 So yesterday, I talked about two-dimensional young melts. This time, I'm going to four-dimensions as planned. So let me first review very quickly the Lagrangian of 4D n equals 2 gauge theories. I'm assuming that you are familiar with n equals 1 superfield notation with n equals 1 supersymmetry. So with n equals 2 supersymmetry, there are basically two types of multiplets. One is the vector multiplets in n equals 2 supersymmetry, and there's a hypermultiplet. So in n equals 1 notation, vector multiplet contains n equals 1 vector multiplet, which we usually denote by w alpha. And the chiral superfield phi, both are in their joint. And of course, the chiral superfield contains the scalar component and fermion component. Vector superfield contains gauge genome and gauge field strength. What happens in this combination is that there is a SU2, r symmetry, which mixes these two n equals 1 multiplet. In a super conformal case, there is also a u1r symmetry. And it is, well, the gauge field cannot have any u1r charge. Then due to the charge of the supercharge, this will have r charge 1, and this will have u1r charge 2. So this is the vector multiplet. So this is some g gauge field. Now, suppose you want to add some matter content. In that case, you want to introduce something called hypermultiplet. So in n equals 1 language, that consists of a chiral superfield in some representation r of g and another chiral superfield, q twiddle, which is in the conjugate representation of the q fields. And this q will contain scalar component q and the fermion component psi alpha. And in order to exhibit n equals 2 supersymmetries, useful to consider the conjugate of this q twiddle field so that this q twiddle dagger is also transforming in this representation r. And that will contain the dotted spinner field and the complex conjugate of the lowest component of this q twiddle. Now, what happens is that, again, this middle component becomes the SU2r doublet. And in this case, the u1r charge assignment is 0, minus 1, and 1. So this is the hypermultiplet. And in the applications I'm going to describe, it is important to also consider something called half hypermultiplet. So this is possible when r is pseudo-real. So for those of you who are not familiar with this terminology, a representation r is called real. If its complex conjugate representation is the same as isomorphic to the original representation, but there are, in fact, two types. So the two types is distinguished by the type of invariant tensor. So if r is a real representation, then it should have either a symmetric invariant tensor or an anti-symmetric invariant tensor. If it has a symmetric invariant tensor, this is called strict-real. And if it has an anti-symmetric invariant tensor, this is called pseudo-real. So the most familiar one for you would be, if you take gauge group to be SU2, 3 would be real representation because delta ij is symmetric. But the doublet representation is a pseudo-real representation because epsilon ij, which is the invariant tensor, is an anti-symmetric one. So half hyper-multiplet is only possible when r is pseudo-real. And this is obtained by imposing a constraint. Impose q to the a equal jabqb. So this is the constraint you impose. So this cuts the number of fields in half. That's why it's called half hyper-multiplet. And the reason why you cannot do this with strict-real representation is that in order to keep this SU2r symmetry, you need to choose anti-symmetric tensor here. So that's what it is. So those are the multiplets. Let's now discuss the Lagrangian. So again, I'm going to use n equals 1 super field notation. Pardon? You cannot have strict-real half-hypers. Half-hyper is only possible for pseudo-real representation. So the Lagrangian has the vector-multiplet part. So you have a standard kinetic term for the adjoint super fields there, plus the gauge kinetic term. Sorry, I forgot trace, plus complex conjugate. So this is the Lagrangian for the vector-multiplet. And you would have a kinetic term for the q fields, q to the dagger, and certain very important coupling, d squared theta, q to the phi q, plus complex conjugate. So very simple. Important thing is that because with n equals 2 super symmetry, gauge field and this phi field are in the same multiplet. Therefore, the coefficient here and coefficient there is related by super symmetry. Maybe I should say that this tau is a certain combination of the gauge coupling constant and the theta parameter. And similarly, this is a standard n equals 1 coupling of the q fields and the vector-multiplet. But the vector-multiplet in n equals 2 not only contains this v, but also phi. Therefore, by applying n equals 2 part of the super symmetry, you obtain this coupling of q and q to the phi starting from this kinetic term. So you see that this part is a superpartner of this gauge-matter-matter coupling. So again, the coefficient here and coefficient there is related by super symmetry and the ratio is fixed. So I'm sloppy about the precise coefficients, but you can find them in the literature. In any case, the precise coefficients are not important in my lectures this time. So this is the Lagrangian. So let's discuss the renormalization. So of course, in a gauge theory, there's a running of this coupling. And the one loop beta function is proportional. It's very easy to compute, as you know. You just count the number of multiplets multiplied by a gauge representation factor. So this has the following form. So there's a contribution from the vector multiplet. And there will be a contribution from the hyper-multiplet. Sorry, plus R i bar. Well, let me give you the notation. So this T of R is basically a quadratic casimir given by TATB in the representation R is given by delta AB T of R. And in any case, we only use the, I only have time to discuss g equals SUn. In that case, T of a joint is n and T of fundamental is 1 half. And this is the contribution from the vector multiplet. And this is a contribution from the hyper-multiplet. And this is the contribution from the Q field. This is the contribution of the Q twiddle field. In the case of half-hyper-multiplet, you need to drop one. So that's why it's useful to split this. I mean, representation theory guarantees that this and that is equal, but this is what it is. So you immediately see that when R is g, it's 0, because this is twice T a joint, T a joint. It's 0. It's 0. So this is the n equals 4 super-animal's case. So this is a very important fact that n equals 4 super-animal. That's 0, 1 of beta function. And another example is the one I referred to yesterday. So if you have SUN as g and two NF copies of fundamental, hyper-multiplet, so it contains Q field in the fundamental. Maybe you cannot read it, so sorry. Let me continue here. So another example is to take g to be SUN. And you take two NF pairs of half-hyper-multiplet, fundamental plus fundamental. Sorry, NF equals 2nc of fundamental plus anti-fundamental. So this gives you 2n for the first term. And fundamental plus anti-fundamental gives you 1 half plus 1 half, which is 1. So in order to cancel 2n, you need to have 2n copies of it. So again, this has 0, 0, 1 loop beta. Here comes a very important fact about and simplifying fact about n equals 2 supersymmetric systems. So this is a very important fact. With n equals 2 SUZ, if 1 loop beta is 0, then it's 0 to all order and even non-perturbitively. So this guarantees that just by checking this group's theoretical fact, this system has a UV coupling constant that's tunable. So let's just quickly show how this fact can be derived. I'm not going into detail, but the rough idea is as follows. So let's use Zyberg's holomorphic trick. So there is a certain renormalization scheme in n equals 1 supersymmetry where holomorphic of the chiral superfields are preserved. So let's assume that. In that case, there are two basic facts. One is that the superpotential term is not renormalized. The second condition is that the gauge coupling constant is only running at one loop. I guess I should write it. So the term runs only at one loop. And the superpotential doesn't run. So this is a generic n equals 1 fact that can be learned in any modern textbook on supersymmetry. So this is a superpotential term. So this coefficient 1 doesn't run. And this coefficient tau only runs at one loop. But we are now assuming that there's no one loop running of tau, so this doesn't run either. In a generic n equals 1 system, there will be some wave function renormalization factor, z phi and z 2 that are the sources of many interesting dynamics of n equals 1 supersymmetric systems. But here we are talking about n equals 2 supersymmetric systems. And as I already told you, the ratio of the coefficient here and the coefficient there is fixed by n equals 2 supersymmetry. Therefore, if this doesn't run, this needs to stay at one. Similarly, if this part doesn't run, because of the n equals 2 supersymmetry, this cannot be non-trivial. Done. So this is the all-order, at least prototype of all-order proof of this fact. So now, so these are the examples of what I referred to at the beginning of my first lecture yesterday. So there is a gauge theory in four dimensions where you can tune the gauge coupling at UV. That can be very weak, but you can choose it to be very strong. So you would like to ask what happens if you tune the gauge coupling very, very strong. So let's concentrate on the particular case of n equals 2 is SU2 and nf equals 4. So in this case, let me just remind you what I told you yesterday. So there are two pairs of quarks in the doublet, but there are four of them. However, as is familiar, doublet and anti-doublet of SU2 are the same. So instead of writing them as this, let's use the notation that we have qA1 and 2, and i goes from 1 to 8. In this notation, this superpotential term has this particular coupling, A, qB, AB. And if you take a symmetric combination of the doublet indices, you get the adjoint index of SU2. And you have i, j, delta i, j. So this is the superpotential term there. So this means that this system has SO8 flavor symmetry. So if you work out the scalar potential from the Lagrangian I just erased, I'm sorry about that, there is a scalar potential. And it's a standard exercise to see that there's a term of the form trace of phi dagger squared plus various terms involving q's. So from this, you easily see that if phi is of this form, A minus A is 0, 0. And if you set both q and q to 0 to be 0, now this is a 0 energy vacua. And that guarantees supersymmetry. So these are supersymmetric vacua. When A is non-zero, this is a wave, non-zero wave in the triplet of SU2. Therefore, this breaks SU2 to U1. And this is a very classic case studied by many people in the past. So first of all, there is a Higgs mechanism going on. So there will be tons of massive excitations. So some fields become massive, non-massive. So some component of phi will have m phi is given by basically 2A, let's say A, absolute value. And some fields, some components of q will have mass of the form 1 half q A. So these are different by a factor of 2 because of the difference in the charge of phi and q. But not only that, when the gauge group SU2 is broken to U1, there is Tohut-Polyakov monopole. And what's the mass of that? What's the mass of that? In any case, the original theory doesn't have any mass scale. The only thing which is introducing the mass scale is this A. Therefore, this should be proportional to A. But monopole is a classical configuration, semi-classical configuration. And the energy can be read off from the Lagrangian. And there was a Lagrangian written here, which I erased. But there's an overall factor 1 over g squared. So typically, you have something of this form. And in fact, you can find an exact formula which just becomes this, tau A absolute value. So when tau is, I mean, when g, coupling concept is very weak, of course, 1 over g is extremely large, which means that this monopole is extremely heavier than these perturbative excitations. So of course, this is important to remember. This is perturbative. These are really quanta, quanta of the original QFT, perturbative quanta. And this is semi-classical soliton. So what would happen if you crank up the gauge coupling? So when g is of order 1, eventually this will become lighter and lighter. Eventually the mass would become comparable. But if you go further and make the coupling constant even larger, what would happen is that eventually the solitonic object would be lighter than the perturbative excitations themselves. So well, in that regime, I shouldn't really call something as a soliton and something as a perturbative excitations, because it's very quantum mechanical. But we can expect that. So the statement of estuality in this case is that when g is very big, there is another weakly coupled frame where Q, quanta, and monopole are exchanged. So in the original description, this has A, and this is tau A, 1 half. And in the dual description, Q, quanta, dual Q, quanta, and the dual monopole would have mass A dual and tau duo A duo. Sorry, tau duo A duo. But they can be exchanged if the ratio if tau d over 2 is 2d, 1 over 2 tau, and if you rescale A, d, and A. So this is the basic property of estuality. This exchanges perturbative excitation and monopole. You might ask, how can it be possible? I mean, I just told you that there are eight fields here in SO8 representation. Very naively speaking, classically, if you are given a system where SO2 is broken to U1, there's just one monopole configuration. So how can you map eight fields here and just one monopole? So this needs to be equal to that. So you just solve one of... I think the equality here is easier to solve. tau A should be 2ad. This means that the ad is 2 tau A. Yeah, sorry for putting additional 2 here. For any course for superman, this 2 doesn't appear. Thank you for the question. So there are, in fact, eight monopole states. Here it is important to perform treat monopole quantum mechanically. So what happens is that in the monopole configuration, there's a nontrivial profile of the gauge field and phi. And then you need to consider Q fields in this nontrivial background. Sorry I should have used I. In particular, this contains a fermion component. And this has zero modes, fermionic zero modes, which I denote as C of I from 1 to 8. So what happens is that if you treat the monopole quantum mechanically, there should be an action of these fermionic operators. If you work out the quantization problem of these fermionic zero modes carefully, you find that CI, CJ satisfy this standard anti-commutation relation. So this is exactly the same as the gamma matrix identities. This means that the monopole states are in the spinar representation of SO8 flavor symmetry. So there are two spinar representations, HS and HC. And if you work more carefully, you can check that one half of the spinar component is not just a monopole, but becomes a diome. So this becomes a diome. And this part becomes a pure monopole. But in any case, we find eight monopole fields. Again, so there are eight fields after quantization. But remember, sorry, what we did is the quantization here, the eight fields. So originally, Q fields were in the vector representation of the SO8 symmetry. And the monopole is in the 8s of the SO8 symmetry. But if you want to do this identification, what happens is that this dual Q quanta should be not the vector representation of SO8, but should be in the spinar representation of SO8. So in this duality, both sides are SU2 gauge theory with four flavors. And they look very, very similar. But you need to remember that the SO8 flavor symmetry representation assignment is different from between this side and the other side. So the reason is that the C field came from this Q field. And Q field is charged under the U1, which is unbroken subgroup of SU2, which means that the C operator also have charge plus minus 1. Therefore, if you act on a monopole state by 1c, it adds an electric charge. And as you know, gamma matrices, if act on a while spinner, that gives you a different while spinner. So if you act by C field once, you get on this side. If you act by another C field, you get back on this side. Therefore, the electric charge of this monopole state and the other monopole state are different by one unit. So that's how you see that they should be diodes. That's a very good question. So you can work out similarly how monopoles transform under the flavor symmetry. But you immediately learn that it is not very easy to do this kind of identification. So this nice way of mapping things only work with SU2. And the SU3 case was only understood in 2007. So this was proposed by Zyberg written in 1994. So it took 20 years or something. No, 10 years. Yeah, yeah. Right, that's right. So I'm skipping over the details. So this monopole with this charge has mass of the form tau A. And the monopole of this other guy has the mass of the form tau plus 1 A. So what happens is that if the gauge coupling is very big with theta angle 0, this guy becomes massless. But if you keep theta angle to be pi and make gauge coupling strong, this guy becomes massless instead. That's how you get to a point where the light degrees of freedom transforms in A to C. So that's how the triality works. So there are three points which are special in the space of complexified tau. That's how the triality works. Right, yeah. Right, so that's why this was originally called a triality. So where should I write? So how should I say? I don't have time to discuss the details. But states in the system can be characterized by the electric charge and magnetic charge. And basically what you find is that odd even combination, if charge is odd and monopole charge is even, this transforms in 8V, well, I mean 1 8-dimensional representation. And if you consider even electric charge and odd monopole charge thing, this transforms in the 8S. And if you consider odd combination, this transforms in the 8C. So they are in a completely symmetric combination because by S duality, you can exchange E and M. And by T transformation, you can mix E and M. So in that way, you can freely mix three types of 8-dimensional representation. Even even. So even even typically, they are in the a joint representation. Yeah. So this is what's going on in SU2 with four flavors. But the arguments I provided so far are very crude, right? We just studied the flavor symmetry content of the excitations in one weakly coupled frame and boldly claim that they should be mapped. So how can we check this more precisely? So in the original paper in 1994, Zyberg and Witten computed something called the Zyberg-Witten curve of the system and checked this triality. That's a fun exercise, but it's rather complicated. So let's do something else. So for this, I introduced the concept of the super conformal index. So the super conformal index I defined for 4D. N equals to super conformal theory. So these theories are examples of these super conformal theory. And you put this system on S3 times R. Well, this R should be straight, but sorry. So you can start from the flat space and perform a vial transformation and conformal transformation to get there. And so let's give a name, so this is T. And very roughly speaking, super conformal index of this theory, T, is just the Witten index of the system on S3. So you consider Hilbert space on S3. And you pick a super charge. Pick a super charge q, cap q. And as always, you put the anti-commutator of q, q dagger here, and you add, you can add some additional objects here, objects commuting with q and q dagger. Sorry for using the same symbol q for the hypermultiple and the super charge, but unfortunately, alphabets have so many letters. We cannot help. So from the standard argument about Witten index, this is independent of beta, first of all. Also independent of deformations of the theory, preserving this super charge q and q dagger. So it's a very nice object. So if this S duality is really true, we should be able to compute the super conformal index on one weakly coupled frame and do the same thing on the dual weakly coupled frame, where the original gauge coupling is extremely strongly coupled, and we should be able to compare the super conformal indices, which should agree, because it should be invariant under the deformation of the gauge coupling constant. So that's what I'm going to do. But first of all, I need to give you a more explicit form of this expression. And the equations I'm going to write would become more and more horrible as the time goes on. So please bear with me for a few seconds, a few 10 minutes or so. But there's something amazing going on in this super conformal index business. So what we need to do is to pick a particular q. So in n equals 2 super conformal algebra, we know that there's SU2R symmetry. So q has this SU2R index and also the spinor index. Spinner of SO4 rotating S3. And also there's a conjugate one in the conjugate spinor, dotted spinor ones. So let's pick our q in the super conformal index to be q1 minus dot, one of this guy. It doesn't really matter. Everything is equivalent. But it helps to pick just one. So I'm following Rastedi et al's convention of using plus dot and minus dot here. Then q, q dagger can be computed. And this becomes delta minus 2j2 minus 2i3 plus this. So this j1, j2 are the spins of SO4. Delta is the scaling dimension, or the energy on S3. And i3 is the carton of SU2. And R is the u1R charge. u1R charge. Yeah, it's really. Then what can be put there? So the thing we can put there depends on lots of parameters. So there are three things we call p, q, and d, and something called xi. And that is trace 3f, q, q dagger, and p2j2 plus j1 minus hono, and qj2 minus j1 minus 2r, and ti3 minus 2r. So these are the combination of the super conformal charges. And finally, you have insertions of something like this. So fi is the generator of carton of flavor symmetry. So this is a horrible expression. Horrible, horrible expression. But it's very explicit. So take a free hypermultiple of u1 charge plus minus 1. So there's a phi psi and phi 2i2 and psi 2i2. So this has charge plus 1 and this has charge minus 1. In this case, this is a free theory. Therefore, the Hilbert space on S3 can be computed very explicitly. It's just a fox space. So in your first lecture of QFT when you enter graduate school, you learned how to do the free field quantization on flat R3. There, you expand in the field in the Fourier modes. Instead here, you expand the modes of phi and psi in terms of a spherical harmonics on S3. It's exactly the same computation conceptually. You can work out this very explicitly. I was thinking of doing the computation, but I don't have time. So let me just say the resulting SCI. SCI free hyper of pqt and x is given by the product of two special functions, gamma pq t1 half x, gamma pq t1 half x inverse, where gamma pq of z is an infinite product of two sets of non-negative integers such that pm plus 1 to n plus 1 and 1 minus z pm qn. So one of the first amazing thing is that somehow mathematicians long time ago came up with this stupid product of tons of factors for their own reasons. But it exactly fits the super conformal index of this free field. I have no idea why mathematicians came up with this. But somehow using this mysterious special function, which is just an infinite product because it's just a focus space, free field contribution is just product of two gamma functions. So this is called elliptic gamma function. So this is something strange. So the next thing, the free fields are OK. How to compute SCI over gauge theory? So these are just choices. You see, super conformal algebra is also a real algebra. Therefore, you have a number of cartoon generators. And what I just erased, delta j1, j2, i3, r, are the generators of the cartoon of the super conformal algebra. I just took all the possible linear combinations that commute with q and q dagger. So the invariant information is that there are just three linear combinations that commute with q and q dagger. And it's just your choice of using p and q and t here. So in fact, when Rastelli et al. first started working this out, they tried various different symbols by doing linear combinations. After two years, they realized that they should have used this particular basis. With that choice, somehow, this relation to the mathematics is very clear. So that's just a convenient choice to compare with the math literature. So SCI of a gauge theory can be computed. Yeah. Yes. U1R symmetry. U1R symmetry will be anomalous for the gauge theory with nonzero beta function. But for zero beta function case, U1R symmetry is guaranteed to be there. Yes. So I guess I don't have time to really do the computation. So the idea is to do the computation, study HS3 in a weakly coupled limit, keeping thinking the gauge group as the flavor symmetry, and then impose gauge invariance. So when you learn quantization of gauge theory, you learn that you need to introduce ghosts. And in the intermediate step, you have tons of unphysical moles in the Hilbert space, which then cut down by the BRST operator. So you just do that. So in a particular case of SU2 gauge theory, let me just skip over the general discussion. So in the case of gauge theory, you pick a Carlton element of SU2. And think of this as an external flavor symmetry variable like this. Then SCI of the gauge theory is given by multiplying a vector-multiple contribution, gamma pqt, gamma pq1, and product of plus minus, gamma pqtz plus minus, gamma pqz plus minus 2, 1 over. So I don't have time to explain, but this is the contribution from the Fox space of the vector-multiple plus ghosts. And you multiply the hyper-contribution, and you need to do this projection down to the gauge invariant part. That is in the case of SU2 just given by a residue integral. So you see there are three factors here, three factors. This is because SU2 is three-dimensional. This is the contribution from the Carlton. Plus comes from the highest weight vector, and minus comes from the lowest weight vector. So this is the expression. So let's consider the case of let's explicitly work out SU2 with NF equals 4. So we need to write down this hyper-contribution. I already gave this general expression for the hyper-contribution, so it's doable. But let's introduce the trick of Gayotto. So hyper, we have been using this expression so far. 1, 2, for the gauge index, and 8, flavor index. Let's consider it as NF equals 2 plus NF equals 2. Then this becomes QA, 1, 2, and I from 1, 2, 4. And another Q, 1, 2. And this time, J is from 1, 2, 4. So there's this SO4 symmetry acting here. Another SO4 symmetry acting here. So what we did is just to artificially reduce the flavor symmetry from SO8 to SO4 times SO4. And I wanted to keep this, but I couldn't. And this can be further decomposed into SU2A times SU2AB from the first one. And to SU2C times SU2D for the second one. So what happens is that 8V we have been talking about becomes, of course, 4 of this and 4 of this SO4. And you know that the composition under this. So this becomes 8V becomes 2A times tensor 2B plus 2C times 2D. And because we have this gauge index, you also have 2 gauge. And this is times 2 gauge. So from this expression, we see that this matter content is transforming under 2A times 2B times 2G, very symmetric. So this is called tri-fundamental. And there's another tri-fundamental. So the good thing is that 8S decomposes under 2A times 2D plus 2B times 2C, for example. And you need to multiply by dual gauge symmetry. So this is the situation. So the original SU2 with 4-flavor duality can be drawn in this way. So in the first setup, you have tri-fundamental ABG, another tri-fundamental GCD. So this is the original setup. SU2 with two flavors and two flavors. In total, there are four flavors. And the dual frame, you have tri-fundamental in ADG prime and another tri-fundamental in BCG. So this is a very nice diagrammatic way of encoding the change from 8V to 8C. This way, you can write this hyper-contribution nicely. So let me do that. Tri-fundamental has three doublets. So we need to multiply tons of gamma PQ, gamma PQ. And the gauge is Z. So t1 half, Z plus minus, A plus minus, B plus minus. And finally, you need to multiply plus, minus, plus, minus, plus, minus, gamma PQ, t1, 2, Z plus minus, C plus minus, D plus minus. So this is it. This is the super-confirmal index of SU2 with four flavors. The variables, so let's call this function I of PQ, t, semicolon AB, semicolon CD. So A minus A, so A inverse is in SU2, A, et cetera. So from this construction, it is clearly clear that the final formula is symmetric under the exchange of A and B. Also, it is symmetric under the exchange of C and D. The question is whether you can replace D and B. So the question is, so if this duality is really true, we should have PQ, t, AB, CD should be equal to PQ, CD, A, C, AD, BC, right? This is not at all obvious. How can you change B and D? So let's check that. So let's see if it works. Hopefully my computer is not dead. Come on. Ah, here it comes. Ta-da. Can you read it? So here is this horrible function. This is just tons of infinite product. And I'm assuming that PQ, t are all of the same order. And I want to do the formal series expansion in terms of t. So I just redefine P and Q to be t times capital P and t times capital Q. Gamma PQ is just an elliptic gamma function. I just typed in the gamma formula into mathematics this morning. And this is gamma PQ1. And this is just a trick of not having to write gamma PQ lots of, lots of things. So this is a tri-fundamental contribution, this one, right? So I just wrote t to 1 half, AB, C, t to 1 half, AB over C, t to 1 half, AC over B, da, da, da. So there are eight terms. This is a vector-multiple contribution, this one, this guy, right? So this is the integration. So it's just a residue integral, very easy. And this is the final SCI, AB, CD. So let's just start from the lowest order computation, right? Yeah, so this is the lowest order computation, 1 plus, blah, blah, blah, blah, blah. Is it symmetric under the exchange of B and D? Well, this time I said B and D. Yeah. So the difference is 0 up to the order we specified. Let's go to the next order. Let's do the computation. Of course, it takes time. Let's hope it ends before Simone needs to start his talk. But we should work. Ah, here it comes. So it's a horrible, horrible expression. Oh my god. Horrible, horrible, horrific expression. But you just take the difference. Ah, and it's not. It's beautiful, isn't it? So this is what Rasteli et al did in the summer 2009. They did much more work to expand it into higher order. But he realized that there should be a better way, because gamma PQ is a special function some mathematicians love. So Rasteli, Leonardo Rasteli told me an interesting story. He found a certain expert on this elliptic gamma functions. He emailed him, asking whether he know whether that math experts know the formula like this. The answer, the reply email, was amazing. Apparently, a PhD student of that math expert had just completed the proof of exactly this equation, not just as a formal power series, but as a meromorphic function. And he was writing up it as a thesis. How can that happen? I mean, this was coming from, I mean, physics-motivated computation of the superconfirmative index. But somehow, mathematicians, with their own motivation, came up with exactly the same combination of these tons of horrible elliptic gamma functions. And they are brave enough to really prove the equation. So I really don't know how that happened. Maybe human beings as a whole has some kind of collective consciousness. So that maybe we are just part of just like neurons in our head, right? I guess neurons inside my head don't know each other. So just like this, maybe there's something going on on the earth trying to do something. Anyway, so this is the story I wanted to tell today. Thank you very much.