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Home » News » Scythians already practiced the genetic improvement of horses 2,300 years ago
Scythians already practiced the genetic improvement of horses 2,300 years ago
A new study published in Science unveils the secrets of horse breeding by Iron Age Scythian nomads. The genomes reconstructed from 14 archaeological horses also provide important insights into the process of animal domestication.
Nomad Scythian herders roamed vast areas spanning the Central Asian steppes during the Iron Age, approximately from the 9th to the 1st century BCE (Before Common Era). These livestock pastoralists, who lived on wagons covered by tents, left their mark in the history of warfare for their exceptional equestrian skills. They were among the first to master mounted riding and to make use of composite bows while riding. A new study published in Science now reveals the suite of traits that Scythian breeders selected to engineer the type of horse that best fit their purpose. This study has been led by Professor Ludovic Orlando in Copenhagen and has involved 33 international researchers from 16 universities, including Tomas Marques-Bonet, from the Institute of Evolutionary Biology (CSIC-UPF) and the CNAG-CRG and Inna Povolotskaya, Aitor Serres and Lukas Kuderna from the same institute.
The study took advantage of exceptionally preserved horse remains in royal Scythian burials, such as the site of Arzhan, Tuva Republic, where over 200 horses have been excavated but also at Berel', Kazakhstan, where no less than 13 horses were preserved in a single, permafrozen funerary chamber. Applying the latest methods in ancient DNA research, the researchers could sequence the genome of 13 Scythian stallions. These were 2,300-2,700 years old and included 11 specimens from Berel' and two from Arzhan. The researchers also sequenced the genome of one 4,100 year-old mare from Chelyabinsk, Russia belonging to the earlier Sintashta culture, which developed the first two-wheeled chariots drawn by horses.
The DNA variation observed at key genes revealed a large diversity of coat coloration patterns within Scythian horses, including bay, black, chestnut, cream and spotted animals. Scythian horses did not carry the mutation responsible for alternate gaits, and as a consequence, were not natural amblers. However, some but not all individuals carried variants associated with short-distance sprint performance in present-day racing horses. This indicates that Scythian breeders valued animals showing diverse endurance and speed potential.
"With the exception of two horses, none of the animals were related. It fits with Herodotus's depiction of Scythian funerary rituals, whereby sacrificed horses represented gifts from allied tribes spread across the steppes", says Dr. Pablo Librado, post-doctoral researcher at the Centre for GeoGenetics, University of Copenhagen, Denmark, and co-leading author of the study.
Scythian horse breeding unveiled: lessons for animal domestication
Importantly, none of the ancient horses analyzed in the study were inbred, which suggests that Scythian breeders succeeded in maintaining natural herd structures and did not perform selection through a limited number of valuable lineages. This contrasts with modern management practice where single stallions can be used to father hundreds of offspring. Patterns of genetic variation along the genome also revealed a total of 121 genes selected by Scythian breeders, most of which are involved in the development of forelimbs. This is reminiscent of the morphological indices measured on bones, and indicates that Scythian breeders selected horses showing more robust morphologies.
"In this study we wanted to go beyond the myth of Scythians being aggressive warriors, drinking the blood of their enemies in skull mugs. We wanted to reveal the many facets of the exceptional relationship that these people developed with their horses," says Orlando, at the Centre for GeoGenetics of Copenhagen.
The genome data set generated in the study also reveals important lessons for the history of horse management, which started some 5,500 years ago, and animal domestication as a whole. By contrasting patterns of genetic variation in ancient and present-day horses, the authors found support for a significant demographic collapse during the last 2,300 years, which resulted in an important reduction of genetic diversity within horse domesticates. During the same time period, reproductive management has involved an increasingly reduced number of stallions, up to the point that, today, almost all domesticates virtually carry the same, or highly similar, Y-chromosome haplotype(s).
The authors also found that the demographic collapse and loss of Y-chromosomal diversity observed within the last 2,300 years was mirrored by a significant accumulation of deleterious mutations in the genome of the horse. As these mutations reduce the fitness of their carriers, it shows that the last two millennia of horse management have negatively impacted the horse. However, early domestication stages, as represented by the Sintashta and Scythian genomes, did not have such an impact. This contrasts with the Cost-of-domestication hypothesis, which posits a negative impact starting from early domestication stages. In the case of horse domestication, it is likely that the demographic collapse within the last 2,000 years reduced the efficacy of negative selection to purge out deleterious mutations, which could then accumulate in the horse genome.
Finally, the researchers also found that the genomic regions showing the most extreme signatures were involved in the development of the neural crest, and expressed within tissues derived from the neural crest. "The neural crest hypothesis proposes a unified model for the origin of similar traits commonly found in most domestic animals. As the neural crest represents a temporary group of cells during development which gives rise to many tissues and cell lineages, selection for genetic variants affecting the neural crest can almost in one go co-select for a range of traits. The over-representation detected in our study supports the neural crest as key to animal domestication and to the rise of common domestic traits in independent animal lineages," concludes ICREA Professor Tomas Marques-Bonet.
Work of reference:
Ancient Genomic Changes Associated with Domestication of the Horse


The period of the final exams is the most stressful for all students. And every student is going through a final exam week with difficulties. But there is a way in which you can avoid so stressful situations. All you need is preparing for the exams in advance.
All students postpone all the work until the last moment and this can play a nasty trick with them. Relaxing all the time students don't think about the great number of course works, essays and other writing assignments which can gather like a huge snowball. And when students realize that they have problems with exams they try to find some magic ways how to solve all these problems.
But, in fact, there is no magic and it is impossible to go through the final exams successful without hard work. But there are some tips which can make the level of the stress lower and can help the students to make sure that they are on the right way. Here you can find several scientifically proven techniques which will help you to prepare and to pass the exams as well as you can.
Cramming is not actually a good way to master the material. You shouldn't sit with the book in your hands from early morning till the late at night and repeat the same things every now and again. It's better to study in the intervals. What does it mean? You should study for 40-50 minutes and after that take a break of 5-10 minutes. This way of studying is more effective than usual cramming.
Everybody knows that even short 15 min cardio workout can make your memory better. You can choose the type of workout which you like the most. It can be walking, jumping, jogging, dancing. All these exercises may easily increase the level of your energy and at the same time, it will definitely reduce the level of stress.
Remember that you should eat a good food not only when you start preparing for the exams. Even the products which you have eaten 5 days ago can influence.
While studying, the brain cells are consuming glucose, so it is important to take short breaks every 40-50 minutes. It will allow your body to produce glucose as a special fuel.
Don't make the preparing for the finals routine. Use some alternative places for studying. The New York Times says you can just change the room in which you study and it can improve your retention.
Our brain has an amazing ability to draw parallels and to establish associations. So if you have to study, for example, many foreign words, study them in different rooms and the brain will connect the associations between the places and words that were studied there.
In fact, cramming and a big number of writing assignments can cause a great anxiety during the exam session. The anxiety in its turn can lower the person's ability to study the information and retain it in the memory. In order to rid yourself of stress and anxiety, you should create a schedule which will be well balanced. If you have strict time limits you can ask for the help with your writing assignments. It will allow you to prepare for the exams in advance and pass the final exams without physical and psychological difficulties.
All students who don't prepare for the exams in advance try to study all night long before the exam. But it should be mentioned that it is not a good idea. The studies have shown that all-nighters usually have a bad memory and get lower grades than the students who study during the whole semester and prepare to pass the exams timely.
Be focused and attentive when you are preparing for the exams. The scientific studies have shown that it is more difficult to study the information and retain it if you are watching television, texting using your computer or the smartphone or listening to music. So put away all distractions and focus on studying.
There are different ways to study the information and retain it while preparing for the exams. Some students may highlight the main statements, other students may re-read some paragraphs. Sometimes one can think that summation is a good way to remember all the information. But you should understand that all those techniques cannot boost the performance of the students. Practice is the most powerful technique. Do a rehearsal of how you will take the exam, use different flashcards and your preparing for the final exams will be effective.


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Obituary of Burton Bodie
It is with profound sadness the family of Arthur Burton Bodie announce his passing on January 22, 2023. Burton was born March 31, 1926 on the farm in Mazenod. He passed away peacefully at the age of 96. He leaves behind his loving wife, Clara with whom he celebrated 75 years of marriage in November 2022. He is survived by his one sister Lila Bernardin; four grandchildren, Sean (Andrea) Connors, Angie Connors, Trevor (Tabatha) Bodie, and Kelly (Dale) Reed; seven great grandchildren, Riley Connors, Tara Rousell, Brendan, Rin and Ariana Bodie, Haylee and Chloe Reed; two brothers-in-law Herb Fox and Don Fox; one sister-in-law Madeline Fox; numerous nieces and nephews.
Burton was predeceased by his parents Arthur and Helen; one sister Audrey; his three children Terry Connors, Robert Bodie, and Jack Bodie; one son-in-law Jim Connors.
Burton farmed until he retired. During his retirement he and Clara enjoyed traveling south for the winters. He also truly appreciated his family, fishing trips to Anglin Lake each summer for many many years. Burton loved fishing, family time, camping and spending time on his computer. He was a great community supporter and spent decades on the following boards: Palmer United Church, Palmer Geneva Cemetery, Palmer Town Council, and Palmer School Board.
Burton dedicated his life to his family and 'his special angel', Clara. They always went to bed each night holding hands. He is now in the hands of the Lord.
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Butler University's Visiting Writer's Series bring talented writers to the Indianapolis all year, poet Mark Strand is no exception. As a premier contemporary American poet, Strand brings an unparalleled grasp of language to his work. This Indianapolis education event will be held in the Reilly Room of Atherton Union on Butler University's campus Tuesday, April 13 at 7:30 pm. As always the Visiting Writer's Series at Butler University is FREE to all Indianapolis people. Hear Mark Strand read his remarkable poetry in a voice all his own, then pick his poetic brain with your questions.
As well as being a talented poet, Mark Strand is also an accomplished editor, translator and prose writer. In 1990, he was named to United States Poet Laureate. Just nine years later his collection Blizzard One garnered him the prestigious Pulitzer Prize for Poetry. His road to this type of success spans more than four decades, as Strand saw the country deteriorate and rebuild itself during the turbulent 1960s.
His first collection of poetry, Sleeping with One Eye Open (1964), introduced Strand's unique approach to writing. His writing is shaped by an overall sense of anxiety and restless foreboding. The title poem alone reveals a great deal about the entire volume as Strand explains that it "speaks to a certain anxiety I experienced back in the early '60s. I was afraid the United States would go to war with Russia or the USSR. I think it's a poem surrounded by a great deal of silence." His early poetry expresses a deep concern for recovering identity and self. This theme would come to permeate all of his work, as he grew as a writer.
Strand wrote seven more volumes of poetry through the sixties and seventies, before his writing no longer made sense to his self. He took a decade long break from poetry to explore other avenues of literature and writing. During that time he wrote three children's books. As in his poetry, these stories focused on loss, using the genre to connect with children about common childhood worries.
In 1990, he returned to poetry, publishing eight more collections of work. Mark Strand has been published in countless collections of poetry, literary magazines and other publications. He uses his many talents to share a universal sense of void, as he (and we) search for meaning in our short little lives.
Don't miss Mark Strand at Butler University's Visiting Writer's Series, this Tuesday, April 13. Before this FREE Indianapolis event, head to the nearby Indianapolis cultural district, Broad Ripple Village, for a bite to eat. With so many Indianapolis restaurants and Indianapolis bars to choose from, there is something for everyone in Broad Ripple Village.
Stay tuned to Indianapolis News, Events and Information on Fun City Finder.com for all the latest on fun things to do in Indianapolis. We cover Indianapolis arts to Indianapolis culture and more. Get out in Indy and find some fun!


Quattro has officially engaged with Mike Shutty, a Chicago-based agency services consultant and longtime advisor to Quattro's leadership team. Shutty is working with Quattro's leaders and account managers to strategically drive client growth by leveraging Quattro's data and analytics capabilities.
Shutty will work cross-functionally at Quattro as a strategic advisor to review select clients, identify what drives their business and KPI's, and elevate Quattro's data and analytics capabilities to drive our clients' business growth.
"Mike is a long-term friend and former coworker of mine, and I've seen him in action – which is why we are thrilled to partner with him as we look to implement increasingly complex marketing programs for our clients," said Tom McNamara, Quattro Partner and Co-Founder.
Over his 35+ years in the industry, Shutty has worked for the country's top data and analytics companies and marketing agencies before successfully branching out on his own to provide consulting services.
"Mike's robust data and analytics experience will help us become more effective at growing our clients' businesses," said McNamara.
Quattro is excited to partner with Shutty, and is looking forward to a prosperous relationship.
Want to know how leveraging data and analytics can help improve your business? Get in contact with us today, and let's spark up a conversation!


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GA Training Institute
Blythe Keeler Robinson
CEO and President
Blythe Keeler Robinson is an innovative leader with 20 years of non-profit management experience and expertise in educational programming, early childhood education, program design and implementation, grant writing and contract review and negotiation.
Robinson is the President and CEO of Sheltering Arms Early Education and Family Centers in Atlanta, Georgia. Founded in 1888, Sheltering Arms is one of Georgia's largest nonprofit organizations and one of its most respected. Sheltering Arms serves nearly 2,500 children daily, ages six weeks to five years old, and their families in 13 early care and education centers across metro Atlanta. As a recognized leader in early education, Sheltering Arms is a National Head Start Center of Excellence, as well as a member of the Educare network. Under Robinson's leadership, Sheltering Arms has increased its national impact and visibility as a member of the Annie E. Casey Foundation's Family Economic Success – Early Childhood Initiative Action Learning Network and Ascend at the Aspen Institute.
Prior to joining Sheltering Arms, Robinson served as the Senior Vice President and Chief Programs Officer of the Early Learning Coalition of Miami‐Dade/Monroe, Florida. She was responsible for the management and implementation of early care and education programs providing services to more than 50,000 children and their families annually. In her role as the Vice President of Strategic Planning and Initiatives at Reading Is Fundamental, she directed programs and strategic initiatives across the country and designed and oversaw the implementation of the Multicultural Literacy Campaign, a national effort to promote and support early childhood literacy in targeted communities.
As a dedicated servant leader, Robinson believes in giving back to the community. She volunteers with local organizations focused on children and the homeless population, and mentors girls and women. Robinson serves on the Board of Directors of the National Black Child Development Institute, the board of Directors of the International Women's Forum-Georgia Chapter, the Atlanta Speech School Advisory Board and the Healthy Beginnings System of Care for Young Children Advisory Board. She is also a member of Delta Sigma Theta Sorority, Incorporated.
In addition to receiving a Women Extraordinaire Award from Business Leader Magazine, Robinson was recognized as one of South Florida's "40 Under 40 Leaders of Today and Tomorrow" and "50 Most Powerful Black Professionals." She has also been named to the Atlanta Business Chronicle's list of "Atlanta's 40 Under 40 Leaders" and is a member of the Leadership of Atlanta Class of 2015.
Robinson holds a bachelor's degree in Government and Politics with a certificate in African American Studies from the University of Maryland, College Park, and a juris doctorate from The Georgetown University Law Center.
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How Fragrances Were Born: A Short History of Perfumery
The history of perfumes dates back thousands of years. Not only the Ancient Egyptians, but also the Romans and Persians discovered various types of fragrances. Fragrances were a symbol of cleanliness and hygiene in ancient times rather than an accessory as they are today. Over time, the use of perfumes evolved, but they remained expensive and difficult to obtain. For nobility, who were often the only people who could afford it, perfume was a way to distinguish oneself. One of the first recorded perfume factories was the Tapputi chemist, whose story was discovered on a clay slab from Mesopotamia dated back to the second millennium BC.
Fragrant Greetings from Ancient Egypt
Ancient Egypt's famous perfumes were of great importance. Even in their mythology,there exists the god Nefertem, the master of perfume. The Egyptians produced fragrances using the distillation of natural ingredients with essential oils. The most frequent ingredients were, of course, flowers and fruits. The scents were not only used on the skin, but also as home fragrances and in public festivities. Fragrant baths and households played a significant role in the history of Egypt. The rulers even took their favourite perfume with them to the grave.
Oil-based Perfumes
The Persian kings were famous for their distinctive scent. Perfume was like a personal signature for them; no one else was allowed to use the same scent. The Persians even ruled the perfume trade and are considered the inventors of oil-based perfumes. According to ancient Persian documents, many workshops existed for the production of perfumes, where perfumers experimented with distillation processes and the creation of new fragrances. Ancient Persia was another huge milestone in developing perfumery as a craft.
Greco-Roman Scents
Thanks to the thorough documentation of perfume production by the ancient Romans and Greeks, you can still find Greco-Roman scents today. One of them dates back to 1850 BC, from the world's oldest perfume factory. Perfumes were associated with Aphrodite, the goddess of love, and with services and temples. They often used frankincense and myrrh to scent temples and public baths, or in the production of body balms and perfumes. Despite their love of fragrances, some Romans condemned the use of perfumes as a wasteful indulgence for the wealthy. As Rome fell, the use of fragrances also fell. As a result, the perfume market in Europe stagnated for hundreds of years to come.
Perfumes in Europe
During the Crusades around the 11th century, the Crusaders brought to Europe from the Far East the materials and techniques to produce various fragrances. The popularity of perfumes rose, and it was perfected by the Italians in the 13th century, who replaced liquid perfumes with solid ones. Marco Polo imported unique, aromatic substances from his travels, and so Venice soon became a trading venue for fragrances. The spread of perfume to the rest of Europe is often attributed to Catherine de Medici, the Italian noblewoman and wife of King Henry II of France. Her Italian perfumer Rene le Florentin created for her a signature perfume of orange blossom and bergamot. From that moment, perfumes became personal accessories. European men and women wore it on their bodies, and in their clothes and wigs. Perfumers began to incorporate more complex ingredients into their products, such as ambergris, civet, and musk derived from animals. Scents were a way to mask body odours, as personal hygiene as we know it today was still not widely practiced. Perfumes were also a way to distinguish between the social classes.
The changing tastes and developments of modern chemistry later laid the foundations of perfumery as we know it today. Even the French Revolution did not reduce the appetite for perfumes; a fragrance called Parfum a la Guillotine was born from it. In France, under the post-revolutionary government, people again dared to express a taste for luxury, including perfumes. Thanks to its shops, the city of Grasse in Provence established itself as the largest production centre of raw materials for the production of perfumes.
Fragrance Creation, Then and Now
Although perfumery is thousands of years old, the methods of production have changed over time. Some fragrant materials were initially turned into balms, while, for example, flower ingredients were part of scented oils. During frequent trade trips, perfumers also began to use imported exotic raw materials, which they often added to water. The first perfume bottles were wooden and clay-made, and with time they gradually evolved into orante, coloured glass bottles.
Today, perfumes are everyday personal accessories, made from both natural and synthetic ingredients. They are no longer only for nobility. In recent years, however, the emergence of small and exclusive brands of fragrances has reappeared on the market. They are known as niche perfumes, and are made by perfumers who often believe perfumery is an art form. These fragrances contain the best oils from around the world and have revived a passion for this ancient craft. If you crave such a perfume, don't forget to read our article that explains how to choose the perfect perfume for you. Like a piece of jewellery made just for you, your perfume is more than just a scent.


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Jenny's Unexpected New Job
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Jenny sighed as she turned over the page in the text book in front of her. She had been studying in the library for over 3 hours now and even she needed a break. She ran her hands through her light brown hair, hair that she had kept in the same plain, chin length bob since she had been about five years old. While the rest of class mates had become interested in the latest fashions, even at the age of 19, fashion just did not interest her. Most days she dressed in shapeless clothes that seemed to have the sole intention of trying to hide her body and so help her blend into the background. Most of the other girls at the college hardly seemed to notice her when she was there, and so Jenny found it easier to keep to her own company. While the popular girls were out partying together, or hanging around down at the local mall, she would usually be found in the library or working on an assignment in the computer lab'.
Packing her books and notes away for the day, Jenny left the library and started to walk across the college car park back to her dorm'. Suddenly a car came screeching around the corner and she had to leap out of the way to avoid being hit. The car swerved and caught the rear fender of another car parked in the lot. The door of the car swung open and the driver leapt out screaming at Jenny.
"What the hell are you doing?", She screamed at Jenny, "You made me hit that car you stupid loser. Why don't you look the hell where you're going?".
Jenny was too stunned to speak and just stood there staring at the angry woman in front of her. Although she had always thought of herself as being quite tall, this woman must have been easily five feet ten, maybe even taller. The high-heeled leather knee boots she was wearing only added to her towering height. But it was the woman's body that really caught her attention. She had never seen such a perfectly proportioned and well-toned physique before, and the tight-fitting, all black outfit she was wearing only served to accentuate it. The leather trousers were pulled in tightly by a thick leather belt, accentuating her slim waist. The black polo neck drew attention to her firm breasts, stretched tightly whenever she moved. It wasn't just her build that grabbed your attention though, it was the way she held herself. There was an air of confidence about her. This was a woman with total self-belief.
"Well?", the woman – Danielle – demanded of Jenny, "What have you got to say?".
Jenny tried to speak but couldn't make the words come out, so she just stood there as a lone tear ran down her cheek.
"Jesus, what a wimp!", the woman spat at her, "Why the hell are you crying baby girl? You've been an idiot and caused me to crash but there's no need to cry about it." She stood looking at Jenny waiting for an answer. Looking at Jenny more closely a plan began to form in Danielle's mind. Maybe this plain, little coward was just the girl she had been looking for. She smoothed down her shiny black hair, pulled back into a tight pony tail and came to a decision.
"I reckon the damage to my car is going to come at least a couple of thousand dollars little girl and I don't think you have got that kind of money have you?".
Jenny shook her head.
"I didn't think so. Well, you can work off the debt by working part-time in my hair salon. Come on. Get in", she commanded pointing towards her now dented car.
Almost in a dream Jenny walked towards the dented car and climbed into the passenger seat, buckling her seat belt automatically. Danielle slipped into the driving seat slamming the driver's door behind door. Slipping the car into drive she roared away from the college heading for the city center. After a short drive they pulled up outside a very fashionable looking hair salon. Danielle slid out of the car; "Hurry up you, I haven't got all day and we have to get you ready to face my clients".
Jenny followed her new boss into the salon, taking in the black and white decor. Danielle threw down her bag, locked the door of the salon and sat down in one of the styling chairs.
"Right then let's see what we have got to work with here. Turn around, what's your name?".
"It's Jenny".
"Far too dull little one. Let's see, hmm, I know, your name when you're in the salon is Serena. Okay Serena, take off the gypsy clothes", she instructed.
Jenny (Serena) hesitated, embarrassed at taking her clothes off in front of another woman. But a quick glance at Danielle told her that this was not something to take her time about. She reluctantly slipped out of her long floral print skirt and the pale pink long-sleeved blouse, leaving standing before her new boss in just her plain white bra and white panties.
"Everything!" Danielle commanded.
Jenny started to protest but then thought better of it. She slowly unhooked the plain bra, dropping it to the floor, then slid her white panties down her slim legs and stepped out of them.
"Nice figure you have there, you should dress better and show it off more", Danielle replied.
"Stay there Serena and I will back with your new uniform". She disappeared through the door at the back of the salon and reappeared a few moments later carrying a small pile of clothes. She handed Jenny a matching bra and panties in black satin. Jenny quickly slipped on the unfamiliar lingerie unused to the shiny material and the way that it felt against her skin.
She showed Jenny a black pleated leather mini-skirt. The skirt was short, very short and she guessed would only fall to about mid thigh. If she wasn't careful people would catch a glimpse of her new panties. Surely she wasn't expected to wear this in front of the clients. Danielle sat still looking at her. She reached down and showed Jenny a black latex T-shirt. Jenny wondered how she would manage to pull the tight-fitting top down over her body. The tight material would cling to her body and emphasize her breasts. The many lights in the salon reflected on the shiny surface of the top.
"And finally", said Danielle pointing to a pair of patent leather ankle boots with a three inch stiletto heel.
"Okay Serena, while you are working in my salon these are the rules for your clothing. If it's not black, then you're not wearing it in here. If you wear a skirt that's fine, but it has to be no longer than the one I have here. All of your tops must either be tight or low-cut at the front. With shoes or boots it's high heels only and preferably no lower than three inches. I will check every day to make sure that your clothes are right. If you break the rules you will have to be punished. Any questions?".
Jenny shook her head, still in shock at she was supposed to wear.
"Right, it's time to sort out your hair", Danielle announced, "Come and sit here".
Jenny walked slowly over to the styling chair trying to cover her body as she felt so exposed in her new bra and panties. She sat down nervously feeling the coldness of the seat against her exposed ass cheeks. She was covered quickly by one of the salon's shiny black satin capes that wrapped around her body and almost reached to the floor. A second, heavy black rubber cape was wrapped around her shoulders and then tied very tightly around her neck. Danielle ran her fingers through Jenny's plain bob.
"Obviously a lot of this length has to go", Danielle announced as she ran her hands through Jenny's hair, pulling it back from her face.
Danielle picked up a water spray bottle and dampened Jenny's hair and then combed it all straight down. She then reached down and picked up a pair of sharp scissors from the nearby shelf. Placing the scissors against the right side of Jenny's head, she slowly started to cut a straight line through her hair across the top of her ear. Jenny gasped as she watched the lengths of her hair slide down her shoulders, onto the cape, and then fall silently to the floor of the salon. The scissors continued to cut the blunt line around her head, carving their way easily through her bob until Jenny was left with a short cap of ha
ir sitting on top of her head like a mushroom. Danielle went back and checked the line she had just cut, re-combing and snipping until she was satisfied that the line was perfect.
Tears started to fall down her cheeks as Jenny sat immobile in the chair in shock, unable to believe the dramatic change that had just been wreaked on her hair. Hoping that perhaps the worst was over she was surprised when she heard the sharp click behind her, and then the buzzing of a pair of electric clippers. She had seen these used before when taking her younger brother to be cropped and knew that it could only mean she would be getting a very short cut.
"Head down!" Danielle instructed, "It's time to clean up the back and sides."
Danielle couldn't help a slight smile spreading across her face as she pushed the young girl's head even further forward. How many times now had she been here before? It always surprised her how they always gave in once the clippers started stripping away their hair. Many times now she had "helped" her female customers towards a short haircut. In they came to her salon with their timid, boring requests for a "just a trim" or "just take the ends off". The end result was always the same however. Out would come the clippers and there would be another unwilling convert to Danielle's little club.
Depending on her mood that day sometimes it would be just a little crop, or sometimes maybe a flat top. Or if she was in a bad mood then it would be time for her big black Oster clippers and a very severe clipping for the poor woman in the chair. Once a girl had complained that it was too short. Big mistake! To teach her a lesson for being ungrateful Danielle had to take her down to a traditional high `n' tight, even going so far as to white-wall the sides of her head with her trusty razor. Then for a month afterwards the girl was made to return every Friday to have the back and sides freshly shaved. Although as severe as that haircut had been, she had yet to shave one of her customers completely with her razor.
Returning to her current new trainee and gripping the clippers tightly, Danielle started to move them gradually up Jenny's neck, pressing them firmly against her skin until they met the hair left below the line she had recently cut. The note of the clippers changed as they started to bite into the hair. With no guard on the clippers Jenny's hair was easily and quickly stripped down to bare skin, revealing a pale strip of just stubble as they climbed higher and higher towards the line cut by the scissors.
"Please don't shave my head", Jenny pleaded.
"Don't be such a baby, I'm just tidying you up", Danielle replied, trying hard not to laugh, as she started a second pass up the back of Jenny's head.
Jenny gripped the arms of the chair tightly as she felt the clippers stripping the hair from the back of her head. The back of her head felt strangely cooler as more and more of her hair met the sharp blades of the clippers. With each fresh path cut by the clippers more hair tumbled down onto the silky black cape. With her head forced forward, Jenny could only look down at the floor and watch through her tears as more and more of her hair joined the long strands already lying there. Suddenly the clippers stopped and, for a moment, she thought her ordeal was over, only to feel her head now pushed painfully over to one side. Again the clippers started to peel the hair from her head as Danielle roughly bent her left ear down, and pressing the blades against her scalp, started to brutally carve a wide strip free from hair above each ear.
After another ten minutes of clipping and buzzing Jenny's head was finally released and she could raise her head. Steeling herself for the worst Jenny looked at her unfamiliar image in the salon's large mirror. She couldn't believe what had happened to her once beautiful hair. Had she ever seen such an ugly haircut? She turned her head to the side and saw in the reflection the brutally clippered outline disappearing around the back of her head. Lifting a hand from under the cape she ran her hand across the nape of her neck feeling the stubble, all that was left now of her hair. With growing horror she raised her hand higher up the back of her head feeling the extent of the newly buzzed hair.
"Don't worry my child, we will soon get rid of that horrible roughness", Danielle laughed, "But first we have some styling to do".
She lead Jenny over to the washbasins at the back of the salon, and quickly rinsed off her hair, removing any bits left clinging to her head. Then Jenny was taken back to the styling chair.
"Time to put some body in this", Danielle told her as she walked away and then came back pushing a trolley in front of her. She picked a large curler from the trolley and started to wind it very tightly around Jenny's hair before pinning it firmly to her head. Danielle continued to wind more large curlers into the hair until Jenny had a halo of curlers surrounding her head. Her hair was then soaked in a very strong perming lotion. She was then taken over to the row of dryers lining the back wall, and she pushed down into one of the chairs and one of the hoods lowered over her head. Danielle turned the dryer up high and switched it on. She knew that the large curlers and the permanent setting lotion would leave Jenny with a very full cap of hair that would make the contrast between the shaved sides and back even more prominent. Danielle knew it wouldn't be long before Jenny would be begging her to get rid of the hairstyle. Then it would be time to bring out her clippers again, and maybe even time for her first full head shave.
After an hour under the dryer Jenny was starting to really worry. Not only was she sure that she could smell burning hair, she couldn't help worrying what the final result would look like. Danielle turned off the dryer and she was taken back once more to the styling chair.
"Before I take your curlers out I need to get rid of that horrible stubble", Danielle informed her.
Jenny sat still in the chair dreading what she knew was going to come next. She jumped as she felt the cold shaving gel being spread over the back and sides of her head below the curlers.
"Hold still, I would hate to cut you", Danielle warned as she picked up the old-fashioned open razor and moved Jenny's head firmly forward again.
Jenny felt the sharp razor scraping against the nape of her neck, stripping away what little hair now remained there. She almost held her breath, scared to move. She felt the razor climbing the back of head, rising up and up stripping the stubble. Danielle worked slowly and confidently as she shaved the back of Jenny's head very close, going over and over to make sure that not a trace of stubble remained. This was perhaps the part that she enjoyed most. As much as it turned her on to have a pretty girl seated in her chair as her clippers sheared away their hair, there was something about shaving with a razor. So clean. So brutal. So hot!
Danielle took her time working with the razor, not stopping until she was certain that no trace of hair remained below the curlers. Wiping Jenny's newly shaved skin roughly with a towel she ran her fingers over the clean, smooth skin. Danielle slowly started to unwind the curlers, smiling to herself as she saw how full the remaining hair was, how far it stood away from the sides and back of Jenny's head. The freshly shaved skin in sharp contrast to the wide, bouncy hair above. Picking up a large canister of hair spray, Danielle coated Jenny's remaining cap of hair with a thick coating of lacquer.
Picking up a rectangular mirror from the shelf, Danielle held it behind Jenny's head so that she could finally see the finished result.
"Well, what do you think"?, Danielle asked.
Jenny tried not to look but felt herself drawn to the reflection in the mirror. Fresh tears started to roll down her cheeks as she took in the startling transformation. Her hair had now been shaved an inch clear of the top of her ears and she could see in the second m
irror the harsh line extended all the way around the back of her head. The curlers had done their work and what hair she had left stuck out all around, the shaved skin below exaggerating the width of bowl.
"We call this a chili bowl cut", Danielle informed her, "Although usually I don't shave the back quite so high", as she ran her hand over the newly shaved skin, "In fact I may shave it a bit higher when I trim it next time".
"You will need to come in half an hour early each day so that I have time to shave you", Danielle instructed Jenny. She removed both of the hair capes leaving Jenny sitting in the chair in just her new red satin bra and panties.
"Get dressed!", Danielle barked.
Jenny picked up the leather mini-skirt and slid it up her long legs. She looked in horror at just how much of her legs were left exposed by the shortness of the skirt.
"Just a minor adjustment I think", Danielle said, as she turned the waistband of the skirt over so lifting the hem of the skirt even higher, "And now put the top on".
Jenny struggled to pull the tight-fitting latex top over her body, feeling it cling to every curve of her young body.
"That's better, you are fit to face my customers now young lady", Danielle suggested, "Now sweep all this hair up and tidy the place up, we open in 20 minutes".
As she busied herself tidying up Jenny couldn't help catching glimpses of her new image in all the reflections in the salon's mirrors. The shortness of her skirt, the hem rising still higher every time that she turned around or stretched to reach for something. The way that the latex top outlined her breasts. And worst of all – her hair. The stark shaved back of her head and the ridiculously wide cap of hair left perched on top of her head. Every so often she stopped to stroke the back of her head, hoping in vain that it wasn't shaved as high as it looked. Of course it was. How could she go outside looking like this? What would her friends say?
How long will Jenny put up with her new hairstyle and her new image? What will Danielle do to Jenny's hair next time….?.
December 24, 2006 clippedOneEnglish, Fantasy, Forced, PG, SalonNo Comment
«Erin's Discovery
This Is For You Shaanu»
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Hair Stories ↑


Quiz: 91% of People Can't Identify All of These Canines. Can You?
91% of People Can't Identify All of These Canines. Can You?
Can you tell whether that canine hanging out in your backyard is safe to pet? Sure it might be a neighbor's dog -- but it could also be a coyote, wolf or some other critter that you definitely don't want to tangle with. Take our quiz to see if you can tell one member of this animal family from another!
The Canidae family includes more than 30 different species, known as canids. They include domestic dogs, like retrievers and poodles, as well as slightly more feral critters, including wolves, coyotes, jackals, foxes and even wild dogs, like dingos. They're found on every continent except for Antarctica and live everywhere from the mountains or desert, to icy tundra to your very own living room.
Canids all share the same basic features -- so if you see one in the wild, you'd probably know it was related to the dog -- but there's also a surprising level of variation between species. Some have wiry hair and raccoon-like faces, while others have layered coats that allow them to survive in some of the coldest spots on Earth. They range in length from less than a foot to nearly 7 feet and can weigh less than a typical housecat or more than an average-sized man.
Think you can name the various members of the canine family from just a single photo? Take our quiz to find out!
What is the name of this canine?


When you look at the current statistics, you will notice that the escape rooms have been growing in popularity. It is also easy to notice that these escape rooms have been growing in number too. Since people are beginning to show interest in them, that's the reason why they have been growing in popularity.
When you visit this escape room, you will get an opportunity of playing different games. Among them is the kidnapping game. You and your colleagues will be placed in a room by the kidnappers with nothing but your will to survive. You will be required to free yourselves before the kidnappers return in less than one hour.
Another game that is interesting is the Hostage game. It follows two people who are kidnaped on a flight. You are captured while in a plane, and you lose consciousness. Your task is to get out of your cuffs before time hits zero.


Since 1989, The Daniel Group has helped industrial manufacturers and value-added dealers and distributors that serve the B2B end user. Our expertise includes a deep understanding and industry focus on the pressures you face and the complexity of your channels.
We work with you, your management team and the people in the customer trenches. Through a combination of LinkConnect, our Web-based portal, surveys, data analysis, and strategy consulting, we help you make the small changes that over time lead to profitable growth.
If you'd like to learn more about how The Daniel Group can help your business gain a competitive advantage in the market by improving customer loyalty and trust, call us – 877-967-4242. Or, send email to information@thedanielgroup.com.


text:
Søkort er en film instrueret af Ib Dam efter eget manuskript.

Handling
Effektiv og nøjagtig søopmåling er afgørende for skibsfartens trygge sejlads. Denne vigtige opgave varetages af søværnet i samarbejde med søkortarkivet. Opmålingsarbejdet såvel i danske som grønlandske farvande vises, foruden fremstillingen af søkort.

Referencer

Eksterne henvisninger

Danske film fra 1967
Film fra Danmark (administrationskategori)
Dokumentarfilm fra Danmark

redpajama_set_name: RedPajamaWikipedia


To be sure, this does not necessarily suggest that all religious studies departments will be gone in 5 years. I don't know any of the particulars regarding the situation at UC Berkeley and how this decision was reached. This move, though, serves to highlight the increasingly difficult predicament of religious studies departments across the country (and elsewhere, like the UK). Further, this move comes in the midst of a time when little seems as important to local, national, and international conversations as religion. Here's how Carrie Schroeder put it after tweeting about the canceling of the major at UC Berkeley.
Indeed, in a time when more and more news organizations are devoting more resources to covering religion - and often doing a less than stellar job - now seems to be exactly the time to double down on our commitment to the academic study of religion at the university level. Emily Vork, a Religious Studies major at Alabama, gets it.
We religionists are often quite poor at explaining why what we do matters to the "real" world (I took a stab at it when discussing the new Noah movie here). The headlines of late should make that an easy enough conversation, but we often fall back on the amorphous ideal of "critical thinking." Yet we fail to adequately explain just what that means. Vork has done that here as well as I've seen from many long-time religion professors.
The academic study of religion does not just teach good reading and writing skills (though it certainly does that), it does not just teach content about various "religions" and traditions (though you'll certainly pick up more than your fair share along the way), it does not just offer the opportunity to learn new languages (though it does and you should seize those opportunities without hesitation), the academic study of religion teaches you that questions exist which you never knew existed. The academic study of religion pulls the veil back on the world around us and examines its attempts to present as "natural" that which is arbitrary and contested. It tells us that history is never objective or neutral. It gives us the tools for recognizing the rhetorical work being done in political speeches, viral illustrations, and the latest sci-fi movie.
We need religious studies - not to tell us what is true or what is the meaning of life, but rather to show us how such claims work, how they naturalize themselves, and what they hope to gain.
There is a phrase over the main door to Dodd Hall on Florida State's campus, the building in which our religion department is housed, which says, "The half of knowledge is to know where to find knowledge." We cannot expect our citizenry, our journalists, or our politicians to understand religion if we do not know where to find those best suited to explain it to us. If we continue this nationwide attack on the liberal arts in general and on the study of religion specifically, there will be nowhere left to look for this knowledge that we so badly need.
Edit: An earlier version of this post incorrectly stated that the Center for the Study of Religion at UC Berkeley was solely concerned with public theology.


text:
\section{Introduction}
\label{section:introduction}

The {\em mapping class group} of a closed oriented $3$-manifold $M^3$, denoted $\Mod(M^3)$, is the
group of isotopy classes of orientation-preserving diffeomorphisms of $M^3$.  In this paper,
we study the mapping class group of the connect sum $M_n$ of $n$ copies of $S^2 \times S^1$.
The fundamental group $\pi_1(M_n)$
is the free group $F_n$ on $n$ letters.  Since
diffeomorphisms of $M_n$ do not fix a basepoint, the action of $\Mod(M_n)$ on $\pi_1(M_n)$ is only
defined up to conjugation, so it gives a homomorphism
\[\rho\colon \Mod(M_n) \longrightarrow \Out(\pi_1(M)) \cong \Out(F_n).\]
A classical theorem of Laudenbach \cite{LaudenbachSpheres, LaudenbachBook} says that $\rho$ is surjective
but not injective.

\paragraph{Sphere twists.}
The kernel of $\rho$ is the subgroup generated by sphere twists, which are defined as follows.  Let
$M^3$ be a closed oriented $3$-manifold and let
$S \subset M^3$ be a smoothly embedded $2$-sphere.  Fix a tubular neighborhood $U \cong S \times [0,1]$
of $S$.  Recall that $\pi_1(\SO(3),\text{id}) \cong \Z/2$ is generated by a loop $\ell\colon [0,1] \rightarrow \SO(3)$
that rotates $\R^3$ about an axis by one full turn.  Identifying $S$ with $S^2 \subset \R^3$, the {\em sphere twist}
about $S$, denoted $T_S$, is the isotopy class of the diffeomorphism $\tau\colon M^3 \rightarrow M^3$ that
is the identity outside $U$ and on $U \cong S \times [0,1]$ takes the form $\tau(s,t) = (\ell(t) \cdot s,t)$.
The isotopy class of $T_S$ only depends on the isotopy class of $S$.  In fact,
Laudenbach \cite{LaudenbachSpheres, LaudenbachBook}
proved that if $S$ and $S'$ are homotopic $2$-spheres in $M^3$ that are non-nullhomotopic, then
$S$ and $S'$ are isotopic, so $T_S$ actually only depends on the homotopy class of $S$.

\paragraph{Actions of sphere twists.}
Since $\pi_1(\SO(3),\text{id}) \cong \Z/2$, the mapping class $T_S$ has order at most $2$.  It
follows from Laudenbach's work that in the case $M^3 = M_n$, the sphere twist $T_S$ is trivial if
$S$ separates $M_n$ and has order
$2$ if $S$ is nonseparating.  Showing that $T_S$ is ever nontrivial is quite subtle since $T_S$ fixes
the homotopy class of any loop or surface $Z$ in $M^3$, and thus cannot be detected by
most basic algebro-topological invariants.  To see this, let $U \cong S \times [0,1]$ be the tubular
neighborhood used to construct $T_S$ and let $p_0 \in S$ be one of the two points of $S$ lying on the axis
of rotation used to construct $T_S$.  Homotope $Z$ to be transverse to $S$.
The intersection $Z \cap S$ is then either a collection
of circles (if $\dim(Z)=2$) or points (if $\dim(Z)=1$).  As is shown in Figure \ref{figure:gather}, we can
then homotope $Z$ such that $Z \cap U \subset p_0 \times [0,1]$, so $T_S$ fixes $Z$.

\Figure{figure:gather}{Gather}{On the left is $S$ along with $p_0$ and $Z \cap S$ (which here
is $1$-dimensional, so $\dim(Z)=2$).  On the right we show how to homotope $Z$ such
that $Z \cap S = \{p_0\}$ -- choose a point $q \in S \setminus (\{p_0\} \cup Z)$, and homotope
$Z$ so as to push its intersection with $S$ along paths from $q$ to $p_0$ until it is entirely
contained in $p_0$.}{95}

\paragraph{Twist subgroup.}
The {\em twist subgroup} of $\Mod(M^3)$, denoted $\Tw(M^3)$, is the subgroup generated by all sphere twists.
For $f \in \Mod(M^3)$ and a sphere twist $T_S \in \Tw(M^3)$, we have
\[f T_S f^{-1} = T_{f(S)}.\]
This implies that $\Tw(M^3)$ is a normal subgroup of $\Mod(M^3)$.  Also, if $T_S$ and $T_{S'}$ are sphere twists,
we saw above that $T_{S'}(S)$ is homotopic to $S$.  Since a sphere twist only depends on the homotopy class
of the sphere along which we are twisting, setting $f = T_{S'}$ in the above relation we get
\[T_{S'} T_S T_{S'}^{-1} = T_S.\]
In other words, $\Tw(M^3)$ is abelian.

\paragraph{Laudenbach sequence.}
Laudenbach's theorem can thus be summarized as saying that there is a short exact sequence
\begin{equation}
\label{eqn:laudenbachseq}
1 \longrightarrow \Tw(M_n) \longrightarrow \Mod(M_n) \stackrel{\rho}{\longrightarrow} \Out(F_n) \longrightarrow 1.
\end{equation}
He also proved that $\Tw(M_n) \cong (\Z/2)^n$ and is generated by the sphere twists about the core
spheres $S^2 \times \ast$ of the $n$ different $S^2 \times S^1$ summands of $M_n$.
This theorem raises two natural questions:
\begin{compactenum}
\item Does the extension \eqref{eqn:laudenbachseq} split?
\item The conjugation action of $\Mod(M_n)$ on its normal abelian group $\Tw(M_n) \cong (\Z/2)^n$ induces
an action of $\Out(F_n)$ on $\Tw(M_n) \cong (\Z/2)^n$.  What action is this?
\end{compactenum}

\paragraph{Main theorem.}
Our main theorem answers both of these questions.  It says that the extension \eqref{eqn:laudenbachseq}
does split, and in fact the image of the splitting $\Out(F_n) \rightarrow \Mod(M_n)$ has a simple geometric
description: it is the stabilizer of the homotopy class of a trivialization of the tangent bundle of $M_n$.
A precise statement is as follows.

\begin{maintheorem}
\label{maintheorem:split}
Let $[\sigma_0]$ be the homotopy class of a trivialization $\sigma_0$ of the tangent bundle of $M_n$
and let $(\Mod(M_n))_{[\sigma_0]}$ be the $\Mod(M_n)$-stabilizer of $[\sigma_0]$.  The following then hold:
\begin{compactitem}
\item $\Mod(M_n) = \Tw(M_n) \rtimes (\Mod(M_n))_{[\sigma_0]}$.
\item $\Tw(M_n) \cong \HH^1(M_n;\Z/2)$ as a $\Mod(M_n)$-module.
\item $(\Mod(M_n))_{[\sigma_0]} \cong \Out(F_n)$.
\end{compactitem}
\end{maintheorem}

\begin{remark}
Before Laudenbach's work, Gluck \cite{GluckOriginal} proved that $\Mod(M_1) \cong \Z/2 \times \Z/2$.  The first factor
was a sphere twist, and the second factor was $\Out(F_1) = \Z/2$.  This is of course a special case of
Theorem \ref{maintheorem:split}.
\end{remark}

\paragraph{Sphere complex and $\Out(F_n)$.}
Laudenbach's exact sequence \eqref{eqn:laudenbachseq} plays an important role in the study
of $\Out(F_n)$.  In his seminal paper \cite{HatcherOriginal}, Hatcher defined the {\em sphere complex}
$\Sphere_n$ to be the simplicial complex whose $k$-simplices are sets $\{S_0,\ldots,S_k\}$ of
isotopy classes of non-nullhomotopic smoothly embedded $2$-spheres in $M_n$ that can be realized
disjointly.  One of his main theorems says that $\Sphere_n$ is contractible.  The group $\Mod(M_n)$
acts on $\Sphere_n$, and since sphere twists fix the isotopy class of any smoothly embedded $2$-sphere
the twist subgroup $\Tw(M_n)$ acts trivially.  By \eqref{eqn:laudenbachseq}, we thus get
an action of $\Out(F_n)$ on $\Sphere_n$.

The space $\Sphere_n$ is also sometimes called the {\em free splitting complex}
and has played an important role in a huge amount of subsequent work (see, e.g.,
\cite{AramayonaSouto, ClayQingRafi, HamenstadtHensel, HandelMosher, HatcherVogtmann, HilionHorbez, Horbez, KapovichLustig}).
It is unsatisfying that the original construction of the action of $\Out(F_n)$ on
$\Sphere_n$ was so indirect: you first construct an action of $\Mod(M_n)$, then
notice that $\Tw(M_n)$ acts trivially, and only then get an induced action
of the quotient group $\Out(F_n) = \Mod(M_n)/\Tw(M_n)$.  It follows from
Theorem \ref{maintheorem:split} that $\Out(F_n)$ can be embedded as a subgroup
of $\Mod(M_n)$, so there is no longer a need to perform this indirect
construction.

\paragraph{Nontriviality of sphere twists.}
Our proof also gives a new and easier argument for seeing that the sphere twists $T_S$ about nonseparating
spheres in $M_n$ (and other $3$-manifolds) are nontrivial (cf.\ Corollary~\ref{corollary:identifytwist}).  This is not as easy as one might expect.  The usual
way that one studies a group like $\Mod(M_n)$ is via its action on homotopy classes of submanifolds of $M_n$.  However, the
sphere twists $T_S$ fix the homotopy classes of all loops and surfaces in $M_n$, so a new idea is needed.  Laudenbach used
an argument involving framed cobordism and the Pontryagin--Thom construction, while we study
the action of $\Mod(M_n)$ on trivializations of the tangent bundle.

\begin{remark}
The idea of using the Pontryagin--Thom construction to study sphere-twists in $3$-manifolds goes
back to early work of Pontryagin; see the example at the end of \cite[\S 4]{PontryaginThree}.
\end{remark}

\paragraph{Automorphisms vs outer automorphisms.}
Let $M_{n,1}$ be $M_n$ equipped with a basepoint $\ast \in M_{n}$, and define $\Mod(M_{n,1})$ to be
the group of isotopy classes of orientation-preserving diffeomorphisms of $M_{n,1}$ that fix $\ast$.
The group $\Mod(M_{n,1})$ then acts on $\pi_1(M_{n,1},\ast) = F_n$, so we get a homomorphism
$\Mod(M_{n,1}) \rightarrow \Aut(F_n)$.  It follows from Laudenbach's work that we also have a short
exact sequence
\[1 \longrightarrow \Tw(M_{n,1}) \longrightarrow \Mod(M_{n,1}) \longrightarrow \Aut(F_n) \longrightarrow 1\]
and that $\Tw(M_{n,1}) = (\Z/2)^n$.  Our work shows that this sequence also splits, and a result
identical to Theorem \ref{maintheorem:split} holds.  This can be proved either by adapting our proof (which
needs almost no changes), or by using the exact sequence
\begin{equation}
\label{eqn:birman}
1 \longrightarrow F_n \longrightarrow \Mod(M_{n,1}) \longrightarrow \Mod(M_n) \longrightarrow 1
\end{equation}
arising from the long exact sequence in homotopy groups of the fiber bundle
\[\Diff^{+}(M_n,\ast) \longrightarrow \Diff^{+}(M_n) \longrightarrow M_n.\]
The $F_n$ in \eqref{eqn:birman} is the image of $\pi_1(M_n)$ in $\Mod(M_{n,1}) = \pi_0(\Diff^{+}(M_n,\ast))$ and
maps to the inner automorphisms in $\Aut(F_n)$.  We leave the details to the interested reader.

\paragraph{Other $3$-manifolds.}
Let $M^3$ be an arbitrary closed orientable $3$-manifold and let $\pi = \pi_1(M^3)$.  The twist subgroup $\Tw(M^3)$ is still
an abelian normal subgroup of the mapping class group $\Mod(M^3)$, and it turns out that
there is still an exact sequence
\begin{equation}
\label{eqn:generalsequence}
1 \longrightarrow \Tw(M^3) \longrightarrow \Mod(M^3) \longrightarrow G \longrightarrow 1,
\end{equation}
where $G < \Out(\pi)$ is the image of $\Mod(M^3)$ in $\Out(\pi)$.  See \cite[Proposition 2.1]{HatcherWahl}
for how to extract this from the literature; in fact, this reference also gives an appropriate
but more complicated statement for $3$-manifolds with boundary.  It is often the case that
$G = \Out(\pi)$; see \cite[Proposition 2.2]{HatcherWahl} for some conditions that ensure this.

In light of Theorem \ref{maintheorem:split}, it is natural to wonder whether \eqref{eqn:generalsequence}
splits.  Let $\TwNosep(M^3)$ be the subgroup of $\Tw(M^3)$ generated by sphere twists about
nonseparating spheres. Our proof of Theorem \ref{maintheorem:split} can be generalized to show
that we have a splitting
\[\Mod(M^3) = \TwNosep(M^3) \rtimes \Gamma,\]
where $\Gamma$ is a subgroup of $\Mod(M^3)$.   However, we cannot show that $\Gamma$ can be taken
to be the stabilizer of the homotopy class of a trivialization of $T M^3$ (c.f.\ the
proof of Theorem \ref{theorem:weakmain} below).

Unfortunately, we do not know how to deal
with separating sphere twists.  For $M^3 = M_n$, separating sphere twists are always trivial,
so $\TwNosep(M^3) = \Tw(M^3)$ and separating twists can be ignored when studying
\eqref{eqn:generalsequence}.  However, for general $3$-manifolds the situation is very complicated.  Let $S \subset M^3$
be a $2$-sphere that separates $M^3$.
\begin{itemize}
\item In \cite{Hendriks}, Hendriks gives a remarkable characterization of when $T_S$
is homotopic to the identity.  Namely, $T_S$ is homotopic to the identity if and only if for one
of the two components $N$ of cutting $M^3$ open along $S$ the following strange condition holds.
Let $P$ be a prime summand of the result of gluing a closed $3$-ball to $\partial N = S^2$.
Then either $P = S^2 \times S^1$, or $P$ has a finite fundamental group whose Sylow $2$-subgroup
is cyclic.
\item However, this is not the whole story.  In \cite{FriedmanWitt}, Friedman--Witt show that in
some cases the separating sphere twists that Hendriks showed were homotopic to the identity
are not {\em isotopic} to the identity, and thus still define nontrivial elements of $\Mod(M^3)$.
\end{itemize}
What happens in the general case is unclear.  See \cite[Remark 2.4]{HatcherWahl} for some
further discussion of it.

\paragraph{Outline.}
The outline of our paper is as follows.  We start in \S \ref{section:extension} by constructing the
exact sequence \eqref{eqn:laudenbachseq}.  To make our paper self-contained, we give a mostly complete proof
of this, simplifying some details from Laudenbach's original paper.  We then have a preliminary
algebraic section \S \ref{section:crossed} on crossed homomorphisms and their relationship
to splitting exact sequences.  We then construct the crossed homomorphisms we need in
\S \ref{section:derivative} and \S \ref{section:twisting} before closing with
\S \ref{section:completing}, which takes care of a few final details.

\paragraph{Notational conventions.}
It will be important for us to distinguish between a diffeomorphism and its isotopy class.  For $f \in \Diff^{+}(M_n)$ we will write $[f] \in \Mod(M_n)$ for the isotopy class of $f$.  More generally, we will use square brackets frequently to indicate that something is being taken up to homotopy/isotopy, though we will try to be explicit about this whenever it might be confusing to the reader.

\paragraph{Acknowledgments.}
We would like to thank Jake Landgraf for helpful conversations and Allen Hatcher for some useful
references, historical remarks, and corrections.  We would also like to thank Oscar Randal-Williams
for pointing out some related work of Crowley \cite{Crowley}, Krannich \cite{Krannich}, and Krylov \cite{Krylov}
on mapping class groups of high-dimensional manifolds (see Remark \ref{remark:high} below). We additionally
thank Diarmuid Crowley for referring us to work of Baues \cite{Baues} on automorphisms of simply
connected connect sums of $S^p \times S^q$.  Finally, we would like to thank Dan Margalit and Richard Wade for some
corrections.

\section{Constructing the extension}
\label{section:extension}

This preliminary section discusses some aspects of Laudenbach's work we will need for our proof.

\paragraph{What is needed.}
Recall from the introduction that Laudenbach \cite{LaudenbachSpheres} proved that
$\Mod(M_n)$ is an extension of $\Out(F_n)$ by $\Tw(M_n)$ and that $\Tw(M_n) \cong (\Z/2)^n$.
Theorem \ref{maintheorem:split} strengthens this and its proof will give as a byproduct that
$\Tw(M_n) \cong (\Z/2)^n$, but it will depend on the following piece of Laudenbach's work.

\begin{theorem}[Laudenbach, \cite{LaudenbachSpheres, LaudenbachBook}]
\label{theorem:laudenbach}
The map
\[\rho\colon \Mod(M_n) \rightarrow \Out(\pi_1(M_n)) = \Out(F_n)\]
is surjective with kernel $\ker(\rho) = \Tw(M_n)$.  Also, $\Tw(M_n)$ is generated
by the sphere twists about the core spheres $S^2 \times \ast$ of the $n$ summands of $S^2 \times S^1$ in $M_n$.
\end{theorem}

\begin{remark}
\label{remark:laudenbachnontrivial}
Theorem \ref{theorem:laudenbach} does {\em not} assert that the indicated sphere twists
are nontrivial.  As we discussed in the introduction, Laudenbach proved that they are and
that $\Tw(M_n) \cong (\Z/2)^n$, but we will only establish this (in a stronger form) later; see
Corollary \ref{corollary:identifytwist}.
\end{remark}

To make this paper more self-contained, this section contains a mostly complete sketch of a proof of
Theorem~\ref{theorem:laudenbach}.  We will follow the outline of Laudenbach's original proof,
but we will simplify one key step (see Theorem \ref{theorem:pi2} below).

\paragraph{Homotopy vs isotopy for spheres.}
Our proof of Theorem \ref{theorem:laudenbach} will depend on three preliminary results.  The first
is the following:

\begin{theorem}[Laudenbach, \cite{LaudenbachSpheres, LaudenbachBook}]
\label{theorem:sphereisotopy}
Let $M^3$ be a closed oriented $3$-manifold and let $\iota,\iota'\colon \sqcup_{i=1}^k S^2 \rightarrow M^3$ be
homotopic embeddings of disjoint smoothly embedded spheres.  Assume that none of the components of
the images of $\iota$ or $\iota'$ are nullhomotopic.  Then $\iota$ and $\iota'$ are ambient isotopic.
\end{theorem}

We omit the proof of Theorem \ref{theorem:sphereisotopy} since it is lengthy and its details do not
shed much light on our work.

\paragraph{Action on second homotopy group.}
The second preliminary result is the following theorem of
Laudenbach.  Our proof is much shorter than his proof.
We remark that Hatcher-Wahl have given a different (but somewhat
longer) simplified proof in \cite[Appendix]{HatcherWahl}.

\begin{theorem}[Laudenbach]
\label{theorem:pi2}
Let $M^3$ be a closed oriented $3$-manifold equipped with a basepoint $x_0$
and let $f\colon (M^3,x_0) \rightarrow (M^3,x_0)$ be a basepoint-preserving diffeomorphism
such that $f_{\ast}\colon \pi_1(M^3,x_0) \rightarrow \pi_1(M^3,x_0)$ is the identity.  Then
$f_{\ast}\colon \pi_2(M^3,x_0) \rightarrow \pi_2(M^3,x_0)$ is the identity.
\end{theorem}
\begin{proof}
Let $(\tM^3,\tx_0) \rightarrow (M^3,x_0)$ be the universal cover of $(M^3,x_0)$.  Let
$\tf\colon (\tM^3,\tx_0) \rightarrow (\tM^3,\tx_0)$ be the lift of $f$.  To prove that
$f$ acts trivially on $\pi_2(M^3,x_0)$, it is enough to prove that $\tf$ acts trivially
on $\pi_2(\tM^3,\tx_0) = \HH_2(\tM^3)$.  By Poincar\'{e} duality, we have
\begin{equation}
\label{eqn:identifyh2}
\HH_2(\tM^3) \cong \HH^1_c(\tM^3) = \lim_{\substack{\rightarrow\\ K}} \HH^1(\tM^3,\tM^3 \setminus K) = \lim_{\substack{\rightarrow\\ K}} \RH^0(\tM^3 \setminus K).
\end{equation}
Here the limit is over compact subspaces $K$ of $\tM^3$ and the final equality comes from the long exact sequence of the
pair $(\tM^3,\tM^3 \setminus K)$ and the fact that $\tM^3$ is $1$-connected.  Elements of $\RH^0(\tM^3 \setminus K)$ can
be interpreted as locally constant functions
$\kappa\colon \tM^3 \setminus K \rightarrow \Z$ modulo the globally constant functions.  Fix such a
$\kappa\colon \tM^3 \setminus K \rightarrow \Z$, and let $K'$ be a compact subspace of $\tM^3$ containing
$K \cup \tf(K)$ such that no components of $\tM^3 \setminus K'$ are bounded (i.e.\ have compact closure).
The image under the homeomorphism $\tf$ of the element of
$\HH_2(\tM^3)$ represented by $\kappa$ under \eqref{eqn:identifyh2} is represented by the function
\[\kappa \circ \tf^{-1}\colon \tM^3 \setminus K' \rightarrow \Z.\]
We must prove that $\kappa = \kappa \circ \tf^{-1}$ on $\tM^3 \setminus K'$.  The key observation
is that since $f$ acts trivially on $\pi_1(M^3,x_0)$, the lift $\tf$ fixes each point in the
$\pi_1(M^3,x_0)$-orbit of the basepoint $\tx_0$.  This orbit will contain points in each component
of $\tM^3 \setminus K'$, so $\kappa$ and $\kappa \circ \tf^{-1}$ agree on at least
one point in each component of $\tM^3 \setminus K'$.  Since they are locally constant,
we conclude that they are equal everywhere on $\tM^3 \setminus K'$, as desired.
\end{proof}

\paragraph{Mapping class groups of punctured spheres.}
The third and final preliminary result we need is as follows.  For a $3$-manifold $M^3$ with boundary, we define
$\Mod(M^3)$ to be $\pi_0(\Diff^{+}(M^3,\partial M^3))$, i.e.\ the group of isotopy classes of
orientation-preserving diffeomorphisms of $M^3$ that fix $\partial M^3$ pointwise.

\begin{lemma}
\label{lemma:spheretwists}
Let $X$ be the $3$-manifold with boundary obtained by removing $k$ disjoint open balls from $S^3$.  Then
$\Mod(X)$ is generated by sphere twists about embedded spheres that are parallel to components of $\partial X$.
\end{lemma}

\begin{remark}
Just like Theorem \ref{theorem:laudenbach}, Lemma \ref{lemma:spheretwists} does {\em not} assert
that these sphere twists are nontrivial in the mapping class group.  In fact, one can show that they
are trivial if $k=1$ and nontrivial if $k \geq 2$, and the twist subgroup
of the manifold $X$ in Lemma \ref{lemma:spheretwists} is isomorphic to $(\Z/2)^{k-1}$.  Here $(k-1)$
appears instead of $k$ since the product of all the boundary twists is trivial;
see \cite[p.\ 214--215]{HatcherWahl}.
We will not need any of this, so we will not prove it.
\end{remark}

\begin{proof}[Proof of Lemma \ref{lemma:spheretwists}]
The proof will be by induction on $k$.  The base case $k=0$ simply asserts that $\Mod(S^3) = 1$, which is
a theorem of Cerf \cite{CerfConnected}.  We remark that even more is true: the $3$-dimensional Smale Conjecture
proved by Hatcher \cite{HatcherSmale} says that $\Diff^{+}(S^3)$ is homotopy equivalent to $\SO(4)$.  Assume
now that $k>0$ and that the lemma is true for smaller $k$.  Let $X'$ be the result of gluing a closed $3$-ball
$B$ to a component of $\partial X$.  We thus have $\Diff^{+}(X,\partial X) = \Diff^{+}(X',\partial X' \sqcup B)$.
There is a fiber bundle
\[\Diff^{+}(X',\partial X' \sqcup B) \rightarrow \Diff^{+}(X',\partial X') \rightarrow \Emb^{+}(B,X').\]
Identifying $\Mod(X)$ and $\Mod(X')$ with $\pi_0$ of the relevant diffeomorphism groups,
the associated long exact sequence in homotopy groups contains the segment
\begin{equation}
\label{eqn:inductiveemb}
\pi_1(\Emb^{+}(B,X')) \longrightarrow \Mod(X) \longrightarrow \Mod(X') \longrightarrow \pi_0(\Emb^{+}(B,X')).
\end{equation}
Fix oriented trivializations of the tangent bundles of $B$ and $X'$.  For an orientation-preserving
embedding $\iota\colon B \rightarrow X'$, these
trivializations allow us to identify the derivative $D_0 \iota\colon T_0 B \rightarrow T_{\iota(0)} X'$ with
a matrix in the subgroup $\GLp_3(\R)$ of $\GL_3(\R)$ consisting of matrices whose determinant is positive.  The map $\Emb^{+}(B,X') \rightarrow X' \times \GLp_3(\R)$ taking
an embedding $\iota\colon \rightarrow X'$ to $(\iota(0), D_0 \iota)$ is a homotopy equivalence, so
\[\pi_0(\Emb^{+}(B,X')) = \pi_0(X' \times \GLp_3(\R)) = 0\]
and
\[\pi_1(\Emb^{+}(B,X')) = \pi_1(X' \times \GLp_3(\R)) = \pi_1(\GLp_3(\R)) = \Z/2.\]
Plugging these into \eqref{eqn:inductiveemb}, we get an exact sequence
\[\Z/2 \longrightarrow \Mod(X) \longrightarrow \Mod(X') \longrightarrow 0.\]
The image of $\Z/2$ in $\Mod(X)$ is a sphere twist about a sphere parallel to a component of $\partial X$, and
by induction $\Mod(X')$ is generated by sphere twists about spheres parallel to components of $\partial X'$.  The
lemma follows.
\end{proof}

\paragraph{The proof.}
We now have all the ingredients needed for the proof of Theorem \ref{theorem:laudenbach} above.

\begin{proof}[Proof of Theorem \ref{theorem:laudenbach}]
Recall that
\[\rho\colon \Mod(M_n) \rightarrow \Out(\pi_1(M_n)) = \Out(F_n)\]
is the natural map.  We must prove the following two facts.

\begin{claims}
The map $\rho$ is surjective.
\end{claims}

\Figure{figure:glueup}{GlueUp}{On the left hand side, $M_3$ is obtained by gluing the $6$ boundary components
of $X$ together in pairs as indicated.  The generators for $\pi_1(M_3) = F_3$ are $\{a_1,a_2,a_3\}$.  In the
middle, we indicate a diffeomorphism $\phi\colon M_3 \rightarrow M_3$ that drags one boundary sphere of $X$ along
a closed path.  As is shown on the right, on $\pi_1(M_3)$ the diffeomorphism $\phi$ takes $a_1$ to $a_1 a_2$.}{95}

Let $X$ be the result of removing $2n$ disjoint open balls from $S^3$.  As in
Figure \ref{figure:glueup}, we can identify $M_n$ with the result of gluing the boundary
component of $X$ together in pairs.  Let $\{a_1,\ldots,a_n\} \in \pi_1(M_n) = F_n$ be
the generators indicated in Figure \ref{figure:glueup}.  It is standard that the group $\Aut(F_n)$ is generated
by the following elements:
\begin{compactitem}
\item For distinct $1 \leq i,j \leq n$, elements $L_{ij}$ and $R_{ij}$ defined via the formulas
\[L_{ij}(a_k) = \begin{cases}
a_j a_k & \text{if $k=i$}\\
a_k & \text{if $k \neq i$}
\end{cases}
\quad \text{and} \quad
R_{ij}(a_k) = \begin{cases}
a_k a_j & \text{if $k=i$}\\
a_k & \text{if $k \neq i$}
\end{cases}
\quad \quad (1 \leq k \leq n).\]
\item For $1 \leq i \leq n$, elements $I_i$ defined via the formula
\[I_i(a_j) = \begin{cases}
a_j^{-1} & \text{if $k=i$},\\
a_j & \text{if $k \neq i$}.
\end{cases}\]
\end{compactitem}
It is enough to find elements of $\Diff^{+}(M_n)$ realizing these automorphisms.  This is an easy exercise; for
instance, as we show in Figure \ref{figure:glueup} we can realize $R_{12}$ as a diffeomorphism that drags
one boundary sphere of $X$ through another.

\begin{claims}
The kernel of $\rho$ is the twist subgroup $\Tw(M_n)$, and $\Tw(M_n)$ is generated
by the sphere twists about the core spheres of the $n$ summands of $S^2 \times S^1$ in $M_n$.
\end{claims}

Clearly $\Tw(M_n) \subset \ker(\rho)$, so it is enough to prove that every element of $\ker(\rho)$ is a
product of sphere twists about the core spheres of the $S^2 \times S^1$ summands of $M_n$.  Consider
some $[f] \in \ker(\rho)$, and let $\iota\colon \sqcup_{i=1}^n S^2 \rightarrow M_n$ be the embedding
of those core spheres.  Fix a basepoint $x_0 \in M_n$.
Isotoping $f$, we can assume that $f(x_0) = x_0$ and that $f_{\ast}\colon \pi_1(M_n,x_0) \rightarrow \pi_1(M_n,x_0)$
is the identity.  Theorem \ref{theorem:pi2} then implies that $f$ also induces the identity on $\pi_2(M_n,x_0)$.
It follows that $\iota$ is homotopic to $f \circ \iota$, so by Theorem \ref{theorem:sphereisotopy} we can
isotope $f$ such that $\iota = f \circ \iota$.  Let $X$ be the result of cutting $M_n$ open along
the image of $\iota$, so $X$ is diffeomorphic to the result of removing $2n$ open balls from $S^3$.  Since
$\iota = f \circ \iota$, the mapping class $[f]$ is in the image of the homomorphism $\Mod(X) \rightarrow \Mod(M_n)$
that glues the boundary components back together.  Lemma \ref{lemma:spheretwists} says that $\Mod(X)$ is generated
by sphere twists about spheres parallel to its boundary components.  These map to the $[T_{S_i}]$ in $\Mod(M_n)$, and
the desired result follows.
\end{proof}

\section{Crossed homomorphisms and exact sequences}
\label{section:crossed}

As preparation for proving Theorem \ref{maintheorem:split}, this section reviews the connection between crossed homomorphisms and split exact sequences.
Let $G$ and $H$ be groups such that $G$ acts on $H$ on the right.  We will write this action
using superscripts: for $g \in G$ and $h \in H$, the image of $h$ under $g$ will be denoted
$h^g$.
A {\em crossed homomorphism} from $G$ to $H$ is a set map $\lambda\colon G \rightarrow H$
such that
\[\lambda(g_1 g_2) = \lambda(g_1)^{g_2} \lambda(g_2).\]
This implies in particular that
\[\lambda(1) = \lambda(1^2) = \lambda(1)^1 \lambda(1) = \lambda(1)^2,\]
so $\lambda(1) = 1$.
If the action of $G$ on $H$ is trivial, then this reduces to the definition of a homomorphism.
Just like for an ordinary homomorphism, the kernel $\ker(\lambda) = \Set{$g \in G$}{$\lambda(g)=1$}$
is a subgroup of $G$; however, it is not necessarily a normal subgroup.

As the following standard lemma shows, these are closely related to splittings of short
exact sequences.

\begin{lemma}
\label{lemma:crossedsemidirect}
Let $G$ be a group and let $A \lhd G$ be an abelian normal subgroup, so $G$ acts on $A$ on the right via the formula
\[a^g = g^{-1} a g \quad \quad (a \in A, g \in G).\]
Letting $Q = G/A$, the short exact sequence
\[1 \longrightarrow A \longrightarrow G \longrightarrow Q \longrightarrow 1\]
splits if and only if there exists a crossed homomorphism $\lambda\colon G \rightarrow A$ that restricts to the identity on $A$.
Moreover, if such a $\lambda$ exists, then we can choose a splitting $Q \rightarrow G$ whose image is $\ker(\lambda)$, so
$G = A \rtimes \ker(\lambda)$.
\end{lemma}
\begin{proof}
If the exact sequence splits, then there exists a subgroup $\oQ$ of $G$ projecting isomorphically to $Q$, so we can uniquely
write all $g \in G$ as $g = q a$ with $q \in \oQ$ and $a \in A$.  This allows us to define a set map
$\lambda\colon G \rightarrow A$ via the formula
\[\lambda(q a) = a \quad \quad (q \in \oQ, a \in A).\]
This restricts to the identity on $A$, and is a crossed homomorphism since for $q_1,q_2 \in \oQ$ and $a_1,a_2 \in A$ we have
\begin{align*}
\lambda(q_1 a_1 q_2 a_2) &= \lambda(q_1 q_2 a_1^{q_2} a_2) = \lambda(q_1 q_2 a_1^{q_2 a_2} a_2) = a_1^{q_2 a_2} a_2 = \lambda(q_1 a_1)^{q_2 a_2} \lambda(q_2 a_2).
\end{align*}
We remark that the second equality is where we use the fact that $A$ is abelian.

Conversely, assume that there exists a crossed homomorphism $\lambda\colon G \rightarrow A$ that restricts to the identity on $A$.
Define $\oQ = \ker(\lambda)$, so $\oQ < G$ satisfies $\oQ \cap A = 1$.  To prove the theorem,
we must prove that the surjection
$\pi\colon G \rightarrow Q$ restricts to an isomorphism $\pi\colon \oQ \rightarrow Q$.  Since $\oQ \cap A = 1$, the projection
$\pi\colon \oQ \rightarrow Q$ is injective, so we must only prove that it is surjective.  Consider $q \in Q$.  We
can find some $g \in G$ such that $\pi(g) = q$.  Since $\lambda(g^{-1}) \in A$, we have
$\pi(\lambda(g^{-1}) g) = \pi(g) = q$, so it is enough to prove that $\lambda(g^{-1}) g \in \oQ = \ker(\lambda)$.  For
this, we compute
\[\lambda(\lambda(g^{-1}) g) = \lambda(\lambda(g^{-1}))^g \lambda(g) = \lambda(g^{-1})^g \lambda(g) = \lambda(g^{-1} g) = 1.\]
Here the second inequality uses the fact that $\lambda$ restricts to the identity on $A$.
\end{proof}

\section{The derivative crossed homomorphism}
\label{section:derivative}

By Lemma \ref{lemma:crossedsemidirect}, to prove that the exact sequence
\[1 \longrightarrow \Tw(M_n) \longrightarrow \Mod(M_n) \longrightarrow \Out(F_n) \longrightarrow 1\]
splits, we must construct a crossed homomorphism $\Mod(M_n) \rightarrow \Tw(M_n)$ that restricts to the
identity on $\Tw(M_n)$.  We will do this in two steps, the first in this section and second in the next.

\begin{remark}
\label{remark:high}
We construct our crossed homomorphism using the action of $\Mod(M_n)$ on the set of homotopy classes
of trivializations of $TM^3$.  We would like to thank Oscar Randal-Williams for pointing out that
a similar idea has been used to study the mapping class groups of high-dimensional manifolds.  Its
earliest appearance seems to be in a paper of Krylov \cite{Krylov}, and these ideas were later developed by
Crowley \cite{Crowley} and by Krannich \cite{Krannich}.
\end{remark}

\paragraph{Frame bundle.}
What we do in this section works in complete generality, so let $M^3$ be any closed oriented $3$-manifold.
Let $TM^3$ be the tangent bundle of $M^3$ and
let $\Fr(TM^3)$ be the principle $\GLp_3(\R)$-bundle of oriented frames of $TM^3$.  Here recall
that $\GLp_3(\R)$ is the subgroup of $\GL_3(\R)$ consisting of matrices whose determinant is positive.
The points of $\Fr(TM^3)$
thus consist of orientation-preserving linear isomorphisms $\tau\colon \R^3 \rightarrow T_p M^3$, where
$p \in M^3$ is a point.  The group $\GLp_3(\R)$ act on $\Fr(TM^3)$ on the right in the usual way: regarding
elements of $\GLp_3(\R)$ as isomorphisms $\R^3 \rightarrow \R^3$, we have
\[\tau \cdot M = \tau \circ M \quad \text{for $\tau\colon \R^3 \rightarrow T_p M^3$ in $\Fr(TM^3)$ and $M \in \GLp_3(\R)$}.\]
This action preserves the fibers of the projection $\Fr(TM^3) \rightarrow M^3$, and its restriction
to each fiber is simply transitive.

\paragraph{Trivializations.}
Since $M^3$ is oriented, its tangent bundle $TM^3$ is trivial.  An {\em oriented trivialization} of $TM^3$
is a section $\sigma\colon M^3 \rightarrow \Fr(TM^3)$ of the bundle $\Fr(TM^3) \rightarrow M^3$.
Let $\Triv(M^3)$ be the set of oriented trivializations of $TM^3$ and let
$C(M^3,\GLp_3(\R))$ be the space of continuous maps $M^3 \rightarrow \GLp_3(\R)$.  The group structure
of $\GLp_3(\R)$ endows $C(M^3,\GLp_3(\R))$ with the structure of a topological group, and $C(M^3,\GLp_3(\R))$
acts continuously on $\Triv(M^3)$ on the right via the formula
\begin{align*}
\sigma \cdot \phi = \Big(p \mapsto \sigma(p) \cdot \phi(p)\Big) \quad &\text{for
$\sigma\colon M^3 \rightarrow \Fr(TM^3)$ in $\Triv(M^3)$}\\
&\quad\quad\quad\text{and $\phi\colon M^3 \rightarrow \GLp_3(\R)$ in $C(M^3,\GLp_3(\R))$.}
\end{align*}
The action is also simply transitive.

\paragraph{Diffeomorphism actions.}
For $f \in \Diff^{+}(M^3)$, the derivative $Df$ of $f$ induces a map
$(Df)_{\ast}\colon \Fr(TM^3) \rightarrow \Fr(TM^3)$ defined via the formula
\[(Df)_{\ast}(\sigma) = \Big(\left(D_p f\right) \circ \sigma\colon \R^3 \rightarrow T_{f(p)} M^3\Big) \quad
\text{for $\sigma\colon \R^3 \rightarrow T_p M^3$ in $\Fr(TM^3)$}.\]
Using this, we can define a right action of $\Diff^{+}(M^3)$ on $\Triv(M^3)$ via the following formula, where
we use superscripts to avoid confusing this action with the above action of $C(M^3,\GLp_3(\R))$:
\[\sigma^f = (Df^{-1})_{\ast} \circ \sigma \circ f \quad \text{for $f \in \Diff^{+}(M^3)$ and $\sigma\colon M^3 \rightarrow \Fr(TM^3)$ in $\Triv(M^3)$.}\]
The group $\Diff^{+}(M^3)$ also has a right action on $C(M^3,\GLp_3(\R))$ defined via the formula
\[\phi^f = \phi \circ f \quad \text{for $f \in \Diff^{+}(M^3)$ and $\phi\colon M^3 \rightarrow \GLp_3(\R)$ in $C(M^3,\GLp_3(\R))$.}\]
These three different actions are related by the formula
\begin{align*}
\left(\sigma \cdot \phi\right)^f = \sigma^f \cdot \phi^f \quad &\text{for $f \in \Diff^{+}(M^3)$ and $\phi \in C(M^3,\GLp_3(\R))$ and $\sigma \in \Triv(M^3)$.}
\end{align*}
Both sides of this formula are the element of $\Triv(M^3)$ whose value at a point $p \in M^3$ is the linear
isomorphism $\R^3 \rightarrow T_p M^3$ given by
\[\big(D_{f(p)} f^{-1}\big) \circ \big(\sigma\left(f\left(p\right)\right)\big) \circ \big(\phi\left(f\left(p\right)\right)\big).\]

\paragraph{Derivative crossed homomorphism.}
Our next goal is to construct a crossed homomorphism
\[\cD\colon \Diff^{+}(M^3) \rightarrow C(M^3,\GLp_3(\R))\]
that we will call the {\em derivative crossed homomorphism}.  In a
suitable sense, it encodes the action of $\Diff^{+}(M^3)$ on $\Triv(M^3)$.
The derivative crossed homomorphism depends on a choice of a base trivialization
$\sigma_0 \in \Triv(M^3)$ that we fix once and for all.  Now consider $f \in \Diff^{+}(M^3)$.
We have $\sigma_0^f \in \Triv(M^3)$, and as we noted above the topological group $C(M^3,\GLp_3(\R))$ acts
simply transitively on $\Triv(M^3)$.  It follows that there exists a unique $\phi_f \in C(M^3,\GLp_3(\R))$
such that
\[\sigma_0^f = \sigma_0 \cdot \phi_f.\]
We define $\cD(f) = \phi_f^{-1}$.  Here the inverse refers to the group structure
on the space $C(M^3,\GLp_3(\R))$ induced by the group structure on $\GLp_3(\R)$.
The inverse will be needed to make $\cD$ a crossed homomorphism -- if you
examine the formulas below, you will see that without it $\cD$ would be a crossed anti-homomorphism.

To check that $\cD$ is indeed a crossed homomorphism, note that for
$f_1,f_2 \in \Diff^{+}(M^3)$ we have
\[\sigma_0^{f_1 f_2} = \sigma_0 \cdot \cD(f_1 f_2)^{-1}\]
and
\begin{align*}
\sigma_0^{f_1 f_2} &= \left(\sigma_0^{f_1}\right)^{f_2} \\
&= \left(\sigma_0 \cdot \cD\left(f_1\right)^{-1}\right)^{f_2} \\
&= \sigma_0^{f_2} \cdot \left(\cD\left(f_1\right)^{f_2}\right)^{-1} \\
&= \sigma_0 \cdot \cD(f_2)^{-1} \cdot \left(\cD\left(f_1\right)^{f_2}\right)^{-1}.
\end{align*}
Here again the inverses refer to the group structure on the space $C(M^3,\GLp_3(\R))$ induced by the group structure on $\GLp_3(\R)$.  We thus have
\[\cD(f_1 f_2) = \cD\left(f_1\right)^{f_2} \cdot \cD(f_2),\]
as desired.

\paragraph{Homotopy classes.}
We now pass to homotopy.  Let $[\sigma]$ denote the homotopy class of $\sigma \in \Triv(M^3)$ and let
\[\HTriv(M^3) = \Set{$[\sigma]$}{$\sigma \in \Triv(M^3)$},\]
and let $[\phi]$ denote the homotopy class of $\phi \in C(M^3,\GLp_3(\R))$ and let
\[[M^3,\GLp_3(\R)] = \Set{$[\phi]$}{$\phi \in C(M^3,\GLp_3(\R))$}.\]
Finally, let $\Mod(M^3) = \pi_0(\Diff^{+}(M^3))$ denote the mapping class group of $M^3$, and
for $f \in \Diff^{+}(M^3)$ let $[f] \in \Mod(M^3)$ denote its isotopy class.
The group structures of $\Diff^{+}(M^3)$ and $C(M^3,\GLp_3(\R))$ induce group
structures on $\Mod(M^3)$ and $[M^3,\GLp_3(\R)]$, and the right actions
of $\Diff^{+}(M^3)$ and $C(M^3,\GLp_3(\R))$ on $\Triv(M^3)$ induce right actions
of $\Mod(M^3)$ and $[M^3,\GLp_3(\R)]$ on $\HTriv(M^3)$ that we will continue
to write with superscripts and $\cdot$'s, respectively.  For $[f] \in \Mod(M^3)$ and $[\phi] \in [M^3,\GLp_3(\R)]$
and $[\sigma] \in \HTriv(M^3)$, we still have the relationship
\[\left([\sigma] \cdot [\phi]\right)^{[f]} = [\sigma]^{[f]} \cdot [\phi]^{[f]}.\]
Finally, the derivative crossed homomorphism $\cD\colon \Diff^{+}(M^3) \rightarrow C(M^3,\GLp_3(\R))$
descends to a derivative crossed homomorphism
\[\fD\colon \Mod(M^3) \rightarrow [M^3,\GLp_3(\R)]\]
whose characteristic property is that
\[[\sigma_0]^{[f]} = [\sigma_0] \cdot \fD([f])^{-1} \quad \text{for $[f] \in \Mod(M^3)$}.\]

\section{The twisting crossed homomorphism}
\label{section:twisting}

Just like in the last section, let $M^3$ be a closed oriented $3$-manifold.  Fix some $\sigma_0 \in \Triv(M^3)$, and let $\fD\colon \Mod(M^3) \rightarrow [M^3,\GLp_3(\R)]$
be the associated derivative crossed homomorphism.

\paragraph{Twisting crossed homomorphism.}
The group $\GLp_3(\R)$ deformation retracts to its maximal
compact subgroup $\SO(3) \cong \RP^3$.
We thus have $\pi_1(\GLp_3(\R)) \cong \Z/2$.  The $\pi_1$-functor therefore induces a group homomorphism
\[[M^3,\GLp_3(\R)] \longrightarrow \Hom(\pi_1(M^3),\Z/2) = \HH^1(M^3;\Z/2).\]
Composing this with the derivative crossed homomorphism $\fD$, we get a crossed homomorphism
\[\fT\colon \Mod(M^3) \longrightarrow \HH^1(M^3;\Z/2)\]
that we will call the {\em twisting crossed homomorphism}.  Since the twist subgroup $\Tw(M^3) < \Mod(M^3)$ acts trivially
on $\pi_1(M^3)$, the restriction of $\fT$ to $\Tw(M^3)$ is a homomorphism (not just a crossed homomorphism).

\paragraph{Effect on sphere twists.}
The following lemma shows how to calculate $\fT$ on a sphere twist:

\begin{lemma}
\label{lemma:spheretwist}
Let $S$ be an embedded $2$-sphere in $M^3$.  Then $\fT(T_S) \in \HH^1(M^3;\Z/2)$ is the cohomology class
that is Poincar\'{e} dual to $[S] \in \HH_2(M^3;\Z/2)$.
\end{lemma}
\begin{proof}
Identify $S$ with $S^2 \subset \R^3$.
Recall that $T_S$ is constructed from a loop $\ell\colon [0,1] \rightarrow \SO(3)$ with $\ell(0) = \ell(1) = \text{id}$
that generates $\pi_1(\SO(3),\text{id}) \cong \Z/2$.  This generator rotates $S$ by a full
twist about an axis, and $T_S$ is represented by a diffeomorphism $\tau$ that is the identity outside a tubular neighborhood
$U \cong S \times [0,1]$ of $S$, and on $U$ is defined by $\tau(s,t) = (\ell(t) \cdot s, t)$.  Let $p_0 \in S$ be one
of the two intersection points of the axis of rotation defining $\ell$ with $S$.

Consider a smoothly embedded closed curve $\gamma\colon S^1 \rightarrow M^3$.  Homotoping $\gamma$, we can assume that it only intersects $U$ in segments of the form $p_0 \times [0,1]$ (which it might traverse in either direction).  It follows that
$\tau$ fixes $\gamma$ pointwise.  Let $\cD\colon \Diff^{+}(M^3) \rightarrow C(M^3,\GLp_3(\R))$ be the derivative
crossed homomorphism that descends to $\fD\colon \Mod(M^3) \rightarrow [M^3,\GLp_3(\R)]$ when we pass
to homotopy.  The composition
\[[0,1] \xrightarrow{\gamma} M^3 \xrightarrow{\cD(\tau)} \GLp_3(\R)\]
is a loop whose image in $\pi_1(\GLp_3(\R)) \cong \Z/2$ represents $\fT(T_S)([\gamma])$.  Examining the definitions, we
see that this element of $\pi_1(\GLp_3(\R)) \cong \Z/2$ simply counts the number of times $\gamma$ traverses
$p_0 \times [0,1]$, which equals the $\Z/2$-algebraic intersection number of $\gamma$ with $S$.  The lemma follows.
\end{proof}

\paragraph{Connect sums of $S^2 \times S^1$.}
We now specialize this to the connect sum $M_n$ of $n$ copies of $S^2 \times S^1$.  Recall from
Theorem \ref{theorem:laudenbach} that $\Tw(M_n)$ is generated by the sphere twists about
the core spheres $S^2 \times \ast$ of the $n$ summands $S^2 \times S^1$ of $M_n$.  These clearly commute
with each other and are Poincar\'{e} dual to a basis for $\HH^1(M^3;\Z/2)$, so
Lemma \ref{lemma:spheretwist} implies the following:

\begin{corollary}
\label{corollary:identifytwist}
The twisting crossed homomorphism $\fT\colon \Mod(M_n) \rightarrow \HH^1(M_n;\Z/2)$ restricts to an isomorphism
$\Tw(M_n) \cong \HH^1(M_n;\Z/2)$.
\end{corollary}

In particular, we recover Laudenbach's theorem \cite{LaudenbachSpheres} saying that $\Tw(M_n) \cong (\Z/2)^n$.
We actually get more: since $\fT\colon \Mod(M_n) \rightarrow \HH^1(M_n;\Z/2)$ is a crossed homomorphism,
the isomorphism $\Tw(M_n) \cong \HH^1(M_n;\Z/2)$ in
Corollary \ref{corollary:identifytwist} is an isomorphism of $\Mod(M_n)$-modules, where
$\Mod(M_n)$ acts on its normal subgroup $\Tw(M_n)$ by conjugation.

\paragraph{Summary.}
Combining Corollary \ref{corollary:identifytwist} with Lemma \ref{lemma:crossedsemidirect} and the
exact sequence
\[1 \longrightarrow \Tw(M_n) \longrightarrow \Mod(M_n) \longrightarrow \Out(F_n) \longrightarrow 1\]
from Theorem \ref{theorem:laudenbach}, we conclude the following:

\begin{theorem}
\label{theorem:weakmain}
Let $[\sigma_0]$ be the homotopy class of a trivialization $\sigma_0$ of the tangent bundle of $M_n$
and let $\fT\colon \Mod(M_n) \rightarrow \HH^1(M_n;\Z/2)$ be the associated twisting crossed homomorphism.  The
following then hold:
\begin{compactitem}
\item $\Mod(M_n) = \Tw(M_n) \rtimes \ker(\fT)$.
\item $\Tw(M_n) \cong \HH^1(M_n;\Z/2)$ as a $\Mod(M_n)$-module.
\item $\ker(\fT) \cong \Out(F_n)$.
\end{compactitem}
\end{theorem}

This is almost Theorem \ref{maintheorem:split}.  All that is missing is the fact that
$\ker(\fT) \cong \Out(F_n)$ is the $\Mod(M_n)$-stabilizer of $[\sigma_0]$, which we will
prove in the next section (see Corollary \ref{corollary:stabilizer}).

\section{\texorpdfstring{$\Out(F_n)$}{Out(Fn)} acts trivially on homotopy classes of trivializations}
\label{section:completing}

In this section, we prove the following.

\begin{lemma}
\label{lemma:fixtrivial}
Let $[\sigma_0]$ be the homotopy class of a trivialization $\sigma_0$ of the tangent bundle of $M_n$
and let $\fT\colon \Mod(M_n) \rightarrow \HH^1(M_n;\Z/2)$ be the associated twisting crossed homomorphism.  Then
$\ker(\fT)$ fixes $[\sigma_0]$.
\end{lemma}

Since the $\Mod(M_n)$-stabilizer of $[\sigma_0]$ is clearly contained in $\ker(\fT)$, this implies the following:

\begin{corollary}
\label{corollary:stabilizer}
Let $[\sigma_0]$ be the homotopy class of a trivialization $\sigma_0$ of the tangent bundle of $M_n$
and let $\fT\colon \Mod(M_n) \rightarrow \HH^1(M_n;\Z/2)$ be the associated twisting crossed homomorphism.
Then $\ker(\fT)$ is the $\Mod(M_n)$-stabilizer of $[\sigma_0]$.
\end{corollary}

As we noted at the end of \S \ref{section:twisting}, Corollary \ref{corollary:stabilizer} together with
Theorem \ref{theorem:weakmain} implies Theorem \ref{maintheorem:split} from the introduction.

\begin{proof}[Proof of Lemma \ref{lemma:fixtrivial}]
Let $G = \ker(\fT)$, so by Theorem \ref{theorem:weakmain} we have $G \cong \Out(F_n)$.  Let
$\fD\colon \Mod(M_n) \rightarrow [M_n,\GLp_3(\R)]$ be the derivative crossed homomorphism
associated to $[\sigma_0]$.  Recall that $\GLp_3(\R)$ is homotopy equivalent to
$\RP^3$.  Let $\tGLp_3(\R)$ be the universal cover of $\GLp_3(\R)$, so $\tGLp_3(\R)$ is homotopy
equivalent to $S^3$.  For $[f] \in G$, we know that $\fD([f]) \in [M_n,\GLp_3(\R)]$ induces
the trivial map on $\pi_1$, so we can lift $\fD([f])$ to an element of $[M_n, \tGLp_3(\R)]$.
Though there are two distinct lifts to $\tGLp_3(\R)$ of a map $M_n \rightarrow \GLp_3(\R)$ that induces
the trivial map on $\pi_1$ (the two lifts correspond to a choice of a lift of a basepoint), these two lifts
are homotopic via a homotopy corresponding to right-multiplication by a path in $\tGLp_3(\R)$ from
the identity to the other element of $\tGLp_3(\R)$ projecting to the identity in $\GLp_3(\R)$.  From
this, we see that in fact $\fD$ lifts to a crossed homomorphism
$\tfD\colon G \rightarrow [M_n,\tGLp_3(\R)]$.

Since $\tGLp_3(\R)$ is homotopy equivalent to $S^3$, elements of $[M_n,\tGLp_3(\R)]$ are classified
by their degree, i.e.\ as a group we have $[M_n,\tGLp_3(\R)] \cong \Z$.  What is more, since orientation-preserving
diffeomorphisms of $M_n$ act by degree $1$, the action of $\Mod(M_n)$ on $[M_n,\tGLp_3(\R)] \cong \Z$ is trivial.
It follows that the crossed homomorphism
\[\tfD\colon G \rightarrow [M_n,\tGLp_3(\R)] \cong \Z\]
is an actual homomorphism (not just a crossed homomorphism).  Since the abelianization of $G \cong \Out(F_n)$
is torsion, we deduce that in fact $\tfD$ is trivial.  This implies that $\fD|_G$ is also trivial.  Since
$\fD$ encodes the action of $\Mod(M_n)$ on $[\sigma_0] \in \HTriv(M_n)$ (see \S \ref{section:derivative}), this implies that
$G$ acts trivially on $[\sigma_0]$, as desired.
\end{proof}

redpajama_set_name: RedPajamaArXiv


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