So, let me first wrap up or slightly review where we ended up yesterday, and before doing so, I should have mentioned yesterday which paper exactly I've been using for the transformation rules and the actions, so it's this one in case you want to see the explicit details written down somewhere. The authors, Hamai and Hosomichi. So, we ended up yesterday starting the localization procedure for n equals 2 theories on the force sphere. So, we argued, and Francesco argued this morning as well, that we should really be interested in studying the zeros of the bisonic part of the variation of some fermionic functional V, and for this fermionic functional V, I chose at the moment the canonical expression, which is a sum over all the fermions in the theory of the supersymmetry variation of the fermion complex conjugate times the fermion itself. So, in particular, that means that if you would like to study the localization locus, it's sufficient to study the BPS equation subject to this reality property. So, we should study these BPS equations, and as I mentioned yesterday, there are two classes of solutions if you do so on the force sphere for the particular supercharge I chose. So, apart from choosing V, I also chose a particular supercharge which had the property that it squares to a sum of two rotations and an SU2R transformation. So, J12, J34 were two rotational generators acting on the force sphere. So, we chose this supercharge, we chose this deformation action, and we would like to study these things, and as I mentioned yesterday, there are two classes of solutions, they're the smooth solutions, which are characterized as follows. All fields in the hypermultiplet are set to zero, and inside the vector multiplet, we find that the field strength is equal to zero, which you can take to mean in a particular choice of gauge, that the gauge field itself is equal to zero. But in the scalar sector of the vector multiplet, we find nonzero stuff, namely the two complex scalars, they are equal and they are equal to some constants, which I will write as A0. So, A0 is just some constant, the algebra valued object, which you can always choose to diagonalize, and the auxiliary fields, so there's a triplet under SU2R of auxiliary fields in the vector multiplet. That object is proportional to this very same constant, and the SU2R indices are soaked up by this object W, which I defined yesterday. Okay, so these were the smooth solutions. On top of these smooth solutions, at the north pole, we found solutions to the instanton equations, and at the south pole, additionally we found point-like anti-instantons. So given this localization locus, as we described yesterday, or this morning by Francesco, we're supposed to evaluate the classical action on these locus, so since the hyper-multiplet fields are all equal to zero, we're not going to get a contribution from there, but the Yang-Mills action does give something, and it is some constant divided by the Yang-Mills coupling squared of the trace of A hat, where A hat is really the same as this A0 over here, but with some dimension of all parameters soaked up. Does the theta term contribute? Yes, I will come to that in a second. The theta term will, of course, talk to the instanton configurations, since they carry non-trivial topological charge. So this is the evaluation of the classical action on the smooth solutions, only on the smooth ones, not yet on the instantons, and it also indicated the result for the one-loop determinants, the determinants that describe the quadratic fluctuations around the localization locus, and that was given by a product over all the positive roots of the Li algebra associated to the gauge group we're considering of the UPSLON function times another UPSLON function, while for the hyper-multiplet we find the following result. It's a product over all the weights in the representation under which the hyper-multiplets transform, so this is a gauge group representation, and we find another expression in terms of the UPSLON function, and this has an inverse. So this is sort of where we left off yesterday, so putting these things together in a in the result, at least the result which you obtained by just starting the smooth part of the localization locus, we find an integral over this constant matrix A hat over there. It is weighted by the classical action, and now we find this product of one-loop determinants, sorry where? E to the minus 8 pi squared, trace A0 squared, yes, oh I'm sorry, there should be a square there. Sorry, what was your question? Okay, let me write it here. So remember that the conformal Killing-Spinner equation looked like, like so, and we said that to define a massive algebra we're going to restrict these guys, and that restriction was given as follows, okay, and then Wij is defined in terms of this S, the matrix as follows, like so. So you can similarly, if you look at the conformal Killing-Spinner equation for XI tilde, you find XI prime here, XI prime is related to XI something like this, it should be like this, or let's do this. So this is W tilde, but it's equal to a similar object, which you can define in terms of S and the Killing-Spinner XI, and these two are actually equal on the force here, yeah, so remove the tilde, so you get W, but it's really equal on this particular super charge we're looking at. So this is the definition of W, if I didn't make any typos, okay, so this is the result we found by just starting these smooth solutions, of course we need to take into account these point-like instantons, so we should expect that there is some additional stuff carrying the information of those non-perturbative configurations. So let me make a few comments, and then we'll start studying this piece in more detail. So first of all, so far I haven't quite mentioned the masses for the hyper-multiplet yet, so remember that yesterday I briefly mentioned that Poincare algebra has a central extension, if you tune the central extension, the central charge to be precisely equal to the mass, then the representation remains short, and if you do so, then you can turn on masses essentially by shifting in the one loop determinant over there, by shifting a hat zero by the mass of the hyper-multiplet. So another way of thinking of turning on masses for the hyper-multiplet is just by introducing a background vector-multiplet, and then giving a BPS configuration to that background vector-multiplet, and as described over here, we can give constant values to the scalars in the background vector-multiplet, so if you do that for a background vector-multiplet for the flavor symmetry, we will precisely turn on the mass. So this result can be easily generalized to include masses for the hyper-multiplet by doing what I just mentioned, a second, so masses can be introduced, a second comment I would like to make is about which observables you can actually introduce additionally in the path integral within this localization scheme. So we chose a particular queue, and the name of the game is to just study which configurations are queue closed, which observables are queue closed, because then I can insert them in the path integral, and I can see how my localization computation changes with those extra observables present. So as far as local operators go, you can insert n equals to chiral operators at the north pole, and similarly anti-chiral guys at the south pole. So concretely what I mean by n equals to chiral operators or anti-chiral operators is trace of phi to some power, and anti-chars trace phi bar to some, possibly other power. So these objects, you can check for yourself that if you introduce those at the north pole and south pole, then they are queue closed. And queue closed stuff can be introduced in the path integral, and then you can run the localization machine, and at the end of the day you will just have to evaluate these local operators on the localization locus. It's described over there, phi and phi bar are just essentially these constants. So what it means to have these local operator insertions is just that you address this integral with additional trace of the matrix variable to these powers. Just a second. And the study of these types of correlators will be analyzed in great detail next week by Zohar. So now you just know that you can do it, and how to relate the objects you find by inserting these things on the force sphere. The use of the partition function will be affected by this. Yes, it will. So yeah, I haven't talked yet about this part, but you can introduce it, and this will be modified if you introduce these objects. So if you don't want to modify them, maybe you should look at A1, and then the only relevant powers are phi squared, and then the instanton part is not quite affected. So for this part, I'll refer you to Zohar's lectures. So these are the local operators. As far as non-local operators go, you can introduce Wilson loops. The Wilson loop insertion is why it will introduce trace of E to the power of the matrix variable. You can introduce Toft loops. So Toft loops are disorder operators. And as Francesco explained this morning, one way to introduce them is to not consider the path integral over the space of fields as you normally would, but instead look at the space of fields with a prescribed singularity at the location of the Toft loop. This object is still Q closed with respect to the Q I described over there, so you can run the localization. But the result is not quite as intuitive as for the Wilson loop. It's a complicated computation. If you are interested, you can look at the paper by Gomes Pestun and Okuda. I don't know if he's present. And finally, you can also introduce surface operators. So surface operators, you can either describe them as a disorder operator or as a 2D4D system. And, well, anyway, you can do it. You can also consider intersecting surface operators just so you have in mind what this localization will allow you to compute as far as observables go. Okay, so let's now start thinking about this non-perturbative part. So let's start by just considering Yang-Mills theory. So the action of Yang-Mills theory, it reads something like this. It's essentially f squared. And now we're interested in studying configurations within Yang-Mills theory which have finite action and which solve the equations of motion. So we want finite action configurations solving the equations of motion. So let's see if we can find configurations with finite action solving the equation of motions. And I will do so by using the standard Bogomolny trick.     So we need to correct this by adding another integral of precisely f wedge f. So this is a trivial rewriting of this left-hand side, but now we find something interesting given that this thing is positive definite, or semi-positive definite, we find that this result is always larger than or equal to integral of trace f wedge f. And this bound is of course saturated precisely when this object is equal to zero. So we find an equality here if f is either self-dual or anti-self-dual, depending on the signs I choose. OK, now I indicated that I would like to have configurations of finite action. If the action is supposed to be finite, it better be that this quadratic term dies off sufficiently fast at infinity. And in particular, right at infinity, that means that the gauge field should become something that looks like a pure gauge. So if you compute the field strength of this object, you will find zero. But notice that the information we have here to describe the gauge field at infinity is precisely a map g from the three-sphere at infinity into the gauge group. So for example, I'm looking at SUM theories, then this g will be a map that takes any point at the sphere at infinity into the g group. And such maps are precisely characterized by the third homotopy group of an SU2 subgroup inside SUN. So what this means is just I have this three-sphere at infinity and the SU2 is also a three-sphere. So you can have a map of the three-sphere into the three-sphere which has some non-trivial winding. In the non-trivial windings, they are parametrized by an integer z. And this integer in fact is precisely computed by this type of an integral. So this is a topological charge and it takes precisely the value of this integer that describes the behavior of the gauge fields at infinity. So we're discussing it for the national flat space or what we're discussing? So at the moment I went back to for the, well, yeah, flat space. And what for, sir? Euclidean. Sorry, what was the second question? And what for we are discussing? Because we talked about the sphere. Right, so I told you there are instant on configuration at North Pole and South Pole. Locally around the North Pole, things look pretty much flat. This is the flat space I'm trying to describe now. But you got some, like, boundary condition in the first case. It was that at minimum zero everywhere apart from the pole or what? Yes, so I'm trying to describe configurations which have F plus equal to zero on a flat space, which is the local environment on R4. You'll see where it goes. If it's not clear, ask me again later. So this type of integral computes precisely this topological charge. In fact, to be totally precise, we should have some coefficients. 1 over 8 pi squared. So this integer number is computed by an integral of this, where F is the configuration satisfying, well, satisfying this type of equation. And a configuration that satisfies this type of equation, it's fine at action. We make sure by imposing this, but it also solves the equation of motion. This is trivial by what I have told you so far. These configurations are precisely the minimal configuration within a given topological sector. So by definition, they must satisfy the equation of motion. But we can just check it. If F satisfies an equation like this, we should just make sure that d mu F mu nu is equal to zero. This is the equation of motion. But we know that F mu nu is really equal to its dual. And this is trivially true. This is just a Bianchi identity. So indeed, these configurations satisfy the equation of motion. And they do so trivially. So I would like to study these equations, because eventually I would like to get a handle on the configurations at North Pole and South Pole. And typically, you should expect that the solutions to these types of self-duality equations, they depend on various what goes under the name of collective coordinates. So the collective coordinates, you can think of as continuous parameters on which the most general solution to the equations depend. And given that we have these configurations in our path integral on S4, we should be integrating over these types of collective coordinates. And that is the task we have ahead of us. In the path integral we had over there, we were supposed to integrate over all the BPS configurations. If the BPS equations tell me that I have to solve for configurations satisfying these equations, and if these equations have various collective coordinates or just various continuous parameters which you can take to be any random number, we should be integrating over these additional collective coordinates. And that isn't really the question, how to integrate over these things, how to integrate over the collective coordinates. So to get a bit of an intuition, let me just present the standard example. I'm sure most of you have seen this, and in that case, well, I will just review it for you. So the most standard example is to look at a k equals 1 where k is this integer, or it's k is this topological charge, is this integer computed by this integral over there. So let's look at the k equals 1 instant solution for gauge group SU2. I'll just write the answer. I will write it in singular gauge, and the answer looks like so. So this is a function of x. It looks like x is x nu divided by, so this is a solution where this eta bar symbol is defined as eta bar mu nu i is the coefficient of the Pauli matrix in the decomposition of sigma mu nu, and this thing is, of course, nothing else than sigma mu sigma bar nu minus sigma nu sigma bar nu. Sorry, where does this curly x come from? I haven't explained yet anything, so let me explain now. So I defined this symbol. So the gauge configuration depends on position x. So this position enters here and here and also here. But as I was saying, the most general solution of the instant on equations depends on various continuous parameters, the so-called collective coordinates over there. So this capital X is one of these collective coordinates. This capital X is precisely x mu parameterizes the positions of the instanton. That should be sort of clear from the explicit expression I have over there. So these are four free parameters, one for each axis. There's another parameter, which I called rho here. Rho parameterizes the size of the instanton. So that's another collective coordinate. And finally, I have included here this G and G inverse. G and G inverse are matrices in SU2.  And since there are SU2 matrices, there's really three variables, which you need to specify. So in total, this particular solution for the SU2 gauge group, instanton number equals to 1, has eight collective coordinates. And if you were to introduce this one instanton solution in our localization, we are supposed to integrate over these four coordinates, these eight coordinates, these four positions, this one size, and these three global gauge thingies. So an eight-dimensional integral, maybe that sounds reasonable, although I haven't quite told you how exactly the space looks like parameterized by these eight coordinates. But really, we're doing SUN gauge theory in the localization computation. So let's try to think about what happens in terms of SUN instantons. So an easy way to obtain SUN instanton solutions is just to take my SU2 instanton solution and embed it inside SUN. So SUN, we can think of as an N by N matrix. At least the gauge field over there, it's an adjoint variable. So we can think of it as an N by N matrix. We can embed the SU2 solution I have over here, for example, in the top corner over here. We put zeros everywhere else. And then this will be a self-dual solution in SUN. But of course, I could have put it everywhere. So in fact, I should be conjugating this type of a solution with a generic SUN matrix. I don't know how to denote it, maybe H. So where H is some SUN matrix, but not quite all SUNs, we should mod out by the stabilizer subgroup that is the subgroup of SUN that leaves this configuration as it is. So in particular, we have this N minus 2, N minus 2 block over here. We can act as much as we want here. It's not going to change anything. So we should mod out by those transformations. And we should also mod out by an additional SU2, because in this block, we already included the global gauge transformations. Is that G over there? We don't want to over count. So we should remove that again. So inside here, we had eight collective coordinates. If you do the counting by reorienting that particular solution inside SUN by this conjugation, you will find another for N minus 8.  So in total, we find four N collective coordinates. So depending on what the value of N is, this becomes a big space of collective coordinates. But of course, we just described some solution. What tells me that these are all solutions? And in fact, what happens if I would like to try to increase the instanton number? Can I still write such explicit solutions? The answer is not quite. But I can describe at least a number of collective coordinates pretty easily, since from an index theorem, which I will not explain, but you will learn all about next week, from an index theorem, namely the index of the Dirac operator in the instanton background, you can deduce that in general, the dimension of a k-instanton configuration in gauge group SUN has, so this is the modelized space, or the space of collective coordinates, with labeled by cotopological charge k and gauge group SUN is given by 4kN. So in particular, when k is 1, we find four N here. This is an exact result, which you get by some advanced technology. So indeed, over here, we at least find all the solutions, which we care about, to describe k equals 1 instantons in SUN. So just take that one and perform the procedure over here. You find all possible collective coordinates. But for a larger k, we don't have an explicit description, but at least we know that the modelized space will have this many collective coordinates. So this index theorem, the way I described it, it's like counting collective coordinates, but what it really does for you is count zero modes. So it's a slightly different problem, which I will relate in a second to the counting of collective coordinates. So what are zero modes? Zero modes are nothing else than normalizable solutions. So it's important that they are normalizable. The normalizable solutions to the linearized field equations, where the linearization is, of course, around the instanton background for fluctuations. So this index theorem counts the dimension of the space of zero modes, which is defined like such. So to be a bit more concrete, let us consider a gauge field, which is given by a classical solution to the instanton equations. It will therefore, let me omit the position dependence, it will depend on a variety of collective coordinates, which I will collectively denote, maybe x. So it's not just a position, but I just mean all eight in that particular solution, or all 4kN, if you weren't able to solve the instanton solutions for non-trivial cases. So we have this classical solution plus some fluctuations around the solution. And what the index theorem now teaches us is what is the space of solutions to the linearized field equations for fluctuations. In particular, the linearized field equations for fluctuations will be equivalent to the linearized self-duality equations. So they will look like so. So I'm trying to solve this equation, linearized self-duality equation. But of course, if this fluctuation around the classical background is a gauge transformation, then we're doing nothing, essentially. So we should make sure that this zero mode is not gauged. So we impose that it is orthogonal to all possible gauge transformations. And we do that by demanding exactly what I said, that it is orthogonal to all possible gauge transformations, where lambda is something arbitrary, which is, of course, equivalent to demanding that it satisfies this equation. So the number of zero modes is really the dimension of the space of solutions to these two equations. And that is what the index theorem tells you all about. Now good, this is, naively speaking, a totally different problem than counting the number of collective coordinates. But in fact, these two problems are intimately related simply because if you have the solution, depending on all of its collective coordinates, if you vary with respect to one of these collective coordinates, so this is the variation with respect to the alpha collective coordinates, that is, sorry, I said it incorrectly. If you want to look at a zero mode that parametrizes the variation of the alpha collective coordinates, you can compute that zero mode from the solution by simply taking the derivative with respect to the corresponding collective coordinates. But you should always make sure that this thing actually obeys this type of a condition, so maybe we need to address it with some gauge transformation. So the logic, again, we're looking for this normalized solution to the linearized field equations. That means we're looking for solutions to this equation. We don't want the solution to be a gauge transformation, so we additionally impose this condition. If you find a solution to these two conditions, then in fact, you have found the collective coordinates because such a solution is precisely related to the derivative of the classical solution with respect to the collective coordinates. Is this clear to everyone?  Maybe I should do so. So why is such a statement true? Why if I have a solution to the linearized field equations, should I expect that the derivative of the classical solution with respect to a collective coordinate precisely correspond to that type of zero mode? And the logic is as follows. We know that the classical solution, of course, satisfies the following equation, that if you vary the classical action with respect to that solution, you find zero. That's by definition, it's solution to the equation of motion. But if I now take this thing and I vary it with respect to some collective coordinate, I still find zero. But I can also write it alternatively. Let me be a bit schematic. As a double variation of the classical action with respect to two gauge field configurations, one at x, one at y, off. So I'm just using the chain rule of the variation of the derivative of the classical solution with respect to precisely that collective coordinate. And then I still have an integral to perform over y. OK, so this is a true equality. Zero is equal to this. But now you can recognize here that this will be some particular matrix. It will depend on the inches mu and nu on the positions x and y. But that operator is precisely the operator that describes the quadratic fluctuations. So it's precisely the operator that tells you which field equations describe the fluctuations. So this thing acts on this beast. And this beast is this thing acting on the beast is equal to zero. That means it's solution of that thing. So indeed, this variation precisely corresponds to this type of zero. So here, OK, maybe as I sweep something under the rock, you should really take the classical action plus its gauge fixing term. And the gauge fixing will precisely correspond to that extra piece. Or said differently, you can trivially see that this thing is satisfying that condition. Is that clear to everyone? OK, so we found these zero modes. You still may find that it's not a very useful concept. We could have maybe just stuck to collective coordinates. But it will become clear in a second as soon as I affirm the odds around that the notion of zero modes is really useful. But let me first make another observation for why these zero modes are useful. They provide a metric on the modelized space of collective coordinates. So say we have two zero modes, delta alpha, a mu, delta beta, a nu. But let me contract the space time indices. Let me take a trace. Let me integrate this thing over all of space. This will define an object which depends on two indices, which is obviously symmetric. And in fact, this thing is precisely the metric on the space of collective coordinates. This is just a side comment. But just now, I'd said we have an eight-dimensional space of collective coordinates. I didn't describe the metric yet. But in fact, this is how you describe the metric for that space. In fact, there may be an exercise later that asks you to precisely compute this metric for that particular solution. Good. So so far, I have just talked about pure Young-Mills theory. Of course, in our final theory of interest, and it goes to theories on the force theory, we have fermions. So let me talk briefly about what we should be doing when fermions are around. And what we should really be doing is also study their zero modes. So their zero modes are defined while it's gone. But they're also just normalizable solutions to the field equations. And since the fermions are, of course, classically set to zero, there's no, I mean, the field equations for the fermions already describe the quadratic fluctuations. So the zero modes of the fermions are defined just as before. In particular, that means that we should try to solve the Dirac equation in the background of an instanton. So in here, there is a gauge field that gauge field takes, for example, for SU2, that type of a profile. And we should be solving equations like this, where d bar slash is, of course, nothing else than sigma bar mu d mu, and similarly for d slash. Now, our first observation you can make as soon as you would like to study fermionic zero modes. So solutions to either one of these field equations in the instanton background is that you will never find any zero modes from lambda tilde in an instanton background. So let me do that little exercise. So imagine that d lambda d slash lambda is equal to zero. That in particular implies that d slash bar d bar lambda tilde is equal to zero. Let me write what these two symbols mean. This is equivalent to, say, in a sigma bar mu sigma nu d mu d nu lambda tilde is equal to zero. Now, sigma bar mu times sigma nu. It has a symmetric piece and an anti-symmetric piece. The symmetric piece is just delta mu nu. The anti-symmetric piece is sigma mu nu. So when acting with a delta function, you just get d squared acting on lambda tilde. When sigma mu nu is in its anti-symmetric in its mu nu indices, you may as well replace this with an anticomitator at the cost of introducing a factor of half. When you have an anticomitator of covariant derivatives, of course, you get the field strength acting on lambda tilde. This thing should still be zero. But now you should remember that we're in an instanton background, which means that f is self-dual. But sigma is anti-self-dual. So an anti-self-dual thing contracted with self-dual thing is equal to zero. So this requirement implies that d squared lambda tilde should be zero. And now you should recall that d squared is a positive, well, actually a negative definite operator. So this will never have any solution except for the trivial solution. So this little argument shows that there will be never any zero modes coming from anti-carol spiners over here in an instanton background. But the same argument doesn't tell you anything about solutions to this equation, simply because when you reach this point, you will not have sigma, but sigma tilde, which is self-dual just like f and this thing is what it is. It's not zero, so you cannot make any conclusions about whether there are solutions. And in fact, again, using an index theorem, you can figure out that the number of solutions to this type of an equation, if, for example, we take lambda to be fundamental in the fundamental representation of the gauge group, then the number of solutions is absolute value k n. Well, in fact, since k should be positive, just k n. So we found zero modes corresponding to the bisonic coordinates. These zero modes we could translate into collective coordinates. Now you also find zero modes on the fermions. And the next step should, of course, be to go even one more step closer to our final go, studying instantons in Enneco's two super young mills theories. So we should study what supersymmetry has to say about these zero modes. So remember, from yesterday, I wrote the supersymmetry transformations of the fermions explicitly. It used to be on that blackboard there. And these variations started off as follows, plus and so forth, the variation of the anti-carrel guy was similar. But notice, in fact, I already noticed it over here, sigma mu nu times f mu nu is zero in an instanton background. So here we see sigma mn f mn that will be zero in an instanton background. So in an instanton background, the variation of the fermion lambda is simply equal to zero. So this means that the supersymmetry is parametrized by psi are preserved. Let me just denote it like this. Supersymmetry is parametrized by psi are just preserved. But over here, we have the same story. This is not equal to zero in instanton background. So the supersymmetry is parametrized by psi till that they're broken. In fact, this broken supersymmetry is a little trick which you may use to construct some of these fermionic zero modes. Since the variation of this fermion the supersymmetry used to be a symmetry of the action. But now I see that it is explicitly broken by the instanton background. But we know it also didn't change the action. So this thing, if you insert it instead of the fermion, it should be a solution of the field equations. You can maybe do that as an exercise. So these broken supersymmetries, they gave me some of the zero modes. But not all, not necessarily all. But what is more interesting are these preserved guys. Can we do the counting a little bit slower? Can I push them? Or how many? I mean, you want to compare with the RISCAN which you get there, right? Oh, no. I'm not claiming that you get all solutions by looking at the broken supersymmetries.  And then the rest you need to find in other ways. So some of the supersymmetries are preserved on the instanton background. Now what that means is that there are some supersymmetries around which will pair for me bosonic zero modes with fermionic zero modes. So the supersymmetry is still there in the instanton background. We know that the instanton has all these zero modes. And then I'm just using the statement that if I would like to study, well, if I have symmetries preserved in some vacuum where the vacuum in my case is the instanton, I should expect the fluctuations above this vacuum to be organized in the preserved symmetry algebra. In this case, we have the supersymmetry algebra. So I should expect that zero modes, which are in this case the fluctuations of interest, are organized in supersymmetry multiplets. So bosonic zero modes will be paired with fermionic zero modes in supersymmetry multiplets. So this is useful. We started off starting bosonic zero modes. We studied separately fermionic zero modes. But in fact, if you have supersymmetry, these two types of zero modes, they will come together in particular supersymmetric multiplets. So the question now is, of course, well, which Susie and which multiplets? Excuse me. All of this KM from my zero modes how is it like in other 4KM bosonic zero modes? Right. All the fermionic zero modes pair up with all the bosonic zero modes in supersymmetric multiplets. How do you mean all of them, the 4KM bosonic zero modes? Well, the 4KM bosonic zero modes, yes. So here I was just describing zero modes of the fundamental lambda. But say in an n equals 2 theory, there are also adjoint guys around. So anyway, all of them, at the end of the day, are paired. So to figure out the questions I raised over there, which Susie and which multiplets, let me do the following. Let me uplift the entire discussion by one dimension. So let me consider an R4 and add an extra direction which maybe you can think of as time, although I'm still in Euclidean signature. So as you can imagine now, imagine that we have an instant on sitting there in R4. And if it moves forward in time, it may want to consider roaming around in its own modelized space. So this guy is parametrized by all these collective coordinates, or if you like, by all these zero modes. And as you move forward, except for the kinetic theory, it doesn't cost any energy to move around in a modelized space. So only the kinetic theory of the motion in time gives you energy, but just moving in the modelized space. These are all flat directions, essentially, by definition. So the question then becomes, in this type of a picture, what is the dynamics of the motion of this instant on as it moves forward in time? And this dynamics is, well, it's easily described in words at least, this dynamics is described by a Lagrangian that starts off as, well, let me write the action. It's the integral along this time of a sigma model into the modelized space of the zero modes of under consideration plus dot, dot, dot. So here I just wrote the bisonic part. We introduced this metric earlier. We just have the kinetic term in the target space. And now we need to complete it supersymmetrically, possibly add some interactions. So I haven't really gained much by this picture, because I still don't know what this thing actually is.  What for? Because I have a trick where I can figure out this theory. I can figure out what the world volume theory is on the instant on world volume. Well, that was, I can figure out what the theory is on the world volume of the instant on. As soon as I know what the world volume theory is, I can try to study the dynamics of that theory. And that dynamics will be related precisely to the integral over the modelized space, which I'm after to complete this computation is for. So. And you will explain that later, yeah? I'm heading there, yes. So let me at least describe the supersymmetry. I described somewhere here, no, somewhere there, that there were some number of preserved supersymmetries. You can study how those supersymmetries should act on the world volume. From there, you see it. We're only size, no xi tilde. So it's some chiral supersymmetry. And indeed, it is two-dimensional 0,4 Suzie. I know. It's two-dimensional 0,4 supersymmetry, dimensional reduced to 1D. So I explicitly tell you what the two-dimensional version of the supersymmetry algebra is. So you know what the r-symmetry is, explicitly. So there's an SU2 r-symmetry acting. And another dimension reduced to 1D. So this is the supersymmetry that I should expect to live on this world volume theory. So this completion should have this type of a supersymmetry. And now the question is, what is the completion? So what is that completion? Well, to find that completion, let me talk a bit about brain realizations. It's OK. We would like to study brain realizations of instantons in four-dimensional Nicos II theories. So let me first engineer the 40 Nicos II theory as follows, where the brains are oriented along the directions. 1, 2, 3, 4, 5, 6. And the D4 sits along 1, 2, 3, 4, and 7. So in particular in this picture, this direction is the 7 direction. And this direction is the 5, 6. OK, so I claim that this is a good description. You have two NS5 brains. Or well, I'm just drawing two NS5 brains. You could have more if you like. And I have a stack of ND4s here, ND4s here, ND4s here. So I choose them all equal for reasons I will not try to explain. But let's just stick to the fact that I'm describing a super conformal field theory. So I will be describing an Nicos II super conformal field theory with gauge group SUn and 2n fundamental hyper multiplets. So let me explain this picture a little bit. I hope that most of you have seen this picture before. If not, let me try my best to at least intuitively indicate why this indeed describes an Nicos II theory with gauge group SUn and 2n fundamental hypers. So let's first start by just considering a stack of ND4 brains, draw them like this. Let's imagine we have this stack of ND4 brains. We used to have 32 supersymmetries in the absence of the brains. So now, including the brains, we have broken half the supersymmetry. So you have 16 supercharges left. And the quantum field theory living on the world volume of these D4 brains is just a dimensional reduction of 10-dimensional super Yang-Mills to five dimensions. So in particular, this is 5D maximally supersymmetric super Yang-Mills that lives on this world volume with gauge group Un, but I will ignore the U1. So that's step number one. Now step number two, I take this stack of D4 brains and now I sandwich it between NS5 brains. And I think of the NS5 brains as sitting very close to each other, so the vibrational mode in this interval is very heavy, so it will decouple. So essentially, we will have a dimensional reduction to four dimensions. And of course, introducing these extra brains again broke half of the supersymmetry. So I have eight supercharges left. So I have a four-dimensional N equals 2 supersymmetric theory. And moreover, the NS5 brains have imposed boundary conditions such that going from this 5D N equals 2 super Yang-Mills to this 4D N equals 2 super Yang-Mills, you lose what in 4D N equals 2 language would be the adjoint hyper-multiplot. So this is really just super Yang-Mills. OK, so far so good. And now I can address this picture by having semi-infinite stacks of additional D4 brains where the strings from this 4D N equals 2 super Yang-Mills stretching to these additional semi-infinite stacks of D4 brains give me additional hyper-multiplots. So these guys produce additional hyper-multiplots, N from here and N from there. So indeed, this picture describes a theory, as I described over here. Good. No, I'm not interested in 4D N equals 2 theories as a brain picture. I'm really interested in instantons in this theory. So I can address this picture with yet another stack of brains over here, a stack of D0 brains, K of them, which threads along the seven direction. So this D0 brain is a co-dimension 4 object with respect to these D4 brains, just like instantons were co-dimension 4 in my four-dimensional space. Moreover, they have BPS objects because, again, another stack of D4 brains with directions chosen well with respect to the directions of the D4 brains preserve half the supersymmetry. So this is really describing a half BPS object in my four-dimensional theory of co-dimension 4 or of dimension 0. And now that is all the ingredients of an instanton. I hope you believe me. If you do not, you should definitely read the paper by Michael Douglas. He has beautiful string words to explain this in string language. But at least intuitively, it's a half BPS object of the right dimension. And I claim it describes an instanton. In fact, if I have K of these brains, I describe it describes a K instanton, an instanton with topological charge K. Which paper is it? The brains we have on the right. Yeah, I can tell you later. So from this picture, we can, in fact, confirm what I was saying earlier about the supersymmetry over there. You can just study all these breakings of supersymmetry I've been telling you. So we had a half from having the D4 brains. We had another half from having the Ns5 brains and another half from having the D0 brains. And you can study what the corollity of that, of the remaining four supercharges is. So if I T-dualize this picture twice, so the world volume of the instantons is two-dimensional, then you can convince yourself that indeed, the preserved supersymmetry on the world volume of the instantons is precisely two-D N equals 0,4. So in this case, well, in this case, I have a zero-dimensional object. But by T-dualization, you can convince yourself that the world volume theory has this type of supersymmetry. Or in my picture over there, the dimensional reduction to one dimension. So now we're in good shape. We should just analyze what the brain picture tells you about the world volume theory of these D0 brains to find this description for the completion of these quantum mechanics I started writing. So to find the quantum mechanics, you just T-dualize once. So let me draw the picture again. So this is my picture. And I would like to study what is the world volume theory living on these D0 brains. That will be the world volume theory I've been after. So we should study the quantization of open strings stretching between all these different brains. We know that since the world volume theory has two-dimensional 0,4, they should organize in such multiplets. So again, at random, I'm changing dimensions just because I can. So I'm describing things in terms of 2D and equal 0,4 multiplets. So the D0, D0 strings, just if you think of the stack of D0 brains in and of themselves, they would preserve 16 supercharges. If you additionally have the NS5 brains, they would preserve eight supercharges. They would just be a vector-multiplet of that amount of supersymmetry. And in 0,4 language, that means that you have a vector-multiplet and a hyper-multiplet. So these strings give me a vector-multiplet and a hyper-multiplet. Well, a vector-multiplet of what? Since I have KD0 brains, it's a vector-multiplet of gauge group UK to UK vector-multiplet. So I have three stacks of D4 brains, stack number one, stack number two, stack number three. So the D0, D4, one strings, they will give you, without much explanation, a Fermi-multiplet. Similarly, the D0, D4, number three strings, they give me a Fermi-multiplet. And finally, the D0, D4, two strings. So these strings, they give me something else. They give me a hyper-multiplet. Maybe shape it with a Fermi-multiplet. Yes, I will do so, but in, yeah, I will do so. Zero comma four, two D zero comma four. But I will reduce them now to, in terms of zero comma two language, just because it's easier to describe the multiplets. So a vector-multiplet decomposes into a vector-multiplet plus a Fermi-multiplet, which I will describe in a second, plus a Chara-multiplet and a Chara-multiplet. So the hyper-decompose into two chirals. This is very much the same as we had in 40 N equals two. The hyper-multiplet decomposes into two chirals. And the vector decomposes into a vector in a Fermi. So let me describe these three multiplets, these three types of multiplets. So the chiral-multiplet contains, well, never mind the symbol. The chiral-multiplet contains scalars, phi and phi bar. It contains a left-moving Fermion and a second left-moving Fermion. The Fermi-multiplet, as the name says, it contains a lot of fermions. It's a right-moving one, and it's complex conjugate. And some auxiliary fields. And finally, the vector-multiplet, of course, contains the gauge field and some right-moving super partners in addition to an auxiliary field. So these are multiplets under the 0,2 supersymmetry. So there's only supersymmetry in the right-moving sector, which you can see reflected in these multiplets. In particular, the variation of the gauge field gives you only some right-moving stuff. OK, so at least the field content should be sort of clear. And these decompositions are what they are. So very good. This took some time, but there it is. This is the field content of the world volume theory of these KD0 brains, which describe an instanton in my four-dimensional N equals 2 theory. And in fact, I can do a bit better than just describing the field content. I can give you the precise quiver gauge description of this theory. So as I said, we have a UK gauge group that's still preserved over here. We have a UK gauge group together with an adjoint fermi multiplet and two adjoint chiral multiplets. So the chiral multiplets are oriented, but since they're adjoint, it doesn't really matter much. So this is for so far I've drawn this piece. Then there are two fermi multiplets. You see that they were fundamental under the UK gauge group and under the UN from these stacks of D4 brains. So we have additional fermi multiplets, like so. And finally, we have these two chiral multiplets, which came from the D0, D4 strings. So again, they're fundamental under UK, fundamental under another UN, and they look like so. So this is the quiver gauge description. And given that we started off with 0,4 supersymmetry, this is not going to mean much, perhaps. But there is a j-type superpotential, which you need to turn on for this adjoint fermi multiplet. And well, let's just keep in mind that you can also write down the interaction term. So this type of superpotential determines the interactions of all these multiplets among each other as described by this quiver. So do you mean that the 0,4 fermi multiplet is the same as the 0,2 fermi multiplet? Yes. I see. It's exactly the same number fermi multiplet. It's the same. It's sort of maybe somewhat similar to a half hyper multiplet, which is a multiplet of n equals 1 and n equals 2 simultaneously in four dimensions. It's somewhat similar in spirit. So with some amount of effort, we found this description for the world volume theory of the k-instantan theory in four-dimensional n equals 2 super Yang-Mills theory with gauge group SU n and 2n fundamental flavors. Now, why did I put all this effort? Because now comes a pretty beautiful statement. The Higgs branch, appropriately defined, of this theory of my theory, is precisely equal to the k-instantan modelized space in gauge group SU n if you have 2n fundamental hypers. So this has to be true simply because we're describing the world volume theory of these instantans. Their vacuum space should describe precisely what three motions you can do, whereby three I mean zero-mode motions. So actually, as one of the exercises, I suggest you to try to verify this statement. So this is a very good statement because now we have a concrete handle on the world volume theory and we know that its vacuum manifold is the manifold we were supposed to be integrating over all along. But now we can actually do the integral over this thing. We should just. And what about the Coulomb branch of the same theory? No, the Coulomb branch is not relevant for this story. So is this different from the ADHDM type of construction? No, this is exactly the ADHDM quiver. OK, so you can explain it in a few seconds. If you like, yeah. But I hope Nikita will do that next week. So one of the exercises is precisely to get the ADHDM constraints by just checking this type of statement. So when doing that, you can just ignore film markers, right? In the one-off film markers? Yes, of course. I mean, you need to be slightly careful because of the superpotential coupling, perhaps. But essentially, you can ignore the film markers. Here, we have to compute the potential from the plane construction. I don't know. Here, I just knew it had to be there because I knew from the brain construction there is a 0,4 theory actually living on this world volume theory. And if I describe a 0,4 theory in terms of 0,2 language, there are canonical interactions which need to be present for the 0,2 descriptions still to have 0,4 supersymmetry. So this is why I knew that this J lambda had to be there. And is it unique to come in and find the same thing? You need to compute all these things scattering them to 3 levels and you will get it there. But, well, fine. You can do that computation or you just realize it's mandatory by the reduction of the Susie. OK, so let me now go back to my computation on the squashed force sphere. Remember that the localizing supercharge I have been considering. I think I wrote it at the beginning of the lecture. Let me do it yet another time. But let me be a little bit more precise. So these are the three pieces I've been telling you about all along. The two rotations, the SU2R carton. But now I also included the masses. So as I said earlier, you can think of masses by giving background value to vector multiplied coupled to the flavor symmetry. Or you can think of them as being central extensions of your algebra. This is how they enter here. They're somewhat, they look like a central extension because, of course, flavor symmetries commute with everything. And over here I made explicit the gauge transformation. Well, at least explicit enough for the computation I'm about to do. So phi is one of the vector multiplied scalars. In fact, what sits here is slightly more complicated, but I already evaluated on the localization locus. So localization locus just had phi and phi bar turned on and the auxiliary field. So I evaluated the gauge transformation parameter. In words, this is great. But how do you really show this? Show what? The fact that you get the m times f there and the phi. Oh, you just compute this on any multiple that you like. But should I use the sub-actionary transformation to give us nothing else? Well, to get this without the masses, you will get this. And what about the phi? So the phi is some gauge transformation which you will get as soon as you act with Suzy on non-gauge invariant stuff. So if you just try to close or compute the algebra with the transformation rules that used to be here, you will automatically get gauge transformations. And these gauge transformations will be precisely of this form. It's, yeah. And then to get the mass, I need to modify somehow the super-gear. Right. But the other point of view I gave you to turn on the mass is precisely to turn on a background vector multiplied coupled to flavor symmetry. So you need to turn on some BPS configuration for that vector multiplied. The BPS configurations were precisely described by phi. So the fact that it exactly mirrors this is precisely because of this statement. Another way of thinking is that you just shift phi by m. OK, so this is my supersymmetry. Let me slightly rewrite it as follows. I will group some rotations in slightly more convenient combinations. OK, so I just rewrote this trivially. But now I have a combination here, which I will start calling J. I have a combination here, which I will start calling J left. And as I mentioned before, well, I haven't mentioned it. But this supercharge, you remember that at the north pole it precisely preserved the, well, as a localization locus at the north pole, we found instantons. In particular, that means that this supercharge acts on the modelized space of instantons, just like we had the preserved supersymmetry just now. The same argument I gave somewhere over there. The preserved supersymmetry, they still act on the modelized space. In particular, this guy will act on the modelized space. So this supersymmetry really corresponds to a supersymmetry on the world volume, on the world volume of the instantons located at the north pole. And over there, it will look like, well, like this. So all these actions in the brain picture, it should be somewhat obvious. All these actions, they can be realized geometrically in the brain picture. And then you can figure out that they also act in some particular way on the world volume theory, which I described over here. And they act as flavor symmetries. So J and J left a particular flavor symmetries of this theory. I will describe for you the charges in a second. So these are just particular flavor symmetries from the point of view of the world volume theory living on the d0 brains, or alternatively of this quiver. I still have these pieces. And additionally, I have a gate transformation of the zero dimensional degrees of freedom. So this is now a supersymmetry on the world volume. And what we're supposed to do is take that supersymmetry and localize this theory. Once we find the partition function of that theory, computed with respect to the supersymmetry, which squares to this object, I will have one. Because then I have described for you the world volume theory partition function, which is precisely the missing ingredient in my localization formula on the force here. So I hope now it became clear why I did all this. Well, maybe I should just stop here. I will reconsider the computation of the partition function of that thing first thing tomorrow.


Well, by now, I'm sure that you've heard but the house voted on the so-called bipartisan infrastructure bill before the Senate voted for Reconciliation, which means that Build Back Better is probably dead And you might think that that's just a little bit too presumptuous because we don't know and the president is still promising a vote on it But I think that the writing is on the wall Corporate Democrats like Joe Manchin and Kirsten Sinema, they don't care about Build Back Better They don't care about the budget reconciliation bill. All they wanted was their corporate toll road bill That's what their donors wanted. There's a reason my Republicans supported it as well And so you gave them exactly what they wanted if you are a progressive in the House of Representatives It was really important that these bills Get linked together Decoupling them was the worst thing ever. I mean it shouldn't ever have been passed as two separate bills But the second you waver and you say all right, I'll vote on one without the other That's when you lose all leverage So if Build Back Better even passes at this point, which we don't know then Manchin can do anything he wants to it But you're at even further and there's nothing you can do because you gave him what he wanted now Thankfully not all progressives in the House caved there were six individuals who voted against it all of them are members of Specifically the squad and that includes representative Alexandria Ocasio-Cortez, Ilhan Omar Jamal Bowman, Corey Bush, Rashida Tlaib, Iyanna Presley They all voted no But unfortunately there weren't enough no votes to sink bipartisan infrastructure because other progressives Supported that all other progressives supported it including Ro Khanna, Pramila Jayapal, Marie Newman, Mark Pokan, Katie Porter and others So now what I say is good luck getting what you want in Build Back Better because It's not looking great right now. Now Joe Biden is promising a vote on Build Back Better before Thanksgiving and Now Apparently progressives are confident the progressive leadership They're confident that this is all gonna happen because I kid you not The moderate corporate Democrats in the House looked her in the eye and they swear they're gonna support it now The problem is that Biden's entire Timeline of voting on this before Thanksgiving is shaky because there's a little bit of a caveat, right? So the moderate Democrats or I shouldn't call them moderates. They're they're corporate right-wing Democrats They don't really have an ideology other than to deliver for their donors But the corporate Democrats who supposedly looked Pramila Jayapal and other progressives in the eye and swear to support Build Back Better They are waiting on a budget analysis to be conducted now whether or not that will be finished before Thanksgiving It looks really unlikely But there's an out for them even with their support if the budget analysis yields results that they don't like Then they can they cannot vote for it, but apparently this is a win So let me read to you what? CPC leader Pramila Jayapal said she got from them as political reports in a sign of how much trust has eroded Congressional Progressive Caucus leader Pramila Jayapal Asked each of the centrists who signed the statement to look her in the eye as they Publicly and privately to vote for the broader spending deal after they've seen cost estimates according to multiple Democrats Familiar with the exchange so I have to laugh because this is ridiculous They don't care their word is shit. They already broke their promise to you They said they want to support both bills at the same time This is what Biden said he said I'm not gonna sign one without the other and He's gonna be signing one very soon without the other So I just feel like you have to know that their word is shit, right?  What are you gonna do? You can't threaten to torpedo the bill that they wanted because you already voted for that So what are you gonna do? Sadly declared that they promised you and this is malarkey. I mean, they don't they don't care But yet Pramila Jayapal after this was passed Claimed that they didn't cave after they literally just caved and went back on what they said they weren't gonna do Um, and she she again assured us that what they promised her by looking her in the eye That was really that that meant something we feel really proud of what we were able to get and how far we've come In just four weeks because we held the line over and over again. We also made the determination that The country needs to continue to move forward. And so we feel like we got the best of all worlds we got a commitment on This vote which and and every single one of those individuals looked us in the eyes and said they are voting for it and so You know look we got a we got a we kind of move things forward Look from a policy perspective Pramila Jayapal is one of the best legislators She sponsored the Medicare for all Act, which is phenomenal legislation That being said under her leadership something like that would never ever have a chance of passing because she very clearly Doesn't know how to play hardball. I mean to say oh well You know each of those corporate Democrats looked us in the eyes and we got the best of both worlds. I Mean it's just it comes off as extremely naive to me and I don't think that progressives are ever going to have Much leverage so long as she's the leader But look I could be proven wrong and if she disproves me and build back better gets passed and it doesn't get butchered further Great, I want to be proven wrong here, but I know how politics works.  She may have gotten this pinky promise because each Corporate Democrat looked her in every progressive in the eye and they said we're gonna vote for this so long as the budget And then also doesn't make us upset but There's nothing that she got from Joe Manchin and Kirsten Sinema They can still say, you know what that $550 billion investment into climate Cut it in half cut it in more than half. What are you gonna do? You can't say anything So what's to stop them from butchering it even more because your corporate Democrat colleagues in the house Might be making you this promise and you might trust them But what's to stop corporate Democrats and the Senate from tanking everything because you voted for the bill that they wanted You to vote for Well in an interview with Jen Ugar of TYT Ro Khanna answered this question before caving and voting for this bill and As you're gonna see he admits there's not really anything that's gonna stop Manchin from From butchering it but we do have something. That's really great. It's gold as a currency in DC and that is a Promise to the president. We think we're not sure but we think This is an important news point in the question here because you're saying that Biden has some sort of implicit promise from Manchin and Presumably Sinema that they're going to vote for this version of the bill with 1.75 the framework Whatever that means and and Manchin is saying no, I'm not giving any promises. So What's actually happening here? Does Biden have Manchin and Sinema's promise or doesn't he? I think publicly Manchin doesn't want to make those commitments But our belief is that the president has a private Commitment from 50 senators that he could pass this and that he Didn't come to us for many many months and he only came to the progress is once he had that commitment I personally believe that he will pass the bill back better agenda that Manchin ultimately will be a Yes vote, but I acknowledge that they're the most popular provision and this was the Bernie Sanders increase dental vision hearing and you know dental and and vision are out So I I great you that the bill it has lost popular provisions that it should have it should have a much more robust prescription drug negotiation it should tax the the billionaires in the Corporations it's not raising corporate tax rates because of senator Sinema So I'm not I'm not here to sort of sugarcoat things what I'm saying is that we have a moderate president We didn't win with Bernie Sanders or or or Elizabeth Warren in light of that He's adopted a lot of progressive policies like childcare universal preschool and most importantly climate and this is Significant progress in light of the circumstance and that's where the disagreement is Yeah, not just us but but amongst people who've been critical of the of the caucus So represent kind of we're not even talking about this agreement. I think that what's remains of the bill is I Think the negotiations went terribly. I agree. So basically what progressives got is A lot of promises by people who are liars who represent Nobody but their corporate donors and now they gave away their leverage. There is nothing stopping Joe Manchin and Kirsten Sinema from Completely destroying and dismantling what is left of build back better and The fact that we're still being told that this is a victory for progressives and that we got the best of all worlds It's it's delusional. So I am incredibly disappointed in the progressives who caved But for the progressives who stood their ground and did not cave like AOC they explain how now this is kind of an issue because if we Pass the bipartisan infrastructure, which we did and that signed into law But we don't get build back better than we're doing more harm than good overall So she explains this very eloquently in a thread on the issue of climate think of the bipartisan Infrastructure as a lock and build back better as the key locks only open with keys Biff climate benefits only unlock with BBB's provisions If we message Biff as good on climate alone when it's not we stop the pressure for BBB's passage Do not let this happen the desire to pass both together isn't the un-nuanced stance some pundits think it is Biff has a lot of bad stuff in it. That's how we got GOP votes It also has a lot of good stuff in it what you're hearing on TV Passing BBB unlocks Biff's climate goals so the good prevails if BBB isn't delivered with Biff's oil and gas locked in We're in trouble that is what makes it a huge gamble So again, we need pressure to deliver the promise on BBB for anyone middle-aged and younger Let's hope this ain't Kyoto all over again because we're going to live with the outcome So she raises a really important point if Biff passes, but BBB fails ultimately we're worse than we were before and Bypassing the so-called bipartisan infrastructure bill, which is really just the corporate toll road bill well, I Mean all the pressure is gone what pressure what urgency is there to pass BBB because you just got a major win And you could sell this as a win if you're a corporate Democrat and really this Momentum is is gone for BBB because you're claiming it's a win And there's an example as to how Democrats are already doing this so Biden claimed on Twitter that What this would do with the bipartisan infrastructure bill would do is it would allow every kid in this country to Be able to turn on the tap and drink clean water, but AOC explains how this is factually incorrect This bill is being oversold. She writes the cost to replace every lead pipe in the United States is 45 to 60 billion Biff only gives 15 billion without BBB many communities historically denied clean water will continue to be denied Build back better has led money for disadvantaged communities. We must keep pushing for build back better It's not just that we made these promises before look at how the infrastructure bill is being messaged now I respect the president and the legit feat he just accomplished, but this is simply wrong We did not fund the replacement of every child's pipe and we shouldn't tell people we did but they're doing it And now Joe Manchin can say look we just spent 1.5 trillion on infrastructure So why should we spend another 1.75 trillion on build back better? It's just because progressives caved Now they have no say they're completely Legislatively irrelevant their input is not valued They caved six didn't though those members of the squad should be commended But the other members like Primera Jayapal and Ro Khanna who are part of the Congressional Progressive Caucus who did cave They need to reevaluate whether or not They're actually good at fighting for their priorities because you might support the correct policies But strategically if you're doing everything that is kind of working against you Then you need to reevaluate how you fight for things because what we saw overall from the Congressional Progressive Caucus was strength at first but ultimately they caved and The corporate wing of the Democratic Party banked on them caving and they were right to make that guess They were right to assume that they were going to cave because they did So I am Going to be watching closely to see what happens But I would not be surprised if build back better is not passed before Thanksgiving And I wouldn't be surprised if it was watered down even further and mansion in cinema still don't support it This is is bad progressives look weak But the six individuals who are imperfect They did the right thing now and those members of the squad should be commended for that


Israel Apologia is one big fat appeal to emotion fallacy. Notes from the edge of the narrative matrix. The Israeli hostages deserve far, far less interest and attention than the hundreds of thousands of Gazans who are being starved and murdered as we speak. The only way to believe otherwise is to believe Israeli lives are worth thousands of times more than Palestinian lives. The PR campaign for Israel's destruction of Gaza revolves around encouraging everyone to focus on the feelings of the people on the pro-Israel side instead of focusing on facts and evidence. The way Israelis are feeling about October 7th. The way Israelis are feeling about the hostages. The way pro-Palestine demonstrations make Western Jews feel inside. What feelings Biden is privately feeling toward Netanyahu?  Meanwhile, the feelings of the two million Palestinians being starved, displaced, dismembered, traumatized, terrorized and having to watch their loved ones die horrific deaths right in front of them never features. We're discouraged from thinking about their emotions at all. Only the feelings of the people who support these atrocities shall draw our attention and sympathy thereby generating support for the continuation of those atrocities. There's a tweet by Ahmed M. Sadaq. It's a video of a British doctor who treated Palestinian victims of Israeli airstrikes in Gaza. The doctor says, I will never forget it. A child's face was burned so much that you could see her facial bones. We didn't even have morphine to give her to ease the pain. She died painfully in the corridor of the emergency room. Comment from Caitlin. Palestinians in Gaza have been experiencing unfathomable suffering for nearly six months now. But the entire Western political media class has been insisting since October 7 that we center the feelings of Israelis and Western Jews at all times. Palestinian anguish doesn't count. In December, a former legal advisor to the State Department named Brian Finucane warned the Washington Post that the first Israeli attack on al-Shifa Hospital back in November could be used as a precedent to quote, pre-excuse future operations against hospitals. Which is exactly what ended up happening. Israel has since been obliterating hospitals throughout the enclave and has attacked al-Shifa three more times since then. The propagandists who have been manufacturing public consent for this mass atrocity are just as guilty as the ones dropping the bombs. Israel apologists will always have excuses and justifications for every horrible thing that comes out about what Israel is doing in Gaza. There is no testimony, investigative report, statistic or raw video clip that could come out showing horrific criminality on the part of the IDF or the Israeli state that would cause Israel supporters to go, oh, okay, now this one really is bad. Israel needs to hold those responsible to account and make some dramatic systemic changes to make sure it never happens again. This is because the function of Israel apologia, also known as Hasbara, has never ever been to promote truth or justice. Its purpose has always been to promote and defend the information interests of the state of Israel. This is why no matter what Israel is exposed as having done, you will always see prominent Israel apologists in politics and media as well as ordinary Israel apologists on social media providing excuses and justifications for it. They're not there to tell the truth or to help the public understand what's happening or to help create a more just and ethical world. Their sole aim is to advance the information interests of their favorite ethno-state to ensure its continued support. But the fact that so many thousands of people need to be working at this every single day shows how depraved Israel really is. Every genocidal regime in history has always had reasons and justifications for its behavior and that's all Israel apologists are providing today. Don't take their manipulations seriously. When I'm criticizing US aggressions toward Russia, I get called a Kremlin agent a lot. When I'm criticizing US aggressions toward China, I get called a CCP propagandist. Now I'm criticizing US-Israeli aggressions in Gaza and they say I'm an anti-Semite. At a certain point you figure out that there are just a lot of bootlickers out there who don't like it when you criticize the world's most powerful government so they call you names. That's all it ever is. Vapid name calling. It's what people do when they have no argument when facts and morality are not on their side.


What's going on everybody? Got Callan's Sean here relating sports cards and we are ripping 22-23 Contenders Optic Basketball 10 box case 27. Pick your team style. Got Alexandros S with the Lakers and T-Wolves. Got Andy F with the Bulls. Bojia B with the Suns. Brett P has the Kings. Chong T has the Pacers. Chun Wei has the Grizzlies. Got Eric S with the Nuggets. Gregory M has the Magic Spurs Wizards. How UW has the Nets. Hornets. Heat Nix. Jason T has the Celtics. We got Jesus T with the Warriors. Joseph M has the Clippers. Got Joshua D with the Mavs and Thunder. Keith B with the Hawks. Kenneth W with the Pistons. Reese A. Cavs. Robert P has the Blazers. Robert S has the Jazz and Pelicans. Shelly E with the Bucks. Thomas C. Sixers. Rockets. And Thomas R with the Raptors. Alrighty guys. Good luck. Good luck. It seems like it is a lot John. Those Thursday night games have not been the best. Past couple years they really have not been very good. Oh they got a penalty. Yonte Johnson. Where's Shite? I said Johnson. I don't know how good these games happen but that should help. Alrighty guys. Contenders on to Case 27. Good luck here. In just three spots left for the Nights. It is Flawless. Flawless got the Falcon Steelers and the Niners. Yonte Johnson. A lot of time though. Four minutes left. A lot of time. Hollow playing the Numbers game is Jason Tatum. And Blue Ice. Uniformity is Palo. Nice one there. Magic. It's Mr. Greg with the Magic. 26 is 75 on the Uniformity Palo. Go Magic. Rookie of the Year last year. Got a Base of Kyle Kuzma season ticket. Redknot numbered is Anthony Edwards on the season ticket. And Hollow Bojan Bogdanovich. And Otto is going to be a Silver Rookie ticket of Isaiah Mobley for the Cavs. Evan's little brother there on the Silver Ticket. Steelers hold on. That's one more down on me. Jesus. On the Pickham man. I am just lost right now man. I'm lost. Hoop Dreams is Jane Ivey. Nice red Rookie. And a Hoop Dreams Rookie of Matherin. Spot us a few bucks. There's a Shea. Gil just got a Blue. That's going to be Nicola Jovic on the Rookie. It's a 99 for the Heat. And a season ticket Red Shea. Not numbered there. And this one is to 25. Nice one there. Orange up and coming Rookie is Jeremy Sohan for these Spurs. I believe that's Greg. I believe Greg closed us out with the Spurs. He did. Thank you sir. Nice one there. Jeremy Sohan to 25. Very nice. Up and coming Orange. Oh boy. There's Jaker. There's Jaker. How are we doing tonight Jaker? How are we doing tonight? Yeah I know. I feel you Joshua. What's the score? What's the score of that one? Say Red Ice. Tari Eason. We choked very very hard last two days. So that could happen. Oh Shea Bajie first class ticket Hollow. There's a season ticket Hornets. Got a Hollow Scotty Barnes. Got Pink. That one is BAM out of bio to 75 out of bio. And Otto is going to be Red Isaiah Mobley on the Rookie ticket. Isaiah Mobley variation to 99. Yeah Greg you got a Sohan Rookie to 25 for the Spurs. Nice one there. Yeah and a Palo to 75. So good start for your couple teams man. Palo in foul trouble. Are we up 90? 90 to 86? Oh yeah you wanted to choke. Wait then we are up. All right. No we uh we've been terrible in the second half Joshua. Really really bad. Really really bad. Like the Clippers game were up three a half and we lost by 18 I believe. 18 or 16. And I only watched the second half so that was not fun. That was not fun. Turning on we're up and just go straight down from there. I watched it too. It was brutal. Superstar Giannis on the red and Green Pulsar, Legendary Tandems Paul Pierce and Ray Allen to 25. This little color match there is Celtics. Yeah that game was not fun to watch. Oh we got a Redemption. Anthony Edwards got a blue. That one is going to be a Kevin Porter Jr. It's a 99. And Hollow is Josh Hart. Redemption is Dirk Nowitzki on the 1985 Tribute Red. That's a nice one there. Nice one Mavs. The Hall of Famer. Where is the Mavs? Mavs are Joshua D. That's a nice one on the 85 Tribute Autumn. Old Dirk. Nice one there Joshua.  Those 85 throwbacks are really nice design if you want to uh if you haven't seen them definitely check them out on like eBay or something. I know Magic Johnson's live in that set on the 85s but a lot of uh Redemption's on those. The set is really nice though. It's a red ice playing the Numbers game in bead and a Legendary Tandems Giannis and Drew Holiday. Season ticket Chris Paul. Got a Zion and Shortprint. It's going to be rookie Nicola Jovic on the Black Pandora. Not numbered but Shortprint about one per case on these. Jovic on the Black Pandora. And Otto is going to be Brian Scalabrini on the back all-time contenders. Celtics. I haven't seen him in a while. 275 Scalabrini. I don't know Greg. Shouldn't be a buzzing noise. Oh I got you. Yeah that's that uh that's that Tequila. That's Tequila and uh Nicola going uh getting the buzzing going. The Whistlies. It's called the Whistlies. I refracted that one's D-Book. Legendary Tandems Duncan and David Robinson. Red Ice. Season ticket Sabonis. Got a Red Devon Booker and Hollow Yanis. Pornio Contenders is blue. Going to be a Herb Jones. Herb Jones Palakins. Blue to 75 on the odd. Hollow Hoop Dreams that one is Marjan Bochamp. Working for the Bucks. Got a Blue Ice playing the numbers game to 75 AD. AD's been till it this year so far. Staying healthy. It's pretty early still but Darius Garland. There's a Hollow uh CJ McCullum and Chris Middleton on the blue to 99. I heard that Greg. I heard that. Yeah that's South Florida man. South Florida is interesting. Oh there we go. Greg again this time we got an up-and-coming Otto of Palo. Palo Bancaro. Boom there we go Greg. Palo Bancaro to um 75. We'll mag that for you Greg. Palo Otto. Got a base to 75 and the Otto. Take that. Really nice break so far. Fourth and four.  That's interesting play call. Interesting decision I guess. That's not good. Red Ice is Jason Kidd and Vince Carter on the Tandems for the Nets. And this one is the 25 Green Pulsar of Fred Van Bleet. Fred Van Bleet. He's been hurt a bit already this year. That's not good. Tyrese Maxie. Got a red not numbered PG. PG-13 and a Jaren Jackson. Triple J. And this one is the 49. It's going to be a blue rookie ticket Otto of AJ Griffin. AJ Griffin. It's the Steel of the Draft. AJ Griffin as Sam would say. S.O.D. Griff. What's the Rangers score? Rangers are winning. Rangers are winning. They're going to be winning for a couple months. This one is a Optic Aleutian Hollow and Ant Edwards. Got a Legendary Tandems Shack and D-Wade to 25. Wade County on the green. It's a Josh Geedy. Hollow is Tatum and red AJ Griffin. And on the back is a Christian Leightner Otto to 75. That one's going to the T-Wolves. T-Wolves on the Leightner All Time Contenders. Christian Leightner. Last one here, guys. Box 10. Case 27. We'll get number 28 going next. And we're on to some Bowman U football. Yeah. No, I don't remember. Yeah, I was too young, but I remember who he was. Yeah, I know. I know. Yeah, not the most popular for sure. Hoop Dreams, Peyton Watson. Good. Yeah. And Green Pulsar is Tarry Eason to 25. Rockets Rookie. A Kevin Porter Jr. Got blue John Morance. That one is to 99. John and a red not number Trey Young. Otto on the back is to 75. Going to be blue up and coming Otto Coloco. Christian Coloco for the Raptors. Alrighty there, guys. That'll do it for Contenders Optic 10 Box Case 27. Do a quick auto recap and then get moving on to 28s. Had Coloco to 75. Isaiah Mobley on the base rookie ticket Otto Capps. There's Isaiah Mobley to 99 on the red variation Otto. Brian Scalabrini to 75 and Leightner to 75. Had 49. This one is a rookie ticket AJ Griffin. Perennial Contenders Herb Jones to 75. To 25, Jeremy Sohan on the up and coming. Dirk Nowitzki on the 1985 tribute for the Mavs. Really nice one there. And on the back we had Palo. Palo on the up and coming Otto to 75. For the magic. Go magic. Alrighty there, guys. That will do it for Contenders Optic Case 27.