Introduction to Feynman Amplitudes
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Brown, Francis
In this talk, aimed at master's and Ph.D students, I will explain how to assign integrals to certain graphs representing physical processes. After discussing the standard integral representations and their underlying geometry, I will give an overview of what is presently known and not known about Feynman integrals, and indicate why they are of interest to mathematicians.
 So first of all I should say that almost everything in the first half of this talk is in standard textbooks in quantum field theory. I would explain how to assign integrals to graphs and in the second half I'll say give a sort of informal survey of what is known and what is not known about these integrals. The problem in this subject is knowing where to start. So typically a quantum field theory course will talk about the Feynman-Parth integral up to a certain point and then by analogy they'll make some noises about wick expansion, expansion in graphs and then sort of start again perturbatively and there's a sort of leap of faith that connects the two parts. So what I will do is start the story then with perturbation theory that seems as good a place as any. So as I just said we want to put graphs at the center of the story. So in perturbative quantum field theory interactions between graphs are represented, sorry not a good start, interactions between particles are represented by graphs. Always feel slightly awkward as a mathematician talking about these things. Here's an example of a graph representing two particles scattering and exchanging a photon. Here's for example another graph, the same two particles exchange a photon which spontaneously becomes an electron-positon pair and later which later annihilate the form of photon again. So every graph that we'll draw like this represents some kind of interaction between fundamental particles and the game is that to every such graph physicists associate an amplitude, i.g. which is given by some integral and it's going to be a function of certain data which are masses and momenta of particles. That's a function and the final answer that you compare with experiment obtained by summing over all graphs. And of course there are infinitely many diagrams of this type which represent the same outcome, the same observed outcome. And so one has to do something. You can only in practice some over finitely many graphs and so at this point you cross your fingers and hope that somehow this expansion converges which it doesn't. So I just, I'm not going to say anything about this, there's some very difficult unsolved problems relating to this but we have to, we run into some divergent series and we don't worry too much about that. Practice it. This all seems to work. Okay, so today to make some simplifications, G, I work in a scalar field theory and I'm going to work in Euclidean as opposed to Minkowski spacetime. So spacetime will be R to the D where the dimension will be, is a positive even integer. So it's important that it's even and so we'll work with a Euclidean norm. So X squared is the sum of the square of its components of X equals, okay. And so the challenge that I want to bear in mind throughout this talk is how we think about amplitudes. So how do we make sense of the zoo of amplitudes high G? So this won't make much sense now but once, when we've seen some examples, I want you to think about how, whether the problem is the challenges to try and find some order in this huge morass of examples. So of course I should say that calculating these, these Feynman amplitudes is big business. Nearly all the predictions for particle collider experiments are obtained by calculating amplitudes high G and comparing them with experiment. So there are, I don't know how many people but many teams around the world, banks of super computers running night and day calculating amplitudes. This is, and if you open any particle physics phenomenology paper, journal, it will be filled with huge expressions for Feynman amplitudes. So this is a very important problem and we are very far from having a mathematical theory of amplitudes. Okay, so let's actually begin. A Feynman graph, for me then, will be a graph. So a graph has a certain number, certain set of vertices, certain set of internal edges and a set of external edges. Okay, so this is a graph that satisfies the usual rules and the internal edges, each edge is connected to two vertices. So this is just a subset of pairs of vertices. The external edges on the other hand are connected to a single vertex. We represent that by a map from external edges to the vertex to which they are attached. So here's an example of a Feynman graph. Just a triangle has three vertices, three internal edges, three external edges. For convenience I'll number the edges for later use, one, two, three. And the external edges represent particles which I will typically represent as incoming but not always in fact. Okay, so Feynman graph also has some extra data, namely the data of... No, it's not an infusion. Ah, yes, thank you. So the graph is not necessarily oriented. No, but I'm good at orientating just a minute to write down the integral, Feynman integral but it won't depend on the orientation. So the graph also has an extra data of a mass, a particle mass Me which is a real number. If you're a paranoid you can take it to be positive but as we'll see it's only the squares of the masses that come into the formulae. So there's a mass for every internal edge is to every external edge we have a momentum, qi which is a d vector for every external edge. So here let's put q2 here, q1 and q3. So the convention will be almost always, sometimes I might change this, the convention will always be to have momenta coming inwards if you reverse the... If you want to represent an outgoing particle that's the same as an ingoing particle with a negative momentum. So we can always assume that all the particles are incoming if we allow the qi's to be negative. So that's totally fine. No vectors, yes. So that's a very small detail. The next remark is that this data is subject to momentum conservation and momentum conservation sum qi equals zero for all external edges. Okay. So now we want to write down an integral associated to such a Feynman graph. So Feynman graph is the graph with this data. And to do that we need some integration variables. So to every edge, e, internal edge, actually, yep, to every internal edge we assign a momentum variable, momentum variable ke, which is a d vector. Now sort of temporarily I just, I'm going to choose an orientation on the graph just to write down the integral but it won't depend on this. The integral won't depend on this orientation. So choose an orientation, you'll see why in a minute. And let me write p e to be either ke if e is an internal edge and the incoming momentum if e is external edge. And then the Feynman integral in momentum space form is, so this depends on the external momentum qi and the particle masses me. So it's an integral of real space of d dimensions, product of internal edges, default integral. So remember ke is a d vector, so integrate, this is the component by component. And here in the denominator we have ke squared plus me squared times the product of all vertices and here a delta function p e where the sum is over all edges both internal and external which meet the vertex v. So this makes sense because you've chosen the orientation, the meaning of this, this, oh thank you very much, r d to, let me give that a name now actually, n g, thank you, equals the number of e n g. Yeah, thank you, so this is r d times n d. So this factor here represents, if you like, momentum conservation at every vertex so that the sum of the momenta at every vertex sums to zero. Okay, so this of course, so this is the momentum space Feynman integral as it's most commonly seen. Of course, it doesn't necessarily make sense, it may diverge. Sorry, a small d, oh this is delta, delta function. D, sorry d as in, yeah, delta is the, is the Dirac delta function. Okay, so yeah, mathematicians tend to prefer a, sorry, yeah, you're assuming that at each vertex your convention is such that for momentum are either in constant. No, I'm going to chew, well you could do that, you could do that, but I, what I've, I'll show you in an example just a minute, but I said we choose an orientation on g and then when we assign a momentum to every edge, so we'll have, let's do it, we'll have k1, k2 and k3 and so at this vertex we'll get the equation k2 plus k3 equals q1 and then at this vertex we'll get k1 plus q3 equals k2 and at this vertex we'll get k1 plus k3 equals q2. So, so this delta function means integrate over, over this region. Okay. So it's sum of plus minus, yeah. I found it's because if you choose a orientation. Yeah, I, I, it's clear what it means, right? Yeah. You change the variation, you change the, you can do the opposite. Yeah, it doesn't change. No, but street is the, what? This, this integral doesn't make, I have to say this integral doesn't make a huge amount of sense in the first place and that's why the first thing I'm going to do is change it into an integral which doesn't make sense. So this is what you see in, in physics textbooks I've kept it exactly as it's written. And exactly as I'm going to say is that, that in fact in the, in the good old days it was common practice to use parametric representations, that's classical and in fact it's after a long hiatus it's now coming back into fashion. So classical and modern approaches use the parametric form, the parametric representation which is much more satisfactory from many points of view. So I'm not going to derive that from this and all should start to make sense. Yeah. Do you need to introduce coupling constant for each vertex? Do I need to introduce what? Coupling constant. Yeah, but I'm working in a very, in principle yes. Absolutely. But I'm working in a very simplified, this is a simplified theory where I can just take it to be one. I could put a parameter. You can, yeah. But I'm only going to consider one graph at a time so we know how many vertices it has. Doesn't make much difference, yeah. Of course in a, in a genuine quantum field theory the, the true final rules are much, much more complicated. But at the end I'll say something very briefly about how that changes. Just to make the idea as a, I mean the, the sum of the idea as a formal power series. Yeah, I think of it as a formal power series exactly here. Okay, so to get from this to parametric form, we first apply the Schringer trick. So this is the identity 1 over x equals into the goal of e to the minus alpha x dx, I had d alpha. So this is valid for x positive. So the idea is to introduce a new variable alpha e for every edge, every internal edge. And then write, and in this integral replace this factor here with integral from zero to infinity e to the minus. Okay so it's something, it looks as if we're going in the wrong direction instead of doing the integrals, we're adding more integrals. But as we'll see where there's a lot to be gained from this. Why am I kidding? So what happens to this integral, it becomes, it will now be the integral, a bunch of integrals from zero to infinity for every alpha parameter assigned to every edge. R to the d and g x minus sum e k, oops, k squared plus me squared alpha e. And then the same stuff, that's the same as before, I won't write it out again. And product d alpha e. Right, so we made the integral more complicated apparently by introducing new variables. But now the point is that we can actually do the momentum integrals now. So this involves Gauss's formula for this integral over the reels, e to the minus pi x squared dx equals one. So you can multiply integrals of this form together and to get a higher dimensional version. And in general, the d dimensional analog of Gauss's integral is as follows. If we have q of x is minus x transpose A x plus 2 b where, so here x is an rd, A is a symmetric positive definite matrix, and b is any vector in r to the d. So this is a nice quadratic form. Then the integral over r to the d, e to the q x dx. Oh, sorry, thank you very much, b scalar x. Let me put b transpose, yeah, thank you, sorry. Of course, if we add a constant, this will come up later, if we add a constant plus c here, it's just going to factor straight out of the integral because it does not depend on x. Okay, yeah, thank you for that. So the upshot is that this is pi to the d over 2 over the square root of the determinant of A times e to the b transpose A inverse b. Right, so, and this can easily be deduced from the previous integral by diagonalizing the matrix A. And it'll break into a product of d copies of this integral after a change of variables. Right, so now the idea is to apply this identity to the integral at the top and to actually do the ke integrals in this expression in the above. And the upshot will be that we will obtain a new integral only in, there'll no longer be any k's, there'll only be the alpha's, the masses and the qi, well only the alpha's, the m's and the q's. Yeah. Okay, so this can be done in general, but it's quite tedious. So I will just do illustrate on one example and then state the general answer. Okay, so let's do a very simple example. So let's take this graph. We assign momentum k1 and k2 to these two internal edges. So if you prefer we can, if I should stick to my conventions, minus q. So the delta function in the integral is momentum conservation at each vertex that reduces, they're both the same condition, so it reduces to the equation k1 plus k2 equals q. So if you like that's our domain of integration. So now I'm going to write down the exponent, so the exponent in that integral, so it was minus k squared e plus m e squared alpha e. So what is it in this case? It is alpha 1 k1 squared plus m1 squared plus alpha 2. Okay, so the first thing to do is to substitute in the momentum conservation condition. So this is alpha 1 k1 squared plus alpha 2 q minus k1 squared, this time on this term. And then the mass terms are going to accumulate here. So then the first thing you have to do is to take this part and complete the square. Of course you can always do this. In this case it gives the main coefficient is alpha 1 plus alpha 2 and then times k1 minus q alpha 2 alpha 2 squared. And now you need to add q squared, sorry this is a plus, let me write, get rid of the minus, that's better, over alpha 1 plus alpha 2. Okay, so we've completed the square. So let's write this. So the next thing is to change variables, or the shift variables. I can call this k. So let's put k equals k1 minus q alpha 2 over alpha 1 plus alpha 2 and then this is just alpha 1 plus alpha 2 k squared plus the rest. And changing variable, of course the we're integrating over r to the d, so the integral is translation invariant. And from this representation we can then plug in Gauss's formula and write down the answer. So what we get is, so this is just the term in the exponential, we have exponential of minus this integrated over k. All the stuff on the right just does not depend on k at all, it's just going to factor out of the integral. And we just have to integrate e to the minus alpha 1 plus alpha 2 k squared. And that's easy using this formula. So I'll just write down the answer. We get pi to the d over 2 integral from 0 to infinity d alpha 1 to alpha 2. The alpha 1 plus alpha 2 comes downstairs. And then all this stuff over here is unchanged, it's factored out of the integral in the first place. Okay. So in the general case, so you can, this works in general. You can complete the square and put in the integrals. It's a bit of bookkeeping to work out what the answer is. At the end you get a, here you get the determinant of a certain matrix. And then you have to apply something called a matrix tree theorem to express that determinant in terms of some graph-threatic quantities. But that's rather straightforward. And the final answer is that the, we get that the integral is some trivial factors which I'm going to ignore. From powers of pi and so on times the integral from 0 to infinity. So here the product is over all internal edges. Then here you get a certain polynomial, or psi g. And then you get exponential minus pi g q over psi g minus the sum. So that is the parametric form of the Feynman integral. And I now need to explain to you what these psi and phi are. So psi g and phi g q are what are called semantic polynomials. Okay. Graph polynomials. So first of all psi g. So psi g is a polynomial first discovered by Kirchhoff I think in 1853. So it's a polynomial in just the alpha parameters. And depends in no way on the external momentum or the masses. No dependence on qi. It's an important fact that it has integer coefficients. But that won't play any role in what I say today. So here's a formula for it. Actually there are many different ways to interpret this graph polynomial. I'll just give one. It is the sum of spanning trees in the graph. And then you take the product over the edge variables not in each spanning tree. So t is a spanning tree. So t is a spanning tree. t is connected. Simply connected. So that's it. That means that it's a tree. And spanning means that it meets every vertex of the graph g. Or perhaps I should say, I should actually say my apologies. Here g is, I'm assuming g is connected. Throughout this I'm assuming g is connected. I forgot to say that. Otherwise this is going to vanish. So let's do an example. Actually, the best of probably to do the example I did earlier. So let's do this example. It's very simple. There are only two spanning trees. There's one. It's a tree which meets every vertex. And there's two. And so the graph polynomial is the product of all the edges not in each spanning tree. So there's only one in this case. And it gives alpha two plus alpha one. And alpha two plus alpha one is indeed the term that we found in the denominator here. You should say that tree doesn't contain an external edge. Yes, yes. I'm sure you're right. So I put no dependence on the QIs. Yeah, spanning tree. So think of t as a subset of the internal edges. Yeah, so the external momentum plate, the external legs play no role in this at all. We just ignore them. Perhaps I should do another example. So this graph which we had earlier, the spanning trees are one, three. It's going to be linear again. That's not a good example. I don't know. Let's do this example for later on. So what are the spanning trees here? It's the same one, two, and three. And so psi g is alpha two, alpha three plus alpha one plus alpha one, alpha two. Okay. So that's one of the terms in the integral. And then we want the second semantic polynomial, phi g. Now now it does depend on the external momenta that I represent just by the letter Q. So it's a polynomial again with integer coefficients in the alphas. And in fact in dot products scale products of all the external momenta. So the formula is that's minus the sum of spanning two trees. And so now you take Q to the t one. So the dot product is the Euclidean dot product. And then here you have the product of the edges not in t alpha e. So what is the spanning two tree? So t spanning two tree if and only if. So it's a forest with two components. So t has exactly two components t one and t two. It's simply connected. Each component is simply connected. And it spans. So spans means that it meets every vertex of the graph. Oh yeah. And then so what is Q to the t one? Q to the t two. So Q to the t i equals total the sum of all momenta entering the tree t i total momentum entering t i. Okay. So an example. So let's do the example we did earlier.. So spanning two trees. Well there's only one. It's this. It's the only spanning two tree t one and t two. And so we get minus Q dot minus Q times the product of all the absent edges. That's alpha one alpha two. And that is exactly Q squared alpha one alpha two. Which was exactly hopefully the term up there minus alpha one alpha two Q squared. It's exactly the same polynomial here. No I didn't get minus sign. Sorry? It's something like this. No no no because it's polynomial skewed. Yeah I put a minus sign here. Maybe you didn't see it here. There's a minus sign here. Yeah yeah. Yeah so the two minus is cancelled to give a plus. Okay let's do a more interesting example. The triangle graph we had earlier. No no I think it's fine. This is the definition of phi. But this is minus phi. So here so here. So this is minus phi. So it's minus alpha one alpha two Q squared from that blackboard over psi which is alpha one plus alpha two. It's exactly this term. Two times minus is plus yeah. So I think it's great. So there's a minus here. So often I don't put this in the formula. I write it differently. There's a minus here and I think it all works. And there's a minus here and then there's another minus here. Three minuses. There is a minus here on the left. No no so this dot product always produces a minus. So this is a minus. Because Q2 power Q2 is minus Q2 power Q1. Yeah thank you. Yeah I was just about to say that. So Qt one plus Qt two is zero by momentum conservation. So the way I normally write this is plus Qt one squared and that equals plus Qt two squared. But if you write it this way it's not clear it's symmetric in t1 and t2. But if you write it this way it's obviously symmetric. But the price you pay is that you need a minus sign. But it's not this minus sign isn't that this is completely positive. This has entirely positive coefficients. So this minus gets absorbed in the stock. Absolutely yeah. Well otherwise t would have a single connected component. It's going to be connected. So one I can't remember how I labeled this but up to rotation it's this. So what are the spanning two trees? Here they are. So let me label them Q1. This is edge one Q3 Q2. That's one spanning tree. Q2 two. So what are the two trees? So what are the two trees? So what are the two trees? Q3 Q1 and then the last one Q3. Okay so then so what do we do? We take the dot product of the moment entering the first tree. So it's minus Q1 dot Q2 plus Q3 and we take the product of the edges which are not present. So that's our first tree. So what do we do? We take the moment entering the first tree. So it's minus Q1 dot Q2 plus Q3 and we take the product of the edges which are not present. So that's alpha two alpha three. Same here. So minus Q2 dot Q1 plus Q3 alpha one alpha three and the last one gives minus Q3 dot Q1 plus Q2 alpha one alpha two. And then of course by momentum conservation which I'll write up here. Q1 plus Q2 plus Q3 is zero. So this is just equal to Q1 squared alpha two alpha three. Q2 squared alpha one alpha three plus Q3 squared alpha one alpha two. And of course in the final answer all the coefficients are positive. That's actually very important. And that's because we're working in Euclidean space and that's what guarantees that these integrals make sense somewhere for at least some values of Q and M. And so these changes of variable formula and so on will be justified in some region of Q and M and for some value of D not necessary which may be a real number as I'll explain in a minute. So that's very important that these polynomials should have all positive coefficients. Okay so oh yeah let me write. So this is not a standard notation but I like to package this, give it a symbol and consider this polynomial phi GQ plus sum M E squared alpha E. So that's not a standard notation. And now. Thank you very much. Yeah thank you. Thank you. Thank you. Thank you very much. Yeah absolutely. So psi G is homogeneous of degree HG. So let me write HG is what this is called the number of loops of the graph. So it's the dimension of H1 of the graph. And on the other hand so phi GQ and hence psi GQ comma M is the number of loops of the graph. So that's the number of loops of the graph. So that's the number of loops of the graph of H1 of the graph. And on the other hand so phi GQ and hence psi GQ comma M are both homogeneous of course in the alphas they depend on other things as well but of degree exactly one more HG plus one. So that follows from some combinatorial property of spanning trees. It's very easy to check. Okay. Right so now we want to get to projective integrals. We want to get actually so there are two things you can do at this stage. We have these Feynman integrals in these form as an exponential integral. You can either do what's called a minimal subtraction which I somehow prefer but you can also do a little bit of a subtraction. So we have a little bit of a subtraction. We have a little bit of a subtraction which we can do as an exponential integral. We can do a minimal subtraction which I somehow prefer but we can also do a dimensional regularization. And the latter is slightly quicker and gets us straight to the arta so I'm going to do that. And so the And so what the basic idea is that we want to get rid of this exponential factor. So there's one that can be done by doing one more integration. So recall that we had i integral from 0 to infinity. So let me rewrite it in the new notation, x over psi g. Now change variables. So write alpha e equals lambda beta e, where the sum of the beta e is equal to 1. Lambda is a positive real number. And then this differential form here, product d alpha e can be written. So we call that ng is the number of internal edges, d lambda wedge omega g, where omega g equals sum minus 1 to the i. Okay, so that's a simple manipulation. And so this can be rewritten in the region where the beta is all positive. And of course, don't forget that the sum is equal to 1 of omega g over psi g. So now I view psi g as a function of the betas. It's homogeneous. So but I'm not going to write beta e in here because it will clutter the integral. So this clearly means it's viewed as a function of the betas instead of the alphas. And here we get e to the minus lambda. So from here over here, because of the degree of the psi polynomial is exactly one more than the degree of psi, when you scale all the alpha parameters by lambda, it will produce exactly one lambda coming out because the degree here is exactly one more than the degree there. So we get lambda to the minus, to the ng minus d over 2 hg, d lambda over lambda. So this integral here we can do, and it's the final step, the final integral that we need to do. And it produces a gamma value, gamma of what I'm going to call SDG times psi g over psi g qm to the minus SDG, where, so SDG is essentially this quantity which pops out, which is minus the number of edges plus d over 2 times the number of loops. And this is called, this is going to measure something about the convergence of this integral. And so it is called the superficial degree of divergence of the graph. It can be positive or negative. So essentially when it's positive, this integral is going to diverge, and when it's negative it's going to converge. And it's called, the graph is called logarithmic, overall logarithmic divergent when it equals zero. Yeah, so everything's a function of beta, but I don't want to write beta everywhere because it will become very cluttered. But from now on, let me write the next integral and everything is a function of beta. So the conclusion of all this is that we can write our Feynman integral up to maybe some trivial factors in the form. So we're going to acquire a gamma function, an integral over this hyperplane. But in fact, you notice that the integrand that comes out is actually homogeneous of degrees zero. So it's slightly nicer to write this as an integral over a projective space. So this is the sign of the argument of gamma. You want the opposite side. You want the superficial degree of divergence. No, I think this is correct. So it's superficial degree of divergence. So this is very positive when the graph is divergent. So this should be negative. Yes, it is. It's not what you wrote. Because you wrote gamma of SDG. Oh, sorry. Yeah, yeah. Thank you. Yeah, this is correct and that is not correct. Yeah, yeah. Okay, thank you. Yeah, so it measures how much it diverges. When it's very big, it means we're going to get a pole here, if that's the point. So we get psi g d over 2 psi g over psi g qm to the power of. So I didn't put superficial degree of divergence here because I want to stare at it in a minute. So this is a projective integral. It's not all projective space, but it's simple. Oh, yeah, sorry. Thank you. Sigma equals the set beta 1. Thank you. So this is a projective integral. And it makes sense if you plug in these degrees here, the integrand is homogeneous of degree 0. Okay. Okay, so now the problem is, of course, that this can have a pole. This gamma factor can have a pole, particularly when the superficial degree of divergence is non-negative. So very briefly, the idea of dimensional regularization. So the idea is that you replace, and this is what is done the most commonly in practice, though I'm not a huge fan myself, is to replace d with d minus epsilon, some small positive epsilon, and perform a Taylor expansion in epsilon. And so what physicists do in practice is they compute the coefficients of epsilon to the i, where i can be both positive and negative. Excuse me. So you can view it as an integral of this hyperplane if you prefer. But the point is that you could have chosen any other hyperplane. We didn't have to choose wherever it is, the sum of the beta i's equals 1. You could choose any other hyperplane. It still works. And this, if you like, is a nice way to codify that. What does a projective integral mean? It means you restrict to an affine chart and calculate the integral on that affine piece. And that's exactly the same as choosing a hyperplane at infinity and integrating over affine real space. So for physicists, this is a way to think of this in integral over r to the ng with a delta function given by this equation. And it contains the fact that you can replace this with any hyperplane, any non-trivial hyperplane. Okay, so that's the end of the first half. As I said, almost everything is in textbooks. So I was informed that instead of having a pause, I should just press on. And so now in the second half, I want to give a sort of panorama of what we know about these integrals. In this way, dimensional regularization doesn't need to invent a defensive geometry in the end. No, absolutely not. No, no, no. Absolutely not. No, absolutely not. That's the point. That's the key point. Okay, so a panorama. Okay, so what do we know about fine integrals? Of course, it's impossible for me to give a completely exhaustive list, but let me try to give a picture of the landscape. So the first remark is that the general one loop diagram. Oh, sorry, let me make a remark before I do this remark. So all these, this procedure, the Schringer trick in Gaussian integrals and all this works in much greater generality. It works for a gauge theory. And the upshot is that we can always write fine integrals in the general form. So what will happen is if you do this, is you get higher, you maybe get different powers of psi and psi in the denominator. So here A and B are z integers and P alpha is some polynomial in the alpha E's. Maybe it has coefficients in some Clifford algebra. And the whole integrand is homogeneous, such that this is homogeneous at a degree zero. So that's the general form of a parametric form. Of course, the P could be extremely complicated. So back to these scalar diagrams. The one loop diagram can be expressed using a single function, well, two functions rather. The dialog with them is defined by the following sum. n of n squared converges for mod x less than or equal to one and has an analytic continuation to C, so multi-valued function, and the logarithm. So any graph like this, you can always express the amplitude using just l2 and log, where the x, it'll have a complicated expression in these with different arguments x, but where the x will be some function of the masses and momenta in some possibly quite complicated way. In any case, there's a single function essentially that describes all one loop amplitudes. It was the initial discovery of Feynman, Long-Long with Lueckum. Feynman introduced the dialog with Lueckum and Lueckum. I'm skeptical. We can discuss this. So the general two-loop diagram, two-loop amplitude is not known. So there isn't a function that you can look up in the canon of special functions. It means, this also means that the singularities and the analytic behavior of the general two-loop amplitude is not understood. But does it reduce the functions like two variables or three variables? So here there's a theorem that if you take an arbitrary large number of external legs, it can always reduce to at most a certain number of legs. In two-loop, the corresponding result is not known. But I think it should follow immediately if one knew something about, from Hodge theory, about weights. So it should follow. No, no, no. Yeah. So it typically involves square roots of a quadratic form in the masses and momenta. So even in a simple example like the triangle, it's, in general, it's a long formula. Let's just not write it down. It's quite complicated. It's not something. So two-loop amplitude is not known in general. There's a special case that's become very fashionable at the moment that's actually been studied for many years. So it's called the Sunset or for optimists, it's the sunrise diagram. So this is a particular two-loop diagram. So of course, faced with these difficulties, what people typically do is make some assumption that certain masses are zero and so on and so forth. This has a long history and it was recently solved for the general case with all non-zero masses and momenta. And the key names are Bloch, Vanhover and McCurr and Adams, Bogner and Weintzeel. So they have very different approaches, but these have the sort of final word on the status of this integral. I recommend, if you want to know about history, you can look at these papers. So I forgot to say something. The dialog is an iterated integral. So I can write it in a very simple integral form, zero less than t1 less than t2 less than x. So let me take x between zero and one for the sake of argument. Then it can be written as an integral just involving two differential forms. And so this is an example of an iterated integral on c minus zero and one. And for some bizarre reason, it's not fully understood, this space seems to, the functions on this space, describes a vast number of Feynman amplitudes for reasons which are not completely clear. This in fact is an exception, but what it does involve are iterated, or rather twice, in twice iterated elliptic integrals on some family of elliptic curves. And one way to write down, to get at the amplitude is by average, is this function and in z. So here q has nothing to do with this q here. So q is, let me call it q naught, q nothing, is nothing to do with the q here. This is some function, it doesn't make sense, it's for many reasons, first of all, we're averaging a multi-valued function, and it's going to have singularities when n is either very big or very small depending on this. So you have to regularize it in some way, but if you do that you get some function, and this function essentially describes the amplitude here completely, where q naught here will be some function of the masses and q. So I won't say more about that. Next there is a, so this is the first sort of difficult case. A huge class of amplitudes are no, and are expressible, using a finite class of functions called multiple polylogarithms, about which I think we'll hear a lot more this afternoon. So part of my job is to set up the afternoon speakers. So here's the definition of the multiple polylogarithm, some generalization of the dialog with them. So it depends on r variables, and it is the sum 1 less than or equal to kr. x1 to the k1, xr to the kr. So that defines some analytic function on the region where the x i's, let's say, are strictly less than 1. And by analytic continuation it extends to a multivalued function on c to the r minus, and then you have to take out a certain class of diagonals, given by the successive products. So this is actually c, c minus, c minus the origin to the r minus, this class of diagonals. And this is nothing other than the moduli space of r minus 3 points on a sphere. So we think of this as r minus 3 particles on a sphere. And it's not entirely clear why a priori. Oh yeah, thank you very much. So this is just consecutive. Oh yeah, r plus 3, thank you. Absolutely, the other points are 0, 1 and infinity. Okay. So for some reason this particular moduli space, which in the case r equals 1, is just c minus 0, 1. It plays a special role, and it's not understood what the analog should be to describe all higher order Feynman integrals, if there indeed is such a thing. Okay, so something which is used a lot in physics, which I'll emphasize, but it is not valid in general at all. So I'm going to explain what it is and then say why one shouldn't extrapolate beyond its range of legitimacy, is that first of all they are iterated integrals. There's no reason to think that general Feynman integrals should be iterated integrals. And there's something very special that they have a weight grading. So what that means is that the weight of, you can attach a weight to such a function, it's the sum of the indices, and the property of this weight is that if you differentiate the multiple polyloga of a certain weight, and there's an explicit formula for this, I just am not going to write it now, but I'm going to write this as a linear combination of functions of lower weight, again of multiple polyloga of strictly lower weight, in fact weight one less. So there's some recursive structure in these functions, and this weight, if you keep differentiating in a certain way you get to weight zero, and that reflects the fact that iterated integrals. But I really want to emphasize because this is often expected, in physics literature it's said that this holds in general, it's not true, this is a very special property of this particular class of functions. Next, so the sunrise is not of this type, it's something else, there should exist a class. I will leave that to the speakers this afternoon, I believe that that will be discussed in detail. So yeah, I'm just really setting up the afternoon talks. There should exist, so there is a well-defined class that we understand, infinite families of graphs of this type. Beyond that, there should exist a class of multiple elliptic type, such that IG is multiple elliptic. So this is a sort of nascent theory, one of two papers defining what such functions are. So this is not known, it's not been worked out at all. So the first class beyond the polylogarithmic is completely unknown, say for one or two examples like this, maybe that and one other. And... No, so it would mean something like this lead to, it would mean sort of forced it to be elliptic on the Jacobi uniformization of the elliptic curve by averaging over Q. You can do the same thing for multiple polylogarithms, it's slightly more subtle, and that's a good, that's a perfectly valid definition of a multiple elliptic polylogarithm. Alternatively, you could look at iterated integrals on the universal elliptic curve. That's a whole class of functions, that's in some sense the genus one analog of this story, and they should describe another large class of diagrams, and that class is not known. And so the boundary, again the boundary where the polylogarithmic ends, is not known. Something is known about it, but it's not known precisely. Okay, so now let me focus on some more specific examples. I'm sure I have interest in number theorists. So I have half an hour, okay. So typically these examples are very difficult because there are lots of masses and momenta, and the amplitude is a function of many complex variables, and that's where a lot of the difficulty comes from. But often in practice you don't need such generality, and the things get easier if you make some restrictions on the momenta and the masses. So the other extreme we have single scale processes. So these are where i,g,q,m is a function of a single variable. So it could be, for example, all the masses are zero, and there's just one incoming momentum, or it could be where you just have one non-trivial mass. And what typically happens is that the dependence on that variable is completely trivial. And it will just factor out of the integral, out of the integrand. So what you're already saying is that the amplitudes are giving numbers instead of functions. So it factors out. And the coefficient is an interesting number. So what can we say about the numbers that come out of quantum field theories? So here's one family of examples, which are in some sense the most exotic. So it's the analog of this sunrise. It's diametrically opposed to the polylogarithmic class where we feel we understand quite a lot. So these are the banana graphs, which have a lot of symmetry and are very interesting. And these were first studied by David Broadhurst. So let me do one example in complete detail. So we have the same picture like this. And what we're going to do is impose, first of all, let me set all the masses equal to the same mass. Any number of masses, or just three? Just one here. So I'm going to put all masses equal. And I'm going to call it equal to n. Now, with the n1 and 2 and 3, all of them, n4 and 5. The bananas will be this family of graphs here. That's a banana. I don't know why it's called a banana. It looks nothing like a banana. Yeah, sunrise, banana. So the graph polynomial is this, alpha 1, alpha 2, plus alpha 1, alpha 3, plus alpha 2, alpha 3. The second graph polynomial, so these are all good exercises to revise the definitions of the graph polynomials I gave earlier. This is q squared, alpha 1, alpha 2, alpha 3. And this polynomial psi gives this equation, q squared, alpha 1, alpha 2, alpha 3, plus m squared, alpha 1, plus alpha 2, plus alpha 3, alpha 1, alpha 2, plus alpha 1, alpha 3, plus alpha 2, alpha 3, which is very nice and symmetric. And defines a beautiful family of qubits. Okay, so the next thing now. Yes. So the next thing, so now Brorow says, okay, we let Nallet work in d equals two dimensions. And put m squared. So this is not quite Euclidean space, I've been slightly dishonest here, but we'll put m squared equals q squared equals, so m squared equals minus q squared equals one. And now, so now we set S, so using Broadhurst's notation so you can look at his paper and directly compare with his conjectures. The integral is omega g. So if you look at the number of loops and number of edges, what happens is the psi drops out of the integral. So we get psi g. And so this m here in his notation is this graph where m minus one is the number of internal edges. So I don't know why he shifted the indices, but I'm going to stick with his indexing to make it more consistent. So if you do stick q squared equals minus m squared, of course the m squared just factors out of the integral. So we can just set m squared equals to one. Yeah, m is not m. Thank you, yes, that's unfortunate. Yeah, of course the m counting number of edges is not the same. But m is, that m is one, so that's okay. So S bar four is what we've just computed. It's this integral, it's omega three. It's this projective integral alpha one plus alpha three, alpha two plus alpha three, very beautiful integral. And to work it out, it's a projective integral. So to work it out, we just work on some affine chart. So let's put alpha three equal to one, and it's just d alpha one d alpha two, alpha one plus alpha two, one plus alpha one, one plus alpha two. So it's a very nice, completely convergent little integral there. And here's a list of what's known about these integrals. Now S bar three is two pi over three squared three. S bar four, which is this one, is pi squared over four. S bar five is conjectually, I'm not sure if this has been proved actually since I looked this up, but it should be four pi square root of 15 times the L function of a modular form. So this is some Dirichlet series that's beloved to number theorists. S bar six is experimentally given by 48 Riemann's ETH of two times special value of the L function of a modular form, where F four is a weight form modular form. It's an explicit product of dedicated ETH of values. So this is something, I won't dwell on this too long, but it's something that number theorists care very much about. And it's not at all understood how and if at all this pattern should continue. Okay, and in general it's not known. After this, there's no conjecture for what these integrals should be. So you see that some very simple family of diagrams is producing some numbers relating to many active topics in number theory and we really do not understand how to make sense of this. Okay, so that's in some sense the worst possible family of examples. Now let me turn to a completely different family of single scale examples. So these will be residues in massless fight the fourth theory. And these are the examples that got a lot of us mathematicians interested. So now D is four again and all masses will be zero. So I'll say that a graph is in fight the four turns out to be equivalent to the fact that all the vertices of the graph of degree four. So here's an example. So what we have in this situation is that the number of edges is twice the loop number. So the superficial degree of divergence is zero and we're in the divergent case. So this quantity we had earlier has a pole here and we're in trouble. But what we're doing is we're looking at the coefficient of one over epsilon in the final integral in dimmer egg. And it turns out that this does not depend on the external momenta. It's just a number. Yeah, so that's exactly what I was going to say. It's the residue. It's called a residue and by total abuse of notation. I'm going to call it I again because I'm running out of letters. So I'll just call it I. It's not the actual formula to call it the residue. And it's just given by the integral of omega g of psi g squared. So in the previous case, the integrals were special in that the psi dropped out of the integrand. These are special because the psi drops out of the integrand. So these integrals don't always converge. A criterion due to Weinberg is that this integral converges if and only if g is what is called primitive. He didn't certainly use that language. Primitive refers to some hop-valje structure that came much later. The condition is that for all strict subgraphs of g, the superficial degree of divergence of the subgraph should be negative. And that means that the number of edges should be at least twice the number of loops. So the first example we saw already. So I leave you to work it out from the definitions that this is alpha 1 to alpha 1 over the graph polynomial squared. So these graphs should have 4 over 4 external edges, yes, of course. Yeah, it does. No, in general, alpha is the original degree. Yeah, yeah, yeah, it will always have 4 external edges, absolutely. It will always have 4 external edges. That's the consequence of this and this. My oil is forming up. Absolutely. So to work this out, we work on an affine chart, let's say alpha 1 equals 1. So that's alpha 2 goes from 0 to infinity d alpha 2 over 1 plus alpha 2 squared, and that is the number 1. So the residue of this graph is just 1. That's easy enough. And after that, they get a lot more difficult as we'll probably hear more about this afternoon. So let me just give the first few examples, some of the first few examples. So the next graph which satisfies all these conditions is this graph. So here the residue is 6 times zeta of 3. This gives 20 zeta of 5. This here gives 36 zeta of 3 squared, which is the square of this amplitude. So there's some identity looking, several identities looking behind the scenes. And finally, this graph with 6 loops, which took many years to first compute numerically, which is done by Bort-Hurston-Kreimer. This with those external legs here. But it can now be done symbolically, because this is now known exactly. And it is 27 over 5 zeta 5, 3 plus 45 over 4. So what are these numbers here? And they're given by a nested sum. So this converges when nr is greater than or equal to 2, and these are called multiple zeta values. So this graph, this 6 zeta 3 turns up in many different quantum field theories. Essentially for the reason that was on this board before, that you can always write a general amplitude by putting numerators in. And the sort of numbers, you only get a finite family of numbers coming out. And it's this same zeta 3 that shows up in quantum chromodynamics, n equals 4 Cp young mills. It shows up again and again. So there's nothing particularly specific. This theory seems very specific, but actually it gives a good indication of the quantities that are describing more general quantum field, or different quantum field theories. Okay, so multiple zeta values have a kind of weight, but now what I'm saying is totally abusive. You want to assign a weight to a number, you can't do that. It doesn't necessarily make any sense. And in fact, we don't know whether assigning this integer to these real numbers, MZVs, makes any sense at all. But conjecturally, these multiple zeta values are graded by the weight. That means that there are no relations, no algebraic relations with rational coefficients between them, between multiple zeta values of different weights. Okay, so I'll come back to that later. You should say that the weight is half of the weight of the unit. Yeah, I'll say that later as well. But so this is how physicists understand the notion of weight. And it's used a lot in modern as a sort of guide to various things in the physics literature. So it is known that there is an infinite class of graphs, of such graphs in phi 4 primitive, such that IG, the residue, is multiple zeta value, or linear combination of multiple zeta values. So there's a criterion. So there's a common at all criterion. I won't say much about that. There are eight loops. There are so-called modular graphs, whose amplitudes, whose residues we do not expect to be of this type. So somehow these numbers describe a vast class of these amplitudes. And something goes badly wrong at eight loops. So we can get some graphs whose residues are expected to be something else. I don't know what to call them, but for want of a better name, they should be multiple modular values, certain types of numbers relating to modular forms. They're absolutely not of this class. Another comment worth making is that not every multiple zeta value seems to occur. So that means if you take the vector space generated by the residues of these graphs, they don't seem to fill the space of multiple zeta values. There's some very specific subspace with some very interesting properties. And finally, there's only one family of graphs which we actually know the amplitudes. And these are the zigzag graphs. So you take a bunch of triangles like this and you connect them from head to toe. And so zigzag five in this case, the index counts the number of loops. So theorem I proved with Oliver Schnetz, which was a conjecture by Bordersen Kramer for a long time, gives a formula for this amplitude. So let me write it down because I think there's some interesting corollaries. So it depends on the parity of n. So this family of graphs gives every odd value of the Riemann zeta function. So we know in particular that all the odd zeta values definitely occur as amplitudes. On this you can deduce that the products of odd zeta values occur. But unfortunately, there is no other family of graphs whose amplitudes are known to all orders in this theory. And there is no family whose amplitude is even conjectured. So there's not much is known. And here you mean it's a vacuum diagram, it's not a residue. So from now on, i always means residue. So in this section, so I've been a bit sloppy here, in this section when I'm talking about residues, I'm calling the residue i, which is an unfortunate abusive notation. But it has no q and m in there, so I suppose. D equals 4, yes, D equals 4 in this here. You can have more in there, 5 to the 4. I'm being sloppy and they don't play any role because of this comment here. The coefficient of one of epsilon does not depend on the external momenta, so I can just ignore it. So this is for external legs, and this is closed graphs, this degree 4, which is 2 upon vectors. And probably 6 does depend on each. So there are two things you can do. You can do what you just said. Or you can also look at two loop functions like this in 5-4 theory. So here you put 0 coming in here, for example, and you have just a single momentum coming in. And the amplitude of this is some trivial dependence on q times the integral where you close up the external legs of a vacuum diagram. So these families of integrals are actually computing, if you take any integral like this, and graph and you break a leg, it's computing the amplitude of a graph like this. And it's also computing the residue of the corresponding 4-point function. So these graphs are encoding lots of integrals at the same time. So here's a conjecture that's an interesting challenge for a young person. It's in my paper with Schnetz. And so let G be a primitive graph of the type we're looking at in 5-4. So of course it satisfies n equals 2Hg in this section with Hg loops. Then we conjecture, in fact, that the amplitude is strictly less than the zigzag graph. So the zigzag is somehow the most, is the biggest possible Feynman amplitude in the theory. It's somehow reminiscent of volumes in hyperbolic geometry where you measure some complexity. These are pure numbers. So this is a positive number. So we conjecture that this graph is the biggest, and if this were true, it would imply a bound. It would imply that I of G over 4 to the H is at most asymptotically, by applying Sterling's formula to this, 2 over root 4 pi Hg to minus 3 over 2. So it gives you some bound on the size of Feynman amplitudes. And of course this is the first step in trying to understand whether you have the convergence properties or the resumability properties of your divergent series. And I think this conjecture is far better, far stronger than what can be proved at the moment. Of course you need to understand all the divergent graphs and the renormalization. But I thought that was a curious feature of this theory. And what it's tempted to try to show that doing elementary operations on the graphs somehow always increases the amplitude or something. There may be something to explore there. Sorry, but why would this bound be important then? Well if you want to understand, if you have the sum of all graphs, you have some divergent series, you need to know very precisely how many graphs there are and how big, how fast they're growing. I meant that usually this is removed by some renormalization, right? Yes, no. So these residues are contributions to the beta function which are independent of all renormalization schemes. So these numbers, no matter how you renormalize, they will always be in there. Of course there will be other graphs which have sub-divergences that need to be renormalized. And there I do not know of a good, I mean there's choices, but I don't know of an analog of this sort of thing. But if you guessed all graphs of this nature, it would tell you something about the radius of convergence. Okay, so some final remarks in the last five minutes. It should be rather open-ended. So I want to talk about this notion of weight that's been creeping into the picture. So the majority of the known amplitudes, which are, so for example, the polylogs and the MZVs, which is, forms the bulk of what we know about at the moment, they had a grading, a notion of weight. So this grading is highly conjectural in the case of multiple zeta values. And we can ask where on earth this weight comes from and if it's got anything to do with physics. So recall that a V, a vector space over Q is graded if and only if it admits the action of a group, where so lambda in Q star, lambda acting on V is lambda to the N, V, if and only if V of degree N. So we can think of this grading as a group action. And this suggests, among other things as well, suggests the possible action of a group on amplitudes. Which I probably doesn't make much sense. So here's an important comment. So there should be a weight as well, but the weight cannot be a grading. If there is a notion of weight on all amplitudes, it certainly can't be a grading. We should be careful of this. And it will be a filtration. It will be an increasing filtration. And if we stick with this, the normalization of the weights we've used up to now, which is the physicist's one, then we must expect half integer weights. So this is a fact that's well known to algebraic geometries, but I don't think it's been fully assimilated into the physics community. So what we do in mathematics, we tend to do is multiply all the weights by two, as Cati pointed out earlier. So what is a conjecture then? So conjecturally, this is a highly conjectural, there should exist a very large algebraic group of matrices. I should really say pro algebraic, so it's like an infinite, a group of infinite matrices. C, which acts on the space of generalized amplitudes, which is the space which certainly contains, let's call it F, F of Feynman or something. So this is the vector space over which field I won't say, of regularized, because these integrals are possibly divergent, but these are generalized, generalized integrals of the form of the general shape, some numerator, some a and b integers, and p a polynomial, which is homogeneous such that the whole integrand is homogeneous to degrees zero. So in some sense, this is like a vector space of all possible amplitudes you'd ever want to consider in any quantum field theory. Yeah, thank you, yes, absolutely, that's a very good point. So this is the space of all amplitudes, and there should be a weight filtration, not a grading. So this is the space of amplitudes of weight at most something. So C acts on F, and the action of C should preserve, preserved by the action of this group. There should be a huge group of symmetries acting on all possible amplitudes. So C would be called a cosmic Galois group. You can include any other integrals such as jump here, what's special about these guys? There's a very, very small subclass of integrals, and what was surprising? For any collection of integrals, you can do that. Absolutely, you can certainly do that. In all of the conjecture, it's kind of conjecture whatever, nor emotives. No, but if you have to replace, not actual numbers, it's a transcendence. Sure, sure, sure. Sure, this is... I agree, you can do this. The point, which I won't have time to mention, but the point is that amplitudes such as these you're finding are closed under the group. It's totally extraordinary. Yeah, absolutely extraordinary. The reason why the Galois conjugates of some crazy number like this should still be an amplitude. This seems to be the case. So in some sense, it's somehow the periods of some operad in the category of motives. It's a very strong condition. It's like the material fundamental group are the periods of a group in the category of motives. It has some very, very strong... So they're very special formally for the action. And here there's some sort of operad structure that's motivic. And so this is a useful concept. It really tells you something about amplitudes. So it says that the action of this group on an amplitude will be expressible in terms of smaller amplitudes in some controllable way. That means the product will tell you. Yeah, yeah, yeah. So as I've said, it's an empty statement if set up appropriately. But it will be a very powerful tool to study precisely this recursive structure. So this notion of cosmic Galois group is due to Pierre Cartier. And really I say this because they were the abstract of one of the talks that was mentioned of a co-product. So let me explain where this comes from. So we should have this group acting on the space of 5 min equals the weight n. So one theorem, for example, is that this space is finite dimensional. So this is extraordinarily strong fact you have infinitely many graphs, infinitely many possible integrands, but this vector space is finite dimensional. That's some magical feature of these integrals. If you pick a random family, that's just not true. So quickly, the dual of an action is a co-action. So if OC is the ring of functions, and I'll finish here on this group, then the group law is equivalent to a co-product on the functions. So I think this may come up again this afternoon. And this action here, star is equivalent to a co-action. So f goes to f times OC, and the game is to compute this co-action on graphs and to use that to get information about amplitudes. And so that's one way, in this challenge I mentioned at the beginning, of how do we think about all amplitudes and how do we put some structure in this, all this information. So the only way to do that is to use this action of a group or a co-action, as I think we'll hear more about this afternoon. So I'll run slightly, so I'll stop exactly there. Any questions? Do you have an example of the general action on a special class of just size? Yeah, so this doesn't make sense on numbers because you run into transcendence conjectures. We don't know that z to five could be rational, so we don't know that it makes sense. But indeed in this family, the conjecture due to Panzer and Schnitz is that this space is closed under this Galois action. So the group will send this amplitude to a linear combination of this and this. So it will map it, so this will generate a representation of the group which involves z to five and z to three, which are two smaller diagrams. So the fact that you're allowed to have z to five three is allowed to appear at this place and only in this place because you previously had a z to three and a z to five in the right places before. You're not allowed to have a z to three times z to two because that would give a z to two and there's no graph giving z to two. So it somehow rigidifies, you can see on these examples exactly how the group is acting or conjectural. I have a question about the non-comitative graph. Yes. You said that when the graph is non-comitative, the residue, the integral giving the residue is infinite? Yes, so you need to, yeah, it diverges and you have to renormalize it. So I didn't have time to explain. So we did an overall regularization for the overall divergence because the superficial degree of divergence was zero. So that's a regularization, but all the subgraphs which have superficial degree of divergence zero need to be regularizing some sense as well and that needs to be done consistently, so that's renormalization. So there's a way to do that. I didn't have time to explain that unfortunately. One more question? Alright. Thank you.
 So first of all I should say that almost everything in the first half of this talk is in standard  textbooks in quantum field theory. I would explain how to assign integrals to graphs  and in the second half I'll say give a sort of informal survey of what is known and what  is not known about these integrals. The problem in this subject is knowing where to start.  So typically a quantum field theory course will talk about the Feynman-Parth integral  up to a certain point and then by analogy they'll make some noises about wick expansion,  expansion in graphs and then sort of start again perturbatively and there's a sort of  leap of faith that connects the two parts. So what I will do is start the story then  with perturbation theory that seems as good a place as any. So as I just said we want  to put graphs at the center of the story. So in perturbative quantum field theory interactions  between graphs are represented, sorry not a good start, interactions between particles  are represented by graphs. Always feel slightly awkward as a mathematician talking about these  things. Here's an example of a graph representing two particles scattering and exchanging a  photon. Here's for example another graph, the same two particles exchange a photon which  spontaneously becomes an electron-positon pair and later which later annihilate the  form of photon again. So every graph that we'll draw like this represents some kind  of interaction between fundamental particles and the game is that to every such graph physicists  associate an amplitude, i.g. which is given by some integral and it's going to be a function  of certain data which are masses and momenta of particles. That's a function and the final  answer that you compare with experiment obtained by summing over all graphs. And of course  there are infinitely many diagrams of this type which represent the same outcome, the  same observed outcome. And so one has to do something. You can only in practice some  over finitely many graphs and so at this point you cross your fingers and hope that  somehow this expansion converges which it doesn't. So I just, I'm not going to say anything  about this, there's some very difficult unsolved problems relating to this but we have to,  we run into some divergent series and we don't worry too much about that. Practice it. This  all seems to work. Okay, so today to make some simplifications, G, I work in a scalar  field theory and I'm going to work in Euclidean as opposed to Minkowski spacetime. So spacetime  will be R to the D where the dimension will be, is a positive even integer. So it's important  that it's even and so we'll work with a Euclidean norm. So X squared is the sum of the square  of its components of X equals, okay. And so the challenge that I want to bear in mind throughout  this talk is how we think about amplitudes. So how do we make sense of the zoo of amplitudes  high G? So this won't make much sense now but once, when we've seen some examples, I  want you to think about how, whether the problem is the challenges to try and find some order  in this huge morass of examples. So of course I should say that calculating these, these  Feynman amplitudes is big business. Nearly all the predictions for particle collider  experiments are obtained by calculating amplitudes high G and comparing them with experiment.  So there are, I don't know how many people but many teams around the world, banks of  super computers running night and day calculating amplitudes. This is, and if you open any particle  physics phenomenology paper, journal, it will be filled with huge expressions for Feynman  amplitudes. So this is a very important problem and we are very far from having a mathematical  theory of amplitudes. Okay, so let's actually begin. A Feynman graph, for me then, will  be a graph. So a graph has a certain number, certain set of vertices, certain set of internal  edges and a set of external edges. Okay, so this is a graph that satisfies the usual rules  and the internal edges, each edge is connected to two vertices. So this is just a subset  of pairs of vertices. The external edges on the other hand are connected to a single vertex.  We represent that by a map from external edges to the vertex to which they are attached.  So here's an example of a Feynman graph. Just a triangle has three vertices, three internal  edges, three external edges. For convenience I'll number the edges for later use, one,  two, three. And the external edges represent particles which I will typically represent  as incoming but not always in fact. Okay, so Feynman graph also has some extra data,  namely the data of... No, it's not an infusion. Ah, yes, thank you.  So the graph is not necessarily oriented. No, but I'm good at orientating just a minute  to write down the integral, Feynman integral but it won't depend on the orientation. So  the graph also has an extra data of a mass, a particle mass Me which is a real number.  If you're a paranoid you can take it to be positive but as we'll see it's only the squares  of the masses that come into the formulae. So there's a mass for every internal edge  is to every external edge we have a momentum, qi which is a d vector for every external  edge. So here let's put q2 here, q1 and q3. So the convention will be almost always,  sometimes I might change this, the convention will always be to have momenta coming inwards  if you reverse the... If you want to represent an outgoing particle that's the same as an  ingoing particle with a negative momentum. So we can always assume that all the particles  are incoming if we allow the qi's to be negative. So that's totally fine. No vectors, yes. So  that's a very small detail. The next remark is that this data is subject to momentum conservation  and momentum conservation sum qi equals zero for all external edges. Okay. So now we want  to write down an integral associated to such a Feynman graph. So Feynman graph is the graph  with this data. And to do that we need some integration variables. So to every edge, e,  internal edge, actually, yep, to every internal edge we assign a momentum variable, momentum  variable ke, which is a d vector. Now sort of temporarily I just, I'm going to choose  an orientation on the graph just to write down the integral but it won't depend on this.  The integral won't depend on this orientation. So choose an orientation, you'll see why in  a minute. And let me write p e to be either ke if e is an internal edge and the incoming  momentum if e is external edge. And then the Feynman integral in momentum space form is,  so this depends on the external momentum qi and the particle masses me. So it's an integral  of real space of d dimensions, product of internal edges, default integral. So remember  ke is a d vector, so integrate, this is the component by component. And here in the denominator  we have ke squared plus me squared times the product of all vertices and here a delta function  p e where the sum is over all edges both internal and external which meet the vertex v. So this  makes sense because you've chosen the orientation, the meaning of this, this, oh thank you very  much, r d to, let me give that a name now actually, n g, thank you, equals the number  of e n g. Yeah, thank you, so this is r d times n d. So this factor here represents,  if you like, momentum conservation at every vertex so that the sum of the momenta at every  vertex sums to zero. Okay, so this of course, so this is the momentum space Feynman integral  as it's most commonly seen. Of course, it doesn't necessarily make sense, it may diverge.  Sorry, a small d, oh this is delta, delta function. D, sorry d as in, yeah, delta is  the, is the Dirac delta function. Okay, so yeah, mathematicians tend to prefer a, sorry,  yeah, you're assuming that at each vertex your convention is such that for momentum  are either in constant. No, I'm going to chew, well you could do that, you could do that,  but I, what I've, I'll show you in an example just a minute, but I said we choose an orientation  on g and then when we assign a momentum to every edge, so we'll have, let's do it, we'll  have k1, k2 and k3 and so at this vertex we'll get the equation k2 plus k3 equals q1 and  then at this vertex we'll get k1 plus q3 equals k2 and at this vertex we'll get k1 plus k3  equals q2. So, so this delta function means integrate over, over this region.  Okay. So it's sum of plus minus, yeah. I found it's because if you choose a orientation.  Yeah, I, I, it's clear what it means, right? Yeah.  You change the variation, you change the, you can do the opposite. Yeah, it doesn't change.  No, but street is the, what? This, this integral doesn't make, I have to say this integral doesn't  make a huge amount of sense in the first place and that's why the first thing I'm going to  do is change it into an integral which doesn't make sense. So this is what you see in, in  physics textbooks I've kept it exactly as it's written. And exactly as I'm going to say  is that, that in fact in the, in the good old days it was common practice to use parametric  representations, that's classical and in fact it's after a long hiatus it's now coming back  into fashion. So classical and modern approaches use the parametric form, the parametric representation  which is much more satisfactory from many points of view. So I'm not going to derive  that from this and all should start to make sense. Yeah.  Do you need to introduce coupling constant for each vertex? Do I need to introduce what?  Coupling constant. Yeah, but I'm working in a very, in principle  yes. Absolutely. But I'm working in a very simplified, this is a simplified theory where  I can just take it to be one. I could put a parameter. You can, yeah. But I'm only going  to consider one graph at a time so we know how many vertices it has. Doesn't make much  difference, yeah. Of course in a, in a genuine quantum field theory the, the true final rules  are much, much more complicated. But at the end I'll say something very briefly about  how that changes. Just to make the idea as a, I mean the, the  sum of the idea as a formal power series. Yeah, I think of it as a formal power series  exactly here. Okay, so to get from this to parametric form, we first apply the Schringer  trick. So this is the identity 1 over x equals into the goal of e to the minus alpha x dx,  I had d alpha. So this is valid for x positive. So the idea is to introduce a new variable  alpha e for every edge, every internal edge. And then write, and in this integral replace  this factor here with integral from zero to infinity e to the minus.  Okay so it's something, it looks as if we're going in the wrong direction instead of doing  the integrals, we're adding more integrals. But as we'll see where there's a lot to be  gained from this. Why am I kidding? So what happens to this integral, it becomes, it will  now be the integral, a bunch of integrals from zero to infinity for every alpha parameter  assigned to every edge. R to the d and g x minus sum e k, oops, k squared plus me squared  alpha e. And then the same stuff, that's the same as before, I won't write it out again.  And product d alpha e. Right, so we made the integral more complicated apparently by introducing  new variables. But now the point is that we can actually do the momentum integrals now.  So this involves Gauss's formula for this integral over the reels, e to the minus pi  x squared dx equals one. So you can multiply integrals of this form together and to get  a higher dimensional version. And in general, the d dimensional analog of Gauss's integral  is as follows. If we have q of x is minus x transpose A x plus 2 b where, so here x  is an rd, A is a symmetric positive definite matrix, and b is any vector in r to the d.  So this is a nice quadratic form. Then the integral over r to the d, e to the q x dx.  Oh, sorry, thank you very much, b scalar x. Let me put b transpose, yeah, thank you,  sorry. Of course, if we add a constant, this will come up later, if we add a constant plus  c here, it's just going to factor straight out of the integral because it does not depend  on x. Okay, yeah, thank you for that. So the upshot is that this is pi to the d over 2  over the square root of the determinant of A times e to the b transpose A inverse b.  Right, so, and this can easily be deduced from the previous integral by diagonalizing  the matrix A. And it'll break into a product of d copies of this integral after a change  of variables. Right, so now the idea is to apply this identity to the integral at the  top and to actually do the ke integrals in this expression in the above. And the upshot  will be that we will obtain a new integral only in, there'll no longer be any k's, there'll  only be the alpha's, the masses and the qi, well only the alpha's, the m's and the q's.  Yeah. Okay, so this can be done in general, but it's quite tedious. So I will just do  illustrate on one example and then state the general answer. Okay, so let's do a very  simple example. So let's take  this graph. We assign momentum k1 and k2 to these two internal edges. So if you prefer  we can, if I should stick to my conventions, minus q. So the delta function in the integral  is momentum conservation at each vertex that reduces, they're both the same condition,  so it reduces to the equation k1 plus k2 equals q. So if you like that's our domain  of integration. So now I'm going to write down the exponent, so the exponent in that  integral, so it was minus k squared e plus m e squared alpha e. So what is it in this  case? It is alpha 1 k1 squared plus m1 squared plus alpha 2. Okay, so the first thing to  do is to substitute in the momentum conservation condition. So this is alpha 1 k1 squared plus  alpha 2 q minus k1 squared, this time on this term. And then the mass terms are going to  accumulate here. So then the first thing you have to do is to take this part and complete  the square. Of course you can always do this. In this case it gives the main coefficient  is alpha 1 plus alpha 2 and then times k1 minus q alpha 2 alpha 2 squared. And now you  need to add q squared, sorry this is a plus, let me write, get rid of the minus, that's  better, over alpha 1 plus alpha 2. Okay, so we've completed the square. So let's write  this. So the next thing is to change variables, or the shift variables. I can call this k.  So let's put k equals k1 minus q alpha 2 over alpha 1 plus alpha 2 and then this is  just alpha 1 plus alpha 2 k squared plus the rest. And changing variable, of course the  we're integrating over r to the d, so the integral is translation invariant. And from  this representation we can then plug in Gauss's formula and write down the answer.  So what we get is, so this is just the term in the exponential, we have exponential of  minus this integrated over k. All the stuff on the right just does not depend on k at  all, it's just going to factor out of the integral. And we just have to integrate e to  the minus alpha 1 plus alpha 2 k squared. And that's easy using this formula. So I'll  just write down the answer. We get pi to the d over 2 integral from 0 to infinity d alpha  1 to alpha 2. The alpha 1 plus alpha 2 comes downstairs. And then all this stuff over here  is unchanged, it's factored out of the integral in the first place. Okay. So in the general  case, so you can, this works in general. You can complete the square and put in the integrals.  It's a bit of bookkeeping to work out what the answer is. At the end you get a, here  you get the determinant of a certain matrix. And then you have to apply something called  a matrix tree theorem to express that determinant in terms of some graph-threatic quantities.  But that's rather straightforward. And the final answer is that the, we get that the  integral is some trivial factors which I'm going to ignore. From powers of pi and so  on times the integral from 0 to infinity. So here the product is over all internal edges.  Then here you get a certain polynomial, or psi g. And then you get exponential minus  pi g q over psi g minus the sum. So that is the parametric form of the Feynman integral.  And I now need to explain to you what these psi and phi are. So psi g and phi g q are  what are called semantic polynomials.  Okay. Graph polynomials.  So first of all psi g. So psi g is a polynomial first discovered by Kirchhoff I think in 1853.  So it's a polynomial in just the alpha parameters. And depends in no way on the external momentum  or the masses. No dependence on qi. It's an important fact that it has integer coefficients.  But that won't play any role in what I say today. So here's a formula for it. Actually  there are many different ways to interpret this graph polynomial. I'll just give one.  It is the sum of spanning trees in the graph. And then you take the product over the edge  variables not in each spanning tree. So t is a spanning tree. So t is a spanning tree.  t is connected. Simply connected. So that's it. That means that it's a tree. And spanning  means that it meets every vertex of the graph g. Or perhaps I should say, I should actually  say my apologies. Here g is, I'm assuming g is connected. Throughout this I'm assuming  g is connected. I forgot to say that. Otherwise this is going to vanish. So let's do an example.  Actually, the best of probably to do the example I did earlier. So let's do this example.  It's very simple. There are only two spanning trees. There's one. It's a tree which meets  every vertex. And there's two. And so the graph polynomial is the product of all the  edges not in each spanning tree. So there's only one in this case. And it gives alpha  two plus alpha one. And alpha two plus alpha one is indeed the term that we found in the  denominator here. You should say that tree doesn't contain an external edge. Yes, yes.  I'm sure you're right. So I put no dependence on the QIs. Yeah, spanning tree. So think  of t as a subset of the internal edges. Yeah, so the external momentum plate, the external  legs play no role in this at all. We just ignore them. Perhaps I should do another example.  So this graph which we had earlier, the spanning trees are one, three. It's going to be linear  again. That's not a good example. I don't know. Let's do this example for later on.  So what are the spanning trees here? It's the same one, two, and three. And so psi g  is alpha two, alpha three plus alpha one plus alpha one, alpha two. Okay. So that's one of  the terms in the integral. And then we want the second semantic polynomial, phi g. Now  now it does depend on the external momenta that I represent just by the letter Q. So  it's a polynomial again with integer coefficients in the alphas. And in fact in dot products  scale products of all the external momenta. So the formula is that's minus the sum of  spanning two trees. And so now you take Q to the t one. So the dot product is the Euclidean  dot product. And then here you have the product of the edges not in t alpha e. So what is  the spanning two tree? So t spanning two tree if and only if. So it's a forest with two  components. So t has exactly two components t one and t two. It's simply connected. Each  component is simply connected. And it spans. So spans means that it meets every vertex  of the graph. Oh yeah. And then so what is Q to the t one? Q to the t two. So Q to the  t i equals total the sum of all momenta entering the tree t i total momentum entering t i.  Okay. So an example. So let's do the example we did earlier.  So spanning two trees. Well there's only one. It's this. It's the only spanning  two tree t one and t two. And so we get minus Q dot minus Q times the product of all the  absent edges. That's alpha one alpha two. And that is exactly Q squared alpha one alpha  two. Which was exactly hopefully the term up there minus alpha one alpha two Q squared.  It's exactly the same polynomial here. No I didn't get minus sign. Sorry? It's something  like this. No no no because it's polynomial skewed. Yeah I put a minus sign here. Maybe  you didn't see it here. There's a minus sign here. Yeah yeah. Yeah so the two minus is  cancelled to give a plus. Okay let's do a more interesting example. The triangle graph  we had earlier. No no I think it's fine. This is the definition of phi. But this is minus  phi. So here so here. So this is minus phi. So it's minus alpha one alpha two Q squared  from that blackboard over psi which is alpha one plus alpha two. It's exactly this term.  Two times minus is plus yeah. So I think it's great. So there's a minus here. So often  I don't put this in the formula. I write it differently. There's a minus here and I think  it all works. And there's a minus here and then there's another minus here. Three minuses.  There is a minus here on the left. No no so this dot product always produces a minus.  So this is a minus. Because Q2 power Q2 is minus Q2 power Q1. Yeah thank you. Yeah I  was just about to say that. So Qt one plus Qt two is zero by momentum conservation. So  the way I normally write this is plus Qt one squared and that equals plus Qt two squared.  But if you write it this way it's not clear it's symmetric in t1 and t2. But if you write  it this way it's obviously symmetric. But the price you pay is that you need a minus sign.  But it's not this minus sign isn't that this is completely positive. This has entirely  positive coefficients. So this minus gets absorbed in the stock. Absolutely yeah. Well  otherwise t would have a single connected component. It's going to be connected. So one I can't  remember how I labeled this but up to rotation it's this. So what are the spanning two trees?  Here they are. So let me label them Q1. This is edge one Q3 Q2. That's one spanning tree.  Q2 two. So what are the two trees? So what are the two trees? So what are the two trees?  Q3 Q1 and then the last one Q3. Okay so then so what do we do? We take the dot product  of the moment entering the first tree. So it's minus Q1 dot Q2 plus Q3 and we take the  product of the edges which are not present. So that's our first tree. So what do we do?  We take the moment entering the first tree. So it's minus Q1 dot Q2 plus Q3 and we take  the product of the edges which are not present. So that's alpha two alpha three. Same here.  So minus Q2 dot Q1 plus Q3 alpha one alpha three and the last one gives minus Q3 dot  Q1 plus Q2 alpha one alpha two. And then of course by momentum conservation which I'll  write up here. Q1 plus Q2 plus Q3 is zero. So this is just equal to Q1 squared alpha  two alpha three. Q2 squared alpha one alpha three plus Q3 squared alpha one alpha two.  And of course in the final answer all the coefficients are positive. That's actually very important.  And that's because we're working in Euclidean space and that's what guarantees that these  integrals make sense somewhere for at least some values of Q and M. And so these changes  of variable formula and so on will be justified in some region of Q and M and for some value  of D not necessary which may be a real number as I'll explain in a minute. So that's very  important that these polynomials should have all positive coefficients. Okay so oh yeah  let me write. So this is not a standard notation but I like to package this, give it a symbol  and consider this polynomial phi GQ plus sum M E squared alpha E. So that's not a standard  notation. And now. Thank you very much. Yeah thank you. Thank you. Thank you. Thank you  very much. Yeah absolutely. So psi G is homogeneous of degree HG. So let me write HG is what this  is called the number of loops of the graph. So it's the dimension of H1 of the graph.  And on the other hand so phi GQ and hence psi GQ comma M is the number of loops of the  graph. So that's the number of loops of the graph. So that's the number of loops of the  graph of H1 of the graph. And on the other hand so phi GQ and hence psi GQ comma M are  both homogeneous of course in the alphas they depend on other things as well but of  degree exactly one more HG plus one. So that follows from some combinatorial property of  spanning trees. It's very easy to check. Okay. Right so now we want to get to projective  integrals. We want to get actually so there are two things you can do at this stage. We  have these Feynman integrals in these form as an exponential integral. You can either  do what's called a minimal subtraction which I somehow prefer but you can also do a little  bit of a subtraction. So we have a little bit of a subtraction. We have a little bit  of a subtraction which we can do as an exponential integral. We can do a minimal subtraction  which I somehow prefer but we can also do a dimensional regularization. And the latter  is slightly quicker and gets us straight to the arta so I'm going to do that. And so the  And so what the basic idea is that we want to get rid of this exponential factor.  So there's one that can be done by doing one more integration.  So recall that we had i integral from 0 to infinity.  So let me rewrite it in the new notation, x over psi g.  Now change variables.  So write alpha e equals lambda beta e, where the sum of the beta e is equal to 1.  Lambda is a positive real number. And then this differential form here, product d alpha e can be written.  So we call that ng is the number of internal edges, d lambda wedge omega g, where omega g equals sum minus 1 to the i.  Okay, so that's a simple manipulation.  And so this can be rewritten in the region where the beta is all positive.  And of course, don't forget that the sum is equal to 1 of omega g over psi g.  So now I view psi g as a function of the betas.  It's homogeneous.  So but I'm not going to write beta e in here because it will clutter the integral.  So this clearly means it's viewed as a function of the betas instead of the alphas.  And here we get e to the minus lambda.  So from here over here, because of the degree of the psi polynomial is exactly one more than the degree of psi,  when you scale all the alpha parameters by lambda, it will produce exactly one lambda coming out  because the degree here is exactly one more than the degree there.  So we get lambda to the minus, to the ng minus d over 2 hg, d lambda over lambda.  So this integral here we can do, and it's the final step, the final integral that we need to do.  And it produces a gamma value, gamma of what I'm going to call SDG times psi g over psi g qm to the minus SDG,  where, so SDG is essentially this quantity which pops out,  which is minus the number of edges plus d over 2 times the number of loops.  And this is called, this is going to measure something about the convergence of this integral.  And so it is called the superficial degree of divergence of the graph.  It can be positive or negative.  So essentially when it's positive, this integral is going to diverge, and when it's negative it's going to converge.  And it's called, the graph is called logarithmic, overall logarithmic divergent when it equals zero.  Yeah, so everything's a function of beta, but I don't want to write beta everywhere because it will become very cluttered.  But from now on, let me write the next integral and everything is a function of beta.  So the conclusion of all this is that we can write our Feynman integral up to maybe some trivial factors in the form.  So we're going to acquire a gamma function, an integral over this hyperplane.  But in fact, you notice that the integrand that comes out is actually homogeneous of degrees zero.  So it's slightly nicer to write this as an integral over a projective space.  So this is the sign of the argument of gamma.  You want the opposite side.  You want the superficial degree of divergence.  No, I think this is correct.  So it's superficial degree of divergence.  So this is very positive when the graph is divergent.  So this should be negative.  Yes, it is.  It's not what you wrote.  Because you wrote gamma of SDG.  Oh, sorry. Yeah, yeah. Thank you.  Yeah, this is correct and that is not correct.  Yeah, yeah. Okay, thank you.  Yeah, so it measures how much it diverges.  When it's very big, it means we're going to get a pole here, if that's the point.  So we get psi g d over 2 psi g over psi g qm to the power of.  So I didn't put superficial degree of divergence here because I want to stare at it in a minute.  So this is a projective integral.  It's not all projective space, but it's simple.  Oh, yeah, sorry. Thank you.  Sigma equals the set beta 1.  Thank you.  So this is a projective integral.  And it makes sense if you plug in these degrees here, the integrand is homogeneous of degree 0.  Okay.  Okay, so now the problem is, of course, that this can have a pole.  This gamma factor can have a pole, particularly when the superficial degree of divergence is non-negative.  So very briefly, the idea of dimensional regularization.  So the idea is that you replace, and this is what is done the most commonly in practice,  though I'm not a huge fan myself, is to replace d with d minus epsilon, some small positive epsilon,  and perform a Taylor expansion in epsilon.  And so what physicists do in practice is they compute the coefficients of epsilon to the i, where i can be both positive and negative.  Excuse me.  So you can view it as an integral of this hyperplane if you prefer.  But the point is that you could have chosen any other hyperplane.  We didn't have to choose wherever it is, the sum of the beta i's equals 1.  You could choose any other hyperplane.  It still works.  And this, if you like, is a nice way to codify that.  What does a projective integral mean?  It means you restrict to an affine chart and calculate the integral on that affine piece.  And that's exactly the same as choosing a hyperplane at infinity and integrating over affine real space.  So for physicists, this is a way to think of this in integral over r to the ng with a delta function given by this equation.  And it contains the fact that you can replace this with any hyperplane, any non-trivial hyperplane.  Okay, so that's the end of the first half.  As I said, almost everything is in textbooks.  So I was informed that instead of having a pause, I should just press on.  And so now in the second half, I want to give a sort of panorama of what we know about these integrals.  In this way, dimensional regularization doesn't need to invent a defensive geometry in the end.  No, absolutely not. No, no, no. Absolutely not.  No, absolutely not. That's the point. That's the key point.  Okay, so a panorama.  Okay, so what do we know about fine integrals?  Of course, it's impossible for me to give a completely exhaustive list, but let me try to give a picture of the landscape.  So the first remark is that the general one loop diagram.  Oh, sorry, let me make a remark before I do this remark.  So all these, this procedure, the Schringer trick in Gaussian integrals and all this works in much greater generality.  It works for a gauge theory. And the upshot is that we can always write fine integrals in the general form.  So what will happen is if you do this, is you get higher, you maybe get different powers of psi and psi in the denominator.  So here A and B are z integers and P alpha is some polynomial in the alpha E's.  Maybe it has coefficients in some Clifford algebra.  And the whole integrand is homogeneous, such that this is homogeneous at a degree zero.  So that's the general form of a parametric form.  Of course, the P could be extremely complicated.  So back to these scalar diagrams.  The one loop diagram can be expressed using a single function, well, two functions rather.  The dialog with them is defined by the following sum.  n of n squared converges for mod x less than or equal to one and has an analytic continuation to C, so multi-valued function, and the logarithm.  So any graph like this, you can always express the amplitude using just l2 and log, where the x, it'll have a complicated expression in these with different arguments x,  but where the x will be some function of the masses and momenta in some possibly quite complicated way.  In any case, there's a single function essentially that describes all one loop amplitudes.  It was the initial discovery of Feynman, Long-Long with Lueckum. Feynman introduced the dialog with Lueckum and Lueckum.  I'm skeptical.  We can discuss this.  So the general two-loop diagram, two-loop amplitude is not known.  So there isn't a function that you can look up in the canon of special functions.  It means, this also means that the singularities and the analytic behavior of the general two-loop amplitude is not understood.  But does it reduce the functions like two variables or three variables?  So here there's a theorem that if you take an arbitrary large number of external legs, it can always reduce to at most a certain number of legs.  In two-loop, the corresponding result is not known.  But I think it should follow immediately if one knew something about, from Hodge theory, about weights.  So it should follow.  No, no, no.  Yeah. So it typically involves square roots of a quadratic form in the masses and momenta.  So even in a simple example like the triangle, it's, in general, it's a long formula.  Let's just not write it down. It's quite complicated.  It's not something.  So two-loop amplitude is not known in general.  There's a special case that's become very fashionable at the moment that's actually been studied for many years.  So it's called the Sunset or for optimists, it's the sunrise diagram.  So this is a particular two-loop diagram.  So of course, faced with these difficulties, what people typically do is make some assumption that certain masses are zero and so on and so forth.  This has a long history and it was recently solved for the general case with all non-zero masses and momenta.  And the key names are Bloch, Vanhover and McCurr and Adams, Bogner and Weintzeel.  So they have very different approaches, but these have the sort of final word on the status of this integral.  I recommend, if you want to know about history, you can look at these papers.  So I forgot to say something. The dialog is an iterated integral.  So I can write it in a very simple integral form, zero less than t1 less than t2 less than x.  So let me take x between zero and one for the sake of argument. Then it can be written as an integral just involving two differential forms.  And so this is an example of an iterated integral on c minus zero and one.  And for some bizarre reason, it's not fully understood, this space seems to, the functions on this space,  describes a vast number of Feynman amplitudes for reasons which are not completely clear.  This in fact is an exception, but what it does involve are iterated, or rather twice,  in twice iterated elliptic integrals on some family of elliptic curves.  And one way to write down, to get at the amplitude is by average, is this function and in z.  So here q has nothing to do with this q here. So q is, let me call it q naught, q nothing,  is nothing to do with the q here. This is some function, it doesn't make sense,  it's for many reasons, first of all, we're averaging a multi-valued function,  and it's going to have singularities when n is either very big or very small depending on this.  So you have to regularize it in some way, but if you do that you get some function,  and this function essentially describes the amplitude here completely,  where q naught here will be some function of the masses and q.  So I won't say more about that.  Next there is a, so this is the first sort of difficult case.  A huge class of amplitudes are no, and are expressible,  using a finite class of functions called multiple polylogarithms,  about which I think we'll hear a lot more this afternoon.  So part of my job is to set up the afternoon speakers.  So here's the definition of the multiple polylogarithm,  some generalization of the dialog with them.  So it depends on r variables, and it is the sum 1 less than or equal to kr.  x1 to the k1, xr to the kr.  So that defines some analytic function on the region where the x i's, let's say, are strictly less than 1.  And by analytic continuation it extends to a multivalued function  on c to the r minus,  and then you have to take out a certain class of diagonals,  given by the successive products.  So this is actually c, c minus, c minus the origin to the r minus, this class of diagonals.  And this is nothing other than the moduli space of r minus 3 points on a sphere.  So we think of this as r minus 3 particles on a sphere.  And it's not entirely clear why a priori.  Oh yeah, thank you very much.  So this is just consecutive.  Oh yeah, r plus 3, thank you.  Absolutely, the other points are 0, 1 and infinity.  Okay.  So for some reason this particular moduli space, which in the case r equals 1,  is just c minus 0, 1.  It plays a special role,  and it's not understood what the analog should be to describe all higher order Feynman integrals,  if there indeed is such a thing.  Okay, so something which is used a lot in physics, which I'll emphasize,  but it is not valid in general at all.  So I'm going to explain what it is and then say why one shouldn't extrapolate beyond its range of legitimacy,  is that first of all they are iterated integrals.  There's no reason to think that general Feynman integrals should be iterated integrals.  And there's something very special that they have a weight grading.  So what that means is that the weight of, you can attach a weight to such a function,  it's the sum of the indices,  and the property of this weight is that if you differentiate the multiple polyloga of a certain weight,  and there's an explicit formula for this, I just am not going to write it now,  but I'm going to write this as a linear combination of functions of lower weight,  again of multiple polyloga of strictly lower weight, in fact weight one less.  So there's some recursive structure in these functions,  and this weight, if you keep differentiating in a certain way you get to weight zero,  and that reflects the fact that iterated integrals.  But I really want to emphasize because this is often expected,  in physics literature it's said that this holds in general, it's not true,  this is a very special property of this particular class of functions.  Next, so the sunrise is not of this type, it's something else,  there should exist a class.  I will leave that to the speakers this afternoon,  I believe that that will be discussed in detail.  So yeah, I'm just really setting up the afternoon talks.  There should exist, so there is a well-defined class that we understand,  infinite families of graphs of this type.  Beyond that, there should exist a class of multiple elliptic type,  such that IG is multiple elliptic.  So this is a sort of nascent theory,  one of two papers defining what such functions are.  So this is not known, it's not been worked out at all.  So the first class beyond the polylogarithmic is completely unknown,  say for one or two examples like this, maybe that and one other.  And...  No, so it would mean something like this lead to,  it would mean sort of forced it to be elliptic on the Jacobi  uniformization of the elliptic curve by averaging over Q.  You can do the same thing for multiple polylogarithms,  it's slightly more subtle, and that's a good,  that's a perfectly valid definition of a multiple elliptic polylogarithm.  Alternatively, you could look at iterated integrals on the universal elliptic curve.  That's a whole class of functions,  that's in some sense the genus one analog of this story,  and they should describe another large class of diagrams,  and that class is not known.  And so the boundary, again the boundary where the polylogarithmic  ends, is not known.  Something is known about it, but it's not known precisely.  Okay, so now let me focus on some more specific examples.  I'm sure I have interest in number theorists.  So I have half an hour, okay.  So typically these examples are very difficult because there are lots of masses  and momenta, and the amplitude is a function of many complex variables,  and that's where a lot of the difficulty comes from.  But often in practice you don't need such generality,  and the things get easier if you make some restrictions on the momenta and the masses.  So the other extreme we have single scale processes.  So these are where i,g,q,m is a function of a single variable.  So it could be, for example, all the masses are zero,  and there's just one incoming momentum,  or it could be where you just have one non-trivial mass.  And what typically happens is that the dependence on that variable is completely trivial.  And it will just factor out of the integral, out of the integrand.  So what you're already saying is that the amplitudes are giving numbers instead of functions.  So it factors out.  And the coefficient is an interesting number.  So what can we say about the numbers that come out of quantum field theories?  So here's one family of examples, which are in some sense the most exotic.  So it's the analog of this sunrise.  It's diametrically opposed to the polylogarithmic class where we feel we understand quite a lot.  So these are the banana graphs, which have a lot of symmetry and are very interesting.  And these were first studied by David Broadhurst.  So let me do one example in complete detail.  So we have the same picture like this.  And what we're going to do is impose, first of all, let me set all the masses equal to the same mass.  Any number of masses, or just three?  Just one here. So I'm going to put all masses equal.  And I'm going to call it equal to n.  Now, with the n1 and 2 and 3, all of them, n4 and 5.  The bananas will be this family of graphs here.  That's a banana. I don't know why it's called a banana.  It looks nothing like a banana.  Yeah, sunrise, banana.  So the graph polynomial is this, alpha 1, alpha 2, plus alpha 1, alpha 3, plus alpha 2, alpha 3.  The second graph polynomial, so these are all good exercises to revise the definitions of the graph polynomials I gave earlier.  This is q squared, alpha 1, alpha 2, alpha 3.  And this polynomial psi gives this equation, q squared, alpha 1, alpha 2, alpha 3, plus m squared, alpha 1, plus alpha 2, plus alpha 3, alpha 1, alpha 2, plus alpha 1, alpha 3, plus alpha 2, alpha 3, which is very nice and symmetric.  And defines a beautiful family of qubits.  Okay, so the next thing now.  Yes.  So the next thing, so now Brorow says, okay, we let Nallet work in d equals two dimensions.  And put m squared.  So this is not quite Euclidean space, I've been slightly dishonest here, but we'll put m squared equals q squared equals, so m squared equals minus q squared equals one.  And now, so now we set S, so using Broadhurst's notation so you can look at his paper and directly compare with his conjectures.  The integral is omega g.  So if you look at the number of loops and number of edges, what happens is the psi drops out of the integral.  So we get psi g.  And so this m here in his notation is this graph where m minus one is the number of internal edges.  So I don't know why he shifted the indices, but I'm going to stick with his indexing to make it more consistent.  So if you do stick q squared equals minus m squared, of course the m squared just factors out of the integral.  So we can just set m squared equals to one.  Yeah, m is not m.  Thank you, yes, that's unfortunate.  Yeah, of course the m counting number of edges is not the same.  But m is, that m is one, so that's okay.  So S bar four is what we've just computed.  It's this integral, it's omega three.  It's this projective integral alpha one plus alpha three, alpha two plus alpha three, very beautiful integral.  And to work it out, it's a projective integral.  So to work it out, we just work on some affine chart.  So let's put alpha three equal to one, and it's just d alpha one d alpha two, alpha one plus alpha two, one plus alpha one, one plus alpha two.  So it's a very nice, completely convergent little integral there.  And here's a list of what's known about these integrals.  Now S bar three is two pi over three squared three.  S bar four, which is this one, is pi squared over four.  S bar five is conjectually, I'm not sure if this has been proved actually since I looked this up,  but it should be four pi square root of 15 times the L function of a modular form.  So this is some Dirichlet series that's beloved to number theorists.  S bar six is experimentally given by 48 Riemann's ETH of two times special value of the L function of a modular form,  where F four is a weight form modular form.  It's an explicit product of dedicated ETH of values.  So this is something, I won't dwell on this too long, but it's something that number theorists care very much about.  And it's not at all understood how and if at all this pattern should continue.  Okay, and in general it's not known. After this, there's no conjecture for what these integrals should be.  So you see that some very simple family of diagrams is producing some numbers relating to many active topics in number theory  and we really do not understand how to make sense of this.  Okay, so that's in some sense the worst possible family of examples.  Now let me turn to a completely different family of single scale examples.  So these will be residues in massless fight the fourth theory.  And these are the examples that got a lot of us mathematicians interested.  So now D is four again and all masses will be zero.  So I'll say that a graph is in fight the four turns out to be equivalent to the fact that all the vertices of the graph of degree four.  So here's an example.  So what we have in this situation is that the number of edges is twice the loop number.  So the superficial degree of divergence is zero and we're in the divergent case.  So this quantity we had earlier has a pole here and we're in trouble.  But what we're doing is we're looking at the coefficient of one over epsilon in the final integral in dimmer egg.  And it turns out that this does not depend on the external momenta.  It's just a number.  Yeah, so that's exactly what I was going to say.  It's the residue. It's called a residue and by total abuse of notation.  I'm going to call it I again because I'm running out of letters.  So I'll just call it I. It's not the actual formula to call it the residue.  And it's just given by the integral of omega g of psi g squared.  So in the previous case, the integrals were special in that the psi dropped out of the integrand.  These are special because the psi drops out of the integrand.  So these integrals don't always converge.  A criterion due to Weinberg is that this integral converges if and only if g is what is called primitive.  He didn't certainly use that language.  Primitive refers to some hop-valje structure that came much later.  The condition is that for all strict subgraphs of g, the superficial degree of divergence of the subgraph should be negative.  And that means that the number of edges should be at least twice the number of loops.  So the first example we saw already.  So I leave you to work it out from the definitions that this is alpha 1 to alpha 1 over the graph polynomial squared.  So these graphs should have 4 over 4 external edges, yes, of course.  Yeah, it does.  No, in general, alpha is the original degree.  Yeah, yeah, yeah, it will always have 4 external edges, absolutely.  It will always have 4 external edges.  That's the consequence of this and this.  My oil is forming up.  Absolutely.  So to work this out, we work on an affine chart, let's say alpha 1 equals 1.  So that's alpha 2 goes from 0 to infinity d alpha 2 over 1 plus alpha 2 squared, and that is the number 1.  So the residue of this graph is just 1.  That's easy enough.  And after that, they get a lot more difficult as we'll probably hear more about this afternoon.  So let me just give the first few examples, some of the first few examples.  So the next graph which satisfies all these conditions is this graph.  So here the residue is 6 times zeta of 3.  This gives 20 zeta of 5.  This here gives 36 zeta of 3 squared, which is the square of this amplitude.  So there's some identity looking, several identities looking behind the scenes.  And finally, this graph with 6 loops, which took many years to first compute numerically, which is done by Bort-Hurston-Kreimer.  This with those external legs here.  But it can now be done symbolically, because this is now known exactly.  And it is 27 over 5 zeta 5, 3 plus 45 over 4.  So what are these numbers here?  And they're given by a nested sum.  So this converges when nr is greater than or equal to 2, and these are called multiple zeta values.  So this graph, this 6 zeta 3 turns up in many different quantum field theories.  Essentially for the reason that was on this board before, that you can always write a general amplitude by putting numerators in.  And the sort of numbers, you only get a finite family of numbers coming out.  And it's this same zeta 3 that shows up in quantum chromodynamics, n equals 4 Cp young mills.  It shows up again and again.  So there's nothing particularly specific.  This theory seems very specific, but actually it gives a good indication of the quantities that are describing more general quantum field,  or different quantum field theories.  Okay, so multiple zeta values have a kind of weight, but now what I'm saying is totally abusive.  You want to assign a weight to a number, you can't do that.  It doesn't necessarily make any sense.  And in fact, we don't know whether assigning this integer to these real numbers, MZVs, makes any sense at all.  But conjecturally, these multiple zeta values are graded by the weight.  That means that there are no relations, no algebraic relations with rational coefficients between them,  between multiple zeta values of different weights.  Okay, so I'll come back to that later.  You should say that the weight is half of the weight of the unit.  Yeah, I'll say that later as well.  But so this is how physicists understand the notion of weight.  And it's used a lot in modern as a sort of guide to various things in the physics literature.  So it is known that there is an infinite class of graphs, of such graphs in phi 4 primitive,  such that IG, the residue, is multiple zeta value, or linear combination of multiple zeta values.  So there's a criterion.  So there's a common at all criterion.  I won't say much about that.  There are eight loops.  There are so-called modular graphs, whose amplitudes, whose residues we do not expect to be of this type.  So somehow these numbers describe a vast class of these amplitudes.  And something goes badly wrong at eight loops.  So we can get some graphs whose residues are expected to be something else.  I don't know what to call them, but for want of a better name, they should be multiple modular values,  certain types of numbers relating to modular forms.  They're absolutely not of this class.  Another comment worth making is that not every multiple zeta value seems to occur.  So that means if you take the vector space generated by the residues of these graphs,  they don't seem to fill the space of multiple zeta values.  There's some very specific subspace with some very interesting properties.  And finally, there's only one family of graphs which we actually know the amplitudes.  And these are the zigzag graphs.  So you take a bunch of triangles like this and you connect them from head to toe.  And so zigzag five in this case, the index counts the number of loops.  So theorem I proved with Oliver Schnetz, which was a conjecture by Bordersen Kramer for a long time,  gives a formula for this amplitude.  So let me write it down because I think there's some interesting corollaries.  So it depends on the parity of n.  So this family of graphs gives every odd value of the Riemann zeta function.  So we know in particular that all the odd zeta values definitely occur as amplitudes.  On this you can deduce that the products of odd zeta values occur.  But unfortunately, there is no other family of graphs whose amplitudes are known to all orders in this theory.  And there is no family whose amplitude is even conjectured.  So there's not much is known.  And here you mean it's a vacuum diagram, it's not a residue.  So from now on, i always means residue.  So in this section, so I've been a bit sloppy here, in this section when I'm talking about residues,  I'm calling the residue i, which is an unfortunate abusive notation.  But it has no q and m in there, so I suppose.  D equals 4, yes, D equals 4 in this here.  You can have more in there, 5 to the 4.  I'm being sloppy and they don't play any role because of this comment here.  The coefficient of one of epsilon does not depend on the external momenta, so I can just ignore it.  So this is for external legs, and this is closed graphs, this degree 4, which is 2 upon vectors.  And probably 6 does depend on each.  So there are two things you can do. You can do what you just said.  Or you can also look at two loop functions like this in 5-4 theory.  So here you put 0 coming in here, for example, and you have just a single momentum coming in.  And the amplitude of this is some trivial dependence on q times the integral where you close up the external legs of a vacuum diagram.  So these families of integrals are actually computing, if you take any integral like this, and graph and you break a leg,  it's computing the amplitude of a graph like this.  And it's also computing the residue of the corresponding 4-point function.  So these graphs are encoding lots of integrals at the same time.  So here's a conjecture that's an interesting challenge for a young person.  It's in my paper with Schnetz.  And so let G be a primitive graph of the type we're looking at in 5-4.  So of course it satisfies n equals 2Hg in this section with Hg loops.  Then we conjecture, in fact, that the amplitude is strictly less than the zigzag graph.  So the zigzag is somehow the most, is the biggest possible Feynman amplitude in the theory.  It's somehow reminiscent of volumes in hyperbolic geometry where you measure some complexity.  These are pure numbers.  So this is a positive number.  So we conjecture that this graph is the biggest, and if this were true, it would imply a bound.  It would imply that I of G over 4 to the H is at most asymptotically,  by applying Sterling's formula to this, 2 over root 4 pi Hg to minus 3 over 2.  So it gives you some bound on the size of Feynman amplitudes.  And of course this is the first step in trying to understand whether you have the convergence properties  or the resumability properties of your divergent series.  And I think this conjecture is far better, far stronger than what can be proved at the moment.  Of course you need to understand all the divergent graphs and the renormalization.  But I thought that was a curious feature of this theory.  And what it's tempted to try to show that doing elementary operations on the graphs somehow always increases the amplitude or something.  There may be something to explore there.  Sorry, but why would this bound be important then?  Well if you want to understand, if you have the sum of all graphs, you have some divergent series,  you need to know very precisely how many graphs there are and how big, how fast they're growing.  I meant that usually this is removed by some renormalization, right?  Yes, no. So these residues are contributions to the beta function which are independent of all renormalization schemes.  So these numbers, no matter how you renormalize, they will always be in there.  Of course there will be other graphs which have sub-divergences that need to be renormalized.  And there I do not know of a good, I mean there's choices, but I don't know of an analog of this sort of thing.  But if you guessed all graphs of this nature, it would tell you something about the radius of convergence.  Okay, so some final remarks in the last five minutes.  It should be rather open-ended.  So I want to talk about this notion of weight that's been creeping into the picture.  So the majority of the known amplitudes, which are, so for example, the polylogs and the MZVs,  which is, forms the bulk of what we know about at the moment, they had a grading, a notion of weight.  So this grading is highly conjectural in the case of multiple zeta values.  And we can ask where on earth this weight comes from and if it's got anything to do with physics.  So recall that a V, a vector space over Q is graded if and only if it admits the action of a group,  where so lambda in Q star, lambda acting on V is lambda to the N, V, if and only if V of degree N.  So we can think of this grading as a group action.  And this suggests, among other things as well, suggests the possible action of a group on amplitudes.  Which I probably doesn't make much sense.  So here's an important comment.  So there should be a weight as well, but the weight cannot be a grading.  If there is a notion of weight on all amplitudes, it certainly can't be a grading.  We should be careful of this.  And it will be a filtration.  It will be an increasing filtration.  And if we stick with this, the normalization of the weights we've used up to now, which is the physicist's one,  then we must expect half integer weights.  So this is a fact that's well known to algebraic geometries, but I don't think it's been fully assimilated into the physics community.  So what we do in mathematics, we tend to do is multiply all the weights by two, as Cati pointed out earlier.  So what is a conjecture then?  So conjecturally, this is a highly conjectural, there should exist a very large algebraic group of matrices.  I should really say pro algebraic, so it's like an infinite, a group of infinite matrices.  C, which acts on the space of generalized amplitudes,  which is the space which certainly contains, let's call it F, F of Feynman or something.  So this is the vector space over which field I won't say,  of regularized, because these integrals are possibly divergent,  but these are generalized, generalized integrals of the form of the general shape, some numerator,  some a and b integers, and p a polynomial, which is homogeneous such that the whole integrand is homogeneous to degrees zero.  So in some sense, this is like a vector space of all possible amplitudes you'd ever want to consider in any quantum field theory.  Yeah, thank you, yes, absolutely, that's a very good point.  So this is the space of all amplitudes, and there should be a weight filtration, not a grading.  So this is the space of amplitudes of weight at most something.  So C acts on F, and the action of C should preserve, preserved by the action of this group.  There should be a huge group of symmetries acting on all possible amplitudes.  So C would be called a cosmic Galois group.  You can include any other integrals such as jump here, what's special about these guys?  There's a very, very small subclass of integrals, and what was surprising?  For any collection of integrals, you can do that.  Absolutely, you can certainly do that.  In all of the conjecture, it's kind of conjecture whatever, nor emotives.  No, but if you have to replace, not actual numbers, it's a transcendence.  Sure, sure, sure.  Sure, this is...  I agree, you can do this.  The point, which I won't have time to mention, but the point is that amplitudes such as these you're finding are closed under the group.  It's totally extraordinary.  Yeah, absolutely extraordinary.  The reason why the Galois conjugates of some crazy number like this should still be an amplitude.  This seems to be the case.  So in some sense, it's somehow the periods of some operad in the category of motives.  It's a very strong condition.  It's like the material fundamental group are the periods of a group in the category of motives.  It has some very, very strong...  So they're very special formally for the action.  And here there's some sort of operad structure that's motivic.  And so this is a useful concept.  It really tells you something about amplitudes.  So it says that the action of this group on an amplitude will be expressible in terms of smaller amplitudes in some controllable way.  That means the product will tell you.  Yeah, yeah, yeah.  So as I've said, it's an empty statement if set up appropriately.  But it will be a very powerful tool to study precisely this recursive structure.  So this notion of cosmic Galois group is due to Pierre Cartier.  And really I say this because they were the abstract of one of the talks that was mentioned of a co-product.  So let me explain where this comes from.  So we should have this group acting on the space of 5 min equals the weight n.  So one theorem, for example, is that this space is finite dimensional.  So this is extraordinarily strong fact you have infinitely many graphs, infinitely many possible integrands, but this vector space is finite dimensional.  That's some magical feature of these integrals.  If you pick a random family, that's just not true.  So quickly, the dual of an action is a co-action.  So if OC is the ring of functions, and I'll finish here on this group, then the group law is equivalent to a co-product on the functions.  So I think this may come up again this afternoon.  And this action here, star is equivalent to a co-action.  So f goes to f times OC, and the game is to compute this co-action on graphs and to use that to get information about amplitudes.  And so that's one way, in this challenge I mentioned at the beginning, of how do we think about all amplitudes and how do we put some structure in this, all this information.  So the only way to do that is to use this action of a group or a co-action, as I think we'll hear more about this afternoon.  So I'll run slightly, so I'll stop exactly there.  Any questions?  Do you have an example of the general action on a special class of just size?  Yeah, so this doesn't make sense on numbers because you run into transcendence conjectures.  We don't know that z to five could be rational, so we don't know that it makes sense.  But indeed in this family, the conjecture due to Panzer and Schnitz is that this space is closed under this Galois action.  So the group will send this amplitude to a linear combination of this and this.  So it will map it, so this will generate a representation of the group which involves z to five and z to three, which are two smaller diagrams.  So the fact that you're allowed to have z to five three is allowed to appear at this place and only in this place because you previously had a z to three and a z to five in the right places before.  You're not allowed to have a z to three times z to two because that would give a z to two and there's no graph giving z to two.  So it somehow rigidifies, you can see on these examples exactly how the group is acting or conjectural.  I have a question about the non-comitative graph.  Yes.  You said that when the graph is non-comitative, the residue, the integral giving the residue is infinite?  Yes, so you need to, yeah, it diverges and you have to renormalize it.  So I didn't have time to explain.  So we did an overall regularization for the overall divergence because the superficial degree of divergence was zero.  So that's a regularization, but all the subgraphs which have superficial degree of divergence zero need to be regularizing some sense as well and that needs to be done consistently, so that's renormalization.  So there's a way to do that.  I didn't have time to explain that unfortunately.  One more question?  Alright.  Thank you.
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