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(AFP/Getty Images)
Badaling Pass, 1972
by John K. Fairbank
John K. Fairbank
John K. Fairbank (1907-1991) was a highly influential scholar of Chinese history. He is largely credited with founding the field of Chinese Studies in the United States. After graduating from Harvard...
The Red Guard: A Report on Mao's Revolution
by Hans Granquist, translated by Erik J. Friis
Praeger, 159 pp.
China in the Year 2001
by Han Suyin
Basic Books, 267 pp.
China Looks at the World, Reflections for a Dialogue: Eight Letters to T'ang-lin
by François Geoffroy-Dechaume, Translated from the French by Jean Stewart, with an Introduction by Paul Mus, a Foreword by the Rt. Hon. Philip Noel-Baker
Pantheon, 237 pp.
This article was first published in the March 28, 1968 issue of The New York Review of Books.
China is so distant, big, and complex that each Marco Polo nowadays tells a different tale. The authors of the three books under review—a cool Swedish journalist, a passionate Chinese true-believer, and a philosophical Frenchman—give very different impressions of Chairman Mao's revolution. Unencumbered by first-hand contact ourselves, we Americans can judge only at second-hand whether Mao's China is working or in deep trouble, expansionist or merely defensive. These questions will become even more interesting if China tries to keep us over the barrel in Vietnam by coming into the war herself.
Hans Granquist has been for many years a correspondent for Swedish newspapers, radio, and television in Hong Kong, and was in China in November and December, 1964, and in April and May, 1966. He finished this book at the beginning of January but has added new material to the English translation from the Swedish, bringing his account up to May of last year. He sees the issues of Mao's Cultural Revolution as a personal competition for power among interest groups. Following Stuart Schram and others, he sees Mao as "a revolutionary romantic, a utopian," "an old man in a hurry." Mao has the idea (which is probably his chief contribution to Marxist-Leninist theory) that "contradictions among the people"—the "non-antagonistic" kind that must be settled by teaching and persuasion—are perpetual and in the nature of things. Therefore they can be resolved only by "permanent revolution," in effect a continual series of campaigns mobilized to reform one or another aspect of social conduct in one or another part of society.
Granquist proceeds to analyze the various kinds of opposition to Mao's nostalgic ideal of a decentralized, self-sufficient "guerrilla communism." Against this opposition, Mao, through Lin Piao, first turned the Army into a political base from which to battle the post-1949 "administrative generation" in the Chinese Communist Party. There then ensued the successive phases of the Great Proletarian Cultural Revolution: the effort, in 1964, to train an elite of "revolutionary inheritors," youths who could carry on the cause; the creation, in August 1966, of the Red Guards as extra-party shock-troops, millions of them, emerging from the youthful masses (more than half of China's 750 million people are in their twenties or younger). These, in turn, met the passive resistance of the local Party bureaucrats, who sabotaged efforts to purge them by putting into the field their own rival Red Guards, raising workers' wages, and generally compounding the confusion. "Mao was stuck in the mire of his own Cultural Revolution…. No one could differentiate reliable Maoists from those who were not." The result of this confusion has been the emergence of the Army as the only available group with sufficient power to restore order.
But Granquist concludes that "Mao's final goal is still the creation of a society in which it is impossible for a certain group of people—be they experts, intellectuals, or Party members—to regard themselves as superior to the masses. We see here the principle of equality pushed to its utmost limits…the most extreme experiment in utopianism ever attempted…based on the assumption that man can be re-educated to learn both self-discipline and unselfishness."
Mr. Granquist's book is as concrete and perceptive a study as we have in brief compass, but like any observer he brings his own feelings and his own responses to the scene. In appraising his stress on egalitarianism in China today, his readers should keep in mind the cultural and political setting within which Maoism has had to operate: the supreme authority is still the personal teaching of a great individual, a hero-sage who leads by concrete example. There is no transcendent yet personal deity to invoke in the background, and the doctrines of Marxism-Leninism-Maoism seem less abstract principles or laws than immanent natural tendencies or operational norms. Mao's anti-Confucian egalitarianism may be a modern infusion of Western origin. But whereas the American myth now expanding over the world is one of individualism and affluence, Chinese egalitarianism entails conformity and poverty as far ahead as one can see. The struggle to remake the old China from the inside may be appreciated by Europeans historically accustomed to violent social revolutions, but it is like nothing Americans have ever experienced. It begins with a new orthodoxy even more all-embracing than the old Confucian one.
Mao's black-and-white orthodoxy is sharply described by the true-believer, Han Suyin. Her book is so full of hopeful exaggerations and bitter distortions that a reader acquainted with the facts of Chinese history would find it shocking. However, although many Americans lack even a rudimentary knowledge of China's history, Dr. Han's message is so shrill that it will prove salutary even to them. Dr. Han's book seems to reflect the elemental emotions of pain, hunger, fear, and rage which Peking itself expresses: pain over China's cruel victimization by imperialism and colonialism, hunger for requital and recognition as the greatest nation on earth, fear of the inevitable American imperialist attack, and rage at enemies in many forms.
Han Suyin is of course a gifted writer. As her autobiographical books—The Crippled Tree, A Mortal Flower, and even her first success, A Many Splendoured Thing—all attest, she was caught between two cultures. Daughter of a Chinese father and a Belgian mother, she grew up in the 1920s and 1930s at a time when privileged foreigners, as part of China's ruling class, could still enjoy the best of both worlds, but being neither the one nor the other, she got the worst. She found her salvation by determining from an early age to study medical science and get the secure status of a doctor. She seems to have few happy memories of the warlord and Kuomintang China of her childhood, but after eleven visits to the mainland in recent years has given up medical practice and become an eloquent interpreter of China's revolution to English readers—in Peking's corner, but not under Peking's control and thus free to express herself. I attempt this crude analysis because Dr. Han's sense of grievance over China's raw deal at the hands of America, like her Marxist-Leninist dramatization of modern history, seems to echo that of the Maoists in general.
Her simplistic view of the 2500 years of "feudal bureaucratism" and "antiquated exploitation" before 1949 denigrates China's past and sells the old culture short. The old "literocracy" had its roots "in the very language, the ideograms (many of which are based on feudal concepts), the system of education, the (still very active) notions of 'face' and many other attitudes. Two millennia of government by the Divine Right of Reading are not easily abolished in less than two decades; the Cultural Revolution of 1966 is attempting once more to uproot the deep stalks of this oligarchic mentality." Han Suyin gives the back of her hand to Western Sinology and shows little interest in the facts, including even the dates, of history. Yet her main complaint is quite intelligible: with the nineteenth century, China's traditional civilization let her down. The old ways were indeed inadequate to modern times.
We face here a pattern that leads toward wishful thinking and unreality: China was a civilization, and proud accordingly. The civilization collapsed, the pride has not only remained but intensified: demanding at least equivalence with the West, superiority if possible. This requires militant struggle to destroy the old China and build it anew. Mao's success in this makes him the leader of the world revolution, and thus responsible for China's renewed importance in the world.
Han Suyin's account of this is full of pathos and self-pity. The century 1840-1949 was an "accumulation of anguish…bitter struggle against occidental exploitation…bitter servitude, misery and exploitation…repetitive violence…occupation by foreign troops, massacres, and the sacking of Chinese cities," and so on. Japan's success in modernizing under rather similar handicaps in the same period is simply brushed aside. Self-pity also fosters the fatalistic conviction that America will destroy Mao's creation. This theme crops up repeatedly: Vietnam is "the prelude" to an attack on China; the decentralized communes will be, "in the event of massive bombing, unconquerable"; "the grand design to 'bomb China back to the Stone Age' " will be thwarted by locating industry so as "to mitigate the bombing that is to come"; industrial planning "takes the coming attack by the USA into account"; scientific research has been "overshadowed by the…almost certain attack, within the foreseeable future, by the USA." She says that "ideological education is essential" against "the threat of nuclear annihilation"; and "nuclear and bacteriological attacks to put down the population"; and also "to prepare the masses for the coming onslaught." "In the ever-deepening shadow of a largescale attack on China…that…would contain and keep down the growth of China"; "China has become the…beacon of revolution for the whole world. China is ready, therefore, for 'any national sacrifice' to fulfill this great duty to humanity"; "to shoulder 'the heaviest burden'…that is, the burden of being attacked and fighting a long war of attrition alone…direct collision between China and the USA is now almost inevitable…the Cultural Revolution today is an expression of…total preparation"; "for a long protracted war upon Chinese soil"; "with nuclear strikes on densely populated areas to 'kill off' as many millions as possible"; an "almost inevitable war that is being imposed upon China through escalation in Vietnam." These references are scattered through chapters on agriculture, industry, science, education, ideology, and Mao's program generally, leading to a chapter on "War and Peace."
Thus Han Suyin finds Mao's leadership valiant and his success almost superhuman in view of China's victimization in the past and her victimization in the future. Dr. Han pictures a resurgent self-reliant China that is nevertheless victimization-prone, resigned to the worst, expecting it.
We Americans say of course that we have no intention of bombing China. But before 1965 we had no intention of bombing North Vietnam either. One can only conclude from Dr. Han's book that our bombing south of China's border—believe it or not—has had psychological repercussions to the north. Outwardly prudent and non-provocative, the Chinese response has been inwardly one of fear, humiliation, rage, and, finally, undying hatred. It is difficult to think of 750 million people storing up a slow-burning resentment of us day by day, because the resentment is so intangible. Even so perhaps such intangibles should begin to enter our calculations. In the end this enmity may indeed give the Chinese world leadership, against a people who find themselves called upon to live by the sword.
François Geoffroy-Dechaume spent the years between 1937 and 1943 as an officer in Indo-China training Vietnamese infantry. He visited China in 1937, and "in 1943 he did consular work in Peking for the French Foreign Service." He subsequently worked for the United Nations and has recently revisited China. He views China's revolution as a "revulsive crisis," a rejection of the West. He urges us to head off a world conflict by trying to understand China's culture in the broadest and deepest sense, seeing ourselves and China in perspective; this would make Sino-Western relations more manageable. This excellent aim he pursues in a number of essays on aspects of Chinese life and civilization, written during the years between 1963 and 1966. Unfortunately they are too discursive and over-reflective to have much interest for those American readers who do not keep Descartes, Pascal, Buffon, Lévi-Strauss, Fénelon, Renan, Teilhard, Brunetière, and the Taoist classics at hand for bedside reading.
The book concludes with the following sentence, which summarizes its eight chapters:
China is undergoing a revulsive crisis (I) because this Civilization-State (II) in order to overcome its internal crisis and its humiliation (III) wants to invent (IV) and thus recover its cohesion (V) and its inspiration (VI) thorugh its educators (VII) who mobilize it against cultural oppression (VIII).
To call Mao a mere "educator" suggests the degree of the author's general imprecision; but to see our problem as culture-wide is surely the beginning of wisdom. There is no other way to comprehend what we and the Chinese are up against.
All three studies leave one wondering. How can two peoples of such self-assurance and righteousness—America's all-conquering image, China's image of invincibility—ever manage to avoid eventual conflict? A Gallup Poll published last February 6 asked a national sample of American adults how much they liked twenty-eight countries. On a scale of 100, Canada was rated the highest at 94, Japan 72, Russia 19, North Vietnam 7, and, at the very bottom, Communist China 5. Before we express this disenchantment in any further action, prudence and our Vietnam experience suggest that we seek some perspective on China and ourselves. We must ask why there is so much mutual animosity?
We may ruefully conclude that we were in some degree sucked into Vietnam because the enemy's cultural differences provoked our virtue: we saw him as the heathen who might be converted. But hostilities come from interaction: both sides contribute. We might well consider now how we may by-pass the role in China which Mao assigns us, and which observers like Han Suyin foresee. I suggest, in short, that the Sino-American confrontation—"anti-imperialism" and "containment" each begetting the other—is likely to be stopped only if we Americans stop it. This we can do only if we understand the roots of this confrontation.
The story of Sino-American love and hate, or toleration and rejection, has been little studied and less understood. Quite aside from the as yet minimal use history has made of social psychology, we know too much about ourselves and too little about the Chinese. Richard Solomon has noted how China's hostility is powered by Mao's need for a foreign oppressor. America, he suggests, is actually in a "revolutionary alliance" with Communist China "in the sense of being one of a series of hated oppressive authorities used by Mao Tse-tung as a foil in his efforts to mobilize support" (Asian Survey, December 1967). We have all seen how necessary we have become as a target of China's own peculiar revolutionary process. It is almost like being victims of a madman's fixation, except that we have responded by playing our assigned role all too realistically.
Each of the three observers—none of them American—touches on this psychological pattern: the oppressiveness of the old regime in China, and the Westerners' support of it to the point of becoming foreign oppressors. Sober historical fact and analysis might suggest the opposite: that the old regime collapsed less from its oppressiveness than from mere inanition, and the Westerners did far more to tear it all down than they ever did to prop it up. Never mind: revolutionary fanaticism and foreign guilt feelings propagate the myth, thereby meeting a need in Maoist cosmology.
The starting point of this fixation is the old, authoritarian Chinese social structure. This was the structure that survived into the twentieth century after 4,000 years of evolution. It put each Chinese in his proper hierarchic place (pen-fen), using all possible means to differentiate between superior and inferior, literate and illiterate, mental labor and manual labor, landlord-scholar-official and artisan-peasant, the privileged governors and the unprivileged governed. At the top of this structure of privilege, the old ruling class carried on the great tradition of Chinese culture, its role sanctified by the Confucian ethical teachings.
This ruling class was a joy to be part of for those who made it. The Mongols and Manchus who conquered China in the thirteenth and seventeenth centuries found upper-class life irresistible. So did the Westerners of modern times, whether missionaries or merchants, diplomats or educators. Privileged status was thrust upon them by extraterritoriality. The American love of old Peking and the treaty ports can hardly be divorced from the personal service, polite deference, freedom from police, and psychological satisfaction that went with honorary membership in the Chinese upper class. For the privileged few in old China consciously floated on the huge sea of peasants, who might be hard-working and intelligent, but nevertheless went through life illiterate and immobile in the village, absorbed in family relationships, with a rich lore of superstition in their heads and, all too often, worms in their bowels.
To remake a two-level social structure, imbedded for centuries, in a country so enormous and self-contained, so walled off by its peculiar writing system, so proud of its long history, has naturally required a great destructive effort. So far as social revolution is concerned, the distance between Confucius's hierarchy of unequal relationships and Mao Tsetung's politics of mass participation is far greater than the distance between European feudalism and modern multiclass democracy. China since 1895 has had to attempt political changes even more drastic than those achieved in Europe since 1789. On our side, Americans have had no real experience of the frustrations and hatreds, the destructiveness and the utopianism, involved in social revolution. Never having been peasants, we have trouble understanding Mao. From the beginning his movement has been fueled by resentment of the restraints imposed and the symbols used by the old order. Strict filial obedience had always entailed some conscious fear and some suppressed hatred of the parent, and the old virtues of loyalty, chastity, and duty-before-self had been preserved only at the price of harsh self-discipline. Amid the altered circumstances and new options of modern times, these old restraints and their symbols could be destroyed and thrown away.
The first target was the monarchy, keystone of the ruling class, and then the study of the Confucian classics, which had for so long produced the scholar-official rulers. Yet in every village, the same system of social harmony enforced through inequality was inherent in the family and clan, with their subjection of youth and women. Once started, the revolution found no stopping place. Three generations have attacked the images of emperor, father, and foreigner as their oppressors. K'ang Yu-wei and Liang Ch'i-ch'ao undermined the classics at the turn of the century. Sun Yat-sen and Yuan Shih-k'ai between them brought down the Manchu monarchy. The May Fourth Movement of 1919 attacked the family. Mao Tse-tung has tried to finish the job, including his attack on the last segment of the old ruling class: foreign participation in Chinese life. Yet, having destroyed the oppression of "colonialism and imperialism" within China, he still encounters a policy of "containment" outside.
It is not easy to remake a people's traditional ideals and habits. The desire to wear a long gown and escape manual labor, to read Chinese characters and give orders to others, to study for examinations and rise in the world, to "be an official and get rich" (tso-kuan fa-ts'ai), is seductive and natural, as natural as our tendency to accumulate personal property. (Making use of a foreign menace is also an old custom in Chinese politics.) These inherited antisocialist tendencies are still Mao's real enemy, though to call them "capitalist tendencies" is unenlightening jargon to American readers. Revival of ruling-class privilege, personal favoritism, bureaucratic cliquism, and nepotistic corruption is a more intelligible way to put it. We can also see the revival of regionalism against too tight central controls and of pragmatism against policies too drastically utopian. (Reliance on Soviet aid has been one form of pragmatism.)
As a result of all this, we find ourselves "American imperialists," caught in a role which has been partly thrust upon us and which we do not want: that of the outside oppressor, last target of Mao's revolution, tied in with the evils of China's past, cause of her inadequacies and obstacle to her rise.
Instead of responding to this Chinese sense of wrong with our usual countervailing emotions of outraged indignation or guilt-ridden sympathy, we must now take an interest in the facts and cut our fear of China down to realistic size. We saw in 1966 how crazy Mao was to close all Chinese schools for a year—how can a country possibly modernize without education? Now in 1968 we ourselves have abolished draft deferment for, among many others, the few hundred talented and carefully selected American graduate students specializing in Chinese studies. Who's crazy now?
History, Politics
Hans Granquist, Lin Piao, Red Guards, Han Suyin, Vietnam, the Great Wall, Cultural Revolution


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Prasocuris junci is een keversoort uit de familie bladkevers (Chrysomelidae). De wetenschappelijke naam van de soort werd in 1790 gepubliceerd door Brahm.

Bladkevers

redpajama_set_name: RedPajamaWikipedia


A stunning large 5 bedroom Townhouse in Elviria with outstanding sea views. The property also features a second living area on the lower floor along with spacious utility area, and direct but private access straight to the pool.
Townhouse - Terraced, Elviria, Costa del Sol.
5 Bedrooms, 3.5 Bathrooms, Built 0 m².


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IIROC Trading Halt - DMR
SLMZF
November 19, 2021, 8:52 a.m. ·1 min read
VANCOUVER, BC, Nov. 19, 2021 /CNW/ - The following issues have been halted by IIROC:
Company: Damara Gold Corp.
TSX-Venture Symbol: DMR
All Issues: Yes
Reason: At the Request of the Company Pending News
Halt Time (ET): 8:43 AM
IIROC can make a decision to impose a temporary suspension (halt) of trading in a security of a publicly-listed company. Trading halts are implemented to ensure a fair and orderly market. IIROC is the national self-regulatory organization which oversees all investment dealers and trading activity on debt and equity marketplaces in Canada.
SOURCE Investment Industry Regulatory Organization of Canada (IIROC) - Halts/Resumptions
View original content: http://www.newswire.ca/en/releases/archive/November2021/19/c5278.html
VAPERS TAKE CENTRESTAGE DURING NATIONAL NON-SMOKING WEEK
The vapers of Canada are claiming their spot in National Non-Smoking Week. Under the theme of #SmokeFreeWithVaping, vapers are speaking out to demand that governments acknowledge that vaping is less harmful than smoking and can help smokers quit.
Government of Canada continues support for clean energy opportunities in Saskatchewan
As Canada pursues a net-zero carbon emissions future, the Government of Canada, through Prairies Economic Development Canada (PrairiesCan) is investing in communities affected by the transition from coal-generated power. Transitioning to a low-carbon economy is a necessary step to protect the climate and create a more sustainable economy in Saskatchewan and across Canada. The Government of Canada is ensuring that this transition is fair for workers and communities with a strong economic connecti
2 Bank Stocks That Could Rocket Higher in 2022
TD Bank (TSX:TD)(NYSE:TD) is a severely undervalued bank stock that Canadian investors may wish to stash atop their watchlists in 2022. The post 2 Bank Stocks That Could Rocket Higher in 2022 appeared first on The Motley Fool Canada.
Moody's places all Citibanamex ratings, assessments on review for downgrade
Ratings agency Moody's Investors Service on Sunday placed all ratings and assessments of the consumer banking arm of Citigroup in Mexico on review for downgrade. In a statement, Moody's said the review was triggered by Citigroup's announcement that it would sell its Citibanamex https://www.reuters.com/business/citi-exit-mexican-consumer-banking-business-strategy-revamp-2022-01-11 consumer banking operations, which will end a two decade retail presence in Mexico.
UK manufacturers feel the pay pressure from rising inflation
British manufacturers have offered higher pay deals to staff but many others are holding off on settlements as they monitor fast-rising inflation and get ready for higher minimum wages and a tax hike, industry group MakeUK said on Monday. Most pay increases ranged between 2% and 3%, below the recent pace of inflation, but went as high as 14% in some cases. The Bank of England is keeping a close eye on the impact of rising inflation on pay deals, something it fears could lead to longer-term inflationary pressures.
If playing catch up with Tesla is what everyone in the auto industry is about then Stellantis, the company formed from the merger of Fiat Chrysler and Peugeot, has had a good start – its shares have far outpaced its U.S. rival in its inaugural year. Fixing its business in China and overcapacity in Europe are just two areas where analysts want to see Stellantis making progress when Chief Executive Carlos Tavares unveils his detailed business plan on March 1.
Human Horizons' HiPhi X Tops China Luxury Electric Vehicle Sales for the 2nd Half of 2021
According to data released by the China Automotive Technology & Research Center, HiPhi X was the best-selling luxury electric vehicles with MSRP more than 500,000 RMB in both December and the second half of 2021. Designed and manufactured by industry-leading new mobility and intelligent driving company Human Horizons, the HiPhi X is equipped with the latest in technological innovation and all of the trimmings of a luxury vehicle to create a new segment of cars, TECHLUXE®.

redpajama_set_name: RedPajamaCommonCrawl


I'm having trouble downloading a picture. I get a message saying that I have exceeded my quota by 1.32 MB. What do I do? Thank you.
Also a quick note for all. If you have the option of uploading a smaller file size, use that because, like all forums, there are limits on the total size of attachments someone can upload. The limits are set high here, but you don't want to waste space on unnecesarily large files. For example, image sizes are best at between 400 - 600 px width. Anything larger will just waste your space. There are image hosting sites, however if you delete your image account, your pictures are gone off the forum. We've all seen that on some of our favorite sites and it is disappointing, so if you use a hosting site, make sure you don't delete your account.


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Q: Content Decoding error on my website when I add a UTM to the URL I set up an ad campaign for my website and added a UTM to the links. So far so good. The problem is, everytime I enter to said link and reload  the page I get one of the following errors in the console:
Failed to load resource: net::ERR_CONTENT_DECODING_FAILED
---
GET https://example.com/?utm=test 200

Also, the loading times are way slower.
Without the UTM I never had any issues with the loading of my website.
Any Ideas of what could be causing this?
The website runs on wordpess.

redpajama_set_name: RedPajamaStackExchange


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Executive Boardroom ideal for private meetings with 33 sq.m.


Fill your tum with yummy grub – we have eight distinctly different cafés & restaurants to satisfy every taste bud!
We are proudly 100% free-range in every one of our eight cafés & restaurants at Cardrona.
Captain's Pizzeria is our home away from home in Captain's Basin – ski here on-piste from the Main Basin via A Real Journey to try the famous pizza & other scrumptious menu items. It has been a Cardrona icon since it opened in 1986, & there's still no better place to enjoy a spring afternoon in the sun than on the Captain's Pizzeria deck! It's also the best spot on the mountain for a great brew, with local beer & cider on tap, including our collabs with Altitude Breweries & Zeffer Ciders. Try them with the scrummy tiramisu for an afternoon après session.
Offering: Self-service, freshly-made convenience food. Base Café is your 1-stop shop for food on the run!
NEWLY REFURB'D FOR 2018! With plenty of sweet things to keep your energy levels up & a wide array of fresh cabinet food, Base Café is a great stop to fill up on the go. Fill up with an award-winning Wanaka Bakery pie, or a home-made sandwich. Check out our famous Mountain Meals– a reasonably-priced meal that changes daily. Base Café is the home of our Courtyard dining – pull up a picnic table, have a cold drink & a hot meal in the sun!
NEWLY REFURB'D FOR 2018, The Mezz is the crème da la crème of Cardrona's dining options! Mezz has breakfast, lunch & dinner a la carte dining, amazing coffee & a fully-stocked bar for those après ski afternoons. The Mezz menu proudly uses locally-sourced ingredients for mouth-watering meals. Ask one of our friendly servers about today's Mezz specials.
Offering: Allpress coffee, selection of teas, beer & giant pretzels!
Juice & Java is our on-mountain brew bar & is the ultimate stop to quench your thirst, no matter what time of day! Grab your fresh pastries, locally brewed beer or mulled ginger bear by Crimson Badger Brewing ordered straight from the main base café tills!
Noodle Bar is the place to be if you have a hankering for Asian-inspired food. With a wide range of noodle dishes cooked right in front of your eyes, as well as freshly-made sushi, there's something at Noodle Bar for every hungry tummy! Ski in from Skyline Ridge for an amazing slope-side dining experience. Noodle Bar has the best views of the Beginners' Area, & with its latest refurb is also the hub for our Conferences & Venue Hire.
Pro tip: Noodle Bar is worth the trek up the stairs for your morning coffee, with much shorter waits than the other cafés. While you're up there, try our new breakfast waffles!
At the bottom of Valley View Quad is the Little Meg Café – perfect for a snack on the go. If you have your lift pass already, park at the bottom of Valley View, grab a coffee from Little Meg & get on the lifts before anyone else! It's the perfect spot to get your early morning caffeine fix. Throughout the day, cruise down to Little Meg for some great tunes, fresh coffee, & cold locally-brewed beer at Cardrona's golden nugget of a café!
At the top of McDougall's Express Chondola is the brand new Vista Bar! Grab a cheeky glass of Mumm champagne, pull up a beanbag & watch the world go by! Take a photo with the Wakatipu Basin as your backdrop on the old Captain's swing chair. Whether you're skiing, snowboarding, or just up to check out the view, Vista Bar is your spot for a little bit of indulgence at the top of the mountain!
Location: Base Area, right-hand side of the Archway as you're looking down from the top of the mountain.
The old Kids' Club has been transformed into the Lounge – a sophisticated slopeside bar & aprés. Sit on the slopeside deck in the sun & have a quiet one with your mates, ski in for coffee at the slopeside window, or send a few emails from the mezzanine floor – the Lounge is the ultimate spot to catch up & get some refreshments.


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JASON EDWARD CHARLES
Jason Edward Charles has spent the greater portion of the past two decades making the world his classroom. His post-collegiate years found Charles taking photographs of everything from street life in Mexico City to the gorgeous Spanish countryside. After Europe, Charles moved to New York and waited tables in the Manhattan restaurant industry, where he was hostage to the good fortune of being surrounded by passionate wine experts. His interest ablaze, Charles landed a harvest internship and promptly moved to Northern California, where he worked the fields to learn every possible aspect of the winemaking craft. From there, Charles headed to France to further his studies at Chateau Haut Goujon in Lalande de Pomerol.
After serving the soil, grape and sun in France, Charles returned to work at some of Napa's most renowned wineries, formally establishing Vinca Minor in 2013.
EMILY O'HARA CHARLES
Emily has fallen into wine making by proxy of her husband. It is not her craft, but business and design is. By day, Emily is Director of Global Brand Operations at eBay. When she's not at her day job, you can find her with their daughter Ava or fully immersed in Vinca Minor, managing the programming and events for the tasting room and wine club.
AVA ELIZABETH CHARLES
Ava was born into it. At three years old she isn't involved in the winemaking, but loves a vineyard visit, specifically for the dogs. In the tasting room, you can find her curating all aspects of the kids activity corner.
Soil grape & sun
Natural wines made with love in Berkeley California.
© Vinca Minor


FOXBOROUGH, Mass. — Jamal Adams reduced the state of the Jets to simplest terms moments after punctuating a humiliating 38-3 loss to the Patriots on Sunday.
"We need more players and we need more dogs. It's as simple as that," Adams said. "We have a lot of talent on the team, but we don't have a lot of dawgs on this team. We just don't."
God Bless the Jets safety who stripped away the B.S. and gave it to everyone straight after this miserable 4-12 season. Adams offered fair criticism to all the culprits in the wake of this lost campaign.
"You got to go get players, man," Adams said. "You got to go get big-time players. You got to go get names. It's simple. You look at the top teams. You look at the Saints, you look at the Rams, you look at the Chicago Bears. You look at all these teams, man. Come on, man.
The unvarnished truth is that the Jets don't have anywhere close to the talent required to be a viable playoff contender. Should they have squeezed out more than four wins this season? Absolutely.
That's part of the reason why Todd Bowles will be fired after a 24-40 four-year run, but Adams is smart and honest enough to admit that massive changes must take place if this franchise has any chance of reversing its fortunes.
He's also fully aware that the time is now to strike with a franchise quarterback and more than $100 million in salary cap space.
Adams isn't bashful of openly recruiting some of those big-time players, either.
Although Adams maintained that "everybody in that building is together," the general manager and head coach are going on divergent paths. Only Bowles will be fired.


text:
Q: Is one of the numbers in this list in between the two given integers? I have a list with barline ticks and midi notes that can overlap the barlines. So I made a list of 'barlineticks':
barlinepos = [0, 768.0, 1536.0, 2304.0, 3072.0, 3840.0, 4608.0, 5376.0, 6144.0, 6912.0, 0, 576.0, 1152.0, 1728.0, 2304.0, 2880.0, 3456.0, 4032.0, 4608.0, 5184.0, 5760.0, 6336.0, 6912.0, 7488.0]

And a MidiFile:
{'type': 'time_signature', 'numerator': 4, 'denominator': 4, 'time': 0, 'duration': 768, 'ID': 0}
{'type': 'set_tempo', 'tempo': 500000, 'time': 0, 'ID': 1}
{'type': 'track_name', 'name': 'Tempo Track', 'time': 0, 'ID': 2}
{'type': 'track_name', 'name': 'New Instrument', 'time': 0, 'ID': 3}
{'type': 'note_on', 'time': 0, 'channel': 0, 'note': 48, 'velocity': 100, 'ID': 4, 'duration': 956}
{'type': 'time_signature', 'numerator': 3, 'denominator': 4, 'time': 768, 'duration': 6911, 'ID': 5}
{'type': 'note_on', 'time': 768, 'channel': 0, 'note': 46, 'velocity': 100, 'ID': 6, 'duration': 575}
{'type': 'note_off', 'time': 956, 'channel': 0, 'note': 48, 'velocity': 0, 'ID': 7}
{'type': 'note_off', 'time': 1343, 'channel': 0, 'note': 46, 'velocity': 0, 'ID': 8}
{'type': 'end_of_track', 'time': 7679, 'ID': 9}

And I want to check if the midi note is overlapping a barline. Every note_on message has a 'time' and a 'duration' value. I have to check if one of the barlineticks(in the list) is inside the range of the note('time' and 'duration'). I tried:
if barlinepos in range(0, 956):
    print(True)

Of course this doesn't work because barlinepos is a list. How can I check if one of the values in the list results in True?

A: Simple iteration to solve the requirement:
for i in midifile:
    start, end = i["time"], i["time"]+i["duration"]
    for j in barlinepos:
        if j >= start and j<= end:
            print(True)
            break
    print(False)


ASHLAND, Ohio - Bret Wackerly hit a 3-pointer with 11 seconds remaining to lift Ashland to a 74-70 come-from-behind victory over Saginaw Valley State in GLIAC men's basketball Saturday.
SVSU led 70-62 with 5:14 remaining before Ashland closed the game on a 12-0 run aided by five Cardinal turnovers and seven missed shots.
Avery Stevenson led the Cardinals with 24 points and 12 rebounds. Greg Foster had 14 points and six assists, while Chris Johnston had nine points and six rebounds.
Kale Richardson led Ashland with 20 points and five rebounds, while Wackerly had 19 points.
SVSU (1-6, 1-4 GLIAC) hosts Indianapolis University on Tuesday at 7 p.m.


Viewing thread 24th Dec ; The eve of festives!
Viewing thread WTSell: Lens - Canon Canon EF24-70 f/4L IS USM. Details below.
Viewing thread Getting correct color from scanning color negatives.
Viewing thread D80 which prime lens to consider for my usage??


There was a smear [=smudge] on the mirror.
His opponent did a smear job on him.
The children smeared the window with fingerprints.
She smeared jam on her toast.
Butter was smeared all over the counter.
The axle was smeared [=covered] with grease.
The ink smeared [=smudged] when she folded the letter.
Water smeared [=smudged] the ink so I could barely read the letter.
They deliberately tried to smear him by telling lies about his war record.
They tried to smear his reputation.
What made you want to look up smear? Include any comments and questions you have about this word.


Join NPT, Vanderbilt University, the Nashville Symphony, the Jewish Federation of Nashville and Middle Tennessee; and the Tennessee Holocaust Commission for a free screening of Violins of Hope: Strings of the Holocaust, on Tuesday, Feb. 27, at Sarratt Student Center. The film will be shown at 7 p.m. and will be followed by a Q&A with James A. Grymes, author of the book that inspired the documentary.
RSVPs are requested; seating is first come, first served.
Violins of Hope, a haunting film by Lance K. Shultz, reveals a bright spot following one of the darkest chapters in history. Narrated by actor Adrian Brody, the documentary follows the journey of Israeli violinmaker Amnon Weinstein in his efforts to restore violins that were played by Jewish prisoners in concentration camps as well as instruments connected to the Klezmer musical culture.


text:
Posted in Broadway, New York Theater, Theater Reviews
Annie Review: Sandy Is Sunny on Broadway
New York Theater November 9, 2012 4 Comments
You could not count me an "Annie" aficionado before I saw its third Broadway production, which has opened tonight. "Annie" to me conjured up nothing but "Tomorrow" sung badly by generations of cloyingly precocious girls. But then again Sandy meant something different then as well.
Now, when Annie tells a cop that weather doesn't bother her and then she sings to Sandy (!) that the sun will come up, well…as a nearby theatergoer Tweeted during intermission: "I can't believe some sentimental clown kept crying during the first act of Annie, or that it was me."
"Annie" is so old that its original star, Andrea McArdle, is 49 years old, yet it has never had a chance to fade away into nostalgia: It's been done not just on Broadway (1977-1983; again in 1997), not just in Hollywood (1982 movie; 1999 TV movie) but this year alone there are more than 100 licensed productions, everywhere from Sao Paolo to Denmark to Wilson West Middle School in Sinking Spring, Pennsylvania… including three "Annies" in New York and five in New Jersey.
Still, the timing does seem ideal for its Broadway revival, and director James Lapine is attempting a timely approach to this story of optimism triumphing in a bleak era. Lapine's "Annie" is as always shamelessly sentimental but it is also actually affecting; dopey but endearing; full of Broadway razzmatazz but also blunt in its depiction of tough times.
"Annie" begins now with a black-and-white newsreel that shows the musical's time and place – New York City in 1933, with its unemployment and breadlines and housing foreclosures, some of it feeling uncomfortably close to the present day. A curtain made up of clotheslines rises to reveal a tenement, and the orphans all asleep in one bed, including Annie, whose sole possessions are the dog-eared note her parents left 11 years ago when they left her at the doorstep, and the half-locket that she wears around her neck; her parents promised to bring the other half when they came back for her.
Some of the later scenes as well are grittier than you might expect, showing people down on their luck, living in Hoovervilles beneath a bridge that looks like the one in "In The Heights" and talking like the street urchins in "Newsies."
This is an "Annie" that has traveled some ways from its feel-good tap-dancing origins, and even further from Little Orphan Annie, the comic strip that inspired it.
Is it an "Annie" for adults? Not really. It's still largely a fantasy for little girls – though their parents are offered the mild social commentary as a reward for their guardian duties. The story remains in the heightened colors of a fantasy: Annie is rescued from the orphanage by a billionaire industrialist Daddy Warbucks (were there any billionaires in the 1930's?), who brings Annie along to a meeting with FDR, where the red-headed orphan teaches the President and his cabinet to be optimistic and inspires them to come up with the New Deal.
The songs by Charles Strouse and Martin Charnin are sturdy, tuneful and familiar: Besides "Tomorrow," there are "It's the Hard-Knock Life," (a favorite of Jay-Z), "Easy Street," "Maybe" and "You're Never Fully Dressed Without a Smile." The production values are pleasingly expensive-looking, especially David Korins' big inventive sets, such as the lush interior of Warbucks mansion. Andrew Blankenbuehler's choreography is at its best not during the conventional numbers, but in what amount to mute intervals telling their own stories in-between the major scenes – when Annie goes from apple seller to hobo in search of her parents, for example, or when she's buying a coat at Bergdorf Goodman with the help of the staff.
Katie Finneran as Miss Hannigan in Annie
The little girls in the show are still cute and sassy, or (depending on your W.C. Fields quotient) a gaggle of polished pre-pubescent performers whose memoirs I do not plan to read in 20 years. The adults are largely one-dimensional, by design. The apparent attempt to make them more than that doesn't get much traction. Others with a background in Annieania are sure to explain how Katie Finneran, who plays the drunken slut of a orphanage mistress Miss Hannigan, differs from the Miss Hannigans of Dorothy Loudon or Carol Burnett or Nell Carter. I don't get what she's doing, other than reprising the drunken slut of a barfly in "Promises, Promises," where she was the best thing about that revival. She is not the best thing about this revival, though her drunken pratfalls are still funny.
The stand-out in a cast that largely doesn't stand out is Anthony Warlow, an Australian opera singer making his Broadway debut as Oliver Warbucks, who has a deep voice and a winning manner; despite his name, he is such a warm, kind killer industrialist that anybody would want him to adopt them.
Lilla Crawford, who plays Annie, has a strong voice, and seems smart, confident and grounded, and that's as far as I'm going in critiquing an 11-year-old. Sandy the stray male dog whom Annie adopts is played by a female terrier who, according to her bio in Playbill, was rescued from a dog pound 24 hours before being euthanized, and whose name is Sunny. Now, a cynic might suspect this story. But who living in New York City these days wouldn't prefer a Sandy who turned out to be Sunny?
At The Palace (1564 Broadway at 47th Street)
Book by Thomas Meehan; music by Charles Strouse; lyrics by Martin Charnin; based on the comic strip "Little Orphan Annie" by Harold Gray; directed by James Lapine; choreography by Andy Blankenbuehler; sets by David Korins; costumes by Susan Hilferty; lighting by David Holder; sound by Brian Ronan; projections by Wendall K. Harrington; hair design by Tom Watson; animal trainer, William Berloni,
Cast: Katie Finneran (Miss Hannigan), Anthony Warlow (Oliver Warbucks), Lilla Crawford (Annie), Brynn O'Malley (Grace Farrell), Clarke Thorell (Rooster Hannigan) and J. Elaine Marcos (Lily).
Buy tickets to Annie
Tagged Annie, Broadway, Daddy Warbucks, Katie Finneran, Miss Hannigan, Sandy
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text:
State Highway 21 (SH-21), also known as the Ponderosa Pine Scenic Byway, is a state highway in Idaho. It runs from Boise to Stanley, primarily as a two-lane road. With two-thirds of its length in Boise County, it passes by historic Idaho City and the village of Lowman to the western edge of the Sawtooth Mountains, then along their northern boundary to Stanley.

The road is designated as one of Idaho's scenic byways and provides access to Sawtooth National Recreation Area from Boise and the Treasure Valley. It primarily follows the Boise River and its tributary Mores Creek to the Boise Basin and beyond, and then the upper South Fork of the Payette River and a tributary from Lowman to Banner Creek Summit.

Route description

SH-21 begins at exit 57 of Interstate 84 southeast of Boise, exactly  above sea level. The highway is an eastern extension of Gowen Road, which runs along the southern and western perimeter of the Boise Airport (and the Air National Guard's Gowen Field) to connect with Orchard Road at exit 52 of I-84.

After  eastbound from I-84, Highway 21 descends and crosses the New York Canal and Boise River on a bridge constructed in 1996, and connects with Warm Springs Avenue, old SH-21, from east Boise. The highway heads upstream, briefly as a four-lane divided highway, along the north bank of the river in a basalt canyon. The route passes the century-old Boise River Diversion Dam (1909), and several miles later the Lucky Peak Dam, completed in 1955.

Above the Lucky Peak Dam, SH-21 climbs over the Highland Valley Summit (), then descends to just above the reservoir. It then crosses the Mores Creek tributary and ascends with it to the Boise Basin gold mining area centered around Idaho City, leaving the sagebrush to enter the Ponderosa pine forest of the Boise National Forest. After Idaho City, the highway continues to climb with Mores Creek for another  to the Mores Creek Summit at , and with Beaver Creek to the Beaver Creek Summit at   further. Following this summit, SH-21 descends  in switchbacks to the village of Lowman at  above sea level.

At Lowman is the intersection with the Wildlife Canyon Scenic Byway, which descends with the South Fork of the Payette River through a canyon popular for whitewater rafting, westward to Banks at Highway 55. A devastating wildfire ravaged the area around Lowman in 1989; it destroyed  and 26 structures, but without injuries or fatalities.

North of Lowman, SH-21 ascends the South Fork of the Payette River to Grandjean, on the west side of the Sawtooths, and climbs the Canyon Creek tributary to the Banner Creek Summit at , the route's maximum elevation, over  above Lowman.

At the Banner Creek Summit, SH-21 enters Custer County and the Challis National Forest, and descends northward with Banner Creek and Cape Horn Creek, tributaries of the Middle Fork of the Salmon River. At "Cape Horn" the highway turns southeast to gradually climb with Marsh Creek to the border of the Sawtooth National Recreation Area, where SH-21 enters the drainage of the main Salmon River, and gently descends to Stanley. About  west of Stanley is the turnoff for the  spur road to the photogenic Stanley Lake at , framed by the jagged McGown Peak at .

In Stanley, SH-21 terminates at  at the junction with SH-75, the Sawtooth Scenic Byway from Galena Summit and Ketchum to the south, originating in Shoshone. North of the intersection in Stanley, Highway 75 becomes the Salmon River Scenic Byway, and continues due north for a mile, then veers east with the twisty main Salmon River as it descends to Clayton, then north towards Challis, where the route terminates and rejoins US-93 just south of the city limits.

The upper elevations of SH-21 are often closed during the winter months.

History
Portions of the highway originated in the 1860s, as a toll road connecting Boise with the gold mining areas near Idaho City.

SH-21 originally terminated in Downtown Boise by following Warm Springs Avenue along the north side of the Boise River. The state government approved a realignment that truncated the highway at the Gowen Road interchange on I-84 in 1980. A temporary routing was used while a new four-lane highway between the freeway and SH-21 near Lucky Peak Dam was planned in the 1980s. The realigned SH-21 would also serve an expanded Micron Technology facility and residential development on land owned by the J.R. Simplot Company. The new bridge over the Boise River began construction in March 1995 and was opened the following year. The former alignment was transferred to the Ada County Highway District on February 1, 1997.

Major intersections

See also

 List of state highways in Idaho

References

External links

 Visit Idaho - Ponderosa Pine Scenic Byway

021
Transportation in Ada County, Idaho
Transportation in Boise County, Idaho
Transportation in Custer County, Idaho


our official sponsor for two FCA Chapter Shows each year!
Opening reception 2PM on November 25th.
The Centre For Creative Arts will host the first PRFCA Show of the new year in Grande Prairie, AB. Abstract works will be on view. Look for more details soon.
The Beaverlodge and Area Cultural Centre in Alberta is the host of this exhibit featuring a wide variety of themed works by many diversely talented artists in the region. Inspiration from the sites, scenes, flora fauna and life activities of the Peace River Region of BC and Alberta will be on view. The morning opening will be followed by an afternoon workshop, Three Perspectives on Colour" delivered by artists Marjorie Henn, Suzanne Sandbow and Patricia Peters.
Peace Gallery North and the North Peace Cultural Centre in Fort St. John are proud to present the 6th juried exhibition of the Peace River Chapter of Federation of Canadian Artists (PRFCA), DIMENSIONS OF REALITY. You are invited to the Gala Opening on Friday, May 3rd from 7 - 9.


text:
\section{Introduction}
Given a compact oriented manifold $M$ of dimension $n$, Chas and Sullivan define a number of operations on the homology of the free loop space $LM = \op{Map}(S^1, M)$ \cite{ChasSullivan, ChasSullivan2}. The most prominent ones are the string product, which is an operation of the type
$$
\star \colon H_\bullet(LM \times LM) \to H_{\bullet - n}(LM),
$$
the string coproduct
$$
\Delta \colon H_\bullet(LM, M) \to H_{\bullet - n + 1}(LM \times LM, M \times LM \cup LM \times M),
$$
and the circle action
$$
B \colon H_\bullet(LM) \to H_{\bullet + 1}(LM).
$$
The string product and coproduct are defined in terms of intersections of chains satisfying certain transversality condition. In particular, it is not a priori clear whether or not they depend on the manifold structure beyond its homotopy type. Or said differently, one can ask whether a homotopy equivalence $f \colon M_1 \to M_2$ that preserves the orientation classes, induces a map $f \colon H_\bullet(LM_1) \to H_\bullet(LM_2)$ that intertwines all the above operations, i.e. $(\star, \Delta, B)$. The operator $B$ is clearly homotopy-invariant. For the string product $\star$, it is shown in \cite{CohenKleinSullivan, crabb2008loop, gruher2008generalized} (or could be deduced from \cite{CohenJones02}) that it is homotopy-invariant. We show in this short note that this is \emph{not} true for the string coproduct. To that extent, we compute enough string coproducts on lens spaces to show that it is sensitive to Reidemeister torsion and transforms with respect to the Whitehead torsion. In particular, string topology can tell $L(1,7)$ and $L(2,7)$ apart. Moreover, we verify (in a certain range) the transformation formula
\begin{equation}
\label{eqn:mainconj}
\Delta f(x) = f(\Delta(x)) + f(x \star d\log\tau(f) ),
\end{equation}
where $\Delta$ is the string coproduct, $\star$ is the string product and $\tau(f)$ is the Whitehead torsion under the Dennis trace map, which we denote by $d\log$. Naturally, one is led to conjecture that this formula is true in full generality (i.e. for all closed manifolds $M$ and all $f \in \pi_*(\operatorname{aut}(M))$). Such a transformation formula is not entirely unexpected considering the following. In \cite{NaefWillwacher} it is shown that the natural comparison map (over the reals) between loop space cohomology and Hochschild homology of the cochain algebra $C^*(M)$ can be made to intertwine coproducts. The description of the coproduct on the algebraic side, however depends on the 1-loop contributions of the partition function of a Chern-Simons type field theory. It is moreover easy to see that not every $\mathrm{Com}_\infty$-automorphism of $C^*(M)$ (the algebraic analogue of a homotopy equivalence) preserves the coproduct since it might change the 1-loop part. In particular, the algebraic analogue of the above transformation formula is true, where the Whitehead torsion term is defined as the action on the 1-loop part. Stretching the analogy a bit we would like to think that this 1-loop part merely computes (a certain expansion of) the Reidemeister torsion as in the cellular model in \cite{cattaneo2020cellular}.

The structure of the paper is as follows. First we compute the integral homology of the free loop space of a lens spaces $M = L(k,7)$ and give generators. We proceed to compute all the string coproducts of generators in $H_3(LM)$ in terms of these generators. We then show that after quotienting out certain "inconvenient"
classes we can write particularly succinct formulas for the previous calculation, and that even after "forgetting" these classes, we can still detect Reidemeister torsion and get the correction terms as in \eqref{eqn:mainconj}. Finally, we show for one particularly striking example that the transformation formula \eqref{eqn:mainconj} is also true with the "inconvenient" classes intact.

\subsection*{Acknowledgements}
I would like to thank Pavel Mnev and Konstantin Wernli, the discussion with whom inspired this note, and Pavel Safronov for helpful discussions. I also would like to thank Nathalie Wahl for discussions, suggestions and support.

\tableofcontents

\section{Lens spaces}
In the following $M$ will be a lens space of the form $L(k,7)$. That is, let $t = e^\frac{1}{7}$ where $e = \mathrm{e}^{2\pi \sqrt{-1}}$ and consider the $\mathbb{Z}_7$-action on $S^3 = \{ (z_1, z_2) \ | \ |z_1|^2 + |z_2|^2 = 1 \}$ generated by
$$
(z_1, z_2) \mapsto (e^\frac{1}{7}z_1, e^\frac{k}{7}z_2).
$$
There is a residual two-torus action given by
$$
(z_1,z_2) \mapsto (e^{t} z_1, e^{kt+s} z_2),
$$
for $(s,t) \in [0,1] \times [0, \frac{1}{7}]$. Note that this action is free away from the two circles
$$
K_1 = \{ z_1 = 0\}
$$
and
$$
K_2 = \{ z_2 = 0\}.
$$
Let $r$ denote the inverse of $k \in \mathbb{Z}_7^{\times}$.

\subsection{Homology}
The free loop space $LM$ decomposes into $7$ connected components corresponding to elements in $\mathbb{Z}_7$. Let $L_l M$ denote the component corresponding to $l$. There is a fibration
\begin{equation}
\label{fibr}
\Omega_l M \to L_l M \to M,
\end{equation}
where $\Omega_l M$ is the component of the based loop space corresponding to $l$. Since $\Omega M$ is group-like, the component $\Omega_l M$ is homotopy-equivalent to $\Omega_0 M$, that is the component of contractible loops. Comparing homotopy groups we see that the map
$$
\Omega S^3 \to \Omega_0 M
$$
is a homotopy-equivalence. Hence the (integral) homology of $\Omega_l M$ is given by
$$
H_*(\Omega_l M) =
\begin{cases}
\mathbb{Z} & \text{for $* = 0,2,4,\dots$} \\
0 & \text{else}.
\end{cases}
$$
The differential on the $E_2$-page of the Serre spectral sequence associated to the fibration \eqref{fibr} is zero by degree reasons.
The $E_3$-page is
$$
\begin{tikzcd}
&&&\\
\dots &&& \\
\mathbb{Z} & \mathbb{Z}_7 & 0 & \mathbb{Z}\\
0 & 0 & 0 & 0 \\
\mathbb{Z} & \mathbb{Z}_7 & 0 & \mathbb{Z} \ar[uulll]\\
0 & 0 & 0 & 0 \\
\mathbb{Z} & \mathbb{Z}_7 & 0 & \mathbb{Z} \ar[uulll]
\end{tikzcd},
$$
where the only possibly non-zero differentials are indicated.
From the residual torus action one can see that the map $H_*(L_l M) \to H_*(M)$ is onto (for the component of the contractible loop it is onto for any space). We will see this in more detail below. Hence the differential on the $E_3$-page also vanishes and we obtain for the homology
\begin{align*}
    H_0(L_l M) &= \mathbb{Z} \\
    H_1(L_l M) &= \mathbb{Z}_7 \\
    H_2(L_l M) &= \mathbb{Z} \\
    H_3(L_l M) &= \mathbb{Z} \oplus \mathbb{Z}_7 \\
    H_4(L_l M) &= \mathbb{Z} \\
    \dots.
\end{align*}

Let us be more precise about these identifications and give more explicit descriptions for $H_0$, $H_1$ and $H_3$. We identify $H_0(LM)$ with $\mathbb{Z}[\mathbb{Z}_7]$. Moreover, let $\Omega^1 = \Omega^1(\mathbb{Z}[\mathbb{Z}_7]) = \mathbb{F}_7[t]/(t^7-1) \frac{dt}{t}$ denote the vector space of formal de Rham 1-forms and identify
\begin{align*}
    \mathbb{F}_7[t]/(t^7-1) \frac{dt}{t} &\longrightarrow H_1(LM) = \oplus_{l \in \mathbb{Z}_7} H_1(L_l M) \cong \oplus_{l \in \mathbb{Z}_7} H_1(M) \\
    ( c_0 + c_1 t + \dots c_6 t^6 ) \frac{dt}{t} &\longmapsto (c_0, \dots, c_6).
\end{align*}
\begin{Rem}
The reason we identify $H_1$ with $\Omega_1$ and not with $\mathbb{Z}[\mathbb{Z}_7]$ is first and foremost to make the formulas later more appealing. One can justify this identification at this point by appealing to the fact that $H_1(LM) = HH_1(\mathbb{Z}[\mathbb{Z}_7]) = \Omega_1$, where $HH_1$ is Hochschild homology. Thus the identification is in particular natural with respect to automorphisms of $\pi_1 = \mathbb{Z}_7$. Furthermore, the circle action $H_0(LM) \to H_1(LM)$ can now be written as the "de Rham differential"
\begin{align*}
\mathbb{Z}[\mathbb{Z}_7] &\to \Omega^1(\mathbb{Z}[\mathbb{Z}_7]) \\
c_0 + c_1 t + \dots + c_6 t^6 &\mapsto c_1 t + 2 c_2 t^2 + \dots + 6 c_6 t^6 \frac{dt}{t}
\end{align*}
\end{Rem}
Let us also define $\bar{\Omega}^1 = \Omega^1 / \mathbb{Z} \frac{dt}{t}$, such that we can identify
$$
\bar{\Omega}^1 \cong H_1(LM, M).
$$
Similarly $\bar{\mathbb{Z}}[\mathbb{Z}_7] = \mathbb{Z} [ \mathbb{Z}_7] / \mathbb{Z} 1$ such that
$$
\bar{\mathbb{Z}}[\mathbb{Z}_7] \cong H_0(LM,M).
$$
For $H_3(L_l M)$, the spectral sequence gives us a short exact sequence
$$
0 \to H_1(M, H_2(\Omega_l M)) \to H_3(L_l M) \to H_3(M) \to 0.
$$
From the residual two-torus action, we can construct a number of classes in $H_3(L_l M)$ that map to the fundamental class in $H_3(M)$. Consider the following $S^1$-actions. For given integers $(l,m)$ we define
$$
(t, z_1, z_2) \mapsto \rho_{l,m}(t, z_1, z_2) :=  (e^{lt}z_1, e^{(kl+7m)t}z_2) \quad \text{for $t \in [0,\frac{1}{7} ]$}.
$$
We view $\rho_{l,m}$ as a map $M \to LM$ and denote the image of the fundamental class by $[\rho_{l,m}] \in H_3(LM)$. The class of $\rho_{l,m}$ lies in the component corresponding to $l$,
$$
[\rho_{l,m}] \in H_3(L_lM),
$$
for all $m$, as can be seen for instance by setting $z_2 = 0$. We will argue below that these classes span all of $H_3(LM)$.

\subsection{String coproduct}
\subsubsection{Definition of the string coproduct}
Let us briefly recall how the string coproduct is computed. Given a homology class $\alpha \in H_p(LM)$ let us for simplicity assume that it is represented by a map $N \to LM$, where $N$ is an oriented $p$-dimensional manifold. In particular, we are given a map
\begin{align*}
    \alpha \colon S^1 \times N &\longrightarrow M \\
    (t,n) &\longmapsto \alpha(t,n).
\end{align*}
The self-intersection locus is defined by
$$
V = \{ (t,n) \ | \ \alpha(t,n) = \alpha(0,n) \ , \ t \neq 0 \} \subset S^1 \times N,
$$
and is (under appropriate transversality assumptions) an oriented manifold of dimension $p + 1 - \dim(M)$. Splitting the loops at intersection points we obtain
\begin{align*}
    \Delta(\alpha) \colon V &\longrightarrow LM \times LM \\
    (t,n) &\longmapsto ( s \mapsto \alpha(st,n) , s \mapsto \alpha( t+ (1-t)s, n) ).
\end{align*}
The resulting class is only well defined in $H_{p-n+1}( LM/M \times LM/M ) := H_{p-n+1}( (LM,M) \times (LM,M) ) := H_{p-n+1}( LM \times LM, LM \times M \cup M \times LM )$ and is the string coproduct of the class $\alpha$. One can furthermore show that it only depends on the image of $\alpha$ in $H_p(LM,M)$.
\subsubsection{String coproduct on $L(k,7)$}
In the case of $M = L(k,7)$ the string coproduct is a map $H_*(LM) \to H_{*+1-3}(LM/M \times LM/M)$.
We consider only the component
$$
\Delta\colon H_3(LM) \to H_{3+1-3}(LM/M \times LM/M) \to H_1(LM,M) \otimes H_0(LM,M).
$$

Under the identifications from the previous section the string coproduct gives a map
$$
H_3(LM) \to \bar{\Omega}^1 \otimes \bar{\mathbb{Z}}[\mathbb{Z}_7] = \mathbb{F}_7[t, t_2]\frac{dt}{t} /\left((t^7-1, t_2^7-1) \oplus \Omega^1 \cdot 1 \oplus \frac{dt}{t} \mathbb{Z}[t_2] \right) .
$$
More concretely, we identify a monomial $t^p t_2^q \frac{dt}{t}$ with the class in $H_1(L_p M) \otimes H_0(L_q M)$ whose image under $H_1(L_p M) \otimes H_0(L_q M) \to H_1(M) \otimes H_0(L_q M) \cong \mathbb{Z}_7$ is the canonical generator.

We now compute the coproduct of classes $[\rho_{l,m}]$ for $(l,m)$ positive coprime integers.

To compute the coproduct we have to find $(t,z_1,z_2)$ where
$$
\rho_{l,m}(t,z_1,z_2) = (z_1,z_2).
$$
Away from the circles $K_1 = \{ z_1 = 0 \}$ and $K_2 = \{ z_2 = 0 \}$ the action is free, since in that case the above equation reads as
$$
\begin{pmatrix}
l \\ kl +7m
\end{pmatrix} t \in \mathbb{Z}^2 + \frac{1}{7}\begin{pmatrix}
1 \\ k
\end{pmatrix}\mathbb{Z} = \begin{pmatrix}
0 \\ 1
\end{pmatrix}\mathbb{Z} \oplus \begin{pmatrix}
1 \\ \frac{k}{7}
\end{pmatrix} \mathbb{Z},
$$
whose solutions are $\frac{1}{7}\mathbb{Z}$ since $l$ and $m$ were coprime. Hence we
only need to consider the self-intersection loci on the circles $K_1 = \{ z_1 = 0 \}$ and $K_2 = \{ z_2 = 0 \}$.
For $K_2 = \{ z_2 = 0 \}$ we chose the (orientation-preserving) coordinates $(\alpha, z)$ around $K_2$ via the assignment
$$
(\alpha, z) \mapsto (e^\alpha \sqrt{1 - |z|^2}, z).
$$
In these coordinates the action reads as
$$
\rho_{l,m}^c(t, \alpha, z) = (\alpha + lt, e^{kl+7m}t z).
$$
We obtain that the self-intersection locus is
$$
\{ (t,\alpha,z) \ | \ \rho_{l,m}^c(t, \alpha, z) - \rho_{l,m}^c(0, \alpha, z) = 0 \} =
\{ (t,\alpha,0) \ | \ t= \frac{1}{7l}, \frac{2}{7l}, \dots, \frac{l-1}{7l} \},
$$
that is a disjoint union of circles. The derivative of $\rho_{l,m}^c(t, \alpha, z) - \rho_{l,m}^c(0, \alpha, z)$ at $(\frac{n}{7l}, \alpha, 0)$ (computed on its cover) is given by
$$
\begin{pmatrix}
l & 0 & 0\\
0 & 0 & 0\\
0 & 0 & e^{(kl+7m)\frac{n}{7l}} - e^\frac{kn}{7}
\end{pmatrix},
$$
from which we see that if $l > 0$ then all the circles are oriented such that $s \to (e^s,0)$ is an orientation-preserving map. Each component of the self-intersection locus gives a term in $H_1(LM) \otimes H_0(LM)$. To identify these terms we only need to know which connected component it belongs to and what the image under $H_1(LM) \to H_1(M)$ is. The term belonging to $(\frac{n}{7l}, \alpha, 0)$ lies in the connected component associated to $t^n t_2^{l-n}$ and as we saw, the coefficient is given by the element in $H_1(M)$ that corresponds to $s \mapsto (e^s,0)$ which is the generator. Thus we get that the contribution from $K_2$ is
\[
(t t_2^{l -1} + t^2 t_2^{l -2} + \dots + t^{l-1} t_2) \frac{dt}{t}.
\]
Similarly, we obtain the contribution coming from $K_1 = \{ z_1 = 0 \}$, here the equation to solve is
$$
(0,e^{(kl+7m)t}z) = (0,z)
$$
intersection-locus is thus
$$
\{ (t,0,z_2) \ | \ t=\frac{1}{7(kl+7m)}, \frac{2}{7(kl+7m)}, \dots, \frac{kl+7m-1}{7(kl+7m)} \}.
$$
The contributions can again be expressed in terms of the class $\alpha \to (0, e^\alpha)$ which in $H_1(M)$ corresponds to $r$ where $r$ is the multiplicative inverse of $k$ mod $7$. Thus the contribution is
$$
r ( t^r t_2^{(kl+7m -1)r} t^{2r} t_2^{(kl+7m -2)r} + \dots + t^{(kl+7m -1)r} t_2^r) \frac{dt}{t}.
$$
Finally, we conclude that for any $(l,m)$ coprime
\begin{align}
\label{eqn:rhocop}
\begin{split}
\Delta([\rho_{l,m}]) =& (t t_2^{l -1} + t^2 t_2^{l -2} + \dots + t^{l-1} t_2) \frac{dt}{t} \\
&+ r ( t^r t_2^{(kl+7m -1)r} + t^{2r} t_2^{(kl+7m -2)r} + \dots + t^{(kl+7m -1)r} t_2^r) \frac{dt}{t}
\end{split}
\end{align}

It thus follows that
$$
\Delta([\rho_{l,m+n}] - [\rho_{l,m}]) = r \, n \, t^{kl+7m} (t \, t_2^6 + \dots + t^6 t_2 + t^7) \frac{dt}{t}.
$$
Recall that we are working in $H_1(LM,M) \otimes H_0(LM,M)$.
In particular, we see that we have found all lifts along $H_3(L_lM) \to H_3(M)$. More precisely, any two classes $[\rho_{l,m_1}]$ and $[\rho_{l,m_2}]$ for $m_1 \neq m_2 \text{ mod } 7$ and both coprime to $l$, are non-zero and not equal and hence span $H_3(L_lM)$. Alternatively, we actually see that the classes $[\rho_{l, 1 + nl}]$ for $n = 1, \dots, 7$ are all the lifts of the fundamental class along $H_3(L_lM) \to H_3(M)$ if $l \neq 0$. For $l = 0$ take the classes $[\rho_{7,n}]$ for $n = 1, \dots, 6$ and $[\rho_{0,0}]$.

\subsection{More convenient notation}
We wish to write to above formulas in a more convenient way. As we have seen $H_3(L_i M)$ is an extension of $H_3(M) = \mathbb{Z}$ by a $\mathbb{Z}_7$. As we have seen in the calculation above there is not much variation in the coproduct of the $\mathbb{Z}_7$ summand, so we are modding it out to simplify notation.
To that effect, let us denote the kernel of the map
$$
H_3(LM) \to \oplus_{i\in \mathbb{Z}_7} H_3(M)
$$
by $K$. Thus we can identify $H_3(LM)$ with $\mathbb{Z}[\mathbb{Z}_7]$ and our the formulas define a map
$$
H_3(LM)/K \to H_1(LM,M) \otimes H_0(LM,M) / \Delta(K),
$$
which we identify with
$$
\mathbb{Z}[\mathbb{Z}_7] \to \bar{\Omega}^1 \otimes \bar{\mathbb{Z}}[\mathbb{Z}_7] / \Delta(K).
$$
Our formulas actually lift to a map
$$
\mathbb{Z}[\mathbb{Z}_7] \to \Omega^1 \otimes \mathbb{Z}[\mathbb{Z}_7] / \Delta(K),
$$
which we will describe and at the very end project to $\bar{\Omega}^1 \otimes \bar{\mathbb{Z}}[\mathbb{Z}_7] / \Delta(K)$.
The target can be identified with the quotient of $\mathbb{F}_7[t,t_2, \frac{dt}{t}]$ by the subvector space spanned by
\begin{gather*}
t^l \frac{dt}{t}, t_2^l \frac{dt}{t}, t^l (t \, t_2^6 + \dots + t^6 t_2 + t^7) \frac{dt}{t}
\end{gather*}
Call that quotient $Q_{\mathbb{F}_7}$.

\subsection{Relation to Reidemeister and Whitehead torsion}
To rewrite the formulas in a more convenient way let us introduce a rational version of above target space. Introduce the algebra $\mathbb{Q}[t,t_2]$ with relations $t^7=1$, $t_2^7=1$. Consider the ideal $I = (t_2^6 + \dots + t^7)$ and define $A = \mathbb{Q}[t,t_2] / I$. This has the convenient effect that now $t^l - t_2^l$ are units in $A$, since $(t-t_2)(t^6 + 2 t^5 t_2 + \dots + t_2^7) = 7 + (t \, t_2^6 + \dots + t^6 t_2 + t^7)$. Let then $Q_{\mathbb{Q}}$ denote the vector space obtained by taking the quotient of $A$ by the subvector space spanned by the elements $t^l$ and $t_2^l$ and formally adjoint a symbol $\frac{dt}{t}$. Similarly there is an integral version of said space $Q_{\mathbb{Z}} \subset Q_{\mathbb{Q}}$. The above formulas define a map
$$
\Delta: \mathbb{Z}[\mathbb{Z}_7] \to Q_{\mathbb{Z}} \subset Q_{\mathbb{Q}}.
$$
After reduction mod $7$ this the component of the string corpoduct (the rationalization is merely to write down the formulas in a more conveninet way). The map $\Delta: \mathbb{Z}[\mathbb{Z}_7] \to Q_{\mathbb{Z}} \subset Q_{\mathbb{Q}}$ is defined by formulas
\begin{align*}
t^l \mapsto &(t t_2^{l -1} + t^2 t_2^{l -2} + \dots + t^{l-1} t_2) \frac{dt}{t} \\
&+ r ( t^r t_2^{(kl+7m -1)r} + t^{2r} t_2^{(kl+7m -2)r} + \dots + t^{(kl+7m -1)r} t_2^r) \frac{dt}{t} \\
=&(t t_2^{l -1} + t^2 t_2^{l -2} + \dots + t^{l-1} t_2 +t^l) \frac{dt}{t} \\
&+ r ( t^r t_2^{(kl+7m -1)r} + t^{2r} t_2^{(kl+7m -2)r} + \dots + t^{(kl+7m -1)r} t_2^r + t^{(kl+7m)r}) \frac{dt}{t}\\
=&(t_2^{l -1} + t t_2^{l -2} + \dots + t^{l-2} t_2 +t^{l-1}) dt \\
&+  ( t_2^{(kl+7m -1)r} + t^{r} t_2^{(kl+7m -2)r} + \dots + t^{(kl+7m -2)r} t_2^r + t^{(kl+7m-1)r}) dt^r \\
=& \frac{t^l - t_2^l}{t - t_2} dt + \frac{t^{r(kl+7m)} - t_2^{r(kl+7m)}}{t^r - t_2^r} dt^r \\
=& \frac{t^l - t_2^l}{t - t_2} dt + \frac{t^{l} - t_2^{l}}{t^r - t_2^r} dt^r \\
&= (t^l - t_2^l) d\log( (t^r-t_2^r)(t-t_2) ) \\
&= (t^l - t_2^l) d\log( R ),
\end{align*}

where $R \in Q$ is the homogenized Reidemeister torsion
$$
R = (t^r - t_2^r)(t-t_2).
$$
We refer to the lecture notes \cite[equation (58)]{MnevLecture} or \cite{milnor1966whitehead} for the fact that this is indeed the Reidemeister torsion (our convention differs slightly).
We summarize our findings in the following
\begin{Prop}
\label{prop:main1}
The string coproduct descends to
$$
\begin{tikzcd}
K \ar[r, "\Delta"] \ar[d] & \Delta(K) \ar[d] \\
H_3(LM) \ar[r, "\Delta"]\ar[d]  & H_1(LM,M) \otimes H_0(LM,M) \ar[d] \\
\oplus_{i\in \mathbb{Z}_7} H_3(M) \ar[r, "\mathcal{R}"] & \oplus_{i \neq 0 \in \mathbb{Z}_7} H_1(M) \otimes H_0(LM,M) / \Delta(K),
\end{tikzcd}
$$
where $\mathcal{R}$ is the map $t^l \mapsto (t^l - t_2^l) d\log( R )$ where $R$ is the homogenized Reidemeister torsion and the term $(t^l - t_2^l) d\log( R )$ is evaluated as explained above.
\end{Prop}

\begin{Rem}
For us Reidemeister torsion is merely an expression of the form $(t^p - t_2^p)(t^q - t_2^q)$. We do not fully explain here what the exact space of these expressions is. We will only need that the Whitehead group acts on these expressions faithfully.
\end{Rem}

\begin{Ex}
Let us give the calculation of $\mathcal{R}$ for $L(1,7)$ and $L(2,7)$
$$L(1,7), \quad k = 1, r = 1$$
\begin{align*}
t^0 \mapsto & 0\\
t^1 \mapsto & 0\\
t^2 \mapsto & 2 t^1 t_2^1 \frac{dt}{t}\\
t^3 \mapsto & 2 t^1 t_2^2 + 2 t^2 t_2^1 \frac{dt}{t}\\
t^4 \mapsto & 2 t^1 t_2^3 + 2 t^2 t_2^2 + 2 t^3 t_2^1 \frac{dt}{t}\\
t^5 \mapsto & 2 t^1 t_2^4 + 2 t^2 t_2^3 + 2 t^3 t_2^2 + 2 t^4 t_2^1 \frac{dt}{t}\\
t^6 \mapsto & 2 t^1 t_2^5 + 2 t^2 t_2^4 + 2 t^3 t_2^3 + 2 t^4 t_2^2 + 2 t^5 t_2^1 \frac{dt}{t}\\
\end{align*}

$$L(2,7), \quad k = 2, r = 4$$
\begin{align*}
t^0 \mapsto & 0\\
t^1 \mapsto & 4 t^4 t_2^4 \frac{dt}{t}\\
t^2 \mapsto & 5 t^1 t_2^1 + 4 t^4 t_2^5 + 4 t^5 t_2^4 \frac{dt}{t}\\
t^3 \mapsto & 5 t^1 t_2^2 + 5 t^2 t_2^1 + 4 t^4 t_2^6 + 4 t^5 t_2^5 + 4 t^6 t_2^4 \frac{dt}{t}\\
t^4 \mapsto & 5 t^1 t_2^3 + 5 t^2 t_2^2 + 5 t^3 t_2^1 + 4 t^5 t_2^6 + 4 t^6 t_2^5 \frac{dt}{t}\\
t^5 \mapsto & 2 t^1 t_2^4 + 5 t^2 t_2^3 + 5 t^3 t_2^2 + 2 t^4 t_2^1 + 4 t^6 t_2^6 \frac{dt}{t}\\
t^6 \mapsto & 2 t^1 t_2^5 + 2 t^2 t_2^4 + 5 t^3 t_2^3 + 2 t^4 t_2^2 + 2 t^5 t_2^1 \frac{dt}{t}\\
\end{align*}
In particular, we see that they cannot possibly be isomorphic. In the first one there are two $i$'s such that $H_3(L_iM) \to H_1(LM/M) \otimes H_0(LM/M)$ has rank one (or rank zero after quotienting out $\Delta(K)$). In the second one there is only one such $i$. Since all the $H_3(L_i M)$ have images in different components we see that the ranks of the maps $H_3(L_i M) \to H_1(LM,M) \otimes H_0(LM, M)$ differ.
\end{Ex}

We summarize the result of the above example in
\begin{Prop}
The string coproduct coalgebras on $L(1,7)$ and $L(2,7)$ are non-isomorphic. More precisely, they are told apart by the dimension of the kernel of $\Delta \colon H_3(LM, M) \to H_1(LM/M \times LM/M)$. For $M = L(2,7)$ the coproduct is injective on $H_3(LM, M)$, while for $M = L(1,7)$ the kernel is spanned by the class $[\rho_{1,0}]$.
\end{Prop}

\subsubsection{Whithead torsion}
Let $f: L(1,7) \to L(2,7)$ be a homotopy equivalence. Let $\tau(f) \in Wh(\mathbb{Z}_7) = (\mathbb{Z} [\mathbb{Z}_7])^\times / \mathbb{Z}_7$ be its Whitehead torsion. We denote by the same symbol its image under the map
$$
Wh(\mathbb{Z}_7) \to HH_1(\mathbb{Z}[\mathbb{Z}_7])/ HH_1(\mathbb{Z}[\mathbb{Z}_7], \mathbb{Z}) = H_1(LM/M) \to H_1(LM/M \times LM/M) \overset{1 \times \sigma}{\longrightarrow} H_1(LM/M \times LM/M),
$$
where $\sigma\colon LM \to LM$ is given precomposing with the orientation-reversing diffeomorphism of $S^1$. Let us recall the definition of the Dennis trace map in our case. The Hochschild homology computes as $HH_1(\mathbb{Z}[\mathbb{Z}_7]) = \Omega^1$ and $HH_1(\mathbb{Z}[\mathbb{Z}_7], \mathbb{Z}) = \mathbb{Z}_7$. Under these identifications the Dennis trace map is then given by
\begin{align*}
    Wh(\mathbb{Z}_7) &\to HH_1(\mathbb{Z}[\mathbb{Z}_7])/ HH_1(\mathbb{Z}[\mathbb{Z}_7], \mathbb{Z}) \\
    \alpha &\mapsto \alpha^{-1} d\alpha = d\log \alpha.
\end{align*}

The calculation in the previous section partially verifies the formula
$$
\Delta f(x) = f(\Delta(x)) + f(x \star d\log \tau(f) ),
$$
where $\star$ is the string product (applied to both factors as a derivation).
Namely, recall that $R_{2,7} = f(R_{1,7} \tau(f))$ and that moreover $\tau(f) \in \mathbb{Z}[\mathbb{Z}_7]^\times$. We then have
\begin{align*}
\Delta f(t^l)  &= (t^{f(l)} - t_2^{f(l)}) d\log R_{2,7} \\
&= (t^{f(l)} - t_2^{f(l)}) (f(d\log R_{1,7}) + d\log f(\tau(f))) \\
&= f(\Delta(t^l)) + (t^{f(l)} - t_2^{f(l)}) d\log f(\tau(f))) \\
&= f(\Delta(t^l)) + f((t^l - t_2^l) d\log \tau(f))),
\end{align*}
where we used the calculation of the string product from the Appendix.
\begin{Rem}
We used the following in the previous calculation. Let $I \to \mathbb{F}_7[\mathbb{Z}_7] \to \mathbb{F}_7$ be the augmentation ideal. Then $(t-1) \in I$ is not a zero-divisor in the algebra $I$, hence neither is $R_1 = R_{1,7} = (t-1)^2$ nor $R_2 = f(R_{2,7}) = (t^2-1)(t^4-1)$. Let $u \in \mathbb{F}_7[\mathbb{Z}_7]^\times$ be such that $R_1 = R_2 u$. Then to show the identity
$$
(t^l-1)d\log(R_1) - (t^l-1)d\log(R_2) = (t^1-1)du u^{-1} \quad \text{mod $\Sigma$},
$$
it is clearly enough to show that it is true after multiplying with $R_1 R_2 = R_2 R_2 u$. Doing this we obtain
\begin{align*}
    (t^l-1)(dR_1 R_2 - dR_2 R_2 u) &= (t^l-1)(R_2 R_2 du) \\
        &= R_2 R_2(t^l-1)du u^{-1}.
\end{align*}
\end{Rem}
Specializing to $l = 1$ we obtain
$$
\Delta f(t) = f( (t - t_2) d\log \tau(f) ).
$$
\begin{Ex}
It is known that there exists a homotopy equivalence $f\colon L(1,7) \to L(2,7)$ that sends the preferred generator $t$ to $t^2$. Its Whitehead torsion (in our convention) is thus
\begin{align*}
\tau(f) &= \frac{(t^4-1)(t^2 -1)}{(t-1)^2} = (t^3 + t^2 + t + 1) (t + 1) - \Sigma \\
&= t + t^2 + t^3 - t^5 - t^6 \\
\tau(f)^{-1} &= \frac{(t^8-1)(t^8-1)}{(t^4-1)(t^2-1)} = (1 + t^4)(1 + t^2 + t^4 + t^6) - \Sigma \\
&= t^4 - t^5 + t^6,
\end{align*}
where $\Sigma = 1 + t + t^2 + \dots + t^6$.
Its image under the Dennis trace is
\begin{align*}
d\log(\tau(f)) &= (1 + 2t + 3t^2 - 5 t^4 - 6 t^5)(t^4 - t^5 + t^6) dt \\
&= (6 +  5 t + 6 t^2 + t^3 + 2 t^4 + t^5 + 2 t^6) dt,
\end{align*}
and hence (after homogenizing again to match notation from above)
\begin{align*}
    (t - t_2) d\log \tau(f) &= (4 + 2 t t_2^6 + 6 t^3 t_2^4 + 2 t^5 t_2^2) dt - ( 1 + t t_2^6 + t^2 t_2^5 + t^3 t_2^4 + t^4 t_2^3 + t^5 t_2^2 + t^6 t_2) dt
\end{align*}
and finally
\begin{align*}
    f((t - t_2) d\log \tau(f)) &= (4 + 2 t^2 t_2^5 + 6 t^6 t_2 + 2 t^3 t_2^4) d(t^2) \\
    &= (4 + 2 t^2 t_2^5 + 6 t^6 t_2 + 2 t^3 t_2^4) 2 t dt \\
    &= (t^2 + 4 t^4 t_2^5 + 5 t t_2 + 4 t^5 t_2^4) \frac{dt}{t},
\end{align*}
where we dropped multiples of $\Sigma$.
\end{Ex}

Summarizing our findings we conclude with
\begin{Prop}
The string coproduct on the lens spaces $L(k,7)$ detects Whitehead torsion. More precisely, the restriction of the string coproduct to $H_3(LM)$ after taking the quotient described in Proposition \ref{prop:main1} transforms according to formula \eqref{eqn:mainconj} and two elements in $Wh(\mathbb{Z}_7)$ give the same correction term if and only if they are equal under the Dennis trace map.
\end{Prop}

\subsection{More details on an example}
Let us make the previous example more concrete and show that the formula in the introduction is still true even without modding out $K$ (i.e. dropping the multiples of $\Sigma$). To that extend recall that the homotopy equivalence $f$ is constructed as
$$
L(1,7) \to S^3 \vee L(1,7) \overset{\operatorname{id} \vee (z_1^2, z_2^4)}{\longrightarrow} S^3 \vee L(2,7) \overset{\Phi}{\longrightarrow} L(2,7),
$$
where $\Phi\colon S^3 \to L(2,7)$ is any map of degree $-7$ (see \cite[section 6.4]{MnevLecture}.
\begin{Lem}
$$
f([\rho_{1,0}]) = [\rho_{2,3}]
$$
\end{Lem}
\begin{proof}
Let us first try to compare the maps $f\circ \rho_{1,0} \colon S^1 \times L(1,7) \to L(2,7)$ and $\rho_{2,3} \circ (\operatorname{id} \times f) \colon S^1 \times L(1,7) \to L(2,7)$ using obstruction theory. Away from a neighborhood of a point in $L(1,7)$ the two maps are given by
\begin{align*}
f\circ \rho_{1,0} \colon (t, z_1, z_2) &\longmapsto (e^{2t} z_1^2, e^{4t} z_2^4 ) \\
\rho_{2,3} \circ (\operatorname{id} \times f) \colon (t, z_1, z_2) &\longmapsto (e^{2t} z_1^2, e^{(4 + 7 \cdot 3) t} z_2^4 )
\end{align*}
Recalling the standard cell decomposition of $L(1,7)$ as
\begin{align*}
    e_0 &= \{ (1,0) \} \\
    e_1 &= \{ (e^s,0) \ | \ s \in (0, \tfrac{1}{7}) \} \\
    e_2 &= \{ (z_1 ,r) \ | \ r \in (0, 1) \} \\
    e_3 &= \{ (z_1, z_2) \ | \ z_2 = e^s r \text{ for $s\in (0, \tfrac{1}{7})$} \},
\end{align*}
we see that the two maps already coincide on $A := \{ 0 \} \times L(1,7) \cup S^1 \times e_0$. The first obstruction for these two maps being homotopic relative to $A$ lies in
$$
H^3(M/A ; \pi_3( L(2,7) ) = \mathbb{Z}_7.
$$
The obstruction is computed by comparing the two maps on the $3$-cell $I \times e_2$. Since the maps coincide on the boundary of that cell, they fit together to a map $S^3 \to L(2,7)$, i.e. an element in $\pi_3(L(2,7)) = \mathbb{Z}$, where the identification is by computing the degree and dividing by $7$. Thus it is enough to show that the degrees of the two maps restricted to $I \times e_2$ are equal mod $49$. For the map $f\circ \rho_{1,0}$, we note that $\rho_{1,0}$ maps the cell $I \times e_2$ homeomorphically onto $e_3$. Since $f$ has degree 1, we get a contribution of $1$. For the map $\rho_{2,3} \circ (\operatorname{id} \times f)$, we note that $f$ is given by $(z_1^2, z_2^4)$ on the cell $I \times e_2$ and hence we are computing the degree of the map
\begin{align*}
    (0, \frac{1}{7}) \times e_2 &\longrightarrow L(2,7) \\
    (t, z_1, r) &\longmapsto (e^{2t} z_1^2, e^{(4 + 7 \cdot 3) t} r^4 ).
\end{align*}
This map has degree $2 \cdot 25 = 50$ (it has the same degree as its $7$-fold cover $S^1 \times e_2 \to S^3$ given by the same formula but now $t \in [0,1]$). We see that the obstruction vanishes since $1 \cong 50 \text{ mod } 49$. We conclude that the two maps in question are homotopic at least up to the $3$-skeleton of $S^1 \times L(1,7)$. They could still potentially differ on their $4$-cell $I \times e_3$ by an element in
$$
H^4(M/A ; \pi_4(L(2,7)) = \mathbb{Z}_2.
$$
We can view $\rho_{1,0}$ as an element in $\pi_1(\operatorname{aut}_1(L(1,7))$, where $\operatorname{aut}_1(L(1,7)$ is the monoid of self-equivalences homotopic to the identity. Under this identification, the action of $H^4(M/A ; \pi_4(L(2,7)) = \mathbb{Z}_2$ corresponds to multiplication by the element
\begin{align*}
&S^1 \times L(1,7) \to (S1 \times L(1,7)) \vee S^4 \to (S^1 \times L(1,7)) \vee S^3 \\
&\to L(1,7) \vee L(1,7) \to L(1,7) \in \pi_1(\operatorname{aut}_1(L(1,7))),
\end{align*}
which is an element of order $2$. However, it follows directly from the definition of the string product that
$$
\pi_1(\operatorname{aut}_1(L(1,7))) \to (H_3(LL(1,7)), \star)
$$
is a morphisms of monoids. Moreover, the image is contained in $\bigcup_{i \in \mathbb{Z}_7} H_3(LM)$ and maps to the fundamental class $[M]$ under $H_3(LM) \to H_3(M)$. We also saw that all these classes are of the form $\rho_{l,m}$. Thus we conclude that the image of $\pi_1(\operatorname{aut}_1(L(1,7)))$ is a group of order $49$ and hence any element of order $2$ gets sent to zero. This shows that indeed
$$
f([\rho_{1,0}]) = [\rho_{2,3}] \in H_3(L L(2,7)),
$$
where $M = L(2,7)$.
\end{proof}
We are now ready to evaluate equation \eqref{eqn:mainconj} for $x = [\rho_{1,0}]$. The left hand side is given by
\begin{align*}
\Delta f([\rho_{1,0}]) =& \Delta [\rho_{2,3}] \\
=& t t_2 \frac{dt}{t} \\
& + 4 \cdot 3 ( t + t^2 t_2^6 + t^3 t_2^5 + t^4 t_2^4 + t^5 t_2^3 + t^6 t_2^2 + t_2) \frac{dt}{t} + 4 (t^4 t_2^5 + t t_2 + t^5 t_2^4 ) \frac{dt}{t} \\
=& 12 ( t + t^2 t_2^6 + t^3 t_2^5 + t^4 t_2^4 + t^5 t_2^3 + t^6 t_2^2 + t_2) \frac{dt}{t} + (4 t^4 t_2^5 +5 t t_2 +4 t^5 t_2^4 ) \frac{dt}{t}
\end{align*}
reading off formula \eqref{eqn:rhocop}.
For the right-hand side we obtain
$$
f(\Delta [\rho_{0,1}]) + f( [\rho_{1,0}] \star d\log\tau(f)) +  f( [\rho_{1,0}] \star d\log\tau(f)) =  f( (t - t_2) d\log\tau(f) ),
$$
using that $[\rho_{0,1}]$ has no self-intersections and the calculation of the string product in the appendix and introducing a Koszul sign. We already calculated the image of the Whitehead torsion under the Dennis trace map in the example above, that is
$$
d\log\tau(f) = (6 +  5 t + 6 t^2 + t^3 + 2 t^4 + t^5 + 2 t^6) dt \in HH_1(\mathbb{Z}[\mathbb{Z}_7]) = H_1(LM),
$$
the map $H_1(LM) \to H_1(LM \times LM) \to H_1(LM) \otimes H_0(LM)$ given by the diagonal and reversing the circle on the second factor sends a monomial $t^l \frac{dt}{t}$ to $t^l t_2^{-l} \frac{dt}{t}$ and hence is homogenization. We hence compute as in the above example (this time without dropping $\Sigma$ terms)
\begin{align*}
    f((t - t_2) d\log \tau(f)) &= (4 + 2 t^2 t_2^5 + 6 t^6 t_2 + 2 t^3 t_2^4) d(t^2) \quad + \quad 6 ( 1 + t t_2^6 + t^2 t_2^5 + t^3 t_2^4 + t^4 t_2^3 + t^5 t_2^2 + t^6 t_2) d(t^2) \\
    &= (4 + 2 t^2 t_2^5 + 6 t^6 t_2 + 2 t^3 t_2^4) 2 t dt \quad + \quad 12 ( 1 + t t_2^6 + t^2 t_2^5 + t^3 t_2^4 + t^4 t_2^3 + t^5 t_2^2 + t^6 t_2) dt\\
    &= (t^2 + 4 t^4 t_2^5 + 5 t t_2 + 4 t^5 t_2^4) \frac{dt}{t} \quad + \quad 12 ( t + t^2 t_2^6 + t^3 t_2^5 + t^4 t_2^4 + t^5 t_2^3 + t^6 t_2^2 + t_2) \frac{dt}{t}\\.
\end{align*}
Thus we see that the two sides of equation \eqref{eqn:mainconj} coincide up to the term $t^2 \frac{dt}{t}$ which corresponds to an element in $H_1(LM) \otimes H_0(M)$ and is hence zero in $H_1(LM, M) \otimes H_0(LM,M)$.

\section*{Appendix}
\subsection{String product}
We compute the string product
$$
H_3(LM) \otimes H_1(LM) \to H_1(LM).
$$
Since the string coproduct is compatible with the projection to $M$, the only thing to check is which component we land in. It is then clear that
$$
[\rho_{l,m}] \star \omega \mapsto t^l \omega
$$
for any $\omega \in \Omega^1$.

\subsection{Transverse string topology}
We show that the transverse calculation of the string coproduct is indeed the invariantly defined string coproduct by comparing it with the definition in \cite{NaefWillwacher}. Recall that in loc. cit. the string coproduct is defined by the following zig-zag of spaces
\begin{equation}
\label{eqn:defredcop}
\begin{tikzcd}
\frac{LM}{M} \ar[dashed]{r}{\text{suspend}} &\frac{I \times LM}{\partial I \times LM\cup I\times M} \ar[r, "s"] & \frac{\op{Map}(\bigcirc_2)}{F} \ar[r] & \frac{\op{Map}(\bigcirc_2) / \op{Map}^\prime(\bigcirc_2)}{ F/ F|_{UTM}} && \\
&&& \frac{\op{Map}(8) / \op{Map}^\prime(8)}{F / F|_{UTM}} \ar[u, "\simeq"] \ar[r, dashed, "Th"] & \frac{\op{Map}(8)}{F} \ar[r] & \frac{LM \times LM}{LM \times M \cup M \times LM},
\end{tikzcd}
\end{equation}
where dashed arrows are only defined on homology and we used the following notations.
\begin{itemize}
    \item $UTM$ is the unit tangent bundle.
    \item $FM_2(M)$ is the compactified configuration space of two points, namely it is obtained from $M \times M$ by a real oriented blowup along the diagonal. It is a manifold with boundary $UTM$ and homotopy equivalent to $M \times M \setminus M$ and fits into the following commuting diagram
    \begin{equation*}
        \begin{tikzcd}
            UTM \ar[r] \ar[d] & FM_2(M) \ar[d] \\
            M \ar[r] & M \times M.
        \end{tikzcd}
    \end{equation*}
    \item $\op{Map}(\bigcirc_2)$ is simply $LM$ thought of as a fibration over $M \times M$ given by evaluating the loop at time $0$ and $\frac{1}{2}$.
    \item Taking the pullback of $\op{Map}(\bigcirc_2)$ along the above square we obtain
    \begin{equation*}
        \begin{tikzcd}
            \op{Map}^\prime(8) \ar[r] \ar[d] & \op{Map}^\prime(\bigcirc_2) \ar[d]\\
            \op{Map}(8) \ar[r] & \op{Map}(\bigcirc_2).
        \end{tikzcd}
    \end{equation*}
    \item $F$ is $LM \sqcup LM \to M$.
    \item The map $s$ reparametrizes a loop. It takes a parameter $t \in I$ and a loop $\gamma$ and reparametrizes it in such a way that the path $\gamma_{[0,t]}$ is run through on the interval $[0, \frac{1}{2}]$ and the path $\gamma_{[t,1]}$ is run through on the inverval $[\frac{1}{2}, 1]$.
    \item The map $Th$ is capping with the Thom class in $H^n(M, UTM)$.
\end{itemize}

Let us first formulate the following
\begin{Lem}
Let $\alpha \colon N \to LM$ be transverse in the sense that
\begin{enumerate}
    \item [i)] $\frac{\partial}{\partial t} \alpha(u,t)$ is non-zero at $t=0$.
    \item [ii)] The map $\bar{\alpha} \colon N \times (0,1) \to M \times M$ given by $(n,t) \mapsto (\alpha(n,0), \alpha(n,t))$ intersects the diagonal transversely in a compact submanifold $V \subset N \times (0,1)$,
\end{enumerate}
then there is a unique map $\hat{\alpha} \colon \widehat{N \times I}^V \to FM_2(M)$ from the real oriented blowup of $N \times I$ at $V$ denoted by $\widehat{N \times I}^V$ to the compactified configuration space of two points such that
\begin{equation*}
    \begin{tikzcd}
    \widehat{N \times I}^V \ar[r] \ar[d] & FM_2(M) \ar[d] \\
    N \times I \ar[r] & M \times M
    \end{tikzcd}
\end{equation*}
commutes. Moreover, $\hat{\alpha}$ identifies the unit normal bundle of $V$ in $N \times (0,1)$ with $\bar{\alpha}|_V^* UTM$.
\end{Lem}
\begin{proof}
In local coordinates the map $M \times M \setminus M \to FM_2(M)$ looks like
\begin{align*}
    \mathbb{R}^n \times \mathbb{R}^n \setminus \mathbb{R}^n &\to \mathbb{R}^n \times S^{n-1} \times [0,\infty) \\
    (x,y) &\mapsto (x-y, \frac{x-y}{|x-y|}, |x-y| ).
\end{align*}
Composing with $\bar{\alpha}$ we readily see that condition i) is sufficient (and necessary) to lift the map $N \times I \to M \times M$ to $FM_2(M)$ in a neighborhood of $N \times \partial I$. To obtain the statement away from the boundary we observe that the function (in coordinates)
$$
(n,t) \to \frac{\bar{\alpha}(n,t) - \bar{\alpha}(n,0)}{|\bar{\alpha}(n,t) - \bar{\alpha}(n,0)|}
$$
smoothly extends from $N \times I \setminus V$ to $\widehat{N \times I}^V$.
\end{proof}

Such a transverse map $\alpha: N \to LM$ naturally defines a map
\begin{align*}
    \Delta(\alpha) \colon V &\to LM \times LM \\
    (n,t) &\mapsto (s \mapsto \alpha(n, st), s \mapsto \alpha(n, (1-t)s + t)).
\end{align*}

The following is a special case of Proposition 3.13 in \cite{HingstonWahl} adapted to our notation.
\begin{Prop}
Let $\alpha: N \to LM$ be transverse in the sense of the previous Lemma. Then
$$
\Delta(\alpha_*([N])) = (\Delta \alpha)_*([V]) \in H_*(LM \times LM, M \times LM \cup LM \times M)
$$
\end{Prop}
\begin{proof}
One checks that the following diagram commutes, where all the maps are the "obvious" ones.
\begin{equation*}
\begin{tikzcd}[row sep=scriptsize, column sep=scriptsize]
&\frac{N}{\varnothing} \ar[dashed]{r}{\text{suspend}} \ar[ddl] &\frac{I \times N}{\partial I \times N} \ar[r, equal] \ar[ddl] & \frac{I \times N}{\partial I \times N} \ar[r] \ar[ddl] & \frac{I \times N / \widehat{I \times N}^V}{\partial I \times N /\partial I \times N} \ar[ddl] & \\
&&&& \frac{V / \bar{\alpha}|_V^* UTM}{\varnothing / \varnothing} \ar[u, "\simeq"] \ar[r, dashed, "Th"] \ar[ddl] & \frac{V}{\varnothing} \ar[ddl] \ar[dd] \\
\frac{LM}{M} \ar[dashed]{r}{\text{suspend}} &\frac{I \times LM}{\partial I \times LM\cup I\times M} \ar[r, "s"] & \frac{\op{Map}(\bigcirc_2)}{F} \ar[r] & \frac{\op{Map}(\bigcirc_2) / \op{Map}^\prime(\bigcirc_2)}{ F/ F|_{UTM}} && \\
&&& \frac{\op{Map}(8) / \op{Map}^\prime(8)}{F / F|_{UTM}} \ar[u, "\simeq"] \ar[r, dashed, "Th"] & \frac{\op{Map}(8)}{F} \ar[r] & \frac{LM \times LM}{LM \times M \cup M \times LM},&
\end{tikzcd}
\end{equation*}
The only thing left to show is that after taking homology the upper zig-zag sends the fundamental class of $N$ to the fundamental class of $V$. Namely, we have to show that under
\begin{equation*}
    \begin{tikzcd}
    H_d(N) \ar[r] & H_{d+1}(I \times N, \partial I \times N) \ar[r] & H_{d+1}(I \times N, \widehat{I \times N}^V) && \\
    && \ar[u, "\simeq"] H_{d+1}(V, \bar{\alpha}|_V^* UTM) \ar[r, "Th"] & H_{d+1-n}(V)
    \end{tikzcd}
\end{equation*}
where $d = \operatorname{dim}(N)$, the class $[N]$ gets sent to $[V]$. First note that the Thom isomorphism here is given by capping with a Thom class that is the pullback of the Thom class on $H_n(M ,UTM)$ along $\bar{\alpha}|_V$. This Thom class is also the natural Thom class by considering $\bar{\alpha}|_V^* UTM$ as the oriented normal bundle of $V$ in $I \times N$. Hence, apart from our insistence on avoiding tubular neighborhoods, we obtained the standard description of the intersection pairing from which it follows that $[N]$ is sent to $[V]$. To see this more concretely, we note that it is enough to show that composing with $H_{d+1-n}(V) \to H_{d+1-n}(V, V \setminus \{ x \} )$ sends $[N]$ to the generator in $H_{d+1-n}(V, V \setminus \{ x \} )$. Thus the situation is local and we can assume that $N = \mathbb{R}^d$ and $\bar{\alpha} \colon \mathbb{R}^{d+1} \to \mathbb{R}^n$ is a linear projection. In this case the statement follows directly from the definitions.
\end{proof}

\bibliographystyle{hplain}

redpajama_set_name: RedPajamaArXiv


" The IGP's Intelligence Response Team (IRT) accomplished this task when it arrested a three man notorious kidnap gang on the 23rd June, 2016", the police said today.
According to the police, the Principal suspect, one Ikechukwu Daniel, a 28 year old indigene of Imo State, who is the mastermind of the gang, was rusticated from Ahmadu Bello University Zaria due to his cult related activities on campus in 2009/2010. He also doubles as the IT guru of the kidnap gang with mastery of Computer applications.
"He made damning revelations of their criminal exploits in the South West of the country. Another member of the gang, 29 year old Adeyemi Kayode, a serving officer of the Nigeria Security and Civil Defence corps and a native of Ojoo in Ibadan, Oyo State serves as a Personal Assistant to the Commandant, NSCDC Oyo State Command.
"The suspect took undue advantage of his office to obtain the GSM number and location of their would be victim. He hatched the plan on how to kidnap the business magnate to make a demand of one billion naira ransom.
"The third suspect, Ayodele Temitayo, a native of Oyo in Ibadan, Oyo State who is the marksman and armourer of the gang, claimed to have been dismissed from 213 Battalion Maiduguri of Nigeria Army as a Private in 2015.
" All the suspects made confessional statements revealing that they had successfully carried out several high profile kidnappings and two AK47 rifles loaded with live ammunition were recovered from them.


text:
"Blood Wolf" is a song performed by the American rock band Dance Gavin Dance. It was released as a single for digital download and streaming on October 11, 2019, by Rise Records. It is a b-side to the band's 2019 single "Head Hunter". It was produced by Kris Crummett. Following its release, it charted at number 24 on the Billboard Hot Rock Songs chart, the group's first career charting single.

Background
On June 8, 2018, Dance Gavin Dance released their eighth full-length studio album, Artificial Selection. The band toured extensively in support of its release, including tours with the American rock band Underoath and embarking on The Artificial Selection Tour from March to May 2019. In January 2019, the group decided to record new material with Kris Crummett at Interlace Audio Recording Studios in Portland, Oregon. On March 22, 2019, the group released a stand-alone single, "Head Hunter", accompanied with its music video.

Recording
On January 22, 2019, Dance Gavin Dance's lead guitarist Will Swan announced on Twitter that he was recording new material in Portland at Crummett's recording studio Interlace Audio. "Blood Wolf" was produced, mixed and mastered by Crummett with audio engineering by Ricky Orozco assisted by GiGi Zimmer.

Music video
The music video for "Blood Wolf" was released on the same day as the single's digital and streaming release, on October 11, 2019. It was directed by Mount Emult, who had previously directed music videos for the band's 2017 single "Summertime Gladness" and the 2018 single "Count Bassy".

Credits and personnel

Dance Gavin Dance
 Tilian Pearson – clean vocals, rhythm guitar
 Jon Mess – unclean vocals
 Tim Feerick – bass guitar
 Will Swan – lead guitar, backing vocals
 Matthew Mingus – drums, percussion

Production
 Kris Crummett – production, mixing engineer
 Ricky Orozco – audio engineer
 GiGi Zimmer – assistant engineer

Chart performance
Following its release, "Blood Wolf" debuted at number 24 on the Billboard Hot Rock Songs chart on October 26, 2019.

References

2019 singles
2019 songs
Dance Gavin Dance songs
Rise Records singles


Erin Dollar, the owner and maker behind Los Angeles-based (and so very chic) Cotton & Flax, recently made her first foray into paper goods, and I'm totally jazzed about it! The new Cotton & Flax notebook collection includes six pocket-sized notebooks in light blue and gold ochre, with a different pattern printed on each.
The notebooks were created in collaboration with Scout Books, and they're made from 100% recycled paper and vegetable-based inks, so they're stylin' AND eco-friendly. Gotta love that. And, if you love getting a peek behind the scenes (I do!), be sure to head on over and have a look at the making of these seriously snazzy notebooks.


If dedicated programs mode is used the child runs ChildCentre using the label on the Windows Desktop.
Then he has to pick up a preferred program that has to be run. Then he pushes Start.
If absolute time counting mode is used the child just works with the computer. He can see the left time by placing the mouse cursor over the System Tray icon of ChildCentre.
If the child does not work with the computer there is a good way to close Windows rather than to shutting down the computer. For that he can press right mouse button on the icon and then push "Close Windows" menu item.While Windows is closed the time is not counted.
The program was designed for automatic control of the time a child spends playing his games during a day. Even if an adult is not near the child, ChildCentre controls the total time and closes the program that is running (or Windows depending on the defined mode) if the time limit is expired. Then ChildCentre does not allow to start dedicated programs (or to work with Windows) during the rest of the day.
What parents have to do?
A parent has to run ChildCentre. Then he has to choose Settings in the System Menu.
It is necessary to define at least one program to be run by the child if you do not use mode Counting absolute time of working in Windows (see Options below).
If you want to add a program push Add button. Then enter the title (that you see proper for the program) and the path to the EXE (or PIF for DOS files) module.
You can use <...> button to activate standard window for picking files up. In that window find the module and push Open.


This is another one of my table gracing flowers. Flowers are my indulgence, whether they are daisies, roses, or orchids. Whether it is a single stem or a bounteous bouquet, I love them all. And now that I have become a phoneography fanatic, I can keep all of my flowers for as long as I wish. They don't smell quite as good, but they are still beautiful!