Topological Field Theory / 1-Loop Amplitudes in N=4 Super Yang-Mills

0:00 Maybe from the outset, of course, that no new thing will be said. I simply want to convey, in a talk designed largely to all of the power, twisted theories, some of the ideas behind some of the computers. So, it's a big subject, biggest subject. I learnt it from a nice paper of Wiccan. So, this isn't necessarily the best reference, but it's certainly very close to the spirit of what I'm going to say. Ok, so there's a paper by Witton on Manifolds and then topological field theory And it's in a conference per TV, so it's in essays on the manifolds. everything right now. So the idea is to sum up with an N equals 2, 2 super symmetric. Sigma model. So this is a model where we consider, we take a Riemann surface, sigma, a simple case is g of zero, but not necessarily g of zero, a Riemann surface, sigma, and we consider mapping it into a Calabi-Dowell manifold M. One could be more general, but we could take n to be a 3-fold column. So, we consider that from sigma into n.

2:30 So, we put coordinates, z, z bar, mod, sigma. We have coordinate x mu, x mu bar goes straight away to complex coordinates. Straight away to complex coordinates on the color of the outline of the hole. And the mass is expressed by giving x at the time. And the action for the second model is written out in this way and the obvious kinetic terms with x's. I really, in here, there ought to be a term h for the half, h, z, z, r, where h is the metric of sigma, but if we're going to conform the gauge, we write h in the way in which it's conforming flat, then the product of these two things is one. So just to simplify, we've got the tau. So there are the obvious terms involving the derivatives of x. Then there are all sorts of thermionic terms involving spinners, which are world sheet spinners, and also vectors, or take values of the cotangent bundle of m. So, phi mu, for example, takes values of the cotangent bundle of m, but is also a well-cheated symbol, similar to the sign. There are terms here, there are derivative terms here. this term contains a partial dp z bar, and it also contains Aristotle symbols to take care of. So then, for reasons to do with supersymmetry, one has to include these four Fermi terms, for the agreement tensor of M, and then

5:00 there are also these extra terms involving the auxiliary field F. F hat is short-fant for this combination. here which involves a subtle symbol. These X can be eliminated, the action involves the X quadratically, there are no derivatives of X, they could be eliminated, it is easier however to carry it on, write things in this first order form so to speak, carry the X around, because it simplifies the simplification of the transformation of this. Now, this action of another thing, I've been so damned, and I've kept track of the transformation properties of the figures. So phi plus, for example, phi mu plus, the plus indicates that this, times dz to the half, is invariant under changes of z. Similarly, phi, psi mu minus plus d z bar to the half is very much more than all the changes on C. I need you to put the plus and minus in here, but it helps, it technically helps with the foot keeping in here. So everything, if you think of a half as being equivalent to half you see that everything in the curly brackets has a z and a z-bar subscripts. Every term has a z-bar subscript. Here you see a z-bar and two pluses together. Now, I had a horrible shock as I was starting to write this up. So, in my notes, I have interchanged what's later going to be the A and B, or half of the A and B, of course. So, I've tried to unscramble that, and I believe that the conventions I want are the ones here. I reserve the right to change half the conventions and begin to try later.

7:30 later. So, the claim is that this action is invariant under these transformations. It's actually invariant on a slightly higher than the four-parameter on transformations, but I want to concentrate on these two. These two parameters were over. So, these are supersymmetry transformations. For example, x rotates on the left. The alphas and the alphas are anti-commuting. Parameters, and again I've tried to put in the pluses and minuses to make it clear how everything transforms as variables on the world shift. So, if I've done it correctly the plus is the minus is the z, to all the boundaries across the equation. this is a two-parameters worth of transformation. It's a little bit of work. It's not supposed to be obvious. It's a little bit of work that if you take these in order, say you take the alpha transformation, make this transformation, then this S is invariant. That also is another reason for including the Fs. If you eliminate the S, then it's invariant up to the total divergence. For the Brahmins, it's up to the total divergence, but if you include the S, then I think it's generally... So, we do that. And we can think of this delta, this consummation, as delta is QI, QI, QI, QI, QI, QI, QI, QI, QI, and QI generally, these kinds of issues.

10:00 So, there's some fields in this 2-2 model, so let me make a table. There's x nu, 5u plus, sin nu minus, 5u plus, and then also x nu bar, 5u bar, plus sine mu-1 minus f mu-bar mu-1. Then we do something that's called the a-twist, which is we mess it out with the spin assignments of the piece. So, x mu gets left alone. But, we now declare this 5 mu plus to be a world-sheet scalar. So it becomes 5 mu. This becomes sinu z bar, so it becomes a one-fourth of the energy, this becomes f mu z bar. That's a new bar, that's a new bar, that's a new bar, that's it. Now, that's called the A-twist, and it makes sense, at least it makes sense in this action, because, for example, the phi's always appear, possibly the derivatives, but the phi's always appear with a phi mu plus and a phi mu plus, say here and here, phi, we need to check it.

12:30 So, for example, phi nu plus phi nu by plus goes under this to phi nu, phi nu by z. So you see that, in the sense that two pluses are equivalent to the Z, the term you get is still something that you can increase in the actions of the bacteria now. and probably these x balance out. So the x mu plus minus becomes the f mu z bar and the combination that we have there So, bi kappa plus and psi lambda minus, if everything works properly, becomes a bi kappa psi lambda minus, yes, psi lambda plus. So, again, this transforms in a way that balances in a way that this transforms, this becomes an fz bar, the other one becomes an fz, and again things cut itself. So, the spin assignments have been adjusted, but in such a way that the resulting theory, the resulting action, still makes sense as something that can integrate over a signal. There's another way of doing this, as it's a p-tree. This is F2, this becomes F2, same, so I'm going to set F2, F2, so we can set F2,

15:00 Next new bar. Sign new bar. Sign new bar like that. My checks have been the same way, with a B-twist, but also with the spin scheme assignment cut away. Now, if one checks through now, another nice thing about this is that under the A twist, This phi mu plus loses its index, so this becomes i alpha, or goes to something like that, i alpha phi mu. And the alpha, the parameter which was previously had some sort of spin-up, the alpha is now escaped. Similarly, the beta becomes scales. So, under this, I should put twiddles on things or something like that, but I won't. This becomes alpha, if you want, plus beta, qb, where alpha and beta are both scales. And this being so, it makes sense now it is consistent to take alpha theta to be constant. So since the scalars, we can adjust the curve to be small, anti-q, lots. We now have the theory in which one checks from these that q a squared is zero, and that's the same thing, and the q a squared is zero. You see that easy enough because, for example, under an A transformation, so if we set B to zero, just look at the A, the transformation,

17:30 go and transform alpha, then for example, under such a transformation, x turns into phi, but phi turns into zero. And similarly, psi turns into f, but f turns into zero. So they all have this property that under double transformation, you get zero. Something turns into something else, and something else is always set up. So we have these generators, qa and qb, and we're going to consider one at a time, consider the a model and the b model separately, so you have, say, start with the a model, you have a Q that squares to zero, and therefore you cannot stop yourself thinking of a cosmology. You're going to think of this as a Q and as an ERS operator or an operator that generates cosmology. So the idea is, when we pont I, we have states I, and we want to consider states invariant under these transformations, so states sidetracked to QA with pont sine in zero, and since Q squared is zero, it's now to try and formulate a theory of the cohomology groups, So we want to identify psi with psi. That's q5 ring pi. And the theory will contain operators, O, and in order that the operators should act nicely on the cohomology groups, we want to consider operators such that Q with O vanishes, and moreover, of course, it's easy to find operators that commute with Q. One way of doing that is it generates trivial operators, and that is if the O is

20:00 is a commutator of Q with some other, what should we call a beta C. If O is itself a commutator or anti commutator of Q with C, then that's trivially zero because QA times times the anti-commitator of QA squared C is a virtual Jacobi identity this is the commmitator of QA squared C which vanishes so we'll consider such operators trivial and we identify operators So in this way we get a cohomology theory which has many fewer degrees of freedom than the original physics. That's the usual story that you go from a theory of forms down to cohomology groups. originally you had infinitely many forms, but under reasonable assumptions you would have a finite number of dimensions. So, in the path integral, In the plot interval, so we'll be concerned with, so the theory is about computing expectation Which is going to be the eggs. It's fine. It's fine. You have to use the weight of all the fields. The times is greater. There you go. That means the time arrived. It's fine. But the theory is about, the body weight is observed, in this term. And in this identical language, O acts as an exterior, sorry, Q acts as an exterior derivative. So, the functional integral of Q of something will vanish, and since these

22:30 If O is trivial, in that way, then the party will vanish. And also, because of this, we can see that this is a variant under a changing action. So if we, if you replace S by S plus Q of anything, then that also will be the value of our charge. So in this way, everything reduces to a continuous ionic set of Q homology. So, this gives a theory different from the original. This gives a theory which, if you want to give it a name, we could call it a name. It's different. It's not the same theory as we have before. It's interesting. So, it's useful for calculating certain observables of the full theory, because sometimes you can show that the observable you're interested in, in the full theory, corresponds to an observable that's reduced in. So, that's the statement that sometimes you're interested in calculating quantity, which which just depends on the chronological properties of theory, and so it can be calculated by some operation which is reminiscent of a sort of topological operation, such as a calculator. So it might reduce, say, to an intersection calculation in chronology, and that, of course, means that Jesus does, if it applies. So, this won't calculate for you all the observations, but it will calculate for you many inches, but it will calculate for you many inches, but it will be. There's a question.

25:00 Now, S can be written as Q or something, because Q is just a sort of symmetry there. So if you write S in superspace, then you write it as 2. I don't think that you can write times. I don't think so. So the claim is, certainly you can't write it as a cue of something or something about one of these cues. Certainly. Let me carry on, I'm going to give you more bits of this. Now, what you can do, at the very next line, is that you can... write parts of this theory as material. So for example, in the lay model, you can take the test, and you can write it a ventricle B to the lid. XQMu bar and here we can have XMu bar divided

27:30 Plus the cube of something. And something big is the end of the day. 2 plus 1 2 plus 1 2 plus 1 2 plus 1 2 plus 1 2 plus 1 2 plus 1 2 plus 1 OK, up to plus and minus is like it's a 2 and so I think you need to change that this is true Now these This is just the integral of the caliform over sigma. So you can write it as the integral of the caliform over sigma plus the q plus something. And so, here, in this integral, this integral of x of the K-a-form will just give you a winding number. So, if J is the E where E is the, say there's only, say H2 is one dimensional, there's only one generator, the Taylor form will be some multiple of the generator, and then this integral will give us S is minus, S is, sorry, S is pi. any winding number, t, plus 1. And so, these digits will break half as the sum of the winding numbers.

30:00 n equals bx taken over our sectors of definite finding number and the action is here, the only bit of the action that provides is e to the minus 2 pi nt that comes outside reaction is too trivial, and so it goes out, and we're just left with this product. So But, let me just say, now let me simplify some more, and that is, so we're left with a simplified integral to evaluate it, and we've got the sound. And now, it's possible to say what happens next in a number of ways, but there's a very pretty way, which I want to pay attention to, which is that this integral, you're evaluating an integral which has a symmetry, because all these operators are invariant of Q, and Q generates function. So this is like saying, so if you break off for a minute and do a quick two-dimensional example, this is like evaluating an integral in a plane where the integrand is invariant, say, on the location. So you have a cylindrical and symmetric integrating. And there what you would do is go into polar coordinates. And you would write this integral as integral vr 2 pi r r. And the 2 pi r, of course, is just the volume of the two orbits. So you have a group of multiplications. you integrate first over into four bits, and then you do the deranged integrals. And so, we want to apply this idea here. The parameters, the analogue of the beta integrals, here you have

32:30 the group parameter, beta, the analogue of the beta parameter is this parameter, alpha, here. And so first, we're going to integrate over alpha, and then there's this wonderful planet that since alpha is anti-commuting, the result of integrating over alpha is zero. So these group orbits give you zeros, and so you think the answer is zero, but the answer isn't zero, but what it is, is it's concentrated on the invariant subsets. So this way, the integral concentrates on the variable subspace On the subspace, where qx is q5, qc2 is 0, so you have to look at the table and see what it is that these equations tell you. And they tell you that pi is zero, that very very important. They tell you that d is able to mu is zero. That must be, since that's important, that comes from here. The delta phi mu tells us that d is able to mu is zero, the two-ticks is real, the two-ticks is real. I said we get into trouble with options, and we can.

35:00 Ok, so if I fix this now I'll get into trouble later on. Thank you. So, we get that condition out of here, Now, dzx is 0, and x being real, we also have dzx is 0, and there's more of the conditions on that side. So, this is a pretty plot. It tells you that the x's are all over. So, in fact, this integral concentrates on the homomorphic embeddings. So we were integrating all embeddings of the region surface into M. But the invariant embeddings are in fact the homomorphic embeddings. So the integral concentrates just on the homomorphic. So it's in this way that things reduce to a character over time incident. One of the observables...

37:30 So we want to write down operators that are closed under Q, and we know more or less what the answer is, it's going to be related to the tomology of M. So, you start by saying next omega n1 to mk, dx n1 to mk, your k4 on n. And then you consider the associated operator of omega, which is going to be omega to the to the mk Right, there's something inconsistent in what I'm doing. I'm sorry, I don't know how to fix it at the point. Chi n1 to chi nk and chi n is, I think, phi nu, sorry, nu bar. So you have a set of thermions with a real value index, chi n, which the holomorphic ones are phi nu, and the anti-holomorphic ones are the same. If I go quickly, then my inconsistencies won't be apparent. If you do it properly, then what you discover is under A, delta of chi, G0, and delta of X, so when you, if we commute this with Q, right, so commute this with QA of omega, that's like different, that's like taking the delta derivative. Then, the delta derivative, it can either act here, and gives you zero, or it acts here, and differentiates the x, and this is the equation of x, differentiates the x, and then there's a relation that says that delta x is chi, and so, in this way, you get the derivative, this time's an extra chi, and so what you get is minus o, d over delta, d over delta, d over delta, d over delta, you get the exterior derivative.

40:00 So, this says something rather interesting. This says that if omega is a closed form, so t omega is zero, then this says that Q commutes with this O of omega. And so there's a map from the Durham cohomology classes of omegas to these operators O of omegas. And so the observables of this theory are these O of omegas, and so this integral here just boils down to a classical intersection that just boils down to a concentration over the Durand's thermology group. So this involves just a classical intersection calculation of the real thermology weighted by the numbers of instruments. So that's the story for the A model. The B model is similar in theory but sort of opposite. So, the A-model, what we do is both sort of the A-model, we sort of said that S was kinetic, was the kinetic term, roughly speaking what we said was the kinetic term, plus exact. And for the B-model, it's sort of the opposite way round. So we said that this bit, if the count and everything concentrated on these first two terms, and that just turned out to be a winding number which you could pull out of the half-integral,

42:30 or some separate base, and the rest didn't matter. With the B model, the opposite is true. And so the nicest way to see that is to do some more work, which I won't do. But there are two ways of writing this sigma model. That's a nice. One of these is intrinsic to the manifold. So, we're talking about the Calabi-Yau manifold M, and we can draw a slightly different picture. So now, we could think of the, say, we could think of M, say, the quintic threefold. Doesn't matter what it is, the important thing is that there's an equation. So, there's, say, there's a CP4, and there's a quintic equation, whose zero locus gives us an atom. So, now this figure looks like P4, and in this P4, here we have N, which corresponds to an equation, W of x is 0, x being the coordinates of P4. So, you can either write the signal model intrinsically involving the metric of a thing, or you can lift this description to the people. So, what we can do is introduce a delta function, different delta of course, a delta function of x, which restricts the integral to the hypersurface w, to the hypersurface in b4. And in order to do that, you would introduce an extra coordinate, so you'd introduce an extra coordinate, dx0, in fact dx0, dx0 bar, and you'd have a term e to the i integral, x0, w, x, e is the minus sign, x0. x bar, in addition to your other terms, which would enforce this delta function.

45:00 And you want more, in fact you really want to do this in terms of superfields, because as well as having an x nought, you also want a phi nought and a psi nought and an f nought, to the finor and the f and the sinor will be the grand multipliers for these other fields that occur and will force them to lie on the hypersurface and similarly for the f. You will have to force, f is a vector, you have to force it to be tangent to the hypersurface and to do that you need a multiplier. So the story about that is that you go from 3x is 3 pi's 3 pi's 3x to 4x is 4 pi's 4 pi's and you write the whole thing in P4 where you have these extra terms. Now, which enforce these terms, plus their fermionic friends, which enforce the fact that the intervals in factories are sticking to hyperspace. And so, in here, we have this, and in here we have these terms of air cylindrical operations. like this. And what you can show is that this, once you've done this, it's a different S now, once you've made this transformation, you can show that this is exact, and this is exact, And all you're left with is this as the only non-triple piece. So in the A-model, you show that everything is in fact restricted to the k-la part of on the action, everything was restricted to the k-metric, and the complex structure, in fact, disappeared. There was no reference in the answer to the complex structure. The observables were related to the real cohomology of metaphors or mention of the complex structure.

47:30 Here, the opposite is true. You could show that all reference to the metric, which is the place where metric occurs, action in kinetic terms, so all reference to that is gone because these transformations distinguish between x and x bar. It's possible for this to be exact without this being exact, so this goes out and the only non-trivial piece that's left is just the piece that it depends homomorphically on the complex structure. And here the observables are related, so essentially in the same way as the other observables were related to the one kind of homology groups, now we get observables which are basically... New bar, you end up evaluating quantities zero P forms, not, no, they're sort of PQ forms, so we're not PQ cohomology groups, and so we get an answer that in fact depends on the conflict structure of N, but not on the chaos size of N. And then just one final comment is that if you trace through how mirror symmetry works with this, then mirror symmetry intersects with two. So the statement is, what's believed to be true is that A of N is B of the mirror. So a different value, and so A of the family will be used with B. that sort of calculations to do with the Kaler class of M turn into the calculations to do with the structure of W. Thank you.

50:00 I've got a paper copy. What's next? Oh, okay. Well, let's see. She's gone. She's gone. She's gone. Oh, no, no, no. Yeah. First I'll describe the work of ours. And then I will describe a better method that we found. And finally, I will discuss the key level of recursion relations that I already mentioned. This is work done with Freddy Cachazzo and Bill Chang, and also with Tim Witton for the group of the key level of recursion relations. We've been interested mostly in one-loop matrices and to use n equals 4 supersymmetry, for which a tree-level is not as much. At one loop, from a particle physics point of view, this is more of a practice problem use the N equals four supersymmetric contributions to populate a particular QCB bite, including

52:30 subtracting that N equals four and contributions with scalers on the nose. And there's no review of progress in these long-winded galaxies in the past ten years. As Bern mentioned, it is taught 4 QCB efforts at one point, and then we'll live to five parts. For the N equals 4, in 1994, the solution was given for MHP efforts and for non-MHP cases with six new ones. and then just last fall the results were reported for seven gluons and next Bern has also discussed already how if you try to do this computation from great vitamin diagrams you get very low expressions with far too many diagrams and it's prohibitive so you need to find something And one clever idea is that the supersymmetric amplitudes of the blue line are more dimensional constructible, which means if you understand the branch cuts and discontinuities, you can be a constructible amplitude. For any close core amplitudes, we need to understand scalar box signals. You can express the whole amplitude in your combination in terms of scalar box integrals for gradual coefficients. And this is what they are called. As you can see, we grouped the gluons cyclically ordered into four groups, sum over all those combinations. And for each of those combinations, we get what's called a scalar box integral, which is just defined as given there. And then, this relation here, with expressing the emphasis in terms of coefficients times these variables, which are known, and these coefficients, the A, B, C, D, or rational

55:00 So, I'm just going to talk about computing these coefficients, and our convention in this talk is to label these coefficients by four indices, which are the first four at each corner of the box. Okay, so first it was interesting to use cooling your ecoplanar operators in twister space. And first I could already describe how scattering amplitudes are naturally localized on certain curves and these operators help us see how that is. So we have a first order differential operator that tells us when three points are collinear and second order differential operator which tells us when PowerPoints are a complainer. And so if a linear operator annihilates an amplitude that he's supported, where a twister space and Hulons are all . And now we need to use the uniformity. The amplitudes are cost-constructible. And there are two ways this is done, and the way we do our calculation is by comparing the two. So first, you construct this integral. You take your whole amplitude, and here are two propagators in the loop, from a climate diagram. And in the integral, we're going to place these propagators in the delta function. The propagator has a principal part and then a delta function. So just take the part with the delta function. Then you get what we call a cut integral, which is contribution to the entire system. So that means we're just placing these two propagators on shell, and then essentially just taking the product, the very product, of these two on shell. Well, what unitarity cuts means is that this is equal to the discontinuity of the amplitude

57:30 across a certain branch, a corresponding branch, and in understanding of unitarity, the discontinuity of the imaginary part of the amplitude. So this is the same cut now, unitarity cut. And here I've written in terms of the expansion, in terms of box-integrals. Here are the coefficients. These are rational. We don't have any discontinuities. And then here is expressed the discontinuity of the box-integral, which you can find from the linear expressions. So we do these two calculations. We do the integral, and then we look at this. And from the integral, we see that we can find information about certain coefficients, the ones that are participating in that. And by participate in the cut, I mean that these two propagators chosen can be the cup propagators' corresponds to some of the edges of the bar. So here's the following observation. If you have a differential operator, which in particular we're going to be using full linear and full planar operators, and you acted on the cut, then if the result happens to be a rational function, then the differential operator annihilates all the coefficients that dissipate in them. And the reason for this is that the coefficients are rational, and this should say, actually, that it's not three-boxy-girls themselves, but they're discontinuities as given here. All have logarithms, and they all have special reference logarithms, and there's no way these logarithms can conspire and cancel them from the boxy-girls. They find that each won't be separately. So that's good. So then all we need to find is how to pick the operator such that the operator is acting on the cuts or rational functions. And the way to do this is by exploiting the full morphine anomaly. So, for example, here's a collinear operator acting on a vector.

1:00:00 Suppose that one of the factors, one of the two amplitudes, is . That means, as we have seen, for example, as we have talked about, that the amplitude is localized on a line. So you'd expect that the collinear operator annihilates this MHPM. But the thing is, we're still doing an integral. And there's a point in phase space at which that's not quite true. There's a delta function, which is the whole work of an integral. But the good thing about a delta function is a localized integral. So instead of just getting zero, you localize the integral and still get a rational function. So you can use this, so then you can pick an operator that would ordinarily annihilate what it would be able to use based on its cluster space work, and find that when you rational function, and therefore, it annihilates all the coefficients that are indicated in their class. Also, as a byproduct, you find it, I mean, we want to find the coefficients, but if we're finding them by using equations like the f of b equals zero, then that tells So one application was to prove that all the next-to-MHB coefficients are co-planar. For next-to-MHB amplitudes, whenever you factorize into these two pieces with a cutting I mean, unless it's zero, one of these is what you're going to be able to be, and another is going to be the next branch. MHP tree amplitudes are collinear, and next to MHP tree amplitudes are always co-planar. And you can just show that any co-planar operator can be shown to annihilate both these factors of the monomorphic anomaly, which means that it annihilates all the coefficients, all the coefficients are different. This result was also given in the paper of five neurons, which is the Dixon concept, by different approaches.

1:02:30 So we also used that previous observation to find actual coefficients, as opposed to for next to MHP amplitudes, again, in those kind of MHP factors. And it turns out that for any coefficient you want, you can find some appropriate class and some appropriate collinear operator such that you get an equation of this form, including coefficient in a train. Now, we know what these box functions are, and we can do this in a... So this is just an algebraic function for these coefficients. This is described in a paper by Pachasso. And he found a whole family of coefficients for amplitudes, with three adjacent line spaces. And for all next-to-MHG amplitudes, We described the systematic method for unpacking this equation and obtaining each coefficient separately. And we gave, as our example, all the 35 coefficients for this practical analysis. But that was still a little bit cumbersome because, as you saw, you did an equation with many coefficients coming together. what we would like, what would be most direct, is just to pick a given co-fiction to the client to ask, what is a co-fiction, and just calculate it, and I'll pass around the other ones. And so our result was good for next-to-MHB amplitudes. And so then naturally we wondered how to go to the next one. And we had four minuses. And then it wasn't so clear. But while working on this, we discovered how to target a given box to get a co-fiction. And what we found is a simple procedure in finding any probes, and this is a quadruple cut. So I've described unitarity cuts before, where you take a belt function contribution for each of the four propagators from climate dynamics. And that had a nice interpretation in terms of unitarity and theory. Well, there's no sort of generalized unitary, where you just consider all the other singularities that NASA's map will choose might have.

1:05:00 And if you want to, you can pick four propagators and replace them by their problem. And this is called reading singularities. What's good about this is that it picks out a given box uniquely, because if you specify these four groups, you already specify a given box. Also, since we're working in four dimensions, we've got four delta functions, the integral is completely localized. So it's very easy. You just have to solve the condition for the loop momentum that puts all these four propagators on shell, and then just plug it in to the product in four three-level languages, and that's it. Finally, we have to mention that for n equals four, this was approved on n equals four here, there are only box functions. So you can get everything left for your looper class. It's very easy to do. If you wanted to go to n equals 1, there are triangle diagrams and bubble diagrams for 3 and 2 sides as opposed to 4, so it's not s-trick. So this is a formula you derive. The product of the four trees on latitude, the corners of the box, subject to S, where S is the solution of the Amstel condition, which is where the term lines are the elements of the loop propagator in the four conditions. You just have to make sure now that these four equations have a simultaneous solution. And in fact, if you just try this at first naively, you run into problems at three gluon vertices where there's just one external gluon at a corner of the box. But it turns out this is only a problem if you insist on doing the whole calculation This is a square very nicely in Whitton's paper. Okay, so the three Goulin vertex is very important. We have three Goulins, and we want them all to be on the shelf.

1:07:30 condition, for example, that R squared equals zero means that P dot Q is different. Well, P dot Q, we can write in terms of our spinner products, this has two factors, so either one or the other is P zero, which means either the polomorphic or anti-polomorphic spinners are proportionate. Now, if you're working in Minkowski's signature with real momentum, lambda tilde is lambda conjugate so you can't have them all together which means you have to go so when you write down the expressions for the three gluon amplitudes they're just zero because it's a higher power but what you can do is you can take complex momenta get non-natural amplitudes or another natural way is to work in two comets for signature and lambda-tilda are real and independent. So you can have these conditions about together. In that case, this plus-plus-minus amplitude is supported where all the lambdas are proportional. And the minus-minus-plus amplitude is supported where all the lambatilas are proportional. And you can have those separately. And I guess what was fortunate for us was that since our whole path of research had run out of Wooden's paper, we were using this and it was very natural to us to work in 2,0-2 signature, and that's what we've always been doing for our numerical experiments. We've always found a convenient case that must be their own. And it turned out to be . So here's our formula again. And now, by going to an appropriate signature, or by taking complex momentum behind it, there are two solutions to these equations. I haven't written them here because they're a bit long, but I don't think we're using this one. So, as I mentioned, the thing is, if you're working in Jankowski's signature with real momenta, the thing is, it just doesn't have the same fuses in the limit. So, once you go away, then you find a clean signature. Okay, adding the two solutions together produces a rational function.

1:10:00 And, you know, this sum also includes the sum over possible internal validity assignments and the whole end equals 4 months. The 1, 2, 3 mass coefficients, these are the ones which have any degree to 1, 2, 3, 2, 2, 2, 2, 3, 2, 3, 2, 1, 2, 3, 2, 1, 2, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3. So this is an illustration, since I haven't written down the actual solution, from a long equation, S plus and S minus, the two solutions for Lucia Mementa. We gave them the terms of vectority. But they actually determined the possible holistic assignment of the cringeline vertices. So, as I was describing, they're either supported the two kinds of cringeline amplitudes and their support-wise are all the landers are proportional or all the landers are proportional. And the two solutions, S plus and S minus, determine what the possibilities are. So here's an example we can do. So here you see the four factors. It's just the amplitude at the corners of the box is MHD . So you just get more simple factors for the amplitude. And then you have to plug in the solutions for L, which I haven't done here. I mean, you can do it and you get a one-night answer. And I haven't done it here because this is one of the examples that we are executed by the whole one anomaly originally. So we just check the base of g. But here, I'm including this example too. This is a full n-blown example. A whole family of coefficients which have the property, when the boxes have the property, you've got one of the three minuses in each one, three separate quarters. And then one, three, one, versus the right here. And the point of giving this is that in the previous slide I left it L in terms of the one left it L, which I didn't plug it in to get the final answer.

1:12:30 But you can do it. It's not that hard. We found some identities that are useful to do, to get these expressions. and so here I take it out with a full anthem for this whole class. Here's an example of a Formascovision, one of the ones that originally inspired us to, Well, as I said, we were looking for an experiment, and we were trying to perfect our method. And this seemed to be . And it turned out that we could calculate all of the coefficients by the whole Morphine-Hoplin method, except for the two four-mass coefficients, which means this one and the one where every index is shifted by one. So it was just this one that was elusive, and by the quadruple cut, I mean, here you could even do this in the LCC picture, because as I mentioned, we don't have a problem unless you want to free the 100s. So we could have done this. And it turns out that the other Pong-Ascope The formula we found from the quadruple cuddle is to end the self-fisher space analysis because we understand So the example I just gave is to find this twister space like each of these lines corresponds to one of the parameters of the bar. Again, this is HB. I mean, these are not all the parameters. So, we found that it's easy to compute N equals 4 will move amplitudes from knowledge of 3-level amplitudes as well. And the interesting thing is that this particular formula inspired the recursion relation for tree-level amplitude, which I'm about to describe,

1:15:00 which, in fact, is the most compact formula for tree-level amplitude that we have, which means it makes this formula easier to use again. So a few words on how that fever occurs came about. There are infrared divergences for the end of this morning. And the infrared divergent behavior encodes the tree level. You can recover the tree level off the fever. Previously, these infrared relations were used to obtain hard-compute coefficients, we did that ourselves, or as consistency checks on coefficients that we compute as well. Because there are many equations. But you can also get further linear combinations of them, and find new compact representations of three-level amplities as linear combinations across all different combinations. And this was illustrated for Seven Bulans in his paper that learned how to do the Dixie-Costellar and discuss the program in the Dixie-Costellar. And Brian Strathley-Volovich began to use our formula for an quadruple class to discuss the possibility of recurring relations as Cedar and Strathley-Costellar. But there's still an issue here, and this is a bit funny, because these box function coefficients are computed for total line equals 4 multiple. Now, what we seem to be getting is a relation for tree-level amplitudes, in terms of tree-level amplitudes, that relies on n equals 4 superturning, which is very strange. And more than strange, it's also tedious and acute, because you'll have these contributions If you want to do an actual computation, you can put them together.

1:17:30 And it turns out you can do better. And we can get here to new recursion relations in all of the only fields. And here it is. Just a picture on the next page. So what you've got here is, well, spread the point of the formula, the form of this recursion, where you've got two trigonal latitudes by a proper gainer. What we're doing here is that you see in this sum, we're singling out two of the guan, which we call reference guan. Here we picked n and n-1. And what you have to do to use this formula is you have to shift their momenta as follows. you can see that you're really only shifting either lambda or lambda. And moreover, you do a compensating shift in this propagator, such that you still have momentum conservation in each of these terms. And all these three are still on the shelf at this time. Here is a... So here is our amplitude. These two terms correspond to the two Felicity assignments for the internal property user. And for each of these you just have n-3 diagrams. And you keep the n-1 gluons fixed over here and the n-gluons fixed over here. And then you sum over all the secret orders and pass gluons from one vertex to the other. I just took this figure from our paper and apparently said we forgot that we only have n-gluons. So the first comment I'd like to make is, is this relation equally valid for any other choice of two external gluons? I picked n-1 here and n, and it had a particular helicity relation, or plus or minus. But you can pick any two, and it can have any helicity. But there are a few reasons that I'm still choosing to write a helicity. First of all, if you choose adjacent gluons, you get the most compact expansion.

1:20:00 Otherwise, you'll have a lot more diagrams if you pick non-adjacent gluons. And you see if you're adjacent to a lot of numbers, it's right here, it's just linear. If you're non-adjacent to a lot of numbers, it's just linear. Another reason we are sticking with minus and plus here is that's a perfectly generic configuration. You can always find adjacent to a lot of the multiplicity, unless they're all the same, but we know that those are zero. And finally, this case where they have opposite multiplicity is the one for which our proof is most complete, meaning that we did it in terms of Feynman diagrams without reporting those three diagrams. Anyway, it's still nice to know for a set of reasons that you should have to do one. Okay, next, this curtain is like building an empathy from various factorization limits. We computed many previously known examples. This was before we had a proof, too. Everyone wanted to check that this was right. So we computed many previously known examples. And we found that in every case that we computed, the formula that we got just from this recursion relation is the most compact one available. In a couple cases we got a new formula that many others are distributed to those contracts and contracts available. And for that, of course, you do have to use Jason Jones. As I pointed out, before momentum is still conserved in each projects and contracts and HB documents. Finally, here, if you iterate the recurrence and take it back to its natural starting point, you can express any amplitude in terms of 3-to-1 amplitude only. You can span the whole thing. And, uh, I hope you saw Ms. Jose's talk yesterday. Perhaps there's a useful way to look at this. So what is the fundamental formulation underlying this? We speculated that there's some connection to string theory and quadric, which is node, deinstatons, and parity invariant. So now we have our two kinds of interaction. That would be interesting to study for you. So here's a proof. And for this we consider a general k and n. And that shift in momentum I was describing earlier, we now just call z. And we think of z as a complex variable.

1:22:30 And then we study the amplitude of a complex function. So here we're only taking momentum The physical amplitude is a of zero. a of z is a rational function with only simple poles, and the only possible singularities comes from those propagators, which now takes us from them. And now here when I write ij as a subscript, I mean ij that separate the elements otherwise there's no z-dependence. So because of this property, you can write the amplitude expanded in terms of its poles, plus the contribution from infinity. And the poles are just given by some of the equations there. So it would show that A of infinity equals zero. Just by tracing for the . Well, this is where the plus-min is. The plus-min is the case where you can do it. You can manage to do it from . In the other case, we use . The C is still being used. The default is . A of infinity is . So consider now a particular hole. and it comes from this diagram. So you've got B-I-J represents the sound of all this dementia as a function of Z. Here you've got K and N, the reference to the ones. This is the contribution of the diagram. Two amplitudes divided by a propagator. H is labeling the internal velocity. Yeah, I wrote the formula on the previous page. This thing is just linear, so it's pretty straightforward to extract what the contribution is from this whole. And you find the numerator becomes the product in terms of the shifted momentum. And the propagator is just what it should be, and you recover exactly the form of our recursion.

1:25:00 Here's an example of one hand. Six gluons of alternating multiplicities. So each of these terms corresponds to one in the diagram. What I've done here is I'm taking two and three to be the reference gluons. And by a bar, what the split is between the T-tree on the left. So we have three diagrams, and each diagram gives one term. The original formula that we were using before had four terms, and now we have three. And what's nice here, unexpectedly nice, is that all the symmetry in this amplitude is manifest by looking at these two terms. We didn't necessarily expect that, just because there's nothing symmetric about our recursion. And here's another example. This is not pretty straightforward. This is long. This is the 81K. So, encouraged by that, we thought that we were going to look at the 80-word case of our training. I haven't read all the terms, even if there's some symmetry here, which which is explicit in our recursion, so we didn't have to calculate explicitly, but there are three terms. So we calculated explicitly, and here they are. But again, we're interested in how much of the symmetry of this amplitude, which is log-symmetry, is with the mathematics. There are eight rotational symmetries and an overall reflection. Now, our terms are not all related by this. All except three of them were. There are three terms that, oh, I should also mention here, that there is still a current typo in the the version I'm going to take for this on one hand. It's correct. So, there are three types of terms that are all related by these symmetries.

1:27:30 But there are three that didn't fit. But there are also three that were missing. That you would expect from the symmetry. And if you just compare the sum of the three that we're missing with the three that we found, they're the same. So, although it was not fully manifestly symmetry, we were able to infer the symmetric plot. And these are the three sort of families that operate the symmetries on these three terms, except where you get the whole capital. So, to conclude, using the Holomorphic Anomaly and some of the operators of Twister Space, you can get configuration of the Twister Space structure of box function coefficients. All the box coefficients in one loop n equals four amplitudes are given by this formula in terms of tree-level amplitudes using political cuts. And finally, we have this recursion relation for tree amplitudes with a linear number of factors. That gives the most complex expression strength. I don't know. Right? Mr goes down. Yes brother? So you're helping out... And since you can formally write down the algebra, then you can write down a single word and then move out, and you need your symmetric ear.

1:30:00 For, uh, yes, for formats. I think it has to be complicated because one of the favorite solutions to real prostitutes you have to choose one of the prostitutes that you can invest in three years in this morning. I'll allow it on the shelf. You'll get an eye in there. It's in your paper. Right. Oh, it's okay. Okay. I guess you can do it. I have a question, it's a sort of general question, which I think may be helpful by my colleagues and co-workers, which is that they start to look very much like the analytic So in the bad old days there was a program that somehow all scouting attitudes were going to be determined by inconsistency. They're going to be incorporated. And this is starting to look very much like that. Does everyone have a comment? I don't know. So you know that you can express your own self-consciousness, because none is overseas, or what you say is very generally. So the amplitudes know you know their point of view, so you know that you can express your own self-consciousness, because none is overseas or what you mean. Thank you.