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Eos Bioreactor uses AI and algae to combat climate change
Written by Bonface Landi
Environment Science Technological Advancements
A new artificial intelligence invention by Hypergiant Industries could prove to be the solution to the world's carbon dioxide problem. The company is launching the second generation Eos Bioreactor, currently still a prototype, that can be used to absorb excess carbon dioxide from the atmosphere and give out oxygen. Besides its ability to reduce environmental pollution, the new AI-based bioreactor also improves health.
The excessive presence of carbon dioxide in the atmosphere has led to a steady rise in the average global temperatures over the years. A National Geographic report states that ocean levels will rise by up to 2.3 feet by 2050 due to melting glaciers. This is just one of many problems that are brought about by excessive carbon dioxide in the atmosphere. Terrestrial radiation, which is supposed to be absorbed by the ozone layer, is also retained in the atmosphere. This leads to the greenhouse effect, where the globe is overheated.
Related: New map exposes secrets of Antarctica's green snow
The Eos Bioreactor seeks to reduce the level of carbon dioxide in the atmosphere to address climate change. Traditionally, the world relies on forests to absorb excess carbon dioxide in the atmosphere and produce more oxygen. However, deforestation in major forests across the world has greatly affected the effectiveness of this approach. For instance, deforestation of Amazon increased by 34% in 2019. Such challenges make it unrealistic for the world to continue relying solely on forests to combat climate change. Technology like the Eos Bioreactor could help address these issues.
According to the manufacturer, the AI-based technology is more effective because each boosted algae bioreactor is 400 times faster in capturing carbon dioxide than trees in the same unit area. Simply put, a single 3-foot by 3-foot bioreactor can absorb the equivalence of the carbon dioxide captured by an acre of forested land.
Besides absorbing carbon dioxide, the bioreactor also monitors airflow, bio-density, pH, type of light and harvest cycles. Because it can be used in a home or office setting, the Eos Bioreactor can completely monitor and purify the quality of the air you breath.
Why use the Eos Bioreactor
According to the CDC, climate change has an effect on human health. Climate change disrupts the quality of natural air, resulting in respiratory and cardiovascular complications. Extreme weather changes can lead to serious cardiovascular injuries and even death. The effects of climate change can also contribute to stress in food production and lead to malnutrition. According to Hypergiant Industries, Eos Bioreactor technology can help reduce such effects.
How the Eos Bioreactor works
Algae require high levels of carbon dioxide to thrive. The bioreactor provides the right environment to grow algae, which can consume most of the carbon dioxide in the atmosphere. However, the system is much more complex than that. Besides exposing algae to the atmosphere for carbon dioxide absorption, the system uses artificial intelligence to control the lighting, airflow, temperature and other factors of the environment. Such factors facilitate the accelerated rate of carbon dioxide absorption and processing.
The bioreactor works in 5 key processes:
Air intake: The air intake absorbs open air in a room or can be connected to a building exhaust. Once absorbed, the air is bubbled into the bioreactor tank, where it combines with algae.
Growing algae: For the algae to grow, it needs carbon dioxide and light. Once carbon dioxide has been pumped into the bioreactor tank, the algae have to be exposed to light. The algae and water are pumped through tubes to maximize exposure to light. They mix with carbon dioxide in the bioreactor tank for the process to commence.
Biomass accretion: Once the algae and carbon dioxide are mixed, the algae consume carbon dioxide to produce biomass. The biomass is harvested to create fuel, oils and high-protein foods and fertilizer.
Harvesting and separation: The Eos Bioreactor uses AI to control the harvesting process. The harvesting system allows the reactor to retain the maximum amount of algae to suck up carbon dioxide.
Clean air exhaust: Once the system uses carbon dioxide to produce biomass, it also consumes all the impurities in the air. As a result, 60% to 90% of the carbon dioxide input is consumed. The resulting oxygen-rich, clean air is released to the environment.
The shape and appearance of the bioreactor
The Eos Bioreactor measures 3-feet-by-3-feet-by-7-feet and is designed to fit in small spaces, including offices and homes. The bioreactor has options for solar power connections, which will make it usable in remote regions. The power used in running the system is minimal, and the waste produced can be utilized for other purposes.
About Hypergiant Industries
Hypergiant Industries is a company that focuses on providing solutions to current humanitarian challenges. One of the biggest challenges that humans face today is climate change. The development of the AI-powered bioreactor is one of many projects spearheaded by the company. Hypergiant Industries is working on several environment-focused products and solutions for clients including governments and Fortune 500.
+ Hypergiant Industries
Images via Hypergiant Industries
Eos Bioreactor
A new artificial intelligence invention by Hypergiant Industries could prove to be the solution to the world's carbon dioxide problem. The company is launching the second generation Eos Bioreactor, currently still a prototype, that can be used to absorb excess carbon dioxide from the atmosphere and give out oxygen. Besides its ability to reduce environmental pollution, the new AI-based bioreactor also improves health.


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Yuheng Hu
I am an Assistant Professor in the Department of Information and Decision Sciences at College of Business Administration of University of Illinois at Chicago. Prior to joining UIC, I was a Research Staff Member at IBM Research. I obtained my Ph.D in Computer Science from Arizona State University, where I studied under Dr. Subbarao Kambhampati. I also did a few fun and fruitful internships at Microsoft Research and Avaya Labs ( A good getaway of Arizona's heat!).
My research interest include Machine Learning and Social Computing and their applications in Online Markets, User-Generated Content, and Finance. My current work focuses on analyzing large-scale and heterogeneous human behaviors on various online platforms at both individual and collective levels. My research has been published in premier journals such as Information Systems Research, Management Information System Quarterly, Journal of Management Information System, INFORMS Journal of Computing, IEEE Transactions on Knowledge and Data Engineering and top CS conferences such as ACM SIGCHI, ICWSM, IJCAI, and AAAI.
My research has received several honors, including 6 Best Paper Awards (and nominations) at WITS (2020 and 2018), HICSS (2021), INFORMS CIST conference (2017), INFORMS eBusiness conference (2017), the ACM CHI conference (2013) , and a Letter of Commendation from the President of Arizona State University. My research has been widely covered by national and international media outlets such as ABC, PBS, WIRED, the Seattle Times, FastCompany, the Sydney Morning Herald, etc. I also hold two patents granted by USPTO.
E-Mail: yuhenghu at uic.edu
Google Scholar: See here
© 2010 - 2021 Yuheng Hu - All rights reserved.

redpajama_set_name: RedPajamaCommonCrawl


This item in Indian Movies category was first added to our catalog on 04-30-2017.
Buy Bahubali 2 & Latest Other Hits Telugu Audio MP3 , sold & shipped by Movie Time Video where you can shop online for Pooja Mandir, Bajot, Gifts, Hindi Movie DVD / CD, Music Video.
This Regional Movies collection has been tagged as 2016 telugu movie, tollywood, south indian music, Tollywood Latest Songs.


She was a southern belle with big hair and a husky voice, and for a short time she was the biggest thing in country music. Born in Mississippi in 1944 to dirt-poor farmers, Roberta Lee Streeter managed to get to college, where she studied philosophy and music. She was good at both, and adopting the stage name Bobbie Gentry, she had a smash hit in 1967 with the swamp-gothic story song "Ode to Billie Joe".
Later she went more mainstream, recording covers albums and duets with Glenn Campbell, hosted some fairly bland television shows, then retired from performing. But in between, she recorded an album, The Delta Sweete which was, writes Dorian Lynskey in the Guardian, "her second record and her masterpiece: a multi-faceted quasi-concept album about Gentry's Mississippi delta roots".
I discovered this astonishing work via one of the singles released from it. The A-side is a vivacious version of the Doug Kershaw song "Louisiana Man": Gentry's take leaps out at the listener, fresh as a kicking catfish. But it is the B-side, "Courtyard" which dazzles, even on the scratchy disc I found. A Scots folk ballad meets Astral Weeks in a graveyard on a sticky summer's day: understated, lovely, chilling.
most of The Delta Sweete's innovative, sophisticated sound is down to Gentry herself, who played piano, guitar, banjo, bass and vibes. Swampy southern grooves mingle with the latest Nashville trends, blue-eyed soul [and] whispered intimations of psychedelia … each track blurs, dream-like, into the next … the earlier tracks chime with her public image as a husky, sensual southern belle but [elsewhere] her voice enters … like sun filtering though curtains.
We are archaeologists of sound, at a dig. We value the pot shards and door knobs and belt buckles, and we sometimes find riches unimagined.
Bobbie Gentry's The Delta Sweete is one of those.


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Events By Week
Interview with Veena Azmanov
March 24, 2015 - 6:45am
by Terri
Veena Azmanov shares her creative cake decorating ideas with hundreds of thousands of cake artists every day through Facebook, YouTube and her website Veena's Art of Cake. Her cakes feature inspiring stunning floral sprays and eye-catching details.
A talented cake artist living in Israel and owner of Veena's Art of Cake, Veena was born in Bombay, India, and began her career working as a flight attendant for Singapore Airlines. Following graduation from college where she majored in English literature, she wanted to see the world and spent the following ten years doing just that with the airline.
"I loved traveling and those were definitely the best years of my life," Veena says.
She attained her second degree in marketing through the University of Ireland while living in Singapore. Later, Veena pursued her MBA in a similar fashion from the University of Chicago.
While traveling as a flight attendant, Veena met her husband Ziv, an American Israeli living in San Francisco. Together, they moved to Israel to be closer to his family where they have resided for the last decade.
Initially, Azmanov pursued a degree in English literature out of a love for teaching and mentoring. Upon traveling to Singapore, she adopted a fondness for marketing. Given that her original intention was to work as a flight attendant for just a few years, she pursued a degree marketing which eventually led to her attaining a MBA.
Upon her arrival in Israel, Veena sought a job in marketing, but she and her husband ultimately decided that having a full-time parent at home for the first five years was the best plan for raising their children. By the time her daughter Rhea reached two years of age, Veena had already fallen in love with decorating cakes.
Veena has been decorating cakes since 2009, when she made her very first cake for her daughter. She instantly fell in love with fondant.
"Before that I baked lots of cakes and desserts, but never really decorated with fondant. It took me almost two years to finally realize that I would love to do this for a living and, in July 2011, I decided to do this professionally," she said.
Because Veena has lived and worked in such diverse cultural environments, one of the things I was curious about was how she handled servicing a community where a large population would require specific ingredients and kitchen conditions.
"There are many who eat only kosher cakes here. Most ingredients found in the local super markets are of course Kosher. For those who need Parve, meaning dairy-free, I am able to accommodate them but not those customers who require food coming out of a Kosher Kitchen. My home kitchen is not Kosher Certified and I personally am not Jewish and do not follow any Shabbat regulation. It does effect business to some extent but, in general, I do attract a lot of customers that are liberal and do not have any such restriction," Veena said.
Yet, Veena did face some challenges in her cake business initially:
"I must say the most difficult challenge has been creating a need for cake where I am. Three years ago, most wedding, birthdays and celebrations had no cakes. It's not a part of the culture to have a wedding cake and because the cost of living is high, it's easily scrapped from the budget. But times are changing and people are becoming more and more aware of cakes. Brides want it, kids want it and parents are also excited to have something special for their kids," she explains.
"When I came to Israel, things were different. I found it very difficult to live and be here. I personally could not blend in. I tried to learn the language but, honestly, it was too difficult for me. The culture was different, too. So, yes, I had to make many adjustments but, on the whole, I think I did well. I try my best to blend in when possible," she adds.
Although she's been immersed in so many different cultures while traveling, Veena's cake decorating style is truly her own. "I am lucky to have many overseas clients who come here to celebrate their wedding or send cakes to loved ones here."
Veena's favorite types of cakes to make are simple, clean cakes, "I'm still young at this profession and finding my way. I love sugar flowers and molds. I truly believe in the concept "less is more," so you will generally never get a cake with too much on it from me."
Here are some of Veena's favorite cakes she's made over the years:
Veena's specialty is wedding cakes but she still does enjoy creating the random novelty or sculpted cake when she has the opportunity.
Despite her popularity online, Veena considers herself a very shy person. "I'm really am a very boring person, self-conscious and shy, so I usually keep to myself and only speak up or say something when spoken to. Most often people perceive this as being proud, but I'm just shy."
Regardless of shyness, Veena maintains a large social presence thanks to her popular blog and video tutorials.
"Not many people know me as a person because I barely ever talk or comment much. I am now trying my best to comment on posts and other pages because some perceive my quietness as pride and unfriendliness. But, I've been lucky to meet a few really lovely people who I can call friends and they now know me better. Personally, I think if I weren't so shy, my work would have more exposure, but it's not easy to just come out and start chatting, so it's something I have to accept."
Veena's short term goals include refining her skills in order to take pride in every cake she makes. Ultimately, she hopes to set an example that others in her profession will want to follow. "I hope that someday I will be as good as the people that I truly admire and aspire to be," she says.
Veena's greatest artistic influences have been Ron Ben Israel and Mercedes Strachwsky. "I have been admiring Ron's work even before I started cake decorating and every time I feel down on my work, I go admire his cakes which somehow gives me the drive to go back and do better. Mercedes Strachwsky gave me the confidence to take up cake decorating as a profession despite no formal cake decorating background or knowledge. She actually started self-taught and won the Oklahoma Grand National Cake Competition in 2009. I figured if working hard and passion could get me to even 1/10 the skill she possesses, it would be worth trying. She has truly inspired me," she shared.
Veena's work has indeed gained worldwide recognition and has led to her being asked to participate in many collaborative projects.
To those who wish to follow their dreams, she suggests, "Don't be afraid to try and don't be afraid to start over. You can take a lot of classes and spend hours watching online videos but practice is the best way to learn. Whenever possible, practice a new technique or cake on a dummy if you can before you do it on a real cake."
The love of cake decorating has become an international passion emanating from all corners of the planet. Veena's cake artistry and worldwide following are an exceptional example of the influence and encouragement in others experienced by sharing our own designs and tips.
If you'd like to see more of Veena's work, visit her website Veena's Art of Cake.
Interview with Master Cake Artist, Elisa Strauss
Interview with Marilyn Bawol
Dawn Parrott
Interview with Mari Senaga
Don Donneruno
See Even More...
Gilles Leblanc wrote on June 20, 2015 - 10:14am
Beautiful interview, loved it.
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Pasquale Pavone represented NOMAD at the recent GO FAIR event "GO BUILD - GO TRAIN - GO CHANGE" in Berlin, Germany, 19 September 2018. At the event, representatives of various institutions met to tackle topics related to the fair provision of research data. NOMAD is involved in the GO FAIR initiative as an implementation network.
Picking up this invitation Klaus Tochtermann, director of the ZBW – Leibniz Information Centre for Economics and strategic director of GO FAIR, explained the relation of the GO FAIR initiative to national efforts in research data management, to the European Open Science Cloud (EOSC) and to the Internet of FAIR Data and Services. Community-driven and bottom-up, the GO FAIR initiative creates an environment in addition to the largely funded EOSC projects.


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Noragami on crack 1988
Part 1 you know its true poor yukine yato beautiful i on october 2nd pink eyed piece of trash fujisaki kouto noragami aragoto episode 2 noragami aragoto so far sorry not sorry thougher than though. Jason liebrecht english yato is the titular stray god of the series noragami who wants to build his own shrine. And we got to learn that vents open because kofuku also, as a bonus. Noragami internet movie firearms database guns in movies. Its one of the millions of unique, usergenerated 3d experiences created on roblox. Not really like noragami at all, but you asked for any other anime you might like, and i wanted to sneak baccano into the list because its such an amazing gem. The story is the kind that pieces itself together to form the big picture at the end, and the characters are really memorable. Ladies and gentlemen, almost 1 year later, i present to you noragami crack part 2.
The following is a list of characters for the noragami manga and anime series. Feb 21, 2017 thank you guys for watching our crack videos, we hope you enjoy this one as well, and hey if you want you can leave a comment or give us a like if you did. Thank you guys for watching our crack videos, we hope you enjoy this one as well, and hey if you want you can leave a comment or give us a like if you did. Like the human to weapon thing, even though in noragami they dont eat souls, good or corrupted ones, like in soul eater. Stray god is an anime television series adapted from the manga of the same title by adachitoka. Zerochan has 43 kazuma noragami anime images, androidiphone wallpapers, fanart, facebook covers, and many more in its gallery. It is produced by bones and directed by kotaro tamura with character designs by toshihiro kawamoto. An archive of our own, a project of the organization for transformative works. When the downonhisluck yato is saved by an unfortunate soul called hiyori, hes forced to. This is noragami episode 1 indoakatsuki by yap kokcien on vimeo, the home for high quality videos and the people who love them. User recommendations about the anime noragami noragami.
Noragami manga goes on extended hiatus due to adachitokas health posted on 20170501 01. Noragami started as a manga series, written and illustrated by adachitoka and published by kodansha. Yato is an unemployed stray god who battles phantoms and solves humans problems for only 5 yen. Stray god on myanimelist, the internets largest anime database. Noragami was first published in monthly shonen magazine in 2010, and is localised in the west by kodansha comics. Access our huge library of subs and dubs, featuring a deep catalog of big hits, fan favorites and alltime classics, as. I do not own this song, it belongs to respective owner. Stray god 18 has been added to your cart add to cart.
Listen to noragami soundtrack soundcloud is an audio platform that lets you listen to what you love and share the sounds you create tokyo. Unfortunately, things just dont seem to be going his way. Also, i apologize for the use of the english dub in some scenes. The series has been collected into twentyone tankobon volumes as of march 2020. Watch noragami episode 4 online where happiness lies shiawase no arika. I think that seeing hiyori being hurt might actually have effect on yukine. Worst of all, his aloof and somewhat goofy personality has just cost him his newest partner and best holy weapon. There is no specific anime for this game so you can roleplay almost anyone you would like.
An anime television series adaptation by bones aired in japan beginning from january 5, 2014 to march 23, 2014, with a. Yato may be a minor god, but hes got a plan to make it big. Yukine makes a new friend, but soon learns what it means to be a regalia. In the past, he was a god of calamity, magatsukami. Bishamon continue uses it at the second season, before yato destroyed it by his sword. It starts with yukine getting lost in the woods, it gets dark and yato has to find him and comfort him. Free shipping get free shipping free 58 day shipping within the u.
Stray god 18 paperback june 6, 2017 by adachitoka author. Noragami original soundtrack 1, anima musica, noragami original soundtrack 1. Noragami ost by ris dominic free listening on soundcloud. The ninth volume was released simultaneously with a limited edition, bundled with a.
In 2012, 88% of imprisonments from crack cocaine were african american. Divine intervention for sale yato is a homeless god. By kittymcwane ongoing updated jan 28, 2016 embed story share via email read new reading list. Season 1 consists of 12 episodes while season 2 has. Noragami, noragami aragoto, mirai nikki, tokyo ghoul, my hero. Children a list of 49 titles created 4 months ago my anime series a list of 40 titles. The series follows yato, a homeless shinto god without a shrine of his own. D it is still workinprogress so there are more stuff to be build and more stuff to be scripted so please. Noragami will tug violently on your heartstrings and have you weeping at the same time as youre hiding every set of eyes in your home out of fear. Noragami aragoto yaboku yato at bishamontens proposal part 1 raboh my god credits to attack on onepiece tumblr crowd genius hiyori how to. Stray god 19 by adachitoka 9781632364395 from boomerang books, australias online independent bookstore. Noragami stage plays entire main cast shown in costume jan 20, 2016 j. He doesnt have a single shrine dedicated to him, his partner has just quit, and now hes got to find a new divine weapon. The series follows a poor deity named yato, who desires to become a famous god, and his adventures with a human girl named hiyori, whose spirit frequently leaves her.
Another mamoru comedy another mamoru meme tuxedo kamen sama. Assuming a regalia could be made from any phantom, hiyori goes in search of one, only for it to turn out to be a really hostile one. He wears a tracksuit and scarf and refers to himself as yato god yatogami. Staff makes flying witch tv anime for april jan 6, 2016 animelab announces more titles for fall simulcast lineup oct. Kazuma remains confused by bishamons declining health. Both styles of art arent bad at all, though they are pretty. Stray god is a japanese manga series written and illustrated by adachitoka. Here are some of my ships and some random photos that i found hope you enjoy it. Soul eater and noragami are pretty similar surprisingly at least to me. The series premiered in monthly shonen magazine s january 2011 issue, released on december 6, 2010, and has been compiled into twentyone tankobon volumes between july 15, 2011 and october 25, 2019. This is a montage of noragami crack from different crack videos. This game will be an anime role playing game in tokyo. Bishamonten or bishamon shinto god of war possess shinki yokai japanese demon, named kazuha, that can transforms to the m1911a1 pistol during the battles.
D make sure to subscribe,like and comment for more. Yato, the supposed god of calamity, is so far down on the shinto totem pole that he doesnt even have any followers or a shrine to call his own and has to do odd jobs to make ends meet. If only he hadnt said all those stupid things to yato, he wouldnt be in this situation now. Both anime have spiritual elements, and the characters have similar powersabilities. Noragami aragoto yaboku yato at bishamontens proposal part 1 raboh my god credits to attackononepiece tumblr crowd genius hiyori how to. May 01, 2017 noragami manga goes on extended hiatus due to adachitokas health posted on 20170501 01. Top 25 dark animes with vampires demons and monsters.
Stream tracks and playlists from noragami soundtrack on your desktop or mobile device. But little does he know that a search for a missing cat is about to lead to a fateful encounter. He doesnt even have a shrine, not to mention worshippers. So to achieve his ambitious goals, hes set up a service to help those in need for a small fee, hoping hell eventually raise enough money to build himself the lavish temple of his dreams. Despite being reminded by hiyori about the job she paid him for, yato insists that he cant do anything until he gets a new regalia. Bishamon aww fuck now shes getting seriouskofuku is actually serious about this its good to see another side of the carefree goddess, if anything happens to yatty. Its not that he has really had much respect for her in the past think skateboard, but i think that if theres one person that would be able to influence him, itd be her. The blond boy was trembling, his hair and clothes had been soaked by the rain and he felt so cold.
The near shore is the place where people reside, while the far shore is the place for deities and ayakashi, which are emotional fragments that can posses humans and cause trouble. When the down on hisluck yato is saved by an unfortunate soul called hiyori, hes forced to. As hiyori ends up drifting out of her body more often than not, she takes advantage of her new agile form. Access our huge library of subs and dubs, featuring a deep catalog of big hits, fan favorites and alltime classics, as well asthe latest shows out of japan. Noragami ost by ris dominic published on 20140506t14. Yato may just be a minor god now, but hes determined to make it big and hes got a plan. I wanted to mix both seasons but it was just too many episodes to download. Looking for episode specific information noragami on episode 2. It began serialization in kodanshas monthly shonen magazine in january 2011s issue. Anime up to date a list of 29 titles created 02 aug 2018 anime a list of 47 titles. Further, the data shows the discrepancy between lengths of sentences of crack. Yato is a war god who wants to build his own shrine.
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Love of Life Orchestra (LOLO) is an experimental music group formed by Peter Gordon in New York in April 1977. The band is associated with the 1970s New York downtown music scene.

A number of the players that would form the band came together in 1976, to perform Peter Gordon's Symphony in Four Movements at The Kitchen. Gordon was then asked to perform at a benefit concert for the New York non-profit Performing Artservices in Spring 1977, and so formed the Love of Life Orchestra with Arthur Russell, Peter Zummo, Jill Kroesen, "Blue" Gene Tyranny, David Van Tieghem, Scott Johnson, Ernie Brooks, Ed Friedman, and Kathy Acker. The group has had a widely varying lineup since, and other notable past members include Laurie Anderson, Rhys Chatham, and Ned Sublette.

In 2007, two tracks by the group — "Another Heartbreak" and "Don't Don't Redux" — were prominently featured on the FabricLive.36 compilation curated by James Murphy and Pat Mahoney of LCD Soundsystem, generating a resurgence of interest in the band.

Partial discography
 Star Jaws (Lovely Music) 1978, reissued on CD 2008
 Extended Niceties EP (Infidelity) 1980
 Geneva (Infidelity) 1980
 Casino EP (Expanded Music/Antarctica) 1982
Love of Life (DFA Records) 2010, retrospective.

References

American post-punk music groups

redpajama_set_name: RedPajamaWikipedia


For years now, Wal-Mart and I have had a sort of love-hate relationship. I remember in the 80's when it was all the rage and I lived in my small town; it was the biggest (and only) store that seemed to be open on a Sunday – oh sure, there was Hastings, but Hastings didn't have the zillion choices I craved in a shopping experience. Where I was typically able to avoid cruising the Dairy Queen for entertainment, I couldn't always resist the siren call of Wal-Mart, especially when I had a small allowance, an overwhelming desire to be a consumer and absolutely had to have a pre-molded plastic something to get me through the next semester.
The thing I absolutely hate about Wal-Mart is it induces in me some sort of shopping daze where I go in empty handed and leave with a basket full of things I'm not sure I remember actually buying. Things I didn't realize I wanted until they were sitting in my trunk.
Yesterday's purchases: camp chairs (ok, I planned on those for April's Thanksgiving in the park), Pyrex portable (it's got a lid for the dish and thermal lining! and I am supposed to bring something to Thanksgiving that's best served warmish – another planned purchase), and then I went into a small frenzy. I "needed" a cover for the splattery food within the microwave – sure, I didn't want one before, but it was only $1.99! Then, looking at this thing that looked like a lid, I realized I needed a pie carrier for all those pies I never make. I find one and it has these nifty inserts – one insert can convert the carrier into a deviled egg carrier (for all the deviled eggs I never make) and one will make it a cupcake holder (again… don't really make cupcakes), but I was absolutely fascinating by it and had to own it. I even eyed a new blender, a rice cooker and espresso machine (because I don't drink coffee, but I did imagine it would make a great present and people like espresso, right?) Thankfully, I managed to keep my hands at my side and not walk out with these as well. Overall, it could have been worse. We also hit the grocery store side, but didn't stay there long thanks to HEB holding a bigger sway over me.
When I checked-out, snatching a copy of People because it had Barack Obama's face on the cover (look, there wasn't a copy of Time or Newsweek there – it was that or OK! magazine and I was still in "consume" mode) I stood back in awe as I saw over the lanes that I could also bank there, get new glasses, get a manicure, have my taxes done, have a family portrait and do all of this while enjoying a Big Mac. I swear, it was like country come to town – boy howdy, it was a consumer's dream come true in one single shopping experience.
Now, if I want to have any money ever again, I need to make a vow to stay away from that place or I'll end up tithing to it without even being aware.
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text:
\section{Introduction}

Let $(M,\omega)$ be a compact connected symplectic manifold, and let $G$ be a compact connected Lie group acting on $M$ in a Hamiltonian fashion with moment map $\mu:M \to \gg^*$. Assume that the equivariant cohomology class $[\omega + \mu]$ is integral. Let $L\to M$ be a complex Hermitian line bundle and let $\nabla$ be a Hermitian connection on $L$ whose equivariant curvature form is the equivariant symplectic form $\omega + \mu$. Suppose that the $G$-action on $M$ lifts to a $G$-action on $L$ which preserves $\nabla$. Let $J$ be a $G$-equivariant almost complex structure on $M$. This almost complex structure gives us a totally complex distribution $\Delta$ in the complexified tangent bundle to $M$, such that $TM\otimes \mathbb{C}= \Delta \oplus \overline{\Delta}$. This gives us a splitting $\Lambda^k(TM\otimes \mathbb{C}) = \sum_{i+j=k}\Lambda^i(\overline{\Delta})\otimes\Lambda^j(\Delta)$, and thus a bigrading on the space of $L$-valued differential forms. We will take $J$ to be compatible with $\omega$; that is, $\omega$ is a $(1,1)$-form and $\omega(v,Jv)>0$ for all $p \in M$ and $v \in T_pM$.

Let $D:\Omega^k(M;L)\to \Omega^{k+1}(M;L)$ be the operator $D(s\otimes \alpha)=\nabla s \otimes \alpha + s \otimes d\alpha$, where $s \in \Gamma(L)$ and $\alpha \in \Omega^k(M)$. Define $\overline{\partial}$ to be the $(0,k+1)$ component of $D$. Fix a Hermitian metric on $M$. Together with the Hermitian inner product on $L$, this gives a Hermitian inner product on $L \otimes \Lambda^{0,k}TM$. Define operators $\overline{\partial}^t:\Omega^{0,k}(M;L)\to \Omega^{0,k-1}(M;L)$ which are $\mathcal{L}^2$-adjoint to $\overline{\partial}$. Then the operator $\overline{\partial} + \overline{\partial}^t:\Omega^{0,even}(M;L)\to \Omega^{0,odd}(M;L)$ is elliptic; the \emph{quantisation} $Q(M, \omega, \nabla, L; J)$ is defined as the virtual $G$-representation on $\ker(\overline{\partial}+\overline{\partial}^t)\ominus \coker(\overline{\partial}+\overline{\partial}^t)$; that is, the equivariant index of the $\overline{\partial}+\overline{\partial}^t$ operator on $L$. (See chapter 6 of Ginzburg, Guillemin and Karshon's book \cite{ggk}.)

In their paper \cite{gp}, Guillemin and Prato prove a formula for the multiplicity with which each irreducible character of $G$ appears in the character of this representation $Q(M, \omega, L, \nabla; J)$. In the special case of a torus $T$ acting on a coadjoint orbit $G/T$ by left multiplication, their formula becomes the Kostant multiplicity formula that Kostant obtained in \cite{kostant}.

Let $G$ be semisimple, let $K \subset G$ be a Lie subgroup of equal rank, choose a common maximal torus $T\subset K \subset G$, let $\lambda$ be a dominant integral weight for $G$, and let $M$ be the coadjoint orbit $G\cdot \lambda$. The choice of positive roots for $G$ determines a complex structure on $G\cdot \lambda$; take $J$ to be the corresponding almost complex structure on $G\cdot \lambda$. Let $L_\lambda=G\times_T \mathbb{C}_{(\lambda)}$ (where $T$ acts on the line $\mathbb{C}_{(\lambda)}$ with weight $\lambda$). Then the quantisation $Q(M, \omega, \ll, \nabla; J)$ is a $G$-representation on the space of holomorphic sections of $\ll$ (see \cite{ggk} for details). The Borel-Weil theorem tells us that the quantisation of $(M, \omega, \ll, \nabla, J)$ is the irreducible representation of $G$ of highest weight $\lambda$, and that all irreducible representations arise in this way, as described by Bott in \cite{bott}. In this case, Gross, Kostant, Ramond and Sternberg provided in \cite{gkrs} a formula for the character of this $G$-representation as a quotient of the alternating sum of a multiplet of $K$-characters. Their formula has its origins in String Theory and is the motivation for our work which provides a generalisation. In the special case where $K=T$, their formula becomes the Weyl character formula.

In this paper, we extend the result of Gross, Kostant, Ramond and Sternberg by replacing the coadjoint orbit $G\cdot \lambda$ with any compact connected symplectic Hamiltonian $G$-manifold $M$, and relate the resulting character formula to the Guillemin-Prato multiplicity formula. In section two we obtain, for arbitrary compact connected symplectic manifolds $M$ with Hamiltonian $G$-actions, a formula for the character of $Q(M, \omega, \nabla, L; J)$ as a quotient of $K$-characters. In section three, we derive from our formula a generalisation of the Guillemin-Prato multiplicity formula.

\paragraph{Acknowledgements} I would like to thank Jonathan Weitsman for suggesting the problem and for his advice and encouragement throughout. I am also grateful to the anonymous reviewer for making valuable corrections and suggestions, and to Nate Bade, Barbara Bolognese, Victor Guillemin, and Ryan Mickler for helpful discussions.

\section{Character Formula}

Let $G$ and $K$ be compact connected Lie groups of equal rank with $K \subset G$, and choose a common maximal torus $T\subset K \subset G$. Write $\tt$, $\mathfrak{k}$ and $\gg$ for the Lie algebras of $T$, $K$, and $G$ respectively. Let $\mathcal{N}_G(T)$ denote the normaliser in $G$ of $T$ and let $W(G) = \mathcal{N}_G(T)/T$ be the Weyl group of $G$. Choose a set $\Phi^+(G)$ of positive roots of $G$, and let $\mathcal{W}_G \subset \tt^*$ be the positive Weyl chamber for $G$. Let $\Lambda \subset \tt^*$ denote the weight lattice. For $\phi \in \Phi^+(G)$, let $H_\phi$ denote the hyperplane orthogonal to $\phi$ in $\tt^*$, and let $w_\phi \in W(G)$ be the reflection in this hyperplane. Let $(M,\omega)$ be a compact connected symplectic manifold, and let $G$ act on $(M, \omega)$ in a Hamiltonian manner. Then $T$ acts on $(M, \omega)$ in a Hamiltonian fashion with $T$-equivariant moment map $\mu: M \rightarrow \tt^*$; we will assume that the fixed points of this torus action are isolated. Suppose the equivariant cohomology class $[\omega + \mu]$ is integral, choose a prequantisation line bundle $(L,\nabla)$, and let $J$ be a $G$-equivariant almost complex structure on $M$ that is compatible with $\omega$. Let $Q(M, \omega, \nabla, L; J) = ind (\overline{\partial}+\overline{\partial}^t)$ be the quantisation of $(M, \omega, L, \nabla; J)$, and let $\chi$ denote its character. In this section we will give an expression for the character $\chi$, as a sum of quotients of (virtual) $K$-characters. We begin by setting up the equivariant cohomology we need, and by recalling the equivariant index theorem and the localisation theorem which will be our main tools.

\subsection{Review of Equivariant Cohomology}

Let $G$ be a compact Lie group acting on a manifold $M$. Let $EG$ be a contractible space on which $G$ acts freely, so that $M\times EG \simeq M$ and the diagonal action of $G$ on $M \times EG$ is free. Form the homotopy quotient $M_G := (M\times EG)/G$.

\begin{defn} The \emph{equivariant cohomology ring} $H^*_G(M)$ is the ordinary cohomology ring $H^*(M_G)$. \end{defn}

Let $H\subset G$ be a subgroup. Then $H$ also acts freely on $EG$, so we can take $EH=EG$ and thus $M_G = M_H/G$. If $p:M_H\to M_G$ denotes the projection, we can pull back classes in $H_G^*(M)$ along $p$ to $H^*_H(M)$.

An alternative approach to defining equivariant cohomology is known as the Cartan model. Define an \emph{equivariant differential form} to be a $G$-equivariant polynomial on $\gg$ taking values in $\Omega^*(M)$. More precisely, the \emph{equivariant differential $k$-forms} are elements of $\Omega^k_G(M)=\bigoplus_{k=2i+j}(S^i(\gg^*)\otimes \Omega^j(M))^G$. The \emph{equivariant exterior differential} $d_G: \Omega^k_G(M) \rightarrow \Omega^{k+1}_G(M)$ is given by $$(d_G\alpha)(X) = d(\alpha(X)) - i_{X_M}\alpha(X),$$ where $\alpha \in \Omega^k_G(M)$, $X \in \gg$, and $X_M$ is the vector field defined by the infinitesimal action of $X$ on $M$. Note that $d_G^2=0$ by invariance.

Hence, an equivariant form $\alpha \in \Omega^*_G(M)$ is closed if $d(\alpha(X)) - i_{X_M}\alpha(X)=0$ for all $X \in \gg$.

\begin{example}[Equivariant symplectic form]
Let $\omega + \mu$ be the equivariant symplectic form. Then $d_G(\omega+\mu)(X) = d(\omega+ \mu)(X) - i_{X_M}(\omega + \mu)(X).$ But $d\omega=0$ since $\omega$ is the non-equivariant symplectic form, and $i_{X_M}(\mu)=0$ since $\mu$ is a 0-form. So $d_G(\omega+\mu)(X) = d\mu(X)-i_{X_M}\omega(X)$, which is zero by definition of the moment map $\mu$. So the equivariant symplectic form is equivariantly closed.
\end{example}

\begin{thm}[Equivariant de Rham Theorem] Let $G$ be a compact connected Lie group acting on a manifold $M$. Then the equivariant cohomology is given by $$H^*_G(M) = \frac{\ker d_G}{\Imm d_G}.$$ \end{thm}

Suppose $S \subset T$ is a subtorus with Lie algebra $\mathfrak{s}$. A $T$-equivariant form $\alpha$ is an $\Omega(M)$-valued polynomial on $\tt$; by restriction we can view this as an $\Omega(M)$-valued polynomial on $\ss$ and hence as an $S$-equivariant form. If $\alpha$ is $T$-equivariantly closed then $d(\alpha(X)) - i_{X_M}\alpha(X)=0$ for all $X \in \tt$, in which case we certainly have $d(\alpha(X)) - i_{X_M}\alpha(X)=0$ for all $X \in \ss$ and so $\alpha$ is also $S$-equivariantly closed.

Let $c_{1,T}$, $e_T$ and $Td_T$ denote the $T$-equivariant first Chern, Euler and Todd classes respectively. For a detailed discussion of these equivariant characteristic classes, see \cite{gs} or \cite{ggk}.

We are now ready to state the following two theorems, which are the key components of our proof. For more details about these theorems we refer the reader to \cite{ggk}.

\begin{thm}[Equivariant Index Theorem] \label{eqind}Let $(M, \omega, L, \nabla; J)$ be a $G$-manifold, and let $\chi$ denote the character of the quantisation $Q(M, \omega, \nabla, L; J)$. Then, in a neighbourhood of $0\in \gg$, $$\chi \circ exp = \int_M e^{c_{1,T}(L)}Td_T(M).$$

\end{thm}

\begin{thm}[Atiyah-Bott-Berline-Vergne Localisation Formula] Let $M$ be a compact, oriented manifold and let the torus $T$ act on $M$ with fixed point set $F$, and let $\iota:F\rightarrow M$ denote the inclusion. Let $\alpha$ be an equivariantly closed form on $M$. Let $NF$ denote the normal bundle to $F$ in $M$. Then $$ \int_M \alpha = \int_F\frac{\iota^*_F \alpha}{e_T(NF)}.$$ In particular, when $F$ is a finite set of isolated points, $$\int_M \alpha = \sum_{p\in F} \frac{\alpha \vert_p}{e_T(T_pM)}.$$ \end{thm}

\subsection{Actions of the Torus and of a Subtorus}

The Equivariant Index Theorem and the Localisation Theorem together give an expression for the character of $Q(M, \omega, \nabla, L; J)$ as a sum of contributions from each fixed point of the torus action on $M$. The fixed point data involved is the $T$-equivariant Chern classes of the line bundle $L$, and the equivariant Todd and Euler classes of the tangent bundle $TM$, all restricted to each fixed point $p$. These classes are all determined by the weights of the torus action on the fibres of the vector bundles above $p$. So we now discuss what the fixed points are and how the torus acts on the fibres above them.

\begin{lem} \label{image} The equivariant first Chern class $c_{1,T}(L)\vert_p$ of the fibre of the line bundle $L$ above the point $p \in M$ is the moment map image $\mu(p)$. \end{lem}

\begin{proof} We chose our prequantisation line bundle $L \rightarrow M$ and connection $\nabla$ such that its equivariant curvature form was equal to the equivariant symplectic form $\omega + \mu$. Restricted to a point $p$ this becomes the image $\mu(p)$.
\end{proof}

\begin{lem} \label{wactsonp} The Weyl group $W(G)$ of $G$ acts on the set of fixed points of the $T$ action on $M$. \end{lem}

\begin{proof} Let $p\in M$ be a fixed point of the torus action, and let $wT \in W(G)$. We define the action of $W(G)$ on the fixed point set by $wT \cdot p = w \cdot p$. Since the torus action fixes $p$, this doesn't depend on the coset representative $w$, and so the action is well-defined. We would like to show that $wT \cdot p$ is a fixed point of the torus action. By definition of the Weyl group, if $wT\in W(G)$ and $t \in T$, then $tw = wt'$ for some $t' \in T$. So $t\cdot (w \cdot p) = tw \cdot p = wt' \cdot p = w \cdot (t' \cdot p) = w \cdot p$, so $w \cdot p$ is a fixed point as claimed. \end{proof}

\begin{cor} The fixed points of the torus action can be partitioned into $W(G)$ orbits. \end{cor}

\begin{lem} \label{wactsonweights}Let $\alpha_1, \alpha_2, \ldots , \alpha_n$ be the weights of the representation of $T$ on the tangent space $T_pM$ to the fixed point $p$, and let $w \in \mathcal{N}_G(T)$ be a representative for $wT \in W(G)$. Then the weights of the $T$-representation on $T_{w\cdot p}M$ are $w\cdot \alpha_1, \ldots, w\cdot \alpha_n$.\end{lem}

\begin{proof} Let $\alpha$ be a weight of the torus action on the tangent space $T_pM$, and let $V_\alpha \subset T_pM$ be the weight space of $\alpha$. Let $v\in V_\alpha \subset T_pM$, and let $X \in \tt$ with $\exp X = t$. By definition this means $t \cdot v = e^{\langle \alpha, X \rangle}v$. We wish to know how $T$ acts on $w\cdot v$. Since $w \in \mathcal{N}_G(T)$, we know $w^{-1}tw = t'$ for some $t' \in T$. Suppose $t' = expY$. So $t\cdot (w \cdot v) = (tw)\cdot v = wt' \cdot v = w \cdot (t' \cdot v) = w \cdot (e^{\langle \alpha, Y \rangle}v) = e^{\langle \alpha, Y \rangle}w\cdot v $. But $\exp Y = w^{-1}\exp X w$, so $Y = w^{-1} X w$ since the exponential map commutes with the adjoint action. So $\langle \alpha, Y \rangle = \langle \alpha, w^{-1} X w \rangle = \langle w\cdot\alpha, X \rangle$, so $t\cdot w \cdot v = e^{\langle w \cdot\alpha, X \rangle}w\cdot v $. That is, $w\cdot \alpha$ is a weight of the $T$ action on the tangent space $T_{w\cdot p}M$, as required. \end{proof}

If we apply localisation to the equivariant index theorem, we get a formula for the character of the $G$-representation in terms of torus characters. We combine these torus characters into $K$-characters. Geometrically, this arises as follows.

Let $\ss \subseteq \tt$ be the algebra $\{\zeta \in \tt \vert \langle \alpha , \zeta \rangle = 0 \forall \alpha \in \Phi(K)\}$, and let $S = \exp(\ss)$. Then $S\subseteq T$ is the maximal torus of the centraliser of $K$ in $G$.

\begin{lem} The fixed manifolds of the $S$-action are preserved by the $K$-action.
\end{lem}

\begin{proof} Suppose $x \in M$ is fixed by the $S$ action. We claim that $k\cdot x$ is also fixed by $S$, for all $k \in K$. But $S$ is contained in the centraliser of $K$, so indeed $s\cdot k \cdot x = k \cdot s \cdot x = k \cdot x$. Since $K$ is connected, the $K$-action preserves connected components of the fixed set of $S$. \end{proof}

\begin{cor} \label{fixedmfldsofsarekspaces} Each fixed manifold $F$ of the $S$-action is a Hamiltonian $K$-space. In particular, $W(K)$ acts on the set of fixed points $p_i$ of the torus action that are contained in $F$, and on the weights of the torus action on $T_{p_i}F$. \end{cor}

Let $p\in M$ be a fixed point of the torus action, and consider the action of $T$ on the tangent space $T_pM$. The tangent space can be broken up into a direct sum of 2-dimensional weight spaces. The point $p$ is contained in some manifold $F$ fixed by the $S$-action. Denote by $NF$ the normal bundle to $F$ in $M$; since the torus action preserves $F$, the decomposition $T_pM=T_pF \oplus N_pF$ respects the weight decomposition.

Recall that an element $\xi$ of a Lie algebra $\mathfrak{k}$ is \emph{regular} if the dimension of its centraliser is minimal among all centralisers of elements of $\mathfrak{k}$. For Lie algebras of compact Lie groups this means that the centraliser of $\xi$ is a maximal torus of $\mathfrak{k}$. Write $\mathfrak{k}^*_{reg}$ for the set of regular elements of $\mathfrak{k}^*$, and $\tt^*_{reg}$ for the intersection of $\mathfrak{k}^*_{reg}$ with $\tt^*$.

Since the action of $K$ preserves $F$, if $\mu(p)$ is in $\tt^*_{reg}$ then $T_pF$ contains a copy of $\mathfrak{k}/\tt$; in particular the set of weights of the representation on $T_pF$ contains either $\phi$ or $-\phi$ for all positive roots $\phi$ of $K$. Note however that unlike Guillemin and Prato in \cite{gp} we are not assuming all of the $\mu(p)$ are regular; this allows us to consider actions on non-generic coadjoint orbits.  If $\mu(p)$ is not in $\mathfrak{t}^*_{reg}$ then it's fixed by some of the $w_\alpha$, where $w_\alpha \in W(K)$ is the element of the Weyl group that acts on $\tt^*$ by reflection in the hyperplane orthogonal to $\alpha$. These $\alpha$ are then not necessarily weights of the torus action on $T_pF$.

\subsection{Some Notation}

Let $P=\{p_1, \ldots, p_n\}$ be the set of fixed points of the torus action on $M$. For the fixed point $p_i$, let $A_i$ be the set $\{\alpha \in \Phi^+(K) \vert \langle \alpha , \mu(p_i) \rangle =0\}$ of roots $\alpha$ of $K$ such that $\mu(p_i)$ lies on the hyperplane orthogonal to $\alpha$ (and so $w_\alpha$ fixes $\mu(p_i)$). Let $\mathcal{A}_i$ be the group generated by the $w_\alpha$, for $\alpha \in A_i$, and let $U_i$ be the set of left cosets of $\mathcal{A}_i$ in $W(G)$. If $X$ is an $\mathcal{A}_i$-invariant set then we define the action of $U_i$ on $X$ by $[w]\cdot x = w \cdot x$, where $w \in W(G)$, the coset of $\mathcal{A}_i$ containing $w$ is denoted $[w]$, and $x \in X$; this action does not depend on the choice of coset representative and so is well defined. For ease of notation we will often drop the square brackets and refer to elements $w \in U_i$.

 If $w\cdot p_i = p_j$, we will also write $w\cdot i = j$. Let $P^+$ be the set of fixed points which map into the closed positive Weyl chamber $\overline{\mathcal{W}_K}$ under the moment map.

Given a fixed point $p_i \in M$, let $F$ be the fixed manifold of the $S$ action containing $p_i$. Let $\{\alpha_{ij}\}^{a_i}_{j=1}$ denote the weights of the torus action on the tangent space to the $K$-orbit of $p_i$ (these form a subset of $\Phi(K)$), let $\{\beta_{ij}\}^{f_i}_{j=1}$ denote the remaining weights of $T_{p_i}F$, and let $\{\beta_{ij}\}^{n_i}_{j=f_i+1}$ denote the weights of $(NF)_{p_i}$. We write $B_i = \{\beta_{ij} \vert 1 \leq j \leq n_i \}$. We will polarise the weights as follows. Choose a $\xi \in \tt$ such that $\langle \alpha , \xi \rangle = 0$ for all $\alpha \in W(K)$, and $\langle \lambda , \xi \rangle \neq 0$ for all other weights $\lambda$ of the torus action on tangent spaces to fixed points. So $W(K)$ preserves $\xi$. Note that there may be some weights $\beta_{ij}$ of the torus action on the normal bundle to the $K$-orbit of $p_i$ which also happen to be roots of $K$; such $\beta_{ij}$ will have $\langle \beta_{ij}, \xi \rangle = 0$. Define \emph{polarised} weights

\[ \beta_{ij}^+ = \begin{cases}
    \beta_{ij}, & \beta_{ij}(\xi) \geq 0\\
    -\beta_{ij}, & \beta_{ij}(\xi) < 0
  \end{cases},\]
and for fixed $i$, let $s_i$ be the number of $j$ with $\beta_{ij}(\xi) <0$. Notice that we've chosen $\xi$ such that $s_i = s_{w\cdot i}$ for all $w \in W(K)$.

For the fixed point $p_i$, write $\beta_i = \frac{1}{2}\sum_j \beta_{ij}$. Let $\mathcal{B}_i$ be the set $\{w \cdot \beta_{ij} \vert w \in W(K), \beta_{ij} \in B_i\}$, and let $\mathcal{B}^+_i$ be the set of polarised weights $\{\beta^+ \vert \beta \in \mathcal{B}_i\}$. Let $\overline{\beta_i}=\sum_{\beta \in B_i: \beta = -\beta^+}\beta$. Let $C_i$ be the set of weights $\beta \in \mathcal{B}_i^+$ that are not equal to $\beta_{ij}^+$ for any $\beta_{ij} \in B_i$; that is, polarisations of all weights in the orbit except for those at the point $p_i$.  Let $m_i(\eta)$ be the multiplicity of $e^\eta$ in $\prod_{\gamma \in C_i}(1-e^{-\gamma})$.

\subsection{The Main Theorem}

\begin{thm} \label{main} Let $G$ be a compact Lie group with maximal torus $T$. Let $(M, \omega, L, \nabla; J)$ be a Hamiltonian $G$-space such that the $T$-action has isolated fixed points. Let $K \subset G$ be a closed connected subgroup with maximal torus $T$.  Write $V^K_\lambda$ for the irreducible $K$-representation of highest weight $\lambda$. Then the character $\chi$ of the quantisation $Q(M, \omega, L, \nabla; J)$ is given by
\begin{equation}
\label{mainthm}
\chi(Q(M, \omega, \nabla, L; J)) = \sum_{F \in \pi_0(M^S)} \sum_{p_i \in F \cap P^+} \frac{\sum_{\eta \in \Lambda}(-1)^{s_i}m_i(\eta) \chi(V^K_{\mu(p_i)+\overline{\beta_i}+\eta})}{ \prod_{\gamma \in \mathcal{B}^+_i} (1-e^{-\gamma})}.
\end{equation}
 \end{thm}

\begin{proof} The equivariant index theorem (Theorem \ref{eqind}) tells us that $\chi \circ \exp = \int_M e^{c_{1,T}(L)}Td_T(M)$. By localisation with respect to the $T$ action we get
\begin{equation}
\chi \circ \exp = \sum_{p_i \in M^T} \frac{\bigl(e^{c_{1,T}(L)}Td_T(M)\bigr)\big|_{p_i}}{e_T(T_{p_i}M)}.
\end{equation}
Let $M^S$ denote the fixed set of the $S$ action on $M$. Since every fixed point $p_i \in M^T$ is contained in $M^S$, we can write $M^T$ as the disjoint union over all $F \in \pi_0(M^S)$ of the fixed sets $F^T$ of the $T$ action on $F$. So
\begin{equation}
\chi \circ \exp = \sum_{F \in \pi_0(M^S)} \sum_{p_i \in F^T} \frac{\bigl(e^{c_{1,T}(L)}Td_T(M)\bigr)\big|_{p_i}}{e_T(T_{p_i}M)}.
\end{equation}
At a fixed point $p_i \in F$, we can write $TM\vert_{p_i}$ as $T_{p_i}F \oplus (NF)_{p_i}$. Hence $Td_T(TM)_{p_i} = Td_T(T_{p_i}F)Td_T(NF)_{p_i}$, and $e_T(TM)_{p_i} = e_T(T_{p_i}F)e_T(NF)_{p_i}$. So
\begin{equation} \label{allbrokenup}
\chi \circ \exp = \sum_{F \in \pi_0(M^S)}\sum_{p_i \in F^T}\frac{e^{(c_{1,T}(L))_{p_i}}Td_T(T_{p_i}F)Td_T((NF)_{p_i})}{e_T((NF)_{p_i})e_T(T_{p_i}F)}.
\end{equation}
Recalling the definitions of the equivariant characteristic classes and lemma \ref{wactsonweights}, we can write
\begin{equation}e_T(T_{p_i}F) = \prod_j \alpha_{ij}\prod_{k=1}^{f_i} \beta_{ik},
\end{equation}
\begin{equation}
Td_T(T_{p_i}F) = \displaystyle \frac{e_T(T_{p_i}F)}{\prod_j (1-e^{-\alpha_{ij}})\prod_{k=1}^{f_i} (1-e^{-\beta_{ik}})},
\end{equation}
\begin{equation}
e_T((NF)_{p_i}) =\displaystyle \prod_{l=f_i+1}^{m_i} \beta_{il},
\end{equation}
\begin{equation}
Td_T((NF)_{p_i})= \frac{\displaystyle \prod_{l=f_i+1}^{m_i} \beta_{il}}{\displaystyle \prod_{l=f_i+1}^{m_i} (1-e^{-\beta_{il}})},
\end{equation}
\begin{equation}
e^{(c_{1,T}(L)\vert_F)_{p_i}}=e^{\mu(p_i)}.
\end{equation}
And so our formula becomes
\begin{equation}\chi \circ \exp = \sum_{F \in \pi_0(M^S)}\sum_{p_i \in F^T} \frac{e^{\mu(p_i)}}{\prod_j (1-e^{-\alpha_{ij}})\prod_{k=1}^{m_i} (1-e^{-\beta_{ik}})}.
\end{equation}

We now organise this formula into characters of representations of $K$. By Lemma \ref{wactsonp}, $W(K)$ acts on the fixed points of the fixed manifolds $F$. This action partitions the fixed points contained in $F$ into $W(K)$-orbits; each orbit contains $\vert U_i \vert$ points, and has a unique representative $p_i$ whose image $\mu(p_i)$ lies in the closed positive Weyl chamber $\overline{\mathcal{W}_K}$. For given $F$, the set of such orbit representatives is $F \cap P^+$.

Recall from Lemma \ref{wactsonweights} that the action of $w \in W(K)$ takes weights of the tangent space at $p_i$ to weights of the tangent space at $w\cdot p_i$. So we can write our formula more suggestively as
\begin{equation}
\label{suggestive}
\chi \circ \exp = \sum_{F \in \pi_0(M^S)} \sum_{p_i \in F \cap P^+} \sum_{w \in U_i}\dfrac{\frac{e^{w(\mu(p_i))}}{\prod_j (1-e^{-w\cdot\alpha_{ij}})}}{\prod_{\beta \in B_i} (1-e^{-w\cdot\beta})}.
\end{equation}
Recall that an irreducible $K$-character has the form $\sum_{w \in U_i}\frac{e^{w\cdot \lambda}}{\prod_\alpha(1-e^{-w\cdot \alpha})}$, where the $\alpha$ are the positive roots of $K$ whose inner product with $\lambda$ is non-zero. (When $\lambda$ is regular, this is all the positive roots and this becomes the more familiar $\sum_{w \in W(K)}\frac{e^{w\cdot \lambda}}{\prod_{\phi \in \Phi^+(K)}(1-e^{-w\cdot \phi})}$.) So to get irreducible $K$-characters, we would like to rewrite
\begin{equation}
\label{rewritethis}
\sum_{w \in U_i}\dfrac{\frac{e^{w(\mu(p_i))}}{\prod_j (1-e^{-w\cdot\alpha_{ij}})}}{\prod_{\beta \in B_i} (1-e^{-w\cdot\beta})}
\end{equation} as
\begin{equation}\frac{\sum_{w \in U_i}\frac{e^{w\cdot \mu(p_i)}w\cdot R}{\prod_j(1-e^{-w\cdot \alpha_{ij}})}}{H},
\end{equation} where $R = \sum_{\tau \in E}c_\tau e^\tau$, for $E$ a finite subset of the integral weight lattice, and $c_\tau \in \mathbb{Z}$. Equivalently, we wish to rewrite $(\prod_{\beta \in B_i}(1-e^{-\beta}))^{-1}$ as $R/H$ with $H$ required to be $W(K)$-invariant. One $W(K)$-invariant set we have is the set $\mathcal{B}_i$ of all weights of the torus action on the spaces $T_{w\cdot p_i}M/T_{p_i}(K\cdot p_i)$ for $w$ ranging over $W(K)$. In fact, since we chose $\xi$ to be fixed by $W(K)$, we know $\langle w\cdot \beta, \xi \rangle = \langle \beta , \xi \rangle$ and so $W(K)$ also preserves the set $\mathcal{B}_i^+$ of polarised weights. Let us therefore take
\begin{equation}
H = \prod_{\gamma \in \mathcal{B}^+_i}(1-e^{-\gamma}).
\end{equation}
Thus we find that
\begin{equation*}
R = \dfrac{\prod_{\gamma \in \mathcal{B}^+_i}(1-e^{-\gamma})}{\prod_{\beta \in B_i}(1-e^{-\beta})}.
\end{equation*}
Consider $\beta \in B_i$. If $\beta = \beta^+$, then $\beta \in \mathcal{B}^+_i$, so the corresponding terms cancel. If $\beta = -\beta^+$, then $-\beta \in \mathcal{B}^+_i$, so we can replace the corresponding terms with a $-e^\beta$. By definition the number of $\beta \in B_i$ with $\beta = -\beta^+$ is $s_i$, and so we are left with
\begin{equation}
R = (-1)^{s_i}e^{\sum_{\beta \in B_i:\beta=-\beta^+}\beta}\prod_{\gamma \in C_i}(1-e^{-\gamma}),
\end{equation}
where $C_i$ is the set of all polarised weights $\gamma \in \mathcal{B}_i^+$ of the torus action on the tangent spaces to the whole orbit of fixed points such that neither $\gamma$ nor $-\gamma$ is in $B_i$; that is, neither $\gamma$ nor $-\gamma$ is a weight of the torus action on the tangent space $T_{p_i}M$ at the particular fixed point $p_i$. Write $\overline{\beta_i}$ for $\sum_{\beta \in B_i: \beta = -\beta^+}$. So we have rewritten \eqref{rewritethis} as
\begin{equation}
\label{rewritten}
\frac{\sum_{w \in U_i}w\cdot\left(\frac{e^{\mu(p_i)}(-1)^{s_i}e^{\overline{\beta_i}}\prod_{\gamma \in C_i}(1-e^{-\gamma})}{\prod_j (1-e^{-\alpha_{ij}})}\right)}{\prod_{\gamma \in \mathcal{B}^+_i}(1-e^{-\gamma})}.
\end{equation}

\begin{lem} \label{trivrepn} Let $A_i \subset \Phi^+(K)$ be the set $\{\alpha \in \Phi^+(K) \vert \langle \alpha, \mu(p_i) \rangle = 0\}$, and let $\mathcal{A}_i$ be the subgroup of $W(K)$ generated by the reflections $w_\alpha$ for $\alpha \in A_i$. Then $$\sum_{w\in \mathcal{A}_i}\frac{1}{\prod_{\alpha \in A_i}(1-e^{-w\cdot \alpha})}=1.$$
\end{lem}
\begin{proof} Let $s_w$ denote the number of $\alpha$ in $A_i$ such that $w\cdot \alpha$ is a negative root of $K$, and let $\rho_{A_i}$be half the sum of the elements of $A_i$. Then $$\sum_{w\in \mathcal{A}_i}\frac{1}{\prod_{\alpha \in A_i}(1-e^{-w\cdot \alpha})}=\frac{\sum_{w\in \mathcal{A}_i}(-1)^{s_w}e^{w\cdot \rho_{A_i}}}{\prod_{\alpha}(e^{\alpha/2}-e^{-\alpha/2})}.$$
We claim that this is equal to $1$. Let $\Psi \subset A_i$ be a set of simple roots (so that each $\alpha \in A_i$ can be expressed uniquely as a non-negative integer combination of the $\psi \in \Psi$). Let $x=(x_1, \ldots, x_{\vert A_i \vert}) \in \{-1,1\}^{\vert A_i\vert }$, and consider the term $X=e^{\sum x_i \alpha_i/2}$ in the expansion of the product in the denominator.

If for all $i, j, k$ for which $\alpha_i = \alpha_j+\alpha_k$ and $x_j = x_k$ we also have $x_i =x_j$, then $\sum x_i \alpha_i/2 = w\cdot \rho_{A_i}$ for some $w \in \mathcal{A}_i$, and the coefficient of $X$ in the denominator is $\prod x_i = (-1)^{s_w}$. Every $e^{w\cdot \rho_{A_i}}$ appears once in the denominator.

Otherwise, suppose there are $i, j, k$ with $\alpha_i = \alpha_j + \alpha_k$ and $x_j = x_k$ but $x_i = -x_j$. Then there's another term $\overline{X} = e^{\sum \overline{x_l}\alpha_l/2 }$ where $\overline{x_l}=-x_l$ for $l = i, j, k$ and $\overline{x_l}=x_l$ otherwise. Then $X$ and $\overline{X}$ both appear once in the expansion of the denominator, each with opposite sign and so they cancel.

Thus we are left with $\prod_\alpha(e^{\alpha/2}-e^{-\alpha/2})=\sum_w (-1)^{s_w}e^{w\cdot \rho_{A_i}}$, which proves the lemma.
\end{proof}

By multiplying \eqref{rewritten} by $\sum_{w\in \mathcal{A}_i}(\prod_{\alpha \in A_i}(1-e^{-w\cdot \alpha}))^{-1}$, we get
\begin{equation}
\frac{\sum_{w \in W(K)}w\cdot\left(\frac{e^{\mu(p_i)}(-1)^{s_i}e^{\overline{\beta_i}}\prod_{\gamma \in C_i}(1-e^{-\gamma})}{\prod_{\phi \in \Phi^+(K)} (1-e^{-\phi})}\right)}{\prod_{\gamma \in \mathcal{B}^+_i}(1-e^{-\gamma})},
\end{equation}
which is a quotient of $K$-characters, where the $K$-character of highest weight $\mu(p_i)+\overline{\beta_i}+\lambda$ appears with multiplicity equal to the multiplicity of $e^\lambda$ in $(-1)^{s_i}\prod_{\gamma \in C_i}(1-e^{-\gamma})$. Write $V^K_\lambda$ for the irreducible $K$-representation of highest weight $\lambda$, and $m_i(\eta)$ for the multiplicity of $e^\eta$ in $\prod_{\gamma \in C_i}(1-e^{-\gamma})$. Then summing over all orbits we get

$$\chi(Q(M, \omega, \nabla, L; J)) = \sum_{F \in \pi_0(M^S)} \sum_{p_i \in F \cap P^+} \frac{\sum_{\eta \in \Lambda}(-1)^{s_i}m_i(\eta) \chi(V^K_{\mu(p_i)+\overline{\beta_i}+\eta})}{ \prod_{\gamma \in \mathcal{B}^+_i} (1-e^{-\gamma})},$$
which is equation \eqref{mainthm}, as claimed. \end{proof}

\begin{remark} There are at most $2^{\vert C_i \vert}$ points $\eta \in \Lambda$ where $m_i(\eta)$ is non zero.
\end{remark}

\begin{remark} Although the proof of our main theorem only requires the $T$ action, considering the $S$ action as well helps with the geometric intuition. Consider one connected component $F \in \pi_0(M^S)$ of the fixed set of the $S$-action. By Corollary \ref{fixedmfldsofsarekspaces} $F$ is a Hamiltonian $K$-space, so the character of the quantisation of the restriction of $L$ to $F$ is given by $\int_F e^{c_{1,T}(L)\vert_F}Td_T(F)$. This is an integral of $S$-equivariant forms, so we can apply localisation with respect to the $S$ action on $F$ to get $\sum_{p \in F^T} \frac{e^{c_{1,T}(L)_p}Td_T(T_pF)}{e_T(T_pF)}$. These terms, for each $F \in \pi_0(M^S)$, appear in the formula \eqref{allbrokenup} for the full character $\chi \circ \exp$, each multiplied by $Td_T(NF)_p / e_T(NF)_p$. Geometrically, this tells us that the way a few $K$-characters combine to form a $G$-character is a consequence of the way that the fixed manifolds of the $S$-action -- each of which is a $K$-space -- are embedded in the $G$-space. \end{remark}

\subsection{Example}
\begin{example}
Let $G = SU(3)$ and let $K\subset G$ be $S(U(2)\times U(1))$. Denote by $\alpha, \beta$ and $\gamma$ the positive roots of $G$, where $\beta = \alpha + \gamma$, and let $\alpha$ be the positive root of $K$. Choose $\xi$ such that the polarised roots are the positive roots. Let $\nu = \beta + \gamma$, and let $M$ be  the coadjoint orbit $G.\nu\cong G/K$. In this case the moment map is the inclusion $i: G\cdot \nu \hookrightarrow \gg^*$ composed with the projection $\pi: \gg^* \rightarrow \tt^*$.

The centraliser of $K$ in $G$ is the subtorus $S$ consisting of matrices of the form \[ \left( \begin{array}{ccc}
e^{i\theta} & 0& 0 \\
0 & e^{i\theta} & 0 \\
0 & 0 & e^{-2i\theta} \end{array} \right),\] for $\theta \in [0,2\pi)$. The $T$ action on $G/K$ has three fixed points: their moment map images are $\nu$, $w\cdot \nu$, and $w^2\cdot \nu$ where $w$ is one of the rotations in $W(G)$. Let's assume $w\cdot \nu$ is in the dominant Weyl chamber $W_K$. These fixed points lie in the two fixed manifolds of the $S$-action: the point $\nu$, and the copy of $K/T\cong \mathbb{P}^1$ which maps to the edge between $w\cdot \nu$ and $w^2\cdot \nu$ under the moment map. Each fixed manifold contains just one $W(K)$-orbit of fixed points.

For the fixed manifold $F \cong \mathbb{P}^1$, the weights of tangent spaces are $-\gamma$ (at $T_{w\cdot p}G/K$) and $-\beta$ (at $T_{w^2\cdot p}G/K$). So the denominator will be $(1-e^{-\beta})(1-e^{-\gamma})$. Each fixed point in this $W(K)$-orbit has one negative weight on its tangent space that's not a root of $K$, so $s_i = 1$. Here $\overline{\beta_i}$ is the sum of the negative weights at $w\cdot \gamma$, which is $-\gamma$. Recall that $C_i$ is the set of weights $\delta$ of the $T$ action on tangent spaces to $M$ at fixed points of the orbit of $p_i$ such that neither $\delta$ nor $-\delta$ is a weight of the tangent space at $p_i$, where $p_i$ is the fixed point in the dominant Weyl chamber for $K$; in this case $p_i$ is $w\cdot \nu$ and $C_i$ is $-\beta$. Thus $m_i(\eta)$ is the multiplicity of $e^\eta$ in $1-e^{-\beta}$, so

\[ m_i(\eta) = \begin{cases}
    1, & \eta = 0\\
    -1, & \eta = -\beta\\
    0, & \text{otherwise.}
  \end{cases}\]
At the point $p$, the weights of the tangent space are $\beta$ and $\gamma$, so the denominator is again $(1-e^{-\beta})(1-e^{-\gamma})$. Both weights are positive so $s_i=0$ and $\overline{\beta_i}=0$. This time there are no weights on tangent spaces in that orbit that are not weights of the tangent space at that point, so $C_i$ is empty and $m_i(\eta)$ is non-zero (and equal to one) only when $\eta = 0$.

So our theorem tells us that

$$\chi(V^G_\nu) = \frac{\chi(V^K_{\nu})-\left(\chi(V^K_{w\cdot\nu - \gamma})-\chi(V^K_{w\cdot\nu -\beta -\gamma})\right)}{(1-e^{-\beta})(1-e^{-\gamma})}.$$

\end{example}

\subsection{Relation to the GKRS formula}

In their paper \cite{gkrs}, Gross, Kostant, Ramond and Sternberg consider equal rank Lie algebras $\mathfrak{k} \subset \gg$ with $\gg$ semisimple and $\mathfrak{k}$ reductive, and give a formula for the characters of irreducible $\gg$-representations in terms of certain irreducible $\mathfrak{k}$-characters. Since representations of Lie groups are also representations of their Lie algebras, in the case where $G$ is semisimple and the quantisation of the $G$-action on $M$ is an irreducible $G$-representation, we expect our formula to agree with the GKRS formula. We now show that the two formulae do indeed coincide.

At the end of the proof of Theorem \ref{main}, we had expressed the contribution from one orbit of fixed points as

\begin{equation}\label{toconvert}
\dfrac{\sum_{w \in W(K)}w\cdot \left(\frac{e^{\mu(p_i)}(-1)^{s_i}e^{\overline{\beta_i}}\prod_{\gamma \in C_i}(1-e^{-\gamma})}{\prod_{\phi \in \Phi^+(K)}(1-e^{-\phi})}\right)}{\prod_{\gamma \in \mathcal{B}^+_i}(1-e^{-\gamma})}.
\end{equation}
We may write $\mathcal{B}^+_i = \{\beta \in B_i \vert \beta = \beta^+ \} \sqcup \{-\beta \vert \beta \in B_i, \beta = -\beta^+ \}\sqcup C_i$. So multiplying numerator and denominator of \eqref{toconvert} by $\prod_{\gamma \in \mathcal{B}^+_i}e^{\gamma/2}$ (which is $W(K)$-invariant), we get

\begin{equation}\label{withhalves}
\dfrac{\sum_{w\in W(K)}w\cdot \left( \frac{e^{\mu(p_i)}(-1)^{s_i}e^{\beta_i}\prod_{\gamma \in C_i}(e^{\gamma/2}-e^{-\gamma/2})}{\prod_{\phi \in \Phi^+(K)}(1-e^{-\phi})} \right)}{\prod_{\gamma\in \mathcal{B}^+_i}(e^{\gamma/2}-e^{-\gamma/2})}.
\end{equation}
Since $\gamma/2$ is not necessarily a weight of $T$, the numerator of \eqref{withhalves} is a character not of $K$ but of a covering of $K$; it is however a character of the Lie algebra $\mathfrak{k}$. Thus if we work at the Lie algebra level we have the alternative expression
\begin{equation} \label{liealgversion}
\chi(Q(M, \omega, \nabla, L; J)) =\sum_{F \in \pi_0(M^S)} \sum_{p_i \in F \cap P^+} \dfrac{\sum_{\eta \in \Lambda}(-1)^{s_i}\tilde{m_i}(\eta)\chi(V^\mathfrak{k}_{\mu(p_i)+\beta_i+\eta})}{\prod_{\gamma \in \mathcal{B}^+_i}(e^{\gamma/2}-e^{-\gamma/2})},
\end{equation}
where $\tilde{m_i}(\lambda)$ is the multiplicity of $e^\lambda$ in $\prod_{
\gamma \in C_i}(e^{\gamma/2}-e^{-\gamma/2})$.

Let $K \subset G$ be compact connected Lie groups of equal rank, and let $G$ act on $G/T$ (by left multiplication). Choose positive roots for $G$, let $\rho_G$ be half their sum, and let $\rho_K$ be half the sum of the positive roots of $K$. Recall that the choice of positive roots determines a complex structure on $G/T$; let $J$ be the associated almost complex structure. Choose $\xi$ such that the polarised roots are the positive roots. Let $\lambda$ be a dominant integral weight, and let $\ll = G \times_T \mathbb{C}_{(\lambda)} \to G/T$ be the prequantisation line bundle, where $T$ acts on $\mathbb{C}_{(\lambda)}$ with weight $\lambda$ (as in \cite{bott}). In this case the quantisation is the irreducible representation of $G$ of highest weight $\lambda$. The images of the fixed points of the $T$ action are the $w\cdot \lambda$, for $w \in W(G)$. Write $p_w$ for the fixed point whose image $\mu(p_w)$ is $w\cdot \lambda$. Let $\mathcal{W}_G$, $\mathcal{W}_K$ denote the positive Weyl chambers of $G$, $K$ respectively (so $\mathcal{W}_G \subset \mathcal{W}_K$). There is one fixed manifold of the $S$ action containing each of the fixed points whose images lie in the positive Weyl chamber $\mathcal{W}_K$, and each of those fixed manifolds contains a single $W(K)$-orbit of fixed points. As in Gross, Kostant, Ramond and Sternberg's paper \cite{gkrs}, let $C$ be the subset of $W(G)$ mapping $\mathcal{W}_G$ into $\mathcal{W}_K$; the fixed manifolds are indexed by $C$. The weights of the torus action on $T_{p_w}M$ are $\{w\cdot \phi \vert \phi \in \Phi^+(G) \}$. The orbit $K\cdot \lambda$ is a generic coadjoint orbit (of $K$) and so the set of weights of the $T$ action on the tangent space to the orbit is $\{w\cdot \phi \vert \phi \in \Phi^+(G)\}\cap \Phi(K)$, while the remaining weights make up the $\beta_w$. For every fixed manifold $F$, $\mathcal{B}^+_i$ is exactly the set $\Phi^+(G / K)$ of positive roots of $G$ that are not roots of $K$. For each weight $\beta\in B_w$ of the torus action on the tangent space to a fixed point, either $\beta$ or $-\beta$ appears as a weight of the torus action on the tangent space at each other fixed point in the $W(K)$-orbit; thus $C_i$ is empty, and so $\tilde{m}_i(\eta)$ is non-zero only when $\eta = 0$ in which case $\tilde{m}_i(\eta) = 1$. By definition $s_c$ is the number of positive roots changed into negative roots under $c$; that is, $(-1)^{s_c} = \epsilon(c)$. The weights at $T_{p_c}G/T$ are $\{c\cdot\phi \vert \phi \in \Phi^+(G)\}$. Note that if $c\cdot\phi \in \Phi(K)$ then $c\cdot\phi \in \Phi^+(K)$ since $C$ maps $\mathcal{W}_G$ into $\mathcal{W}_K$. So $\beta_c = \frac{\sum_{\phi \in \Phi^+(G)}c\cdot\phi - \sum_{\alpha \in \Phi^+(K)}\alpha}{2} = c\cdot\rho_G - \rho_K$. Substituting this all into equation \eqref{liealgversion}, we get

$$\chi(V) = \frac{\sum_{c \in C}\epsilon(c)\chi(V_{c(\lambda + \rho_G)-\rho_K})}{\prod_{\phi \in \Phi^+(G / K)}(e^{\phi/2}-e^{-\phi/2})},$$ which is exactly the GKRS formula (equation 5 in \cite{gkrs}).

An alternative approach to obtaining a GKRS-like formula that applies to Lie groups is discussed in Landwebber and Sjamaar's paper \cite{ls}.

\section{Multiplicity Formula}

In this section, we will give a formula for the multiplicity in $Q(M, \omega, \nabla, L; J)$ of $V^K_\lambda$,the irreducible representation of $K$ of highest weight $\lambda$. We will compare our formula with the Guillemin-Prato formula from \cite{gp}.

\subsection{Derivation of the Multiplicity Formula}

Our starting point is the equation

\begin{equation}
\label{currentsuggestive}
\chi \circ \exp = \sum_{F \in \pi_0(M^S)} \sum_{p_i \in F \cap P^+} \sum_{w \in U_i}\dfrac{\frac{e^{w(\mu(p_i))}}{\prod_j (1-e^{-w\cdot\alpha_{ij}})}}{\prod_{\beta \in B_i} (1-e^{-w\cdot\beta})}.
\end{equation}
from the previous section. Let us start by only considering fixed points belonging to one Weyl orbit. The contribution to formula \eqref{currentsuggestive} from such a set of points is

\begin{equation}
\label{oneorbit}
\sum_{w \in U_i} \frac{\frac{e^{w\cdot\mu(p_i)}}{\prod_j (1-e^{-w\cdot\alpha_{ij}})}}{\prod_k(1-e^{-w\cdot\beta_{ik}})}.
\end{equation}

We polarise the $\beta_{ij}$ as above. If $\beta_{ij}=\beta_{ij}^+$, we have $(1-e^{-\beta_{ik}})^{-1}=(1-e^{-\beta_{ik}^+})^{-1}$; if $\beta_{ij}=-\beta_{ij}^+$ then rewrite $(1-e^{-\beta_{ik}})^{-1}$ as $(-e^{-\beta^+_{ik})}(1-e^{-\beta^+_{ik}})^{-1}$. By definition the number of $\beta_{ij}$ with $\beta_{ij}=-\beta_{ij}^+$ is $s_i$; thus $(\prod_k(1-e^{-\beta_{ik}}))^{-1}$ becomes $((-1)^{s_i}e^{-(\beta^+_i - \beta_i)})(\prod_k (1-e^{-\beta^+_{ik}}))^{-1}$.
As discussed earlier, we have $s_{w(i)}=s_i$, and $w\cdot(\beta^+_i - \beta_i) = \beta^+_{w(i)} - \beta_{w(i)}$. Thus we can rewrite \eqref{oneorbit} as

\begin{equation}
\sum_{w \in U_i} \frac{\frac{(-1)^{s_i}w\cdot(e^{\mu(p_i)-\beta^+_i + \beta_i})}{\prod_j (1-e^{-w\cdot\alpha_{ij}})}}{\prod_k(1-e^{-w\cdot\beta^+_{ik}})}
\end{equation}
Expanding $(1-x)^{-1}$ as $1+x+x^2+\cdots$, this gives us

\begin{equation}
\sum_{w\in U_i}\frac{(-1)^{s_i}w\cdot(e^{\mu(p_i)-\beta^+_i + \beta_i})}{\prod_j (1-e^{-w\cdot\alpha_{ij}})}\prod_k(1+e^{-w\cdot \beta^+_{ik}}+e^{-2w\cdot \beta^+_{ik}}+e^{-3w\cdot \beta^+_{ik}}+\cdots).
\end{equation}
For $\zeta \in \tt^*$, write $P_i(\zeta)$ for the number of ways to write $\zeta$ as a sum $\sum_jc_j\beta^+_{ij}$, where the $c_j$ are non-negative integers. So the contribution of this $W(K)$-orbit of fixed points is

\begin{equation} \label{orbitcont}
(-1)^{s_i}\sum_\zeta P_i(\zeta)\sum_{w \in U_i}\frac{w\cdot(e^{\mu(p_i)-\beta^+_i + \beta_i + \zeta})}{\prod_j (1-e^{-w\cdot\alpha_{ij}})}.
\end{equation}
Recall that an irreducible character of $K$ of highest weight $\lambda$ takes the form $$\sum_{w \in W(K)}\frac{e^{w\cdot \lambda}}{\displaystyle \prod_{\alpha \in \Phi^+(K)}(1-e^{-w\cdot\alpha})}.$$
Note that if $\mu(p_i)$ is regular and $w^{-1}\mu(p_i)$ is in the positive Weyl chamber $W_K$, then the $\alpha_{ij}$ are exactly the $w\cdot \alpha$, for $\alpha \in \Phi^+(K)$. If $\mu(p_i)$ is not regular and lies on a hyperplane orthogonal to the root $\alpha$ of $K$, then neither $\alpha$ nor $-\alpha$ will be a weight on $T_{p_i}(K\cdot p_i)$. As before let $A_i$ be the set $\{\alpha \in \Phi^+(K) \vert \langle \alpha, \mu(p_i) \rangle = 0\}$; notice that this is exactly the set of positive roots $\alpha$ of $K$ such that neither $\alpha$ nor $-\alpha$ is a weight of the torus action on $T_{p_i}(K\cdot p_i)$. Hence $\Phi^+(K) = \{\alpha_{ij}\} \sqcup A_i$. By lemma \ref{wactsonweights} we know that $A_{w\cdot i}=\{w\cdot \alpha \vert \alpha \in A_i\}$ for $w \in W(K)$. Let $\mathcal{A}_i < W(K)$ be the subgroup generated by the reflections $w_\alpha$ for $\alpha \in A_i$. Note that $\prod_{\alpha \in A_i}((1-e^{-\alpha})^{-1}+(1-e^\alpha)^{-1})$ is $\mathcal{A}_i$-invariant, so the action of $U_i$ on this product is well defined.

Using Lemma \ref{trivrepn} we can rewrite \eqref{orbitcont} as
\begin{equation} \label{wv}
(-1)^{s_i}\sum_\zeta P_i(\zeta)  \sum_{w \in U_i}\left(\frac{w\cdot(e^{\mu(p_i)-\beta^+_i + \beta_i - \zeta})}{\prod_k (1-e^{-w\cdot\alpha_{ik}})} \sum_{v\in \mathcal{A}_i}\frac{1}{\prod_{\alpha \in A_i}(1-e^{-v\cdot \alpha})} \right).
\end{equation}
But since $\Phi^+(K) = \{\alpha_{ik}\}\sqcup A_i$, this can be simplified to
\begin{equation} \label{wholewk}
(-1)^{s_i}\sum_\nu P_i(\nu) \sum_{w \in W(K)}\frac{w\cdot(e^{\mu(p_i)-\beta^+_i + \beta_i - \nu})}{\prod_{\alpha \in \Phi^+(K)} (1-e^{-w\cdot\alpha})}.
\end{equation}
\begin{remark}
In the previous formula \eqref{wv}, the sum over $U_i$ corresponded to a sum over the fixed points in one $W(K)$-orbit. In the formula \eqref{wholewk}, we instead sum over the set of pairs $(p,w)$ where $p = w\cdot p_i$ for $w\in W(K)$.  Geometrically, what we have done is replace each point $p$ with several copies of itself, one for each Weyl chamber on whose boundary $\mu(p)$ lies.
\end{remark} Let $Z_i$ denote the set of pairs $(p,w)$ where $p = w\cdot p_i$ for $w\in W(K)$. Note that $\vert Z_i \vert = \vert W(K) \vert$.
 We let $Z$ be the union over all $p_i \in P^+$ of the $Z_i$; that is, let $Z$ be the set of pairs $(p,w)$ where $p$ is a fixed point of the $T$ action on $M$ and $w\in W(K)$ is such that $w^{-1}(\mu(p))$ is in the closed positive Weyl chamber $\overline{\mathcal{W}_K}$. (So each regular point appears once; the non-regular points appear once for each Weyl chamber in whose boundaries they lie.)

Equation \eqref{wholewk} expresses the contribution of this orbit as a sum of irreducible $K$-characters; we would now like to know the multiplicity with which the $K$-character of highest weight $\lambda$ occurs, for each $\lambda \in \tt^*$.
Let $\rho_K$ denote half the sum of the positive roots of $K$. We can rewrite \eqref{wholewk} as
\begin{equation} \label{qi}
(-1)^{s_i}\sum_\zeta P_i(\zeta)\sum_{(p_i,w)\in Z_i}\epsilon(w) \frac{e^{\mu(p_i)-\beta^+_i+\beta_i-\zeta-\rho_K+w\cdot \rho_K}}{\prod_{\alpha \in \Phi^+(K)}(1-e^{-\alpha})}
\end{equation}
The contribution of one $W(K)$-orbit to the multiplicity of $\lambda$ is given by the multiplicity in \eqref{qi} of terms that look like $\frac{e^\lambda}{\prod_{\alpha \in \Phi^+(K)}(1-e^{-\alpha})}$. So write $\lambda = \mu(p_i)-\beta^+_i+\beta_i-\zeta -\rho_K+w\cdot\rho_K$. Then we can rewrite the contribution from one $W(K)$ orbit as
\begin{equation}
(-1)^{s_i}\sum_\lambda \sum_{(p_i,w)\in Z_i} \epsilon(w) P_i(-\lambda + \mu(p_i)-\beta^+_i +\beta_i-\rho_K +w\cdot\rho_K)\frac{e^\lambda}{\prod_{\alpha \in \Phi^+(K)}(1-e^{-\alpha})}
\end{equation}
and hence, summing over all Weyl orbits, we get the following formula:
\begin{thm} Let $K \subset G$ be compact Lie groups of equal rank with a common maximal torus $T$. Let $(M, \omega, L, \nabla; J)$ be a compact Hamiltonian $G$-space, and let $Q(M, \omega, \nabla, L; J)$ be its quantisation. Let $\lambda$ be a dominant weight for $K$. Then the multiplicity of the irreducible $K$-representation of highest weight $\lambda$ in the $G$-representation $Q(M, \omega, \nabla, L; J)$ is given by $$\#(\lambda, Q(M, \omega, \nabla, L; J)) = \sum_{(p_i,w)\in Z} (-1)^{s_i}\epsilon(w)P_i(\mu(p_i)+w\cdot\rho_K +\beta_i -\lambda - \rho_K - \beta^+_i).$$ \end{thm}
\begin{remark} This is almost the Guillemin-Prato multiplicity formula from \cite{gp}, generalised to allow fixed points whose images under $\mu$ are not regular. The two formulae do not quite agree; the formula appearing in \cite{gp}, in our notation, is the following: $$\#(\lambda, Q(M, \omega, \nabla, L; J)) = \sum_{(p_i,w)\in Z} (-1)^{s_i}\epsilon(w)P_i(\lambda +w\cdot\rho_K +\beta^+_i -\beta_i - \rho_K -\mu(p_i)).$$ \end{remark}

\subsection{Examples}
We will illustrate our formula using the same examples as in the previous section.

\begin{example}[Generic coadjoint orbits] Suppose $M=G/T$, and denote by $\nu$ the moment map image of the fixed point $p$ that lands in the dominant Weyl chamber. So we can identify $G/T$ with the coadjoint orbit $G\cdot \nu$. Then the set of fixed points is $\{u\cdot p \vert u \in W(G)\}$, and their moment map images are the $u \cdot \nu$. Write $p_u = u\cdot p$ for $u\in W(G)$. So the set $Z$ is $\{(p_u, w)\vert u \in W(G), w \in W(K), w^{-1}(u\cdot \nu) \in \mathcal{W}_K\}$. The weights of the $T$ action on the spaces $T_pM/T_p(K\cdot p)$ are the roots of $G$ that are not roots of $K$. So $\{\beta_{uj}\}=\Phi^+(G / K)$, and at the point $p_u$ the $\beta_{uj}$ are $\{u\cdot \phi \vert \phi \in \Phi^+(G / K), u\cdot \phi \notin \Phi(K)\}$. For each $u$, $\beta_u = u\cdot \rho_G - w \cdot \rho_K$, and $\beta_u^+=\rho_G - \rho_K$.  Since the $\beta_{uj}^+$ are the same at each fixed point, we will write $P(\lambda)$ for the number of ways to express $\lambda$ as a sum $\sum_{j}c_j\beta_j$, where the $\beta_j$ are the positive roots of $G$ that are not roots of $K$, and the $c_j$ are non-negative integers. Finally, $(-1)^{s_u}\epsilon(w)=(-1)^n$, where $n$ is the number of roots of $G$ that change sign under $u$, and so $(-1)^{s_u}\epsilon(w) = \epsilon(u)$. So in this case, our formula becomes
\begin{equation} \label{coadj}
\#(\lambda, Q(M, \omega, \nabla, L; J)) = \sum_{u \in W(G)}\epsilon(u)P(u\cdot \nu+u\cdot\rho_G-\lambda-\rho_G).
\end{equation}
\begin{remark}
Goodman and Wallach proved a branching formula which gives the multiplicity of the irreducible representation of $H$ of highest weight $\lambda$ in the irreducible representation of $G$ of highest weight $\nu$ (Theorem 8.2.1 in \cite{gw}). They do not require $G$ and $H$ to be of equal rank, only that the chosen maximal torus of $H$ is contained in that of $G$, but in the case where the two groups do have a common maximal torus their formula coincides with our \eqref{coadj} above.
\end{remark}
\end{example}

\begin{example}

Consider the same non-generic coadjoint orbit we discussed in the previous section. That is, take $G=SU(3)$ and $K=S(U(2)\times U(1))$. Let $\alpha, \beta$ and $\gamma$ be the positive roots of $G$ (with $\alpha + \gamma = \beta$), and take $\alpha$ to be the positive root of $K$. Let $\nu = \beta + \gamma$, and let $M$ be the coadjoint orbit $G\cdot \nu \cong G/K$. Then the torus action has three fixed points: $\nu$, $w_+\nu$ and $w_-\nu$ (where $w_+$ and $w_-$ are the rotations in $W(G)$ that map $\nu$ into the chambers $\mathcal{W}_K$ and $-\mathcal{W}_K$ respectively).
\begin{center}
\begin{tikzpicture}
\draw [white, fill=gray, fill opacity =0.2](-4,0) --(4,0) --(4,3.46410) --(-4,3.46410);
\node [above] at (3,2) {$\mathcal{W}_K$};
\node [below] at (3,-2) {$-\mathcal{W}_K$};
\draw [gray, fill=gray, fill opacity =0.6](-1.5,2.598) --(3,0) --(-1.5,-2.598) --(-1.5,2.598);
\node [above] at (-1.5,2.598) {$w_+ \nu$};
\node [below] at (-1.5,-2.598) {$w_- \nu$};
\node [below] at (3,0) {$\nu$};
\draw [gray](-4,0) --(4,0);
\draw [gray, rotate=60](-4,0) --(4,0);
\draw [gray, rotate=120](-4,0) --(4,0);
\draw [->][black](0,0) --(0,1.732);
\node [below right] at (0,1.732) {$\alpha$};
\draw [->][black, rotate=-60](0,0) --(0,1.732);
\node [below] at (1.5,0.866025) {$\beta$};
\draw [->][black, rotate=-120](0,0) --(0,1.732);
\node [below] at (1.5,-0.866025) {$\gamma$};
\draw [fill] (3,0) circle [radius=0.1];
\draw [fill] (-1.5,2.598) circle [radius=0.1];
\draw [fill] (-1.5,-2.598) circle [radius=0.1];
\draw [fill] (0,1.732) circle [radius=0.1];
\draw [fill] (1.5,0.866025) circle [radius=0.1];
\draw [fill] (0,0) circle [radius=0.1];
\draw [fill] (1.5,-0.866025) circle [radius=0.1];
\draw [fill] (0,-1.732) circle [radius=0.1];
\draw [fill] (-1.5,0.866025) circle [radius=0.1];
\draw [fill] (-1.5,-0.866025) circle [radius=0.1];
\end{tikzpicture}
\end{center}
Let $u$ be the non trivial element in $W(K)$. Since $w_+\nu$ is in the positive chamber $W_K$, $w_-\nu$ is in the other chamber and $\nu$ is on the boundary wall, the set $Z$ consists of the four elements $(w_+\nu, e), (w_-\nu, u), (\nu, e),$ and $(\nu, u)$. The data at each of these points is as follows:
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
$(p,w)\in Z$ & Weights of $T$ action on $T_pM/T_p(K\cdot p)$& $s_i$ & $\epsilon(w)$  & $\beta_i$ & $\beta_i^+$ \\
\hline
$(w_+\nu,e)$ & $-\gamma$ & 1 & 1 & $-\gamma/2$ & $\gamma/2$ \\
$(w_-\nu,u)$ & $-\beta$ & 1 & -1 & $-\beta/2$ & $\beta/2$ \\
$(\nu,e)$ & $\beta, \gamma$ & 0 & 1 & $(\beta + \gamma)/2$ & $(\beta +\gamma)/2$ \\
$(\nu,u)$ & $\beta, \gamma$ & 0 & -1 & $(\beta + \gamma)/2$ & $(\beta +\gamma)/2$ \\
\hline
\end{tabular}
\end{center}
In this case $\rho_K=\alpha/2$. Putting this into our formula, we get $$-P_{\gamma}(w_+\nu-\lambda-\gamma)+P_{\beta}(w_-\nu - \beta - \alpha-\lambda)+P_{\beta,\gamma}(\nu-\lambda)-P_{\beta,\gamma}(\nu - \alpha - \lambda),$$ where $P_A(\psi)$ denotes the number of ways to write $\psi$ as a sum $\sum_{a \in A}c_a a$ where the $c_a$ are in $\mathbb{N}$. Notice that in this example, each $A$ is a linearly independent set of roots, so $P_A$ will always be one or zero. For each point $(p_i,w)\in Z$, we will shade the region of $\tt^*$ containing those $\lambda$ where $P_i(\lambda + \beta^w_i - \beta_i + \rho_K - w\cdot\rho_K - \mu(p_i))=1$ with red lines of positive slope if $(-1)^{s_i}\epsilon(w)$ is $+1$ or blue lines of negative slope if $(-1)^{s_i}\epsilon(w)$ is $-1$. (Thus the regions where cancellations occur will appear hashed in red and blue.) The point $(w_+\nu,e)$ contributes a multiplicity of $-1$ for every $\lambda$ that can be written as $w_+\nu -\gamma - c\gamma$ for $c \in \mathbb{N}$, so we illustrate this with a blue dashed halfline starting at $w_+\nu - \gamma$ and travelling in the $-\gamma$ direction. The point $(\nu, e)$ contributes a multiplicity of $+1$ for every $\lambda$ that can be written as $\nu -b\beta -c\gamma$ for $b, c \in \mathbb{N}$, so this is illustrated with a red shaded cone with vertex $\nu$. The contributions from the other points are shaded in the same way.
\begin{figure}
\begin{center}
\begin{tikzpicture}
\draw [fill] (3/2,0) circle [radius=0.1];
\draw [black] (3/2,0) circle [radius=0.2];
\draw [fill] (-1.5/2,2.598/2) circle [radius=0.1];
\draw [black] (-1.5/2,2.598/2) circle [radius=0.2];
\draw [fill] (-1.5/2,-2.598/2) circle [radius=0.1];
\draw [fill] (0,1.732/2) circle [radius=0.1];
\draw [black] (0,1.732/2) circle [radius=0.2];
\draw [fill] (1.5/2,0.866025/2) circle [radius=0.1];
\draw [black] (1.5/2,0.866025/2) circle [radius=0.2];
\draw [fill] (0,0) circle [radius=0.1];
\draw [fill] (1.5/2,-0.866025/2) circle [radius=0.1];
\draw [fill] (0,-1.732/2) circle [radius=0.1];
\draw [fill] (-1.5/2,0.866025/2) circle [radius=0.1];
\draw [fill] (-1.5/2,-0.866025/2) circle [radius=0.1];
\draw [fill] (-3/2,0) circle [radius=0.1];
\draw [fill] (-3/2,-1.732/2) circle [radius=0.1];
\draw [fill] (-3/2,1.732/2) circle [radius=0.1];
\draw [fill] (-3/2,3.4641/2) circle [radius=0.1];
\draw [fill] (-3/2,-3.4641/2) circle [radius=0.1];
\draw [orange](-1.5/2,2.598/2) --(3/2,0) --(-1.5/2,-2.598/2) --(-1.5/2,2.598/2);
\node [above] at (-1.5/2,2.598/2) {$w_+ \nu$};
\node [below] at (-1.5/2,-2.598/2) {$w_- \nu$};
\node [below] at (3/2,0) {$\nu$};
\draw [red, pattern color=red, pattern = north east lines](-6/2,5.19615/2) --(3/2,0) --(-6/2,-5.19615/2) --(-6/2,5.19615/2);
\draw [blue, pattern color=blue, pattern = north west lines](-6/2,3.461/2) --(3/2,-1.73205/2) --(-6/2,-6.928203/2) --(-6/2,3.461/2);
\draw [blue, dashed, ultra thick](-3/2,3.46410/2) --(-6/2,5.19615/2);
\draw [red, dashed, ultra thick](-3/2,-3.46410/2) --(-6/2,-5.19615/2);
\draw [white, fill=white, fill opacity =0.6](-6/2,0) --(4/2,0) --(4/2,-3.46410/2) --(-6/2,-6.928203/2);
\draw [yellow, ultra thick](-7,0) --(4,0);

\end{tikzpicture}
\caption{The circled points are the only points in $\overline{\mathcal{W}_K}$ with a non-zero net multiplicity, showing that they are the highest weights of the four characters of $K$ that appear in $Q(M, \omega, \nabla, L; J)$ with multiplicity one.}
\label{colour}
\end{center}
\end{figure}

The portion of the picture we are interested in is the top half, corresponding to the closed positive Weyl chamber $\overline{\mathcal{W}_K}$, since our formula tells us, for an element $\lambda \in W_K$, the multiplicity of the irreducible $K$ representation of highest weight $\lambda$. We can see that most of this region is either unshaded or hashed in red and blue; in fact the only integral weights in $\mathcal{W}_K$ that lie in the positive region are the four lying along the top edge of the triangle. Thus we see that our $G$ character breaks up into the four $K$ characters which have the circled weights as their highest weights (see figure~\ref{colour}).
\end{example}

\bibliographystyle{plain}

redpajama_set_name: RedPajamaArXiv


The Golden Circle is a logical way of thinking that questions the legitimacy of the brand. It encourages us to review our relations with customers, prospects and employees. Why is Apple such an incredible success? Why has Keytrade Bank outstripped the major Belgian banking players? Why did Kodak almost disappear? Ultimately, why has the consumer become the consumactor (and changed from consumer to prosumer)?
Probably because some companies have decided to ask themselves a single question. Why?
Why do it? Why say it? Why join? This is what we will demonstrate in this article, also explaining how this question can be objectified via Emolytics' emotional statistics.
And the last is Why? Why maintain a presence in a particular market, for what purpose and above all with what legitimacy? In other words, to make what and for whom, to say what and to whom, to join what and for what reason?
These three questions follow each other logically. They can be answered by going from what to why. Or, conversely, by going from why to what. Sinek demonstrates that the direction in which you ask the questions is fundamental.
In effect, a vast majority of companies go from what to why. And this is precisely their fundamental error, because in this way they are not in a purchasing process (focused on their client), but a sales process (centered on the company). They don't care about their customers' desire to buy what they have to offer (products and services), so much as their need to sell such products and services to the customers. They haven't adopted the "The Golden Circle" approach.
We are going to illustrate this error with an example, linked to the banking sector, that of Keytrade Bank.
However, first we are going to take a brief detour via our benchmark study "Banking the Customer Experience Dividend" by KPMG. It highlights a revealing insight, which is actually a new paradigm that will require banks to move from a "sell more" position to a "deliver more" position.
Each point within this paradigm defines exactly why Keytrade Bank is so successful because each one is an expression of "The Golden Circle" questions.
In 1998, Keytrade Bank (originally VMS Keytrade) became the first online bank in Belgium. At first glance it was a crazy idea, because investing in the stock market was something already offered by a lot of other more legitimate and more powerful players. And yet, they were immediately successful. Thousands of Belgians agreed to give a young bank come out of nowhere the responsibility for their stock portfolio (in whole or in part). The question is obviously: Why?
The answer is very simple. Keytrade Bank had nothing to sell that was better than the others, but on the other hand, it positioned itself in a way that was totally different to the rest of the market. They gave access to the stock exchanges of the whole world to amateurs and people with small portfolios. They streamlined processes (everything could be done online), supported customers and, above all, made opening an investment account with drastically reduced brokerage fees accessible to everyone.
Why: Keytrade Bank, a bank for people who dare (still their current claim). To be daring is to be uninhibited, listening, open, available. Exactly what a lot of small portfolio owners were waiting for in a closed environment aimed at experts.
What: The classic products and services of the stock market but accessible to all, and cheaply.
Keytrade had nothing to sell that was better than its competitors – it was a powerful digital tool, but it was not the only one. On the other hand, right from the start, its positioning strategy focused on the why (why they existed). And consequently was centered on the user, their clients.
The bank made an appeal to the emotions. Those of envy and the desire to belong. Nowadays, Keytrade customers are significantly emotionally attached to their bank. Brand loyalty is far higher than that associated with the Belgian Big 4 (ING, BNP Fortis, KBC / CBC, DEXIA), who are, nonetheless, far more powerful. In only 20 years they have gone from being the challenger to being the reference. The founders of Keytrade Bank clearly adopted the logic of Simon Sinek's, The Golden Circle.
How can you use Emolytics to objectify the why, how and what?
Emolytics is a statistical tool that allows you to work these three dimensions continuously, in a simple, scientific and objective way.
We will illustrate this through the data of a typical web site (see tables below). They are exemplative of the different insights you can gain when using the logic of "The Golden Circle". They also show the need to question the legitimacy of the communication channels used. Emolytics' platform allows you to make this reflection.
"What ": Which page is responsive, effective, ineffective, which day of the week causes a particular reaction, what is the socio-demographic profile of respondents, what are the most effective channels, what content generates a lot of reactions, etc.
The tables above show which emotions are primarily expressed in terms of age for a given site (top left table top). The same exercise can be done for gender (top right table).
When the two results are superimposed, some interesting insights are revealed. The site in question is obviously very efficient (high Emoscore) for ages 20-29 and 30-39. Whichever gender. On the other hand, beyond 39 some sensitive areas appear, which increase with age, especially for women (25% are irritated, confused or bored).
So now we have an answer to "What" and "Who". But still not to "Why". We'll get this thanks to an analysis of the insights given by the users.
2. "How": which digital channel (or not) do they come from, how long do they stay, within a defined channel how do they express themselves, etc.
So let's continue the analysis – how do my users come to my site? And how is the Emoscore distributed depending on the channel (top left chart) and the device (top right chart)?
We can quickly see that the most satisfied are those that come from organic searches, direct traffic and brand campaigns, on the other hand, the most irritated come from social networks and paid campaigns (SEA type). Similarly, the most irritated are found among desktop users and the most interested in the tablet users. As the differences are small, these interpretations need to be qualified.
But, why don't paid campaigns and social networks produce satisfactory results? We still do not know that. Moreover, is it an effective channel or is it simply useful for this brand?
3. "Why": The Emoscore (an emotional performance indicator) is what best expresses the balance between the legitimacy of the company (the reason for its presence) and the positive / negative emotions experienced by users (their response to this presence).
Let's go back to our example: We saw sensitivities with regard to certain age groups, gender and some of the digital channels used to communicate. The questions that need to be asked now is: Why do users come to this site? What are their objectives? And, is the site is in line with these objectives?
If the site's priorities are to enable the discovery of products, services or pricing, then the objective is clearly achieved (table above, left). This is also confirmed when looking at the digital channels used (table above, right). So, we know what visitors are looking for on the site, and which channels are the most effective, based on the aforesaid objectives.
But is the content strategy, which is obviously strategic, up to user expectations (table below)?
This is at the heart of "Why". We know exactly why visitors endorse content (simple, clear, relevant, high quality, …) but also the weaknesses which need to be corrected (content hard to find, too much content, content not always relevant…). On this basis, it is possible to improve the information available, even though we see that the majority of visitors are satisfied.
If we now superimpose the goal of the visit (the user's "why") and its source (where did they come from), we have a meaningful picture of the strengths (in green) and sensitivities (in red) of the logic of the site content and the digital communication tools used (see table below).
Using Simon Sinek's concept of The Golden Circle, we wanted to demonstrate the importance of focusing upstream on "why" and then going toward "what".
This approach should prevail in any search for meaning, because the legitimacy of the presence of a brand or institution primarily depends on the answer to this question: to make what and with whom, to say what and to whom, to join what and for whom (as a consumer, partner, employee)?
Now that this methodology has been established, in the next part of our dossier we will develop the concept of a digital ecosystem, which is implicitly linked to Sinek's concept because it is intimately linked to the users' experience. It is central to the debate, the cornerstone of any digital ecosystem. And yet … it is often overlooked.
Etienne Gonsette is a Psychologist, analyst and digital auditor for over 20 year, he worked as a digital strategist for numerous Belgian and international companies. He works as Business Developper at Emolytics.