 At seven, you know, is it dinner? No, it's just to get together at which drinks and food happen to be served. I mean, not served, you kind of serve yourself. It's buffet. And also, most of you have to pay for your stay and then get reimbursed, right? So don't run, try not to run away without paying. Yeah, okay, I cheated on all of you. Okay, so yesterday, the point of yesterday is lecture to make some propaganda for gauge theory and tell you that it's exciting for a reason that you already know, because you can use it to study for many faults, and for reasons that maybe you were not aware with, often that sort of through connections to physics, there are some conjectural gauge theories that should give us modular forms and that should be interesting. And I actually did discuss this and I only mentioned this, but it's discussed in the lecture notes a little bit and you will learn more about this in tomorrow's talk by Alfred Holmes and their references in the lecture notes about the Kapustin-Witton equation, which has connections with the Geometric Languages Program, the Jones polynomial, some five-dimensional version of this also has connections with Hovanoff homology. But you will hear about that tomorrow, okay, so. And there are many other gauge theories that you can be interested in. And even I talked to some of you after the talk yesterday and you did not agree that they all have sort of a similar shape, I claim that they all sort of come from the same kind of equation. So the curvature, the safety part of the curvature is equal to something that comes from the spinner. Here there's no spinner, so it must be zero. Here it's some quadratic form in the spinner and the spinner itself satisfies some Dirac equation. This also is a sort of Dirac equation and this is some quadratic form in the spinner. So at least from this perspective, some schematic similarity should be clear. Okay, so and today I want to present to you a framework that covers all of this and then I want to tell you what equations come out of this and then I will say a little bit about what we know about this, what the problems are and what one maybe can do with this. Okay, anyway, so as a warmup, today I want to recall hopefully some aspects of spin in dimension four. Okay, so it turns out that spin in dimension four is the same as SP1 times SP1 and the reason for this is that there's a natural map from spin four to SO4 and it's given from SP1 times SP1 to SO4, which we think of as SO of the quaternions and we must write down the map that a way of making two unit length quaternions act on the vector and one way to do it is to say we call the first one P plus, the second one P minus and we act with this on V by left multiplying with P minus and right multiplying with P plus conjugate. There are other conventions, they lead to slightly different things but it doesn't really matter. This is the convention I want to stick to for concreteness. This is the vector representation, right? This is a two to one cover. You can check if you've never done that. It's good exercise. If you already know this, then you know this and then there are two spinner representations. They are the obvious ones. So if you can project the first factor, so I just map P plus comma P minus to P plus and then multiply with P on the left hand side. Okay, so these are the spinner representations and then there's another representation in dimension three that just comes by taking the adjoint of one of those. I call this one the vector representation. They're called delta, the spinner representation, I call sigma plus minus and this adjoint version of the spinner representation I call delta plus minus because we need to give some name to this. Okay, so this is hopefully, you've seen this before. Let me erase this, already started. Okay, so and then you can check that with, if you write it in this way, in this way, you can check that add, the act with add plus, add of P plus P minus on V, yeah? Yes, yes. But that should be a map. Yes, there's also a map. There's also a map this way. Everything corrects, right? And I will tell you what one of those maps is in a second. So this one here is obvious. This one is not totally obvious to get from there to there but I will show you where it comes from. Okay, so and the reason to introduce this is that if you left multiply V, so you want this Clifford multiplication, right? You want to be able to multiply V with V on the left-hand side and it's supposed to map something that has a plus representation and something with a minus representation. This is how this is engineered. So if you do this, you do this, then you can do, I mean this might be obvious to you or you have to do a little computation but I guess it's obvious. So it means that this multiplication operation here is equivariant if we act on the vectors with the add representation and spinners with sigma plus minus. Okay, so this is very good. But then there's another thing that happens that I also sort of alluded to yesterday and it comes from, there's another map that you can see that goes from the imaginary quaternions to wedge plus say this is the one we need of H. So I give you an imaginary quaternion and you give me an imaginary quaternion and I make for you self dual two form on H. And the formula for this is for concreteness, let's say, I call this one here Q. Actually, I don't want to call this Q, call this C because I'm already using Q. And yesterday I wrote down the special form, this DP, sorry, DQ, which DQ par. This is a self dual two form that takes values in imaginary quaternions. I'm just taking the inner product on the imaginary quaternions with that. And you can check that if you make delta plus act on this, and there's an induced action of add on this, this is equivariant, this is delta plus and add equivariant. And there's something similar for minus that you have to put the bar over here. Okay, good. This is some linear algebra of spin four. But the consequence of this is that if S was to be a moth track S, if this is a spin structure, favorite oriented four manifold, remaining four manifold, then you're positive. This is actually a negative spinner bundle, but let's do this. The negative spinner bundle corresponds to the adjoint vector bundle that comes from the sigma plus representation, then positive one comes from the sigma minus representation and the tangent bundle of your manifold itself just comes from the vector representation. So these representation give you all of the bundles and you can also see that the wedge plus T star X, the bundle of self dual two forms comes from this representation delta plus on the imaginary. And then you get all of the structure comes out of this. The spin structure automatically has a connection. There's no choice in lifting it. There's this two to one cover and then this identity here gives you the Clifford multiplication from TX to say homomorphisms from S plus sorry S minus in this convention S plus and then you have a Dirac operator and so on. So this is the usual. This is sort of what's going on in spin geometry in dimension four. This is an explicit description of the spin representations without going through Clifford algebra so anything like this. Everyone happy with this? Yeah? Which one, this one? Sorry, I said again. Okay, so the question is with respect to what action this map is equivalent, is that right? So this group here acts on this by a delta plus and delta minus. I'm taking the delta plus action on here. And this group also acts on H and therefore it has an associated action on wedge plus of H star and this is the action I'm taking on here. Any other questions? Yeah? Aha, these, okay. So V, yeah. So in all of this, P plus P minus some element of SP one times SP one. V is an element of H but I think of this as a vector because it's supposed to sort of transform in the vector representation and phi is also an element of H but I think of this as a spinner because it transforms in the spinner representation. Sorry? Now this map, so okay, the question is how does this give this map? When you have map, equivalent maps between representations, they give maps between the vector bundles associated to these representations. Is that sufficient answer? Do you want me to write down an explicit formula? If you try to just write down an explicit formula, the obvious thing is the answer to this. Equivalence classes of a pair in the thing, of a pair consisting of one thing in here and one thing in here and just make it act on this in the obvious way and equivalence means everything is well defined. Okay, good. So okay, so this is, I mean, this is supposed to be classical stuff that hopefully you know and now we want to make some cyber-written equations knowing this. Okay, so we're going to make a group that's more complicated than this. It's going to act on some stuff. The stuff somehow has some geometry. It has some representation theory and some geometry. This is a cyber-written equation. It comes from this. And all of the equations that I just wrote, erased, or just erased, come from this and many other things. Many, many other things. And I forgot to say at the very beginning why we do all of this. So one reason to do all of this is that you already see there's a whole host of these equations and you don't want to write a hundred page compactness paper on all of them. You just want to write a slightly longer one on one of them. Okay, so this makes your life easier. Then the other thing is that we will see that actually the geometry of this will give us more insight into what is actually going on. And the reason I personally care about this is that I care about some problem that actually involves a whole infinite dimension, infinite family of these guys. And if you want to know more about this, ask me later. But for now let's just press on and write some generalized Zeppelin equations. So there's some input into this construction and it consists of algebraic data and geometric data. So, I mean, what's the algebraic data? There's three pieces of data. One is sort of easy, it's just a structure group. The compact, connected, semi-simple if you want. Doesn't have to be a group. Structure group. So this is the gauge theory will really be a gauge theory for this structure group. If you only have this, then your spinners are always actual spinners and very often you don't want that. You want to be able to twist the situation a little bit. And for this, you need some auxiliary group. I just call it K. It's just some extraly group. I've previously called this a flavor group, but I've also been told that this is really not a good name. So, auxiliary space, I guess. And then we want the representation. So we take the product of these league groups and the representation of this on some quaternionic vector space. This is the quaternionic representation. Representation. So, as there's some vector space, it has an inner product, it has a quaternionic left multiplication, I, J, and K act by, they preserve the metric. And then I can look at all linear transformations of S that preserve the metric and commute with I, J, and K. This is called SP of S. This is a symplectic group or sometimes unitary symplectic group. And I just want the representation of this. This is really just, it's a matrix representation really. And I want the property on this. I also want an element, epsilon, supposed to be in the center of this group, such that rho of epsilon acts as minus one. So, if this element is absent in your group, you can always add this back in. But usually there's one that's already given. Oh, and I want that this squares to, that this squares to the identity. So, I want some square root of the identity that acts as minus one in, and now we'll use this to generalize this setting here. So, at first we have to generalize what the quaternions mean. We try to replace them by S. We define some Clifford multiplications and so on. Then afterwards I'll tell you what the groups are. We fix one of those data sets, right? And yeah, if the center is trivial, then it's not going to happen. Then you have to make the center larger. So, if your group really does not have an element that satisfies this, you can take the product with plus minus one, and just force minus one to act like this. This is one way of doing it. But we will see that usually there is one. So now we define a Clifford multiplications. I'm going to define some maps now. So, what do we need? So, we define gamma. This is supposed to be the Clifford multiplication. I'll define a bunch of maps. In addition, three maps, no two maps first. And I also define this map called gamma underline, which is sort of a slightly more complicated version of this. So we multiply not just by a vector, but also by a Lie algebra element, but just of G, and we define them sort of in a simple way. I multiply a vector with a spinner by, well, this is a left-edge module, so I know how to multiply an element. So this is just to give a name to this operation of left multiplying the quaternion. And this, I have to tell you how to multiply a vector and the Lie algebra element. Well, the Lie algebra representation acts on S, so I make that act, and the vector also acts. And since it's an SP1, it doesn't matter if I multiply before or after, so I just, this is Lie of rho is the Lie algebra representation that comes from this Lie group representation, like C into this gives me endophobic morphism of S, and I multiply this, okay. So these are some maps. It turns out that there's another map sort of comes naturally out of this. So, you wouldn't actually like to take a quotient of this. And there's a theory of taking quotients of this, just a special case of taking hypercalic quotients, and if you remember, we have seen that before, you know that it's a moment map and there's a natural moment map that arises in this case, which we can write down the formula. Tinguished moment map, hypercalic moment map. I call this mu. It's going to take the role of this quadratic map that I wrote in all of these cyber-githnic and these equations over there. This thing eats a spinner, and then it's supposed to spit out an element in the dual space of the imaginary quaternions tends to the Lie group, the algebra. But all of these things have been a product, so you might as well not do a lie, this is sort of a matter of, you don't need to do this, but we will do it here for now. This is defined by the following formula. Okay, so what can you do? So you have this Lie algebra representation, which allows you to go, can go back with this map. So we can certainly produce an endomorphism of S. It's given by phi, phi star, and we can think of this as an endomorphism of S or in a dual of endomorphisms, that doesn't really matter. And then we take the adjoint map of this map to go back, technically speaking, this ends up being an H tensor G dual, but we only take the impart of that. I'm forgetting some stuff here, right? I'm restricting to, I'm not putting in any full quaternion, just an imaginary quaternion, this is okay. I have not used K in this. That was a question, I have not used K in this, it was a statement, and this is important, right? I have not used K in this. K plays no role in this. Okay, so this is the algebraic. These are the maps. We need something that replaces the groups. Write the groups. This might look complicated now, but we do some examples. This stuff becomes extremely concrete and fun to do some computations and so on. You will see it's not as complicated as it might initially seem. So now I define bin G times K of four is just bin four. There is a natural element called minus one in there. Times G times K, and then I can quotient. So I have an element that's called minus one in here. Also an element epsilon over there, I quotient by the action of this. This is called spin G times K. This thing has a natural map down to the same group, but where I only use K. And I think I gave this map a name. I called it pi. This is defined in just the same way. And then I can completely forget about this group and spin four still has a map to SO4. I call this map kappa. I'm sorry for all of this notation. There's just a lot of maps flying around that they all need a name. So, aha, okay. So this thing here is a component and G has one and K. I just take the component in K. That might go to one. Yeah, yeah. And in fact, yeah, yeah. There are cases where it's minus one here. It's one here and get away around so everything happens. But so this is where I have to say it is like this. It might go to one, that's right. So it's also, yeah, you will see it. There will be examples. Okay. So, and then this thing acts on the spinner bundle. So we also define some actions tau plus or minus of spin G times K four. This acts on, let's call it just acts on S. And the formula is the following. This defined by tau plus, tau minus. Let's call this one little K, this one little, this one little G, this one little K. Okay, this acts on phi in the following way. So I certainly know how G and K are going to act. They're going to act by the representation. This is going to act by the spinner representation and I can make this act because this has this module structure. So it's all, everything is fine. I can actually write this down. And then the other one is called row. Good. Okay, so not much more. Not much more. This is the last algebraic data. Yeah, I'm sorry. This is a setup, it's a little bit elaborate. But if you recall, the setup for the set of equation is not that much more complicated. But we'll get there and just do it enough. Okay, so you can see with this representation, the usual Clifford multiplication is still linear and so on, so we're in good shape. So now we can define some geometric data and then we write the equation. It's certainly easier to digest than the definition of an A infinity category. So it doesn't say much. Okay, good, geometric data. I mean, it's easier to come up with as well. Okay, so geometric data. Okay, so we need, as you see, we have these two groups. So the auxiliary group is really, this is supposed to be there to make, put some extra structure on the manifold. That's right. There's the auxiliary group, the purpose of this is to put some structure on the manifold and this can be all sorts of things, right? It can be sort of an auxiliary vector bundle that you want to twist by. It can be a complex structure, whatever, an almost complex structure, whatever you want. Okay, so we need, we fix the geometric data. That's only two things. It's a reduction. I call this Fr star. This is supposed to be a slightly larger frame bundle of the frame bundle and it's a reduction along this map kappa. So just put some extra structure. It puts some extra structure on. And you see if k, for example, is a trivial group, this is just S of four and just means doing nothing, right? And there will be some equations where we basically do nothing. And then the second thing is a reduction called this S of this reduction along pi. So this is the spin structure, right? This is the thing we would choose in sub-eviton theory. Okay, so we choose a set of those things. Okay, so now I want to define, I want to finally get to being able to write the equations. So out of this, you construct the positive and the negative spinner bundle. You construct a relative adjoint bundle, denote by S slash frame star. The action of this structure group on the Lie algebra and that's it. Okay, good, okay, so I'll write the equation over there in just a second. Okay, so what do we need as input for our equation? We need to write the Clifford multiplication. There's obviously going to be some Clifford multiplication. So we get some Clifford multiplication, gamma. What else do we get? So this, let me just write this. This is going to be from Tx to homomorphisms S plus to S minus. We get an analog of this map mu, takes a spinner in S plus and spits out two-form values in this relative adjoint bundle. And then the last piece of notation is that if I have a connection on S, that induces the given connection on this frame bundle, the extended frame bundle, when I only want to consider the curvature with values in here, I denote this by F A slash F R star. So just remove, it just removes a certain component. It just takes values in this. This is a two-form of values in this bundle. So finally, if you have chosen all of this stuff, you get a Dirac operator out of A and this and we can define the generalized algorithm equation. Sorry, this is the definition of the project. So the question is, what the hell does this mean? This is the projection of the usual curvature. Sorry, this, yeah. The question was what this ad S slash F R star means. It's defined over here. It's this group here acts on the little algebra by the adjoint representation and constructed from this. There's another question. Yeah, this is the linear algebra. The question is there are two gammas, are they the same? This is the linear algebra map that underlies this and this gamma underline here, we need it to write this formula. Okay, the generalized sub-barbiton equation. Finally, for the setup is an equation for a connection on this S that induces a given connection on this extended frame bundle and a spinner, so a section of S plus. And we demand that the spinner satisfies the Dirac equation and that the curvature of this connection when I project it onto the sub-bundle, take the self to a part of this is equal to mu of F. Okay, so this is called the generalized sub-barbiton equation. Okay, there we have it, okay. And it looks like all of the other ones, sort of. Okay, good, okay. So this is a lot to digest and so maybe we'll do some examples to see that recovers what we get, what we already have. Yeah, there's more. Yeah, it makes the bundle larger but it's the language. Sorry, the question is by reduction I mean lift. Yeah, in this case it makes the bundle larger but nevertheless the language in differential geometry for this is still called a reduction. Ah, okay, okay. We'll do some examples and we see what's going on. I think it will be clearer. We'll do some examples and see what's going on. So let's start with an example that we didn't cover so far. Okay, so this is examples. I have a whole bunch of examples in the lecture notes and we'll not discuss all of them. I will discuss just a few. So what's the simplest thing you can do? Okay, so you can make these groups as small as possible. This is example one. Let's take this to be trivial gauge theory. Okay, this is not a good gauge theory example. Okay, you can't do just this, right? You need to at least have an element plus minus one in the group so we'll force it to be in by saying that K is plus minus one. Okay, sorry, yeah. Yeah, okay, okay. Four, phi being a section of S plus. Yes, that's okay. And A is a connection on S but I demand that it induces a certain fixed connection on this extended frame bundle. Yeah, sure. D, A, phi, find this as follows. You can take this connection on the spinner bundle. It's induced by this connection A. I differentiate in the direction of some vector field in the frame and then do the Clifford multiplication. Some of all of these things, A goes from one to four. And here, E one up to E four is some local of a normal frame. Same definition that I used yesterday. Okay, so this is the simplest thing you can do, right? The epsilon is uniquely determined by this. Epsilon has to be one minus one. And the representation is also uniquely determined because we know, I mean, this must be a quaternary vector space. Let's just take H, the obvious one. This thing must act as minus one, this element. It's the only non-trivial element in the group G times K and this F R star I don't have to choose. Sorry, F R star is a spin structure. Choice of spin structure. S I don't have to choose because this group is trivial. S is this. And the equation is an equation on some connection with trivial structure groups. So this equation disappears and this just becomes the equation for harmonic spinners. So it's a Beguin equation. In this case, it's just a Dirac equation for a spin bundle. So this is one of the simplest cases that you can write down. Is everyone happy with this? This is really the simplest instance that you can write down because it's not even a gauge theory. Okay, so here's another. It's an exercise. The next one is an exercise. So you can also take K to be... They want to say K is going to be... K is the trivial group. My G is whatever I want it to be. It's an arbitrary group. Epsilon is the element one comma one. Okay, so if I do this, then there's only one representation that is allowed because this has to act as minus one. For minus one to be one, the representation has to be the representation on the trivial vector space. Okay, so this gives you A is the instantons. An example slash exercise to figure out why this is. Okay, so now let's go to the usual cyber-written equation. Where does the usual cyber-written equation come from? So for the usual cyber-written equation, we know that there's nothing that we have to choose to be able to define this. So this K must be equal to the trivial group. G, the structure of the gauge theory is U one. Epsilon is minus one one. The spin representation underline vector space has H, but we need to figure out how this acts on this, but there's only one way to make this act. That is reasonable, well, up to equivalence. So E to the I of three is simply given. So this is supposed to be quaternionic linear. So I must multiply on the right-hand side. And I mean this is a quaternion, so it's multiplied by that. And it's commutative, so don't need to conjugate. It's already fine. Okay, so you can do another exercise to figure out that this really recovers the cyber-written equation. The exercise is solved in the lecture notes. Classical cyber-written. How do you want to mess with it? Sorry, the question was how can you mess with Epsilon? I guess you can try. Oh, no, you cannot mess it with it so hard. So you get the wrong science of compactness and it becomes false, no, no, no, no. No. It's not part of this framework. No, yeah, yeah, yeah. This framework is made so that the equation is as good as possible. So you can't, you can't, right? If you write the usual cyber-written equation, there's an opportunity for you to mess up royally by just putting the wrong sign in the quadratic form. All right, and then really, I think also, okay, maybe I'll tell you later. Okay, I mean we all make mistakes. So yes, maybe do this as the next example then, okay. Yeah, so let me say one more thing. So you can play with this, for example. If you replace this by, for example, S-U-N, you get cyber-written with N spinners and so on. Maybe let me write this, let me write this and we do what I've written. Because I think tomorrow I want to talk about cyber-written with multiple spinners and I guess not today. We do the same with K is equal to, let's say for concreteness, S-U-N, this makes our life a little bit easier. Then this gives cyber-written N spinners. I think I will maybe say more about this tomorrow and write a more concrete equation than that. Okay, let's do what I've written. Yeah, let's do U-N for today, why not? It doesn't really, I mean you can, yeah, U-N is also fine. There's something you're doubling up. If you put, if you make this, have a U-N sitting inside it, you're sort of over-parameterizing because you have two U-Ns that act in the same way. It's sort of two U-1s that act in the same way. That's why it's a little bit better to not put that in there. Yeah, let's do Waffa Witten. Where does Waffa Witten come from? Yeah, it's a good exercise. If you have your favorite representation and it's already quaternionic, then go with that. If not, then quaternionify. So for Waffa Witten, this is going to be good for Waffa Witten. We take our favorite representation of G on its Lie algebra. Of course, this is not a quaternionic vector space, but we can make it one. Answering with H, okay. So this is now a quaternionic representation. This is called rho, all this rho. I will tell you what the K is and so on later. But let's stick with this first. So for this, you can do little exercise and compute the moment map. So moment map of this is... So you put an element of this. Let's write this as XC0 plus XC1I plus XC2J plus XC3K. And then you can do a little computation and you will see that this comes out to be the commutators of these elements. Namely, you get XC0, XC1 plus XC2, XC3. And set with I. Here, identify the Lie algebra and the imaginary quaternions all with their duos. This is the I component. Then here, I go through IJK and here I cyclically rotate these indices one, two, three. See that these are the same dual two forms. IJ plus XC0, XC3, it's a little computation. So this is going to be the C field. You see that this multiplies with this and this is going to be this cross product that showed up inside the Waffa-Witton equation. So now to tell you what K is and so on, how do we get K? And I will tell you how to get this FR star as well. K in our case is SP1. K in our case is SP1. And so now this group, SOK, sorry, spin 4K, spin K4, sorry. This is SP1 times SP1 times SP1 divided by plus minus 1. And our SO4, of course, sits inside here and we just map it inside in some obvious way. So SO4, SP1 times SP1 mod plus minus 1, map this in here. And this is going to define me in the frame bundle by saying that P comma Q maps to PQP. And this defines FR star. This, in fact, there's no, the structure is not really extended or reduced. It's just I'm embedding this inside here in some trivial way. There's really nothing going on. This is just to make the spinner bundle something else. So I still need to tell you what the choice of S is. The choice of S is equivalent to a G principal bundle over X. Just in mind. I need to think a little bit about what the spinner bundle is. And then what happens comes out of this. So what's the spinner bundle? What's S plus? So this is going to be the K. So S plus has the action of this. I hope I didn't mess this up to this the wrong way around. Yeah, yeah, this is right. So this is the first one I call plus. So we act with this. And with this, the spinner bundle, in this case, is therefore going to be the adjoint representation of SP1 on the full quaternions. It fixes the identity. So we get a trivial bundle plus the self-dual two forms on X. And then this whole thing, the spinner bundle also has this G. This is going to give us the adjoint bundle of P. Give this bundle a name. It's called this P over X. The spinner bundle is going to be this. Positive spinner bundle. The negative spinner bundle is going to be just the quaternion bundle and so at P. And then, so it has to do some computations to check that the maps that I wrote down yesterday and the maps that come from this actually are the same. But you can sort of see structurally that they're going to be the same. So if you do this with, I mean, A is the connection on this. This is OK. But the spinner field, we call the first component comes from this I call C. The other one I call B. If you use this notation, then this is the same as the Waffa-Witton equation that I wrote down yesterday. And then there are many more things one can do with this. These are just recovering the equations that we have. And we have only seven more minutes left. Maybe I'll tell you some other equations that you can. Are there any questions? Let me tell you some more exciting things one can do with this. So probably know that there are many infinite dimensional hyperkiller representations. So for example, you can do. I'm not going to make these examples very explicit. But so one thing you can do is you can take, this is these are infinite dimensional examples. You can take S to be just maps from your favorite interval. Let's just make this from 0 to 1. But you can change this to be other things in some Lie algebra. It's called as Lie algebra H. This is a Lie algebra of some group. Quaternionify, this is an infinite dimension. Quaternionify is based somewhat formally. This is acted upon by group G. Sorry? Yeah, H is some Lie group. H is a Lie group. Smurf frag H is the Lie algebra of H. This is the quaternionic representation. This is one parameter families of afferent equations. So you can formally do this. You can compute the moment map for this action. You can see the moment map for this action is more or less the NAM equation. Moment map is more or less NAM's equation. And the equation that comes out of this, this is the Heides-Witton equation in dimension of, looks like a five-dimensional equation if you wanted to. But it comes from this construction. But this leads to the Heides-Witton equation, which I think you will see more about tomorrow. We'll spare. Yeah? Sorry, did I write K? Oh, this is supposed to be an H here. Are you asking about this? Oh, K! Oh, K in this case, take it to be, what did I say? Yeah, take it to be trivial in this case. No, sorry, sorry. You do the same thing as for Waffa-Witton. K is in Waffa-Witton. There are modifications of this. So you take it as K to be SP1, and you pick the same structure as here. Then you get the thing that we call the Waffa-Witton equation. But there are modifications of this that you could get out of this as well. You can get spinorial versions of this as well. Good point, thank you. And then let me maybe... One more thing that one can do with this, the final example, which connects us with gauge theory in higher dimensions, which is sort of why I'm secretly interested in this, is the following. So you can consider connections of S is connections on the trivial G bundle over the quaternions or R4. We identify this with one forms on H with values, sorry, H. G is always the same G. We identify this with one forms on H with values in G. Of course, I can act on this with quaternions. In the usual way, I take G to be the gauge group of this. Well, this in this case turns out to be the same as, let's call this the gauge group of this trivial bundle H. And then the moment map in this case, this was observed by Tia many years ago. That the moment map for this, you formally compute the moment map for this action. So A is an element in here is a connection on the trivial bundle. You can compute the moment map of this and this turns out to be the self-dual part of the curvature of A. If you think about this correctly, and then you can compute the generalized Saber-Gutten equation for this and this turns out to be the equation for spin seven instantons. So then it depends on how you twist it but always gives you spins of an instanton equation. I have not told you what this is, but it's in the lecture notes. You can take a look there. So you can even, this is an eight dimensional gauge theory. It comes out of this construction as well. Okay, I'm finished here. Maybe you can, more or less, yeah? Yeah, I mean, you're going to construct this bundle somehow. It's going to be the gauge theory, for example, on the spinner bundle of a four manifold that would require a spin bundle. But it could be a gauge theory, for example, where the eight manifold is the cotangent bundle or something like this. Yeah, yeah, yeah, there are no, yeah, yeah. Are the metrics not constrained? No, no, no. Yeah, the question was if I can do this on any four manifold. Yeah, in principle I can do this on any four manifold and I need to choose some extra data, but the extra data is going to be some rank four bundle. I mean, I can tell you more precisely what the extra data is, but it's not a rare thing. You will have this data very, very often. Okay, there are three questions. Is there a function such that the cyber-contact equations are critical points? Yes, there's a function. What's the second question? What's the relation to the positivity of scalar curvature? This is more complicated. So this K has the effect of changing the spin bundle to something else. If you take the K to be trivial, then you see the scalar curvature appearing. But if K is something else, other curvature tensors appear. So for example, Ricci can show up. It really depends. I mean, we can write down some generalized for some generalized cyber-contact, generalized Weizberg formulas where you see exactly what replaces the scalar curvature. So you have the corresponding vanishing theorems as well. Question three. Do they exist in some four dimension larger than four? If you really want to know, maybe I can say something in private. Yeah, yeah, you can. I mean, the question is, if an arcala manifold reduces to holomorphic things, this that depends a little bit on what your representation is. But so the Nakajima has this paper where he also talks about this, about towards the mathematical definition of the Coulomb branch. Look in there. You will find some conditions that guarantee that you can do it.