 So yesterday, we discussed the ADHDM equations, which describe the modulate space of framed instantons. So I didn't introduce the terminology yesterday. So instantons are the solutions of the equation, the self-dual part of the curvature is equal to 0. And framed means that we identify the gauge equivalent solutions of these equations with the requirement that this gauge transformations approach 1 as x goes to infinity. So the resulting space, which I denoted by m plus k, I should probably introduce the second label for the rank. So we'll be talking about the u-alansantons. So this space has an SUN symmetry, which comes from the possibility of performing a constant gauge formation. So I claimed that there is a certain compactness theorem, which one can prove in this case, which is useful in analyzing the partition function. So let me state this theorem. So in addition to the SUN symmetry, we also have the action of the group u2. So this is the group of rotations. In fact, let me work with this spin cover. So these are the rotations of R4. And the equations, the original equations are invariant under all symmetries of R4. In fact, they're invariant under the conformal symmetries. But we introduced certain deformation when we discussed the compactification of the modular space, which was adding to the so-called real map equation, a constant. So this deformation breaks the rotational symmetry from a spin forward down to u2. This deformation uses a specific complex structure on R4. So in order to single out one equation out of three, so the other two were combined into the homomorphic equation, I have to distinguish a specific pair of homomorphic coordinates. And so this is the result of the symmetry. So it basically acts on b1 and b2 as a doublet. i is invariant. And j transforms as a determinant representation. So if t is in u2, you act. So t is a 2 by 2 matrix. You act with this matrix on b1, b2. And you multiply j by the determinant of this matrix. So the commutator of u1, b2 will be multiplied by this determinant. And that's why this equation is compatible with the symmetry. Now, so this is a big group. And in applications, we only use its maximal torus. So let me denote it as the torus of rotations. So if you look at the fixed point, set of fixed points, under the maximal torus and the maximal torus of the group of the conjugation formations, so those fixed points were classified yesterday. And they were simply isolated. There was a finite number of those points. T rotation cross t gauge fixed points and tuples of partitions whose total size is equal to k. So each partition is a young diagram. So this is lambda alpha for one of the values of alpha. And the size of the partition is the number of squares. So each square, remember, each square was a vector in the k dimensional vector space. And so if you have a vector, if you have a square with coordinates i and j, j goes here, i goes here, then this vector, so alpha ij, this is the vector b1 to the power i minus 1, b2 to the power j minus 1, i applied to the one-dimensional vector, the eigenvector of the transformation from the global gauge rotations corresponding to the eigenvalue e to the i alpha, e to the alpha. So this is the situation when you have the full maximum torus at your disposal. But it may happen that the problem you are solving forces you to consider special specific values of the equivalent parameters. So if these parameters, epsilon 1, epsilon 2, are not generic, you may not have the full torus at your disposal. So remember, OK, so this is my two-dimensional torus. And epsilon 1, epsilon 2 give me a direction, a specific slope for the one-parametric group, which is generated by this transformation. So if I start iterating it, put some coefficient s. I will generate one-parametric subgroup. So if the ratio epsilon 1, epsilon 2 is not rational, then the subgroup will actually fill the torus densely. And that means that you have the whole torus at your disposal. And so the set of fixed points is precisely that. But if this ratio is rational, then you actually have only one-dimensional symmetry group. So then this spans s1 with u1 embedded into u1 cross u1 at some rational slope. And so since this group is strictly smaller than the maximum torus, the set of fixed points will be strictly larger and need not be isolated. So these points are isolated. So let's say this is my modular space. So this is the picture of these fixed points. So these are just examples of these fixed points for the case of general epsilons. But if the epsilons are non-generic, then what happens is that these points become continuous submanifolds. So this is what happens for non-generic epsilons. In fact, if you further reduce the symmetry group, namely you start playing with Coulomb parameters, the parameters which the generators of the global symmetry transformations, your set of fixed points becomes even larger and actually can go off to infinity. So this is what happens if A alphas are non-generic. And that's kind of bad for the partition function. That means that you actually have integrated out something which is massless, can run away to infinity. And so that means that that signals actual singularity of the partition function and requires further analysis. But I claim that if the differences between the Coulomb parameters, so the different eigenvalues, are not of the form epsilon 1 times a positive integer plus epsilon 2 times a positive integer, then this never happens. The fixed point set is always. So let me prove it. Yes. Yes. So there is a case when you specify A's to the value that it front-case number of data. Say what? So you specify A's parameter to front-case number of data or are these choices? No, no, no, not in this situation. Not in this case, but in general, the rest of it. So this is, so if you, when we'll consider a theory with matter, we'll have other types of parameters which are the masses. And so those correspond to, those can be used to truncate the size of young diagrams. So those correspond to the possibility of adding new equations to the set of equations I've written before. And so when you add more equations, you restrict the modular space. So what happens then is that your modular space becomes strictly smaller. So you go to vertices. Vertices? Vortices. It could be, sometimes it could be interpreted as vortices, sometimes it could be interpreted as something else. But that's what you do. So you have extra equations from extra matter. We'll do that momentarily because we will need to consider theories which correspond to asymptotically super conformal theories. And for those theories, the modular space which we're integrating over will be of virtual dimension 0. So ultimately, you actually want to count points, some points. But right now, these quantities appear in the denominator of the localization formula. And so they lead to the poles if something is non-generic. What Maxim is asking is in theories with matter, you have numerators in the localization contribution. And so those can vanish, which simplify life sometimes. OK, so the reason I want to present this compactness theorem is because it's a kind of baby version of a more complicated story, which leads to the non-perturbative Dyson-Schwinger equations. So let me assume from the very beginning that the parameters epsilon 1, epsilon 2, so these are infinitesimal rotations of spacetime that they are integers. So I can rescale them. If the ratio is rational, I can rescale them so that they're integers. And so what we are looking at, we are looking at the solutions of the following equations. This is the infinitesimal version of the equations which I wrote yesterday, hopefully without the typos which I had yesterday. OK, we'll see that in a moment. So what this equation says is that if I do infinitesimal rotation of v1 and v2, I can undo that by the infinitesimal UK symmetry. And now, remember, there was a statement that instead of solving this real moment by equation with zeta and then dividing by the group UK, you can actually forget about this equation. You impose a stability condition, which is phrased in holomorphic terms. And then you divide by the group of general linear transformations. So that allows you to think of phi as a general just k by k matrix. So then i is acted on by the, so let me write this like this. This is phi a. So a is the diagonal matrix with eigenvalues a1 an. i is the map from n to k. So phi is this compensating transformation which acts from the space k to itself. And finally, epsilon 1 plus epsilon 2 j minus j a is equal to, sorry, minus aj is equal to j phi. So this sum epsilon 1 plus epsilon 2 is the linearization of, it's a logarithm of this determinant. So what this equation says is that the infinitesimal rotation and the global gauge transformation can be undone by the local GLK transformation. So this defines the fixed points, the set of fixed points, which we want to study. And so what we said yesterday was that from this equation, well, we know that i cannot be zero because if i were zero, then this condition would not be satisfied. So it means that at least one component of phi is non-zero. So it means that at least one of the eigenvalues of phi has to be equal to the eigenvalue of a. And so the spectrum of phi has the form. So this part comes from acting on the eigenvector of phi with b1 and b2. So this, like in supersymmetry, by looking at this equation, if psi is the eigenvector of phi with eigenvalue lambda, then b1 times psi will again be an eigenvector using this equation. So I'm using that the commutator of phi with b1 is equal to epsilon b1. So I can pull it like here. And so what I get is lambda plus epsilon 1 times b1 psi. So it means that if this vector is not killed by b1, then it is an eigenvector of phi with the new eigenvalue, which is lambda plus epsilon 1. And so by acting with b1 and b2, you generate these shifts. And from this it follows that j has to vanish because for j, the eigenvalues go in a different direction. You see, j wants to subtract from some combination of epsilon 1 plus epsilon 2. So this argument works when the signs of epsilon 1 and epsilon 2 are the same, so if they are of equal sign. So for p and q both, let's say positive or negative. I mean, by scaling I can always assume, let's say, that q is positive. And I'm assuming that both p and q are not equal to 0. So that's important for compactness theorem, otherwise it will not work. So this is an important assumption. So if this is the case, then by these arguments you conclude that j is 0. If the signs on p and q are opposite, then you cannot use this argument because, well, epsilon 1 plus epsilon 2, for example, could even vanish. And so it wouldn't matter. So that will not be enough to conclude that j is equal to 0. We'll need to argue differently. But let's first dispose of this case. Actually, it's more complicated than others. So the idea is the following. Well, first of all, if a's are generic, we can actually split the space k into the subspaces such that phi restricted on the subspace k alpha is simply. And under my assumption that epsilon 1, epsilon 2 are integers, positive integers, well, this is some non-negative integer. So the spectrum of phi consists possibly of several arithmetic progressions, which start at different complex numbers a alpha. And so we can separate the space into the orthogonal summands corresponding to different values of alpha. So it's actually enough to understand it for a fixed value of alpha because they don't talk to each other. And therefore, so what I want to do, I want to find the bound on the norm squared to find a bound on the norm of all these matrices. Since we just proved that j equals to 0, this makes life easier. And in fact, the norm of i, you can compute from my equation, but we'll not do that. What we will do is the following trick. So now I work at fixed alpha. And so by shifting a, I can assume that a alpha is equal to 0. It doesn't really matter. So my equations now have the following form. This is 0. I forget about this. This is p and this is q. So I want to find some bound on this norm. So we now can split the space k alpha into, in turn, some of the subspaces on which phi has eigenvalue k alpha n is the kernel of phi minus n. So n is non-negative integer. So the question is, well, so this integer can be arbitrarily large, not too large, because in the dimension of k is fixed. So we cannot have more than k, little k values for this n. But the danger is to have the matrices be with matrix elements which are arbitrarily high, arbitrarily large. So what we want to show is that even though these are non-isolated fixed points, they actually sit inside some big ball in this modular space. So this is what we want to find. We want to find the bound on the size of this ball. So let's compute. Let me define the following quantity, delta sub n. It's a trace over the subspace k alpha n of b1, b1 dagger, plus b2, b2 dagger, plus i, i dagger. So you see, b1, when acting on the space k alpha n, sends it to the space k alpha n plus p. And b2, this is exactly the repetition of this argument. And b dagger acts in the opposite direction. So the norm which I wanted to compute is just a sum of these partial norms. So now we need to find some identity for these norms. And what we can do, we can, using the real moment map equation. So here, for this normal analysis, I'm using the unitary picture. So I'm only devising by the unitary to summations UK and using the real moment equation. So from this equation, I'm using the j is equal to 0. So what you can see is that this guy is almost everything that you see on the left-hand side. What you see on the left-hand side is bb dagger plus i, i dagger minus b dagger b. So we can rewrite this as 1 over zeta trace k alpha n zeta times 1 plus b1 dagger b1 plus b2 dagger b2. This is just the algebraic manipulation. And now using this property, well, you see that now b1 takes this space and pushes it further into the space with eigenvalue n plus p. b2 does it with the map set to the space n plus q. So you can estimate, use the fact that the trace is cyclic. So this is the dimension of the space k of n from this unit. And then you can estimate it by delta of n plus p plus delta n plus q. So you get some kind of sequence of telescopic inequalities. And eventually, so you use this inequalities, you effectively increase the value of n until you get to the point when this space k alpha n is simply 0. So k alpha n is 0 for n much bigger, let's say, than k. Just bigger than k, yes. So from that, you can get an estimate. And the conservative estimate is rather high, but it doesn't matter. It's finite. Could you repeat again how you found b1 dagger b1? OK, so this is the definition. So this is the real moment equation. So now I want to, I'm claiming the following. So the trace in the space k alpha n of b1 dagger b1, I want to rewrite as a trace of the space k alpha n plus p of b1, b1 dagger. So normally under the trace, you can exchange, but you have to be careful in which spaces the operators act. So b1 acts from k alpha n to k alpha n plus p. And so now, since in this formula, k alpha n plus p appears outside, that's the space at which I'm taking the trace. And so this is less than or equal to the same quantity where I'm adding something which is non-negative, namely b2 b2 dagger plus ii dagger. So this is a very conservative estimate. I mean, this is small, and I'm increasing it by, which is, so if I divide by zeta, this is delta n plus p. So you can introduce generalized Fibonacci numbers for this problem to package this. So usually Fibonacci numbers are solving the equation of that kind, I think. So the generalized Fibonacci, so you start with f1 equals 1, f less than 0 equals to 0, and then you impose this condition. So just like for ordinary Fibonacci numbers, you can write a formula using the eigenvalues of the two-by-two matrix. So here you solve the Fermi equation lambda to the p plus lambda to the q is equal to 1. So it has a max of p and q solutions. And so in terms of, so you can write the formula for f of n as a sum over all solutions of this equation, lambda to the n, and some prefactors, initial conditions. The point is that it shows that these numbers, they grow, but at most exponentially. So there is some constant by which you can bound them. And from this inequalities, so the inequalities of that type, delta n is less than k plus delta n plus p plus delta n plus q. Again, I'm making conservative estimates. And increasing the right-hand side, it follows that delta n is less than equal to the sum k n plus prime fn plus prime plus 1, which means that this can be bound by k exponential. Say it again? Should we see the minus delta? Yes, but that's probably the same. You can multiply by, oh, OK, maybe I won't. Right, so it follows that the sum over delta n has some finite bound. OK, so that's the way this argument works. So a little bit of work, but not too much. And for now when p and q have opposite signs, then the following trick works. If pq is less than 0. Well, first of all, you need to prove that j equals to 0, because now you cannot use the eigenvalues. I claim it nevertheless. So how do I prove that? Well, let me assume now that n is one-dimensional. So this is what we did here. We split. So the problem splits into n-Abelian problems when the Coulomb parameters are generic. And so let me do it now in the Abelian case. So the equation B1, the commutator of B1 B2 plus ij equals to 0, let me take the trace of this equation. So it will imply that ji is equal to 0. So now, you see, normally I would have to put a trace here. But trace of the 1 by 1 matrix is equal to the matrix itself. So this is because i was acting from n to k. So this is the n to n operator, which is one-dimensional. So it's a number. So j annihilates the image of i. Now, if I multiply this equation by B1 or B2, in fact, if I multiply it by any power of a linear combination of B1 and B2 and take a trace, so this is 0 because of the moment of my equation. But now, if you look carefully, you can rewrite this as the commutator of alpha B1 plus beta B2 gamma B1 plus delta B2, provided that alpha delta minus beta gamma is equal to 1. Now you see, you have the power of the matrix times the commutator of this matrix with something, and then you take a trace. So that means that that part of the trace is 0. And so you are left with j multiplying alpha B1 plus beta B2 with power n for any alpha, beta, and n. So if you knew that B1 and B2 were commuting, that would imply that j vanishes polynomial of B1 B2 applied to i, which is by the stability assumption is equal to k, which means that j vanishes on k, which means that j vanishes. But B1 and B2 may not commute, but that's OK, because now, if you try to insert a commutator of B1 B2 into some word involving B's i and j here, you can use the moment of my equation to replace it by as a product of two words. And so by induction, if you prove that j annihilates all words in letters B1 and B2 acting on i up to the, let's say, depth k, then by this argument, if you have a longer word, and then inside you have a commutator of B1 and B2 that vanishes. That means that inside this, when you close any word with B's by j, inside B's have actually commute. And that means that this argument works, j vanishes. And then it means that actually they do commute, because if j is 0, then B1 commutator of B2 is equal to 0. OK, so that's nice, but that's still not the full story. But actually, the full story is almost over. I claim that if this is the case, if epsilon 1, epsilon 2 are in rational relation with opposite signs, then actually the corresponding fixed point is invariant under the full two-dimensional torus. So if pq is less than 1, actually it follows that any epsilon 1, epsilon 2 fix this configuration. And this is how we prove it. Look at the following operator. So suppose p less than 0, q is greater than 0. So I will raise B1 to the power q, B2 to the power minus p. If p is negative, this is positive, so it's an allowed operator. Well, from these fixed point equations, it follows that phi come used with n, because the weight of this combination is actually 0. So B1 has weight p, and B2 has weight q. So p times q minus p times q is equal to 0. And so n, so curly n. So curly n is a nilpotent operator which commutes with u1, which is generated by this epsilon 1, epsilon 2. Therefore, so if I decompose k as a sum of the eigenspaces, so these are the eigenspaces of phi. So u1 is generated by phi, in the sense. Each space kn is invariant under the action of this nilpotent operator. And then there is a theorem by Jacobson and Morozov, which says that a nilpotent operator can be included into the SL2 subalgebra. So it follows that there exist operators h and n star such that the commutator of nh with n is n, commutator with h with n star is minus n star, and n with n star is twice h. And so this h becomes the second grating, so that kn and this further split into the knm. And that means that now you have the u1 cross u1 symmetry, so h and phi. And then we are back in business as yesterday. So we have the full two-dimensional torus of symmetries. So that proves that as long as both epsilon 1 and epsilon 2 are non-zero, so this is what I used, the set of fixed points will be compact. OK, so now let me say something new. So so far, we discussed pure n equals 2 d equals 4 theory. Let's add some matter. It is convenient to, in some sense, so if you want to add matter while preserving the asymptotic freedom of theory, and so if we only add matter in the fundamental or by fundamental presentation if you have a product of unitary groups as a gauge group, then the choices are actually limited. So you only can work with quivers, quiver theories. And the condition that the theory is asymptotically free or asymptotically conformal means that these quivers have to be of the ad type, either a fine linking diagrams of ad type or finite linking diagram of ad type. And that means in turn that the matter content of the theory can be obtained by certain orbital truncation from the n equals 4 theory. So it's all can be gotten by Orbe faults of n equals 4 superannual theory with maybe much bigger gauge group, possibly with some massive deformation. So it could be n equals 2 star theory. And so let me simply explain what happens to the adHM data when we pass from n equals 2 theory to n equals 4 theory, because everything else will be straightforward orbital truncation. Now, unfortunately, unlike the pure instant on case where there was this nice reciprocity theorem mapping adHM data to gauge fields and back and forth, in the n equals 4 case, I'm not aware of such theorem, so what I'm going to present will be really construction by analogy, not systematic derivation. So n equals 4 gauge theory has, in addition to the gauge field, it has six scalars, which we can organize into the complex scalar sigma and its conjugate as n equals 2 case. And the other four, I will twist, let's say in the Waffa-Wieden way, and so they will become a scalar and the cell dual 2 form. And so n remains the gauge field. So this is done in order to give a certain mathematical structure to the localization equations, which we will be analyzing. Again, I'm working on flat space on R4. So in fact, I'm not changing my theory in any way. You can actually just use some flavor indices to label your fields. But I will single out one of the supercharges out of 16 supercharges of this theory. And from the point of view of that supercharge, it's convenient to write the theory in this way. So the equations in which we will be localizing our theory deformation of the instanton equations, I think it starts like this. So you add the commutator of the cell dual 2 form with the scalar, and then there is another term where with some gymnastics you can produce a bilinear map from cell dual 2 forms into cell dual 2 forms. And then there is an equation that da. So this is just cross-project. Yeah, it's some kind of cross-project. It's the, so if you think about the cell dual 2 forms as a triplet of SU2, there is a, well, it's a triplet of SU2 times the adjoint representation of the Lie algebra. So you have the, that's not three-dimensional, so three times the Lie algebra, right? There is a Lie algebra structure there. There is also a structure, there is also a structure constant. Anyway, but, and then gosh, so what is the, so I guess, so this, here I want to write an equation of this type, I think. So this equation is valued in, again, in cell dual 2 forms, valued in adjoint bundle. And this equation is valued in one forms. Yes, well, on R4 it's kind of tricky because it's not compact, so you have to be careful. Presumably with some vanishing conditions at infinity, these equations imply that B and C are actually equal to 0. But the point is that now that you have more fields and more equations, and you produce the supersymmetric, I mean, the path integral over these fields and their superpartners, even though physically it will reduce to the same, to the integral of the same modulus space of instantons because on solutions B and C vanish. So these equations imply that B and C equals to 0. But since you use more fermions to impose these equations, the actual measure of integration over the modulus space of instantons will be not 1, as in n equals to 2 case, but it will be essentially the Euler class of the tangent bundle or cotangent bundle. So you upset the sigma and sigma bar to 0? Sigma and sigma bar are the, so they are in the equivalent multiplet. So if you like, there is also the equation of this form. This part is not changed. I'm not, this is, so, I mean, I'm working in the paradigm of fields equations and symmetries. And so this is the part of the symmetry multiplet. These are the equations, and these are my fields. So what's nice about these equations is that they actually kind of match the set of fields. Previously, we had the fields were one forms, but the equations were self-dual two forms, and the symmetry were scalars. So there was some mismatch. You were comparing one forms to, so in n equals 2 case, you're comparing one forms to self-dual two forms and scalars. But here, we have one forms here, one forms here, and we have self-dual two forms here, self-dual two forms here, and we have a scalar here and a scalar in the symmetry. And so this is an example of what digraph Moore called a balanced topological theory. So we can mimic, right, sorry, one more thing. So these equations, you can actually derive from some kind of super potential. Namely, if you look carefully, you'll see that you can write some kind of functional, which is like integral trace B plus wedge F A plus some sort of change assignments action plus 1 third B plus B. So if I vary this functional W with respect to B, I'll get the first equation. And if I vary it with respect to A, and I didn't make a mistake, maybe I didn't make a mistake, I will not get the second equation, but the second equation means that the gradient of, so maybe I made a mistake in the first equation. So the gradient, so the variation of W with respect to A is equal to the gauge of summation generated by C. So maybe I don't need this part. And the variation of W with respect to B plus is the gauge of summation of B generated by C. So these equations are kind of generalized gradient flow equations. So that's what we will mimic on the ADHM construction. So I claim that in order to produce the same measure, which I just erased, the Euler class on the module space of instantons, the ADHM data will have to be supplemented by two more matrices. And the equations will be now, so we'll have more equations. So the first equation, so I start with the ADHM equation as a complex map, but then I add to it the conjugate of the commutator of B3 and B4. So somehow you should see part of it in this structure. You see, if I choose complex structure on R4, then B plus can split as B2 0 plus B11 omega plus B02, which is conjugate. And so B2 0 is like B3, and B11 omega plus IC is like B4 in this notation. And then I will have equations. B1 commutator would be 3 plus B4 commutator would be 2 dagger is equal to 0. B1 B4 plus B2 B3 dagger is equal to 0. And finally there, or not finally. So then there is equation B1 B1 dagger plus B2 B2 dagger. And so I can size with the sum of all four commutators plus i i dagger minus j dagger j is equal to zeta. And finally B3 i minus plus j B4 dagger is equal to 0. And B4 i minus j B3 dagger is equal to 0. So you can compare these equations to the equations which follow from the superpotential. So the superpotential will not respect the symmetry of rotation of four matrices. It will be something like B commutator B1 B2 plus ij. And the equations will be that the gradient of w respect to the field commission conjugate is equal to the gauge variation of this field x generated by B4. So for example, if x is B3, the derivative of w respect to B3 is the first part of the addition equation. And then the gauge transformation of B3 is a commutator of B3 with B4. And so you take commission conjugate. So that's the first equation. And if you take the variation respect to j, you'll get B3 times i. So B3 times i. And this is minus j B4 is the gauge variation of j. So that's the logic here. Again, so here is a vanishing theorem which is easy to prove that these equations imply that B3 and B4 are equal to 0 when zeta is positive. But now if you set up the calculation, the integration of 1 over the modulus space of solutions of these equations. And now we have more symmetry. So the symmetry is u4 is SU4. Sorry, it's not quite SU4. It's the maximal torus of SU4. So you can multiply matrices B a by any phases provided that the product of these phases is equal to 1. So the first equation, you multiply the commutator of B1, B2 by T1, T2. And this commutator of B3 and B4 is multiplied by T3, T4, which is inverse of T1, T2 by this condition. So it means that compared to the n equals 2 case where we had two epsilon parameters, the parameters of rotation, here we have three parameters. So we have epsilon 1, epsilon 2, epsilon 3, and epsilon 4, which is negative epsilon 1 plus epsilon 2 plus epsilon 3. The physical meaning of these parameters is, so epsilon 1, epsilon 2 correspond to the rotations of spacetime. And epsilon 3 is the mass of the adjoint hypermultiply. So this is the mass of adjoint. So now if you do the localization computation for the integral of the module-specific solutions to these equations, which means literally what? It means that for every variable here, so you have six types of complex variables, or eight Hermitian matrices and four rectangular matrices, you introduce the fermionic variable of the same type. And then for each equation, you introduce a pair of boson and fermion, the Lagrange multiplier for the equation and its fermionic partner. And then for the symmetry, so the symmetry is UK, you also introduce a small multiplet like sigma, sigma bar and eta. And then you write the standard homological field theory Lagrangian, which enforces these equations and takes care of the symmetry, which would deform using this global symmetry and the global framing symmetry, which still acts on i and j. Then the result would be the partition function, which is again the sum over all n-tuples of partitions. And then you'll get the product of the eigenvalues of phi acting on the tangent space to the modular space of instantons at the corresponding n-tuples of partitions. Let me call them lambda. So these lambdas have the differences of the Coulomb parameters plus integral linear combinations of epsilon 1, epsilon 2. And now you'll have in the numerator the same eigenvalues shifted by epsilon 3. And even though it's not obvious, this expression is actually symmetric under exchanges of epsilon 3 and epsilon 4. So these equations were completely symmetric. But there was no preferred, I mean there was no, well I can exchange b3 with b4 and b4 with negative b3. The equations remain the same. And so there should be symmetry which exchanges these parameters. And that means that when I call epsilon 3 a mass of the adjoint hypermultiple, I could equally well call epsilon 4 the mass of the adjoint hypermultiple. The notion of mass is slightly ambiguous in the mega background. So you say that in this case the proof is not known. But if you can construct the explicit form of the gauge field, et cetera, from given this solution. No, this is what I don't know. I mean, oh sorry. Of course, the equations, since the equations imply that b3 and b4 equals to 0, the equations reduce the old adjoint equations. So of course, at the level of, at the set theoretical level, of course, you have the correspondence. So you get, it's the old instant which you get from the old b1, b2, and g. But these equations have the meaning themselves. Even though the solutions are simple, but the equations are more important than the solutions. Somehow, I would like to have a way of connecting the equations to the equations without assuming that the equations imply this thing. Because also, if you, instead of r4 you work on some other manifold, the vanishing theorem does not hold. So then you have, anyway. So from this, by looking at these equations and observing that these equations have a symmetry, let's say, SU2 symmetry which rotates b1 into b2, you can perform some orbit faults which will replace this r4 spacetime by the ALE space. That's one option. There is another SU2 symmetry. So the symmetry which these equations actually have is the product of SU2 cross u1 cross SU2. So this SU2 acts on b1, b2 as a doublet. This SU2 acts on b3 and b4 as a doublet. And this u1 scales simultaneously b1, b2, and scales in opposite way b3 and b4. So if you have a discrete subgroup of this SU2, or if you have a discrete subgroup of this SU2, and so there are AD classifications of such subgroups, you can impose additional equivalence condition under the subgroups and produce other equations which will produce the ADHM data for quiver gauge theories. So from this, you get for free, essentially. You get, so these are quiver gauge theories on r4, or if you also impose the orbital projection by gamma 1, it will be quiver gauge theories on r4 mod gamma 1. But now you can say, you can observe one thing, that here you have some kind of symmetry between 1, 2 space and 3, 4 space. And these two procedures look symmetric, but the equations are not quite symmetric because you have ij, which couples to b1, b2, but nothing couples to b3, b4. And so you can try to fix that symmetry by introducing one more set of fields, more data. So we introduce new space and tilde. And maps i tilde, which maps n tilde to k, j tilde, which maps k to n tilde. And modify these equations. So here we add additional terms into the real moment map. And then now symmetrically, so b3 wants to annihilate i. So b1 wants to annihilate i tilde. So now life looks truly symmetric. So what these equations describe, they describe the instantons, which live on the intersecting worlds. So in the brain world picture, in the brain world picture, you have n brains, which span r4 with complex coordinates z1 and z2. And then there is n tilde brains, which live on c2 with coordinates z3 and z4. And these two complex planes intersect at one point. And this is the point where there is some interaction between this data. So these equations imply that separately b1, b2 plus ij equals to 0. So this follows from these equations. And separately b3, b4 plus i tilde, j tilde is equal to 0. So this is the adhm equation for instantons. So you get un instantons on c2, 1, 2. And you get un tilde instantons on c2, 3, 4. But you don't know in advance what is the instanton charge of the solutions. So you may have the solution where you have charge k1 on this plane, k1, 2, and charge k3, 4. On that plane, complex plane, so on that four dimensional space. And the sum of these charges is greater than or equal to k. So some of these instantons are actually shared by both planes. And so in what sense are they shared? Well, this is algebraic condition. So what we need to do, we need to look at the space generated by applying to the image of i polynomials in b1, b2. So this is the space k1, 2. And similarly, with b3, b4, i tilde, and tilde. And so these two spaces, the sum should be equal to k. So this is the stability condition. This is what follows from this equation when z is positive. But these two spaces may intersect. And so the dimension of intersection, this is the number of instantons trapped at the intersection. So I call the intersection the cross. And this modulate space is the modulate space of crossed instantons. So we live on a cross, on a coternioni cross. Now the game which you can now play, so there's a partition function for this cross instanton configuration. You can first integrate out instantons living on one plane. And that will produce an observable in the theory on the second plane. And then the compactness theorem, which again holds, will imply that there's a function of certain equivalent parameters, which from the point of view of the observer living on the horizontal plane are just some spectral, zero spectral parameters. The correlation functions of those observables have no poles. And from that, follow Dyson-Schminger equations. So compactness theorem implies regularity of correlation functions, implies word identities of safety, actually. So from these Dyson-Schminger equations, one derives B, P, Z, and K, Z equations of two-dimensional conformal field theories. So let's just say that I'm done, finished. Thank you. No more questions. Sorry, sorry. The fact that in some instant the charges shared by 1, 2, and 3, 4 seems to mean that they kind of make a bounce data or interact with the other. But still you're solving the actual solution you're constructing seems to be the self-dual solution, not the more complicated solution. So talk with each other, but the profile is not looking good. Well, you see the point is that, well, they talk to each other, in particular, through the real moment map equation. So in order to produce the instant on-gauge field via usual edichem construction, you need this equation and the real map equation, which involves only B1 and B2 and INJ, not B3 before. So I don't know how to produce something like a gauge field using solutions of these equations. Because now once you added second brain, second set of brains, it is not true that B3 and B4 is 0. So this is no longer 0. So qualitatively, there are solutions for which these two parts are very far away, separated. So then approximately, you can construct instantons of some charge k1, 2 leaving here and instantons leaving here. But what happens differential geometrically when they all come together, nobody knows. This here is only the algebraic geometric description which works, only in terms of the holomorphic data, not in terms of the differential geometric data. So the fixed points on this modular space consist now of two types of young diagrams. So some young diagrams grow in directions epsilon 1, epsilon 2. And other young diagrams grow in directions epsilon 3 and epsilon 4. And so generically, they have separated in complex plane. But sometimes they touch. And so when they touch is when the possible poles may appear, but then they come back in a theorem that the poles actually cancels. So this configuration I call the butterfly. When two young diagrams start touching, these are like wings of a butterfly. And to the observer who lives on the horizontal plane, things look in a slightly mysterious way, because it appears that the instanton can actually disappear. So you start with the configuration of instanton charge k. And then one box of young diagram can disappear, gets chipped off. It doesn't disappear in the combined system, because it just goes off into the other direction. And so this is what happens when these two diagrams kiss each other. So they kiss. And then the box disappears in one diagram and gets attached to another diagram. And so the poles cancel between these two contributions. Maybe I should describe the solutions of these equations in the case of k equals 1. Maybe it's helpful. So if k goes to 1 and n equals to 1 and n until this equal to 1, so everything is one dimensional. Then all matrices are just numbers. All commutators vanish. So you have very simple equations to deal with. Oh yes, I forgot one more equation. j tilde i minus i tilde dagger j equals dagger equals to 0. So one proves very quickly that j actually vanishes. And so the equations which you have are, you only need to look at these equations. So b3 i equals to 0, b4 i equals to 0, b1 i tilde is equal to 0, b2 i tilde is equal to 0. And these are all numbers. So bA is just a complex number. i is a complex number. i tilde is a complex number. And then you have a real moment equation which says that the absolute value squared of i plus absolute value squared of i tilde is equal to zeta. So this is a three-dimensional sphere. But now we have several branches. So if at least one of the numbers b3 or b4 is non-zero, then i equals to 0. And so you have a branch which is asomorphic to c2. So it's parameterized by b3 and b4. On this branch, i equals to 0. Then the norm of i tilde is fixed. And up to location of formation, i tilde can be set to be just square root of zeta. So that's one branch. So inside this branch, there is a special point when b3 and b4 vanish. If they vanish, then i may be non-zero. Similarly, there is a branch when b1 and b2 are non-zero. So then i tilde is 0, and i is equal to square root of zeta. And then there is a special point, 0 when i tilde may become non-zero. And then there is a branch when all b are equal to 0. So these equations are solved trivially. And here you have S3 mod u1. So this is Cp1 parameterized by the ratio i to i tilde. And so what this module space is, you have one instanton which can sit either on the vertical plane, so this is that branch, or it can sit somewhere away from the origin, or it can sit somewhere on the horizontal plane, or when it reaches this point, so this is the position of this instanton, reaches the intersection point, then it actually starts thinking whether it wants to stay or it wants to go, and has the whole two-dimensional sphere worth of choices. And so that's the extra component of this modular space. And so now in the classification of the fixed points, they remember that we were supposed to have n-tuples of young diagrams, so n plus n-tuple, n plus n tilde, tuple of young diagrams of total size 1. And so we have two choices, either one box empty or empty one box, and so these two choices are these two points. So these are these two fixed points in this case. One of them is visible to the observer in the horizontal plane, and another one is not visible. So when instanton has gone to a different branch, to us it seems that we lost instanton charge. Once everything is a bit in string theory, nothing is lost. Go string theory. Thank you.