Live from Las Vegas, Nevada, it's theCUBE. Covering AWS re-invent 2016. Brought to you by AWS and its ecosystem partners. Now, here are your hosts, John Furrier and Stu Miniman. Okay, welcome back everyone. We are live here in Las Vegas for Amazon re-invent, Amazon Web Services re-invent 2016 user conference. 32,000 people, the center of the technology industry of the universe now. I'm John Furrier with Stu Miniman. It's been packed, housed, our next guest. We're excited to have the CTO of Rackspace, John Engates back on theCUBE here at re-invent. Incredible milestone status to talk about. Congratulations. Welcome back to theCUBE. Thanks for having me back. It's good to see you guys. 19 interviews yesterday. My, our voices are going, but we're going to go on and on today. So much going on here. I call the center of the universe because it's a consolidation of innovation. And it's not a land grab per se. It's cloud. You know a lot about that. You guys served customers going from hosting to cloud. Now Amazon, you guys have some special news. What's the big news? Well, this week we announced that we are part of Amazon's premier partner tier of their partner program. We're at the top tier and we did it, I think in record time. We went from basically launching about a year ago here. This was sort of the place where we announced fanatical support for AWS. A year ago and today we're in the top tier. We also have now over 500 certified engineers and architects on the AWS platform. So we're really building up a breadth of expertise on this platform or helping customers make that migration to the cloud and really creating great outcomes for customers. Take us to the journey of Rackspace within relationship to Amazon because what I find very interesting is that the innovation that you guys are doing really comes down to the heart and soul of Rackspace. You guys have been doing infrastructure for a long time. Talk about your journey vis-à-vis this new cloud world that's been spun to mainstream enterprises now. What is it all about? I mean, what's the big... Well, look, I get excited about cloud because it feels like the early days of Rackspace. Back in the early days, Linux was the new thing. The web was the new thing. Customers were trying to get to those technologies and those platforms were the foundation for the next new thing for them. The web was going to be this Wild West and Frontier. And today, cloud is sort of representing that. It's the thing that companies feel pressure and they feel excitement and they want to get onto these new platforms. I mean, all the announcements here are around things like machine learning and IoT and analytics and really cool technologies that many enterprise companies don't know how to take advantage of. They don't have the skills. They don't have expertise to take advantage of it. But they feel pressure to because of the fact that it's where they differentiate nowadays. Data is the true differentiation for companies. And so Rackspace is in a position to help companies do just that. For many years, we've helped them get up and running on infrastructure, use Linux, take advantage of open source. And today, we're translating that heritage into services and support for AWS. We got enough inbound questions from customers like, hey Rackspace, we love what you do. We love your fanatical support, but can you do it on other platforms like AWS? We're using those platforms. We're going to use multiple platforms, so it's not just AWS, but can you help us across these different platforms? John, Rackspace is a really interesting place to look at kind of the hybrid cloud, multi-cloud world. We're here to AWS, but you've got lots of different services you offer. You work with, I think, most all of the cloud providers out there. What do you hear from customers? How do they, when you talk to your average customer, do they use those terms? What do their situations look like? And maybe compare to that as to what you hear from Amazon. Well, look, I think the word multi-cloud is certainly on the mind of companies nowadays. This is something that they feel, again, a little bit of pressure to make sure that they have a couple of vendors in the race. This is like multi-vendor of a number of years ago. Everybody used to have an HP and a Dell or an EMC and a NetApp. And today, I think it makes sense to have multiple cloud providers. Also, it depends on the layer of the stack that you're thinking about, because certainly they're going to be multi-cloud up at the SaaS layer. I mean, they're using Salesforce and different service offerings at that layer. And so multi-cloud is kind of becoming sort of a given. And so when you start to think about the different layers of the cloud stack, they may have private cloud, they may have public cloud, they may have more than one public cloud, and they may want to keep the options open for where they place their workloads, especially in a geographic, global kind of a perspective. Yeah, offering those various offerings, do you help customers with kind of migrations or choosing or give them that decision tree? How do you help them move forward? Well, sometimes they come in with an opinion. You know, oftentimes they do it. It's like I tell people, I never convince somebody to move from Linux to Windows or Windows to Linux. They usually come in with an opinion about what they're going to do. And it's usually based on their skills and where they're coming from. And so oftentimes they come in with an opinion about which cloud provider or platform they want to use. But we do help them really navigate once they've sort of gone down that path, you know, what should I place where, which kind of workloads are best suited for which cloud, which particular application, you know, or which services within a particular cloud or the appropriate ones, which ones are mature, which ones should I, you know, adopt now or later. And I think those are hard things to know unless you've sort of been there and done it a number of times. And we do that with customers across the spectrum. John, I want to get your take on some of the themes here at the show. Obviously Andy Jassy's on stage, Verner today. The theme is pretty much old guard versus the new guard. New way old guard references to Oracle and others. But really the buyer, Verner Vogels, like, hey, I don't want to buy it the old way.  And I was talking to Peter Burris, our head of research earlier today. And the new guard and the new persona making decisions are changing too because the capabilities are different. You're seeing scale, speed, AI, machine learning, virtual reality, internet of things, whether it's consumer enterprise, the new game has changed. So when you're out engaging with customers, who's in the room when you're talking about this new way? Who's sitting at the table? Does that defines kind of the new architecture? That's true. I mean, look, in the past, the CIO had all the spend. He had all the, you know, he held the pocketbook for where IT spend went. But today, business units tend to sometimes drive the decisions, especially for net new applications. Things that are greenfield, things that are consumer facing. We hear a lot of times that the CMO or the chief marketing officer is getting more and more wallet share, you know, of the IT spend. And that to me is an indicator that, you know, changes going on within organizations. Big changes are going on in terms of what they want to do. I mean, IT is now a differentiator. It's not just a keeping the lights on behind the scenes kind of thing, but it's a thing that feel, you know, companies feel pressure to go adopt cloud technologies because they feel like it helps them have an advantage or at least keep pace with the other companies out there that are disruptive. And so if you think about who's driving that, it's marketing folks, it's business units, it's product line leaders within organizations.  In the old days of when I grew up in the business in the 80s and 90s, you bought the big software and platforms because as a customers didn't have the expertise to build it and come to Facebook, Google error, they build their own stuff. They build their own stuff.  Now we're kind of going back to the world where, damn, Amazon's growing their scale so damn fast, speed of new services, competition's coming in, I'm a customer, I can't build, I can't build those services as fast but yet I need to have some control. How do you reconcile that new shift now that's happening where I won't say it's multi-purpose cloud or general purpose cloud, it's general purpose sets of tools that can be composed. So it's kind of a different animal but the world is shifting where Amazon's shipping more capabilities than the customer's sales can build. So it's kind of shifting. It's true, you're drinking from a fire hose when you're thinking about consuming Amazon's services because they're just coming at you left and right and they're really changing very rapidly. Even if you look beyond just the launches of new services, the iterations within existing services are going very rapidly too. And a lot of our customers are excited but at the same time they feel a little bit of concern about like if we adopt these too rapidly what happens if our expert gets poached to another company down the street that's got a bigger budget or more interesting tech and so they're reluctant sometimes to go too fast, too deep into these technologies. So I think that's a balance that every company has to take. The intellectual capital with people is huge. Yeah, that's right. A lot of times companies are heavily invested in the existing expertise they have, the people that have been certified on technology platforms for their entire career are now having to sort of rethink how they approach IT and so we're helping companies navigate that. We're giving them a little bit of a safety net when it comes to that. So do you de-risk that? You're looking at de-risk it. We certainly do because we give them a backstop in terms of expertise on all these different platforms.  It used to be one engineer could keep the whole stack in his head or her head and now it's not so easy anymore because you've got this matrix of all these different services, all these different ways of composing applications, serverless and microservices and really complex components and that's hard to really be an expert on all of that and so that's a way that Rackspace also really takes the risk out of it. John, I want you to give us a little insight into kind of the enterprise mindset.  Is it public cloud? Is it safe? Is it secure? Today I try to follow, it's the data and the applications and it's a little bit less of a discussion about like where it lives and more about, you know, where are the tools I need? Where can it have, you see a broad spectrum is that kind of infrastructure versus applications? Is that a good way to look at it or how do you see it? What are the conversations you have? Well, I think, again, you're- And in any commentary on OpenStack I'd kind of be interested to. Sure. I do think that the person writing the application or owning the data owner, the person trying to build a new product is oftentimes in control of where something goes and their needs have to be met for them to be comfortable or, you know, really be effective in building an application. I mean, they really have to kind of take into account all the things they need. And I had this analogy the other day in my head. You know, we oftentimes use the utility grid as the analogy for cloud, but I thought instead of electricity, how about water, right? Water is something we usually get from the public utility. It's easy to get, you know, everybody accepts that. But then there's companies and organizations that dig their own well. Why do they dig their own well? Because sometimes geographic reasons dictate that they do. Sometimes they have a use case that's bigger or different than, you know, a typical consumer. And so I think that same thing applies in public and private cloud. I think we're going to have users that continue to see use cases where private cloud makes a lot of sense. I've had discussions here where companies are using both of those. They're using them for different applications. Sometimes they're using even together. And I just don't see, I don't maybe see it as big of a debate as it was in the early days of cloud where like it was this religious war of public versus private. I think it's all of the above. I'm going to, you know, if I'm a big enough company, I'm probably going to have all of the above. So wait, is private cloud the artisanal bottled water that I pay a lot for? And public cloud is good clean drinking water that you get out of the cloud. I don't know if I've taken the analogy that far, but it's certainly something that, you know, companies, you know, they certainly know that there's a lot of options out there, they're going to use every option that available and they're going to make the best use out of it. And I think now the world's changed to the top line drivers really are impacted to your business unit driving, some of the budget. That's right. And also choice. I mean, multi-vendor in the old world really came out of lock in. And so choice became a buzzword. Now choice is ultimately a toolbox decision for customers. I don't think that makes it a, you know, win or take all environment, but certainly gives the customer hell of a lot of choice. They've got a lot of control, a lot of choice, a lot of options, you know, competition is a good thing. Having people building these competitive tools around AI and machine learning and IoT. I mean, we're going to have tons of opportunity to go build really compelling application. Sometimes even mixing and matching capabilities from different clouds. I imagine that that will happen more in the future than in the past. Stu and I are competitive strategy junkies. We think Amazon strategy, I think Amazon strategy is scale and speed faster than the competition. Leverage open source, go hard, run hard and get the economies of scale. And people who try to replicate that won't have those, will have this economy. So it's interesting to see that play out. Yeah, John, you know, Rockspace is now a private company. So you've got the ability to kind of, you know, move towards where it needs to be. How important is public cloud in general and AWS specific for the future of Rockspace? Well, public cloud is a huge investment for our company. I mean, we have put a lot of resources into this. A lot of our new marketing campaign that you'll see out on the Las Vegas strip and in news media, it's basically about, you know, what we're doing with the public cloud. And Amazon is a great partner. They're the fastest growing cloud out there. They're the biggest cloud out there. They're the one with the most mind share, I would say, in the market. And so certainly they are going to be a big focus for us. But again, you know, people are going to have a lot of choices in cloud and they're going to need help on all of those. Yeah, you're one of the, probably the largest service provider for VMware too though. What's your take then? Do you have a play in the VMware on AWS? Or, you know, how does that impact what you guys do? I don't know yet about the VMware. You know, the VMware, I mean, it's a relatively new product. And I think that's just getting started. You know, we do have a ton of VMware customers hosted in Rackspace data centers. A lot of those companies are looking to Rackspace to help them take steps to move some of those workloads to the cloud over time. We may end up, you know, working with the technologies that are available in the AWS cloud to make that easier. If they work for us, if they don't work for us, we'll, one way or the other, we will get these companies to where they want to be and how, you know, however that makes sense. Well congratulations on the fastest partner to achieve a premier status on the certification. Absolutely, very excited about that. And no surprise, you guys have a great team of people. I mean, we've worked with Rackspace since you started cloud back in the, when the whole open stack, pre-open stack days, you built your own cloud, you guys done a lot of work. And I think it's going to pay off for you guys. Congratulations. Thank you so much, yeah, thank you guys. All right, John and Gay, CTO of Rackspace, now a privately held company, which by the way, when that happens, innovation usually comes out of that. So they're not on the 90 day shot clock, as Michael Dell will say as well. This is theCUBE, back with more live coverage. With Stu Miniman, I'm John Furrier, you're watching theCUBE.


All right. Justice Compensation Fund it is. So we have anything that people had just going through to in the meeting, anyone didn't anticipate? Yes. So I want to announce that our wonderful MS Sunrise Youth Group, they are planning know your rights training. And we also promise that there will be a second public comment tonight before we close up. And there are still some audience in this meeting. So yes, I just wanted to make sure there weren't any other things or announcements like you just shared. AMR Sunrise is a youth group. And this is a training. Know your rights training. So excited that the young people are bringing this to the community. October 9th. October 9th at 3 o'clock, 3 to 5. And I've been part of the plan. I've been receiving their email. So I'm kind of like CSSJC rep. If people don't mind, because we'll be talking about the Know Your Rights. So we are co-sponsoring the event as well. Great. And so the human rights commission is also sponsoring. I'm sorry.  So you said the Human Rights Commission is also sponsoring. So it's not working yet. Great. So if folks want to know more about it, they can contact you all. Great. Yeah. Anyone else? I just wanted to say that the budget hearing, the first budget hearing is scheduled for November 7th. So I imagine that at our next meeting, I'd like to propose pending some sort of formal letter to the town manager and town council about what we would like to see in the budget, specifically for Crescent DEI, drawing most likely from CSWG and the recommended budget that they had put forward to use it. I also have one announcement. The CDBG request for proposal is on. It's something in our next meeting. We may want to discuss how we can try to see if the DEI department, we can support the department to apply for grants. And I'm not remembering the deadline, but it's something that we should think about, put it in our next agenda. Great. The deadline is November 4th. And I just want to point out that you do have one person in the attendees with a hand raise for this last public comment. Yes. Yes. Thank you. OK, I think we are ready to move on to public comments. Yeah. Thank you. Vera? Hi, this is Vera Duongmini Cage, a 12-long meadow drive in Amherst. I want to thank each member of the committee and DEI officer Pamela Young for staying so long at this extended meeting once again. I appreciate the need for a more robust DEI department. And I want to raise that when I was serving on the Amherst School Committee, we wanted to establish the participatory action research project involving youth and families. Unfortunately, that never got funded from the school level. And I hope the town will take this seriously to consider that methodology to engage community. And I think we've heard a lot from white people in this town. And we need to be intentional about reaching BIPOC communities. I am aware that the Crest Program has launched, but it is missing the voice of Asian-Americans in that body. And I want to see the town be able to acknowledge that and seek ways to rectify that. Additionally, I applaud this committee for spearheading something that is an alternative to the criminal justice system, to the legal system. We know that that process can be re-traumatizing for individuals and very threatening for families. And that's why some people do not resort to seeking that type of resolution.  But what is important is what it means to the children and families involved to be able to put this behind them in a way that this is not a topic of conversation that involves them directly. But they can be part of saying that this may have happened to me and my friends. But I did everything I could to stop it from happening to my younger siblings and to the other generations that follow me. Thank you very much. Thank you, Vera. Thank you, Vera. I just want to, since we mentioned the budget real quick, I'd like to know where we are in terms of translations of town council meetings and Zoom translation ability for folks that speak languages other than English. So I think that's part of the work that we're committed to and inclusion. So I just want to be mindful and remind ourselves that there's still a need for that. And funding should go towards that. Thank you all. We have to, did we set the next meeting? You did not. OK, this is still a good time for everyone, these Wednesday nights. OK. So we need this for me as long as we don't go on the third Wednesday because that is on the human rights commission. Oh, right, right. That's our member that. So the next meeting, are we going to try to meet twice in October? There isn't been if the town council invites us, that I think it would be way too many meetings for us in October. Good point. Yeah. So the 17th, right, is that meeting? So we should meet after the 17th or before. That's, I guess, the critical question. So the Wednesday before is the 12th. And then the Wednesday afterwards is the 19th or the 26th. I'm sorry, say that. The 26th. Yes. So the one immediately following the 17th is the 19th. And then the 26th is at the end of the month. So do we want to put it to the 26th? Can I make a motion that we meet on the 12th? And then on that meeting, then we decide kind of what's the next step? Because if we could meet on the 17th, we will know that by hopefully the 12th. And then if not, then we can just move the meeting to the 26th, because the 19th is the third Wednesday. OK. Yeah, because 12th can't meet on the third Wednesday. So that's why I was saying the 26th. Yeah, I think we should try to meet on the 12th just because too, so we can prep for the 17th. Let's see if they do give us the 17th. Yeah. And we can prepare for it too. OK, let's do that then. So we're doing the 12th. What are we doing? That's the 12th. And it's the Wednesday in October. OK. So put that on calendar still at 6 o'clock. And I think that's a good suggestion so we can prepare. Hopefully if they put us in. Yes, well, we'll find out. OK. We'll find out. And that's where media comes in. So we'll find out before then when we have a meeting, we should be inviting the media. Right. Yeah. So the other thing is, Allegra and I will write the press release. We'll have our names as the contact for that press release. So just in case, I don't know, Scott Smersbach or whoever will contact us potentially. Do you have your hand up, Allegra? Are you over? No, I just might. All of a sudden the volume got really loud. OK. But I mean, but I think even though you all will have your contact, shouldn't it come from all of us though? Oh, absolutely. But usually on a I'm just thinking. They got you. No, no, no, that's perfect. I just wanted to make sure that all our names will be on it. It will be everybody's name will be on it as the committee most certainly. OK. Anything else? And if we forget something in the draft, please feel free to send comments. So they would send comments, Pam, to Jen and to you. And then it would circle back to us. Is that correct? Yeah. OK. All right. Great. Yeah, that's how you do it. All right. Thank you all for serving. So when we plan the next agenda items, if we can have the DEI and press update first, like we did it the last time. We didn't do it today. If we can put that as the first thing that, you know, because I want to hear more about the Cress program. I think I read somewhere about the Cress responders, you know, in the school, CSWG never, never, never, never recommended that. It's a no, no, no, no, no. But anyway, it's for next meeting. OK. So if we could have that, Pam, already on the agenda, I'm sure we'll be we'll remind you. But yeah, the updates of that and that's helpful. OK. Thank you, everyone. See you next time. Take care. Yeah. Good night, everyone. Night.


OK, well then, good morning. So my plan for this first lecture is to introduce various equations which would have come to us via the way of physics. And it turns out that they're useful in mathematics. For some of them, we know this. For some of them, we expect this. And towards the end of the lecture, I want to explain how they all sort of fit in one framework, and in some sense, are all the same equation. And I want to start with something that hopefully you all know. Let's just do some review of anti-serve duality and Youngville's theory. So the setup for this is some sort of minimal setup that you always have to do for gauge theory.  And for convenience, I will take it to be semi-simple. So this thing is called the structure group of the theory. And I want semi-simple so that the Lie algebra has a canonical inner product that comes from the killing form, or the negative of the killing form. And so this is the group in this theory. Of course, this all happens on some space. The space I call x comma g. x is some manifold smooth. Of course, g is a Riemannian metric on this. I want this to be oriented, because you want to integrate. And then you need a g-principle bundle over this space to do some gauge theory on it. So this is some other manifold p, has some map down to x. It's the right action of g. Let's be pedantic and give this a name. And this quotient here, this group acts freely. And this is the quotient of this action. So this is a g-principle bundle. And the gauge theory sort of happens on this thing. And then just want to fix some notation in case you use different notation. I denote by script a of p. This is the set of connections on p, actions on g-principle connection. And this is a quite boring space. This is an affine space modeled on one forms on x of values in the adjoint bundle of p. And here adjoint bundle of p. This is the bundle that comes from the representation of g on its Lie algebra, the adjoint representation. It's mafthrag g is supposed to denote the Lie algebra of g. So this is the setup. And all of this is acted upon by the gauge group, g of p, gauge group, or the group of gauge transformations. This acts on everything that you can see over there. Well, it doesn't act on everything on this blackboard. The very left doesn't act. So this is the basic setup. And there's a map that associates to a connection that's curvature. So we give that a name as well. So if we have a connection, we can associate to it its curvature. And we denote the curvature by capital F of a. This is a two form also on x. And it takes values, the adjoint bundle. You should think of this as taking the derivative of a in some sense. Good. OK, so all of these are things that were naturally used in differential geometry. But we don't want to use them. We want to sort of study this space, but equivalently with respect to the gauge group. We need to have some more structure than just this. And it turns out that there's a natural functional on the space, which is just the L2 norm of this. And it's called the Young-Mills function. This is the, let me write this maybe more explicitly. Can I repeat the definition of at p, please? So when you have these principal bundles, they have a right action of g. So now if you have anything else that g acts on, you can make a bundle out of p and that space by making p times the space that you want. In our case, the space is the Lie algebra. And we can act on p with g in whatever way this is given. And we can act on g in the way given by this representation. And this quotient is the adjoint bundle. This is a vector bundle, because this is a linear. This is a representation, exactly. This is a vector bundle. Oh, this is omega 1. This is the one form that takes values in this vector bundle. The vector, these one forms are sections of the quotient bundle. We take sections of the quotient bundle, tensor at p. This is what this means. OK, good. So back to the Young-Mills functional. This is a natural energy functional on the space of connections. This norm here uses the norm that comes from the killing form, and I think the other stuff is reasonably sensible. OK, this is a natural energy functional. In the generality, I have defined that it could be infinite, because the space is not assumed to be compact. The second, I want a non-compact space. OK, so this is the Young-Mills functional. And you can formally, at least, compute the equations satisfied by a critical point, the so-called Euler-Lagrange equation. And this is the Young-Mills equation. If you compute this, you see it's the d star of fA, simple looking equation, but maybe not completely harmless. OK, and one can say some stuff that I will maybe not say in detail, but sort of Maxwell's equations can be formulated in this. This is an important equation in physics, but we don't really care about what the physics says. OK, so if you're an analyst, you would try to solve this equation, right? And then this is not easy to solve this. Maybe try direct minimization. This doesn't really work if the dimension is reasonably high. But if you're a geometer, maybe you observe that there's a special solution to this equation.  So an easy way to solve this, let's give this a name. A is flat. A flat means that the curvature of this connection vanishes. I mean, then certainly it satisfies this equation, because if I apply something to 0, but I stay 0, this implies the Young-Mills equation. And this actually gives us an interesting class of solutions, and they're completely characterized by the fundamental group of the space. So if you want, maybe there are some exercises in what I've discussed so far already for you, if any of that doesn't really make sense to you. But here's maybe the first exercise that you should think about. You can look at all principal bundles of structure group G up to isomorphism. And then you can look at all flat connections on these bundles, modulo gauge. And I claim that this space has a natural bijection to space that you all know, hopefully. This is a bijection. And the bijection is with representations of the fundamental group in the group G, modulo conjugation. So this G acts on G by conjugation. So this is the first exercise. And the map just comes from doing parallel transport. This is the map comes from monodromy. It's a good exercise in covering theory, if you understand why this is true. So this is, in some sense, an extremely satisfactory answer if you're an analyst, because it becomes a topologist's problem. And you're done. You go to maybe one of you in the audience, and then they have to solve it. But it turns out that this is maybe not that easy to compute. And you can actually use this both ways. So let me not say more about this. But this is useful in both ways. This is not just the way to make your problem someone else's problem. But this is also not very interesting, because this really just depends on the topology. So yeah, there's a question. Yeah, DA star. OK. So this connection on the principle bundle, so the question was, what is DA, or DA star? The connection on the principle bundle associates covariant derivatives or connections on any bundle that you associate with it. So in particular, you have this adjoint bundle here. And this inherits a covariant derivative. And then in the usual way that you extend the derivative on functions to a derivative on differential forms, it extends to differential forms valued in your vector bundle. So this gives you a map from differential forms on x with values in adp, degree 1. And all of these spaces have inner products, so you can take the formal adjoint. And this is the formal adjoint. Hope that most of the gauge group related to G. So the gauge group is the group. This was the question in the chat, I think, maybe. How's the gauge group related to G? The gauge group is the group of automorphisms of this bundle, so these G-equivariant maps from P to P. And this is how this affects the gauge group. One can say something more, but maybe I will not say more about this. OK. Good. So we have these solutions that come from flat connections. Here's another interesting solution. So I don't know who first discovered this. Maybe this was an obvious observation immediately when people developed the theory of maybe Erismann already noticed this. But the next observation I think I can pin down to precisely who figured this out. So far there's no physics except for maybe the young who is functional, but very natural thing to discover. In 1975, four physicists, Berja Wien, Poljakow, Schwarz and Chupkin, they made the following observation. So I'll abbreviate their names as BPST because their names are quite long, and it's four of them. They observed the following. So there's something special that happens in dimension four. It doesn't happen in other dimensions quite in the same way. So the dimension of x is four. Then the Hodge star operator maps two forms to themselves, and there are two eigenvalues. So you can compute that the square is one. So the eigenvalues must be plus or minus one. And they observed the following. If your connection is anti-safe dual, so anti-safe dual means that when you write this over there, it means if you take the curvature and you apply the Hodge star to it, you get minus the Hodge star. You can also write this as saying that the safe dual part, one-half of FA plus, if this vanishes, then this connection must automatically satisfy the Young-Mills equation. Because the Hodge star in this dimension is star D star. The curvature connection is self-dual, anti-safe dual, so you just get this minus sign here. But this expression, the exterior derivative of the curvature, this always vanishes. This is the Bianchi entity, the basic structure property of a curvature, that satisfies this equation. So somehow, we have been able to solve this equation which differentiates F by just imposing an algebraic condition on F. This should be much easier to solve. So this means that in particular these solutions, these anti-safe dual connections, they satisfy the Young-Mills equation, but in fact, something much better is true than this. You can actually write down precisely what the Young-Mills energy is in terms of a topological invariant if you have an anti-safe dual connection. So in general, you have this energy identity, the Young-Mills energy of A is given by 1 fourth integral, while it's basically the norm squared of this thing, which let me write it in this form. It's this plus something else, and the something else is topological. I'm going to write this in some really obnoxious way, because the way that I'm writing it, there's going to be an expression that has a meaning later. So this is written as 8 pi squared. And then some number H check of G. You're not supposed to know what that means. It's some number that depends on G. Actually, we will sort of take this number out in a second times k of A. I mean, this is certainly true until I tell you what k of A is. This is the definition of k of A. This k of A is an important thing. It's called the instanton number of the connection. And it turns out on a closed manifold to be a topological thing. This holds with k of A. So I'm now taking this number back out again, because this has to be normalized for specific reasons that minus 1 over 2, then I'm taking this number back out. In general, it's the integral of fA which fA over x. This is a 4 form on x that takes values in a tensor product of the Lie algebra, a joint bundle with itself. But we can compose this with this contraction map that comes from the killing form. And this is an actual 4 form that we can integrate. And if you remember some Chen-Wei theory from a differential geometry class, then you identify this as, I'm sorry, I'm even taking the 8 pi squared out again. If you remember some Chen-Wei theory from your differential geometry class, then you will recognize this as 1 over 2 times this number. The Pontriagin class of the adjoint bundle paired with x, the fundamental class of x. So this is if x is closed. And in that case, this number really only depends on p. If it quite generally only depends on p, even if x is non-compact, but one has to say more technically precise things about this. Anyway, this thing has a special significance. This is called the instanton number. This is another exercise to prove this formula. You know that the formula doesn't actually contain this number, so I don't need to tell you to prove what this number is to do the exercise. I will tell you, however, that this is some number that comes from representation theory. If you do this for SUn, let's do it for SU2 instead of the most important case. For SU2, this number is just 2. For other ones, you can look at the paper of Atiyah Hitchin and Singer from 1978. And they will have a table with these numbers. Any questions? Am I stopping now? Yes, yes, yes, only depends on the principle. Yeah? Very good. OK, so I repeat that, because I would have said exactly the same thing next. So from this formula, we see that we can write, this was your question, so all of the credit goes to you. The Young Mills Functional, you can decompose into two parts. One of them is just the square of the positive part of the curvature, and then a topological number. That depends on the principle bundle and the structure group. In particular, this tells you that the Young Mills Functional is always bounded below by whatever this number is, because this is whatever it is. The connection has no influence on this. If the x is close, and this is non-negative. So you can make this as small as possible by making a anti-self dual. So the lesson you learn from this is that they don't just satisfy, ace the incidence, don't just satisfy the Young Mills equation, they're absolute minima for the equation. You haven't just found the critical point of this function, it's really the bottom. You can't go any deeper. So a fancy way to say is that they saturate the topological lower bound. This is what it is. OK. Very good. So they didn't just observe that this is true. They also found an explicit solution of this equation. Yes, that's right. So your question or your statement was that the number k a, in general, it depends on a. But under reasonable assumptions, like x being closed, it only depends on the bundle. Yeah, henceforth, I will sometimes write this if the space is closed. And one time more, I will write this, because we will be on a non-compact space. OK. Good. I mean, this is already good observation. But in hindsight, it's probably obvious. If you, even obvious observations have to be made somehow by someone at some point. OK, so but they also found the solution. This thing is called the BPST Instanton. And this thing is called Instanton number, because it tells you how many instantons can be in your bundle. What would be obvious? I mean, so if you think about connections and principle bundles and so on, and you know that you have this equation, and you know. So this is an equation that you actually have to solve. But there's another equation which looks very similar, which is always true. This is this Bianchi identity. So if you meditate over this for long enough, I guess you should be able to discover that if you can somehow relate star of FA to FA itself, that would be a good idea. And it would help you to solve this equation. But someone had to observe this first. And I guess it was these guys.  So if someone tells you something is true, once you digest it, it's hard to understand why it once was complicated. So they also found a solution. And the solution they found on R4. And to write the solution, it's very convenient to identify R4 with the quaternions.  But in case you don't, I would just briefly say that it's a normed division algebra. Similar to the complex numbers, but doesn't just have i. It also has j and k. And they satisfy the relations that i squared is equal to j squared is equal to k squared is equal to minus 1. And this is also the same as i times j times k. So these are the quaternions. And the quaternions naturally relate to a special league group, namely the league group SP1, the same as SU2. This is just the unit elements of the quaternions. The quaternions have a norm. This is the basis. And you take this basis to be orthonormal. And then the condition is just p is equal to 1. This is SUP. This is a group. And there's a very easy formula for them. So let me say, this is the league group. The league algebra, therefore, little SP1 is just the imaginary part of the quaternions. And the solution is given by following formula. It's the imaginary part of q. Sorry, I should say something before. So q is sort of the identity function on the quaternions. Every quaternion, if I'm somewhere on the quaternions, I can say the point where I'm at is a point in here. And it's just the identity. So you can take the function q. You can multiply it by dq conjugate. It's just q conjugate by flipping the signs on i, j, and k. This is some one form that takes values in the quaternions. It should take values in this. So we take the imaginary part. And then this is not an instanton, but it's an instanton if you divide it by q squared plus 1. This is a one form on h with values on SP1. Because this is not quite what I told you connection is. But this is defined on h. There's only the trivial bundle. The trivial bundle has a canonical connection. It's just product connection. And if you add this one form to the product connection, you get the BPST instanton. And on trivial bundles, one usually identifies the connections with just the one forms for this reason. So it's a good exercise to check that this thing actually is anti-self-duel. And q, it just maps, it's just a coordinate function. It's the identity. It's a name for the identity function. Does that make sense? It's like when you write the coordinate on the complex plane, you just write z. It's just on quaternions, people write q. So it's also a good exercise, actually, to try to find if you know that the connection must be of this form times a function to find this function. It's a good exercise to do that. Yeah? Yeah? It does, because, I mean, OK, yeah, it doesn't, yeah. Oh, OK, OK, OK, OK, OK, OK. But Paul asked, the Yang-Mills function looks like it doesn't depend on the orientation. But this seems to say that something depends on the orientation. But I mean, when you define, so this is actually differential geometry question, when you define the integral of a function, you have to say how to define the integral, and you have to say what the volume form is. And if you flip the sign on both, it flips the sign on both. So it does, let me be, I'm not quite sure. Does this answer your question? OK, good. So there's a similar identity with, you can also put a minus here. There's a similar identity that goes on. You flip the sign here and here. Good, so it's a good exercise to check that this is actually. So there are two exercises. The first exercise is to understand, if you actually mean understand what it means to compute the Yang-Mills function with respect to the killing norm of the inner product that comes from the killing form, is to compute that the instanton number of this thing is 1. This is not, I mean, this is an integral from 0 to infinity. You can do that. And the second exercise is to check, so this is 1, 2, to check that f plus of i actually is 0. And for this, the hint, compute dq, which dq bar, vary explicitly and see the expressions that show up there. And from that, I think you can do it. OK, good. OK, so this is some physics. This, they do this in a two and a half page paper. And I guess if a normal person would have just read that, they would have said, OK, it's an interesting thing. But at here, I really realized that this is a deep observation about the connection between geometry and physics. And he probably instructed all of his graduate students to work on problems related to this, connect this to algebraic geometry, and so on. They wrote this famous paper with Hitchin and Singer that computes the index formula of the modular space of this. They computed with Hitchin and separately Manning and Dreenfeld computed all solutions of this equation on R4 when the structure group is SUN. This is the ADHM correspondence. And then, of course, in 82, Simon Donaldson realized that all of the modular spaces of solutions of this equation and their compactification that was explained how to do by Uhlenbeck before, you get invariance of four manifolds and so on. So I hope you all understand that this is a significant equation, profound impact in mathematics that probably cannot be. I mean, the impact of this simple observation that it had on all of mathematics I think is very hard to underestimate. Overestimate, sorry. So then probably all of you who have ever actually tried to follow the proof of anything in Donaldson theory, you've seen that this is really complicated. Start by the analysis is quite complicated. And then, is there any question? If there are no questions, then we will come to the second equation that comes to us by the way of physics. So a few years after the introduction of the Donaldson and variance, the polynomial invariance, Witten realized he wanted to understand what this means physically. And he realized that from a physicist's perspective, you can somehow understand this as a topological from the certain physical limit of the n equal to supersymmetric Young-Mills theory, whatever that means. This is some physics theory. And then together with Cybeck, they studied this theory in a different limit. And they observed that you can describe the same physical theory by a different gauge theory. In particular, so this is I think a development from 1994. 1994 is that Cybeck Witten sort of observed by some physics magic that the SU2 Donaldson theory. By this, I mean the polynomial invariance that we extract out of the modular spaces. If you don't know what that means, it's not going to be important. But these are some numbers, basically, that come out of the modular space. So this theory, you should be able to compute them in terms of a U1 gauge theory. So this is a physics conjecture. Butable by U1 gauge theory. And this U1 gauge theory, I'll explain what this is in a second. This is the Cybeck Witten theory. This is the theory that gives us the Cybeck Witten equation. And you would expect that to be much simpler. This is an Abelian group. If someone says U1 gauge theory, expect it can't be much more complicated than Hodge theory. Maybe it is, in fact, much more complicated than Hodge theory, as you will see in a second. And it's much richer as well. I've heard that mathematicians have written down the same equations, but didn't realize the significance of that without this. But they've never actually seen someone's old notes where this was written down. Anyway, so this is the so-called Cybeck Witten theory. So in mathematics, this is called Cybeck Witten theory. In physics, they have a Cybeck Witten theory, which doesn't, where these equations don't show up. So what is this? OK. For the setup, this is a little different from the setup that we studied for the ASD instantons. I mean, it's somehow slightly more complicated. But at the end, it's simpler. So here's how the setup works. So if someone gives you Riemannian manifold, you can always construct the frame bundle. I'm sorry. Bear with me, OK? This is going to explain this to you in a way that maybe it's not the way that you usually see, because we will have to do something like this later on. Something called the frame bundle. So I denote this of the tangent bundle with the metric. I denote this by FR. So what this is, is simply at every point over x, we choose concrete, orthonormal basis, positive orthonormal basis, in fact, of the tangent space. If you have a different orthonormal basis, they're related by a transformation. In this case of SO4, so this is an SO4 principle bundle, simple example. I guess if you learn differential geometry from Czern's book, he tells you to work on this space throughout, never work on the manifold anyway. OK, so this is a natural thing that always comes with any Riemannian manifold. And then when you want to put structure on a manifold, it means changing this bundle to something different. So if you want to put a complex structure on it, it means that you want to change this group from SO4 to U2 and something. So you would like to modify this group to a different group. So if you have studied some spin geometry, then you know that the spin structure is basically changing this group from SO4 to spin 4. What you need to do for the cyber-written theory is to change this to spin C. What we need to do is we pick a spin C structure. So I will now finally tell you what a spin C structure is. I will also later tell you what the Dirac operator is. Maybe today, OK, pick a spin C structure. So what does this mean? I mean, this is something that's not much more complicated than this. This is spin C4. I will tell you what spin C4 is in a second if you don't know what this is. You pick a spin C4 principle bundle. This is supposed to be a Muffrack W. I'm sorry. Can't do much better than this over X. So this is not a spin C structure. This is just a spin C4 principle bundle together with. And it's important to distinguish this also for spin structures. Sometimes people get confused about this and then together the map from W to the frame bundle. And we demand that this W is equivariant under the action of spin C variant. We're not going to tell you what spin C is. The spin C4 has a very explicit description. So spin C in general is spin N. In our case, it's going to be spin 4 times U1. So spin N is the double cover of SON. N is sufficiently large, so 4 is fine. And it contains an element minus 1. So if you quotient by the element, you map to SO4. And of course, U1 also has an element called minus 1. So you just quotient by that. So and then this acts on SO4 because I can just forget this factor and map down to SO4. Call this. I don't think we need the name for this. So in general, whenever you have a homomorphism from one Lie group to another, there's a notion of a reduction of structure group of a set principle bundle along this homomorphism. Spin C structures are one example of this. Any geometric structure is basically an example of this that you usually use. OK, so you fix this data. And then you get a lot of structure out of this. Out of this comes the following. So out of this come two bundles, W plus and minus. They're the complex. So these are the, I denote this by W plus. I think it's what Francesco called S plus and S minus. These are the complex spinner bundles, they're positive or negative. There's a Clifford multiplication with some map from TX. You can use this either to make the elements of this are called spinners or sections of this generally are called spinners. You can make a positive spinner into a negative spinner by multiplying with a tangent vector. And you can do the same way the other way around. Thing is called the Clifford multiplication. This is sort of the globalized version of these matrices that Francesco introduced in his very first talk that he was writing over there on the little board. What else? The line this. So let me maybe say right away that sort of important thing that you have to do to write down the usual type of equation is you can see that you can extend this to the exterior algebra. So if you have two vectors, you just multiply them in order. So we only need the extension to wedge two. And then you observe that since they flip the chirality or whatever design you want to call, when you do this with a two form, it doesn't anymore. So you can, for example, make a positive spinner, take a positive spinner, make a new one out of this by multiplying it with a two form. And then I have not defined what these are, but I will tell you what they are maybe later or tomorrow. You can check that when you do this with a self dual two form, on these bundles you actually get zero. So this vanishes when you do it on wedge minus. Let's also make this on covectors. This is these bundles. They have metrics, so they're isomorphic to each other. If you do this on negative two forms, you get zero. And in fact, so this is an injection. You can prove this as well. And it identifies the subspace of self dual two forms with a special subspace in here. So you can check, in fact, you can check that induced map on self dual two forms. Doesn't just go to arbitrary endomorphisms, but it goes to skewer joint endomorphisms, which are trace-free. So I denote this by little su of w plus. So this means skewer joint, trace-free. This is an isomorphism. But this is going to be important because we will be able to write down an element of this and we want an element of this. So any questions? Yeah? Oh, yeah. Thank you. The question was how I go from gamma defined on tx to gamma defined on wedge two of tx star. OK, yeah. So when I have an element of this, I can decompose this into wedge w. I can make sure that they are orthogonal to each other. And it doesn't matter if I think of them as vectors or covectors because they have a metric that identifies them. So let's not bother with this. So then I make this thing act on the spinner. So the spinners are usually called phi or something fancy to make everyone scared. And then this is just defined as gamma of v first and then gamma of w second acting on phi. If you assume that they're already perpendicular to each other, if you don't do this, then you anti-symmetrize this. Oh, this is an element of this w plus. This phi is an element of w plus. And I do this. So I assume already that v and w are perpendicular. Otherwise, this is wrong. This is why Paul got scared. OK, good. OK, so we're now at page three of 15 pages I wrote for the first lecture. So OK, well, let's try to press on anyway. OK, so more structure comes out of this than this. OK, I told you what the spinners are. OK, so now we need to differentiate these guys. There was another question. Sorry, can you repeat the question? I don't think I understood. Ah, OK, OK, OK. So this is a good question. The question is, what is the structure of these bundles? OK, good. So these are complex vector bundles. They have a Hermitian metric, and they have rank two. Hermitian of rank two. That's right. That's right. OK, so the question is about the plus minus. There's some larger bundle called the full spinner bundle, and that has a natural operation on it with eigenvalues plus and minus one. And this is the plus eigenspace. I mean, one of the pluss, the other one is the minus eigenspace. But tomorrow, I will tell you more concretely what they are in terms of really elementary stuff, right? Just quaternions, multiplying, quaternions, and so on. You can also axiomatically just define a spin structure as something like this with some properties, but that doesn't generalize very well. Any more questions? The volume form gives the opposite splitting. So this confused me until the spring. Well, it didn't confuse me until the spring. I just didn't bother with it until the spring, when I had to explain why there's a difference. I will explain this to you tomorrow why there's a difference. Volume form gives you the opposite splitting. It's absolutely bizarre. Anyway, so it's better not to worry about these things and just, there are also choices that you think don't actually exist that you have to make. But if you are interested in this, I can tell you hours about this later, maybe not today in the lecture. Okay, so we have this. Okay, so now we also want to differentiate things because ultimately we want to define some Dirac operator. So how do you differentiate things? You need a connection, and you don't need an arbitrary connection on this. We want the one that induces Levi-Civita on this. They have a name that calls spin connections. So spin connections, W, so just connections. Let's give them a name A on W, which induce Levi-Civita on the frame bundle. Or if you want to think of this in terms of the induced connection on the spinner bundle, it just means that this operation here is parallel. It's the same if you have seen that before. Okay, so the space of these guys, I want to denote as connections on W that induce Levi-Civita on the frame bundle, so I put comma Levi-Civita. This is absolutely not the usual notation, but later we will need some notation of this nature. Okay, so now I can write the equation, so finally. So what's the Cyberg-Witton equation? This is an equation for connection, a spin connection, and one of those positive spinners, and it has two components. The first component says that you want the Dirac operator defined by A and gamma applied to phi to be zero. So now we'll tell you what this means. So you pick some point, choose your favorite orthonormal frame. Some that I call the frame E1 up to E4 and differentiate using the connection A and the direction of EA. Spinner, this gives you a new spinner, at least defined locally where this frame is defined, and then you just multiply one of those vectors. And then this comes out to be a negative chirality spinner field, and I want this to be zero. This is the definition of the Dirac operator. Should remind you of the formulas that Francesca wrote. You differentiate, you multiply it by these matrices. It's just this. So the second equation is a little more complicated. It's of the form, so. It's the save to a curvature of something, but it's not, so this connection here, the structure, if you take the curvature of this, the curvature takes values in some huge bundle whose fibers look like the Lie algebra of spin C4, but we don't want all of that, right? At the beginning I said it should be an abelian gauge theory. So a convenient way to say this is that you look at the connection of the determinant bundle, so this is the connection on the exterior, the second exterior power of W plus, for example. This is the same, by the way. I mean, now you cannot solve this exercise, but if you know what this is, you can solve this exercise. See that these actually are the same. So you want this to be equal to something that comes out of the spinner. Okay, so one thing you can do. So okay, here we see I can make a self dual two form if I have a skewer joint thing, right? Okay, I cannot really make a skewer joint thing, so well, I can, I can, so. But let's make first a self-adjoint thing. So you can take the spinner and you can form this quadratic form, then sort of you take the spinner, sort of takes inner product with the spinner, but with the spinner again. This is a self-adjoint endomorphism of W plus. It's not trace-free, you can make it trace-free by subtracting the trace identity. So now it's trace-free, it's not skewer joint, but I can make it skewer joint by multiplying it with I, but this actually takes values not just in wedge plus, it takes wedge plus times I because this is a connection on the unitary bundle. So I can just take gamma inverse of this directly and it's going to give me an element of this. And then for reasons, I want to put a one-half in front of this, okay. So this is the Cyberg-Witton equation and this looks complicated at first sight, but it turns out to be much easier to work with. Much easier to work with. And so now that I've defined this equation and maybe let me say that immediately after this was introduced in Witton's 1984, 1994 paper, people reproved basically everything they used to be able to prove. Donaldson's theory, the Cyberg-Witton theory, many new things were approved with this and it's technically just a much easier theory to work with. And there are two big open conjectures left from these early days that one of them is a simple type conjecture. We're not saying anything about that. And the other one is the conjecture by Witton that, so out of this you can construct invariance and you can put them in some complicated generating series and you can do the same thing with Donaldson variance and there's coordinate transformation as opposed to do, that makes them the same. This is Witton's conjecture and there have been approaches to, towards this and a lot of work, source of initial suggestion by Pitzley-Gadzian Turin and work by Fian and Linness but as far as I know it's not completely settled yet even though people have worked on this for 28 years. Anyway, so this is the Cyberg-Witton equation, okay. So far this is all supposed to be classical stuff and I think anyone who, yeah, there's a question. The definition of a spin connection, okay. So can I repeat the definition of a spin connection? I can repeat the definition of a spin connection. It's a connection on this principle bundle W with the property that the induced connection on the frame bundle agrees with Levi-Civita. There's another question? Yeah, it's, that's right. So what you, the question or statement was if I just think about this as pushing down the horizontal distributions, yes, that's one way of doing it. But there are like a million ways to think about this. This is one of them that I think geometrically is clear. Okay, so, so it turns out that there are many more equations that physicists have proposed all of which have some interesting, fascinating connections with gauge theory. I mean, they come from quantum gauge theories that are not as well explored as ASD and Cyberg-Witton and so one of them is the Waffa-Witton equation. That was actually discovered, I think, just a few months before the Cyberg-Witton equation. I think the paper just predates the Cyberg-Witton paper slightly, but who knows, I think Witton was just generally working on sort of ideas around this. And there's a second equation called the Kapustin-Witton equation which sort of has fascinating connections which we will hear more about on Friday to the Jones polynomial. It was originally discovered because Witton wanted to understand the Geometric Langlands program, sort of very, very complicated conjecture in algebraic geometry. Comes out of a more complicated conjecture in number theory and anyway. So I think I cannot reasonably say anything about the Kapustin-Witton in the remaining 22 minutes, but let me say something about Waffa-Witton. Okay, so in physics there's some conjecture that goes back to the work of Monton and Olive. It's called electromagnetic duality. It's some conjecture. So I think for Maxwell's equations, we have all seen that you can exchange the electric in the magnetic field if you flip the sign on one of them, the equation and the vacuum are still satisfied, some sort of simple symmetry. But they observe that this should be true in sort of a much more general framework. And this is really, even in physics, this is really a conjecture, right? This is not something that they know how to argue for rigorously, physically as far as I understand. And let me try to say what it says. So this is something that's supposed to happen for a large class of gauge theories. And apparently the best gauge theory, the best chance of this being true comes from N equal four, whatever this means, super symmetric youngest theory, right? This is supposed to have, this is a physical theory. And in this context, it's called S duality. So let me explain what this means. So this physical theory here, this depends on two parameters. The first parameter is what there's a general parameter, which is the structure group, right? This tells you what kind of a gauge theory this is. And then there are two constants. One of them is the coupling constant, which is called E, coupling constant. And then there's another parameter, which is called theta. This is supposed to be the angle. And it turns out that this coupling constant, this is always positive. They actually like to normalize this in some sort of bizarre way where the way that the coupling constant actually appears is four pi squared over E, sorry, four pi over E squared. Don't ask me why they do this, but that's what they do. So, and then the conjecture says, so S duality conjecture says that the theory is invariant under doing the following. So you don't do anything to the angle, but you change the coupling constant. You invert that thing. You invert this to one over four pi, okay? So this is easy. We can all invert this. There's a complicated thing that you also have to do. You have to replace this group by its dual group. This goes to the Langlands dual group. Let me not explain what this means in any detail. But this is an operation that you can do to a compact leak group. If you do it twice, you get the original group back. If you do it to SU2, you get SO3. In general, to understand this, you have to understand how to associate to a group. It's root datum. If you know what that means, you will have observed that there are two things that you can flip around. If you flip them around, you get a new group. This is the Langlands dual group. Yes? So the question is what this means. This is physics talk. I'm not sure what this means precisely. I have not spent an extraordinary amount to figure out what this means. No, no, here you don't square. As far, oh, sorry, no, no, no, sorry, yeah. Okay, no, no, you also square this. Okay, thank you. I don't know. I mean, we will combine this. Yeah, I'm sorry, sorry. Definitely the square has to be there. Otherwise, it's not true. Otherwise, it doesn't give the right thing. Okay, so, okay. So this is a very old conjecture in physics. It's a very important conjecture, sort of, I think this was one of the first sort of deep dualities they conjectured in physics, sort of remarkable. Actually, also cyber-written, sort of one part of cyber-written argument, also seems to use this electric magnetic duality. Okay, so in 1994, also, Waffa and Witten, they tried to test this conjecture. Okay, so this is extremely difficult to test because even physicists don't know how to compute what this means. So then physicists have this thing that they can do called a topological twist, and then you can do some magic, and then if you're lucky at the end of this, you get some PE, and then you can imitate what people usually do, right, when you do Donaldson theory or cyber-written theory, and the equation they arrive at is the following. I promise it's much more simpler than this, right? So I can write it down probably in two minutes. So this is the Waffa-Witten equation now. So this is, again, the setup is like for ASD. You have a principle bundle P, you take a connection on this. They try to engineer a particular structure of this equation, which you will see. It's like doubling up the ASD equation, or it's a little bit like taking the cotangent bundle of the ASD equation. So there's a second field called B. This is a two-form, sorry, this is a safe dual two-form on X with values in the I've got a bundle. And then there's another field, which is the most often is zero, but let me write it anyway. This is just a section of the adjoint bundle over X. So then you write down something that looks a little bit like this, okay? So if you wanted to guess this equation, it would not be too difficult to guess this. So a kind of Dirac equation that you can write in this setting, where we know how to differentiate this, gets us a one-form. We know how to differentiate this. We can get a three-form or one-form if we take the dual, so let's do that. Take dA star B plus dAC. If you just had to guess some reasonable equation, I would write this down. And then, well, FA plus, right, that worked well for ASD, so let's do that again. A little bit suspicious about the sign that I wrote here, I think that's wrong, I think it goes on the other side. Okay, so FA plus is supposed to be some quadratic form in B and C. So something easy we can do is we can take the commutator between C and B on the adjoint part that gives us a two-form, so let's do that. So, but this somehow, this doesn't seem like it's enough. So there's actually something you can do to these two-forms to make another two-form. We may be in the interest of time, not tell you what this is, but there's a natural bilinear form that you have on this that basically says, this basically comes about as follows. You think of these, of a self-dual two-form, you think of this as an endomorphism of the tangent bundle, two endomorphisms you can compose, and this is basically this composition. So, yeah, good. So this is the Waffa-Witton equation, and then you can be, you can sort of throw all reason away and say, well, this looks like the cyber-Witton or Donaldson variance, surely there must be four manifold invariants that come out of this, right? So then there's a wild guess. There's four manifold invariants, which are just defined by counting the modular space of these Waffa-Witton solutions on P. There's a modular space of solutions of these guys, you count them, and they're supposed to give you integers, and this is supposed to give you invariants that are called the Waffa-Witton invariants. I mean, none of this makes sense. The way I just said this, because we all know that to actually do this you need to do some actual work, you need to compactify the space for cyber-Witton, it's no problem because it's automatically compact, to do some transversality that's maybe not so hard, and then you can do this. But if this really is an invariant, it really depends on what your compactification is, it's not, I mean, some actual thought has to go into this. But let's sort of not worry about whether, yeah? No, so if the Waffa-Witton equations are not related to the geometric Langland's conjecture, there's a different one that is, so the appearance of this Langland's dual is very mysterious, and I think also Attia observed that, so there was a paper before in physics that sort of explained that the magnetic charge should correspond to roots, the electric charge to core roots, or the other way around, and you should swap them and that corresponds to Langland's duality for the groups, and Attia realized that this is, mathematicians already know what this is. Anyway, yeah, no, this equation apparently cannot explain geometric Langland's, or physically explain, but there's another one that, that, I mean, whether it can explain geometric Langland's, or not, this is a different question, but you can sort of phrase the concepts in the language of physics. Interesting mathematical problems come out of this. Okay, so, good. So you're supposed to be able to do this. So in algebraic geometry, people know how to do this for a project of surfaces. They have some technique of doing this. This is the work of Tanaka and Thomas from 2017. I guess I think you will hear more about this next week in Götzsch's talk. In differential geometry, we really don't know how to do this. I will say maybe more about this tomorrow, why we don't know how to do this, but let's just press on and pretend that these numbers exist. So, what the wild guess is that out of this four manifolds, the question is what is the wild guess? The answer is the wild guess is that out of this four manifold variance arise. But how precisely, it's not so clear. I mean, this is a very, very successful topic in algebraic geometry now. Okay, so let me tell you, this itself would not be interesting. So these invariants are supposed to not give you new invariants. What is interesting is the connection with this s-duality over there. So let me say, let's pretend we know what these are, and then Waffa Witten say you are supposed to write down this thing. So you're supposed to do your first combine, you combine the coupling constant and the angle into one number. So the real part we take to be the angle and the coupling constant we take to be the imaginary part B squared. You observe that this gives you something in the upper half space. Imaginary part of z is positive. This is the upper half space. And that might already make you a little bit suspicious about what's supposed to come out of this. Okay, so now you define a function in the upper half space. But the function depends on all sorts of things. It depends on the manifold, but it also depends on this parameter and the group. So then you're supposed to define this function as follows. Me, sure, I'm not messing this up. So you take the sum of these Waffa Witten invariants whatever they are, and you somehow want to put this in a generating series. Okay, so you do this in a sort of more complicated way than you would maybe naively guess. And I'm pretty sure when you read the Waffa Witten paper you see that they first made the naive guess and then sort of fiddled around with it until it works. Because the outcome is also not what they expected, I think at the very beginning of this. So natural thing would be, I mean, if you see this upper half plane here, the natural thing would be to say to write e to the two pi i tau times the instanton number. The instanton number, this is why we find the instanton number before. It would be a natural guess, but it turns out this is not quite right. You have to shift this down by some number that depends on x, which I will give you in a second. So here, S of x is supposed to be the rank of your group plus one divided by 24 times the Euler characteristic of x, why this is true? There's no physical argument. They computed some cases and then this one works. But that's fine. The sum over all isomorphism classes of P's, thank you. So this is some function, you have the half space. It's not obvious if this can purchase anything, but it's something you can write. Oh, this is the center of the league group and you just count how many elements it has. Thank you. Thank you. The question was what this is. It's the center of the league group. Okay, so then the prediction of S duality. This is not actually the prediction of S duality, I would say, but it's what they say. So if this is really true, S duality, if this is true on the nose, you would say, okay, inverting this number means doing this transformation. You do this merbius transformation, minus one over tau. You can see it preserves this, but it flips the sign on this. Sorry, it inverts this. And then you change this to the length of the dual group. And you would guess if, I mean, if you say the physics is invariant, you would say it should be the same, but it's not the same, there's some factor. The factor is very interesting. It's plus minus tau over i times w of x over two, and then times this function. If you have not seen anything like this, it might seem bizarre to you. If you have seen similar things, this might sort of seem obvious that something like this should appear. Let me tell you what these numbers are, plus or minus one are given by minus one to the rank of g over four, times the order characteristic of x plus the signature of x, and then maybe looks familiar to you, and w of x is just minus the order characteristic, times x, okay. So this is a complicated conjecture, right, okay. So if this is true, some magic happens, right? I mean, if this cannot be true, just by accident. Now, we know that on the upper half plane, there's a bigger group that acts the modular group. SL2Z acts on the upper half plane, and SL2Z has two generators. One generator does this, this generator is called, this is S of tau, and the other generator is T. It just shifts tau by one. So if this, for some reason, is an integer, certainly it's invariant under this, right? If not, then it's still the shift by tau plus one and it's a predictable thing. Okay, so this is the S-tuality conjecture. This is a very complicated thing, yeah? Yeah? Oh, sorry, sorry. This is plus minus one, plus minus. This is the sign. This is the sign that shows up here. Yeah, sorry, it's not upside, yeah. Yeah, it's a weird thing to write, just plus minus is equal to. As sum of all, I take, for every isomorphism class of principle, I take one representative, and I sum of all of them. So for example, if this is SU2, I sum over all principle bundles. I take, for every second-gen class, I take one principle bundle, yeah? Oh, this is the upper half plane. Yeah, it's the usual action by Mibius transformations. Okay, and then this is supposed to remind you of modular forms, right? So this is like a modular form, but the space is not the upper half place. The space is sort of two copies of the upper half space. One is labeled by G, the other one is labeled by L. There's some tricks that one can do to get, ultimately get only a sort of a modular form, but maybe of some higher level. Out of this equation, if you want, I can tell you more about this later. The question is, is the coincidence that this is called S duality and this is S over there? I don't know, but I suspect not.  The physicists, you never know. I suspect that this is why it's called S duality. I mean, they usually also call this electric magnetic duality. So I have only one minute left, so let me say. My plan was also to introduce a framework that generally explains how to think about these. I will do this tomorrow, and then I will tell you that this is a reasonable framework to study all of these equations and similar equations and maybe give you some other reasons why these equations are interesting.  And even though it's hard, I will tell you some things that we can do and I will tell you what we can't do. And maybe on Friday, I will tell you why we can do the things we can do. Yeah, that's it. No, no, no, no, no, no, no. No, they should not be, they should not be, they're supposed to have the same information as even less information maybe than Donaldson variance. You can do this for arbitrary, the question was what kind of structure groups? You can do this in principle for arbitrary compact connected D groups. This is also important because otherwise it's a little bit boring if you only do it for one, right? I mean the really the fascinating thing is that the Langland's dual shows up here. So if you do it only for UN, the Langland's dual is itself. It's not so interesting when that's what algebraic geometry is to a care. Hi, this is a very good, do you want me to write this down? I'll show you later. So Waffa and Witten, okay, they write this down, right? And then they set out to actually compute this in some cases. And they compute it in particular for the K3 surface. I can show you the formula for the K3 surface later. And so they compute for the K3 surface for SU2 and SO3 and they check that this works. And it involves Dedekind eta functions and so on. It's, I mean, it's good evidence that something is going on. Yes, the question was if for that form of S duality G is simply less, and this is a good point. So in Waffa Witten original paper, they don't mention this, but it's right. If S, if G is not simply laced, apparently, you don't do this transformation, but you multiply with some number N that depends on the difference between the long and the short roots. Can be two or three. And apparently there's also some correction to this statement when the league group and the Langlands dual league group don't have the same league algebra. Two questions. The first question, if there's heuristic how to go between S cyber Britain and ASD, I mean, yes, there's this program by Pitstree Gutsch and Turin. There's some other equation that has some modular space, contains both of the modular spaces. You should get some co-bordisms. So this should give you some relation. The physical intuition, I cannot say anything about this. I mean, I have 15 pages of lecture notes for the first lecture. Maybe I cut it down to what I actually said and I put it on my web page. So maybe try to go to my web page later today. If you don't find it there, email me and I can send you something.


Hello and welcome to the session. In this session, we will discuss properties of binomial coefficients. Now, the binomial expansion of 1 plus x will raise to power n is equal to nC0 plus nC1 into x plus nC2 into x square plus so n up to nCR into x raise to power r plus so n up to plus nCn minus 1 into x raise to power n minus 1 plus nCn into x raise to power n where n is a positive integer. Here nC0, nC1, nC2, nCR and so on up to nCn are called the binomial coefficients. And when n is known, that is the index of the binomial is known, then we can write these binomial coefficients as C0, C1 and so on up to these binomial coefficients as C0, C1 and so on up to C. Let us discuss the properties of these binomial coefficients. And the first property is the coefficients from the beginning, the terms which are equidistant from the beginning and the end equal coefficients. Now, let us put this property. Now, in the expansion of 1 plus x whole raise to power n which is equal to nC0 plus nC1 into x plus nC2 into x square plus so into x raise to power r plus so n up to minus 1 into x raise to power n minus 1 plus nCn into x raise to power n. Now, the total term, so the adorn expression has n plus 1 terms on the beginning coefficient of r plus 1 th term from the beginning plus 1 th n plus 1 the whole minus r plus 1 the whole terms before it.  Therefore, the coefficient from the beginning of binomial coefficients of the binomial coefficients is equal to the expansion of 1 plus x whole raise to power n is equal to nC0 plus nC1 into x plus nC2 into x square raise to power n. Now, put 1 plus 1 that is 2 raise to power n is equal to dot plus C1 plus C2 plus Cn that is when n is the sum of the binomial coefficients is equal to 3 raise to power n. Now, as we know that C0 is equal to 1 and this equation is equal to 1 it will be 2 raise to power n is equal to 1 plus C1 plus C2 plus C3 plus C1 plus C3 plus Cn is equal to 2 raise. Now, let us discuss the next property which is in the binomial expansion of 1 plus x whole raise to power n the sum of the coefficients is equal to the coefficients even terms. Now, we have the binomial expansion of 1 plus x whole raise to power n is equal to nC0 plus nC1 into x plus nC2 into x square plus so on n into x raise to power n. Now, putting x is equal to 1 in this expansion we get 1 plus 1 that is 2 raise to power n is equal to nC0 plus nC1 plus nC2 plus so on the equation number 1 equal to minus 1 in the given expansion we get 0 is equal to nC0 minus nC1 plus nC2 minus nC3 plus so on raise to power n into nCn. Now, let this be equation number 2. Now, adding power n is equal to 2 into nC0 plus nC2 plus nC4 plus so on raise to power n is equal to 2 into C0 plus C2 plus C4 plus so on the whole.  But in the binomial expansion of 1 plus x whole raise to power n the sum of the coefficients is equal to the sum of the coefficients of the even terms n minus 1. Therefore, we can write that the sum of the coefficients of odd terms is equal to 2 raise to power n by 2 which is equal to 2 raise to power n minus 1. The coefficients of even terms is also 2 raise to power n by 2 which is equal to 2 raise to power n minus 1. Now let us discuss the next property which is summation minus 1 whole raise to power into nC r where r varies from 0 to n C0 minus C1 plus C2 minus C3 plus C4 minus C5 plus so on up to minus 1 whole raise to power n is equal to 0. We know the binomial expansion of 1 plus x whole raise to power n which is equal to C0 plus C2 x C3 x cube plus so on up to Cn into x raise to power n equal to minus 1 will be equal to 1 plus x whole raise to power n will be 0 which is equal to C0 minus C1 plus C2 minus C3 plus so on up to 1 whole raise to power n. Therefore, we have minus 1 whole raise to power r into nC r where r varies from 0 to n is equal to this property which is equal to n over r r the whole into r minus 2 is equal to now n factorial is equal to n into n minus 1 the whole over r factorial is equal to n factorial is equal to n minus 1 the whole minus r minus 1 the whole is equal to n over r the whole into n minus 1 factorial that is n minus 1 whole factorial minus 1 the whole minus r minus 1 the whole is complete whole factorial this is equal to n over r the whole into now this complete expression will be equal to n minus 1 C r minus 1 is equal to n over r the whole into n minus 1 C r minus 1 then is equal to minus 1 whole upon r minus 1 the whole into r the whole into 1 whole upon r minus 1 the whole into we have 1 so in this session we have learnt about properties of binomial coefficients so this completes our session but we don't have enjoyed the session.