Yang-Mills Amplitudes from Twistor String Theory

0:00 I think I'll be using those. Is there a way? The middle one. Do you want the board also? Okay, so I'll be talking about Yang-Mills attitudes from Twister's string theory, and this will be based on a couple of papers written in collaboration with Roy Vaughan and Bolovich. And I should start by saying that the calculation of Yang-Mills attitudes from Twister's string theory is not just some incidental byproduct of the theory, but rather it's so far been the primary result or application of Quister's string theory, and moreover, it was one of the prime motivations for Witten's original formulation of the theory. So in his talk yesterday, Witten explained his motivation for trying to find a string theory description of weakly coupled Yang-Mills theory, just as we have a string theory description of the strongly coupled theory. But more specifically, I think, certainly it's evident in his paper that one of his motivations was to try to understand some of the remarkable simplicity that certain Yang-Mills amplitudes already at that time were known to have. And we heard a lot about this already in Steve Byrne's talk. So the primary example are the so-called MHV, or maximal echolicity-violating amplitudes, which also played a role in my ears talk. So really, Twister-String theory is really predicated on this idea of understanding simplicity against those amplitudes. And we've also heard, of course, that once you start understanding the deep mathematical structure, that also has the pleasant byproduct of making the calculations easier. And that helps our particle physics colleagues. So let me start by giving a brief outline. And that is that there have been several versions of Twister-String theory proposed. at least two of them, Whitney's original topological B model, and also the Berkovich OpenString Formalisms, and I think we might hear a couple more in future talks.

2:30 And at least two of these formalisms, the ones I've indicated here, have so far been used to recover in slightly different ways that I'll discuss later in my talk, but they've both been used successfully to recover tree-level luan scattering amplitude in a mill series. And in fact, even within the context of these string theories, there are several ways of doing this calculation, and each one of these ways has, I must confess, some poorly understood twists and turns, some of which I'll get to later in my talk, but also each of the calculations reveals new and different mathematical structures in the end of amplitudes. So the primary focus of my talk would be the first calculation of, I should say, of course, that in his first paper Witten did show that his topological B models could be used to reproduce the MHB amplitudes, so I'll be discussing the first calculation of non-MHB amplitudes from the B model based on my work with Roy Barman-Volovich, and it will be the one that's usually called the connected prescription, and the reason for that terminology will So, as I said, the subject of this talk will be the calculation of gluon scattering amplitudes, and as Steve Byrne explained, it doesn't really matter now whether we're talking about QCD or N equals 4-Yang-Mills theory, if we stay at tree level, because there are no superpartners, we're not having any loops, so it doesn't matter what other content our theory has. And he also reviewed the fact that instead of labeling each gluon by its momentum, p, and its polarization, epsilon, which one must confess are rather inefficient variables, so it looks like you have to specify eight real degrees of freedom for each external particle, but in fact you have some nonlinear conditions. Each momentum obviously has to be null. The momentum has to be orthogonal to the polarization, And furthermore, you have the gauge equivalence, which identifies the polarization vectors which differ from each other by shifting by the momentum. So it's a pretty inefficient use of your kinematic variable. So instead, we use the spinner helicity notation, where we decompose a null momentum 4 vector into a product of commuting spinners, and we furthermore use the fact that it's possible

5:00 to write down a formula, as Byrne did, which shows that you can canonically determine the polarization tensor epsilon from lambda and lambda tilde if you furthermore specify a plus or minus solicity for each neuron. And this determination is canonical up to a gauge transformation, which, of course, doesn't affect the amplitude. So to summarize that statement in the picture, we consider some tree-level amplitude with n-luons specified by lambda i, lambda tilde i, and the callicity plus or minus i for each p. And the only remaining redundancy variable is that there's a scaling switch. So obviously, because p equals lambda times lambda tilde, the sigma u is how we make your case. Obviously, if you scale lambda and lambda tilde oppositely by any non-zero complex number, you need to keep the momentum unchanged. And furthermore, you can check that this introduces a pretty simple scaling on the overall amplitudes. We've heard a lot about how things, well, okay, obviously. And now I have to apologize that I'm going to leave out a lot of mathematical, I'm going to be rather non-rigorous in my talk today until I get to the very end, And so I have to apologize here, my use of, so I put in quotes, this one-half Fourier transform as being operationally what we're going to do in order to implement the transformed cluster space. So what we do, we have the amplitude as a function of lambda and lambda-thilbic, and we just Fourier transform all of the lambda-thilbic variables into variables that I'll call mu. And now what happens is that this object becomes a function of endpoints on projected Pusser's base P3. Function is in yellow because, of course, it's not actually a function, it's a section of some line bundle over P3, and this particular bundle depends on the helicity of each external nuance. So we heard in several talks yesterday how there's a connection between the helicity of a particle and how its associated wave function, it will scale homogenously to the surface. So that's what's going on here. And the coordinates on the projected, well, the homogeneous coordinates on projected solar space are going to be labeled by capital D I. So capital I runs from one to four, and it encompasses the two lambdas and the two

7:30 use. Now, so far this is just a notation, but the magic is that in Witten's paper of 2003, he checked in several cases, and conjectured in general, that if you have an n-luon scattering amplitude with q-phalicity gluons and n-q-phalicity gluons, that this object should in fact be On whole work, it curves in P3 of a specific degree, degrees equal to Q minus 1. So that in particular, the very simple, the simplest amplitudes, which are maximally helicity-violating, they have two negative helicity neurons, Q equals 2, so they're supported on lines in twister space, in projected twister space, which is one way to explain their remarkably simple structure. Now, of course, by itself, this remarkable observation is not enough to allow you to explicitly calculate any amplitudes. All it tells you is that the amplitude is, say, an integral over the moduli space of degree d curves of something, some integrand, and you need to know what that integrand is. Well, fortunately, Witten further proposed that the annual, well, he proposed a specific way to calculate this integrand. So he proposed that the Yang-Nose amplitudes could be calculated from a particular string series. The Olka topological B model on P3-4. So my goal in this talk will be to show you in quite a bit of detail, although hopefully not too much to bore everybody, how specifically one could do this calculation, and we'll obtain thereby new and interesting formulas for tree-level Luan scattering amplitudes in Yang-Nose So before I begin the calculation, are there any questions at this point? Okay, so the next transparency is unfortunately the worst, sort of contains the most information of any transparency in my talk, so let's spend a few minutes on it. It's basically a cartoon of what's going on in this B model. So I hopefully have introduced all of the characters here. So, we take this twisted sigma model on P3 slash 4, and it's a six-real-dimensional

10:00 manifold. The bosonic manifold P3 is six-real dimensions, so we can fill it with D5 brains, which have a six-dimensional world volume. So we start by filling the bosonic dimensions with N D5 brains. Now, of course, when you have D brains, you have open strings which end on them. And the open strings ending on these D5 grains will be the blue ones. I wouldn't explain that in this talk, how you see that, but I'll have a little bit more later. And the effective action for these open strings, so just intrinsically from themselves, from quantizing these open strings, you find that the effective action is not that of Yang Yang-Nos theory, but rather only the self-dual part of Yang-Nos theory. And in order to supplement those interactions, Witten argued, in order to reproduce all of Yang-Nos theory, Witten argued that you should consider additional contributions from instantons. So the instantons in this theory take the form of certain V1 grains, which have the totology of the sphere, and they can wrap some holomorphic curves inside P3-4. And whenever you have an instanton, their additional physical degrees of freedom. In this case, the additional degrees of freedom are the strings, the strings colored purple here, which run between the D5 brains and the D1 brains. So the point is that what you do is you want to integrate out these effective degrees of freedom, well, I'm sorry, integrate out the degrees of freedom in this instanton background, and doing that will give you contributions to the effective action for the gluons, which which supplements the effective action so that you're promoted from just-self-dual-yang-nose theory to the full-yang-nose theory. Well, that's the idea. Let's now do the calculation in gory detail. So using the background field method, we can pretty much write down a cartoon of what the formula should look like for a contribution to the gluon scattering method. So what we have is, we start with what I've called number one there, which is an open Numbers 1, 2, and 3, I'm going to go into more details later, so this is just a cartoon at this point. But 1 here is the open string correlation function, so I have n of these d1, d5 strings ending on the instantons, so ending on the sphere.

12:30 So they give me some open string correlation function. I have an integral over n points on the sphere, n points on p1. That would just tell me that I have to integrate over the points where the vertex operators are inserted. I have to integrate over the moduli space of curves, because this D1 instanton can wrap any holomorphic curve inside P3 slash 4, where equivalently that just means I'm summing over, well, summing, or in this case integrating, over all possible instantons. And finally, there's, you know, whenever you calculate a scattering amplitude, you need to specify wave functions for the external particles that you're scattering. So in this case, I need to specify n wave functions for my blue eyes. And so those will just be functions of the embedding. So here, this capital Z, as a function, so little z is a coordinate on the sphere. Capital Z is the embedding of that sphere into P3 slash 4, and I'm integrating over the space of all such curves. Okay, so now let me go through numbers 1, 2, and 3 in a little bit more detail. The point of this whole exercise is to show you that there's really nothing scary going on. It's nice to be able to use sheep cohomology and all that wonderful mathematics, but really it is the case that you can just sit down and write out a formula and follow your notes and see where it takes you. So let's start with the open string correlations. So there are two kinds of D1, D1, D5 strings are oriented. So we have the D1, D5 strings alpha and the D5, D1 strings beta, which are respected in the fundamental or anti-fundamentals of the UN. It's UN, if I take N, V5 range, each of the open streams, there's a Shannon-Patton factor. So I combine those following link into the form of the LJ. It's fundamental times anti-fundamental, so I take that using the D up in the adjoint of GLN. And so the idea now is that if we specify the gauge group part of the wave function, so we specify a wave function for a gluon, it not only has a space-time wave function,

15:00 which will usually be a plane wave, but also carries a gauge index. So we can specify where each gluon points in the gauge group. So we do that just by saying that the wave function of blue on i will be proportional to t x of i. That will be some generator of the gate here. And so then the correlator of this open string, well, the open string correlation function takes the following form. There are many terms, so you basically write out j equal to alpha beta, and then look at all the possible contractions you can get between the alphas and the betas. And there are many terms. So there's sort of a leading term, which is a single trace over a usual kind of denominator, plus different possible permutations, depending on how you... There are many ways of doing the contraction. But then you can also have double trace terms, where the contractions split into two factors. And of course you can have even higher trace terms. So you get all that. And now the thing is that at tree level, gluon scattering amplitudes are always single-trace interactions, right? So here I've just drawn a four-point scattering, well, one contribution to the four-point interaction of the gluon, and the purple line traces the color index, if you will, and everything is nicely connected and it's proportional to trace TA1, TA2, TA3, TA4. It's a nice single-trace interaction. Whereas multi-trace interactions indicate that you've exchanged some particle which is not charged under the gauge group. So in this example, for example here, maybe you have a scalar field propagating in the intermediate step there, and so you see that the trace splits into a double trace. And so this, the second kind of interaction you don't have in Peoria-Ing-Mills theory at tree level, nevertheless, here we do seem to have that in the topological B model. So as we've noticed, this already indicates that the B model will have more than just Peoria-Ing-Mills theory in it. In fact, it will have conformal supergravity. But for the purpose of the present calculation, we're going to ignore that, since you can consistently ignore that kind of thing at tree level, and we're just going to focus on the coefficient of the single trace term. So we're going to ignore all of this and just

17:30 look at the coefficients of this single trace term. And that will correspond in Yang-Mills theory to, as Speedburn also indicated in his talk, how you look at specific color factors. You look at the coefficient of a specific trace structure in Yang-Mills theory. That's frequent trace that you can do. Okay, so that will be, okay, so to summarize, the blue box upstairs will be replaced just by this factor of one, this will be one over V1 minus V2 times dot dot dot of VN minus V1. So let's look at the second object that I promised to explain in a little bit more detail, which is this integral over the modulite space of curves. So at genus zero, all our instantons go at genus zero, a curve of degree V in P3-4 is most conveniently parametrized by writing an explicit embedding as a degree V polynomial. So here, I continue to use capital Z as the four bosonic homogeneous coordinates on P3-4, and I'm also using psi A to be the corresponding for fermionic homogeneous coordinates. And so each of those, I'm just writing an explicit embedding map from little z, which is the point on P1, into P3 slash 4, as an arbitrary degree B polynomial. So here I'm going to have 4 times D plus 1, those aren't in coefficients, and similarly 4 times D plus that specify the exact shape of this curve. Now, to integrate over all possible curves in P3 slash 4 means, practically, that you should just integrate over all of these parameters, A and theta. And so, when we argue that the natural measure, or the polymorphic measure, I should say to be more precise, on the space of curves is given in this parameterization as the obvious the product of the differentials of all of these variables. But that's a little bit of an over-counting. So let me do the U1 first. It's obviously over-counting by a scale factor because capital Z and capital Psi are homogeneous

20:00 coordinates. So obviously, if I just do an overall rescaling simultaneously of all the A's and all the where I'm going to in P3 slash 4, so I don't care about the overall scaling of all of those moduli, so I have to divide that by a U1 factor. And furthermore, I have to divide out by an SL2 factor, which is familiar whenever you do a tree-level calculation of string theory, because you don't care about how the actual parametrization of that curve, of that P1 as it sits in P3 slash 4, you only care about sort of the, you don't care how it's parameterized, you don't care about, basically you could do an SL2 transformation on the little z's here, and that's not going to change the overall shape of the curve sitting inside P3 slash 4. So modding out by those symmetries gives us the holomorphic measure on the space of curve. Now, the third ingredient in the formula is the choice of wave functions for the external nuance. Now, the physical open string stage of the B model corresponds to the cohomology group H1 of P3 minus a small set, so it's P3 prime. I tried at the last minute to correct some minus signs. So I wrote, I had to change O of minus K. Okay, so there may be some inconsistencies with previous thoughts regarding O of K versus O of minus K, but I hope I haven't done anything too drastic. So anyway, so Witten showed yesterday that physical open string states of the B model correspond to these particular cohomology groups with coefficients in some line bundles over P3. And we also heard in several talks yesterday that by the correspond precisely to the space of solutions of the conformally invariant free-massive wave equation in the galaxy space. So in other words, this spectrum of bundles gives us exactly the field content of any case-story angles. Okay, that's wonderful. So here we have to choose some cohomology elements to do this integral. Now this is the... We've had some e-mail of correspondence with some mathematicians, and this is always the point where we lose them, because we say, oh, we need to choose this cohomology element.

22:30 And they never want to choose a cohomology element, right? Because they're all equivalent. That's the whole point of cohomology. Why would you ever want to choose a particular cohomology element? It's dirty to them. You mean you're choosing a class or you're choosing a representative? I'm sorry, you're choosing a representative. Right. So which representative do we want to choose? So our particle physics colleagues, of course, they always evaluate scattering amplitudes evaluated with plane waves. So that's, of course, the most natural thing to do. So what we want to do is we want to choose the cohomology, the representative, which corresponds to a plane wave in Minkowski space. So we want a wave function that has a definite four-momentum, pi. Pi would be the momentum of muon number i. So now what does this look like in twistering space? Well, There's one way we can do this. It turns out a pretty convenient way. We'll see. I'm looking at one thing. If you are looking at these rational maps that you multiply by a real state of a that it's not going to be changing the map again, there's a better that things show up. Is that serious? I'm sorry. So the re-canning that you're considering is just a phase, right? No, it's an arbitrary complex. Okay, this actually combines... That's right. So this will combine the GL2, C. So how do you normalize the measurements? It is. It is because the So the fermions and the bosons transform opposites. Oh, I... Yeah, that's crucial. So this descend... So this modulite space is actually itself a Pallavi-Yau manifold. And this is the... Because this is binary, it can descend into the... ...morphic measure. Okay, so back to the ray function. So we started by writing something which is essentially a delta function on projected twister space. So, well, a delta function, we have four homogeneous variables, right, lambda 1, lambda 2, mu 1 and mu 2. You can't write a delta function for them all four simultaneously because you need to, that would be too restrictive of a condition.

25:00 You need to allow two points to be identified with each other even if the homogeneous coordinates only agree with each other up to an overall scale. So one really simple, trivial way to implement that homogeneity is to write a formula like the following. You have the four delta functions you want, but they don't tell you necessarily that mu i should equal mu and lambda i should equal lambda. It just tells you that these forms equal each other up to the same number, which we call C. So, by integrating over C with the proper normalization, this gives us the properly normalized delta function on P3. So now, now we want to go from projective twister space back to the original, well, momentum variables, ideally. We've written momentum as lambda times lambda tilde, so we just Fourier transform back, pardon my line technicality. So that gives us this formula here in the yellow box. This will be, so it's basically just what you expect. It's a delta function in the lambda variables times a plane wave in the lambda-silda variables. I just wanted to, I have a line in this thing, but I just wanted to say how that formula would be interpreted in context with this, because it's really good. That's right, please. Are you about to? No, actually I won't say anything. So every investigator would see that it would be a one-dimensional vector. And to make it D bar closed, delta would be both a ZZ bar and a ZZ bar. So the complex system states that would be a zero-one point, which would bring us a dependence on the other, and the complex system, the wave functions, like one-dimensional vector. So the point of my derivation, yeah, so many of the steps that I'm doing a little bit fast and loose can be properly rigorized with the requisite mathematical machinery, but some of the steps I'm going to do can't be, well, it's not yet quite known exactly how to rigorize So, at the end, when we have the final formula, I'll give a completely honest assessment of what's rigorous and what's not. So now all we need to do is put together all three of the ingredients that I've hopefully spelled out for you in quite detail and will last several transparencies.

27:30 So here's what we get. This is the twister string amplitude. I'm suppressing here the choice of helicity, so what happens is, because I'm going to suppress all the fermions just for the moment in this formula, and the choice of helicity is basically a choice of, when you expand the superfield in terms of its fermionic components, the different component fields, as we reviewed yesterday, correspond to the different helicity n equals four supermultiple. So depending on which fermionic component you're looking at, essentially, that tells you which particular helicity assignments you're looking at. So all of that comes in the fermionic determinant. So I'm not written out on this transparency. It's roughly speaking just a similar copy of some of this will come along to the first. And it's essential that it be there, but I just haven't drawn it on this transparency So we have the integral over the bosonic modulite space, the A's, the integral over the n points on the sphere, we have the integral over these n parameters I've introduced to enforce the homogeneity, we have this factor which comes from the correlation function of the open-stream vertex operators, and then we have the n weight functions for our gluons, which again are delta functions and the lambda variables and plane weights and the lambda-tilda variables. And capital Z is, as we saw before, this polynomial embedding of P1 into P3. Okay, so now there are many things one might imagine doing with this formula, but let's try one thing which I know will give us a somewhat pleasing answer. And that's to notice that what we're going to do is we're going to do the integral over half of these moduli. So remember that A carries an upstairs index, which runs over 1, 2, 3, and 4. And 1 and 2 correspond roughly to the alpha spinner index, and 3 and 4 correspond to the alpha dot spinner index. So what we're going to do is we're going to do the integral over half of the A's, namely those that correspond to an alpha dot index. And those only appear up here in this term, in the exponential, and so doing this integral will just give us even more delta functions, because they appear linearly here in this

30:00 exponential. So let's see how that works on the next transparency. So here's the formula after doing that integral. We have these delta functions which we kept from over there, and then we have a new series of delta functions from integrating out half of the bosonic bond. Now this is kind of an interesting formula for the following reason. So, if we count how many integration variables there are, we have 2b, 2b plus 2 from the remaining bosonic monduli, n from the points of this sphere, another n from this homogeneity parameter, minus 4, and we have 6, so that's 2n plus 2b minus 2 integration variables. Now, if you stare at a count of how many delta functions there are, there are 2n in front of 2b plus 2 delta functions. So, for any amplitude, there are always precisely 4 extra delta functions. But that's great, because that's exactly what we want. We know that any scattering amplitude and angular theory should have a delta-4 momentum conservation sitting in front of it. So, can we see that? Yeah, it turns out it's really pretty simple. So you just take this quantity, which is enforced to be zero by that delta function, and sum it in this combination, and you would rearrange it a little bit, and recall that this was the formula here, which gives the embedding, and then you use this equation here, and you find that, basically you find that the equations which these delta functions imply necessarily imply momentum conservation. Or in other words, if you don't have momentum conservation, you're never going to solve those equations. So accounting for this overall delta function of four momentum, which you can imagine pulling out of the integral, possibly at the expense of the Jacobian factor, so after accounting for that, there are precisely as many delta functions as there are integration variables. So the integral localizes onto a discrete set of points. So this leads to the main result for the connected prescription for the tree-level S matrix of the A-Mils-D, which goes as follows. Given lambda and lambda tilde, which specify the

32:30 data for the n-luons we want to scatter, you start by finding all solutions to these polynomial polynomial equations, so let me collect all of the equations into hr, so it will denote the union both of this kind of equation and that kind of equation. So you solve these polynomial equations in terms of these 2n plus 2d plus 2 variables, being the z's, the size, and the half of the moduli that we haven't integrated over yet. And then the scattering amplitude is given by this formula. You have the delta function momentum conservation, over the roots of these polynomial equations times this Jacobian factor. And the Jacobian factor has J, so these are the factors we had before. Additionally, there's a determinant which you introduce when you fix the GL2 symmetry. So, commonly you fix the SL2 part of that symmetry by, you can fix three of the Zs, say, to 0, 1, and infinity. common choice, and doing so introduces some determinants. And similarly, it's convenient to, say, fix the overall scaling symmetry of the moduli by fixing one of the bosonic moduli to be one, for example. So, but I should say that's, I mean, there are different choices one could make to fix the GL2 symmetry, and they would give you different Jacobians, but this turned out to be a pretty convenient choice, and that's what I mean. Finally, you have And it actually takes a remarkably simple form. It's the fourth power of the determinant of a d plus 1 times d plus 1 matrix. And the matrix is formed by taking xi prime times xi prime to the k power. And k, recall, runs from 0 to d. It was basically the number of bosonic moduli you had appearing in this degree d polynomial. And I prime denotes that you only let I run over that subset of the end muons that corresponds to the negative full of speed muons. So that's a pretty elementary matrix, and you just tap that in there. And so this is the formula for the tree-level S matrix that you get from the connected prescription.

35:00 And the final comment that I'm going to go into more detail quite a bit is that there are no absolute value signs here. So, had this really been an integral over the real axis of these delta functions, you should have expected that when you do the integral of a delta function, you get it to turn 1 over its Jacobian matrix with the absolute value symbol. And so, the part of the derivation which I can't justify at the moment, other than to say that this is correct, is that you definitely don't want to put the absolute value symbol. So, I think I'm going to leave that transparency on the board while I make six comments about that formula. So, are there any questions? Well, I'm a little hesitant to ask questions because I'm now going to make six comments about that formula. So, but if there are any questions, I'd be happy to take them. I'd be happy to take them or to tell you that I'm about to get to them. So comment number one is this formula is correct. Now remember, I guess it was just a conjecture of Whitten's that the B model should correctly reproduce Yang-Mills scattering amplitudes, but at least that formula works. So for the mostly plus MHB amplitudes, which have two negative for the 3d1s, and for the mostly minus MHD amplitude, it's reasonably straightforward to check that the equations always have precisely a unique root, and that if you plug that root into this equation, you reproduce the simple form for these amplitudes. And this, basically in this case, it reduces to the calculation Witten did in his paper that in turn essentially reduces to a similar calculation that you had done over a decade I should say, for the MHV amplitudes, mostly plus. For the mostly minus MHV amplitudes, it was perhaps a little bit striking that it worked, that we found that it worked, but nevertheless did it work. Now, for other amplitudes, unfortunately, these polynomial equations have more than one solution, and they're really quite tricky. Actually, if anybody in the audience is an expert on numerically solving polynomial equations, love to hear more input on how to do this. The problem is that the computer programs

37:30 such as Mathematica and Maple are completely useless on these equations because certain subsets of these equations have just awful intersections which don't localize on points but on all sorts of higher degree lines and intersections. It's pretty nasty to separate out the discrete points that is the common intersection locus of all those equations. But in any case, in all the cases where we can't numerically attack that with the equations and compare with no-yang-mills amplitudes, one can check that the agreement is 100% correct. And more importantly, there are a number of consistency conditions you can check. For example, it's known on general grounds how Yang-Gil's amplitude should behave under when you take a soft limit of some gluon momentum or as you take the momentum of two gluons to be collinear with each other. And these properties, one can check, is satisfied by this procedure. And there's some other checks as well that I'm going to do with them. So certainly, we're confident that this formula is correct. It certainly would be very interesting to have, well, it would be interesting to have a direct proof of that procedure within, either directly from Yang-Mills theory, and or a more rigorous derivation from the B model. So I mentioned that one can do a number of consistency checks of this formula, this procedure. And one of those is parity invariance. So of course, parity invariance is completely obscured in Twister's space. whether you're reminding parity invariance, says that an amplitude should be invariant when they're interchanging lambda, lambda, tilde, and simultaneously exchanging the holistically of all particles. So, nevertheless, it's elementary to prove, and by elementary, I mean elementary in the mathematical sense. I don't mean that it's easy to find, but I mean it's just an algebraic manipulation. It doesn't require sheet cohomology or anything. You can show that there's a one-to-one correspondence between the roots of those equations the corresponding parity-flift equations, which is a little bit interesting because you have a different number of variables because the dimension of the moduli-like space of curves is changing. You're going from degree-D curves to degree-D tilde curves. And the proof is constructed, so if you take a root of those equations,

40:00 you can write down a formula that will produce for you a root of the parity-flift equation. And furthermore, you can prove quantity in brackets, kind of forms in precisely the right way to account for the switch between plus and minus. So we conclude from that that the connected prescription, again defined by this transparency, is parity-symmetric. And there's a related, well I assume it seems related as far as I can tell, but it involves some heavier math proof of parity and variance So my third comment, so I mentioned how hard it is to solve even numerically these equations, so one interesting question you might have is, what's the number of groups as a function of n and d? I remind you, n is the number of Guelan's you're scattering, and d is equal to the degree of the curve, and is equal to the number of negative solicited Guelan's minus one. So this is all we know at the moment. I told you that MHB and grouply MHB amplitudes unique group. I told you the parity invariants, and five is this interesting identity which reminds you of your symmetry or something. And then there's this other case which you can verify numerically, and you find that they're always precisely four groups. So it might be interesting, for example, if you're curious about how the computational complexity of calculating in those amplitudes grows as you go to higher and higher amplitudes, you might wonder, it might be interesting to determine, for example, whether this number grows polynomially or exponentially. At the moment, I can fill it up. But I should emphasize, of course, this question of how many roots there are only makes sense within the context of how we chose our particular way of fixing the GL2 symmetry. So I remind you, we fixed three points on the sphere and one overall scale of a moduli. There are other ways to fix the GL2 symmetry, but they're more complicated. You could imagine fixing four of the So, some other research groups have looked into this, and what they find, so obviously you're going to get a different set of equations if you fix the moduli differently, and they could very well have a different number of roots, and they do. And unfortunately, it always seems to be the case that fixing the GL2 symmetry in a different way leads to even a nastier set of equations than these. So, it might be interesting to see if there's

42:30 some non-obvious way of fixing the GL2 symmetry that's a nicer set of equations. I just don't know. Comment number four has to do with felicity permutations. So it's an interesting fact that almost all of the currently known calculational techniques, and I think certainly all of the calculational techniques which have been motivated by string theory, become harder as you have more plus and minus neurons interspersed. So here I'm using a cyclic ordering of the gluons appearing as the external legs on the two-level amplitude. So all of the calculational techniques suggest that, for example, that this amplitude should be much more complicated than this amplitude, even though they have the same number of minus listed gluons, so they come from the same degree curve in Twister's state. However, this formula for the connected prescription treats all the elicity orderings democratically because the only place the elicity enters is in the determinant of this matrix F. So once you solve for the roots of this equation, it's a simple algebraic, you just evaluate different determinants on the same set of roots and you immediately can determine the So, I don't know if that has any profound meaning or if it's just a curiosity, but I'd certainly find it interesting. So now I'm going to explain why this is called a connected prescription. So the answer to that is that we consider it only connected in some times, meaning that we had our green V1 brain here, here at Drone Young. Yeah, this is supposed to roughly indicate a qubit curve, approximately, schematically. So we considered the case where the instanton is purely connected. But of course, we could also have instantons which are disconnected. So here you can have a multiple instanton of degree 1, or you can have an instanton of degree 2 sitting over here, and then some other instanton of degree 1 sitting over there. So, and at first it was thought that maybe you would have to add together all of these contributions in order to reproduce the correct A-nose amplitude.

45:00 But remarkably, we found from this formula that the connected, it seemed like the connected instantons themselves reproduced the complete A-nose amplitude. And about the same time, Kachalda, Skirczek, and Witten formulated these MHB rules motivated by a study of the completely disconnected instantons, which separately give you, by themselves, these give you the complete dang-nose amplitudes. So once you know these two, it's perhaps not surprising, and it was soon thereafter shown by Benaburn and Kolfsauer that you could use intermediate, what I call, or what's frequently called intermediate prescriptions, to calculate the amplitudes also. So there's an interesting story here, which is still... So I've explained some of, none of these, well as Wooden said in his talk yesterday, the derivation of these rules from the topological B model is not 100% nailed down as well as we would like to understand it. So we really don't, so since that's even the case, maybe we can't yet be at the stage where we completely understand exactly what's going on here, but There's a very interesting argument by Hupalt Modlin-Misky, who argued that the equivalence of these prescriptions, and I don't have time to go into this in more detail, and I only recently learned that they probably wouldn't be talking about this, unfortunately. So they argued that the equivalence could be understood by the fact that there are singularities in each of these modulite spaces. So for example, in this modulite space, there's a singularity as the curve is becoming degenerate. And similarly, there's a singularity in this modulite space as the two lines approach one each other. And they argue that in each case, the amplitude is sort of received, well, that it would be possible to choose a contour such that the integral expression which reproduces these various amplitudes only picks up the poles associated with the degeneration of these curves. And furthermore, those points in the moduli space are common. So what I mean by that is, if you have a degree 2 curve, which is the hyperbola, there's a limit in this moduli space where it becomes two intersecting lines. And similarly, if you have two disconnected lines, there's a limiting point in this moduli space where the lines touch.

47:30 So both of these moduli spaces share a common singular point. And the argument was that the integral can be written in a form where the contribution comes entirely from that point. So again, it would be very interesting to rigorize that argument, and if the primary instructions doing that is the problem that we don't really understand what this integration contour should be. So there's a puzzle here. So, the measure on the moduli space of curves was really a holomorphic measure. So in order to honestly do an integral, we need to choose a contour or a real cycle in the moduli space and integrate this holomorphic form over that real cycle. But it's not really understood, for example, how one should make this choice or whether different choices differ. So, pragmatically speaking, perhaps what I'm saying is this, that all of the integrals and all of the formulas that I wrote down, all of the integration variables were complex. All of the moduli for the curves in P3 slash 4 are complex moduli. Points on P1 are obviously complex variables. So, when I did those integrals, I must have chosen some contour in a complex plane. And indeed, I did. I implicitly chose the most-maid contour, which is the one where all the integration variables go along the real axis. And this makes sense in signature plus, plus, minus, minus, where len and len are independent real variables. Now, okay, so this was a trick, a slight at hand, And even this didn't quite work. So here's the following puzzle. These equations, it can happen... Okay, so one element of that fact about the conch we're not working is that one might have expected absolute values here, but the formula is correct without the absolute values. But another aspect, completely related, is the fact that even when you choose the lambdas and lambda-tildas to be real, signature two comma two, it can happen generically that the solutions to these equations are complex. You can have, you know, they can become complex in pairs. And nevertheless, it's the case, and so, so suppose you're in a situation where all of the roots are complex, for example, and this does happen. Naively, you're thinking that if you're integrating

50:00 these delta functions over the real axis and all the roots are complex, that you would think you should get zero, but zero is not the correct Yang-Mill's amplitude. Nevertheless, you do get the correct Yang-Mill's amplitude. If you plug in all the roots here, complex or not, and as you add this up over the roots, the complex phases will cancel out, so the answer here will always be real. So that's a little bit of a puzzle still, how to think about that. Okay, so I have about seven minutes left, and I was planning to change gears a little bit at this point in the talk, so let me ask first if there are any questions about this connected prescription for the G-level S. That's the justification for the intermediate It's the justification for the If it was correct This formula It's generated from the string field So if you know the cubic vertex Then the string field theory essentially will generate Oh I see So it doesn't I mean One thing I did forget to mention Let me come back Of course I even had a note here I forgot to mention it If you use the open string formulation of Vertex, what you get is, when you first write down the formula for the scattering emphasis, you get a formula which looks basically exactly like this description. But so now what you're saying, I guess, is that, so I have been under the impression that using your formalism, you would get a formula which looks like an integral. Yes, you have the same structure. But this is not an integral. Is that the point you wanted to make? Well, my point was that suppose you know the cubic version. Yeah. And you get the image, the image, the image, the image. Yeah. Yeah. Yeah. And you have a field theory generalization, a field theory which generates the cubic vertex, the higher vertex. Yeah. Then you essentially guarantee to get all the right ingredients just by sewing the cubic vertex together. Yeah, so right, so what you have is basically, that's right, it's like the MHB rules, except you have both kinds of vertices, the plus plus minus, and the minus plus plus.

52:30 It's a strange field here, but it's a different field. Right, that's right, and it reproduces this, that's right. Okay, so now I'm going to switch directions a little bit. So one of the things which C. Byrne emphasized in his talk that's been one of the primary outputs of the recent twist-through-stream developments is the realization that scattering amplitudes are a lot simpler than we had any right to expect. So you can sum up an expression which involves thousands or millions of Feynman diagrams, and magically there's some structure there which needs to be explained. And so the question, and we're going to hear a lot more about this in future talks. So the point is, twister string theory has inspired a number of insights into the remarkably rich structure of tree and loop amplitudes in N equals 4 in those theories. We see these talks, possibly for a variety of results and instructions. And now the question is, you know, how do we know when we reach the end? How do we know when we've distilled all the possible simplicity down to the very bare minimum? How do we know when we can go no further? Well, I'm now super optimistic. The reason is that we recently discovered the following formula for an 8-luon amplitude, which, first of all, would be completely impossible to calculate using Feynman diagrams. But even if you had used the previously known twister-inspired constructions for this amplitude, a mess. It would involve, for example, summing together 44 different CSW diagrams. And so, and yet we found a formula for this which is much simpler, even than might have been guessed possible using previously known to a certified construction. So, I don't know how to answer the question of how we know we've reached the end. But let me give you of brief introduction into how we stumbled across this formula. It's kind of a roundabout way. There's a more direct way that's known now, and I'll get to it in a second. So we looked at the infrared behavior of one-loop amplitudes and n equals Fourier in those two. So, and this was largely motivated by, I mean, of course this has been studied by many people

55:00 over the years. But our particular study was largely motivated by the paper and also were known to be the fixed term process. So basically, when you evaluate loop-amplitude in any post Fourier-Mills theory and their infrared divergence so you evaluate them in 4 minus 2 epsilon dimensions, it's known on general grounds that the coefficient of the 1 over epsilon pole in that one loop-amplitude should be proportional to a tree amplitude in some particular combination with some kinematic variables appearing. And so what you can do, you can basically read off tree-level amplitudes by looking at the residue of the 1 over epsilon 4 in a loop amplitude. And loop amplitudes are known, it's known, there's a complete classification of all the possible types of integrals which can appear in dimensional regularization. And those are called, I'm sure you'll hear a lot more about this. So those are called box integrals, and so it's known that any one-loop amplitude man equals four-angle theory can be expressed as a linear combination of those box integrals, with coefficients that we call box coefficients. So I'm not, I don't have time, unfortunately, going into all the details, but here's certainly much more about this, but I guess this discussion now is mostly for the experts, so. So, by looking at the coefficients of various poles of this form, where t is some kinematic variable, you can derive a series of n times n minus 3 over 2 linear equations, which relate these box coefficients to the corresponding tree amplitude. And what we found is a really remarkable particular combination of these equations, which takes the following form. So, we basically showed, so the first line just tells you which particular linear combination to look at. But the essential result is labeled equation 1. We showed that any tree amplitude, and remember that tree amplitudes can be the same in QCD or N equals 4 angles theory. So on the left-hand side, it's n equals 4 QCD. But on the right-hand side, I, of course, mean the coefficient of a one-loop box function in the n equals 4 QCD. So you can use this formula to calculate QCD tree amplitudes from n equals 4 loop amplitudes, for example. So we found that the tree amplitude could be expressed as this particular linear combination of box coefficients.

57:30 And this is nice for two reasons. short equation. The right-hand side of this equation only has n minus 3 box coefficients in it. And at least for n equals 8, we prove that there's no shorter linear combination. So you always have at least five terms appearing on the right-hand side. You can have more, of course, if you add zero in an unclever way. But five terms was the shortest you could get. And we conjecture, we haven't proven that it's probably true, that this is the equation you can get for any n. And furthermore, the particular box coefficients which appear here, they have the property that they have two massless legs. So the legs marked i and i plus 1, you hold fixed. And then you sum over all possible distributions of the other legs. They can be on the upper corner or the upper right corner or the lower right corner. And these particular box coefficients are not that complicated to calculate. If we use the recent technology of Brito, Pachazzo, and Fang, who showed that one could use generalized quadruple cuts, quadruple cuts were brief, or generalized cuts were briefly mentioned in Bern's talk, to compute these coefficients, roughly speaking, and again, I'm sure you'll hear much more detail about this in Brito's talk, but this coefficient is, roughly speaking, going to be this tree amplitude sitting in that corner times the tree amplitude sitting in the bottom form. So if you take these two ingredients together, this equation we found in this new technology, it obviously implies, and I should say, so suppose you want to use this equation upstairs to calculate an n-point tree amplitude. Then here, the most you'll ever get is an n-1 point tree amplitude. Can you please see? So that shows you that there There must exist a quadratic recurrence relation for tree amplitudes, which roughly speaking has a form like that. So, you know, within a week of having our paper, this rough expectation that there should exist this quadratic recurrence was realized in a paper by Brito Kachalzman-Feng, who had written down a precise formula for such a quadratic recurrence, which has recently, and by recently I mean yesterday, if you check SDH, been proven in collaboration with Whitton.

1:00:00 Okay, so my summary, that's basically all I wanted to say. So first of all, if there's one thing I want you to learn from this talk, it's that the connected prescription is not an integral. It's a sum over root. But more generally, so Twister-String theory has led to many fascinating insights into scattering of the tooth in the Amos theory, revealing rich mathematical structure, which has the pleasant side effects of making things easier to compute. And this structure is completely obscured in the traditional Feynman diagram calculations. So in my talk, I explain in a little bit of detail the connected prescription, the derivation of which is, I should emphasize again, the derivation of which is not 100% rigorous, but the final formula is 100% correct, we believe. to indicate, as best we currently understand, how it can be derived from the B-model, but of course it would be better to patch up that a little bit. And I've also demonstrated that there must exist a quadratic occurrence of tree amplitudes, which has indeed recently been realized. So that's it. Any questions? You described the real contrast you were expecting based on why is it a surprise that it doesn't work because the signature is really Minkowski? I mean, shouldn't there be some different reality from the Minkowski signature? Right, there should be, but in the Minkowski signature everything is a little bit more complicated in that slide. So for example, even this idea that you do the Penrose transform by half Fourier transform, that already becomes problematic in the calcium space because then you have a Fourier integral in the complex plane. So I really am doing the thing that the twister people don't like, which is trying to work in plus, plus, minus, minus signature and and then analytically continuing at the end of the day back to, but it's very possible that if one were to sit down and try to do everything completely, precisely, correctly in the correct signature from the beginning, using the correct sheet for homology, then maybe one would get the correct prescription at the end of the day. That would be very interesting.

1:02:30 But what you explained to us, it's unlikely that this can be captured by an integral under a frequency level. Right, so... So it's not more to go to the top of your head. Right, so I don't know what to... You had a framework which gave something like this in a different context. Any more questions? Thank you very much.