 So it's a basic fact that we haven't used at all. We haven't used this because we've used crossing symmetry to view it as a symmetrical function of all sort of our momenta, our label to be incoming so that the s matrix is determined by symmetrical functions of four momenta. But it really is a matrix and that can be used for some things. What I want to apply this to is to the renormalization of local operators. So in this lecture we'll relate this matrix when viewed as a matrix to another very natural matrix which occurs in gauge theory, which is the matrix of anomalous dimensions. And that there's a connection is a relatively recent realization. So this lecture will build on the work by Willem which was published just last fall and building on previous work by Zwiebel. And this work discussed the case of Ennipot's four super angles and it really used, but it was motivated with Triggered originally by integrability of the theory and so on. But the general relation between those two things I think is a very deep and important idea and it has nothing to do with Ennipot's four, nothing to do with planarity and so on. So I will present it in general and give some example in QCD at first. And then we'll make contact with this integrability. And the relation occurs in, it's very, the map first of all, to understand the map we basically have to understand what we want this matrix to act on. So the first ingredient is that they are very natural states to act on which are polynomials. The first ingredient is that a local operator in the weekly couple theory is in one-to-one correspondence with a polynomial state or what they call more precisely polynomial form factor. And the idea is simple. Let's look in a free theory. If we write down some local operator, anything, trace five square, we can compute its matrix element to, let's call it yeah, general or anything. We can compute the matrix element for O to decay to a bunch of unshell particle. We can compute this matrix element and that is the form factor for O to decay into this particle, okay? I will denote it. So it's a very simple thing and diagrammatically I draw this O air, for example, at the origin and then this particle come out to infinity. At loop level, you get diagram, sorry, when you turn on interactions, you get diagrams like that, you don't get a polynomial, but the first thing you get, if you just do it in a free theory, in the free theory, this form factor will always be just a polynomial. If this is a local operator, because you get derivatives and derivatives turn into D mu, just goes into P mu. Actually, there's an I air. And a local operator, for example, in the scalar field theory, if you have the mu phi, but yeah, if you have phi, it just turns into, the form factor is turned into one. If you have the mu phi, where I then mu phi, form factor just turn to P one mu, if the scalar decays to N and so on. In gauge theory, we can do the same, but now the form factor for F to decay for F now, it's very useful again to use the spin or less denotation. So F is a spin one object, there's a, can either be a symmetrical combination of spin or indices or dotted spin or indices. And this has three components, because it's symmetrical. This has three, because this, this may be a four. So this could include F one one, F one two, and F two two. It's symmetrical, so there's no F two one. This, and you add these guys, you get the three, you get the six, Lorentz components of F mu. This is the self-dual part and this self-dual part. This notation is useful, because this F alpha beta can decay to one negative elastic neuron. And the form factor is just lambda alpha, lambda beta. So again, it's a polynomial in the lambdas. And similarly, F bar decay to a plus neuron. So the form factor is equal, so this is an equality. So the form factor for this operator to decay into a minus elastic neuron is this polynomial. So in general, there's a one to one correspondence between polynomials in lambdas and lambdatildes and local operator. And you make this correspondence by looking at just at the at the form factor. You just overlap the operator with on-chain states. The, one important thing about this, let's look at more complicated operator. For example, we can look at something like F alpha beta. Let's put a trace, look at something gauge invariant. And look at this, for example, F. Let's compute the form factor for this, just at three level again. So, and, sorry, this is dot, this is F bar. So this guy decays to a minus neuron. This guy decays to a plus neuron. Let's call this guy one, two. The form factor is we have lambda one alpha, lambda one beta for this guy. For this guy, we get lambda two beta dot, lambda two gamma dot. And for this derivative, the derivative act on this guy. So we get well, lambda two, lambda two tilde, which is the momentum in that guy. That's P two. And this beta, beta dot. However, you see that this contraction here, that's this Lorentz-Envarnian contraction, which is anti-symmetric, anti-vanished. So this gives zero. This particular form factor is zero. Which of course, we could have guessed because this is the Yang-Mill's equation of motion. So in general, this form factor map throws to zero everything, which is an equation of motion. So it give us local operators modulo equation of motion. Which is good because this is what we actually care about. This is much more useful than our local operators of shell. So this map automatically projects on projects of the equation of motion. One can do more example. So as another illustration, let's classify before just to familiarize ourselves more. Let's classify possible dimension five or six operators in Yang-Mill's theory. Say with spin zero. The total spin zero means that we're looking at form factors where all the Lorentz index are and these are contracted. Let's go through this exercise now. So let's look at operators which can decay to two particles first. So what are the form factor we can write? So let's say we decay to minus minus. Let's compute some. So we have to list all polynomials. So the dimension four guys who already know it decays to each particle, each F decays to lambda one, lambda one, lambda two, lambda two. And if we get the spin zero these get constructed to one, two square. So that's what we get at dimension four. And that comes from that can be recognized trace of F alpha beta, F alpha beta. But we didn't need to know that. We just this emotional polynomial and at the lowest dimension we need two lambdas for each guy because of the elicity of the gluon and the first thing we get is this. It's dimension four. This object is dimension two but remember each state is dimension one. So the dimension is what looks like the naive dimension plus the number of states. So the dimension of O is equal to the dimension of the polynomial plus the number of legs. So this is a dimension four. This polynomial is dimension four. At dimension five. What can we write? Well actually we cannot write down anything for two things because if we want to have the correct weight we need to add a lambda and a lambda tilde. Oh no sorry that brings us to dimension five. That's very good. So let's start from this and add lambda lambda tilde. Let's put index one here for example. We have to contract this into a singlet. Can we? We should be able to. No we can't because the lambda tilde we cannot contract against anything. Okay so sorry. That was a kind of trivial example. There cannot be any spin zero operator of dimension five. You need to add two derivatives. So that's trivial. Dimension six. Lambda one, lambda one, lambda two, lambda two. Let's add for example this square. We have to contract the indices in some way now. And well actually the dotted index here we'll have to contract against itself. That doesn't exist. So we have to put the other guy under two under the tilde. And then if we contract the indices there's only one way to contract the indices and it goes plus one, two, cube, one, two. That's the only form factor we have. Which we recognize as because one, two times one, two. Angle times square that's P one plus P two. Square times one, two, four. So this operator is recognized as del square of trace. F alpha beta, F alpha beta. So the only dimension six operators which decay to two particles are total derivatives. So if we're interested in writing down a grand gene we don't need to bother about them. But we can also look at operators which can decay to three particles. Let's look at minus, minus plus for example. Then we need on the tilde is like that two, two, three, three. We need to contract the indices in some way. If we want dimension six we cannot add anything because the dimension is already six. This polynomial is dimension three and it's three days. So that's the dimension six operator already. And we cannot contract this into a singlet because there's no tilde to contract this guy. Doesn't exist. We can look at this minus, minus, minus. And that now we get our lambdas. Now we can make a contraction. That's the spin zero dimension six. There's one of these guys. And that guy we could recognize if we wanted to as F alpha beta, F beta gamma, F gamma alpha. It's antisymmetric. And if you want it to be gauge invariant it will just be proportional to F A. So that's the only dimension six operator which is not at all the derivatives. So this is just an example for how you can use these rules to list physical operators in theory. So it's just a labeling system that I wanted to use you to. And it's a very convenient, it's a very powerful system. For example, people are interested in the standard model in listing all dimension six operators because they classify possible small deviations from the standard models in precision measurements. And there's actually a lot of dimension six operator in the standard model. There's a lot because there are many fields. And if you have to mod out by equation of motion it becomes a very non trivial game. But yeah, this problem has been solved anyway by brute force but probably would be interesting to apply this method to revisit this. But yeah, so that's the first ingredient. So we know if we want to act, the S matrix is a matrix and these polynomials give us very natural states on which to act. So we're gonna take this S matrix and act on this polynomial. So now why would we expect this to have anything to do with renormalization? So the second ingredient is unitarity. That's a very simple idea. If we look at this form factor at one loop I'll be first sloppy and then less sloppy. If you look at this form factor at one loop O to decay into something by dimensional analysis this should scale like P to the dimension of the operator. And the dimension is the canonical dimension plus the anomalous dimension. Okay. One actually has to be more careful. There are infrared divergences also. So that's the total scaling respect to P. I will describe this gamma. In general, that's the behavior. If we expand this in perturbation theory these things are order G square. They're small at recoupling. If we expand this, well this looks like F. Zero times a log. A log of P square with some pace. And then times gamma plus gamma over two. It's very important that this log as usual you have to take the correct branch of this log. And we're looking at the physical operator which decays to on shell states. So the total center of mass energy is time like. So this log has an imaginary part. So a quick way to compute the question of the log is to take the imaginary part of this thing. So the imaginary part of the form factor. Yeah, let me rewrite the equation the other way. So gamma plus gamma infrared times three level form factor is equal to minus two over pi times the imaginary part of the one loop form factor. And this is just, you can generalize the statement while loop but the one loop statement takes this form. You see if you take the imaginary part because this log is log P square minus i pi. So if you take the imaginary part it's equivalent to computing the quotient of the log. One can be a bit more rigorous about that. Because the M2 does other contributions. I claim that all the contributions are real. Because all the other finite parts are real. And the simple way to see that is to use the RG equation. Which essentially where this follows is that, because we have many momenta, but we have the statement that if we rescale all the momenta on F, we get gamma plus gamma IR times F. So this is a general RG equation which we can use. And using this equation, if we look in the, you have many, many momenta, but the imaginary part, the amplitude is defined. Remember that amplitude of branch cut on the real axis. And the amplitude is evaluated here. And the imaginary part is the amplitude X and Y. But you can go from this point to this point by exponentiating the RG equation. If you take, so the imaginary part of the amplitude is equal to one minus E to the minus 2 pi I P dot DRD. If you do that, that's the way of computing the amplitude. And now you can use the RG equation to relate that to the anomalous dimension. So it's completely general that the imaginary part of the amplitude is related to the anomalous dimension. Any loop orders, the duration is just slightly more complicated than that. So let's continue with that equation. So the anomalous dimension is related to an imaginary part. Now we have to actually use the uniterity. We evaluate this imaginary part. And recall the uniterity relation I gave yesterday. That the imaginary part, two times the imaginary part of the T matrix is T dagger T. For the form factors, there's something similar. If we look at this form factor at one loop, you have a graph like that and the Kutkowski rules tell us the imaginary part is obtained by cutting it. So the form factor, two times the imaginary part of the form factor is equal to the T matrix acting on the form factor. Hey, I'm actually, I wasn't able to, well, I don't properly derive where the dagger goes here. Perhaps someone can help me after the talk. So just to be conservative, let's average the question mark. It will not matter for the one loop check because T is real at one loop. It will not matter for this talk, but it's next to her. But yeah, if someone can derive the correct equation, please tell me. It will not be needed for A. But that, this is the main relation. So let's just call this Tf. So we have the enormous dimension is equal to minus one over pi times Tf. And this T matrix is, now is really, we think of it as a matrix. It's the, at three level, this would be just, at leading order, sorry, this would be the two to two scattering amplitude. So here's a situation where we really have to think of the amplitude as a matrix. And when it's going to act on polynomials, its eigenvalue when it acts on polynomials will give us this quantity. Which is not exactly the, which is not quite the enormous dimension because of this infrared stuff. But this IR here doesn't depend on the operator. So we'll find a way to subtract it. And this occurs basically, so this enormous dimension comes from UV divergences and this comes from IR divergences. So we still have to disentangle these two kind of divergences. And the scaling respect to P doesn't tell us. Yes. Yes, that's why I put a little one here. Yes, yes. You can generalize this, but it's not really being tested and here I would just discuss this case. It's really an open problem to really concretely do it at IR reporters. So just to stay concrete, I've discussed. So yeah, so we have this equation. Let me box it. So this is the basic equation. So let's look at examples. I was planning to jump directly to QCD, but maybe a simple example in 5-4 would be useful. So in 5-4, the matrix element, well, T222, if I ignore the delta function and on shell stuff, but let's just call it the 222 amplitude is minus lambda. If I have the Lagrangian, let's say that if you put one over four factorial in the action, five over four factorial, then the amplitude is just that. This, what is this? Let's act on the simplest operator, which has coefficient one, which has a form factor one. So let's look at five square. The form factor for five square to decay to two articles is just one, very simple operator. And in that case, the formula here to be useful to put one more equality, this T matrix for 222 scattering is integral over angle. So it's defined as, well, T222f, it's by definition integral over Lorentz and Viren, phase space of P1 prime, P2 prime, times the form factor to decay to P1 prime, P2 prime. And P1 prime plus P2 prime is equal to P1 plus P2, as in this diagram, P1, P2, P1 prime. This D-lips can just go to a center of mass frame and write it as one over eight pi integral over angle. And then here we need, sorry, here there's a 222 amplitude here, then there's a 222 amplitude for 12 goes to one prime, two prime times F of P1 prime. That's the basic formula I would use. If you combine the factors, that's a third more explicit version of this relation. We have to integrate over the four point amplitude over angles. So for the, in this special case of five fourths, this is a constant, this form factor is a constant. So this integral just gives one, and we get that the inverse dimension is minus one over eight pi square, times minus lambda. So get the inverse dimension of pi square is plus lambda. And you can cross check this against computing the UV divergence of this graph, unless I made a mistake because I didn't prepare that specific example to work. Yeah, and which is actually, yeah, that's a good point. There are no, there's never any IR divergence in five four anyway. There are coordinate divergences which do occur in this gamma IR, but they first appear at two loops through this. So at two loops, one would have to worry about this in five fourths, not yet. Yeah, so this IR contains coordinate stuff also. But in QCD, we won't be so lucky and this gamma IR will be important. So, okay, I can move here. So now I need some room. There's never anything which is completely easy in QCD. So I will need some room. But the starting point would be simple. So what do we need in QCD? I've already described, or at least let's just do it preangles and I already described to you what the polynomials are in preangles. Now let's discuss this guy, which was the content of our first lecture. And so the amplitude, the formula we'll use is, now I actually have to be careful about signs and factors of two. And there are simple ways to check this. I will not go into that. So there's this thing that we add. And then, if we keep all color and disease and everything, there's the FAB, sorry, if the colors are A, B, C, D, we have F, this guy, F, C, D, F, B, D. And downstairs, one, two, two, three, three, four, four, one, plus, this is the formula we derive in the first lecture. This delta function and this notation introduced yesterday. Yeah, it was just one, two, three, four in the first lecture. And okay, this amplitude is fully permutation environment, which is, did I get it right? Sorry, AD. This amplitude can be checked to be permutation environment, which is not so obvious maybe, but let's just call this thing, if we call this thing, this factor, one over one, two, three, four, an exercise, check both symmetry. Check that this is permutation environment. And the identity which guarantees this is, so this term involves this, this second term involves this. And if you do a permutation, you think you'd get a different term, but it turns out this is an identity. And if you use this identity, you can check that this is permutation and symmetric. That's called U1 decoupling. I will not explain why. There's a good reason for existing. So that's the formula for the triumphant zoom. The second thing we need is a useful way to do this angular integral in terms of spinors. Because you see this formula and the form factors are all expressed in terms of lambdas and lambda tildes. We have to express the angular integral in terms of these measures. And the way to do that, just remember, when you have a spinner, if you make a rotation of it, if you start from spinner one zero, you make a rotation, you get something like cos theta over two. You can get the motion of spinors parameterized in this way for some, and theta is the angle with respect to the z-axis, phi is the azimuthal angle. We can prioritize a spinner like that. And here, we have two particles, p1 and p2. So they're labeled by, so natural, naturally want to work in a frame in which they are opposite. And we can do all that coherently just by rating that lambda one prime is equal to lambda one cos theta over two plus lambda two sin theta over two. Lambda two prime is equal to lambda two cos theta over two minus lambda one sin theta over two. So this kind of representation is derived in the paper by William that I referenced, but I just wanted to motivate it. It's very simple to see why you would write that down by thinking about rotations of spinors. So the integral then becomes integral d2 omega over four pi. And these ones are really annoying. Let me rescale all angles by two. Then it's an integral over zero to pi over two because it's rescaling d theta sin two theta and then integral zero to two pi d phi over two. So now we can do everything on these spinors. I don't need to erase, I have plenty of room. The equation will be long, it should fit here. Don't get scared, it's simple building blocks. And hopefully I'm writing big enough that everybody can read. So let's look, we look at the more, some of the most important operators which is f square. So that's the one of the operator we look at. And the enormous dimension of f square is essentially the better function. So that's the very natural guy to look at. And the form factor for f square is, as we discussed, one to square, where it goes to minus, minus. But then this will only give us gamma of f square plus gamma IR. So we get some integral. But that's not useful yet until we find some way of computing gamma IR. We could compute this from the infrared divergence by computing infrared divergence, some cross-section and so on. Our shortcut is to use the fact that the stress answer has zero enormous dimension. So the other thing we calculate is the, that's f, form factor, to go to stress answer. And the stress answer can indicate to minus plus. And it's matrix element, so it's spin two, it's conserved, you can write it as t alpha. Sorry, it's traceless at three level. You can write it in spinner indices like that. It's lambda one, lambda one, lambda two. So we'll look at these two form factors. And by looking at these two, this second one will give us gamma T mu nu plus gamma IR. And because this is zero, the difference of the two equation will give us gamma f square. Oh, so this is a very good point. It's a, they are very simple, they are factorization theorem, shall prove. And the IR divergence, it's simple. If you look at this form factor here, the IR divergence comes from very far away. They don't care at all about what's happening there. So they depend on the angles at which the particles are going. It depends on what the particles are. They don't depend on the operator. So this gamma IR is the same, and it cancels when we take the difference. That's very important. Yeah, the infrared divergence are completely factorized from the UV one. So what integral do we get here? Okay, let's look at the first, let's compute the first one first. So I have a four point amplitude here. I have this cut, and here I have this, and this cut, and this cut, I have minus, minus, plus, plus. So the amplitude here, okay, let me, the amplitude here will be given by one, two, the fourth over, say this is one, two, one prime, two prime. The amplitude here, okay, and I missed, I forgot the, looking at the gauge environment case. So this delta AB. If I put this delta AB into this, I just get Nc times delta AB. Let me put some space. Then I get this 2G square, and then this parenthesis gives me one, two to the fourth, or one, two, and then two, two prime, two prime, one prime, one prime, one. Okay, and then there'll be the other terms. The other terms is simple permutation. So let's just evaluate this using the explicit spinors. You give us some experience. So there's a one, two, two, two prime. So that's minus, so that's minus two, one, that's plus one, two, sign. Things will simplify momentarily. The next guy is two prime, one prime. That one is harder to compute, but it's just two, one. As you sort of expect, it's this, because the center of mass energy here and here is the same, and one, two is essentially the center of mass, or square center of mass energy. So I want the right, I want, check that one. And the last one, one prime, one. One prime, one is proportional to two, one, two, one times sine theta. It's a sine, and now there's the opposite phase. The phase cancels, the one, two's all cancel. So we just give one over sine square theta. The other guy, the color structure gives the same, and it's rated by symmetry, it really just gives this. And that is one over sine square cos square theta. So we computed that matrix element. So this here, let's collect the terms. There's a minus one over I pay pi square. There's a two g square and c. Then this, okay, let's ignore the delta eb for a second. There's a symmetry factor of two we have to include, because these two particles are indistinguishable. And then there's integral from zero pi over two, d theta sine two theta. Okay, then we have this thing here, sine square, cos square. And now we just have to evaluate this operator or the shifted place. And that is actually just one, two again. So we get one, two square delta eb, which, so we get the left-hand side and right-hand side are proportional to each other. Yeah, so the endless dimension is just that. Is that what I, and yes, very good. So that's our first guy. Notice the integral is colonial divergence. That's not an issue, because we expected, we know that the amplitude is soft and colonial divergence. So this gamma IR, that we're subtracting, actually is colonial divergence. But it will, this colonial divergence will cancel in a second when we take, when we cancel gamma IR. So the other guy give basically the same thing. Let's simplify this. This is sine two theta is two sine cos. So that's just two, and now that the squares are gone. Two d theta over sine theta, cos theta. And the bracket, now we have to evaluate again. Okay, let me draw the, now the integral is not trivial. What's the integral? We have to evaluate, now the amplitude is a different one. We have two cases. One case as, we're still looking at one plus here, can have one plus on the inside, or we can have plus minus on the inside. Let's do, sorry, let me, let's do this case first. It's simpler. This case, plus minus case, give us, and the plus minus case, the numerator is one, two prime. The factor is one, two prime to the fourth, which differs from the one, two to the fourth, that we've included previously. That's the numerator from the amplitude. And then there's a form factor, which is this. So we have to put, evaluate it at lambda prime, and lambda prime is essentially lambda one, cos plus lambda one, sine EI phi, sorry, lambda two. Is this guy square? Then it's lambda tilde two minus lambda tilde one, sine EI phi. And the tilde is a complex conjugate. So the phase is, is the opposite of that one here. You see that when we do the angular integral, something very simple happens is that all these terms just get killed. The angular integral just give us, there's a cos here, sorry. Just give us lambda one square, lambda two square, which is what we had originally. So if we divide by this, we get one, but we get cos to the fourth. And this thing, so this whole thing here, just give cos four theta. This thing, this is one two prime, one two prime, so that's one two times cos. So this thing is also cos the fourth. So the whole thing is just cos four, cos to the eighth theta. And the other guy related by symmetry is just sine square, sorry, sine eighth theta. So we're done. If we take the difference, we have that gamma phi square. Let's collect factors. These two cancel with a two here. So that gives minus alpha strong over pi, G square over four pi square times integral zero, pi over two d theta, sine theta, cos theta. One minus sine eight minus cos eight. Now the divergences have canceled. That's just a number. Can compute this number. That's minus alpha strong over two pi. I'm sorry, there's an NC. So it's in front times 11, 30. So this, what have we learned? This thing here knows about the QCB better function. But that was expected in general ground. So what is the lesson in general? So I've checked this on a couple of examples. I don't have time to have some yet unpublished notes on this, hopefully this would be out. But I think this is a very simple idea. And what's the simple idea here? Which is the one used by Willem, is that S matrix acting on polynomial is equal to the dilatation operator. And that, well, the strict equality here is at one loop. But I think it is clear that in the future this would be generalized wire loops. Gang Yang has been working on this among others. I'm sure there's a long and nice story awaiting here. Another thing is, we obtain formulas for you can, once we have computed this gamma IR. Which we've done here, because yeah, this is zero. So we've just computed gamma IR, in terms of this cos eight plus sine eight. Now we can subtract it to any operator we want. So now this gives the, that's the full, that's the full one loop dilatation operator in QCB. So I think it's a very, very, very beautiful idea that was discovered last fall. And yeah, it worked, has nothing to do with N course four. It has nothing to do with symmetry or integrability. It's just unitarity. And the 222 amplitude gives the renormalization. I have five minutes left, I think. This is enough to discuss the N course K, N course four Ks briefly. So in N course four, historically, what people looked at are, the simplest operator in N course four are involved scalars. And people, historically, following Minahan and Zarembou in 2001, look at chain of operators involving two scalar fields, where Z would be phi one two and X would be phi one three, these are two SU four indices. But look at trace of operators like that. And what Minahan and Zarembou found is that the dilatation operator acting on this kind of chain can be mapped to an integrable spin chain. And it's interesting to see this here. So we've just, I've just said that the, the dilatation operator at one loop is the S matrix. So let's just hit this guy with the S matrix. So we need four things. So we need the S matrix for, we need the amplitude for ZZ goes to ZZ. So let's draw a diagram here, ZZ X and so on. And the planar limit at one loop would just get pairwise interaction between nearest neighbor. We'll have to integrate over these cuts. And this amplitude here can have, we don't have to scalars, but of course, of course scalars, because of, yeah, ZZ can only go to ZZ by a symmetry. This ZZ goes to ZZ amplitude can be computed from this factor. And just record the answer up to a constant minus S over T. Then the next case is ZX goes to ZX, which turns out to be equal to plus U over T. And the next thing is a ZX goes to XZ, which turns out to be equal to plus one, very good. So these are the basic S matrices. Now the dilatation operator. Now there's a nice thing. We don't have to look at the stress answer in this theory because we know that gamma ZZ equals zero. It's a protected coward operator. But let's just call it the BPS operator. The center's dimension is zero. And we have this other equation, which shows that gamma ZZ plus gamma IR is equal to integral of, integral D to omega of AZZ goes to ZZ. So essentially this amplitude gives us gamma IR. So now we can subtract it from the other one so that the gamma, the matrix element for gamma ZX goes to ZX. Or in general gamma would be equal to integral A222 minus AZ, ZZ goes to ZZ times the identity. So that's what we get here. So if you compute this here, the matrix element for ZZX goes to ZX, involves the integral of U over T minus S over T. And because S plus T plus U equals zero, that's minus one. Integral of minus one and gamma for ZX goes to XZ equal to plus one, because the identity doesn't contribute. So that gives us the action of dilatation, so how does this class of operator renormalize? What happens is that this is the amplitude for the operator to go to itself when this is ZX. It's the amplitude for it when the ZX exchange. So we can write that the gamma one on this chain of operator, right, the chain, yeah. Gamma one is a sum over adjacent pairs of the amplitude for them to exchange minus the amplitude for two particles to go straight, which is the famous Hamiltonian that was discussed by Hans Beti in 31 in the very early on. It was diagonalized explicitly, so it's a integrable. So yeah, just to finish it extra, I will rewrite this object again. So this is delta eight on eta over one, two, two, three, three, four, four, one. So it's a very simple object, but it knows about many important things. It knows about the QCD beta function, which is the preangles, but this has fermions in it also, and you can get the NF terms also if you want. And it knows about, well, Eisenberg's spin chain, the spin chain, and it generalizes it to a, yeah. So, and there's a starting point for this N-course four theory being integrable. But yeah, this, I think I've said enough. Thank you.