 All right, okay, so today I give you last lecture. And that's good for me. And so I don't know if my message came across because sometimes I'm confused, but one of the thing is that, you know, I'm not concentrating supersymmetry here, but so let me summarize what I did last time and then I will switch to 5D, otherwise I would not have. So we were discussing last time, Chen-Simon theory, Chen-Simon theory, Simon's theory. And I even didn't bother to write for you all supersy, all fields, all transformation, et cetera. And if you look up, so for example, it's written if you wanna look it up. So I think it's paper from 2012 by my former student, Shalene. So it's called co-homological blah, blah, Chen-Simon theory. So what is important that if you write everything, all transformations, et cetera, I mean, they look complicated, but in fact, they're not that complicated. So let me just try to write to you. So this is connection, then I have my field Chi, so connection is at linearized level, it's a one form, these guys are horizontal two forms, then I have a ghosts, this is zero forms, odd, then I have this field Sigma, sort of then there is Psi, then there is H, then there is C bar, then this is zero form, odd, and there is B, it's a Lagrangian multiplier. So if I use notations again, it's very symbolically. So in this direction, that's what I call our operator, and in this direction, this is the operator. So any complicated things actually done like this. So this guy, so this part, again, it's not realized, it's not a common knowledge, but this part is actually a transversal lactic problem. And these guys is correspondently where you localize, it's Lv plus whatever joint of your Sigma. So it's hard to realize, you have to play with these things, et cetera. And for example, when you write supersymmetry, of course, original, if you write supersymmetric theory, it's even more obscure. But in a way, the structure, which I told you right away, it's almost done like this. So we always, so I will now for you write 5D, but we always will have these spaces, so we always have two type of complexes, E0, E1, E2, for example, that can be good to zeros. Then there will be some spaces, F1, F2, sorry, F0, F1, F2, I mean, they can go for some theories, they go longer, the higher dimensions you go, and then there are these errors. So again, this is typically different Ds, this is R, and at linearized level, I always have as determinant of R and whatever cohomology on the kernel of operator D. That's always the answer, how it's organized. So for example, here's a natural problem, so this basically was suggested, all the things was written in 97 in the paper by Losev, Nikrasov, and Bollet, and that's a lift of two-dimensional problem, and two-dimensional problem is elliptic problem. So here's actually, I mean, this is a DRAM, this is DRAM projected to horizontal guys, et cetera. So this is exactly a transversal elliptic problem, and it comes from the lift of two-dimensional problem. Anything coming from supersymmetry always has a structure. That was one of my message, I don't know if I got across, of course it's maybe a bit complicated, et cetera. So again, a few words about your assignment that you understand what we are trying to do. So we are trying to build all fields, so we are trying to space of connection, then we put some bundle over this, and you know, we put more fields. Important thing that we wanna localize over Lv plus I joint of thing, which is constant gauge transformations, constant gauge transformations. If you do everything properly, it's a subtle, it's has been done in paper and by a pestle, because you also have to treat zero modes. I mean, there are some subtleties. So the fixed points, he is very simple. So the fixed points, he is just, so if I'm looking at S3, so my student did it on General Seifert, and then story becomes more complicated, but on S3 is just F equals zero, sigma equal to constant, okay. And then eventually what I told you, so the integral on S3 would become just integral with d sigma. So you basically integrate over the algebra, so let me write answer in two ways, minus some number, trace of sigma square, and then he will, basically if you cancel everything, this is S determinant of a horizontal zero. That forms Lv plus adjoint, sigma. And that's why you use index theorem here. So try to do it yourself if you don't get it. So you basically do a mode expansion over your hopf vibration, and index theorem tells you about constellations. Okay, and there are some subtleties which I'm not telling you, they're written in this paper. I'm taking square roots here, and I don't care about details, but it's very subtle. It will have effect on this coefficient. As you know, in Chin Simon's level gets shifted by basically dual coseter number. So if you do take square roots correctly, everything works correctly. And eventually the answer, if you wanna look it up, it's integral over carton over d sigma, integral minus again some number trace of sigma square. And then also, by the way, here's this determinant cells. So don't forget that I always have here whatever my manifold as three into the algebra valid. So determinant also assumed that I have a little algebra valid guys. And here's integral of whole algebra. Here you reduce over carton, and then this is over all roots not equal to zero sine of i beta sigma. And that's coincides with what Witton found many years ago. It's integral representation which Marinio found to study large N. So it's very convenient to study large N, okay? So that was just repetition of 3D story. So my main message was for you to say that again, I basically wanted to stress this point of view, which is of course a bit formal, but this is actually ability for us to calculate it. Now again, supersymmetry can be mapped. You've heard a lot of lectures about supersymmetry. There is a map to map supersymmetry there and, okay? Any questions about this stuff? So now we'll switch to 5D. So in 5D, everything will work the same way. The things just will get more complicated. The paper you refer to was this, or you also mentioned one by Pestun. Do you have the reference? Pestun, it's an origin of everything, okay? That's like a Bible. But that's, I mean, commentary to the Bible, okay? So this is exactly paper where my student, former student, dealt with co-homological Chan-Simon theory, I mean, this concrete theory on cypher manifolds. Right, any other questions about this? 3D, before we go to 5D? So again, all theories basically, just to give you idea, then you have all this 2D, 3D, 4D, 5D, well, I mean, there is 6D, 7D. There are good reasons to stop at 7D. And so typically, quite often, if you look up, so for example, if you look at, I mean, theory which you localize here, it's an equal two, and they're related about these conditions, which is good elliptic problem. Then actually, here is the same theory, an equal two. It's, you map it here. So it's basically around the lift of this problem. So which becomes transversely elliptic problem here. Okay. So then the standard problem about which Nikita was talking here, so this is again an equal two, so an equal two here and an equal two here mean different things. It's a different theories, of course. Because if you leave this theory up here, it's an equal one, which you don't know how to localize. So an equal two here originally is related about this problem. That's what Nikita was talking about. If you lift it up here, then that's what I will discuss now, this problem. So the role of Pestun here, what he did, it's actually a reduction of this problem from 5D, and this is transversely elliptic problem. So I don't wanna talk about this that much. They're supposed to come, hopefully, this is here. Our paper with Fistuchia, John Q., and Jacob Winden. So I don't know, hopefully we'll come this year where we explain all the things. But his problems, whatever operators he gets, it's transversely elliptic, it's just tilted reduction of this problem. And then you can keep going, there is elliptic problem here, there is it's lift here, et cetera. So my now interest will be here. So supersymmetry always basically picks you either elliptic problem or transversely elliptic, depending on context. Okay, right. So the idea what, so just let me give you some words, and that's original idea goes back to 97 basically, 97, by this Losev, Nekrasov, Berlin. And what they did is that if you have a full manifold, you have naturally this F plus problem. And if you try to put, it's very natural, there is a way to lift this problem to five dimensional manifold. And what you have to do, so you have to introduce a vector field, DT. I mean, so T is along this direction. And then what would be natural to do DTF equal to zero and F horizontal plus equal to zero. Well, horizontal mean that I have legs only here. Again, this problem in fact is not, I mean, don't analyze it's whatever, transversely elliptic, it's too many solutions. So actually there, well, I will tell you. So the idea is now in five D what I have to do, I mean, this is of course, that's what this guy suggested was local picture, what we have to co-orientize it. So I have to discuss for you a bit of geometry. So this would be now crash course on a contract geometry. So let me discuss two minus N manifold. So in fact, 3D falls down to the same class as just, I mean, 3D is more boring, five Ds are infinity, many examples. So I have to N minus manifold and you call it contact manifold, contact manifold if this exists the contact form such that kappa D kappa N is nowhere to zero. So contact manifold sets generalization of symplectic manifolds. So the simple thing is that if you take, I mean, typically usual system NRN, you pick up some Hamiltonian reasonable and you look at constant energy levels, this has become the contact manifold. The idea is such a way that there is exist a plane where D kappa is actually invertible. So it's like a symplectic form. So what is important to hold the story that there is exist and it's unique, what's called rib vector field, field V, which satisfies the following conditions, IV kappa equal to one, IV D kappa equal to zero. So if you fix kappa with this condition, so if kappa does not satisfy this condition, then it's not granted, otherwise it's unique. So for example, you know, that's what I kept telling you on hop vibration. For example, you can choose the things always that they satisfy all these conditions. Right, then the robot we will need is the following, that the peak chief you wanna look visually or dimensional manifolds. So by the way, so just tell you that any three dimensional manifold is contact manifold, for example, there is no obstruction for rentable manifold to be contact. In 5D, well, there are some non-contact manifold, but most of them are contact. So the picture is the following, that you actually have some plane, you have direction R, and this is what's called contact plane, contact plane. And here you have D kappa, and this is basically non-degenerate. It plays the role of symplectic form. So there is again exist actually a metric always exist, such that, such that what? G, R GV is equal to kappa. So basically what I'm telling you, I always can choose a metric that this plane is orthogonal. So by this plane, you understand distribution. And you can make them basically orthogonal and you can make this choice. And moreover, on this plane, you can choose almost complex structures that there is exist almost complex structure. It's a point like thing. So any basically any or dimensional manifold upon choice of contact structure, you choose one directions, you have a plane and this plane, I mean locally looks like a standard symplectic manifold. Again, there are no obstructions for these structures to exist. Okay, so what's a catch? In 3D, in 5D, et cetera. So the catch is the following that if I'm on such manifold, so if I choose this structure, so I choose V contact form and metric compatible, then it means that for any manifold, I can do the following decomposition of P form. I can decompose this P vertical form plus P horizontal form. So this is a form which is given by this projector and this is given by this projector. So please prove, so exercise. So prove that these two spaces are orthogonal. Two spaces, spaces are orthogonal. So to prove it actually, think what GV it means. So basically what the hint is that you have to use conditions that IV star is up to sign will be star, kappa, wage, whatever your form. That's type of conditions and that basically follows from this thing. So that's a way. So this is orthogonal space. So now we go to five dimensions, so 5D. So in 5D if I look, so I'm in particularly interested in looking at two forms on five manifold, okay? And on five forms, I can decompose them as vertical two forms, okay? Plus of course horizontal two forms. And on horizontal two forms, there is exist what you would expect because effectively, I mean, this is linear algebra. And we are not discussing any integrability. So these guys, I mean, as a space, it looks exactly as a two forms in four dimensions. So you should not be surprised that there is exist further decomposition. So actually there is exist the following further decomposition in horizontal cell dual plus horizontal anti cell dual. The operator which does for you this is one over two plus IV star. And the operator here is one over two, one minus IV star. So you see what is going on. So you take a star on two form, you get a three form, you contract with V. So this would be necessary in horizontal space, et cetera. Again, please prove that this is projectors. The only warning is this is projectors not here, not on the whole space. This is projectors on horizontal forms, okay? So they actually, this operator squares to itself. So to prove that it's projector, you have to take this guy, take it square, and it squares to itself, then it's a projector. So it's important thing that if you try to do this calculation in the whole space, you will fail. So it's actually projectors on this subspace. So then you have this decomposition and again, you can prove that this is orthogonal spaces. And this is true for any contact manifold actually. This compatible metric and metric always exist. So here if I start to think, so if I have the structure, there is very natural lift of instanton. So basically what I would try, it's natural to do the following. So I will take a horizontal guy and put to zero. And then I will have a F vertical guy put to zero. So again, I'm warning you, this is not transversely. I mean, it has nothing to do with ellipticity. There are too many conditions because this is three, this is four, this is seven conditions. This problem can be embedded in elliptic problem, this Wafa written type of problem. That's what Nikita was talking yesterday about in 4D, in 5D it's exist similar story. Okay. So this in fact thing you can really write in totally equivalent way. You can write that star of F equal to minus kappa wage F. Prove it. Again, it's a simple algebra. So these things imply this and vice versa. So we call this contact instanton. Instanton. So just one comment, that supersymmetry actually prefers one particular sign here. I mean, formally you can write when you look here, you can also equally write so anti-cell dual, et cetera. So of course, if you change here minus to plus or plus to minus, then you get here different sign. One sign is good and others is not good. One sign actually relates to supersymmetry and also it's important thing that with this choice of sign actually the Young-Mills equation is automatically satisfied. There's another choice of sign that's not satisfied. So supersymmetry actually takes you one contact instanton. So there is no reason to differentiate them, okay? So I'm writing what supersymmetry actually is picking up. So this is later on what we will try to localize. And I mean, this is rather complicated the question and we have very limited idea how it's solved. And again, it's, yep. So let me just, before I will switch to construction of the theory, I need a bit more to tell you about contact geometry and give some examples and we will concentrate of course on S5. So all this contact stuff is very much related to supersymmetry because there are the following thing. So when I have a contact manifold, then there is exist a very canonical way. You just can put a plus line here. And this would call simplectization, which is the following thing that I can introduce for u omega, which would be d r square kappa. So r, it just coordinate here. So this would be symplectic form. In principle, you can think this is a cone. So this is like a conic manifold. So this is a cone. So your manifold is like this. This is your m to n minus one. So this is symplectic. So then of course what has been very natural, let's introduce a metric on the cone, which would be exactly d r square plus r square metric on my odd dimensional manifold. So then, you know, so this is manifold, a second it becomes even dimensional. And in even dimensions, we know a lot of nice, you know, geometry. So first thing is that if your cone is Keuler manifold, then these guys call this Sasaki manifold. So this is a definition. I mean, you can write this intrinsically in terms of odd dimensional manifold, but you can just basically construct the cone and require this. Now, if cone is Calabi Yao, then two n minus one is called Sasaki times Sasaki. Einstein manifold. There is very fundamental theorem. That's where supersymmetry pops up. And that's why 5D becomes incredibly rich. And 4D, we don't know much. Because there are very fundamental theorems, which is saying that if you wanna, if you have a metric cone and you wanna find killing spinners on the base of the cone, it's equivalent to finding correctly constant spinners on the whole cone. Now, Calabi Yao, it's exactly where we know there are currently constant spinners. So it means that Sasaki Einstein manifold admits killing spinners. So there is automatically supersymmetric cangmills. So in 5D, I mean, we automatically get millions of examples. Not just millions, but we get simply a lot of examples with historic geometry, et cetera. In 5D nothing, actually, sorry, in 4D nothing. And the reason is very simple because if you try to apply this construction, I mean, there is, we actually, I mean, we don't have any good class of manifolds in 5D with currently constant spinners. But in 6D, it's Calabi Yao's with all this geometry. And so to give you a concrete example, so again, my interest in this, so S5, right? So S5, this is this condition in C3. So the cone of S5, right? It's just a spherical coordinate. This is C3 by itself. So C3 is obviously Calabi Yao. So all killing spinners related to currently spinners on C3. And so there are plenty of killing spinners. I will talk to you about this geometry in a moment, but let me just give you a hint why Hall's sense, I mean, in this particular respect becomes very nice, because I can produce for you millions of examples of manifolds. If I have time, I can tell you what's going on there. Of course, I will concentrate for the sphere mainly. But that's why 5D story is interesting, because I mean, we do know how to produce Toric-Calabi Yao's cones. It's very, very simple, right? So what you have to do, I mean, thanks for the logical string theory, people studied a lot Toric-Calabi Yao's. And exactly the same things. The only thing we have to require that they are cone and we have to just look at the base of the cone, et cetera. So for example, I mean, famous example, conifold, right? You take a C4 and you do simple equation one, one, minus one, minus one, then you get conifold. If you look at the base conifold, so the base, this is called T11. It's a sake Einstein manifold, so you can write supersymmetric theory there. So topologically, it's S2 times S3. So I mean, it's automatically, it's they're killing spinners, killing spinners related, all this related to contact geometry, the supersymmetric Young Mills, et cetera. So then of course, I mean, you can keep going. So for example, another example would be the following. If you take P minus Q, P plus Q minus P minus P and divisor, so this is called the base. This gives me, so if I do this quotient, simple electric quotient, so P and Q, just to integer's positive, let's say they have co-prime. Then if I do this quotient, I get Calabi-Yau cone. If you look at the base, this is this famous YPQ spaces. Again, topologically, S2 times S3. So in this example, the horizontal four-manifold, it exists only as a transverse structure. So it's not there as a manifold. But still, is it okay to say that solve the antiseptic equation on that thing? It's not a manifold. It's a good thing. No, first of all, you don't care about if it's manifold or not. So what you are talking about, it's a regular. So for example, this is a regular thing. This is not regular. So let me give you, that's a good point that you raised. Absolutely, but the equation by itself makes sense. What you would suspect that most of the time, there will be no smooth solutions. So they will be singular. But in a way, if you look at Necrase's story, it's exactly, I mean, the only interesting solutions, if you put omega background, everything only singular. Of course, it's a purely speculative fact. I don't know how to look at this. So let me, that's a good question you ask. So let me give you example at the level of the sphere. So I have sphere, and then what is important that I have a T3 octane on S5. And this is basically done, the zeta i goes to e alpha i zeta i. Okay, so if I rotate by face, every three zetas, of course, this condition is preserved. So then, of course, there is one case for which I can choose. So this is when V is equal. So let me say that for every guy, I would have a diagonal rotation. So I will have a vector field E1, E2, E3. So this is a vector fields corresponding, correspondingly if I write my T3, it's just S1, S1, S1. So of course, if I take diagonal guy, so first choice, if I take my V to be just E1 plus E2 plus E3. So it means that all zetas are rotated by the same face, identical face. This corresponds to hopf vibration. And this is what's called regular contact structure. So you can actually write kappa very explicitly. So let me give this exercise. Write kappa explicitly. This is basically a problem about writing connection on hopf. So S5 is S1 bundle over CP2. So by the way, many exercises I told you, you can upgrade them instead of doing S3 over CP1. Now you can do S5 over CP2. There are more coordinates, et cetera, about four year modes, everything. It just, it requires more guts and CP2, I mean it requires not two patches, but three patches actually, okay? So now this, if you look at orbits of these guys, they're regular. So this is just regular S1 sitting over CP2, okay? So let's come from these other cases. So actually, so what I can do in other choice is the following. Let me choose omega one, omega two, omega three, some numbers, real. Well, in fact, it's better to choose them positive if I wanna say that it comes from. So then, E1 plus E2, omega two, E2 plus omega three, E3. And let me choose them generic. So generic, that's, you know, so Nikit spent yesterday a lot of time discussing epsilon when they're not generic. So they're typically rational. So I'm actually requiring for these guys to be irrational. I wanna avoid rational points. So then in fact, you also can construct copper, there is a contact structure. So this is all examples of what's called toric contact geometry. So whenever your rib vector field is related as a combination of torus things, then this is a toric contact geometry. And in all these examples, it's an example of toric contact geometry. So now you can ask the question, how the structure looks like. So in principle, how can you visualize a sphere if you wanna think about T2 vibrations? Well, very easily, right? So you have this equation. So what I would like, so my torus is just basically a face of every this guy. If I look at this equation, it's basically a triangle. And then over this triangle, I have a generically vibration of T3. On the edges, it degenerates on T2. And on the vertices, it degenerates to T1. So this H, for example, would be Zeta1 equal to zero. This is Zeta2 equal to zero. This is Zeta3 equal to zero. So basically it's a, you can think of S5 as a T3 vibration over triangle, when you go to different things, there is different degenerations. So analogous story, just for S3. So here's a story that's falling. It's much easier to visualize. It's just a vibration of interval. And then when you come here and degenerates one thing, it comes here, degenerates another cycle. So here's very similar story. It's a bit, you know, I cannot draw for UT3, over et cetera, but you sort of understand, so it's degenerate. And the important thing that it generates now here is one. So then you can ask which vector field this has a closed reborbits. And for rational omegas, the only reborbits, so the closed orbits, it sits here. So in fact, the conjecture, of course, you know, we are not mathematicians, but this is a conjecture that if you would try to solve this equation here, then there are no smooth solutions and there will be only singular solutions sitting in these closed reborbits. But this is purely conjecture. But generically, most of the situation, the interesting context structure, they're in fact not regular. So there is no actually any manifold, et cetera. So the situation with S5, I mean S5 written as one over CP2, this is very non-generic. I have to choose actually coefficients one, one, one, but I can choose them in rational numbers. And in fact, it would be very important when I will try to write a general result and relate to necrosis story, okay? So this is about geometry. Any questions? Because by absence of the questions, I understand that you completely decoupled. So everything is crystal clear, right? Conjecture that it has no smooth solution for a generic choice. Right. But have you any insight if it's less generic than one, one, one, like if it's rational? Well, if it's rational, then basically you have a torus. So it's the same thing, you know, necrosis was writing yesterday that it will close on the torus. You will have more orbits. By the way, there is cold Weinstein conjecture. So anybody who wants to get a million dollars or whatever. So this is one of the outstanding problems in contact geometry to prove that any contact manifold has at least one closed orbit. It has been proven only in three dimensions. And the thing is that you actually use gauge theory. It's related to cyber grit and equations. In five digits remains to be open problem. So actually every manifold, whatever context structure you write, they will be at least one closed orbit. So for these things you will have, for example, I mean for this Toric type, you will have three closed orbits on S5, okay? So that's about geometry. Now, what I actually would like to introduce a theory, supersymmetric theory. So I will write for you this theory in homological terms. But this exists theory for Sasaki Einstein manifolds. It exists a supersymmetric theory and there is a well-defined map one to one with the things I'm writing. And again, I'm concerned only about physicists called vector multiplet and I'm not discussing any matter multiplet but everything can be added. And my main example of course will be S5. I may comment maybe later on about other things. So I have the story looks very much similar as before. So let me just write for you this theory. Delta A goes to psi. Delta psi goes to IVF plus Ida sigma. So delta sigma goes to minus IV psi. Delta chi H plus goes to H H plus. Delta H H plus goes to LV A chi H plus minus I think I sigma chi H plus. So let me, let me see that I'm not making, yep. So what's of the field? So A is a connection of course. Then psi is a odd one form in adjoint. Sigma is a even bosonic zero form in adjoint. So then kappa H plus is element of horizontal cell dual forms. And this is odd, odd and again it's in adjoint. So all fields accept connection and adjoint. And then H plus is in horizontal to cell dual form. It's even and it's in adjoint. So now on Sausage Einstein manifolds, on Sausage Einstein manifolds. So there is exist map between n equal one vector multiplied to this homological field theory is all this transformations field theory. So this map is invertible, very nice, et cetera. I mean, we wrote it explicitly. Typically, you know, to do this maps and everything to derive the answer looks nice, but you have to do some nasty things. Fiercing all these gamma matrices. So it's not something I particularly like, but some people do like. Okay, so in a way, this is just to write in a vector multiplet and here also what is important if you actually start to count, for example, degrees of freedom. So odd guys here are two objects. So this is where supersymmetry is extremely smart because I will tell you, I will get very good problems, et cetera. But you see, I cannot by hand add some fields because this guy is odd, this is odd, right? So psi. So this has five components because this is a five form and one form in five dimensions. This guy is a horizontal cell dual guy, right? So this has three component. So five plus three, this is exactly eight. This is how many Dirac spinners should have in four dimensions. So in a way, you know, the action which you will write have a Dirac operator, et cetera, and supersymmetry. So everything will work nicely. So when you're, again, I'm repeating this statement I told you before. I cannot keep writing some other fields out of blue. My problems will be very badly defined. So all things are matched. Okay. So this is just vector multiplet, et cetera. Of course, in reality, what you will do if you wanna do a calculation, you have to add more fields. So the missing fields, missing fields is C, C bar B. So ghost, anti-ghost, ghost, and Lagrangian multiplier. So this is related to gauge fixing. So the story here, again, it's not always, again, it requires some work, but I have to write actually all supersymmetry because I didn't write for your ghosts. For example, if I start to write ghosts, they will be like plus, you know, dsigma, et cetera, et cetera. So I mean, the complex becomes bigger. So I wrote for your supersymmetry part. So one of important things of localization that I have to actually gauge fixed, write everything, and I have a non-trivial mixture of supersymmetry with BRST. So my BRST guys also have a super partners. So this is very important. So if I write everything sort of as a scheme, I told you before, so I will have C. So this is zero form odd. So then I would have here connection. So this is one form even. Then I will have chi field, which is two plus form horizontal. Typically here I will have another field, which is C bar, which is just zero form odd. And here I will have, I'm a bit lying, it's not quite sick with some combination part of the fields, but let's not care about this psi. Then you have H, and then you have here B. So this would be DRAM, this would be D horizontal plus, plus another operator, which projects me here. I don't want to actually write it to IV, the dagger. So this is plus, this is a dagger. And here, then the errors go slide here. So also, persimmetries can be written like this. And now what I told you, so the problem I was telling you, so 4D translation of this problem, that was elliptic complex. Now it becomes a transversely elliptic complex. So if you reduce this complex, not along the V direction, but over other directions, you will get complex, which Peston has in his paper, which is transversely elliptic. That's the nature of this complex. Again, just believe me, I'm not writing for transformations. If I write everything, it's pretty messy. You can look it up in our papers. It's basically generalization of Peston paper, but it's important that that's a complex we have. And again, basically the idea that this is your manifold, this is differentials, this is part of tangent bundle. So the story exactly the same. So you have this, you have this. So if I would write for you the answer, we'll discuss it now, but if I would write the answer around A equals zero. So my answer, by just looking at this, I can tell you exactly what it should be, because what it should be is the following. So I look at this object, I supposed to have a determinant, because it's odd determinant comes upstairs, determinant over zero for one over two Lv plus adjoint of sigma. Then I'm going here, it's downstairs. It's a determinant of one over two of one form Lv plus adjoint. Then I go here, then it goes, it's odd, it goes up. It's a determinant one over two of a horizontal cell dual forms of Lv plus adjoint of sigma. Then this is odd forms, it again goes up. It's zero forms, so I just write this guy. So I mean, everything is built in. Of course, this is not the full answer. I'm basically looking at linear things, but that's what is important. So if you take this picture, write everything, but then you organize the things. So actually, I mean, this is coordinates on your super manifold. This is, so this is an alex of x. So this is like x, this is like size. Okay, so I will come back to this determinant in a moment, but let me now try to write for you the exact terms. Any questions? Sir, probably it's such a good question, but you write that you're acting by D on ghost and guess what you're doing. Yes. D is the run differential. Yes. What does it mean that I'm acting by D on differential ghost? What's the Birsty transformation or Getsch transformation? Do you remember from your childhood when you studied Birsty? If, I mean, then everybody should go back to Birsty symmetry if you don't remember. I mean, I assume, so the thing is the following. So this is, I mean, this is something you should remember. This is very basic stuff. So this is a Birsty transformation of connection. So you take a ghost and, I mean, that's you act by Getsch transformation. So everything, how do you act ghosts here? You just look at Getsch transformations and you write these things. So this acts in adjoint, so you add C psi. So this acts in adjoint, you write C sigma. This acts in adjoint, you write C H plus, et cetera, et cetera. So then, I mean, there is non-trivial thingies, I mean, you have to see how C transform, et cetera, you have to add. But the idea is that, I mean, this is all related, this is very basic stuff. So when I was telling you that, you know, this complex, for example, in 2D, this is equivalent to the same problem as when I was writing for F plus equal to zero, D diagram of A equal to zero. I mean, at the level of determinants, it's related to the fact that I have a ghost here. So I mean, this is dualization of the thing. So this is a problem when I look at elliptic complex, I equivalently can do Hodge theory and I can map it here. So instead of writing this, I would write omega plus to omega two plus plus zero form. And then my operator here will be D plus plus D. Plus and dagger. So in theory, all gauge symmetries, you forget about them. They're not there, you fix them. Ghosts take care of this. So ghosts sit here. So this is ghost, this is A, this is Chi. Again, this is once Witton wrote his Donaldson Witton theory, I think all this mathematical aspects were understood by I.T.I. Jeffrey. So it's a very nice paper when they discuss all the things. But in a way, I mean, this is just, I mean, original Birsty. So you do Fadeev Popov trick and that's if you write Birsty without super symmetry, this is what you will write. Any other questions? Stupid questions are okay. What did I, okay, we write Birsty exact terms. So what I will do again, I'm writing for a simpler version. I mean, this is what is sort of convenient in the field. People do localization. You have to do with all fields, but a lot of things you sort of forget about ghost, et cetera, et cetera. So I'm writing for a sort of gauge invariant part only. So I will write Delta Psi Wage star Delta Psi bar plus then I will have Chi Wage star H. So H plus plus H plus minus F. Horizontal plus, okay. So if you work out the thing, then what you would get, you will get the following. So this would be F vertical star F vertical plus F horizontal star F horizontal, sorry, plus plus, plus DA sigma star DA sigma. So this I actually related that we are working in the Euclidean theory. So typically if you will get your 5D theory from reduction from 6D and you would like to keep a reality condition for spinners, then you will reduce the long time directions and your scholar will have a wrong kinetics, I mean, sign and then you just have to do analytical continuation. That's why you have I there. So this what you get. So first of all, one fact which you can work out that this is the same as terms F star F plus, I don't have to write integrals, plus kappa F Wage F. So of course my localization locus will be F vertical equal to zero, F horizontal plus equal to zero. So this is exactly this contact instanton. And then there will be of course D of A sigma equal to zero. So that's a localization locus. Now what I wrote there is the following thing. So of course, I mean, and that's where life becomes complicated. So in 3D we just had a flat connection and the flat connection on S3 is trivial. So there is nothing to worry about this. Now the problem is that we have to solve this equation and we have no idea about this equation actually. But as first thing one can do, and that's for example, related to many large checks of whatever we did is that there is a following solution of this equation when F equal to zero and sigma equal to constant. So F equals zero on S, it's the same as equal to zero up to gauge transformations because it's simply connected guy. And moreover, this is isolated point there are no deformations of this, isn't this a question? So this is actually isolated point. So you may have more solutions, but it makes sense to isolated point. So this we call perturbative thing. And if you would assume that all other guys will be suppressed in large chain, that's basically the answer at least which is relevant in large chain. So that's what I was writing for you. So let me go back to this determinants. So this actually, so this determinant if I'm writing things more explicitly, this is integral over d sigma. I have this and then I have to evaluate my action. So I didn't write it, there will be some terms with sigma square, et cetera, which I didn't write it. So this would be some number, sigma square term, basically times volume of my manifold. Okay, so that's what I will get. And of course here will be further corrections related to non-trivial solutions of this equation. So let me first tell you about this determinants because you have to use exactly the same thing as before. So let's just go to simplest case where in principle we have enough technology to do everything. So my V is just hope vibration, right? So actually what I can do here further on, I can do these decompositions. Two plus, this is the same as a horizontal two zero plus horizontal zero two plus basically omega zero times omega. So omega is just a color form. Does it sounds familiar? Let me check it. Did you hear already during this lectures about this? Good, yeah. So he already mentioned this a few times during lectures. So I'm doing exactly the same thing. So I'm doing this in 5D, but it's adjusted basically. It's a Cp2 part direction, right? So now I'm looking this. So now one guys can be decomposed in the following way. So one forms a decomposed SM. One zero, horizontal plus horizontal zero one plus basically a zero form. This is just a vertical part. So I'm doing this just to look for you for these determinants, right? So I'm looking here. And so now just look what is up there, what is down. So if I write everything up, so I will have upster there. I will have determinant. Again, I'm not caring about face horizontal two comma zero. Oh, I don't know, zero comma two LV plus adjuvant. So because I had here two copies, I take a square root. Then I have another of zero forms and I have a zero form here. I'm sure your face is. I'm sure your face is. There are no trivial faces, but unlike, so we did calculate them originally, but it seems they do not appear in 5D. There are no any physics behind it. This is problem with physics. You're right when you have other checks to check that you're right. Otherwise, so you do identical calculation like in 3D, in 3D, you know it's correct. In 5D, you have no clue. So presumably there is some reason that it's irrelevant, okay? So then here determinant of zero form. So this is in power three over two. So one I had here and one comes from here. LV plus adjuvant. So now look here. So I have one zero zero one. So it's the same thing. Let me take square root. I'm ignoring the faces. So I will have horizontal zero one LV plus adjuvant. And then I would have a zero form one over two, zero form. So actually you can see that this guy cancel and you just have here one. So what you have here, you have basically as determinant of a horizontal forms zero bullet LV. So this complex, which has very natural operator. So I mean all zero forms horizontal. So there is this operator DH bar, horizontal zero one. Horizontal zero two. So outside of this kernel of DH bar, actually everything will cancel. So I can actually go to the homology of this operator, which is infinite dimensional. And then here I have to use index theorem. So I can tell you, I will not derive it although I have written it somewhere. So the idea is the following that in this setting on, so I can do it. So there are many ways of calculating things, but since we are talking about index theorems, we are talking. So here I can do exactly the same story. I can expand in modes and all my modes. So what I can do, I can do the following thing. It's exactly like in 3D. Just like it becomes a bit more complicated. So if I have a horizontal form, zero P on S5, this is the same as the sum over N over zero P forms on CP two with the values in O and bundle. So the calculation I told you, you can think of this exercise, which is a bit more complicated. So you have to do all the CP two stuff very, very explicitly, which is of course, more complicated, but otherwise. And the thing is, for example, you can derive that index of this double operator for N bundle twisted. It's just one plus three over two N plus one over two N square. So this is a different. So the answer will look very much the same as before, but that's a factor you will have there. So you remember what I will actually have. I will have a product of N's not equal to zero. And I will have whatever two pi IN plus a joint of sigma. And then here would have exactly this coefficient. So this is for round sphere one over two N square. This is a special function. So actually it's related. It's a special function. And it's related. It's a triple sign at special values. But if you take a log of this, it involves logs, dialogues and tree logs. So this is this type of special function. But again, since I don't have time, I'm just suggesting for you play with this around. So the logic of playing and doing calculations is exactly the same as I told you before. So you have to prove this guy. You have to prove this formula. Again, you can do this formula using some details of characteristic classes or doing the current things over CP2. So this is for dull boot, twisted bio and bundle. You have to write everything explicitly. CP2 has a three fix points. So you have to do this, the same exercise. I was very briefly telling you before. Okay, and that's what you get. Okay. Let me write you the module answer. So of course we were more general answer. And there is actually much more powerful trick to do it. I can mention for you. So if you would ask me to, if I choose this V to be omega one plus omega two plus omega three generic. So this is just written when all omegas equal to each other and equal to one. So this is what's called round sphere. Sorry, I meant here one E one plus omega two E two plus omega three E three. Okay. Then the answer will look the following. So I can switch integral to carton. I will have minus classical piece, sigma square. Then I will have a product over roots, not equal to zero. Then I will have this function S three of I X omega one, omega two, omega three. So this function is called triple sign. Sorry, I wrote not very precise. I sigma with roots pairing omega one, omega two, omega three. So this is called triple sign. So the function sign fact very nice, which appear there. So just to give you idea, if you do on S one things, you get a sign function. If you get things on S three, you get what's called double sign function. And if you do an S five, you get triple sign function. Question how it's defined sign it's periodic. This function is defined in such a way that if you shift things by one of the omegas. So in a way it's double thing, right? So it has X and this would be omega one, omega two. So it has two periods. If you shift this by either omega one, omega two, you will get sign. So double sign is periodic up to sign. Triple sign is periodic up to double sign, et cetera. So there are whole hierarchy functions. So they're very nice functions. If you wanna have explicit formula of S three on X omega, this is the following, this is N one, N two, N three from zero. And then you have X plus N on omega. And then you have another product N one, N two, N three. Now starting from one minus X plus N on omega. So that's explicit formula for triple sign. Again, as a physicist, you understand this as a regularized thing. So this is entire function with a specified zeros. Again, there are purely index theorems calculations, calculators for generic omega, but it requires a bit more work. That's why exactly you have to use full equivalent index theorem for transversal elliptic operator. So this, I mean, this trick I told you works only with, I mean, for round sphere. Now, let me, so this is a function. And for example, this is what we really want for large N. If you basically believe that non-trivial solutions to, I mean, suppressed in large N, then you can study asymptotics of these things and large N and study and see that for example, it agrees with whatever we would expect, things like ADS, CFT, et cetera. Right. So the question is, okay, very good, but what's the general answer? I mean, how can we expect, et cetera? The thing is the following, that things come from localization and actually, so my omegas for geometry, when I choose geometrically, they are supposed to be real numbers. If I want actually to stick to contact structure to geometry. If I write this function, the function actually defined for any complex omega. And in a way, this is a story very much like a necrosis. I mean, it's the most natural thing. It's actually to assume your parameters to be, I mean, epsilon parameters, omega background to be complex. It's very natural. So in this story, it's also very natural to assume parameters to be complex. So when they're actually complex, there is the following factorization of this answer. So you can decompose this in the following things. So it will be two pi i x over omega one, two pi i omega two of omega one, two pi i omega three over omega one, and cyclic permutation. So what's this symbol means? So I can tell you in a moment. So this symbol means the following thing. So for this factorization to work, there is some polynomial thing. Yes, I'm Bernoulli. This works actually, we need to make sure that these things are convergent. So let me write this explicitly. So if I would write for zeta q one q two infinity, this is just nm from zero to infinity, one minus zeta q one n q two m. And this is a good function if q one is less than one and q two is less than one. If it's another region, there are other things to write. So you see for these convergences, for me I have to guarantee that these guys actually modulus less than one. It means that ratio of omega two of omega one should have imaginary parts. So if you, I mean, this is purely analytical results. So you can take the special function and decompose upon this analytical continuation. But what you can see if you stare at this guy. So this is actually a perturbative answer for necrosis partition function, function on R four times S one. And the beta, the radius of this is related to one omega one and then corresponding epsilon, it's omega two over omega one and omega three over omega one. And then you have three pieces like this. So you have one piece, you have second piece and three piece. You just do cyclic permutation and all these guys. So next thing it will be instead of omega one, there will be omega two, et cetera, et cetera. So you have this decomposition result and then also you can stare at purely at the local geometry because it's exactly. So I told you about this picture of toric vibration of S five. And if I look at this corner exactly, I told you that here in this corner, I have my circle. So for general values of omega, so what is important here that all these factorizations, I have to analytically continue in omega and my omega should be generic. Otherwise there will be extra problems, et cetera. But the thing is that you can also look at geometry and this is of course agrees because around on every point, this part looks locally like R four of ball times S one. And if you identify parameter, so sphere is actually five sphere glued from three pieces like this and toric geometry exactly gives you this parameter. So I mean, you should not be surprised that analytical continuation agrees with geometry. Then after thinking of these that's okay, you calculated perturbative answer, it factorizes so nicely, et cetera. Then we know any cross-off partition function for, this is a five D version what Nikita was describing. So five D version, if in four D version on ADHM construction or more or less space, you calculate the volume, you put one. If you put instead of one a roof genus, it's not a characteristic class. It has interpretation as a quantum mechanics of model space. So it gives you five D theory, but it's a trigonometric version of what Nikita was discussing. So these objects are defined. So from this point of view, it's very natural to make the following conjecture that the full partition function on S five is equal to D sigma whatever some classical term sigma square. And then there will be any cross-off partition function of epsilon four times S one, times another partition function, times another partition function. And the parameters of course will be exactly what postulate to omega one, omega three of omega one, one, one. Then for example, here would be omega two, omega one, omega three of omega two, one of omega two. So this is analog. This is epsilon one, epsilon two, this is beta. And then here would be, so now remaining thing is omega three, omega one of omega three, omega two of omega three, one of omega three. And then of course you integrate here over sigma, so it depends on this sigma. So I didn't write explicitly perturbative part because you can put in there in the cross-partition function. So it means again, if you believe in this, you would basically say that your solutions, your own point like instantons is sitting exactly around. So your instantons, I mean instantons, in 5D it's not, I mean instantons, it's the particles like, because in 5D the action is actually related to Hamiltonian 4D action. So there are particles there. So they sit around these things. So now this is a story, it's very conjectural, but let me stress this is very important point about localization. And this is presumably, again physicists don't care, but this is the most important problem in localization compact spaces. So in 2D and 3D our localization locus is very small and we understand this very well, et cetera. So I mean when I was telling you about Chin Simons I didn't lie anything. When you go to Pestun result it's conjectured because his configuration actually singular. He has no means to regularize them. He say they're there, let's assume they're there. In Ecrasa story you don't have to worry because on non-compact space you have a lot of tools. He actually has well-defined modular spaces. He looks at fixed points. He regularizes, he calculates the current volumes. And for example there is a story related to non-commodative regularization. And all the stories agree. So in doing this story on R4, despite the fact that are in singular configurations which you don't worry, because actually singular configurations is just reinterpretation of these fixed points which is interpreted by this n-tuples of yantablo. Now when we go and find a dimensional space we have no modular space. We have nothing. So this is pure conjecture. So Pestun conjecture thinks, and again everybody, I mean in community including Vasily is aware of the things. And the thing is that it's not that easy mathematically to introduce any non-commodative regularization, et cetera. So I mean there is this problem. So this result is conjectured and it's not even clear how to make it more precise, et cetera. And it's not clear if actually makes sense to study the things. So for example I can prove there are no smooth solutions and I can say that presumably if there are singularity around reborbits but it doesn't mean anything. But of course in the community this type of answer is accepted. So we studied higher dimension, the problem just get worse and worse. Of course PDs becomes more and more complicated. In 3D and 2D there are no problems. Even if you have vertices you can get them not by localizing on the vertices but you can localize on Coulomb branch and then do contour integrals and then go to Higgs branch, et cetera. But here I wanted to tell you that this is presumably the biggest challenge. I mean nobody questions the answer but if actually passed on result can be made more precise in the long the line. So there is this friction between compact and non-compact examples that you have to be aware. And in principle it would be good to resolve it but I don't know. But again there is many, many indirect checks and 4D it's related to AGT, et cetera. So this is a correct answer. Yep. What's the classical action that you're localizing? Classical action. I'm localizing, well it's young mills, right? I didn't write there is a chance I'm in store and I can write, et cetera, 5D. Oh so the sigma squared. Yeah, yeah, sigma squared. Yeah, you get mills true. Absolutely. So this conjecture formula includes contribution from the trigger solution sometimes. Right, I mean conjectured, yeah, absolutely, yeah. So if I want to take the further limit I have to send all the omegas to infinity. How do you take, I mean, 4D limit here? But then, well, I guess the PQP camera should tend to some... No, but let's ask questions geometrically. I give you five sphere. What do you want to get? Four sphere from this, how? Yeah, but I mean... No, it doesn't work on that. I mean sometimes you can do 4D limits but I mean here I'm on a very hard to get out of these things, I mean any 4D limit. So the thing is that if you tried for the level of one guy to do something it blows up on another way. But geometrically it's very clear because if you have a sphere and you start to squash it here then it goes like this. So five sphere does not have any good 4D limit and it's purely obvious geometrically. So this formula C also cannot do. At the same time if you think about the sensor at something very amazing because in Necrasa formula, 5D formula, there are epsilon one, epsilon two and there is a beta. There is a democracy between these parameters but beta stays sort of outside. It's instant on counting parameter. What you actually do, you're symmetrizing over these guys and taking integral. So in a way it should be very wonderful object. But I mean, even studying this explicitly, this object, it's already takes time. Yeah, five, I can stop it now if you want. And I don't know, I think I don't wanna tell you anything more and I feel like I failed actually to deliver the message because it's, I mean, last subject. But I tried to tell you that there is a very nice structural things and at least I was trying to stress on mathematical side but there is a physical side supersymmetry everything works but Guido told you everything last week. So it was not my job. So you don't worry, there are formulas, everything works. I was actually trying to stress mathematical part. So for example, in general for Toric guys, we conjecture the same thing that for more complicated Toric guys, for example, Sasai Kainstang, you will have more complicated Toric diagrams and for every closed report, but you just put a copy of Necrasa partition function and exactly which U epsilon, et cetera, you can read from Toric data around fix, I mean around reborbit, et cetera. Again, it's a conjecture to answer. And, but I think I will stop here and if you have questions, I may answer them remaining four minutes. Questions. You're wishing you have some good reason to stop and save on the machine? Yeah. So what is the reason? It's the same reason why you stop with super conformal series in 60. It's the same reason as Pilarov. You cannot construct this complex. The thing is that you cannot actually put so the reason related to supersymmetry, you cannot put supersymmetric young mills on a dimensional sphere preserving, I mean sort of nice amount of, I mean symmetries, that's the main problem. It's related to super groups, but in a way 70, it's related to the fact that seven minus one, it's equal to six. So it's the same thing like num classification. So num classification, you cannot have super conformal. I mean theory beyond, I mean, because there is no unitary representation. So there are similar things. Of course you can put still eight D theory on some manifold, et cetera. But the problem is that it will not be related to supersymmetric theory with good symmetries. That's what I mean. It's basically related to, so this argument is stated in our paper with John Minahan. So you have to look at super groups with correct asymmetry, with correct symmetry of the sphere and beyond seven there are nothing. Of course you can still try to put seven eight dimensional young mills on eight sphere, but you have to make bigger sacrifices. That's the only thing I mean. But in a way, at least from 2D to 70s, the story looks very uniformly and nice. Other questions? So we started with real generic omega. So the function seems to not be equal or does it only go up attractionally? I mean, this guy, I mean, if you look at these things, et cetera, you cannot have real omega, it's divergent. If omega is real, this has a more or less one. It doesn't make sense. Yes, but I mean, for the same epoch conversable, people say that we don't blow up rational points and it doesn't make sense for irrational ratios. So does something similar hold for that moment? I think so, yeah. I mean, the story is very much the same thing. Once you go outside of a rational points, I mean, once you go outside of generic points, you have to think more. So that's what Nekrasov was telling you yesterday, I mean, and I mean, there are other representations of these partition functions which we can have when omegas are rational. If they become rational, then I mean, many things just fell down, et cetera. Other questions? 45, I'm done. Okay, thank you.