4/4 Index theorems and 5d localization
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Algebraic structureNumerical analysisTransformation (genetics)Positional notationGauge theoryModulformSupersymmetryDimensional analysisFiber bundleDeterminantPhysical lawEnergy levelAlgebraHomologieDuality (mathematics)Connected spaceVibrationGroup actionPrice indexComplex (psychology)MereologyMultiplicationTheorySigma-algebraStandard errorTheoremStochastic kernel estimationThermal expansionEllipseGoodness of fitLogical constantOperator (mathematics)RootMusical ensembleSquare numberPoint (geometry)Körper <Algebra>Symmetric matrixTwo-dimensional spaceMany-sorted logicCoefficientDirection (geometry)Student's t-testCorrespondence (mathematics)Multiplication signSpacetimeTransverse waveCohomologyChi-squared distributionLecture/Conference
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33:08
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49:11
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01:06:18
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01:29:52
Diagram
Transcript: English(auto-generated)
00:15
All right, okay, so I today give you last lecture, and that's good for me.
00:24
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.
00:42
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 fields, all transformation, etc.
01:01
And if you look up, so for example, it's written if you want to look it up. So I think it's paper from 2012, by my former student, Shalin, so it's called cohomological blah, blah, Chen-Simon theory.
01:21
So what is important that if you write everything, all transformation, etc., 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.
01:40
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, well, sort of, then there is psi. Then there is H, then there is C bar.
02:11
Again, this is zero-form, odd, and there is Cb. It's a Lagrangian multiplier. So if I use notations again, it's very symbolically.
02:24
So in this direction, that's what I call our operator. And in this direction, this is the operator. So any complicated things are actually done like this. So this guy, so this part, again, it's not realized. It's not a common knowledge, but
02:41
this part is actually a transverse problem. And this guy is correspondently where you localize. It's LV plus whatever joint of your sigma.
03:02
So it's hard to realize. You have to play with these things, etc. 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 always done like this.
03:24
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 go to zeros. Then there will be some spaces, F1, F2.
03:44
Sorry, F0, F1, F2. I mean, they can go for some theories, they go longer, the higher dimensions you go, then there are these errors. So again, this is typically different Ds, this is R. And at linearized level, I always have S determinant of R and
04:05
whatever cohomology on the kernel of my 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 by,
04:23
in the paper by Losev, Nikrasov, and Boleh. 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 deram. This is deram projected to horizontal guys, etc.
04:43
So this is exactly transversally 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, etc.
05:04
So again, a few words about Chen-Simon 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 we put more fields.
05:21
Important thing that we wanna localize over LV plus adjoint of thema, which is constant gauge transformations. Constant gauge transformations. If you do everything properly, it's a subtle. It has been done in paper and by Pesto, because you also have to treat zero modes.
05:43
I mean, there are some subtleties. So the fixed points here is very simple. So the fixed points here is just, so if I'm looking at S3. So my student did it on general ciphered, and then story becomes more complicated. But on S3, it's just f equals zero sigma cubed to constant.
06:04
Okay, and then eventually what I told you, so the integral on S3, what becomes just integral with d sigma. So you basically integrate over the algebra.
06:23
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.
06:42
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 hop vibration. And index theorem tells you about cancellations. Okay, and there are some subtleties which I'm not telling you.
07:03
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 Simon's level gets shifted by, basically, dual number.
07:21
So if you do take square roots correctly, everything works correctly. And eventually, the answer, if you want to 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 these determinants.
07:40
Also, don't forget that I always have here whatever my manifold test tree and the algebra valid. So determinant also assume that I have a real algebra valid, guys. And here's integral of the whole algebra. Here, you reduce over carton. And then this is over all roots not equal to 0 sine of i beta sigma.
08:08
And that coincides what Witten found many years ago. It's integral representation which Marino found to study large n. So it's very convenient to study large n, okay?
08:21
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.
08:41
Now again, supersymmetry can be mapped. You 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.
09:01
The things just will get more complicated. The paper you referred to was this, or you also mentioned one by Peston. Do you have the reference? Peston, it's origin of everything, okay? That's like a Bible. But that's, I mean, commentary to the Bible, okay? So this is exactly a paper where my student,
09:24
former student dealt with cohomological transignment theory. I mean, this concrete theory on cypher manifolds. Right, any other questions about this 3D before we go to 5D?
09:48
So again, all theories, basically, just to give you an idea, then you have all this 2D, 3D, 4D, 5D,
10:01
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 n equal two, and they're related about these conditions, which is good elliptic problem.
10:22
Then actually, here is the same theory, n 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?
10:40
So then the standard problem about which Nikita was talking here, so this is again n equal two. So n equal two here and n equal two here mean different things. It's a different theories, of course. Because if you leave this theory up here, it's n equal one, which you don't know how to localize. So n equal two here originally is related about this problem. That's what Nikita was talking about.
11:03
If you lift it up here, then that's what I will discuss now, this problem. So the role of Pastern here, what he did, it's actually a reduction of this problem from 5D, and this is transversely elliptic problem.
11:26
So I don't wanna talk about this that much. They're supposed to come, hopefully, this year, our paper with Fis-Tuche, John Q, and Jacob Winding, so I don't know. Hopefully, they'll come this year
11:41
where we explain all the things. But his problem, 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 a research lived here, et cetera. So my now interest will be here.
12:02
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,
12:23
and that's original idea goes back to 97, basically, 97, by this Lossiv-Nikrasov-Burley. And what they did is that if you have a four manifold, you have naturally this F plus problem. And if you try to put, it's very natural,
12:42
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, D-T. I mean, so T is along this direction. And then what would be natural to do, D-T F equal to zero
13:03
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 ellipticity, it's too many solutions. So actually there, well, I will tell you.
13:23
So the idea is now in 5D, what I have to do, I mean, this is of course, that's what this guy suggested was local picture, what we have to cover and ties it. So I have to discuss for you a bit of geometry. So this would be now crash course on a contact geometry.
13:45
So let me discuss 2-N manifold. So in fact, 3D falls down to the same classes, this is just, I mean, 3D is more boring, 5D, there are infinitely many examples. So I have 2-N minus manifold
14:01
and you call it contact manifold. Contact manifold if this exists contact form such that kappa d kappa n is nowhere to zero.
14:27
So contact manifolds is a generalization of symplectic manifolds. So the simple thing is that if you take, I mean, typically usual system in Rn, you pick up some Hamiltonian, reasonable, and you look at constant energy levels,
14:41
this becomes contact manifold. The idea is such a way that there exists a plane where d kappa is actually invertible. So it's like a symplectic form. So what is important to hold this story is that there exists and it's unique,
15:00
what's called the Rieb vector field, field V, which satisfies the following conditions, I V kappa equal to one, I V d kappa equal to zero. So if you fix kappa with this condition, so if kappa does not satisfy this condition,
15:21
then it's not guaranteed, otherwise it's unique. So for example, that's what I kept telling you on Hopf vibration, for example, you can choose the things always that they satisfy all these conditions, okay? Right, then the robot we will need is the following.
15:41
The picture, if 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 any three dimensional manifold to be contact. In 5D, there are some non-contact manifold, but most of them are contact.
16:02
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.
16:23
So there is again, exists actually a metric, always exists, such that, such that, what, g v is equal to kappa.
16:45
So basically what I'm telling you, I always can choose the 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.
17:01
And moreover, on this plane, you can choose almost complex structures, there exists almost complex structure. It's a point-like thing, so any, basically any odd dimensional manifold upon choice of contact structure, you choose one direction, you have a plane,
17:20
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 the catch in 3D, in 5D, et cetera? So the catch is the following, that if I'm on such manifold,
17:41
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.
18:01
So this is the form which is given by this projector, and this is given by this projector. So please prove, so exercise,
18:21
so prove that these two 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 i, v star
18:42
is up to sign, will be star, kappa, h, whatever you form. That's type of conditions, and that basically follows from this thing. So that's the way, so this is orthogonal space. So now we go to five dimensions, so 5D.
19:05
So in 5D, if I look, so I'm particularly interested in looking at two forms on the five manifold, okay? And on five forms, I can decompose them as a vertical two forms, okay?
19:24
Plus, of course, horizontal two forms. And on horizontal two forms, there exists what you would expect, because effectively, I mean, this is linear algebra, and we are not discussing any integrability. So these guys,
19:42
I mean, as a space, it looks exactly as the two forms in four dimensions. So you should not be surprised that there exists further decomposition. So actually, there exists the following further decomposition in horizontal cell door plus horizontal anti-cell door.
20:00
So 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.
20:21
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 whole space. This is projectors on horizontal forms, okay?
20:43
So actually, this operator squares to itself. So to prove that it's a 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 on whole space, you will fail. So it's actually projectors on this subspace.
21:01
So then you have this decomposition, and again, you can prove that this is orthogonal space. And this is true for any contact manifold, actually. This compatible metric, and metric always exist.
21:26
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,
21:43
and then I will have a vertical guy put to zero. So again, I'm warning you, this is not transverse. I mean, it has nothing to do to ellipticity, there are too many conditions. Because this is for the seven conditions.
22:07
This problem can be embedded in elliptic problem, this type of problem, that's what Nikita was talking yesterday about, in 4D, in 5D, it's a similar story. Okay, so this in fact,
22:22
thing you can write in totally equivalent way. You can write that star of F equal to minus kappa, which prove it. Again, it's a simple algebra. So these things imply this and vice versa.
22:45
So we call this contact instanton. So just one comment, that supersymmetry actually prefers
23:01
one particular sign here. I mean, formally you can write when you look here, you can also equally write anti-cell dual, et cetera. So of course, if you've changed your minus to plus or plus to minus, then you get here different sign. One sign is good, and others is not good.
23:22
One sign actually relates supersymmetry and also it's important thing that with this choice of sign, actually the Yank-Mills equation is automatically satisfied. There's another choice of sign that's not satisfied. So supersymmetry actually pins, I mean, takes you one contact instanton. So there is no reason to differentiate them, okay?
23:43
So I'm writing what supersymmetry actually is picking up. So this is later on what we will try to localize. This is a rather complicated equation and we have very limited idea how it's solved. And again, it's, so let me just,
24:13
before I will switch to construction of the theory, I need a bit more to tell you about contact geometry
24:23
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,
24:41
then there exists 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 you omega,
25:01
which would be dr square kappa. So r is just coordinate here. So this would be simplectic 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.
25:21
This is your m to n minus one. So this is simplectic. So then of course, what has been very natural, it's introduced symmetric on the cone, which would be exactly dr plus r square
25:44
metric on my odd dimensional manifold. So then, you know, so this is manifold as a cone, it becomes even dimensional. And in even dimensions, we know a lot of nice geometry.
26:05
So first thing is that if your cone is color manifold, then these guys call this Sasaki, Sasaki manifold.
26:25
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 color BL,
26:44
then two n minus one is called Sasaki Einstein, Sasaki Einstein manifold. There is a very fundamental theorem. That's where supersymmetry pops up.
27:02
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 have a metric cone, and you wanna find killing spinners
27:20
on the base of the cone, it's equivalent to finding currently constant spinners on the whole cone. Now, Calabi, 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 Yang-Mills.
27:42
So in 5D, I mean, we automatically get millions of examples, not just millions, but we get simply a lot of examples of historic geometry, et cetera. In 5D, nothing, actually. Sorry, in 4D, nothing.
28:01
And the reason is very simple, because if you try to apply this construction, I mean, 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-Yau's with all this geometry. And so to give you a concrete example,
28:34
so again, my interest is this, so S5, right? So S5, this is this condition in C3.
28:49
So the cone of S5, right, it's just a spherical coordinate. This is C3 by itself. So C3 is obviously Calabi-Yau.
29:01
So all killing spinners related to covariate 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 whole science,
29:22
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 this here mainly.
29:42
But that's why 5D story is interesting, because, I mean, we do know how to produce toric Calabi-Yau's cones. It's very, very simple, right? So what you have to do, I mean, thanks to photopological string theory, people studied a lot toric Calabi-Yau's.
30:01
And exactly the same things. The only thing you 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 simply a quotient,
30:21
one, one, minus one, minus one, then you get conifold. If you look at the base conifold, so the base, this is called T1-1. It's a Sake-Einstein manifold, so you can write supersymmetric theory there. So topologically, it's S2 times S3.
30:45
So I mean, it's automatically, they're killing spinners, killing spinners related, all this related to contact geometry, this is to be symmetric Yang-Mills, et cetera. So then, of course, I mean, you can keep going. So for example, another example would be the following.
31:04
If you take p minus q, p plus q, minus p minus p, and divisor.
31:22
So this is called the base. This gives me, so if I do this quotient, symplectic quotient, so p and q are just two integers positive, let's say they are co-prime. And then, if I do this quotient, I get Calabi-Yaukon. If you look at the base, this is this famous YPQ spaces.
31:43
Again, topologically, S2 times S3. So in this example, the horizontal form manifold, it exists only as a transverse structure, so it cannot vary as a manifold. But still, is it okay to say that
32:01
solve the antiseptal expression on that thing? It's not a manifold, is it? 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.
32:21
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 Nikrassa 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.
32:43
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 acting on S5.
33:01
And this is basically done, the zeta i goes to e alpha i zeta i. Okay? So if I rotate by phase, every three zetas, of course, this condition is preserved. So then, of course, there is one case for which I can choose.
33:23
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 correspondently if I write my T3, it's just s1, s1, s1.
33:46
Okay? 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 phase,
34:01
identical phase. This corresponds to Hopf vibration. And this is what's called a regular contact structure. So you can actually write kappa very explicitly. So let me give this exercise.
34:21
Write kappa explicitly. This is basically a problem about writing connection on Hopf bundle. So s5 is s1 bundle over cp2. So by the way, many exercises I told you,
34:40
you can upgrade them instead of doing s3 over cp1. Now you can do s5 over cp2. There are more coordinates, etc., about Fourier modes, everything. 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,
35:03
they're regular. So this is just regular s1s sitting over cp2, okay? So let's confront with other cases. So actually, so what I can do, another choice,
35:48
is the following. Let me choose omega-1, omega-2, omega-3, some numbers, real. Well, in fact, it's better to choose them positive if I wanna say that it comes from.
36:02
So then, e1 plus e2 omega-2 e2 plus omega-3 e2. e3, e3, e3. And let me choose them generic. So generic, well, that's, you know,
36:21
so Nikita spent yesterday a lot of time to discuss in 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 kappa. There is a contact structure. So this is all examples
36:40
of what's called toric contact geometry. So whenever your RIB vector field is related as a combination of torus things, then this is 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.
37:01
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,
37:22
so my torus is just basically a face of this guy. If I look at this equation, it's basically a triangle. And then over this triangle, I have a generic vibration of T3. On the edges, it degenerates on T2,
37:42
and on the vertices, it degenerates to T1. So this edge, for example, would be zeta one equal to zero. This is zeta two equal to zero. This is zeta three equal to zero. So basically, you can think of S5 as a T3 vibration over triangle.
38:02
When you go to different things, there is different degenerations. So analogous story, just for S3. So here's the story of the following. It's much easier to visualize. So it's just a vibration of interval.
38:25
And then when you come here, it degenerates one thing. It comes here, it degenerates another cycle. So here's a very similar story. It's a bit, I cannot draw for you T3 over et cetera, but you sort of understand. So it's degenerate. And the important thing that it generates now here is one.
38:43
So then you can ask which vector field this has a close reward with. And for rational omegas, they're only re-orbits. So the close orbits, it sits here. So in fact, the conjecture, of course, we are not mathematicians, but this is a conjecture
39:02
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 this closed re-orbit. But this is purely conjecture. But generically, most of the situation, the interesting context structure,
39:21
in fact, not regular. So there is no, actually, any manifold, et cetera. So the situation with S5, I mean, S5 written S1 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 the general result
39:42
and to relate to Nikrasov story, okay? So this is about geometry. Any questions? Because by absence of the questions, I understand that you're completely decoupled. So everything is crystal clear, right?
40:08
The 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?
40:21
Well, if it's rational, then basically you have a torus. So it's the same thing Nikrasov was writing yesterday, that it will close in the torus. You will have more orbits. By the way, there is Cole-Weinstein conjecture, so anybody who wants to get million dollars or whatever, so this is one of the outstanding problems
40:42
in contact geometry to prove that any contact manifold has at least one close orbit. It has been proven only in three dimensions, and the thing is that you actually use gauge theory. It's related to Seiberg-Witten equations.
41:01
In five digits, it remains to be open problem. So actually, every manifold, whatever context structure you write, there will be at least one close orbit. So for these things, you will have, for example, I mean, for this toric type, you will have three close orbits on S5. So that's about geometry.
41:21
Now, what I actually would like to introduce a theory, supersymmetric theory. So I will write for you this theory in cohomological terms, but there exists a theory for Sasek-Einstein manifold. It exists a supersymmetric theory, and there is a well-defined map, one-to-one with the things I'm writing.
41:42
And again, I'm concerned only what physicists call 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 the story looks very much similar as before,
42:00
so let me just write for you this theory. Delta A goes to psi, delta psi goes to I V F plus I D sigma. So delta sigma goes to minus I I V psi,
42:22
delta chi H plus goes to H H. Delta H H plus goes to L V A chi H plus, minus I sigma chi H plus.
42:42
So let me, let me see that I'm not making, yep. So what's out of the field? So A is the connection, of course. Then psi is the odd one-form in a joint.
43:06
Sigma is a even bosonic zero-form in a joint. So then kappa H plus is element
43:22
of horizontal self-dual forms, and this is odd. Odd, and again, it's in a joint. So all fields except connection are not joint. And then H plus is in horizontal two self-dual forms.
43:42
It's even, and it's in a joint. So now on Sasseke-Einstein manifolds, on Sasseke-Einstein manifolds, so there is exist map between n equal one vector multiplet to this cohomological field theory
44:07
with all this transformation field theory. So this map is invertible, very nice, et cetera. I mean, we wrote it explicitly. Typically, you know, to do these maps and everything to derive, the answer looks nice,
44:21
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.
44:45
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,
45:02
because this is a one form in five dimensions. This guy is a horizontal self-dual guy, right? So this has three components. So five plus three, this is exactly eight. This is how many Dirac spinors should have in four dimensions. So in a way, the action which you will write
45:20
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, right.
45:44
So this is just vector multiplied, et cetera. Of course, in reality, what you will do if you wanna do a calculation,
46:00
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.
46:23
So this is related to gauge fixing. So the story here, again, it's not obvious, again, it requires some work, but I have to write
46:42
actual supersymmetry because I didn't write for your ghost, for example. If I start to write ghost, it will be like, plus the sigma, 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,
47:02
and I have a non-trivial mixture of supersymmetry with BRS-T. So my BRS-T guys also have a superpartners. 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.
47:23
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.
47:44
Typically here we'll 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 sigma, it's some combination of other fields, but let's not care about this psi.
48:03
Then you have H, and then you have here B. So this would be deram, this would be D horizontal plus, plus another operator, which projects me here.
48:24
I don't want to actually write it as D dagger. So this is plus, this is a dagger. And here, then the arrows go slide here.
48:44
So also, presymmetries can be written like this. And now what I told you, so the problem I was telling you, so for the truncation of this problem, that was elliptic complex. Now it becomes transversally elliptic complex.
49:04
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 transversally elliptic. That's the nature of this complex, okay? Again, just believe me, I'm not writing for transformations.
49:22
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.
49:40
So the story exactly is 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 0. 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.
50:02
I suppose to have a determinant, because it's odd determinant comes upstairs, determinant of a zero form, one over two, LV plus adjoint of sigma. Then I'm going here, it's downstairs. It's a determinant of one over two,
50:23
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 horizontal cell dual forms of LV plus adjoint of sigma.
50:46
Then this is odd forms, it again goes up in zero form. So I just erase this guy. So I mean, everything is built in. Of course, this is not the full answer. I'm basically looking at linearized things, but that's what is important.
51:02
So if you take this picture, write everything, but then you organize the things. So actually, I mean, this is coordinates in your supermanifold. This is so, this is analogs of x. So this is like x. This is like size.
51:22
Okay, so I will come back to this determinant in a moment. But let me now try to write for you the exact terms.
51:47
Any questions? Sir, probably, you took a question, but you read that you're writing by D on
52:01
ghosts and ghosts go to, how do you call it? D is the wrong differential. Yes. And what does it mean that I'm not going to write the wrong differential in ghosts? What is the B-R-C transformation, gauge transformation? Do you remember from your childhood when you studied B-R-C?
52:24
No, I mean, then everybody should go back to B-R-C 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.
52:42
So this is a transfer, maybe R-C transformation of connection. So you take a ghost, and I mean, you act by gauge transformation. So everything, how do you add ghosts here? You just look at gauge transformations and you write these things. So this acts in adjoint, so you add C psi.
53:01
So this acts in adjoint, you write C sigma. This acts in adjoint, you write C H plus, etc., etc. So then, I mean, there is non-trivial thing is, I mean, you have to see how C transforms, etc., you have to add. But the idea is that, I mean, this is all related.
53:23
This is very basic stuff. So when I was telling you that 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 dagger of A equal to zero.
53:42
I mean, at the level of determinants, it's related 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, omega two plus, plus zero form.
54:01
And then my operator, he will be D plus, plus D, again, 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.
54:26
So this is ghost, this is A, this is sky. Again, this is once Witten wrote his Donaldson-Witten theory, I think all this mathematical aspects were understood by Atiyah Jeffrey.
54:43
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 BRS-T. So you do five-day pop-off trick and that's, if you write BRS-T without supersymmetry, this is what you will write. Any other questions?
55:00
Stupid questions are okay. Okay, we write BRS-T exact terms. So what I will do again, I'm writing for you a simpler version. I mean, this is what is sort of convenient in the field.
55:20
People do localization, you have to do with all fields, but a lot of things you sort of forget about, ghost, etc., etc. So I'm writing for 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.
55:45
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.
56:01
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.
56:25
So typically, if you will get your 5D theory from reduction from 6D and you would like to keep a reality condition for spinors, then you will reduce along time directions and your scholar will have a wrong sign and then you just have to do analytical continuation.
56:42
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 by h f.
57:06
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.
57:23
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.
57:41
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.
58:05
But as first thing one can do, and that's for example related to many large n checks of whatever we did, is that there is the following solution of this equation when f equal to zero and sigma equal to constant.
58:24
So f equals zero on S, it's the same as a 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 within this equation. So this is actually isolated point.
58:41
So you might 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 n, that's basically the answer, at least which is relevant in large n.
59:02
So that's what I was writing for you. So let me go back to these determinants. So this actually, so this determinant, if I'm writing things more explicitly,
59:20
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 squared, etc., which I didn't write it. So this would be some number, sigma squared term, basically times the volume of my manifold, okay? So that's what I will get.
59:43
And of course, here will be further corrections related to non-trivial solutions of this equation.
01:00:05
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.
01:00:22
So my V is just hope vibration, right? So actually what I can do here further on, I can do these decompositions. 2+, this is the same as a horizontal. 2, 0 plus horizontal 0, 2 plus basically omega 0 times omega.
01:00:48
So omega is just a color form. Does it sound familiar? Let me check it.
01:01:01
Did you hear already during this lecture about this? Good, yeah. So he already mentioned this a few times during the 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?
01:01:20
So now I'm looking at this. So now one guys can be decomposed in the following way. So one forms are decomposed as 1, 0. Horizontal plus horizontal 0, 1 plus basically a 0 form. This is just a vertical part.
01:01:44
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.
01:02:03
So if I write everything up, so I will have upstairs there. I will have determinant. Again, I'm not caring about face. Horizontal 2, 0, or I don't know, 0, 2, lv plus adjoint.
01:02:21
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. And I'll show you faces, I'll show you faces. There are non-trivial faces, but unlike, so we did calculate them originally,
01:02:42
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.
01:03:00
In 5D, you have no clue. So presumably, there is some reason that it's irrelevant, okay? So then here, determinant of 0 form. So this is in powers 3 over 2. So 1 I had here, and 1 comes from here. lv plus adjoint.
01:03:23
So now look here, so I have 1, 0, 0, 1. So it's the same thing. Let me take square root, I'm ignoring the faces. So I will have horizontal 0, 1, lv plus adjoint.
01:03:40
And then I would have a 0 form, 1 over 2, 0 form. So actually, you can see that this guy cancels, and you just have here 1.
01:04:01
So what you have here, you have basically s determinant of a horizontal forms 0, bullet, lv. So this complex, which has very natural operator, I mean, all 0, 0 forms horizontal.
01:04:24
So there is this operator dh bar, horizontal 0, 1, horizontal 0, 2. So outside of this kernel of dh bar, actually everything will cancel.
01:04:43
So I can actually go to cohomology 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.
01:05:10
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.
01:05:23
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.
01:05:45
So if I have a horizontal form, 0p on S5, this is the same as the sum over n, over 0p forms on Cp2 with the values in on bundle.
01:06:04
So the calculation I told you, you can think of this exercise, which is a bit more complicated. So you have to do all this Cp2 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
01:06:25
double operator for n bundle twisted, which is one plus three over two. And plus one over two n square.
01:06:50
So this is the difference. 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 will have. I will have a product of n not equal to zero, and
01:07:04
I will have whatever 2 pi i n plus adjoint of sigma. And then here we'll have exactly this coefficient. So this is four rounds here, one over two n square.
01:07:26
This is a special function, so actually it's related. It's a special function. And it's related, it's a triple sign of special values. But if you take a log of this, it involves logs, dialogs, and tree logs.
01:07:44
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, exactly the same as I told you before.
01:08:03
So you have to prove this guy, you have to prove this formula. Again, you can do this formalizer using some details of characteristic classes or doing the current things of a CP2. So this is for the old boot twisted by one bundle.
01:08:21
You have to write everything explicitly, CP2 has three fixed points. So you have to do this same exercise I was very briefly telling you before. Okay, and that's what you get, okay? Let me write you the more general answer.
01:08:41
So of course, we're aware of more general answer. And there is actually much more powerful trick to do it, I can mention for you.
01:09:00
So if you would ask me to, if I choose this v to be omega1 plus omega2 plus omega3 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,
01:09:21
omega1 plus omega2 plus omega3 is 3, okay? Then the answer will look as the following. So I can switch integral to carton. I will have minus classical piece sigma square. And I will have a product over roots not equal to zero.
01:09:44
Then I will have this function s3 of ix omega1, omega2, omega3. So this function is called triple sine. Sorry, I wrote not very precise. I sigma these roots pairing omega1, omega2, omega3.
01:10:05
So this is called triple sine.
01:10:22
So the function sine fact very nice which appear there. So just to give you idea, if you do on s1 things, you will get a sine function. If you get things on s3, you get what's called double sine function. And if you do an s5, you get triple sine function.
01:10:41
Question how it's defined. Sine is 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 omega1, omega2. So it has two periods. If you shift this by either omega1 or omega2, you will get sine.
01:11:03
So double sine is periodic up to sine. Triple sine is periodic up to double sine, etc. So there are whole hierarchy of functions. So they're very nice functions. If you want to have explicit formula of s3 on x omega.
01:11:22
This is the following. This is n1, n2, n3 from 0. And then you have x plus n on omega. And then you have another product, n1, n2, n3. Now starting from 1 minus x plus n on omega.
01:11:43
So that's explicit formula for triple sine. Again, as a physicist, you understand this as a regularized thing. So this is entire function is a specified zeros. Again, there are purely index theorems, calculations,
01:12:01
calculators for generic omega, but it requires a bit more work. That's why exactly you have to use full equivalent index theorem for transverse elliptic operator. So this trick I told you works only with, I mean, for round sphere. Now, let me, so this is a function.
01:12:25
And for example, this is what's really one for large n. If you basically believe that non-trivial solutions, 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 what we would expect with things like AD, CFT, etc.
01:12:46
Right. So the question is, okay, very good, but what's the general answer? I mean, how can we expect, etc.? The thing is the following, that things come from localization and actually, so my omegas for geometry,
01:13:04
when I choose geometrically, they're supposed to be real numbers. If I want actually to stick to context 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.
01:13:21
I mean, it's the most natural thing. It's actually to assume your parameters to be, I mean, epsilon parameters on the ground 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.
01:13:41
So you can decouple this in the following things. So it will be 2 pi i x over omega 1, 2 pi i omega 2 over omega 1, 2 pi i omega 3 over omega 1,
01:14:04
and cyclic permutation. So what's this symbol mean? 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,
01:14:24
e s and 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, q1, q2, infinity. This is just n m from 0 to infinity,
01:14:41
1 minus zeta, q1 and q2 m. And this is a good function if q1 is less than 1 and q2 is less than 1. If it's another region, there are other things to write. So you see for this convergence, for me, I have to guarantee that this guy actually models less than 1.
01:15:04
It means that the ratio of omega 2 over omega 1 should have imaginary parts. So if you, I mean, this is purely analytical result. So you can take this special function and decompose upon this analytical continuation. But what you can see if you stare at this guy.
01:15:22
So this is actually a perturbative answer for Nikrasov partition function on alpha times s1.
01:15:41
And the beta, the radius of this is related to 1 omega 1. And then corresponding epsilon, it's omega 2 over omega 1 and omega 3 over omega 1. And then you have three pieces like this. You have one piece, you have second piece, and three pieces. You just do cyclic permutation and all these guys.
01:16:01
So next thing it will be instead of omega 1, they will be omega 2, etc, etc. So you have this decomposition result and then also you can stare at purely the local geometry because it's exactly.
01:16:39
So I told you about this picture of toric vibration of s5.
01:16:45
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 an omega and my omega should be generic.
01:17:03
Otherwise, there will be extra problems, etc. But the thing is that you can also look at geometry. And this, of course, agrees because around in every point this part looks locally like alpha of bold times s1.
01:17:20
And if you identify parameters, so c 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 this, that's okay, you calculated perturbative answer.
01:17:42
It factorizes so nicely, etc. Then we know new class of partition function for, this is a 5D version what Nikita was describing. So 5D version, if in 4D version on ADHM construction or you calculate the volume, you put one.
01:18:00
If you put instead of one, a roof genus, it's not a characteristic class. It has interpretation as quantum mechanics of modal space. So it gives you 5D 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.
01:18:21
The full partition function on S5 is equal to d sigma, whatever some classical term, sigma square. And then there will be any cross of partition function of epsilon 4 times S1, times another partition function, times another partition function.
01:18:48
And the parameters, of course, will be exactly what postulated omega 1, omega 3 of omega 1, 1 over omega 1. Then, for example, here would be omega 2, omega 1,
01:19:01
omega 3 of omega 2, 1 over omega 2. So this is another, this is epsilon 1, epsilon 2, this is beta. And then here would be, so now remaining thing is omega 3, omega 1 over omega 3, omega 2 over omega 3, 1 over omega 3.
01:19:24
And then, of course, you integrate here over sigma, so it depends on this sigma. So I didn't try to 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
01:19:40
solutions, your own point like instantons, is sitting exactly around. So your instantons, instantons in 5D, it's not instantons, it's particles.
01:20:00
Because in 5D, the action is actually related to Hamiltonian for the 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 a very important point about localization.
01:20:22
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, etc. So when I was telling you about Chen-Simons, I didn't lie anything.
01:20:43
When you go to Peston result, it's conjectured because his configuration actually is singular. He has no means to regularize them. He says they're there, let's assume they're there. In the Krasov story, you don't have to worry, because on non-compact space, you have a lot of tools.
01:21:01
He actually has well-defined model 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 in R4, despite the fact that are in singular configurations, which don't work.
01:21:24
Because actually, singular configurations are just reinterpretation of these fixed points, which is interpreted by this n-tuples of Yantablot. Now when we go and find the dimensional space, we have no modulus space. We have nothing. So this is pure conjecture.
01:21:40
So Peston 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, etc. So I mean, there is this problem. So this result is conjectured, and
01:22:01
it's not even clear how to make it more precise, etc. And it's not clear if actually it makes sense to study the thing. So for example, I can prove there are no smooth solutions. And I can say that presumably, is there a singularity around reorbids, but it doesn't mean anything. But of course, in the community, this type of answer is accepted.
01:22:25
So we studied higher dimension, the problem just get worse and worse. Of course, PDEs 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,
01:22:40
but you can localize on Coulomb branch, and then do contour integrals, and then go to Higgs branch, etc. But here, I wanted to tell you that this is presumably the biggest challenge. I mean, nobody questions the answer. But if actually Peston result can be made more precise in the longer line. So there is this friction between compact and
01:23:01
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's related to AGT, etc. So this is a correct answer. Yep?
01:23:20
What's the classical action that you're localizing? Classical action, I'm localizing, well, it's Young Mills, right? I didn't write the research on Simon's term, I can write, etc., 5D. So the sigma squared- Yeah, yeah, sigma squared. Yeah, no square. Absolutely. So this conjecture formula includes contribution from the contribution from that is.
01:23:46
Right, I mean, conjecture, yeah, absolutely, yeah. So if I want to take the 4D limit, I have to send all the omegas to infinity. No, how do you take, I mean, 4D limit here? But then, well, I guess the PQ per camera should tend to some-
01:24:06
No, but let's ask question geometrically. I give you five sphere. Well, do you want to get four sphere from this, how? Yeah, but, I mean- No, it doesn't work, no. I mean, sometimes you can do 4D limits, but I mean, here, I'm unaware how to get out of these things, I mean, any 4D limit.
01:24:25
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
01:24:43
it's purely obvious geometrically. So this formula, see, also cannot do. At the same time, if you think about this answer, it's something very amazing, because in the cross of 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.
01:25:02
It's instant on counting parameters. What you actually do, you're symmetrizing over these guys and taking integral. So in a way, it should be a very wonderful object, but, I mean, even studying this explicitly, this object, it already takes time.
01:25:21
Yeah, 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, a vast subject.
01:25:41
But I tried to tell you that there is a very nice structural thing, and at least I was trying to stress on the 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 the mathematical part.
01:26:01
So for example, in general, for toric guys, we conjectured the same thing. For more complicated toric guys, for example, Sasaki and Sanghi will have more complicated toric diagrams. And for every closed report, but you just put a copy of Nikrasov partition function, and exactly which your epsilons, etc.
01:26:22
You can read from toric data around fixed, I mean, around the orbit, etc. Again, it's a conjecture to answer. And, but I think I will stop here, and if you have questions, I may answer them in the remaining four minutes.
01:26:41
Questions? You mentioned you have some good reason to stop and saving the machine. Yeah. So, what is the reason? It's the same reason why you stop with super conformal series in 6D. It's the same reason as P.
01:27:00
You can help construct this complex? The thing is that you cannot actually put, so there isn't related to supersymmetry. You cannot put supersymmetric Yang-Mills on a dimensional sphere preserving, I mean, sort of nice amount of symmetries. That's the main problem.
01:27:21
It's related to supergroups. But in the way 7D, it's related to the fact that 7 minus 1 is equal to 6. 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.
01:27:41
Of course, you can put still 8D theory on some manifold, etc. 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.
01:28:03
So you have to look at supergroups with correct asymmetry, with correct symmetry of the sphere, and beyond 7 there are nothing. Of course, you can still try to put 8 dimensional Yang-Mills on 8 sphere, but you have to make bigger sacrifices.
01:28:21
That's the only thing, I mean, but in a way, At least from 2D to 7D, the story looks very uniform and nice. Other questions? So we started with real generic omega, so the function seems to not commit to, or does it only go up a rational point?
01:28:43
I mean, this guy, I mean, if you look at these things, etc., you cannot have real omega, it's divergent. If omega is real, this has a modulus one, it doesn't make sense. Yes, but I mean, often for the same people, people say that it only blows up at rational points,
01:29:01
and it doesn't make sense for irrational ratios. So does something similar hold for the double? I think so, yeah. I mean, the story is very much the same thing. Once you go outside of irrational points, I mean, once you go outside of generic points, you have to think more.
01:29:20
So that's what Nikrassa was telling you yesterday, I mean, 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, etc.
01:29:40
Other questions? 45, I'm done. Okay, thank you.