text:
s1 — E10
s1 - E10
Heather: Today, I'm speaking with Kate Barney—entrepreneur, Harvard grad, champion of female leadership, new mom and head of HR for TikTok's global business solutions in the US and EMEA. TikTok was the fastest app in the history of social media to reach a billion user downloads, and it spurred a revolution in self-expression while changing the landscape of music, comedy, activism and so much more. In the last two years, TikTok has gone from 300 to 3,000 employees. I'm so excited to sit down with Kate today and talk about the war for talent, how TikTok is attracting the next generation of professionals and how it's actually becoming a platform for others to recruit talent. So welcome, Kate.
Kate: Thanks for having me, Heather.
Heather: A special thanks for doing this while on maternity leave, nonetheless. Some might think that you have more time on your hands when you're on maternity leave; but, in fact, you have less.
Kate: I will say that I have spent a lot of time on TikTok during maternity leave, getting me through the late-night feedings and sleeplessness. There is an amazing tribe of moms on TikTok, so I found a whole new community since going on mat leave.
Heather: That's fantastic. And it really has been stunning to see how explosive the growth has been and how it's taking on so many new forms. I think, for many people working at TikTok, it would be the dream job. What attracted you to TikTok?
Kate: It definitely is my dream job. I started my career in China. I studied Mandarin in college and, right after college, I went to Beijing. I had a fellowship at the Beijing Film Academy and lived in China for another 11 years doing startups, working in tech and finance, fashion, pretty much everything… moved back to New York, worked in startups…was in London…and we moved back to New York in 2019 with one-year-old twins. I was looking for a new job, coming off of maternity leave and moving back to the US, and a recruiter from ByteDance, the parent company of TikTok, reached out. It was actually a LinkedIn InMail. Always check your InMail, always respond to things, because it is genuine—and there are really great opportunities there. It just sounded too good to be true for me. It was an opportunity to work in a global capacity to support the US and Europe.
I now support global teams throughout Asia and Latin America and Australia. And I work with China, having spent so much time there, I loved the working culture there. I loved the people there. It's so entrepreneurial…and then to be able to do it in New York and to really build up a company that had already seen pretty good market fit. So, it wasn't like a small startup. I definitely cut my teeth at lots of small startups that were trying to sell people on the importance of the product, but TikTok had already hit the sticky factor. When I joined, there were only 35 people in the US, and it was really, really fresh. I was excited about the work culture and being on a global team, which truly is global. Every day, I have calls all over the world. And for me also as a new mom, it was really convenient. I might start my day at 5:30 or 6 in the morning with calls to Europe or China and might end at midnight with calls to Australia or Latin America—but I also have a lot of flexibility in the middle of the day, which was really important for me to be able to have dinner with my kids, put them down for bed and spend an hour or so with them in the morning. And by having a global role, having these bookend-kind of hours, meant that I had a lot more flexibility during the little-kid waking hours.
The people who TikTok seeks out are incredibly open-minded. They're very, very strong in their experience—in their professional career to date—because, again, it's not a tiny startup that was started in, like, a college dorm and is hiring friends of friends. It's really out there trying to find the best there is in the market. I love being able to spend my time with these very smart, very passionate people who are really excited about building something new. And that's what we encourage in the culture. So, for me, it's my dream job. It's amazing.
Heather: Attracting talent is hard in any time, but I think the time we're in now feels particularly hard. There's a lot of opportunity out there, and then you have a lot of people that are questioning what they want to do with their lives. How are you seeing that pan out from your perspective? Are you seeing similar trends to the rest of us? And what are some of the ways that you're keeping talent?
Kate: We have gone from about 300 to 3,000 during COVID, and this is just in the US. So, 90% of our team have never met; we are still all remote. And it's been such a journey for us over the past almost two years now; because at first, with COVID, we needed to hire. We were growing. We got incredibly lucky that we were a pandemic-friendly product. And we want to be very reflective of how lucky we are that, because people were at home and looking for community and looking for joy, TikTok found even more users and even faster acceleration. We had to hire like crazy, but we had to do it all virtually. So, most people have never been in an office, never been in a TikTok client meeting in person, never met each other. People were really unsure of the situation with COVID, so there wasn't a lot of job-hopping.
I think this resignation that we're talking about is very, very new in the past six to eight months. I think it's wonderful that everyone is being a little bit self-reflective of what matters most in their life. And do they love the job they're at? Should they look somewhere else? And because everything's virtual, I think people are really questioning: Do I love the company I work for? Do I relate to this product that we're selling? Do I see myself in this, or is it just a job? And I think, for a lot of us, we put 150% into work and want to see ourselves in it. And if we don't relate with the product, we have to look at what else is there. Hopefully, then, it's the team, it's your work family that keeps you there. What we're trying really hard to do is to create that soft network, those family ties within TikTok, so that people feel like they can come and be their authentic selves—really show who they are and have a lot of fun. We're being as innovative as we can to create that soft network, because I think that that's what is going to keep people at companies rather than moving on.
Heather: Tell me more about how do you actually create those soft networks in a way that is authentic and doesn't necessarily feel forced on people. I think it has been hard for a lot of us to figure out the right way to engage.
Kate: We have an incredible head of diversity and inclusion at the company, who has stood up ERGs [Employee Resource Groups] virtually. They have speakers, they have events, we bring in TikTok creators…it's open to the whole company. You actually get more attendance than if it were in-person. We have seen incredible community over Zoom chat. There's just so much love going on in our group over the Zoom chat: "That's so great. I totally relate to that. That's amazing." And you see this going on, and it feels very authentic, and it feels live. I think that's where we're trying to harness the greater community. So my HR team has grown from 20 to 120 in the last six months.
Heather: Oh, my goodness.
Kate: It's just exploded, as well. How do we get to know each other? How do the newcomers feel just as plugged in as the old people? And I put coffee talk on the calendar. I'll just pick four people two or three times a week. We don't talk about work, but we'll pick a topic. It can help to have these funny conversations that you might have; for example, if you were the first three people in a meeting room before the meeting started and you're looking to small talk, how do you create small talk virtually? And that's how you really feel a bond and get to know funny things about each other. I think you do have to put in a little extra effort when you're virtual. And maybe when we go back to work, we'll have to put in extra effort, as well; because we're getting to know each other again.
Heather: I think a little bit of the extra effort goes a long way. What does your return to office look like here in the States?
Kate: We're very conservative and want to put the health of our employees first, so we keep pushing it back. But for us right now, we're looking at three days in the office, two days at home—trying to recognize that people are used to a more flexible in-office situation. A lot of people really value being at home and not commuting and not spending the extra time blow-drying their hair and getting dressed and just, like, working through the day. A lot of other people are desperate to get back in the office, because their apartment or home feels very small. Or they share an apartment or a home, and they don't have great Wi-Fi; they don't have great resources, so they really want to get back to the office even more.
Right now, I think a great benefit of being online and on Zoom is that it's really accessible. Everybody's there on the screen. Everyone has the same-size square. It's very, very equal. And my worry about going back to the office is a hybrid situation; while it gives employees flexibility, there could be an in crowd and out crowd—people who show up at the office and have a great side conversation over coffee and then say, "Don't even worry about the meeting, guys. We figured it out. I chatted with Heather in the hall. We've got it all down. We're not even going to bother with that meeting anymore." And you're sitting at home waiting to log in. You have your great discussion points. And you're like, "But I want to be there. My opinion's important too."
So, I think a lot of it is about setting the right norms. Perhaps we have some meetings that are all virtual; even if you're in the office, you'd log into Zoom and sit at your laptop so that everybody is fully present for it. Those are the different layers that we're trying to figure out so that, when we return the office, it's a great experience for them.
Heather: There had been, I think, a lot of cultures that were very much about being physically in the office. And if you weren't, you really didn't feel like you were part of the team. Hopefully, this shift actually helps create greater access and equity. A lot of people will talk about Millennials or Gen Z, and what they are looking for in a career has shifted. As you have looked at different positions that you filled and rising talent, what do you think matters most to people? And do you see it changing?
Kate: I think what matters most is feeling like they have a voice, that they're adding impact. I think, regardless of generation cohort right now, people want to see their individual impact added and want to hear their voice. Even if it's somebody who's fresh—it's their first job ever right out of college—they want a seat at that table. It's not an "I want to work my way up." But, to be fair, I've also never wanted to work my way up. I relate to that. I've always thought I have a voice, I have a lot of opinions, I've done my homework, I'm here working hard. Don't you want to hear what I have to say, too? And so, I respect that. I think that if we're going out and we're trying to find really talented, ambitious, hardworking people, then we should be able to let everybody have a voice at the table to get involved. And I think at a company like TikTok, where we're growing really, really fast and there's so much to do, that there's an incredible opportunity for anybody who joins.
TikTok, itself, is not hierarchical at all. We don't use titles. It's very open, and it's really what impact you're having. So, it blends nicely with this desire for individuals to have a lot of impact. I think for someone who is starting off their career, that can be really gratifying to be able to move on to really challenging work much quicker. And I think that also blends well with, perhaps, a generation that doesn't really care about the titles. I think it's, "I want to prove how smart I am with the impact of my work."
So that's exactly what we're looking for, too—people who want to have impact and work hard. For anybody who is looking to rise quickly, find a company that's growing fast and then just keep saying, "Yes, that sounds great. I'll dive into that project." And just keep learning and take on more and more projects, more opportunities. Work with teams that are outside of your team, because you get this incredible exposure very fast. And then you can find yourself in a really interesting role within a company—because, even if you've only been there for a year or two, you can be seen as an expert.
I think that that blends nicely with the aspirations of a new generation of workers who are looking to come in, make a splash, have that seat at the table. We're saying, "Great, come on in and do it. You have to work hard." We're very, very collaborative as a company and want people to work hard together and to be creative together and to be open-minded and listening to each other. I think that is really important, but it's also important for this new generation who wants to be heard and seen.
Heather: Let's talk about TikTok Resumés. Over the summer of 2021, this soft launch and now TikTok is kind of helping other companies recruit talent. Talk to me about that, what the response has been and how you see that effort growing over time.
Kate: I think it is a really interesting new idea and product. For us, it started incredibly organically. We had a fresh college graduate apply for our account manager role, and she happened to make a TikTok after she applied and said, "Hey, TikTok. Look at me. I applied. You should hire me." And one of the women on my HR team happened to see the video in her "for you" feed. The "for you" feed is just so amazing in how random it shows you things but that, at the same time, it really knows you.
She was hired and is doing incredibly well. And then we looked at this as a way to say, "Hey, this is an opportunity to get to know people much more authentically, much more 3D." I think a lot of us realize that the CV is very stale. Your resumé…we edit it down. We use all these action words. Everybody uses the same words. It's the same bullet. You look online for a template.
Having a video that's also super short—15 seconds, 30 seconds—shows your personality. You might be able to say, especially if you are applying to a sales or marketing role, "I'm capable at selling something. This is how I would talk to you. This is how I would pitch a brand or a product." It's really unique, and it allows people to be themselves in a way that might present a lot better than just the CV. And especially for fresh graduates or people who are newer in their career, where your CV is relatively slim, it's an opportunity to shine.
So, where we started this summer was partnering with a few different companies—HBO was one of them—to advertise for their internship program on TikTok. And interns applied only through TikTok Resumes. We are very excited for it to keep growing; it is going to continue to evolve. And like most of the things on TikTok, it started out organically. It was a user that said, this is how I want to use the product. I want to use TikTok, as a platform for myself to pitch myself as a resume. And I think we looked at that and said, that's a great idea; let's build off of that.
And so right now, in partnering with different brands, we're learning: What would they like to see? How will this enhance the applicant process? It still needs to be really useful, so that it's not an added step for recruiters—because we all know that the recruitment team is one of the hardest-working teams in every company right now, trying to screen tons of applicants, as well as do outreach and hire high volumes. We want to make sure it's a real value-add, and also something very innovative and creative.
Heather: Talk to me about your role. What would people be surprised to hear?
Kate: I'm laughing, because I'm saying this incredibly humbly and also just joking. No offense, please. No one take offense at this…
I don't think there are that many people who aspire to work in HR when you're in college. I think, as a field, it is not appreciated enough for what it does. I know that when I first started working in HR, I also did not understand the full breadth of what the team does. I think the misunderstanding of what HR does is just hire and fire. Read resumés and throw them out. You're just kind of like there shredding resumés. Okay. You have a job? No, I don't like you now. Fine. And then you're the police: It was an HR violation. Watch out! HR is in the room. You can't make a joke, or don't do anything! And then you get fired by HR. I think that's an old vision of HR. A lot of great companies have worked hard to change it. I think Google has done a ton around starting a People department.
But I love the field, and I love all the people who work in it. I think that, at its core, it's trying to understand people. It's trying to understand what motivates us. So, it's a really interesting role when you're hiring. If you're a recruiter, you're trying to understand the applicant at its core. Who is this person? What motivates them? Will they be happy in this role? And then there are the HR business partners, who are trying to help teams work together and help people find their next career move and find their self-satisfaction or help leaders really bond while with a team or a leader makes a really difficult decision.
If you do the job right, you understand the business really well, you have the trust of everyone; and people come to you for advice, because they trust you as the third party who knows them, who knows the intentions, who will give them advice to help them succeed at their job. And you're there to make a difference in someone's life, you're there to make someone get the promotion or you're there to help give someone a job. HR executives are really about building trust, understanding human emotions, understanding the motivation for people and helping people excel. We're here to amplify. We amplify culture. If you're a people person, if you're the fixer of your friends, if you're the matchmaker or the one who's always organizing things for people, I think HR is a great fit. I think that's maybe where it's misunderstood—and that's what I really love about it.
Heather: I think that the tide is turning in the recognition that HR is a hugely strategic important role in a company, especially as we're grappling with new technologies, with the importance of inclusion and belonging and diversity. All of those things have risen to the top, and they're not just the HR issues but are company issues. You're kind of in the center of that, which I think is so exciting. As I mentioned earlier, you're currently on maternity leave—and I, have to say, very bright-eyed and awake and not sleep-deprived at all! Tell me how your second leave has been different from your first leave and advice that you would give to new moms or dads that are going to be returning after taking some time off.
Kate: We're seeing a huge baby boom right now at TikTok. I imagine lots of other companies are, as well. And for me, this second leave is so much calmer. The first time I was out on leave, I had a lot more anxiety around being out. "I hope I'm not replaced. Am I going to be seen as working hard enough? Should I take all of this leave? What else could I be doing?" I was always wanting to check in.
This time around, it goes so fast. I know it goes so fast, and I realized I have an incredible team. They don't need me to micromanage while I'm out on leave. It's actually a really great opportunity for people to step up and take on the responsibility or share responsibilities and have exposure to things that they might not have, because they're covering for me while I'm gone. So, I am appreciating every moment. I haven't felt any anxiety over being out in terms of was I seen as taking too long of leave, was I missing out on important projects? Even at TikTok, which moves at this lightning pace. And I joke with my boss that it might be a different company when I come back, because going out for four and a half, five months is dog years. But it'll be really fun to go back and see what everyone else has accomplished while I've been gone. If you're fortunate to have it—and I know not everyone does but, if you're at a company that has it—I say take it and appreciate it, because it goes by really quickly and you have to trust in your teams.
Heather: This is the time that you deserve and that the baby deserves, and then you come back, hopefully, as energized as ever before.
Kate: I will say, it was very interesting being pregnant over Zoom, because no one saw it. So, it wasn't a topic of conversation. Even after I told people, it wasn't first and foremost—whereas with the first pregnancy, it was the first thing people said when I walked into a meeting at work. That was my personality for so many months at work: I was the pregnant HR director at work. How would I be able to do this project while pregnant? Could I do this after the babies? I found it incredibly refreshing to not have it be the first point of conversation in every single meeting just because it wasn't visibly there.
Heather: I can completely appreciate that. Being on a panel, let's say, of one woman and three men, the question that the woman is asked is, "How do you balance being a mom and work?" I appreciate that as a question, but the men don't get asked that. So, it becomes something that, yes, should be part of your persona but becomes almost too much to a certain extent.
The podcast is all about showcasing rising leaders, iconic brands and people who I think are iconic. So, my question to you is, do you have an icon?
Kate: I have a couple of work mentors. I think the person right now, who I'm learning from most, is our global head of sales, Blake Chandlee. He's one of the reasons why I took the job. When I met him, everything clicked; and he's the person whom I spend all of my time with now from an HR partnering perspective—planning the path of the business, planning the path of the team. I remember our first conversation over Zoom—he has a three-year-old and a five-year-old, and we talked about our family and our kids the whole time. He was saying he would want to work really, really hard—and he knows exactly how much effort goes into building a rocket ship, having done it at Facebook and previously at Yahoo—but it was important not to lose sight of putting in your family time and putting in your own time. And I thought, "Wow, that was just a really powerful thing to hear from someone who's going to be the global head of sales at this new company." He is incredibly empathetic as a leader and looks at the people at the core of every business decision. So he's an incredible partner from an HR perspective—somebody who's really on the same wavelength.
Heather: It sounds amazing; and not just saying, "Yes, family is important," but really living it that I think is amazing and something you don't always see.
I want to thank you so much, Kate, for spending time with us today. And, your daughter did not make her podcast debut—but maybe in a couple of months.
Kate: I can go wake her up if you want to see her.
Heather: You're doing amazing work, and the passion that you bring to the table in talking about this brand that is on everybody's minds and phones and lips is really exciting. I can't wait to see what happens next.
Kate: Well, thank you, Heather.
TikTok itself is not hierarchical at all. We don't use titles. It's very open and it's really what impact are you having?
Starting a brand from scratch
Kristen Tomlan, Lippincott alum and founder and CEO of DŌ Cookie Dough Confection, shares how she pivoted from a career in branding to founding a company that quickly went viral. Listeners will be inspired by Kristen's endless optimism and her mission to spread joy—one sweet bite at a time.


I knew this kitchen would turn out wonderful when I first met the client. She was open to exploring the options available in new materials and was more than willing to "go that extra step" with the design that took it from great to amazing. The tall and upper cabinetry is a high gloss gray laminate and the base cabinetry is a composite rift cut white oak veneer with a custom stain on it. This was a difficult kitchen to plan for, as we had to order the cabinetry from two different manufacturers, yet make it look seamless. I think we accomplished this feat to perfection. These photos were taken by yours truly so the quality isn't fantastic. We plan to professionally shoot this project in about a month though, so I'll post those photos as soon as they're available!
Cooking appliances: Wolf induction cooktop, island hood, and oven/microwave. Sub Zero refrigeration. All integrated and all very sleek. We raised the dishwasher in a custom cabinet to elevate it 12" higher than normal. It is very easy to get into and out of. That sort of configuration is not always doable and is based on the space available to us. In this case, it worked perfectly - so why not??
The perimeter stone is quartz and the island is honed granite. Stainless steel upstands and back-painted glass doors also add to the character.


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Q: Trust issue while sending a post to my API since DST Root CA X3 Expiration I have a C# api running on a aws S3 with ubuntu.
This API is use by a website, a windows application and a xamarin app deployed on Samsung android devices.
Since today 16:00 (paris time), the android part is not working anymore, I have a "trust issue".
Clearly it seems to be related with DST Root CA X3 Expiration (No release on my side and the timing is perfect).
But I don't understand why...

*

*SSL certificate
I checked my SSL certificate and regarding let'sencrypt forums, I have one of the path base on  "ISRG Root X1". The second one is base on "DST Root CA X3" (expired).
I renew them anyway to be sure, but still the same certificate path. (and no problem for chrome to contact them).

*Internet with https is working
I can reach internet with a webview inside the app (to my website in https)

*Can't connect using restsharp
When I use RestSharp to contact my server, I have the trust issue.
My android devices are all the same: Samsung A7 tab, half up to date, the other half was update in august, all of them with Android 11. So theorically they are "not concerned" with this certificate expiration.
Can the problem come from Xamarin or RestSharp ? Maybe my server certificate ?
EDIT
Ok half resolved....
If I go to the "Trusted Root Certificates folder" in my android device (don't know the exact name), If I disable the "Digital Signature Trust Co. - DST Root CA X3", it's working again !
Not a "real solution" since I need to update something like 150 devices...
2 options in my mind

*

*Can I force RestSharp to use a certificate more than another ?

*Is it just because Android know the expiration date is 30/09 and still use it because we are still the 30/09 and everythin will work Tomorow ?

EDIT 2 resolved.
Thx to all of you, sorry I should have been able to validate this answer before some post, but stackoverflow was on readonly mode this night and I fall asleep after that.
What I did (not sure if all step are mandatory).
1/ I updated the certbot since mine was < 1 (check with certbot --version)
sudo apt-get remove -y certbot python3-certbot-apache
sudo snap install certbot --classic
sudo ln -s /snap/bin/certbot /usr/bin/certbot

Procedure from @dmaehler from Letsencrypt forum
2/ As @hamzaElHanna and @ThatCoderGuy said, I updated the certificates
update-ca-certificates

3/ I made a cert-sync to synchronise Ubuntu and Mono certificates (if it's working as i undertand it)
4/ I renew my SSL certificate to remove the CA X3 since it's no longer in my server certificates
sudo certbot renew --force-renewal --preferred-chain "ISRG Root X1"

As proposed by @dmaehler and @jsoetewey
SSlabs to check, the path with the old certifcate is removed and RestSharp work as previously.
Thx for the help everything were usefull !

A: If you are using certbot:
certbot renew --preferred-chain "ISRG Root X1" --force-renewal

A: We faced the similar problem, we fixed it by adding ! before mozilla/DST_Root_CA_X3.crt in the /etc/ca-certificates.conf file , it's for disabling the DST Root CA X3:
!mozilla/DST_Root_CA_X3.crt
After that, you need to update ca-certificates:

$ update-ca-certificates

A: We've had similar issues today, unfortunately we were using older Amazon Linux on elasticbeanstalks. Upgrading to the latest Ubuntu build in your case should fix your issues.
The issue we had was the Amazon Linux version trusted certificate service was always adding the expired root certificate.
The reason restsharp is having problems is probably because it's trying to do something like a curl request behind the scenes and is doing a handshake to verify the validity of the ssl cert when sending a request. The way it does this is checks it against certs that are trusted on the server, which includes the expired certificate.
See here for Ubuntu builds that have the latest certs upgrade https://ubuntu.com/security/notices/USN-5089-1

A: I made an example of the workaround 1  on Cent OS 7 / openssl 1.0.2f.
https://gist.github.com/kujiy/67ef342170c4b0a36bb4bd9615ae2916
# fix the issue
cd /etc/pki/ca-trust/source/blacklist
wget -O dst.pem https://crt.sh/?d=0687260331A72403D909F105E69BCF0D32E1BD2493FFC6D9206D11BCD6770739

cd /etc/pki/ca-trust/source/anchors
wget --no-check-certificate https://letsencrypt.org/certs/isrgrootx1.pem

update-ca-trust

DOMAIN=$YOUR_DOMAIN
openssl s_client -connect $DOMAIN:443 -servername $DOMAIN | grep verify

DESIRED STATE
Files
$ ls -al /etc/pki/ca-trust/source/blacklist
-rw-r--r--  1 root root 1200 Oct  1 16:48 dst.pem

$ ls -al /etc/pki/ca-trust/source/anchors
-rw-r--r--  1 root root 1939 Oct  1 16:45 isrgrootx1.pem

Trust stores
$ trust list | grep -C3 ISRG
pkcs11:id=%79%b4%59%e6%7b%b6%e5%e4%01%73%80%08%88%c8%1a%58%f6%e9%9b%6e;type=cert
    type: certificate
    label: ISRG Root X1
    trust: anchor                <----------
    category: authority

$ trust list | grep -C3 DST
pkcs11:id=%c4%a7%b1%a4%7b%2c%71%fa%db%e1%4b%90%75%ff%c4%15%60%85%89%10;type=cert
    type: certificate
    label: DST Root CA X3
    trust: blacklisted              <----------
    category: authority

redpajama_set_name: RedPajamaStackExchange


text:
\section{Introduction}
Quantized affine algebras are pseudo-triangular Hopf algebras \cite{Drinfeld-QG};
as such, given any pair
of ``generic'' representations $V_1$ and $V_2$, there exists an intertwiner between $V_1\otimes V_2$ and $V_2\otimes V_1$,
the so-called $R$-matrix. $R$-matrices are the central object in quantum integrable systems,
and for most algebras and many low-dimensional representations, they are known explicitly
\cite{KRS-YBE,ZamFat,IK-SM,Bazh-Lie,Jimbo-Rmat,Jimbo-YBE,Ma-YBE,Kuniba-G2,DGZ-YBE,Manga-R}.
As far as the author knows, the only
case that has not been studied in detail is that of (the quantized affine algebra of) $\e$.
The goal of this short paper is therefore
rather modest: it is the explicit computation of the $R$-matrix associated
to the quantized loop algebra $\Uq(\ee)$ in its lowest nontrivial representation, which is of dimension
$249$ (and is {\em not}\/ irreducible for the nonaffine algebra $\Uq(\e)$, which is a source of complication).
In order to to do so, we develop a natural diagrammatic language for the theory of invariants of $\Uq(\e)$
based on tensors of the adjoint representation, which we then apply to the computation of the $R$-matrix.
The latter is fixed (up to normalization)
by solving the equations that require it to be an intertwiner \cite{Jimbo-qg-YBE}.

The main challenge of this paper is computational: for instance, the $R$-matrix is a matrix
of size $249^2=62001$, so altogether an array of approximately 4 billion entries (albeit a sparse array, since only
$0.05\%$ of these are nonzero). This means that many computations in this paper cannot be performed by hand,
and the help of a symbol computation program is necessary. (The author's code is available on request.)
In particular, quite a few results are presented below without proof; if so, the reader should assume that
they are the result of a computer-assisted calculation.

The immediate motivation for this paper came from the work of the author in collaboration with
A.~Knutson \cite{artic71,SchubII} where the $R$-matrix of $\Uq(\ee)$ appeared unexpectedly
in the computation of structure constants of the $K$-theory of 4-step flag varieties.
Another possible source of interest is the fact that the scaling limit of the Ising model at the
critical temperature in a magnetic field
is known to be related to an $\e$ integrable field theory \cite{Zamo-e8,BNW-e8}, and a vertex lattice model
based on $\e$ might be desirable.

The paper is organized as follows. \S\ref{sec:prelim} contains basic definitions related
to the Hopf algebra $\Uq(\ee)$. \S\ref{sec:nonaff} presents the diagrammatic calculus for $\Uq(\e)$
invariants. $\S\ref{sec:aff}$ is the core of the paper: the $249$-dimensional
representation of $\Uq(\ee)$ is defined, the
trigonometric $R$-matrix introduced and its main properties discussed.
Finally, appendix \ref{app:mult}
contain some cumbersome diagrammatic identities,
appendix \ref{app:main} is the full expression of the $R$-matrix,
and appendix \ref{app:rat} discusses the rational limit
of the $R$-matrix.

Intentionally, this paper has very concrete formulae which we hope will be useful in future
studies of the quantum integrable systems associated to $\e$.

\vfill\eject

\section{Preliminaries}\label{sec:prelim}
\subsection{Dynkin diagrams}\label{sec:dynkin}
The Dynkin diagram of $\ee$ is
\begin{equation}\label{eq:dynkin}
\tikz[baseline=0,every node/.style={circle,fill=blue,draw=black,inner sep=3pt},background rectangle/.style={}]{
\draw (0,0) -- (7,0) (5,0) -- (5,1) node[label={above:$8$}] {};
\foreach\i in {1,...,7} \node[label={below:$\i$}] at (\i,0) {};
\node[fill=green,label={below:$0$}] at (0,0) {};
}
\end{equation}
Its adjacency matrix $A$ uses the index set
$I=\{0,1,\ldots,8\}$ as on the figure. Let $C=2-A$ be the corresponding Cartan matrix.

If one removes the (green) vertex labeled $0$ in \eqref{eq:dynkin}, one obtains the Dynkin
diagram of $\e$.

\subsection{The $\e$ root system}\label{sec:rootsys}
Let $\Phi=\Phi_+\sqcup \Phi_-$ be the root system of $\e$, divided into
positive and negative roots; $|\Phi|=240$.
Let $Q$ be the root lattice (generated by $\Phi$).
Let $\alpha_1,\ldots,\alpha_8\in \Phi_+$ be the set of simple roots.

There is a nondegenerate scalar product $\braket{\cdot}{\cdot}$
on $Q\otimes_{\ZZ} \RR \cong \RR^8$ (inverse of the Killing form on the Cartan subalgebra of $\e$),
such that $\Phi$ is exactly the subset
of $Q$ of vectors of square length $2$. More generally, one has, for $\beta,\gamma\in\Phi$,
$\braket{\beta}{\gamma}\in \{-2,-1,0,+1,+2\}$ with $\braket{\beta}{\gamma}=\pm2$
iff $\beta=\pm\gamma$.

The dual basis of the simple roots $\alpha_i$ is that of fundamental weights $\omega_i$:
$\braket{\alpha_i}{\omega_j}=\delta_{i,j}$.

Define the sequence of integers
\begin{equation}\label{eq:n}
(n_0,n_1,\ldots,n_8)=(1,2,3,4,5,6,4,2,3)
\end{equation}
The highest root of $\e$ (highest weight of the adjoint representation,
and therefore a fundamental weight) is given by $\omega_1=\sum_{i=1}^8 n_i\alpha_i$.
Alternatively, define $\alpha_0=-\omega_1$ to be the lowest root of $\e$; it then satisfies
\[
\sum_{i=0}^8 n_i \alpha_i=0
\]
The Cartan matrix of \S\ref{sec:dynkin} encodes the scalar products: $C_{ij}=\braket{\alpha_i}{\alpha_j}$
for $i,j\in I$ (including the value $0$).

We need three more definitions.

Given $\beta\in\Phi$, define $\height(\beta)$ to be the sum of its entries in the basis of simple roots,
that is, $\height(\beta)=\braket{\rho}{\beta}$ where $\rho=\sum_{i=1}^8 \omega_i=\frac{1}{2}\sum_{\beta\in \Phi_+}\beta$.

Denote $\Phi^3_0$ the set of triples $(\beta,\gamma,-\beta-\gamma)\in\Phi^3$ that
sum to zero.
Define the binary relation $\to$ on $\Phi^3_0$ by
$(\beta,\gamma,-\beta-\gamma)\to (\beta+\alpha_i,\gamma-\alpha_i,-\beta-\gamma)$,
$(\beta,\gamma,-\beta-\gamma)\to (\beta,\gamma+\alpha_i,-\beta-\gamma-\alpha_i)$ and
$(\beta,\gamma,-\beta-\gamma)\to (\beta+\alpha_i,\gamma,-\beta-\gamma-\alpha_i)$
for $i=1,\ldots,8$. Given $x\in\Phi^3_0$, write $\ell(x)$ for the maximum of lengths $k$ of chains
$x\to x_1 \to\cdots\to x_{k}$ starting from $x$.
\rem[gray]{is there a better definition?}

Finally, define a map $\epsilon: \Phi^2 \to \{-1,+1\}$ by
\[
\epsilon(\beta,\gamma)=\prod_{\substack{1\le i<j\le 8\\[1pt]
C_{ij}
=-1}} (-1)^{\braket{\alpha_i}{\beta}\braket{\alpha_j}{\gamma}}
\]
Note that it satisfies
\[
\epsilon(\beta,\gamma)\epsilon(\gamma,\beta)=(-1)^{\braket{\beta}{\gamma}}
\]

\subsection{The quantized affine algebra $\Uq(\ee)$}
In all that follows, $q$ is a ``generic'' nonzero complex number, i.e., not a root of unity.

Introduce the notation for $q$-number
\[
[n]=\frac{q^{n}-q^{-n}}{q-q^{-1}}
\]

The quantized affine algebra $\Uq(\ee)$ is given by
generators $\{e_i,f_i,k_i^{\pm},\ i\in I\}$ and relations (see e.g.~\cite[Chap.~12]{CP-book})
\begin{align}\notag
&k_i k_j=k_jk_i
&&
[e_i,f_j] = \delta_{ij} \frac{k_i-k_i^{-1}}{q-q^{-1}}
\\
&k_i e_j k_i^{-1} = q^{C_{ij}} e_j
&&
k_i f_j k_i^{-1} = q^{-C_{ij}} f_j
\\\notag
&e_ie_j=e_je_i&& f_if_j=f_jf_i&&\text{if }C_{ij}=0
\\\notag
&e_ie_j^2-[2]e_je_ie_j+e_j^2e_i=0
&&
f_if_j^2-[2]f_jf_if_j+f_j^2f_i=0
&&
\text{if }C_{ij}=-1
\end{align}
for all $i,j\in I$.

The coproduct, antipode and counit are defined by
\begin{align}\notag\label{eq:co}
\Delta(e_i) &= e_i \otimes 1 + k_i \otimes e_i
&
S(e_i)&=-k_i^{-1}e_i
&
\varepsilon(e_i)&=0
\\
\Delta(f_i) &= f_i \otimes k_i^{-1} + 1 \otimes f_i
&
S(f_i)&=-f_ik_i
&
\varepsilon(f_i)&=0
\\\notag
\Delta(k_i) &= k_i \otimes k_i
&
S(k_i)&=k_i^{-1}
&
\varepsilon(k_i)&=1
\end{align}

The Cartan matrix being degenerate has two consequences. Firstly,
there is a central element $\prod_{i=0}^8 k_i^{n_i}$ where the integers $n_i$ were defined in \eqref{eq:n}.
In all representations that we consider, this element is sent to $1$.

Secondly, there is a nontrivial
gradation on $\Uq(\ee)$: we choose
$e_0$ to have degree $+1$, $f_0$ degree $-1$ and all other generators degree $0$.

If one restricts the index set of the generators to
$\{1,\ldots,8\}$, one obtains the quantized algebra
$\Uq(\e)$; it is naturally a (horizontal, or degree $0$) sub-Hopf algebra of $\Uq(\ee)$.

\section{\texorpdfstring{$\Uq(\e)$}{Uq(e8)} invariants}\label{sec:nonaff}
We first study the non-affine algebra $\Uq(\e)$. We shall be interested in two representations of it: the trivial representation,
which is simply given by the counit in \eqref{eq:co}, and the 248-dimensional representation (a deformation of the adjoint representation
of $\e$), which we discuss now.

\subsection{The $248$-dimensional representation}\label{sec:248}
The definition is a straightforward $q$-deformation of the adjoint action of $\e$.
$V:=V_{\omega_1}$ is a $248$-dimensional vector space with basis $v_\beta$, $\beta\in\Phi$ and $u_j$, $j=1,\ldots,8$.
The action of $\Uq(\e)$ on it is given by
\begin{align*}
k_i v_\beta &= q^{\braket{\alpha_i}{\beta}} v_\beta
&
k_i u_j &= u_j
\\
e_i v_\beta &= \begin{cases}
\epsilon(\alpha_i,\beta) v_{\beta+\alpha_i}&\beta+\alpha_i\in\Phi
\\
u_i&\beta+\alpha_i=0
\\
0&\text{else}
\end{cases}
&
e_i u_j &= -[C_{ij}] v_{\alpha_i}
\\
f_i v_\beta &= \begin{cases}
\epsilon(-\alpha_i,\beta) v_{\beta-\alpha_i}&\beta-\alpha_i\in\Phi
\\
-u_i&\beta-\alpha_i=0
\\
0&\text{else}
\end{cases}
&
f_i u_j &= [C_{ij}] v_{-\alpha_i}
\end{align*}
for $i=1,\ldots,8$.

Because the representation ring of $\Uq(\e)$ (for generic $q$) is isomorphic to
that of $\e$, various results are known. For example, the adjoint representation
of $\e$ has an invariant bilinear form (the Killing form) and an invariant trilinear
form (built out of the bracket). The same is true of $V$:

\begin{lem}
The following bilinear form $B$
is an $\Uq(\e)$ invariant:
\begin{align*}
B(v_\beta\otimes v_{-\beta})&=\epsilon(\beta,\beta)q^{1-\height(\beta)}
\\
B(u_i\otimes u_j)&=[C_{ij}]
\end{align*}
all other entries being zero (as implied by their nonzero weight).
\end{lem}
\begin{proof}
The $k_i$ invariance is trivial. Let us check the $e_i$ invariance (the $f_i$ invariance works similarly, and is in fact implied by $k_i,e_i$
invariance in finite dimension). There are three nontrivial cases to consider: either one starts with $v_\beta\otimes v_\gamma$,
in which case for its image under $Be_i$ to be nonzero one must have $\gamma=-\beta-\alpha_i$; then
\begin{align*}
B(e_i (v_\beta\otimes v_{-\beta-\alpha_i}))&=
B(\epsilon(\alpha_i,\beta)v_{\beta+\alpha_i}\otimes v_{-\beta-\alpha_i}
+q^{\braket{\alpha_i}{\beta}}\epsilon(\alpha_i,-\beta-\alpha_i)v_\beta\otimes v_{-\beta})
\\
&=\epsilon(\alpha_i,\beta)\epsilon(\beta+\alpha_i,\beta+\alpha_i)q^{1-\height(\beta+\alpha_i)}+q^{\braket{\alpha_i}{\beta}}\epsilon(\alpha_i,-\beta-\alpha_i)\epsilon(\beta,\beta)q^{1-\height(\beta)}
\\
&=\epsilon(\alpha_i,\beta)\epsilon(\beta,\beta)q^{-\height(\beta)}(1-q^{1+\braket{\alpha_i}{\beta}})
\\
&=\epsilon(\alpha_i,\beta)\epsilon(\beta,\beta)q^{-\height(\beta)}(1-q^{1+\frac{1}{2}(|\beta+\alpha_i|^2-|\beta|^2-|\alpha_i|^2)})
\\
&=0
\end{align*}
where in the last line we used the fact that all roots of $\Phi$ have square length $2$.

Similarly, the other two cases are
\begin{align*}
B(e_i(u_j\otimes v_{-\alpha_i}))
&=
B(-[C_{ij} v_{\alpha_i}\otimes v_{-\alpha_i}) + u_j\otimes u_i)
\\
&=
-[C_{ij}]q^{1-\height(\alpha_i)} + [C_{ij}]
\\
&=0
\end{align*}
and
\begin{align*}
B(e_i(v_{-\alpha_i}\otimes u_j ))
&=
B(u_i\otimes u_j - q^{-\braket{\alpha_i}{\alpha_i}} v_{-\alpha_i}\otimes [C_{ij}] v_{\alpha_i})
\\
&=[C_{ij}]-q^{-2} q^{1-\height(-\alpha_i)} [C_{ij}]
\\
&=0
\end{align*}

\end{proof}

The trilinear form is more complicated to define:
\begin{lem}
Define $T\in V^*\otimes V^*\otimes V^*$ as follows.

Given a triple $(\beta,\gamma,-\beta-\gamma)\in\Phi^3_0$, write
\[
T(v_\beta\otimes v_\gamma\otimes v_{-\beta-\gamma})=\epsilon(\beta,\beta)\epsilon(\gamma,\gamma)\epsilon(\gamma,\beta) q^{-41+\ell(\beta,\gamma,-\beta-\gamma)}\qquad \beta\in\Phi_+,\ -\beta-\gamma\in\Phi_-
\]
and then use
\[
T(v_\beta\otimes v_\gamma\otimes v_\alpha)=T(v_\alpha\otimes v_\beta\otimes v_\gamma)q^{2\height(\alpha)}
\]
to reduce to the case above.

Also write
\[
\left.
\begin{aligned}
T(u_i\otimes v_{\alpha_i}\otimes v_{-\alpha_i})&=q
\\
T(u_i\otimes v_{-\alpha_i}\otimes v_{\alpha_i})&=-q
\\
T(v_{\alpha_i}\otimes u_i\otimes v_{-\alpha_i})&=-q^{-1}
\\
T(v_{\alpha_i}\otimes v_{-\alpha_i}\otimes u_i)&=q
\\
T(v_{-\alpha_i}\otimes u_i\otimes v_{\alpha_i})&=q^3
\\
T(v_{-\alpha_i}\otimes v_{\alpha_i}\otimes u_i)&=-q
\end{aligned}
\right\}
(q^{d(i)}+q^{-d(i)})
\]
and
\[
T(u_i\otimes u_j\otimes u_k)
=(q-q^{-1})
\begin{cases}
[2]\left(q^{d(i)}+q^{-d(i)}\right)
&i=j=k
\\
-[d(k)]&i=j\leftarrow k\text{ or permutations}
\\
[d(k)]&i=j\rightarrow k\text{ or permutations}
\end{cases}
\]
where $i,j,k$ are viewed as nodes of the Dynkin diagram,
$d(i)$ is the distance to the central node $5$,
and the Dynkin diagram is oriented away from the central node.

All other entries of $T$ are zero.

Then $T$ is an $\Uq(\e)$ invariant.
\end{lem}
The proof is similar to that of the previous lemma, and is skipped.

\medskip

More generally, the
decomposition of $V^{\otimes 2}$ is known: one has
\begin{equation}\label{eq:dec}
V_{\omega_1}\otimes V_{\omega_1}
\cong \bigoplus_{\omega\in \Omega} V_\omega\qquad
\Omega=\{0,\omega_1,\omega_7,2\omega_1,\omega_2\}
\end{equation}
where $V_\omega$ is the irreducible representation of $\Uq(\e)$ of highest weight $\omega$;
it is also the unique irreducible representation of dimension $1,248,3875,27000,30380$
for $\omega=0,\omega_1,\omega_7,2\omega_1,\omega_2$, respectively.

This implies that $\text{End}(V^{\otimes 2})^{\Uq(\e)}$,
the space of $\Uq(\e)$ intertwiners of endormorphisms of $V^{\otimes 2}$,
is of dimension $5$. Out of the invariants above we can build a basis for
this space. It is convenient to do so using a diagrammatic language.

\subsection{Diagrammatic depiction of invariants}\label{sec:diag}
We use here the graphical calculus which is standard in mathematical physics, in the particular form
which is adapted to quantum groups \cite{Kassel-QG}, namely our diagrams are {\em planar}.
See also Cvitanovic \cite{Cvita-book} for a discussion of the non $q$-deformed case (for all
simple Lie algebras), so without the planarity requirement.

In general, to a graph embedded in the plane of the form
\[
\tikz{
\foreach\i in {1,...,3} \draw (\i+0.5,1) node[vertex] {} -- ++(0,-0.5);
\draw[decoration=brace,decorate] (1.5,1.2) -- node[above] {$m$} (3.5,1.2);
\foreach\i in {1,...,4} \draw (\i,-1) node[vertex] {} -- ++(0,0.5);
\draw[decoration=brace,decorate] (4,-1.2) -- node[below] {$n$} (1,-1.2);
\draw[pattern=dots] (0.5,0.5) rectangle (4.5,-0.5);
}
\]
we will associate an $\Uq(\e)$ intertwiner from $V^{\otimes m}$ to $V^{\otimes n}$,
where tensor products are read from left to right.
As usual, vertical concatenation corresponds to multiplication, whereas horizontal
juxtaposition corresponds to tensor product.

The simplest graph is the identity from $V$ to $V$, which is
\[
\tikz[baseline=-3pt]{\draw (0,1) node[vertex] {} -- (0,-1) node[vertex] {};}
\]

Next is the bilinear form $B$, which we draw as a cup:
\[
\mycup
\]
Because it is nondegenerate, it possesses an inverse, drawn as a cap:
\[
\mycap
\]
with the identities
\[
\tikz[baseline=-3pt]{
\draw (-0.5,1) node[vertex] {} -- (-0.5,0) [bend right=75] to (1,0) [bend left=75] to (2.5,0) -- (2.5,-1) node[vertex] {};
}
=
\tikz[baseline=-3pt,xscale=-1]{
\draw (-0.5,1) node[vertex] {} -- (-0.5,0) [bend right=75] to (1,0) [bend left=75] to (2.5,0) -- (2.5,-1) node[vertex] {};
}
=
\tikz[baseline=-3pt]{\draw (0,1) node[vertex] {} -- (0,-1) node[vertex] {};}
\]

Gluing together cup and cap in the opposite way, we obtain our first
nontrivial identity:
\begin{equation}\label{eq:rela}
\tikz[baseline=-3pt]{\draw (0,0) circle[radius=0.75];} = \frac{[20][24][31]}{[6][10]}
\end{equation}
(we shall not prove this identity, or any that follow, but mention in passing that this is simply
the principal specialization of the character of the adjoint representation of $\e$).

Similarly, the trilinear form $T$ can be drawn as
\[
\tri
\]
Combining it with caps, we can get three more intertwiners, namely
\[
\U\qquad\D\qquad\triinv
\]
In principle, there are several ways to glue caps to $T$ to produce these diagrams,
but by Schur's lemma and normalization condition these all produce the same result.
\rem[gray]{is there a better to say this?}

One also has the trivial identities:
\[
\tikz[baseline=-3pt]{
\draw (0,0) node[vertex] {} -- (0,0.5); \draw (0,1) circle[radius=0.5];
}
=
\tikz[baseline=-3pt]{
\draw (0,0) node[vertex] {} -- (0,-0.5); \draw (0,-1) circle[radius=0.5];
}
=0
\]
as well as two nontrivial ones:\footnote{Note that these identities {\em do}\/ depend
on the choice of normalization of $B$ and $T$, which is determined by imposing that it agree with
the standard $\e$ convention at $q=1$, that it
be the simplest possible (entries are coprime polynomials)
and that constants occurring in the identities be palyndromic in $q$
(which fixes the remaining freedom of multiplying by a power of $q$).}
\begin{gather}\label{eq:relb}
\tikz[baseline=-3pt]{\draw (0,-1) node[vertex] {} -- (0,-0.5) (0,0.5) -- (0,1) node[vertex] {};
\draw (0,0) circle[radius=0.5];}
=
\frac{[10][15]^2[18][32]}{[5][9][16][30]}
\ \tikz[baseline=-3pt]{\draw (0,-1) node[vertex] {} -- (0,1) node[vertex] {};}
\\
\label{eq:relc}
\tikz[baseline=-3pt]{
\draw (-1.2,1) node[vertex] {} -- (-0.5,0.45) (0.5,0.45) -- (1.2,1) node[vertex] {};
\draw (0,-0.35) -- (0,-1) node[vertex] {};
\draw (0,-0.35) -- (-0.5,0.45) -- (0.5,0.45) -- cycle;
}
=\frac{[6][10]^2[15]}{[2][5][30]}
\left(
\frac{[32]}{[3][16]}+\frac{[36]}{[9][12]}
\right)
\U
\end{gather}
(the mirror image of the last identity follows from it).

At this stage we have all we need to produce a basis of intertwiners
of endomorphisms of $V^{\otimes 2}$; we propose
\[
\hspace{-6mm
\full
\qquad
\cupcap
\qquad
\vertical
\qquad
\horizontal
\qquad
\mysquare
\]
It is a consequence of what follows that these intertwiners are indeed linearly
independent.

Now let us work out the multiplication table in this basis. Note that the last
basis element is simply the square of the fourth.
Thanks to \eqref{eq:rela}--\eqref{eq:relc}, we can compute all products except
the product of the last two (which itself allows to compute the square of the last).
This identity is not particularly nice, and is given in Appendix~\ref{app:mult} for
the sake of completeness. We present in the next section a more pleasant
alternative.

Before we proceed, let us point out that the trivial representation, being
the same as $V^{\otimes 0}$, is diagrammatically\dots invisible. However, in what follows
it is sometimes useful to emphasize that in a tensor product some trivial factors
are included, in which case we draw an empty vertex; e.g.,
\[
\tikz{
\draw (1,-1) node[vertex] {} -- (-1,1) node[vertex] {};
\node[empty] at (1,1) {};
\node[empty] at (-1,-1) {};
}
\]
is the natural map from $V\otimes \CC$ to $\CC\otimes V$.

\subsection{The spectral parameter-independent $R$-matrix}\label{sec:R0}
As mentioned before, our diagrams are required to be planar. Let us informally discuss this point.
In the undeformed case of $\e$, it is natural \cite{Cvita-book} to consider
arbitrary diagrams (i.e., not embedded into the plane); however,  one can always
turn them into planar diagrams by projecting them onto the plane, thus creating new ``virtual crossings''
where two formerly nonintersecting lines are now on top of each other.
These crossings simply correspond to the natural
permutation of tensors $x\otimes y\mapsto y\otimes x$, which commutes with the $\e$ action. The point is of
course that such a permutation does not commute with the $\Uq(\e)$ action; instead one must use
the $R$-matrix (of $\Uq(\e)$).

With a bit of foresight, we therefore define
\begin{multline}\label{eq:defR0}
\crossing=f(q)\
\full
+
f(q^{-1})
\cupcap
\\
+g(q)
\vertical
+g(q^{-1})
\horizontal
+h(q)
\mysquare
\end{multline}
where
\begin{align*}
f(q)&=\frac{[6][10][15]^2}{[3][5][30]^2}\left({q^{-24}}-{q^{-22}}-{q^{-18}}+{q^{-16}}-{q^{-14}}+q^{-8}-1+q^6-q^{12}\right)
\\
g(q)&=\frac{[15]}{[3][30]}\left(
-q^{-16}-q^{-14}-q^{-12}-q^{-8}+q^{-6}+q^{-2}-1+q^2+q^6-q^8-q^{14}-q^{16}-q^{18}
\right)
\\
h(q)&=h(q^{-1})=\frac{[5]}{[6][10]}
\end{align*}
Rather than deriving this ($\Uq(\e)$, i.e., spectral parameter-independent) $R$-matrix now, we postpone
its justification to \S\ref{sec:pties}, where the full ($\Uq(\ee)$, i.e., spectral parameter-dependent)
$R$-matrix will be obtained.
One can also define a similar diagram where overcrossing and undercrossing
are switched, by a simple 90 degree rotation; equivalently, it corresponds
to switching $q\leftrightarrow q^{-1}$ in the coefficients of the right hand side.

These two diagrams are inverses of each other so that one has the relations
\[
\tikz[rounded corners=7mm,baseline=-3pt]{
\draw (-1,-1) node[vertex] {} -- (1,0) -- (-1,1) node[vertex] {};
\draw (1,-1) node[vertex] {} -- (0.1,-0.55) (-0.1,-0.45) -- (-1,0) -- (-0.1,0.45) (0.1,0.55) -- (1,1) node[vertex] {};
}
=
\tikz[rounded corners=7mm,baseline=-3pt]{
\draw (1,-1) node[vertex] {} -- (-1,0) -- (1,1) node[vertex] {};
\draw (-1,-1) node[vertex] {} -- (-0.1,-0.55) (0.1,-0.45) -- (1,0) -- (0.1,0.45) (-0.1,0.55) -- (-1,1) node[vertex] {};
}
=
\full
\]
which one can recognize as Reidemeister move II. We also mention for the sake of completeness
the Reidemeister move III (or braid relation)
\[
\tikz[baseline=-3pt]{
\draw (1,-1) node[vertex] {} -- (-1,1) node[vertex] {};
\draw[bend left=60,draw=bgcolor,double=black,line width=2pt] (0,-1) node[vertex] {} to (0,1) node[vertex] {};
\draw[draw=bgcolor,double=black,line width=2pt] (-1,-1) node[vertex] {} -- (1,1) node[vertex] {};
}
=
\tikz[baseline=-3pt]{
\draw (1,-1) node[vertex] {} -- (-1,1) node[vertex] {};
\draw[bend right=60,draw=bgcolor,double=black,line width=2pt] (0,-1) node[vertex] {} to (0,1) node[vertex] {};
\draw[draw=bgcolor,double=black,line width=2pt] (-1,-1) node[vertex] {} -- (1,1) node[vertex] {};
}
\]

\rem[gray]{is there a useful knot theory application?}

We can now use the alternative basis of
$\text{End}(V^{\otimes 2})^{\Uq(\e)}$:
\[
\full
\qquad
\cupcap
\qquad
\vertical
\qquad
\crossing
\qquad
\invcrossing
\]

In order to complete the multiplication table, we need the following additional relations:
\begin{align*}
\tikz[baseline=-3pt,rounded corners=5mm]{
\draw (-1,0) node[vertex] {} -- (-0.1,-0.9) (0.1,-1.1) -- (1,-2) -- (0,-2.5) -- (-1,-2) -- (1,0) node[vertex] {};
}
&=
q^{60}\ \mycup
&
\tikz[baseline=-3pt,rounded corners=5mm]{
\draw (-1,0) node[vertex] {} -- (1,-2) -- (0,-2.5) -- (-1,-2) -- (-0.1,-1.1) (0.1,-0.9) -- (1,0) node[vertex] {};
}
&=
q^{-60}\ \mycup
\\
\tikz[baseline=-3pt,rounded corners=5mm]{
\draw (-1,0) node[vertex] {} -- (1,2) -- (0,2.5) -- (-1,2) -- (-0.1,1.1) (0.1,0.9) -- (1,0) node[vertex] {};
}
&=
q^{60}\ \mycap
&
\tikz[baseline=-3pt,rounded corners=5mm]{
\draw (-1,0) node[vertex] {} -- (-0.1,0.9) (0.1,1.1) -- (1,2) -- (0,2.5) -- (-1,2) -- (1,0) node[vertex] {};
}
&=
q^{-60}\ \mycap
\end{align*}
which is nothing but Reidemeister move I, as well as
\begin{align*}
\tikz[baseline=-0.4cm,rounded corners=5mm,yscale=0.6]{
\draw (-1,1) node[vertex] {} -- (-0.1,0.1) (0.1,-0.1) -- (1,-1) {[sharp corners] -- (0,-1.5)} -- (-1,-1) -- (1,1) node[vertex] {};
\draw (0,-1.5) -- (0,-2.5) node[vertex] {};
}
&=
-q^{30}
\U
&
\tikz[baseline=-0.4cm,rounded corners=5mm,yscale=0.6]{
\draw (-1,1) node[vertex] {} -- (1,-1) {[sharp corners] -- (0,-1.5)} -- (-1,-1) -- (-0.1,-0.1) (0.1,0.1) -- (1,1) node[vertex] {};
\draw (0,-1.5) -- (0,-2.5) node[vertex] {};
}
&=
-q^{-30}
\U
\\[2mm]
\tikz[baseline=0.4cm,rounded corners=5mm,yscale=0.6]{
\draw (-1,-1) node[vertex] {} -- (1,1) {[sharp corners] -- (0,1.5)} -- (-1,1) -- (-0.1,0.1) (0.1,-0.1) -- (1,-1) node[vertex] {};
\draw (0,1.5) -- (0,2.5) node[vertex] {};
}
&=
-q^{30}
\D
&
\tikz[baseline=0.4cm,rounded corners=5mm,yscale=0.6]{
\draw (-1,-1) node[vertex] {} -- (-0.1,-0.1) (0.1,0.1) -- (1,1) {[sharp corners] -- (0,1.5)} -- (-1,1) -- (1,-1) node[vertex] {};
\draw (0,1.5) -- (0,2.5) node[vertex] {};
}
&=
-q^{-30}
\D
\end{align*}
and
\begin{multline*}
\tikz[rounded corners=7mm,baseline=-3pt]{
\draw (-1,-1) node[vertex] {} -- (1,0) -- (0.1,0.45) (-0.1,0.55) -- (-1,1) node[vertex] {};
\draw (1,-1) node[vertex] {} -- (0.1,-0.55) (-0.1,-0.45) -- (-1,0) -- (1,1) node[vertex] {};
}
=(q^{-2}-q^{10}+q^{12})\full
+q^{31}(q-q^{-1})^2 \frac{[6][10][60]}{[20][30]} \cupcap
\\+
q^{18}(q-q^{-1})\frac{[5][30]}{[10][15]}\vertical
+(q^{-2}-q^{10}+q^{12})\invcrossing
-q^{10}\crossing
\end{multline*}
(and the same identity with undercrossings $\leftrightarrow$ overcrossings,
$q\leftrightarrow q^{-1}$).

\subsection{Primitive idempotents}
Because $V^{\otimes 2}$ is multiplicity-free,
$\text{End}(V^{\otimes 2})^{\Uq(\e)}$ is a commutative algebra, and multiplication
operators can be diagonalized simultaneously, with eigenvectors the primitive
idempotents (projectors $P_\omega$ onto irreducible subrepresentations $V_\omega$).
The eigenvalues give the coefficients of the expansion:
\begin{align*}
\full&=P_0+P_{\omega_1}+P_{\omega_7}+P_{2\omega_1}+P_{\omega_2}
\\
\cupcap&=\frac{[20][24][31]}{[6][10]}P_0
\\
\vertical&=\frac{[10][15]^2[18][32]}{[5][9][16][30]}P_{\omega_1}
\\
\crossing&=q^{60}P_0-q^{30}P_{\omega_1}+q^{12}P_{\omega_7}+q^{-2}P_{2\omega_1}-P_{\omega_2}
\end{align*}
and the expansion of the inverse crossing is obtained by inverting all coefficients, i.e.,
$q\leftrightarrow q^{-1}$ as usual.
\rem{are these the correct eigenvalues of the $R$-matrix?}

The multiplication rule for the diagrams is equivalent to the statement
\[
P_{\omega}P_{\omega'}=\delta_{\omega,\omega'} P_{\omega}
\qquad
\omega,\omega'\in\Omega
\]

We also mention the expansion:
\begin{multline*}
\horizontal=\frac{[10][15]^2[18][32]}{[5][9][16][30]}P_0
+\frac{[10]^2[15]}{[30]}\left((q-q^{-1})^2\frac{[7][12]}{[4]}+\frac{[6]^2}{[2]^2[3][5]}\right)
P_{\omega_1}\\
+\frac{[6][10][15][32]}{[5][16][30]}P_{\omega_7}-\frac{[10][15][18]}{[5][9][30]}P_{2\omega_1}+\frac{(q-q^{-1})^2[6][10][15]}{[30]}P_{\omega_2}
\end{multline*}
and the expansion of the ``square'' diagram is obtained by squaring all the coefficients above.

\section{\texorpdfstring{$\Uq(\ee)$}{Uq(e8(1))} and its $R$-matrix}\label{sec:aff}
\subsection{The $(248+1)$-dimensional representation}
In this section we present explicit formulae for the one-parameter
family of representations $W_z$ of dimension $249$ of $\Uq(\ee)$. Note that the parameter $z\in\CC^\times$,
called ``spectral parameter'' comes automatically from the $\ZZ$-grading of
the algebra.

The action of $\Uq(\e)\subset \Uq(\ee)$ on $W_z$ is as follows:
it is the direct sum of the $248$-dimensional
representation $V$ defined in \S\ref{sec:248} and of the trivial representation. We use the same
basis $\{v_\beta,\ \beta\in\Phi\}\cup\{u_i,\ i=1,\ldots,8\}$ for the former and fix some
nonzero vector $w$ in the latter.

The action of the remaining generators of $\Uq(\ee)$ on $W_1$ is given by
\begin{align*}
k_0 v_\beta &= q^{\braket{\alpha_0}{\beta}} v_\beta
&
k_0 u_j &= u_j
&
k_0 w &= w
\\
e_0 v_\beta &= \begin{cases}
\epsilon(\alpha_0,\beta) v_{\beta+\alpha_0}&\beta+\alpha_0\in\Phi
\\
-c_0\kappa^{-1}w-\sum_{i=1}^8 c_i u_i&\beta+\alpha_i=0
\\
0&\text{else}
\end{cases}
&
e_0 u_j &= -[C_{0j}] v_{\alpha_0}
&
e_0 w &=\kappa\, v_{\alpha_0}
\\
f_0 v_\beta &= \begin{cases}
\epsilon(-\alpha_0,\beta) v_{\beta-\alpha_0}&\beta-\alpha_0\in\Phi
\\
c_0\kappa^{-1}w+\sum_{i=1}^8 c_i u_i&\beta-\alpha_i=0
\\
0&\text{else}
\end{cases}
&
f_0 u_j &= [C_{0j}] v_{-\alpha_0}
&
f_0 w &=-\kappa\,v_{-\alpha_0}
\end{align*}
where the $c_i$ are constants to be determined, and $\kappa$ is a parameter related to the freedom
in the relative normalization of $w$ and of the basis of $V$, which we do not fix for now.

Imposing the relations of the algebra is
equivalent to the following system of equations:
\begin{align*}
\sum_{j=1}^8 [C_{ij}] c_j  &= \delta_{i,1} \qquad i=1,\ldots,8
\\
c_0+c_1&=[2]
\end{align*}
which can be readily solved:
\begin{align*}
c_i&=\frac{1}{\det [C]} \widetilde{[C]}_{i1}
=\frac{[6][10][15]}{[2][3][5][30]}
\left(
\frac{[2][3][18]}{[6][9]},
\frac{[2][3][12]}{[4][6]},
\frac{[2][8]}{[4]},
[5],
[2][3],
[2]^2,
[2],
[3]
\right)
\qquad
i=1,\ldots,8
\\
c_0&=(q-q^{-1})^2\frac{[6][10][15]}{[30]}
\end{align*}

We then restore the $z$-dependence to obtain the general action on $W_z$ by
transforming generators according to their degree: $e_0\mapsto z\,e_0$, $f_0\mapsto z^{-1}f_0$,
all others unchanged.

\subsection{The spectral parameter-dependent $R$-matrix}
Because $\Uq(\ee)$ is pseudotriangular, we know
that for generic $z,z'\in \CC^\times$, $W_z\otimes W_{z'}$ and $W_{z'}\otimes W_z$ are isomorphic.
Up to normalization, the (unique) intertwiner is given by the universal $R$-matrix in these representations;
since the universal $R$-matrix is of degree $0$, this intertwiner only depends on $z'/z$.

Here we follow Jimbo's approach \cite{Jimbo-qg-YBE} which is to directly compute the intertwiner from the
intertwining relations. The normalization that we obtain may be different from the one coming
from the universal $R$-matrix.

We thus look for an operator $\cR(z'/z): W_z\otimes W_{z'}\to W_{z'}\otimes W_z$ that satisfies
\begin{equation}\label{eq:inter}
\cR(z'/z)(\rho_z\otimes \rho_{z'})(\Delta(x))=(\rho_{z'}\otimes \rho_z)(\Delta(x))\cR(z'/z)
\end{equation}
for all $x\in \Uq(\ee)$, where in this equation only, we write explicitly the representation map
$\rho_z: \Uq(\ee)\to \text{End}(W_z)$.

First we consider the commutation with $x\in \Uq(\e)$ only. We know that $W_z\cong V_{\omega_1}\oplus V_0$
as a $\Uq(\e)$ representation; using \eqref{eq:dec}, we obtain\footnote{Note in particular
that the ``multiplicity-free'' property that is crucial in e.g.\ \cite{DGZ-YBE} fails here.}
\[
W_z\otimes W_{z'}\cong \CC^2\otimes V_0 \oplus \CC^3\otimes V_{\omega_1}\oplus V_{\omega_7}\oplus V_{2\omega_1}
\oplus V_{\omega_2}
\]
This means that the space of $\Uq(\e)$ intertwiners from $W_z\otimes W_{z'}$ to $W_{z'}\otimes W_z$
is of dimension $2^2+3^2+1^2+1^2+1^2=16$.
Using the diagrammatic language of \S\ref{sec:diag}, it is easy to produce
a basis of this space, namely
\[
\begin{array}{cccc}
\full
&
\cupcap
&
\vertical
&
\horizontal
\\[13mm]
\mysquare
&
\myempty
&
\cupa
&
\capa
\\[13mm]
\leftid
&
\diagida
&
\diagidb
&
\rightid
\\[13mm]
\Yaa
&
\Yba
&
\Yab
&
\Ybb
\end{array}
\]
Recall that filled (resp.\ empty) vertices correspond to $V_{\omega_1}=V$ (resp.\ $V_0=\CC$), so that
when one concatenates diagrams, if the type of the vertices does not match, the resulting product is
zero.

$\cR(z)$ must therefore be a linear combination of these 16 intertwiners. Imposing \eqref{eq:inter}
for the remaining generators $e_0,f_0,k_0$ of $\Uq(\ee)$ results in a very large number of linear equations
on the entries of $\cR(z'/z)$. For practical purposes, since we know the system has a nontrivial solution,
it is sufficient to extract a subset of $15$ independent equations among these and then solve them.
Such a subset can be found for any $z,z'$ (as opposed to, for generic $z,z'$), implying uniqueness of the
intertwiner. Finding and solving these equations
is best performed by computer, and we present the explicit solution in Appendix~\ref{app:main},
where the chosen normalization is
\begin{equation}\label{eq:norm}
\cR(z) v_{\omega_1}\otimes v_{\omega_1}=v_{\omega_1}\otimes v_{\omega_1}
\end{equation}
(i.e., the highest-weight-to-highest-weight entry is $1$).

We now discuss properties of $\cR(z)$.
These properties should be checked using the diagrammatic algebra itself (which is only of dimension $16$,
so $16\times16$ matrices in the regular representation), rather than using the actual
operators acting on copies of $W_z$. It is sometimes useful to represent graphically the $R$-matrix as
\[
\check R(z'/z)=
\tikz[baseline=-3pt]{
\draw[dashed] (-1,-1) node[halfvertex] {}  -- (1,1) node[halfvertex] {} node[right] {$z'$};
\draw[dashed] (1,-1) node[halfvertex] {} -- node[Rmat] {} (-1,1) node[halfvertex] {} node[left] {$z$};
}
\]
where half-filled vertices correspond to the whole of $W_z\cong V_{\omega_1}\oplus V_0$,
and correspondingly, dashed lines to identity operators on $W_z$,
namely
$\tikz[baseline=-3pt]{\draw[dashed] (0,1) node[halfvertex] {} -- (0,-1) node[halfvertex] {};}
=\tikz[baseline=-3pt]{\draw (0,1) node[vertex] {} -- (0,-1) node[vertex] {};}
+\tikz[baseline=-3pt]{\path (0,1) node[empty] {} -- (0,-1) node[empty] {};}$.

\subsection{Basic properties of the $R$-matrix}\label{sec:pties}
First, from Schur's lemma one knows that $\cR(z) \cR(z^{-1})$ must be proportional to the identity.
With the chosen normalization \eqref{eq:norm}, one in fact has the so-called unitarity equation
\begin{align}\label{eq:unit}
\tikz[rounded corners=7mm,baseline=-3pt]{
\draw[dashed] (-1,-1) node[halfvertex] {} -- node[Rmat] {} (1,0) -- node[Rmat] {} (-1,1) node[halfvertex] {};
\draw[dashed] (1,-1) node[halfvertex] {} -- (-1,0) -- (1,1) node[halfvertex] {};
}
&=
\full+\leftid+\rightid+\myempty
\\\notag&=\identity
\end{align}
As a check of the methods of this paper, the reader is invited
to verify this equation by direct computation.

Similarly, one can check the Yang--Baxter equation
\begin{align}\label{eq:YBE}
(\cR(z)\otimes 1)
(1\otimes\cR(wz))
(\cR(w)\otimes 1)
&=
(1\otimes\cR(w))
(\cR(wz)\otimes 1)
(1\otimes\cR(z))
\\\notag
\tikz[baseline=-3pt,dashed]{
\draw (1,-1) node[halfvertex] {} -- node[Rmat] {} (-1,1) node[halfvertex] {};
\draw[bend left=60] (0,-1) node[halfvertex] {} to node[Rmat,pos=0.3] {} node[Rmat,pos=0.7] {} (0,1) node[halfvertex] {};
\draw (-1,-1) node[halfvertex] {} -- (1,1) node[halfvertex] {};
}
&=
\tikz[baseline=-3pt,dashed]{
\draw (1,-1) node[halfvertex] {} -- node[Rmat] {} (-1,1) node[halfvertex] {};
\draw[bend right=60] (0,-1) node[halfvertex] {} to node[Rmat,pos=0.3] {} node[Rmat,pos=0.7] {} (0,1) node[halfvertex] {};
\draw (-1,-1) node[halfvertex] {} -- (1,1) node[halfvertex] {};
}
\end{align}

Next we come to the so-called crossing symmetry. Consider the 90 degree counterclockwise rotation of
the $R$-matrix. Then we expect it to be related to the original $R$-matrix up to shift of the
spectral parameter. The 90 degree rotation simply permutes our basis of $\Uq(\e)$ intertwiners,
so it easy to compute the result. More specifically, because the dual representation of $W_z$
satisfies $W_z^*\cong W_{q^{30}z}$, we expect the two diagrams below to be proportional:
\[
\tikz[baseline=-3pt,rounded corners=3mm]{
\draw[dashed] (-1.5,-1) node[halfvertex] {}  -- (-1,0.5) -- (-0.5,0.5) -- node[pos=0.1] {$q^{30}z'$} (0.5,-0.5) -- (1,-0.5) -- (1.5,1) node[halfvertex] {} node[right] {$z'$};
\draw[dashed] (0,-1) node[halfvertex] {} -- (-0.5,-0.5) -- node[Rmat] {} (0.5,0.5) -- (0,1) node[halfvertex] {} node[left] {$z$};
}
\propto
\tikz[baseline=-3pt]{
\draw[dashed] (-1,-1) node[halfvertex] {}  -- (1,1) node[halfvertex] {} node[right] {$z'$};
\draw[dashed] (1,-1) node[halfvertex] {} -- node[Rmat] {} (-1,1) node[halfvertex] {} node[left] {$z$};
}
\]
For this to happen, it is clear from the expression of the $R$-matrix
in appendix~\ref{app:main} that one must define dashed cups and caps as follows:
\begin{align}\label{eq:dashcap}
\tikz[baseline=-3pt]{\draw[bend left=60,dashed] (-1,0) node[vertex] {} to (1,0) node[vertex] {};}
&=
\mycap+\frac{(q-q^{-1})^2[6][10][15]}{[30]\kappa^2}
\ \tikz[baseline=-3pt]{
\node[empty] at (-1,0) {};
\node[empty] at (1,0) {};
}
\\\label{eq:dashcup}
\tikz[baseline=-3pt]{\draw[bend left=60,dashed] (1,0) node[vertex] {} to (-1,0) node[vertex] {};}
&=
\mycup+\frac{[30]\kappa^2}{(q-q^{-1})^2[6][10][15]}
\ \tikz[baseline=-3pt]{
\node[empty] at (-1,0) {};
\node[empty] at (1,0) {};
}
\end{align}
Of course, it is very natural at this stage to set the free parameter $\kappa$ to the value
\begin{equation}\label{eq:kappa}
\kappa=(q-q^{-1})\left(\frac{[6][10][15]}{[30]}\right)^{1/2}
\end{equation}
\rem[gray]{i.e., $\kappa=\sqrt{c_0}$, which kinda makes sense}
and we do so in the remainder of this paper. The exact crossing relation is
\begin{equation}\label{eq:crossing}
\tikz[baseline=-3pt,rounded corners=3mm]{
\draw[dashed] (-1.5,-1) node[halfvertex] {}  -- (-1,0.5) -- (-0.5,0.5) -- node[pos=0.1] {$q^{30}z'$} (0.5,-0.5) -- (1,-0.5) -- (1.5,1) node[halfvertex] {} node[right] {$z'$};
\draw[dashed] (0,-1) node[halfvertex] {} -- (-0.5,-0.5) -- node[Rmat] {} (0.5,0.5) -- (0,1) node[halfvertex] {} node[left] {$z$};
}
=q^4\frac{(1-z)(1-q^{10}z)(1-q^{18}z)(1-q^{28}z)}{(1-q^2z)(1-q^{12}z)(1-q^{20}z)(1-q^{30}z)}
\tikz[baseline=-3pt]{
\draw[dashed] (-1,-1) node[halfvertex] {}  -- (1,1) node[halfvertex] {} node[right] {$z'$};
\draw[dashed] (1,-1) node[halfvertex] {} -- node[Rmat] {} (-1,1) node[halfvertex] {} node[left] {$z$};
}
\end{equation}
Note that the prefactor can be absorbed in the redefinition
\[
\check R_{\mathrm{poly}}(z)=z^{-2}(1-q^2z)(1-q^{12}z)(1-q^{20}z)(1-q^{30}z)\check R(z)
\]
(which turns entries into Laurent polynomials of $z$),
but then it is the unitarity equation \eqref{eq:unit} which acquires a prefactor.\footnote{%
Requiring both unitarity and crossing relations without a prefactor would take us beyond the realm
of rational functions, and we do not pursue this here.}

There are three special values of the spectral parameter. First, at $z=1$,
one has
$
\cR(1)=1
$.
The other two values are $0$ and $\infty$:
\begin{align*}
\cR(0)&=q^2\left(\crossing+\diagida+\diagidb+\myempty\right)
\\
\cR(\infty)&=q^{-2}\left(\invcrossing+\diagida+\diagidb+\myempty\right)
\end{align*}
\rem{why these annoying factors of $q^{\pm2}$?}
where the crossing is defined in \eqref{eq:defR0}.
This gives an {\em a posteriori}\/ justification for the introduction of the crossing in
\S\ref{sec:R0}.

At this stage it is natural to switch to the alternative basis which involves these crossings,
and the expression of the $R$-matrix simplifies. We find, using \eqref{eq:kappa} to further simplify,
\begin{align*}
\cR(z)&=
q^3(q-q^{-1})^2[6]\frac{z}{(1-q^2z)(1-q^{12}z)}
\full[0.85]
\\
&\
+q^3\frac{\kappa^2}{[5]}\frac{z(1-z)(1-q^{10}-q^{20}z+q^{30}+q^{40}z-q^{50}z)}{(1-q^2z)(1-q^{12}z)(1-q^{20}z)(1-q^{30}z)}
\cupcap[0.85]
\\
&\
-q^{13}(q-q^{-1})^2[5]\frac{z(1-z)}{(1-q^2z)(1-q^{12}z)(1-q^{20}z)}
\vertical[0.85]
\\
&\
+q^7\frac{1-z}{(1-q^2z)(1-q^{12}z)}
\left(
q^{-5}\crossing[0.85]
-q^{5}z
\invcrossing[0.85]
\right)
\\
&\
+q^2(q-q^{-1})\kappa^2 \frac{z(1+q^{30}z)}{(1-q^2z)(1-q^{12}z)(1-q^{20}z)}
\left(
\leftid[0.85]+\rightid[0.85]
\right)
\\
&\
+\left(\cf{\leftid[0.28]}+\frac{q^2-z}{1-q^2z}\right)\left(\diagida[0.85]+\diagidb[0.85]\right)
\\
&\hspace{-2cm}
+q^{17}(q-q^{-1})\kappa\frac{z(1-z)}{(1-q^2z)(1-q^{12}z)(1-q^{20}z)}
\left(
\Yaa[0.85]+\Yba[0.85]
+\Yab[0.85]+\Ybb[0.85]
\right)
\\
&\hspace{-2cm}
+q^{32}(q-q^{-1})\kappa^2\frac{z(1-z^2)}{(1-q^2z)(1-q^{12}z)(1-q^{20}z)(1-q^{30}z)}
\left(\capa[0.85]
+\cupa[0.85]
\right)
\\
&\hspace{-2cm}
+\left(\cf{\full[0.28]}+\cf{\cupcap[0.28]}
+\cf{\crossing[0.28]}+\cf{\invcrossing[0.28]}
-(q-q^{-1})\frac{[6][15]}{[5]\kappa}\cf{\Yaa[0.28]}
\right)
\myempty[0.85]
\end{align*}
where we have reused certain coefficients by denoting them $\cf{\cdot}$.

\subsection{Fusion channels}
In the three multiplicity-free channels, one can directly rewrite the $R$-matrix in terms
of the idempotents:
\begin{multline}
\cR(z)=P_{2\omega_1}
+\frac{z-q^2}{1-q^2z}P_{\omega_2}
+\frac{(z-q^2)(z-q^{12})}{(1-q^2 z) (1-q^{12} z)}P_{\omega_7}
\\
+\frac{1}{(1-q^2 z) (1-q^{12} z)(1-q^{20})} \cR_{\omega_1}(z)
+\frac{1}{(1-q^2 z) (1-q^{12} z)(1-q^{20})(1-q^{30})}\cR_0(z)
\end{multline}
where the last two terms are the restrictions of $\cR(z)$
to $\CC^3\otimes V_{\omega_1}$ and $\CC^2\otimes V_0$, respectively, in which we have taken out the LCM of
denominators (and the GCD of numerators is $1$). Diagonalizing $\check R(z)$ in these remaining
blocks results in an eigenvalue
$\frac{z-q^2}{1-q^2z}$ in the $V_{\omega_1}$ block, and then pairs of eigenvalues that are
solutions of quadratic equations in each block.
\rem{why the extra degeneracy between $V_{\omega_2}$ and $V_{\omega_1}$? some non QG invariant extra symmetry?}

As a function of the spectral parameter $z$, $\cR(z)$ is a rational function with
poles at $z=q^{-2},q^{-12},q^{-20},q^{-30}$. Conversely, due to unitarity equation~\eqref{eq:unit},
at $z=q^2,q^{12},q^{20},q^{30}$, $\cR(z)$ is noninvertible. More precisely, define
\[
\begin{aligned}
f_a&=\mathrm{Res}_{z=q^{-a}} \cR(z)
\\
g_a&=\cR(z=q^a)
\end{aligned}
\qquad
a\in \{2,12,20,30\}
\]

Then from \eqref{eq:unit} we have $f_ag_a=g_af_a=0$, and in fact, expanding this equation around
$z=q^a$, diagonalizing $\cR(z)$ and noting that each zero eigenvalue of $\cR(z)$ at $z=q^a$ corresponds to a
pole eigenvalue of $\cR(z)$ at $z=q^{-a}$, we conclude that their ranks are complementary,
so that $\textrm{Im}(f_a)=\textrm{Ker}(g_a)=:X_a$ and $\textrm{Im}(g_a)=\textrm{Ker}(f_a)=:Y_a$.
$X_a$ is a submodule of $W_{z}\otimes W_{q^a z}$ with quotient isomorphic to $Y_a$, and vice versa
for $W_{q^a z}\otimes W_{z}$. Because $\cR(z'/z)$ is the {\em only}\/ intertwiner from
$W_z\otimes W_{z'}$ to $W_{z'}\otimes W_z$ even at $z'/z=q^{\pm a}$,
$W_z\otimes W_{q^{\pm a}z}$ is indecomposable. \rem[gray]{can we say more? irreducibility of $X$?}

If $a=30$, $X_{30}$ is of dimension $1$: we recognize it as the trivial representation of $\Uq(\ee)$.
As an exercise, let us recover this by looking at possible embeddings of the trivial representation into
$W_z\otimes W_{z'}$. From the $\Uq(\e)$ invariance, we know that this embedding must take the form
\begin{equation}\label{eq:triv1}
\alpha\ \mycap + \beta\
\tikz[baseline=-3pt]{
\node[empty] at (-1,0) {};
\node[empty] at (1,0) {};
}
\end{equation}

Imposing the invariance under $e_0$ results in the relations
\[
z'=q^{30}z\qquad
\alpha=\beta
\]
which is consistent with the above. Of course, we recognize that \eqref{eq:triv1} is (up to normalization)
the dashed cap \eqref{eq:dashcap},
and its invariance is once again nothing but the statement that $W_z^*\cong W_{q^{30}z}$.

Dually, the invariance of the  $W_z\otimes W_{z'}\to \CC$ diagram
\begin{equation}\label{eq:triv2}
\alpha\ \mycup + \beta\ \tikz[baseline=-3pt]{
\node[empty] at (-1,0) {};
\node[empty] at (1,0) {};
}
\end{equation}
leads to
\[
z'=q^{-30}z\qquad
\alpha=\beta
\]
(compare with the dashed cup \eqref{eq:dashcup}).
Up to normalization, $f_a$ is just the product of \eqref{eq:triv1} and \eqref{eq:triv2}.

If $a=20$, $X_{20}$ is of dimension $249$, and so it must be isomorphic to
$W_{z''}$ for some $z''$. Again, we look for an embedding of $W_{z''}$ into $W_{z}\otimes W_{z'}$;
by $\Uq(\e)$ invariance it is of the form
\begin{equation}\label{eq:fus1}
\alpha\ \D+\beta\ \tikz[baseline=-3pt]{
\draw[bend left=60] (-1,-1) node[vertex] {} to (1,-1) node[vertex] {};
\node[empty] at (0,1) {};
}
+\gamma\
\tikz[baseline=-3pt]{
\draw[bend right=40] (-1,-1) node[vertex] {} to (0,1) node[vertex] {};
\node[empty] at (1,-1) {};
}
+\delta\
\tikz[baseline=-3pt]{
\draw[bend left=40] (1,-1) node[vertex] {} to (0,1) node[vertex] {};
\node[empty] at (-1,-1) {};
}
+\epsilon\
\tikz[baseline=-3pt]{
\node[empty] at (1,-1) {};
\node[empty] at (-1,-1) {};
\node[empty] at (0,1) {};
}
\end{equation}
and requiring that it commute with the $e_0$ action leads to
\[
z'=q^{20}z\qquad z''=q^{10}z\qquad
\frac{\alpha}{\epsilon}=
-\frac{[5]}{[20]}\sqrt{\frac{[10][30]}{[6][15]}}
\qquad
\frac{\beta}{\epsilon}=
\frac{\gamma}{\epsilon}=\frac{\delta}{\epsilon}=\frac{[10]}{[20]}
\]
so that $W_z\otimes W_{q^{20}z}\supset X_{20}\cong W_{q^{10}z}$.

Dually, one finds
\begin{equation}\label{eq:fus2}
\alpha\ \U+\beta\ \tikz[baseline=0cm,yscale=-1]{
\draw[bend left=60] (-1,-1) node[vertex] {} to (1,-1) node[vertex] {};
\node[empty] at (0,1) {};
}
+\gamma\
\tikz[baseline=0cm,yscale=-1]{
\draw[bend right=40] (-1,-1) node[vertex] {} to (0,1) node[vertex] {};
\node[empty] at (1,-1) {};
}
+\delta\
\tikz[baseline=0cm,yscale=-1]{
\draw[bend left=40] (1,-1) node[vertex] {} to (0,1) node[vertex] {};
\node[empty] at (-1,-1) {};
}
+\epsilon\
\tikz[baseline=0cm,yscale=-1]{
\node[empty] at (1,-1) {};
\node[empty] at (-1,-1) {};
\node[empty] at (0,1) {};
}
\end{equation}
and requiring that it commute with the $e_0$ action leads to
\[
z'=q^{-20}z\qquad z''=q^{-10}z
\]
and the same ratios of parameters.
Again, $f_a$ is proportional to the product of \eqref{eq:fus1} and \eqref{eq:fus2}.

For $a=2,12$, the spaces $X_a$ are higher-dimensional, and are not so simple to describe in our diagrammatic
language. In terms of $\Uq(\e)$, one has:
\begin{align*}
X_{12}&=V_{\omega_7}\oplus V_{\omega_1}\oplus V_0
\\
X_2&=V_{\omega_2}\oplus V_{\omega_7}\oplus \CC^2\otimes V_{\omega_1}\oplus V_0
\end{align*}
\rem[gray]{in the first case, it has to be irreducible, otherwise sub/quotient would be one of the previous
two cases}
By obvious complementation, one can get similar decompositions for the spaces $Y_a$, but again, their description
is more complicated. For instance, we have
\[
Y_2\cong V_{2\omega_1}\oplus V_{\omega_1}\oplus V_0
\]
and we can describe its embedding into $W_{q^2z}\otimes W_z$ as
\begin{multline*}
\alpha\dbl[1]
+ \beta
\left(
\D[1]
+
\frac{[32]}{[16]}\sqrt{\frac{[6][10][15]}{[30]}}
\left(
\tikz[baseline=-3pt,scale=1]{
\draw[bend right=40] (-1,-1) node[vertex] {} to (0,1) node[vertex] {};
\node[empty] at (1,-1) {};
}
+
\tikz[baseline=-3pt,scale=1]{
\draw[bend left=40] (1,-1) node[vertex] {} to (0,1) node[vertex] {};
\node[empty] at (-1,-1) {};
}
\right)
\right)
\\
+\gamma
\left(
\tikz[baseline=-3pt,scale=1]{
\draw[bend left=60] (-1,-1) node[vertex] {} to (1,-1) node[vertex] {};
\node[empty] at (0,1) {};
}
+ [31]\
\tikz[baseline=-3pt,scale=1]{
\node[empty] at (1,-1) {};
\node[empty] at (-1,-1) {};
\node[empty] at (0,1) {};
}
\right)
\end{multline*}
where $\dbl[0.5]$ is the embedding of $V_{2\omega_1}$ into $V_{\omega_1}^{\otimes 2}$ (with some normalization).
However, the coefficients $\alpha,\beta,\gamma$ can only be fixed by independently specifying the
$\Uq(\ee)$ action on $V_{2\omega_1}\oplus V_{\omega_1}\oplus V_0$.

redpajama_set_name: RedPajamaArXiv


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alwaysdata | blog » 2019: more than ever here to guide you!
We're already nearby end January, so before it's too late, we want to wish you the best for the upcoming year! 2019 started faster than ever, so remove the last glitters, put on your helmet, ride your scooter, here's our roadmap!
All of this was a lot of fun to do alongside you.
We plan to upgrade our Debian based distribution during the year, and thanks to our migration process, this will be totally transparent.
We spent a long time on it, here's a new design for our website! A new identity to present the alwaysdata's magic to everyone.
More than ever, we're a hosting provider: our servers allow you to get your Web services online and running, for your websites, web apps, databases… also e-mails, backups, and more. However, we're not just a hosting provider. We built a platform that embeds and runs everything you need, from programming languages to brokers, and we manage all of them for you, regardless of your offer, from shared hosting to dedicated servers. From this point of view, we're a PaaS provider1).
So it's time to change that identity a little bit — new terms: environments instead of plans, PaaS along with Hosting provider, etc. There's also a new Gold environment available.
All details in an upcoming blog post about the brand new but still very same alwaysdata.
We want to continue to give you feedback and right inputs about new technologies. So we continue our Tech Evangelism Tour, with new events, and new topics!
We cover a lot of topics with new talks about blockchain, containers, timezones, and REx about our backups and migration tools. Get in touch by waving at community@alwaysdata.com if you want to see us at your event!
We pledge in favor of well tested, secured environments. That's why we decided to handle a Bug Bounty Policy. Whether you're a security researcher or a bug hunter, feel free to report your bug accordingly to our program.
More than ever, we build our platform with you, for you, to help you to power the best services ever. Welcome again to our great community, 2019 is already running on a rainbow!


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Anne of Great Britain (6 February 1665 - 1 August 1714) became Queen of England, Scotland, and Ireland on 8 March 1702. On 1 May 1707, when England and Scotland combined into a single kingdom, Anne became the first sovereign of the Kingdom of Great Britain.
This page was last changed on 11 September 2018, at 22:53.


text:
Trevor Noah is a big fan of Patek Philippe, Audemars Piguet, Richard Mille and of course Rolex
His jokes are crazy funny but The Daily Show host doesn't play when it comes to watches.
May 28, 2019 Words By Celine Yap
LOS ANGELES, CA - MAY 16: Trevor Noah is seen at 'Jimmy Kimmel Live' on May 16, 2019 in Los Angeles, California. (Photo by RB/Bauer-Griffin/GC Images)
JOHANNESBURG, SOUTH AFRICA - DECEMBER 02: Trevor Noah speaks on stage during the Global Citizen Festival: Mandela 100 at FNB Stadium on December 2, 2018 in Johannesburg, South Africa. (Photo by Kevin Mazur/Getty Images for Global Citizen Festival: Mandela 100)
NEW YORK, NY - SEPTEMBER 29: Comediann Trevor Noah performs onstage at Stand Up LIVE! during Advertising Week 2015 AWXII at the Gotham Comedy Club on September 29, 2015 in New York City. (Photo by Andrew Toth/Getty Images for AWXII)
SAN FRANCISCO, CA - JUNE 01: Trevor Noah speaks onstage during 'The Daily Show Live: A Conversation with Trevor Noah and The World's Fakest News Team' in the Larkin Comedy Club during Clusterfest at Civic Center Plaza and The Bill Graham Civic Auditorium on June 1, 2018 in San Francisco, California. (Photo by Jeff Kravitz/FilmMagic)
WEST HOLLYWOOD, CALIFORNIA - FEBRUARY 22: Omar Dorsey, Common and Trevor Noah speak during Common's 5th Annual Toast to the Arts at Ysabel on February 22, 2019 in West Hollywood, California. (Photo by Arnold Turner/Getty Images for Freedom Road Productions)
NEW YORK, NY - FEBRUARY 14: Trevor Noah and Jordyn Taylor attend the New York Knicks vs Washington Wizards game at Madison Square Garden on February 14, 2018 in New York City. (Photo by James Devaney/Getty Images)
NEW YORK, NY - NOVEMBER 13: Trevor Noah attends The Children's Monologues at Carnegie Hall on November 13, 2017 in New York City. (Photo by Walter McBride/WireImage)
TORONTO, ON - DECEMBER 02: Trevor Noah signs copies of his new book "Born A Crime" at Indigo Eaton Centre on December 2, 2016 in Toronto, Canada. (Photo by GP Images/WireImage)
JOHANNESBURG, SOUTH AFRICA - DECEMBER 02: Trevor Noah speaks on stage during the Global Citizen Festival: Mandela 100 at FNB Stadium on December 2, 2018 in Johannesburg, South Africa. (Photo by Michelly Rall/Getty Images for Global Citizen Festival: Mandela 100)
He was spectacular in Afraid Of The Dark, and he totally killed it in Son Of Patricia, and Trevor Noah is coming to Singapore this August for his latest show Loud & Clear. If you didn't manage to score any tickets, well don't cry. You'll see it soon enough on Netflix.
Colorado! Thank you to everyone that braved the cold weather and came to the show last night! An amazing night, I won't forget ??#LoudAndClearTour ?: @marklashark
A post shared by Trevor Noah (@trevornoah) on May 20, 2019 at 2:30pm PDT
It's been 10 years since Noah started touring; his was the first one-man stand-up comedy show from—of all places—South Africa. Yes, they have comedy there, as Noah would have you know. His meteoric rise to international fame is based purely on sharp wit and intelligent commentary regarding race, gender and politics.
Sometimes it's good to think out loud but silently. Sometimes you need to speak and sometimes you need to listen and sometimes you need to do both but not exactly at the same time because that's hard. Hard is how we get better. Better is more than good. And that's how we become great! ?: @imseangallagher #motivationalquotes #motivation #motivationalspeaker #motive #nomakeup
A post shared by Trevor Noah (@trevornoah) on Jan 22, 2019 at 12:27pm PST
Most recently he tackled the 5G race between China and the US, specifically the strategies of President Donald Trump, who provides much comedy gold 5G or not.
So as all eyes and clicks are on Trevor Noah, man of many strengths, we zero in on his wrist to discover what is surely his one weakness. Watches! The man goes around with such gorgeous wrist candy we can't even.
Audemars Piguet Royal Oak
Patek Philippe Aquanaut Travel Time
Vintage Rolex Paul Newman Daytona
Montblanc Star Nicolas Rieussec
Patek Philippe Nautilus in steel
Patek Philippe Nautilus in rose gold
Panerai Luminor Base Logo
Richard Mille RM67-01
Take a gander at the gallery above and you'll see exactly what we mean. Also, don't forget to check out his wrist at the show come August.
Ice blue watches to take you through the coldest time of the year
By Celine Yap
Spruce up your fall/winter wardrobe with the latest men's jewellery launches
The best new watches to take you through summer and beyond

redpajama_set_name: RedPajamaCommonCrawl


Dr. Elmi Zulkarnain Osman is an award-winning educator, motivational speaker, top rated mega event emcee, and a highly recommended corporate trainer with over 15 years serving government, grassroots organisations, non-profit, private and public sectors. Dr. Elmi has proven himself having an innate ability to connect with diverse audiences across disciplines.
Prior to becoming CEO of Elemantra Training and Event Consultancy, Dr. Elmi Zulkarnain held middle management and leadership positions in various government schools and grassroots organisations. Currently, Dr. Elmi oversees Elemantra Consultancy's training and business development practices. Dr. Elmi has personally trained 5,000+ participants including volunteers and staff of various government agencies such as the People's Association.
Although Elemantra Consultancy's clients include government agencies, non-profits, and publicly traded companies, Dr. Elmi's most passionate about leveraging the best practices of multinational enterprises to be a catalyst, compass, and connector of new business development and expansion for entrepreneurs and the small business community.
Dr. Elmi is a husband and father of two toddlers. They are currently residing in Singapore but commute to Malaysia regularly. In addition to serving as a youth mentor in his local community, Dr. Elmi is a passionate advocate of the language arts and he aspire to build and equip businesses and individuals globally using corporate training, education, and coaching tools.
Contact Dr. Elmi Zulkarnain Osman today to find out more about how he can help you with your organisation'slearning needs.


The revised rule, effective December 1, 2016.
This revised rule increases the minimum salary threshold for exempt employees from $455/week (approximately $23,660/year) to $913/week (approximately $47,476/year). For campus employees paid monthly, the threshold to remain exempt is $3,956.34/month.
All staff and academic personnel, including rehired retirees and part-time employees, who do not qualify for the professional teaching exemption and who earn less than $913 per week, will be reclassified as non-exempt and paid on an hourly basis. This is because an employee must earn at least $913 in each week in which they perform any work regardless of the number of days or hours worked to qualify for the exemption under the FLSA.
Due to certain exemptions under FLSA, it will remain the case that some University employees will remain overtime-exempt even if they earn a salary that is below the new salary level. For example, University employees whose primary duty is teaching will be unaffected by this rule no matter how much they earn, because the minimum salary threshold requirement does not apply to bona fide teachers. This exemption from the salary basis test also applies to many University coaches, who have the primary duty of teaching or instructing student athletes. If a coach is primarily engaged in administrative duties, then the salary threshold must be applied. Additionally, the DOL does not consider undergraduate and graduate students engaged in research or extracurricular activities as employees, and therefore they are not subject to the minimum wage or overtime requirements of the FLSA.
The transition to non-exempt biweekly pay will take place on November 20, 2016, the start of the biweekly pay cycle covering December 1, 2016.
Departments are being provided with lists of affected employees that will need to be updated in PPS from monthly, exempt appointments to biweekly, non-exempt appointments.


Continuing the countdown to the 84th Academy Awards, we're still gearing up for a fabulous Oscars party at home! Now that you have your Oscar party invitations sent out, it's time to think about some glamorous, Hollywood-inspired decor ideas to set the mood. You don't have to erect towering, gold statues in your front hall to achieve the perfect setting. All it really takes are a few fun accessories. Here are five ideas to inspire your own Oscar party decor for this Sunday. It's just not a star-studded night without a red carpet! Make your guests feel like celebrities when they enter your home with this textured red runner. Horchow, $299. Add a little glitz and glam with some sequined decorative throw pillows. West Elm, $24.99. Keep your guests refreshed with freshly shaken cocktails. Williams-Sonoma, $40. Whether you're setting up a table or passing around the hor d'oeuvres, serve your appetizers on gorgeous silver trays for an elegant and chic touch. Pottery Barn, $99. And what Oscars night would be complete without the paparazzi? Capture the memories of the evening with a lo-fi camera for that stylish, vintage look. Indigo, $89.99. Which party accessory do you like best?


text:
Tips.Net
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> Dogs
> Dog Breeds
> Terrier Dog Breeds
by Doris Donnerman
(last updated August 17, 2009)
The Skye Terrier is beautiful with his smooth cascades of shining groomed coat, the tender softness of eyes obscured by a thick veil of hair. His attributes earned him the name of the "heavenly breed" as a tribute from admirers.
This purely Scottish breed had his early home in the mist-shrouded island of Skye, but he was also found in other islands of the Hebrides as well as on the mainland of Scotland. It is said by a well known doctor of the time, returning from a visit to the Islands in 1773, remarked that otters and weasels were plentiful on the island of Skye and that the foxes were numerous and were hunted by small dogs.
The earliest descriptions of the breed present a Terrier much smaller in size than the type of today, a size more adapted to burrowing and taking to water. It had a profuse, hard coat, with short legs, a body long in proportion to its height, and with ears that were neither erect nor drooping, but half erect and rose to alertness when aroused. In the early days the Highlander cared far less for beauty than for the sporting ability of his dogs. It seems difficult to credit the languishing-beauty appearance of the Skye Terrier of today with bloodthirsty tendencies. The breed was originally used and developed for hunting among the rough rocks and crevices of his native shores. He needed the face fringe to protect his eyes, his lithe, long body for agility in getting into burrows and dens.
When the Skye first made his appearance in England, it was taken up by the nobility. Queen Victoria greatly admired the breed and owned several Skyes starting in 1842. In 1896 a Skye Terrier was presented to the then Princess of Wales who later became Queen Alexandria. Sir Edwin Landseer's paintings in which the Skye was introduced helped to draw public attention. It was still considerably smaller in size than today's type.
The Skye Terrier has been surpassed in popularity by many of the more modern breeds of terriers, but it is unrivalled in length of time of popularity because it has existed over a period of long duration. Before the 1900s it was an important breed in American dog shows. It retains the greatest popularity in Scotland and England.
Two varieties of Skye Terrier are recognized, the prick-eared and the drop-eared (pendant). In other respects both are identically the same in all points. The Skye Terrier has great working capacity despite his small frame. He displays great stamina, strength and agility. Although cautious with strangers, his demeanor is friendly and loyal to those he knows.
As a general description of the Skye Terrier has the color of either a dark or light blue or grey. The coat is a double coat with the under coat being short, close, soft and woolly. The height at the shoulder for the male dog is nine inches, and 23-1/2 inches from the skull to the root of the tail while the female is half and inch lower and 2-1/2 inches shorter.
Doris Donnerman
Doris is a jack of all trades, writing on a variety of topics. Her articles have helped enlighten and entertain thousands over the years. Learn more about Doris...
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