 Today's discussion would be very different than yesterday in the following sense. So yesterday I discussed some list of problems which were, let's say, open for a while, and now there are some new techniques that allow to solve them. Today what I'm going to do is to talk about a problem that is interesting for a while and there is not much progress, but there are some hints that one can make progress. And so I would just like to explain the subject a little bit and why I think one can make progress. And then maybe some of you will be able to push it forward and see how much further we can go. Also it's an interesting subject on which there are already partial results and these partial results may be interesting in their own right. So let me start by quickly reminding you about what happens if you study Young-Mills theory at large N, so there is no, I'm not talking about some supersymmetric theory, I'm talking about ordinary SUN Young-Mills theory which is known to confine, so this theory confines. There is a mass gap, so there is, so the spectrum of the theory, if this is the mass squared axis, there will be some global, so this would be zero. There are no massless particles, this theory confines. So there would be some global, it's a mass lambda, and then there would be many other states. There would be many other resonances in this theory. And, analogically, we know that these other resonances are unstable particles that can decay to the lightest globals with the right quantum numbers. But one result that is of importance is that if you take the limit of N going to infinity, then all these resonances, at least from analogically, they become exactly stable. So the resonances become exactly stable. And a related fact, a related fact is that the three-point functions of resonances are very, very small, and that's why their decay probability goes to zero, the large N limit. So, let's say the amplitude for some resonance to go to some other two resonances goes like one over N squared. In canonical, when everything is canonically normalized, then this is the right answer. And so the decay probability goes to zero. And what you get to know logically is a bunch of stable resonances with some masses and some coupling constants that characterize how resonances interact with each other. And these coupling constants are small. They're one over N squared, which means that the resonances are stable. I just wanted to, in five minutes, to review the argument for why you get infinitely many stable resonances and then talk about how could we possibly try to understand if there is an internal structure in the spectrum of resonances. So the problem that I would like to discuss, this problem would not have an answer in the end of the lecture. So unlike the three previous lectures where I posed the problem in the end I solved it, in this problem I would not solve it. There would be only very partial results, but I think maybe they can be pushed further. So the problem is to determine PN, to determine this masses. So we can call them MN squared, maybe more physically. Determine the masses of this resonances squared, of this exactly stable resonances. Since they're exactly stable, then these numbers make sense. It's some mathematical series of numbers and it makes sense to ask what they are. Now even partial results could be interesting. For example, if you could show, one of the things that you could try to show is that asymptotically MN squared looks like some AN plus B. So this is one conjecture that you could try to prove. That as N going to infinity, these resonances sit on a straight line. The resonances sit on a straight line, at least the resonances with the highest pin. At least those, at least, so we can say that this would, we could at least expect that this is true. At least for those MN that correspond to maximal spin. I'll explain what I mean by that. So those particles that spin very, very fast could sit on a straight line. And that intuition kind of follows if you believe in the string picture, that at least in some naive string pictures, you would expect that there would be asymptotically linear projectories of particles. Okay, so there are some things of that sort that you could try to show. And in fact, there's a lot of data in this sense. There's lots of Monte Carlo simulations of large and young mill theory and people have gone pretty far. We have lots of data about the low line resonances, some partial data about higher resonances. And so if you can make concrete predictions about this theory, then it would be very worthwhile. There would be many people who could contrast your predictions with experiment. I mean, Monte Carlo simulations and there may also be connections to string theory and so on. Okay, so anyway, that's the problem that I would like to discuss. I just want to start with some reminder of the large n limit of how we take the large n limit. That's just for the beginner students among you that haven't seen it. So the idea is that you take g and mills to zero and n to infinity in such a way that a combination g and mill squared n is kept fixed. Now, this makes sense. This makes sense because let's recall the beta function. The beta function for g and mills, it looks like g young mills cube times n squared. So, plus higher order corrections. Let me see. And so this is g squared and this is right. So this is the one of beta function of young mill theory. Now, let's multiply it by another g young mills. So we get squared and here we get to the power four. Now, we multiply it by n and we have n here and n squared here. And so we see that for this final combination g squared, but for this final combination g squared n, which is called lambda, the beta function doesn't have factors of n. So we have lambda cubed. Sorry, we have lambda squared plus higher order corrections. These higher order corrections are functions of lambda and there are terms which are one over n times lambda. So these higher order corrections could be lambda cubed, lambda to the four, but they could also be one over n times lambda cubed and so on. So these corrections would be small if lambda is small and there would be also corrections that die away when n is taken to infinity. Famously, the sign here is negative, of course. And so what this beta function says is that the lambda explodes at some scale. So if this is the energy scale and if this is lambda, if you start off from some small lambda at very high energies, it would naively explode at some scale. At least this is what the leading term predicts. Of course, we can't trust the leading term near where it explodes, but it doesn't matter. This theory clearly flows to strong coupling. It confines. And so the large n limit in this way where g squared times n is kept fixed is a good limit because in this limit you expect confinement with the resonances having finite mass, which is independent of n. So because of this property, the beta function doesn't depend on n, you would expect that the resonances of Yang-Mills theory that we've observed in Monte Carlo simulations and even in some sense in nature, the resonances of this Yang-Mills theory would stay the masses, the masses of this resonances. The masses of this resonances in Yang-Mills theory would stay finite in this larger limit. That's because the strong coupling dynamics kicks in at some fixed energy scale. It doesn't scale with n to a leading quarter, okay? So you expect that these resonances would have some finite masses in the larger n limit. Furthermore, you expect that their couplings would be one over n, simply because if you just draw a diagram of three resonances. So let's assume we have three mesons or let's say three globals. We're just doing Yang-Mills theory. If we try to draw a diagram of three globals, then this is of order n squared just because we have these two loops here. This is how that kind of diagram would look like in the large n limit, in the truth notation. So there would be two loops here and you would get a factor of n squared. But the propagators are also of order n squared. So the two point functions of power of order n squared. But also the three point functions, let's say of globals. I keep writing mesons, but I mean globals. So the two point functions of globals and the three point functions of globals scale like n squared. And so if you just canonically normalize, canonically normalize the two point functions to be one, therefore you divide that squared by n, then you get the canonically normalized three point function. The canonically normalized three point function would look like one over n. I meant here the amplitude squared. Yes, yeah, there could be, there would be one over n squared in general. So we have a, so the canonically normalized three point functions go like one over n. And the canonically normalized four point functions go like one over n squared. And that's how it goes. So higher correlation functions are more and more suppressed when they're canonically normalized. And so you can think about the theory at least at larger gen as a bunch of mesons with very, very, sorry, bunch of globals with very, very weak interactions. And the mesons are almost stable because the interactions are so weak that the probability of decay goes like one over n squared. So therefore you can imagine that at infinite n, there is some effective Lagrangian for globals. So you could imagine that there is a bunch of globals, G and with various spins. So there could be rotating globals, there could be spinless globals. And they have some kinetic terms which are normalized like something of this or it, very schematically, they could be spinning particles. They would be massive, so all of them would be massive. So the mass of the global n with spin s, you can denote by m and s squared. And then there are weak interactions which could possibly connect any three globals. The only constraint being Lorentz invariance. And there could be derivatives in these interactions. They don't have to be so simple. There could be various derivatives at various points with some powers. But these interactions would be one over n suppressed. And then there could be four point functions, you know, with G, G, G, G and so on. Like infinite n, we expect the theory to basically consist of infinitely many globals with various spins. So we expect the theory to consist of infinitely many globals with various spins and various masses, various coefficients for three point interactions, four point interactions and so on. This is the leading interaction, this is somehow in some sense, smaller. If you want to compute four point functions of globals, let's say you want to compute a four point function, then you can either exchange a resonance like this or you can just pick a contact term from here. They would be both of order one over n squared. But you don't need to care about higher order terms because they would be further suppressed. Okay, so this is just a reminder about the phonology of large n, that we expect a bunch of globals with various masses and spins. Now, one interesting observable that emerged in the 60s is a relation between the masses of these globals and there's, are there any questions? Yes? Are those scales like the energy in units of number? No, this, this scaling doesn't scale with n. This, this part doesn't scale with n because the masses are finite as. Suppose I think the energy to be some power of n, some point to this. Ah, yeah, I mean, if you, I mean, there's a series of globals with various masses and spins and we don't consider this masses to be so high that they are of order n. For us n is the largest parameter. And we just care about, you know, the low line, the globals that are not that massive that are comparable with n. Let's say if n is 100, we care about the first 100 globals. They weren't an estimate for how long, how far we can go before they start becoming very wide. That's a, a good question. I wouldn't be able to give you an answer right away. But indeed, for high enough masses, you would expect that the phase space would be so huge that they could decay to so many particles that. Do I think it's an exponential? Yes, yes, it could be even an exponential. It depends on the density of states. It depends on how much phase space you've got to decay to. Indeed, it's an interesting question you could try and solve. Okay, so some interesting thing that emerged in the 60s, I think it's called something like the Chui Farucci plot. I'm not exactly even sure how to spell it. But it's a diagram where you plot the mass squared as a function of spin. Okay, so we have a bunch of particles with very spins. And for a given spin, there could be many particles of that spin. However, it turns out, so let's start from spin zero, one, two, and so on. However, it turns out that it's interesting to consider for a given spin you could try to consider. Let me just do it the opposite. It would be slightly nicer. So this would be, so for each particle, you could try to consider, for each mass, you could try to consider the particle that spins the fastest with that mass. And so what people have found is that at least in QCD, this kind of spectra can be approximated by straight lines and then there are sub-leading straight lines. So to some phonological approximation, it seems that these resonances lie on straight lines, approximately straight lines. Of course, in real life, these resonances are wide and you can't even place exactly where they sit. But it seems that to some precision, they sit on straight lines. This would be for globals. Let's, this is the mass squared of, this is the mass squared. So these lines are straight in the plane S, M squared. And these mass squareds are for globals. And this came from X, that? This came from Monte Carlo. For me, Monte Carlo experiment is the same thing. They have a statistic, when you have errors, it's experiment. But it's also true for mesons. It's also, for mesons, it's a fact of, it's a fact that you can see in nature. That it seems to be well approximated by straight lines. For globals, it comes from Monte Carlo simulations where people have gone pretty far. Both in four and three dimensions. Especially the group of tapper. So if you want to look at the data, you should look up papers by Michael Tapper. He's done amazing work in three and four dimensions simulating these globals for a bunch of gauge tiers. And he has these plots. And you can see that things line up extremely nicely. In fact, it's so good that one can't even with the naked eye see differences between what he finds in exactly straight lines. Which is somewhat mysterious. But anyway, one expects this kind of analogy. So for a given spin, you would eventually expect that there would be infinitely many particles of the same spin. For a given mass, you would expect that there would be a maximal spin. A maximal spin for which there would be finitely many particles. And there would be a maximal spin that a particle of that mass can have. For a given spin, you expect infinitely many particles. So that's what we expect from large gen physics. We expect that for a given spin, there would be infinitely many particles. For a given mass, there would be finitely many particles with a maximal spin. And I think that there is an expectation that perhaps these lines are not exactly straight. There could be some interesting curves. But I think there is a reasonable expectation that they at least asymptotically look linear. As an exact statement about large gen. There is a long debate about that in the literature. There is sort of a proof that they have to be exactly straight by Mandelstamm. I'll just say a quick historical overview. So Mandelstamm came up with an argument that's called the Mandelstamm argument. I won't review the argument. It requires a little bit too much background, but Mandelstamm had an argument to the effect that these lines are exactly straight at large gen. This argument I think is known to be false. In fact, from ADS CFT, we have examples of large gen theories that can be solved via the ADS dual. And deviations from straight lines have been observed. So people have seen that these lines, if you can analytically continue them to a negative M squared and they don't look straight. They look something like this. So this is Polchinski-Straszler. So if you look up in the Polchinski-Straszler papers, you will see that in their example, they have a controlled regime where they could compute these lines and you will see deviations from straight lines. Small deviations, they are asymptotically linear in Polchinski-Straszler. But they're not exactly linear. So I think that the Mandelstamm argument is clearly wrong and if you look at the argument, I can send you the link if you send me an email, you'll be able to see that there is an assumption that is most likely false. There is an implicit assumption. But I think that there is no reason not to expect that these lines do become asymptotically linear at large mass squared, at least for the leading guide. These sub-leading lines may or may not exist in large NQCD. I don't know. It's hard to say what happens to these sub-leading lines, but at least for the leading line, I think that there is a good expectation that they would be asymptotically linear. And the question is, can we prove it or understand why this has to be true? This is like in some sense showing that there is a string that indeed in four-dimensional Yang-Milster, there is a hidden string description of some sort. This is the sort of thing that you would expect from a string description. Now, of course, due to the, after the ADS-CFT correspondence, we learned that this string description might not be a string in flat space. It might be a string in ADS and expectations from strings in ADS might be different from the expectations from strings in flat space. But having said all that, it seems that in all the known examples, it is asymptotically linear. So it's still asymptotically linear in all the known examples, even in those that have been studied using the ideas of ADS-CFT. Yes. Well, my, for me, capital N is always the largest parameter. So once N is a very large parameter, you have lots of resonances to study before these issues that Shiraz mentioned become important, like before they become wide. There are lots of very stable resonances to study at large N. And so you can discuss what happens when the mass squared is large, but not so large that it would be of order capital N. It's a completely well posed question. Roughly speaking, you would expect that when the mass squared becomes of order N times the QCD scale, which is finite, maybe things would break down. But if N is very large, you have like a thousand resonances to study. Or if N is a million, you have a million resonances to study. And so this question makes complete mathematical sense. And it's also something that can be settled with future Monte Carlo simulations that would go further than before. Now there it's been, I think that they can go up to spin eight or 10, but it's completely reasonable that with the advent of computing and maybe even other methods, they could go even further. And so this question is something that will be settled in the future. Okay, any other questions? Now I'm going to describe some way to think about these questions, which allows you to prove two things. So what I'm going to do now is that I'll set up a certain framework that allows you to study these resonances. And I will prove for you two results. These are not results about asymptotic linearity, but these are nevertheless two results that are correct. And maybe this can be pushed further. So the first result that I'll prove is that there have to be infinitely many resonances. There have to be infinitely many resonances at large n, where n is the biggest parameter. And the second result is that for every spin, for every spin, there must be infinitely many resonances. Okay, so these are some results in some well-defined mathematical framework that you can prove. And it begins to sound like some sort of hagedorn density of states, linear trajectories, but these are speculations. These things are pretty easy to prove, but I think that the framework is so constrained that it's obvious that you can go much further. It's just that it hasn't been done yet. So what's the framework? The framework is the S matrix. So the S matrix bootstrap, all right? So it's very similar to what you heard about in the context of conformal filters. So I'll define the problem of the S matrix bootstrap. It's a well-defined problem in mathematics. And at some point you can forget about the underlying physics. It becomes a completely well-defined problem in mathematics. One of the first consequences that you can prove, well, the two first consequences that you can easily prove by these two. So I'm going to define what I mean by the S matrix bootstrap program. This program has not been as successful as the CFT bootstrap program, but it might be as successful in the future. There is no reason why not. So the idea is that we study two-to-two scattering. We study two-to-two scattering of the lightest global, okay? You can study any other. You could pick other states. It doesn't matter. So we studied two-to-two scattering. It's an exclusive scattering amplitude. So the outstates are the same globals that went in, but they can be with other momentum. So there would be P1, P2, P3, P4. And these are the same globals. So Pi squared is given by some M naught squared. This is the mass of the lightest global. And since this is two-to-two scattering, this is of order one over N squared. So this amplitude is small. Most of the time these globals would just fly across each other, because it's a large N theory. So this amplitude is small, but we're going to subtract the component which describes the process in which they just fly through each other. And we're going to concentrate on the actual scattering amplitude. So the actual scattering amplitude is a function of two complex variables, S and T. S is defined to be P1 plus P2 squared. And T is defined as P1 minus P3 squared. Let me just make sure that these are the same conventions that I have in my notes. Yeah, okay. There is a third variable that you could define which is called U, which is P1 minus P4 squared, but it's redundant because this is minus T, minus S plus four times M naught squared. You can just check that by momentum conservation. And so since there is momentum conservation and since each of the P-squards is M naught squared, this is a linear combination of this and that, okay? So this is a check you can do. Now in the center of mass, you can view this process in the center of mass, where P1 comes from here, the P1, P2 comes from here. So in the center of mass, this momentum are the same. In the center of mass, the spatial momenta are opposite to each other and the energies are the same and they collide here. And then some particles fly out. Okay, this is the center of mass picture in which P2 is almost the same as P1. P1 would have a momentum to the left and P2 to the right and they have the same energy. And then they fly out and you can define an angle theta with which they fly out. The process has an azimuthal symmetry, so it doesn't depend on the angle on the cylinder, but it depends on theta. So you can define the angle theta and you can just do a small computation and show that this is one plus two T over S minus four M naught squared. So this is the cosine of the angle of scattering. So if T vanishes, if T vanishes, then we're talking about forward scattering or theta vanishes. And this also defines the physical regime. So I allowed these parameters T and S to be two complex parameters, but there are some configurations for these parameters that don't describe any physical process. You see that this has to be for this angle to be real. This object has to be between one and minus one. So that means that if S is bigger than four M squared, this has to be negative. So there are some configurations of parameters that describe physical processes. There are some configurations that describe unphysical processes. And so we do some analytic continuation. So these processes make sense for some T and S and then we just analytically continue this to be a function of two complex variables. Yeah, so I'm assuming that the lightest texture asks questions to which he knows the answer so that I would tell you. So yeah, I'm indeed assuming that these lightest particles are scalar. So the lightest globals are scalars. This is known from Monte Carlo simulations. The lightest global is a scalar. And where have I used that? I have used that in assuming that there is an azimuthal symmetry. If this particle carries spin, then there would not be an azimuthal symmetry and the process would depend not just on theta, but also on phi. It would be more complicated. But in this way, it's very simple. Okay, so now we have this function A S of T, A of S and T and it has to satisfy some axioms. And I'm going to list now the axioms that this function has to satisfy. And then you have to look for functions that satisfy this kind of things. So remember that this function A is of order one over N squared. So this function is always of order one over N squared in the large N limit. And I'm going to throw away this factor of one over N squared because it's an overall factor. It won't be important for me. So I'm going to list all the axioms that this function has to satisfy. First of all, it has to be symmetric in its two variables. Why? This is what's called crossing symmetry. It's very much like in CFT. I mean, you can choose to look at this process from this point of view, but you can also choose to look at this process from this point of view. They are the same by analytic continuation because these are the same particles. If they were different particles, it would not hold. But since these are all the same particles, you can just move this state, you can just think about this state as an out state and about this state as an in state by flipping some signs. And the amplitude has to be the same. So A S of T is the same as A T of S. And this is the analog of the bootstrap equation that comes from the associativity of the operator product expansion. That's one constraint. That's an easy constraint to satisfy. You can write lots of symmetric functions. Okay, that's not an interesting. Now let's fix T for a second. Then we have a function of one complex variable. This function of one complex variable is a meromorphic function. So it's a meromorphic function with certain poles. The poles are when there are particles that can be exchanged. So the poles correspond to exchanging some of the heavier resonances. So we have our light global outside, but here there could be any resonance. M squared, N of S. You could exchange resonance with spin and some mass. And there will be some coefficients for the interactions here. These coefficients are of order one over N times some pure number. We throw away the factor of one over N squared. We just keep the pure numbers. So there would be poles to these meromorphic functions that to this meromorphic function when S hits M squared and spin. So let me now denote the spin by L because S stands for something else. So whenever S hits one of those numbers, you get a pole. Now in some applications of this problem, like in string theory, some of the M squareds are not actually positive. Like there could be tachyons. That's one example that I'll analyze. But in Young-Mills theory, we expect all of these things to be positive. So the poles would only be on the positive axis, positive real axis. Now away from the real axis, this function cannot have any singularities. There cannot be poles. There cannot be branch cuts. It's a meromorphic function. And the only poles are here. Forget about the U-channel. I'm simplifying it a little bit. Forget about the U-channel poles. But they exist. Yeah, yeah, but we forget about them. Yeah, it doesn't change anything. We just assume that it's a function of two variables and the poles are just here for fixed t. What Shiraz is saying is that there could be also poles on the negative axis due to U-channel. There will be, yeah, but you can forget about it. It doesn't change anything. It's just a technical complication. Okay, so there are poles at these points. And now comes the most important constraint. Now, these two constraints are natural for mathematicians. I mean, they have studied functions of that sort. But now comes a constraint which is kind of pesky from the mathematical point of view. And this constraint, maybe I'll write it on this board so it will be more visible. This is the hardest constraint to satisfy. It leads to lots of interesting consequences. And it's not very natural mathematically, which is why, I mean, you can't just go and open some book in complex analysis and read out the properties of these functions. So the hardest constraint comes from unitarity, okay? So on the one hand, you would think that unitarity poses no constraints at large. Because the amplitudes for scattering are so small that the probabilities are always much smaller than one. So the probability for scattering would never exceed one, right? And that's correct. But still, unitarity imposes some non-trivial constraints because what it says is that this coupling times this coupling is positive. Because we have the same particles here as here and we exchange some resonance. The coefficient that comes with this resonance must be positive because the coupling here and the coupling here is the same. So it's exactly the same sort of constraint that you've seen in the CFT bootstrap program where the coefficients of conformal blocks were positive. So here the coefficients of resonances are positive, yes? So here the coefficients of these resonances have to be positive. What it means is that if you look at the residue, so let's look at the residue of this function A of S and T. For fixed T, we look at the residue near S equals M squared N S. So we just look at the residue that corresponds to one of the resonances. So this has to be a positive definite sum of Legendre polynomials of, well, more generally, Gaggenbauer polynomials. So it has to be a positive definite sum. Let's do four dimensions. So I'll just stick for a second, three plus one dimensional physics. So then these Gaggenbauer polynomials are just Legendre polynomials. So you have to have something like, you should be able to decompose the residue as a sum of Legendre polynomials of theta where theta was defined there. So cosine theta is one plus two T. But now we're looking at the residue. So we can replace the S in the denominator of cosine theta by M squared N L. So this was L and L minus four times M naught squared. So this formula looks a little bit weird. If you assume that these masses are completely distinct, that all the masses are completely distinct, then the residue would be just one Legendre polynomial with a positive coefficient that corresponds to this particular guy. So it would be just P L. So this index here tells you about the spin. So if there are the generacies, then a certain pole in the S channel would encompass lots of resonances of different spins if there are the generacies. Then you would have a sum of our Legendre polynomials where this index runs over all the possible spins that sit at that mass squared. But if there are no degeneracies, there would be just one Legendre polynomial. And the important constraint is that these coefficients have to be non-negative. Okay, this is what unitarity implies at large head. Not more, nothing more, nothing less. If you satisfy these constraints, then you have a good scattering amplitude. Yeah. Does unitarity also, I think something also give you some, past this amplitude, guys of energy at large? Yeah, we'll discuss that in a second. Are there any other questions about that? Okay. So you have to find solutions to this problem. Which consists of three different constraints. One well-known solution is what string theory gives, namely the Veneziano amplitude, which I'll discuss a little bit in a few minutes. But I just want to discuss a little bit the general consequences of these three different constraints. Okay, so one technical comment, one technical comment, this is something that one has to realize is that, in fact, it's easy to construct amplitudes that satisfy these three different constraints. Here is an example. So you could have one pole here at M naught and one here, that's it. So this is a good solution to this problem. Why? It's obviously symmetric. It's obviously meromorphic when you fix TRS. And the residue at the pole for fixed T is just one. And one is a good Legendre polynomial with a positive coefficient. So this is an example of a solution. But the problem with this thing is that for any fixed T, for any T, A S going to infinity T is non-zero. So this is order one. So this amplitude does not decay for any T. So this is a cheap solution. It's unphysical. The physically interesting solutions to this problem have yet another constraint, which I'll write and then I'll take the question. So there is some, there should be some T naught or maybe many T naughts for which A S T naught would go to zero as S going to infinity. And then now the combination of all of these constraints is extremely hard to solve. You wouldn't be able to guess a function that satisfies all of them. It's extremely hard to solve. And in fact, we just know, we know that young mills should solve this. And we know that many other gauge tiers at large and should solve these four constraints. String theory solves these four constraints, but we can't write functions that do that except for one example. Yes. So we do some analytic continuation. There is the physical sheet, which is the space of configurations for T and S, which describes actual physical processes. And there is then the rest, which is an analytic continuation. Yeah. It's just obtained from the physical sheet by analytic continuation. Yeah, okay. I'll explain where does this condition come from. So this condition cannot be derived rigorously. This condition comes from looking at all the known examples that emerge from the ADS-CFT correspondence and extrapolating wildly experimental data. This condition is something that always has asymptotic freedom. So if you are happy with arguments of the sort that BFKL has to be right and therefore, I mean, something follows, then you could argue that that is true. In fact, you could argue, there is a very nice argument that if you assume asymptotic freedom, the trajectories cannot be exactly linear because BFKL for some negative T forces it to curve. And so if you are, you could say that this is a property of asymptotically free theories. It's definitely satisfied in string theory. This is just a mathematical condition. There is some T naught. Typically, T naught is negative. Typically, the T naught that you need is negative to satisfy that. This doesn't have to be some physical process that has some meaning. It's just a property of this analytic function that is needed for consistency. It's needed for, like, if you have asymptotic freedom, then you need to impose it. In string theory it follows. If you look at other examples that emerge from the ADS-CFT correspondence, then it's true. It's a condition that is basically meant to eliminate these cheap examples that don't describe physical situations. It could be, you could try to prove it. It's not unfeasible that it can be proved, but here I'm just imposing it as an axiom. Okay. Now, from these four conditions, you can prove a lot of things. You can prove that there are infinitely many resonances for every spin. That's what I want to do. And then we'll discuss one example, and I'm running out of time. So I'll just give you the argument that there must be infinitely many poles, okay? So let me just prove that there are infinitely many poles, and you'll prove at home that for every spin there are infinitely many poles. Yes. Say again, U, you said the pole in U. Yeah, that's, so I throw away all the U-channel poles. And this is not something crazy to do because you can arrange for, you can arrange for a scattering process in which the TNS channels are the same, but the U-channel is different, and it may not even have any poles. It's a self-consistent thing to do to throw away this U-channel poles. It's something that is good to do for simplicity. It doesn't change anything. It's something that I'm doing just for simplicity so that the poles are on the positive axis. We don't have to worry about poles on the negative axis. It's just simply. It doesn't change any of the qualitative conclusions. Okay, so here I'll give you an example of how an argument could, you could, you know, devise an argument that will prove that under these restrictions, there must be infinitely many resonances. And then you can try to quantify how big is this infinity? Do we get hugged during density and so on? So this could start, this could be the beginning of a systematic mathematical analysis of which functions satisfy that. Okay, so I'm going to do a five minutes of the technical proof, proof of infinitely many resonances. Okay, so how would the argument go? It's an argument from complex analysis. So we start from this condition where S goes to infinity, and T-naught is some fixed number that satisfies this. Typically there would be infinitely many T-naughts. There would be some interval where this would be true. So starting from this condition, we can make a contour argument. So let's write first an equation that's always true for T-naught. So we do a small contour argument around some S prime. Okay, so this means that we have some contour which is very small around S prime. And I'm just writing the Cauchy theorem, okay? Then we have S prime T-naught over S prime minus S, and then we have the S prime. So we pick some S and we do a small contour argument around S. So we have a contour that's centered at S, and the contour runs over S prime, which is some contour like this, okay? So this is a true equation because we assume that this function A has only poles on the real axis. So this is a true equation. We can choose a small enough contour that it would not include any of the other poles. But now since this is true, since it decays at infinity for this particular T-naught, we can deform this contour in an interesting way to pick up all the poles. So we can imagine deforming this contour continuously to look like this. So you see that this can be deformed into this kind of thing. And since the amplitude vanishes at infinity, this allows us to write A S of T-naught as an integral over this minus this. The rest goes to zero because A goes to zero, and therefore A over S goes to zero faster than one over S. So you don't pick up anything from infinity, but you pick up something from here minus here. Yes? A pole at infinity? Yeah, so if the amplitude was of order one, then there would be a pole at infinity and then the argument wouldn't work. But since I'm picking something out where it vanishes, then there is no pole at infinity. That is not the first order pole. There might be a second order pole, but we don't care or something of that sort. Okay, I didn't hear the question very well, but you're asking what happens if I don't pull it all the way to infinity or what? But there isn't any pole at infinity. If there was a pole at infinity, then this would pick up the pole at infinity plus the term that I'm about to write. You could try to rerun the argument if there is a pole at infinity, but we know that if there is a pole at infinity, you won't be able to conclude there are infinitely many resonances because they just gave you a counter example where there was just one resonance. Okay, so what if what dies off? Like one of ours? Oh, even better. No, because this has one over S in the denominator. You would get something that dies off like one over S squared and therefore this contour contribution would vanish. So as long as eight dies off, it doesn't matter how fast, you don't pick up anything. Okay, so this can be written as an integral over the difference between the amplitude here and the amplitude here. What's the difference? The difference is called the imaginary part of eight and this integral is now over the real S prime is on the real axis now, okay? Because we assume that all the poles are the real axis, so we just have to be careful around the real axis. Now, if you have a pole, the imaginary part that it induces is a delta function, okay? This is a fundamental equation that you should know. The imaginary part of the function S minus S naught is delta S minus S naught, okay? It's a fundamental equation that you need to know. There are no factors in this lecture, or there are one factors are omitted, okay? This is a fundamental fact that you need to know. And so this can be now written as, so I can finish now this little exercise with the following representation for S T naught. It's going to be a sum over all the resonances which are picked up when you pick up these delta functions. So we have to sum over all the N's and L's over all the possible resonances. And then in the top, we have some positive definite combinations of the Legendre polynomials of one plus two T four M naught. And these FK are positive definite. So this is a representation which you find. To be true, if this assumption is satisfied. So this representation makes sense for the amplitude. It looks like just a bunch of resonances that you have exchanged with some spins. Now, this function can, sorry, this is T naught. This function cannot be symmetric in T and S unless there are infinitely many poles because this is a polynomial in P, okay? So, right, but T naught is some range of parameters T naught is some finite region in the complex plane where this condition is satisfied, okay? And therefore this cannot be, if this were true in some region of T naught, this could never be a symmetric function. It's a very simple argument, but you can push it very, very far, very far from starting from here. You can prove that for every spin there are infinitely many particles. You can start to make estimates on what happens to the high resonances. So this is just the starting point, but I think it's a good starting point. It's very simple, but it's a good starting point. So the conclusion is that this function, this cannot be symmetric. Symmetric in ST because it's a, because unless there are infinitely many poles, why? Because if there were finitely many poles, the numerator, the fact that the dependence on T would be polynomial. There would be no singularities anywhere. Otherwise polynomial in T. So if the theory is asymptotically free, this is known to be true. So there have to be infinitely many resonances. Fine, it's a simple result, but it can be improved. It can be refined. Okay, now I want to give you an example. This is the only example that we know of an amplitude that satisfies all of these constraints. This is the Veneziano amplitude, which I'm going to write down. And I'll finish with an open question about the Veneziano amplitude that you could try to solve at home. So I'll go over time five minutes. Just want to make this point. So this is the only example of a function that is known to satisfy all of these constraints. The only example that we know, up to a tiny, tiny dressing that you can do that appears in the heterotic string theory. So up to some tiny, tiny dressing that is allowed, this is the only known function that satisfies all of these constraints, which is amazing. There is only one. So let's analyze this function. Let's see how this function works. Yeah, yeah, yeah, that's true. So these axioms would be satisfied. Yeah, but they're all the same. I mean it's, basically, you just change these parameters here. Is that right? Yeah. Well, they're very, very, very similar. They don't really give new examples. It's of the same sort. But what I wanted to emphasize, what I wanted to emphasize is that this is a solution to these four constraints, but it hasn't been proven that it is a solution. We just, you can check it on the computer, but it hasn't been proven. And I think it would be really, really nice and potentially very important to try to prove that it satisfies the four constraints. So one constraint that's obvious is that it's symmetric in S and T. The other constraint that's obvious is there are only poles on the real axis. So let's fix T. The poles are at minus one, yes, minus one, zero, one, two, all the way to infinity. So the poles are at integer mass quarts. Okay, so this is exactly linear. Well, you don't see it yet, but the poles are on the real, the mass quarts are just integer numbers and there is a tachyon. This guy means that one of the mass quarts is negative, so there is a tachyon. But nevertheless, you can proceed. Yeah, for what? Well, I don't, you could, I mean, there are tiny modifications of this function that also solve the constraints, but then you have a massless gravity tone. Here you just have a massless, let's see. Here you have a tachyon, but they're all the same. It doesn't matter. This attitude admits a little bit of modifications in the intercept and some tiny dressing factor that you can put, but they're all the same. These are not really different examples. So I'm just giving this one example for concreteness. So there are poles on these numbers. This is where this function has poles for any t, okay? And it does satisfy this constraint. It does satisfy this constraint. So starting from, I think that t naught here is minus one. So if t naught is smaller than minus one, if the real part of t naught is smaller than minus one, this is true. You can check. It's the sterling formula. It follows from the sterling formula. So everything is, this is obviously satisfied. The symmetry is obviously satisfied. The meromorph is obviously satisfied. But nobody has ever proven that the residues are given by positive definite sums of the gender polynomials. It's a fact of life. We just know it by putting this function on the computer to very high orders, but it hasn't been proven mathematically. And in fact, okay, so let me just explain what I mean by that. So let's compute the residue. Okay, so the residue is given by, let me just take, so the residue at S equals N where N is in minus one, zero, one, all the way to infinity. The residue for S equals N is given by a polynomial of order N plus one of t over, okay. So I already took the residue. So this would be a polynomial of order N plus one of t. It's a polynomial that you can compute by taking ratios of these gamma functions. And this polynomial, so the question is whether this polynomial of N plus one of t can be decomposed as a positive definite sum over the gender polynomials that run from spin zero to spin N plus one. So in this Veneziano amplitude, there is a huge degeneracy, huge degeneracy that at a given pole, there are particles of spin zero, one, two, three, up to N plus one. So at the pole N, there are particles of spin up to N plus one. For example, at minus one, there's just a scalar. At zero, there is a vector. At one, there is a spin two, okay. So there is a huge degeneracy. This is a property of this solution which we don't expect from QCD. So the question is whether we can write this polynomial as a positive definite sum, A N L of P L of one plus two t over N plus four, yeah. N plus four, okay. And the question is whether the A and L are positive definite. So these polynomials are known. I can write them down. So let me just write down this polynomial. It's very simple. This polynomial is given by a t plus two, two t plus three times t plus three all the way to t plus N plus two. And this is what the polynomial is equal to. And the case of N equals, sorry, yes. And the case of N equals minus one, this polynomial is just one, okay. So the polynomial for zero is just one. So it obviously satisfies everything. And for N plus one, where N is bigger than minus one, this is the answer. So you have to decompose this product in terms of the gender polynomials of these arguments. And this is generally not known. You can check it on the computer that it works up to very high orders, but there is no proof. In fact, some of the coefficients become very, very small. If you go to very high orders, some of these coefficients remain positive. They never cross zero, but they become very, very, very small, exponentially small. So the fact that it's unitary is not obvious. It's not like there is a simple decomposition that we haven't discovered. Some of these coefficients do become very, very, very small when you go to very high little N. And they remain positive, but it seems that there is some conspiracy that this works. Now another fun thing about this amplitude, this is the last thing that I'm mentioning and I'm done. This is the last fun fact. So this Legendre polynomials can be defined in any number of dimensions. They just become Gaggenbauer polynomials. So PL can be extended to, so PL mathematically can be understood as the zonal function of the group SOD minus one. And when D is equal to four, this is the Legendre polynomial, but if D is equal to 26 or any other D, then this is a Gaggenbauer polynomial. So in general, we get Gaggenbauer polynomial, polynomials, Gaggenbauer polynomials. Now, Gaggenbauer polynomials are known. They can be obtained by some hypergeometric function that generates them all. And you can do a small exercise that's in my notes, in my lecture notes, you'll find the solution to this exercise, but you can also do it yourself. That if you take D that's bigger than 26, then A and L is not positive, is not positive. That's a very fun fact that this stupid function knows about 26 dimensions. So this decomposition is positive for every number of dimensions that's smaller or equal to 26. But if you take D equals 27 at the second order, meaning that at N equals two, so I mean at, sorry, here. Oh, yes, yes, good. Thank you so much. Okay, so what I wanted to say is that, this is what I, the last thing that I wanted to say. So the last thing that I wanted to say is that actually it doesn't fail at the zeroes order. You see, this is always a positive Gigan-Bauer polynomial. It doesn't fail at N equals zero. It fails here. So if you take D equals 27, then the composition of this quadratic polynomial would not have positive coefficients. So it fails when the polynomial is quadratic. So, and that's it. But for D, 26 and smaller, it just works. You can just put it on the computer and it works. But it hasn't been proven. I think it would be very nice to directly prove that. It will teach us about why these functions exist, like we expect in QCD or Young-Bills theory, how constrained these functions are. And I think a constructive proof would be useful because it might teach us how to modify the Veneziano amplitude. So the Veneziano amplitude has this crazy feature that there is a huge degeneracy in the masses. So for a given mass, there are many, many particles with different spins that have exactly the same mass. And this is the only solution that we know to these four constraints. But we don't expect this degeneracy in QCD. We know from Monte Carlo simulations that it doesn't exist. There is no such degeneracy in QCD. So maybe if we could understand why this is true, we could understand how to modify the Veneziano amplitude in a consistent way. So to preserve these four axioms, but remove this degeneracy. And then we can get closer to QCD. So this is a well-defined analysis that I think can be done to look for consistent modifications of this function that removed the degeneracy. This can be set up as a perturbative exercise, very far on the rigid trajectory, I think. So I think this is an interesting open problem that would be nice to solve.