 Today, we'll be studying instant terms in gauge theory. And we'll start with an exercise of sorts. We'll take the spacetime to be of the form of a product over three manifold. And the timeline, let's say it's compact. And we'll try an ansatz for the gauge field of the following simple form. So it's going to be some function of time. So time is the coordinate on the real line times a connection. So theta is going to be a flat connection on M3. So it means that it solves this equation. This is the derivative of M3. So theta does not depend on time. The time dependence is in the scale factor. So if you compute the curvature of this connection, it's going to be something like f dot dt w h theta plus f d theta plus f squared theta w h theta, which you can rewrite using this equation as in this form. So this is the only component which has time component, time direction. And this is purely a spacelike. So when you compute the Young-Middle's action with the ansatz, you'll get the integral over time. And then some constant times f dot squared. Let me assume, actually, for simplicity that f is a scalar function. So this is simply, and then some other constant times f squared f minus 1 squared. So this constant knows something about this connection theta. So c1 is the integral over M3 of theta star theta. And c2 is the integral theta w h theta w h star theta w h theta. So I'd like to point out that this connection theta may be a pure gauge. Actually, in fact, if M3 is, let's say, three sphere, then it has to be a flat connection. But it may be a flat connection with non-trivial, winding number. So you can actually estimate these quantities using your third constant, which is the chancems invariant, which is the integral of theta cube. I'm skipping some factors of 2 pi here. It's not important. And so actually, if I edit at the topological term, the integral of trace f by gf, this will also appear. At any rate. So for this ansatz, the action reduces to the classical mechanical action of a particle. So the particle coordinate is f. And what you can see is that this particle is moving in the potential of the kind which we started yesterday. So this is the f coordinate, and this is the potential. So this potential looks like the one just accepted is shifted. So it's a double well potential. It has two minima. One is at f equals to 0. So a is trivial. And another is at f equals to 1, where a is flat. And so from yesterday, we know what the classical solutions look like. So the classical solutions, which are in general complex, will be the rational windings of an elliptic curve, which corresponds to a complex f in general. But there is one solution which is very easy to construct. Namely, it's the one which is half of the full periodic solution. So this is for periodic solutions where we replace r1 by s1. But if we don't replace it so we consider the timeline to be truly infinite, then there is a solution which is a tunneling between 0 and 1. And so that's a solution that will be solving the first equation. f dot is proportional to f minus 1. So that's the instanton in the sense of quantum mechanics. And that's actually going to be the instanton which was discovered by Belavin, Polyakov, Schwartz, and Tupkin in 1975, I believe, but in a slightly unusual form. So this is the BPST instanton in radial coordinates. So what Sasha Belavin et al. did, they were, of course, they were studying solutions of the, well, they were studying a self-duality equations. Today we will be replacing change in orientation, so it will be an unself-duality equation on R4, which have a rotational symmetry. Actually, it's S of 4 symmetry with gauge group SU2. So here I slightly generalize, but if you keep the gauge group to be SU2 and take, so R4, if you write the metric in the radial spherical coordinates, you see that this metric is actually conformally equivalent to the metric on the direct product of S3 times the real line. So it's R squared times R squared over R squared plus, so this is the round S3 metric. And in the exercises, you're supposed to find that these equations are conformally invariant. So the solution of the equations on flat space is the same solution of these equations on that space. And now by going to the coordinate, which is a logarithm of the radius, with now time changing from minus infinity to plus infinity, that becomes simply the metric on the product R1 across S3. And so the ansatz of BPST corresponds to taking theta to be G inverse dG, where G is the map from S3 to SU2, which is simply the isomorphism. So when you identify the group SU2 with three dimensional sphere, so you can view it as a map which identifies the sphere in space with the group SU2, which is kind of isotopic sphere. And so the F, which you will get by solving this equation, I will not do it for you, it's a neat exercise, will precisely correspond to the scale factor which BPST found. Of course, then once you found the solution here and you map it here, you can produce more solutions by applying translational invariance in conformal transformations. And so that gives you the modular space of solutions. Unfortunately, that method is hard to generalize to higher instanton charges. So here the instanton charge is the number of times you tunneled between 0 and 1, which is only 1. And if you look at the solutions which I described yesterday, they actually tunnel back and forth. So they net instanton charge for those solutions is 0. And so they correspond to the true smooth instanton and instanton configurations. But today, I'm not going to be interested in those. I want to generalize this construction and describe all instanton solutions on R4, which actually extend to the conformal compactification of R4. So it's another conformal story. So you can add one point to R4 and produce force field topologically. And you can do some conformal rescaling of the metric here to produce this sphere isometrically. And so the solutions which we're interested in will extend to the four-dimensional sphere, which is compact and on which you can define the topological number. So we will be interested in the solutions of general charge and for the group SUn. So it turns out that the solutions can be described more explicitly than naively you would think. So the equations which we're solving are partial differential equations. But it turns out that instead of solving the PDEs, you can solve algebraic equations on matrices. This is known as the ADHM construction for T-adrinfield, Hitchin, and Mannend, which in turn grew out of the study of twisters and which in turn go back to the inverse scattering method in integrable equations, Zakharev-Shabat, and those techniques. So there are two ways. I can start by just stating the result or I can try to motivate it. And let me try to motivate it a little bit. So the idea is that suppose you have an instanton, so a solution of this equation with finite action, then what you can do you can study the Dirac equation with spinors which are supposed to be L2 normalizable and which have one of the positive and negative corollity. So here E will be the vector bundle. So it's rank n complex vector bundle over S4. So I will be thinking about this equation actually on R4 with flat metric. So of course, over R4, this bundle is trivial. But I want it to extend to a bundle on a sphere with the trivial Pantrain class. Say it again. It's not hobb bundle. It's not hobb bundle. There is no, it's a vector bundle. Vector bundle rank n, n is an arbitrary number. So let's not get confused. And so the spinors, so I will be working on R4. So I will actually pick a complex structure on R4. That's for simplicity because so that it's not, you have a whole two sphere of choices here, but let's just make one choice. And then you can identify the spinors with something which is simpler, namely, so for me, S minus will be identified with the 0, 1 forms. So it's things like dZ1 bar or dZ2 bar. And S plus will be identified with 0, 0 and 0, 2 forms. So it's 1 or dZ1. And then the Dirac operator under these identifications is the sum of the Dolbor differential couple to the gauge field plus its conjugate. So d bar maps 0i forms into 0i plus 1 forms. And d bar star in the opposite direction. So this star is a respect to Hermitian metric. So I'm identifying, so here I have a flat, I have an euclidean space, and I identify it with Hermitian complex vector space. So the metric will be dZ1 dZ1 bar plus dZ2 dZ2 bar, where dZ1 and dZ2 are holomorphic coordinates on C2. So the nice thing about this identification is that if you rewrite the anticellularity equations in terms of the complex geometry, this equation is equivalent to the equation that the 0, 2 component of the curvature is equal to 0, which is equivalent to saying that the operator d bar A squares to 0, that's a fundamental equation. So this is what used to be the integrability equation, the zero curvature condition in the first approaches to this problem. And then the, so this is worth two real equations. It's one complex equation. And here we have three real Li-algebra valued equations, 3 is 6 over 2. So you have the rank of the space of two forms is 4, choose 2 is 6, and the cell dual forms, and add the cell dual two forms, three dimensional spaces. So this is three real real equations. This is worth two equations, because it's one complex equation. And so there's one equation missing, which is the component of the curvature along the scalar form. So the scalar form is, so this equation is invariant under complexified gauge transformations. This equation is not, so sometimes people say that, well, you can use this equation as to fix the non-compact part of the complex gauge transformations. So instead of imposing this equation and dividing by complex gauge transformations, you impose only this equation and divide by complexified gauge transformations. So this correspondence is slightly subtle, but in the first approximation, the space of solutions to this equation is the space of anti-halomorphic connections, which have flattened complex sense up to complexified gauge transformations. So because of this condition, you can actually form a certain chronology problem. And this chronology problem, so namely, you look for forms which are annihilated. So these are E-valued, let's say, L2 normalizable E-valued 0 comma i forms, which are killed by d bar a up to exact forms. And now instead of dividing by exact forms, you can impose a gauge condition on these forms. And this gauge condition will involve one of the choices is to use the conjugate operator. So in the end, this problem with the gauge condition will be equivalent to solving Dirac equation. So long story short, solving the equation, so we have two choices. So either for S plus, so we are looking for the positive chirality spinor is a sum of a scalar and 0, 2 form. So let me call the scalar eta, the 2 form, let's say, will be called chi. So we want to impose this condition. So this equation takes values in 0, 1 forms. So I'm applying the del bar operator to the scalar. So I'll produce 0, 1 form. And then I add the conjugate operator acting on chi. So that's my equation. So if you write this in components, we actually get two equations, d1 bar eta plus, let's say, d2 chi equals to 0 and d2 bar eta minus d1 chi equals to 0. Now let's act on the first equation with d1 and the second equation with d2. So now we'll get d1 d1 bar plus d2 d2 bar acting on eta. It's equal to 0 because d1 with d2 commute, that's the conjugate of the equation of 0, 2 equals to 0. And also the second equation, this equation, allows me to rewrite. I can simultaneously flip the order in the first term and the second term. And in fact, I can average. So I can write this as an anti-commutator. So this is the covariant Laplacian acting on eta. So these equations imply that eta is covariantly harmonic, from which you conclude that eta is actually equals to 0. So you multiply by eta, integrate by parts. Derived from this, this eta is covariantly constant. And then because of the L2 condition, this constant will have to be 0. Otherwise, it will not be normalizable on R4. So from this, it follows that the Dirac equation for positive chirality spinors has no solutions in the instant on background. On the other hand, the difference of the dimensions of the kernels of in mind normalization, this is computed by the index theorem, which I'm sure you learned from the lectures of Professor Zabzin. And so this is equal to k. And so this is being 0. We conclude that we actually have a k-dimensional space of solutions to the Dirac equation in the space of 0, 1 forms. So this is k-dimensional complex vector space, which I've all denoted by the capital letter k. And now having this space, we can play with it. Namely, suppose psi belongs to k. So it means that psi is the L2 normalizable 0, 1 form valued in a vector bundle such that it solves two equations. So this is what you get by. So when you apply d bar to psi, you get 0, 2 form. And when you apply d bar star to psi, you get 0, 0 form. So that's the first condition is this closeness condition. And the second condition is like a gauge for the identification. So this is really the homology problem. And so we can do the following. We can multiply the coordinate functions on psi. So we'll produce some 0, 1 form valued in a vector bundle. And it will not be a solution of a Dirac equation, but we can actually project it on the space of solutions of Dirac equation. And so we can re-expand it, how to say this, in a strange form. OK, I can project p. So this is orthogonal projection. Namely, this space you can decompose as a kernel of Dirac operator plus the h, plus h perpendicular, where it's perpendicular in the sense of the L2 norm. L2 norm. And so if you like, you can form the kind of Laplace operator. And so on this space, this Laplace operator will be strictly positive. And now you can define the projection projector by 1 minus d slash star, 1 of the Laplacian d slash. So what this operator does, if you act on a spinner, if the spinner is annihilated by Dirac operator, you'll get the spinner itself. But if it's not then, well, it will subtract from it some portion which is in the space. And so if you act with Dirac operator on anything on which you acted with this operator, you'll get 0. So the image of this operator is the subspace k, which we're interested in. So what we did, we took, so for every vector in the space k, we defined mu 4 vectors by multiplying it by 4 coordinate functions and then projecting. And so in this way, we've defined 4 complex operators. In fact, they are, so they're two complex operators and they conjugate by these formulas. So the conclusion is that this abstract vector space k, which is Hermitian because of the L2 norm, carries an action of four matrices. So that's one structure. And another structure is that if you take, so let's look at the asymptotic behavior of the solutions. So let me write psi now, psi is right in components. So each vector, when r goes to infinity, at infinity the gauge field approaches the pure gauge because the curvature should vanish at infinity in order for the energy of the solution to be finite. And so you can approximate the Dirac operator coupled to the instant gauge field up to this gauge summation g by a flat Dirac operator. And so psi should approach the solution of flat Dirac equation, which you can find in the following way. So this first equation, so we have two equations on the components of our spinor. And remember that separately d bar and d are flat connections. So you can solve one of these equations by, let's say, writing psi alpha as d bar alpha of some chi. And then this chi will have to solve the Laplace equation. Of course, it cannot be solved globally because then we will get again that the solution is 0. But far away to infinity, this is possible. All you need to do, you need to choose the harmonic function chi. So chi should be harmonic as r goes to infinity. And the simplest harmonic function on r4, which is well-defined near infinity, which is rotationally symmetric, is what? What is the simplest harmonic function on r4? No, 1 is not good because 1 will give 0. So you don't want something which is still not non-trivial. Green's function of Laplace. Yes, Jan knows. No? Yes, perfect. So this is one possibility. And you can multiply it by a constant factor, which I will denote, I think, by i. So i dagger, actually. So i is going to be a map and is the n-dimensional vector space, which is the fundamental representation of our gauge group. OK, so let me write it here. So the possible asymptotic behavior, so psi alpha as r goes to infinity, behaves either as d alpha bar 1 over r squared. And this d is really just its usual d dressed with a flat gauge transformation, with a gauge transformation. So it's i dagger. And the second option is to write epsilon alpha beta again 1 over r squared times j. So i and j are constant. So i is the map from n to k. And j is a map from k to n. So what I wrote here is kind of a fundamental solution. So this psi has two labels. Psi is valued in the n-dimensional vector space, which I denoted by e, so the fiber of the vector bundle. And also psi is valued in k, because we have k solutions. So k, and so that's the neat way of packaging this. So by looking at the solutions of Dirac equation, we required the following algebraic structure. We have a vector space k, k and n. And we have maps i and j. And we have the operators b1 and b2, which act on k. And they're commission conjugates. Now, when you multiply coordinate functions on anything, of course, these operations commute. You can multiply them in arbitrary order. But here, after we multiply, we project it. And so after you project, what used to be commuting doesn't commute anymore. So this, of course, should be familiar from the studies of the quantum Hall effect, where the coordinates on a two-dimensional plane, where the electrons move, become non-commutative coordinates when you project in the first Landau level. So this is exactly what's going on here, except that it's in four dimensions. So the matrices b1, b2 do not commute in general. But what's remarkable, and that's been shown by Corrigan and Goddard, was that the commutation relation between b1 and b2 close up to the matrices i and j. So the only thing which you need to know is this quadruple, this quiver of which you can summarize in the quiver diagram, modern notations. So this is i, this is j, this is b1 and b2. And so it is a computation, which is straightforward, but rather long, to show that the commutator of b1 and b2 fails to be 0 by this finite rank matrix ij. And then the commutator of b1 with b1 dagger plus b2. And this is modular. The choices of bases in this k dimensional vector space, we didn't have any specified bases, so we were allowed to conjugate the matrices b1, b2 and multiply i on the left and j on the right. So on one hand, from the instanton data you produce these matrices which solve these equations. These are called the GHM equations. And they have the hypercalor mean. So these are hypercalor moment maps. In some sense, they are kind of infinite dimensional Fourier transforms of the equations at 0, 2 equals to 0 and f11 omega a equals to 0. So this is the analog of the first equation and this is the analog of the second equation. And this analogy is actually helpful in designing the integration formulas. Now, so I described how given a, you produce b1, b2, i and j, there is an inverse construction, which is actually the GHM construction, which is given the solution of these equations. You can produce the gauge field. You can also produce the solutions of Dirac equation. So this is done as follows. So you consider now, again, kind of a Dirac operator which will be acting from k tensor c squared plus n to this space. So explicitly, this is this rectangular matrix. It's easy to remember how to construct it. So first you go b1, b2, and i. And then here you take the Hermitian conjugate of this guy with minus sign, Hermitian conjugate of this guy, and minus j dagger. The property of this map, so it's a family of maps which are parameterized by point on c2. So for each point on our space, this is going to be our space R4. We have a linear operator. In other words, you have a family. You have a vector bundle. You have two vector bundles with these fibers over R4. They're trivial bundles. But then you have a map bundle map, which depends on where you are. And this map has an interesting, has nice property that if you look at its conjugate and so compute the product. So this is going to be a map from k times rc squared to k times rc squared. Well, when you compute it and use the dHM equations, you'll find that this map actually commutes. So it's actually identity on c squared times some map on k. So this is delta times 1 on c squared, where delta is a Hermitian operator on k, which is non-negative. And then one actually imposes a non-singularity condition of the dHM data, which guarantees that this operator is strictly positive. So delta is strictly positive for good quadruples of matrices. And then the original vector bundle, e, is recovered as the kernel of this map. So you see what I'm doing, I'm doing the following. I take first the trivial vector bundle with the fiber n, and then add to it something. And then map result to the same vector bundle. So I'm trying to add some redundant data. But then I'm twisting the isomorphism in a way which depends on z. And so the kernel of the separator is my vector bundle. And the connection is just a projection, it's just a projected connection. So it's where, so you solve these equations. Psi is just an identifier. So psi is the map from n to this space. So you solve the equation d dagger psi is equal to 0. You can write it again explicitly in this form. And then you just differentiate and project. And the claim is that this solves these internal equations. And moreover, the claim is that that gives all solutions. So if you repeat this procedure twice, so start with b, i, and j, produce a. Look at the solutions of Dirac equation. So the Dirac equation is solved in this very explicit form. So look at the solution of Dirac equations and recover the operators b1, b2, i, and j. You'll get what you start with up to maybe UK transformation. So that's the claim. OK, so this is nice. The infinite dimensional, the PDE problem is reduced to the algebraic problem. So let's see how can we use it. So we are interested in computing the path integral. The simplest path integral, which one can compute, is that in the context of n equals 2, d equals 4, superannual theory. So it has eight supercharges, of which I will single out one supercharge, call its action delta. So it acts on the gauge field, produces something which I will call psi m. So this is twisted green. It's twisted in the same sense in which I replace spinors by forms. Previously, I'm working on R4. So actually, I'm not doing anything to my theory, just using more mathematical notation, if you like. And so the variation of the fermion is the covariant derivative of the scalar. So the field content of n equals 2, minimal theory, minimal superannuals is you have a gauge field and you have a complex scalar in that joint. And it's conjugate. And the fermions, these notations correlate with what I used for my spinors before. So this is one form. This is a scalar, and this is a self-dual two form. It's all adjoint valued, and these are fermions. Sir, why do you say you don't know anything? This is weak and topological, so I'm not making my theory topological like this. So well, if you like super symmetric theories in general on flat spaces, you have topological subsectors. You're not making a theory topological. You're defining a subset of operators, which are topological in a certain sense, but that's always the case. So with the super symmetric word identities, which say that the Hamiltonian is the commutator of the supercharge, imply that if something is analyzed by the supercharge, the expectation, the correlation functions don't depend on time. Things like sigma bar varies into eta. This is on show. This is after some auxiliary field has been integrated out. And so the whole action of the superannuals can be written as a sum of the topological term. Again, I'm skipping the factors of 2 pi plus delta of chi plus times f plus, well, maybe I should write the auxiliary field here. So this is the auxiliary bosonic field, plus psi t star. So when you compute the delta variation of the right-hand side and then you integrate out the field h, you'll recover the standard superannual section written in a slightly different notation for the fermions. But it's identical to the standard action. So tau here is a complex combination of the theta angle and the angles coupling. So the reason I wrote is that you can actually write the parallel, similar multiplet for the five-dimensional model of the horizontal modular space. So here the point is that the localization in this path integral on the locus of fields which are fixed by delta will reduce the infinidimensional path integral to the integral of the five-dimensional modular space, for instance. So delta localization gives you the integral over the modular space of instantons. And let's find dimensional space. Yes, I should have said that just like a computer index for a drug operator coupled to the fundamental valued spinors, you can also compute it for the adjoint value spinors and that will compute the tangent, the virtual dimension of the modular space solution of this equation, which happens to be the actual dimension, which is matched by the dimension of this five-dimensional space you get from matrices. So that space has a real dimension 4 and k. So you have four k by k matrices and you have two complex k by n matrices. And now these four k by k matrices are killed by imposing three equations and divided by the group u k. And so that's the number of degrees of freedom you are left with. So now instead of starting the infinidimensional problem, you can do the single-fine dimensional problem where you would introduce the super partners for these matrices and then for the corresponding multiplet for the u k gauge symmetry. And so you can compute the same integral using the five-dimensional model. But of course, it's going to be something kind of ill-defined because actually this modular space is non-compact. And if you naively compactify it, it's singular. So these integrals by themselves are not very well-defined. What we're interested in gauge theory, actually, it's not in computing the partition function per se. We want to fix the asymptotic value of the scalar sigma. So which is, again, it's allowed to have non-zero expectation value because as long as it computes with conjugate. And compute the effective action for the small fluctuations around this configuration. And so the effective action will also be equals to supersymmetric. So you can write it also in a way as a sum of something topological and delta exact. But now the coupling tau will start depending on the expectation value of this field sigma. So instead of the SUM gauge group, when you expand your theory around the configuration where this field sigma has generic expectation value, you only see the maximum torus as a gauge group. So instead of the SUM valued gauge field, you will now have the maximum torus valued one. And so now the gauge symmetry is less restrictive. And supersymmetry allows more general couplings where the curvatures of different gauge fields are coupled to the scalars through some matrix where it's interaction plus. And then supersymmetry tells you that this has to be completed. So there are further terms like the derivative with respect to ak tau ij. And then there is something like psi k wedge psi i wedge j plus, so this is 2 thirds, maybe 1 third. Maybe this is 1 fourth second derivative ak l tau ij. And then there are four fermions, psi, psi, psi, psi ij k l. And then when you look at the symmetry structure of these various couplings, you conclude that the coefficients have to be completely symmetric. So tau actually should come from some pre-potential, which should be a locally homomorphic function of A. And so determining this function is equivalent to determining the effective action, at least up to delta exact terms. But then remember, we singled out arbitrarily one supercharge, which was null potent, out of 8. So you can do repeat this exercise for different choices of supercharges. And then requiring the consistency of this structure will fix the effective action at the level of two derivatives and four fermions, all in terms of this single function. So it's this function which you want to compute. And now this function you can compute by designing a trick, which is to deform the original theory in such a way that instead of the integral of the modular space of all instantons, you'll be dealing with the localized version of this integral, where it is saturated by contribution of some special insertion configurations, the fixed points of certain symmetry. And so that's what omega deformation does. So I'm deforming. So the idea is to recognize in this supercharge, the equivalent differential, the RAM differential acting on the space of gauge fields, which is equivalent to respect to the action of the gauge group. Now, in gauge theory, the gauge group is something we divide by. So it's not a symmetry. It's a redundancy. But we can slightly help ourselves by saying that when we fix the vacuum expectation value of the gauge field of the scalar, it actually means that we single out, single doubt, a subgroup of constant gauge transformations. And we fix it as a global group. So we decide we will not divide by it. We'll only divide by gauge transformations, which vanish at infinity. And so we can separate sigma into the part which vanishes at infinity plus something which will be the expectation value. So this guy goes to 0 as r goes to infinity. And this will be keeping fixed. And now we can think of another global symmetry group, which we can use to our advantage, which will deform the supercharge. And that symmetry group is a group of rotations. So what follows is, if you like, you can view it as a kind of regularization of the theory. So in the end, we will be interested in removing this regularization, at least for the purposes of the computing effects of action. So I'm deforming delta to delta with parameters epsilon. So these are the parameters of the infinitesimal SO4 rotation of R4. So this is where my twisted notation comes handy, because if you want to do the same in the untwisted language, then you'll have to say that in addition to the R4 rotation, you also do the SU2R symmetry rotation, which will preserve part of the supercharges. So this delta epsilon acts on the gauge field in the old way. But the fermion now, let me skip the indices. So it used to be given by the covariant derivative of sigma. And now I'm adding a new term, which is the contraction of the curvature of the gauge field A with the rotational vector field. So V epsilon is epsilon 1 z1 d by d z1 minus z1 bar epsilon 2. I should stress here that just as with gauge transformations, the actual gauge transformations are generated by anti-commission matrices, the supersymmetry algebra contains the complexified parameters of the transformations. And so here, the parameters epsilon 1 and epsilon 2 are complex parameters. And in fact, the story is holomorphic in them. Even though the actual rotation of spacetime is generated by the vector fields when epsilon 1 and epsilon 2 are real. Finally, d epsilon of sigma now is the contraction of psi with V. So it used to be that any function, any gauge invariant function of sigma was delta closed. So that gave rise to the curl ring of local operators, which were built out of sigma, which you could insert everywhere in spacetime. Now, this is no longer possible once epsilon is in place. If your operator sits somewhere away from the origin of my rotation, it is no longer supersymmetric. So now the homology of this supercharged delta, so delta squared is equal to 0 up to rotation and gauge and global symmetries. So the only operators which are annihilated by delta, these are the gauge invariant functions of sigma inserted at the origin. So this is where z is equal to 0. So the homology has been reduced, so to speak. It actually has not been reduced in size, because previously, even though it looked like you could insert an operator of a function of sigma at any point in spacetime, if you move that point, you didn't get new homology. The result was homologous to the previous one. And so in particular, you could move everything which you liked to the origin of spacetime. And so this is where this homology sits. So the homology actually didn't change in size, but it's changed in kind of presentation. Now, you can ask whether, so I deform the supercharge, but of course, I want my theory to be invariant under this uniform supercharge. And so I need to modify the action in an epsilon-dependent way. But as its notation suggests, it's a small deformation. So in the end, we can take epsilon to 0 and recover the original theory. Now, the advantage is that we no longer have kind of propagating fields, which will spoil the Wilsonian integration procedure. So here, in the standard approach, you cannot integrate out all the fields in your theory because you have massless fields which you cannot integrate out. When you try to integrate them out, you get infrared divergences. Now, with the epsilon deformation, with the omega deformation in place, you can integrate out everything except for the zero modes of sigma. And so that's what we will do. So in particular, if you first reduce onto the module space of instantons and look at the expression at the differential form, which the path integral measure will produce here, well, it will produce something like exponential of delta of psi. OK, how should I write this? So you see, the rotational symmetries of R4 on this B, I, and J data, namely, the rotation generated by epsilon 1, epsilon 2 acts, well, simply multiplies B1 by phase B2 by phase and J by the product of these phases. So you can check that these equations are covariant under these transformations. And so it means that to every, so I have a vector field with parameters epsilon. So it's a vector field on the module space of instantons, which for real epsilon would be even an isometry. And now, so if G denotes a metric which comes from, so it descends from a two metric on gauge fields, so it's an integral over our four trace delta a new squared with some choice of gauge, same thing. It's a hypercard. So now if I substitute the vector field V with the parameter, which I will call epsilon bar into the metric, that defines, so it gives me one form on the module space. In components, this is gmn vm epsilon bar dxn. And so I have the same, I have a similar differential, delta epsilon, which acts on differential form, equivalent differential forms on M plus. And so I can use this differential to produce an integration measure on M plus of this form. So this is, so this you can trace, so you can trace to the original omega deformed N equals 2 action. And now by the usual tricks, so if you send epsilon bar to infinity, the answer doesn't change since it's delta exact. This is equal to the sum over the fixed points of rotational symmetry, 1 over the product of the weights. So at each fixed point, you look at the tangent space. So P is in the modular space, which is, so this is where the vector field vanishes. It preserves the point, but it acts on the tangent space. So this tangent space becomes a representation of the rotational group. If you like the derivatives, so this is an upper, it's like a matrix, which depends linearly on epsilon. It's a matrix which acts in the tangent space and you take its eigenvalues. So this is the, so eigenvalues, it's just the determinant of this matrix. So this is what you put in denominator. Nikita, can I ask, do you send epsilon bar to infinity? Yes. Or you put the sum and keep in epsilon fixed? Yes. This is just a way to compute this integral. All right, but you can also put a real parameter in front and send it, I mean, formula to give you. What's the difference? It's linear in epsilon bar. I mean, certain equations will decouple. I mean, a question d sigma plus IV epsilon bar will decouple if you send it. This is epsilon. This is not epsilon bar. This is not epsilon bar. This is epsilon. This is finite. It's only here that you send epsilon bar to infinity. I didn't write the full delta differential, so you didn't see where epsilon bar. You don't have these things times complex conjugate, right? Not if epsilon to epsilon bar. So once epsilon bar becomes independent of epsilon, they're not complex. Right, but what I'm trying to say that when you send epsilon bar to infinity, you will send a real parameter in front infinity. It will get slightly different conditions. You'll be stronger. What you are going, you are going for stronger, so I'm asking. Maybe, yes. Whatever works. So I will derive this result, except that, unfortunately, unfortunately, this modular space is not a space at which this rotational symmetry has isolated fixed points with well-defined tangent space. And so this is where we need to do a couple more tricks. So further regularization. OK, Google. We introduced a fire post from. So at this point, this looks completely arbitrary, but I don't have time to explain what it really means. But so once you make this deformation, which is, in some sense, it's a delta exact deformation, but the outcome is drastic. So instead of the space which has some conical singularities, you now will get a space which is smooth. And now what used to be a bad point where the instantons were point-like and sitting on top of each other is now getting resolved into a multi-dimensional space on which the rotational symmetry actually acts with isolated fixed points. So in this picture, so this is the simplest model where you had, let's say, a1 singularity. And then by introducing zeta, these a1 singularities gets replaced by a small two-sphere. And then the rotational symmetry acts in particular on this two-sphere with two isolated fixed points. And so each fixed point now has a tangent space on which you can compute this quantity. Here, the parameter zeta, which I introduced, doesn't appear here. So you can formally say, well, you can send zeta to 0. And the answer will be still the same, except that you will not have an interpretation of the individual term. So when zeta is non-zero, you can actually make sense out of each term separately. But when zeta is 0, only the whole sum makes some sense. OK, so what are these fixed points? Let's do a little bit of algebra. Maybe I should say something about what happens to these equations when you introduce zeta. So what does it really mean, algebraically? So suppose zeta is positive. Then I claim that these equations are equivalent to the following equations. But first of all, well, you keep the first equation. It doesn't change. And the second equation, we replace by the stability condition, which says that when you, so you have this operator i, which maps n to k. So if I apply i to n, I get a subspace in k on which I can further act with matrices B1 and B2 in arbitrary order. So I act with all polynomials in B1 and B2. They are not commuting in general, but I don't fear that. And the conditions that when you act with all possible combinations of B1 and B2 on the image of n, that should generate all of the space k. So that's the stability condition. So I claim that this is equivalent to imposing the second equation and dividing by the group uk. In this way of formulating things, I can divide by the group glk. So it acts on, that's the way it acts on these matrices. So here, there is no complex conjugation involved. Everything is holomorphic. It's a neat exercise to prove that stability is equivalent to that, but I don't have time to do this. Now, the fixed points. So let me introduce the parameters of rotation. So t1 is like e to the epsilon 1, t2 is e to the epsilon 2. And let me use the matrix B. So this is the second. So the group, so this will be the group elements of the maximum torus of the global symmetry group. And so these guys act on my adhm data. In this way, I be inverse c1, c2, bj. So what does it mean to have a fixed point? It means that you're looking for quadruple b1, b2, i, and j, such that this is equivalent to that up to the transformation of the symmetry group. So to be at a fixed point, it means that for each c1, c2, and b, you find the compensating transformation compensator, h, which depends on c1, c2, and b, such that this is equal to h, c1, c2, b acting on b1, b2, i, and j. OK, so t1, b1 is equal to h inverse b1, h. t2, b2 is h inverse b2, h. i, b inverse, is equal to h inverse i. And t1, t2, bj is equal to jh. Very good, so let's classify the solutions of this equation. Provided b1, b2, provided we solve this equation and we have a stability condition. Well, first of all, our space n splits as a sum of the, so these are eigenspaces. So it's an eigenspace of b with the eigenvalue e to the a alpha. So these are one-dimensional spaces for generic choices of a's. And now I have a vector i alpha, which is the image of i. And now this equation says that h acting on i alpha is equal to b alpha. So this matrix h has eigenvectors. We already know n of its eigenvectors with its eigenvalues. And moreover, if I act with some polynomial, let's say, binomial of b1 and b2 on the vector i alpha, let me call this vector ij alpha. It is an eigenvector of h with the eigenvalue b alpha, t1 to the power i minus 1, t2 to the power j minus 1. So you use these equations to show that when you act with b1, you raise the eigenvalue of h by, the eigenvalue is multiplied by t1 or by t2 depending on these things. And finally, so what you see, you see that these vectors, which should span the vector space k. So here I'm not yet careful about the ordering. The ordering might matter, but irrespective of how you order the matrices, the eigenvalue will be the same. So for generic t1, t2, and b, these are different eigenvalues. So these vectors cannot be linearly dependent for different ij and alpha. So the eigenvalues have this form. They grow, if you take a logarithm, so they have the form epsilon alpha a alpha plus epsilon 1. So they grow into some kind of lightest way with certain preferred directions. So the direction is given by the vectors epsilon 1, epsilon 2. On the other hand, the operator j, if you look carefully at what's written here, if I didn't make a crucial mistake, which I might have, would want to have an eigenvalue which grows in different direction. And so from this, you actually conclude pretty fast that j vanishes on all these vectors. So actually, j vanishes on k and b1, b2 commute. And the structure of this solution is such that the space k splits into the sum of n subspaces. And each subspace has a structure of young diagram where the box with the coordinates i and j, and so the block number alpha, is this factor here. So the fixed points, n-tuples of young diagrams of total size equals to k. So the total number of vectors. So each block is a basis vector, and the total number of vectors is equal to k. So that's the condition. And then the algebraic calculation is to calculate the linearization of the symmetry transformations on the tangent space at such a fixed point. So once you found such b1, b2, and j, which I just described, I described this as complex operators. So b1 acts by moving from one box to the right, and b2 acts by moving this box one step up until you get out of the young diagram when it becomes zero. So then what you need to look at, you need to look at the variations of b, which should solve the linearized equation up to linearization of the symmetry. So on this linearization, you act with a symmetry group. Namely, so t1, t2, b act in the following way. Well, first of all, there is this naive transformation, which is what you did. But then you compensate with the same compensator h, which you had before. So delta i, b inverse h, t1, t2, b delta j, h inverse. So the idea is that on the space of all matrices, you act simultaneously with the fixed global symmetry transformation, t1, t2, b. And then you compensate by h, which was designed in such a way that that quadruple was kept intact. So it was fixed not as an orbit, but as actually representative. And so then for the nearby matrix, which is b plus delta b, you act with this combination of the symmetry, global symmetry and local symmetry. And so when you undo this, this is going to be b1 plus what I wrote here. So this is an explicit linear transformation of these variations, and you need to find the eigenvalues. And so these eigenvalues will be the generalized arm-leg formula, which you get into the denominator. And then in this way, you produce the partition function, the case of contribution of the partition function, then you sum it up, and then explore the limit epsilon goes to 0. So unfortunately, I didn't get far. OK, let me just make one remark, which is kind of a beginning of an interesting story, is that in this analysis, I assumed that my parameters were generic. So there was no accidental relation between these eigenvalues. So all these factors were forced to be linearly independent because the eigenvalues were different. Nevertheless, sometimes one is forced to consider non-generic values of parameters. And in particular, if epsilon 1 over epsilon 2 is a rational number, and it's a positive rational number, then, well, you see, it means that epsilon 1 times i minus 1 plus epsilon 2 times j minus 1 is some epsilon times, let's say, p. So you have a bunch of integers, and it may happen that for different i and j, you get identical eigenvalues. And so what it amounts to is that these fixed points are no longer isolated. So they are non-isolated. And so in this localization formula, which I raised, which was a sum over fixed points with some denominators, these denominators will vanish. And so the individual contribution will be infinite, and so the formula will not make sense. However, if you can prove that the set of fixed points, even though no isolated, is compact, then that will guarantee that, well, the formula will be slightly more complicated, but it will be still finite. The limit will be finite. So the limit exists. This is important because if you are not doing this as the way I did when I mean my epsilon's were at my disposal, but you try to find these supercharges, the way these supercharges act as specific supergravity backgrounds, then sometimes the values of epsilon are fixed for you by supergravity. And so in this later work by Peston, for example, when he computed the partition function of n equals 4 and s4, the abstinence which he got were equal. So he could not use literally my formula because individual contributions contain things like epsilon 1 minus epsilon 2. So the expression is infinite. And so the actual formula is more complicated because instead of the points, you have now some sub varieties in the Hilbert scheme of points which are compact but complicated. But anyway, this is the beginning of the story of compactness theorems on the module space of instantons and the modifications which lead to various powerful word identities and Dyson-Schwinger formulas from which one can derive most of what is known about the relation of this story to confirm a few theories and make good deformations. So I apologize for not covering all the material, but you might have access to the notes and try to run from there. Thank you.