This year I've really wanted to start making an effort to get fit. So many of us say it in January, but it can be so hard to stick to. This month, I've actually started making progress and it's actually been quite easy (not to speak too soon!).
When Fenetic Wellbeing, a mobility company, got in touch about a blog post about getting back into fitness, I realised it would be a good opportunity to let you know what I've been doing to feel fitter and lose weight.
I realised where I usually fail is by doing too much at once. I make a great plan where I'll be going to the gym 4 times a week, and doing exercise at home for the rest of it. I'll eat like the healthiest person that ever existed, and only drink water and herbal tea. It doesn't work. We know this. It gets overwhelming, and it always feels like I've let myself down before I've even started.
For years I have steadily gained weight and now I am at a really uncomfortable place fitness wise. I live a sedentary office life and am glad I've started changing that.
I thought sharing what I've been doing might help somebody else get back into exercise.
The first thing to do is get your mindset right. You can write all the shopping lists, buy all the workout clothes, but if you're heart is not fully set on making lifestyle changes, it likely won't work. It's good to have a goal, but just doing it for an event or to impress somebody means outside factors affect your progress.
Despite what I said above, once your mind is set on getting fitter, stylish workout clothes can be a great boost to your confidence when going to the gym.
I've still eaten some junk food, I've still been out for food, but at home, I've made some small changes.
I have less carbs and have focused on eating better food at regular times.
I've also been having smaller portions, which feels normal within a couple of days.
Meal planning has been the biggest game changer, because I know what I'm having each day and don't have the after-work worry of wondering what to eat.
I only planned to have 2 days of exercise a week, and it's made it much more practical.
I don't always have a chance to go to the gym, so like to find other ways to exercise.
For as long as I can remember, I've had a bad knee, meaning running is not an option really as I'm worried I'll do more damage. That's why I prefer to cycle. If you find it tough, or are recovering from an injury, this can be made even easier with the use of an electric bike. It's just like a normal bike, but with an extra bit of help. Even if it's a little bit of movement, it's better than none and a great, easy way to explore your local surroundings.
Some of us have quite high hopes for our fitness and goals can be overwhelming. It's worth breaking up into smaller pieces to make your goal more manageable.
For example, if you want to lose a certain amount of weight in a year, work out how much that is monthly, and work towards the smaller goal each month instead.
To help me track my progress, Fenetic Wellbeing also sent me a MISFIT Shine 2 activity tracker in Rose Gold and it's so handy. Activity trackers are a great way to make sure you're keeping on track day by day. It's such good motivation for when you are on the way to hitting your daily goals.
On the MISFIT Shine 2, there is a setting to make it vibrate when you haven't moved for a while, which is so useful because I really get in the zone when working. You don't have to wear it as a wristwatch as the face pops out which is super handy, and something else I love is not having to charge it, which I would definitely forget to do. The replaceable battery lasts 6 months.
Thanks to Fenetic Wellbeing, you can own one of these activity trackers (worth £79.99) too! Just enter in the Rafflecopter competition below.
Do you have any extra tips to share? Have you got any fitness goals to work towards this year?
Disclaimer: As mentioned within the post, the activity tracker was sent as a gift and this post was written in collaboration with Fenetic Wellbeing.
Baby steps are always important to remember. You can't make huge leaps and bounds when it comes to working out in the first day or week. So break up that big goal and work at it. Also I'm the type of person who loves music while working out so a good playlist is important!
Since finding out I have chronic kidney disease stage 3 just last month, I have cut right down on my alcohol intake. Prior to the news, I was drinking half a bottle of wine every night – now I have just a couple of glasses at the weekend.
I'm so sorry to hear that, but it's great to have made such a positive change!
Wow that's amazing! Good luck preparing, I'm sure you'll do great. Likes like such a great adventure!
I'm decluttering and tidying my house room by room. Sorting into rubbish/charity/sell and any money I make is going in our family 'holiday fund' so everyone benefits!
Aw that's a great idea! I plan on having a massive declutter too, but I get so attached to things haha! I love the idea of a reward though, that might get me through.
I'm planning to take a trip to the mountains,probably in Morocco.
I booked our first family holiday in 4 years… this is going to be a great year for us!
I'm planning to get the conservatory revamped! I've also never dyed my hair but have got it booked in for highlights, eek!
I've gotten so much fitter this year. So far so good!
getting married in May which will make 2017 my best year yet!
I plan on giving birth next month and that will make 2017 amazing! Also looking forward to getting back to working out once baby is here.
Learning to swim & moving to our 'forever' home!
I plan to walk somewhere different each week with friends.
Meet my 1st grandchild when she is born in around 4 weeks time.
We are planning to move house, to a smaller one so we benefit from spending our time together, doing what we love, rather than maintaining the house and garden. So looking forward to more time together.
A big family summer holiday.
I plan (and hope!) to graduate uni with a first!
I've booked a holiday to Cornwall and we've never been there.
I'm planning to buy my first home this year – very exciting being a proper grown-up!
I've got a new job which is great and I'm hoping to increase my fitness by running.
I have created a lot more balance in my life as far as work goes as last year i was working every day except sunday and very long days.Being self employed i have cut my hours down and can now spend more time with my family and am getting more done at home and start some hobbys too.
I'm doing more housework; a neater house is so calming.
I've made a vow to smarten up. So far it's going well.
I plan to exercise more and declutter.
I have bought a bike and started cycling!
Be more mindful! To myself and others around me, im a bit of a control freak in my house and my house mates get really annoyed with me (its annoying when people leave lights on, ha!) but going to try and let it NOT get to me as much when they do things that annoy me, haha! wish me luck!
I'm gonna celebrate my 40th birthday and my son's 21st, so its a big year for celebrations in our house!!
I'm planning to look after myself more! Admittedly I've neglected my body over the past few years!
I've put the awfulness of 2016 into a box, buried it in the garden and am moving on – Inca Trail done, next stop London Marathon then a triathalon in the Summer….
I've cut out sugary snacks, no more sweets or chocolate for me. I've stopped craving them now too, so that's great.
I have a mission to lose weight to make my 2017 great! I would love to be lucky! This prize would be perfect for me! Thank you!
Ironically – itss pending less time online!!
Hopefully going to finish converting our farm into holiday cottages.
I try and do one random act of kindness every day, I think if we can all do this, it makes the world a better place.
I am joining the gym and I am going to get FIT!
I'm also trying to make an effort to get fit this year, but also to be more active. There are so many walking routes nearby that my boyfriend and I have really loved getting out, getting fresh air and exploring new parts of Yorkshire. Plus, we sleep like babies after those long walks, which isn't too bad either haha.
We are in the process of buying our first home! Which is super exciting.
I want to start going to meditation classes or yoga….. I think that they would help me relax a bit more and control my breathing too.
I plan to lose some weight to I can have stomach surgery. This will put an end to the health problems I've been experiencing and therefore make the rest of this year great!
Umm I think I am going to drink more beer and eat more cheese & onion crisps in 2017.
Haha this is my favourite! Cheers to this!
I'm making a big effort this year to eat nutritious meals and snacks, so this fitness tracker would really help the journey!
We've entering an IVA to try and sort out all of our debt problems so hopefully this year will be a lot easier than the last.
I have been walking more rather than using my car.
Take the time to enjoy the simple things, take a gentle stroll, enjoy a picnic in the the fresh air, relax and appreciate what you already have.
I plan on eating healthier and keeping me and my family fitter in 2018 to make it great!
Help my partner recover after his stroke last year!
The plan is to lose weight and get super fit.
Get married!! which i will be doing in July!!
I'm going to qualify as a yoga teacher!
I'm striving for my goal of education and a good job. Only 3 years to go!
Simply to spend more time with family members, today's society means we are often rushed off our feet but life's too short to not spend as much quality time with loved ones as we can.
Wow, that's incredible, well done!
I'm planning to run my first ever half marathon!! eeeeeek!! LONG way to go!!
Ohh wow, good luck! Sounds great!
I am trying my best to overcome my anxieties and not to live in fear anymore!
I'm getting married this year to my amazing fiancé – so we are trying to lose weight and get fit for our big day!
Oh wow amazing news! Congrats and good luck!
get as much out of every day as I can!
I plan to create a little patio where we can sit outside during the evenings. It will be another room in an otherwise tiny home.
Im going back to work after a long spell off good luck to me.
I'm no natural athlete (which is a bit of an understatement!) but recognise I need to get more active so a friend and I have pledged to try a new activity every month. We've been trampolining (which was great fun) and this month we're trying climbing.
I've set myself a challenge to do a half marathon in 2018. Have a long way to go with weight loss and fitness but given myself plenty of time to do it!
I plan on taking more pictures of my son with his aunties/nan and grandad!
Try and be more patient with the kids!!
When the weather gets a bit more comfortable, we as a family aswell as the children and dogs have planned to walk the 3 peaks. Snowdon, Ben Nevis and Scafell Pike.
Aswell as more general walking throughout the year to keep us all more healthy and our adventurous minds happy!
I've ditched the car and try to walk as much as possibly. I don't feel much better for it physically, but do mentally and am saving £60 a month in petrol!
I try and walk my 10,000 steps a day after having my third in December.
I plan to learn to crotchet and make one of those 60's frocks.
Get fit and lose my mummy tummy! Spin class here I come!
We are having a new baby so there's lots of planning going on at the moment. We are having a good clear out and decorating the baby's room. I plan to start a keep fit routine after the baby is born.
My aim is to get down to a 30″ waist by the end of 2017! I have gone from 42″ to 36″ since November, so watch this space….
Wow that's incredible! Doing so well already! Good luck.
I'm adding ingredients such as teff into my diet to replace foods removed due to allergies. This should make me even healthier and make 2017 great.
I plan to do more hiking over the Summer when the weather is nice.
After spending 8 years as a mum at home (I've always worked a few hours a week but no more), I completely lost myself, my personality. There's something about New year's resolutions and I've decided it's time for change; it's time for me. I am in the middle of a Nutritional course and I am determined to lead healthier and more fulfilling life, but also, to help others. I am so excited and, I hope that this year, is my year :)!
Spend more time outside with family!
The only goal I have is to be me, but a better version of me.
I'm training for my first marathon this year – so excited/nervous!
GET FIT BY DOING MORE EXERCISE + EATING HEALTHILY!
I plan to try out new things. So far have started a course it is on Community Building. Forming groups to take action with support to make positive change for the Community. Interesting so far.
I am training to become a quantity surveyor with my work.


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Olsim: The Ibaloy Today
VALRED OLSIM
OUR national tourism policy is more than numbers that help generate employment and pump the economy but also includes the preservation of culture and heritage.
As government workers who are dealing with these issues, we are often invited to related conferences as part of our work. One of the most memorable, perhaps, is the Ibaloy Studies Conference entitled "The Ibaloy Today" which was organized by UP (University of the Philippines)-Baguio a couple of years ago.
I remember that on the said event, I was compelled to stand up and address the room's white elephant after many confusing queries pertaining to my generation.
"How do we, the multi-cultural children- the one-halves, and one fourths, address our identity? What is the implication of being a multi-cultural child in the Ibaloy's advocacy of cultural preservation?" I nervously rose. After I let "the" question out, the speaker, Dr. Julie Camdas-Cabato, heaved a sigh not knowing that the inquisitive person in front of her is just one of the many babies she ushered to this world via caesarian section.
Dr. Cabato's story of the "vanishing Ibaloys of Baguio City" echoes the many fears of the Ibaloys in this age – the displacing of its people, the fading of its culture, and the disappearance of its identity.
"That is indeed a very difficult question" she continued. The questions actually broke the room - each participant looking side and up, asking the ceiling the same thing.
There was no real answer.
Earlier before that, messages were given asking the obvious question: "Who are the Ibaloys?"
One politician, who admitted that he has no Ibaloy blood but can speak fluent Ibaloy, shared that he is considered by many as one of them simply because he lived as an Ibaloy, and speaks the Ibaloi language. Does that mean that those who do not know how to speak Ibaloi, despite having Ibaloy parents, like most from the later generation, are not considered Ibaloys? Is being an Ibaloy by blood? Or by cultural orientation (as being able to speak Ibaloi)?
NCIP (National Commission on Indigenous Peoples) commissioner Zenaida Hamada-Pawid (a one-half, one- fourth herself) answered my query in less than three parts. She started her story by recalling her experience as a young Anthropology professor of UP who is tracing her ancestry by collecting the genealogy of the five biggest clans of the Benguet villages. What she and other researchers found out was that all of the people in the southern Cordillera can trace their roots to only one people, the Kalanguyas (Ikalahans), a distinct sub-group of the Ifugaos.
The Kalanguyas (from "Keley ngoy iya" a term used to pacify misunderstandings), in turn, are blood brothers of many different ethnic groups, not only from the Cordillera region, but also of the people from Region I and II. "Enshi nai-afafil" I smirked.
If we go deeper, we will find out that our different tribes in the Cordillera region belong to the same Austronesian peoples in Southeast Asia and Oceania. This means that we belong to the same family with the Taiwanese Aborigines, the majority ethnic groups of Malaysia, east Timor, Indonesia, Brunei, and Micronesia, as well as the Polynesian peoples of New Zealand and Hawaii (and many other groups).
"So how can the young Ibaloy today understand how is it to become an Ibaloy of the past?" She asked. "The Ibaloy is a culture and people in constant change… they are not frozen in time"
"But if you want to know the core value of the Ibaloys, go ask the Ibaloys who perished in the Battle of Tonglo. Our ancestors who spent hundreds years of fighting to protect the ownership of our lands, properties, and resources." She almost raised her voice. "The Ibaloys are not shy… they are strong and empowered!"
The room almost became somber. For some of the Ibaloys who were there, the obvious implication of her last lines hit them like a brick.
-o0o-
My late father, Alberto Ingosan Olsim Jr., descended from the Ingosan-Gabol Clan of Irisan, Baguio City who can trace their roots to Buguias (and areas near Ahin) and Kabayan, and the Olsim and Bacquian clan of Buguias, Benguet. He is a "Kanibal" (Kankana-ey – Ibaloy). He, however, grew up in Mt. Province because of his father's choice to farm at nearby Bauko and Sabangan, Mt. Province.
It is a different case with my Mother, Marcelina Dulay Elwas, who is predominantly a "Bontokis" from Sabangan and, Samoki, Can-eo, and Gonogon, Bontoc, Mt. Province. She, however, grew up in the Ibaloy mining village of Itogon, Benguet where she lived and spoke like a true blue Ibaloy.
This (comically) means that I have an i-Benguet father who grew up in Mt. Province, and a Mt. Province mother who grew up in Benguet.
I was born in Baguio City, and was raised in La Trinidad, Benguet. I lived in a multi-cultural neighborhood who speak most of the Cordillera region's "neutral" language – the "Ilokano" or "Baguio-Ilokano", or perhaps our washed-out version of it. Our parents did not use the Kankana-ey or Ibaloi language in our home, just like many parents today.
So, how do we, the part Ibaloys, address our identity, especially in the Ibaloy's call for cultural preservation in this modern multi-cultural society? Does it mean that one has to marry an Ibaloy to 'promote and continue the blood-line'? Does it mean that we have to discourage the entry of other cultural groups? How can I compromise the preservation of the different cultures that I belong to (which are all equally wonderful)?
For us hybrids, we can only imagine in our silence.
In this multi-cultural generation in which the young Ibaloys have learned to love without the issue of tribe, language, or colors, they may hardly understand the importance of such "preservationist" call.
They can only learn the cultural constructs, the tools; language, cultural dances, symbols. They can only wear their names and their ethnic costumes. But beyond that, they have changed... just like this ever-changing world.
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National Commission on Indigenous Peoples
Ibaloy
vanishing Ibaloy

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Perspectives about the american dream essay conclusion
Without the rights enshrined in this document, the American dream wouldnt be able to persevere. Since then, many aspects of Martin Luther Kings Dream for the American people have come true , but some remain a dream. It may refer to any type of paper. However, the American dream for any person and for any epoch has always been based on the following pillars: freedom, equality, control over ones destiny and an incessant pursuit of ones dream. This American Dream essay example will focus on the ways Americans have in the past and present, attempted to achieve a life of happiness and fulfilment in the United States. People did not have a chance to enjoy life, much less try to better themselves for the future of their families and themselves. Joanne Rowling, the books author, agrees with my opinion. The verdict from the King trial was seen as a direct attack on the civil liberties of African-Americans and their ability to live a life free from oppression and persecution. Example #1 According to the.S. For the parents, the American Dream is dying, but for their children it is still within reach if they join the exodus away from rural life. It goes without saying, that the perception of the American dream vastly differs if we compare views of immigrants perspectives about the american dream essay conclusion and people who have lived in this country for their whole life.
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Some Americans felt discouraged concerning the future of the countrys job market and decided to take a step towards what they thought would bring true prosperity. It has perspectives about the american dream essay conclusion many differences with the original legend but is understandable for students. Economic desperation can lead to violence. Those Americans who have achieved the Dream are afraid to lose it and are prepared to protect it at all costs. Poorly funded schools cannot provide the level of education of those in wealthy areas, thus affecting the educational and economic future of the students and consequently the ability to achieve the Dream. On reflection, the American Dream lives on, even if it remains just that, a dream, for the majority of the population. Conclusion Many say the American Dream is gone. For the large landowners, be it Ted Turner in Montana or the Cattle Kings in Texas, the American Dream is alive and well. Those seeking religious freedom came to North America and formed the thirteen colonies so they may pursue a life filled with happiness and upward mobility. In this book, the phrase appears for the first time in the Preface, when Adams refers to the American dream of a better, richer, and happier life, adding that that dream or hope has been present from the start (Kochan, 2007,.
That is because jobs do not pay as well or offer as many hours as they used. In many peoples eyes people did not have to come from wealth to be perspectives about the american dream essay conclusion wealthy. dead or alive? How to Write a Conclusion for an Essay? Topics, the Modern Day American Dream, why Do People Pursue the Elusive American Dream? The purpose is to take a specific position on the topic. The History Behind the American Dream. Not necessarily the dream of a house with a picket fence, but the dream that they too could be the next Michael Jordan. Even though not all immigrants want to achieve the American Dream, the ones that do, do so keeping in mind the pursuit of a fresh start and a better life. Streets have become a battleground where the elderly are beaten, terrified women are viscously attacked and raped, teenage gangs shoot it out for a patch of turf to sell their illegal drugs and, innocent children are caught in the crossfire of drive-by and school-yard shootings. The Ideals Behind the American Dream.
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Washington, DC: Potomac Books, Inc. Did you enjoy conclusion sentence examples for essays shared in our article? However, that does not mean it has died or it has been forgotten. Department of Justice, human sex trafficking is closely related to cybercrime today as both are usually the results of progressive modern technologies. One great thing about the American way of life is that people are in control of their destinies. Never say someone that you cannot do something this is also a peculiar example of the American dream. The New Immigrants The recent immigrants of today and the last few decades have been Hispanic and Latino people from the Americas. Finally, we have the dream of novelty. Many that crossed the border into America whether legally or illegally has experienced hardships that make them question if the move to this seemingly prosperous nation was a good idea.
The New Immigrants, iII. Meatless meals might improve the health conditions of patients suffering from various disorders, and the observed study proves. Places offered through affirmative action help overcome the economic divide and, by hopefully opening career opportunities to a greater cross-section of the community, help to widen the attainment of the American Dream. Stern, K; A Force Upon The Plain: The Militia Movement and the Politics of Hate, Simon perspectives about the american dream essay conclusion Schuster, USA, 1996 (Reading Pack). This is the ability of everyone to purchase the goods of America, regardless of where they come from and who they are. Start with paraphrasing a thesis.
Students can get both separate parts of the papers and fully written works from scratch to use as templates for their future assignments. Some decided to head towards the east coast of North America to practice their beliefs and make money for themselves and their families. The American Dream is usually defined in terms of financial security, homeownership and higher education perspectives about the american dream essay conclusion (Currier, 2015). The image of Hercules remains the same popular and recognized through centuries, and it never changes. The only way for them to get out of the ghetto, move their families out of poverty and receive a decent education was through basketball. Both characters were natural leaders who could create a better relationship tension than Ron and Hermione.
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Feagin, J Vera, H; White Racism, The Basics, Routledge, USA, 1995 (Reading Pack). The acquittal of the two lapd officers involved in the beating sparked outrage in the LA black community and widened the rift of racial inequality throughout the entire country (Klein, 1997:3). One may be born in an indigent family but become a billionaire 40 years later. Example #2, euthanasia should be legalized everywhere. For perspectives about the american dream essay conclusion the British colonists, they wanted not only a new start, but a new identity, away from the ones they left on the British shores. The difference today is young people may go to college instead of an apprenticeship to go about their pursuit of the American dream. A chance at Upward Mobility: The American Dream. He waited for the day they could walk down the street and hold hands with a child of a different race. The rags to riches story is what many people in the United States who have little money, hope to achieve. Klein, J; The Race Course, Mother Jones, Sept/Oct, 1997 (Reading Pack). Such effort is reminiscent of the British colonists of the 17th and 18th centuries. It can hardly be imagined that the American dream cherished, for example, by George Washington, Martin Luther King or by a fictional character Jay Gatsby, were ideologically similar.
Such essay requires research, but the main goal is to explore and analyze the problem in-depth to describe it in details. The Modern Day American Dream, people who have grown up looking towards America, thinking as adults about achieving the American Dream, have seen an evolution of such a notion throughout the years. To fulfil this part of the constitution, the dream of a democracy of goods has to exist. It may have changed throughout the decades, but the core of the American Dream has always remained. The US may be a welfare state for the rootless, or it may be a country with strong cultural power to unite people.
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In reality, whites and blacks still lead separate lives. That hope is the underpinning of every immigrant struggle and of every effort made to come to the United States and achieve something meaningful. Years of economic downturn have led many in rural communities to despise and distrust the US Government, remote in Washington,.C (Stern, 1996). In 1963, Martin Luther King Jnr said that he too had a dream that on the red hills of Georgia the sons of former slave owners will be able to sit down together at a table of brotherhood Video: The Legacy). It means the current skills in demand are forever changing. But many Americans say they remain doubtful about economic opportunity and the ability to move up the ladder. But theres a more elemental calculation: Whether you achieved more than the generation that came before you (Lee, 2014). The kind of American Dream that persists even during the difficult economic times is the dream of starting with nothing and becoming rich and successful. However, there is no direct or absolutely clear definition of the American dream, as it is viewed by people on the basis of their different worldviews, mental perception, sociological status and time background. Those that came, did so because they wanted the liberty of being their own person. To sum up, the American Dream is freedom to act, freedom to take responsibility for ones actions, freedom to pursuit ones dreams and life goals, freedom to choose ones destiny.
Therefore, despite any possible limitations and obstacles, a strong desire to give up, insurmountable challenges, unbearable hardship, one should perspectives about the american dream essay conclusion always strive for his dreams as it is the main drive of humans life. In the inner cities of Chicago and Los Angeles, riots fuelled by racial and economic inequality flared in the 1960s. Critical thinking is required! Such a sentiment could be for many reasons. Yes, it is definitely heterogeneous as far as different epochs are taken into account. Todays job market is filled with competition for even low-wage jobs, and the need to multi-task. Citizens Militia groups have grown up with their own ideas on how to seek redress for years of perceived neglect and the destruction of their Dream, most famously in the bombing of the Federal Building in Oklahoma City in April 1995.
Essay, what is the, american, dream?
The perceived threat to a citizens achievement of the Dream and the reason for possession of a firearm is the fear generated by Americas alarmingly high crime rate, including an average 20,000 homicides committed annually with firearms (Sugarmann, 1992). Thirty years on, the Rodney King beating in Los Angeles in 1992 sparked more rioting. Retrieved from Smith,. The rural crisis has changed family farmers expectations of the American Dream. We recommend involving a paraphrased thesis statement and adding a powerful hook sentence to attract the readers attention again and leave an unforgettable impression. Include a couple of general sentences. They are graduating high school and earning college degrees. View All Our Example American Dream Essays. The course of the United States has taken many turns and zig zags. In cities and towns, affluent areas are mainly white, poor areas black or Hispanic. Example #1 I believe Harry Potter and Hermione should have made a couple.
It is an exception from the rules. Example #3 I have analyzed the given piece of literature to prove that the author uses ghostwriting services. Many of them face writers block in the beginning. While the symbolism of the American Dream came from the settlers to what is now the United States, the actual phrase existed centuries later. Swifts pamphlet gained an appreciation and influenced next generations. That is not the case, at least for the new immigrants that come to the United States. Video Hoop Dreams, video perspectives about the american dream essay conclusion Troublesome Creek. The tiny percentage of farms owned by African-Americans are mainly in the south-east of the country. Yet true prosperity has evaded but a few million people in the United States. For these people the Dream is not dead, yet. It all leads to a nation of debt filled constant loss and instability. Introduction, the, american Dream began as a journey from oppressive Great Britain to a new land. Kids graduated and acquired gainful employment right after graduation, continuing the cycle of stability and upward mobility with their children.
A conclusion is made of 3 different parts. Thesis Statement, the past and the present remind Americans what the American Dream consisted and consist of; that is where this will essay will go to understand why such a notion has persisted since the first settlers came to shore. I have realized the ways meat impact our life and health. The Great Gatsby and the American Dream. She changed perspectives about the american dream essay conclusion the plot because of the comments of famous colleagues and fans who wished to have a less predictable plot.
Perspectives on the, american, dream
An essay conclusion is the next most important part after the introduction. People have noted in recent years, to pursue any form of upward mobility in the United States is to hit a brick wall. Now, this seems like an impossible task for many that live in the United States, especially for recent immigrants. However, not everyone got what they wanted and soon the American Dream became an American Nightmare for some. This leads to the idea of the American Dream disappearing. While the disadvantaged continue to see others around them moving towards a Dream that they can never hope to achieve themselves, material gain by illegal or violent means continues to be a problem. Its no surprise, then, that for many the American Dream may feel out of reach (Currier, 2015). A lot of it has to do with the continued economic instability of the United States. In conclusion, the American dream is about both choice in the consumer industry and unlimited freedom of choice. Is economic reality wiping out the American dream?
That is doing better than ones parents. In Hoop Dreams, the two young basketball players were both attempting to fulfil the criteria of the American Dream. Rockefeller came from humble beginnings and quickly became one of the richest men of America thanks to hard work and good business acumen (Chernow, 2010). The answer is not yet. So whilst King today would see black and white children learning and playing together, discrimination still exists. Rather, it is withering away with the rest of their crops, taken over by an economical weed that is rooted too deep even for a Federal fertiliser to kill. While it was something that existed in the earlier part of the 20th century, the Great Depression and the World Wars brought tough times and a harsh, economic reality. For some, the answer to their economic problems is simply to take by force from others what they themselves do not have. The latest immigrants, Hispanics perspectives about the american dream essay conclusion and Latinos from South, North, and Central America, come to the United States to escape persecution, an unstable government, and lack of employment. Narrative Essay Conclusion Example A purpose of the narrative essay conclusion example should simply sum up everything described and discussed in the essay. So is the American Dream dead or alive? Interest rates for savings accounts are near zero, discouraging people from saving. More recently, opposition to immigration has been rising (Daniels, 1991:400).
The, american, dream, essay, example - Academic Writing
High profile, high earning professional sports stars keep the Dream alive in the hearts of many. In The Legacy, the Dream as Martin Luther King Jnr saw it was that all people would be treated equally regardless of race, religion, or sex. It is the closing part of any type of an academic assignment, no matter whether it is a school perspectives about the american dream essay conclusion essay or college research paper. People will always want a chance at upward mobility. We have come up with essay conclusion examples and useful tips to help students master the art of academic writing. While many still suffer once they relocate to the.S., others have achieved a new start, a better life for their themselves and their children. A new poll by The New York Times found the public is more pessimistic than it was right after the financial collapse (PBS, 2014). The main reasons to make education of all levels free is the fact adolescents who fail to enter college remain unemployed and get engaged in criminal activities more often than their more successful peers. Guns are seen as an easy solution and gun ownership is justified by people as protecting their Second Amendment rights I have a right to own my gun for my protection and to protect my family (ibid:22).
Cynicism and the evolution of the American dream. For years hidden discrimination has hampered possibility of their economic success (ibid). Is the American Dream Gone? They may not reach their final aims, but many have provided for their children enough so they could become doctors, lawyers, and even a Supreme Court Justice of the United States. Today, the American dream is still relevant. That is reminiscent of the original American Dream, bringing everything, full circle. There were so many that sought to reap from the new land a chance at wealth, stability, upward mobility, and most importantly hope. Have A Custom Example Essay Written.
The American Dream was built on the rough foundation of American individualism, a revolutionary substance unlike anything the world had seen before (Caldwell, 2011,. To prevent the expansion of such problem, it is perspectives about the american dream essay conclusion necessary to limit access to social media. These entrepreneurs took the risk and had the talent to rise above everyone else and achieve amazing success. To someone like James Truslow Adams, the American Dream meant a strong sense of hope and desire that fueled the long journey to a better life. For some, belief simply leads to disappointment and bitter feelings. Essay Hook, although the versions of the American Dream have changed throughout the centuries, people still maintain a strong desire to make it and come to the United States to see what life there can offer. The last thing to observe is how to write a conclusion for an expository essay. The ones that make it in America do so because of hard work, some luck, and a lot of persistence. Example #3, my position is 100 correct. The rest of the plot reflects the myth better than other related shows. Poverty too can stand in the way of the American Dream. Several factors show that. Both Steven and his daughters teacher are right.
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The Numbers Show Rags-to-Riches Happens Only in Movies. When one boy is not selected you see the heartache in his eyes and the fear that his Dream may not be fulfilled. Jeffrey Klein states.that although American society is more diverse than ever before, a charged division still exists; namely the line between blacks and whites (1997:3). Often, they come with very, little money or belongings and work difficult jobs to ensure a future for themselves and their loved ones. The stagnant economy, the multiple recessions could be one reason. They could gain the home and the education that they wanted. That is how they can become the victims of several categories of victimization crimes: identity theft, sexual abuse, and property theft. It has to be measured on its principles and how they apply to society. Looking at countries like England where classism was and is intrinsically connected to culture and society, America was the opposite. Example #2, speaking of the article, I should perspectives about the american dream essay conclusion say that the most complicated dilemma recalled by the author is the lack of time versus storing resources and not the rest of the ideas. They sought freedom from religious persecution. The epic TV series known as Hercules: The Legendary Journeys, is the best start in the process of learning Roman and Greek mythology.
With the global economy shrinking as well, the once rich and powerful country has begun to seem like a disappointment more than an opportunity. The ones that do not find work, feel their economic prospects shrink as each year passes. People have lost their hope for a better life. Skolnick, j Fyfe, J; Above The Law: Police The Use of Excessive Force, Free Press, USA, 1993. Lifes best moments do not come from everyday routine, but the dreams and the wishes that come true through hard work and determination. In this essay, we will explore the American dream and just what. The concept of the American Dream is sold to every citizen along with the belief in the might and power of the US, patriotism, the American flag and the all-American apple-pie. Any buyer will be better off purchasing a modest automobile. They come to the United States to escape from the dangers and instability of their home country. Have you ever heard the expression American Dream? Gone are the days where people can apply for an entry level job right after high school or college.
What happened to the American dream? It is possible to find more examples online. America has traditionally been a land of immigrants and the home to people of many nationalities and cultures, all aspiring to achieve the Dream. I have explored several studies to prove that. On the one hand it could be argued that the easy accessibility of guns in America is killing the Dream (especially in the ghettos) as death by gunshot wound is the most perspectives about the american dream essay conclusion common way for young black males to die (Skolnick Fyfe, 1993:65).
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Poverty, seclusion and total destruction of traditional life has seen their Dream wither. The essay proves that violent video games may motivate adolescents to take part in school bullying and even commit crimes. Analysis Essay Conclusion Example, an analysis essay conclusion example should sum up the analysis. However, one article notes, the real modern American Dream is about doing better than ones parents. Add a warning of the outcomes of not following the thesis and a general statement of the way society will benefit from using offered ideas. One more pillar of the American Dream is based upon the assumption that one should pursuit ones dream without ever giving up, without ever doubting oneself despite any obstacles or sufferings life can put forth. But there still may be hope for the American Dream. Great Britain (Caldwell, 2011). For immigrants, or for people who have just received American citizenship, this notion may be directly related to a chance of earning more than they could in countries where they were born, a chance of giving proper and substantial education for their children. Retrieved from Kochan,. It is this perception that draws people to the idea of the great American Dream and the land of the free, a perception that for the right person at the right time, everything is possible. Although it is now mainly economic, unfortunately it tends to follow a racial divide, lessening the access of the American Dream to minority groups.
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Miami Cruise Destinations
Miami Cruise Guide
Sandy September 18, 2017
Carnival is the world's largest cruise line and has more cruises departing from Miami than any of the other cruise lines.
Carnival's "Fun Ships" are known for providing cruises packed with lots of entertainment options. Onboard activities vary by ship, but may include everything from mini-golf to water parks and sky-rides.
Carnival Cruise Lines ships sailing from Miami include Carnival Celebration, Carnival Conquest, Carnival Freedom, Carnival Horizon, Carnival Magic, and Carnival Sunrise.
Carnival Cruise Lines Ships Sailing from Miami
Max Passsengers: 5,374
The Carnival Celebration makes its debut in the Carnival Cruises lineup in November of 2022 and will sail five, six, and seven-day cruises from Miami to the Eastern and Western Caribbean.
The Carnival Conquest is all about fun. This ship specializes in short 3 & 4 night cruises to the Bahamas and Mexico.
Lots of dining options, plenty of bars and entertainment, WaterWorks slides, and kids' programs like Camp Ocean, make Carnival Freedom fun for cruises of all ages.
Carnival Horizon features innovative amenities such as the Dr. Seuss WaterWorks, a huge Seaside Theatre poolside LED screen, a bike-ride-in-the-sky attraction called SkyRide, an IMAX Theatre, and a new aft-view category of balcony staterooms.
The Carnival Magic is scheduled to sail from Miami in 2023 and 2024, with 7 & 8-night cruises to the Eastern and Southern Caribbean. One of Carnival's Dream-class ships, the Carnival Magic is a large ship packed with entertainment and dining options for cruisers to enjoy.
Originally named Carnival Triumph, this ship had a $200 million refurbishment in 2019 and was relaunched as Carnival Sunshine. Carnival Sunrise sails 4 to 5-night cruises from Miami to The Bahamas and the Eastern and Western Caribbean. The ship has a guest capacity of 2,984.
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When you choose to cruise from Miami, you're departing from the world's busiest cruise port. Over 4 million passengers leave from Miami for the Caribbean and other destinations every year. All of the biggest cruise lines have trips leaving from Miami, including Norwegian, Carnival, Celebrity, Royal Caribbean and Crystal, as well as companies with smaller fleets and smaller ships, such as Azamara and Oceania.
Miami Cruise Guide provides you with the information you need about cruising from Miami, including info about ships sailing from Miami, cruise schedules, deals and special offers, and the best hotels to stay in before or after your cruise.
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Kenny Chesney, Brett Young Reign on Albums, Singles Chart
Kenny Chesney debuts at the top of Billboard's country albums chart this week with Songs for the Saints. It's his 12th studio album to reach this pinnacle -- and his 16th if you count in his greatest-hits compilations.
Brett Young's "Mercy" rolls in as the week's most-played song. It is Young's 3rd No. 1 single, and it took 26 weeks to get there.
Cody Jinks' Lifer pops in at No. 2. Three albums re-enter the charts — Chesney's Live in No Shoes Nation (back at No. 37), Jordan Davis' Home State (No. 49) and Songs You Know by Heart: Jimmy Buffett's Greatest Hit(s) (No. 50).
There are two new singles — Danielle Bradbery's "Goodbye Summer," with Thomas Rhett(No. 49) and Brandon Lay's "Yada Yada Yada" (No. 58).
Aaron Watson's "Run Wild Horses" gallops back on at No. 40.
The No. 3 through No. 5 albums, in that order, are Luke Combs' This One's for You, Jason Aldean's Rearview Town and Dan + Shay.
Trooping in directly behind Young's "Mercy" in the Top 5 songs array are Chesney's "Get Along," Keith Urban's "Coming Home," featuring Julia Michaels, Dan + Shay's "Tequila" and Aldean's "Drowns the Whiskey," featuring Miranda Lambert.
Jake Owen's "I Was Jack (You Were Diane)," last week's No. 1 slips to No. 12.
Last week, I ended this column by asking, "Any questions?"
"Yes," a reader responded. "Where are the women?"
The answer is not a happy one: They're mostly missing.
Among the Top 50 country albums this week, only seven are by women. And it's not much sunnier within the Top 60 airplay rankings — just seven songs by women as principal artists and five more with women "featured" as secondary artists.
To whom does this imbalance matter? Obviously, to the women singers and songwriters who seek to be heard and to earn a living from their art. But do buyers and listeners care if the voices and personalities that entertain them mirror the gender makeup of the population? If so, how can they most effectively assert their preferences and discontents?
Damned if I know. What do you think?
Read more http://www.cmt.com/news/1797738/kenny-chesney-brett-young-reign-on-albums-singles-chart/


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text:
\section{Introduction}

A typical feature of quantum theory is that we have many layers of descriptions, differing by their amount of details.
Usually the choice of the descriptions depends on our needs or on available information.
We can use highly complete or total description of some device or setting, but also less detailed description of the same thing if that is more suitable for our purposes or we simply lack information.
A prototypical example is joint states versus reduced states.
A joint state describes a state of a composite system, whereas a reduced state describes a state of one of the subsystems.
A generic joint state contains more information than the related reduced states, and even if we know the reduced states of all subsystems, we may not know the joint state.

In a typical case lower layer description (e.g.\  reduced states) does not uniquely determine higher layer description (joint state), since variation is greater in the latter and even big differences can lead to the same coarser descriptions.
However, in some special cases pieces can be combined into a whole in a unique way.
In this work we identify one such condition to be extremality of a piece and we unify several results that fall under this theme.
By an extremal object we mean an extremal point in the convex set of all similar objects. (This total set can be e.g.\  the set of states, observables, instruments or channels.)

We will consider objects consisting of two pieces.
We show that, in a well specified setting,
\begin{quote}
\emph{when one of the pieces is an extremal object, then pieces uniquely determine the whole}.
\end{quote}
We will further prove that
\begin{quote}
\emph{when both of the pieces are extremal objects, then the whole is an extremal object}.
\end{quote}

We recall the following well-known results that exemplify the previously sketched ideas.

\begin{itemize}

\item[(a)] \emph{Joint state with a pure marginal state}: Suppose that $\varrho$ is a state of a composite system $\mc H_1\otimes\mc H_2$. If one of the reduced states $\mr{tr}_{\mc H_2}[\varrho]\equiv\varrho_1$ or $\mr{tr}_{\mc H_1}[\varrho]\equiv\varrho_2$ is pure, then $\varrho=\varrho_1\otimes\varrho_2$.
If both $\varrho_1$ and $\varrho_2$ are pure, then also $\varrho$ is pure.

\item[(b)] \emph{Joint observable with a sharp marginal observable}: Suppose that $\ms{M}$ and $\ms{N}$ are jointly measurable observables (POVMs).
If $\ms{M}$ or $\ms{N}$ is sharp (i.e.\  projection valued measure), then their joint observable $\ms{J}$ is unique and it is determined by the condition $\ms{J}(X\times Y)=\ms{M}(X)\ms{N}(Y)$ for all outcome sets $X,Y$.
(See e.g.\  \cite{HeReSt08} for a proof of this fact.)

\item[(c)] \emph{Instruments related to a sharp observable}: Suppose that an observable $\ms{M}$ is sharp and $\Gamma$ is an instrument such that $\Gamma(X,\id)=\ms{M}(X)$ for all outcome sets $X$.
Then $\Gamma(X,A)=\ms M(X)\mc E(A)$, where $\mc E=\Gamma(\Omega,\cdot)$ is the total state transformation and $\Omega$ is the total set \cite{Ozawa84}. Hence the instrument $\Gamma$ is completely determined by its total state transformation $\mc E$.

\item[(d)] \emph{Variant of `No Cloning Theorem'}: Suppose $\mc F:\trh\to\trhh$ is a quantum channel such that $\mr{tr}_{1}[\mc F(\varrho)]=\varrho$ for every state $\varrho$.
Then $\mr{tr}_{2}[\mc F(\varrho)]\equiv\sigma$ for some fixed state $\sigma$, hence the attempted copy $\mr{tr}_{2}[\mc F(\varrho)]$ contains no information on the input state $\varrho$.

\end{itemize}

\noindent Our main result contains all these statements as corollaries and identifies the common source behind the uniqueness claims  as being extremality of a marginal map.
We will also demonstrate some new applications of this result.
Our main theorem implies the following:

\begin{itemize}

\item[(e)] Suppose that $\ms{M}$ and $\ms{N}$ are jointly measurable observables (POVMs).
If $\ms{M}$ or $\ms{N}$ is extremal, then their joint observable is unique.
If both $\ms{M}$ and $\ms{N}$ are extremal, then their unique joint observable is extremal.

\item[(f)] Suppose that an observable $\ms{M}$ and a channel $\Ec$ are parts of a single instrument $\Gamma$, i.e., $\Gamma(X,\id)=\ms{M}(X)$ for all outcome sets $X\subseteq\Omega$ and
$\Gamma(\Omega,\cdot)=\Ec$ for the total set $\Omega$.
If $\ms{M}$ or $\Ec$ is extremal, then the instrument $\Gamma$ is unique.
If $\ms{M}$ and $\Ec$ are both extremal, then $\Gamma$ is extremal.

\end{itemize}

Our investigation is organized as follows.
In Section \ref{sec:prel} we fix the notation and recall some standard results.
In Section \ref{sec:ext} we derive a general criterion for extremality.
Our main result, sketched here in the introduction, is proved in Section \ref{sec:joint}. In Section \ref{sec:comp} we concentrate on quantum channels on compound systems and prove that causal channels are local if one of the channels reduced to one of the subsystems is extremal.
Finally, in Section \ref{sec:end} we summarize our conclusions and make some final remarks.

\section{Notation and preliminaries}\label{sec:prel}

We denote by $\mc A$ a von Neumann algebra with the unit $\id_{\mc A}$ and $\mc H$ is a complex Hilbert space with the inner product $\sis{\cdot}{\cdot}:\mc H\times\mc H\to\mb C$.
We denote the set of bounded linear operators on $\mc H$ by $\lh$ and the unit of $\lh$ by $\id_{\mc H}$.
The subscripts of the unities may be omitted if there is no ambiguity of the algebra or Hilbert
space.
Although we concentrate on von Neumann algebras, it should be noted that Theorem \ref{dilat} and Proposition \ref{RNd} of this section also hold in the more general
setting of $C^*$-algebras. We follow the convention $\nat=\{1,\,2,\,3,\ldots\}$.

We recall that a linear map $\Phi:\mc A\to\mc L(\mc H)$ is {\it completely positive} (CP) if for any $n\in\mb N$, $a_1,\ldots,\,a_n\in\mc A$ and $\varphi_1,\ldots,\,\varphi_n\in\mc H$ we have
\begin{equation}
\sum_{j,k=1}^n\sis{\varphi_j}{\Phi(a_j^*a_k)\varphi_k}\geq0.
\end{equation}
We denote the set of all CP maps $\Phi:\mc A\to\mc L(\mc H)$ by ${\bf CP}(\mc A;\mc H)$.
We recall the following fundamental result for CP maps \cite{Stinespring55}.

\begin{theorem}\label{dilat}
For any $\Phi\in{\bf CP}(\mc A;\mc H)$ there is a triple $(\mc M,\pi,J)$ where $\mc M$ is a Hilbert space, $\pi:\mc A\to\mc L(\mc M)$ is a unital *-representation and $J:\mc H\to\mc M$ is a linear map
such that
\begin{itemize}
\item[{\rm (i)}] $\Phi(a)=J^*\pi(a)J$ for all $a\in\mc A$ and
\item[{\rm (ii)}] the linear span of the set $\{\pi(a)J\varphi\,|\,a\in\mc A,\ \varphi\in\mc H\}$ is dense in $\mc M$.
\end{itemize}
If $(\mc M',\pi',J')$ is another such triple, there is a unitary operator $U:\mc M\to\mc M'$ such that $U\pi(a)=\pi'(a)U$ for all $a\in\mc A$ and $UJ=J'$.
\end{theorem}

\begin{definition}
{\rm
Suppose that $\Phi\in{\bf CP}(\mc A;\mc H)$ is associated with a triple $(\mc M,\pi,J)$ satisfying the conditions (i)--(ii) of  Theorem \ref{dilat}.
The triple $(\mc M,\pi,J)$ is called a {\it minimal Stinespring dilation for $\Phi$}.
}
\end{definition}

If the Hilbert space $\mc H$ and the algebra $\mc A$ are separable, the dilation space $\mc M$ of a minimal Stinespring dilation of a CP map $\Phi\in{\bf CP}(\mc A;\mc H)$ can be chosen to be
separable \cite[Section 9.2]{QTOS76}. This is the case in the typical physical applications; see the end of this section.

Suppose that $\Phi,\,\Psi\in{\bf CP}(\mc A;\mc H)$. We denote $\Psi\leq\Phi$ or $\Phi\geq\Psi$ when $\Phi-\Psi\in{\bf CP}(\mc A;\mc H)$. This relation is a partial order in ${\bf CP}(\mc A;\mc H)$.
The following result has been proved in \cite{Raginsky03}.
We provide a proof for the reader's convenience.

\begin{proposition}\label{RNd}
Suppose that $\Phi,\,\Psi\in{\bf CP}(\mc A;\mc H)$ and $\Psi\leq\Phi$. Also assume that $(\mc M,\pi,J)$ is a minimal Stinespring dilation of $\Phi$. There is a unique operator $E\in\mc L(\mc M)$ such
that $[E,\pi(a)]=0$ for all $a\in\mc A$ and
\begin{equation}\label{deriv}
\Psi(a)=J^*\pi(a)EJ  \qquad \forall a\in\mc A \, .
\end{equation}
\end{proposition}

\begin{proof}
Suppose that $(\mc M',\pi',J')$ is a minimal Stinespring dilation of $\Psi$. Pick $n\in\mb N$, $a_1,\ldots,\,a_n\in\mc A$ and $\varphi_1,\ldots,\,\varphi_n\in\mc H$ and set
\begin{equation}\label{eq:eta}
\eta=\sum_{j=1}^n\pi(a_j)J\varphi_j\in\mc M,\qquad\zeta=\sum_{j=1}^n\pi'(a_j)J'\varphi_j\in\mc M'.
\end{equation}
Using $\Psi\leq\Phi$ we may evaluate
\begin{eqnarray*}
\|\zeta\|^2&=&\sum_{j,k=1}^n\sis{J'\varphi_j}{\pi'(a_j^*a_k)J'\varphi_k}=\sum_{j,k=1}^n\sis{\varphi_j}{\Psi(a_j^*a_k)\varphi_k}\leq\sum_{j,k=1}^n\sis{\varphi_j}{\Phi(a_j^*a_k)\varphi_k}\\
&=&\sum_{j,k=1}^n\sis{J\varphi_j}{\pi(a_j^*a_k)J\varphi_k}=\|\eta\|^2.
\end{eqnarray*}
Using property (ii) of $(\mc M,\pi,J)$ in Theorem \ref{dilat} we define an operator $D:\mc M\to\mc M'$ such that $D\pi(a)J\varphi=\pi'(a)J'\varphi$ for all $a\in\mc A$ and $\varphi\in\mc H$; by the above calculation $D$ is well-defined and bounded.
It follows that
\begin{equation}
\Psi(a)=(J')^*\pi'(a)J'=J^*D^*D\pi(a)J
\end{equation}
for all $a\in\mc A$.
Moreover for all $a,\,b\in\mc A$ and $\varphi\in\mc H$ one has
\begin{equation*}
D\pi(a)\pi(b)J\varphi=D\pi(ab)J\varphi=\pi'(ab)J'\varphi=\pi'(a)\pi'(b)J'\varphi=\pi'(a)D\pi(b)J\varphi \, .
\end{equation*}
The minimality of $(\mc M,\pi,J)$ yields thus $D\pi(a)=\pi'(a)D$ for all $a\in\mc A$.
Hence also $\pi(a)D^*=D^*\pi'(a)$ for all $a\in\mc A$.
Define $E:=D^*D$ and, using the results above, we see that $E$ is as in the claim.

Suppose now that $E_1,\,E_2\in\mc L(\mc M)$ are as in the claim.
Suppose that $\eta\in\mc M$ is the vector defined in \eqref{eq:eta}.
We obtain
\begin{eqnarray*}
\sis{\eta}{(E_1-E_2)\eta}&=&\sum_{j,k=1}^n\sis{J\varphi_j}{\pi(a_j^*)(E_1-E_2)\pi(a_k)J\varphi_k}\\
&=&\sum_{j,k=1}^n\sis{J\varphi_j}{\pi(a_j^*a_k)(E_1-E_2)J\varphi_k}\\
&=&\sum_{j,k=1}^n\big(\sis{\varphi_j}{\Psi(a_j^*a_k)\varphi_k}-\sis{\varphi_j}{\Psi(a_j^*a_k)\varphi_k}\big)=0 \, .
\end{eqnarray*}
Again, from the minimality of $(\mc M,\pi,J)$ follows that $E_1=E_2$.
\end{proof}

Any $\Phi\in{\bf CP}(\mc A;\mc H)$ is guaranteed to be norm continuous. If $\Phi$ is additionally ultraweakly continuous, i.e., is continuous with respect to the ultraweak topologies of
$\mc A$ and $\mc L(\mc H)$, we say that $\Phi$ is {\it normal}. We denote the set of normal CP maps $\Phi\in{\bf CP}(\mc A;\mc H)$ by ${\bf NCP}(\mc A;\mc H)$. Whenever $\Phi\in{\bf NCP}(\mc A;
\mc H)$, there is a unique predual map $\Phi_*:\mc T(\mc H)\to\mc A_*$, where $\mc T(\mc H)$ is the set of trace class operators on $\mc H$ (the predual of $\mc L(\mc H)$) and $\mc A_*$ is the predual of
$\mc A$, which is continuous with respect to the trace norm topology of $\mc T(\mc H)$ and the norm topology of $\mc A_*$. The maps $\Phi$ and $\Phi_*$ are related by
\begin{equation}
\tr{T\Phi(a)}=\langle\Phi_*(T),a\rangle,\qquad T\in\mc T(\mc H) \, , \quad a\in\mc A \, ,
\end{equation}
where $\langle\cdot,\cdot\rangle:\mc A_*\times\mc A\to\mb C$ is the canonical bilinear form (dual pairing).
A typical physically relevant CP map $\Phi$ is normal and hence has the predual $\Phi_*$.
These two maps $\Phi$ and $\Phi_*$ are considered as different representations of the same object, and $\Phi$ is said to be in the {\it Heisenberg picture} while $\Phi_*$ in the {\it Schr\"odinger picture}.

Since $\mc A$ is a von Neumann algebra, Theorem \ref{dilat} can be augmented with the following additional result \cite[Section 9.2]{QTOS76}:
{\it For any $\Phi\in{\bf NCP}(\mc A;\mc H)$, there is a minimal Stinespring dilation $(\mc M,\pi,J)$ where the *-representation $\pi$ is normal.}

\begin{definition}
{\rm
For each positive operator $P\in\mc L(\mc H)$, we denote by ${\bf CP}_P(\mc A;\mc H)$ the set of maps $\Phi\in{\bf CP}(\mc A;\mc H)$ satisfying $\Phi(\id_{\mc A})=P$.
We further denote ${\bf NCP}_P(\mc A;\mc H)={\bf CP}_P(\mc A;\mc H) \cap {\bf NCP}(\mc A;\mc H)$.
}
\end{definition}

The set ${\bf NCP}_P(\mc A;\mc H)$ has different functions in quantum mechanics depending on the choices for the algebra $\mc A$ and the operator $P$.
In the following we recall the most relevant ones. The Hilbert spaces (input and output spaces; typically denoted by $\mc H$ or $\mc K$) appearing in the definitions of the sets of quantum
mechanical CP maps are always assumed to be separable.
In most of the cases the dilation spaces of these CP maps are also automatically separable.

\begin{itemize}

\item \emph{States}:
We say that an operator $\varrho\in\mc L(\mc H)$ is a {\it state} if it is a positive trace class operator and $\tr{\varrho}=1$.
We denote the set of states on $\mc H$ by $\mc S(\mc H)$.
From the duality $\lh_\ast=\trh$ follows that $\mc S(\mc H)={\bf NCP}_1(\mc L(\mc H);\mb C)$.

\item \emph{Channels}:
Suppose that $\mc K$ is a separable Hilbert space.
An element in the set ${\bf NCP}_{\id}(\mc L(\mc K);\mc H) \equiv \mc C(\mc K,\mc H)$ is called a \emph{channel}.
A channel corresponds to a procedure where a system described by the Hilbert space $\mc H$ transforms
into a system described by the space $\mc K$.
The dilation space appearing in a minimal Stinespring dilation of a channel is separable since $\hi$ and $\hik$ are separable.

\item \emph{Observables}:
Suppose that $(\Omega,\Sigma)$ is a measurable space, i.e., $\Omega\neq\emptyset$ and $\Sigma$ is a $\sigma$-algebra of subsets of $\Omega$. Also assume that $\nu:\Sigma\to[0,\infty]$ is a
$\sigma$-finite measure.
An element in the set ${\bf NCP}_{\id}(L^\infty(\nu);\mc H)\equiv\mc O_\nu(\mc H)$ is called an \emph{observable}.
Observables that are *-homomorphisms are called sharp observables \cite{OQP97}.
Usually observables are identified with normalized positive operator valued measures (POVMs) in the following way: for any $M\in\mc O_\nu(\mc H)$ and $X\in\Sigma$, we denote $\ms M(X)=M(\chi_X)$, where $\chi_X$ is the characteristic function of $X$. Now $\ms M$ is a POVM and $M(f)=\int f\,d\ms M$ for all
$f\in L^\infty(\nu)$.
Conversely, every POVM is of the above form for some measure $\nu$ and $M\in \mc O_\nu(\mc H)$ \cite{Part1}. From now on, we view observables as POVMs and we denote POVMs defined on $\Sigma$ by $\mc O_\Sigma(\mc H)$. In this view, sharp observables are projection-valued measures. The minimal
dilation of a POVM is often called a minimal {\it Na\u{\i}mark dilation}. The separability or non-separability of the dilation space of a minimal Stinespring dilation of a POVM depends on the properties of the
value space $(\Omega,\Sigma)$.
For instance, if the $\sigma$-algebra $\Sigma$ is numerably generated, then the algebra $L^\infty(\nu)$ with any $\sigma$-finite measure $\nu:\Sigma\to[0,\infty]$ is separable and hence the
dilation space in a minimal Stinespring dilation is guaranteed to be separable.

\item \emph{Instruments}:
Suppose that $(\Omega,\Sigma)$ is a measurable space and $\nu:\Sigma\to[0,\infty]$ is a
$\sigma$-finite measure.
An element in the set ${\bf NCP}_{\id}(L^\infty(\nu)\otimes\mc L(\mc K);\mc H)\equiv\mc I_\nu(\mc K;\mc H)$ is called an \emph{instrument} \cite{QTOS76}.
An instrument describes the conditional state changes
experienced by a quantum system in a measurement with the value space $(\Omega,\Sigma,\nu)$. Instruments $G\in\mc I_\nu(\mc K;\mc H)$ are usually treated as maps $\Gamma:\Sigma\times
\mc L(\mc K)\to\mc L(\mc H)$ such that $\Gamma(X,B)=G(\chi_X\otimes B)$ for all $X\in\Sigma$ and $B\in\mc L(\mc K)$. As in the case of observables, instruments can be defined in a broader
context as maps $\Gamma:\Sigma\times\mc L(\mc K)\to\mc L(\mc H)$, where $\Gamma(\cdot,B)$ is an operator measure for all $B\in\mc L(\mc K)$, $\Gamma(\cdot,\id)\in\mc O_\Sigma(\mc H)$,
$\Gamma(X,\cdot)$ is a normal linear CP map and $\Gamma(\Omega,\cdot)\in\mc C(\mc K;\mc H)$.
We denote the set of such maps by $\mc I_\Sigma(\mc K;\mc H)$. The dilation space of a minimal
Stinespring dilation of an instrument $\Gamma$ is separable if and only if the dilation space of a minimal Na\u{\i}mark dilation of the associated POVM $\Gamma(\cdot,\id)$ is separable, hence the
separability of the dilation space depends on the properties of the value space $(\Omega,\Sigma)$.

\end{itemize}

\section{Extremal points}\label{sec:ext}

The set ${\bf CP}_P(\mc{A};\mc H)$ is convex; if $\Phi_1,\Phi_2\in {\bf CP}_P(\mc{A};\mc H)$, then $t\Phi_1 + (1-t) \Phi_2\in{\bf CP}_P(\mc{A};\mc H)$ for all $0\leq t \leq 1$.
An element $\Phi\in{\bf CP}_P(\mc{A};\mc H)$ is called  \emph{extremal} (or \emph{extreme} or \emph{pure}) in ${\bf CP}_P(\mc{A};\mc H)$ if it cannot be written as a mixture $\Phi=t\Phi_1+(1-t)\Phi_2$ for two different elements
$\Phi_1,\Phi_2 \in{\bf CP}_P(\mc{A};\mc H)$ and $0<t<1$.
It is obvious that when deciding whether $\Phi$ is extremal or not, it suffices to study mixtures with $t=1/2$.
The following result characterizes the
extremal elements in ${\bf CP}_P(\mc{A};\mc H)$ \cite{Arveson69}. We provide a proof for the reader's convenience.

\begin{theorem}\label{ext}
Suppose that $\Phi\in{\bf CP}_P(\mc{A};\mc H)$ and let $(\mc M,\pi,J)$ be its minimal Stinespring dilation. The map $\Phi$ is extremal in ${\bf CP}_P(\mc{A};\mc H)$ if and only if for any $E\in\mc L(\mc M)$ the
conditions $[E,\pi(a)]=0$ for all $a\in\mc A$ and $J^*EJ=0$ imply $E=0$.
\end{theorem}

\begin{proof}
Suppose that $\Phi\in{\bf CP}_P(\mc{A};\mc H)$ is not extremal, i.e., there are $\Phi_\pm\in{\bf CP}_P(\mc{A};\mc H)$ such that $\Phi_+\neq\Phi_-$ and $\Phi=\frac12\Phi_++\frac12\Phi_-$. Hence
$\Phi_\pm\leq2\Phi$ and according to Proposition \ref{RNd} there are operators $F_\pm\in\mc L(\mc M)$ that commute with $\pi$ and
\begin{equation}
\Phi_\pm(a)=2J^*\pi(a)F_\pm J=J^*\pi(a)E_\pm J \qquad \forall a\in\mc{A} \, ,
\end{equation}
where $E_\pm=2F_\pm$.
Define $E=E_+-E_-$.
The operator $E$ is nonzero since $\Phi_+\neq\Phi_-$.
Moreover, $E$ commutes with $\pi$ and
\begin{equation}
J^*EJ=J^*E_+J-J^*E_-J=\Phi_+(1)-\Phi_-(1)=P-P=0 \, .
\end{equation}

Suppose then that $E\neq0$ satisfies $[E,\pi(a)]=0$ for all $a\in\mc{A}$ and $J^*EJ=0$.
We may assume that $E$ is selfadjoint; if it is not selfadjoint we may redefine $E'=i(E^*-E)$.
We may also assume that $\|E\|\leq1$; otherwise we can redefine $E''=\|E\|^{-1}E$.
Thus we are free to assume that $-\id_{\mc{M}}\leq E\leq \id_{\mc{M}}$ and we can define the positive operators
$E_\pm=\id_{\mc{M}}\pm E$ that commute with the representation $\pi$ and the CP maps $\Phi_\pm\in{\bf CP}(\mc{A};\mc H)$ through
\begin{equation}
\Phi_\pm(a)=J^*\pi(a)(\id_{\mc{M}}\pm E)J=J^*\sqrt{\id_{\mc{M}}\pm E}\pi(a)\sqrt{\id_{\mc{M}}\pm E}J \qquad \forall a\in\mc{A} \, .
\end{equation}
Using the properties of $E$ we obtain
\begin{equation}
\Phi_\pm(1)=J^*(\id_{\mc{M}}\pm E)J=J^*J\pm J^*EJ=J^*J=\Phi(1)=P \, .
\end{equation}
Thus $\Phi_\pm\in{\bf CP}_P(\mc{A};\mc H)$.
Since $E\neq0$ and the dilation $(\mc M,\pi,J)$ is minimal, it follows that for some $\varphi,\,\psi\in\mc H$ and $a,\,b\in\mc{A}$ we have
\begin{equation}
0\neq\sis{\pi(a)J\varphi}{E\pi(b)J\psi}=\frac12\Sis{\varphi}{\big(\Phi_+(a^*b)-\Phi_-(a^*b)\big)\psi} \, .
\end{equation}
Thus $\Phi_+\neq\Phi_-$ and $\Phi$ has a nontrivial convex decomposition $\Phi=\frac12\Phi_+
+\frac12\Phi_-$.
Therefore, $\Phi$ is not extremal in ${\bf CP}_P(\mc{A};\mc H)$.
\end{proof}

Theorem \ref{ext} can also be applied to the convex set ${\bf NCP}_P(\mc{A};\mc H)$.
Namely, {\it a map $\Phi\in{\bf NCP}_P(\mc{A};\mc H)$ associated with a minimal Stinespring dilation $(\mc{M},\pi,J)$ is
extremal in ${\bf NCP}_P(\mc{A};\mc H)$ if and only if for an operator $E\in\mc L(\mc M)$ the conditions $[E,\pi(a)]=0$ for all $a\in\mc{A}$ and $J^*EJ=0$ yield $E=0$.} This is due to the fact that when
$\Phi\in{\bf NCP}_P(\mc{A};\mc H)$, then any map $\Psi\in{\bf CP}_P(\mc{A};\mc H)$ appearing in a convex decomposition of $\Phi$ is necessarily normal.

Theorem \ref{ext} gives extremality conditions for the usual physically relevant CP maps.
We recall the following characterizations.

\begin{itemize}

\item \emph{States}: Since the set of states can be associated with ${\bf NCP}_1(\mc L(\mc H);\mb C)$ the extremal points are
characterized by Theorem \ref{ext}; {\it a state $\varrho$ is extremal in $\mc S(\mc H)$ if and only if $\varrho$ is a one-dimensional projection.}
This is, of course, a classical and well-known result.
An extremal state is usually called {\it pure}.

\item \emph{Observables}: Let $T$ be a trace class operator satisfying $\sis{\varphi}{T\varphi}>0$ for all $\varphi\in\mc H\setminus\{0\}$. Suppose that $\ms M\in\mc O_\Sigma(\mc H)$ and define a finite measure $\nu_T^{\ms M}=\tr{T\ms M(\cdot)}$. One sees that the complex measures $X\mapsto
\sis{\varphi}{\ms M(X)\psi}$, $\varphi,\,\psi\in\mc H$, are absolutely continuous with respect to $\nu_T^{\ms M}$, or briefly $\ms M\ll \nu_T^{\ms M}$. Hence, $\ms M\in\mc O_{\nu_T^{\ms M}}(\mc H)$.
If $\ms M\in\mc O_\Sigma(\mc H)$ and $\ms M\ll \nu$, where $\nu:\Sigma\to[0,\infty]$ is a $\sigma$-finite measure, it follows that whenever $\ms N\in\mc O_\Sigma(\mc H)$ appears in a convex
decomposition of $\ms M$, then $\ms N\ll \nu$. Thus, we may apply Theorem \ref{ext} to the larger set $\mc O_\Sigma(\mc H)$: {\it An observable
$\ms M\in\mc O_\Sigma(\mc H)$ with a minimal Na\u{\i}mark dilation $(\mc M,\ms P,J)$, where $\mc M$ is a Hilbert space, $\ms P\in\mc O_\Sigma(\mc M)$ a projection-valued measure and $J:\mc H\to
\mc M$ is an isometry so that $\ms M(X)=J^*\ms P(X)J$, is extremal in $\mc O_\Sigma(\mc H)$ if and only if for an operator $E\in\mc L(\mc M)$ the conditions $[E,\ms P(X)]=0$ for all $X\in\Sigma$ and
$J^*EJ=0$ yield $E=0$.}
This result has been also proved in \cite{Pellonpaa11}.

\item \emph{Channels}: Let us consider a channel $\mc E\in\mc C(\mc K;\mc H)$.
Assume that $(\mc M,\pi,J)$ is a minimal Stinespring dilation of $\mc E$. The *-representation $\pi$ of $\mc L(\mc K)$ is normal and hence it is unitarily
equivalent with a direct sum of the identity map. This means that we may choose $\mc M=\mc K\otimes\mc M_0$, where $\mc M_0$ is a separable Hilbert space (often called as an {\it ancilla})
and $\pi(B)=B\otimes \id_{\mc M_0}$ for all $B\in\mc L(\mc K)$. We often identify the triple $(\mc M,\pi,J)$ with the pair $(\mc M_0,J)$. The set of those operators in $\mc L(\mc M)$ that
commute with the representation $\pi$ are of the form $\id_{\mc K}\otimes D$ with $D\in\mc L(\mc M_0)$. We may characterize the set of extremal channels in the following way: {\it A channel
$\mc E\in\mc C(\mc K;\mc H)$ with the minimal dilation $(\mc M_0,J)$ is extremal in $\mc C(\mc K;\mc H)$ if and only if for any $D\in\mc L(\mc M_0)$ the condition $J^*(\id_{\mc K}\otimes D)J=0$
implies $D=0$.}

We may characterize the extremality of a channel $\mc E\in\mc C(\mc K;\mc H)$ also in another way, first proved in \cite{Choi75} in the case of finite Hilbert spaces.
Suppose that $(\mc M_0,J)$ is a minimal dilation of $\mc E$ as above.
We fix an orthonormal basis $\{\xi_j\}_{j=1}^{\dim{(\mc M_0)}}$ and define the operators $R_j:\mc K\otimes\mc M_0\to\mc K$, such that $R_j(\varphi\otimes\psi)=\sis{\xi_j}{\psi}\varphi$ for all $\varphi\in\mc K$ and
$\psi\in\mc M_0$.
For all $B\in\mc L(\mc K)$, we may write
\begin{eqnarray*}
\mc E(B)&=&J^*(B\otimes \id_{\mc M_0})J=\sum_{j=1}^{\dim{(\mc M_0)}}J^*(B\otimes\ktb{\xi_j}{\xi_j})J\\
&=&\sum_{j=1}^{\dim{(\mc M_0)}}J^*R_j^*BR_jJ=\sum_{j=1}^{\dim{(\mc M_0)}}K_j^*BK_j,
\end{eqnarray*}
where we have defined $K_j=R_jJ$. The operators $K_j$ are said to set up a {\it Kraus decomposition} for $\mc E$, and when the operators are obtained as above from a minimal dilation, the
decomposition or the set $\{K_j\}_j$ is said to be {\it minimal}. Suppose that $D\in\mc L(\mc M_0)$ and define $d_{j,k}=\sis{\xi_j}{D\xi_k}$. Inserting the identity operator $\sum_j\ktb{\xi_j}{\xi_j}$
on both sides of $D$, one finds that
\begin{equation}
J^*(\id_{\mc K}\otimes D)J=\sum_{j,k}d_{j,k}K_j^*K_k \, .
\end{equation}
A set $\{B_{j,k}\}_{j,k}$ of bounded operators on $\mc K$ is said to be {\it strongly independent}
if $\sum_{j,k}\alpha_{j,k}B_{j,k}=0$ implies $\alpha_{j,k}=0$ for all $j,\,k$ whenever $(\alpha_{j,k})_{j,k=1}^{\dim{(\mc K)}}$ is a matrix representation of a bounded operator on $\mc K$. We may
rephrase the extremality condition of channels: {\it A channel $\mc E\in\mc C(\mc K;\mc H)$ with a minimal set $\{K_j\}_j$ of Kraus operators is extremal in $\mc C(\mc K;\mc H)$ if and only if the
set $\{K_j^*K_k\}_{j,k}$ is strongly independent.}
This result can be found in \cite{Tsui96}.

\item \emph{Instruments}:
A complete characterization of extremal instruments is presented in \cite{Part1}.
It can be stated in a similar form as for channels by using minimal pointwise Kraus decompositions of instruments.
Actually, all the above characterizations are special cases of this general characterization due to the fact that states, observables, and channels can be viewed as instruments \cite{Part1}.

\end{itemize}

\section{Marginal maps and joint maps}\label{sec:joint}

In this section $\mc A$ and $\mc B$ are von Neumann algebras so that we may define their von Neumann tensor product $\mc A\otimes\mc B$.
We also assume that $P\in\mc L(\mc H)$ is fixed.
We study maps belonging to ${\bf CP}_P(\mc{A};\mc H)$, ${\bf CP}_P(\mc{B};\mc H)$ or ${\bf CP}_P(\mc{A}\otimes\mc{B};\mc H)$.
We give our main result (Theorem \ref{extmarg}) without assuming normality.
According to the discussion in the end of the previous section, all results apply also in the case of normal maps.

\begin{definition}
{\rm
Suppose that $\Psi\in{\bf CP}_P(\mc A\otimes\mc B;\mc H)$.
The maps
\begin{equation}
\Psi_{(1)}:\mc A\to\mc L(\mc H) \, , \quad \Psi_{(1)}(a)=\Psi(a\otimes \id_{\mc B}) \quad \textrm{for all $a\in\mc A$}
\end{equation}
and
\begin{equation}
\Psi_{(2)}:\mc B\to\mc L(\mc H) \, , \quad \Psi_{(2)}(b)=\Psi(\id_{\mc A}\otimes b) \quad \textrm{for all $b\in\mc B$}
\end{equation}
belong to ${\bf CP}_P(\mc A;\mc H)$ and ${\bf CP}_P(\mc B;\mc H)$, respectively.
We call the maps $\Psi_{(1)}$ and $\Psi_{(2)}$ as the {\it marginals of $\Psi$}. Suppose that $\Phi_1\in{\bf CP}_P(\mc A;\mc H)$ and $\Phi_2\in{\bf CP}_P(\mc B;\mc H)$. If $\Phi_1$ and $\Phi_2$
are marginals of some $\Psi\in{\bf CP}_P(\mc A\otimes\mc B;\mc H)$, i.e., $\Phi_1=\Psi_{(1)}$ and $\Phi_2=\Psi_{(2)}$, we say that $\Phi_1$ and $\Phi_2$ are {\it compatible}. Moreover, we say that $\Psi$ is a {\it joint map} for $\Phi_1$ and $\Phi_2$.
}
\end{definition}

To demonstrate the content of this definition, let us recall two common instances of compatibility.
Suppose that $\mc K$ is a Hilbert space, $(\Omega,\Sigma)$ is a measurable space, and $\nu:\Sigma\to[0,\infty]$ is a $\sigma$-finite measure.
We set $\mc A=L^\infty(\nu)$, $\mc B=\mc L(\mc K)$ and $P=\id_\mc H$.
If an observable $\ms M\in\mc O_\nu(\mc H)$ and a channel $\mc E\in\mc C(\mc K;\mc H)$ are compatible, we say that their joint
map $\Gamma\in\mc I_\nu(\mc K;\mc H)$ is their {\it joint instrument}.
Clearly, every instrument is a joint instrument of some observable and channel.
For instance, let $\Omega=\nat$ and $\Gamma(X,B)=\sum_{j\in X} K_j^\ast B K_j$ for some set of operators $\{K_j\}_j$.
Then the marginals of $\Gamma$ are $\ms M(X)=\sum_{j\in X} K_j^\ast K_j$ and $\mc E(B)=\sum_{j\in\Omega} K_j^\ast B K_j$.

For another example, we set $\mc A=L^\infty(\nu)$ and $\mc B=L^\infty(\nu')$, where
 $(\Omega',\Sigma')$ is a measurable space and $\nu':\Sigma'\to[0,\infty]$ is a $\sigma$-finite measure.
If two observables $\ms M_1\in\mc O_\nu(\mc H)$ and $\ms M_2\in
\mc O_{\nu'}(\mc H)$ are compatible, we say that their joint map $\ms M\in\mc O_{\nu\times\nu'}(\mc H)$ is their {\it joint observable}.
Compatible observables are usually called {\it jointly measurable}.
According to the remarks made in the previous section, all the results concerning observables and instruments in this section also apply to the entire sets $\mc O_\Sigma(\hi)$,
$\mc O_{\Sigma'}(\hi)$ and $\mc I_\Sigma(\mc K;\hi)$. It should be noted that a joint observable $\ms M$ of $\ms M_1\in\mc O_\Sigma(\mc H)$ and $\ms M_2\in\mc O_{\Sigma'}(\mc H)$ is in $\mc O_{\Sigma
\otimes\Sigma'}(\mc H)$, where $\Sigma\otimes\Sigma'$ is the product $\sigma$-algebra of $\Sigma$ and $\Sigma'$.
The fact that $\ms M$ is a joint observable means that
$\ms M(X\times \Omega)=\ms M_1(X)$ and $\ms M(\Omega' \times Y)=\ms M_2(Y)$ for all  $X\in\Sigma$, $Y\in\Sigma'$.

The main result of our investigation is the following.

\begin{theorem}\label{extmarg}
Suppose that $\Phi_1\in{\bf CP}_P(\mc A;\mc H)$ and $\Phi_2\in{\bf CP}_P(\mc B;\mc H)$ are compatible.
\begin{itemize}
\item[{\rm (a)}] If $\Phi_1$ is extremal in ${\bf CP}_P(\mc A;\mc H)$ or $\Phi_2$ is extremal in ${\bf CP}_P(\mc B;\mc H)$, then they have a unique joint map.
\item[{\rm (b)}] If $\Phi_1$ is extremal in ${\bf CP}_P(\mc A;\mc H)$ and $\Phi_2$ is extremal in ${\bf CP}_P(\mc B;\mc H)$, then their unique joint map is extremal in ${\bf CP}_P(\mc A\otimes\mc B;\mc H)$.
\item[{\rm (c)}] If $\Phi_1$ or $\Phi_2$ is a *-representation, then $\Phi_1$ and $\Phi_2$ commute and the unique joint map $\Psi\in{\bf CP}(\mc A\otimes\mc B;\mc H,P)$ is of the form
\begin{equation}\label{tahtiesit}
\Psi(a\otimes b)=\Phi_1(a)\Phi_2(b),\qquad a\in\mc A,\quad b\in\mc B.
\end{equation}
\end{itemize}
\end{theorem}

Before we present a proof of Theorem \ref{extmarg}, we need some auxiliary results.

\begin{lemma}\label{submin}
Suppose that $\Psi\in{\bf CP}_P(\mc A\otimes\mc B;\mc H)$ and $(\mc M,\pi,J)$ is a minimal Stinespring dilation of the first marginal $\Psi_{(1)}$. There is a unique map
$E\in{\bf CP}_\id(\mc B;\mc M)$ such that $[\pi(a),E(b)]=0$ for all $a\in\mc A$, $b\in\mc B$ and
\begin{equation}\label{tulo}
\Psi(a\otimes b)=J^*\pi(a)E(b)J,\qquad a\in\mc A,\quad b\in\mc B \, .
\end{equation}
\end{lemma}

\begin{proof}
Fix $b\in\mc B$ and denote the map $\mc A\ni a\mapsto\Psi(a\otimes b)$ by $\Psi_b$. Clearly, if $b\geq0$, then $\Psi_b\in{\bf CP}(\mc A;\mc H)$. Furthermore
\begin{equation}
\|b\|\Psi_{(1)}(a)-\Psi_b(a)=\Psi\big(a\otimes(\|b\|\id_{\mc B}-b)\big),
\end{equation}
and hence $\Psi_b\leq\|b\|\Psi_{(1)}$ for any positive $b\in\mc B$.
Thus, according to Proposition \ref{RNd}, for any positive $b\in\mc B$, we have an operator $E(b)\in\mc L(\mc M)$ that commutes
with $\pi$ such that $\Psi_b(a)=J^*\pi(a)E(b)J$. We may uniquely extend this construction into a linear map $E:\mc B\to\mc L(\mc M)$ such that $[\pi(a),E(b)]=0$ and \eqref{tulo} holds.
The unitality of $E$ is obvious since
\begin{equation}
J^*\pi(a)J=\Psi_{(1)}(a)=J^*\pi(a)E(\id_{\mc B})J
\end{equation}
and the choice of $E(\id_{\mc B})$ is unique by the minimality of the dilation.

Pick natural numbers $n$, $n_j$ and vectors $\varphi_{jk}\in\mc H$ and $a_{jk}\in\mc A$, $k=1,\ldots,n_j$ $j=1,\ldots,n$ and $b_1,\ldots,b_n\in\mc B$ and define
\begin{equation}
\eta_j=\sum_{k=1}^{n_j}\pi(a_{jk})J\varphi_{jk}\in\mc M.
\end{equation}
We may evaluate
\begin{eqnarray*}
\sum_{j,k=1}^n\sis{\eta_j}{E(b_j^*b_k)\eta_k}&=&\sum_{j,k=1}^n\sum_{l=1}^{n_j}\sum_{m=1}^{n_k}\sis{J\varphi_{jl}}{\pi(a_{jl}^*)E(b_j^*b_k)\pi(a_{km})J\varphi_{km}}\\
&=&\sum_{j,k=1}^n\sum_{l=1}^{n_j}\sum_{m=1}^{n_k}\sis{J\varphi_{jl}}{\pi(a_{jl}^*a_{km})E(b_j^*b_k)J\varphi_{km}}\\
&=&\sum_{j,k=1}^n\sum_{l=1}^{n_j}\sum_{m=1}^{n_k}\sis{\varphi_{jl}}{\Psi\big((a_{jl}\otimes b_j)^*(a_{km}\otimes b_k)\big)\varphi_{km}}\geq0
\end{eqnarray*}
using the complete positivity of $\Psi$. Again, the minimality of $(\mc M,\pi,J)$ implies the complete positivity of $E$.
\end{proof}

\begin{definition}\label{subminma}
{\rm
Suppose that $\Psi\in{\bf CP}_P(\mc A\otimes\mc B;\mc H)$ is associated with a quadruple $(\mc M,\pi,E,J)$ where $(\mc M,\pi,J)$ is a minimal dilation of the first marginal $\Psi_{(1)}$ and
$E\in{\bf CP}_I(\mc B;\mc M)$ as in Lemma \ref{submin}.
We call the quadruple $(\mc M,\pi,E,J)$ as an {\it $\mc A$-subminimal dilation of $\Psi$}. We define the $\mc B$-subminimal dilation in
the same way.
}
\end{definition}

Let $\Psi\in{\bf CP}_P(\mc A\otimes\mc B;\mc H)$ and $(\mc M,\pi,E,J)$ be an $\mc A$-subminimal dilation of $\Psi$.
By extending the map $a\otimes b\mapsto\pi(a)E(b)$ into a (unique) CP map
$F\in{\bf CP}_\id(\mc A\otimes\mc B;\mc M)$, we have $\Psi(c)=J^*F(c)J$ for all $c\in\mc A\otimes\mc B$. Typically, the map $F$ is not a *-representation; hence the name `subminimal dilation'.

\begin{proof}[Proof of Theorem \ref{extmarg}.]
(a) Suppose that $\Psi$ and $\Psi'$ are joint maps for $\Phi_1$ and $\Phi_2$ and assume that $\Phi_1$ is extremal in ${\bf CP}(\mc A;\mc H,P)$ and
$(\mc M,\pi,J)$ is a minimal dilation of $\Phi_1$. Assume that $(\mc M,\pi,E,J)$ and $(\mc M,\pi,E',J)$ are $\mc A$-subminimal dilations for $\Psi$ and respectively for $\Psi'$.
We may write
\begin{equation}
0=\Phi_2(b)-\Phi_2(b)=\Psi(\id_{\mc A}\otimes b)-\Psi'(\id_{\mc A}\otimes b)=J^*\big(E(b)-E'(b)\big)J
\end{equation}
for all $b\in\mc B$. Since the operator $E(b)-E'(b)$ commutes with the representation $\pi$ for all $b\in\mc B$, the extremality of $\Phi_1$ yields $E=E'$. This means that $\Psi(a\otimes b)=
\Psi'(a\otimes b)$ for all $a\in\mc A$ and $b\in\mc B$ and thus $\Psi=\Psi'$.

(b) Suppose that both $\Phi_1$ and $\Phi_2$ are extremal. Item (a) yields the uniqueness of the joint map $\Psi\in{\bf CP}(\mc A\otimes\mc B;\mc H,P)$. Let us
make a counter assumption: there are $\Psi^\pm\in{\bf CP}(\mc A\otimes\mc B;\mc H,P)$ such that $\Psi^+\neq\Psi^-$ and $\Psi=\frac12\Psi^++\frac12\Psi^-$. Hence also $\Phi_r=
\frac12\Psi^+_{(r)}+\frac12\Psi^-_{(r)}$, $r=1,\,2$. Suppose that $(\mc M,\pi,J)$ is a minimal dilation of $\Phi_1$. Since, especially, $\Psi^\pm_{(1)}\leq2\Phi_1$, combining
the result of Proposition \ref{RNd} and the beginning of the proof of Lemma \ref{submin}, we obtain CP maps $E^\pm:\mc B\to\mc L(\mc M)$ such that $\Psi^\pm(a\otimes b)=J^*\pi(a)
E^\pm(b)J$ for all $a\in\mc A$ and $b\in\mc B$. Since $\Psi^\pm$ are characterized by the quadruples $(\mc M,\pi,E^\pm,J)$, it follows that $E^+\neq E^-$. Suppose that
$b\in\mc B$ is such that $E^+(b)\neq E^-(b)$. Since $\Phi_2$ is extremal, it follows that $\Psi^+_{(2)}=\Psi^-_{(2)}$ and thus
$$
0=\Psi^+_{(2)}(b)-\Psi^-_{(2)}(b)=\Psi^+(\id_{\mc A}\otimes b)-\Psi^-(\id_{\mc A}\otimes b)=J^*\big(E^+(b)-E^-(b)\big)J \, .
$$
The extremality of $\Phi_1$ thus yields that $E^+(b)=E^-(b)$, contradicting our assumption on $b$.

(c) Suppose that $\Phi_1$ is associated with a minimal dilation $(\mc M,\pi,J)$ yielding the $\mc A$-subminimal dilation $(\mc M,\pi,E,J)$ so that (\ref{tulo}) holds.
Let us assume that $\Phi_1$ is a *-homomorphism. Then $J$ is unitary and we may simply choose $\mc M=\mc H$, $\pi=\Phi_1$ and $J=I$. It is clear that $\Phi_1$ is extremal and hence the joint
map $\Psi$ is unique. From (\ref{tulo}) it immediately follows that $E=\Phi_2$ and thus $\Phi_2$ commutes with $\Phi_1$ and equation (\ref{tahtiesit}) holds.
\end{proof}

The statements (a)--(f) sketched in the introduction are now consequences of the previous theorem.
Note that, when viewed as CP maps, any two states $\varrho_1\in\mc S(\mc H_1)$ and $\varrho_2\in\mc S(\mc H_2)$ have a joint state $\varrho_1\otimes\varrho_2 \in \mc{S}(\mc H_1\otimes\mc H_2)$.

\begin{cor}
Suppose that $\mc K$, $\mc H$, $\mc H_1$ and $\mc H_2$ are Hilbert spaces, $(\Omega,\Sigma)$ and $(\Omega',\Sigma')$ are measurable spaces and $\nu:\Sigma\to[0,\infty]$ and $\nu':\Sigma'\to[0,
\infty]$ are $\sigma$-finite measures.
\begin{itemize}
\item[{\rm (a)}] Suppose that $\varrho\in\mc S(\mc H_1\otimes\mc H_2)$.
If $\mr{tr}_{\mc H_1}[\varrho]=\varrho_2$ or $\mr{tr}_{\mc H_2}[\varrho]=\varrho_1$ is pure, then $\varrho=\varrho_1\otimes\varrho_2$.
\item[{\rm (b)\&\rm (e)}] Suppose that $\ms M_1\in\mc O_\nu(\mc H)$ and $\ms M_2\in\mc O_{\nu'}(\mc H)$ are jointly measurable. If $\ms M_1$ is extremal in $\mc O_\nu(\mc H)$ or $\ms M_2$ is extremal in
$\mc O_{\nu'}(\mc H)$, then their joint observable $\ms M\in\mc O_{\nu\times\nu'}(\mc H)$ is unique. If both $\ms M_1$ and $\ms M_2$ are extremal, then the joint observable is extremal in
$\mc O_{\nu\times\nu'}(\mc H)$. If $\ms M_1$ or $\ms M_2$ is sharp, then $\ms M_1$ and $\ms M_2$ commute and their unique joint observable $\ms M\in\mc O_{\nu\times\nu'}(\mc H)$ is determined by the condition
$$
\ms M(X\times Y)=\ms M_1(X)\ms M_2(Y),\qquad X\in\Sigma,\quad Y\in\Sigma'.
$$
\item[{\rm (c)\&\rm (f)}] Suppose that $\ms M\in\mc O_\nu(\mc H)$ and $\mc E\in\mc C(\mc K;\mc H)$ are compatible. If $\ms M$ is extremal in $\mc O_\nu(\mc H)$ or $\mc E$ is extremal in $\mc C(\mc K;\mc H)$,
then their joint instrument $\Gamma\in\mc I_\nu(\mc K;\mc H)$ is unique. If both $M$ and $\mc E$ are extremal, then this $\Gamma$ is extremal in $\mc I_\nu(\mc K;\mc H)$. If $\ms M$ is sharp or $\mc E$
is a *-representation, then $[\ms M(X),\mc E(B)]=0$ for all $X\in\Sigma$ and $B\in\mc L(\mc K)$ and the unique joint instrument $\Gamma$ for $\ms M$ and $\mc E$ is given by
$$
\Gamma(X,B)=\ms M(X)\mc E(B)=\sqrt{\ms M(X)}\mc E(B)\sqrt{\ms M(X)},\qquad X\in\Sigma,\quad B\in\mc L(\mc K)
$$
\item[{\rm (d)}] Suppose that $\mc F:\mc T(\mc H)\to\mc T(\mc H)\otimes\mc T(\mc H)$ is a Schr\"odinger channel such that the partial trace $\mr{tr}_1[\mc F(T)]=T$ for all $T\in\mc T(\mc H)$. There
is a state $\sigma\in\mc S(\mc H)$ such that $\mr{tr}_2[\mc F(T)]=\tr T\,\sigma$ for all $T\in\mc T(\mc H)$.

\end{itemize}
\end{cor}

\begin{proof}
Items (b)\&(e) and (c)\&(f) are direct consequences of Theorem \ref{extmarg}.
Let us first concentrate on item (a). Given any Hilbert space $\mc H$, the set $\mc S(\mc H)$ of state operators on $\mc H$ is in
bijective affine correspondence with ${\bf NCP}(\mc L(\mc H);\mb C,1)$ when one defines the map $\mc S(\mc H)\ni\varrho\mapsto\Phi^\varrho$, $\Phi^\varrho(C)=\tr{\varrho C}$, $C\in\mc L(\mc H)$. Suppose that
$\varrho\in\mc S(\mc H_1\otimes\mc H_2)\simeq{\bf NCP}(\mc L(\mc H_1)\otimes\mc L(\mc H_2);\mb C,1)$. The marginals of the corresponding positive map $\Phi^\varrho$ are $\Phi^\varrho_r$, $\Phi^\varrho_1(A)=
\Phi^\varrho(A)=\tr{\varrho(A\otimes I_{\mc H_2})}=\tr{\varrho_1A}$ and similarly $\Phi^\varrho_2(B)=\tr{\varrho_2B}$ for all $A\in\mc L(\mc H_1)$ and $B\in\mc L(\mc H_2)$, where $\varrho_1=\mr{tr}_{\mc H_2}[\varrho]$
and $\varrho_2=\mr{tr}_{\mc H_1}[\varrho]$. It now follows from Theorem \ref{extmarg} that if either one of the marginals, i.e.\ partial traces, of a state $\varrho\in\mc S(\mc H_1\otimes\mc H_2)$ is pure, then
there is no other state $\varrho'\in\mc S(\mc H_1\otimes\mc H_2)$ that has the same marginals. Since especially the state $\varrho_1\otimes\varrho_2$ has the marginals $\varrho_1$ and $\varrho_2$, the claim
follows.

Let us prove item (d). Denote the transpose (Heisenberg channel) of $\mc F$ by $\mc E$, i.e.\ $\mc E\in\mc C(\mc H\otimes\mc H;\mc H)$. We may write
$$
\tr{\mc E_{(1)}(A)T}=\tr{\mc E(A\otimes I)T}=\tr{(A\otimes I)\mc F(T)}=\tr{\mr{tr}_1[\mc F(T)]A}=\tr{TA}
$$
for all $A\in\mc L(\mc H)$ and $T\in\mc T(\mc H)$. Hence $\mc E_{(1)}$ is the identity map and especially a *-homomorphism. Thus, according to item (c) of Theorem \ref{extmarg}, the second
marginal $\mc E_{(2)}$ must take values in the center of $\mc L(\mc H)$, i.e.\ there is a state operator $\sigma\in\mc S(\mc H)$ such that $\mc E_{(2)}(B)=\tr{\sigma B}\id$ for all $B\in\mc L(\mc H)$.
One easily sees that this means $\mr{tr}_2[\mc F(T)]=\tr{T}\sigma$ for all $T\in\mc T(\mc H)$.
\end{proof}

\section{Causal channels on bipartite systems}\label{sec:comp}

In this section we study a special consequence of the results obtained in the previous sections to the structure of channels on bipartite systems.
Before we switch to the channel terminology, we present the essential result in a general form.

Suppose that $\mc A$ and $\mc B$ are von Neumann algebras, $\mc H_1$ and $\mc H_2$ are Hilbert
spaces, and $P_1\in \lin{\mc H_1}$ and $P_2\in \lin{\mc H_2}$ are positive operators.
Let $\Phi_1\in{\bf CP}_{P_1}(\mc A;\mc H_1)$ and $\Phi_2\in{\bf CP}_{P_2}(\mc B;\mc H_2)$.
There is a single CP map
$\Psi\in{\bf CP}_{P_1\otimes P_2}(\mc A\otimes\mc B;\mc H_1\otimes\mc H_2)$ such that $\Psi(a\otimes b)=\Phi_1(a)\otimes\Phi_2(b)$ for all $a\in\mc A$ and $b\in\mc B$, and we denote $\Psi\equiv\Phi_1\otimes\Phi_2$.
We denote the marginals of $\Phi_1\otimes\Phi_2$ by $\tilde\Phi_1$ and $\tilde\Phi_2$, i.e.\ $\tilde\Phi_1(a)=\Phi_1(a)\otimes P_2$ and $\tilde\Phi_2(b)=P_1\otimes\Phi_2(b)$.

\begin{proposition}\label{laajext}
Retain the notations defined above. If $\Phi_1$ is extremal in ${\bf CP}_{P_1}(\mc A;\mc H_1)$, then $\tilde\Phi_1$ is extremal in ${\bf CP}_{P_1\otimes P_2}(\mc A\otimes\mc B;\mc H_1\otimes
\mc H_2)$.
\end{proposition}

\begin{proof}
We prove the claim by showing that if $\tilde\Phi_1$ is {\it not} extremal in ${\bf CP}_{P_1\otimes P_2}(\mc A\otimes\mc B;\mc H_1\otimes\mc H_2)$, then $\Phi_1$ is not extremal in ${\bf CP}_{P_1}(\mc A;\mc H_1)$.
Assume that $(\mc M,\pi,J)$ is a minimal dilation of $\Phi_1$.
Further assume that $\mc M_2$ is a Hilbert space and $J_2:\mc H_2\to\mc M_2$ is a linear map such that the image space $J_2(\mc H_2)$ is dense in
$\mc M_2$ and $P_2=J_2^*J_2$.
(A possible choice is, for instance, $J_2=\sqrt{P_2}$ and $\mc M_2$ is the closure of the range of $\sqrt{P_2}$.)
We denote by $\tilde\pi$ the map defined by $\tilde{\pi}(a)=\pi(a)\otimes\id_{\mc M_2}$, $a\in\mc{A}$.
Then $(\mc M\otimes\mc M_2,\tilde\pi,J\otimes J_2)$ is a minimal dilation for $\tilde\Phi_1$.
For each $\psi\in\mc M_2$, we define the linear operator
$R_\psi:\mc M\otimes\mc M_2\to\mc M$ through
\begin{equation}
R_\psi(\zeta\otimes\xi)=\sis{\psi}{\xi}\zeta
\end{equation}
 for all $\zeta\in\mc M$ and $\xi\in\mc M_2$.
 Pick any $\zeta\in\mc M$, $\psi\in\mc M_2$ and
$a\in\mc A$. We have
\begin{equation}
R_\psi^*\pi(a)\zeta=\big(\pi(a)\zeta\big)\otimes\psi=\tilde\pi(a)(\zeta\otimes\psi)=\tilde\pi(a)R_\psi^*\zeta.
\end{equation}
This yields
\begin{equation}
R_\psi^*\pi(a)=\tilde\pi(a)R_\psi^* \, , \qquad
\pi(a)R_\psi=R_\psi\tilde\pi(a)
\end{equation}
for all $\psi\in\mc M_2$ and $a\in\mc A$.

Suppose that $D\in\mc L(\mc M\otimes\mc M_2)$, $D\neq0$, is such that $\tilde\pi(a)D=D\tilde\pi(a)$ for all $a\in\mc A$ and $(J\otimes J_2)^*D(J\otimes J_2)=0$. Since $D$ is nonzero,
there are $\chi,\,\psi\in\mc M_2$ such that $R_\chi DR_\psi^*\neq0$; this is obvious because $R_\psi^*R_\psi=\id_{\mc M}\otimes \ktb{\psi}{\psi}$. Using the properties of $R_\psi$ and $R_\chi$ established above, one obtains
\begin{equation}
R_\chi DR_\psi^*\pi(a)=R_\chi D\tilde\pi(a)R_\psi^*=R_\chi\tilde\pi(a)DR_\psi^*=\pi(a)R_\chi DR_\psi^* \, ,
\end{equation}
i.e.,\ the operator $R_\chi DR_\psi^*$ commutes with $\pi$. We also obtain
\begin{equation}
\sis{\varphi}{J^*R_\chi DR_\psi^*J\varphi}=\sis{J\varphi\otimes\chi}{D(J\varphi\otimes\psi)}=0 \, ,
\end{equation}
where the last equality follows from the last defining property of $D$ and the fact that $\chi$ and $\psi$ can be approximated with vectors from the image space $J_2(\mc K_2)$. Hence
$D':=R_\chi DR_\psi^*\in\mc L(\mc M)$ is a nonzero operator that commutes with $\pi$ and $J^*D'J=0$. According to Theorem \ref{ext}, $\Phi_1$ is not extremal in ${\bf CP}_{P_1}(\mc A;
\mc H_1)$.
\end{proof}

Suppose that $\mc H_1,\mc H_2$ and $\hik_1,\hik_2$ are separable Hilbert spaces.
Let us study channels $\mc E\in\mc C(\mc K_1\otimes\mc K_2;\mc H_1\otimes\mc H_2)$.
This kind of channel acting on a compound system may have independent action on the first system in the
sense that there exists a channel $\mc E^1\in\mc C(\mc K_1;\mc H_1)$ such that the equivalent conditions
\begin{eqnarray}\label{2rton}
\mc E(A\otimes \id)&=&\mc E^1(A)\otimes \id \, , \quad A\in\mc L(\mc K_1) \qquad \mr{or}\nonumber\\
\mr{tr}_{\mc K_2}[\mc E_*(S\otimes T)]&=&\tr{T}\mc E^1_*(S) \, , \quad S\in\mc T(\mc H_1),\ T\in\mc T(\mc H_2)
\end{eqnarray}
hold.
On the other hand, there could be a channel $\mc E^2\in\mc C(\mc K_2;\mc H_2)$ such that
\begin{eqnarray}\label{1rton}
\mc E(\id\otimes B)&=&\id\otimes\mc E^2(B),\quad B\in\mc L(\mc K_2)\qquad \textrm{or equivalently}\nonumber\\
\mr{tr}_{\mc K_1}[\mc E_*(S\otimes T)]&=&\tr{S}\mc E^2_*(T) \, , \quad S\in\mc T(\mc H_1),\ T\in\mc T(\mc H_2) \, .
\end{eqnarray}
With the terminology used in \cite{EgScWe02}, we say that a channel satisfying the conditions \eqref{2rton} and \eqref{1rton} is {\it causal}.

A special class of causal channels are local channels.
A channel $\mc E\in\mc C(\mc K_1\otimes\mc K_2;\mc H_1\otimes\mc H_2)$ is \emph{local} if there exist channels $\mc E^1\in\mc C(\mc K_1;\mc H_1)$ and $\mc E^2\in\mc C(\mc K_2;\mc H_2)$ such that the equivalent conditions
\begin{eqnarray}\label{localchannel}
\mc E(A\otimes B)&=&\mc E^1(A)\otimes\mc E^2(B) \, , \qquad A\in\mc L(\mc K_1) \, , \   B\in\mc L(\mc K_2)\quad\mr{or}\nonumber\\
\mc E_*(S\otimes T)&=&\mc E^1_*(S)\otimes\mc E^2_*(T) \, , \qquad S\in\mc T(\mc H_1) \, , \  T\in\mc T(\mc H_2)
\end{eqnarray}
hold.
Given channels $\mc E^1\in\mc C(\mc K_1;\mc H_1)$ and $\mc E^2\in\mc C(\mc K_2;\mc H_2)$, we can always combine them to the joint local channel $\mc E^1\otimes\mc E^2$, but there might be
other joint channels with the same marginals $a\mapsto\mc E^1(a)\otimes \id_{\mc H_2}$ and $b\mapsto \id_{\mc H_1}\otimes\mc E^2(b)$.
However, Proposition \ref{laajext} implies that, in the case of an
extremal marginal, the only joint map is the local one $\mc E^1\otimes\mc E^2$.
In other words, a causal channel with an extremal marginal is local.

\begin{cor}\label{causallocal}
Assume that a channel $\mc E\in\mc C(\mc K_1\otimes\mc K_2;\mc H_1\otimes\mc H_2)$ is causal, i.e.,\ there are channels $\mc E^1\in\mc C(\mc K_1;\mc H_1)$ and $\mc E^2\in\mc C(\mc K_2;\mc H_2)$ such that the conditions
(\ref{2rton}) and (\ref{1rton}) hold. If $\mc E^1$ is extremal in $\mc C(\mc K_1;\mc H_1)$ or $\mc E^2$ is extremal in $\mc C(\mc K_2;\mc H_2)$, then $\mc E$ is local, i.e.,\ $\mc E=\mc E^1\otimes
\mc E^2$.
\end{cor}

It is easy to see that convex combinations of local channels are causal.
If we restrict to the case where $\mc E$ is assumed to be in the convex hull of local channels, then the above result
could be proved without the results on extremal marginals developed in this paper.
However, the result of Corollary \ref{causallocal} is more extensive since there are causal channels that cannot
be represented as convex combinations of local channels; concrete examples are given in \cite{BeGoNiPr01}, \cite{DaFaPe11}.

\section{Conclusions and Remarks}\label{sec:end}

We have identified extremality of a marginal map as a sufficient criterion for the uniqueness of the joint map.
This result directly implies several well-known results.
For instance, if either one of two jointly measurable observables is sharp, then they commute and their joint observable is unique
and of the product form.
Another well-known result states that if a sharp observable $\ms M$ and a channel $\mc E$ are parts of a single instrument, then this instrument $\Gamma$ is unique and $\Gamma(X,B)=\ms M(X)\mc E(B)$.
Our main result states that in the previous results the uniqueness part remains valid also in the case of a looser condition that one of the marginals of the joint map is extremal.
This is a true expansion of previous results since, for instance, there are extremal observables that are not sharp observables but are physically relevant \cite{HePe09}.

Finally, we emphasize that the extremality of a marginal is not a necessary condition for the uniqueness of the joint map.
Namely, a joint map of two compatible maps can be unique and extremal even if the constituents are not extremal.
To give an example of this kind of occurence, we consider binary observables on a qubit system.
As proved in \cite{Busch86}, the observables $\pm 1 \mapsto \half (I \pm  t \sigma_x)$ and  $\pm 1 \mapsto \half (I \pm  t \sigma_y)$ are compatible exactly when $0\leq t \leq 1/\sqrt{2}$.
It is clear from the proof of \cite[Theorem 4.5.]{Busch86} that in the boundary case $t=1/\sqrt{2}$ the observables have a unique joint observable (the four spheres used in the proof have a single point in their intersection).
A qubit observable is extremal if and only if each effect is rank-1 and the set of effects is linearly independent \cite{DaLoPe05}.
Therefore, it is easy to see that the unique joint observable is extremal while the marginal observables are not.

\section*{Acknowledgements}

Authors wish to thank Takayuki Miyadera and Michal Sedlak for comments on the manuscript and acknowledge financial support from the Academy of Finland (grant no. 138135). E. H. is also
funded by the Finnish Cultural Foundation.

redpajama_set_name: RedPajamaArXiv