Ok. So, we were discussing exprance localizacon. And so, in particular we added one more Q-exact term to the action, one new localization term, these of course is not gonna change the result, however you remember this term was not poslyzded definite? Well, I didn't show it, but I told you. in nekoraz sem ni da vsej dojelima, da je to krha ciljev, počke je skolja za pevnega reginta, in zato je zadajte ima, besej bps soluzije in bpsa rekinta, kjel je počkemo na ker ne bo nohnone, na krežnih, da samo to izvajemo, da je vse počent. In včetno, asešelj, njih ne videlem vse deteži, ker ne so več nekaj, ali vsešelj je tudi podobno to, da je tukaj na s4, in vsešelj, da v bolj dela tukaj, tako da je tukaj spesiali pojnje, nekaj nekaj nekaj pojnje, tukaj nekaj nekaj pojnje, nekaj nekaj pojnje, nekaj nekaj pojnje, nekaj nekaj pojnje, Tudi tudi in tudi tudi tudi in here we chose one here and one here.  So in the bulk the solutions, well, I remember also this was some, let me call it Q exact parameter. And it says this was a parameter that appears in the Q exact term. So this is not a parameter that will affect the answer, but it might affect the presentation of the answer. V tem, kaj sem tudi ga svoj, je vse je zelo neko zelo, v ko ta parametru pojala na infinity. Moj se poznala, da je sistem na naša vse kaj, in tudi pa nekaj priješal nekaj evočnih. Čekaj, kaj to neki parametru, tako da je zelo na koncu, da je vse naši začil v tem storiji. Tako, nekaj sem tako nekaj vse, da ni tu vse počil na pappri. V nekaj bolj je na hradi, So, let me stres. So let me put this in quotation mark. I mean, this is an analogy of course. So the theories on S2, okay, there is no time, there is no Hamiltonian associated to S2. So the is not that the theory is physically put on the X-pranz. This is just an analogy to understand what type of VPS configurations contribute to this pat integral. Čekaj, zelo da sem tukaj na kotacij. Zelo, da je zelo, da je zelo, da je zelo, in karal multiplice završila tukaj sem zelo, z tukaj nekaj parametrem, nekaj nekaj parametrem, in tukaj sem tukaj nekaj parametrem, tukaj sem tukaj nekaj parametrem, tukaj nekaj parametrem, Zelo da sem bila, da sem後rečnikočno povsem. Zelo da sem Chocolate Hot Cheeswanja. Aha. Yo rej. kakoardo je, vključa, nič kot, tudi povolj. Eni taj ta dolик, bi bila d услiknit, ta daj, ki smo, prvo izpočušila, kako bilo, tako, tudi naredil. So bi malo z trk, štup. V intervenciju nekaj začajno zepleaner na zbe 40, dolik, kečke vse, vsega tako zelo. To je razlih vsega. Ta je vsega, the BPS locus, and so now we have to integrate or sum over this different set of BPS equations, but apart from that, the story is always the same. We have to evaluate classical action, we have to evaluate one loop determinants, and then we have to sum or integrate. Okay, and this is interesting because vertices are essentially the instantons in two dimensions. Okay, so first of all, we have these branches, and as I told you, in general, it's just one, but it's some finite number, so this part, so there will be a sum over this, this is not particularly interesting, it's just a statement that if this is the would-be Coulomb branch, so if you want this parameterized, what yesterday we called A, I think that was sigma one, and then depending on the twisted masses that you turn on, and I think also the R charges and so on, there are some special points where some component of this phi, so this phi is both charged under the gauge group and under the flavor group, so it contains a lot of components, and the special points are points where one of these components is massless, and so this would be x branch opens up, but there are many of these points. In fact, essentially there is one point for each, essentially for each component, more or less speaking, of this one. Okay, there is more than one. Okay, and once again, these points were identified by the equation sigma one plus m phi equal to zero, and that is the corresponding, when one component of this matrix, this is a matrix in gauge space, this is a matrix in the adjoint of the gauge group, this is in the adjoint of the flavor group, and when this big matrix has a zero again value, then the corresponding component can be different from zero. Okay, is this so far everything is clear? So, the interesting part now is this part here, because now we have to integrate over the modular space of solutions to these equations. Sir, can you explain here, you said that at these points phi is massless, the one component of the phi is massless. What is the component supposed to mean? Oh, it's supposed to mean that now that component now can be different from zero, and so the cone, or maybe I should write it as a complex line, it's the fact that you can choose different expectation value for this phi. But in fact, this expectation value is fixed, because we also have the equation, we also have this other equation, which is telling you that if you have just one component, this is completely fixing the modulus, and the phase will be fixed by gauge transformation. So, this cone opens up, but in fact there is a single, or maybe I should draw it as a line, but there is a single point, which is selected by this equation. So, in fact this part is pretty boring, it's just a finite sum over these various components. There is no integral there. So, what about these solutions? So, these are very similar to the instanton equations in many respects, so if you know something about instantons, many of the things go through in this case, essentially the modulus space of solutions to these equations is separated into disconnected branches, and each branch is characterized by a number, essentially, which is the vortex number, the equivalent if you want of the instanton number. And, well, for instance, if you take u n, which will be our example, then you just, so we are in two dimensions, so what you can do is you integrate this f, either on r2, essentially on neighborhood of the point, and this is going to be the vortex number, in particular is an integer, and so you have these disconnected components in which you change the vortex number, and then for each vortex number, you have some modulus space of vortices. So, you have these guys here, and then for each of the points in this modulus space, what you are supposed to do is to compute the classical action, compute the one loop determinant and put everything together. So, of course, in this background it is more complicated to compute the one loop determinants and one needs index theorem, so one cannot compute the spectrum by hand. So, I am not going to explain how to do that, but one can do that. I just want to give you the general idea. And so the final form of the formula, if you want a little bit abstractly, is the following. So, there will be this finite sum and the stress, this is finite. So, it is not an infinite series or something complicated. And then there will be the classical action. There will be some one loop determinant. And then there will be the contribution from vortices at one pole and the anti-vortices at the other pole. So, the one loop determinant, well, it turns out that the one loop determinant in each of these vortex backgrounds is the same as the one that we computed yesterday. The only difference is that you have to, essentially, some of the fields will be massive. So, in particular, the component of phi, which is fixed, that will be massive and that will be massive. That will be massive and, in particular, will make massive also one component of the gauge field by its mechanism. And so for massive fields that one loop determinant just is equal to one and you can remove it. Essentially, because as we discussed various times, this expression is independent of the RG scale of the gauge coupling and so you expect that everything which is very massive does not contribute. So, here you have to believe in my, essentially, the one loop determinant is the one that we already computed. You just remove the fields that get a vev. So, the interesting part that you would like to understand is this guy's queue. So, what are these guys? Well, this is precisely the result of what contains two factors. This is the result of the integration of the vortex modular space and also contains a piece of collection that depends on this vortex number. So, somehow here I have taken out everything that does not depend on this vortex number. So, this contribution here and what still depends on this k is contained in these pieces. So, what this z vortex is? So, essentially, if you want, this is the integral. So, as we said, this is the integral on this modular space. So, essentially, it's volume and this is weighted by a factor which is the vortex action which depends on this k. If you do things a little bit more carefully you discover that, in fact, the action close to the pulse is not just the flat action because there is still some curvature and because of this curvature the action looks like the theory in the omega background. So, really, this vortex partition function corresponds to vortices in the omega background. And what is this omega background? Well, I'm just gonna give you some words because I don't have time to go into full details. But, essentially, this omega background give you some potential, some trapping potential that keeps the vortices around the origin. Because if you're starting vortices in R2 and this is the same with the instantons now, ok, we are looking at the modular space of these vortices and these vortices are some lamps of energy or action more precisely but they can move and, in particular, they can go to infinity so this modular space is non-compact and it has infinite volume. So, what the omega background does is to add some trapping potential such that these vortices can still move but there is some cost in action and so at the end of the day they are all trapped in the middle and so this somehow the volume of this modular space becomes finite. I thought that these vortices only contributed to the point in the north and south that is not completely correct in the sense that, as I said, so here I'm writing the equation in the limit in which this kaj goes to infinity but when kaj is finite you have some smooth equation on the sphere and these smooth equations essentially allow only the vortices around the point but not just at the point. So in some region which scales inverse kaj depends on kaj and so if you write down the action it looks around the point which is what matters in the limit kaj to infinity it looks like the omega background. So in R2 you need the omega background but on the sphere it's just done automatically by this q-exact term. And as I said, this should be the equivalent of the story of Pesto minus 4 if you do the non-commutative deformation but that is not been done. So essentially we have to compute the volume of this of the modular space of vortices in the omega background weighted by essentially the classical action of the vortex k times the classical action of the vortex and so this object here will be a sum over the vortex numbers which are positive and then this q is the weight associated to the that comes from the classical action so it's a power of the I mean there's one factor for each vortex and then we have the integral over the modular space of k vortices and this so this is essentially the volume however the volume is infinite because it's non-compact and what this potential does is to transform this volume form in the equivalent volume form such that now this this is finite and so what this q is where this q is e to the minus for xi minus i theta so in the instanton so in the case of instantons instanton action depends on the gauge inverse gauge coupling in the theta angle in two dimensions what plays the role is the theta angle and this phi etilopoulos parameter so this just come from integrating the classical action on this background sorry if we weren't under what what's the group action yes I'm gonna describe is there some k factorial in the denominator typically that's not yes you don't want to over count yeah I mean here you have to count the volume of the modular space where you don't over count so it's not just I mean this volume is not just k time the volume of a single because they also interact I mean they can have relative rotations but at infinity if you wish you could look like a orbifold maybe ok I don't want to say things I'm not sure about so what about this modular space so this modular space is scalar is not hypercaler as the instanton modular space but it still is scalar in particular is simplatic so the scalar form is a simplatic form and so there is a simplatic form omega is a two form and it's closed and if you take the maximal possible power of this omega give you the volume form however we said the volume so we could try to take to the volume would be equal to would be equal to omega to the power so this is scalar and simplatic so the dimension of this space is to L and ok we could determine what this L in terms of k and the gauge group but ok I don't remember the formula but it's even so the volume form would be this but as you say this volume is infinite so we don't want to compute this because it's just going to be infinite and so but in fact what you have to compute is the volume so equivalent with respect to what so first of all we have rotations of R2 and moreover we have flavor rotations so in this equation we have this phi which transforms in some in some under some flavor group and so we have flavor rotations and so ok this is already a billion in the flavor rotations group we take the maximal torus and we do the equivalent and so we want to do equivalent homology if you want under the maximal torus in here so the maximal torus Is L equal to k no I mean as I say it depends on the gauge group and it depends on the so it depends on the gauge group and the representation of phi so here I am being very general as you see I mean the example that we consider I mean if phi will be in the fundamental but you could consider any representation for sure there is a formula I just don't know ok so we have these two guys and so for each of these guys we should introduce a vector field once again since I don't have much time let me just keep the discussion in the notation light so let's call v the various vector fields really there is one vector field for each of the u1 so we should introduce a parameter for these vector fields or more precisely one should make these vector fields taking values in the algebra in the carton algebra we have not done this properly in the first lecture but ok of course if you are interested you can take for instance the lecture of Alex Save or any other review and you can see the details so we just write v and now what happens that ok so first of all this volume form is invariant so is equivariant under this action essentially because this is a symmetry of the problem however it is not equivariantly closed and so we construct we construct the equivariant volume form let me say that we can also write this as e to the omega restricted to the top component and so this equivariant volume form is just equal to e to the omega plus mu where omega is the very same sympathetic form we had before but now mu is the moment map for this u1 action so it is a moment map but a moment map is a function and they say more precisely it should take value in the cartana algebra of the group that acts but we will pretend there is just so if you want what we say is just you can see that we just do equivariant in respect to r2 in this discussion key and so what this does so the differential of mu which is a one form is must be equal to the contraction of the symplatic form by v so this is the equation that you impose and the function that satisfies this equation is a moment map so this moment map exists if you have a u1 action and we have a symplatic manifold and in fact so you can think of this as a Hamiltonian if you want in the rational mechanics context as a Hamiltonian for this time evolution and now you can check that this is equivariantly closed with respect to our equivariant differential and so we can use this equivariant volume form to define the equivariant volume and the integral of this will be finite maybe to slow you down a bit this is a moment map I mean focus lecture we show that this is basically the triplet measure from the hypermultiplet how should I understand the looking bit more well here there is no triplet there is no SU2 we are not in the hyper hyperkeller context we are just in the keller context I don't understand the question no, I'm just trying to understand the looking bit more in terms of fields rather than keller inflective models but where does it come from I can't understand in terms of the Higgs branch physics probably surely yes but I don't know I will give you the HM construction in a moment so maybe we will be more clear but for the mathematical or even physical point of view so first of all the moment map is fixed so if you have the U1 action and you have the symplatic form this equation fixes mu up to a constant and this constant is not important and as I say just from the point of view of this modular space you can think of it as a Hamiltonian and in fact this so Hamiltonian for this action and in fact the motion of this action will be at constant mu so if you want this U1 action so sometimes this is called a height function because if you have some space by the way one exercise is to compute that the equivalent volume of C so just R2 just a complex plane is 1 over epsilon so if you do this exercise maybe you qualify better these concepts but so essentially if you have some manifold suppose that you have S2 so it will not be that example but of course very close and your U1 action rotation of the S2 then the moment map would be literally the height in this picture of the z component if you embed in R3 and then this motion is so you can think of a map of mountains there are these lines of equal height and precisely the U1 action will be along these lines of equal height which means that this is a conserved quantity along the flow generated by V this probably doesn't answer to your question but say for example if you have a fundamental model it will be like a Q squared it is one of the three of the I think it is the D term equation so is it different from the D term? I think it is the D term but so to be 100% sure as you think about it because I guess in Wolfian case it was the D term and this complex D term which is an F term so that is probably the D term what is the parameter epsilon in your answers? so we are doing so we want to compute the equilibrium volume of C so we do dotations ok, maybe I should say it but it is the only U1 action and yes and then in the first lecture we say that if we want to make something homogeneous we should introduce a parameter epsilon in front of the contraction with V so if you want it is a parameter that we need to insert here and then this is the epsilon that will appear there otherwise you will get one I think yes essentially I mean it depends how you parameterize V what is the the length of the circle ok, so maybe here in this exercise we can say that we use D maybe I had a minus in my ok so now that we can get so this defines a finite volume for this space and so now what we would like to compute is this and in order to do this integral I will take a physicist approach I will not do it mathematically but essentially what we can use is the A to HM construction that Wolfgang explained so this integral is the zero-dimensional integral so essentially it is a matrix model of a theory whose modular space is precisely the modular space that we want to compute the volume of so that has this space modular space and this this way of computing it is called A to HM construction of course A to HM and then precisely do this I mean there were mathematicians and they did it in a mathematical way so you want this is a physicist interpretation of the HM construction so you already saw essentially how it works in Wolfgang's lecture and so and now of course the problem here is that we need to know what this theory is and this is no, there is no algorithm to do that as far as I know and so one can discuss some examples in which we know the A to HM construction but we cannot apply it in general if you don't know the A to HM construction the same comment applies to instantons of course there is no nice brain picture here that you can always follow that you can always follow, no in examples usually you can so A to HM HM doesn't use the brain construction but it can be interpreted with brains I think that all the examples for which we know the HM construction we also know the brain derivation but as you know if I give a generic group and generic representation we don't have a brain construction so that's a problem so let me consider then a specific example and so we take N super QCD so this will be the 2D theory so this is the 2D theory of which we want to understand the vortex modular space so the 2D theory is UN super QCD with NF fundamentals and NF tilde anti-fundamentals and we take NF bigger or equal to N and generic NF tilde so this is the 2D theory so a special case was this exercise that Bruno gave you now you had U1 with 2 fundamentals ok and so ok of this theory we would like to understand the vortex modular space and one can have a brain construction then you need a brain construction for the vortices in this theory and then you read off what is the theory on the object that realizes the vortices and this was done by Ananian Tong in entry and so ok so now we go to the k-vortex sector and I will give you the answer so this will be a Z so this was a 2D 2.2 theory and so the zero-dimensional theory whose modular space is the vortex modular space in this theory so this is a zero-dimensional theory which is the dimensional reduction of 2D 0.2 so if you want this is a theory with two real supercharges I don't know if we should call it N equal 2 but essentially it is the dimensional reduction of this and it contains the following the following ingredients so there is a UK vector ok let me give names to these fields I don't know if this is important or not let's see then there is one adjoint chiral that I call X and chiral so this is a scalar, fermion-fermion scalar this scalar fermion then there are N fundamental chiral are Nf minus N anti-fundamental chiral and then there are Nf tilde fundamental fermi a fermi multiplet starts with a fermion and then it contains an auxiliary field which is a scalar so in particular notice that in this theory gauge group is not the gauge group that you are into D somehow the gauge group depends on the number of vortices it does not depend on the gauge group that you have here but this is ok, this is a different theory this is a zero dimensional theory which you want is precisely constructed in such a way that if you compute the modular space of this theory it's equal to the modular space of solutions to the vortex equation ok, it's a non trivial fact no one can check or one can use a brain construction but ok, I will not go into the full details I just want to give an example can you comment on this a symmetry between Nf and Nf tilde and is there maybe some side that will do all to this that has the opposite yes, why there is a symmetry because there is a fatalopulosk here xi and I have chosen Nf bigger or equal to N but this is something I should have said I am choosing xi to be I think positive so this construction is for positive fatalopulos term so because the fatalopulos term is essential to have vortices and since this is real there is a there is a transition because at zero is a special point you can distinguish positive or negative so this is for positive and so in particular when xi is positive this is essentially the fundamental fields are the ones that can take the expectation value because the term equation is phi phi dagger minus or chi let me call it xi so this is the D term equation and so in particular if and here here you have the charge of the field in the D term equation here you have the charge of the field and so if the fatalopulosk is positive then positively charged fields can get an expectation value which is the situation described here if it is negative it is the opposite so you have this nice construction and then and so essentially the way to compute this this equivalent volume is to do the patin integral of this theory the patin integral of this theory computes for us the equivalent volume so how do you compute the patin integral of this ok, first of all this is a matrix model so it's simple but ok in fact we can apply once again the localization but now to this problem instead of the big problem so this is somehow the idea that I would like to convey that we started with a patin integral of the 2D theory we apply localization to that this involves some model space which is complicated we have to compute this volume but we have a description of it in terms of a lower dimensional theory and now we apply localization to this theory of course the localization of a zero-dimensional patin integral in the end of the day is essentially a covariant localization that we discussed in the first lecture and so again I'm not going into the full details but essentially what we have to do while we have to compute the one loop determinants I mean we have to solve the bps equations in one loop determinants for instance one loop determinants are very simple because we are in zero dimensions so for instance they take a chiral multiplet now in this chiral multiplet there is this x that I had here and there is chi however in the action there are no derivatives this is zero dimensional it's a matrix model and so for instance the piece quadratic action for them is just this why is this? so it is quadratic in the fields but once again there are no derivatives this action is just the dimensional reduction of this theory but there is a mass term which is controlled by the scalar in the vector multiple of course in the vector multiple there is no vector we are in zero dimensions there is only this scalar but this scalar is a sort of twisted mass as in two dimensions so if you take this scalar and plug into the action so if you want this is the dimension when you do the dimensional reduction from two dimensions you are left with a twisted mass this is but this is a Gaussian integral that you can do so you get one over phi squared from this you get phi from this and so this may be up to some constant and not important in just one over phi so you can think of this as the one loop determinant for the carol multiplet so you go through all the steps you have to do the gauge fixing and you have to do everything carefully and let me write the result and I will not explain exactly how to get to these results because so that is what I wanted to explain in the second part of the lecture so the final result is that you get some integral along some contour so essentially this is once again the flavor of the I mean this follow the zena discussion that we have so we have contribution from the one loop determinant of all the fields it turns out that there is no classical action in this particular problem and then we have to integrate somehow over the BPS configuration that I have not described and what you obtain in some contour integral in this space and I am not going to explain this right now because this is, as I said, something that I want to explain in a different context in a moment so let me just say that this contour I mean a contour integral is essentially a residue and so this contour is called Geoffrey Kiman residue in a moment what this is and how it arises so since I didn't give you any detail I mean it's useful at this point that I described what are the fixed points and so on but one can do this and one can get some expression so maybe let me just write the final expression and I write the final expression just because you had an exercise so you maybe want to compare with the exercise so so those are masses so I didn't explain essentially anything so these equivalent parameters they appear as twisted masses in this ADHM construction and you saw the same thing in the case of instantons because this adjoint chiral is massive and it has a mass so if you want the U1 action that rotates spacetime this R2 corresponds to a U1 that rotates this chiral because the expectation value of this chiral roughly speaking describes the position of the vortex it has many components it is in the adjoint so it has K and so each component describes the position of one of the vortices and so rotating spacetime corresponds to rotating this guy so there is a U1 action here and to each U1 action you can associate a twisted mass and so this mass is epsilon and then there are the flavor rotations and the flavor rotations act on these chirals and fermi multiplets and so these guys here have masses m with an index and these guys here have masses that they called m tilde and when you connect this problem with the two dimensional problem this epsilon has to be fixed to 1 over R the radius of the sphere that is because I told you that if you look at the action very close to the pole you see that it looks like that you are in the omega background and what plays the role of epsilon is curvature and these guys here are precisely corresponds to the twisted masses that you had originally in this theory ok so so ok, so one does this procedure computing these residues very well, is this expression independent from independent of chi? yes because you said to be pinky not the 7 chi now yeah, I understand which chi yes, why is that? because so we are looking at the vortex modular space and ok, what we are computing is this let's see if I still have the expression ok, I just erased the problem in the expression so the expression that we have into d is so we are looking at zs2 this is the guy that we are trying to compute in a different way and this was ok, there was this sum over higgs branches there was there was this classical action this one loop and then there was the z vortex and the anti vortex ok, and now we are trying to compute this z vortex we know that chi it was in a q exact term so chi is not gonna affect the answer so it should not so this z v should not depend on it in fact z v is the equivalent volume of the vortex modular space and this is the equivalent volume it depends on the equivalent parameters it depends on epsilon and this m it does not depend on chi if you want chi in the vortex equation you have a chi but that modular space does not depend at least it's equivalent volume does not depend on it ok so at the end of the day you get some expression that once again the details are not important I'm just writing it because you will be able to compare so ok, so this will be the so we do this computation for each k and then as I wrote at some point then we have to sum over k we have all these vortex sectors and we have to sum over them so here this will be the sum over all the vectors and in general if you have many components for the gauge group when we break when you go to the x branch it's broken to a bunch of u1s these are the numbers in each of the directions but ok, this will arise from doing that computation in particular looking what are the vps solutions that I didn't describe so when I get some expression ok, let me write it I'm not sure so in this expression depends on epsilon depends on these twisted masses and in this object here the pochamer symbols so in particular these are some finite products there is a lot of notation that I have not explained here so there is a lot of notation probably there is no point for me to explain everything so let me just make some comments so what does this depends on so it depends on epsilon and it depends on these masses there is also this dependence on q but this comes when we sum over the vortex of the vortex sectors because we say that there are vortex sectors they are weighted by the vortex action and so only when we sum we get this contribution here these pochamer symbols they are essentially products, finite products and so this expression is relatively simple there is just some number of monomials in these various parameters above and some number of monomials below and some rational function and you have to do this series and essentially this hyper geometric series is a high dimensional version of it but essentially this is a hyper geometric function so once again it is a relatively simple object and why I am writing this so first of all because in your exercise so you had to compute this S2 partition function by doing some residue some residues and so you want to compare with this expression but so the point that I want to stress is that so we started so we started yesterday so yesterday we obtained some expression for this 2 partition function there was this Coulomb-Brensch formula this Coulomb-Brensch formula essentially was a sum over some magnetic fluxes and it was an integral over some real lines of some integrand and now so yesterday and today we found an alternative formula in the localization in a different way which is this X-Brensch formula so this is a Coulomb-Brensch formula now we found some different expression in which there was this finite sum and then in here we had up to some classical or one loop terms we have this Z-Vortex and this is the anti-Vortex so this formula looks very different but it should be equal because we are just computing the same path integral in two different ways and in fact one can at least in this example check very explicitly that these are the same precisely doing what I described yesterday and you are supposed to do in the exercise so you start with this integral which is along a then you close the contour on one of the two sides and which side depends on the value of the Phi etiopoulos term in such a way that the contribution here is zero and then you have to sum over the poles and as I described there is some wedge of a lattice of poles so you have to sum over all these residues and if you do that you discover that each of these residues is precisely one of the terms in this sum of course there is a double sum where we have vortex and anti-vortex so this double sum precisely corresponds to this, the fact that this lattice has two directions and each of the terms in here corresponds to one of the terms these times the other term in here and so this you can do very explicitly and you can check that this is really true at least in an example in which you know how to compute this explicitly this superqcd example is one such example ok what happens if you take a pure superior meals in that case there is no heat strike so I don't expect any heat strike variation but you can still close the contour for chrome bright let's see pure superior meals so don't you break super symmetry I think you get zero in fact the constraint that I put there is that the number of flavors is equal or larger than n is in such a way that you get something non-zero otherwise you get zero because you break super symmetry so you don't even have a problem maybe it's divergent to zero ah it's zero and if it's not large enough it's always zero even if you have say one favorite ph3d yeah you see it it's easy to see it at positive phyethiopulos and so let's say we do as you are doing un so if you are doing un at positive phyethiopulos there is just no solution we go on flat space we study the Lorentzian theory on flat space there is no solution to do the determine equation ok so yeah so this so what this computation shows besides the fact that it's interesting that we can write this in two different ways is that in fact this formula that we wrote that apparently didn't contain the instantons because this was just classical in one loop and you might say ok there is just an integral here but in fact it does contain all the instanton corrections which are manifest in this other way of writing the partition function so the dimension of the lattice so so yeah so this is for u1 so this was for you essentially for each of these integrations you have this sorry how u1 should be u2 then it's a two dimensional lattice so let's try to yeah so if you have so if your 2d theory is u1 like it was in the exercise then this integral is a single integral it's only integral of da so you really have just one line and then when you close the contour you get this two dimensional lattice so where does it come from one of the directions of the lattice come from this sum and the other comes from the fact that the integrand was a ratio of gamma functions but so gamma function does not have zeros so you don't get poles from zeros here but you get poles from poles here and gamma function as poles for when the argument is zero or negative integer so that give you the other direction but you have to combine them but if you do the excess as you will see how it works so this concludes what I wanted to say about this are there any other questions ok so in the second part so I wanted to talk about elliptic genus and how one gets this thing here I don't have much time so probably I will just give you some ideas because I have half an hour so so far we talked about so I want to remain in the same context so I want to remain into 2d n equal 2,2 teoris gauge teoris and so far we discussed this untwisted S2 so it was just one example but as I described there are other backgrounds the one can use to preserve supersymmetry and compute things about this very same theory and these other backgrounds will give access to other observables so in particular one interesting example is to take t2 and compute this partition function of t2 and and this is interesting for two reasons because ok, we compute something about these two dimensional teoris and for instance if this is a sector of the string we compute something about the string theory and so on it's also interesting because in doing this computation we will get an example in which we have fermion zero modes and it turns out that the system of zero modes that one obtaining this example appears in many other examples so learning how to deal with the system of zero modes one learns how to do many other examples Maybe the supersymmetry can be lower Yes, so the setup in which one can do this is 2d n equals 0,2 so it's a special case so with this lower amount of supersymmetry still everything I will say works I just prefer to do it here because we already discussed the multiplets and the actions I don't want to introduce more notation and stuff but everything works essentially the same so yeah there is no point in doing this there is nothing really new that one learns I think so first of all what is the object that we are computing so as we say as we change the manifold we get access to different observables so what are the observables that we obtain if we study the theory on t2 but these are super conformally index and so essentially now we make contact with what Guido described yesterday and the day before he was talking about an index in for dimension n equal to 1 and he describes what an index is you can either count states on the sphere or operators in flat space this is the very same store in two dimensions so we can write this index this is called the elliptic genus but in fact is a super conformal index in two dimensions and this will be a function of various parameters or chemical potential so what this is well instead of putting the theory on we put the theory on a sphere but since we are in two dimensions the special sphere is just the circle if you want we have this cylinder so this is an S1 and now we have two sectors so we go in the Ramon-Ramon sector so periodic boundary condition for fermions now we want to compute a protected index so we put minus 1 to the fermion number and then we start well we count states this is a trace of a liber space so it's a sum of the states in the liber space and we put weights for these states and so now we have so of course we have emiltonian and momentum along S1 and it is useful in two dimensions to rewrite them as left moving emiltonian and right moving emiltonians so so we can have two weights that they call q and q bar for left moving emiltonian and right moving emiltonian so we are in 2,2 so we need to preserve some super symmetry if you want to construct any index but here we have so I will insist on theories that have full r symmetry so for instance super conformal theories like here and so the r symmetry is u1 left times u1 right so now we will use the full r symmetry I will insist that there is the full r symmetry and in order to compute these index I only need to preserve one half as is apparent from the fact that it would work in this context so the other one I can turn on a fugacity for this and I can do refinement and weight states according to what is the r charge here so this is this other ok so let me write it essentially these these are the chemical potentials and these are the fugacities this is standard there will be some sign and then the theory in general can have some flavor symmetry that commutes with the supercharges and so we can put fugacities for these flavor symmetries so this is the object that we can try to define and but this object has some nice properties so first of all it does not depend on q bar and the argument is a standard with an argument that if you have a state whose rise moving in Newtonian is above zero then there must be a... so we use the supercharges and we construct a partner of this state which has all the same quantum numbers because all these other operators commute with the right moving in Newtonian but the fermion numbers since these are partners opposite statistics and so we get another contribution which is all the same but for the sign and so the two cancels and so only the states for which hr is equal to zero can contribute to this index and so in particular means now the same argument does not work for the left moving in Newtonian because now we are also waging the state by the left moving r symmetry and so when we act with the... if you have a state with non-zero left moving in Newtonian we act with the supercharges, we get the partner but the partner will have a different r charge and so now the two things do not cancel and this is why there is actually dependence on q okay is this point clear? so it does not depend on q bar because it is anamorphic function or meromorphic or analytic of q or of tau so this is something which is interesting of course we can send y to 1 and then we remove this parameter and then if we don't have this parameter we can apply the same argument so this also not depends on q bar on q and in fact this reduces to a number which is the written index so now we are just really counting ground states of the system on S1 where both are 0 and so okay, so these are nice properties and let me stress, so what this means is that what we are counting are some special states whose L0 L0 bar quantum numbers are p0 so we are counting so in terms of doing radial quantization in to relate these some of our states to some of our operators in flat space we are counting essentially allomorphic operators operators whose that only have dimension is equal to the spin or in other words the right moving Hamiltonian is 0 okay, any question? maybe the Ramon Ramon is new now that we are into the okay well essentially so we are on the circle we are on the cylinder and so here we are studying the theory on the cylinder and well there are there are two spin structures on the cylinder Ramon or Nevel Schwartz and this just amount of we have to specify is fermions are periodic or anti-periodic and here we are taking the Ramon sector so we are taking periodic fermions but you can take the other one what are we counting? yeah so in general we cannot take the other one so if you are doing 0,2 we cannot and the reason is and this is a general fact I mean we want to preserve supersymmetry so if we have multiplets and fermions we want that all fields in the multiplet behave in the same way so as we go around the circle we want that all the fields have the same boundary conditions and if you are taking periodic boundaries for the condition for scalars then we should also take periodic boundary conditions for fermions so if you want the Ramon is the natural boundary condition that you have to choose in a width and index the other one in general will correspond to some thermal this is not the this is not the time circle but now 2,2 is special because there is more supersymmetry and in fact the two choices Ramon, Ramon and Naversovac and Naversovac are related by spectral flow now I am not going to explain spectral flow you can take Polchinski but the upshot is that you can compute the one in the Naversovac sector but the two are just related by the definition of the variables essentially you just shift the chemical potentials and this reproduces one in terms of the other so there is no new information if you compute the one in the Naversovac sector ok any other question? in the two points the spectral flow could be applied separately on the last column down by there is no special point down by there is no special point well I mean if you have let's see I mean if you have an actual theory where you know in positive spin is quantized then you apply you have the spectral flow just in the combined direction no, I mean in 2x2 case you can apply and change in the spin of the formula separately for the whole of your part wow, but the full states are really a combination of some left moving part and some right moving part so if you want something which is well defined in the full theory I think you do have to do it simultaneously ok so this is an interesting objekt and the point is that this object which here is written as a trace of a some inverse space can be computed by a part integral and this is a standard statement that if you want to compute so if you have some manifold and you want to compute states on this manifold what you do you go to Euclidean so this is Lorentian and this is time and you want to compute states in this manifold what you can do you go to Euclidean you take the same manifold but then you compactify time so now this is an S1 and so the part integral of this Euclidean compact manifold is counting for you the number of states on this special manifold and so in particular so we can compute this index in T2 because you already have a special circle we have to make the other cycle a long time and we get a T2 now when we do that how do all these fugacities appear well the fugacities for left moving and right moving emiltonian they precisely corresponds to the fact that well ok in general we have fugacities for the emiltonian and they just corresponds to the size of the circle ok, because the part integral in this setup this is computing a partition function so this is a sum over all the energies of E to the minus beta H of all the states but here we also have on S1 we also have a momentum and so one can reorganize the sum over energy and momentum left moving emiltonian, right moving emiltonian if you want this just a sum over L0 and L0 bar the other fugacities will appear as some flat connections for vector multiplets so now we have our T2 and on T2 there are two cycles and so every time we have a symmetry we can couple the current to an external vector multiplet and then we can turn on a flat connection for this vector multiplet for this vector so in particular we can impose that so we can fix we can fix the integral of this flat connection along each cycle of the torus and so one can introduce some complex parameters for instance the parameter Z that appears there will be nothing else that the Wilson line for the vector field for the r symmetry along say the temporal cycle minus tau the integral along the space cycle and so turning on these two parameters which are real one gets a complex variable and this complex variable corresponds to to this chemical potential this is standard so what are these background fields for the r symmetry so in the index we argue that what we can do and still have a protected index is a weight for the left moving r symmetry and a weight for flavor symmetries so this corresponds to the fact that if you go on the torus we can turn on so if you on this is the left moving r symmetry you can turn on a Wilson line for a background gauge field that comes to the left moving r symmetry and this does not spoil the right moving r symmetry and then we can do a similar thing for the flavor for the flavor symmetries so similarly nu A would be equal to the integral or some flavor A along the temporal cycle minus tau ok so this observation allows us to translate to rephrase or setup this sum of all these states that we want to do in terms of a pat integral so now we have to compute a pat integral over supersymmetric theory and so we can apply localization to this pat integral this is essentially the idea now of course the steps are the same as before ok one has to find a q exact action now it turns out that in this particular setup the full action is q exact so we can use the original action we don't have to introduce anything else this is just a manifestation of the fact that an index as Guido explained yesterday is independent of continuous variations in your theory this is why it is a protected object it is interesting and so obviously cannot depend on any parameter in the action so q exact and it is you can check so the full action is q exact so in particular there is no ok you can use the full action as our localization term there is no classical action because it is q exact and give you zero on the bps locus so you need to find the bps locus and compute one loop determinants now the bps locus I will be sketchy because I have seven minutes so the bps locus ok essentially I will just be able to give you the answer so the bps locus is given precisely by flat connections so in the same way as this flat connection for background fields preserves supersymmetry so do flat connection for the dynamical gauge fields and so it is the very same thing so flat gauge dynamical connections on t2 and so well this space is essentially also a torus so for each component is also a torus because so you can parameterize in this way but of course a flat connection is gauge invariant but large gauge transformation can shift it by one in some normalization and so this lives on a circle and this lives on a circle on torus with the very same tau so each of the cartons and moreover there is a condition the flat connection should commute on the torus just because of the structure of the cycles and so this bps locus is given essentially by r, this is the rank of the gauge group copies of the spacetime torus in this map and now you go and compute your one loop determinants of course again the problem is trivial because it's t2 so once again you do just three modes and you find two interesting features so the first feature is that these one loop determinants now have divergencies on the space where you are supposed to integrate this t2 because the form of this one loop determinants is for chiral multiplets up to factors which are not very important so they are a ratio of Jacobi teta functions so for instance for a chiral multiplet you will get something like let me call this just a suppose it's just a u1 so you get this ratio of Jacobi teta functions and these Jacobi teta functions have zeros on the torus this is a function of y and of sorry I should call yx is the same thing so x is 2 pi i I guess some u and this u is this parameter and the parameter has a flat connection so this u lives on the torus yes so but the important point is that there are divergencies because this Jacobi teta function has zeros on the torus and so there are some poles and you are supposed to integrate over this x so you have these torus where you are supposed to integrate and there are poles in what you have to integrate which is not very nice so this is one feature the other feature is that now there are fermion zero modes and these fermion zero modes come from the gay genie in particular the gay genie along the carton directions and the point is that this gay genie do not have any fugacity associated with them because of course gay genie cannot be charged under flavor symmetries more specifically the right moving gay genie is not charged on the left moving gar charge so I am not charged under this and the gay genie along the carton are not charged under the maximal torus in the gage group so there is no fugacity that can leave these zero modes and so one is left with zero modes from left moving gay genie in particular there are n if n is the rank of the gage group and so we are in this more complicated situation where are these fermion zero modes and we have to deal with them but in fact it turns out that these two problems solve one solved the other so they take care of each other and the way in which essentially this works and it will be very schematic here is that these zero modes form can be organized into super multiplets from zero dimensional super multiplets and so schematically so schematically one can organize the zero modes into u let me call them u bar lambda plus lambda plus bar and d zero and so these are our fermion zero modes these are our bosonic zero modes so these are in fact the bosonic coordinates that we are integrating over so it's not so bad that they appear in the same multiplet and this is a zero mode for the auxiliary field it's also a constant and these zero modes form an off shell multiplet so in particular one can set up so q is the supercharger we can preserve and one gets something like the following so one gets this action and in fact both q and q tilde square to zero and so the formula now that we are supposed to compute is the following so schematically should be an integral over the bosonic zero modes but now it's also an integral over the fermionic zero modes and let's restore at this point also the integral around these d zero modes for d and the object that we have to compute to integrate is the product of all the one loop determinants but since there are these zero modes as I said we should expand so the first leading term does not contribute because of this so we really have to write we really would have to write these as a full dependence on the fermion zero modes and now so the question is how to compute this well we know how to compute the standard one loop determinant this is what we have discussed so far and in fact if we set to zero this fermionic zero modes and d this is just the one loop determinant that we have computed many times so the u u bar with the zeros this is just the standard one loop determinant that we have computed so it does not depend on u bar so how do we compute this more general object and here the nice thing is that supersymmetry comes to the rescue because this object here now this is written in off shell formulation and so this must be closed under supersymmetry so in particular for instance q tilde of this z and I am just two minutes just to finish so this is supersymmetric so let's see what does it mean well essentially we just apply this superalgebra this action because there are two so if you want there is one complex super charge which corresponds to the zero two so q and q tilde we could use q one gets the same answer so what do we have well q acts on u bar and in lambda plus bar sorry q tilde transforming minus lambda plus so this is minus lambda plus in the derivative of z we respect to u bar and then q tilde so on u doesn't act I give you zero this give you zero but this doesn't give you zero and so I give you d plus sorry zero ok so this relation come from supersymmetry and now if we take this equation which is equal to zero and we differentiate once again once more with respect to lambda plus we get the second derivative of z with respect to lambda plus lambda bar plus is equal to what is equal to one over d zero d z over d u bar and and of course since we have differentiated twice with respect to already lambda plus and lambda bar plus this object is just that is just a component of lambda plus lambda plus bar equal to zero because we cannot have two lambda plus these are anti commuting so we get this nice relation and this is precisely what we need because the object that we need that contributes to this integral it precisely the second derivative it's not the first term in the expansion it's the one that contains lambda plus and lambda plus but this relation is telling us that this is related to the component that we know because this component here is the one in lambda plus and lambda plus is equal to zero but differentiated with respect to u bar and so using this and I write this it's just the integral in d2u and the integral in d0 of this object here but now you see that this is a total derivative with respect to u bar and so combining with this we can apply Stokes theorem and this gives us a contour integral and of course there is still this d0 integration that I don't have time to explain all the details so you can just get the main idea but it is that you get now a contour integral along some specific contour of z and what this is is this z1 loop because it's the one evaluated at zero fermion with no zero modes so this is just the z1 loop which we know how to easily compute and so first of all we have gotten rid of the zero modes and moreover you see a contour integral solves the problem of the divergence because we have converted an integral of all this space to a contour integral and it turns out that this contour integral is at the boundary of the region and the boundary are precisely around the singular point so you get some nice contour integral so ok, I didn't have time to explain what precisely this contour is because it depends that comes out of the computation and as I advertised before this contour is called Jeffrey Kevan residue ok, so my time is completely over so I think I will stop here


How's it going everyone? Caleb here with Layton Sports Cards. We're working on Chronicles football next tonight. This is gonna be a six box break this time. It's number 30 random teams. Here are the results from that random that we just did. Thanks again for getting in guys. Here we go, here we go. See we got good luck on it's Chronicles time. We're gonna be working on select football after this then tonight. And then a 10 box Tuesday. Let's do it Chronicles.  Let's buy that pack a little bit. We have Autograph, OZ's, Ojalari to 50, Giants, T-Law, Trellance, Cadarius, Tony, Prism Black. That'd be sick. Yep, see another Prism Gold Auto. That'd be big time. Clear vision is Sam Ellinger there. There he is. Clear vision of 199, Jalen Waddle, and Russell Wilson. We have a Crown Royale Autograph of Pat Fryermuth number 73 and 99. Solid, there you go Steelers. Not a bad start. Pat Fryermuth there. Come on, clear vision. There we go. Kyle Pitts, clear vision of Ellinger. Ellinger's the only backup QB hanging on. Just like that. Chronicles patched, Deontay Johnson to 199. That's not very good. No, definitely not. Kyle Pitts, Prism Black. That's correct. Chronicles is a mix of all the products that are not a standalone product. Yep. So anything that does not make it as a standalone product is in this set. So here for example is Score Update Chrome. And then you've got right here Luminance. You've got Panini. And then this is going to be Prism Black. And that's going to be purple to 199. Rondale Moore. Cardinals. This is Legacy. This would be like Update Legacy as it has all the NFL jerseys. And then Rashad Bateman, which is a playoff acetate perk. Yeah, exactly. I forgot about that, Sonny. You got the gold final. Gotcha, gotcha. Okay, okay, okay, okay. To 399 to Andre Hopkins. There's a Tiwa nice on the Dynagon. And then Deontay Smith, Magnitude. But exactly. Exactly, man. It is just a mix of everything that is not its own product. Let's keep going. Like Chronicles. Yeah, it's cool. It's a cool product. Oh, here's going to be a good one. Nice. Mac Jones on the Timeless Treasures Patch to 399. Very solid Patriots. Take that. Deontay Smith, Zach Wilson, Eli Mitchell, Prism Black, and Davis Mills. Clear vision. Here we go. Stealing all the brakes again. Try to tell him not to. That's amazing. Nice Dynagon, Silver of Trevor Lawrence. That's a nice one. There you go. Mac Jones. Ooh, Najee Harris, Prism Black, Blue. 23 of 25. There you go. Nice hit. Congrats Steelers. Jamar Chase. Micah. Oh, nice, Sonny. That's awesome, man. Let's go. The Gold Vinyl, Trevor Lawrence, Prism Black Auto. So that's the five. Gin and Waddle. And Michael Carter there to 199. Jets. Najee Harris, Score Update. Najee Harris, Playoff Lamentum. Correct. Yes, that or the Treasures. Treasures RPAs would also be up there. But that is correct. Nice. It's Ryan Shays here now. Five at 25 on the auto. There you go, Steelers. Awesome hit there. Nasty. Very cool. Great card. Justin Fields. Najee Harris. Jamar. Yeah, Gold Vinyl's a heck of a tough pull out of that. Yeah, to five. Yeah, especially out of this. Silver Flux of Najee Harris. Ewinger. That is red on the playoff there. Magnitude to 199. Kyle Pitts. Nice to 199. Trace Sermon. Zach Wilson. Clear vision. Yeah, Treasures RPAs would be like the next best thing all-star. So very nice, man. That was nasty, dude. We have this time number to 20. Kellen Mond. Rookie Autograph Red Zone. Eight to 20. Red Najee Harris there to 399. Shimon T. Williams. Dynagon. And Jalen Waddle. Next one up. Do the Ravens sign Kenyan Drake? I did not. That's a good sign. Yeah. It's a sweet signing. When was that? What's up? When was that? This morning or yesterday? I think it was this morning. Ravens. It's going to be Patch Marquise Brown to 199. Yeah, that's a nice signing for sure. I like that. Kenneth Gainwell, Dynagon, Jamar Chase. Magnitude. I woke up and I was like, oh, nice. Kenyan Drake. I'll take that. Yeah, that's really solid. Devonte Smith, Score Update Silver. It is John. Yep. For sure, John. Nico Collins. Blue. Legacy. Jamar Chase. Nice. One of 25 jersey number. There you go. Congrats. Awesome hip angles. Yeah, we'll mag that for you. Traskin Zach Wilson. Very cool there. Lee Bay Franks, Kyle Pitts. Jamar Chase Rookie to 99. Let's go, Bengals. Justin Fields and Jalen Waddle. Marquise Colston there. That's going to be Ascension autograph. Heck yeah. I love the Ascension dude. Yeah. It's great names. Marquise Colston. Black Ope. I'm thinking Rondale Moore. I think they made it in like 14 and 15 and then it switched over to just Panini Black after that. I was going to say I think it's just called Panini Black now. I was going to say that might be the same like thoughts there. Rashad Bateman on the Flux Silver. Tray Lance. Derek Carr, that's going to be number to 99. Magnitude. Mac Jones, Prism Black. There you go. Thanks. I remember in 14 for sure. Yeah, I was going to say that's an older, older products. Turn of the century is going to be Cadarius Tony on the auto to 75 giants. Not bad there. Kellen Monds. Jalen Hurts this time to 399. Eagles. I remember Naji. I remember buying random team break spots on eBay for one boxes of Panini Black gold. I remember hitting that. Probably like 499 a spot. Oh yeah. It's probably like actually like 1199. Yeah, that's a great shade dude. I remember I pulled, I mean at the time it was nasty, a Blake Bortles auto and I was like oh that's lit and then now it's like yeah. I hear you. I also. Oh, it's going to be a branding coax to 199. There he is. Texans Patch. T-Law. Bateman. What were you saying? I also blew like 75 bucks on a Blake Bortles NT Bookweight RPA. That's a W. Like way back like 2015. That's a W. That's still a W today. Zach Wilson-Silver. Score update. Panini. Justin Fields there. Devonte Smith. Purple Flux. Not bad. That is to 199. 122-199. Kaderis Tony. Rondale Moore. I mean granted I was 14 at that time too so like all the money I was making on like chores was going toward cards. Good choice. Yeah exactly. Jamar Chase. To 199. Justin Hurts. That's when everything was cheap. Right. Amandra. That's why Blake, that's why hitting a Blake Bortles in a group is a big deal. That's what I mean by that. That's a big thing then. It's like let's go. That's big time. Made my money back this week. Correct. Tyler Wallace. Crown Royale. Sillow XRPA. Nice. Raymonds Kinkratz. Nice card. Very solid. Bateman. Micah Parsons. Silver-Kellen Von Flux. Jamar Chase to 199. Nice one there on the clear vision. On a numbered Jamar Chase. Yeah. Bangles having a great break. Congratulations. Loads of them. Kaderis Tony this time. It's going to be Timeless Treasures RPA. Number 299. Nice hit there. Giants again. We'll take that one. Beautiful patch. James D with the Bangles. You go James to 99 on the Sam Ellinger Colts. You're crushing it with the Bangles. That's second Kaderis. Yeah Giants are also doing solid. Correct. Alex C. Alex doing well with the Giants. Let's keep going. Let's find a QB Rookie Auto now. Good luck guys. Six here. Six. Here we go. Solid stuff so far. Keep going though. We have a patch card of Marshawn Lattimore to 199. Saints. Jamar Chase Flux. Lamar. Brady Base. We have Lance Briggs Autograph. Bears Ascension. Not numbered. You go. Heck yeah. The funniest thing about the Ascension set is for the one of ones they don't do anything different besides just stamping one of one of your cards. Really? Yes. Right next to the signature. Waddle. I had like two of the same person in a break and the only difference was the one of one stamp. I was like eh I guess it works. Scoropted. Zach Wilson. Seems the Ravens have any weird Ascension players. Shuba Hubbard to 99. Here you go. 2-2. That's a good checklist dude. There's some like. Yeah there's some cool ones. Cool guys in there. Silver Prism Blacks. Sam Elinger. Silver there. Davis Mills. Purple Dynagon. Devante Smith. Numbered. 166 to 199. Nice. There you go Dean. You go Eagles. Nice one. Najee. I like Patrick Queen. I guess that's cool. There you go. Ron Dale Moore this time. That's red to 199. Najee Hairspace. Greg Rousseau. This guy's a machine. Already right onto the next break. It's going to be Ron Dale Moore this time. 44 or 49. Crown Royale Auto for the Cardinals. Ron Dale Moore. You go Braden Kay. Nice man. Teantre Hopkins to 25. Chronicles Base. UI Mitchell and Kellen Monds. Last box of Chronicles. We got Select after this and then the 10 box Tuesday. Empty. Good luck all. See if we got anything else here. Come on rookie QB Inc. Justin Fields. We're going to start out with a patch of Justin Fields at 3.99. Bears. Nice one there Bears. Pull on Andrew Locke Auto right now. Is he in there? Okay that's sick. Burrow and Herbert. Nice. The Ascension actually. Nice. Yeah. Gridiron Kings Autograph of Michael Carter. 17 to 25 Jets. Call on Andrew Locke. Here we go Jets. That's an incredible call out if you can get that right. Mac Jones. Caleb will do an IRL back flip if people's names are left and right. I will just like lean back on this chair until it falls over. You heard it everyone. Silver, Jalen Wilde, Tynagon, Naji Harris, and bronze Michael Carter there. Nice. 41 to 75. Yo Jets. Michael Carter, Elijah Moore. Battle going on right now. Oh geez. Silver score update. Jalen Wilde in book. That's what I'm saying Libby. The people need to see it right here. The people need to see it. Oh man. Let's keep her going. To 3.99 Clyde Everett-Hallaire. Zach Wilson. My baby. Last one. Last patch motor. Come on Andrew. Oh man. Kenan Allen. And oh nice. 18 to 35. Auto color match. PS2. Patrick Sertain the second. Nice hit there. Awesome looking legacy card. Congrats. That was cool. That was pretty nasty. Javon Holland to 1.99. I think he's going to be good for like forever. Yeah dude. Forever. He's going to be locked down. All right guys. Recap. Thanks again for joining. I think Sauce Gardner is going to be nasty too. All right. Patch cards here of Justin Fields Bears. Marshawn Lattermore Saints. Branding Cook's Texans. Marquis Brown Ravens. Mack Jones. Patriots. And Deontay Johnson Steelers. That's Bradley's boy. Yeah. Autographs. Patrick Sertain the second to 35. Michael Carter Jets. Rondale Moore Cardinals. Lance Briggs. Bears. Cadarius Tony. Giants. Talon Wallace. Ravens. Got another Cadarius Tony. Giants. Ascension is Marquis Colston there. Saints. Kellen Mond. Vikings. Ryan Shazir for the Steelers. Nice one there. Pat Friar Youth. Crown Steelers. Ozeez Arjalari Giants. And the Jamar Chase there. This is numbered 1 of 25. Jersey number. Thanks again everyone. I do appreciate it. We have Select coming up next tonight. Three Boxer and then a Six Boxer. Thanks all.


Hey guys, it's Jen from I Create Crafts. In today's video, I'm gonna show you how to make these car decals. These are my number one sellers in the stores are my car decals. I have a bunch of video tutorials on this. So go back and check out my other ones. This one is going to be start to finish, how to create this, how to package it, where to get your little decal instructions from. So stay tuned, you don't wanna miss this one.  So stay tuned, guys. So I'm starting in design space and I'm gonna go to upload here and I'm gonna grab the image that I'm going to be working with today. So I wanna do this specific one here, the skull. I'm gonna add to canvas. I absolutely love making car decals. They are my number one seller that I sell at the stores that I'm in. So if you are interested in making some extra money, I would highly suggest trying to make some car decals because they are going crazy for me. I guess it all depends where you live and where you sell them at, but you can also do it on the Facebook marketplace. But I think that's a little bit full of crafters on there right now. So your best bet would be looking into selling at a craft store or a consignment shop or just outright selling to somebody. I've sold in a few stores where I just completely sold outright where I just sold $500 worth of signs and then they bought them directly from me, which was really awesome. So talking about this skull really quick, I wanna talk about where I got it. So I'm gonna pull this up really quick and show you. So this Etsy seller is amazing, very easy to work with, has tons and tons of files. And as you can see, you get all of these for a dollar. So every single one of these, you get them all for a dollar. So I'm gonna just talk about it really quick. I'm not affiliated with this. I just really love working with really good creators and he's definitely one. So as you can see, all of these files are for a dollar. I've gotten several of them from him already and they're very easy to work with, very easy to cut out. I'm actually looking, I see some more than I wanna get. So I'm going to be doing maybe another video tutorial on this. I wanna do a few different video tutorials for you guys, but in case you're interested, I will leave the link down below. His store is Atkins Productions Art and honestly, really, really great. So he was working with me and he said that he would give away, do a giveaway for five of these files. So all you have to do is go into his store, favorite his, you gotta heart his shop up here and then leave a comment down below, which one would you like? Why do you like it? Just some sort of comment down below. Let me know that you hearted his shop and even if you purchase a few, but he's going to be doing five giveaways on my site. So you can get one of these, I'm gonna do five different giveaways. So whoever we're randomly gonna pick, but just let us know that you hearted this page or his store and you can pick one of these. So going back, I'm gonna show you really quick. This is his design. I love it. I love this creator because he makes it so easy. You can see on the side here, all of these are different pieces. So you can go in and you can change the color if you wanted, but I'm just going to leave it just plain old black. The only thing I wanna do to it is change the size and attach it. So with this selected, I'm gonna go up here to the size and I'm gonna type in six. And I'm not gonna unlock it. All I'm gonna do is push a tab button and let it automatically resize for me. If I do a six by six, it would be really skewed and it would not look good at all. So I just want this roughly six by five, six by four. So I love how this looks. The only other thing I'm gonna do is actually go down here and push attach. If I don't push attach, all these little pieces would cut out separately and you would have a heck of a time trying to put this together. So this is really awesome file. I think I'm actually gonna duplicate this a few times. I'm gonna do maybe a black one, a white one, and then even a holographic one. So I'm gonna duplicate this two more times. So with this selected, go up to the duplicate button and then once it duplicates, I'll just push it one more time and then we'll have three. So you can go ahead and change the color to this. I'm going to leave one black, like I said, I'm gonna do one white. So go up to color box, change it to white, whatever color you want. You don't have to choose what I'm choosing. And then this one I'm gonna do, let's say this peach pink color, just so I know that this is going to be the holographic one when I cut it out. That was that simple. So all I have to do is cut this out and weed it out. So I'm gonna push make it and I'm gonna show you which setting I use. If you guys liked this tutorial, give me a thumbs up, hit that subscribe button, leave some comments down below. Do you like this file?  I mean, I have so many people requesting videos on how to create things. I hope this is helping you guys out. So here's what it looks like. Here's the white one, the black one, and then this peach color one or pink color you wanna call it. So the only thing you have to do is go to continue and then select which setting you want to use. And there are two specific settings that I always use. So these are my favorite materials that I always use. I use either the stencil vinyl here or the washi sheet here. You can go into browse on materials and you can select one of those or use whatever you do. When you are cutting this out, it might be different than mine. So I'm going to be actually using a wallet cut adhesive vinyl from Heat Transfer Warehouse. I've never used this vinyl before. So we're gonna try to see together how it cuts out and how it weeds out. So again, I'm going to use this stencil vinyl here and then I'm gonna cut this out and I'll weed it out and I'll kind of show you and tell you how it weeds out. So stay tuned for this. I'm really curious and excited to see how this cuts out. All right, so I finished weeding out my files and they were actually a lot easier to weed than it looked. Sorry, you can't really see this one, but this one's actually white on here. I'm glad I did the black one so you can see. So I'm actually gonna trim the edges a little bit because we don't need this. So I'm gonna trim the edge just a little bit because this one is a little bit bigger and it's not gonna fit in the bag that I normally put it in. And this one I'm just gonna trim just the bottom part a little bit and then I'm gonna go back in with my paper cutter and make a nice straight cut. So the first thing you're gonna wanna do is take your transfer tape and put it on here. So I'm gonna show you a really quick trick. So I have this transfer tape, I have a huge roll of this. So what I found to be easiest is actually have your transfer tape flipped upside down, which means the sticky side up. This way you're not trying to find where it's supposed to go on here, you're not getting ear bubbles in it and you're not getting little creases in it like it is down here. So then you're just gonna take your file and flip it upside down. And you can go all the way up to the top if you need to to get a nice straight angle or you can just lay it on here like I am and I'm just giving myself enough space in between each one so that I can go in later with my cutter and cut them easily. So there, just like that. So again, this is just the sticky side here so it makes it a little bit easier. Then I'm gonna take my scissors and trim off this excess here. And these decals are my number one sellers. They are going like crazy. So that's why I'm doing all different designs. I do have a bunch of videos on creating decals if you're interested in any other ones, like learning how to make your own. I'm showing you start to finish how to create them, how to package them. And hopefully you guys will go out and sell them. You know, like I said, these are my number one sellers and I've never done this one before but I'm really excited to put this one up for sale. So for this one, I'm actually gonna trim around the edges. Normally I would take my paper cutter but because it's so big, I'm not gonna have a bag to put it in. So I'm gonna make it nice to look nice for presentation when I'm selling it. And where I sell them is a craft store but it's also a consignment store. So I'm in three consignment stores right now and hopefully I'm going to expand this summer and reach out to more stores, not just consignment stores. So if you're interested in learning more on how to sell crafts, how to package crafts, how to get started on making your own business, I do have an e-book for sale if you are interested. It's on my website, icreatcrafts.com. I'll leave the link in the description below for you. So for this top part, I'm actually gonna take my paper cutter because I wanna have a nice straight edge. So this is just a regular paper cutter. It comes in really, really handy. It's got a sharp edge up here on the other side here so be very careful. This comes in so handy rather than the taking scissors and trying to figure out what is straight. So I can see my file, I can show you guys, I can see my file. I like flipping it upside down because now it's sticky on here. So I like flipping it upside down and it's not sticky but I'm gonna show you guys what it looks like. So I'm just taking it, setting it in here and then I'm going to get a nice straight cut. And you can see on the middle here where I'm going to cut it, there's a little line on it so I can see where I'm actually cutting. So I'm not cutting any of my file itself. So there, just do this, move it like this and then it is done. So now I have a nice straight cut at the top, a little curved at the bottom so it looks just a little bit nicer when you're gonna go to sell it but that is it for this one. I'm gonna get rid of my tape here so I can show you the next one really quick. One last thing before I finish this one is you wanna take a scraper and just scrape this transfer tape on really hard so that when you give this or you sell this to somebody they don't have to do this. Then I like to go ahead and just peel back the top just to make sure that it does actually stick to the transfer tape like it is. And then just take it and scrape it down again. And I'm not sure if this is gonna stick in my bag or not. Maybe, nope, well, we're gonna cut it pretty close. So it did pretty good that I trimmed it like this. So if you do have to trim them, just make sure you're trimming it to make it look nice. You wanna have a nice presentation when you're selling these. So the next part you're gonna wanna do is take a bag. I bought these bags off of Amazon. I love these bags. They're nice, they're heavy duty. They has the plastic at the top that you can peel off and then you can close it. So what I like to do is flip my design upside down so it's showing in the front part. And then I'll show you the tag that I add in here as well. Hopefully this will fit. We'll see. I don't do really this large of decals. So I might have to grab some new bags eventually but I got these just off of Amazon and I love them. I think they come in like a package of 200 or something. So it's really gonna go a long way. So we'll wiggle this down in here. Make it look nice. And then also what I add into this is just a decal application instruction. I found this on Google. It was free and I just printed it off and then I'll use my paper cutter again to get nice straight lines. But I printed off, I'll show you really quick a whole bunch of these because I run out so quickly because like I said, these are my number one sellers. So I sell out of these really quick. So the last thing I'm gonna do is take one of these and now I'm putting it at the backside so people can see it. And you can also put your business card in here if you want. You can put the price in here if you want. Whatever else you want but this is how I package it. And then here comes the nice part up at the top here. We got this little tab that peels off just like this. Move this out of the way. And then you can just bring it down and it seals it. So it's nice package like this. So there's the front, there's the back with the instructions on it. You can put your price tag on it right here. You could add it to the inside however you want but that's how I do it. So I'm gonna show you one more really quick. I'm gonna show you this other one that I did. This one's a little bit smaller so I can show you how I do this one. So here's the white one I'm sorry is kind of harder to see. But I'm actually gonna trim the edges just a little bit just to get some of that extra transfer tape off because we don't want that going on to our cutting board. So I'm just gonna cut this. And then also like I said, it's very important for presentation to show this that it's really nice and it looks nice so people are gonna wanna buy it. So I'm gonna take my cutting board again. I keep calling it cutting board. It's just a paper cutter. And then I'm gonna take it and put my file in. And then again I can see where it's gonna cut right here. So I'm just gonna make sure I'm not going over any of my lines or my file on here and then just slide it. So now I have a nice straight even cut at the top and I'm gonna do the same at the bottom. These really do come in very handy rather than trying to take a scissors and figure it out. So I'm just gonna do the side and then I'll show you what it looks like just like that. So that is perfect. I think it looks really good. Then you're gonna just wanna take your scraper again, get any bubbles out that may be in here and then just do the test again to make sure it comes off. So there it is. Now you can see the white on there and then fill it back in. So I have two different sizes of these bags. This is what the bag looks like. This is a pack of 200, it's a five by seven. I'll leave this in the description below in case you're interested in it. They really do come in handy and they were pretty inexpensive. So I'm gonna take this, flip it over so that the front goes with the front so you don't see the back because that's where the tape is gonna go. So there's the front and then add your decal information with it.  So here is that part and then you got the nice instructions on the back too. So it's just a nice presentation. Like I said, this is my number one seller if you're interested in learning on how to make your own or do any more. I have tons of videos on how to make card decals. They are great. They don't have to be just card decals.  Like let's say you wanted to put your name tag or something on here or, you know, scissors even or a laptop or, you know, whatever you can put your decal on anything but mine are card decals and they go, they sell like crazy. So I hope this video tutorial helps you out. Give me a thumbs up, hit the subscribe button if you're not subscribed already and go to my channel or go to my website, I create crafts and check out the ebook. You might be interested and it gives you a lot of great information. Hope this helps guys. Happy crafting. Bye.


Professor Health coverage cannot be achieved without involving the people who are supposed to benefit from these services. This was a consensus of speakers at the 2023 Bioneer Conference of the Association of Public Health Physicians of Nigeria Legal State Chapter held in Lagos. Placidly correspondent Love Ikukuo Yedokun tells us more in this report. It was a gathering of public health care physicians and experts. They are here to bring some exchange ideas and learn new concepts in the public health care space. In every two years, the Lagos Chapter comes together to critically look at topical public health issues affecting the state. Chairman of APHPN, Professor Yetunde Kuyinu, in her welcome address encouraged everyone to take issues of their health seriously. She lamented the dirt of doctors and the Jaguar syndrome, which she said has created a crisis. Every day, not a focus of all kids and disciplines living in the choice of this country in such a way that it depends on the push and the boom, factors are never ending. Focuses are projecting a new, enlarged, new flows of animals, removed because of no human resource management. For his part, the chairman of the conference, Professor Ushibogun Akin, noted that universal health coverage is beyond finance. He maintained that involvement of the communities is germane to achieving great results. We cannot achieve universal health coverage without involving people who are supposed to benefit from health services, who must involve the communities. So one of the things that we need to do to achieve universal health coverage is to make sure that our world health communities function, that our local community authorities function and that the citizens themselves have parts of the decision making. Otherwise, you expect all the money, you can mobilize all the money, but you will never get the health outcomes that you desire. The wife of the legal state governor, Dr Ibidyukesawonlu, delivered the keynote address. She was represented by Ola Hibiyoi. She emphasized that the overall well-being of all legosians is of paramount interest to the governor. The health and well-being of all residents is of top priority to the state government under the administration of his excellency, Mr. Baba Gideson Kourishola Insomunu. This explains the reasons for the various initiatives and sustained interventions in the health sector to ensure the well-being and welfare of legosians in line with the things plus the gender of the greater legos. Mr. Simba Belu, CEO of Avan HMO, stressed that collaboration with OC coders could just pull a string to achieve in the sword after universal health care coverage. Early challenging, the current situation we find ourselves in, health care inflation is real. We all know in the press, social media, in recent times, you've seen the huge astronomical jump in cost of medication. So you know, these kind of problems can only be resolved when you have the two arms in health care working together and recognizing that both arms are under this huge financial burden, as well as the people we seek to serve. They are also under this huge financial burden and there's no way of transferring the burden to one side. We all will take a hit together. The theme of the conference is accelerating progress towards universal health care coverage, building resilience, health, system, love and co-coordination. Plus, TV News. So you get notified about fresh news updates.