A}(\rho)\circ\alpha_{(B,Y),A}(\theta)$ is equal to
$$\scalebox{0.935}{$F(\underline{\rho}^Y,A)_{\eta_{(A,M(w)(Y))}}\circ F((v,\id{}),A)\left(\eta_{(\id{A},M(\rho)_Y)}\right)\circ F(\underline{\theta}^Y,A)_{\eta_{(A,M(v)(Y))}}\circ F((u,\id{}),A)\left(\eta_{(\id{A},M(\theta)_Y)}\right)$}$$
By the extraordinary naturality of $\eta$,
$$\eta_{(\id{A},M(\rho\circ \theta)_Y)}=F((\id{A},M(\theta)_Y),A)\left(\eta_{(\id{A},M(\rho)_Y)}\right)\circ \eta_{(\id{A},M(\theta)_Y)}$$
And by the uniqueness of the liftings of $2$-cells through $\groth{M}$, we have that
$$\underline{(\rho\circ \theta)}^Y=\underline{\rho}^Y\circ (\id{A},M(\rho)_Y)\h \underline{\theta}^Y.$$
So, by $2$-functoriality of $F$, it suffices to prove that
$$F(\underline{\theta}^Y,A)_{F((\id{A},M(\rho)_Y),A)\left(\eta_{(A,M(w)_Y)}\right)}\circ F((u,M(\theta)_Y),A)\left(\eta_{(\id{A},M(\rho)_Y)}\right)$$
is equal to
$$F((v,\id{}),A)\left(\eta_{(\id{A},M(\rho)_Y)}\right)\circ F(\underline{\theta}^Y,A)_{\eta_{(A,M(v)(Y))}}.$$
But this is true by naturality of $F(\underline{\theta}^Y,A)$ applied to the morphism
$$\eta_{(\id{A},M(\rho)_Y)}\:\eta_{(A,M(v)(Y))}\to F((\id{A},M(\rho)_Y),A)\left(\eta_{(A,M(w)(Y))}\right).$$
The fact that ${\left(\alpha_{(B,Y),f}\right)}_u$ is a natural transformation is checked with techniques similar to the above ones, noticing that
$$\underline{(f\h\theta)}^Y=(f,\id{})\h\underline{\theta}^Y.$$
Whereas showing that ${\left(\alpha_{(g,\gamma),A}\right)}_u$ is a natural transformation uses that for $(g,\gamma)\:(B,Y)\to (B',Y')$
$$\underline{(\theta\h g)}^{Y}=\underline{\theta}^{M(g)(Y)}\h (g,\id{})\quad \text{and}\quad \underline{\theta}^{Y'}\h(\id{},\gamma)=(\id{},M(v)(\gamma))\h[1.5]\underline{\theta}^{M(g)(Y)}.$$
At this point, it is straightforward to check that $\alpha_{(B,Y),A}$ is $\oplaxn\mbox{-}\lax$ in $((B,Y),A)$. And it is immediatly seen that we obtain a bijection between the $\alpha$'s and the $\eta$'s by construction.
Finally, we extend such bijection to an isomorphism of categories. Given a modification $$\Theta_{(B,Y),A}\:\alpha_{(B,Y),A}\aR{}\beta_{(B,Y),A}\:\HomC{\catfont{A}}{K(B,Y)}{A}\to F((B,Y),A)$$
between $\oplaxn\mbox{-}\lax$ natural transformations in $((B,Y),A)$, we send it to the modification
\begin{cd}[6][7]
1 \arrow[r,"{\id{B}}"]\&[-1ex]\HomC{\catfont{A}}{B}{B}\arrow[r,bend left=27,"{\alpha_{(B,Y),B}}",""'{name=A}]\arrow[r,bend right=27,"{\beta_{(B,Y),B}}"',""{name=B}]\&F((B,Y),B)
\arrow[Rightarrow,from=A,to=B,shift right=0.5ex,"{\Theta_{(B,Y),B}}"{description,pos=0.46},shorten <=0.15ex,shorten >=0ex]
\end{cd}
between extraordinary lax natural transformations (that the latter is such is easily checked). And such assignment is surely functorial.
Given a modification
$$\Gamma_{(B,Y)}\:\eta_{(B,Y)}\aR{}{\eta'}_{(B,Y)}\:1\to F((B,Y),B)$$
between extraordinary lax natural transformations, we construct a corresponding modification $\Theta$. We see that, if we want to reach an isomorphism of categories, we need to define
$${\left(\Theta_{(B,Y),B}\right)}_{\id{B}}\mathrel{\mathrel{\mathop\ordinarycolon}\mkern-1.2mu=} \Gamma_{(B,Y)}.$$
Whence, given an arbitrary $u\:B\to A$ in $\catfont{A}$, since we want $\Theta_{-,A}$ to be a modification, considering $(u,\id{})\:(B,Y)\to (A,M(u)(Y))$ in $\Grothop{M}$, we need to define
$${\left(\Theta_{(B,Y),A}\right)}_{u}\mathrel{\mathrel{\mathop\ordinarycolon}\mkern-1.2mu=} F((u,\id{}),A)\left(\Gamma_{(A,M(u)(Y))}\right).$$
It is straightforward to check that $\Theta$ is then a modification between $\oplaxn\mbox{-}\lax$ natural transformations. And such assignment is surely functorial. At this point, it is immediate to see that the two functors are, by construction, inverses of each other, giving the desidered isomorphism of categories.
\end{proof}
\begin{rem}
We are now ready to show that a pointwise left Kan extension in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$ along a $2$-$\catfont{Set}$-opfibration is always a weak left Kan extension.
\end{rem}
\begin{prop}\label{pointkanisalsoweak}
Consider a diagram
\begin{cd}[7][5]
\catfont{B} \arrow[rr,"{F}"]\arrow[d,"{K}"']\&\arrow[dl,Rightarrow,shift right=0.75ex,"{\lambda}"{pos=0.355, inner sep= 0.35ex},shorten <=1.65ex,shorten >= 3.35ex]\& \catfont{C} \\
\catfont{A}\arrow[rru,"L"']
\end{cd}
in $2\h[2]\mbox{-}\CAT_{\h[-3]\lax}$ with $\catfont{B}$ small and $K$ a $2$-$\catfont{Set}$-opfibration. Assume that $\lambda$ exhibits $L=\Lan{K}{F}$ \pteor{in the sense of \nameit{Definition}\ref{defpointkanextensionlaxthreecat}}. Then $\lambda$ also exhibits $L=\lan{K}{F}$.
\end{prop}
\begin{proof}
Since $L=\Lan{K}{F}$, for every $C$, the $2$-cell
\begin{equation*}
\HomC{\catfont{A}}{K(-)}{A}\aRR{L}\HomC{\catfont{C}}{(L\circ K)(-)}{L(A)}\aRR[\oplaxn]{\HomC{\catfont{C}}{\lambda_{-}}{\id{}}}\HomC{\catfont{C}}{F(-)}{L(A)};
\end{equation*}
is $2$-universal, giving an isomorphism of categories
\begin{equation}\label{eqisomorphismpointkanextnelladimcheeweak}
\HomC{\catfont{C}}{L(A)}{C}\arr{}\HomC{\moplaxn{\catfont{B}\ensuremath{^{\operatorname{op}}}}{\catfont{CAT}}}{\HomC{\catfont{A}}{K(-)}{A}}{\HomC{\catfont{C}}{F(-)}{C}}
\end{equation}
We need to prove that, for every $U\in \mlax{\catfont{A}}{\catfont{C}}$, pasting with $\lambda$ gives an isomorphism of categories
\begin{equation*}
\HomC{\mlax{\catfont{A}}{\catfont{C}}}{L}{U}\cong \HomC{\mlax{\catfont{B}}{\catfont{C}}}{F}{U\circ K}.
\end{equation*}
So consider a lax natural transformation $\varphi\:L\alax{} U$. For every $A\in \catfont{A}$, the component $\varphi_A\:L(A)\to U(A)$ corresponds to an oplax normal natural transformation
$$\alpha_{-,A}\:{\HomC{\catfont{A}}{K(-)}{A}}\aoplaxn{}{\HomC{\catfont{C}}{F(-)}{U(A)}}$$
via the isomorphism of equation~\refs{eqisomorphismpointkanextnelladimcheeweak}. And $\varphi_A$ being lax natural in $A\in \catfont{A}$ precisely corresponds to the oplax normal natural transformations $\alpha_{-,A}$ being lax natural in $\catfont{A}$, with structure $2$-cell on $f\:A\to A'$ in $\catfont{A}$ given by the image of $\varphi_f$ through the isomorphism of equation~\refs{eqisomorphismpointkanextnelladimcheeweak}. This means that the lax natural transformations $\varphi$ precisely correspond to functors $\alpha_{B,A}$ $\oplaxn\mbox{-}\lax$ natural in $(B,A)\in \catfont{B}\ensuremath{^{\operatorname{op}}}\times \catfont{A}$.
\noindent Consider then a modification $\Sigma\:\varphi\aM{}\psi\:L\alax{}U$. The components $\Sigma_A$ with $A\in \catfont{A}$ correspond to modifications $\Theta_{-,A}$ between oplax normal natural transformations $\alpha_{-,A}$ and $\beta_{-,A}$. And the modification axiom for $\Sigma$ corresponds to the modification axiom for $\Theta_{B,-}$ for every fixed $B\in \catfont{B}$. So the modifications $\Sigma$ precisely correspond to modifications $\Theta$ between $\oplaxn\mbox{-}\lax$ natural transformations $\alpha$ and $\beta$. By the functoriality of the isomorphism of equation~\refs{eqisomorphismpointkanextnelladimcheeweak}, we obtain an isomorphism of categories between $\HomC{\mlax{\catfont{A}}{\catfont{C}}}{L}{U}$ and the category of $\oplaxn\mbox{-}\lax$ natural transformations
$$\alpha_{B,A}\:{\HomC{\catfont{A}}{K(B)}{A}}\aoplaxn{}{\HomC{\catfont{C}}{F(B)}{U(A)}}$$
in $(B,A)\in \catfont{B}\ensuremath{^{\operatorname{op}}}\times\catfont{A}$ and modifications between them.
By \nameit{Theorem}\ref{parametrizedyonedaoplaxnlax} (the $\oplaxn\mbox{-}\lax$ parametrized Yoneda lemma), the latter category is then isomorphic to the category of extraordinary lax natural transformations
$$1\to \HomC{\catfont{C}}{F(B)}{U(K(B))}$$
in $B\in \catfont{B}$ and modifications between them, which is isomorphic (for example, by Hirata's paper~\cite{hirata_notesonlaxends}) to $\HomC{\mlax{\catfont{B}}{\catfont{C}}}{F}{U\circ K}$. Therefore we have produced an isomorphism of categories
\begin{equation*}
\HomC{\mlax{\catfont{A}}{\catfont{C}}}{L}{U}\cong \HomC{\mlax{\catfont{B}}{\catfont{C}}}{F}{U\circ K},
\end{equation*}
and we can read that this is given by pasting with $\lambda$.
\end{proof}
\bibliographystyle{abbrv}
| {
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Many aficionados of classic art consider the MET Museum NYC to be the nexus of the art world, for it is the largest museum in the United States, and among the 10 largest in the world. And with a collection of over 2 million works, the MET Museum of Art houses the most important art collection of any museum in the world. But the exhibits at the museum are so diverse, particularly when you consider the ever changing "special exhibits," that no matter how many times you have been to the New York Metropolitan Museum of art, you still never know what to expect.
An example of the special exhibits that are on display at the MET was just announced today. According to the New York Times, the opening of this very special exhibit is scheduled to coincide with the 2013 Major League Baseball All Star Game, which will take place at the home of the New York Mets, Citi Field.
Although MET Museum of Art possesses a collection of more than 31,000 baseball cards, will consist of the 600 rarest and valuable cards. The collection, the size of which is only surpassed by the collection housed in the Professional Baseball Hall of Fame, was donated to the MET during the 1940s by Jefferson R. Burdick. After his generous donation, the MET even temporarily hired Burdick to catalog his massive collection.
The infamous Jefferson R. Burdick Collection is merely one example of the unique exhibits that can be viewed at the museum. While each special exhibit attracts people to the MET Museum who might not consider themselves the "Museum type," some come to witness first hand artifacts from the earliest day of civilizations. Still, others come to the MET Museum in NYC because they never know what surprises await them.
This entry was posted on Friday, July 12th, 2013 at 10:05 am and is filed under Homepage. You can follow any responses to this entry through the RSS 2.0 feed. You can skip to the end and leave a response. Pinging is currently not allowed. | {
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Лаос на летних Олимпийских играх 2012 будет представлен в двух видах спорта.
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\section{Preliminaries}
We now give a formal definition of the discrete Fr\'echet distance and its variants.
Let $A=(a_{1},\ldots,a_{m})$ and $B=(b_{1},\ldots,b_{n})$ be two
sequences of $m$ and $n$ points, respectively, in the plane. Let $G(V,E)$ denote a graph whose vertex set is $V$ and edge set is $E$, and let $\|\cdot\|$ denote the Euclidean norm.
Fix a distance $\delta>0$, and define the following three directed graphs $G_\delta=G(A\times B, E_{\delta})$, $G_\delta^-=G(A\times B, E_{\delta}^-)$, and $G_\delta^+=G(A\times B, E_{\delta}^+)$, where
\begin{align*}
E_{\delta}= & \left\{ \Bigr((a_{i},b_{j}),(a_{i+1},b_{j})\Bigl)\:\middle|\,\|a_{i}-b_{j}\|,\;\|a_{i+1}-b_{j}\|\le\delta\right\} \bigcup\\
& \left\{ \Bigr((a_{i},b_{j}),(a_{i},b_{j+1})\Bigl)\:\middle|\,\|a_{i}-b_{j}\|,\;\|a_{i}-b_{j+1}\|\le\delta\right\},\\
E_{\delta}^{-}= & \left\{ \Bigr((a_{i},b_{j}),(a_{k},b_{j})\Bigl)\:\middle|\, k>i,\,\|a_{i}-b_{j}\|,\;\|a_{k}-b_{j}\|\le\delta\right\} \bigcup\\
& \left\{ \Bigr((a_{i},b_{j}),(a_{i},b_{j+1})\Bigl)\:\middle|\,\|a_{i}-b_{j}\|,\;\|a_{i}-b_{j+1}\|\le\delta\right\}
\end{align*}
\begin{align*}
E_{\delta}^{+}= & \left\{ \Bigr((a_{i},b_{j}),(a_{k},b_{j})\Bigl)\:\middle|\, k>i,\,\|a_{i}-b_{j}\|,\;\|a_{k}-b_{j}\|\le\delta\right\} \bigcup\\
& \left\{ \Bigr((a_{i},b_{j}),(a_{i},b_{l})\Bigl)\:\middle|\, l>j,\,\|a_{i}-b_{j}\|,\;\|a_{i}-b_{l}\|\le\delta\right\}.
\end{align*}
For each of these graphs we say
that a position $(a_{i},b_{j})$ is a \emph{reachable}
position if $(a_{i},b_{j})$ is reachable from $(a_{1},b_{1})$ in the respective graph.
Then the discrete Fr\'echet distance (DFD for short) $\delta_F^{*}(A,B)$ is the smallest $\delta>0$
for which $(a_{m},b_{n})$ is a reachable position in $G_\delta$.
Similarly, the one-sided Fr\'echet distance with shortcuts (one-sided DFDS for short) $\delta_F^{-}(A,B)$ is the smallest $\delta>0$
for which $(a_{m},b_{n})$ is a reachable position in $G_\delta^-$, and the two-sided Fr\'echet distance with shortcuts (two-sided DFDS for short) $\delta_F^{+}(A,B)$ is the smallest $\delta>0$
for which $(a_{m},b_{n})$ is a reachable position in $G_\delta^+$.
\section{Decision procedure for the one-sided DFDS}
\label{sec:DFDS1} We first consider the corresponding decision problem.
That is, given a value $\delta>0$, we wish to decide whether $\delta_F^{-}(A,B)\le\delta$ (we ignore the issue of
discrimination between the cases of strict inequality and equality, in the decision procedures of both the one-sided variant and the two-sided variant, since this will be handled in the optimization procedures, described later).
Let $M$ be the matrix whose rows correspond to the elements of $A$ and
whose columns correspond to the elements of $B$ and $M_{i,j}=1$ if
$\|a_{i}-b_{j}\|\le\delta$, and $M_{i,j}=0$ otherwise. Consider
first the DFD variant (no shortcuts allowed), in which,
at each move, exactly one of the frogs has to jump to the next
point. Suppose that $(a_{i},b_{j})$ is a reachable position of the
frogs. Then, necessarily, $M_{i,j}=1$. If $M_{i+1,j}=1$ then the
next move can be an {\em upward move} in which the $A$-frog moves
from $a_{i}$ to $a_{i+1}$, and if $M_{i,j + 1} = 1$ then the next
move can be a {\em right move} in which the $B$-frog moves from
$b_{j}$ to $b_{j+1}$. It follows that to determine whether
$\delta_F^{*}(A,B)\le\delta$, we need to determine whether there is
a \emph{right-upward staircase} of ones in $M$ that starts at
$M_{1,1}$, ends at $M_{m,n}$, and consists of a sequence of
interweaving upward moves and right moves (see
Figure~\ref{fig:staircase}(a)).
\begin{figure*}[htp]
\centering
\begin{tabular}{ccc}
\includegraphics[scale=0.8]{staircase3_a.pdf} & \hspace{0.5cm}
\includegraphics[scale=0.8]{staircase3_c.pdf} & \hspace{0.5cm}
\includegraphics[scale=0.8]{staircase3_b.pdf} \\
(a) & (b) & (c)
\end{tabular}
\vspace{-0.5cm}
\centering \caption{\small (a) A right-upward staircase (for DFD with no simultaneous jumps). (b) A semi-sparse
staircase (for the one-sided DFDS). (c) A sparse
staircase (for the two-sided DFDS).}
\label{fig:staircase}
\end{figure*}
In the
one-sided version of DFDS, given a
reachable position $(a_{i},b_{j})$ of the frogs, the $A$-frog can
move to any point $a_{k},k>i$, for which $M_{k,j}=1$; this is a
\emph{skipping upward move} in $M$ which starts at $M_{i,j}=1$,
skips over $M_{i+1,j},\ldots,M_{k-1,j}$ (some of which may be 0),
and reaches $M_{k,j}=1$. However, in this variant, as in the DFD variant, the $B$-frog
can only make a consecutive \emph{right move} from $b_{j}$ to $b_{j+1}$,
provided that $M_{i,j+1}=1$ (otherwise no move of the $B$-frog is
possible at this position). Determining whether
$\delta_F^{-}(A,B)\le\delta$ corresponds to deciding whether there
is a \emph{semi-sparse staircase} of ones in $M$ that starts at
$M_{1,1}$, ends at $M_{m,n}$, and consists of an interweaving
sequence of skipping upward moves and (consecutive) right moves (see
Figure~\ref{fig:staircase}(b)).
Assume that $M_{1,1}=1$ and $M_{m,n}=1$; otherwise, we can immediately
conclude that $\delta_F^{-}(A,B)>\delta$ and terminate the decision procedure. From now on, whenever we
refer to a semi-sparse staircase, we mean a semi-sparse staircase
of ones in $M$ starting at $M_{1,1}$, as defined above, but without the requirement that it ends at $M_{m,n}$.
\begin{figure}[htbp]
\MyFrame{
\smallskip
\begin{itemize}
{\small \sf
\item $S \leftarrow \langle M_{1,1}\rangle$ \vspace{-0.3cm}
\item $i \leftarrow 1$, $j \leftarrow 1$ \vspace{-0.3cm}
\item While $(i < m$ or $j < n)$ do \vspace{-0.3cm}
\begin{itemize}
\item {\small \sf If (a right move is possible) then } \vspace{-0.1cm}
\begin{itemize}
\item Make a right move and add position $M_{i,j+1}$ to $S$ \vspace{-0.1cm}
\item $j \leftarrow j+1$ \vspace{-0.2cm}
\end{itemize}
\item Else \vspace{-0.2cm}
\begin{itemize}
\item If (a skipping upward move is possible) then \vspace{-0.1cm}
\begin{itemize}
\item Move upwards to the first (i.e., lowest) position $M_{k,j}$, with $k>i$ and $M_{k,j}=1$, and add $M_{k,j}$ to $S$ \vspace{-0.1cm}
\item $i\leftarrow k$ \vspace{-0.2cm}
\end{itemize}
\item Else \vspace{-0.2cm}
\begin{itemize}
\item Return $\delta_F^-(A,B)>\delta$ \vspace{-0.3cm}
\end{itemize}
\end{itemize}
\end{itemize}
\item Return $\delta_F^-(A,B)\le\delta$
}
\end{itemize}
}\caption{Decision procedure for the one-sided discrete Fr\'echet distance with shortcuts.} \label{alg:semi-sparse}
\end{figure}
The algorithm of Figure~\ref{alg:semi-sparse}, that implements the
decision procedure, constructs a semi-sparse staircase $S$ by always
making a right move if possible. The correctness of the decision
procedure is established by the following lemma.
\begin{lemma} \label{lem:one-sided}
If there exists a semi-sparse staircase that ends at
$M_{m,n}$, then $S$ also ends at $M_{m,n}$. Hence $S$ ends at $M_{m,n}$ if and only if $\delta_F^-(A,B) \le \delta$.
\end{lemma}
\begin{proof}
Let $S'$ be a semi-sparse staircase that ends at $M_{m,n}$. We think
of $S'$ as a sequence of possible positions (i.e., $1$-entries) in
$M$. Note that $S'$ has at least one position in each column of $M$,
since skipping is not allowed when moving rightwards. We claim that
for each position $M_{k,j}$ in $S'$, there exists a position
$M_{i,j}$ in $S$, such that $i \le k$. This, in particular, implies
that $S$ reaches the last column. If $S$ reaches the last column, we
can continue it and reach $M_{m,n}$ by a sequence of skipping upward
moves (or just by one such move), so the lemma follows.
We prove the claim by induction on $j$. It clearly holds for
$j=1$ as both $S$ and $S'$ start at $M_{1,1}$. We assume then
that the claim holds for $j=\ell-1$, and establish it for $\ell$.
That is, assume that if $S'$ contains an entry $M_{k,\ell-1}$, then $S$
contains $M_{i,\ell-1}$ for some $i\le k$. Let $M_{k',\ell}$ be
the lowest position of $S'$ in column $\ell$; clearly, $k'\ge k$.
We must have $M_{k',\ell-1} = 1$ (as the only way to move from a column to
the next is by a right move). If $M_{i,\ell}=1$ then $M_{i,\ell}$
is added to $S$ by making a right move, and $i\le k\le k'$ as
required. Otherwise, $S$ is extended by a sequence of skipping
upward moves in column $\ell-1$ followed by a right move between
$M_{i',\ell-1}$ and $M_{i',\ell}$ where $i'$ is the smallest index
$\ge i$ for which both $M_{i',\ell-1}$ and $M_{i',\ell}$ are one.
But since $i\le k'$ and $M_{k',\ell-1}$ and $M_{k',\ell}$ are both
$1$, we get that $i' \le k'$, as required.
\end{proof}
\paragraph{Running time.}
The entries of $M$ that the decision procedure tests form a row-
and column-monotone path, with an additional entry to the right for
each upward turn of the path. (This also takes into account the
$0$-entries of $M$ that are inspected during a skipping upward move.)
Therefore it runs in $O(m+n)$ time.
\section{One-sided DFDS optimization via approximate distance counting and selection}
\label{sec:tec}
We now show how to use the decision procedure of Figure~\ref{alg:semi-sparse} to solve the optimization problem of the one-sided discrete Fr\'echet distance with shortcuts. This is based on the algorithms provided in Lemma \ref{lem:finding_interval} and Lemma
\ref{lem:searching_in_interval} given below. We believe that the algorithm of Lemma~\ref{lem:finding_interval} is of independent interest, and give another application of it to a different distance-related problem in Corollary~\ref{cor:aprox_selection} below.
First note that if we increase $\delta$
continuously, the set of $1$-entries of $M$ can only grow, and
this can only happen when $\delta$ is a distance between a point of $A$ and a
point of $B$. Performing a binary search over the $O(mn)$ pairwise distances of pairs in
$A\times B$ can be done using the distance selection
algorithm of~\cite{KS97}. This will be the method of choice for the two-sided DFDS problem, treated in Section~\ref{sec:DFDS2}. Here however, this procedure, which takes
$O(m^{2/3}n^{2/3}\log^3(m+n))$ time is rather excessive when compared
to the linear cost of the decision procedure. While solving the
optimization problem in close to linear time is still a challenging
open problem, we manage to improve the running time considerably, to
$O((m+n)^{6/5+\eps})$, for any $\eps>0$.
We first present the two main subproblems that the algorithm uses as independent building blocks, in the following two lemmas.
\begin{lemma}
\label{lem:finding_interval} Given a set $A$ of $m$ points and a set
$B$ of $n$ points in the plane, an interval
$(\alpha,\beta]\subset\reals$, and parameters $0 < L \leq mn$ and
$\eps>0$, we can determine, with high probability, whether
$(\alpha,\beta]$ contains at most $L$ distances between pairs in
$A\times B$. If $(\alpha,\beta]$ contains more than $L$ such
distances, we return a sample of $O(\log(m+n))$ pairs, so that, with
high probability, at least one of these pairs determines an
approximate median (in the middle three quarters) of the pairwise
distances that lie in $(\alpha,\beta]$. Our algorithm runs in
$O((m+n)^{4/3+\eps}/L^{1/3} +m+n)$ time and uses
$O((m+n)^{4/3+\eps}/L^{1/3} +m+n)$ space.
\end{lemma}
\begin{proof}
We partially construct a batched range counting data structure
for representing (some of) the pairs $(p,q)\in A \times B$ whose distance lies
in $(\alpha,\beta]$, as the edge-disjoint union of complete bipartite
graphs. If we build the complete data structure, it will require
$O((m+n)^{4/3+\eps})$ time and $O((m+n)^{4/3+\eps})$ storage (similar
to the parameters of the structure of~\cite{KS97} used in Section~\ref{sec:DFDS2}).
Since this is too expensive, we run the construction
until it reaches a level where the size of each subproblem is at most
$L$ (details will follow shortly), and then stop. The pairwise distances of $A\times B$
that fall in $(\alpha,\beta]$ are now of two types, those recorded in the
complete bipartite graphs that we have constructed, and those where the
two points belong to the same remaining ``leaf'' subproblem. We estimate the
number of distances of the latter type using an appropriate random sample of
points. As we show later, this also allows us to return a sample that contains
an approximate median of the pairwise distances in $(\alpha,\beta]$, with high
probability, if the number of these distances is indeed larger than $L$.
In more detail, we proceed as follows. Let $C$ denote the collection
of the circles bounding the $(\alpha,\beta)$-annuli that are
centered at the points of $A$. We choose a sufficiently large
constant parameter $1 \leq r \leq m$, and construct a
$(1/r)$-cutting for $C$. That is, for a suitable constant $c$, we
partition the plane into $k\leq cr^2$ cells
$\Delta_1,\ldots,\Delta_k$, each of constant description complexity,
so that each $\Delta_i$ is crossed by at most $m/r$ boundaries of
the annuli, and each $\Delta_i$ contains at most $n/r^2$ points of
$B$. This can be done in $O((m+n)r)$ deterministic time, as
in~\cite{Cha93,CF90,Mat91}.\footnote{The construction
in~\cite{Cha93,CF90,Mat91} shows that each $\Delta_i$ is crossed by
at most $m/r$ circles in $C$. To ensure that each
$\Delta_i$ contains at most $n/r^2$ points of $B$, we duplicate each
$\Delta_i$ that contains more than $n/r^2$ points as many times as
needed, and assign to each copy a subset of at most $n/r^2$ of the
points (these sets are pairwise disjoint and cover all the points in the cell).
Then each cell of the resulting subdivision contains at most
$n/r^2$ points, and the size of the cutting is still $O(r^2)$.} We
then dualize the roles of $A$ and $B$, in each cell $\Delta_i$
separately, where the set $B_{\Delta_i}$ of the at most $n/r^2$
points of $B$ in $\Delta_i$ becomes a set of $(\alpha,\beta)$-annuli
centered at these points, and the set $A_{\Delta_i}$ of the at most
$m/r$ points of $A$ whose annuli boundaries cross $\Delta_i$ is now
regarded as a set of points. We now construct, for each $\Delta_i$,
a $(1/r)$-cutting in this dual setting. We obtain a total of at most
$c^2r^4$ subproblems, each involving at most $m/r^3$ points of $A$
and at most $n/r^3$ points of $B$.
In both the primal and dual stages, we output a collection
of complete bipartite graphs, one for each (primal or dual) cell $\Delta_i$.
The sets of vertices of the graph associated with $\Delta_i$ are the
set of points whose annuli fully contain $\Delta_i$ and the set of
points contained in $\Delta_i$ (one of these sets is a subset of $A$
and the other is a subset of $B$). The total vertex-size over all these
graphs is at most $c'(r)(m+n)$, for some constant $c'(r)$ depending on $r$.
We now process each of the $O(r^4)$ subproblems recursively, using the
same parameter $r$, and keep doing so until we get subproblems of size
at most $L$ (in terms of the number of $A$-points plus the number of $B$-points). If this happens at
level $j$ of the recursion, we have (roughly) $(m+n)/r^{3j} = L$, or
$r^j = ((m+n)/L)^{1/3}$. The number of subproblems is at most
$c^{2j}r^{4j} = c^{2j} ((m+n)/L)^{4/3}$. If we choose $r$ sufficiently large,
we can bound $c^{2j}$ by ${(r^j)}^\eps$, where $\eps$ is the positive parameter
prespecified in the lemma.
The total (vertex) size of the graphs output so far is dominated by the
size of the graphs output at the last level, which is at most
$$
c^{2j}c'(r) r^{4j} \cdot (m+n)/r^{3j} \leq c'(r) (m+n)r^{j(1+\eps)} \leq
c'(r) (m+n)^{4/3+\eps}/L^{1/3} .
$$
With some care (again, see~\cite{Cha93,CF90,Mat91}), this also
bounds the cost of constructing the structure.
We count (within the same time bound) the number $N_1$ of edges in the
graphs produced by the algorithm. If $N_1$ exceeds $L/2$, we generate a
random sample $R_1$ of $c_1\log (m+n)$ pairs of points (i.e., edges) from
these graphs, for some sufficiently large constant $c_1>0$. We omit the routine
details of the sampling mechanism, but remark that it ensures that each of the
sampled distances is a random element of the (uniform distribution on the)
set of all distances recorded by the graphs. By construction, all sampled
distances lie in $(\alpha,\beta]$. Moreover, with high probability, the sample
contains an approximate median (in the \emph{middle half}) of these distances;
the routine justification of this claim is provided below.
Each subproblem at the bottom of the recursion may contain additional distances
that lie in $(\alpha,\beta]$; these are distances between centers of annuli
whose boundaries cross the cell of the subproblem and points in the cell that
lie inside these annuli. Denote by $N_2$ the overall number of these distances
at the bottom subproblems ($N_2$ is not known to us). Note that the total number
of pairs in the subproblems is
$$M:= O(((m+n)/L)^{4/3+\eps}\cdot L^2) = O((m+n)^{4/3+\eps}L^{2/3}).$$
Our next step is to determine (approximately) how many of these distances lie inside
$(\alpha,\beta]$; that is, our goal is to estimate $N_2$.
To this end, we generate a random sample $R_2$ of $c_2(M/L)\log (m+n)$ pairs from these subproblems, for some sufficiently large constant $c_2>0$, and check how
many of them lie in $(\alpha,\beta]$. (Again, the sampling mechanism is straightforward, and we omit its details.) Let $R_2'$ denote the subset of $R_2$ of those pairs whose distances lie in $(\alpha,\beta]$. It can be shown, similar to the
analysis of Har-Peled and Sharir \cite{HS11} (which in turn is based on
the work of Li et al.~\cite{LLS01}) that, with high probability, $N_2$ is at most $L/2$ if and only if
the number of distances in $R_2'$ is $O(\log (m+n))$, for an appropriate constant of proportionality.
(Sharir and Shaul~\cite{ShSh}, who also use this tool, call such samples \emph{shallow $\eps$-nets}.)
If $N_1\leq L/2$ and we have determined that $N_2 \leq L/2$ too, then $N_1+N_2$ is at most $L$, and we terminate the algorithm (for the task considered in the lemma), since we have determined that the number of distances of $A\times B$ that lie in $(\alpha,\beta]$ is at most $L$.
Otherwise, with high probability, $R_1 \cup R_2$ contains an approximate median (in the middle three quarters) of the pairwise distances of $A\times B$ in $(\alpha,\beta]$.
In more detail, in the case under consideration we either have $N_1 >L/2$ or $N_2>L/2$ (or both). In the former case, the probability that $R_1$ does not contain a pair of distances in the middle half of the distances recorded in the complete bipartite graphs is $(1/2)^{c_1\log (m+n)} = 1/(m+n)^{c_1}$.
If $N_2>L/2$, then the probability that $R_2$ does not contain a pair at distance in the middle half of the corresponding $N_2$ distances is
$$
\left(1-\dfrac{N_2}{2M}\right)^{c_2(M/L)\log(m+n)} < e^{-\frac{1}{2}c_2(N_2/L)\log(m+n)}<e^{-\frac{1}{4}c_2\log(m+n)} <
\dfrac{1}{(m+n)^{c_2/4}}.
$$
Assume for the moment that both $N_1$ and $N_2$ are greater than $L/2$. Then, with high probability, $R_1$ contains a pair $(a_1',b_1')$ whose distance,
$d_1$, lies in the middle half of the distances recorded in the graphs. Similarly, $R_2'$ contains a pair $(a_2',b_2')$ whose distance, $d_2$, lies in the
middle half of the distances in $(\alpha,\beta]$ that are generated by the leaf subproblems. An easy calculation shows that either $d_1$ or $d_2$ must lie in the
middle three quarters of the overall set of distances in $(\alpha,\beta]$. Similar reasoning applies when either only $N_1$ or only $N_2$ is greater than $L/2$.
We thus return $R_1\cup R_2'$ as the output of the algorithm.
The cost of the algorithm is composed from the following sub-costs:
\noindent{(i)}
We pay $O((m+n)^{4/3+\eps}/L^{1/3})$ for the partial construction of the data structure.
\noindent{(ii)}
We then sample and test
\begin{align*}
& O((M/L)\log (m+n)) = O\left(\left((m+n)^{4/3+\eps}L^{2/3}/L\right)\log (m+n)\right) = & O\left((m+n)^{4/3+\eps}/L^{1/3}\right)
\end{align*}
pairs in
the bottom subproblems, for another, but still arbitrarily small $\eps>0$, at the same asymptotic cost.
\end{proof}
The way it is described, the algorithm does not verify that the
samples that it returns satisfy the desired properties, nor does it
verify that the number of distances in $(\alpha,\beta]$ is indeed at
most $L$, when it makes this assertion. As such, the running time is
deterministic, and the algorithm succeeds with high probability
(which can be calibrated by the choice of the constants $c_1, c_2$).
See below for another comment regarding this issue.
We use the procedure provided by
Lemma~\ref{lem:finding_interval} to find an interval $(\alpha,
\beta]$ that contains at most $L$ distances between pairs of
$A\times B$, including $\delta_F^-(A,B)$. We find this interval
using binary search, starting with $(\alpha,\beta]=(0,\infty)$, say.
In each step of the search, we run the algorithm of
Lemma~\ref{lem:finding_interval}. If it determines that the number
of critical distances in $(\alpha,\beta]$ is at most $L$ we stop. (The concrete choice of $L$ that we will use is given later.)
Otherwise, the algorithm returns a random sample $R$ that contains,
with high probability, an approximate median (in the middle three
quarters) of the distances in $(\alpha,\beta]$. We then find two consecutive distances $\alpha',\beta'$ in $R$ such that $\delta_F^-(A,B)\in (\alpha',\beta']$,
using the decision procedure (see Figure~\ref{alg:semi-sparse}). $(\alpha',\beta']$ is
a subinterval of $(\alpha,\beta]$ that contains, with high
probability, at most $7/8$ of the distances in $(\alpha,\beta]$. We
then proceed to the next step of the binary search, applying again
the algorithm of Lemma~\ref{lem:finding_interval} to the new
interval $(\alpha',\beta']$. The resulting algorithm runs in
$O((m+n)^{4/3+\eps}/L^{1/3}+(m+n)\log(m+n))$ time, for any
$\eps>0$.
Once we have narrowed down the interval $(\alpha, \beta]$, so
that it now contains at most $L$ distances between pairs of $A\times B$,
including $\delta_F^-(A,B)$, we can find $\delta_F^-(A,B)$ by
simulating the execution of the decision procedure at the unknown
$\delta_F^-(A,B)$. A simple way of doing this is as follows. To determine whether $M_{i,j}=1$ at
$\delta_F^-(A,B)$, we compute the critical distance
$r' = \|a_i-b_j\|$ at which $M_{i,j}$ becomes $1$. If $r' \le \alpha$
then $M_{i,j}=0$, and if $r' \ge \beta$ then $M_{i,j}=1$.
Otherwise, $\alpha < r' < \beta$ is one of the at most $L$ distances
in $(\alpha, \beta]$. In this case we run the decision procedure at
$r'$ to determine $M_{i,j}$. Since there are at most $L$ distances
in $(\alpha, \beta]$, the total running time is $O(L(m+n))$. By
picking $L= (m+n)^{1/4+\eps}$ for another, but still arbitrarily
small $\eps > 0$, we balance
the bounds of $O((m+n)^{4/3+\eps}/L^{1/3}+(m+n)\log(m+n))$ and
$O(L(m+n))$, and obtain the bound of $O((m+n)^{5/4+\eps})$, for any $\eps >0$,
on the overall running time.
Although this significantly improves the naive implementation
mentioned earlier, it suffers from the weakness that it has to run
the decision procedure separately for each distance in
$(\alpha, \beta]$ that we encounter during the simulation. Lemma~\ref{lem:searching_in_interval}
shows how to accumulate several unknown distances and resolve them all
using a binary search that is guided by the decision procedure.
This allows us to find $\delta_F^-(A,B)$ within the interval
$(\alpha, \beta]$ more efficiently.
\begin{lemma}
\label{lem:searching_in_interval}
Given a set $A$ of $m$ points and a set $B$ of $n$ points in the plane, and
an interval $(\alpha,\beta]\subset\reals$ that contains at most $L$ distances
between pairs in $A\times B$, including $\delta_F^-(A,B)$, we can find
$\delta_F^-(A,B)$ in $O((m+n)L^{1/2}\log(m+n))$ (deterministic) time using $O(m+n)$ space.
\end{lemma}
\begin{proof}
We simulate the decision procedure (of Figure~\ref{alg:semi-sparse}) at the unknown value $\delta^- = \delta_F^-(A,B)$. During the simulation, when
attempting to retrieve specific entries $M_{i,j}$ of $M$, we encounter comparisons between $\delta^-$ and concrete distances between pairs of points in
$A\times B$. When we need to compare $\delta^-$ with such a distance $r'$, we first check whether $r'$ is in $(\alpha,\beta]$. If not, we know the result of the
comparison (if $r' \leq\alpha$ then $r'<\delta^-$, and if $r'> \beta$ then $r'> \delta^-$). If $\alpha <r'\leq\beta$, we bifurcate, continuing along two
possible paths, one assuming that $r'\leq\delta^-$ and one assuming that $r' > \delta^-$. However, we proceed along
each of these paths for only $s$ steps, for another parameter $s$ that will be specified shortly. (More precisely, we proceed until we
have examined $s$ known entries of $M$ (i.e., entries lying outside $(\alpha, \beta]$), including $0$-entries that we encounter
as we climb upwards in a column.)
If, before examining $s$ entries, we encounter another unknown entry,
we bifurcate again, and keep doing so, until we have examined a total of
$m+n$ entries of $M$, in which case we terminate the current ``phase''.
(It is conceivable that some entries of $M$ are examined more than once
in this procedure, but when such a multiply-visited entry is unknown,
we bifurcate there only once.) The resulting object is a binary tree $T$,
with some number, $x$, of outdegree-$2$ nodes (at which we have bifurcated),\footnote{%
Technically, when we are at some known entry
$M_{i,j}$, check its right neighbor $M_{i,j+1}$, and find that it is (known
to be) 0, we continue the tracing upwards in column $j$. This 2-way exploration
is not considered a bifurcation in the present analysis.}
so that the maximum stretch of consecutive outdegree-$1$ nodes is $s$,
and so that the total number of nodes in the tree is at most $m+n$.
We now sort the set $X$ of the $x=O(m+n)$ critical values at which
we have bifurcated, in $O((m+n) \log (m+n))$ time. We then run a binary search over $X$, using the decision procedure (of Figure~\ref{alg:semi-sparse}) to guide the search.
This step also takes $O((m+n)\log (m+n))$ time. This determines all the $x$ unknown
values that we have encountered, and allows us to choose the lowest
path in $T$ that is still a semi-sparse staircase, as the next portion of the overall lowest semi-sparse path $S$ in $M$
(at the optimal value $\delta^-$). This also allows us to shrink the interval that is known to contain $\delta^-$ to be bounded by two consecutive critical
values of $\{\alpha, \beta\} \cup X$. What we have gained, in a ``successful''
phase, is at least $s$ extra steps of the desired semi-sparse path $S$. Assuming this to be
the case, and since the total length of $S$ is $O(m+n)$, we need
at most $O((m+n)/s)$ successful phases of this kind, whose total cost is thus $O(((m+n)^2/s)\log (m+n))$.
This is only one side of the story, though, because there might be phases where
we do not manage to gain $s$ steps, because we run into ``too many''
bifurcations. If a phase generates $x$ bifurcations, then, continuing
the search in $M$ beyond them, we encounter at most $xs$ entries of $M$.
The reason for not having a ``tail'' of $s$ entries beyond any bifurcation
is that we have exceeded the number of steps per phase, namely $m+n$.
We thus have $m+n\le xs$ or $x\ge (m+n)/s$. In other words, the number of
such ``unsuccessful'' phases is $O(Ls/(m+n))$, and each such phase takes $O((m+n)\log (m+n))$
time, as before, for a total of $O(Ls\log(m+n))$ time. At the end of the simulation, we have two consecutive critical values $\alpha, \beta$ of distances between pairs of $A\times B$, where $\alpha<\delta^- \leq \beta$, so we conclude that $\delta_F^-(A,B) = \delta^- =\beta$.
Overall, the cost is
$$
O\left( \dfrac{(m+n)^2\log (m+n)}{s} + Ls\log(m+n) \right) ,
$$
and we make the
overall cost of this stage $O\left((m+n)L^{1/2}\log(m+n)\right)$, by choosing $s=(m+n)/L^{1/2}$ to balance the terms.
Note that after each phase of the algorithm, we can free the memory
used to process the phase and only remember $\alpha,\beta$ and the
path in $M$ (a prefix of the desired $S$) that we have traversed so
far. Since each phase processes $O(m+n)$ entries of $M$, the space
needed by this algorithm is $O(m+n)$.
\end{proof}
\medskip
To balance the two terms in Lemma~\ref{lem:finding_interval} and
Lemma~\ref{lem:searching_in_interval}, we choose $L = (m + n)^{2/5+\eps}$,
for another, but still arbitrarily small $\eps > 0$. This gives the
following main result of this section.
\begin{theorem}
\label{thm:one_sided}
Given a set $A$ of $m$ points and a set $B$ of $n$ points in the plane,
and a parameter $\eps>0$, we can compute the one-sided discrete Fr\'echet
distance $\delta_F^-(A,B)$ with shortcuts in randomized expected time
$O((m+n)^{6/5+\eps})$ using $O((m+n)^{6/5+\eps})$ space.
\end{theorem}
\begin{proof}
All the details of the proof have already been given, except for the precise statement concerning the running time. As noted earlier, the algorithm of Lemma~\ref{lem:finding_interval} does not verify explicitly that the sample that it generates does contain an
approximate median, nor does it verify that the number of distances in $(\alpha,\beta]$ is at most $L$ when it so asserts.
We can either let things stay as they are, knowing that the algorithm will succeed with
overall high probability. Alternatively, as asserted in the theorem, we can realize that
something went wrong in one of two situations: Either we apply the algorithm of
Lemma~\ref{lem:finding_interval} too many times, or we encounter a total of more than
$L$ bifurcations during the execution of the algorithm of Lemma~\ref{lem:searching_in_interval}.
In these cases we scrap the whole execution and start afresh from scratch.
The expected number of rounds of this kind is $O(1)$, and the theorem follows.
\end{proof}
\paragraph{Remark.}
In principle, our algorithm for the one-sided Fr\'echet distance with shortcuts can be generalized to higher dimensions. The only part that limits our approach to $\mathbb{R}^2$ is the algorithm of Lemma~\ref{lem:finding_interval}. However, this part can be replaced by a random sampling approach that is similar to the one that we use in Lemma~\ref{lem:semi_cont_finding_interval} for the semi-continuous Fr\'echet distance with shortcuts. This will increase the running time of the algorithm, but it will stay strictly subquadratic.
We believe that Lemma~\ref{lem:finding_interval} is of independent interest,
and that it may find other applications in distance-related optimization
problems. Here is one such example.
\begin{corollary}
\label{cor:aprox_selection}
Given a set $A$ of $m$ points and a set $B$ of $n$ points in the plane,
and parameters $0<k<mn$, $0<L<k$, and $\eps>0$, we can find a pair
$(a,b) \in A\times B$ such that, with high probability, $\|a-b\|$ is the
$\kappa$th smallest distance between a point of $A$ and a point of $B$,
for some rank $\kappa$ satisfying $k-L<\kappa<k+L$, in
$O((m+n)^{4/3+\eps}/L^{1/3}+m+n)$ time, using
$O((m+n)^{4/3+\eps}/L^{1/3}+m+n)$ space.
\end{corollary}
\begin{proof}
Consider the following decision problem (already discussed earlier).
Given a set $A$ of $m$ points, a set $B$ of $n$ points, a parameter
$\delta>0$ and a parameter $0<k<mn$, determine whether the number,
$N$, of pairs in $A\times B$ at distance at most $\delta$ is at most
$k$. To solve this decision problem, we use the algorithm of
Lemma~\ref{lem:finding_interval} with $k$ as its parameter $L$, but
we replace the annuli centered at the points of $A$ and $B$ by
respective disks of radius $\delta$ centered at the same points.
Since a point $a\in A$ is at distance at most $\delta$ from a point
$b\in B$ if and only if $a$ is in the disk of radius $\delta$
centered at $b$, and vice versa, the algorithm allows us to
determine, with high probability, whether $N$ is at most $k$.
The cost of this step is $O((m+n)^{4/3+\eps}/k^{1/3} +m+n)$, and
it is subsumed by the cost of the further steps.
Let $\delta^k(A,B)$ denote the $k$th smallest distance between a
point of $A$ and a point of $B$. We now use again the algorithm of
Lemma~\ref{lem:finding_interval}, together with the above decision
procedure, to find an interval $(\alpha, \beta]$ that contains at
most $L$ pairwise distances from $A\times B$, including
$\delta^k(A,B)$. To this end, we repeatedly shrink $(\alpha, \beta]$
using a binary search, starting with $(\alpha,\beta]=(0,\infty)$,
say. In each step of the search, we call the algorithm of
Lemma~\ref{lem:finding_interval} (this time, in its original
setup, with $L$ as the parameter). If it determines that the
number of critical distances in $(\alpha,\beta]$ is at most $L$, we
output $\alpha$ (together with its generating pair) as an
approximation for $\delta^k(A,B)$, in the sense asserted in the
lemma ($\beta$ would do equally well). Otherwise, we have a random
sample $R$ that contains, with high probability, an approximate
median (in the middle three quarters) of the pairwise
distances in $(\alpha,\beta]$. We locate a
consecutive pair $x,y$ of distances in $R$, using the decision procedure, such that the interval
$(x,y]$ contains $\delta^k(A,B)$. Since $R$ contains an approximate
median, the number of distances in $(x,y]$ is, with high probability,
at most ${7/8}$ of the number of distances in $(\alpha,\beta]$. We
then proceed with the next step of the search. The overall
resulting algorithm runs in $O((m+n)^{4/3+\eps}/L^{1/3}+m+n)$ time,
and it uses $O((m+n)^{4/3+\eps}/L^{1/3} +m+n)$ space.
\end{proof}
\paragraph{Remark.}
This should be compared with the near-linear algorithms in~\cite{BS02,HR13}
that approximate the \emph{value} of the $k$th smallest distance in $A\times B$
(rather than its rank as provided in Corollary~\ref{cor:aprox_selection}).
It would be interesting to understand better the relationship between these algorithms and ours.
\section{The two-sided DFDS}
\label{sec:DFDS2}
We first consider the corresponding decision problem.
That is, given $\delta>0$, we wish to decide whether
$\delta_F^{+}(A,B)\le\delta$.
Consider the matrix $M$ as defined in Section~\ref{sec:DFDS1}.
In the two-sided version of DFDS, given a reachable position
$(a_{i},b_{j})$ of the frogs, the $A$-frog can make a \emph{skipping
upward move}, as in the one-sided variant, to any point $a_{k},k>i$,
for which $M_{k,j}=1$. Alternatively, the $B$-frog can jump to any
point $b_{l},l>j$, for which $M_{i,l}=1$; this is a \emph{skipping
right move} in $M$ from $M_{i,j}=1$ to $M_{i,l}=1$, defined
analogously. Determining whether $\delta_F^{+}(A,B)\le\delta$
corresponds to deciding whether there exists a \emph{sparse
staircase} of ones in $M$ that starts at $M_{1,1}$, ends at
$M_{m,n}$, and consists of an interweaving sequence of skipping
upward moves and skipping right moves (see
Figure~\ref{fig:staircase}(c)).
Katz and Sharir~\cite{KS97} showed that the set
$S=\{(a_{i},b_{j})\mid\|a_{i}-b_{j}\|\le\delta\}=\{(a_{i},b_{j})\mid
M_{i,j}=1\}$ can be computed, in $O((m^{2/3}n^{2/3}+m+n)\log n)$
time and space, as the union of the edge sets of a collection
$\Gamma=\{A_{t}\times B_{t} \mid A_{t}\subseteq A, \; B_{t}\subseteq
B\}$ of edge-disjoint complete bipartite graphs. The number of
graphs in $\Gamma$ is $O(m^{2/3}n^{2/3}+m+n)$, and the overall sizes
of their vertex sets are
\[
\sum_{t}|A_{t}|,\sum_{t}|B_{t}|=O((m^{2/3}n^{2/3}+m+n)\log n).
\]
We store each graph
$A_{t}\times B_{t}\in\Gamma$ as a pair of sorted linked lists $L_{A_{t}}$
and $L_{B_{t}}$ over the points of $A_{t}$ and of $B_{t}$, respectively.
For each graph $A_{t}\times B_{t}\in\Gamma$, there is $1$ in each entry $M_{i,j}$ such that $(a_{i},b_{j})\in A_{t}\times B_{t}$.
That is, $A_t \times B_t$ corresponds to a submatrix
$M^{(t)}$ of ones in $M$ (whose rows and columns are not necessarily
consecutive). See Figure~\ref{fig:submatrices}(a).
Note that if $(a_{i},b_{j})\in A_{t}\times B_{t}$ is a reachable
position of the frogs, then every pair in the set $\{(a_{k},b_{l})\in A_{t}\times B_{t}\mid k\ge i,l\ge j\}$
is also a reachable position. (In other words, the positions
in the upper-right submatrix of $M^{(t)}$ whose lower-left entry
is $M_{i,j}$ are all reachable; see Figure~\ref{fig:submatrices}(b)).
\begin{figure}[htb]
\centering \includegraphics[scale=0.6]{submatrices}
\centering \caption{\small(a) A possible representation of the matrix $M$ as a collection of submatrices of ones, corresponding to the complete bipartite graphs $\{a_1,a_2\}\times\{b_1, b_2\},
\{a_1, a_3, a_5\}\times \{b_4, b_6\},
\{a_1,a_3\}\times\{b_7, b_{11}\},
\{a_2, a_3,a_5\}\times\{b_5, b_8, b_9\},
\{a_4, a_7, a_8\}\times\{b_3, b_4\},
\{a_4, a_7\}\times\{b_8, b_{10}\},
\{a_6\}\times \{b_9, b_{11}\},
\{a_8\}\times \{b_9,b_{12}\}$. (b) Another matrix $M$, similarly decomposed, where the reachable positions
are marked with an x.}
\label{fig:submatrices}
\end{figure}
We say that a graph $A_{t}\times B_{t}\in\Gamma$ \emph{intersects}
a row $i$ (resp., a column $j$) in $M$ if $a_{i}\in A_{t}$ (resp., $b_j\in B_{t}$). We denote the subset of graphs of $\Gamma$ that
intersect the $i$th row of $M$
by $\Gamma_{i}^{r}$ and those that intersect the $j$th column by $\Gamma_j^c$. The sets $\Gamma_i^r$ are easily constructed from the lists $L_{A_t}$ of the graphs in $\Gamma$, and are maintained as linked lists.
Similarly, the sets $\Gamma_j^c$ are constructed from the lists $L_{B_t}$, and are maintained as doubly-linked lists, so as to facilitate deletions of elements from them. We have $\sum_{i}|\Gamma_{i}^{r}|=\sum_{t}|A_{t}|=O((m^{2/3}n^{2/3}+m+n)\log n)$
and
$\sum_{j}|\Gamma_{j}^{c}|=\sum_{t}|B_{t}|=O((m^{2/3}n^{2/3}+m+n)\log n).$
We define a 1-entry $(a_k, b_j)$
to be {\em reachable from below row $i$}, if $k \geq i$ and there exists an entry
$(a_\ell, b_j)$, $\ell < i$, which is reachable.
We process the rows of $M$ in increasing order
and for each graph $A_{t}\times B_{t}\in\Gamma$ maintain a
reachability variable $v_t$, which is initially set to $\infty$.
We maintain the invariant that when we
start processing row $i$,
if $A_t\times B_t$ intersects at least one row that is not below the $i$th row, then
$v_t$ stores the smallest index $j$
for which there exists an entry $(a_k,b_j) \in A_{t}\times B_{t}$
that is reachable from below row $i$.
Before we start processing the rows of $M$, we verify that
$M_{1,1}=1$ and $M_{m,n} = 1$, and abort the computation if this
is not the case, determining that $\delta_F^{+}(A,B)>\delta$.
Assuming that $M_{1,1}=1$, each position in $P_1=\{(a_{1},b_{l})
\mid M_{1,l}=1\}$ is a reachable position. It follows that for each
graph $A_t\times B_t \in \Gamma$, $v_t$ should be set to
$\min\{l\mid A_t\times B_t\in \Gamma_l^c \text{ and } (a_1, b_l)\in
P_1\}$. Note that graphs $A_t\times B_t$ in this set are not necessarily in $\Gamma_1^r$.
We update the $v_t$'s using this rule, as follows. We
first compute $P_1$, the set of pairs, each consisting of $a_1$ and
an element of the union of the lists $L_{B_t}$, for $A_t\times
B_t\in \Gamma_1^r$. Then, for each $(a_1,b_l) \in P_1$, we set, for
each graph $A_u\times B_u \in \Gamma_l^c$, $v_u \leftarrow
\min\{v_u,l\}$.
In principle, this step should now be repeated for each row $i$.
That is, we should compute $y_i=\min\{v_t\mid A_t\times B_t\in
\Gamma_i^r\}$; this is the index of the leftmost entry of row $i$
that is reachable from below row $i$. Next, we should compute
$P_i=\{(a_i,b_l)\mid M_{i,l}=1 \text{ and } l\geq y_i\}$ as the
union of those pairs that consist of $a_i$ and an element of
$$
\{ b_j \mid b_j \in L_{B_t} \;\mbox{for}\; A_t\times B_t\in
\Gamma_i^r \;\mbox{and}\; j \ge y_i \}.$$
The set $P_i$ is the set of reachable positions in row $i$.
Then we should set for each $(a_1,b_l) \in P_i$ and for each graph
$A_u\times B_u \in \Gamma_l^c$, $v_u \leftarrow \min\{v_u,l\}$.
This however is too expensive, because it may make us construct
explicitly all the $1$-entries of $M$.
To reduce the cost of this step, we note that, for any graph
$A_t\times B_t$, as soon as $v_t$ is set to some column $l$ at some
point during processing, we can remove $b_l$ from $L_{B_t}$ because
its presence in this list has no effect on further updates of the
$v_t$'s. Hence, at each step in which we examine a graph $A_t\times
B_t \in \Gamma_l^c$, for some column $l$, we remove $b_l$ from
$L_{B_t}$. This removes $b_l$ from any further consideration in rows
with index greater than $i$ and, in particular, $\Gamma_l^c$ will
not be accessed anymore. This is done also when processing the
first row.
Specifically, we process the rows in increasing order and when we
process row $i$, we first compute $y_i=\min\{v_t\mid A_{t}\times
B_{t}\in\Gamma_{i}^{r}\}$, in a straightforward manner. (If $i=1$, then we simply set $y_1=1$.)
Then we
construct a set $P_i' \subseteq P_i$ of the ``relevant'' (i.e., reachable)
$1$-entries in the $i$-th row as follows. For each graph $A_t
\times B_t\in \Gamma_i^r$ we traverse (the current) $L_{B_t}$
\emph{backwards}, and for each $b_j \in L_{B_t}$ such that $j \ge y_i$
we add $(a_i,b_j)$ to $P_i'$. Then, for each $(a_i,b_l)\in P_i'$,
we go over all graphs $A_u\times B_u\in \Gamma_l^c$, and set $v_u
\leftarrow\min\{v_u,l\}$. After doing so, we remove $b_l$ from all
the corresponding lists $L_{B_u}$.
When we process row $m$ (the last row of $M$), we set
$y_m=\min\{v_t\mid A_{t}\times B_{t} \in\Gamma_{m}^{r}\}$. If
$y_m<\infty$, we conclude that $\delta_F^{+}(A,B)\le\delta$
(recalling that we already know that $M_{m,n}=1$). Otherwise, we
conclude that $\delta_F^{+}(A,B)>\delta$.
\paragraph{Correctness.}
We need to show that $\delta_F^{+}(A,B)\le\delta$ if and only if
$y_m <\infty$ (when we start processing row $m$). To this end, we
establish in Lemma~\ref{lem:two_sided} that the invariant stated above
regarding $v_t$ indeed holds. Hence, if $y_m<\infty$, then the
position $(a_m,b_{y_m})$ is reachable from below row $m$,
implying that $(a_m,b_n)$ is also a reachable position and thus
$\delta_F^{+}(A,B)\le\delta$. Conversely, if
$\delta_F^{+}(A,B)\le\delta$ then $(a_m,b_n)$ is a reachable
position. So, either $(a_m,b_n)$ is reachable from below row $m$, or
there exists a position $(a_m, b_j)$, $j<n$, that is reachable from
below row $m$ (or both). In either case there exists a graph
$A_t\times B_t$ in $\Gamma_m^r$ such that $v_t\leq n$ and thus
$y_m <\infty$.
We next show
that the reachability variables $v_t$ of the graphs in $\Gamma$ are maintained
correctly.
\begin{lemma}
\label{lem:two_sided} For each $i=1,\ldots,m$, the following
property holds. Let $A_t \times B_t$ be a graph in $\Gamma_i^r$, and
let $j$ denote the smallest index for which $(a_i,b_j) \in A_t\times
B_t$ and $(a_i,b_j)$ is reachable from below row $i$. Then, when we
start processing row $i$, we have $v_t = j$.
\end{lemma}
\begin{proof}
We prove this claim by induction on $i$. For $i=1$, this claim holds trivially.
We assume then that $i>1$ and that the claim is true for each row $i'<i$,
and show that it also holds for row $i$.
Let $A_t\times B_t$ be a graph in $\Gamma_i^r$, and let $j$ denote the smallest
index for which there exists a position $(a_i, b_j) \in A_t\times B_t$ that is reachable from
below row $i$. We need to show that $v_t= j$ when we start processing row $i$.
Since $(a_i,b_j)$ is reachable from below row $i$, there exists a position
$(a_k,b_j)$, with $k<i$, that is reachable, and we let $k_0$ denote the smallest
index for which $(a_{k_0}, b_j)$ is reachable. Let $A_o\times B_o$ be the graph
containing $(a_{k_0},b_j)$. We first claim that when we start processing row
$k_0$, $b_j$ was not yet deleted from $L_{B_o}$ (nor from the corresponding
list of any other graph in $\Gamma_j^c$). Assume to the contrary that $b_j$
was deleted from $L_{B_o}$ before processing row $k_0$. Then there exists a
row $z<k_0$ such that $(a_z, b_j)\in P_z'$ and we deleted $b_j$ from $L_{B_o}$
when we processed row $z$. By the last assumption, $(a_z,b_j)$ is a reachable
position. This is a contradiction to $k_0$ being the smallest index for which
$(a_{k_0}, b_j)$ is reachable. (The same argument applies for any other graph,
instead of $A_o\times B_o$.)
We next show that $v_t \leq j$. Since $(a_{k_0},b_j)\in A_o\times B_o$,
$A_o\times B_o \in \Gamma_{k_0}^r \cap \Gamma_j^c$. Since $k_0$ is the
smallest index for which $(a_{k_0}, b_j)$ is reachable, there exists an
index $j_0$, such that $j_0 <j$ and $(a_{k_0}, b_{j_0})$ is reachable from
below row $k_0$. (If $k_0=1$, we use instead the starting placement
$(a_1,b_1)$.) It follows from the induction hypothesis that
$y_{k_0}\leq j_0 <j$. Thus, when we processed row $k_0$ and we went
over $L_{B_o}$, we encountered $b_j$ (as just argued, $b_j$ was still in
that list), and we consequently updated the reachability variables $v_u$ of each
graph in $\Gamma_j^c$, including our graph $A_t\times B_t$ to be at most $j$.
(Note that if there is no position in $A_t\times B_t$ that is reachable
from below row $i$ (i.e., $j=\infty$), we trivially have $v_t \leq \infty$.)
Finally, we show that $v_t=j$. Assume to the contrary that $v_t=j_1<j$ when
we start processing row $i$. Then we have updated $v_t$ to hold $j_1$ when
we processed $b_{j_1}$ at some row $k_1 <i$. So, by the induction hypothesis,
$y_{k_1} \leq j_1$, and thus $(a_{k_1}, b_{j_1})$ is a reachable position.
Moreover, $A_t \times B_t \in \Gamma_{j_1}^c$, since $v_t$ has been updated
to hold $j_1$ when we processed $b_{j_1}$. It follows that
$(a_i, b_{j_1})\in A_t\times B_t$. Hence, $(a_i, b_{j_1})$ is reachable
from below row $i$. This is a contradiction to $j$ being the smallest
index such that $(a_i, b_j)$ is reachable from below row $i$. This
establishes the induction step and thus completes the proof of the lemma.
\end{proof}
\paragraph{Running Time.} We first analyze the initialization cost of
the data structure, and then the cost of traversal of the rows
for maintaining the variables $v_t$.
\paragraph{Initialization.}
Constructing $\Gamma$ takes $O((m^{2/3}n^{2/3}+m+n)\log(m+n))$ time.
Sorting the lists $L_{A_t}$ (resp., $L_{B_t}$) of each $A_{t}\times B_{t}\in\Gamma$ takes
$O((m^{2/3}n^{2/3}+m+n)\log^2 (m+n))$ time. Constructing the lists
$\Gamma_i^r$ (resp., $\Gamma_j^c$) for each $a_i \in A$ (resp., $b_j
\in B$) takes time linear in the sum of the sizes of the $A_t$'s and
the $B_t$'s, which is $O((m^{2/3}n^{2/3}+m+n)\log(m+n))$.
\paragraph{Traversing the rows.}
When we process row $i$ we
first compute $y_i$ by scanning $\Gamma_i^r$. This takes a total of
$O\left(\sum_{i}|\Gamma_{i}^{r}|\right)=O((m^{2/3}n^{2/3}+m+n)\log
n)$ for all rows. Since the lists $L_{B_t}$ are sorted, the
computation of $P_i'$ is linear in the size of $P_i'$.
This is so because, once we have added a pair $(a_i,b_j)$ to $P'_i$,
we remove $b_j$ from all lists that contain it, so we will not encounter
it again when scanning other lists $L_{B_{t'}}$. For each pair
$(a_i,b_\ell)\in P_i'$ we scan $\Gamma_\ell^c$, which must contain at
least one graph $A_t\times B_t \in \Gamma$ such that $a_i \in A_t$
(and $b_j\in B_t$).
For each element $A_t\times B_t\in \Gamma_\ell^c$ we spend constant
time updating $v_t$ and removing $b_\ell$ from $L_{B_t}$. It follows
that the total time, over all rows, of computing $P_i'$ and scanning
the lists $\Gamma_\ell^c$ is
$O\left(\sum_{l}|\Gamma_{l}^{c}|\right)=O((m^{2/3}n^{2/3}+m+n)\log n)$.
We conclude that the total running time is
$O((m^{2/3}n^{2/3}+m+n)\log^2 (m+n))$.
\paragraph{The optimization procedure.}
We use the above decision procedure for finding the optimum
$\delta_F^+(A,B)$, as follows. Note that if we increase $\delta$
continuously, the set of $1$-entries of $M$ can only grow, and
this can only happen at a distance between a point of $A$ and a
point of $B$. We thus perform a binary search over the $mn$
pairwise distances between the pairs of $A\times B$. In each
step of the search we need to determine the $k$th smallest
pairwise distance $r_k$ in $A\times B$, for some value of $k$.
We do so by using the distance selection algorithm of
Katz and Sharir~\cite{KS97}, which can easily be adapted to
work for this bichromatic scenario. We then run the decision
procedure on $r_k$, using its output to guide the binary search.
At the end of this search, we obtain two consecutive critical
distances $\delta_1, \delta_2$ such that
$\delta_1 < \delta_F^+(A,B) \leq \delta_2$, and we can
therefore conclude that $\delta_F^+(A,B) = \delta_2$. The
running time of the distance selection algorithm of \cite{KS97}
is $O((m^{2/3}n^{2/3}+m+n)\log^2(m+n))$, which also holds for the bipartite version that we use.
We thus obtain the following main result of this section.
\begin{theorem}
Given a set $A$ of $m$ points and a set $B$ of $n$ points in the
plane, we can compute the two-sided discrete Fr\'echet distance
with shortcuts $\delta_F^+(A,B)$, in time
$O((m^{2/3}n^{2/3}+m+n)\log^3 (m+n))$, using
$O((m^{2/3}n^{2/3}+m+n)\log (m+n))$ space.
\end{theorem}
\input{semi_continuous1.tex}
\section{Discussion}
The algorithms obtained for the discrete Fr\'echet distance with
shortcuts, run in time significantly better than those for the Fr\'echet
distance without shortcuts. It is thus an interesting open question
whether similar improvements can be obtained for the continuous version of
the Fr\'echet distance with shortcuts, where shortcuts are made only
between vertices of the curves. This variant, that was considered by Driemel
and Har-Peled~\cite{DH12}, may be easier than the NP-Hard variant that was
considered by Driemel et al.~\cite{BDS13}. We hope that the techniques that we have developed for the semi-continuous problem will be useful for tackling this harder problem.
Another topic for further research is to find additional
applications of some of the ideas that appear in the optimization technique
for the one-sided variant.
\section{Introduction}
\label{sec:introduction}
\section{Semi-continuous Fr\'echet distance with shortcuts}
\label{sec:semi_cont} Let $f\subseteq\mathbb{R}^{2}$ denote a
polygonal curve with $n$ edges $e_1,\ldots, e_n$ and $n+1$ vertices
$p_0,p_1,\ldots,p_n$, and let $A=(a_{1},\ldots,a_{m})$ denote a
sequence of $m$ points in the plane. Consider a person that is
walking along $f$ from its starting endpoint to its final endpoint,
and a frog that is jumping along the sequence $A$ of stones. The
frog is allowed to make shortcuts (i.e., skip stones) as long as it
traverses $A$ in the right (increasing) direction, but the person
must trace the complete curve $f$ (see Figure~\ref{fig:hdfd}(a)).
Assuming that the person holds the frog by a leash, our goal is to
compute the minimal length $\delta_F^s(A, f)$ of a leash that is
required in order to traverse $f$ and (parts of) $A$ in this manner,
taking the frog and the person from $(a_1,p_0)$ to $(a_m,p_n)$.
\begin{figure}[htb]
\centering \includegraphics[scale=0.8]{HDFD.pdf}
\caption{\small (a) A curve $f$ and a sequence of points $A=(a_{1},\ldots,a_{5})$.
(b) Thinking of $f$ as a continuous mapping from $[0,1]$ to $\mathbb{R}^2$, the $i$th row depicts the set $\{t \in [0,1] \mid f(t) \in D_\delta(a_i)\}$. The
dotted subintervals and their connecting upward moves (not drawn) constitute the lowest semi-sparse staircase between the starting and final positions.}
\label{fig:hdfd}
\end{figure}
Consider the decision version of this problem, where, given a
parameter $\delta>0$, we wish to decide whether the person and the
frog can traverse $f$ and (parts of) $A$ using a leash of length
$\delta$. This problem can be solved using the algorithm for solving
the one-sided DFDS, with a slight modification that takes into account
the continuous nature of $f$. Specifically, for a
point $p \in \mathbb{R}^2$, let $D_\delta(p)$ denote the disk of
radius $\delta$ centered at $p$.
Now, consider a vector $M$ whose entries correspond to the points of $A$. For each $i=1,\ldots,m$, the $i$th entry of $M$ is
$$M_i = M(a_i) = f \cap D_\delta(a_i)$$
(see Figure~\ref{fig:hdfd}(b)). Each $M_i$ is a finite union of connected subintervals of $f$.
We do not compute $M$ explicitly, because the overall ``description complexity'' of its entries might be too large.
Specifically, the number of connected subsegments of the edges of $f$ that comprise the elements of $M$ can be $mn$ in the worst case.
Instead, we assume availability of (efficient implementations of) the following two primitives.
\begin{enumerate}[(i)]
\item \label{enu:1}
{\bf NextEndpoint($x, a_i$):} Given a point $x\in f$ and a point $a_i$ of $A$, such that $x\in D_\delta(a_i)$,
return the forward endpoint of the connected component of $f\cap D_\delta(a_i)$ that contains $x$.
\item \label{enu:2}
{\bf NextDisk($x,a_i$):}
Given $x$ and $a_i$, as in (\ref{enu:1}), find the smallest $j>i$ such that $x\in D_\delta(a_j)$,
or report that no such index exists (return $j =\infty$).
\end{enumerate}
Both primitives admit
efficient implementations. For our purposes it is sufficient to
implement Primitive (\ref{enu:1}) by traversing the edges of $f$ one
by one, starting from the edge containing $x$, and checking for each
such edge $e_j$ of $f$ whether the forward endpoint $p_j$ of $e_j$ belongs
to $D_\delta(a_i)$. For the first $e_j$ for which this test fails,
we return the forward endpoint of the interval $e_j \cap D_\delta(a_i)$. It is
also sufficient to implement Primitive (\ref{enu:2}) by checking for
each $j>i$ in increasing order, whether $x \in D_\delta(a_j)$, and
return the first $j$ for which this holds. To solve the decision
problem, we proceed as in the decision procedure of the one-sided
DFDS (see Figure~\ref{alg:semi-sparse}), except that when we move
``right'', we move along $f$ as long as we can within the current
disk (using Primitive (\ref{enu:1})), and when we move ``up'', we
move to the first following disk that contains the current point of
$f$ (using Primitive (\ref{enu:2})). More precisely, we use the
decision procedure $\Gamma$ given in Figure~\ref{alg:semi-continuous}.
\begin{figure}[htbp]
\MyFrame{
\smallskip
\begin{itemize}
{\small \sf
\item \textbf{Input:} $A, f, \delta$ \vspace{-0.3cm}
\item $S\leftarrow \emptyset$ \vspace{-0.3cm}
\item $a^1\leftarrow a_1$, $x^1 \leftarrow p_0$ \vspace{-0.3cm}
\item If ($x^1 \notin D_\delta(a^1)$) then \vspace{-0.3cm}
\begin{itemize}
\item Return $\delta_F^s(A,f)>\delta$ \vspace{-0.3cm}
\end{itemize}
\item Add $(a^1,x^1)$ to $S$ \vspace{-0.3cm}
\item $k \leftarrow 1$ \vspace{-0.3cm}
\item While ($a^k$ is not $a_m$ or $f$ is not fully traversed) do \vspace{-0.3cm}
\begin{itemize}
\item $x^{k+1} \leftarrow$ {\bf NextEndpoint($x^k, a^k$)} \vspace{-0.1cm}
\item Add $(a^k, x^{k+1})$ to $S$ \vspace{-0.1cm}
\item If ($a^k=a_m$ and $x^{k+1}=p_n$) then \vspace{-0.1cm}
\begin{itemize}
\item Return $\delta_F^s(A,f)\leq\delta$ \vspace{-0.1cm}
\end{itemize}
\item $l \leftarrow$ {\bf NextDisk($x^{k+1}, a^k$)} \vspace{-0.1cm}
\item If ($ l \leq m$) then \vspace{-0.1cm}
\begin{itemize}
\item $a^{k+1} \leftarrow a_l$ \vspace{-0.1cm}
\item Add $(a^{k+1}, x^{k+1})$ to $S$ \vspace{-0.2cm}
\end{itemize}
\item Else
\begin{itemize}
\item Return $\delta_F^s(A,f)>\delta$ \vspace{-0.2cm}
\end{itemize}
\item $k \leftarrow k+1$ \vspace{-0.3cm}
\end{itemize}
\item Return $\delta_F^s(A,f)\leq\delta$
}
\end{itemize}
}\caption{The decision procedure $\Gamma$ for the semi-continuous Fr\'echet distance with shortcuts.} \label{alg:semi-continuous}
\end{figure}
The path $S$ computed by $\Gamma$ is a sequence of reachable
positions $(a^1, x^1), (a^2, x^2),\ldots$, where $a^k$ is a point of
$A$ and $x^k$ is a point on an edge of $f$. Let $P$ be the sequence
of pairs $(a^1,s^1), (a^2,s^2),\ldots$, where $s^k$ is either the edge
of $f$ containing $x^k$ (in its relative interior) or the vertex of $f$
coinciding with $x^k$, for $k=1,2,\ldots$.\footnote{%
Note that we use superscripts $a^k$ and $s^k$ to denote the
sequence $S$ defining the solution produced by the decision
procedure. This is to distinguish them from $a_k$
and $e_k$ or $p_k$, with subscripts, that denote the original input sequence
of points of the frog and the sequence of segments and their endpoints on $f$.}
The correctness of the decision procedure $\Gamma$ is proved
similarly to the correctness of the decision procedure of the one-sided
DFDS (of Figure~\ref{alg:semi-sparse}). More specifically, here
a \emph{semi-continuous semi-sparse staircase} is an interweaving
sequence of \emph{discrete skipping upward moves} and
\emph{continuous right moves}, where a discrete skipping upward move
is a move from a reachable position $(a_i,p)$ of the frog and the
person to another position $(a_j,p)$ such that $j>i$ and $p \in D_\delta(a_j)$.
A continuous right move is a move from a reachable position $(a_i,p)$ of
the frog and the person to another position $(a_i,p')$ where
$p'$, and the entire portion of $f$ between $p$ and $p'$, are contained in $D_\delta(a_i)$. Then there exists a semi-continuous
semi-sparse staircase that reaches the position $(a_m, p_n)$ if and only
if $\delta^s_F(A,f) \le \delta$.
Concerning correctness, we prove
that if there exists a semi-continuous semi-sparse staircase $S'$ that
reaches position $(a_m, p_n)$, then the decision procedure
maintains a partial semi-continuous semi-sparse staircase $S$ that is
always ``below'' $S'$ (in terms of the corresponding indices of the positions of the frog), and therefore $S$ reaches a position where the person is at $p_n$ (and the frog can then jump directly to $a_m$). Intuitively, this holds since the decision
procedure can at any point join the plot of $S'$ using a discrete
skipping upward move. The running time of this decision procedure is
$O(n+m)$ since we advance along $f$ at each step of Primitive
(\ref{enu:1}), and we advance along $A$ at each step of Primitive
(\ref{enu:2}), so our naive implementations of these primitives never back up along the path and sequence, and consequently take $O(m+n)$ time in total.
We thus obtain the following lemma.
\begin{lemma}
Given a polygonal curve $f$ with $n$ edges in the plane, a set $A$
of $m$ points in the plane, and a parameter $\delta>0$, we can
determine whether the semi-continuous Fr\'echet distance
$\delta^s_F(A,f)$ with shortcuts between $A$ and $f$ is at most
$\delta$, in $O(m+n)$ time, using $O(m+n)$ space.
\end{lemma}
\paragraph{The optimization procedure.}
We now use the decision procedure $\Gamma$ to find the optimal value
$\delta_F^s(A, f)$. To make the dependence on $\delta$ explicit,
we denote, in what follows, the decision procedure for distance
$\delta$ by $\Gamma(\delta)$. The path $S$ computed by
$\Gamma(\delta)$, and each element $(a^k,x^k)$ of $S$, depend on
$\delta$, so we denote them by $S(\delta), a^k(\delta)$ and
$x^k(\delta)$, respectively. The sequence $P$ of pairs $(a^k, s^k)$,
and each of its elements, also depend on $\delta$,
so we denote $P$ by $P(\delta)$, and $s^k$ by $s^k(\delta)$.
Of course, $\Gamma(\delta)$ might fail, i.e., report that
$\delta_F^s(A,f) > \delta$. In such a case, the path $S(\delta)$
and the sequence of pairs $P(\delta)$ consist of everything that
was accumulated in them until $\Gamma(\delta)$ has terminated (that is, aborted).
In particular, $S(\delta)$ does not end in this case at $(a_m, p_n)$.
The path $S(\delta_1)$ is {\em combinatorially different} from the
path $S(\delta_2)$, for $\delta_1, \delta_2>0$, if $P(\delta_1) \neq
P(\delta_2)$; otherwise, we say that $S(\delta_1)$ and
$S(\delta_2)$ are {\em combinatorially equivalent}.
We next argue that each critical value of $\delta$ where $S(\delta)$
changes combinatorially must be of one of the following two types:
\begin{enumerate}
\item\label{enum:pv}
The distance between a point of $A$ and a vertex of $f$ (point-vertex distance).
\item\label{enum:ppe}
For two points $p, q \in A$ and an edge $e$ of $f$, the distance between $p$ (or $q$)
and the intersection of $e$ with the bisector of $p$ and $q$ (point-point-edge distance).
\end{enumerate}
See Figure~\ref{fig:critical} for an illustration.
We assume general position of the input, so as to ensure that these
critical distances are all distinct.
\begin{figure}[htb]
\centering \includegraphics[scale=0.8]{critical_distances.pdf}
\caption{\small Two of the critical distances between $f$ and $A$. $\delta_1$ is a point-vertex distance between $a_4$ and $p_2$. $\delta_2$ is a point-point-edge distance between $a_1$, $a_3$ and $e_1$.}
\label{fig:critical}
\end{figure}
\begin{lemma}
Let $\delta$ be such that $S(\delta^-)$ is combinatorially different
from $S(\delta)$, for all $\delta^- < \delta$ and arbitrarily close to
$\delta$. Then $\delta$ is either a point-vertex distance or a
point-point-edge distance.
\end{lemma}
\begin{proof}
In what follows, we use $\delta^-$ to denote an arbitrary point from the neighborhood of $\delta$ mentioned in the lemma. Consider the point at which the executions of $\Gamma(\delta^-)$ and of $\Gamma(\delta)$ add a pair to $P(\delta^-)$ which is
different from the pair added to $P(\delta)$ (this includes the case
in which we add a pair to only one of the sets $P(\delta^-)$, $P(\delta)$).
If $(a_1,p_0)$
is in $P(\delta)$ but not in $P(\delta^-)$ then $\delta$ is the
distance between $a_1$ and $p_0$, a point-vertex distance.
Otherwise, assume that the different pairs arose following a call
to {\bf NextEndPoint($x^k,a^k$)}. Then
$x^{k+1}(\delta) =$ {\bf NextEndPoint($x^k(\delta),a^k(\delta)$)} and
$x^{k+1}(\delta^-) =$ {\bf NextEndPoint($x^k(\delta^-),a^k(\delta^-)$)}
belong to different edges / vertices of $f$. Note that $a^k(\delta) =
a^k(\delta^-)$ since this is the first call that causes a
discrepancy between $P(\delta)$ and $P(\delta^-)$.
A simple continuity argument implies that the disk of radius $\delta$
about $a^k(\delta)$ must contain a vertex of $f$, so $\delta$ is a point-vertex distance.
Finally assume that the first difference in the pairs added to
$P(\delta^-)$ and $P(\delta)$ occured following a call to {\bf NextDisk}($x^{k+1}, a^k$).
Put $a_{\ell(\delta)} =$ {\bf NextDisk}$(x^{k+1}(\delta),a^k(\delta))$ and
$a_{\ell(\delta^-)} = $ {\bf NextDisk}$(x^{k+1}(\delta^-),a^k(\delta^-))$.
As before, $a^k(\delta) = a^k(\delta^-)$ by our assumption.
Moreover, since $x^{k+1}(\delta)$ is not a vertex of $f$ (or else the previous call to {\bf NextEndPoint} would have produced different pairs at $\delta^-$ and at $\delta$), a simple continuity argument shows that $x^{k+1}(\delta^-)\rightarrow x^{k+1}(\delta) $as $\delta^-\uparrow \delta$. Assume that $\ell(\delta^-) \neq \ell(\delta)$. We claim that in this case $x^{k+1}(\delta)$ must lie on $\bd D_\delta(a_{\ell(\delta)})$ and on $\bd D_\delta(a_{\ell(\delta^-)})$, showing that $\delta$ is a point-point-edge distance. The first containment follows from the execution of {\bf NextEndPoint} at $x^k(\delta), a^k(\delta)$ with distance $\delta$. By the same reasoning, we also have $x^{k+1}(\delta^-)\in \bd D_{\delta^-}(a_{\ell(\delta^-)})$. As $\delta^- \uparrow \delta$, $x^{k+1}(\delta^-) \rightarrow x^{k+1}(\delta)$, and $\bd D_{\delta^-}(a_{\ell(\delta^-)}) \rightarrow \bd D_\delta(a_{\ell(\delta^-)})$ (in the Hausdorff sense), which implies that $x^{k+1}(\delta)\in \bd D_\delta(a_{\ell(\delta^-)})$, as claimed.
\end{proof}
Note that the distance between a point of $A$ and an edge of $f$ is
not a critical distance (unless it coincides with a
point-point-edge distance, which is ruled out anyway by our general
position assumption); informally, this is because the appearance or
disappearance of the small interval of intersection between the edge and the
corresponding disk has no effect on $S(\delta)$ (because there is no way to get to this interval that did not exist for $\delta^-$ too). Note also that not all triples
of two points $p,q$ of $A$ and an edge $e$ of $f$ create a point-point-edge
critical event, since the bisector of $p$ and $q$ might not intersect $e$.
The following lemmas lead, in combination with the decision procedure
given above, to an algorithm for the optimization problem that runs in
$O((m+n)^{2/3}m^{2/3}n^{1/3}\log(m+n))$ randomized expected time. The framework
of the proof is similar to that of the discrete case.
Lemma~\ref{lem:semi_cont_finding_interval} shows that, given a
parameter $L>0$, we can find, with high probability, an interval
$(\alpha,\beta]$ such that $\alpha < \delta^s_F(A,f) \leq \beta$
and $(\alpha,\beta]$ contains no more than $L$ critical distances,
in $O(m^2n\log(m+n)/L+(m+n)\log(m+n))$ randomized expected time. Then,
Lemma~\ref{lem:semi_cont_searching_in_interval} shows that we can
find $\delta^s_F(A,f)$ within $(\alpha,\beta]$ in
$O((m+n)L^{1/2}\log(m+n))$ time. Choosing
$L=m^{4/3}n^{2/3}/(m+n)^{2/3}$, we obtain an algorithm that runs in
$O((m+n)^{2/3}m^{2/3}n^{1/3}\log(m+n))$ randomized expected time. The proof of
Lemma~\ref{lem:semi_cont_finding_interval} is different from that of
the analogous lemma for the discrete case (Lemma~\ref{lem:finding_interval}), and uses
a generalization of a random sampling technique by Har-Peled and Raichel
(see~\cite{HR13}). The proof of
Lemma~\ref{lem:semi_cont_searching_in_interval} is similar to but more involved
than the proof of the analogous lemma (Lemma~\ref{lem:searching_in_interval}) for the discrete case.
\begin{lemma}
\label{lem:semi_cont_finding_interval}
Given a polygonal curve $f$ with $n$ edges and a set $A$ of $m$ points
in the plane, and a parameter $L>0$, we can find an interval
$(\alpha,\beta]$ that contains, with high probability, at most $L$
critical distances $\delta$, including $\delta^s_F(A,f)$,
in $O(m^2n\log(m+n)/L+(m+n)\log(m+n))$ randomized expected time.
\end{lemma}
\begin{proof}
We generate a random sample of $cx$ triples of two points of $A$ and
an edge of $f$, where $x=(\binom{m}{2}n+mn)\log(m+n)/L$, and $c>1$ is
a sufficiently large constant. We also sample $cx$ pairs of a
point of $A$ and a vertex of $f$. This will generate at most $2cx$
critical values of $\delta$ (some of the triples that we sample might
not contribute a critical value, as noted above).
We perform a binary search over the sampled critical values that actually
arise, using the decision procedure $\Gamma$ to guide the search.
This takes $O(m^2n\log(m+n)/L+(m+n)\log(m+n))$ time (using a
linear time median finding algorithm).
We claim that the interval $(\alpha,\beta]$ that this procedure generates contains,
with high probability, at most $L$ (non-sampled) critical values of
$\delta$, including $\delta^s_F(A,f)$. To see that, consider the set
$U$ of the $L/2$ (defined) critical values that are smaller than
$\delta^s_F(A,f)$ and closest to it, and denote by $u$ (resp., $v$) the number of point-vertex distances (resp., point-point-edge distances) amond them; thus $u+v = L/2$. The total number of triples and
pairs that potentially (but not necessarily) define a critical value
is $z=(\binom{m}{2}n+mn)$, of which ${m \choose 2} n$ define point-point-edge distances, and $mn$ define point-vertex distances. The probability that none of the $2cx$
triples and pairs that we sampled generate a critical value in $U$ is
at most
$$
\left(1-\dfrac{u}{{m \choose 2}n}\right)^{cx}\cdot
\left(1-\dfrac{v}{mn}\right)^{cx} \leq e^{-cx\left(\dfrac{u}{{m \choose 2} n}+\dfrac{v}{mn}\right)} \leq e^{-cx \cdot\frac{u+v}{z}} = e^{-\frac{c}{2} \log(m+n)} = \dfrac{1}{(m+n)^{c/2}}.
$$
The same argument, with the same resulting probability bound,
applies for the set $U'$ of the $L/2$ (defined) critical values
that are greater than $\delta^s_F(A,f)$ and closest to it.
Hence, the probability that we miss all the $L$ critical values
in $U\cup U'$ is polynomially small (and can be made arbitrarily
small by increasing $c$).
\end{proof}
The following lemma shows that we can find $\delta^s_F(A,f)$
within $(\alpha,\beta]$, in $O((m+n)L^{1/2}\log(m+n))$ time. Notice the high-level similarity with the discrete counterpart in Lemma~\ref{lem:searching_in_interval}.
\begin{lemma}
\label{lem:semi_cont_searching_in_interval}
Given a polygonal curve $f$ with $n$ edges in the plane, a set
$A$ of $m$ points in the plane, and an interval
$(\alpha_0,\beta_0]\subset\reals$ that contains at most $L$
critical distances $\delta$ (including $\delta^s_F(A,f)$),
we can find $\delta^s_F(A,f)$ in $O((m+n)L^{1/2}\log(m+n))$
(deterministic) time using $O(m+n)$ space.
\end{lemma}
\begin{proof}
For an edge $e$ of $f$ and two points $p,q\in e$, let $e(p,q]$ be the
subedge of $e$ starting at $p$ (not including $p$) and ending at
$q$, and let $\ell(e)$ denote the line containing $e$.
We simulate the decision procedure $\Gamma$ at the unknown value
$\delta^* = \delta^s_F(A,f)$. Each step of $\Gamma$ involves a call to
one of the procedures {\bf NextEndPoint} and {\bf NextDisk}.
The execution of each of these procedures consists of a sequence
of tests---the former procedure tests the current disk against a
sequence of edges of $f$, for finding the first exit point from
the disk, and the latter procedure tests the current point
$x^{k+1}(\delta)$ against a sequence of disks centered at the points
of $A$, for finding the first disk (beyond the present disk) that contains the point.
Each such test generates a critical value $\delta_0$, and we check
whether $\delta_0$ lies outside $(\alpha_0,\beta_0]$, in which case we know
the (combinatorial nature of the) outcome of the test (in the procedure {\bf NextEndPoint}, the point
$x^{k+1}(\delta_0)$ itself varies continuously with $\delta_0$), and
we can proceed to the next test. If $\delta_0$ lies in $(\alpha_0,\beta_0]$,
we bifurcate, proceeding along two branches,
one assuming that $\delta^* \le \delta_0$
and the other assuming that $\delta^* > \delta_0$.
These bifurcations generate a tree $T$. At each node of $T$
we maintain a triple $(\tau, a^k, e^k(p,q])$, where
$\tau=(\alpha,\beta]$ is a range of possible values for $\delta^*$ (a subrange of $(\alpha_0,\beta_0]$).
Each such triple satisfies the following invariant. For each
$\delta \in \tau$ there exists a pair $(a^k(\delta),x^k(\delta))\in
P(\delta)$ such that $a^k(\delta)=a^k$ and $e^k(p,q]$
is the set of all points $x^k(\delta)$ for $\delta\in\tau$. In
particular, $q=x^k(\beta)$ and $p$ is the limit of $x^k(\alpha^+)$
where $\alpha^+ > \alpha$ approaches $\alpha$.
The process is initialized as follows.
Let $\alpha'$ be the distance between $a_1$ and $p_0$. Clearly
$\delta_F^s(A, f) \ge \alpha'$. We run the decision procedure at
$\alpha'$, and return $\delta_F^s(A, f) = \alpha'$ if the
procedure finds a path to $(a_m,p_n)$. Otherwise we
initialize the root of $T$ with the triple
$((\alpha', \infty],a_1,e_1(p_0,p_0])$.
For simplicity of presentation, we represent a single cycle of the
decision procedure (consisting of a call to {\bf NextEndPoint}
followed by a call to {\bf NextDisk}) by two consecutive levels of $T$, each catering to the corresponding call.
Let $v$ be a node of $T$ that represents the situation at the beginning
of such a cycle. We now show how to construct the triples of the
children and the grandchildren of $v$ from the triple
$(\tau=(\alpha,\beta], a^k, e^k(p,q])$ of $v$.
To construct the
children of $v$ we simulate {\bf NextEndPoint} assuming that the current pair in $S$ is $(a^k,x^k(\delta))$ for $\delta \in \tau$ and $x^k(\delta)\in e^k(p,q)$. The idea is to compute, for $\delta=\alpha$ and for $\delta=\beta$, the edge containing the forward endpoint of the connected component of $f \cap D_\delta(a^k)$ that contains $x^k(\delta)$. If we obtain the same edge $e$ for $\delta=\alpha$ and for $\delta=\beta$, we conclude that all values in $\tau$ agree that $(a^k, e)$ is the next pair in $P$ and we continue to the next step of the procedure. Otherwise, we have detected a critical value $\delta_0$ in $\tau$ and we bifurcate, proceeding along two paths --- one assuming that $\delta^*\in (\alpha,\delta_0]$ and one assuming that $\delta^*\in (\delta_0,\beta]$.
We now give a more detailed description of the simulation of {\bf NextEndPoint} at $v$.
Let $q_\alpha$ be the first intersection of $D_{\alpha}(a^k)$ with $f$
following $p=x^k(\alpha)$, and let $q_\beta$ be the first intersection
of $D_{\beta}(a^k)$ with $f$ following $q=x^k(\beta)$. Let $e_j$ be the
edge of $f$ equal to $e^k$. We traverse $e_j, e_{j+1},\ldots, e_n$
in order, and for each such segment $e_\ell$, we have three possible cases.
(i) $q_\alpha \notin e_\ell$ and $q_\beta \notin e_\ell$.
In this case, the forward endpoint
$x^{k+1}(\delta)$ of the connected component $f\cap D_{\delta}(a^k)$
containing $x^k(\delta)$ is not in $e_\ell$, for all
$\delta \in \tau$. So we proceed to the next edge $e_{\ell+1}$.
(ii) ${q_\alpha \in e_\ell}$ and ${q_\beta \in e_\ell}$.
In this case, $e_{\ell}(q_\alpha, q_\beta]$ is the set of all the
forward endpoints $x^{k+1}(\delta)$ of the connected components
of $f \cap {D_{\delta}(a^k)}$ containing $x^{k}(\delta)$, for
$\delta\in\tau$. In this case $v$ has a single child $v'$, with
the triple $(\tau=(\alpha,\beta], a^k, e_\ell(q_\alpha,q_\beta])$.
(iii) ${q_\alpha \in e_\ell}$ and ${q_\beta \notin e_\ell}$.
In this case, we encounter a point-vertex critical distance
$\delta_0 \in \tau$, between $a^k$ and $p_\ell$. That is, for each
$\delta\in (\delta_0,\beta]$, the forward endpoint of the connected
component $f\cap D_{\delta}(a^k)$ containing $x^k(\delta)$ is not in
$e_\ell$ (but in a segment beyond $e_\ell$), and for each
$\delta\in (\alpha,\delta_0]$, the forward endpoint of the
connected component $f\cap D_{\delta}(a^k)$ containing $x^k(\delta)$
is in $e_\ell(q_\alpha, p_\ell]$. We generate one child $v'$ of $v$
with the triple $((\alpha,\delta_0], a^k, e_\ell(q_\alpha,p_\ell])$,
and generate the other children of $v$ by proceeding to the
following edge $e_{\ell+1}$ with the smaller range $(\delta_0, \beta]$,
replacing $q_\alpha$ with $p_\ell$, and continuing the process recursively,
generating a child for each consecutive segment in which we need to bifurcate.
Next we generate the grandchildren $v''$ of each child $v'$ of $v$,
which result from the simulation of the call to {\bf NextDisk}.
The idea is to compute, for $\delta=\alpha$ and for $\delta=\beta$, the next point $a_\ell$ of $A$ such that the disk $D_\delta(a_\ell)$ contains $x^k(\delta)$ (note that $x^k(\alpha)=p$ and $x^k(\beta)=q$). If we obtain the same point $a_\ell$ for $\delta=\alpha$ and for $\delta=\beta$, we conclude that all values in $\tau$ agree that $(a_\ell, e^k)$ is the next pair in $P$ and we continue to the next step of the procedure. Otherwise, we have detected a critical value $\delta_0$ in $\tau$ and we bifurcate, proceeding along two paths --- one assuming that $\delta^*\in (\alpha,\delta_0]$ and one assuming that $\delta^*\in (\delta_0,\beta]$.
We now give the details of simulating {\bf NextDisk} at $v'$.
For that we use the following easy observation, whose trivial proof is omitted
(see Figure~\ref{fig:bisector} for an illustration).
\begin{observation}\label{obs}
Let $a$ and $b$ be two points in the plane, let $h$ be their bisector,
and let $s$ be a point on $\bd{D_\delta(a)}$, for some $\delta>0$.
Then $s\in D_\delta(b)$ if and only if $s$ is in the halfspace
bounded by $h$ that contains $b$.
\end{observation}
\begin{figure}[htb]
\centering \includegraphics[scale=0.6]{bisector}
\centering \caption{\small The points on $\bd{D_\delta(a)}$ that are in
$D_\delta(b)$ are in the halfspace bounded by $h$ that contains $b$.}\label{fig:bisector}
\end{figure}
Let $v'$ be a child of $v$ corresponding to the triple
$(\tau=(\alpha,\beta], a^k, e^{k}(p,q])$, where $a^k$ is the point $a_i$. We
simulate {\bf NextDisk} at all possible $\delta \in \tau$ by traversing
$a_{i+1},\ldots, a_m$, distinguishing between the following cases at each such
point $a_\ell$.
(i) ${p \notin D_\alpha(a_{\ell})}$ and ${q \notin D_\beta(a_{\ell})}$.
In this case, each point $x^{k}(\delta) \in e^{k}(p,q]$, for
$\delta \in(\alpha,\beta]$, satisfies
$x^{k}(\delta) \notin D_{\delta}(a_\ell)$. Indeed, by the way we
computed the triple for $v'$, $p$ is a point on $\bd{D_\alpha(a^k)}$
and $q$ is a point on $\bd{D_\beta(a^k)}$. Thus, by
Observation~\ref{obs}, $p$ and $q$ are not in the halfspace $h^+(a^k,a_l)$ bounded
by $h(a^k,a_\ell)$ that contains $a_\ell$, where $h(a^k,a_\ell)$ is
the bisector of $a^k$ and $a_\ell$. Thus, $x^{k}(\delta)$, for any
$\delta\in (\alpha,\beta]$, is also not in the halfspace $h^+(a^k,a_l)$. Since $x^{k}(\delta)$ is a
point on $\bd{D_{\delta}(a^k)}$, again by Observation~\ref{obs},
$x^{k}(\delta) \notin D_{\delta}(a_\ell)$. Hence, in this case, the
$A$-frog cannot jump to $a_\ell$ when the person is at $x^{k}(\delta)$
(for each point $x^{k}(\delta) \in e^{k}(p,q]$, over $\delta \in
(\alpha,\beta]$), so we proceed to the next point $a_{\ell+1}$.
(ii) ${p \in D_\alpha(a_{\ell})}$ and ${q\in D_\beta(a_{\ell})}$.
By a similar reasoning to that of the preceding case,
each point $x^{k}(\delta) \in e^{k}(p,q]$
satisfies $x^{k}(\delta) \in D_{\delta}(a_\ell)$. Hence, for each
point $x^{k}(\delta)\in e^{k}(p,q]$ the $A$-frog can jump to
$a_{\ell}$ when the person is at $x^{k}(\delta)$.
So in this case $v'$ has only one child $v''$ that corresponds to
the triple $(\tau=(\alpha,\beta], a_\ell, e^{k}(p,q])$.
(iii) ${q \in D_\beta(a_{\ell})}$ and ${p \notin D_\alpha(a_{\ell})}$.
By a similar reasoning as in the previous
cases, using Observation~\ref{obs}, $q$ is in the halfspace
$h^+(a^k,a_l)$, and $p$ is not.
Thus, there exists a point $s$ such that $s = e^k(p,q] \cap h(a^k,a_\ell)$.
By the way we constructed the triple of $v'$, it follows that for
each point $p' \in e^k(p,q]$ there exists a $\delta \in (\alpha, \beta]$
such that $p'$ is $x^k(\delta)$ and $x^k(\delta) \in \bd D_{\delta}(a^k) \cap e^k$.
By observation \ref{obs} we have: (a) If
$x^k(\delta) \in e^k(p,s]$ then $x^k(\delta) \not\in
D_{\delta}(a_\ell)$ (so the frog cannot jump to $a_\ell$ when the
person is at $x^k(\delta)$). (b) If $x^k(\delta)\in
e^k(s,q]$ then $x^k(\delta)\in D_{\delta}(a_\ell)$ (so the frog can jump to
$a_\ell$ when the person is at $x^k(\delta)$).
Let $\delta_0=\|a^k-s\|=\|a_\ell-s\|$. Clearly, $x^k(\delta) \in
e^k(p,s]$ for $\alpha < \delta \le \delta_0$ and $x^k(\delta) \in
e^k(s,q]$ for $\delta_0< \delta \le \beta$. Furthermore, $\delta_0$
is a point-point-edge
critical value involving $a^k, a_\ell$ and $e^k$.
We generate a child $v''$ of $v'$ that corresponds to the triple
$((\delta_0, \beta], a_\ell, e^k(s,q])$, and continue to generate the
other children of $v'$ by proceeding to the next point (if there is
one) $a_{\ell + 1}$ with the updated triple $((\alpha, \delta_0],
a^k, e^k(p,s])$. See Figure~\ref{fig:bifurcate}(a).
(iv) ${p\in D_\alpha(a_{\ell})}$ and ${q \notin D_\beta(a_{\ell})}$.
Arguing similarly to the preceding case,
we encounter a point-point-edge critical value $\delta_0$ involving
$a^k, a_\ell$ and $e^k$, where $\delta_0$ is the distance between
$s = e^k(p,q] \cap h(a^k,a_\ell)$ and $a^k$ (or $a_\ell$).
We generate a child $v''$ of $v'$ that corresponds to the triple
$((\alpha, \delta_0], a_\ell, e^k(p,s])$, and continue to generate
the other children of $v'$ by proceeding to the next point (if there
is one) $a_{\ell + 1}$ with the updated triple $(( \delta_0,\beta],
a^k, e^k(s,q])$. See Figure~\ref{fig:bifurcate}(b).
\begin{figure}
\centering
\begin{subfigure}[b]{0.4\textwidth}
\centering
\includegraphics[width=\textwidth]{bifurcate1}
(a) $q \in D_\beta(a_{l})$ and $p \notin D_\alpha(a_{l})$.
\label{fig:bifurcate1}
\end{subfigure}%
\hspace{1cm}
\begin{subfigure}[b]{0.4\textwidth}
\centering
\includegraphics[width=\textwidth]{bifurcate2}
(b) $p \in D_\alpha(a_{l})$ and $q \notin D_\beta(a_{l})$.
\label{fig:bifurcate2}
\end{subfigure}
\caption{\small Situations that cause bifurcation in when simulating a call to NextDisk.
The disks $D_{\delta_0}(a_{i})$ and $D_{\delta_0}(a_{l})$, where $\delta_0$ is the corresponding critical value, are drawn dashed. }\label{fig:bifurcate}
\end{figure}
Note that we may reach the last point $a_m$ without generating
children of $v'$ (i.e., grandchildren of $v$). In this case
$\delta^*$ cannot be in $(\alpha, \beta]$, and we can abandon this branch.
When we reach a node whose triple is
$((\alpha, \beta], a_m, e_n(p,p_n])$ then if $\delta^* \in (\alpha,
\beta]$ then $\delta^* = \beta$.
It is straightforward to show that the triple of each node $v$
that we generate satisfies the invariant mentioned at the beginning
of the proof.
We do not generate the entire tree $T$ but proceed as in
Lemma~\ref{lem:searching_in_interval}. We distinguish between unary
nodes $v\in T$, each having a single child, and nodes $v\in T$ with
more than one child. A node $v$ with $d>1$ children is associated
with $d-1$ critical events that triggered these $d-1$
bifurcations. We construct the relevant subtree $T'$ of $T$ top down,
and we stop expanding a node $v\in T'$ if it has $s$ unary immediate ancestors.
We stop expanding $T'$ altogether when it contains $m+n$ nodes or when each of
its leaves has $s$ unary immediate ancestors.
Note that we might stop the construction of $T'$ in the middle of the expansion of a node $v$ that has too many children. In that case, if $v$ happens to be the node with the range containing $\delta^*$, we either resume the procedure from one of the children already constructed, or abandon them and construct the remaining children.
We then run a binary
search over the set of $O(m+n)$ critical values that we have
accumulated at the bifurcations of $T'$, using the decision procedure
$\Gamma$ to guide the search. This determines the leaf $v$ of $T'$
such that the range $\tau$ of $v$ contains $\delta^*$. The path of $T'$ leading to $v$ is the next portion of the semi-sparse path $S$ produced by the decision procedure (in Figure~\ref{alg:semi-continuous}) at $\delta^*$. We then repeat the whole procedure starting at $v$. We stop when we reach a node that records the last step of $S$, which reaches $(a_m,p_n)$, and the final range $(\alpha,\beta]$ of that node yields $\delta^*=\beta$. An analysis analogous to the one in
Lemma~\ref{lem:searching_in_interval} shows that this algorithms
runs in $O((m+n)L^{1/2}\log(m+n))$ time using $O(m+n)$ space.
\end{proof}
By combining Lemma~\ref{lem:semi_cont_finding_interval} with Lemma~\ref{lem:semi_cont_searching_in_interval} as noted above, we obtain the following result.
\begin{theorem}
Given a set $A$ of $m$ points and a polygonal curve $f$ with $n$ edges in the plane, we can compute the one-sided semi-continuous Fr\'echet distance $\delta_F^s(A,f)$ with shortcuts in randomized expected time $O((m+n)^{2/3}m^{2/3}n^{1/3}\log(m+n))$ using $O((m+n)^{2/3}m^{2/3}n^{1/3})$ space.
\end{theorem}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 5,640 |
Джозеф Філдінг Сміт Старший (; 13 листопада 1838, Фар Вест, Міссурі, США — †19 листопада 1918, Солт-Лейк-Сіті, США) — шостий президент Церкви Ісуса Христа Святих останніх днів, племінник Джозефа Сміта.
Біографія
Син Гайрума Сміта і Мері Філдінг Сміт. Народився, коли його батько був у в'язниці. Потім родина переїхала в Наву, а в 1844 році Джозеф й Гайрум Сміти були вбиті натовпом. У 1846 році овдовіла Мері разом з дітьми переїжджає в Юту і в 1952 році померає.
У віці 13 років Джозеф був хрещений, а в 15 років направлений яко місіонер на Гаваї. Після цього він неодноразово брав участь у проповіді вчення мормонів в різних країнах. У 27 років Джозеф Філдінг Сміт був посвячений в апостоли і служив радником при декількох президентах Церкви. Після смерті Лоренцо Сноу в 1901 році був покликаний стати наступним главою Церкви.
Діяльність як президента церки
Ставши керівником церкви, Джозеф Філдінг Сміт зайнявся роботою з придбання місць, пов'язаних з історією мормонів такі як: в'язниця в місті Картідж, місце, де знаходився храм в місті Індепенденс, ферму сім'ї Джозефа Сміта, а також гай, де у засновника руху були перші видіння.
При Джозефі Філдінгу були закладені кілька нових храмів, започатковано практику обов'язкового щотижневого сімейного вечора.
Приватне життя
У 1859 році Джозеф Філдінг Сміт одружився зі своєю двоюрідною сестрою Лівайрою. Дітей від цього шлюбу не мав. Після того, як у церкві мормонів було введено багатоженство, Лівайра відмовилася прийняти цю практику і розлучилася з Джозефом у 1868 році.
Крім неї Сміт мав ще п'ять дружин, у шлюбах з якими народилося сорок три дитини.
Посилання
Grampa Bill's GA Pages: Joseph F. Smith
Президенти Церкви Ісуса Христа Святих останніх днів
Одержувачі помилування президента США | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 1,568 |
<? if(arg(0) == 'node' && is_numeric(arg(1))) { global $base_url; print ' data-url="'. $base_url .'/node/'. arg(1) .'"'; print ' data-counturl="'. url('node/'. arg(1), array('absolute' => TRUE)) .'" '; } ?>>Tweet
22,000 People Call on Mekong Governments to Cancel Xayaburi Dam
U.S. Senate Committee Calls for Delay in Xayaburi Dam
Bangkok, Thailand – 22,589 people from 106 countries submitted an international petition today to the Prime Ministers of Laos and Thailand, calling for cancellation of the proposed Xayaburi Dam on the Mekong River in Northern Laos. The petition comes one week before the four Mekong governments meet on December 8th in Siem Reap, Cambodia, where they are likely to decide whether to proceed with the project.
The Xayaburi Dam is the first of eleven dams proposed for the Lower Mekong River. The petition expresses grave concern about the future of the Lower Mekong Basin, and urges the Prime Ministers to cancel the project and defer all decisions on Mekong dams for a period of at least ten years, until further studies can be conducted. The petition was presented to Thailand's Government House and the Lao Embassy in Bangkok on Wednesday.
"The people of Southeast Asia and concerned citizens around the world have once again voiced their opposition to the Xayaburi Dam," said Pianporn Deetes, Thailand Campaign Coordinator for International Rivers. "The whole world is watching. We do not want to remember December 8th as the day the Mekong died."
The petition comes a day after the U.S. Senate Foreign Relations Committee unanimously approved a resolution by Senator Jim Webb (D-VA) calling for the protection of the Mekong River Basin and for delaying mainstream dam construction along the river. The resolution calls for the U.S. Government to allocate more funding to help identify sustainable alternatives to mainstream hydropower dams and to analyze the impacts of proposed development along the river.
"The Committee's adoption of this resolution sends a timely signal of U.S. support for the Mekong River Commission's efforts to preserve the ecological and economic stability of Southeast Asia," Senator Webb, chair of the Senate East Asian and Pacific Affairs Subcommittee, said in a statement. "The United States and the global community have a strategic interest in preserving the health and well-being of the more than 60 million people who depend on the Mekong River."
Although Laos is proposing the dam, Thailand is also playing a key role as investor, project developer, and purchaser of 95% of the dam's electricity. The petition calls on the government of Thailand to cancel its plans to purchase electricity from the Xayaburi Dam and any other Mekong Mainstream Dams.
"Laos has a duty under international law to provide enough information about the regional impacts of the Xayaburi project to allow its neighbors to make an informed decision, but it has yet to do so," said Sor.Rattanamanee Polkla, a lawyer for the Community Resources Centre in Thailand and a member of Mekong Legal Network. "Moreover, Thailand, as the primary beneficiary of the dam, should be equally responsible for providing more information about the project's impacts. Under international best practice, Thailand should assess all energy options before deciding to dam a river of such importance for millions of people's livelihoods, in line with the recommendations of the Strategic Environmental Assessment sponsored by the Mekong River Commission."
"Through this petition, the international community has spoken out against the Xayaburi Dam as this is a river of global significance," said Guadalupe Rodriguez, a member of the German-based organization Rettet den Regenwald (Rainforest Rescue), and one of the sponsors of the petition. "We cannot allow a privileged few to trade away the biodiversity and ecosystems that feed millions, as it would spark tension in the region."
At a meeting in April, the governments of Cambodia, Thailand, and Vietnam raised concerns about the Xayaburi Dam's transboundary impacts and recommended further study and public consultations. The four governments could not agree on a solution, and elevated the decision to a ministerial meeting now scheduled for December 7-8.
The full text of the petition is as follows:
To: H.E. Yingluck Shinawatra, Prime Minister of Thailand and H.E. Thongsing Thammavong, Prime Minister of Laos
Copy: H.E. Mr. Choummaly Sayasone, President of Laos
We the undersigned are writing to express our grave concern over the proposed Xayaburi Dam in northern Laos, and the fate of the Mekong River Basin.
The Xayaburi Dam threatens the lives and food security of millions of people. The Mekong River is home to the world's most productive freshwater fishery and is a vital resource for the entire Mekong Region.
Scientists agree that the Xayaburi Dam would have severe environmental, social and economic impacts in the region, impacts which cannot be mitigated.
There are still critical knowledge gaps in understanding the dam's far-reaching impacts and the Mekong River Commission's own Strategic Impact Assessment calls for a 10-year deferment of all decisions about whether to proceed with Mekong Mainstream Dams, allowing time for more studies to be done.
We are aware that a decision about whether to proceed with the Xayaburi Dam will be made in the coming months. With so much at stake in the Mekong Region, we urge the Governments of Laos and Thailand to:
Cancel the Xayaburi Dam and defer all decisions on whether to proceed with Mekong Mainstream dams for a period of at least 10 years, until further studies can be conducted.
Additionally we call on the government of Thailand to cancel its plans to purchase electricity from the Xayaburi Dam and any other Mekong Mainstream dams.
Pianporn Deetes, Thailand Campaign Coordinator, International Rivers, +66 81 422 0111, pai@internationalrivers.org
Sor.Rattanamanee Polkla, Lawyer for the Community Resources Centre, +66 81 772 5843
Ame Trandem, Southeast Asia Program Director, International Rivers, +855 92 569 113, ame@internationalrivers.org
Press release from Senator Webb and full text of Senate Resolution
America Africa South
Asia China Southeast
Asia Policy
Reform Climate
Programs and Campaigns (2)
Mekong Mainstream Dams
Xayaburi Dam
Pianporn Deetes
pai@internationalrivers.org | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 3,912 |
So first rehearsals kicked off today and Twitter was just about exploding with the biggest load of drama queens. I have decided not to watch any rehearsals this year, in order to keep giving fair song reviews and to saviour how special it is watching the performances for the first time. I couldn't help but watch a 15-sec snippet of Slavko though, and I just can't wait for his semi-final performance! The song might be a joke and the talent might be somewhat missing, but this only exacerbates my anticipation for what will be slavtacular.
Hungary have a habit for knocking around the 20th position, and I think whether they break out the top 5 would come down to the draw. | {
"redpajama_set_name": "RedPajamaC4"
} | 4,382 |
{"url":"https:\/\/socratic.org\/questions\/how-do-you-solve-u-u-5-8u-u-u-2-4","text":"# How do you solve u(u - 5) + 8u = u(u + 2) - 4 ?\n\nMay 27, 2017\n\nSee a solution process below:\n\n#### Explanation:\n\nFirst, expand the terms in parenthesis on each side of the equation by multiplying each term within the parenthesis by the term outside the parenthesis:\n\n$\\textcolor{red}{u} \\left(u - 5\\right) + 8 u = \\textcolor{b l u e}{u} \\left(u + 2\\right) - 4$\n\n$\\left(\\textcolor{red}{u} \\times u\\right) - \\left(\\textcolor{red}{u} \\times 5\\right) + 8 u = \\left(\\textcolor{b l u e}{u} \\times u\\right) + \\left(\\textcolor{b l u e}{u} \\times 2\\right) - 4$\n\n${u}^{2} - 5 u + 8 u = {u}^{2} + 2 u - 4$\n\n${u}^{2} + \\left(- 5 + 8\\right) u = {u}^{2} + 2 u - 4$\n\n${u}^{2} + 3 u = {u}^{2} + 2 u - 4$\n\nNext, subtract $\\textcolor{red}{{u}^{2}}$ from each side of the equation to eliminate this term while keeping the equation balanced:\n\n$- \\textcolor{red}{{u}^{2}} + {u}^{2} + 3 u = - \\textcolor{red}{{u}^{2}} + {u}^{2} + 2 u - 4$\n\n$0 + 3 u = 0 + 2 u - 4$\n\n$3 u = 2 u - 4$\n\nNow, subtract $\\textcolor{red}{2 u}$ from each side of the equation to solve for $u$ while keeping the equation balanced:\n\n$- \\textcolor{red}{2 u} + 3 u = - \\textcolor{red}{2 u} + 2 u - 4$\n\n$\\left(- \\textcolor{red}{2} + 3\\right) u = 0 - 4$\n\n$1 u = - 4$\n\n$u = - 4$","date":"2020-08-14 00:41:36","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 15, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.7814702391624451, \"perplexity\": 322.63291701356724}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2020-34\/segments\/1596439739104.67\/warc\/CC-MAIN-20200813220643-20200814010643-00119.warc.gz\"}"} | null | null |
from oslo_config import cfg
CONF = cfg.CONF
class FakeVolume(object):
def __init__(self, **kwargs):
self.id = kwargs.pop('id', 'fake_vol_id')
self.status = kwargs.pop('status', 'available')
self.device = kwargs.pop('device', '')
for key, value in kwargs.items():
setattr(self, key, value)
def __getitem__(self, attr):
return getattr(self, attr)
class FakeVolumeSnapshot(object):
def __init__(self, **kwargs):
self.id = kwargs.pop('id', 'fake_volsnap_id')
self.status = kwargs.pop('status', 'available')
for key, value in kwargs.items():
setattr(self, key, value)
def __getitem__(self, attr):
return getattr(self, attr)
class API(object):
"""Fake Volume API."""
def get(self, *args, **kwargs):
pass
def create_snapshot_force(self, *args, **kwargs):
pass
def get_snapshot(self, *args, **kwargs):
pass
def delete_snapshot(self, *args, **kwargs):
pass
def create(self, *args, **kwargs):
pass
def extend(self, *args, **kwargs):
pass
def get_all(self, search_opts):
pass
def delete(self, volume_id):
pass
def get_all_snapshots(self, search_opts):
pass
| {
"redpajama_set_name": "RedPajamaGithub"
} | 6,559 |
namespace bpt = boost::property_tree;
namespace {
void test_tilejson_fetch() {
char cwd[PATH_MAX];
if (getcwd(&cwd[0], PATH_MAX) == NULL) { throw std::runtime_error("getcwd failed"); }
bpt::ptree conf = avecado::tilejson((boost::format("file://%1%/test/tilejson.json") % cwd).str());
test::assert_equal<int>(conf.get<int>("maskLevel"), 8, "maskLevel");
test::assert_equal<int>(conf.get<int>("maxzoom"), 15, "maxzoom");
test::assert_equal<int>(conf.get<int>("minzoom"), 0, "minzoom");
}
// TODO: this is a pretty awful test, and isn't able to dig any deeper than
// the overzoom, not even to inspect it. might need an upgrade to how the
// built-in HTTP server works so that we can do more with these tests...
void test_tilejson_parse() {
using namespace avecado;
char cwd[PATH_MAX];
if (getcwd(&cwd[0], PATH_MAX) == NULL) { throw std::runtime_error("getcwd failed"); }
bpt::ptree conf = tilejson((boost::format("file://%1%/test/tilejson.json") % cwd).str());
std::unique_ptr<fetcher> f = make_tilejson_fetcher(conf);
fetch::overzoom *o = dynamic_cast<fetch::overzoom *>(f.get());
test::assert_equal<bool>(o != nullptr, true, "is overzoom");
}
// utility function to create TileJSON from Mapnik XML.
std::string tile_json_for_xml(const std::string xml) {
using test::make_map;
using avecado::make_tilejson;
char cwd[PATH_MAX];
if (getcwd(&cwd[0], PATH_MAX) == NULL) { throw std::runtime_error("getcwd failed"); }
mapnik::Map map = make_map(xml, 256, 0, 0, 0);
std::string json = make_tilejson(map, (boost::format("file://%1%/") % cwd).str());
return json;
}
// https://github.com/MapQuest/avecado/issues/54
//
// maxzoom, minzoom & metatile (masklevel, presumably, too) are all
// supposed to be numeric, but are being generated as strings.
//
// the 2.1.0 tilejson spec doesn't specify metatile or masklevel,
// but we'll just assume they're supposed to be integers too.
void test_tilejson_generate_numeric() {
using namespace boost::xpressive;
for (auto &xml : {"test/empty_map_file.xml", "test/tilejson_params.xml"}) {
try {
// this test is kinda half-assed, what with the regex and all...
// but it should work to detect the difference between a JSON
// string and a number.
const std::string json = tile_json_for_xml(xml);
for (auto ¶m : {"metatile", "maskLevel", "minzoom", "maxzoom"}) {
// first, need to check if the key is present at all. if not,
// then it doesn't make sense to check that it's a number.
// NOTE: this is /"${param}" *:/
sregex is_present = as_xpr('"') >> param >> '"' >> *space >> as_xpr(':');
if (regex_search(json.begin(), json.end(), is_present)) {
// if present, we check if the value begins with a valid number
// prefix: an optional minus sign and then digits. this might
// match invalid JSON, but the validity of generated JSON is a
// test we'll have to do elsewhere.
sregex is_number = as_xpr('"') >> param >> '"' >> *space >> as_xpr(':')
>> *space >> !as_xpr('-') >> +digit;
if (!regex_search(json.begin(), json.end(), is_number)) {
throw std::runtime_error(
(boost::format("Parameter \"%1%\" should be a number, but is not "
"in TileJSON: %2%") % param % json).str());
}
}
}
// Some params need to be arrays of integers
for (auto ¶m : {"center", "bounds"}) {
// The same check as above for presense
sregex is_present = as_xpr('"') >> param >> '"' >> *space >> as_xpr(':');
if (regex_search(json.begin(), json.end(), is_present)) {
sregex is_array = as_xpr('"') >> param >> '"' >> *space >> as_xpr(':')
>> *space >> as_xpr('[');
if (!regex_search(json.begin(), json.end(), is_array)) {
throw std::runtime_error(
(boost::format("Parameter \"%1%\" should be an array of numbers, but is not "
"in TileJSON: %2%") % param % json).str());
}
}
}
} catch (const std::exception &e) {
std::throw_with_nested(
std::runtime_error(
(boost::format("while processing XML file \"%1%\"") % xml).str()));
}
}
}
// same as the above test, but we force Mapnik to store strings
// in its parameter list so that we have to convert to numeric
// before outputting to JSON.
void test_tilejson_generate_numeric_force() {
using namespace boost::xpressive;
using test::make_map;
using avecado::make_tilejson;
char cwd[PATH_MAX];
if (getcwd(&cwd[0], PATH_MAX) == NULL) { throw std::runtime_error("getcwd failed"); }
mapnik::Map map = make_map("test/empty_map_file.xml", 256, 0, 0, 0);
map.get_extra_parameters().emplace("maxzoom", mapnik::value_holder(std::string("0")));
std::string json = make_tilejson(map, (boost::format("file://%1%/") % cwd).str());
sregex is_present = as_xpr("\"maxzoom\"") >> *space >> as_xpr(':');
if (!regex_search(json.begin(), json.end(), is_present)) {
throw std::runtime_error(
(boost::format("maxzoom key not in generated JSON: %1%") % json).str());
}
sregex is_number = as_xpr("\"maxzoom\"") >> *space >> as_xpr(':')
>> *space >> !as_xpr('-') >> +digit;
if (!regex_search(json.begin(), json.end(), is_number)) {
throw std::runtime_error(
(boost::format("Parameter \"maxzoom\" should be a number, but is not "
"in TileJSON: %1%") % json).str());
}
}
void test_tilejson_generate_masklevel() {
using test::make_map;
using avecado::make_tilejson;
test::temp_dir tmp;
std::string base_url = (boost::format("file://%1%/") % tmp.path().native()).str();
mapnik::Map map = make_map("test/empty_map_file.xml", 256, 0, 0, 0);
for (int maxzoom = 0; maxzoom < 23; ++maxzoom) {
map.get_extra_parameters().erase("maxzoom");
map.get_extra_parameters().emplace("maxzoom", mapnik::value_integer(maxzoom));
const std::string json = make_tilejson(map, base_url);
const std::string json_file = (tmp.path() / "tile.json").native();
std::ofstream out(json_file);
out.write(json.data(), json.size());
out.close();
bpt::ptree conf = avecado::tilejson((boost::format("%1%tile.json") % base_url).str());
test::assert_equal<mapnik::value_integer>(
map.get_extra_parameters().find("maxzoom")->second.get<mapnik::value_integer>(),
mapnik::value_integer(maxzoom), "maxzoom parameter");
test::assert_equal<size_t>(map.get_extra_parameters().count("maskLevel"), 0,
"maskLevel presence in original parameters");
test::assert_equal<int>(conf.get<int>("maxzoom"), maxzoom, "maxzoom");
test::assert_equal<int>(conf.get<int>("maskLevel"), maxzoom, "maskLevel");
}
}
} // anonymous namespace
int main() {
int tests_failed = 0;
std::cout << "== Testing TileJSON parsing ==" << std::endl << std::endl;
#define RUN_TEST(x) { tests_failed += test::run(#x, &(x)); }
RUN_TEST(test_tilejson_fetch);
RUN_TEST(test_tilejson_parse);
RUN_TEST(test_tilejson_generate_numeric);
RUN_TEST(test_tilejson_generate_numeric_force);
RUN_TEST(test_tilejson_generate_masklevel);
std::cout << " >> Tests failed: " << tests_failed << std::endl << std::endl;
return (tests_failed > 0) ? 1 : 0;
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 3,094 |
\section{Introduction}
The Standard Model (SM) describes globally the existing data on quark-flavour violating processes rather well \cite{Buras:2013ooa} but with the reduction
of experimental errors and increased precision in non-perturbative and
perturbative QCD and electroweak calculations a number of tensions
at the level of $2-3\, \sigma$ seem to emerge in various seemingly unrelated
observables. While some of these tensions could turn out to be the result of statistical
fluctuations, underestimate of systematical and theoretical errors, it is not
excluded that eventually they all signal the presence of some kind of new physics (NP). Therefore, it is interesting to investigate what this NP could be.
In the present paper we will address some of these tensions in
331 models based on the gauge group $SU(3)_C\times SU(3)_L\times U(1)_X$ \cite{Pisano:1991ee,Frampton:1992wt}. As these models have much smaller number of
new parameters than supersymmetric models, Randall-Sundrum scenarios and
Littlest Higgs models, it is not evident that they can remove all present
tensions simultaneously.
Our paper has been motivated by a recent analysis in \cite{Blanke:2016bhf}
which demonstrates that
the new lattice QCD results from Fermilab Lattice and MILC Collaborations
\cite{Bazavov:2016nty} on $B^0_{s,d}-\bar B^0_{s,d}$ hadronic matrix elements imply
a significant tension between
$\varepsilon_K$ and $\Delta M_{s,d}$ within the SM. The authors of
\cite{Bazavov:2016nty} find also inconsistences
between $\Delta M_{s,d}$ and tree-level determination of $|V_{cb}|$. But the
simultaneous consideration of $\varepsilon_K$ and $\Delta M_{s,d}$
in \cite{Blanke:2016bhf} also demonstrates that the tension between these two
quantities cannot be removed for any value of $|V_{cb}|$. Moreover, the situation worsens for other
models with constrained MFV (CMFV), indicating the presence of new flavour- and CP-violating interactions beyond CMFV framework at work.
The question then arises how
331 models face this tension and what are the implications
of new lattice results on other observables for which some departures from
SM predictions have been identified.
In particular, taking the results in \cite{Blanke:2016bhf,Bazavov:2016nty} into account we want to concentrate our analysis on
\begin{equation}
\varepsilon'/\varepsilon, \qquad B_s\to\mu^+\mu^-, \qquad B\to K^*\mu^+\mu^-\,.
\end{equation}
In this context the following facts should be recalled.
\begin{itemize}
\item
Recent analyses in \cite{Blum:2015ywa,Bai:2015nea,Buras:2015yba,Buras:2015xba}
find the ratio $\varepsilon'/\varepsilon$ in the SM to be significantly below
the experimental world average from NA48 \cite{Batley:2002gn} and KTeV
\cite{AlaviHarati:2002ye,Abouzaid:2010ny} collaborations.
The recent analysis in the large $N$ approach in \cite{Buras:2016fys}
indicates that final state interactions will not modify this picture at least
on the qualitative level. The analysis in \cite{Buras:2015yca} shows
that CMFV models cannot cure this problem. In any case models providing
an enhancement of $\varepsilon'/\varepsilon$ should be favoured from present perspective.
\item
The branching ratio for $B_s\to\mu^+\mu^-$ measured by CMS and LHCb \cite{CMS:2014xfa} has been always
visibly below rather precise prediction of the SM \cite{Bobeth:2013uxa}. The most recent result from ATLAS\footnote{$\overline{{B}}(B_s\to\mu^+\mu^-)=(0.9^{+1.1}_{-0.9})\times 10^{-9}$ \cite{Aaboud:2016ire}.},
while not accurate, appears to confirm this picture and models suppressing the rate for this decay relative to its SM prediction appear to be favoured.
\item
LHCb data on $B_d\to K(K^*)\mu^+\mu^-$ indicate some departures from SM
expectation although this issue is controversial. See \cite{ Altmannshofer:2014rta,Descotes-Genon:2015uva} and references to the rich literature therein. Assuming again that statistical fluctuations
or underestimated errors are not responsible for these effects, significant
NP contributions to the Wilson coefficient $C_9$ or $C_9$ and $C_{10}$
are required.
\end{itemize}
These three items have been already addressed by us within 331 models
in the past \cite{Buras:2012dp,Buras:2013dea,Buras:2014yna,Buras:2015kwd}.
In particular in \cite{Buras:2015kwd} the issue of $\varepsilon'/\varepsilon$ anomaly has been
addressed, while in \cite{Buras:2013dea,Buras:2015kwd} the last two items above have been considered. The main result of \cite{Buras:2015kwd} is that among
24 331 models only three (M8, M9 and M16 in the terminology of \cite{Buras:2014yna}) have a chance to survive if an improved
fit to electroweak precision observables relative to the SM is required and the $\varepsilon'/\varepsilon$ anomaly
will be confirmed in the future. Two of them (M8 and M9) allowed simultaneously a suppression of the
rate for $B_{s}\to \mu^+\mu^-$ by $20\%$ thereby bringing the theory closer to the data without any significant impact
on the Wilson coefficient $C_9$. The third model (M16) provided, simultaneously to the enhancement of $\varepsilon'/\varepsilon$, a
shift up to $\Delta C_9=-0.6$, softening the anomalies in $B\to K^*\mu^+\mu^-$, without any significant impact on $B_{s}\to \mu^+\mu^-$. While, before the ATLAS data on $B_s\to \mu^+\mu^-$, M16 seemed to be slightly favoured over M8 and M9, this data and the fact that the theoretical uncertainties in $B_{s}\to \mu^+\mu^-$ are significantly smaller than in
$B_d\to K^*\mu^+\mu^-$ make us believe that at the end models M8 and M9 have a bigger chance to survive.
However, the constraints from $\Delta F=2$ transitions used in \cite{Buras:2015kwd},
prior to the lattice QCD result in \cite{Bazavov:2016nty}, were significantly weaker
and it is of interest to investigate what is the impact of these new
lattice results on our previous analyses and whether the increased tension between
$\varepsilon_K$ and $\Delta M_{s,d}$ within the SM pointed out in
\cite{Blanke:2016bhf} can be removed in these three models. In fact one
should recall that the mixing and CP-violation in $B^0_{s,d}-\bar B^0_{s,d}$
systems play very important roles in the determination of new parameters
in 331 models \cite{Buras:2012dp} and it is not surprizing that our previous results will be indeed
modified in a visible manner.
In this context let us remark that within the SM,
dependently on whether $\Delta M_{s}$ or $\varepsilon_K$ has been used as a constraint, rather different values for $|V_{cb}|$ have been required to fit the data within the SM \cite{Blanke:2016bhf}:
\begin{equation}\label{VCB}
|V_{cb}|=(39.7\pm1.3)\times 10^{-3}\quad (\Delta M_s),\qquad |V_{cb}|=(43.3\pm1.1)\times 10^{-3} \quad (\varepsilon_K).
\end{equation}
This in turn resulted in rather different predictions for rare $K$ and $B_{s,d}$ decays as seen in Table~4 of \cite{Blanke:2016bhf}.
In our most recent analysis in \cite{Buras:2015kwd} we have performed numerical
analysis for $|V_{cb}|$ in the ballpark of the higher value in (\ref{VCB}), that
is $0.042$. In the present paper we will also use this value
in order to see the impact of new lattice data on our previous results, but in addition we will
perform the analysis with $0.040$ which is in the ballpark
of its lower value in (\ref{VCB}). This will tell us whether 331 models can
cope with the tensions in question for both values of $|V_{cb}|$ and whether
the implications for NP effects are modified through this change of $|V_{cb}|$.
The second motivation for a new analysis of 331 models is the following
one. In our analyses and also in \cite{Diaz:2004fs,CarcamoHernandez:2005ka,Promberger:2007py} the $U(1)_X$ factor in the gauge group $SU(3)_C\times SU(3)_L\times U(1)_X$ differed from the hypercharge gauge group $U(1)_Y$.
As various 331 models are characterized by
two parameters $\beta$ and $\tan\bar\beta$ defined through
\begin{equation}\label{QTX}
Q=T_3+\frac{Y}{2}= T_3+\beta T_8+X, \qquad \tan\bar\beta=\frac{v_\rho}{v_\eta}\,
\end{equation}
these analyses dealt with $\beta\not=0$.
Here $T_{3,8}$ and $X$ are the diagonal generators of $SU(3)_L$ and
$U(1)_X$, respectively. $Y$ represents $U(1)_Y$ and $v_i$ are
the vacuum expectation values of scalar triplets responsible for the generation of down- and up-quark masses in these models.
Recently a special variant of 331 models with $\beta=0$ or equivalently
$U(1)_X=U(1)_Y$ has been considered in \cite{Hue:2015mna}. Moreover, these authors
set $\tan\bar\beta=1$ as this choice with $\beta=0$ simplifies the model by
eliminating $Z-Z^\prime$ mixing studied by us in detail in \cite{Buras:2014yna} for $\beta\not=0$. As this is the simplest among the 331 models, the
question arises whether it is consistent with the flavour data in
the setup in \cite{Hue:2015mna} and what are the
implications for quark flavour observables for arbitrary $\tan\bar\beta$
when $Z-Z^\prime$ mixing enters the game. In particular the comparison with our studies for $\beta\not=0$ in \cite{Buras:2012dp,Buras:2013dea,Buras:2014yna,Buras:2015kwd} and in the present paper is of interest. As the authors of \cite{Hue:2015mna} did not address this question, to our knowledge this is the first quark flavour study of this simplified 331 model.
We will see that in the absence of $Z-Z^\prime$ mixing
the choice $\beta=0$ provides a unique 331 model in
which the phenomenologically successful relation
\begin{equation}\label{R1}
C_9^\text{NP}=-C_{10}^\text{NP}
\end{equation}
is satisfied. Here $C_9^\text{NP}$ and $C_{10}^\text{NP}$ stand for the shifts in the
Wilson coefficients relevant in particular for $B\to K^*\mu^+\mu^-$ and
$B_s\to\mu^+\mu^-$, respectively. This is good news. The bad news
is that setting $\beta=0$ modifies the values of all couplings relative to the ones in M8, M9 and M16 models. We find then that NP contributions to $\varepsilon'/\varepsilon$ in
this simple model are at most $1 \times 10^{-4}$ for $M_{Z^\prime}=3\, {\rm TeV}$ and
decrease with increasing $M_{Z^\prime}$.
The effects in $C_9^\text{NP}$ and $C_{10}^\text{NP}$ are at most at the level of a few percent even if $Z-Z^\prime$ mixing is taken into
account. Thus the model fails in solving three anomalies listed above.
But as we will see it is able to remove
the tensions between $\Delta M_{s,d}$ and $\varepsilon_K$.
Our paper is organized as follows.
In Section~\ref{sec:2} we address the tensions between
$\Delta M_{s,d}$ and $\varepsilon_K$ in M8, M9 and M16 models and
we update our analysis of $\varepsilon'/\varepsilon$, $B_s\to\mu^+\mu^-$ and $C_9$ in \cite{Buras:2015kwd} taking new $\Delta F=2$ constraints from \cite{Bazavov:2016nty}
into account and performing the analysis at two values of $|V_{cb}|$ as discussed
above.
In Section~\ref{sec:3} we specify the existing formulae in 331 models to
the case $\beta=0$ but for arbitrary $\tan\bar\beta$ and we derive the results
mentioned above. We conclude in Section~\ref{sec:4}.
\section{M8, M9 and M16 Facing Anomalies}\label{sec:2}
\subsection{Preliminaries}
Let us recall that in these three models new flavour-violating effects are
governed by tree-level $Z^\prime$ exchanges with a subdominant role played
by tree-level $Z$ exchanges generated through $Z-Z^\prime$ mixing. All the
formulae for flavour observables in these models can be found in
\cite{Buras:2012dp,Buras:2013dea,Buras:2014yna,Buras:2015kwd} and will not
be repeated here. In particular the collection of formulae for $Z^\prime$ couplings to quarks and leptons for arbitrary $\beta$
are given in (17) and (18) of \cite{Buras:2013dea}.
New sources of flavour and CP violation in 331 models are parametrized by
new mixing parameters and phases
\begin{equation}\label{PAR}
\tilde s_{13},\qquad\tilde s_{23},\qquad \delta_1,\qquad \delta_2
\end{equation}
with $\tilde s_{13}$ and $\tilde s_{23}$ positive definite and smaller than unity and
$0\le \delta_{1,2}\le 2\pi$. They can be constrained by flavour observables as demonstrated in detail in \cite{Buras:2012dp}.
The non-diagonal $Z^\prime$ couplings relevant for $K$, $B_d$ and $B_s$ meson
systems can
be then parametrized respectively within an excellent approximation through
\begin{equation}\label{vij}
v_{32}^*v_{31}=\tilde s_{13}\tilde s_{23}e^{i(\delta_2-\delta_1)}, \qquad
v_{33}^*v_{31}=-\tilde s_{13}e^{-i\delta_1}, \qquad
v_{33}^*v_{32}=-\tilde s_{23}e^{-i\delta_2} \,.
\end{equation}
$\tilde s_{13}$ and $\delta_1$ can be determined from $\Delta M_d$ and CP-asymmetry $S_{\psi K_S}$ while $\tilde s_{23}$ and $\delta_2$ from $\Delta M_s$ and CP-asymmetry $S_{\psi \phi}$. Then the parameters in the $K$ system are fixed. This correlation tells us that the removal of tensions between $\varepsilon_K$ and $\Delta M_{s,d}$ is not necessarily automatic in 331 models and constitutes an important test of these models.
The remaining two parameters, except for $M_{Z^\prime}$ mass, are as seen in
(\ref{QTX}), $\beta$ and $\tan\bar\beta$. Moreover, the fermion representations
of SM quarks under the $SU(3)_L$ group matter. The three models in question
are then characterized by
\begin{align}
\beta&=\frac{2}{\sqrt{3}},\qquad \tan\bar\beta= 5\, \qquad (F_1), \qquad ({\rm M8}),\\
\beta&=-\frac{2}{\sqrt{3}},\quad \,\tan\bar\beta= 1\, \qquad (F_2), \qquad ({\rm M9}),\\
\beta&=\frac{2}{\sqrt{3}},\qquad \tan\bar\beta= 5\, \qquad (F_2), \qquad ({\rm M16})\,
\end{align}
with $F_1$ and $F_2$ standing for two fermion representations. In $F_1$
the first two generations of quarks belong to triplets of $SU(3)_L$,
while the third generation of quarks to an antitriplet. In $F_2$ it
is opposite.
With the values of $\beta$ and $\tan\bar\beta$ being fixed flavour phenomenology depends only on the parameters in (\ref{PAR}) and $M_{Z^\prime}$.
\begin{figure}[!tb]
\centering
\includegraphics[width = 0.47\textwidth]{DMsvsepsKM8M16.jpg}
\includegraphics[width = 0.47\textwidth]{DMdvsepsKM8M16.jpg}
\caption{\it $\Delta M_{s,d}$ vs. $\varepsilon_K$ in M8 and M16.
Red dots represent central SM values, black dots the central experimental values. $M_{Z^\prime}=3\, {\rm TeV}$ and $|V_{cb}|=0.042$.}\label{MBAJB}
\end{figure}
\subsection{Numerical Analysis}
The difficulty in doing the numerical analysis are tensions between
inclusive and exclusive determinations of the CKM elements $|V_{cb}|$ and $|V_{ub}|$.
The exclusive determinations have been summarized
in \cite{DeTar:2015orc} and are given as follows
\begin{equation}\label{excl}
|V_{cb}|_\text{excl}=(39.78\pm0.42)\cdot 10^{-3},\qquad |V_{ub}|_\text{excl}=(3.59\pm0.09)\cdot 10^{-3}.
\end{equation}
They are based on \cite{Lattice:2015rga,Lattice:2015tia,Glattauer:2015teq,Bazavov:2016nty,Aaij:2015bfa}.
The inclusive ones are summarized well in
\cite{Alberti:2014yda,Gambino:2016fdy}
\begin{equation}\label{incl}
|V_{cb}|_\text{incl}=(42.21\pm0.78)\cdot 10^{-3},\qquad |V_{ub}|_\text{incl}=(4.40\pm0.25)\cdot 10^{-3}.
\end{equation}
We note that after the recent Belle data on $B\to D\ell\nu_l$ \cite{Glattauer:2015teq}, the exclusive and inclusive values of $|V_{cb}|$ are closer to each other
than in the past. On the other hand in the case of $|V_{ub}|$ there is a very significant difference.
Furthermore, after recent precise determinations of hadronic matrix elements
entering $\Delta M_{s,d}$ in $B_{s,d}^0-\bar B_{s,d}^0$ mixing by
Fermilab Lattice and MILC Collaborations \cite{Bazavov:2016nty} there
are significant tensions between tree-level determinations of $|V_{cb}|$ and $|V_{ub}|$
and $\Delta M_{s,d}$ within the SM \cite{Bazavov:2016nty} and also
the tensions between $\varepsilon_K$ and $\Delta M_{s,d}$ \cite{Blanke:2016bhf}
in this model. Moreover, as found in the latter paper, the value of the angle $\gamma$ in the unitarity
triangle extracted from the ratio $\Delta M_d/\Delta M_s$ and the CP-asymmetry
$S_{\psi K_S}$ is with $\gamma=(63.0\pm 2.1)^\circ$ visibly smaller than it
tree-level determination
\cite{Trabelsi:2014}
\begin{equation}\label{gamma}
\gamma = (73.2^{+6.3}_{-7.0})^\circ.
\end{equation}
In the present paper, as in \cite{Buras:2015kwd}, we will
set first the CKM parameters to
\begin{equation}\label{CKMfix}
|V_{ub}|=3.6\times 10^{-3}, \qquad |V_{cb}|=42.0 \times 10^{-3}, \qquad \gamma=70^\circ.
\end{equation}
This choice is in the ballpark of exclusive determination of $|V_{ub}|$ in (\ref{excl}) and
the inclusive one for $|V_{cb}|$ in (\ref{incl}). Moreover, it is in the ballpark of tree-level determination
of $\gamma$. In view of new parameters in 331 models the value of $\gamma$ does not follow
from the ratio $\Delta M_s/\Delta M_d$ and $S_{\psi K_S}$ like in CMFV models
and it is better to take $\gamma$ from tree-level determinations as it is
to first approximation not polluted by NP. Having the same CKM input
as in our previous analysis will allow us to see the impact of new lattice
data on phenomenology.
The choice in (\ref{CKMfix}) is also motivated by the fact that NP contributions to
$\varepsilon_K$ in 331 models are rather small for $M_{Z^\prime}$ of a few $\, {\rm TeV}$ and SM
should perform well in this case. Indeed for this choice of CKM parameters
we find
\begin{equation}
|\varepsilon_K|_{\rm SM}=2.14\times 10^{-3}, \qquad (\Delta M_K)_{\rm SM}=0.467\cdot 10^{-2} \,\text{ps}^{-1}
\end{equation}
and $|\varepsilon_K|$ in
the SM only $4\%$ below the data. Due to the presence of long distance effects
in $\Delta M_K$ also this value is compatible with the data.
While the CKM parameters do not enter
the shift in $\varepsilon'/\varepsilon$ and $\varepsilon_K$, their choice matters in the predictions for NP contributions to $\Delta F=2$ observables in $B_{d,s}^0-\bar B_{d,s}^0$
systems and the rare $B_{s,d}$ decays. This is not only because
of their interferences with SM contributions. The departure of SM
predictions for $\varepsilon_K$ and $\Delta M_{s,d}$ from the data depends on
the CKM parameters, in particular on the value of $|V_{cb}|$, and this has an
impact on the allowed ranges of new parameters extracted from
$\Delta F=2$ observables and consequently on final values of $\varepsilon'/\varepsilon$,
$\overline{\mathcal{B}}(B_s\to\mu^+\mu^-)$ and the shift in $C_9$. We will
illustrate this below by choosing also $|V_{cb}|=0.040$
which corresponds to its exclusive determination in
(\ref{excl}). See (\ref{CKMfix1}).
\begin{figure}[!tb]
\centering
\includegraphics[width = 0.47\textwidth]{epspvsepsKM8.jpg}
\includegraphics[width = 0.47\textwidth]{epspvsepsKM16.jpg}
\caption{ \it $\Delta(\varepsilon'/\varepsilon)$ versus $\varepsilon_K$ for M8 and M16.
Red dots represent central SM values. $M_{Z^\prime}=3\, {\rm TeV}$ and $|V_{cb}|=0.042$.}\label{M8M16}~\\[-2mm]\hrule
\end{figure}
\begin{figure}[!tb]
\centering
\includegraphics[width = 0.47\textwidth]{epspvsBsmumuM8.jpg}
\includegraphics[width = 0.47\textwidth]{epspvsC9M8.jpg}
\includegraphics[width = 0.47\textwidth]{epspvsBsmumuM16.jpg}
\includegraphics[width = 0.47\textwidth]{epspvsC9M16.jpg}
\caption{ \it Correlations of $\Delta(\varepsilon'/\varepsilon)$ with $B_s\to \mu^+\mu^-$
(left panels) and with $ C_9^{\rm NP}$ (right panels) for M8 and M16.
Red dots represent central SM values.
$M_{Z^\prime}=3\, {\rm TeV}$ and $|V_{cb}|=0.042$.}
\label{CORR816}~\\[-2mm]\hrule
\end{figure}
Next, as in \cite{Buras:2012dp,Buras:2015kwd}, we perform a simplified analysis of
$\Delta M_{d,s}$, $S_{\psi K_S}$
and $S_{\psi\phi}$
in order to identify oases in the space of four parameters (\ref{PAR})
for which these four observables are consistent with experiment.
To this end we use the formulae for $\Delta F=2$ observables in
\cite{Buras:2012dp,Buras:2014yna} and set input parameters listed in
Table~3 of our recent analysis in \cite{Buras:2015kwd} at their central values. The only modifications in this input are the recently calculated parameters \cite{Bazavov:2016nty} \footnote{These results are more accurate than ETM results \cite{Carrasco:2013zta}, but compatible with them. We look forward to new FLAG averages on these
quantities.}
\begin{equation}\label{Kronfeld}
F_{B_s}\sqrt{\hat B_{B_s}}=(274.6\pm8.8)\, {\rm MeV},\qquad F_{B_d} \sqrt{\hat B_{B_d}}=
(227.7\pm 9.8)\, {\rm MeV} \,,
\end{equation}
that should be compared with $F_{B_s}\sqrt{\hat B_{B_s}}=(266.0\pm18.0)\, {\rm MeV}$ and
$F_{B_d}\sqrt{\hat B_{B_d}}=(216.0\pm 15.0)\, {\rm MeV}$ used by us in \cite{Buras:2015kwd} . This change implies the modifications in the SM values of $\Delta M_{s,d}$
that now are significantly higher than the data:
\begin{equation}
(\Delta M_s)_{\rm SM}=19.66/{\rm ps},\qquad (\Delta M_d)_{\rm SM}=0.620/{\rm ps},
\qquad S^{\rm SM}_{\psi\phi}=0.037, \qquad S^{\rm SM}_{\psi K_S}=0.688\,
\end{equation}
with CP asymmetries unchanged and compatible with the data. Thus the 331
models are requested to bring the values of $\Delta M_{s,d}$ down to their
experimental values \cite{Amhis:2014hma}
\begin{equation}
(\Delta M_s)_{\rm exp}=17.757(21)/{\rm ps},\qquad (\Delta M_d)_{\rm exp}=0.5055(20)/{\rm ps},
\end{equation}
while being consistent with the data for $\varepsilon_K$, $S_{\psi K_S}$ and $S_{\psi\phi}$.
As we keep the input parameters at their central values, in order to take partially hadronic
and experimental uncertainties into account we require the 331 models
to reproduce the data for $\Delta M_{s,d}$ within $\pm 5\%$
and the
data on $S_{\psi K_S}$ and $S_{\psi\phi}$ within experimental
$2\sigma$ ranges.
Specifically, our search is governed by the following allowed ranges:
\begin{equation}
16.9/{\rm ps}\le \Delta M_s\le 18.7/{\rm ps},
\qquad -0.055\le S_{\psi\phi}\le 0.085, \label{oases23}
\end{equation}
\begin{equation}
0.48/{\rm ps}\le \Delta M_d\le 0.53/{\rm ps},\qquad 0.657\le S_{\psi K_S}\le 0.725\, . \label{oases13}
\end{equation}
The ranges for $\Delta M_{s,d}$ are smaller than used in \cite{Buras:2015kwd}
because of the reduced errors in (\ref{Kronfeld}).
{We also impose the constraint on the ratio $\Delta M_s/\Delta M_d$ using
\cite{Bazavov:2016nty}
\begin{equation}\label{xi}
\xi=\frac{F_{B_s}\sqrt{\hat B_{B_s}}}{F_{B_d}\sqrt{\hat B_{B_d}}}=1.206\pm0.019\,.
\end{equation}
In the spirit of our simplified analysis we will keep this ratio
at its central value in (\ref{xi}) but in order to take into account the uncertainty in $\xi$ we will
require that $\Delta M_s/\Delta M_d$ agrees with the data
within $\pm 5\%$. Specifically we will require that
\begin{equation}\label{RCMFV}
33.3\le \left(\frac{\Delta M_s}{\Delta M_d}\right)\le 36.8\,
\end{equation}
is satisfied.}
In the case of $\varepsilon_K$ and $\Delta M_K$ we will just proceed as
in \cite{Buras:2015kwd} imposing the ranges
\begin{equation}\label{CONE}
1.60 \times 10^{-3}< |\varepsilon_K|< 2.50\times 10^{-3}\,, \qquad
-0.30\le \frac{(\Delta M_K)^{Z^\prime}}{(\Delta M_K)_\text{exp}}\le 0.30\,.
\end{equation}
Having determined the ranges for the parameters (\ref{PAR}) we can calculate
all the remaining flavour observables of interest.
\begin{figure}[!tb]
\centering
\includegraphics[width = 0.47\textwidth]{DMsvsepsKM8M16vcb4.jpg}
\includegraphics[width = 0.47\textwidth]{DMdvsepsKM8M16vcb4.jpg}
\caption{\it $\Delta M_{s,d}$ vs. $\varepsilon_K$ in M8 and M16.
Red dots represent central SM values, black dots the central experimental values. $M_{Z^\prime}=3\, {\rm TeV}$, $|V_{cb}|=0.040$ and $|V_{ub}|=0.0036$.}\label{MBAJBv4}
\end{figure}
\begin{figure}[!tb]
\centering
\includegraphics[width = 0.47\textwidth]{epspvsepsKM8vcb4.jpg}
\includegraphics[width = 0.47\textwidth]{epspvsepsKM16vcb4.jpg}
\caption{ \it $\Delta(\varepsilon'/\varepsilon)$ versus $\varepsilon_K$ for M8 and M16. Red dots represent central SM values. $M_{Z^\prime}=3\, {\rm TeV}$, $|V_{cb}|=0.040$ and $|V_{ub}|=0.0036$.}\label{M8M16v4}~\\[-2mm]\hrule
\end{figure}
\begin{figure}[!tb]
\centering
\includegraphics[width = 0.47\textwidth]{epspvsBsmumuM8vcb4.jpg}
\includegraphics[width = 0.47\textwidth]{epspvsC9M8vcb4.jpg}
\includegraphics[width = 0.47\textwidth]{epspvsBsmumuM16vcb4.jpg}
\includegraphics[width = 0.47\textwidth]{epspvsC9M16vcb4.jpg}
\caption{ \it Correlations of $\Delta(\varepsilon'/\varepsilon)$ with $B_s\to \mu^+\mu^-$
(left panels) and with $ C_9^{\rm NP}$ (right panels) for M8 and M16.
Red dots represent central SM values.
$M_{Z^\prime}=3\, {\rm TeV}$, $|V_{cb}|=0.040$ and $|V_{ub}|=0.0036$.}
\label{CORR816vcb4}~\\[-2mm]\hrule
\end{figure}
In Fig.~\ref{MBAJB} we show $\Delta M_{s,d}$ vs. $\varepsilon_K$ in M8 and M16. Red dots represent central SM values and black dots the central experimental
values. The experimental errors are negligible and the parametric and theoretical errors are represented by the allowed departure from them as explained
above. These results do not depend on the fermion representation up to tiny
effects from $Z-Z^\prime$ mixing and consequently are practically the same
for M8 and M16. As far as M9 is concerned all results presented in our paper
are very similar to the ones in M8 and will not be shown. We observe that
the tensions between $\Delta M_{s,d}$ vs. $\varepsilon_K$ present in the
SM can be easily removed in 331 models for $M_{Z^\prime}=3\, {\rm TeV}$.
In Fig.~\ref{M8M16} we show $\Delta(\varepsilon'/\varepsilon)$ versus $\varepsilon_K$ for M8 and M16
at $M_{Z^\prime}=3\, {\rm TeV}$. Taking the uncertainties due to charm contribution and
CKM parameters in $\varepsilon_K$ into account the maximal shifts in $\varepsilon'/\varepsilon$
in both models amount to
$6\times 10^{-4}$, very similar to what we found in \cite{Buras:2015kwd}.
But NP effects in $B_s\to\mu^+\mu^-$ and $C_9$ are smaller relative
to the ones found in the latter paper by a factor of two.
This is seen in Fig.~\ref{CORR816}, where
we show correlations of $\Delta(\varepsilon'/\varepsilon)$ with $B_s\to \mu^+\mu^-$
(left panels) and with $ C_9^{\rm NP}$ (right panels) for M8 and M16 and
$M_{Z^\prime}=3\, {\rm TeV}$.
In M8 the rate for $B_s\to\mu^+\mu^-$ can be suppressed
by $10\%$ bringing the theory closer to the data in (\ref{LHCb2})
\cite{CMS:2014xfa}. Moreover, this happens
for the largest shift in $\varepsilon'/\varepsilon$. But the shift in $C_9$ is very small. In
M16 the pattern is opposite with only a very small NP effects in
$B_s\to\mu^+\mu^-$ and a shift of $-0.3$ in $C_9$ which brings the theory
closer to the data.
\boldmath
\subsection{$|V_{cb}|$ and $|V_{ub}|$ Dependence}
\unboldmath
It is well known that $\varepsilon_K$ and $\Delta M_{s,d}$ in the SM are sensitive functions of $|V_{cb}|$. Moreover, $\varepsilon_K$ and $S_{\psi K_S}$ depend sensitively on $|V_{ub}|$.
Setting
$|V_{cb}|$ and $|V_{ub}|$ to the values in (\ref{CKMfix}) we have necessarily constrained the allowed range of NP parameters that are consistent with the data on
$\Delta F=2$ observables. Changing $|V_{cb}|$ and $|V_{ub}|$ will necessarily modify
this range and will modify NP contributions to flavour observables even if they do not depend directly on $|V_{cb}|$ and $|V_{ub}|$.
A sophisticated analysis
which would include the uncertainties in both CKM elements from tree-level decays would wash out NP effects and
would not teach us much about the impact of $|V_{cb}|$ and $|V_{ub}|$ on our results.
Therefore,
we prefer to show how our results presented above are modified for
a different value of $|V_{cb}|$ that we choose to be lower so that instead of
(\ref{CKMfix}) we now use
\begin{equation}\label{CKMfix1}
|V_{ub}|=3.6\times 10^{-3}, \qquad |V_{cb}|=40.0 \times 10^{-3}, \qquad \gamma=70^\circ.
\end{equation}
We keep $|V_{ub}|$ and $\gamma$ unchanged as this will allow us to see the
role of $|V_{cb}|$ better. The dependence on $\gamma$ in the observables in question is weak. The dependence of $\Delta M_{s,d}$ on $|V_{ub}|$ is totally negligible. The inclusive value of $|V_{ub}|$ would compensate the decrease of $|V_{cb}|$
in $\varepsilon_K$ but would simultaneously have an impact on $S_{\psi K_S}$
shifting it in the ballpark of $0.80$ within the SM. While in the SM this
problem cannot be cured because of the absence of new CP-violating phases, in
331 models the presence of the phase $\delta_1$ in (\ref{vij}) allows to
satisfy the constraint on $S_{\psi K_S}$ in (\ref{oases13}). We will demonstrate it below.
This assures us that the tension between $\varepsilon_K$ and $S_{\psi K_S}$
for exclusive value of $|V_{cb}|$ present in the SM can be avoided within the
331 models. But as we would like to investigate the impact of the change in $|V_{cb}|$ on our results, we keep first $|V_{ub}|$ at its exclusive value. Moreover
there is some
kind of consensus in the community that in the case of $|V_{ub}|$ one can trust
more exclusive determinations of this parameter than the inclusive ones. This
is based on the fact that the exclusive determinations use formfactors from
lattice QCD, which on the one hand are already rather precise and on the other hand do not require
the assumptions like hadron duality necessary for the inclusive determination
of $|V_{ub}|$.
The
results of this exercise are shown in Figs.~\ref{MBAJBv4}-\ref{CORR816vcb4}.
Comparing Fig.~\ref{MBAJBv4} with Fig.~\ref{MBAJB} it is evident that 331 models
perform better for our nominal choice of CKM parameters in (\ref{CKMfix}) than
for a lower value of $|V_{cb}|$. This is seen in particular in the case of $\varepsilon_K$ for which the maximal values of $\varepsilon_K$ are by $10\%$ below the
data. But, as discussed above, increasing $|V_{ub}|$ towards its inclusive value and taking the uncertainties in the QCD corrections to the charm contribution in $\varepsilon_K$ into account one can bring the theory much closer to data without
violating the constraint on $S_{\psi K_S}$ in (\ref{oases13}). We demonstrate this in Fig.~\ref{epeVubH3TeV} where we use $|V_{ub}|=0.0042$. Indeed $\varepsilon_K$ is now in a perfect agreement with the data. The slight increase of the
maximal value of $\Delta(\varepsilon'/\varepsilon)$ in this case will be analyzed in more details below.
\begin{figure}[!tb]
\centering
\includegraphics[width = 0.47\textwidth]{epspvsepsKM8vcb4VubH.jpg}
\caption{\it $\Delta(\varepsilon'/\varepsilon)$ versus $\varepsilon_K$ for M8.
Red dot represents central SM value.
$M_{Z^\prime}=3\, {\rm TeV}$, $|V_{cb}|=0.040$ and $|V_{ub}|=0.0042$.}\label{epeVubH3TeV}
\end{figure}
Therefore we can claim that 331
models also in this case remove the tensions in question, which is not possible
within the SM. Interestingly, as we will see below for significantly
higher values of $M_{Z^\prime}$ the removal of tensions for this value of
$|V_{cb}|$ will be much easier. We refer to Section~4 in \cite{Buras:2015kwd} for the explanation of this behaviour.
Next Figs.~\ref{M8M16v4} and \ref{CORR816vcb4} should be compared with
Figs.~\ref{M8M16} and \ref{CORR816}, respectively. We observe:
\begin{itemize}
\item
The correlation between $\varepsilon'/\varepsilon$ and $\varepsilon_K$ in Fig.~\ref{M8M16v4}
has a very different shape than in Fig.~\ref{M8M16} but a shift of $\varepsilon'/\varepsilon$ of $(5-6)\times 10^{-4}$ is
possible. In fact this plot is similar to a corresponding plot in \cite{Buras:2015kwd} obtained with CKM parameters in (\ref{CKMfix}) but older hadronic matrix
elements. This similarity is easy to understand. The increase of non-perturbative parameters in (\ref{Kronfeld}) has been roughly compensated by the decrease
of $|V_{cb}|$.
\item
The size of NP effects in $B_s\to\mu^+\mu^-$ and $C_9$ is now larger than
for our nominal value of $|V_{cb}|$ and similar to the ones found in \cite{Buras:2015kwd}: suppression of the rate for $B_s\to\mu^+\mu^-$ by $20\%$ in the case of
M8 and a shift of $C_9$ by $-0.5$ in M16 are possible. But what is interesting is that the decreased value of $|V_{cb}|$ lowers also the SM result for the $B_s\to\mu^+\mu^-$ rate so that with the NP shift central values from CMS and LHCb
\cite{CMS:2014xfa}
\begin{equation}\label{LHCb2}
\overline{{B}}(B_{s}\to\mu^+\mu^-) = (2.8^{+0.7}_{-0.6}) \cdot 10^{-9},
\end{equation}
can be reached.
\end{itemize}
This value should be compared with central SM values
\begin{equation}
\overline{{B}}(B_{s}\to\mu^+\mu^-)_\text{SM} = 3.5 \cdot 10^{-9},\qquad
\overline{{B}}(B_{s}\to\mu^+\mu^-)_\text{SM} = 3.2 \cdot 10^{-9}
\end{equation}
for $|V_{cb}|=0.042$ and $|V_{cb}|=0.040$, respectively.
Thus within 331 models, on the whole, the results for $\Delta F=1$ for $|V_{cb}|=0.040$ appear more interesting than for $|V_{cb}|=0.042$. As we will see below
this is in particular the case for larger values of $M_{Z^\prime}$.
\begin{figure}[!tb]
\centering
\includegraphics[width = 0.47\textwidth]{epsvsVub.jpg}
\caption{\it Maximal values of $\Delta(\varepsilon'/\varepsilon)$ for $|V_{cb}|=0.040$ as
function of $|V_{ub}|$ for $M_{Z^\prime}=3\, {\rm TeV}$ and $M_{Z^\prime}=10\, {\rm TeV}$.}\label{MBAJB10TeV}
\end{figure}
Finally, we show in Fig.~\ref{MBAJB10TeV} the maximal value of $\Delta(\varepsilon'/\varepsilon)$ for $|V_{cb}|=0.040$ as a
function of $|V_{ub}|$. We observe that this value rises approximately linearly with increasing
$|V_{ub}|$ and for $M_{Z^\prime}=3\, {\rm TeV}$ and $|V_{ub}|=0.0044$ that is consistent with
the inclusive determinations could reach values as high as $\simeq 7.7\times 10^{-4}$.
This possibility should be kept in mind even if such high values of $|V_{ub}|$
seem rather unlikely as stated above. For $|V_{cb}|=0.042$ the effects of changing
$|V_{ub}|$ turn out to be smaller.
\begin{figure}[!tb]
\centering
\includegraphics[width = 0.47\textwidth]{DMsvsepsKM8M1610TeV.jpg}
\includegraphics[width = 0.47\textwidth]{DMdvsepsKM8M1610TeV.jpg}
\caption{\it $\Delta M_{s,d}$ vs. $\varepsilon_K$ in M8 and M16.
Red dots represent central SM values and black dots the central experimental values. $M_{Z^\prime}=10\, {\rm TeV}$, $|V_{cb}|=0.042$ and $|V_{ub}|=0.0036$.}\label{epsvsVub}
\end{figure}
\begin{figure}[!tb]
\centering
\includegraphics[width = 0.47\textwidth]{epspvsepsKM810TeV.jpg}
\includegraphics[width = 0.47\textwidth]{epspvsepsKM1610TeV.jpg}
\caption{ \it $\Delta(\varepsilon'/\varepsilon)$ versus $\varepsilon_K$ for M8 and M16.
Red dot represents central SM values.
$M_{Z^\prime}=10\, {\rm TeV}$, $|V_{cb}|=0.042$ and $|V_{ub}|=0.0036$.}\label{M810TeV}~\\[-2mm]\hrule
\end{figure}
\begin{figure}[!tb]
\centering
\includegraphics[width = 0.49\textwidth]{parM83TeV.jpg}
\includegraphics[width = 0.49\textwidth]{parM810TeV.jpg}
\caption{ \it Allowed values of $\sin(\delta_2-\delta_1)$ and
$\tilde s_{13} \tilde s_{23}$ for M8 and
$M_{Z^\prime}=3\, {\rm TeV}$ (left panel) and $M_{Z^\prime}=10\, {\rm TeV}$ (right panel).
The {\it old} ranges
are the ones from \cite{Buras:2015kwd} and the {\it new} ones found here.
$|V_{cb}|=0.042$ and $|V_{ub}|=0.0036$.
}\label{par310}~\\[-2mm]\hrule
\end{figure}
\boldmath
\subsection{$Z^\prime$ Outside the Reach of the LHC}
\unboldmath
\boldmath
\subsubsection{$|V_{ub}|=0.042$}
\unboldmath
We will next investigate what happens when higher values of $M_{Z^\prime}$,
outside the reach of the LHC together with CKM parameters in
(\ref{CKMfix}), are considered. As an example we set
$M_{Z^\prime}=10\, {\rm TeV}$. In Fig.~\ref{epsvsVub} we demonstrate that also in this case
the tension between $\Delta M_{d}$ and $\varepsilon_K$ can be easily removed.
In the case of $\Delta M_s$ 331 models perform much better than
the SM represented by the red point so that the inclusion of the uncertainty
in $F_{B_s}\sqrt{\hat B_{B_s}}$ in (\ref{Kronfeld}), can bring easily 331
models
to agree with data which is not possible within the SM.
The
question then arises what happens with NP effects in other observables for
such high values $M_{Z^\prime}$.
On the basis of our discussion in Section~4 in \cite{Buras:2015kwd}
we expect the effects in $B_s\to\mu^+\mu^-$ and $C_9$ to be smaller
than for $M_{Z^\prime}=3\, {\rm TeV}$, which can be confirmed as seen in Tables~\ref{panoramaM8} and \ref{panoramaM16}. On the other
hand $\varepsilon'/\varepsilon$ was found in \cite{Buras:2015kwd} to be significantly
enhanced for $M_{Z^\prime}=10\, {\rm TeV}$ as can be seen in Fig.~6 of that paper.
Moreover through renormalization group effects it could be even enhanced
slightly more than for $M_{Z^\prime}=3\, {\rm TeV}$. However, as seen in Fig.~\ref{M810TeV}, with new lattice constraints, this is no longer the case and the
maximal allowed shifts in $\varepsilon'/\varepsilon$ are below $1\times 10^{-4}$, far too small to
remove $\varepsilon'/\varepsilon$ anomaly.
\begin{figure}[!tb]
\centering
\includegraphics[width = 0.47\textwidth]{DMsvsepsKM8M1610TeVvcb4.jpg}
\includegraphics[width = 0.47\textwidth]{DMdvsepsKM8M1610TeVvcb4.jpg}
\caption{\it $\Delta M_{s,d}$ vs. $\varepsilon_K$ in M8 and M16.
Red dots represent central SM values and black dots the central experimental values. $M_{Z^\prime}=10\, {\rm TeV}$, $|V_{cb}|=0.040$ and $|V_{ub}|=0.0036$.}\label{MBAJB10TeV4}
\end{figure}
\begin{figure}[!tb]
\centering
\includegraphics[width = 0.47\textwidth]{epspvsepsKM810TeVvcb4.jpg}
\includegraphics[width = 0.47\textwidth]{epspvsepsKM1610TeVvcb4.jpg}
\caption{ \it $\Delta(\varepsilon'/\varepsilon)$ versus $\varepsilon_K$ for M8. Red dot represents central SM values.
$M_{Z^\prime}=10\, {\rm TeV}$, $|V_{cb}|=0.040$ and $|V_{ub}|=0.0036$. }\label{M810TeV4}~\\[-2mm]\hrule
\end{figure}
\begin{figure}[!tb]
\centering
\includegraphics[width = 0.47\textwidth]{parM8BF2vcb4.jpg}
\includegraphics[width = 0.47\textwidth]{parM810TeVBF2vcb4.jpg}
\caption{ \it Allowed values of $\sin(\delta_2-\delta_1)$ and
$\tilde s_{13} \tilde s_{23}$ for M8 and
$M_{Z^\prime}=3\, {\rm TeV}$ (left panel) and $M_{Z^\prime}=10\, {\rm TeV}$ (right panel).
$|V_{cb}|=0.040$ and $|V_{ub}|=0.0036$.
}\label{par3104}~\\[-2mm]\hrule
\end{figure}
\begin{table*}[t]
\centering \caption{Summary of results for M8. $|V_{ub}|=0.0036$.}\label{panoramaM8}
\begin{tabular}{|c | c || c| c |c| }\hline
$M_{Z^\prime}$ & $|V_{cb}|$ & $|\Delta(\varepsilon'/\varepsilon)|_{max} (10^{-4})$ & $\overline{B}(B_{s}\to \mu^+\mu^-)_{min} (10^{-9})$& ${\rm Re}(C_9^{NP})_{min}$ \\
\hline \hline
{$3$ TeV}& $0.042$ & $6.2$ & $3.21$ & $-0.09$
\\ \cline{2-5}
& $0.040$ &$5.9$ & $2.69$ & $-0.16$
\\ \hline {$10$ TeV}& $0.042$ & $0.98$ & $ 3.45$& $-0.02$
\\ \cline{2-5}
& $0.040$ & $ 6.94$ & $3.02$ & $-0.05$
\\ \hline
\end{tabular}
\end{table*}
\begin{table*}[t]
\centering \caption{Summary of results for M16. $|V_{ub}|=0.0036$.}\label{panoramaM16}
\begin{tabular}{|c | c || c| c |c| }\hline
$M_{Z^\prime}$ & $|V_{cb}|$ & $|\Delta(\varepsilon'/\varepsilon)|_{max} (10^{-4})$ & $\overline{B}(B_{s}\to \mu^+\mu^-)_{min} (10^{-9})$& ${\rm Re}(C_9^{NP})_{min}$ \\
\hline \hline
{$3$ TeV}& $0.042$ & $5.55$ & $3.40$ & $-0.28$
\\ \cline{2-5}
& $0.040$ &$5.17$ & $3.01$ & $-0.49$
\\ \hline {$10$ TeV}& $0.042$ & $0.87$ & $ 3.48$& $-0.05$
\\ \cline{2-5}
& $0.040$ & $5.95 $ & $3.13$ & $-0.15$
\\ \hline
\end{tabular}
\end{table*}
In order to understand this drastic change we recall the general formula for
$\varepsilon'/\varepsilon$ for arbitrary $M_{Z^\prime}$ in 331 models
in the absence of $Z-Z^\prime$ mixing which is irrelevant in M8, M9 and M16
\cite{Buras:2015kwd}
\begin{equation}\label{eprimeZPlarge}
\left(\frac{\varepsilon'}{\varepsilon}\right)_{Z^\prime}= \pm r_{\varepsilon^\prime} 1.1 \,[\beta f(\beta)]\,
\tilde s_{13} \tilde s_{23} \sin(\delta_2-\delta_1)
\left[\frac{B_8^{(3/2)}}{0.76}\right]\left[\frac{3\, {\rm TeV}}{M_{Z^\prime}}\right]^2
\end{equation}
with the upper sign for $F_1$ and the lower for $F_2$. $r_{\varepsilon^\prime}$,
$\beta$ and $f(\beta)$ are $\mathcal{O}(1)$ so that only the remaining factors
are of interest to us. Now as discussed in \cite{Buras:2015kwd} with increasing
$ M_{Z^\prime}$ larger values of $\tilde s_{13}$ and $\tilde s_{23}$ are allowed
by constraints from $B^0_{s,d}-\bar B^0_{s,d}$ mixing
\begin{equation}
\tilde s_{13}^{\rm max}\propto M_{Z^\prime}, \qquad \tilde s_{23}^{\rm max}\propto M_{Z^\prime}\,, \qquad (\Delta M_{s,d}~~{\rm constraints})\,.
\end{equation}
In this manner the $M_{Z^\prime}$ suppression in (\ref{eprimeZPlarge}) is
compensated and the fate of $\varepsilon'/\varepsilon$ depends on the allowed values of $\sin(\delta_2-\delta_1)$ that follow not only from $\Delta M_{s,d}$ constraints but also
from
$S_{\psi K_S}$ and $S_{\psi\phi}$ constraints. Our analysis shows that whereas
in our previous analysis values of $\sin(\delta_2-\delta_1)=1$ were allowed
this is no longer the case after new lattice results and maximal values of
$\sin(\delta_2-\delta_1)$ are significantly below unity. While this suppression
appears to be roughly compensated by the increase of the product $\tilde s_{13} \tilde s_{23}$\footnote{These parameters must be larger in order to bring down
the values of $\Delta M_{s,d}$ to agree with data.} for $M_{Z^\prime}=3\, {\rm TeV}$,
this is no longer the case for $M_{Z^\prime}=10\, {\rm TeV}$ and $\varepsilon'/\varepsilon$ is strongly suppressed.
This feature is clearly seen in Fig.~\ref{par310}, where the {\it old} ranges
are the ones from \cite{Buras:2015kwd} and the {\it new} ones found here.
\boldmath
\subsubsection{$|V_{cb}|=0.040$}
\unboldmath
We next consider the CKM input in (\ref{CKMfix1}). The results for
$M_{Z^\prime}=10\, {\rm TeV}$ are shown in Figs.~\ref{MBAJB10TeV4}--\ref{par3104}.
We observe:
\begin{itemize}
\item
The tensions between $\Delta M_{s,d}$ and $\varepsilon_K$ can be much easier
removed than for $M_{Z^\prime}=3\, {\rm TeV}$ because of the increased NP effects in
$\varepsilon_K$. Comparing Fig.~\ref{MBAJB10TeV4} with Fig.~\ref{MBAJB10TeV}
we also observe that the agreement with data is better for $|V_{cb}|=0.040$.
\item
The upward shift in $\varepsilon'/\varepsilon$ up to $(6-7)\times 10^{-4}$ in now possible so that
$\varepsilon'/\varepsilon$ with $|V_{cb}|=0.040$ can probe much higher mass scales than
it is possible for $|V_{cb}|=0.042$ because of other constraints.
\item
The plots in Fig.~\ref{par3104} when compared with those in Fig.~\ref{par310}
explain why the NP effects in $\varepsilon'/\varepsilon$ for $|V_{cb}|=0.040$ have a different structure than for $|V_{cb}|=0.042$. $\sin(\delta_2-\delta_1)$ can for $|V_{cb}|=0.040$
reach unity even for $M_{Z^\prime}=10\, {\rm TeV}$, while this is not possible for
$|V_{cb}|=0.042$.
\end{itemize}
As seen in Tables~\ref{panoramaM8} and \ref{panoramaM16} NP effects in $B_s\to\mu^+\mu^-$ and $C_9$ are suppressed for
$M_{Z^\prime}=10\, {\rm TeV}$ but not as much as for $|V_{cb}|=0.042$.
All our results for M8 for different values of $|V_{cb}|$ and $M_{Z^\prime}$ are
summarized in Table~\ref{panoramaM8}. Very similar results are obtained for
M9. The corresponding results for M16 are summarized in Table~\ref{panoramaM16}.
These tables show
again how important is the precise determination of $|V_{cb}|$ in tree-level
decays.
Finally the red curve in Fig.~\ref{MBAJB10TeV} demonstrates that $\Delta(\varepsilon'/\varepsilon)$ for $|V_{cb}|=0.040$ and $M_{Z^\prime}=10\, {\rm TeV}$ can for large values of $|V_{ub}|$ reach values in the ballpark of
$\simeq 8.8 \times 10^{-4}$. This increase relative to $M_{Z^\prime}=3\, {\rm TeV}$ is related to
renormalization group effects as discussed in detail in \cite{Buras:2015kwd}.
\section{The Simplest 331 Model: M0}\label{sec:3}
\subsection{Preliminaries}
We will next look at the simplest 331 model recently proposed in \cite{Hue:2015mna} in which $\beta=0$. We will denote it by M0. Even if this model fails to remove most of the anomalies
in question, its simplicity invites us to have a closer look at its flavour
structure. We will list $Z^\prime$ and $Z$ couplings in this model and present formulae for
$C_9^\text{NP}$ and $C_{10}^\text{NP}$ as well as $\varepsilon'/\varepsilon$.
The expressions for $\Delta F=2$ processes and for $B_s\to\mu^+\mu^-$
as functions of the couplings listed below can be found in
\cite{Buras:2014yna,Buras:2012dp,Buras:2013dea} and we will not repeat them
here. One only has to set $\beta=0$ in that formulae.
In this manner, in contrast to \cite{Hue:2015mna}, we take $Z-Z^\prime$ mixing in all observables automatically into account.
\boldmath
\subsection{$Z^\prime$ Couplings}
\unboldmath
Setting $\beta=0$ in (17) of \cite{Buras:2013dea} we find for
quark couplings
{\allowdisplaybreaks
\begin{subequations}
\begin{align}
\Delta_L^{ij}(Z') &=\frac{g_2}{\sqrt{3}} v_{3i}^*v_{3j}\,,\\
\Delta_L^{ji}(Z') &=\left[ \Delta_L^{ij}(Z')\right]^\star\,, \\
\Delta_L^{d\bar d}(Z') & = \Delta_L^{u\bar u}(Z')=\Delta_V^{d\bar d}(Z')=\Delta_V^{u\bar u}(Z')=-\frac{g_2}{2\sqrt{3}}
\,,\\
\Delta_A^{d\bar d}(Z') & = \Delta_A^{u\bar u}(Z')= \frac{g_2}{2\sqrt{3}}\\
\label{RH}
\Delta_R^{d\bar d}(Z')& = \Delta_R^{u\bar u}(Z')=0\,.
\end{align}
\end{subequations}}%
with $v_{ij}$ given in (\ref{vij}).
The diagonal couplings given here are valid for the first two generations of quarks
neglecting very small additional contributions \cite{Buras:2012dp}.
For the third generation there is an additional term which can be found in
(63) of \cite{Buras:2012dp}. It is irrelevant for our analysis of FCNCs but
plays a role in electroweak precision tests \cite{Buras:2014yna}. For $\beta=0$ the diagonal $b$ quark couplings differ from $d$ and $s$
couplings only by sign:
\begin{equation}
\Delta_L^{b\bar b}(Z') =\Delta_V^{b\bar b}(Z')=\frac{g_2}{2\sqrt{3}}, \qquad
\Delta_A^{b\bar b}(Z')=-\frac{g_2}{2\sqrt{3}}\,.
\end{equation}
Setting $\beta=0$ in (18) of \cite{Buras:2013dea}
we find for lepton couplings
{\allowdisplaybreaks
\begin{subequations}
\begin{align}
\Delta_L^{\mu\bar\mu}(Z')&=\Delta_L^{\nu\bar\nu}(Z')=\Delta_V^{\mu\bar\mu}(Z')=\frac{g_2}{2\sqrt{3}}\,,\\
\Delta_A^{\mu\bar\mu}(Z') &= \Delta_A^{\nu\bar\nu}(Z')= -\frac{g_2}{2\sqrt{3}}\,,\\
\Delta_R^{\nu\bar\nu}(Z')&=\Delta_R^{\mu\bar\mu}(Z')= 0
\end{align}
\end{subequations}}%
where we have defined
\begin{align}\label{DeltasVA}
\begin{split}
&\Delta_V^{\mu\bar\mu}(Z')= \Delta_R^{\mu\bar\mu}(Z')+\Delta_L^{\mu\bar\mu}(Z'),\\
&\Delta_A^{\mu\bar\mu}(Z')= \Delta_R^{\mu\bar\mu}(Z')-\Delta_L^{\mu\bar\mu}(Z').
\end{split}
\end{align}
These definitions also apply to other leptons and quarks. All these couplings
are evaluated for $\mu=M_{Z^\prime}$ with $g_2=0.633$ for $M_{Z^\prime}=3\, {\rm TeV}$.
\boldmath
\subsection{$Z$ Couplings}
\unboldmath
The flavour non-diagonal couplings to quarks are generated from $Z^\prime$
couplings through $Z-Z^\prime$ mixing
\begin{equation}
\Delta^{ij}_L(Z)=\sin\xi \, \Delta^{ij}_L(Z^\prime),
\end{equation}
where using the general formula (10) in \cite{Buras:2014yna} we find for $\beta=0$
\begin{equation}\label{sxi}
\sin\xi=a\frac{c_W}{\sqrt{3}} \left[\frac{M_Z^2}{M_{Z^\prime}^2}\right]= a\, 4.68\times 10^{-4}\left[\frac{3\, {\rm TeV}}{M_{Z^\prime}}\right]^2 \,.
\end{equation}
Here
\begin{equation}\label{basica}
-1\le a=\frac{1-\tan^2\bar\beta}{1+\tan^2\bar\beta}\le 1, \qquad \tan\bar\beta=\frac{v_\rho}{v_\eta}
\end{equation}
with the scalar triplets $\rho$ and
$\eta$ responsible for the masses of up-quarks and down-quarks, respectively.
Thus for $\tan\bar\beta=1$ the parameter $a=0$ and the $Z-Z^\prime$ mixing
vanish in agreement with \cite{Hue:2015mna}.
On the other hand in the large $\tan\bar\beta$ limit we find $a=-1$ and
in the low $\tan\bar\beta$ limit one has $a=1$.
The flavour diagonal $Z$ couplings are the SM ones and collected in \cite{Buras:2013dea}. We evaluate them with
$g_2=0.652$ and $\sin^2\theta_W=0.23116$ as valid at $\mu=M_Z$.
\boldmath
\subsection{$C_9^\text{NP}$ and $C_{10}^\text{NP}$}\label{RARE}
\unboldmath
The corrections from NP to the Wilson coefficients $C_9$ and $C_{10}$ that weight the semileptonic operators in the effective hamiltonian relevant
for $b\to s\mu^+\mu^-$ transitions
are given as follows
\begin{align}
\sin^2\theta_W C^{\rm NP}_9 &=-\frac{1}{g_{\text{SM}}^2M_{Z^\prime}^2}
\frac{\Delta_L^{sb}(Z')\Delta_V^{\mu\bar\mu}(Z')} {V_{ts}^* V_{tb}}(1+R^V_{\mu\mu}) ,\label{C9}\\
\sin^2\theta_W C^{\rm NP}_{10} &= -\frac{1}{g_{\text{SM}}^2M_{Z^\prime}^2}
\frac{\Delta_L^{sb}(Z')\Delta_A^{\mu\bar\mu}(Z')}{V_{ts}^* V_{tb}}(1+R^A_{\mu\mu})\label{C10}.
\end{align}
As seen in these equations $C^{\rm NP}_9$ involves leptonic vector coupling
of $Z^\prime$ while $C^{\rm NP}_{10}$ the axial-vector one. $C^{\rm NP}_9$ is
crucial for $B_d\to K^*\mu^+\mu^-$, $C^{\rm NP}_{10}$ for $B_s\to\mu^+\mu⁻$ and
both coefficients are relevant for $B_d\to K\mu^+\mu^-$.
Here
\begin{equation}\label{gsm}
g_{\text{SM}}^2=
4 \frac{M_W^2 G_F^2}{2 \pi^2} = 1.78137\times 10^{-7} \, {\rm GeV}^{-2}\,,
\end{equation}
with $G_{F}$ being the Fermi constant. The terms $R^V_{\mu\mu}$ and
$R^A_{\mu\mu}$ are generated by $Z-Z^\prime$ mixing and are given as follows
\begin{equation}\label{DF1}
R^V_{\mu\mu}=\sin\xi \left[\frac{M_{Z\prime}^2}{M_{Z}^2}\right] \left[\frac{\Delta^{\mu\mu}_V(Z)}{\Delta^{\mu\mu}_V(Z^\prime)}\right]\,,
\end{equation}
\begin{equation}\label{DF2}
R^A_{\mu\mu}=\sin\xi \left[\frac{M_{Z\prime}^2}{M_{Z}^2}\right] \left[\frac{\Delta^{\mu\mu}_A(Z)}{\Delta^{\mu\mu}_A(Z^\prime)}\right]\,,
\end{equation}
For $\beta=0$ we find then
\begin{align}
\sin^2\theta_W C^{\rm NP}_9 &=-\frac{1}{g_{\text{SM}}^2M_{Z^\prime}^2}
\frac{g_2^2(M_{Z^\prime})}{6}
\left[\frac{v_{32}^*v_{33}} {V_{ts}^* V_{tb}}\right](1+R^V_{\mu\mu}) ,\label{C90}\\
\sin^2\theta_W C^{\rm NP}_{10} &= +\frac{1}{g_{\text{SM}}^2M_{Z^\prime}^2}\frac{g_2^2(M_{Z^\prime})}{6} \left[\frac{v_{32}^*v_{33}} {V_{ts}^* V_{tb}}\right](1+R^A_{\mu\mu})\label{C100}
\end{align}
with
\begin{equation}
R^V_{\mu\mu}=-0.08\, a, \qquad
R^A_{\mu\mu}=-1.02 \, a\,
\end{equation}
that do not depend on $M_{Z^\prime}$ except for logarithmic $M_{Z^\prime}$ dependence of $g_2$. The numerical factors above correspond to $M_{Z^\prime}=3\, {\rm TeV}$.
We observe then that in the absence of $Z-Z^\prime$ mixing ($a=0$), independently of the parameters $v_{ij}$, the following phenomenologically successful
relation
\begin{equation}\label{R1a}
C_9^\text{NP}=-C_{10}^\text{NP}, \qquad (a=0)
\end{equation}
holds. This should be contrasted with models M8, M9 and M16
for which we found \cite{Buras:2015kwd}
\begin{equation}\label{M8M9}
C_9^{\rm NP}=0.49\, C_{10}^{\rm NP}\quad ({\rm M8})\,, \qquad
C_9^{\rm NP}=0.42\, C_{10}^{\rm NP}\quad ({\rm M9})\,.
\end{equation}
The result in (\ref{R1a}) differs also from
\begin{equation}\label{M16}
C_9^{\rm NP}=-4.59\, C_{10}^{\rm NP}\quad ({\rm M16})
\end{equation}
which is close to one of the favourite solutions in which NP resides dominantly in the coefficient $C_9$.
Thus already on the basis of $B$ physics observables we should be able to
distinguish between the models M0, (M8,M9) and M16.
However in the presence of $Z-Z^\prime$ mixing the relation (\ref{R1a}) does not
hold. While this effect is small in $C_9^{\rm NP}$, it can be large in $C_{10}^{\rm NP}$, in particular for $a=1$, when $C_{10}^{\rm NP}$ becomes very small and the suppression
of the rate for $B_s\to\mu^+\mu^-$ is absent. More interesting is then the
case of $a\approx -1$, corresponding to large $\tan\bar\beta$, as then the simultaneous suppressions of $C_9$ through $C_9^{\rm NP}$
and of $B_s\to\mu^+\mu^-$ rate through $C_{10}^{\rm NP}$ are stronger. We find
then
\begin{equation}\label{R2a}
C_9^\text{NP}\approx -0.5\, C_{10}^\text{NP}, \qquad (a\approx -1),
\end{equation}
that on a qualitative level is still a better description of the data than the results in (\ref{M8M9}) and (\ref{M16}). But the crucial question is whether
the values of both coefficients are sufficiently large when all constraints are
taken into account. Before answering this question let us make a closer look
at $\varepsilon'/\varepsilon$ in this model.
\boldmath
\subsection{$\varepsilon'/\varepsilon$}
\unboldmath
\subsubsection{Preliminaries}
The analyses of $\varepsilon'/\varepsilon$ in 331 models with $\beta\not=0$ have been
presented by us in \cite{Buras:2014yna,Buras:2015kwd}. We want to generalize
them to the case $\beta=0$. Generally in 331 models we have
\begin{equation}\label{total}
\left(\frac{\varepsilon'}{\varepsilon}\right)_{331}=\left(\frac{\varepsilon'}{\varepsilon}\right)_{\rm SM}+\left(\frac{\varepsilon'}{\varepsilon}\right)_{Z}+
\left(\frac{\varepsilon'}{\varepsilon}\right)_{Z^\prime}\equiv \left(\frac{\varepsilon'}{\varepsilon}\right)_{\rm SM}+\Delta(\varepsilon'/\varepsilon)
\,
\end{equation}
with the $\Delta(\varepsilon'/\varepsilon)$ resulting from tree-level $Z^\prime$ and $Z$ exchanges.
Now, as demonstrated by us in \cite{Buras:2014yna}, the shift $\Delta(\varepsilon'/\varepsilon)$
is governed in 331 models
by the electroweak $(V-A)\times (V+A)$ penguin operator
\begin{equation}\label{O4}
Q_8 = \frac{3}{2}\,(\bar s_{\alpha} d_{\beta})_{V-A}\!\!\sum_{q=u,d,s,c,b,t}
e_q\,(\bar q_{\beta} q_{\alpha})_{V+A} \,
\end{equation}
with only small contributions from other operators. This result applies to both
$Z$ and $Z^\prime$ contributions with the latter ones significantly more important as demonstrated in \cite{Buras:2014yna}. Here we would like to point out
that this pattern is not valid for $\beta=0$.
Indeed as seen in (41) of \cite{Buras:2014yna} the important coefficient
$C_7(M_{Z^\prime})$ generated by tree-level $Z^\prime$ exchange for $\beta\not=0$ vanishes for $\beta=0$ and consequently, $Q_8$ operator cannot
be generated from $Q_7$ operator by renormalization group effects. Contributions of other operators are very small
so that $Z^\prime$ contributions to $\varepsilon'/\varepsilon$ can be neglected. This is directly
related to the fact, as seen in (\ref{RH}), that the diagonal right-handed couplings of $Z^\prime$ to
quarks vanish for $\beta=0$. But for $Z$ such couplings are present implying
that tree-level $Z$ exchanges can provide a shift in $\varepsilon'/\varepsilon$.
\boldmath
\subsubsection{$Z$ Contribution}
\unboldmath
The inclusion of this contribution is straightforward as the only thing to be done is to calculate the shifts from NP
in the functions $X$, $Y$ and $Z$ that enter the SM model contribution to $\varepsilon'/\varepsilon$. One finds then \cite{Buras:2014yna}
\begin{equation}\label{DXYZ}
\Delta X=\Delta Y =\Delta Z= \sin\xi \, c_W\frac{8\pi^2}{g_2^3}\frac{{\rm Im}\Delta_L^{sd}(Z^\prime)}{{\rm Im}\lambda_t}
\end{equation}
where $g_2=0.652$ and $\lambda_t=V_{td}V^*_{ts}$. Replacing then the SM functions
$X_0(x_t)$, $Y_0(x_t)$ and $Z_0(x_t)$ by
\begin{equation}
X=X_0(x_t)+\Delta X, \qquad Y=Y_0(x_t)+\Delta Y, \qquad Z=Z_0(x_t)+\Delta Z
\end{equation}
in the phenomenological formula formula (90) for $\varepsilon'/\varepsilon$ in
\cite{Buras:2015yba} allows to take automatically the first two
contributions in (\ref{total}) in 331 models into account.
Inserting $\sin\xi$ in (\ref{sxi}) and the $Z^\prime$ coupling into (\ref{DXYZ}) we obtain for $\beta=0$
\begin{equation}\label{DXYZ0}
\Delta X=\Delta Y =\Delta Z= 47.6\, a\, \left[\frac{M_Z^2}{M_{Z^\prime}^2}\right] \frac{{\rm Im}(v_{32}^*v_{31})}{{\rm Im}\lambda_t}.
\end{equation}
Evidently for $a=0$, as done in \cite{Hue:2015mna}, NP contributions to
$\varepsilon'/\varepsilon$ from $Z$ exchanges vanish and as the ones from $Z^\prime$ can be
neglected, $\varepsilon'/\varepsilon$ is full governed by the SM contribution. This appears
presently problematic
in view of the findings in \cite{Bai:2015nea,Buras:2015yba,Buras:2015xba, Buras:2015jaq} that a significant upwards shift $\varepsilon'/\varepsilon$ of at least $5\times 10^{-4}$
is required to bring the theory to agree with the data from
from NA48 \cite{Batley:2002gn} and KTeV
\cite{AlaviHarati:2002ye,Abouzaid:2010ny} collaborations.
The question then arises whether including $Z-Z^\prime$ mixing
we can obtain the required positive shift in $\varepsilon'/\varepsilon$. But, as seen in
(\ref{R2a}), in order to preserve
at least partly the pattern in (\ref{R1a}) we are interested in
\begin{equation}\label{arange}
-1\le a < 0, \qquad \tan\bar\beta > 1\,.
\end{equation}
In order to answer this question
we insert
(\ref{DXYZ0}) into (90) in Appendix B of \cite{Buras:2015yba}
to obtain first
\begin{equation}
\Delta(\varepsilon'/\varepsilon)=\Delta(\varepsilon'/\varepsilon)_Z= 47.6 \, a\left[\frac{M_Z^2}{M_{Z^\prime}^2}\right][P_X+P_Y+P_Z] {\rm Im}(v_{32}^*v_{31}).
\end{equation}
From Table 5 in \cite{Buras:2015yba} we find then for the central value of
$\alpha_s(M_Z)$:
\begin{equation}
P_X+P_Y+P_Z= 1.52+0.12\, R_6-13.65\, R_8
\end{equation}
where
\begin{equation}\label{RS}
R_6\equiv B_6^{(1/2)}\left[ \frac{114.54\, {\rm MeV}}{m_s(m_c)+m_d(m_c)} \right]^2,\qquad
R_8\equiv B_8^{(3/2)}\left[ \frac{114.54\, {\rm MeV}}{m_s(m_c)+m_d(m_c)} \right]^2.
\end{equation}
Now the results from the RBC-UKQCD collaboration imply the following values
for $B_6^{(1/2)}$ and $B_8^{(3/2)}$ \cite{Buras:2015yba,Buras:2015qea}
\begin{equation}\label{Lbsi}
B_6^{(1/2)}=0.57\pm 0.19\,, \qquad B_8^{(3/2)}= 0.76\pm 0.05\,, \qquad (\mbox{RBC-UKQCD})
\end{equation}
that are compatible with the bounds from large $N$ approach
\cite{Buras:2015xba}
\begin{equation}\label{NBOUND}
B_6^{(1/2)}\le B_8^{(3/2)} < 1 \, \qquad (\mbox{\rm large-}N).
\end{equation}
Using then the results in (\ref{Lbsi})
we find
\begin{equation}
P_X+P_Y+P_Z= -8.78\pm 0.68
\end{equation}
and finally
\begin{equation}
\Delta(\varepsilon'/\varepsilon)=-a\, (0.39\pm 0.03)\left[\frac{3\, {\rm TeV}}{M_{Z^\prime}}\right]^2 {\rm Im}(v_{32}^*v_{31}).
\end{equation}
The question then arises whether for $a$ in the range (\ref{arange}) one
can get sufficient shift in $\varepsilon'/\varepsilon$ while satisfying other constraints.
In particular the ones from $\Delta F=2$ transitions, where as seen in the
previous section the SM experiences tensions in
its predictions for $\Delta M_{s,d}$ and $\varepsilon_K$.
\begin{figure}[!tb]
\centering
\includegraphics[width = 0.47\textwidth]{DMsvsepsKM0.jpg}
\includegraphics[width = 0.47\textwidth]{DMdvsepsKM0.jpg}
\caption{\it $\Delta M_{s,d}$ vs. $\varepsilon_K$ in M0.
Red dots represent central SM values and black dots the central experimental values. $M_{Z^\prime}=3\, {\rm TeV}$ and $|V_{cb}|=0.042$.}\label{MBAJB0}
\end{figure}
\begin{figure}[!tb]
\centering
\includegraphics[width = 0.60\textwidth]{epspvsepsKM0.jpg}
\caption{ \it $\Delta(\varepsilon'/\varepsilon)$ versus $\varepsilon_K$ for M0
for several values of the $Z-Z^\prime$ mixing parameter $a$. $M_{Z^\prime}=3\, {\rm TeV}$ and $|V_{cb}|=0.042$.}\label{M0}~\\[-2mm]\hrule
\end{figure}
In Fig.~\ref{MBAJB0} we show $\Delta M_{s,d}$ vs. $\varepsilon_K$ in M0. Red dots represent central SM values and black dots the central experimental
values. We observe that
the tensions between $\Delta M_{s,d}$ vs. $\varepsilon_K$ present in the
SM can be removed in the M0 model. But as seen in Fig.~\ref{M0} the shift in
$\varepsilon'/\varepsilon$ can be at most $1.1\times 10^{-4}$ which is far too small to be able
to remove $\varepsilon'/\varepsilon$ anomaly. Moreover, this maximal shift can only be obtained
for the maximal $Z-Z^\prime$ mixing. We have checked using the expressions
in \cite{Buras:2014yna} that then the fit to EWPO is significantly
worse than the one in the SM, whereas the three models analysed in the previous
section perform better in these tests than the SM \cite{Buras:2014yna}.
We do not show the results for $C_9$ and $B_s\to\mu^+\mu^-$ as NP effects
are significantly smaller than in M8 and M16. Thus even if M0 can remove
the tensions between $\Delta M_{s,d}$ and $\varepsilon_K$, it fails badly
in the case of other anomalies and therefore cannot compete with models
M8, M9 and M16, unless all anomalies disappear one day.
\section{Summary}\label{sec:4}
Motivated by the recently improved results from the Fermilab Lattice and MILC Collaborations on the hadronic matrix elements entering $\Delta M_{s,d}$ in
$B_{s,d}^0-\bar B_{s,d}^0$ mixing \cite{Bazavov:2016nty}
and the resulting increased tensions
between $\Delta M_{s,d}$ and $\varepsilon_K$ in the SM and
generally CMFV models \cite{Blanke:2016bhf}, we have performed a new analysis of 331 models.
In order to illustrate the sensitivity of the results to the modification of
hadronic parameters we have first used the CKM input of our previous analysis
in \cite{Buras:2015kwd} that is given in (\ref{CKMfix}). In addition, in order
to illustrate the sensitivity of the results to the value of $|V_{cb}|$
we have also performed the analysis with the CKM input in (\ref{CKMfix1}),
where $|V_{cb}|$ is lower than in (\ref{CKMfix}). We also investigated $|V_{ub}|$
dependence.
The most important results of our analysis, summarized in Tables~\ref{panoramaM8} and \ref{panoramaM16} are as follows:
\begin{itemize}
\item
The tensions between $\Delta M_{s,d}$ and $\varepsilon_K$
can be removed in the three 331 models with $\beta\not=0$ (M8, M9, M16) considered by us and this for both CKM inputs. This turns out to be also possible in the model with $\beta=0$ (M0) in the case of the input (\ref{CKMfix}) but it is
much harder when $|V_{cb}|$ is smaller as in (\ref{CKMfix1}).
\item
Models M8, M9 and M16 can provide a positive shift in $\varepsilon'/\varepsilon$ up to $6\times 10^{-4}$ for
$M_{Z^\prime}=3\, {\rm TeV}$ for both choices of $|V_{cb}|$ and $|V_{ub}|=0.0036$. But in contrast to our previous analysis this shift decreases fast with increasing $M_{Z^\prime}$ in the case
of $|V_{cb}|=0.042$ but its maximal values are practically unchanged for
$M_{Z^\prime}=10\, {\rm TeV}$ when $|V_{cb}|=0.040$ is used. We also find that for $|V_{cb}|=0.040$ and the
inclusive values of $|V_{ub}|$ the maximal shifts in $\varepsilon'/\varepsilon$ are increased to
$7.7 \times 10^{-4}$ and $8.8\times 10^{-4}$ for $M_{Z^\prime}=3\, {\rm TeV}$ and
$M_{Z^\prime}=10\, {\rm TeV}$, respectively.
In the
model M0, in which NP contribution to $\varepsilon'/\varepsilon$ is governed by $Z-Z^\prime$ mixing, NP effects are very small even for $M_{Z^\prime}=3\, {\rm TeV}$.
\item
In M8 and M9 the rate for $B_s\to\mu^+\mu^-$ can be reduced by at most $10\%$
and $20\%$ for $M_{Z^\prime}=3\, {\rm TeV}$ and $|V_{cb}|=0.042$ and $|V_{cb}|=0.040$,
respectively. This can bring the theory within $1~\sigma$ range of
the combined result from CMS and LHCb and for $|V_{cb}|=0.040$ one can even
reach the present central experimental value of this rate (\ref{LHCb2}). The maximal
shifts in $C_9$ are
$C_9^\text{NP}=-0.1$ and $C_9^\text{NP}=-0.2$ for these two $|V_{cb}|$ values,
respectively. This is only a moderate shift and these models
do not really help
in the case of $B_d\to K^*\mu^+\mu^-$ anomalies.
\item
In M16 the situation is opposite. The rate for $B_s\to\mu^+\mu^-$ can be reduced for $M_{Z^\prime}=3\, {\rm TeV}$ for the two $|V_{cb}|$ values by at most $3\%$ and $10\%$,
respectively but
with the corresponding values $C_9^\text{NP}=-0.3$ and $-0.5$ the anomaly in $B_d\to K^*\mu^+\mu^-$ can be partially reduced.
\item
In M0 NP effects in $\varepsilon'/\varepsilon$, $B_s\to\mu^+\mu^-$ and $B_d\to K^*\mu^+\mu^-$
are too small to be relevant. Therefore our analysis demonstrates that
in the presence of the anomalies discussed
by us the $U(1)_X$ factor in the gauge group
of 331 models cannot be $U(1)_Y$ .
\item
For higher values of $M_{Z^\prime}$ the effects in $B_s\to\mu^+\mu^-$ and $B_d\to K^*\mu^+\mu^-$ are much smaller.
We recall that NP effects in rare $K$ decays and $B\to K(K^*)\nu\bar\nu$ remain small in all 331 models even for $M_{Z^\prime}$ of few TeV.
\end{itemize}
Even if models M8, M9, M16 still compete with each other and M0
does not appear to be phenomenologically viable from present perspective,
our feeling is that eventually only models M8 and M9 have a chance to survive future tests if the anomalies discussed by us will be confirmed in the future. The point is that with present theoretical uncertainties in
$B_d\to K^*\mu^+\mu^-$ NP effects, even in M16, will be hardly seen in this
decay. The decay $B_s\to\mu^+\mu^-$ is much cleaner and in the flavour
precision era $15-20\%$ effects from NP, which are only possible in M8 and M9,
could in principle be distinguished from SM predictions but this would require
very large reduction in the experimental error on its rate.
Thus the main virtue
of 331 models as opposed to SM and CMFV models is the ability to remove
the tensions between $\Delta M_{s,d}$ and $\varepsilon_K$ and simultaneously
provide a significant upward shift in $\varepsilon'/\varepsilon$ but only for lower values of
$|V_{cb}|$ can this property remain for $M_{Z^\prime}$ beyond the LHC reach.
The possibility of a significant suppression of the rate for $B_s\to\mu^+\mu^-$ in M8 and M9 for $|V_{cb}|=0.040$ is also a welcome feature. In particular, as
it is correlated with the maximal shift in $\varepsilon'/\varepsilon$.
While the NP pattern in flavour physics identified by us in 331 models
is interesting, we should hope that eventually NP
contributions to flavour observables are larger than found in these models and
are also significant in rare $K$ decays which are theoretically very clean and in $B\to K(K^*)\nu\bar\nu$ which are cleaner than $B\to K(K^*)\mu^+\mu^-$ decays.
Most importantly the comparison of our results in \cite{Buras:2015kwd},
prior to the lattice results in \cite{Bazavov:2016nty},
with
the ones obtained using this new input demonstrates clearly how the shifts and increased accuracy
in non-perturbative parameters can have important impact on the size of NP
effects. Similar comment can be made in connection with $|V_{cb}|$.
\section*{Acknowledgements}
This research was done and financed in the context of the ERC Advanced Grant project ``FLAVOUR''(267104) and has also been carried out within the INFN project (Iniziativa Specifica) QFT-HEP. It was partially
supported by the DFG cluster
of excellence ``Origin and Structure of the Universe''.
\bibliographystyle{JHEP}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 4,633 |
Riencourtia is a genus of South American plants in the tribe Heliantheae within the family Asteraceae.
The generic name honors the birth name of Cassini's wife, Riencourt.
Species
Riencourtia latifolia Gardner - Venezuela, Guyana, Brazil
Riencourtia longifolia Baker - Brazil
Riencourtia oblongifolia Gardner - Brazil, Bolivia, Suriname
Riencourtia pedunculosa (Rich.) Pruski - Venezuela, Guyana, Fr Guiana, Suriname, Brazil
Riencourtia spiculifera Cass.
Riencourtia tenuifolia Gardner - Brazil
References
Heliantheae
Asteraceae genera
Flora of South America | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 5,144 |
La Resolució 1836 del Consell de Seguretat de les Nacions Unides fou adoptada per unanimitat el 29 de setembre de 2008. Després d'observar la situació a Libèria, el Consell va decidir ampliar el mandat de la Missió de les Nacions Unides a Libèria (UNMIL) per un any més, fins al 30 de setembre de 2009, tot ajustant el seu desplegament autoritzat. Alhora va ratificar la recomanació del Secretari General d'una reducció de 1.460 efectius militars i la racionalització dels quatre sectors actuals en dos, que es durien a terme entre octubre de 2008 i març de 2009.
També va aprovar l'augment immediat de 240 policials, per tal de proporcionar experiència en camps especialitzats, suport operatiu a la policia regular i reacció a incidents urgents. El president del Consell va donar la benvinguda al nou representant permanent de la Xina, que es va comprometre amb altres membres i esperava amb interès la presidència de la Xina el proper mes.
El Consell va demanar al Secretari General que continués supervisant el progrés en els punts de referència de construcció de pau detallats en els seus darrers informes. Sobre la base d'aquest progrés, li va demanar que recomanés abans del 15 de febrer de 2009 altres ajustaments en el desplegament de la UNMIL, i incloure en el seu informe, en consulta amb el govern de Libèria, escenaris de llarg abast per a la retirada gradual de la tropes de la missió.
Referències
Vegeu també
Llista de resolucions del Consell de Seguretat de les Nacions Unides 1801 a la 1900 (2008 - 2009)
Enllaços externs
Text de la Resolució a UN.org
Resolucions del Consell de Seguretat de les Nacions Unides de 2008
Resolucions del Consell de Seguretat de les Nacions Unides sobre Libèria | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 4,694 |
Q: Remove black borders when converting GeoTiff to MBTile with gdal_translate I am using gdal_translate to convert a fairly large geoTiff to MBTiles.
As the geoTiff is reprojected, it is not square anymore in Google Mercator and the parts that are not covered appear black when I would like them to be transparent.
I think it could have something to do with the -a_nodata parameter and I have tried with "-a_nodata 0", but it did not change things.
I have tried with both PNG8 and PNG. It did not change things either.
Here is what my command looks like:
gdal_translate -a_nodata 0 -of MBTILES -co TILE_FORMAT=PNG8 France.tif france.mbtiles
Any idea what I am doing wrong?
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 5,173 |
ARA is dedicated to developing sustainable and scalable energy technologies. We continually push beyond the boundaries of typical research and development to provide real-world energy savings solutions for our clients. Our focus on expeditionary forces and austere locations has brought rapid and flexible energy security innovations to our service members in operations throughout the world.
To achieve success in this fast-paced and ever-changing technological environment, we employ a diverse set of highly experienced specialists that includes scientists, engineers, and technicians from the electrical, mechanical, chemical, and mathematical fields. These experts form a cohesive group that effectively evaluates clients' needs and develops immediate and cost effective solutions.
Our extensive skill set stems from a synergistic combination of experience in power system design, renewables, electrochemistry, fluid and thermal sciences, high-performance computing, database design, embedded control systems and instrumentation, design for manufacture, design of experiments, modeling and simulation, and developing turnkey applications. We are also able to draw on knowledge from ARA's diverse subject matter experts to provide across-the-board satisfaction to our clients.
We look forward to working with you to solve your toughest power and energy concerns.
ARA is developing energy efficient technologies to meet the 21st century energy vision. | {
"redpajama_set_name": "RedPajamaC4"
} | 5,427 |
pub mod builder;
pub mod context;
pub mod function;
pub mod module;
// pub mod structure;
// pub mod ty;
pub mod tools;
pub use self::builder::Builder;
pub use self::context::Context;
pub use self::module::Module;
pub use self::function::{Function, FunctionType};
// pub use self::structure::Struct;
// pub use self::ty::Type;
| {
"redpajama_set_name": "RedPajamaGithub"
} | 265 |
Thursday 1st February 1990
Friday 2nd February 1990
Saturday 3rd February 1990
Sunday 4th February 1990
Monday 5th February 1990
Tuesday 6th February 1990
Wednesday 7th February 1990
Thursday 8th February 1990
Friday 9th February 1990
Saturday 10th February 1990
Sunday 11th February 1990
Monday 12th February 1990
Tuesday 13th February 1990
Wednesday 14th February 1990
Thursday 15th February 1990
Friday 16th February 1990
Wednesday 21st February 1990
Thursday 22nd February 1990
Friday 23rd February 1990
Exceptional warm weather brought violent storms, torrents of rain, killed 14 in this country and 35 across Europe as flooding inundated some areas of Wales. I badly scalded my foot when working on my new boat in Norfolk and foot injury then dominated my month as it need constant hospital and doctor's examination and re-dressing. Daniel helped drive me around for appointments and I still made some engagements on my own and used rest periods between times to catch up with writing and administration.
All the children went back at school this month after Daniel's and Della's illnesses and Daniel was told he needed to buckle down to gain his grades for university entrance and Dan was preparing for his driving test and Debbie as her Kimbolton Entrance Exam. There were worries this month over Debbie's treatment at the Offord riding stables, which did not seem to welcome them anymore. We sought advice again about Della's urinary infection. Freda now had an idea of buying a P.O./General Stores in this area to be able to look after Mum.
Later this month, after my foot had mostly recovered, we saw our first family holiday aboard the newly named 'Paxton Princess' for which I had also used my lay-up time organising her equipment manuals and administration. We cruised locally and then down the Bure, via Acle, to stay in Great Yarmouth for our normal family seaside fun and then drove home via Mum's in Stanton to discuss future plans.
My political campaigning forged ahead with our new plans for a wider circulation of even better Liberal Democrat Focus newsletters, now half financed, outside election times, by advertising revenue. I was chairing campaigning meetings, helping colleagues from neighbouring areas, but still resisting offers to stand for parliament! There was still the unending round of Parish and District Council meetings , with Little Paxton now getting well-organised and, together with Labour opposition, the District Tories were being subjected to a very effective opposition with development and the costs and complications of Thatcher's Poll Tax being easy prey.
The world's exchanges looked very shaky again, which was the background to Abbey National put their mortgage interest rates up by nearly 1% to 15.5%, with most public professionals getting 9 to 11% increases but no decent money yet for the ambulancemen. There is controversy about the under-cover role of the security services in undermining former Prime Ministers, but Thatcher resisted pressure for a fuller enquiry. She was circumventing sanctions against South Africa against international wishes but the month's news event was of the release of Nelson Mandela in South Africa, with a possible end to the State of Emergency which I saw live on the T.V.
Gorbachev is trying to reform Russia by removing article 6 in the constitution that gives Communism supremacy and agreeing troop reductions in Vienna to make the Warsaw Pact becomes less of a threat to the West. A tanker chartered by British Petroleum has shed 320,000 gallons of oil off of a sensitive part of the Californian coastline. All 14 A320 Airbus aircraft in India have been grounded following safety fears when one crashed, killing 90 passengers
** PRESS "Read More" BELOW for the complete story **
Read more: February 1990
February 1990 daily entries | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 3,355 |
Q: Javascript best practice to use undefined as boolean evaluation? In this simple javascript code, method verify returns a string if conditions do not match, else undefined. I am using undefined value in 'if' clause. Is this common / acceptable javascript programming practice?
if (verify(text)) {
alert(text + " is not bar");
}
function verify(foo) {
if (foo + "" != "bar") return "foo is not same as bar";
}
A: Yes, since you know the verification outcomes will be either truthy (a non-empty string) or falsy (undefined), it's perfectly fine to use it like that.
Here's a list of truthy and falsy values in JS, you might find them handy: http://www.sitepoint.com/javascript-truthy-falsy/
A: It is a common practise but in my experience its better to be explicit. For instance, I prefer:
verify(text) === undefined
The way I would suggest is to make the 'verify' to return a boolean value. The reason for this, is if for whatever reason you have a verification method that returns 0 (zero) and you do if (verify), the condition will be false but 0 is actually a verified value.
I hope that helps.
A: Although this is technically possible and works fine in most cases, I personally find it better style to design functions so that they always return the same type. My IDE warns me about inconsistent functions:
and this warning should be taken seriously. In particular, it helps you avoid unpleasant surprises when you test your app:
// hmm...
function isZero(n) {
if(n == 0)
return true;
}
// let's test it
describe("test", function() {
it("isZero", function() {
expect(isZero(0)).toBe(true);
expect(isZero(1)).toBe(false); // FAILURE!
});
});
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 5,351 |
\section{Introduction}
\label{Sec1}
\setcounter{equation}{0}
The dynamics of tensor metric perturbations in non-trivial geometric backgrounds is a crucial ingredient in the process of amplification of the vacuum fluctuations, and in the related production of a relic background of cosmic gravitational waves (GW) (see e.g. \cite{Muk}). The aim of this paper is to derive such dynamics from an action variational principle for the most general type of background geometry, sourced by arbitrarily given matter-energy distributions.
What we need, to this purpose, is to apply to the action the Hamilton variational formalism expanded up to second order. We will assume that the gravitational interaction is described by the usual Einstein-Hilbert action, but the same procedure can be easily applied to more general (e.g. higher-derivative, non-minimally coupled, scalar-tensor, etc) models of gravitational dynamics.
The first-order variation of the Einstein action with respect to the metric generates, as is well known, non-vanishing boundary terms that are to be cancelled by adding the variational contributions of an appropriate (York-Gibbons-Hawking) boundary action, in order to recover the background Einstein equations. At the second variational order, on the contrary, it will be shown that all boundary terms are automatically vanishing, that the condition of stationary action leads to a dynamical equation for the evolution of the metric fluctuations, and that such equation exactly corresponds to the linear perturbation of the background gravitational equations.
The main result of this paper is that we may expect in principle interesting changes to the dynamics of GW propagation in vacuum (see e.g. \cite{Mis}), even without considering modified theories of gravity. Even for the Einstein theory of gravity, and for gravitational sources minimally coupled to the geometry of a Riemann space-time manifold, we find that non-trivial contributions to the GW propagation equation are possible, in principle, provided the the energy-momentum of the matter sources satisfies some special appropriate condition. It should be stressed, also, that we are perturbing {\it only} the metric tensor appearing in the coupling of the matter fields to gravity, and not the other physical variables (energy-density, velocity, \dots) typical of the matter fields themselves.
The paper is organized as follows. In Sect. \ref{sec2} we derive the most general evolution equation for linear metric perturbations by imposing on the full action to be stationary, up to second order, with respect to the variation of the metric tensor. In Sect. \ref{sec3} we concentrate our discussion on the case of transverse and traceless metric fluctuations and on the general class of non-vacuum background spacetimes compatible with this choice. We give examples with scalar fields and fluids as gravitational sources, and we discuss how the fluid viscosity can affect GW propagation even in a very simple, spatially flat, FLRW background geometry. In Sect. \ref{sec4} we presents our final remarks, and suggest possible applications of the results derived in this paper. Finally, the Appendix A contains
a few explicit computations whose results are used in the main text; in the Appendix B we report a further example of modified GW propagation in a background geometry with the electromagnetic field as the dominant gravitational source.
{Conventions}: in this paper we will use the metric signature $g_{\mu\nu}= {\rm diag} (+,-,-,-)$; the Riemann tensor is defined as $R_{\mu\nu\alpha}\,^\beta = \partial_\mu \Gamma_{\nu\alpha}\,^\beta + \Gamma_{\mu\rho}\,^\beta \Gamma_{\nu\alpha}\,^\rho - \cdots$, the Ricci tensor is given by $R_{\nu\alpha}= R_{\mu\nu\alpha}\,^\mu$, and we define covariant derivatives with the following convention: $\nabla_\mu A_{\nu \cdots}= \partial_\mu A_{\nu \cdots} - \Gamma_{\mu\nu}\,^\alpha A_{\alpha \cdots}+ \cdots$.
\section{Evolution equation for linear metric perturbations in a general background geometry}
\label{sec2}
\setcounter{equation}{0}
Let us start by linearly expanding the metric tensor around a given background metric $g_{\mu\nu}$, by defining
\begin{equation}
g_{\mu\nu} \rightarrow \overline g_{\mu\nu} = g_{\mu\nu} + \delta g_{\mu\nu},
\label{21}
\end{equation}
and let us impose on the action $S$ (describing the geometry and its interactions with the matter fields), to be stationary ($\delta S=0$) with respect to the above background perturbation. By applying the standard variational formalism to first order in $\delta g_{\mu\nu}\equiv h_{\mu\nu}$ we obtain, as is well known, the field equations for the background metric $g_{\mu\nu}$. To second order, the condition of stationary action gives a differential equation governing the dynamical evolution of the metric fluctuations $h_{\mu\nu}$, and exactly corresponding -- as will be shown -- to the linear perturbation of the background gravitational equations.
In order to compute the action variation induced by the infinitesimal transformation (\ref{21}), to any given order in the powers of $\delta g_{\mu\nu}$, it is convenient to apply the formalism of functional differentiation, and expand the action in Taylor series of functional derivatives with respect to the metric. Let us recall, to this purpose, that if we have an arbitrary function of the metric and of its derivatives, $A= A(g, \partial g, \cdots)$, and we perform the metric expansion (\ref{21}), we can expand $A(\overline g, \partial \overline g)$ in power series of $\delta g_{\mu\nu}$ as follows:
\begin{equation}
A(\overline g, \partial \overline g) = A(g, \partial g) + \delta^{(1)}A + \delta^{(2)}A + \cdots
\label{22}
\end{equation}
where
\begin{eqnarray}
\delta^{(1)}A &=& \left(\delta A\over \delta \overline g_{\mu\nu}\right)_0 \delta g_{\mu\nu},
\label{23} \\
\delta^{(2)}A &=& {1\over 2} \left({\delta \over \delta \overline g_{\rho\sigma}}{\delta A \over \delta \overline g_{\mu\nu}}\right)_0 \delta g_{\rho\sigma} \,\delta g_{\mu\nu},
\label{24}
\end{eqnarray}
and where the subscript $``0"$ denotes a function to be evaluated for the unperturbed metric, i.e. for $\overline g_{\mu\nu}=g_{\mu\nu}$.
Finally, we have introduced the symbol of functional derivative, $\delta / \delta g_{\mu\nu}$, defined as usual by
\begin{equation}
{\delta \over \delta g_{\mu\nu}} A\left(g, \partial g, \cdots \right) \delta g_{\mu\nu} \equiv
{\partial A\over \partial g_{\mu\nu}} \delta g_{\mu\nu} +{\partial A\over \partial(\partial_\lambda g_{\mu\nu})} \partial_\lambda \delta g_{\mu\nu}+ \cdots
\label{25}
\end{equation}
(we have used the commutation property $\delta \partial_\lambda g_{\mu\nu}\equiv \partial_\lambda \overline g_{\mu\nu}- \partial_\lambda g_{\mu\nu} \equiv \partial_\lambda \delta g_{\mu\nu}$, due to the fact that the metric transformation (\ref{21}) is performed {\it locally}, i.e. at fixed spacetime position).
As a simple illustrative example of the above formalism we may consider the expansion of the controvariant components of the metric tensor. By setting $A=g^{\alpha\beta}$, using the definitions (\ref{23}) and (\ref{24}), and the property $d g^{\alpha\beta} g_{\beta\nu}= - g^{\alpha\beta}d g_{\beta\nu}$, we have:
\begin{eqnarray}
\delta^{(1)} g^{\alpha\beta} &=&\left(\partial\overline g^{\alpha\beta} \over \partial \overline g_{\mu\nu}\right)_0 \delta g_{\mu\nu} = - g^{\alpha\mu} g^{\beta\nu} \delta g_{\mu\nu}=-h^{\alpha\beta},
\label{26} \\
\delta^{(2)} g^{\alpha\beta} &=& {1\over 2} \left({\delta \over \delta \overline g_{\rho\sigma}}{\delta \overline g^{\alpha\beta}\over \delta \overline g_{\mu\nu}}\right)_0 \delta g_{\rho\sigma} \,\delta g_{\mu\nu}
\nonumber \\
&=&{1\over 2}
\left(g^{\alpha\rho}g^{\mu\sigma} g^{\beta\nu}+g^{\alpha\mu}g^{\beta\rho} g^{\nu\sigma}\right)\delta g_{\rho\sigma} \,\delta g_{\mu\nu}= h^{\alpha\mu}h_\mu\,^\beta.
\label{27}
\end{eqnarray}
The functional expansion of $\overline g^{\alpha\beta}$ to order $h^2$ then takes the explicit form
\begin{equation}
\overline g^{\alpha\beta} = g^{\alpha\beta}- h^{\alpha\beta} +h^{\alpha\mu}h_\mu\,^\beta + \cdots,
\label{28}
\end{equation}
in obvious agreement with the condition
\begin{equation}
\overline g^{\alpha\beta}\,\overline g_{\beta\nu} = \delta^\alpha_\nu + {\cal O}(h^3).
\label{29}
\end{equation}
Let us now apply the above formalism to compute the variation of the action for gravity and its matter sources, under the effects of the transformation (\ref{21}), to the linear and to the quadratic order in powers of the metric fluctuation $\delta g_{\mu\nu}$. Let us consider the Einstein model of gravity, described by the action
\begin{equation}
S= \int d^4 x \left( -{1\over 2\lambda_{\rm P}^2} \sqrt{-g}\, R + \sqrt{-g}\,{\cal L}_m\right),
\label{210}
\end{equation}
where $\lambda_{\rm P}^2=8\pi G$ is the Planck length parameter, ${\cal L}_m$ the Lagrangian density of the matter sources, and let us expand the infinitesimal variation of the action as follows:
\begin{eqnarray}
&&
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\delta S \equiv \delta^{(1)} S+\delta^{(2)} S + \cdots =
\nonumber \\ &&
\!\!\!\!\! \!\!\!\!\! \!\!\!\!\! =
{1\over 2 \lambda_{\rm P}^2} \int d^4 x\left[ \delta^{(1)}\left(- \sqrt{-g} R + 2 \lambda_{\rm P}^2 \sqrt{-g}{\cal L}_m\right)+ \delta^{(2)}\left(- \sqrt{-g} R + 2 \lambda_{\rm P}^2 \sqrt{-g}{\cal L}_m\right)+ \cdots \right].
\label{211}
\end{eqnarray}
\subsection{First-order variation of the action}
\label{sec21}
The application to the above action of the variational principle to first order in $\delta g_{\mu\nu}$ is well known, but it may be useful to briefly report here the results for its subsequent second-order generalization.
Let us start with the standard definition of the dynamical energy-momentum tensor $T_{\mu\nu}$, given by
\begin{equation}
{\delta \sqrt{-g} \,{\cal L}_m \over \delta g_{\mu\nu}}= -{1\over 2} \sqrt{-g} \,T^{\mu\nu}.
\label{212}
\end{equation}
Let us write the Einstein action in terms of the Ricci tensor as $ \sqrt{-g}\,R= \sqrt{-g}\, g^{\alpha\beta} R_{\alpha\beta}$, and recall the well known result
\begin{equation}
d \sqrt{-g}= {1\over 2} \sqrt{-g} \, g^{\mu\nu} d g_{\mu\nu}= - {1\over 2} \sqrt{-g} \, g_{\mu\nu}d g^{\mu\nu}.
\label{213}
\end{equation}
We then easily find that the condition of stationary action to first order in $\delta g_{\mu\nu}$, namely $\delta^{(1)} S=0$, implies:
\begin{equation}
\delta^{(1)} S={1\over 2 \lambda_{\rm P}^2} \int d^4 x \sqrt{-g}\left[ \left( R^{\mu\nu}-{1\over 2} g^{\mu\nu} R- \lambda_{\rm P}^2 T^{\mu\nu} \right) \delta g_{\mu\nu} -
g^{\alpha\beta} \delta^{(1)} R_{\alpha\beta} \right] =0,
\label{214}
\end{equation}
where
\begin{equation}
\delta^{(1)} R_{\alpha\beta} =\left( \delta R_{\alpha\beta} \over \delta \overline g_{\mu\nu} \right)_0 \delta g_{\mu\nu}.
\label{215}
\end{equation}
Let us check now that this last contribution to the action integral represents the well known (non-vanishing) boundary term, to be cancelled by the variational contribution of an appropriate boundary action \cite{York,Gib} which must be added to the Einstein action.
This important aspect of the variational formalism is of course largely understood and well illustrated in the specialistic literature on gravitational theory (and not only in the context of General Relativity, but also for generalized theories of gravity, see e.g. \cite{Dyer, Cosmai}). For later use in this paper, it will be enough to recall here that the first functional derivative (with respect to the metric) of the Ricci tensor can be expressed in terms of the so-called ``contracted Palatini identity" (see e.g. \cite{Gasp}), namely as
\begin{equation}
\delta^{(1)} R_{\nu\alpha}= \nabla_\mu \left( \delta^{(1)} \Gamma_{\nu\alpha}\,^\mu \right)
- \nabla_\nu \left( \delta^{(1)} \Gamma_{\mu\alpha}\,^\mu \right).
\label{216}
\end{equation}
Here the covariant derivative $\nabla_\mu$ is performed in terms the background metric $g_{\mu\nu}$, and $\delta^{(1)} \Gamma_{\nu\alpha}\,^\mu$ is the first-order perturbation (computed as a first functional derivative) of the Christoffel connection. By applying the metric property of the Riemann geometry ($\nabla_\mu g^{\alpha\nu}=0$), and using the Gauss theorem, we then find that the last contribution to Eq. (\ref{214}) can be written as an integral over the boundary hypersurface $\partial \Omega$ of the considered space-time domain $\Omega$:
\begin{eqnarray}
\!\!\!\!
\int_{\Omega} d^4x \sqrt{-g}\,g^{\alpha\beta} \delta^{(1)} R_{\alpha\beta}&=&
\int_{\Omega} d^4x \sqrt{-g}\, \nabla_\mu \left( g^{\nu\alpha} \delta^{(1)} \Gamma_{\nu\alpha}\,^\mu- g^{\mu\alpha} \delta^{(1)} \Gamma_{\nu\alpha}\,^\nu \right)=
\nonumber \\
&=&
\int_{\partial \Omega} dS_\mu \sqrt{-g}\left( g^{\nu\alpha} \delta^{(1)} \Gamma_{\nu\alpha}\,^\mu- g^{\mu\alpha} \delta^{(1)} \Gamma_{\nu\alpha}\,^\nu \right).
\label{217}
\end{eqnarray}
On the other hand, an explicit computation of $\delta^{(1)} \Gamma_{\nu\alpha}\,^\mu$, performed according to the definitions (\ref{23}) and (\ref{25}), provides a result which can be written in compact form as follows:
\begin{eqnarray}
\delta^{(1)} \Gamma_{\nu\alpha}\,^\mu&=&-{1\over 2} \delta g^{\mu\beta} \left( \partial_\nu g_{\alpha\beta}
+\partial_\alpha g_{\nu\beta}-\partial_\beta g_{\nu\alpha} \right)
+{1\over 2} g^{\mu\beta} \left( \partial_\nu \delta g_{\alpha\beta}
+\partial_\alpha \delta g_{\nu\beta}-\partial_\beta \delta g_{\nu\alpha} \right)
\nonumber \\
&=& {1\over 2} g^{\mu\beta} \left( \nabla_\nu h_{\alpha\beta}
+\nabla_\alpha h_{\nu\beta}-\nabla_\beta h_{\nu\alpha} \right).
\label{218}
\end{eqnarray}
By inserting the above result into the boundary integral (\ref{217}) we can immediately check that, besides the terms proportional to the metric perturbation $\delta g$ (whose contribution on the boundary is identically vanishing by assumptions, thanks to the rules of the Hamilton variational principle), there are also terms proportional to the derivatives of the metric variation, $\partial \delta g$. Those terms are {not} automatically vanishing on the boundary, and -- as already stressed -- they are to be eliminated by the addition of an appropriate boundary action. By including such an action into the variational procedure
the overall contribution of the last term disappears from the integral (\ref{214}), and the condition of stationary action, to the first order in $\delta g_{\mu\nu}$, leads to the well-known background equations for the unperturbed metric $g_{\mu\nu}$:
\begin{equation}
R^{\mu\nu}-{1\over 2} g^{\mu\nu} R= \lambda_{\rm P}^2 T^{\mu\nu}.
\label{219}
\end{equation}
\subsection{Second-order variation of the action}
\label{sec22}
Let us now impose on the action to be stationary with respect to the infinitesimal transformation (\ref{21}) up to terms quadratic in $\delta g_{\mu\nu}$. From Eq. (\ref{211}), and from the definition (\ref{24}), we are led to the condition
\begin{equation}
\delta^{(2)} S\equiv {1\over 4 \lambda_{\rm P}^2} \int d^4 x \left[ {\delta \over \delta \overline g_{\rho\sigma}}{\delta \over \delta \overline g_{\mu\nu}}\left(- \sqrt{-g} \,R + 2 \lambda_{\rm P}^2 \sqrt{-g}\,{\cal L}_m\right) \right]_0 \delta g_{\rho\sigma} \delta g_{\mu\nu} =0.
\label{220}
\end{equation}
It is important to note that there is no need of explicitly including into the above equation the additional (York-Gibbons-Hawking) boundary contribution, because the second functional derivative of such term is identically vanishing.
By exploiting the results already reported in the previous subsection, and using in particular Eq. (\ref{214}), we can then explicitly write the second order variation of the action as follows:
\begin{eqnarray}
\delta^{(2)} S&=& {1\over 4 \lambda_{\rm P}^2} \int d^4 x \left[ \delta^{(1)} \left(\sqrt{-g}\, R^{\mu\nu}-
{1\over 2} \sqrt{-g}\, g^{\mu\nu} R\right) - \lambda_{\rm P}^2 \delta^{(1)} \left(\sqrt{-g}\, T^{\mu\nu}\right) \right] \delta g_{\mu\nu}
\nonumber \\
&-& {1\over 4 \lambda_{\rm P}^2} \int d^4 x \left[ \delta^{(1)} \left(\sqrt{-g}\,g^{\alpha\beta}\right) \delta^{(1)} R_{\alpha\beta} + 2 \sqrt{-g}\,g^{\alpha\beta} \delta^{(2)} R_{\alpha\beta} \right]
\label{221}
\end{eqnarray}
(the variational operators $\delta^{(1)}$ and $\delta^{(2)}$ are defined, respectively, by Eqs. (\ref{23}), (\ref{24})).
Let us now separately consider the two integral terms $I_1$ and $I_2$ appearing in the second line of the above equation, namely
\begin{equation}
I_1= \int d^4 x \delta^{(1)} \left(\sqrt{-g}\,g^{\alpha\beta}\right) \delta^{(1)} R_{\alpha\beta},
~~~~~~~~~~~
I_2= 2 \int d^4x \sqrt{-g}\, g^{\nu\alpha} \delta^{(2)} R_{\nu\alpha},
\label{222}
\end{equation}
and show that, in spite of their non-trivial value, they give no contribution to the second-order variation of the action because they exactly cancel each other.
Let us start with $I_1$. By recalling the results (\ref{216}) e (\ref{218}) we can explicitly write $\delta^{(1)} R_{\nu\alpha}$ as
\begin{equation}
\delta^{(1)} R_{\nu\alpha}={1\over 2} \left( \nabla_\mu \nabla_\nu h_\alpha\,^\mu + \nabla_\mu \nabla_\alpha h_\nu\,^\mu - \nabla^2 h_{\nu\alpha} - \nabla_\nu \nabla_\alpha h \right),
\label{223}
\end{equation}
where we have set $h= g^{\mu\nu} \delta g_{\mu\nu} = g^{\mu\nu} h_{\mu\nu}$, and we have denoted with $\nabla^2$ the covariant D'Alembert operator, $\nabla^2\equiv g^{\mu\beta} \nabla_\mu \nabla_\beta$. By using Eqs. (\ref{26}), (\ref{213}), (\ref{223}), by integrating by part, and neglecting the total divergences, we obtain:
\begin{eqnarray}
\!\!\!\!\!\!\!\!\!\!
I_1&=&
\int d^4 x \sqrt{-g} \left( {1\over 2} h g^{\nu\alpha} - h^{\nu\alpha} \right) \delta^{(1)} R_{\nu\alpha}
\nonumber \\ &=&
\int d^4 x \sqrt{-g}\left(-{1\over 2} \nabla_\mu h^{\nu\alpha} \nabla^\mu h_{\nu\alpha}
+\nabla_\mu h^{\nu\alpha} \nabla^\nu h_{\alpha}\,^\mu +{1\over 2} \nabla_\nu h \nabla^\nu h - \nabla_\alpha h^{\alpha\mu}\nabla_\mu h \right).
\label{224}
\end{eqnarray}
We have neglected the contribution of the total divergences because, using the Gauss theorem, they lead to integrals defined over the boundary hypersurface $\partial \Omega$ with an argument proportional to $h_{\mu\nu}= \delta g_{\mu\nu}$: hence, they are automatically vanishing thanks to the assumptions of the standard variational principle.
Let us now compute $I_2$. To this purpose, we need the second functional derivative $\delta^{(2)} R_{\nu\alpha}$ which, by using the general definition of the Ricci tensor and of the covariant derivative with respect to the background metric, can be written as:
\begin{equation}
\delta^{(2)} R_{\nu\alpha}= \nabla_\mu \left( \delta^{(2)} \Gamma_{\nu\alpha}\,^\mu \right)
- \nabla_\nu \left( \delta^{(2)} \Gamma_{\mu\alpha}\,^\mu \right)
+ \delta^{(1)} \Gamma_{\mu\rho}\,^\mu \delta^{(1)} \Gamma_{\nu\alpha}\,^\rho
-\delta^{(1)} \Gamma_{\nu\rho}\,^\mu \delta^{(1)} \Gamma_{\mu\alpha}\,^\rho
\label{225}
\end{equation}
(see Appendix A). The first-order perturbation of the connection is given in Eq. (\ref{218}). At the second order we have
\begin{equation}
\delta^{(2)} \Gamma_{\nu\alpha}\,^\mu =
{1\over 2} \left({\delta \over \delta \overline g_{\gamma\delta}}{\delta \over \delta \overline g_{\rho\sigma}}
\Gamma_{\nu\alpha}\,^\mu
\right)_0 \delta g_{\gamma\delta}\, \delta g_{\rho\sigma},
\label{226}
\end{equation}
and an explicit computation (see Appendix A) gives
\begin{eqnarray}
\delta^{(2)} \Gamma_{\nu\alpha}\,^\mu&=&{1\over 2} h^{\mu\lambda}h_\lambda\,^\beta \left( \partial_\nu g_{\alpha\beta}+\partial_\alpha g_{\nu\beta}-\partial_\beta g_{\nu\alpha} \right)
-{1\over 2} h^{\mu\beta} \left( \partial_\nu h_{\alpha\beta}
+\partial_\alpha h_{\nu\beta}-\partial_\beta h_{\nu\alpha} \right)
\nonumber \\
&\equiv& -{1\over 2} h^{\mu\beta} \left( \nabla_\nu h_{\alpha\beta}
+\nabla_\alpha h_{\nu\beta}-\nabla_\beta h_{\nu\alpha} \right).
\label{227}
\end{eqnarray}
By inserting these results into the integral $I_2$ we then find that the covariant-derivative terms appearing in the explicit expression of $\delta^{(2)} R_{\nu\alpha}$ lead to boundary integrals proportional to $\delta^{(2)} \Gamma$. In that case, the integrand is always proportional to the metric perturbation $\delta g$, so that the corresponding contribution to the action variation is automatically vanishing, as imposed by the assumptions of the Hamilton variational principle.
We are thus left with the contribution of the last two terms of Eq. (\ref{225}). By using the result (\ref{218}) for $\delta^{(1)} \Gamma$, and combining all terms, we obtain:
\begin{equation}
I_2=
\int d^4 x \sqrt{-g}\, \left( {1\over 2} \nabla_\mu h^{\alpha\beta} \nabla^\mu h_{\alpha\beta} - {1\over 2} \nabla_\nu h \nabla^\nu h + \nabla_\mu h \nabla_\nu h^{\nu\mu} - \nabla_\nu h_\alpha\,^\mu \nabla_\mu h^{\nu\alpha}
\right).
\label{228}
\end{equation}
By comparing this result with Eq. (\ref{224}) we can immediately check that $I_1+I_2=0$, so that their contribution of the second-order action variation (\ref{221}) completely disappears. Hence, the condition of stationary action $\delta^{(2)} S=0$ simply implies the vanishing of the first integral of Eq. (\ref{221}). Such a condition, by putting $R^{\mu\nu}= g^{\mu\alpha} g^{\nu\beta}R_{\alpha\beta}$ and $T^{\mu\nu} = g^{\mu\alpha} g^{\nu\beta}T_{\alpha\beta}$, by factorizing the variational contribution $\delta^{(1)} (\sqrt{-g} g^{\mu\alpha} g^{\nu\beta})$, and imposing the validity of the background equations (\ref{219}), can be finally expressed as
\begin{equation}
\delta^{(1)} \left(R_{\alpha\beta}-
{1\over 2} g_{\alpha\beta} R\right)= \lambda_{\rm P}^2 \, \delta^{(1)} T_{\alpha\beta}.
\label{229}
\end{equation}
It exactly corresponds to the condition one would obtain by computing the first functional differentiation of the background equations (\ref{219}).
More explicitly, the above equation can be rewritten as
\begin{equation}
\delta^{(1)} R_{\alpha\beta}-{1\over 2} R h_{\alpha\beta}+{1\over 2} g_{\alpha\beta} h^{\mu\nu} R_{\mu\nu} -{1\over 2} g_{\alpha\beta} g^{\mu\nu} \delta^{(1)} R_{\mu\nu} = \lambda_{\rm P}^2 \, \delta^{(1)} T_{\alpha\beta},
\label{230}
\end{equation}
where, by using Eq. (\ref{223}),
\begin{equation}
g^{\mu\nu} \delta^{(1)} R_{\mu\nu}= \nabla_\mu \nabla_\nu h^{\mu\nu}- \nabla^2 h.
\label{231}
\end{equation}
It may be convenient, also, to rewrite the first two terms on the right-hand side of Eq. (\ref{223}) by applying the commutation rule of the covariant derivatives, which gives:
\begin{equation}
\nabla_\mu \nabla_\beta h_\alpha\,^\mu= \nabla_\beta \nabla_\mu h_\alpha\,^\mu +h_\alpha\,^\nu R_{\nu\beta} - R_{\mu\alpha\beta\nu}h^{\mu\nu},
\label{232}
\end{equation}
where $R_{\mu\alpha\beta\nu}$ is the Riemann tensor. By combining all contributions, and multiplying by $-2$, we finally obtain the following dynamical evolution equation for the linear fluctuations $h_{\mu\nu}$ of a general background metric $g_{\mu\nu}$, with general matter sources $T_{\mu\nu}$, in the form:
\begin{eqnarray}
&&~~~~
\nabla^2 h_{\alpha\beta}+ 2 R_{\mu\alpha\beta\nu} h^{\mu\nu}+ Rh_{\alpha\beta} - g_{\alpha\beta} h^{\mu\nu} R_{\mu\nu}- h_\alpha\,^\nu R_{\nu\beta} - h_\beta\,^\nu R_{\nu\alpha} -
\nonumber \\ && \!\!\!\!\!\!
-\nabla_\beta \nabla_\mu h_\alpha\,^\mu -\nabla_\alpha \nabla_\mu h_\beta\,^\mu + \nabla_\alpha \nabla_\beta h - g_{\alpha\beta} (\nabla^2 h- \nabla_\mu\nabla_\nu h^{\mu\nu})= -2 \lambda_{\rm P}^2 \delta^{(1)} T_{\alpha\beta}.
\label{233}
\end{eqnarray}
In a vacuum geometry ($T_{\mu\nu}=0= \delta T_{\mu\nu}$, $R_{\mu\nu}=0=R$), and for metric fluctuations satisfying the so-called TT gauge conditions, i.e. $\nabla_\nu h_\mu\,^\nu=0= h_\mu\,^\mu$, we recover the well-known vacuum propagation equation (see e.g. \cite{Mis}),
\begin{equation}
\nabla^2 h_{\alpha\beta}+ 2 R_{\mu\alpha\beta\nu} h^{\mu\nu}=0.
\label{234}
\end{equation}
The general result (\ref{233}) -- for a non-vacuum geometry and with no gauge fixing -- was first derived (as far as we know) in a different but equivalent form in \cite{Sci}, and, in the same form as Eq. (\ref{233}), in an unpublished lecture note \cite{Det}. In both cases, however, it was derived without starting from a variational principle but directly perturbing, to first order, the background Einstein equations. It has been recently obtained with the same method in \cite{Brev} (see also \cite{Fig}). In other papers (see e.g. \cite{Cap,Dol}) a similar equation is presented by imposing however the TT conditions from the beginning, so that the last five terms on the left-hand side of Eq. (\ref{233}) are missing. It should be noted, in this regard, that in the presence of gravitational sources it is not possible, in general, to write the full metric perturbations in the TT gauge, as clearly stressed in \cite{Flan}.
Let us conclude this Section by noting that the general propagation equation (\ref{233}) can also be rewritten in terms of another frequently used variable, the so called ``trace-reversed" perturbation $\psi_{\mu\nu}$, defined by:
\begin{equation}
\psi_{\mu\nu} \equiv h_{\mu\nu} -{1\over 2} g_{\mu\nu} h= h_{\mu\nu} +{1\over 2} g_{\mu\nu} \psi, ~~~~~~~~~~~~~
\psi \equiv g^{\mu\nu} \psi_{\mu\nu} = -h.
\label{235}
\end{equation}
In terms of $\psi_{\mu\nu}$, our previous equation (\ref{233}) takes the form
\begin{eqnarray}
&&
\nabla^2\psi_{\alpha\beta}+ 2 R_{\mu\alpha\beta\nu} \psi^{\mu\nu}+ R\,\psi_{\alpha\beta} - g_{\alpha\beta} \,\psi^{\mu\nu} R_{\mu\nu}- \psi_\alpha\,^\nu R_{\nu\beta} - \psi_\beta\,^\nu R_{\nu\alpha} -
\nonumber \\ &&
-\nabla_\beta \nabla_\mu\psi_\alpha\,^\mu -\nabla_\alpha \nabla_\mu \psi_\beta\,^\mu +g_{\alpha\beta} \nabla_\mu\nabla_\nu \psi^{\mu\nu}= -2 \lambda_{\rm P}^2 \,\delta^{(1)} T_{\alpha\beta}
\label{236}
\end{eqnarray}
(useful, in vacuum, to impose the so-called Lorentz gauge condition $\nabla_\nu \psi_\mu\,^\nu=0$, see e.g. \cite{And}).
\section{Evolution of transverse and traceless metric fluctuations}
\label{sec3}
\setcounter{equation}{0}
Let us now recall that, in general, not all the independent components of the metric perturbations $h_{\mu\nu}$ may satisfy the condition of vanishing trace and vanishing covariant divergence when the stress tensor of the matter sources is nonzero. In the rest of this paper, however, we will restrict our discussion to those metric perturbations (and to the associated background geometries) which satisfy such conditions, and which describe radiative degrees of freedom (GW) possibly propagating to infinity (as illustrated in details in \cite{Flan}). Why are we interested in such modes? Because we want to discuss the possible effects of the geometric sources on the dynamics of GW propagation, considering in particular non-vacuum backgrounds of cosmological type, generated by conventional (or more ``exotic") classes of energy-momentum distributions.
We shall thus consider components of the metric perturbations satisfying
the conditions
\begin{equation}
\nabla _\nu h^{\mu\nu}=0, ~~~~~~~~~~~~ h= g^{\mu\nu}h_{\mu\nu}=0,
\label{31}
\end{equation}
which we shall call, for brevity, ``TT gauge" conditions. In that case we have also $\psi=0$, $\psi_{\mu\nu}= h_{\mu\nu}$, and the corresponding propagation equation (\ref{233}) -- or, equivalently, our Eq. (\ref{236}) -- reduces to the simplified form
\begin{equation}
\nabla^2 h_{\alpha\beta}+ 2 R_{\mu\alpha\beta\nu} h^{\mu\nu}+ Rh_{\alpha\beta} - g_{\alpha\beta} \,h^{\mu\nu} R_{\mu\nu}- h_\alpha\,^\nu R_{\nu\beta} - h_\beta\,^\nu R_{\nu\alpha} -
= -2 \lambda_{\rm P}^2 \,\delta^{(1)} T_{\alpha\beta}.
\label{32}
\end{equation}
By computing the trace (with respect to the unperturbed metric $g^{\alpha\beta}$) of the above equation we then find, for consistency with the selected gauge, that
the background geometry and its sources must satisfy the condition
\begin{equation}
2 R_{\alpha\beta} h^{\alpha\beta} = \lambda_{\rm P}^2 \,g^{\alpha\beta} \delta^{(1)} T_{\alpha\beta}.
\label{33}
\end{equation}
Note that for a vacuum, Ricci-flat metric ($T_{\mu\nu}=0=R_{\mu\nu}$), the above equation is always automatically satisfied.
It may be convenient, for the application of this paper, to rewrite Eq. (\ref{32}) in a form which better displays the possible modifications induced {by} the matter sources with respect to the standard GW propagation in vacuum (described by Eq. (\ref{234})). By using the background equations (\ref{219}) and the consistency condition (\ref{33})
to eliminate everywhere the Ricci tensor, we can then rewrite Eq. (\ref{32}) as follows:
\begin{equation}
\nabla^2 h_{\alpha\beta}+ 2 R_{\mu\alpha\beta\nu} h^{\mu\nu} = \lambda_{\rm P}^2 \left( h_\alpha\,^\nu T_{\nu\beta}+ h_\beta\,^\nu T_{\nu\alpha}+{1\over 2} g_{\alpha\beta} \,g^{\mu\nu} \delta^{(1)} T_{\mu\nu} -2 \delta^{(1)} T_{\alpha\beta}\right).
\label{34}
\end{equation}
Let us finally recall that, as previously stressed, the symbol $\delta^{(1)} T_{\alpha\beta}$ represents the second-order perturbation of the matter Lagrangian performed with respect to the metric only (see Eqs. (\ref{220}), (\ref{221})), and it does not imply any direct perturbation of the other specific and intrinsic variables (like density distributions, velocity distributions) typical of the matter sources that we are considering.
In order to display the physical differences with vacuum GW propagation
it may be appropriate, at this point, to write down the modified perturbation equation (\ref{34}) explicitly, for some specific case of non-vacuum background geometry satisfying the consistency condition (\ref{33}). To this purpose we shall provide two examples (typical of cosmological applications), where the sources of the unperturbed geometry are represented by a fluid or by a self-interacting scalar field.
\subsection{Perfect fluid source}
\label{sec31}
Let us consider a perfect fluid with barotropic equation of state, energy density $\rho$, pressure $p$, four-velocity of the fluid element $u^\mu = dx^\mu/ d\tau$, described by the Lagrangian density (see e.g. \cite{For}):
\begin{equation}
\sqrt{-g} {\cal L}_m =-{1\over 2} \sqrt{-g} \,\left[ \left(\rho+p\right) g_{\rho\sigma} u^\rho u^\sigma -\left(\rho+3p\right) \right].
\label{35}
\end{equation}
For the explicit application of Eqs. (\ref{33}) and (\ref{34}) we need both the
energy-momentum tensor of the unperturbed fluid, given (according to Eq. (\ref{212})) by
\begin{equation}
T_{\alpha\beta} = \left( {2\over \sqrt{-\overline g}}
{\delta \sqrt{-\overline g} \,{\cal L}_m \over \delta \overline g^{\alpha\beta}}\right)_0,
\label{36}
\end{equation}
and its first-order perturbation, given by
\begin{equation}
\delta^{(1)} T_{\alpha\beta} = \left( {\delta \over \delta \overline g_{\mu\nu}} {2\over \sqrt{-\overline g}}
{\delta \sqrt{-\overline g} \,{\cal L}_m \over \delta \overline g^{\alpha\beta}}\right)_0 \delta g_{\mu\nu}.
\label{37}
\end{equation}
From the definition (\ref{36}), by using the metric property (\ref{213}) and imposing -- but only {\it after} the functional differentiation -- the standard normalization condition
$ g_{\rho\sigma} u^\rho u^\sigma=1$, we are then led to the well known result
\begin{equation}
T_{\alpha\beta}= - p\, g_{\alpha\beta} + (\rho+p) u_a u_\beta.
\label{38}
\end{equation}
Let us now compute the second functional derivative, by applying the definition (\ref{37}) to the fluid Lagrangian (\ref{35}). We easily obtain:
\begin{eqnarray}
\delta^{(1)} T_{\alpha\beta} &=&
\left( {\delta \over \delta \overline g_{\mu\nu}} \left[ {1\over 2} (\rho+p) \overline g_{\alpha\beta} \overline g_{\rho\sigma} u^\rho u^\sigma - {1\over 2} (\rho+3p) \,\overline g_{\alpha\beta} + (\rho+p) \, u^\rho u^\sigma \,\overline g_{\rho\alpha} \overline g_{\sigma\beta} \right] \right)_0 \delta g_{\mu\nu}
\nonumber \\
&=&-p\, h_{\alpha\beta} +{1\over 2} \left(\rho+p\right) g_{\alpha\beta}\, h_{\mu\nu} u^\mu u^\nu + (\rho+p) \left(u_\alpha u^\nu h_{\nu\beta} + u_\beta u^\nu h_{\nu \alpha} \right).
\label{39}
\end{eqnarray}
On the other hand, by using the background equations (\ref{219}), by inserting the stress tensor (\ref{38}) into the left left-hand side of the condition (\ref{33}), and taking into account that $h=0$, we obtain
\begin{equation}
2 R_{\alpha\beta}h^{\alpha\beta}= 2 \lambda_{\rm P}^2
\left(\rho+p\right)u_\alpha u_\beta h^{\alpha\beta}.
\label{310}
\end{equation}
By inserting the perturbed stress-tensor (\ref{39}) into the right-hand side of Eq. (\ref{33}) we have:
\begin{equation}
\lambda_{\rm P}^2 \,g^{\alpha\beta} \delta^{(1)} T_{\alpha\beta} = 4 \lambda_{\rm P}^2
\left(\rho+p\right) h_{\mu\nu} u^\mu u^\nu .
\label{311}
\end{equation}
It follows that the condition of consistency with the TT gauge, for a perfect fluid source, is satisfied if and only if
\begin{equation}
\left(\rho+p\right) h_{\mu\nu} u^\mu u^\nu =0
\label{312}
\end{equation}
(and in that case both sides of Eq. (\ref{33}) are identically vanishing). The above constraint can be satisfied in two ways.
A first possibility is the case $p=-\rho$ (which describes, in a fluid dynamics language, a background geometry generated by a cosmological constant $\Lambda= \rho=$ cost). In that case we have $T_{\alpha\beta}= - p g_{\alpha\beta}$, $\delta^{(1)} T_{\alpha\beta}= -p h_{\alpha\beta}$, and we can check that the right-hand side of Eq. (\ref{34}) is identically vanishing. It turns out that the linearized propagation equation of the metric fluctuations satisfying the TT gauge condition is exactly the same as that obtained in the context of a vacuum, Ricci-flat geometry, and is given by Eq. (\ref{234}).
A second possibility to be consistent with the TT gauge is to assume that $h_{\mu\nu} u^\mu u^\nu=0$. In that case the right-hand side of Eq. (\ref{34}) is in general non-vanishing, and by using the explicit results (\ref{38}), (\ref{39}) for the fluid stress tensor we find:
\begin{equation}
\nabla^2 h_{\alpha\beta}+ 2 R_{\mu\alpha\beta\nu} h^{\mu\nu} +\lambda_{\rm P}^2
(\rho+p) \left(u_\alpha u^\nu h_{\nu\beta} + u_\beta u^\nu h_{\nu \alpha} \right)=0.
\label{313}
\end{equation}
Interestingly enough, we can then obtain a propagation equation different from the vacuum equation even for the components of the metric fluctuations which satisfy the TT gauge conditions, and which are typical of gravitational radiation.
The above ``non standard" correction terms disappear, however, from the previous equation for metric fluctuations satisfying the condition $h_{\mu\nu}u^\nu = 0$. For instance, they may disappear if the fluid has a ``comoving" energy-momentum distribution, i.e. the fluid element has velocity $u^i=0$, $u^0=1$, and we assume
that the TT gauge is valid for the spatial components $h_{ij}$ of the metric fluctuations, with $h_{\mu0}=0$. For such modes the consistency condition $h_{\mu\nu} u^\mu u^\nu=0$ is also automatically satisfied, and
the propagation equation (\ref{313}) reduces to
\begin{equation}
\nabla^2 h_{ij}+ 2 R_{kijl} h^{kl}=0.
\label{314}
\end{equation}
If we take a spatially flat, homogeneous and isotropic geometry, described by a FLRW metric with $g_{00}=1$, $g_{ij}=- a^2 \delta_{ij}$, we have the following components of the connection and of the curvature tensor,
\begin{equation}
\Gamma_{0i}\,^j= H \delta_i^j, ~~~~~~~~~ \Gamma_{ij}\,^0= - g_{ij}H, ~~~~~~~~~
R_{kijl} h^{kl} = H^2 h_{ij},
\label{315}
\end{equation}
where $H= \dot a/a$ and the dot denotes differentiation with respect to the cosmic time $t$. In such a background, if we explicitly write the propagation equation (\ref{314}) for the mixed components $h_i\,^j$ of the metric perturbations, $h_i\,^j\equiv g^{jk}h_{ik}$, where $\partial_j h_i\,^j=0=g^{jk} h_{jk}$,
we then recover the standard, well-known result (see e.g. \cite{Muk})
\begin{equation}
\ddot h_i\,^j + 3 H \dot h_i\,^j -{\partial^2\over a^2} \,h_i\,^j=0
\label{316}
\end{equation}
(where $\partial^2= \delta^{kl} \partial_k\partial_l$).
\subsection{Minimally coupled scalar field}
\label{sec32}
Another typical source of the spacetime geometry at the cosmic level is possibly represented by a self-interacting scalar field $\phi$, described by the Lagrangian density:
\begin{equation}
\sqrt{-g} \,{\cal L}_m= \sqrt{-g} \left[{1\over 2} g^{\rho\sigma} \partial_\rho \phi \partial_\sigma \phi - V(\phi) \right]
\label{317}
\end{equation}
(which also includes the case of a cosmological constant $V=\Lambda=$ cost). The corresponding energy-momentum tensor, defined by Eq. (\ref{36}), is given by
\begin{equation}
T_{\alpha\beta}= \partial_\alpha \phi \partial_\beta \phi - g_{\alpha\beta} \left({1\over 2} g^{\rho\sigma} \partial_\rho \phi \partial_\sigma \phi - V\right),
\label{318}
\end{equation}
and its first order perturbation, according to Eq. (\ref{37}), leads to
\begin{equation}
\delta^{(1)} T_{\alpha\beta}=h_{\alpha\beta} \left(V- {1\over 2} \partial_\mu \phi \partial^\mu \phi \right) +{1\over 2} g_{\alpha\beta}\, h^{\mu\nu} \partial_\mu \phi \partial_\nu \phi.
\label{319}
\end{equation}
By inserting the two above results into Eq. (\ref{33}), by using the background equations to eliminate $R_{\alpha\beta}$ in terms of $T_{\alpha\beta}$, and taking into account that $h=0$, we then find that the consistency condition with the TT gauge in this case is always automatically satisfied, namely:
\begin{equation}
2 h^{\alpha\beta} R_{\alpha\beta} \equiv 2 \lambda_{\rm P}^2 h^{\alpha\beta} \partial_\alpha \phi \,\partial_\beta \phi \equiv
\lambda_{\rm P}^2 \, g^{\alpha\beta} \delta^{(1)} T_{\alpha\beta},
\label{320}
\end{equation}
for any given type of scalar field dynamics.
This does not imply, however, that there are no modifications to the vacuum propagation equation. By inserting into the right-hand side of Eq. (\ref{34}) the results (\ref{318}), (\ref{319}), we find
\begin{equation}
\nabla^2 h_{\alpha\beta}+ 2 R_{\mu\alpha\beta\nu} h^{\mu\nu} -{\lambda_{\rm P}^2}
\left(h_\alpha\,^\nu \partial_\beta \phi + h_\beta\,^\nu \partial_\alpha \phi \right) \partial_\nu \phi =0.
\label{321}
\end{equation}
If we have a homogeneous scalar source, with $\partial_i\phi=0$, it follows that there are no corrections to the propagation equation for the spatial fluctuations $h_{ij}$ (like in the case of the comoving fluid source). But non-trivial corrections are in general possible, even for the spatial fluctuations satisfying the TT gauge conditions, if $h_\alpha\,^\nu \partial_\nu \phi \not=0$.
\subsection{Fluid source with viscosity}
\label{sec33}
Let us finally consider a possible source of the background geometry which can be described as a fluid with energy density $\rho$, pressure $p$, velocity field $u^\mu$, bulk viscosity $\zeta$, and shear viscosity $\eta$, where $\zeta$ and $\eta$ are constant parameters. By introducing the effective pressure $\widetilde p$,
\begin{equation}
\widetilde p =p - \left(\zeta -{2\over 3} \eta \right) \nabla_\lambda u^\lambda,
\label{322}
\end{equation}
we can put the full energy-momentum tensor of the viscous fluid \cite{Mis,Wei,Gron}
in the following convenient form (see e.g. \cite{Ana,Mon} for recent discussions):
\begin{equation}
\widetilde T_{\alpha\beta} = \left(\rho+ \widetilde p \right) u_\alpha u_\beta - \widetilde p g_{\alpha\beta} - \eta \left[ u_\alpha u^\rho \nabla_\rho u_\beta + u_\beta u^\rho \nabla_\rho u_\alpha - \nabla_\alpha u_\beta - \nabla_\beta u_\alpha \right].
\label{323}
\end{equation}
The first-order perturbation can then be written as
\begin{eqnarray}
\delta^{(1)} \widetilde T_{\alpha\beta} &=& \delta^{(1)} T_{\alpha\beta}(\widetilde p) + \left( u_\alpha u_\beta - g_{\alpha\beta} \right)
\left({2\over 3} \eta - \zeta \right) \delta^{(1)} (\nabla_\rho u^\rho )
\nonumber \\ &-&
\eta \, \delta^{(1)} \left[ \left(g_{\alpha\rho} g _{\beta\sigma} + g_{\beta\rho} g _{\alpha\sigma}\right)u^\rho u^\lambda \nabla_\lambda u^\sigma - g_{\beta\rho} \nabla_\alpha u^\rho - g_{\alpha\rho} \nabla_\beta u^\rho
\right],
\label{324}
\end{eqnarray}
where $ \delta^{(1)} T_{\alpha\beta}(\widetilde p)$ denotes the perfect fluid result ({\ref{39}), written however in terms of the total effettive pressure $\widetilde p$.
By using the result (\ref{218}) for the first-order perturbation of the connection we find, in the TT gauge,\begin{equation}
\delta^{(1)} (\nabla_\rho u^\rho )= u^\alpha \delta^{(1)} \Gamma_{\nu \alpha}\,^\nu = {1\over 2} u^\alpha \nabla_\alpha (g^{\nu\beta} h_{\nu\beta}) \equiv {1\over 2} u^\alpha \nabla_\alpha h =0.
\label{325}
\end{equation}
so that the second contribution to Eq. (\ref{324}) is identically vanishing. Let us explicitly compute the remaining contributions, by using in particular the following results:
\begin{eqnarray}
u_\alpha g_{\beta\sigma} u^\lambda \delta^{(1)} (\nabla_\lambda u^\sigma ) &=& u_\alpha u^\lambda u^\rho \left( \nabla_\lambda h_{\beta\rho} -{1\over 2} \nabla_\beta h_{\lambda\rho} \right),
\label{326}
\\
g_{\alpha\sigma} \delta^{(1)} (\nabla_\beta u^\sigma )&=&{1\over 2} u^\rho \left( \nabla_\beta h_{\rho\alpha} + \nabla_\rho h_{\beta\alpha}- \nabla_\alpha h_{\beta\rho} \right).
\label{327}
\end{eqnarray}
We finally obtain:
\begin{eqnarray}
\delta^{(1)} \widetilde T_{\alpha\beta} &=& \delta^{(1)} T_{\alpha\beta}(\widetilde p)
-\eta \left[ h_{\alpha\rho} u^\rho u^\lambda \nabla_\lambda u_\beta + u_\alpha h_{\beta\sigma} u^\lambda \nabla_\lambda u^\sigma -
h_{\alpha\sigma} \nabla_\beta u^\sigma
\right.
\nonumber\\ &+&
\left.
u_\alpha u^\lambda u^\rho \left( \nabla_\lambda h_{\beta\rho} -{1\over 2} \nabla_\beta h_{\lambda\rho} \right)
+ ( \alpha \leftrightarrow \beta) \right] + \eta \, u^\lambda \nabla_\lambda h_{\alpha\beta},
\label{328}
\end{eqnarray}
where the symbol $\{ \alpha \leftrightarrow \beta\}$ denotes that all the preceding terms inside the square brackets are to be repeated with the index $\alpha$ replaced by $\beta$ and vice-versa.
We can now impose the consistency condition for the TT gauge. By computing the left-hand and right-hand side of Eq. (\ref{33}) for our viscous fluid source, and imposing the equality, we obtain the condition
\begin{equation}
2 \left(\rho+ \widetilde p \right) h_{\mu\nu} u^\mu u^\nu = \eta \left[2 h^{\mu\nu} \nabla_\mu u_\nu + u^\beta u^\lambda u^\rho \nabla_\lambda h_{\beta\rho} \right].
\label{329}
\end{equation}
In the absence of shear viscosity ($\eta=0$), it simply reduces to the same consistency condition of the perfect fluid (\ref{312}) (with the possible contribution of bulk viscosity, contained inside the effective pressure $\widetilde p$).
We are now in the position of discussing whether the presence of viscosity can modify or not the propagation of those metric fluctuations satisfying the TT gauge condition. To this purpose, we have to explicitly evaluate the right-hand side of Eq. (\ref{34}) by using, for the viscous fluid, the results (\ref{323}) and (\ref{328}).
We obtain, in this way, a propagation equation which generalizes the perfect fluid equation (\ref{313}), and introduces many additional new terms depending on the viscosity parameters, on the covariant derivatives of the velocity $u^\mu$, and on the perturbation components $h_{\alpha\beta}$. For the purpose of this paper it will enough to illustrate the case, very simple (but physically relevant for cosmological applications), of a viscous fluid with comoving velocity distribution, $u^0=1$, $u^i=0$, which is source of a background geometry described by a spatially flat FLRW metric (whose relevant connection and curvature components have been already reported in Eq. (\ref{315})).
In that case we can satisfy the consistency condition (\ref{329}) by assuming, as in the perfect fluid case, that $h_{0\mu}=0$, and that the TT gauge is valid for the spatial part $h_{ij}$ of the metric fluctuations. We then find that the source contributions on the right-hand side of Eq. (\ref{34}) are all vanishing, with the only exception of the contribution arising from the last term of the perturbed stress tensor (\ref{328}), generated by the shear viscosity and proportional to
\begin{equation}
u^\lambda \nabla_\lambda h_{ij} \equiv \dot h_{ij} - 2 H h_{ij}.
\label{330}
\end{equation}
By including this contribution, the propagation equation (\ref{34}) then takes the form
\begin{equation}
\nabla^2 h_{ij}+ 2 R_{kijl} h^{kl}+ 2 \eta\, \lambda_{\rm P}^2 \left( \dot h_{ij} - 2 H h_{ij} \right)=0.
\label{331}
\end{equation}
Written in terms of the mixed components, $h_i\,^k= g^{kj} h_{ij}$, we are finally lead to the result:
\begin{equation}
\ddot h_i\,^j + \left(3 H+ 2 \eta\lambda_{\rm P}^2\right) \dot h_i\,^j -{\partial^2\over a^2} \,h_i\,^j=0
\label{332}
\end{equation}
(the same equation was previously obtained, but with a different procedure, also in \cite{Pra}).
For vanishing (or negligible) viscosity, $\eta \rightarrow 0$, we recover the standard equation (\ref{316}). It is important to note, however, that for $\eta \lambda_{\rm P}^2 \sim H$ the presence of shear viscosity can affect in a significant way the evolution and the amplification of tensor metric fluctuations, and the consequent production of a relic cosmic GW background.
It may be useful to note, finally, that the above equation can be rewritten in terms of the conformal time coordinate $\tau$ (related to the cosmic time $t$ by $dt= a d\tau$) as follows:
\begin{equation}
h''_i\,^j +2\left( {\cal H}+ \eta \lambda_{\rm P}^2 a\right) h'_i\,^j -\partial^2 h_i\,^j=0,
\label{333}
\end{equation}
where ${\cal H}=a'/a$, and the prime denotes differentiation with respect to $\tau$.
In other words, we can say that the GW dynamics is affected by a friction coefficient
$\delta$ such that:
\begin{equation}
h''_i\,^j +2{\cal H}h'_i\,^j \left[1-\delta(\tau)\right] -\partial^2 h_i\,^j=0,
\label{334}
\end{equation}
where $\delta(\tau) =- \eta \lambda_{\rm P}^2 a/{\cal H}=-\eta \lambda_{\rm P}^2 a^2/a'$. It follows that the presence of shear viscosity, correctly included into the propagation dynamics of tensor metric perturbations according to Eq. (\ref{34}),
may produce effects very similar to those obtained in the context of modified theories of gravity (see e.g. the models and the examples discussed in
\cite{Mag,Fan}).
\section{Conclusion}
\label{sec4}
\setcounter{equation}{0}
In this paper we have presented a variational method to obtain the equations governing the evolution of linear metric perturbations in a general background geometry, with general sources and general choice of the coordinates. We have considered in particular the Einstein model of gravity, but the method can be applied to any given action assumed to describe the gravitational dynamics. Also, we have applied the action variational formalism to the metric variable only, without perturbing other field variables typical of the matter sources.
We have concentrated our discussion on the metric fluctuations satisfying the transversality and traceless conditions, and on the associated background geometries compatible with such conditions. In that case, we have derived a general equation describing the propagation of gravitational radiation, and the important result is that the effective form of such equation depends (as expected) not only on the background geometry, but also -- for the same given geometry -- on {\it how} the metric couples to the gravitational sources inside the matter part of the action.
We have given explicit examples for standard cosmological geometries generated by fluid sources, with and without bulk and shear viscosity, and we have found that the metric coupling to the shear viscosity is able in principle to leave an imprint on the primordial GW spectrum (an effect that we plan to discuss in detail in a future paper). The case of electromagnetic field sources, presented in the Appendix B, is also worth of further study. Finally, we are planning to apply the method of this paper to generalize the results obtained for metric perturbations in less trivial cosmological scenarios,
like those described by background geometries of Bianchi-type, Lema\^itre-Tolman-Bondy-type, or Light-Cone-type
(see e.g. \cite{Pereira,Fanizza2,Fanizza1,Mitsou}).
\section*{Acknowledgements}
Maurizio Gasperini and Luigi Tedesco are supported in part by INFN under the program TAsP ({\it ``Theoretical Astroparticle Physics"}), and by the research grant number 2017W4HA7S {\it ``NAT-NET: Neutrino and Astroparticle Theory Network"}, under the program PRIN 2017 funded by the Italian Ministero dell'Universit\`a e della Ricerca (MUR). Giuseppe Fanizza acknowledges support by FCT under the program {\it ``Stimulus"} with the grant CEECIND/04399/\\2017/CP1387/CT0026. It is a pleasure to thank Gabriele Veneziano for useful comments and discussions.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 7,812 |
Q: Missing 'Open Source', 'Apps & Software' sections in 'My Profile' on SO Careers For some reasons I don't have the 'OPEN SOURCE' and 'APPS & SOFTWARE' sections under 'My Profile' page on SO Careers.
Just on a side note, I'm not looking for a job so my SO Careers profile is 'Private'. I mainly use my SO Careers profile for answering, suggesting UI changes and reporting bugs. It's only '19% complete' but I used to see the 'OPEN SOURCE' and 'APPS & SOFTWARE' sections even with this level of completion.
Here is what I see now:
As you can see in the above screenshot, all I have now is 'JOB STATUS', 'TECHNOLOGIES', 'EXPERIENCE' and 'EDUCATION' sections below the 'Personal Statement'.
How do I get the 'OPEN SOURCE' and 'APPS & SOFTWARE' sections back in my profile?
P.S.
I can reproduce this in Chrome Ver. 45.0.2454.101 (64-bit), Firefox Ver. 41.0.1 and Safari Ver. Version 9.0 (10601.1.56.2) on my MacBook Pro Retina; OS X Yosemite Ver. 10.10.5
A: This is part of a UI change we're testing out. To consider your profile passing the required minimum for being classified as a non-abandoned profile, we require that you fill out a personal statement, at least two tags, and at least one entry under education or experience.
Until that threshold is met, we're currently trying out hiding all the other fields since for most people the extra fields are just extra noise right now.
The threshold is mostly used for determining if your profile shows up in our candidate search product, and since your CV is set to private this will never be the case, so it's worth looking into and seeing if we should not do this test for profiles that are set to private like yours.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 5,487 |
The 2001–02 Umaglesi Liga was the thirteenth season of top-tier football in Georgia. It began on 28 July 2001 and ended on 22 May 2002. Torpedo Kutaisi were the defending champions.
Locations
First stage
League table
Results
Second stage
Championship group
Table
Results
Relegation group
Table
Results
Relegation play-offs
Top goalscorers
See also
2001–02 Pirveli Liga
2001–02 Georgian Cup
References
Georgia - List of final tables (RSSSF)
Erovnuli Liga seasons
1
Georgia | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 2,973 |
Rita, married Edwin James McEvoy, son of Edwin Robert Joseph McEvoy and Alice Gallah Barrett.
Mrs. R. Adkins, widow of Mr. Norman Adkins, returned this morning after an absence of three and a half years spent at Cungena, west coast of South Australia. She has large property interests here, and intends to once again take up her residence in the city.
Mrs. Adkins, after leaving, intended to remain on the west coast, but the death of her father, Mr. Joseph McEvoy, a farmer of Kildalton, Cungena, altered her plans. She made the trip from Streaky Bay to Adelaide by boat and intended to come on here by 'plane, but owing to the boat being delayed by bad weather she travelled by train.
Mrs. Adkins travelled on the Yandera, and the sea was so rough that the boat sheltered for a day and a half at Flinders Island. It was intended to load wool there, but owing to the rough sea the wool had to be left behind. A horse was put ashore, the animal swimming the last quarter of a mile. This was the second horse taken on ship for Mr. Willie Bascombe of the island. He and his family are the sole occupants. The famous Junkers 'plane, Mount Wedge, which was on the Broken Hill Adelaide run, and was piloted by Mr. J. A. Mollison, was named after Mr. Bascombe's father's station out of Elliston.
Ronald Robert McEvoy was born in March 1929, son of Alexander Dennis McEvoy and Myrtle Cicely North.
Ronald and Jean were divorced. | {
"redpajama_set_name": "RedPajamaC4"
} | 291 |
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\section{Introduction}
Suppose $G$ is an amenable group and $A$ is a finite set, called an alphabet.
Suppose we fixed some F\o lner sequence $\lbrace F_n \rbrace$.
For an infinite element $x \in A^G$ in the discourse of the Kolmogorov complexity it is very natural to consider some kind of mean information along the F\o lner sequence:
\[
hc(c)=\limsup_{n \to \infty} \frac{C(x{\vert}_{F_n})}{\lvert F_n \rvert}
\]
In this way we should make some assertions, to guarantee well behaviour of the defined quantity. If we are working with a group like $\mathbb{Z}^d$ we do not care about computability. Passing to more general case we have to explicitly require our group to be computable. Then, there is a choice of an object to measure the complexity: the string, obtaining by listing of letters, marking elements of given $F_n$ in some order, or the whole object: the map from $F_n$ to $A$. In the second case we get an additional information from the configuration of $F_n$. To avoid such troubles, we introduce the notion of the modest F\o lner sequence, that is such F\o lner sequence, that complexities of its elements are asymptotically negligible to their sizes: $C(F_n)=o(\lvert F_n \rvert)$. This will help us establish equivalence of the two definitions of the asymptotic Kolmogorov complexity.
It is not hard to see, that asymptotic Kolmogorov complexity do not increases under the action of cellular automaton(proposition \ref{asymptotic complexity properties}). The point of interest is to describe those cellular automata, which preserves asymptotic complexity in terms of some well-known properties of cellular automata. The first result of such kind, known to author, contained in the paper \cite{CRI}, there proved, that every invertible cellular automaton over $\mathbb{Z}$ with finite alphabet preserves the asymptotic Kolmogorov complexity, which is defined in the following way:
\[
\limsup_{i\to \infty} \frac{C(x_i)}{2 i + 1}
\]
there $C(x_i)$ is the Kolmogorov complexity of the restriction of $x$ to the segment $[-i,i]$.
The Garden of Eden theorem states, that for the cellular automaton with finite alphabet over an amenable group surjectivity(that is non-existence of Garden of Eden configurations) and pre-injectivity are equivalent(cellular automata is called pre-injective, if every two different configurations with same images under action of cellular automaton, differ on infinite set). We will prove in this paper, that the class of cellular automata, which preserve asymptotic Kolmogorov complexity is exactly the class of surjective and pre-injective cellular automata.
This paper organized in the following way. In section \ref{computable amenable groups} we discuss the notion of a computable group and prove existence of a computable F\o lner sequences in computable groups. In section \ref{kolmogorov complexity} we remind the notion of the Kolmogorov complexity and its basic properties. Also, there is proved a crucial proposition \ref{complexity and size}, which equip us with the lower bounds of the preimages of a computable functions. In section \ref{cellular automata} we remind some definitions, concerning cellular automata. In section \ref{entropy and complexity} we discuss properties of entropy and asymptotic Kolmogorov complexity. At first, we give the definition of the modest F\o lner sequence, Then, we remind the definition of the entropy and define the asymptotic Kolmogorov complexity. After, we prove the basic properties of the asymptotic complexity.
In proposition \ref{asymptotic complexity properties} we prove, that the entropy of effectively closed set bounds the asymptotic complexity of its elements.
After it we remind the Garden of Eden theorem for amenable groups and Curtis-Hedlund-Lyndon theorem. In the end of this section we give the proof of the main result, theorem \ref{main}, which states, that asymptotic Kolmogorov complexity along the modest F \o lner sequence is presserved by the action of the cellular automaton over the computable amenable group if and only if this automaton is preinjective.
\section{Computable amenable groups}\label{computable amenable groups}
\begin{deff}
A pair of a countable or finite set and a bijection of it with an enumerable subset of the natural numbers called a constructible set.
\end{deff}
\begin{deff}
Let $G$ be some some group structure on $\mathbb{N}$, such that $0$ is the identity element. Then, if function $(x,y)\mapsto x \circ y$ is computable, we will call the group $G$ computable.
\end{deff}
It is obvious, that the operation $x \mapsto x^{-1}$ in computable group is computable function.
Throughout this paper we will assume, that $G$ is computable, if not, we could enhance our computability class by a suitable oracle.
\begin{deff}
Suppose $G$ is a countable group. We call a sequence $\lbrace F_n \rbrace$ of its finite subsets a F\o lner sequence for this group, if for every $g \in G$
\[
\lim\limits_{n\to \infty}\frac{\lvert F_n g\Delta F_n \rvert}{\lvert F_n\rvert} = 0
\]
there $\Delta$ denotes the symmetric difference.
\end{deff}
\begin{deff}
We call a countable group amenable, if it has a F\o lner sequence.
\end{deff}
\begin{prop}\label{folner reformulated}
Suppose, $G$ is a countable amenable group, $\lbrace F_n \rbrace$ --- its F\o lner sequence. Then for every finite subset $S$ of $G$ we have
\[
\lim\limits_{n\to \infty}\frac{\lvert F_n S\Delta F_n \rvert}{\lvert F_n\rvert} = 0
\]
\end{prop}
\begin{proof}
Since $S=\lbrace s_1, \ldots, s_k\rbrace$, we have
\begin{multline*}
F_n S \Delta F_n = \bigcup_{i=1}^{k}{(F_n s_i \setminus F_n)} \cup \bigcap_{j=1}^{k}{(F_n \setminus F_n s_j)} \subset \bigcup_{i=1}^{k}{(F_n s_i \Delta F_n)} \cup {(F_n \setminus F_n s_1)} \subset \\ \subset \bigcup_{i=1}^{k}{(F_n s_i \Delta F_n)} \cup {(F_n \Delta F_n s_1)}
\end{multline*}
From which we could infer, using the definition of F\o lner sequence
\[
\lim_{n\to \infty} \frac{\lvert F_n S\Delta F_n \rvert}{\lvert F_n\rvert} \leq \lim_{n \to \infty}\frac{\sum_{i=1}^{k}{\lvert F_n s_i \Delta F_n \rvert} + \lvert F_n \Delta F_n s_1\rvert}{\lvert F_n \rvert} = 0
\]
\end{proof}
\begin{prop}
Suppose $G$ is an infinite countable amenable group, and $\lbrace F_n \rbrace$ is a F\o lner sequence on it, then $\lvert F_n\rvert \to \infty$
\end{prop}
\begin{proof}
Suppose, on the contrary, for some constant $c$ we have infinitely many indices $i$, such that $\lvert F_i\rvert \leq c$. Then, passing to subsequence, we will get a F\o lner sequence $\lbrace F_{n_i}\rbrace$ with $\lvert F_{n_k}\rvert \leq c$. Take some finite subset $S$ of $G$ with $\lvert S \rvert \geq 2 c$ we get a contradiction with the proposition \ref{folner reformulated}:
\[
\frac{\lvert F_{n_i} S \Delta F_{n_i} \rvert}{\lvert F_{n_i}\rvert}\geq 1
\]
for every $i$.
\end{proof}
Every finite subset of the natural numbers could be computably encoded by the natural numbers, so, they forms the constructible set. It is clear, that set-theoretic operations with the finite subsets are computable, and function $(A,B) \mapsto AB$ there $A$ and $B$ are the finite subsets of computable group is computable too.
\begin{prop}\label{existence of folner}
There exists a computable F\o lner sequence , with $\lvert F_n\rvert \geq n$ in every computable amenable group.
\end{prop}
\begin{proof}
Fix some computable enumeration of the finite subsets. Let $F_n$ be the first set, such that \[\frac{\lvert F_n i\Delta F_n \rvert }{\lvert F_n\rvert}\leq \frac{1}{n} \text{ for every
i, }0\leq i\leq n \] and $\lvert F_n \rvert \geq n$.
${F_n}$ is a desired computable F\o lner sequence.
\end{proof}
\section{Kolmogorov complexity}\label{kolmogorov complexity}
We will use the notation $C(x)$ for the plain complexity and $C(x \vert y)$ for the conditional complexity, for the definition see \cite{LV08}, definition 2.1.2, p. 106 . Plain complexity of the pair $(x,y)$ is denoted as $C(x,y)$, for the definition see \cite{LV08}, example 2.1.5, p. 109.
\begin{prop}\label{complexity properties}
For the plain Kolmogorov complexity the following statements hold:
\begin{enumerate}
\item $C(x) \leq \log x + O(1)$
\item if $f$ is computable, then $C(f(x)) \leq C(x) + O(1)$
\item There exist a constant $c$, such that $\left\lvert \lbrace x \vert C(x) \leq n\rbrace \right\rvert \leq c 2^n$
\item $C(x,y) = C(y,x)+ O(1)$
\item $C(x,y) \leq C(x) + C(y) + 2 \log \min(C(x),C(y)) + O(1)$
\item $C(x|y) \leq C(x) + O(1)$
\item $C(x,y) = C(x) + C(y\vert x) + O(\max(\log C(y),\log C(x)))$
\end{enumerate}
\end{prop}
\begin{proof}
For the proofs, see \cite{LV08}:\\
(1) is theorem 2.1.2 on the page 108, \\
(2) and (3) are simple consequences of the definition of plain complexity,\\
(4) is follows from (2) and definition of complexity of pair,\\
(5) is proved in example 2.1.5 on the page 109, \\
(6) is from exercise 2.1.5 on page 113,\\
(7) is a slight reformulation of the theorem 2.8.2 on the page 190.
\end{proof}
\begin{prop}\label{inverse complexity}
Suppose, $f$ is a computable function. If $x=f(y)$, then
\[
\log \lvert f^{-1}(x) \rvert \geq C(y) - C(x) - O(\log \lvert C(y) - C(x) \rvert)
\]
\end{prop}
\begin{proof}
Fix some computable enumeration $t(x,i)$ ($i \leq 0$) of the preimage of the function $f$, such that if $t(x,i)$ does not fail, then $t(x,j)$ does not fail for all $0 \leq j\leq i$ and
$t(x,i) \neq t(x,j)$ (if both do not fail) for $i \neq j$ . Then, there exists $i < \lvert f^{-1}(x)\rvert$, such that $y=t(x,i)$. Using proposition \ref{complexity properties}, we
get
\[
C(y)=C(t(x,i)) \leq C(x,i) + O(1) \leq C(x) + \log \lvert f^{-1}(x) \rvert + \log \log \lvert f^{-1}(x) \rvert + O(1),
\]
so
\[
\log \lvert f^{-1}(x) \rvert + \log \log \lvert f^{-1}(x) \rvert \geq C(y) - C(x) + O(1)
\]
which implies desired estimate.
\end{proof}
\begin{prop}\label{complexity and size}
Suppose, that ${\lbrace V_i \rbrace}_{i \in I}$ is an enumerable family of enumerable sets, there $I$ is some constructible set. Suppose, there exists a function $f: \mathbb{N} \to
\mathbb{N}$, such that for every $i \in I$ we have $\lvert V_i\rvert \leq f(i)$, then for every $x\in V_i$ the following holds:
\[
C(i,x)\leq \log \lvert V_i \rvert + C(i) + O(log C(i))
\]
\end{prop}
\begin{proof}
Follows from proposition \ref{complexity properties}.
\end{proof}
\section{Cellular automata}\label{cellular automata}
Suppose $G$ is a group and $A$ is a finite set. We will call $A$ an alphabet. We will call $A^G$ a configuration space.
There is a natural left action of $G$ on $A^G$ defined in the following way:
\[
(gx)(g')=x(g^{-1}g').
\]
\begin{deff}
We will call a word every map from any finite subset of $G$ to $A$.
\end{deff}
The set of all words is denoted by $A^{\star}$. For a word $w$ we denote $\mathop{dom} w$ its domain. By $\lvert w \rvert$ we will denote the size of its domain. Consider a word $w$, by the
definition it is a map from the finite set $\lbrace g_1, \ldots g_{\lvert w \rvert}\rbrace$ to the alphabet $A$
We will endow $A^G$ with the product topology, assuming that $A$ endowed with the discrete topology. By Tychonoff theorem, $A^G$ is a compact Hausdorff topological set. Sets
\[
U(w)=\left\lbrace \left. x\in A^G \right \vert x{\vert}_{\mathop{dom} w}=w \right\rbrace
\]
are clopen and forms the base of the topology on $A^G$.
Assuming $G$ is computable, it is obvious, that every word could be computably encoded by some natural number.
Since all finite subsets of natural numbers and all words forms constructible sets, it make sense to consider the Kolmogorov complexity of such objects, and we will use the same notation as fo plain complexity: $C(B)$, $C(w)$, there $B$ is a finite subset, $w$ is a word.
Consider a word $w$, by the definition it is a map from the finite set $\lbrace g_1, \ldots g_\rbrace$. Assume, that $g_i < g_j$ for $1\leq i < j \leq \lvert w \rvert $ (remind, that group $G$ is defined on the set $\mathbb{N}$). Then, string $w(g_1)w(g_2)\ldots w(g_{\lvert w \rvert}) $ is called a content of word $w$, and denoted as $\mathop{cont} w$. By the definition, for word $w$ holds $C(w)=C(\mathop{dom} w, \mathop{cont} w)$.
\begin{deff}
Cellular automaton over group $G$ and the finite alphabet $A$ is a map $\tau: A^G \to A^G$ such that there exist a finite set $S \subset G$, and a map $\mu: A^S \to A$, for that the following
equality holds:
\[
\tau(x)(g)=\mu(g^{-1}x{\vert}_S)
\]
Set $S$ is called a memory set, $\mu$ is called a rule.
\end{deff}
\begin{deff}
Suppose $\tau$ is a cellular automaton with a memory set $S$ and a rule $\mu$, Suppose, $w$ is a word, such that $\mathop{dom} w = BS$ for some subset $B$ of $G$. Then we will define
\[
\tau(w)(g)=\mu(g^{-1}x{\vert}_S)
\]
for $g \in B$. So it is no more, than the restriction of the action of the cellular automaton to the word $w$.
\end{deff}
Evidently, if $G$ is a computable group and $\tau$ is a cellular automaton over it, then restriction of $\tau$ on words is computable function.
\section{Entropy and asymptotic Kolmogorov complexity}\label{entropy and complexity}
We want our F\o lner sequence not to affect Kolmogorov complexity
\begin{deff}
We will call a F\o lner sequence $\lbrace F_n \rbrace$ a modest F\o lner sequence, if
\[
\lim\limits_{n\to \infty} \frac{C(F_n)} {\lvert F_n \rvert} = 0
\]
\end{deff}
It obviously follows from the proposition \ref{existence of folner}, that there exist at least one modest F\o lner sequence.
It would be convinient for us to use the following notation for the restriction of configuration $x\in A^G$ to the subset $B$ of G: $x{\vert}_{B}$. If $B$ is a finite set, then this restriction is a word, so, if in addition, $G$ is a computable group, then this would justify the notation $C(x{\vert}_{B})$. We would use the same notation for the restriction of words as well: $w{\vert}_{B}$ if $B$ is a subset of the domain of word $B$.
Let us fix some modest F\o lner sequence $\lbrace F_n \rbrace$.
\begin{deff}
Consider some subset $X$ of $A^{G}$. We will define its entropy as
\[
h(X)=\limsup\limits_{n \to \infty} \frac{\log \left\lvert{ X {\vert}_{F_n}}\right\rvert}{\lvert F_n\rvert}
\]
\end{deff}
\begin{prop}
For the entropy the following holds:
\begin{enumerate}
\item $h(X) \leq \log \lvert A \rvert$
\item $h(A^G)=\log \lvert A\rvert$
\item $h(\tau(X)) \leq h(X)$.
\end{enumerate}
\end{prop}
\begin{proof}
For the proof see \cite{CAG}, propositions 7.7.2 and 5.7.3 on pages 125,126.
\end{proof}
\begin{deff}
For the element $x \in A^G$ we will define its asymptotic complexity as
\[
hc(x)=\limsup\limits_{n \to \infty} \frac{C( x {\vert}_{F_n})}{\lvert F_n\rvert}
\]
\end{deff}
\begin{prop}
For every $x \in A^G$ we have
\[
hc(x)=\limsup\limits_{n \to \infty} \frac{C \left({ \mathop{cont} (x {\vert}_{F_n}) }\right) }{\lvert F_n\rvert}
\]
\end{prop}
\begin{proof}
It is enough to prove, that
\[
\lim\limits_{n \to \infty} \frac{C(x{\vert}_{F_n})-C( \mathop{cont} (x {\vert}_{F_n}))}{\lvert F_n\rvert}=0
\]
Using proposition \ref{complexity properties} we have
\begin{multline*}
\lim\limits_{n \to \infty} \frac{\left\lvert C(x{\vert}_{F_n})-C( \mathop{cont} (x {\vert}_{F_n}))\right\rvert}{\lvert F_n\rvert} = \\
= \lim\limits_{n \to \infty} \frac{\left\lvert C(F_n \vert (x{\vert}_{F_n})) + O(\log (\max (C(F_n), C(\mathop{cont} (x {\vert}_{F_n})))))\right\rvert} {\lvert F_n\rvert} = 0
\end{multline*}
the last equality is true, since
\[
C(F_n \vert (x{\vert}_{F_n})) \leq C(F_n) + O(1) =o(\lvert F_n \rvert),
\]
(by the proposition \ref{complexity properties} and the definition of modest F\o lner sequence), and
\[
C(\mathop{cont} (x {\vert}_{F_n})) \leq \lvert F_n \rvert \log \lvert A \rvert + O(1).
\]
by the proposition \ref{complexity properties}.
\end{proof}
The following proposition allows us to estimate the difference of the complexities of two words in terms of the symmetric difference of their supports.
\begin{prop}\label{complexity difference}
Suppose, $w_1$ and $w_2$ are two words, then
\begin{multline*}
\left\lvert C(w_1)-C(w_2) \right \rvert \leq \lvert \mathop{dom} (w_1) \Delta \mathop{dom} (w_2) \rvert \cdot \log \lvert A \rvert +\\+ O(\max(C(\mathop{cont} (w_1), \mathop{cont}(w_2))))
\end{multline*}
\end{prop}
\begin{proof}
Evidently, there exists an algorithms, which recovers $w_1$ from the triple $w_2$, $\mathop{dom} (w_1)$ and $\mathop{cont}(w_1\vert_{\mathop{dom}(w_1 \setminus w_2 )})$. From the proposition
\ref{complexity properties} follows
\[
C(w_1) \leq C(w_2) + O(C(\mathop{dom}(w_1))) + C(w_1) + \log \lvert A \rvert \cdot \lvert \mathop{dom} (w_1) \Delta \mathop{dom} (w_2) \rvert.
\]
By symmetry we get
\[
C(w_2) \leq C(w_1) + O(C(\mathop{dom}(w_2))) + C(w_2) + \log \lvert A \rvert \cdot \lvert \mathop{dom} (w_1) \Delta \mathop{dom} (w_2) \rvert.
\]
Combining this two we get the desired estimate.
\end{proof}
Let us endow $A^G$ with the Bernoulli measure $\nu$.
\begin{prop}\label{asymptotic complexity properties}
The following holds:
\begin{enumerate}
\item for every $x\in A^G$ we have $hc(x) \leq \log \lvert A \rvert$
\item for almost every $x \in A^G$ we have $hc(x)=\log \lvert A \rvert$
\item for every $x \in A^G$ we have $hc(\tau(x)) \leq hc(x)$.
\end{enumerate}
\end{prop}
\begin{proof}
The following inequality implies the first statement :
\[
C(\mathop{cont}(x\vert_{F_n})) \leq \log \lvert F_n\rvert\lvert A \rvert + O(1),
\]
For the proof of the second statement consider a set
\[
E_{\varepsilon}=\lbrace x\in A^G \vert hc(x) < (1-\varepsilon)\log \lvert A \rvert \rbrace
\]
We have
\[
E_{\varepsilon} = \bigcup_{i>0} \bigcap_{j>i} \lbrace x \in A^G \vert C(\mathop{cont} (x\vert F_j)) \leq (1-\varepsilon) \lvert F_j\rvert \log \lvert A \rvert\rbrace
\]
But since
\begin{multline*}
\nu\left(\left\lbrace x \in A^G \vert C(\mathop{cont}(x\vert F_j)) \leq (1-\varepsilon) \lvert F_j\rvert\log \lvert A \rvert\right\rbrace\right) \leq \\ \leq \frac{\left\lvert \left\lbrace {y \vert C(y) <
(1-\varepsilon)
\lvert F_j\rvert \log \lvert A \rvert} \right\rbrace \right\rvert}{{\lvert A \rvert}^{\lvert F_j \rvert}} \leq {\lvert A \rvert}^{-\varepsilon \lvert F_j\rvert} \to 0
\end{multline*}
We have $\nu(E_{\varepsilon})=0$. Taking a sequence $\varepsilon_n \to 0$, we see, that $\nu(\cup_{i>0} {E_{\varepsilon_i}})=0$, so, almost every $x \in A^G$ has $hc(x)=\log \lvert A
\rvert $, because $hc(x) \leq \lvert A \rvert$ for every $x$.
For the proof of the third statement, consider
\[
hc(x)=\limsup\limits_{n \to \infty} \frac{C( x {\vert}_{F_n})}{\lvert F_n\rvert}
\]
and
\[
hc(\tau(x))=\limsup\limits_{n \to \infty} \frac{C( \tau(x) {\vert}_{F_n})}{\lvert F_n\rvert}
\]
but
\[
\tau(x) {\vert}_{F_n} = \tau(x {\vert}_{F_n S}),
\]
so
\[
C(\tau(x) {\vert}_{F_n}) \leq C(x {\vert}_{F_n S})+O(1).
\]
This mean, it is enough to prove, that
\[
\lim\limits_{n\to\infty}\frac{C(x {\vert}_{F_n S})-C(x {\vert}_{F_n})}{\lvert F_n \rvert }=0
\]
Proposition \ref{complexity difference} implies
\begin{equation*}
\left\lvert C(x {\vert}_{F_n S})-C(x {\vert}_{F_n}) \right\rvert \leq \lvert F_n S \Delta F_n \rvert \cdot \log \lvert A \rvert + O(\max(C(F_n),C(F_n S)))
\end{equation*}
By the definition of the F\o lner sequence,
\[
\lvert F_n S \Delta F_n \rvert = o(\lvert F_n \rvert)
\]
Using proposition \ref{complexity properties}, since $F_n S$ is computable from $F_n$, and by the definition of modest F\o lner sequence we have
\[
C(F_n S) \leq C(F_n) + O(1) = o(\lvert F_n \rvert),
\]
which finishes proof.
\end{proof}
\begin{deff}
An open set $V$ is called an effectively open set, if
\[
V=\bigcup\limits_{w \in W} U(w)
\]
for some enumerable set $W$ of words.
A closed set is called effectively closed, if it is a complement of an effectively open set.
\end{deff}
\begin{prop}\label{entropy and asymptotic complexity}
If $X$ is an effectively closed set, then for every $x \in X$ we have
\[
hc(x) \leq h(X).
\]
\end{prop}
\begin{proof}
It is clear, that $\lvert X {\vert}_{F_n}\rvert \leq {\lvert A \rvert} ^ {\lvert F_n \rvert}$
Let us fix some rational $q>h(X)$. By the definition of the entropy, there exists a constant $c$, such that
\[
\lvert X {\vert}_{F_n}\rvert \leq c 2^{q \lvert F_n \rvert}
\]
If there exists an enumerable family of enumerable sets $\lbrace V_f \rbrace$, there $f \in
\lbrace F_1, F_2, \ldots \rbrace$, such that $X {\vert}_{f} \subset V_f$ and $\lvert V_{F_n}\rvert
\leq c 2^{ q \lvert F_n \rvert}$ for
all $n$, then by the proposition \ref{complexity and size} we would have for $y \in V_n$
\[
C(y, F_n) \leq O(C(F_n)) + q \lvert F_n \rvert
\]
and therefore, for every $x \in X$
\[
hc(x)\leq q
\
But since $q$ is an arbitrary rational, bigger than $h(X)$, we would have $hc(x) \leq hc(X)$ for every $x \in X$.
Let us prove existence of $\lbrace V_f \rbrace$.
$X$ is an effectively closed set, so
\[
A^G \setminus X = \bigcup_{w \in W} U(w)
\]
for some enumerable set of words $W$.
Consider a word $v$ with $\mathop{dom} v = f$. Suppose, that $U(v) \cap X = \varnothing$. Then,
\[
U(v) \subset \bigcup_{w \in W} U(w),
\]
and since it is an open covering of the compact set, we have, $U(v)$ could be covered by some finite subset. This mean, that if $U(v) \cap X = \varnothing$, we could realize it in
a finite time. Therefore, the set of such $v$-s is enumerable(and computably depends on n). Really, assertion, that $U(v)$ is covered by $U(w_1), \ldots, U(w_k)$ is equivalent to the fact, that for every word $t$ with domain
\[
E = \mathop{dom} (v) \bigcup_{i=1}^k{\mathop{dom}(w_i)}
\]
from the fact, that $\left(t{\vert}_{\mathop{dom} v}\right)=v$
follows, that there exists $1 \leq i \leq k$, such that $\left(t{\vert}_{\mathop{dom} w_i}\right)=w_i$
This means, that sets
\[
V_{f,k}=(A^G \setminus \bigcup_{i=1}^{k}{U(w_i)}){\vert}_f
\]
forms the computable family of the finite sets, and we could get $V_f$ to be the first $V_{f,k}$ with
$ \lvert V_{f,k} \rvert \leq c \cdot 2^{q \lvert f \rvert}$.
\end{proof}
\begin{deff}
A cellular automaton $\tau$ over a group $G$ and an alphabet $A$ is called pre-injective, if for every $x,y \in A^G$, such that $x$ and $y$ coincides outside some finite set, and $\tau(x)=\tau(y)$, we have $x=y$.
\end{deff}
\begin{deff}
A configuration $x\in A^G$ is called a Garden of Eden configuration, if its pre-image is empty.
\end{deff}
\begin{theor}[The Garden of Eden theorem]
For a cellular automaton $\tau$ over a group $G$ and an alphabet $A$ the following are equivalent:
\begin{enumerate}
\item There is no Garden of Eden configuration.
\item $\tau$ is pre-injective.
\item $h(\tau(A^G))=h(A^G)$.
\end{enumerate}
\end{theor}
\begin{proof}
See \cite{CAG} theorem 5.8.1 on page 128.
\end{proof}
\begin{theor}[Curtis-Hedlund-Lyndon theorem]
Suppose $\tau$ is a map from $A^G$ to itself, there $G$ is a group and $A$ is a finite set. Then the following are equivalent:
\begin{enumerate}
\item $\tau$ is a cellular automaton.
\item $\tau$ is continuous and shift-invariant, that is for every $x \in A^G$ and $g \in G$ we have $\tau(gx)=g \tau(x)$.
\end{enumerate}
\end{theor}
\begin{proof}
See \cite{CAG}, theorem 1.8.1 on page 20.
\end{proof}
\begin{prop}
Suppose, that cellular automaton $\tau$ has a Garden of Eden configurations. Then for some $x \in
A^G$ we have $hc(\tau(x)) < hc(x)$.
\end{prop}
\begin{proof}
By the Garden of Eden theorem, if there exists a Garden of Eden configuration, then $h(\tau(A^G)) < h(A^G)$. By the proposition \ref{asymptotic complexity properties} there exist an element $x \in A^G$ with
$hc(x)=\log \lvert A \rvert$. The set $\tau(A^G)$ is a closed subset of the set $A^G$, because it is image of the compact set under the action of the continuous map.
Since the statement $U(v)\cap \tau(A^G)=\varnothing$ is equivalent to ${\tau}^{-1}(v)=\varnothing$, we have, that the set $\tau(A^G)$ is effectively closed. So, by the proposition \ref{entropy and asymptotic complexity}, we have, that $hc(\tau(x)) \leq h(\tau(A^G)) < \log \lvert A \rvert = hc(x)$.
\end{proof}
\begin{prop}
Suppose we have $x \in A^G$, such that $hc(\tau(x)) < hc(x)$. Then the cellular automaton is not pre-injective.
\end{prop}
\begin{proof}
Without loss of generality we may assume, that $S=S^{-1}$, and that $S$ contains the identity.
Since $hc(\tau(x)) < hc(x)$, there is a subsequence $\lbrace n_k \rbrace$ and two constants $a<b$,
such that
\[
\frac{C(\tau (x) {\vert}_{F_{n_k}})}{\lvert F_{n_k}\rvert} < a < b
< \frac{C (x {\vert}_{F_{n_k}})}{\lvert F_{n_k}\rvert}
\]
Using the proposition \ref{complexity difference} we could assume(maybe, passing to subsequence and slightly modifying constants $a$ and $b$), that
\[
\frac{C(\tau(x){\vert}_{F_{n_k}})}{\lvert F_{n_k}\rvert} < a' < b' < \frac{C( x {\vert}_{F_{n_k}
S})}{\lvert F_{n_k}\rvert}
\]
then, since
\[
\tau(x {\vert}_{F_{n_k} S})= \tau(x) {\vert}_{F_{n_k}}
\]
we could use the proposition \ref{inverse complexity} and denoting $y=\tau(x)$ we get
\[
\log \lvert {\tau}^{-1}(y{\vert}_{F_{n_k}})\rvert \geq
C(x{\vert}_{F_n S}) - C(y{\vert}_{F_n}) - O(\log \lvert C(x{\vert}_{F_n S}) - C(y{\vert}_{F_n}) \rvert)
\]
which imply, that for some big enough $N$ and positive constant $c$, for every $k>N$ the following holds:
\[
\log \lvert {\tau}^{-1}(y{\vert}_{F_{n_k}})\rvert \geq c \cdot \lvert F_{n_k} \rvert
\]
so
\[
\lvert {\tau}^{-1}(y{\vert}_{F_{n_k}})\rvert \geq 2^{c \lvert F_{n_k}\rvert}
\]
Consider a set $T$ of pairs of words $(v, w)$, there $f(v)=w$, $\mathop{dom} (w) = F_{n_k} S S$ and $v{\vert}_{F_{n_K}}=\tau(x){\vert}_{F_{n_k}}$. Its cardinality is at least $2^{c \lvert F_{n_k}\rvert}$ for $k>N$.
That is, for big enough $k$, by the pigeonhole principle there exists at least two pairs $(v_1,w_2)$ and
$(v_2,w_2)$ in $T$, such that $v_1$ and $v_2$ coincides outside the the set $F_{n_k}$, and $w_1=w_2$.
Really, $w_1$ and $w_2$ coincides on $F_{n_k}$, and we have ${\lvert A
\rvert}^{o(\lvert F_{n_k}\rvert)}$ variants for the filling $F_{n_k} S S \setminus
F_{n_k}$ and $F_{n_k} S \setminus F_{n_k}$ by elements of $A$.
Extending $w_1$ to some $x_1 \in A^G$ and $w_2$ to some $x_2 \in A^G$ in such a way, that they will coincide outside $F_{n_k}$, we will get, that automaton is not pre-injective, since $\tau(x_1)=\tau(x_2)$.
\end{proof}
Combining last two propositions and the Garden of Eden theorem, we get the following result
\begin{theor}\label{main}
Suppose, that $G$ is an infinite computable amenable group, a cellular automaton $\tau$ defined over it, and a modest F\o lner sequence selected(that is, such a F\o lner sequence, that $C(F_n)=o(\lvert F_n \rvert)$). Then the following are equivalent
\begin{enumerate}
\item $\tau$ has no Garden of Eden configurations
\item $\tau$ is pre-injective
\item $h(\tau(A^G))=h(A^G)$
\item $hc(\tau(x))=hc(x)$ for every $x \in A^G$.
\end{enumerate}
\end{theor}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 2,523 |
Hans Zuidersma (Nieuwenhagen, 17 januari 1949) is een voormalig profvoetballer van onder meer Roda JC, FC Wageningen en Fortuna Sittard.
Zuidersma begon met voetballen bij SVN uit zijn geboortedorp. Hij maakt zijn debuut in het betaalde voetbal in de toenmalige tweede divisie bij SV Limburgia. Na de noodgedwongen terugkeer van SV Limburgia naar de amateurs in 1971, stapt Zuidersma over naar Roda JC, dat toen uitkwam in de eerste divisie.
Met de club uit Kerkrade promoveert de rechtsbuiten in mei 1973 naar de eredivisie. Het lukt Zuidersma echter niet om een vaste basisplaats te veroveren bij Roda JC. Halverwege het seizoen 1974/75 haalt Fritz Korbach hem op huurbasis naar FC Wageningen dat tegen degradatie uit de Eredivisie vecht. De komst van Zuidersma mag niet baten, want aan het einde van het seizoen degradeert FC Wageningen alsnog.
Zuidersma keert terug naar Limburg waar hij vanaf het seizoen 1975/76 gaat spelen voor dan nog Fortuna SC. Hij groeit in Sittard uit tot een gewaardeerde vaste kracht en speelt jaarlijks met Fortuna mee voor de bovenste plekken in de eerste divisie.
Uiteindelijk lukt het Zuidersma in het seizoen 1981/82 om de zo gewenste promotie te bewerkstelligen. Na één seizoen met vanaf dan Fortuna Sittard op het hoogste niveau te hebben gespeeld, maakt Zuidersma in het seizoen 1983/84 op 34-jarige leeftijd nog een opmerkelijke rentree bij zijn oude club Roda JC. Een seizoen later beëindigt hij zijn betaald voetbalcarrière bij Roda JC en keert terug bij de amateurs, waar Zuidersma nog 3 seizoenen in de Hoofdklasse speelt bij eerst Limburgia en later SVN.
Profstatistieken
Zie ook
Lijst van spelers van Limburgia
Lijst van spelers van Roda JC
Lijst van spelers van FC Wageningen
Lijst van spelers van Fortuna Sittard
Nederlands voetballer | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 3,279 |
Guus Hiddink was approached to manage in Chinese Super League
Mike Tranter Feb 07, 2016
Mike Tranter
The excitement sparked by the announcement of Pep Guardiola's impending arrival as Manchester City manager has only served to underline the sense of uncertainty hovering over the dug-outs at both United and Chelsea.
On whether Premier League clubs should be concerned about China's willingness to compete for top players, Wenger said, "Yes, of course, because China looks to have the financial power to move the whole league of Europe to China".
He made way for Mathieu Flamini in the midweek goal-less draw against Southampton and Arsene Wenger is now weighing up whether to start Coquelin at Bournemouth on Sunday. They have an fantastic chemistry together and are one of the main reasons that Chelsea have been unbeaten in their last 10 games.
"He will not be in for the upcoming few games because we want to have him fit and we don't want to run the risk of injury in this tough league".
The United manager is hopeful of taking the Red Devils all the way to the Europa League final this season, and van Gaal spoke on Luke Shaw's inclusion into the squad. United are in some of their best form of the season, having scored three goals in consecutive matches.
"Everything is to do with the ambition of the club and loyalty is secured by contracts", said Hiddink.
United has struggled against Chelsea recently and has been even worse at Stamford Bridge, but things are trending in their favor heading into Sunday's meeting.
Guus Hiddink admits he has a professional relationship with Manchester United Louis Van Gaal.
Van Gaal said: "I have the feeling that when we shall beat Chelsea – and that is not simple, that is very hard, since Guus Hiddink has been there they don't lose any more – I think our competition, our league is starting again".
Van Gaal is quoted by the official club website as saying: "Jones is coming back". It's not that he can train with us but, still, he is doing well. Techinically he is very good.
Former Milan striker Pato has not played competitive football since late November and Hiddink says the 26-year-old remains some weeks off his Chelsea debut.
Hiddink is unbeaten in seven Premier League games, winning two, since replacing Jose Mourinho, who has been linked with the manager's job at Manchester United. In the Netherlands they know I am too arrogant to doubt myself, but I also know that such a nonsense is being created about me. | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 1,732 |
{"url":"https:\/\/www.tutorialspoint.com\/line-edge-covering","text":"# Line\/Edge Covering\n\nMathematicsComputer EngineeringMCA\n\n#### Mathematics for Data Science and Machine Learning using R\n\n64 Lectures 10.5 hours\n\n#### Engineering Mathematics - Numerical Analysis & more\n\n6 Lectures 1 hours\n\n#### Advanced Mathematics Preparation for JEE\/CET\/CAT\n\n30 Lectures 3.5 hours\n\nA covering graph is a subgraph which contains either all the vertices or all the edges corresponding to some other graph. A subgraph which contains all the vertices is called a line\/edge covering. A subgraph which contains all the edges is called a vertex covering.\n\n## Line Covering\n\nLet G = (V, E) be a graph. A subset C(E) is called a line covering of G if every vertex of G is incident with at least one edge in C, i.e.,\n\ndeg(V) \u2265 1 \u2200 V \u2208 G\n\nbecause each vertex is connected with another vertex by an edge. Hence it has a minimum degree of 1.\n\n## Example\n\nTake a look at the following graph \u2212\n\nIts subgraphs having line covering are as follows \u2212\n\nC1 = {{a, b}, {c, d}}\nC2 = {{a, d}, {b, c}}\nC3 = {{a, b}, {b, c}, {b, d}}\nC4 = {{a, b}, {b, c}, {c, d}}\n\nLine covering of 'G' does not exist if and only if 'G' has an isolated vertex. Line covering of a graph with 'n' vertices has at least [n\/2] edges.\n\n## Minimal Line Covering\n\nA line covering C of a graph G is said to be minimal if no edge can be deleted from C.\n\n## Example\n\nIn the above graph, the subgraphs having line covering are as follows \u2212\n\nC1 = {{a, b}, {c, d}}\nC2 = {{a, d}, {b, c}}\nC3 = {{a, b}, {b, c}, {b, d}}\nC4 = {{a, b}, {b, c}, {c, d}}\n\nHere, C1, C2, C3 are minimal line coverings, while C4 is not because we can delete {b, c}.\n\n## Minimum Line Covering\n\nIt is also known as Smallest Minimal Line Covering. A minimal line covering with minimum number of edges is called a minimum line covering of 'G'. The number of edges in a minimum line covering in 'G' is called the line covering number of 'G' (\u03b11).\n\n## Example\n\nIn the above example, C1 and C2 are the minimum line covering of G and \u03b11 = 2.\n\n\u2022 Every line covering contains a minimal line covering.\n\n\u2022 Every line covering does not contain a minimum line covering (C3 does not contain any minimum line covering.\n\n\u2022 No minimal line covering contains a cycle.\n\n\u2022 If a line covering 'C' contains no paths of length 3 or more, then 'C' is a minimal line covering because all the components of 'C' are star graph and from a star graph, no edge can be deleted.\n\nUpdated on 23-Aug-2019 11:54:44\n\nAdvertisements","date":"2022-10-06 20:46:39","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.3460344970226288, \"perplexity\": 1571.328828062066}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": false}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2022-40\/segments\/1664030337855.83\/warc\/CC-MAIN-20221006191305-20221006221305-00060.warc.gz\"}"} | null | null |
\section{Introduction}
NGC 1097 is a SBb galaxy with an Active Galactic Nucleus (AGN), located at a
distance of 14.5 Mpc (Tully, 1988; 1~arcsec~$\sim$~70~pc), which places it
among the nearest AGN in the Southern Hemisphere. It has been originally
classified as a LINER and further reclassified as a Seyfert type 1 (Sy 1)
nucleus after the detection of broad, double peaked $H\alpha$ emission by
Storchi-Bergmann et al. (1993). However, it is very moderate AGN with
$L_{2-10 \rm{keV}}\sim 10^{40}$ erg s$^{-1}$ (Terashima et al. 2002). Dust
extinction toward the nucleus, as inferred from the broad Balmer line ratio,
$H\alpha / H\beta = 3-4.2 $ (Storchi-Bergmann et al. 1995), is quite moderate,
$A_v \sim 1$. The ratio of Balmer
narrow components indicates similar extinction (Storchi-Bergmann et al. 1996).
NGC 1097 has a very strong bar at $PA\sim$141$^{\circ}\,$ and a prominent nuclear
star-forming ring of 0.7 kpc radius inside it. Interior to the ring, a
secondary nuclear
bar crosses the nucleus at PA=28$^{\circ}\,$ (Quillen et al. 1995), almost orthogonally
to the primary bar. The nucleus and the star-forming ring are prominent in CO
and HCN molecular emission (Kohno et al. 2003). The CO map shows three main
peaks: at the nucleus and at two diametrically opposite locations in
the
nuclear star-forming ring. The latter correspond to the connecting
points between
principal dust lanes in the outer bar and the star-forming ring. They also
roughly coincide with the ends of the inner nuclear bar, but they slightly
trail the bar (have larger PA). The resolution of Kohno's et al maps, $\sim
7.7''\times 4''$ in CO(1-0) and $10''\times4.5''$ in HCN(1-0), is insufficient
to confirm the presence of cold material between the nucleus and the ring.
The map and radial profile distribution presented by the authors hint to
the presence of some weaker CO emission in this region. First evidence for
the presence of cold material there comes from the Hubble Space Telescope (HST)
optical images analyzed by Barth et al. (1995), who reported on the presence
of dusty filaments in the surroundings of NGC 1097's nucleus.
Regarding the presence of warmer gas in the nuclear region of NGC 1097,
infrared (IR) spectroscopy reveals emission from both $H_2$ and HII at the
nucleus, the star-forming ring and the region in between. Long-slit IR
spectroscopy in several directions across the nucleus (Reunanen, Kotilainen \&
Prieto 2002; Kotilainen et al. 2000) shows strong $H_2$ 2.12 $\mu$m$\,$ line emission
from the
nucleus and the star-forming ring, and weak extended emission from the region
in between. Optical long-slit spectroscopy (Storchi-Bergmann et al. 1996)
shows strong ionized gas emission from both the nucleus (Sy 1 spectrum with
broad $H\alpha$ emission) and the ring (low excitation HII region spectrum),
and faint line emission from the region in between (a LINER type spectrum).
Br$\gamma$ emission is detected in the ring but not at the nucleus (Reunanen,
Kotilainen \& Prieto, 2002; Kotilainen et al. 2000).
Also the $H\beta$ line is
very weak in the nuclear spectra due to the absorption component by
the underlying
stellar contribution.
In this paper we present high spatial resolution images of NGC1097, obtained
with the Very Large Telescope (VLT) and the Adaptive-Optics-assisted NACO
camera/spectrograph in
the 1-2.4 $\mu$m$\,$ range. The physical scales traced by these images are of the
order of 10 pc, and thus they probe with unprecedented detail the presence and
extent of obscuring material in the very proximity of the nucleus.
After describing the observations in Section 2, and results in Section 3, in
Section 4 we attempt to explain the dynamics of the
central filamentary structure that is observed.
\section{Observations}
Images of the NGC~1097 nucleus in $J-$ $H-$ and $Ks-$ bands were collected
with the UT4 unit of the VLT during August 2002. The images were taken with
the NACO camera that provides a field of view of 27$\times$27 arcsec and a
scale of 0.027 arcsec per pixel. The adaptive optics correction was done
with the optical wave-front sensor closing the loop on NGC~1097 nucleus.
The achieved spatial resolution has FWHM of about $0.18''\times 0.15''$
in $Ks-$band, $0.20''\times 0.18''$ in $H-$band, and $0.20''\times0.19''$
in $J-$band. These values were measured on various isolated HII regions in
the star-forming ring, which we assumed to be point-like. These HII regions
are resolved in the HST WFPC1 image at 5500~\AA$\,$ (Barth et al. 1995), having
sizes about 4 pc (0.06 arcsec). In the NACO images,
we get almost constant angular
size for several of the HII regions,
as well as for the nucleus, and thus we believe this to
be the real resolution of the data. It is different from that derived from
a stellar PSF, but
estimating the spatial resolution on the basis of separate images of
stars is not viable because atmospheric correction achieved by the Adaptive
Optics (AO) system
is different depending on the type of source used for monitoring the
atmosphere. NGC~1097 images were corrected using the nucleus of the galaxy,
which means that the contamination by the galaxy light degrades the contrast
of the nuclear light. This leads to a systematically worse AO correction when
compared with that achieved from a separate image of an isolated bright star.
The spatial resolution achieved in the PSF star images
taken for this program turned out in general to be at least 30\% better than
that of the science images.
NACO $J-$ and $Ks-$band images, and the HST WFPC1 optical image of NGC 1097 are
shown in Fig. 1. The NACO $H-$band image is similar, and therefore it is not
shown. Although the resolution of the NACO images is about 4 times worse than
that of the optical HST image, this shortcoming is largely compensated
by the amount of new information revealed in the IR images. At all
wavelengths the AGN and the star-forming ring are detected, but
the IR images show in addition the presence of a central nuclear bar
and newly discovered HII regions in the ring, more than 300 compared
with 88 measured by Barth et al. (1995) in the HST image. The
comparison of the optical and IR images gives evidence for the presence of dust
all over the star-forming ring interlaced with the HII regions. The
ring morphology is more like two spiral arms, with each of the arms connecting
to an end of the nuclear bar. Note however that a different morphology is
disclosed by the $J-Ks$ color image presented in Section 3.3.
\section{Results}
\subsection{The nucleus}
At the resolution of these IR observations, the nucleus of
NGC~1097 remains a point-like source, with an upper limit to the size of
$0.19'' \times 0.14''$ FWHM in the $Ks-$band.
This is about the FWHM measured in
several of the best isolated HII regions in the star-forming
ring from the same $Ks$-band
image (see Section 2), and thus we consider the nucleus unresolved. At the
distance of NGC 1097, that angular resolution implies an upper limit
for the nucleus size of less than 10 pc diameter.
A light-profile decomposition of the central 3 arcsec ($\sim$210 pc)
radius of
NGC~1097 is shown in Fig. 2. It is fitted with a simple
two-component model consisting of an unresolved source -- the nucleus, represented
by a PSF,
and an underlying galaxy component or pseudo-bulge,
represented by a generalized Sersic model.
Surface brightness in the Sersic model follows a formula
$I(r) = I_o \exp [-(r/r_o)^{1/n}]$,
where $I_o$ is the central surface brightness, $r_o$ is a scaling radius,
and $n$ is the Sersic exponent (Sersic 1968). The PSF was estimated as
follows. The useful dynamic
range of the profile from the HII regions in the star-forming ring
is only $\sim5^m$; the
dynamic range reached in the PSF star images observed separately from
the science frames is however $10^m$. The core of PSF star images is
narrower than that measured for the HII regions. Therefore, a PSF
composite profile was produced which includes the profile of an HII
region up to a radius of 0.45 arcsec in $Ks-$band, 0.3 arcsec in $J-$ and
$H-$ bands, complemented with that of the stellar PSF from those radii on.
The Sersic exponent $n$ was allowed to vary between 1 and 4, but the best
fit in any of the bands was obtained for $n=2$. The results of the best fit
are shown in Fig. 2, and the parameters derived from this
fit are given in Table 1. The residuals from the fit are less than 5\%
at all radii. However, within 0.3-0.6 arcsec radius, the residuals
show structure which is most probably due to a still unsatisfactory
PSF, particularly in $J-$ and $H-$ bands.
The contribution of the point-like source, the AGN, to the total light at different
radii is given in Table 2. As expected, the AGN
contribution increases with wavelength. Within a 0.2 arcsec aperture
diameter, which is $\sim$ 1.5 times the achieved spatial resolution,
its contribution in the $Ks$-band is dominant, 90\%, but
it is 20\% less important in $J-$band. The colors of the nucleus are
rather red: $J-H=1.3$, $H-Ks=1$, indicative of dust temperature of 1000
K. However, this temperature should be taken as an upper limit
considering that the measurements still correspond to relatively large
physical scales --- the resolution of these
observations is about 10 pc. Warm and hot gas should also be present
at these distances from the center and therefore
free-free emission from these gas phases might also be
contributing at those scales to the near-IR, in particular if shock
excitation is present. This indeed seems to be the case as indicated
by the large line-widths, $100-300$ km s$^{-1}$, measured in the optical
spectra from regions in between the nucleus and the star-forming
ring (Storchi-Bergmann et al. 1996). Shock
velocities of that range can heat the gas to $T\sim 10^6 K$, and
bremsstrahlung emission from this gas will substantially contribute in
the IR (see e.g. Contini et al. 2004).
\subsection{The pseudo-bulge}
Table 2 gives the colors of the region between the nucleus and the
star-forming ring integrated directly on the images over increasing
concentric radii. This region, 0.7 kpc in radius, is referred to as
the pseudo-bulge of NGC 1097 in Section 3.1. The colors
are subtracted from the point source contribution estimated from the
radial profile fit (Section 3.1). Equivalent colors derived from the Sersic model
used in the fit are given in Table 2 for comparison. There is reasonable
agreement within a few percent
between the two estimates except for radii below 0.4 arcsec where the
nucleus and an additional source of emission
become important. This is discussed below.
The radial variation of the colors in the central 10 arcsec together
with a color-color diagram from the same region are plotted in
Fig. 3. These colors
are measured on the images from
azimuthally averaged concentric rings. By comparing the upper and
lower panels in Fig. 3, one sees that the reddest
colors, i.e. $0.8 < J-H< 0.95 $ and $0.3< H-Ks < 0.5$, are measured
at radii below 1 arcsec. The particular trend
followed by these extreme colors in the $J-H$ vs $H-Ks$ diagram is caused
by contamination by the nucleus light particularly in $Ks-$band (see Table
2). It may also be introduced by the underlying central spiral
structure revealed in the color images (see Section 3.3), which
becomes more dense with decreasing radius. This contamination seems
to fade at about 1 arcsec ($\sim$70 pc) radius as indicated by the
clear color turnover at $J-H \sim 0.8$ and $H-Ks \sim 0.3$ in the color-color
diagram. From that radius outward pseudo-bulge dominates the color,
which becomes monotonically bluer, reaching a minimum $J-H \sim 0.7$,
$J-Ks \sim 0.24$ at $\sim 5$ arcsec (350 pc) radius from the
center. Further out, the colors become redder again due to the
incipient contribution of NGC 1097 nuclear star-forming ring.
With the aim of determining the extinction towards the nucleus,
as a first step we took
those minima, $J-H \sim 0.7$,
$J-Ks \sim 0.24$, as the intrinsic colors of NGC 1097 pseudo-bulge. In doing
so, we note that the progressive reddening with decreasing
radius is difficult to explain by dust extinction alone,
unless very steep IR extinction curves, with $\alpha \sim$ 3.5
($A_\lambda \propto \lambda^{-\alpha}$), are considered. The problem is
better illustrated in Fig. 4 where a detailed view of the color
variations in a selected area of the bulge -- e.g. the South-West
quadrant -- is shown. $J$-$H$ and $H$-$Ks$ are measured in
consecutive boxes of size 0.27$\times$0.27 arcsec, covering the region between
0.5 and 5.5 arcsec ($\sim 40 - 400$ pc) from the center.
Colors measured in the other quadrants show similar behavior. Taking
again $J-H \sim 0.7$ and $J-Ks \sim 0.24$ as the intrinsic colors of the
pseudo-bulge, the progressive reddening at radii larger than $\sim 3$ arcsec
is well explained by dust extinction alone (two extinction laws
with $\alpha=1.5$ and 1.8 are shown in the figure). However further in,
$J-H$ becomes much redder than predicted.
We then compared the observed color variation with the one predicted by
the Sersic model used to fit the pseudo-bulge (Section 3.1). These colors are
plotted as a red line in Fig. 4. It can be seen that the observed
colors follow Sersic predictions down to a radius of $\sim 2.5$ arcsec.
Further in, they depart from Sersic trend, but the departure is now easily
reconciled with a more pure extinction law. Based on this result, we estimate
the extinction toward the nucleus by taking as a face value for the
intrinsic colors of the pseudo-bulge those measured at $\sim$2.5 arcsec
(180 pc) radius from the center: i.e. $J-H\sim0.78$, $ J-Ks\sim$ 0.25
(Fig. 4). The reddest values at about 0.6'' from the nucleus
are $J-H \sim 0.86$, $J-Ks \sim 0.32$. These compared with our fiducial
value lead to a moderate extinction of $A_v\sim 1$ within the
central $\sim 60 - 40$ pc region. This extinction is in the range
measured from the nuclear Balmer decrement estimated from an aperture
larger than 1 arcsec (Storchi-Bergmann et al. 1996).
\subsection{The central spiral network}
In Fig. 5 (bottom-left panel), a $J$-$Ks$ color map of the central
2 kpc region of NGC~1097 is presented. It shows the morphology
of the nuclear
star-forming ring as a complete annulus, filled by a continuous distribution
of HII regions and gas. It differs from two spiral arms morphology
seen in the optical
image (Fig.~1). In the $J$-$Ks$ color map most of the contribution
from both the pseudo-bulge and the nuclear bar cancels out, disclosing a
filamentary structure that spirals around the center. This structure is
similar to those unveiled in nearby galaxies by 'structure maps' constructed
from HST WFPC2 images (Pogge \& Martini 2002), or through visible-near-infrared
color maps from the HST (Martini et al. 2003). The resolution of
our NACO images, when compared to that of HST/NICMOS images,
is similar in $H-$band and
a factor of 2 worse in $J-$band. However, the PSF of the VLT/NACO images
does not show the boxy diffraction pattern often seen in NICMOS images of
strong point-like sources, and therefore nuclear structure at distances
as small as 10 pc from the center can be revealed. An additional advantage of
using a near-IR color image in this study is that this spectral range is
less prone to the contamination by line emission. $V-H$ images often used
to trace dust structures in the surrounding of an AGN may largely be
polluted by strong optical emission lines in the $V-$band.
Most of the central filamentary structure around the nucleus of NGC~1097
cannot be distinguished in the
direct HST optical and NACO IR images, except for the outer parts of the
longer filaments which are seen as obscured fingers in the HST WFPC1/F5500W
and the NACO
$J-$band images (Fig.~1). These filaments are seen in extinction against
background light from the pseudo-bulge, and therefore they are likely to be
tracers of cold material and dust. The low extinction, $A_v \sim 1$,
indicated by the IR colors of the pseudo-bulge (see Section 3.2) suggests that the
filaments are not filling much of the volume within this region, and are likely
distributed in a rather thin disk.
To investigate the central filamentary structure in more detail, a simple
elliptical model was fitted and subtracted from the $J$-band image, with
the fit restricted to the region inside the nuclear star-forming ring.
The residual image, shown in Fig. 5 (bottom-right panel),
looks like the negative of the $J$-$Ks$
color map. The brighter filaments in the color map appear as dark
channels in the residual, yet, with sharper detail and increasing length in
the latter. Our analysis of the VLT/NACO images of NGC~1097 shows that these
filaments, within the achieved resolution of about 10 pc, end up at the very
center of the galaxy. Several filaments can be traced in the residual image,
but most prominent are three spiral arms that wind around the center at radii
below 200 pc. At larger radii, the pitch angles of these
arms increase, and two arms seem to align with the nuclear secondary
bar present in NGC~1097. The northern arm then splits into two at a
radius of about 300 pc. The third arm appears to be unrelated to the
nuclear bar, and it splits into a number of spiral filaments at a similar
radius. Some of the filaments can be followed outward up to the
nuclear
star-forming ring. It can be seen in the HST/ACS image (Fig. 5, top panel) that
this ring also encircles the curving innermost parts of the dust lanes which
run along the primary bar of NGC~1097. The nuclear filaments seem to connect
with the large-scale dust lanes, but there is a clear discontinuity in the
pitch angle at the meeting points.
\section{Dynamics of the central spiral network}
The VLT high-resolution infrared images presented in this paper unveil a
complex spiral network of dust filaments around the nucleus of NGC~1097.
Here we examine how such
a network could have been formed, and whether it could funnel gas inward.
The mass of the molecular gas makes only a fraction of the total
dynamical mass within the inner few hundreds parsecs of the galaxy.
We estimate an upper limit to the central
molecular gas mass within the beam size of the CO observations by Kohno et
al. (2003), $\sim 7.7''\times 4''$ or $\sim 550 \times 270$ pc, to be
$\sim 5 \times 10^{7}$ M$_{\odot}\,$. The dynamical mass within the same area,
inferred from the rotation curve (Storchi-Bergman et al. 1996), is at
least $\sim 5 \times 10^8$ M$_{\odot}\,$.
\subsection{Origin of the central spiral}
The spiral dusty filaments around the nucleus of NGC~1097 are unlikely to be
formed by a self-amplified density wave, analogous to the one that shapes
the classical large-scale spiral arms in galaxies. This is because stars,
which make up most of the mass in this region, mostly settle in the pseudo-bulge
and in the nuclear bar that we observe in the infrared. This spiral pattern
is also unlikely to be of
acoustic origin (Elmegreen 1994), because it consists of a few sharp
spiral filaments, while acoustic spirals, generated by the amplification
of sound waves in the gaseous nuclear disk, are expected to have
flocculent, multi-arm morphology, and to uniformly fill the nuclear disk.
The spiral network around the nucleus of NGC~1097 closely resembles the
nuclear spirals seen in hydrodynamical models (Englmaier \& Shlosman 2000,
Maciejewski et al. 2002, Maciejewski 2004b). Straight principal shocks in
a large-scale bar may end in a nuclear ring, which itself is a tightly wound
spiral, or they can give rise to more open nuclear spirals, which are
an extension of these shocks inward to smaller radii. Hereafter we call
the nuclear spirals that extend to the galactic center the {\it central
spirals}.
There is a strong large-scale bar in NGC~1097, which could
trigger a nuclear spiral,
but there is also a prominent nuclear star-forming ring
there, abundant in gas. Central spirals have not been found in models
inside gaseous nuclear
rings, because shocks triggered by the large-scale bar get
damped in such rings, so they cannot give rise to central spirals
(Maciejewski 2004b). The coexistence of the nuclear star-forming ring
and the dusty central spiral in NGC~1097 seems to contradict these models,
unless principal shocks in the bar have some effect on the gas encircled
by the nuclear ring.
Numerical models of gas flow in bars have so far
only been built in two dimensions, in order to reflect the dynamics of gas
in the disk of a galaxy. On the other hand, shocks generated in gas by a
large-scale bar are intrinsically three-dimensional, and they propagate
also in the off-plane gas. There have been studies of the vertical structure
of galactic spiral arms (Martos \& Cox 1998, Gomez \& Cox 2002), but not
of the shocks in bars. Here we make a qualitative attempt to describe
gas flow in bars in three dimensions. We base our description on two
assumptions: 1) in addition to the gas in the plane of the disk there is
off-plane gas of lower density, 2) larger scale-height of the off-plane gas
is supported by its energy stored in velocity dispersion of the clouds.
Within this framework we can guide our understanding of gas flow in three
dimensions using various two-dimensional models for various heights above
the galactic plane. However, full quantitative description requires
high-resolution three-dimensional hydrodynamical models, which are beyond
the scope of this paper.
The dispersion relation of a wave generated in the gaseous disk implies that
the pitch angle of the wave is proportional to the velocity dispersion in gas
(Englmaier \& Shlosman 2000, Maciejewski 2004a). Thus in gas with higher
velocity dispersion (dynamically warmer), waves generated by the principal
shocks in the bar should have higher pitch angle than in a dynamically colder
gas. When the pitch angle is high enough, waves can spiral freely into the
galactic center (Maciejewski et al. 2002). This may be a scenario
for the extraplanar gas. On the other hand, shocks propagating in the disk,
where gas is dynamically colder, generate waves with smaller pitch angle. These
waves are damped in dense post-shock gas, and a nuclear gaseous ring forms with
excellent conditions for star formation (Maciejewski 2004b). The morphology of
waves, which are expected to form in the disk as well as in the extra-planar
gas, together with gas morphology in the plane of galaxy are shown in the top
panel of Fig. 6. These morphologies should be compared with those in the
bottom-right panel of Fig.5. Although the structure in the model is generic,
and not tuned to reflect particular features in NGC~1097, there are
similarities between the model and the data. The large-scale dust lanes enter
the nuclear star-forming ring along the paths marked by the red lines that
indicate shocks in the cold in-plane gas in the model, whereas the observed
central spiral takes a shape similar to that of the green line marking
shocks in the extraplanar gas. The distribution of the emission from the
star-forming ring resembles in-plane gas density distribution in the model.
The nuclear star-forming ring in NGC~1097 consists mostly of
molecular gas, and therefore it is well confined to the galactic plane. It
damps waves generated by the principal shocks in the bar that are propagating
in the galactic plane. However, in the scenario proposed above, shocks
propagating in the off-plane gas may pass the ring almost unaffected.
Inward from the ring our simple plane-parallel scenario breaks down, and
full three-dimensional approach is needed to determine whether off-plane
shocks couple with the dynamically cold gas in the disk and whether they
can generate spiral density
waves there. It is possible that the gaseous disk inside the nuclear
star-forming ring
is no longer dynamically cold, as indicated by the large pitch angle of the
central spiral, which should facilitate the coupling. If the dispersion
relation of the excited wave allows for its propagation inward from the
nuclear ring, this wave will naturally take a spiral shape. Waves
inward from the ring can propagate freely toward the galactic center,
because they are not hindered by the dense post-shock gas in the galactic
plane, which has settled in the ring outside them. We hypothesize that these
waves form the observed spiral network inside the nuclear star-forming
ring. In this picture the central spiral in NGC~1097 would be a wave in
gas, and a continuation of the principal shocks present in the large-scale
bar.
We note that the innermost parts of the central spiral have a three-arm
symmetry, not an obvious consequence of triggering by bisymmetric
shocks in a bar. Three-arm spirals can form as a response to lopsided ($m=1$)
or $m=3$ terms in the forcing. One source of these odd terms in NGC~1097 may be
the difference in strength of the two principal shocks in the main bar.
It has been observed
that the amount of post-shock gas can differ between the two shocks by a
factor of two (Schinnerer et al. 2002). This can effectively introduce
odd terms in forcing, since in our scenario it is the shock, and not the bar,
that triggers the central spiral. A dwarf elliptical companion to NGC~1097
may also introduce $m=1$ mode in the shocks. Regardless of its explanation,
the three-arm symmetry of the central spiral indicates that there is a
departure from bisymmetry in the forcing that generates this pattern.
\subsection{Interpretation of the polarization maps of NGC~1097}
Our scenario for the origin of the central spiral in NGC~1097, presented
above, is supported by the morphology of the regular magnetic field seen in
the radio polarization maps of this galaxy (Beck et al. 1999). In the lower
panel of Fig. 6, we present the magnetic field vectors observed by Beck et al.
(1999, 2005), which are aligned with the shearing flow, plotted on top of our
processed $J$-band image that shows the dusty central spiral. At radii where
the principal shocks originating in the main bar approach the nuclear
star-forming ring, the magnetic vectors progressively swing from an alignment
with these shocks to a clear spiral pattern that crosses the nuclear star-forming ring under
a rather large angle ($\sim$50$^{\circ}\,$). Directly inward from the ring this
pattern tends to follow central dusty spiral structure that we
observe.
At the position of the nuclear star-forming
ring, the spiral morphology of the magnetic
field is very different from the ring-like morphology of gas and of
star-forming regions. Here we propose that this is because star formation is
confined within the
disk of the galaxy, while radio polarization maps show the regular magnetic
field averaged along the path length that includes the off-plane gas. The
scale-height of the off-plane gas, from which polarized emission is observed
in galaxies, is about 1 kpc, as
indicated by maps of the magnetic field in edge-on spirals (Beck 2000).
Although the total radio signal is largest from the dense in-plane gas,
star-forming activity creates field turbulence there, which lowers the
degree of polarization in the nuclear star-forming ring in the plane of
NGC~1097. If field turbulence in regions far from the galactic plane is much
lower, then most of the polarized emission may be coming from the off-plane
gas, even when most of the total radio emission is coming from the disk.
Then vectors in polarization maps will trace kinematics of the
off-plane gas. Polarization vectors observed in NGC~1097 (bottom
panel of Fig.6) indicate strong ordered magnetic field on top of the
star-forming nuclear ring, as well as a field aligned with the large-scale
dust lanes and with the central spiral. Directions of these vectors
are consistent with our scenario, in which parts of the principal shock that
propagate in the off-plane gas give rise to a spiral pattern,
which crosses the nuclear ring. Note that in this scenario the magnetic
field is just a tracer of gas motions, and it does not play a dynamical role.
The role of magnetic fields in extraplanar gas is most likely important, and
full magnetohydrodynamical approach may be needed in order to understand
the propagation of waves past the nuclear ring.
If the polarized emission comes mostly from the off-plane gas, then another
paradox in NGC~1097 noticed by Beck et al. (1999) can be explained in a
similar way. Beck et al. observe a depolarization zone, which they ascribe to
a shock, {\it upstream} from the straight principal dust lanes. Hydrodynamical
models (e.g. Englmaier \& Gerhard 1997, Maciejewski et al. 2002) indicate that
the distance of the principal shock from the major axis of the stellar bar
decreases with
the velocity dispersion in gas. This is also seen in the top panel of Fig. 6.
If the depolarization zones mark shocks in the off-plane warmer gas, then they
are expected to be located closer to the major axis of the bar than the dust
lanes, which mark shocks in the dynamically cold gas in the plane of the disk.
\subsection{Central spiral in NGC~1097 as a shock in gas. The role of the
inner bar}
Whether the central spiral in NGC~1097 is a shock or a weak wave has direct
implications for the expected inflow of gas to the center. A weak wave
triggers no inflow, while shocks are capable of supplying enough gas to the
galactic center to maintain the AGN activity (Maciejewski 2004b). Nuclear
spirals in barred galaxies can be shocks throughout their extent from the
principal shock in the bar to the galactic center (Maciejewski et al. 2002).
Nuclear spiral shocks in gas tend to have pitch angle larger than what the
linear theory of gaseous density waves predicts (Maciejewski 2004b).
The observed large pitch angle of the central
spiral in NGC~1097 is beyond the linear regime, which is suggestive of
a spiral shock. The optical emission from the region between the
nucleus and the ring reveals a typical LINER spectrum with strong low
excitation lines and large line widths (FWHM of [NII] 6548 \AA\ between
100 and 300 km s$^{-1}$, Storchi-Bergmann et al. 1996), which are signatures of
shock-excited gas (Contini \& Viegas 2001, Viegas \& Contini, 1989). This,
together with the observed lack of star formation in the central spiral,
favour the idea that this spiral is a shock with strong shear.
In Section 3.3 we noted that two of the three most prominent arms of the
central spiral in NGC~1097 unwind at radii above 200 pc to almost straight
filaments aligned with the inner bar. Thus one may think of an alternative
explanation, which ascribes this spiral to the principal shock in the
inner bar. However, principal shocks triggered by bars approach the major
axis of the bar toward the bar's ends, as seen both in nature and in numerical
models, and accounted for by studies of orbital structure. On the contrary,
in NGC~1097 the most pronounced north filament running along the inner
bar departs from the bar's major axis as the radius increases (see NACO
$J$-band image in Fig. 1). This difference already indicates that the gas
dynamics in the inner bar of NGC~1097 is different from the well studied
cases of large-scale bars. In particular the central spiral in this galaxy,
with its three-arm symmetry, is most likely not generated by the inner bar.
Nevertheless, appropriate modelling of this dynamics is necessary before
firm conclusions are to be drawn about the role of the inner bar. The only
models of a spiral density wave propagating through a nuclear bar built so
far (Maciejewski 2004b) indicate that this bar can reshape the spiral shock,
generated there by the outer bar. It does so by straightening parts of the
spiral arms and aligning them with itself, which gives an impression of
straight shocks, and resembles the morphology of the two spiral arms observed
in the center of NGC~1097. However, the nuclear bar in those models is rather
strong, while the inner bar observed in NGC~1097 is weak.
Finally, the curling of the spiral pattern in the innermost 100 pc (its
pitch angle decreasing inward) may point at the presence of a
supermassive black hole in the center of NGC~1097. Such a black hole in
the center of a galaxy causes the nuclear spiral to wind up as it
approaches the center, while in its absence the spiral would be unwinding
with decreasing radius (Maciejewski 2004a).
\section{Conclusions}
The nucleus of NGC~1097 remains a point-like source at subarcsec scales in
the near-IR. An upper limit to its size has FWHM $<10$ pc. Accordingly, any
putative central dusty torus or gaseous disk required by AGN unified
schemes has to be smaller than 10 pc in diameter at
near-IR wavelengths.
High spatial resolution achieved in the NACO/VLT observations
of NGC~1097 presented here was sufficient to reveal in sharp detail an
intricate network of filamentary structure in the central 10'' (700 pc)
of the galaxy. These filaments are seen directly in $J-Ks$ color maps
(Fig. 5, bottom left). They also appear in extinction
against the IR background light from the galaxy,
and therefore they are likely
to be tracers of cold material. On the basis of the IR colors, and assuming
foreground extinction, a very moderate extinction toward the nuclear region
($A_v\sim < 1$) is determined. This is consistent with the extinction values
derived from the Balmer decrement. Thus, if the filaments are tracing cold
dust, they contribute to a very low extinction in the line of sight, and
therefore they are likely to be distributed in a rather thin disk. The
filaments are spiralling around the center from the nuclear star-forming
ring to the nucleus,
which is suggestive of them being channels by which cold dust and gas
are flowing to the center. The CO and HCN maps hint to the presence of cold
material in this region, though their resolution is insufficient to confirm
the exact location of this material. The molecular $H_2$ 2.12 $\mu$m$\,$ emission
is also present between the nucleus and the star-forming ring. We have
recently collected AO-corrected $K-$band spectra of this region
which should resolve spatially the dynamics of this $H_2$ warm gas.
The spiral network in NGC~1097 resembles nuclear spirals found in
hydrodynamical models, where they are density waves driven by the shocks
generated in a large-scale bar. There is a strong large-scale bar in NGC~1097
(Fig. 5, top panel), which can generate the shocks, but the nuclear
star-forming ring may hamper propagation of the waves inward in the plane of
the disk. In this paper we propose a scenario, in which, because of different
dynamical conditions in the off-plane gas, waves propagating there wind into a
spiral, and pass the nuclear ring. We cannot make a decisive statement on how
these waves propagate inward from the nuclear ring -- this would require full
three-dimensional and probably magnetohydrodynamical models -- but if the
dispersion relation of waves excited there allows for their propagation,
they should naturally form the observed central spiral. Radio
polarization maps of NGC~1097 support this scenario, on the assumption that the
polarization vectors trace the motion of the off-plane gas. These
vectors show a progressive swing from an alignment
with the principal shocks along the primary bar to a clear spiral
pattern at the ring and further in. Interior to the ring, they closely
follow some of the nuclear filaments (Fig. 6, bottom panel). Emission lines
in this region, which show all the
characteristics of shock-excited gas, together with large pitch angle of the
central spiral, and the lack of star formation in the spiral arms, make a
suggestive evidence that this spiral is a shock with strong shear, along
which gas and dust can be channelled to the center of NGC~1097.
\section{Acknowledgments}
We are thankful to Rainer Beck for interesting discussions and for providing
us with the radio polarization map of NGC~1097 displayed in Fig. 6, bottom. We acknowledge comments
from John Maggorian on this manuscript. This work was partially
supported by the grant 1 P03D 007 26 from the Polish Committee for Scientific
Research.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 7,764 |
Al Regne Unit:
Trinity College (Cambridge)
Trinity College (Glenalmond), situat a Glenalmond (Perthshire, Escòcia)
Trinity College (Oxford)
Trinity College (Carmarthen, Gal·les)
Irlanda:
Trinity College (Dublín)
Estats Units:
Trinity College, Universitat Duke
Trinity College (Connecticut)
Trinity College (Florida)
Trinity College (Vermont):
Trinity Christian College (Illinois)
Trinity College and Seminary, situat a Newburgh (Indiana)
Trinity College, antic nom de la Trinity University (Washington, DC) fins a 2004
Trinity College and University, antic nom de la Bronte International University, a l'estat de Delaware
Austràlia:
Trinity College (Melbourne)
Canadà:
Universitat de Trinity College, situada a Toronto (Ontario)
Trinity College School, situada a Port Hope (Ontario) | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 5,640 |
ScrapAttic Creations: Allison's Wedding Album - So Far!
I started Alli's album in September as her shower gift.
finishing the album. Here is the album she received.
Family shots, and all the reception!!!!!
I hope to finish by their 1st anniversary. | {
"redpajama_set_name": "RedPajamaC4"
} | 7,937 |
Q: Redirect parent window from iFrame not working in Safari My site is mysite.com. I am iframing othersite.com and having my user complete a task on it.
I can check via API if my user has completed the task on othersite.com. However, I'd like to figure out another way to figure out when the user has completed the task so that I can navigate to the next page on mysite.com.
When the user completes the task on othersite.com, othersite.com redirects them to mysite.com/complete/ -- inside the iframe. I've been taking advantage of this by putting the following JS code on mysite.com/complete/
window.top.location == 'mysite.com/next';
So that the parent window is automatically pushed to the next page. I've got this working on Chrome, but on Safari it redirects the inner iframe, not the parent.
What might be causing the difference between Chrome and Safari? Is there a better way to accomplish this goal?
Note: I've seen this this answer, so it seems like there's something happening with the window.top.location function in Safari, but even trying that fix I'm having trouble escaping the inner iframe.
A: 31
You just cannot do that. Cross-site scripting is not allowed in most browsers.
You can, however, communicate with the other window via cross-document messaging described here: https://developer.mozilla.org/en/DOM/window.postMessage
The most you can to is to send a message from the popup to the opener and listen for such message in the opener. The opener then has to change its location on its own.
Parent Window:
<script language="javascript">
function open_a_window()
{
var w = 200;
var h = 200;
var left = Number((screen.width/2)-(w/2));
var tops = Number((screen.height/2)-(h/2));
window.open("window_to_close.html", '', 'toolbar=no, location=no, directories=no, status=no, menubar=no, scrollbars=no, resizable=no, copyhistory=no, width='+w+', height='+h+', top='+tops+', left='+left);
return false;
}
// opener:
window.onmessage = function (e) {
if (e.data === 'location') {
window.location.replace('https://www.google.com');
}
};
</script>
<input type="button" onclick="return open_a_window();" value="Open New Window/Tab" />
Popup:
<!DOCTYPE html>
<html>
<body onload="quitBox('quit');">
<h1>The window closer:</h1>
<input type="button" onclick="return quitBox('quit');" value="Close This Window/Tab" />
<script language="javascript">
function quitBox(cmd)
{
if (cmd=='quit')
{
window.opener.postMessage('location', '*');
window.open(location, '_self').close();
}
return false;
}
</script>
</body>
</html>
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 96 |
A 3-phase Brushless DC motor control based on a motor control DSP controlling Infineon half-bridge intelligent power modules (IPMs).
In a continuous drive for smaller and higher performing products, a world leader in laboratory pumps developed a new pump that would significantly move the bar. The new product would be powered by an efficient, brushless DC motor supplied from a 24V DC in-cord power supply similar to the ones used to power laptop computers. The combination would allow for a very small pump taking up one-half the bench space compared to their existing products. With product plans in place, Tecnova was called-in to join the team. Our role would be to develop the motor drive electronics, user interface electronics, and firmware inside the product.
The design called for a continuous 75W of output power and 2 to 3 times the continuous torque (current) for starting. The electronics was also tasked to manage analog and contact closure process control inputs and outputs, a membrane keypad, and a graphic LCD display. This all was to be accomplished with a defined footprint (3" X 3.5") and with no external heatsink – and at low cost.
Fortunately, modern and inexpensive DSPs are up to the control challenge. A 16-bit DSP was chosen as the controller. At a blazing speed of 140 MHz, and supporting dedicated hardware for motor position feedback, current sensing, and 3-phase PWM, Tecnova was able to implement an effective vector control using only a small fraction of the processor's available instruction cycles. The product delivers smooth 0-300 RPM performance while maintaining seamless refresh of the 128 X 64 pixel LCD and responsive I/O.
Three half-bridge drivers were chosen to drive the 3-phase bridge powering the motor. The maximum 5.0m Ω (@ 25C) RDS(on) of the half-bridge MOSFETs coupled with the synchronous rectification of space vector PWM permitted operation without a heatsink. The entire power stage including two conductive polymer hybrid Aluminum Electrolytic bus capacitors occupies less than 2 sq-in of PCB area. The assembly is 0.5" high.
The new product is pumping up sales by meeting expectations for ever improved performance in a smaller package and at a lower price. | {
"redpajama_set_name": "RedPajamaC4"
} | 1,090 |
Amazon nears deal to put big distribution center in Pullman
The e-commerce giant has an agreement in principle for a $60 million, 150,000-square-foot facility that will employ up to 300 workers. That's the word from the local alderman.
Amazon has decided to bulk up a little more here, this time on the city's South Side.
According to Ald. Anthony Beale, 9th, the giant internet retailer has an agreement in principle to create a 40-acre distribution center in the Pullman section of his ward, roughly at 104th Street and Woodlawn Avenue.
Update: Amazon says Pullman is one of several sites under review
Amazon eyes South Side for new warehouse
The "wish fulfillment center," as Amazon calls its warehouses, would occupy a 150,000-square-foot structure to be built on former Ryerson Steel property. About 500 construction jobs will be created, with 200 to 300 permanent positions after construction is completed by the end of the year, Beale said.
Pay figures are not available, but at a somewhat larger facility Amazon recently announced it will open on the site of the former Maywood Park horse track, the firm indicated pay would begin at $15 an hour. The Pullman jobs will be located not far from some of the neediest and most job-short neighborhoods in the city.
"We could not be happier," said Beale, reporting that the center will be located immediately north of Amazon's existing Whole Foods distribution center. "It underscores that Pullman is rapidly becoming the new green industry/transit logistics and distribution center of Chicago," with more than $350 million in private investment in recent years.
Beale pegged Amazon's investment at $60 million. The company will seek a county property tax abatement that could slash its property tax by more than half for a decade, Beale said, but those are relatively routine for new industrial projects nowadays. The City Council would have to approve the tax break.
Amazon has been edging closer to the city proper with new distribution facilities after initially focusing on Joliet, Aurora, Waukegan and other outlying suburbs, apparently hoping to get orders faster to city residents. The company does have one small warehouse on Goose Island on the North Side.
Jan. 27 update: Mayor Lori Lightfoot is expressing some skepticism about whether the deal is fully baked.
"I don't think there is anything to talk about yet," she told reporters at an unrelated event today. "We haven't heard anything from Amazon. So I think Ald. Beale got a little ahead of himself."
However, Beale is standing by his story. "I feel fully confident this deal will get done," he said, with some preliminary land clearance already underway.
Officials at Amazon and Ryan Companies, which owns the property, did not immediately return requests for comment.
Lightfoot and Beale have a very hostile political relationship—perhaps one of the reasons he went public with the news himself. But one City Hall source says Amazon actually is looking at several locations, of which the Pullman site is only one under consideration.
Amazon plans new 500-job fulfillment center
This time, Amazon placed the order—a big one
This is the restaurant replacing Tavern on Rush—and it's not a steakhouse
At $15 million, this might be the biggest price cut ever for a Chicago home | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 5,419 |
package org.jenkinsci.plugins.ghprb;
import hudson.Extension;
import hudson.model.AbstractProject;
import hudson.model.UnprotectedRootAction;
import hudson.security.ACL;
import hudson.security.csrf.CrumbExclusion;
import org.acegisecurity.Authentication;
import org.acegisecurity.context.SecurityContextHolder;
import org.apache.commons.io.IOUtils;
import org.kohsuke.github.GHEventPayload.IssueComment;
import org.kohsuke.github.GHEventPayload.PullRequest;
import org.kohsuke.github.GHIssueState;
import org.kohsuke.github.GitHub;
import org.kohsuke.stapler.StaplerRequest;
import org.kohsuke.stapler.StaplerResponse;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.StringReader;
import java.io.UnsupportedEncodingException;
import java.net.URLDecoder;
import java.util.HashSet;
import java.util.Set;
import java.util.logging.Level;
import java.util.logging.Logger;
import javax.servlet.FilterChain;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServletResponse;
import javax.servlet.http.HttpServletRequest;
/**
* @author Honza Brázdil <jbrazdil@redhat.com>
*/
@Extension
public class GhprbRootAction implements UnprotectedRootAction {
static final String URL = "ghprbhook";
private static final Logger logger = Logger.getLogger(GhprbRootAction.class.getName());
public String getIconFileName() {
return null;
}
public String getDisplayName() {
return null;
}
public String getUrlName() {
return URL;
}
public void doIndex(StaplerRequest req, StaplerResponse resp) {
String event = req.getHeader("X-GitHub-Event");
String signature = req.getHeader("X-Hub-Signature");
String type = req.getContentType();
String payload = null;
String body = null;
if (type.toLowerCase().startsWith("application/json")) {
body = extractRequestBody(req);
if (body == null) {
logger.log(Level.SEVERE, "Can't get request body for application/json.");
resp.setStatus(StaplerResponse.SC_BAD_REQUEST);
return;
}
payload = body;
} else if (type.toLowerCase().startsWith("application/x-www-form-urlencoded")) {
body = extractRequestBody(req);
if (body == null || body.length() <= 8) {
logger.log(Level.SEVERE, "Request doesn't contain payload. " + "You're sending url encoded request, so you should pass github payload through 'payload' request parameter");
resp.setStatus(StaplerResponse.SC_BAD_REQUEST);
return;
}
try {
String encoding = req.getCharacterEncoding();
payload = URLDecoder.decode(body.substring(8), encoding != null ? encoding : "UTF-8");
} catch (UnsupportedEncodingException e) {
logger.log(Level.SEVERE, "Error while trying to decode the payload");
resp.setStatus(StaplerResponse.SC_BAD_REQUEST);
return;
}
}
if (payload == null) {
logger.log(Level.SEVERE, "Payload is null, maybe content type ''{0}'' is not supported by this plugin. " + "Please use 'application/json' or 'application/x-www-form-urlencoded'",
new Object[] { type });
resp.setStatus(StaplerResponse.SC_UNSUPPORTED_MEDIA_TYPE);
return;
}
logger.log(Level.FINE, "Got payload event: {0}", event);
// Not sure if this is needed, but it may be to get info about old builds.
Authentication old = SecurityContextHolder.getContext().getAuthentication();
SecurityContextHolder.getContext().setAuthentication(ACL.SYSTEM);
try {
GitHub gh = GitHub.connectAnonymously();
if ("issue_comment".equals(event)) {
IssueComment issueComment = getIssueComment(payload, gh);
GHIssueState state = issueComment.getIssue().getState();
if (state == GHIssueState.CLOSED) {
logger.log(Level.INFO, "Skip comment on closed PR");
return;
}
if (!issueComment.getIssue().isPullRequest()) {
logger.log(Level.INFO, "Skip comment on Issue");
return;
}
String repoName = issueComment.getRepository().getFullName();
logger.log(Level.INFO, "Checking issue comment ''{0}'' for repo {1}", new Object[] { issueComment.getComment().getBody(), repoName });
for (GhprbTrigger trigger : getTriggers(repoName, body, signature)) {
try {
IssueComment authedComment = getIssueComment(payload, trigger.getGitHub());
trigger.handleComment(authedComment);
} catch (Exception e) {
logger.log(Level.SEVERE, "Unable to process web hook for: " + trigger.getProjectName(), e);
}
}
} else if ("pull_request".equals(event)) {
PullRequest pr = getPullRequest(payload, gh);
String repoName = pr.getRepository().getFullName();
logger.log(Level.INFO, "Checking PR #{1} for {0}", new Object[] { repoName, pr.getNumber() });
for (GhprbTrigger trigger : getTriggers(repoName, body, signature)) {
try {
PullRequest authedPr = getPullRequest(payload, trigger.getGitHub());
trigger.handlePR(authedPr);
} catch (Exception e) {
logger.log(Level.SEVERE, "Unable to process web hook for: " + trigger.getProjectName(), e);
}
}
} else {
logger.log(Level.WARNING, "Request not known");
}
} catch (IOException e) {
logger.log(Level.SEVERE, "Unable to connect to GitHub anonymously", e);
} finally {
SecurityContextHolder.getContext().setAuthentication(old);
}
}
private PullRequest getPullRequest(String payload, GitHub gh) throws IOException {
PullRequest pr = gh.parseEventPayload(new StringReader(payload), PullRequest.class);
return pr;
}
private IssueComment getIssueComment(String payload, GitHub gh) throws IOException {
IssueComment issueComment = gh.parseEventPayload(new StringReader(payload), IssueComment.class);
return issueComment;
}
private String extractRequestBody(StaplerRequest req) {
String body = null;
BufferedReader br = null;
try {
br = req.getReader();
body = IOUtils.toString(br);
} catch (IOException e) {
body = null;
} finally {
IOUtils.closeQuietly(br);
}
return body;
}
private Set<GhprbTrigger> getTriggers(String repoName, String body, String signature) {
Set<GhprbTrigger> triggers = new HashSet<GhprbTrigger>();
Set<AbstractProject<?, ?>> projects = GhprbTrigger.getDscp().getRepoTriggers(repoName);
if (projects != null) {
for (AbstractProject<?, ?> project : projects) {
GhprbTrigger trigger = Ghprb.extractTrigger(project);
if (trigger == null) {
logger.log(Level.WARNING, "Warning, trigger unexpectedly null for project " + project.getFullName());
continue;
}
if (trigger.matchSignature(body, signature)) {
triggers.add(trigger);
}
}
}
return triggers;
}
@Extension
public static class GhprbRootActionCrumbExclusion extends CrumbExclusion {
@Override
public boolean process(HttpServletRequest req, HttpServletResponse resp, FilterChain chain) throws IOException, ServletException {
String pathInfo = req.getPathInfo();
if (pathInfo != null && pathInfo.equals(getExclusionPath())) {
chain.doFilter(req, resp);
return true;
}
return false;
}
public String getExclusionPath() {
return "/" + URL + "/";
}
}
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 8,168 |
\section{Introduction}
Notwithstanding the aptness of semiclassical (SC) approximations for uncovering
classical structures underlying quantum evolution,
their use for constructing ready algorithms to deal with increasingly complex experiments
is yet to be established. The practical advantage of integrating Hamilton's ordinary differential equations, rather than dealing with Schr\"odinger's partial differential equation,
is counterbalanced by the need to search for trajectories that are only indirectly
specified by boundary conditions, instead of arising directly from their initial conditions.
This difficulty has led to the development of initial value methods (or {\it inital value representations}, IVR) that substitute the, so called, {\it root search} by an integration
over families of initial valued trajectories \cite{Mil70,Mil01,Mil02,FGrossmann,HerKLuck,Kay}.
Thus, IVR's have been seen as a workable alternative,
in spite of considerable criticism \cite{BarAguiar01}.
One of their main achievements is the evaluation of {\it correlations} for quantum operators, ${\rm tr} \hat{A} \hat{B}(t)$ \cite{Mil12}, even though, if $\hat{A}$ is chosen as a density operator,
this reduces to a single evolving expectation value $\langle \hat{B}(t) \rangle$.
Here we establish general SC approximations for multiple correlations of observables evolved
by various unitary operators: $\langle\opU \hat{B} \hat{V} \hat{C}... \rangle$.
In a recent paper \cite{IVRFVR}, henceforth labled {\bf I}, the IVR approach to SC approximations
was realized entirely within the Weyl representation, that represents the operator $\hat{B}$
by the phase space function $B(\x)=B(p,q)$, its {\it Weyl symbol},
or its Fourier conjugate, $\tilde{B}(\Vxi)$, its {\it chord symbol},
another complete representation \cite{Report}.
A remarkable feature of these particular representations
(including the Wigner function \cite{Wigner}, in the case of the density operator)
is that they are based, respectively, on reflection operators and translation operators
\cite{Grossmann,Royer, Report}. These are themselves unitary,
so one can combine them with the evolution operators which act on each observable
into a single composite entity. Then the expectation of an evolving observable was cast as a phase space integral over the Wigner function multiplied by the nearly classical function that represents the observable.
This procedure is here generalized to the correlations of an arbitrary number, $\nu$, of
observables undergoing general unitary evolutions. The evolving correlation depends
on a single family of {\it compound unitary operators},
labeled by $\nu$ continuous parameters. For each parameter, the required trace of this compound operator
is then obtained from the {\it compound periodic orbits} in the corresponding classical evolution,
according to the standard SC procedure \cite{Gutzwiller,livro}.
Even though this is an important step, the identification of the appropriate compound unitary operator
does not free us from a search for orbits. It is true that continuous families of periodic orbits,
within continuous families of canonical transformations, can be followed by a generalized Newton's method, as the parameters are varied in small steps \cite{Aguiaretal87},
but this is still a formidable task.
Furthermore, such a reliance on continuity is at odds with the use of efficient Monte Carlo methods
at the next stage, where one integrates over the parameters. It is then fortunate that the IVR approach
can be extended to the general evaluation of correlations, by simply freeing one of the parameters:
The corresponding segment of the periodic orbit is then removed,
so that one then deals with a {\it reduced compound trajectory}.
This is still composed of evolutions intercalated by reflections,
but now the trajectory is determined by its initial value. There always exists an extra reflection
which closes such an open orbit, so that its reflection centre can be chosen as the extra free parameter.
Just as in {\bf I}, the IVR algorithm avoids caustic singularities,
transforming them into nodal lines (or surfaces) of the compound propagator. There remains an overall
ambiguity of sign to be determined as such a line is crossed, but the general procedure
presented in \cite{OAI}, henceforth labled {\bf II}, can be immediately generalized for correlations.
In the special case where all the evolution operators are {\it metaplectic}
\footnote{Unitary operators corresponding to classical symplectic transformations,
that is, linear canonical transformations, e.g. those driven by harmonic oscillators
\cite{Peres, Bargmann, KramMoshSel, GuilStern, Voros76, Voros77, Littlejohn86, deGosson06}.},
the semiclassical theory is exact, including its IVR version. This provides scope
for simple applications that illustrate the general features of the method, without
gripping with the difficulties of a full SC calculation, as presented in {\bf I}.
Perhaps, the greatest simplification concerns caustics (or nodal lines in IVR):
The important point is that, even though families of metaplectic propagators
do cross caustics in any representation, as a parameter is varied,
in their case, the final integral for the correlation has no risk of
being divided into regions with different signs that need to be determined,
as was discussed in {\bf II}.
The present paper follows closely on the track of {\bf I}:
The same notations are adopted and we incorporate here many relevant features.
For instance, descriptions of any of the observables to be averaged may be supplied
by the translation operators underlying the chord representation, instead of the reflections
that belong to the Weyl representation, so that here we just focus on the latter.
Again, we shall not develop explicitly the alternative of picking a pair of trajectories
(forming a Final Value Representation, FVR)
the possible advantages being discussed in {\bf I}.
These alternatives shall remain implicit
so as to emphasize the purely original features of the present work.
We shall also rely on the discussion of sign ambiguities associated with crossings
caustics in {\bf II}, since they are readily incorporated into the wider setting
of evolving correlations. The great simplification here is one of scale:
By focusing on {\it mechanical observables} for which the Weyl symbol
coincides with the corresponding smooth classical phase space function except for corrections
that are of first order in $\hbar$, one reduces the expression for the evolving correlation
to a single phase space integral, irrespective of the number of observables.
The following section presents the general construction of the appropriate compound unitary operator
as the kernel for the correlation of evolving operators. Section 3 then interprets its
SC approximation in terms of a compound canonical transformation defined by a sequence of trajectory segments and derives its trace from the periodic orbits. The alternative IVR scenario is then
developed in section 4, whereas section 5 presents the simplifications inherent to the propagation
of mechanical observables. All formulae are valid for an even number of observables. Modifications
that may be required in the odd case are discussed in the appendix.
\section{Compound unitary operators}
The outcome of a standard repeated experiment on a quantum system
is expressed as an average over an (observable) operator,
which may correspond to a standard classical variable, such as position,
a projector, or a POVM. One can also measure correlations between such
observables that have undergone different evolutions.
In the simplest case, these may concern the same operator
traversing coherently the alternative paths of an interferometer, or just
measured at different times, as in the correlations of Leggett-Garg \cite{Leggett-Garg}.
Then, so that the correlation is real, one evaluates some suitable symmetrization of
\begin{equation}
\C = \langle\opA_{\nu}(t_{\nu})...~\opA_2(t_2)\opA_1(t_1)\rangle
= {\rm tr}~\opA_{\nu}(t_{\nu})~...~\opA_2(t_2)\opA_1(t_1)~\oprho,
\label{C1}
\end{equation}
where each of the Hermitian operators $\opA_j(t_j)$ undergoes a Heisenberg evolution
driven by some unitary operator, ${\widehat V}_j$, that is
\begin{equation}
\opA_j(t_j) = {{\widehat V}_j(t_j)}^{\dagger} \;\opA_j \;\; {\widehat V}_j(t_j).
\label{Heisenberg}
\end{equation}
If one defines the intermediate steps as
\begin{equation}
\opU_{j+1} \equiv{\widehat V}_{j+1}(t_{j+1}){\widehat V}_j(t_j)^{\dagger}
\label{intermediate}
\end{equation}
(with $\opU_1 \equiv{\widehat V}_1$ and $\opU_{\nu+1} \equiv{\widehat V}_{\nu}(t_{\nu})^{\dagger}$),
the general form is obtained,
\begin{equation}
\C = {\rm tr}~\opU_{\nu+1}\opA_{\nu}~\opU_{\nu}~...~\opA_2~\opU_2~\opA_1~\opU_1~\oprho,
\label{C2}
\end{equation}
in terms of the original observables $\opA_j$. No longer is one limited to symmetric Heisenberg evolutions, being that each observable can now be sandwitched between arbitrary unitary operators $\opU_j$. Hence, \eref{C2} also includes evolutions such as the fidelity, i.e. the quantum Loschmidt echo \cite{Gorin, IVRFVR}.
\footnote{This is the case of a single observable $\opA_1 = \opI$.}
An example of direct application of such a time evolved correlation arises
in the theory for time-resolved electronic spectra, depending on the evolution of two pairs of transition dipole operators. The Franck-Condon approximation then leads to an expression for
the spectrum in terms of the fidelity, which was obtained in \cite{ZamSVan} using IVR.
The present theory supplies in principle the full correlation witout any supplementary approximation.
Following the same notation as in {\bf I} and {\bf II},
the {\it centre symbol or Weyl symbol} of operator $\hat{A}$ is
\begin{equation}
\label{covW}
A(\x) = 2^N{\rm tr}\;\left[\hat{R}_{\x}\;\hat{A}\right],
\label{Weylrep}
\end{equation}
where $N$ is the number of degrees of freedom, while the unitary operator, $\hat{R}_{\x}$,
corresponding to the (classical) reflection through the phase space point $\x$
\cite{Grossmann, Royer, Report}, plays a fundamental role throughout.
In other words, $A(\x)$ is the Weyl representation of $\hat{A}$.
Alternatively, the {\it chord symbol} of the operator $\hat{A}$ can be defined as
\begin{equation}
\label{covC}
\tilde{A}(\Vxi) = {\rm tr} \;\left[\hat{T}_{-\Vxi}\;\hat{A}\right],
\end{equation}
where $\hat{T}_{\Vxi}$ is the {\it Heisenberg operator} corresponding to a phase space translation
by the vector $\Vxi$. The chord symbol and the Weyl symbol
are related by Fourier transformation. The advantage of both these representations is that
their families of {\it basis operators},
$\{\hat{R}_{\x}\}$ and $\{\hat{T}_{\Vxi}\}$, belong to the group of unitary operators.
An arbitrary operator is then expressed as a superposition of unitary operators
\begin{equation}
\label{conW}
\hat{A} = 2^N\int \frac{d\x}{(2\pi\hbar)^N}\; A(\x) \;
\hat{R}_{\x} ~
= \frac{1}{(2\pi\hbar)^N}\int d\Vxi \;
\tilde{A}(\Vxi)\; \hat{T}_{\Vxi} \,
\end{equation}
though it is convenient to keep the special notation for the density operator,
\begin{equation}
\label{rho}
\oprho = 2^N\int d\x\; W(\x) \;
\hat{R}_{\x} ~
= \int d\Vxi \; \chi(\Vxi)\; \hat{T}_{\Vxi} \,
\end{equation}
in terms of the Wigner function \cite{Wigner}, $W(\x)$,
and the chord function \cite{Report}, $\chi (\Vxi)$.
In the case of the Weyl-Wigner representation, one can now insert these
in the expression for the evolving correlation
\begin{equation}
\fl \C = \frac{2^N}{(\pi\hbar)^{\nu N}}\int d\x_{\nu}... d\x_2d\x_1d\x_0\; A_{\nu}(\x_{\nu})...A_2(\x_2)~A_1(\x_1)~W(\x_0)~~
{\rm tr}~ {\widehat{\mathbf U}}\{\x_0,\x_1,\x_2,...,\x_{\nu}\},
\label{C3}
\end{equation}
where the family of {\it compound unitary operators} is defined as
\begin{equation}
{\widehat{\mathbf U}}\{\x_0,\x_1,\x_2,...,\x_{\nu}\}
\equiv \opU_{\nu+1}~\hat{R}_{\x_{\nu}}~\opU_{\nu}...~\hat{R}_{\x_1}\opU_1~\hat{R}_{\x_0}.
\label{compu1}
\end{equation}
Before any evolution takes place, that is, when all of the $t_j=0$ in \eref{C1},
the compound operator is just a product of reflections, because all the $\opU_j=\opI$,
the identity operator:
\be
{\widehat{\mathbf U}}_0\{\x_0,\x_1,\x_2,...,\x_{\nu}\}
=\hat{R}_{\x_{\nu}}~...~\hat{R}_{\x_1}~\hat{R}_{\x_0}.
\label{compu0}
\ee
The simplest case is when $\nu$, the number of reflections is odd, i.e. an even number
of observables. Then the product is also a reflection and we can identify
${\rm tr}~ {\widehat{\mathbf U}}_0$ with the kernel for the Weyl representation
of a product of Weyl operators, each of which is specified by its Weyl symbol \cite{Report}:
\be
{\rm tr}~ {\widehat{\mathbf U}}_0\{\x_0,\x_1,\x_2,...,\x_{\nu}\}
= 2^{-\nu N} \exp\left[\frac{i}{\hbar} \Delta_{\nu+1}(\x_0,\x_1,\x_2,...,\x_{\nu})\right],
\label{tr0}
\ee
where $\Delta_{\nu+1}$ is the symplectic area of the polygon whose sides are centred on $\{\x_0,\x_1,\x_2,...,\x_{\nu}\}$ as drawn in Fig 1; this is a bilinear function of each pair of variables, which arises in the general product formula for the Weyl representation
of the product of an even number of operators, $\opA_{\nu}...\opA_1$, in \cite{Report}, that is,
\be
\fl \{A_{\nu}...A_1\}(\x_0) = \int \frac{d\x_{\nu}...d\x_1}{(\pi\hbar)^{\nu N}}~ A_{\nu}(\x_{\nu})...A_1(\x_1)~
\exp\left[\frac{i}{\hbar}\Delta_{\nu+1}(\x_0,\x_1,...,\x_{\nu})\right],
\label{Weylproduct}
\ee
so that one retrieves the simple expression for the initial correlation:
\be
\C_0 = \int d\x_0~ \{A_{\nu}...A_1\}(\x_0)~ W(\x_0).
\ee
In the following section, we show how this {\it polygonal scenario} is extended
to evolving correlations within the SC approximation.
\begin{figure}[htb!]
\centering
\includegraphics[height=4cm]{correvol0.eps}
\caption{The initial kernel of the integral for the correlation of an even number of observables
is the complex exponential of $\Delta_{\nu+1}(\x_0,\x_1,\x_2,...,\x_{\nu})$, the symplectic area of the unique polygon with sides centred on $\{\x_0,\x_1,\x_2,...,\x_{\nu}\}$.
For, $\nu=2$, this is a triangle (left), whereas the plane projection of higher polygons may have self-intersections (right).}
\label{Figt0}
\end{figure}
Evidently, one can express evolving correlations equally well in terms of compound operators
with translations in place of reflections,
\begin{equation}
{\widehat{\mathbf U}}\{\Vxi_0,\Vxi_1,\Vxi_2,...,\Vxi_{\nu}\}
\equiv \opU_{\nu+1}~\hat{T}_{\Vxi_{\nu}}~\opU_{\nu}...~\hat{T}_{\Vxi_1}\opU_1~\hat{T}_{\Vxi_0},
\label{compu2}
\end{equation}
or it may be more convenient to keep some reflections and some translations.
The issue appears in the context of a single evolving expectation and it is already
discussed in {\bf I}. Indeed, the product of any even number of reflections,
${\widehat{\mathbf U}}_0\{\x_0,\x_1,\x_2,...,\x_{\nu}\}$, results
in a translation operator, rather than a reflection, so that it is advantageous
to also use the chord-translation basis for the density operator.
For this reason, the results that must be adapted for an odd number of observables,
by resorting to the Fourier transform of the Wigner-Weyl representation,
will be remitted to the appendix.
\section{SC approximation for the compound Weyl propagator}
The key ingredient for the SC approximation of the compound propagator and its trace
is the general SC {\it Weyl propagator} corresponding to an arbitrary
classical canonical transformation that is
generated by a Hamiltonian, $H(\x)$, acting during a time, $t$.
In the simplest case, this is simply \cite{Berry89, Report}
\begin{equation}
U(\x) \approx \frac{2^N}{|\det(\Id+ \M)|^{1/2}}\>\>
\exp \left[ \frac{i}{\hbar}(S(\x)+ \hbar \pi \sigma)\right].
\label{Uweyl}
\end{equation}
The geometric part of the centre or Weyl action, $S(\x)$, is just the symplectic area between
the trajectory and the chord, $\Vxi={\x}^+ -{\x}^-$, joining its endpoints.
From this, one subtracts $-Et$, where $E$ is the energy of the trajectory.
The {\it Maslov index}, $\sigma$, is zero in a neighbourhood
of the identity operator, in which case there is indeed only a single classical trajectory
centred on the point $\{\x=(\vecp, \vecq)\}$ of classical phase space ${\bf R}^{2N}$.
Otherwise, there may be multiple solutions to the variational problem that identifies trajectories
with a given centre, $\x$, so the actions may have many branches
and these branches meet along caustics where the semiclassical amplitude diverges.
(See {\bf II} for the phase $\sigma$ and further details.)
The centre action specifies the classical canonical transformation, ${\x}^- \mapsto {\x}^+$,
corresponding to $\opU$ indirectly through \cite{Report}
\begin{equation}
\Vxi = -\J \frac{\der S}{\der \x}, \>\> {\x}^+ = \x + \frac{\Vxi}{2},
\>\> \x^- = \x - \frac{\Vxi}{2}.
\label{centran}
\end{equation}
The linear approximation of this transformation near the $\x$-centred trajectory is defined
by the {\it symplectic} matrix $\M$. This has the Cayley parametrisation:
\begin{equation}
\M = [\Id + \J\mathbf{B}]^{-1} [\Id - \J\mathbf{B}],
\label{Cayley}
\end{equation}
in terms of the symmetric matrix $\mathbf{B}$, which is just the Hessian matrix of $S(\x)$.
A notable exception for this SC form of the Weyl propagator is precisely
that of a reflection operator, $\opR_\x$. Indeed, its Weyl symbol is simply,
\be
R_\x(\x') = 2^{-N} \delta(\x'-\x).
\ee
It is the chord representation of this operator that has the standard semiclassical form.
To construct the SC approximation of the compound propagator,
${\widehat{\mathbf U}}\{\x_0,\x_1,\x_2,...,\x_{\nu}\}$, we assume that each of the Weyl propagators, ie. the Weyl symbols, $U_j(\x)$, for ${\opU}_j$ can be expressed in the form \eref{Uweyl}.
Then the key point is that the compound unitary operator,
${\widehat{\mathbf U}}\{\x_0,\x_1,\x_2,...,\x_{\nu}\}$,
for any choice of parameters, has its own Weyl symbol,
${\mathbf U}\{\x_0,\x_1,\x_2,...,\x_{\nu}\}(\x)$ or ${\mathbf U}(\x)$ for short.
This compound propagator is determined from the factor propagators, $U_j(\x'_j)$
according to the product formula \eref{Weylproduct}:
\begin{eqnarray}
\fl {\mathbf U}\{\x_0,\x_1,\x_2,...,\x_{\nu}\}(\x)=
\int \frac{d\x'_{\nu+1}d\x''_{\nu}...d\x'_1 \x''_0}{(\pi\hbar)^{2(\nu+1) N}} ~
&& U_{\nu+1}(\x'_{\nu+1})~R_{\x_{\nu}}(\x''_{\nu})...U_1(\x'_1)~ R_{\x_0}(\x''_0) \nonumber \\
&&\exp\left[\frac{i}{\hbar}\Delta_{2\nu+3}(\x''_0,\x'_1,...,\x''_{\nu}, \x'_{\nu+1})\right].
\label{compprop}
\end{eqnarray}
The integrals over the arguments of the reflections merely fix the respective centres, $\x''_j=\x_j$,
whereas all other integrals are evaluated semiclassically by stationary phase.
But the deduction of \eref{Uweyl} in \cite{Report} was itself based on the same product formula,
that is, from a product of infinitesimal small time propagators,
so the result is again a Weyl propagator of the same form,
but it is constructed from a {\it compound trajectory} built up from the sequences
of partial trajectory segments.
The relevant trajectory, which is centred on $\x$, i.e. the argument of the Weyl propagator,
is built up out $\nu +1$ such segments joined by $\nu +1$ reflections.
The appropriate trajectory segments that are generated by each centre generating function,
$S_j({\x'}_j)$, as well the centres themselves, need to be chosen so that
the full compound trajectory is continuous. This is achieved by imposing the requirement \cite{Report}
that the overall action in \eref{Uweyl} is just
\begin{equation}
{\mathbf S}(\x) = \Delta_{2\nu+3} + S_1({\x'}_1) +...+ S_{\nu+1}({\x'}_{\nu+1}).
\label{compaction}
\end{equation}
Here $\Delta_{2\nu+3}$ is the symplectic area of the {\it dynamical polygon} with a side centred on $\x$, as well as $\nu+1$ sides centred on the points $\x_j$ and $\nu+1$ sides centred on ${\x'}_j$.
The stationary conditions,
\begin{equation}
\frac{\partial {\mathbf S}} {\partial{\x'}_j} = 0 ~~~ {\rm or} ~~~
\frac{d{\Delta}_{2\nu+2}}{{d\x'}_j} = -~\frac{dS_j}{{d\x'}_j}= {\Vxi}_j,
\label{statcon0}
\end{equation}
define the variables ${\x'}_j$, so that the trajectory arcs fit precisely the sides of
the dynamical polygon, as depicted in Fig. \ref{Figcom}, which exemplifies the general layout for the case of two observables, i.e. $\nu=2$. It is naturally satisfied by any compound trajectory
that is followed through from its initial value.
\begin{figure}[htb!]
\centering
\includegraphics[height=4cm]{corevol1.eps}
\caption{ Phase space structure corresponding to the compound Weyl propagator for the evolution
of a pair of observables. The arguments of their Weyl symbol are the reflection centres
$\x_1$ and $\x_2$, while the reflection centre for the Wigner function is $\x_0$.
These reflections are joined by the trajectories of the intervening SC Weyl propagators
with centres at ${\x'}_1$ and ${\x'}_2$ so as to form the dynamical polygon.
(These trajectories need not be generated by the same Hamiltonian)}
\label{Figcom}
\end{figure}
As for the compound monodromy matrix, $\M$, this is just the product of the sequence of
monodromy matrices for each step. Indeed, since the matrix for a reflection is just $-\Id$,
we have simply
\begin{equation}
\M = [-\Id] \cdot \M_1 \cdot [-\Id] \cdot \M_2 \cdot ... [-\Id] \cdot \M_{\nu+1}
= (-1)^{\nu+1} \M_1 \cdot \M_2 \cdot ... \M_{\nu+1} .
\label{compmon}
\end{equation}
The kernel for the propagation of the correlation between observables, ${\hat A}_j$,
evaluated at multiple times in terms of the initial Weyl symbol, $A_j(\x)$,
is simply ${\rm tr}~{\widehat{\mathbf U}}\{\x_0,\x_1,\x_2,...,\x_{\nu}\}$.
Just as for other representations, the SC limit of the trace of an unitary operator
singles out the contributions of the periodic trajectories. In the case of the Weyl
symbol, this is derived from
\begin{equation}
{\rm tr}~ \widehat{{\mathbf U}} = \int \frac{d\x}{(2\pi\hbar)^N}~{\mathbf U}(\x).
\label{tru}
\end{equation}
Now, the only explicit dependence on the argument, $\x$ of the compound propagator
is in the symplectic area of the dynamical polygon, $\Delta_{2n+3} = \x \wedge \Vxi + Const$,
where $\Vxi$ is the side centred on $\x$. Like the term $Const$, this depends only on the other centres \cite{Report}:
\begin{equation}
\frac{\Vxi}{2} = ({\x'}_1 - \x_0) + ({\x'}_2 - \x_1) + ...+({\x'}_{\nu+1} - \x_{\nu}).
\end{equation}
Thus,
\begin{equation}
\int \frac{d\x}{(2\pi\hbar)^N}~ \exp\left[\frac{i}{\hbar}\Delta_{2\nu+3}(\x, \x_j,{\x'}_j)\right] =
\exp\left[\frac{i}{\hbar}\Delta_{2\nu+2}(\x_j,{\x'}_j)\right] \delta(\Vxi).
\label{zerochord}
\end{equation}
that is, the integral singles out those polygons where the open side has zero length.
The compound action for the trace is the same as \eref{compaction},
but with $\Delta_{2\nu+3} \mapsto \Delta_{2\nu+2}$, which is the polygon for a periodic orbit.
Such a {\it compound periodic orbit} for the trace is depicted in Fig.\ref{Figtr}.
The corresponding monodromy matrix is just \eref{compmon} without the last factor, $[-\Id]$.
\begin{figure}[htb!]
\centering
\includegraphics[height=4cm]{corevol2.eps}
\caption{ The SC expression for trace of the compound propagator is determined by
the corresponding periodic trajectories: The {\it open side} of the dynamical polygon,
$\Delta_7$ in the previous figure collapses, so that $\x=\x^{-}=\x^{+}$.}
\label{Figtr}
\end{figure}
The stationary condition for the intermediate centres, ${\x'}_j$, of the periodic orbit
is just \eref{statcon0}, so that this merely prescribes each side of the dynamical polygon
to be the chord for the corresponding trajectory segment. However, in practice
one need not search for the solution of each of these equations
because they are automatically satisfied for any given periodic orbit of the compound
canonical transformation, i.e. one then has all the centres and the sides of the $(2\nu)$-sided
polygon for a particular pinning of the reflection centres $\{\x_0, \x_1, ..., \x_{\nu}\}$.
Then one needs only to evaluate \eref{tru} by stationary phase,
so that the contribution of the p'th periodic orbit to the trace is just
\begin{eqnarray}
{\rm tr}~ \widehat{\mathbf U}_p &&\approx \frac{2^N}{|\det {\mathbf B}\det(\Id+ \M)|^{1/2}}\>\>
\exp \left[ \frac{i}{\hbar}({\mathbf S}(\x)+ \frac{\hbar \pi \sigma'}{4})\right] \nonumber\\
&&= \frac{2^N}{|\det(\Id - \M)|^{1/2}}\>\>
\exp \left[ \frac{i}{\hbar}({\mathbf S}(\x)+ \frac{\hbar \pi \sigma'}{4})\right],
\end{eqnarray}
where, following \cite{Report}, one takes the determinant of \eref{Cayley}.
\footnote{The stationary phase evaluation adds a further phase of $\pi/4$
times the signature of $\mathbf B$. This will not be needed in the IVR theory in the next section,
so that it is here just included in the {\it Maslov index} $\sigma'$.}
The parameters $\{\x_0,\x_1,\x_2,...,\x_{\nu}\}$ vary continuously
in the ultimate integration for the correlations,
so that the periodic orbits at each neighbouring parameter can be found
by a generalized Newton's method, such as in \cite{Aguiaretal87}.
The amplitude of the periodic orbit contribution to
${\rm tr}~ \widehat{\mathbf U}\{\x_0,\x_1,\x_2,...,\x_{\nu}\}$ becomes singular at those values of
$\{\x_0,\x_1,\x_2,...,\x_{\nu}\}$ that determine periodic orbit resonances: $\det(\Id - \M)=0$. This problem is avoided by the generalization of the IVR theory of {\bf I} in the following section.
The IVR approach even dispenses with the search for periodic orbits
in the evaluation of evolving correlations.
\section{Initial value representation}
Let us reinterpret the kernel of the evolving correlation \eref{C3} as
\begin{equation}
\fl {\rm tr}~ \widehat{\mathbf U}\{\x_0, \x_1, ..., \x_{\nu}\}
= {\rm tr}~ \opU_{\nu+1}~\hat{R}_{\x_{\nu}}~\opU_{\nu}...~\hat{R}_{\x_1}\opU_1~\hat{R}_{\x_0}~
= ~ {\mathbf U'}\{\x_1, ..., \x_{\nu}\}(\x_0),
\end{equation}
that is, according to \eref{Weylrep}, the Weyl symbol for the {\it reduced compound operator}:
\begin{equation}
\widehat{\mathbf U'}\{\x_1, ..., \x_{\nu}\} \equiv \opU_{\nu+1}~\hat{R}_{\x_{\nu}}~\opU_{\nu}...~\hat{R}_{\x_1}\opU_1.
\end{equation}
In its terms one generalizes the Weyl representation to the evolved product operator
in \eref{Weylproduct} as
\be
\fl \{A_{\nu}...A_1\}'(\x_0) = \frac{2^N}{(\pi\hbar)^{\nu N}}
\int d\x_{\nu}...d\x_1 ~ A_{\nu}(\x_{\nu})...A(\x_1) ~
{\mathbf U'}\{\x_1, ..., \x_{\nu}\}(\x_0),
\label{evproduct}
\ee
so that the correlation is simply
\be
\C = \int d\x_0 ~ W(\x_0) ~ \{A_{\nu}...A_1\}'(\x_0).
\label{Csimple}
\ee
Thus, $\widehat{\mathbf U'}$ has one less reflection than $\widehat{\mathbf U}$, but clearly
its SC approximation has the same form as \eref{Uweyl}, with the corresponding
classical polygonal path given by Fig.\ref{Figtr}, rather than Fig.\ref{Figcom} .
In other words, Fig.\ref{Figtr} is now reinterpreted as an open polygonal line,
with its begining and end points, ${\x_0}^{\pm}$, centred on $\x_0$,
as shown in Fig.\ref{FigIVR}.
It is important to note that, even though the full canonical transformation,
${\x_0}^{-} \mapsto {\x_0}^{+}$, can be decomposed into several
partial canonical transformations, the quantum unitary transformation $\widehat{\mathbf U'}$
corresponds to this single reduced compound canonical transformation. Furthermore,
each branch of its centre generating function , ${\mathbf S'}(\x_0)$, in the SC approximation
to the Weyl propagator, ${\mathbf U'}(\x_0)$, is constructed from those compound trajectories
that satisfy ${\x_0}^{-} + {\x_0}^{+} = 2\x_0$.
\begin{figure}[htb!]
\centering
\includegraphics[height=4cm]{corevol3.eps}
\caption{ The same polygonal trajectory as in the previous figure is reinterpreted
as an open trajectory of the {\it reduced} compound canonical transformation, from which
the initial reflection at $\x_0$ is removed.
This point is now the argument of the Weyl symbol for the reduced compound propagator.}
\label{FigIVR}
\end{figure}
The {\it initial value representation} (IVR) now results from the exchange of the integration variable
from the trajectory midpoint, $\x_0$, that is the argument of the Wigner function in the evolving correlation \eref{C3}, to the initial point, ${\x_0}^{-}$. This generalization of the
procedure in {\bf I} relies on the simple form of the Jacobian of the transformation:
\be
\det \frac{d\x_0}{d{\x_0}^{-}} = \det \left( \frac{\Id+\M'}{2}\right).
\ee
Thus, it brings the SC approximation for the evolving correlation to the form
\begin{eqnarray}
\C \approx \frac{1}{(\pi\hbar)^{\nu N}}\int d\x_{\nu}...d\x_1 d{\x_0}^{-} A_{\nu}(\x_{\nu})...~A_1(\x_1)~W(\x_0({\x_0}^{-})) |\det(\Id + \M')|^{1/2} \nonumber \\
\exp\left[ \frac{i}{\hbar}({\mathbf S'}(\x_0({\x_0}^{-}))+ \hbar \pi \sigma)\right],
\label{C4}
\end{eqnarray}
where $\M'$ is the monodromy matrix for the linearization of
${\x_0}^{-} \mapsto {\x_0}^{+}$ in the neighbourhood of ${\x_0}^{-}$.
It can be decomposed as the product
\begin{equation}
\M' = \M_1 \cdot [-\Id] \cdot \M_2 \cdot ... [-\Id] \cdot \M_{\nu+1},
\label{compmon2}
\end{equation}
a reduced version of \eref{compmon}, but now the sequence of factor monodromy matrices
is directly determined by the initial value ${\x_0}^{-}$.
Likewise, each of the variables ${\x'}_j$ will be just the centre of of the respective
side ${\Vxi'}_j$ of the polygonal path starting at ${\x_0}^{-}$.
The crucial point is that the Jacobian for the switch to the new integration variable,
${\x_0}^{-}$, kills off the caustic singularities in the SC kernel
for the evolving correlation. Thus, in a single step, without increasing the number of integrations,
one does away with the need to search for trajectories while erasing all caustics!
The generating function, ${\mathbf S'}(\x_0)$, for the canonical transformation
is now defined as
\begin{equation}
{\mathbf S'}(\x_0) = {\Delta'}_{2\nu+2}(\x_0, {\x'}_1,..., \x_{\nu}, {\x'}_{\nu+1})
+ S_1({\x'}_1) +...+ S_{\nu+1}({\x'}_{\nu+1}),
\label{compaction2}
\end{equation}
where the requirement
\begin{equation}
\frac{\partial {\mathbf S'}} {\partial{\x'}_j} = 0 ~~~ {\rm or} ~~~
\frac{d{\Delta'}_{2\nu+2}}{{d\x'}_j} = -~\frac{dS_j}{{d\x'}_j}= {\Vxi'}_j,
\label{statcon}
\end{equation}
defining the variables ${\x'}_j$
is automatically satisfied by any compound trajectory that is followed through
from its initial value. Indeed, as discussed in \cite{Report},
the symplectic area of the {\it reduced dynamical polygon} satisfies
\begin{equation}
2 {\Delta'}_{2\nu+2} = {\Vxi'}_1 \wedge {\Vxi}_1 + ({\Vxi'}_1 + {\Vxi}_1)\wedge {\Vxi'}_2 + ...
+ ({\Vxi'}_1 + {\Vxi}_1 + {\Vxi'}_2 + ... +{\Vxi'}_\nu)\wedge {\Vxi'}_{\nu+1},
\label{primepoly}
\end{equation}
where each chord refers to the appropriate reflection (unprimed) or partial evolution (primed).
The discussion of the Maslov phase $\sigma$ in {\bf II} applies direcly to \eref{C4}.
In a full SC calculation in which each trajectory segment needs to be integrated numerically,
the numerical error will build up along the sequence of segments. It may then be preferable
to start somewhere in the middle of the sequence, from where the sequence is taken partly backwards
and partly forwards. This is a direct generalization of the {\it Final Value Representation} (FVR)
for the evolved average of a single observable that was presented in {\bf I}, even though
the forward and backward paths only have the same number of segments if $\nu$ is odd.
In any case, the Jacobian for the exchange of the initial value for an intermediate value,
${\x_0}^{-} \mapsto {\x_j}^{-}$, a canonical transformation, is just unity.
\section{Evolving mechanical observables}
So far no account has been taken of features that distinguish observables from other operators.
Within the Weyl representation, {\it mechanical observables} are represented by real smooth functions
of the phase space variables. Indeed, the Weyl representation of the operator function,
$A(\hat{\x})$, is just the classical phase space function, $A(\x)$, plus corrections of order $\hbar$, which depend on the chosen symmetrization of products of $\op$ and $\oq$. Even more to the point, the Weyl representation of a product of mechanical observables,
${\opA}_{\nu}...{\opA}_1$ is just $\{A_{\nu}...A_1\}(\x)=A_{\nu}(\x)...A_1(\x)$, up to first order terms in $\hbar$,
which again depend upon ordering.
It is important to understand how this simplification arises, starting from \eref{Weylproduct}.
So one adapts the discussion concerning equations \eref{zerochord} and \eref{statcon}:
In the absence of any other phase term beyond the symplectic polygonal area,
stationary phase evaluation of the multiple integral in \eref{Weylproduct}
for all the variables $\{\x_{\nu}...\x_1\}$ collapses each side of the polygon in Fig.1,
$\Vxi_{\nu}=0~...~\Vxi_1=0$, and hence the polygon itself, with $\x_{\nu}=...=\x_1=\x_0$.
The $\hbar$-dependent corrections can be calculated via a generalization \cite{Report} of the familiar
Groenewold product formula \cite{Groenewold}, but they will be only of second order in $\hbar$
for symmetrizations that guarantee a Hermitian product.
This simple classicality of the Weyl representation of a product of mechanical observables,
which is shared by the initial correlation,
\be
\C_0 \approx \int d\x_0~ A_{\nu}(\x_0)...A_1(\x_0) ~W(\x_0),
\label{C0}
\ee
may be destroyed as the observables evolve. Even so, the absence of high period oscillations
in each of the Weyl symbols, $A_j(\x)$, still allows for stationary phase evaluation of \eref{evproduct}, analogous to that of \eref{Weylproduct}.
Indeed, the expression \eref{compaction2} for the reduced action, $S'(\x_0)$,
can now be reinterpreted as the composition of $2\nu+1$ transformations, but with zero action
for the unprimed variables.
\footnote{Note the subtle difference with respect to the multiple integral \eref{compprop}:
There, one integrates over the primed variables, arguments for the Weyl symbols for $\opU_j$.
Here, we integrate first over the unprimed variables, which represent the mechanical observables, $\opA_j$.}
Then the stationary condition \eref{statcon} for each of these centres
collapses the corresponding side of the (reduced) dynamical polygon,
that is $\Vxi_j(\x_j)=0$, as depicted in Fig 5.
\begin{figure}[htb!]
\centering
\includegraphics[height=4cm]{correvol5.eps}
\caption{Stationary phase integration over the unprimed reflection centres collapses the corresponding sides of the dynamical $(2\nu+1)$-sided polygon: These corners of the resulting $(\nu+1)$-sided polygon
now depend on the initial value, ${\x_0}^-$.}
\label{Figmechevol}
\end{figure}
Nonetheless, the {\it evolution sides}, ${\Vxi'}_j(\x'_j)$ are no longer zero, so there is generally a non-zero phase in the integrand that is constructed from the remaining $(\nu+1)$-sided polygon:
Each unprimed variable, $\x_j$ is now a free corner, depending on the initial value ${\x_0}^-$,
instead of being the fixed centre of a side.
In this way there results an enormous simplification of the correlation formula \eref{C4}:
\begin{eqnarray}
\C \approx \frac{1}{(\pi\hbar)^{\nu N}} \int d{\x_0}^{-} && A_{\nu}(\x_{\nu}({\x_0}^{-}))...~A_1(\x_1({\x_0}^{-}))~W(\x_0({\x_0}^{-})) |\det(\Id + \M')|^{1/2} \nonumber \\
&&\exp\left[ \frac{i}{\hbar}({\mathbf S'}(\x_0({\x_0}^{-}))+ \hbar \pi \sigma)\right].
\label{C5}
\end{eqnarray}
Thus one no longer deals with a full family of compound canonical transformations. Instead of this,
the polygonal trajectories are built for each initial value within a single canonical transformation without the intermediate reflections.
In the limit of short times, $\nu$ sides of the remaining $(\nu+1)$-sided polygon shrink to zero.
In this limit the overall phase is zero, so that one retrieves \eref{C0} if $\nu$ is even.
This restriction on $\nu$ follows from the expression \eref{compmon}
for the monodromy matrix in the amplitude of the compound propagator: As each
of the matrices $\M_j \rightarrow \mathbf I$, one obtains $\det(\Id + \M') \rightarrow 2^{2N}$
if $\nu$ is even, or zero if $\nu$ is odd (and hence a caustic).
In the latter case, the Appendix obtains the correlation
as a single integral of the Weyl symbols for the observables weighed by the chord function
instead of the Wigner function.
The case of multiple Heisenberg evolution \eref{Heisenberg} also collapses
the SC phase, but for all time! The easiest way to see this is to propagate directly
the Weyl representation of each observable, $\opA(t)$:
\be
A(\x, t) = \int \frac{d\x_1d\x_2}{(\pi\hbar)^{2N}} [V(\x_1, t)]^* A(\x_2 +\x_1 - \x) V(\x_2, t)
\exp\left[\frac{i}{\hbar} \Delta _3(\x, \x_1, \x_2)\right],
\ee
so that the phase space point representing $\opA$ is placed at the corner of the triangle
oposite the point $\x$ where the evolved observable is evaluated. In \cite{Report} it is shown
that $A(\x,t)= A(\x'(t, \x))$, the classically evolved observable, for a metaplectic evolution,
i.e., for a quadratic driving Hamiltonian. This is also the correct semiclassical
approximation for a mechanical observable, resulting from stationary phase integration:
There is only a single trajectory traversed both forwards and backwards from the initial point, $\x$,
and $\x_1=\x_2$ is its midpoint, so that there is complete phase cancellation.
In the case of the full correlation, the curved polygon in Fig 5 for the correlation collapses
into a thin legged $\nu$-{\it spider} as shown in Fig 6.
Then one can merely obtain the evolved correlation from \eref{C0} with the
classically evolved observables:
\be
\C \approx \int d\x_0~ A_{\nu}(\x'_{\nu}(\x_0))...A_1(\x'_1(\x_0)) ~W(\x_0).
\label{CHeis}
\ee
It is important to note that the collapse of the dynamical polygon into a spider concerns
exclusively the pairing of classical trajectories corresponding to the unitary operators within
the reduced compound operator. Thus there is no restriction on the Wigner function,
that is, the evolution is purely classical even for a highly oscillatory quantum Wigner function.
\begin{figure}[htb!]
\centering
\includegraphics[height=4cm]{correvol6.eps}
\caption{
The dynamical polygon collapses for multiple Heisenberg evolution of mechanical observables,
with each point $\x_j(\x_0, t)$ placed at the tip of a classical trajectory that is retraced with negative time. The {\it spider} shape arises when each observable is propagated by a different Hamiltonian. There is no phase factor in this essentially classical evolution
for the correlation between mechanical observables. Any difference between forward
and backward pair of Hamiltonians fattens the legs and these separate,
even if the same pair propagates all the observables.}
\label{Figspider}
\end{figure}
\section{Discussion}
The elaborate theory underlying SC approximations for the multiple evolution of correlations
between arbitrary numbers of quantum observables leads to deceptively simple results.
The approximate semiclassical scenario presented here becomes exact in the limit
where all the evolution operators are metaplectic, that is, generated by quadratic Hamiltonians.
Correlations are independent from the choice of representation that is employed in their calculation,
but the reliance on Weyl symbols and the Wigner function for the present theory
leads to a rich phase space structure. The evolution kernel is identified
with a compound propagator, which is constructed by a dynamical polygon whose sides
are orbit segments that are combined into a compound classical trajectory
and thence to a quantum phase.
Depending on the choice of symmetrization of the observables, an appropriate
trigonometric function of the polygonal area will dephase the correlation integral.
It is shown in the Appendix that the need to distinguish whether the number of observables
is even or odd dissolves for short times and for the restricted class of {\it mechanical observables},
such that the Weyl representation of their product coincides with the corresponding smooth classical phase space function, except for corrections that are of first order in $\hbar$. The general rule is that independent Heisenberg evolution for each observable leads to \eref{CHeis}
a classical expression of the evolving correlation,
even if the Wigner function employed in their average has quantum oscillations
that are separately though concurrently sampled by each observable.
In contrast, if each observable does not follow its own Heisenberg evolution
so that the intermediate evolution operators are not given by \eref{intermediate},
a phase factor will grow in time within the single phase space integral for the correlation.
This is generalizes the result for the fidelity (or the quantum Loschmidt echo) presented in {\bf I}, a special case within the present framework, that of a single {\it observable}, the identity operator, undergoing different forward and back evolutions. In the language of section 5,
the dephasing grows with the symplectic area of a single slightly fattened spider leg,
i.e. a curvilinear triangle. What about a repeated echo:
On returning, one evolves again with the same pair of forward and back Hamiltonians?
Then the initial value for the second traversal has changed,
so that a new spider leg is drawn which is only initially close to the first leg,
even though it is generated by the same pair of Hamiltonians
(specially if they are chaotic). The correct symplectic area
that determines the dephasing is then that of the full two legged spider,
a curvilinear pentagon, instead of twice the area of the first triangular leg.
For small times, the difference may be small,
but the denominator for the phase is Planck's constant...
Whereas the simple {\it dephasing representation} of Van\'{i}\v{c}ek \cite{Vanicek}
atributes the dephasing factor for the fidelity to a single classical trajectory,
this was shown in \cite{ZamOA11} to result from a first order classical perturbation theory
of an action for a pair of trajectories.
A further generalization in {\bf I} related the evolution of the expectation of a single observable
to a pair of trajectories surrounding a translation or a reflection. Now we find that the only
price to pay for having more mechanical observables in a correlation is to add more segments
to the corresponding compound classical trajectory. In all cases, the relevant trajectory is
completely specified by its initial value, i. e. the integration variable in the average.
Furthermore the general analysis of {\it Maslov phases} in {\bf II}, that is valid for all cases,
guarantees that initially, they are absent and it is only after a first caustic is crossed
that extra phases need to be taken into account.
One should note that relaxing the restriction to mechanical observables does not necessarily
complicate matters. Observables may well be projectors, so that
the correlations become correlations between measurements. For instance,
one may measure the momentum (see \cite{Mil12} for examples), i.e. the projector, $\hat{B}_P$,
rendered in the Weyl representation by the singular distribution,
$B_P(\x)= \delta(p-P)$, which actually simplifies the multiple integral for a correlation.
More generally, positive operator valued measures (POVM, see \cite{Peres})
are also represented by phase space functions in the Weyl representation.
A specially interesting case to study in this setting is that of a parity
projection onto the subspace for either eigenvalue $\pm 1$
of the reflection operators. Reflections are observables as well as being unitary operators
and their Weyl symbol is just a pointwise delta function. Their measurement was
proposed in \cite{Englert,LutterbachDav} and carried out experimentally in \cite{Bertet02}.
Thus, one may readily extend the present results
beyond the restricted class of mechanical observables.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 3,522 |
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