 So, jesrde, we were in the process of performing this localization computation of the partition function of the dimensional and equals to comma 2 gauge theories. And here, we are constantly into vector and还是 multiplets on the untwisted around 2. Pistim, da bi bilo pozorovati srednji, da smo vsega taj viče odvrli vsega. Zato smo odvrli vsega vsega vsega vsega. Nelam ne jazno izgleda, bo da je zelo narednico. To je vsega, da je vsega večo večanje, in je bilo komplexivno. kratične interakcije, ki je standard. In zelo smo zelo skupili materiakcije, zelo skupili kinetično for karan multiplič. In karan multipliči je v nekaj reprezentacij, zato v plizi je, da je kontrakcija z zeločenimi indice. Zeločen R je zeločen R, kako R je zeločen kratične. Vse je Vse. Vse je�o u punic Najlandski sama, kaj se vse začne, tač nemoramo z kavim aktivistim, da vse nas tačne vse je, začne je rank attribute dr. Tako, tako vidim, skoliko, tess, bila, nimi je Slatom vzačne. Tako, tako, tess, bila, tess, bila, tess, začnost. T, bo, bo, bo, to, Tukaj imamo 4 na vse predstavne predstavne, ali posledamo 1,2 tukaj, 1,2 tukaj tukaj. Tako, če so tukaj, v teori in interakciju? Zelo, da imamo super potenšel interakcij, in v sej super potenšel interakcij, tako, da imamo tako, tako, da imamo tako, da imamo nekaj olomorfički funkcij na karelj multipleti. Tako, to je karelj multiplet, tako, da imamo karelj multiplet in tako, da imamo f terma in tako, da imamo vse karelj multipleti v teori in to prijevajte, kar super potenšel interakcij in teori in zelo, tako, in vsej, in tako, in tako, da imamo karelj multiplet v teori in teori in tako, da imamo kaj je tukaj komponent tukaj karamultipleti, je to samo izgleda, da je tukaj v vsej spas, da je vsej derivati superpotenšel v svojimi karalfitih, da je tukaj, in tukaj da je vsej ukawačne interakcije. Tako je, da je tukaj izgleda, da je tukaj vsej tukaj vsej, in tako vsej, da je vsej, da je vsej, Ko ostajremo polinomi nekaj, to ognist愵 za košnje, ali ne pa ne bi tem, ki oblijemo patintečne nakoččne za košnje na superpotenčne. Izdrža in je nekaj, da teorič naredil independent z superpotenčnev. Zato, kako superpotenčni da bo vzelo R2, fpsestimo R-cimetrij, kaj ne kaj ma drugi del, je ni zgledaj, da bi njihovi postanje za vsega, ki je to samo vsega, ki je to dobro potenčenja. Zelo je bilo? Zelo smo tudi tudi tudi potenčenja, in tudi je semila, da je to je zelo vsega, da jelem tudi tudi kali grafik V, tudi tudi karal multiplet. In tudi je to, da so tudi malo morav vsega. Zelo, da smo priči vsega, da je to liniar, Zelo, da je linijačne, nekaj je konstant, nekaj je konstant nekaj je teori. Linijačne termi so kojefitični. Zelo, da sem se parametrizala z nekaj rejalnih part in nekaj imaginaričnih part. Zelo, da se počekajte, nekaj ga je v dela granžan, nekaj ga je komplekcij fajetiljopolis termi. If you wish this real part controls a linear term in proportional to the D term, that you put in the action, this is called the phytelopolis term and imaginary part controls the topological term. The field strength integrated in a two-spher, we are in two dimensions so we have such a topological term. What is it called, the teta angle, because the theory is invariant under shift of this angle by two pi. This is very similar to the term FJGF that you have in 4 dimensions. Of course, we could consider a higher, higher power of the sigma, but let's consider this very simple case. And the interesting thing now that this time is not Q exact. So in fact, our partition function will depend on the parameter, almost generally, on the various parameters so we haveittingijs with super cryptocurrency. Pomembno so bili vzdušnjih mase, in vzdušnjih mase, kaj je zelo svoje vzdušnje, in zelo počusti sva, kaj je vzdušnje mase se postavili na našlih smetri, včasno, ils tega vzdušnja v kartanju, subalge, in vzdušnjih smetrih. Kaj je, če ti je smetri, ali se vse vzdušnje nr. vse, in je šeljena v rovku vektormultipljete, zazve si ga prejsteši in pošliši, od nekaj posluši vzvečenji, a zelo je korisponsov, kaj plači se odvajne. To plače, in tako, je velice in zavečenje vzvečenji vzvečenji vzvečenji vzvečenji. Zato to pličenje vzvečenje v izku, vzvečenji vzvečenji vzvečenji. Na zelo je vzvečenji vzvečenji vzvečenji. In zelo sprem, da je izgled, da se da vse sveti. Zelo prejder, da vse vse vse vse vse, zelo izgleda spetetjče tudi vse vse vse vse vse vse vse. Kako iko tudi vse teori, posebno vsega, kako izgleda toga gravitino in tudi održiv. Vse vse vse vse vse vse vse vse vse. Zelo vse. In je ta vse obtah, da se dajimo, zelo se njimljeznje. Zelo se imamo zapraveno taj kundišen, kaj imamo opravboj do vse. Zato da sem začala povradi, da je vse, kaj je povradi. Ok. Zelo sem mužel njega premačnje v skupi vsej lokalizacij. Zato bi se nekaj dvjih mladov, zelo sem kaj povradi. Svet začak, da smo počkati vzahvali tudi iz veči dobro vzpečne termje, in da se zelo zagravimo hodnje. To bi smo počku, da smo počkati, da so jeli vzpečne nekaj vzpečne. To bi smo jeh tudi nekaj, da je vzpečne začak, da bi je tudi začak zelo nagrače, je to superimetrične, je to na veče, da so vzpečne in pomečite, da se počkati in vzpečne. To je tudi tudi vzpečne, Here comes the obvious contour which is just take real fields, which are complex conjugates, and this is manifestly positive. This is the sum of positive terms Of course let me stress once again if you don't impose a contour, there is no way that this is going to be positive, because we become complex fields, it is getting the generic values. But on the real contour this is a positive, so this is okay. T JOJSOKI Povedajte, da je tko, četeno o dvoreza povedajte bozonicje bozice, ker, zato, četno tako downloadede ili da je to vse pa je tko, tako, tko, četno tko, je to ovoreza povedajte četno da je tko, zamerak tko za tvoje 0 in 2 in tko so dodačimo komputation v tom, četno tko, Vspite sodak, da to neč koputka tega se odpeljezava. If you do the same thing, sum that you sum over the fermions of Q psi, the agru psi in you have to have Q on that. You get precision with this. This one is not in the kanonical form. If you prefer, and this is related to the fact that there is this imaginary part that is not positive, so if you prefer you could instead kronikl form, je veliko izgleda z toga, ali je bilo dobro. Zdaj, da je bilo dobro. Ok. Čekaj smo prišli v zelo tudi površenju, ker smo prišli v zelo tudi površenju bozoničkih. Zelo, da je bilo površen, bozonički površen je površen, zvrst received to zero each of the squares. So the equations are just that F is related to sigma two, d is related to sigma one. These derivatives of sigma one and sigma two are zero, that's valid in commutator. And then once you have fixed this, again, if we are in this region for the charges, In se je zelo, da je to vse odložite. Zelo je zelo, da je vse odložite. Vse je zelo, da je to zelo. Vsih je zelo, da je v sektor. Zelo je zelo, da je vse odložite. Prejde, tako, ta rekojst. Prejde. Takaj rekojst je zelo, da vse zelo, da je skelo. In zelo, da je skelo. In zelo, da je skelo. Tako, da je skelo. Bhejte nekaj dolič, bo se vse zelo, izgleda se, da je tukaj držav, izgleda se, da sigma 1 in sigma 2 je konstant, zelo všeč je konstant. Tako, zelo, da vse bps lokus, kaj smo lokalizati, je ekstremljno zelo. Zelo, da je tukaj, zelo, da je tukaj, zelo, da je tukaj. Zelo, da je tukaj, da je tukaj, da je tukaj, da je tukaj, da je tukaj, da je tukaj, da je tukaj, da je tukaj, doko jo, colluščemski soj, je mentalia, tako, da je tukaj, da je tukaj, cinem. Zelo, da je tukaj, da je tukaj, Tukaj modija začeljne objeljno bi bil, da je zelo, seveda, da je zelo v tukaj začeljno, ko je objeljno zelo pliz. To je ne. Se videlimo, da imamo parametrav, ko je d, nekaj je tukaj, raz od vrstvek, ko je da je tukaj, tako, a zalaj je zelo, F12, kako je tukaj 2 v r, je zelo vrstvek, nekaj je tukaj, tukaj m, Tako tukaj je diagonal, zelo je tukaj, zelo je diagonal. Tukaj zelo je, da je tukaj držav. Tukaj je tukaj kvantizacijen konditon na kompak spas, da je in tegral na tukaj, da je kvantizacija. Tukaj je tukaj gno kvantizacijen konditon, da nekaj kod tukaj k vsega se zoprati, nekaj kod tukaj trkaj. Tukaj be gno kvantizacija. Konditon je, da je tukaj vsega, zelo je tukaj, da je tukaj vsega. Konditon je, da je tukaj kvantizacija. Tako, zelo je, da je tukaj vsega, Tako, da je tukaj početil, da je ta m ozirčil. Načo je, da je izgleda, da je to izgleda o druh, se boš tamo vsega, da je to izgleda. Zelo je to kaj naša vsega, da je to brožem vsega, kaj ne zadači, da je to vsega, da je to vsega, kaj je to, da je to, da je to prist. Mi je ta začel, da je to, z vsega, of magnetic charges, and then what we have to compute, well, we have to compute the classical action and the one loop determinant, so the classical action is very simple. As we said, the only term which is not q exact is this twisted superpotential. So essentially you have to evaluate this very simple action on these configurations, and you get something very simple. And then the next task is to compute one loop determinants. Sorry, small a is diving on magic? Yes. So how do we compute this one loop determinants? So essentially what we have to do, we take for each of these points of these configurations, we take the localization term, we expand that quadratic order around that point. So the feature is that we have the space of the configurations, we have the special set, special configurations for each point, we should expand the quadratic action and quadratic level and multiply all the eigenvalues. And so we could lower the middle part a bit, the shape next to the impossible to see the last thing that was the SFI? Yes, but it's not very important. I kept that one will be more useful. So OK, I can do that, but OK. So how do we do that? Well, the most pedestrian way is that, so it's a ratio of determinants for the bosonic sector and the fermionic sector, and we do it. So we've compute the full spectrum in each sector. Now let me stress one of the advantages of working of shell is that for each multiplet, we have essentially a separate problem, and so we can compute for different multiplets the contribution. Now, in general, this is a very hard problem to compute the spectrum, because these operators, morally speaking, are a Laplacian or a Dirac operator. OK, this is a generalization thereof, but this is what they are. In general, we don't know how to compute the spectrum of these operators, this is a very hard problem. We are able to do that if you have a special situation with a lot of symmetry, like here you have the round S2. So as I was sure, this is doable, but if I already have some squashed sphere or some more general space, in general, we don't know how to compute the spectrum. It's a very hard problem. Luckily, as I said, there are many cancellations between the spectrum of these two operators, and so one is really interesting just in computing the eigenvalues that do not cancel. And there are more sophisticated methods to compute just those eigenvalues that do not cancel, somehow because the second-order equations reduce to first-order equations, morally speaking, because of supersymmetry. But as I said various times, I will not go into that, because I don't have time. So hopefully, someone else will do that. So I will just take the pedestrian approach of computing the full spectrum, which in this case we can do. And in general, so one just has to decompose the fields in harmonics in the space, and since this is S2, we know what the harmonics are. Now, of course, our fields, first of all, they have spin in general. So some of them are scalars, but some of them are not. So we should use harmonics for fields with spin, so sections of either a spin bundle or a vector bundle, sorry, tangent bundle and so on. And they're also charged. So this should also be a bundle and a gauge bundle. But luckily, in two dimensions, the spin group is abelian, and moreover, everything is diagonalized. So all the background is abelian at the end of the day. And so essentially, we only need harmonics for fields, which are in line bundle, essentially. And so these objects are called, in the physics literature, spin-spherical harmonics. So these are called YSJJ3. So S is the spin, but more generally, it's just the charge under the line bundle, and this J and J3 are the angular momentum on the sphere. So these J and J3 are half-integer, depending on S. Moreover, J3 and S are smaller or equal to J. And so essentially, if you want this S, it's a sort of a hectic spin, which contains the z-component of the spin, but also charges under gauge bundle. So this gets a contribution from the magnetic field. And so these are just the eigenfunctions of the Laplacian. So if you take the Laplacian, where you put both the spin connection and the various gauge connections, you have some eigenvalues of this operator. So this is just the standard scalar spherical harmonics. If you set S equal to 0, this is precisely that. OK, and so with this, we can decompose all our fields in these harmonics. And we just have to plug in and take the product of all the eigenvalues. So it's very standard, it's straightforward computation. So let me just give you an example. So let's look what type of thing that one get for the Karan multiplet. So what do we have to do for the Karan multiplet, where we take this matter action. So first of all, we look at, first, the bosonic sector and the fermionic sector. So the bosonic sector I wrote. And so essentially, you have to do the quadratic expansion around the background. And the operator that you get, let me call it O5, is the following, where what I mean here is that you have to plug in the values of the background. This is the quadratic expansion. So in S1 and S2 and D, you plug in what the background is. So in particular, the result will depend on the point, where we are. And you just plug in all the spherical harmonics and you collect the factors. And what you obtain is something like the following. So the details are not important. Just want to give you a flavor of the type of computations and results that one obtain, just that you see it once. So in particular, you don't have to copy these expressions. They're not particularly important. And you can find them in the papers. But the point is that you get some infinite product of eigenvalues like this. And then, OK, one repeats the same thing for the fermion operator, which I'm not going to write because I didn't write it. I wrote it yesterday. Now, in this case, these are two-component spinor. So really, you have to decompose both components in terms of these spherical harmonics. And they have spin 1,5 and minus 1,5. Yes, in this particular computation, phi is a scalar. And so my value for s that I'm using here is really just the magnetic flux. But here, I have two components, 1,5 minus 1,5. And so once again, I plug in and I compute the eigenvalues. And you get some similar expression, which there is nearly no point for me to write. Well, OK, let me just write it to show the feature that I was talking about. So the expression that one obtains is something like the following. So a very similar state. There is some product of this similar j. So as you see, it's very similar, but it's not quite the same. So most of the terms will cancel, but not all of them. And so again, one finds, so we are interested in this super determinant, which is the ratio of the two, the fermionic over the bosonic determinant. And this is, so this will be some simplified expression, but it will still be some, I mean, this is not completely trivial, it's still some infinite product. OK, so this is the expression. Now, well, this expression is not well defined because this product is not convergent. So we need to regularize it. So let me put this in quotation mark. So this is telling us what the function should be, but this per se is not a good, what is not convergent. So we should regularize it. And for instance, we can use zeta function regularization. And so, OK, in this particular case, what do we do where we can use the hord width zeta function? This zeta function depends on two parameters, z and q. And it is defined in some infinite sum. And this is convergent if the real part of z is large enough, but then you can do analytic continuation and you can define it in the whole complex plane up to some poles. And now, OK, so now if you take this expression and you compute a certain derivative of this expression, so you take the derivative respect to z and then you set z to zero. So formally, you obtain the following expression, which, OK, up to taking the exponential is precisely the thing that we want. It's an infinite product over this n for some shift. This shift will be this expression here, this expression here. But in fact, if you do it more properly, what you get is a log of a gamma function. And so this is the regularized expression that we associate to this regular function. We associate to this expression. And so because of this, so let me also say that so here, OK, here it was, OK, and this expression is like we have a single mode. But of course, this current multiple is in some representation of the gauge group. So we have as many components as the dimension of the representation. Each component is associated to a weight of the representation. And so really here there is also a product over the weights of the representation. And so the full thing that I can call z1 loop chiral, because for the chiral multiplet, is first of all a product over all the weights in the representation. And then there is a ratio of gamma functions. So once again, the specific function that we get here is not important in different examples and in different dimensions. We will get different functions. The important thing is that is explicit and relatively simple function. And so in general, it's not a big problem to compute in this one loop determinants. So this is not a hard part in localization. It might be technical, it might take some time to compute it, but usually it's not a big obstacle. OK, so this is the result for the chiral multiplet. Now we want to do something similar for the vector multiplet. Are there questions? OK. So the computation is similar. The only point that I would like to stress, which is different, is that we need to fix the gauge. We are dealing with a gauge theory. And we can use the standard Fadevko-Pov method. However, as this drawing, so let me make this drawing again, so we are expanding around a certain point, a certain background configuration. And so we have to do gauge fixing on a background, not around the zero configuration. And so what we do, so as we said in the general discussion, so we take this gauge field and we separate it into some background, which is essentially this constant magnetic flux that we have on the sphere, plus an oscillatory part. And we should perform gauge fixing on this background. Let me also say that we don't need to do gauge fixing for the background, because essentially we have already done it. So when we wrote the set of BPS configurations, we said that, OK, we have to sum over magnetic fluxes. So we already were using a gauge invariant description of that. So we don't need to do gauge fixing for this, but we need to do gauge fixing for these oscillations. And so we do the standard thing. We add a gauge fixing action. So we introduce ghosts, C and C tilde. These are a scalar. These are fermionic scalar in the adjoint representation. And now this parameter xa, so maybe I should use a different name. Let me call it, OK, let me use xai gauge fixing. So this is now the Fajeti-Lopoulos term. This is the parameter that we are using in arc C gauge. And so in particular, as a check, there should not be a dependence on this parameter at the end of the computation. And this covariant derivative, these are computed on a background. So in this covariant derivative, you only have the background. And so in particular, this action is quadratic in the oscillating fields, in this a hat and the ghosts. And so the only difference is that now we have to do so when you expand a quadratic order and we compute one loop determinant, we also have to include this and the ghosts in the computation. But otherwise, this is the very same thing. So in particular, we can use a superior maze plus this gauge fixing action. This is already quadratic and we do the same thing. And the one loop determinant will have a similar form. So there will be a contribution from the ghosts. Once again, we expand all the fields in harmonics. So there will be a contribution from the ghosts, which are fermionic, a contribution from the gauge genie and a contribution from A and sigma, which are mixed together. Well, here we get a square root because these are really real fields, while all the other ones are complex. And there are some primes here. So we get some zero modes. So we get some zero modes in this operator and these zero modes will be associated to sigma. So there will be some flat direction from sigma. So we remove the zero modes and we'll have to integrate over them. You also get some zero modes from these ghosts. And in particular, zero modes corresponds to constant gauge transformations. These constant gauge transformations, you don't have this problem in flat space because in flat space usually you set to zero your gauge transformation at infinity. But now we are in a compact space, so we do have constant gauge transformations. And we know that we have to divide by them so we can just remove these zero modes. So we know that these zero modes are just related to gauge transformations. So we can either remove them by hand. If you want to be more formal, we could introduce ghost for ghost, but the result will be the same. So since we understand where these zero modes come from, I will just remove them. Okay. And so the expression that we get is the following. So there is a product over the roots, which annihilate the magnetic flux. So of course this answer depends on where we are. And actually probably a more better picture is the following. So if you want each of these slices, it's parameterized by A, but we have many slices because we have this M, this discrete parameter. And the answer depends where we are. So this will be product over the roots that are invariant under this magnetic flux, that commute with this magnetic flux. And then you find some product over the positive roots instead of finding a product over the weights, but these are the weights of the joint representation. And here the function that you get is even simpler than before because it's just a quadratic polynomial. Okay, so we have this very simple expression. Is there any question so far? What's the argument of the first alpha of the input for it? Well, it's not an argument. In the 1 over... Here? So this is 1 over absolute value, let me write it, 1 over absolute value of alpha of A, sorry. Yes, alpha M is 0. There's no argument. You just do the computation and this is what you get. And then the other alphas are also all alpha of A? No, this is alpha positive. So you divide the roots into positive and negative. Like, after that, the exponent minus 1 to the alpha of... Alpha of M. And then it's alpha of M squared plus... Because so this is a product over alpha M equal to 0, this is a product... These are two different products. So this is alpha of M and this is alpha of M squared and alpha of A. Once again, the details are not particularly important. I mean, they are important for this computation, not for the general idea. And so as we say, we have these zero modes and so we should integrate over these zero modes. In particular, these zero modes span the cartanto-balgebra, which commutes with the magnetic flux. So in particular, if the magnetic flux is... The magnetic flux is diagonal. If the magnetic flux is totally generic, all the again values are different, then it's just broken to, if you want... I don't know, U n is broken to U1 to the n and so the zero modes are only corresponds to these n directions. But if the magnetic flux is zero, then there is no breaking at all and so the zero modes span the full U n algebra. Essentially, because these magnetic flux lift the zero modes. So if this is generic, it lifts almost all of them, it cannot lift the ones in the cartan because they are neutral. But if it is special, then there are more zero modes. In particular, if it is zero, it doesn't lift anything. So we should integrate over these zero modes. And so what does it mean? Well, it means that we integrate over these a. So these are many as the rank of g, this a n. And you see essentially, so as I said in the general discussion, these zero modes are nothing else that the continuous direction that you have in this space. So the pat integral reduces to a lower dimensional space and the coordinates around that are precisely the zero modes. So we have this, however, for special values of m, in fact these zero modes are not just the diagonal, as I say, the span, the unbroken algebra, and so we should include a Van der Monde determinant. And moreover, we should also divide by the residual vile group. So when we break un, for any group to the cartan, we have a residual gauge transformation that corresponds to exchange in the cartan, but there is a gauge symmetry, so you should divide by that. So this is the vile group, which leaves the magnetic flux invariant. And so we have this integral to do. In particular, notice that this factor will cancel with this factor. And then finally, we have to sum over all these magnetic fluxes. So we sum over these lattice of magnetic fluxes, which depends on the group, not just on the algebra, as I said. But once again, when we sum, we have to divide by gauge transformations, sorry, by the residual gauge transformations, the residual vile transformations. And these are, so this is the total vile group divided by the vile group that leaves M invariant. So this is just the transformation that actually act on M. And once again, there is a cancellation between this factor and this factor. So we are ready to write the final formula. Is there any question on the, on the general procedure? Sir, from this question, but you only fixed flux of your connection. You only fixed flux, and no flux, yes? But why don't integrate the wrong integration which provides flux, or they are all gauging, they are either grimo, and what what. They are gauge equivalent, because so what are the gauge invariants? So the gauge invariant and the integral of the field strength, and then there are the Wilson lines. But we are on S2, there are no Wilson lines. So the only gauge invariant is the field strength. I mean, in general, the field strength is not the only gauge invariant, you should care about Wilson lines, but you need some cycle. But, okay, if we use the same argument for S4, for example. Yes. And we spent on S4. Yes. For with this argument that we go to the, in the two gauging invariants, it's f and f squared, and we got no Wilson lines in S4. We have not? We don't have any Wilson lines on S4. Yes, there is no extra gauge invariant observable associated to. So the only configurations are enumerated by numbers also? Well, no, I mean, there is not just the integral of the field strength. I mean, the actual field strength is a profile. So the instantons have a modular space, right? You can move these instantons around. Yes. But you only have to tell me on S4, you only have to tell me what is the field strength. You don't have to tell me an extra parameter, which is some Wilson line somewhere, because it's completely fixed. So if you want on S4, so let's say that you tell me what is the field strength. Okay, you specify what is the field strength as f nu nu of x. And then, okay, I can start with a small Wilson line here. This Wilson line is zero, because it is extremely smooth. I am assuming it is smooth. And then I can make this Wilson line larger, and I know how it changes, because it is just the integral of the field strength, right? When I change the Wilson line, I know that the variation in the Wilson line is just the integral of the field strength. And so, if you tell me what is the field strength, I can tell you what is the value of any Wilson line on S4. So there is no extra information. I think the information that may be missing is that f nu nu is set to be constant by the gs equations on S2. Yes, no, absolutely. So, I mean, there are equations that tells us that f will be constant. In general, if you remove equations, I'm saying the only gauge invariant information is in the field strength, but which is a full function. No, any other question? OK, so, OK, we put everything together and we get the following expression. Again, I write it, the details is not super important, because they are specific to this, but then I will make some comments. Yes, the gauge fixing apprehend is qx. It's qx. So you want the result not to be... I'm actually not even using one which is supersymmetric. I mean, I'm not introducing partilence for c. I mean, you could do everything in a supersymmetric way, but I mean, the only thing that I want to do is to fix the gauge. I don't have to do that in a supersymmetric way. I could, then it will be more formal. As I said, you can also introduce gs for gs, but the result is not going to change. The sum of the charge to magnetic bias, is that supposed to be inside of that integral on the line above or is that something separate your writing edges? It's outside, because for each magnetic sector, if one of the zero modes are different. So you should have first... If you think first compute the above integral, then multiply it by that pre-factor and sum over the magnetic fluxes. Sorry, so this integral... So you first take the... OK, let me write the expression, it will be clear. OK, let me write it, and then let's see if this is clear. And you should not copy this. I mean, this is in the papers. There is no point in it. OK, so this is the expression. So let's see if it makes sense. So we have this... So let me repeat today the drawing. We have this modular space of special configuration where we localize. This is a set of submanifolds. So we choose a point on this submanifold for each point we have an integrand, which is this thing here. This is the classical action, and this was coming from this oscillation, so this was the one loop determinant. And then we have to integrate over these submanifolds and then to sub... Over the submanifolds, so we first integrate, and this is our integral over the submanifolds and in fact the zero modes that we integrate over precisely parametrajozati these manifolds. And then we have a sum over the values manifolds, which is a sum over magnetic fluxes. And of course here we are still using a description, which is gauge redundant, because there is the vile group that acts, so we divide by the dimension of the vile group. Does this make sense? Does this answer your question? OK. So let's comment on this. So the first comment is that... OK, I don't know what you think, but this expression is extremely simple. So after all we are computing, so this is an exact evaluation of an unperturbated evaluation of a pat integral. So the fact that we have such a simple expression is quite interesting. And of course this is a common theme in these localization computations. One is able to find exact expression, which is relatively simple. After all, it's just an integral that... OK, I mean, if you want, you can even do it numerically. And then there is a sum. You can do it numerically. So it's a very simple expression. In fact, as we say, so this is an exact unperturbative result. So in particular it should contain... I mean, there should be all the instanton corrections. So you might ask, OK, where are the instantons? And if you wish, this is... In fact, this expression is a bit different from what you get, for instance, on S4, as both can explain, because in that case you also have to include all nonperturbative corrections that you need to compute in some other way. But here we don't find extra nonperturbative corrections. So in some sense it might seem like this is just perturbative. At the end of the day we just put in this one loop determinants which are computed perturbatively. So what about the instantons? And the point is that, in fact, this expression does already contain the instantons. So they are not manifest from here, but it does contain them. And one way to see, but it will be more specific in a moment, is the following. So this is an integral over some real lines. So let me consider the rank one case like it was u1. So this is an integral over the real line. So this is the plane A. Of course there is also this sum over the magnetic fluxes. And we integrate along this real line. So how do we do this integral? One way is to do the Cauchy take. So we close the contour at infinity. Of course we have to make sure that there is no contribution if this circle is large enough. And you can check that this is the case. You will have to choose one of the two sides. That's OK. So then you have this contour integral. And so this contour integral reduces to a sum over residues. And if you... So here there are gamma functions. The gamma function doesn't have any zero, but it has poles. So you get poles from here. And so, OK, take into account onto this sum over the magnetic fluxes, you discover that if you wish at this point you can bring the sum in, because there is no... because this integration contour does not depend on M. So what we discovered, there is a sort of... rank to lattice. It's not really lattice, it's a wedge. But essentially there is a sort of... sort of wedge of poles. And so you can reduce the computation to computing residues around this poles and summing the residues. And, well, as we will see better in a moment, it turns out that, in fact, each of these residues is an instanton contribution. And so we can rewrite this expression as a sum of an instanton contribution. OK? Yes, so if you have pure gauge theory in two dimensions there are no instantons. But if you have a gauge theory with matter and you go on the X branch, there are vortices, and vortices are the instantons in two dimensions. Well, these are non-perturb... these are solutions to the Euclidean equation of motion with finite action. OK, so we will see this in a moment. Well, we will start seeing it. OK. Sorry, but is that... satisfying some... No, it's a vortex equation, which is a different equation. It's just a totally fine. Well, it's... in three dimensions it's a soliton. So if you add time, so this is the standard story. If you have an instanton in some... in dimension d, you can go in d plus one dimensions and becomes a soliton. So the vortex becomes a vortex particle in three dimensions, while it is an instanton in two dimensions. So these are the very same vortices that... I don't know if you are familiar with the particle vortex duality or... Sorry? Yeah. So there are... The vortices are solitons, so there are vortex particles. In four dimensions there are vortex strings and they play a role in superconductivity. They distinguish between type one and type two superconductivity. So they are very... the very same vortices. So let's continue with this comment. So here we computed just the pure partition function, but in fact one might be interested in insertions, operator insertions. And in fact it turns out that there are... So it's easy to make insertions of order operators. For instance we can insert twisted chiral fields at one point of the sphere and antitwisted chiral fields at the other point of the sphere. So here we can include some sigma or more generally some function of sigma gauge invariant and here some other function of sigma tilde at antipodal point and it's easy to include in the localization because essentially you just have these functions appearing in the integrand. And so this is already an example of what I said at the beginning. It's not a holomorphic correlator because this holomorphic is anti-holomorphic. So if you wish this is a type of correlator which you do not access with the topological twist. So this is different from what was done in the past. And in fact there is a lot of... these correlators are quite interesting and I think next week Komar Gotski will talk about this in four dimensions instead of two but essentially there is a story. It's also easy to include the Wilson line operators. So it turns out that you can insert the Wilson line operators along some parallel. So here you can have some Wilson line that depends on some arbitrary representation r. And once again this is very easy to include essentially you have just another exponential in your integrand and so you can compute them and once again this is very similar to the Wilson line. One can also include the other type of operators for instance these order operators of course these are more complicated because then you need to change your boundary condition integrate over a different set of fields because now the fields will be singular points with singular boundary conditions so you have to repeat the computation but this can be done. And for instance Takuja Okuda who is here did these sort of computations so you can ask him if you are interested. Ok. So I will not discuss this. Yes. So you say it's an untested theory and yet in the final expression for the partition function I don't see any explicit dependence on the form of superpotential. Yes. I don't know how you reconcile those two facts because I would imagine that an untested theory would also depend on the form that the superpotential dates. Yes so the full theory of course depends on everything. So if you take a generic operator and you compute correction function of generic operators those correction functions will depend on everything but we are either computing that we just computed the pat integral on the sphere with no insertion so just the partition function that particular computation does not depend on the superpotential. In fact if you and also as I said you can compute more generally these correlation functions for instance of twisted and untrained twisted but these are very special correlation functions so those objects do not depend on the superpotential but a generic correlator does and we don't know how to compute those generic correlators in the strongly coupled theory. Any other question? Similar, yes. If I remember correctly you have to integrate a and then there is also sigma and they have to combine there will be some conditions I don't remember the conditions but there is some conditions. Ok then let me just say that if one study 3D n equal to theory on gauge theory on S3 things are very similar of course the details are different the type of function that you get are different you don't get the sum of magnetic fluxes only the integral but otherwise the same philosophy and the expression is extremely similar so one could leave this as an exercise but you can find it on papers but let me stress once again instead the competition in four dimensions in these respects are different I mean there is a qualitative different that actually in the integrand you need to put all the non-perturbative and non-perturbative corrections but no you have to put the contribution from non-perturbative corrections that you have to compute in some other way so this makes the computations like on S4 or in five dimensions more complicated we can say much more complicated for D S4 requires instantons Ok and then let me stress once again something that we said before that this computation does not depend or my precise does not depend on the scale of the set by the gauge coupling and so in particular something which is a setup which is particularly useful is a setup in which we start with some gauge linear sigma model in the UV so gauge and sigma model is just a gauge theory and for instance we can take a gauge theory of vector and kernel multiplets and we have a formula for the partition function now this gauge linear sigma model may flow in the infrared to a non-linear sigma model and in particular this is particularly interesting if this non-linear sigma model is conformal because then there is no non-unitary deformation and what we compute really knows about the physical theory in the infrared and so this can become a very efficient way in the quantities in non-linear sigma models and but it is very useful in general but in particular for instance it is very useful for the string because the string is a two-dimensional theory and if you compactify the string for instance on some kalabijao there is a sector of the of the worksheet theory which is a non-linear sigma model on the kalabijao and what we compute thinks about the string on these manifolds ok so I'm gonna move to the next topic so if there are extra questions this is a good time so the next thing I would like to talk about is to clarify this story of how the instantons are included in this expression yes so how this expression knows about the instantons and how can we understand this in a more abstract way and so in particular this is related to I don't remember if this question was posed in this in this school but ok natural question one could have is so there are various choices that we have to do in this localization computation in particular we have to choose this localizing term what if we choose a different term are things gonna change or not of course the final result as a function must not depend on it otherwise it means that we are doing something wrong because localization is just a way to compute the original path integral however we might obtain some different expression for the same quantity so let's suppose that we add besides the two terms we were adding before we add a new localization term let me call this LH and I will not write this term with full details because once again the details are not so important but this term looks like the following here probably I should put a tilde in the notation that I am using here plus many other terms and this is a q-exact term so we want to add localization term so it is something q-exact I am not gonna write all the terms but the important thing is that in this term there is a constant so this k is an number and of course since this goes into a localization term there will not be a dependence on this constant and the end of the day it depends on whatever I want this is controlled by some constant here and then you see that this term looks a little bit like a D term sorry like a phyatelopoulos term because the phyatelopoulos term precisely is a D chi so there is a phyatelopoulos term but of course it is not just a phyatelopoulos term because as we commented before the final part integral the partition function does depend on the physical phyatelopoulos term explicit I mean it is written there so you see that the physical phyatelopoulos term was this chi and the final expression does depend on chi this one is a sort of phyatelopoulos term dressed with other stuff in such a way that there is no dependence on it but it looks like a phyatelopoulos term now action is not positive semi-definite and ok, it is not manifest from here so here this is the imaginary part but if you take the full expression that you can find in the papers and you see what is there this is not positive definite so it looks like this is not good however D appears in the action so there is no problem in doing the part integral over D we can do it by hand without localization and if you do the part integral by hand over D the expression that we find is positive definite so if you do this piece of the integral by hand then we can apply the localization argument afterwards and if we integrate out of this D once again since this is auxiliary field just means that we substitute D by the value that it gets by the equation of motion which is just algebraic and this is the reason why we can do the substitution and in fact one gets the following equation which, correctly, does not contain derivatives and D is just linear so this is why we can just substitute the equation of motion to the action and in fact if you look at this precisely it looks like the D term equation but with a phiatilopoulos term and the nice thing is that if we are in two dimensions and we turn on a phiatilopoulos term and we look what happens to the modular space the vacua are moved to the X branch and on the X branch there are vortices so we might expect that now somehow the localization locus is moved to a sort of something that looks like the X branch in flat space and so the configurations that matter are v vortices by contrast if you want the configuration that we are contributing there looks more like a Coulomb branch in flat space because essentially we integrate over this sigma the sigma was the scalar in the vector multiplet and when we turn it on we break the gauge group to the cartan and this is what happens on the Coulomb branch so in fact sometimes this formula is called Coulomb branch localization formula but here it looks like we might localize to something different so I don't have much time so I will not once again go into the full details if you are interested you can ask me or you can look in the papers but now that you impose this constraint now you go and solve the BPS equations and the solution to the BPS equations are different than before and you might ask why the BPS equations are different, the BPS equations do not depend on the action and of course you are correct the point here is that so when we go to Euclidean we have these BPS equations and they are for complexified fields and so there are lots of solutions which are general complex solutions and in general this is complicated however then we restrict to a real contour and then there is just a subset of the solution so in particular when I wrote those solutions those were the zeros of the real part of the actions which if you go to the BPS equations those are the restriction of complexified BPS equations to some real contour so it is a subset of the solutions and somehow imposing these constraints is changing the real contour but it is only changing it for the D term for the auxiliary field so after all it is not a big change because anyway it is non dynamical but still it changes the reality condition and then you get some other subset of the complexified BPS equations that I am going to tell you so this is how you can get different set of BPS solutions the first set of BPS solutions is I don't know what it is it is very large, very complicated or maybe it is not but I am not studying or I don't think anybody has studied it is there any question on this point? so what happens is the following that so we can imagine what would be the Coulomb branch on flat space the classical Coulomb branch so this is parameterized by this A and there are special points on this would be Coulomb branch in flat space where x branches open up and these points are points where one of the current multiplates become massless so these current multiplates have masses because there are twisted masses into the game but twisted masses can be offset by moving on the Coulomb branch and so special points one of the current multiplates becomes massless and those are the points where the x branch open up of course in the x branch you give wave to the sum of the current multiplates and so they better be massless wave and in fact in the BPS equation you find things like sigma 1 minus m phi equal to sigma 2 phi equal to 0 where this m I should use a different letter maybe I can use capital M so where these are the twisted masses and in fact now I realize I didn't tell you how the twisted masses enter in that formula so let me just say that so in this formula I set all the twisted masses to 0 for simplicity but you understand how the twisted masses appear because twisted masses are expectation values for the scalars in background vector multiplate so they are exactly on the same footing as this A so if you turn on the twisted masses you also have some other contribution which is the same as this A the only difference is that A is dynamical and so it's integrated over the twisted masses are not dynamical so they are not integrated over and there will be some parameters at the end of the day but otherwise they appear exactly in the same way so you find this type of equations and so this is what I described before so if you have a mass of course the field should be 0 but on special points of the Coulomb branch some of the components because of course this is a diagonal matrix in the cartano in the algebra while this is if you want this is a diagonal matrix in the flavor algebra but you can have a conspiracy in which some component cancel out and that component then can get a V because this equation allows for phi to get a V so these are precisely the special points where sigma 1 minus m has some vanishing eigenvalue and then you find some other equations now these equations involve so these are differential equations that involve this chi and we once again since there is no dependence on this chi we can take a limit in which chi goes to infinity we can take a limit in which chi goes to infinity and so I will just describe the type of solution that you obtain when chi is very large but I mean there is some differential equations for finite chi and one can study this differential equation so I take the limit and chi goes to plus or minus infinity and the type of solution that you get is the following so we have the s2 and and so outside so there are two special points which are the poles these two special points are selected by the particular q that you chose because this q is made of these epsilon and epsilon tilde of some components of them so there are two special points and outside of the points essentially this equation forces you on the x branch so you have the field strength is 0 and phi, phi tilde is equal to chi and this is precise what you have on the x branch so phi takes a wave and this is we can call it fake phi tilde opolost term in the sense that it is not the physical one however if we go close to the poles then we have some non-trivial differential equation so if you want this is similar to what happens in paston's case on s4 but with the difference that I will remark so we get this equation ok, so this plus or minus I think depends on whether we are at the north pole or the south pole so in one case we get a plus and this d plus minus maybe I should call them the z, z bar these are holomorphic derivatives so maybe I should call either z bar or z depending on the two and so these equations are equations which are known in physics and mathematics and these are called vortex equations so as we said before these equations are so the solutions are essentially instantons so you can study this equation in flat space and you can think of the instantons the difference is that the instantons in four dimensions are equations only for the field strength in two dimensions so if you just write the equation for the field strength you don't get something non-trivial you need the equation for the field strength and some scalar field so this is what you have in these vortex equations and so and so now what you localize on well ok, you will have some finite sum over these x branches but this is just a finite sum so this is a single x branch it's not an important point the important point is that now we have to include all configurations which are solved, yes oh, I was just wondering what's the relation between the off the side those vortex equations so those are the number of the configurations but the equations you wrote under x branch F1 is equal to 0, 5, 5 is equal to 5 is that just giving an ansatz to solve the vortex equations? Yeah, this is the behavior far from the core of the vortex so as we are taking the sphere to be big enough so that you can say ok, it looks like flat space at each pole? Yeah, so the point is that so if you take this equation of course the vortices have modules that corresponds to moving them the core of the vortex but really the vortices here are confined to the poles they are in a omega background essentially and are you saying that all solutions have the core mass topically, or that's kind of an easy class to construct? No, all of them has up to the, so if you go far from the core of the vortex, so the vortex solution looks like the following, so there will be a core the field strength is as a bump around the core so this is the field strength and then the field strength goes to 0 I think exponentially far from the vortex and so if the field strength goes to 0 you see this is constrained to be 0 so this is, so the field strength is 0, this just comes from solving the equation you see that it goes to 0 exponentially yeah and then this follows so if you want phi goes something opposite, phi is 0 at the core essentially because phi winds so it goes to the core so it goes like r essentially this goes like z to the n where n is the let me call it k, the vortex number so phi winds, the phase of phi winds as you go around set infinity if you count the number of windings this is the vortex number it's a topological concept charge and so since it winds it has to go to 0 at the origin to be smooth so this goes to 0 and then when you go far from the origin it goes to the value phi equal to square root of chi which is the so far from the vortex we are on the x branch but then in the core of the vortex I behave like this yes, in fact it's nice how this is related to superconductivity because so when you are in the x branch you know x mechanism is superconductivity essentially the photon is massive and the magnetic field cannot penetrate but when you have a vortex this breaking parameter goes to 0 so the mass of the photon goes to 0 and then the magnetic field can form a tube that goes through the vortex is a tube of magnetic flux that goes into the superconductor so there is this some actual real world physics in this of course also in instantons ok yes, the only difference so here I am a bit simplifying the equations in this limit because the only difference with paston case is that so somehow in paston case you have this point-like instantons at the pulse of the S4 if you would like to be completely let me just finish this sentence so if you would like to be completely precise you would need to do what Necrozov did in flat space you would need to do probably some non-commutative deformation and open up the instantons to smooth configurations still in paston set up these are not smooth these are really point-like and I don't think that this has been done on S4 I think everybody believes that it's gonna work but here there is a parameter that you can tune and when chi is finite you actually have a smooth solutions and if the guys go into infinity they become the standard vortex equations in general there are some smooth deformation of these equations so in some sense here you already have the deformation that makes all the configurations smooth but otherwise it's very similar ok so we will continue tomorrow