 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 will explain how to assign integrals to graphs, and in the second half I'll 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. Typically a quantum field theory course will talk about the Feynman path integral up to a certain point, and then by analogy they'll make some noises about a 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 centre of the story. So in perturbative quantum field theory, interactions between graphs are represented, sorry, not a good start, interactions early in the morning, interactions between particles are represented. I 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-positron pair, and later which later annihilate to form a photon again. So every graph that will 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 are 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 sum 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 won't say anything about this. There are some very difficult unsolved problems relating to this, but we have to run into some divergent series and we don't worry too much about that. Practice it. This all seems to work. OK. 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 is a positive even integer. So it's important that it's even. And so we'll work with the Euclidean norm. So x squared is the sum of the square of its components, so if x equals. OK. 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 I, G? So this won't make more sense now, but when we've seen some examples, we need to think about how 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 finite amplitudes is big business. Nearly all the predictions for particle collider experiments are obtained by calculating amplitudes I, G and comparing them with experiment. So there are, I don't know how many people, but many teams around the world, banks of supercomputers 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 finite amplitudes. So this is a very important problem and we are very far from having a mathematical theory of amplitudes. OK. So let's actually begin. A Feynman graph for me then will be a graph. So a graph has a certain number, a certain set of vertices, a certain set of internal edges and a set of external edges. OK. 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. 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. The 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. OK. So Feynman graph also has some extra data, namely the data of... Is the graph from the other side only one Volley for two points? No, it's a different one. No, so it's not increasing. Yes, thank you. It's an application. And the orientated graph is always... No, not yet. Felly, y graff wedi bod yn ystod o'r ffyrdd o'r ffyrdd. Felly, mae'n fyrdd o'r ffyrdd yn ystod o'r ffyrdd o'r ffyrdd o'r ffyrdd. Mae'r ffyrdd o'r ffyrdd, ond mae'n ddod o'r ffyrdd. Felly, y graff wedi bod yn ymgyrchol o'r mas, y mas ymgyrchol, mE, yn y ffordd o'r ffordd. A y ty ein �faint'u yn am für romance o daithun Girl because he will become the newb like I do but it will be good to you.paranoid dw i fod holl oed dan na telefon o권 yllaf ldeidogaethche. 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 and to every external edge we have a momentum q i Arbeit for every external edge. Here let's put Q2 here, Q1 and Q3. 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 sign, if you reverse the, if you want to represent an outgoing particle, that's the same as an ingoing particle with the negative momentum. So we can always assume that all the particles are incoming if we allow the QIs to be negative. So that's totally fine. No vectors, yeah. So that's a very small detail. And the next remark is that these, this data is subject to momentum conservation. Momentum conservation sum QI equals zero for all external edges. Okay. Right, so now we want to write down an integral associated to such a Feynman graph. So Feynman graph is, is the graph with this, 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. We'll see why in a minute. And let me write PE 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, on the external momenta QI and the particle masses ME. So it's an integral over real space of D dimensions, product of internal edges, default integral. 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 PE, 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... Doesn't make sense to integrate, it's not over RD to follow up. Well, thank you very much. RD to times E. Let me give that a name now, actually. NG, thank you, equals a number of E, NG. Thank you. So this is RD times NG. 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. OK, so this, of course... So this is the momentum space, a five-minute integral, as it's most commonly seen. Of course, it doesn't necessarily make sense. It may diverge. Sorry? A small D. Ah, this is delta, delta function. D is... Libation. It doesn't make perturbation in D direction. Oh, here, D, sorry, D as in... Delta is the Dirac delta function. OK, so... Yeah, mathematicians tend to prefer a... Sorry? You're assuming that at each vertex, your convention is such that momentum... No, I'm going to chew, well, you could do that. But 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... Oh, let's do it. We'll have K1, K2, and K3. And so at this vertex we'll get the equation K2 versus K3 equals Q1. And then at this vertex we'll get K1 versus Q3 equals K2. And at this vertex we'll get K1 plus K3 equals Q2. So this delta function means integrate over this region. Yeah, so it's sum of plus minus... I can't see if you choose a rotation. Yeah, I... I just clear what it means, right? If you change the orientation, you can choose the opposite, and it doesn't change anything. This integral doesn't make... I don't 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 integrate it into an integral which does make sense. So this is what you see in physics textbooks. I've kept it exactly as it's written. And exactly as I'm going to say is that, in fact, in the good old days, it was common practice to use parametric representations. That's classical. And in fact, 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 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 genuine quantum field theory, the two 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... I mean, the sum of the idea has a formal power series. Yeah, I'll think of it as a formal power series, exactly. OK, so to get from this to parametric form, I'll 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 0 to infinity e to the minus. OK, so this is something... It looks as if we're going in the wrong direction. Instead of doing the integrals, adding more integrals, but as we'll see where there's a lot to be gained from this. So what happens to this integral, it becomes... It will now be the integral... a bunch of integrals from 0 to infinity for every alpha parameter, assigned to every edge, r to the d and g, x minus sum e 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 reals, e to the minus pi x squared dx equals 1. So you can multiply integrals of this form together, and to get a higher dimensional version, and that 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 2b, where x is in 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 qx dx. Sorry, thank you very much. B scalar x. Let me put b transpose. 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. 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. This can easily be deduced from the previous integral by diagonalising the matrix A. It will break into a product of d copies of this integral. I have to change the variables. 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. The upshot will be that we will obtain a new integral only in... There will no longer be any k's. There will only be the alpha's, the masses and the qi. Only the alpha is the m's and the q's. This can be done in general but it's quite tedious. I will just illustrate on one example and then state the general answer. Let's do a very simple example. Let's take this graph. We assign momenta k1 and k2 to these two internal edges. If you prefer, we can stick to my conventions. Minus q. The delta function in the integral is momentum conservation at each vertex. They are both the same condition. It reduces to the equation k1 plus k2 equals q. That's our domain of integration. Now I'm going to write down the exponent in that integral. Minus k squared e plus me squared af e. What is it in this case? It is alpha1 k1 squared plus m1 squared plus alpha2. The first thing to do is to substitute in the momentum conservation condition. This is alpha1 k1 squared plus alpha2 q minus k1 squared. This time on this time. The mass terms are going to accumulate here. 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 alpha1 plus alpha2 and then times k1 minus q alpha2 alpha2 squared. Now you need to add q squared. Sorry, this is a plus. Let me write. Get rid of the minus. That's better. Over alpha1 plus alpha2. We've completed the square. Let's write this. The next thing is to change variables. The shift variables. I can call this k. Let's put k equals k1 minus q alpha2 of alpha1 plus alpha2. This is just alpha1 plus alpha2 k squared plus the rest. Changing variable. Of course the we're integrating over r to the d. The integral is translation invariant. From this representation we can then plug in Gauss's formula and write down the answer. What we get is this is just the term in the exponential. We have exponential of minus this integrated over k. All the stuff on the right does not depend on k at all. It's just going to factor out of the integral. We just have to integrate e to the minus alpha1 plus alpha2 k squared. That's easy using this formula. I'll just write down the answer. We get pi to the d over 2 integral from 0 to infinity d alpha1 plus alpha2 comes downstairs. All this stuff over here is unchanged. It's factored out of the integral in the first place. In the general case 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 the determinant of a certain matrix. Then you have to apply something called the matrix tree theorem to express that determinant in terms of some graph-threatic quantities. That's rather straightforward. The final answer is that the integral is some trivial factors that ignore powers of pi and so on times the integral from 0 to infinity. Here the product is over all internal edges. Here you get a certain polynomial of psi g and then you get exponential minus pi g q of psi g minus the sum. That is the parametric form of the Feynman integral. I now need to explain to you what these psi and phi are. Psi g and phi g q are what are called semantic polynomials. First of all psi g Psi g is a polynomial first discovered by Kirchoff I think in 1853. It's a polynomial in just the alpha parameters and depends in no way on the external momenta 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 on the graph and then you take the product over the edge variables not in each spanning tree. So t actually it's a spanning tree so t is a spanning tree means t is connected simply connected so that means that it's a tree and spanning means that it meets every vertex of the graph g. Perhaps I should 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 1, 2 It's very simple there are only two spanning trees there's one 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 2 plus alpha 1 and alpha 2 plus alpha 1 is indeed the term we found in the denominator here. You should say that tree doesn't contain an external edge. Yes, yes, absolutely right. So I put no dependence on the QIs Yeah, spanning tree So think of T as a subset of the internal edges So the external momentum the external legs play no role in this at all if you just ignore them. Perhaps I should do another example So this graph which we had earlier the spanning trees are 1, 3 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 1, 2 and 3 and so psi g is alpha 2 alpha 3 plus alpha 1 plus alpha 1, alpha 2 So that's one of the terms in the integral and then we want the second semantic polynomial psi g Now it does depend on the external momentum 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 scalar products of all the external momentum So the formula is that's minus the sum of a spanning two trees and so now you take Q to the t1 So the dot product is the Euclidean Euclidean dot product and then here you have the product of the edges not in t alpha E So what is a spanning two tree So t if and only if So it's a forest with two components So t has exactly two components t1 and t2 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 t1 Q to the t2 So Q to the ti equals total the sum of all momentum entering the tree ti Total momentum entering ti Okay For example So let's do the example we did earlier So spanning two trees Well there's only one It's this That's the only spanning two tree t1 and t2 And so we get minus Q dot minus Q The absent edges That's alpha 1, alpha 2 And that is exactly Q squared alpha 1, alpha 2 Which was exactly hopefully the the term up there minus alpha 1, alpha 2, Q squared It's exactly the same polynomial here No I didn't get it's minus sign Sorry? I did something wrong with minus sign No no because it's polynomial Yeah I put a minus sign here Maybe you didn't see it here There's a minus sign here Yeah so the 2 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 But this is minus phi So here So this is minus phi So it's minus alpha 1, alpha 2, Q squared from that blackboard over psi which is alpha 1 plus alpha 2 It's exactly this term Two times minus is plus So I think it's great So often I don't put this in the formula I write it differently So minus here 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 there So this dot product always produces a minus Because Q2 power Q2 is minus Q2 power Q1 Yeah thank you, I was just about to say that So Qt1 plus Qt2 is zero by momentum conservation So the way I normally write this is plus Qt1 squared And that equals plus Qt2 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 this minus sign isn't there This has entirely positive coefficients So this minus gets absorbed in the stock market Absolutely, yeah Well otherwise T would have a single connected component It's going to be connected I can't remember how I labelled 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 1 Q3 Q2 That's one spanning tree Q2 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. Q2 plus Q3 And we take the product of the edges which are not present So that's alpha2, alpha3 Same here Sorry minus Q2.Q1 plus Q3 Alpha1, alpha3 And then the last one gives minus Q3.Q1 plus Q2 Alpha1, alpha2 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 Alpha2, alpha3 Q2 squared, alpha1, alpha3 Plus Q3 squared Alpha1, alpha2 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 in Euclidean space And that's what guarantees that these integrals make sense somewhere but at least some values of Q and M And so these changes of variable formulae and so on will be justified in some region of Q and M and for some value of D not necessarily 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 PhiGQ plus some Me squared, alphaE So that's not a standard notation And now Thank you very much Thank you very much So PsiG 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 PhiGQ and hence PsiGQ, M are both homogeneous of course in the alphas they depend on other things as well but of degree exactly one more HG plus 1 So that follows from some combinatorial property of spanning trees it's very easy to check So now we want to get to projective integrals So there are two things you can do at this stage We have these finite 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 dimensional regularisation And I'm going to The latter is slightly quicker and gets us straight to the answer so I'm going to do that And so 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 V side G And 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 lambda Recall that NG is the number of internal edges D lambda wedge omega G where omega G equals sum minus 1 to the I So that's a simple manipulation And so this can be rewritten in the form in the region where the beta 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 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 And 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 the final step, the final integral that we need to do and it produces a gamma a gamma value gamma of what I'm going to call sdg times psi g over psi gqm 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 overall logarithmic divergent when it equals 0 so everything is 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 final integral up to maybe some trivial factors in the form of a gamma function an integral over this hyperplane but in fact you notice that the integrand that comes out is actually homogeneous of degree 0 so it's slightly nicer to write this as an integral of a projective space is the sign of the argument of gamma or you want the opposite sign you want the superficial degree of divergence no I think this is correct 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 that's the point so we get psi G d over 2 psi G over psi G Q M 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 a projective integral 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 that this can have a pole so this gamma factor can have a pole particular when the superficial degree of divergence is non-negative so very briefly the idea of dimensional regularisation 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 and perform a tailor 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 why do you call this a projective integral okay it's a question for physicists so I don't know this vocabulary could you explain it more yeah okay so so you can view it as an integral over 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 betas equals 1 you could choose any other hyperplane it still works and this if you like is a nice way to codify that so what does a projective integral mean it means you restrict to an affine chart and calculate the integral on that affine piece so 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 an 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 nontrigal hyperplane okay so that's the end of the 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 regularisation doesn't need to invent a fancy geometry 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 before let me make a remark before I do this remark so all these this procedure the string of trick and Gaussian integrals and all this works for in much greater generality it works for a gauge theory and the upshot is that we can always write fine and integrals in the general form so what will happen is if you do this as you get higher you maybe get different powers of psi and psi in the denominator so here a and b is integers and p alpha is some polynomial in the alpha is maybe it takes has coefficients in some clifid algebra and the whole integrand is homogeneous such that this is homogeneous of degree zero so that's the general form of a parametric form of course the p could be extremely complicated ok so back to these scalar diagrams the one loop diagram can be can be expressed using a single function well two functions rather the dialog rhythm the dialog rhythm is defined by the following sum actually n of n squared so this converges for mod x less than or equal to 1 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 will have a complicated expression in these with different arguments x but with the x will be some function of the masses and momenta in some possibly quite complicated way but in the case there's a single function essentially that describes all one loop amplitudes it was the initial discovery of Feynman's long normal book that Feynman introduced a dialogary long normal book on that 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 behaviour of the general two loop amplitude is not understood but does it reduce the functions like two variables or three variables so yeah so here there's a theorem that if you take an arbitrary large number of external legs you can always reduce to at most a certain number of legs in two loops 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 Do you need a function x in general it's a rational function? No, no, no It's a general function 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 in general it's a long formula that shows not to 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 0 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 Matt Kerr Adams Bogner and Vinesill sort of 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 sorry I forgot to say something the dialog with them the dialog is an iterated integral so I can write it in a very simple integral form 0 less in T1 less in T2 less in X so let me take X between 0 and 1 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 0 and 1 and for some bizarre reasons 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 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 regularise it in some way but if you do that you get some function and this function essentially describes the amplitude here completely 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 polylogithms 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 polylogithms some generalisation of the dialogithms so it depends on our variables and it is the sum 1 less than equal to k r x 1 to the k 1 x r to the k r 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 the a certain class of diagonals given by the successive products sorry 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 again it's not entirely clear why a priori ah yeah thank you very much yep so this is just consecutive oh yeah r plus 3 thank you absolutely the other points are 0, 1 and infinity ok so for some reason this particular moduli space 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 finite integrals if there indeed is such a thing ok so something which is used a lot in physics which I'll emphasise 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 finite 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 um the multiple polylogithum of a certain weight and there's an explicit formula for this I just am not going to write it out I can write this as a linear combination of functions of lower weight again of multiple polylogithums 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 emphasise because this is often expected in physical 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 okay next so the sun rise was 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 I'm just really setting up the afternoon talks there should exist so there is a well defined class that we understand in families of graphs of this type there should be on that there should exist a class 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 this is not known it's not been worked out at all so the first class beyond the polylogithmic 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 le2 we sort of forced it to be elliptic on the jackaby uniformisation of the elliptic curve by averaging over Q you can do the same thing for multiple polylogithms it's slightly more subtle and that's a good that's a perfectly valid definition of a multiple elliptic polylogithm 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 analogue 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 polylog regime ends is not known something is known about it but it's not known precisely ok so now let me focus on some more specific examples which are of interest to number theorists 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 at the other extreme we have single scale processes so these are where IGQM is a function of a single variable so that 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 really saying is that the amplitudes are giving numbers instead of functions so it factors out and the coefficient is an interesting number and then the question is 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 analogue of this sunrise it's diametrically opposed to the polylogithmic 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 M just one here so I'm going to put all masses equal and I'm going to call it equal to M so 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 but yeah sunrise banana so 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 xi 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 cubics okay so in the next thing now yes so the next thing so Narm Brwdaw says okay we let now let work in D equals two dimensions and 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 Borthhurst'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 and 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 M squared equals to one yeah M is not M yes that's unfortunate yeah of course the M counting number of edges is not the same but 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 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 conjecturally I'm not sure if it's been proved actually since I looked this up but it should be four pi square root of fifteen times the L function of a modular form so this is some Dirichlet series beloved to number theorists S bar six is experimentally given by forty eight Riemann's eto 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 eta values so this is something I don'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 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 we really do not understand how to make sense of this 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 four 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 are 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 dimreg and it turns out that this does not depend on the external momenta it's just a number so that's exactly what I was going to say it's the residue it's called a residue and by total abusive 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 final integral it's 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 but out of the integrand these are special because the psi stops out of the integrand so these integrals don't always converge so a criteria and 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 but primitive refers to some hopfalgic structure that came much later but 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 loops should be bigger than twice sorry 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 it does no, in general the output is the original degree because it's 3% of the polynomial yes yes yes it will always have 4 external edges absolutely it will always have 4 external edges that's the consequence of of this and this about all this formula 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 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 this 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 Borthurst and Crimer this for 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 ok so what are these numbers here so the zeta and r are the values at 1 of the multiple polylogithms 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 ok 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 and n equals 4 Cp young mills 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 different quantum field theories ok so multiple zeta values have a kind of weight but now it's what I'm saying is totally abusive you want to assign a weight to a number you can't do that it's it doesn't necessarily make any sense and in fact we don't know whether assigning this integer to these real numbers mzv makes any sense at all but conjecturally these multiple zeta values are graded by the way that means that there are no relations no algebraic relations relations with rational coefficients between them between multiple zeta values of different weights ok 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 i g 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 ok I won't say much about that there are at eight loops there are so-called modular graphs which 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 we get some graphs which are 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 a theorem I proved with Oliver Schnetz which was a conjecture by bought her some crime for a long time gives a formula for this amplitude so let me write it down because I think there's some interesting crawleries that 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 in fact from 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 is no not much is known and here you mean it's a vacuum diagram it's not a residue yeah this is a residue here so from now on I always means a 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 abuse of notation but it has no Q and M in there so I suppose D equals 4 yes, D equals 4 and it's here I'm being sloppy and they don't play any role because of this comment here the coefficient of 1 of epsilon does not depend on the external momenta so I can just ignore it this graph is for legs external legs can this close graphs this degree 4 which is 2 upon brackets 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 a 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 I 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 ng 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 amplitude in the theory it's somehow reminiscent of volumes in hyperbolic geometry 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 Stirling's formula to this 2 over root 4pi hg minus 3 over 2 so it will give you some bound on the size of finite amplitudes and of course this is the first step in trying to understand wether 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 renormalisation but I thought that was a curious feature of this theory and what is tempted to try to show that doing elementary operations on the graphs somehow always increases the amplitude or something and there may be something to explore there sorry but why would this bound be important then if 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 how many graphs there are and how big, how fast they are growing I meant that usually this will be moved by some renormalisation so these residues are contributions to the beta function which are independent of all renormalisation schemes so these numbers no matter how you renormalise they will always be in there of course there will be other graphs which have sub divergences that need to be renormalised and there I do not know of a good answer to their choices but I do not know of a nonlog of this but if you guessed all graphs of this nature it would tell you something about the radius of convergence so some final remarks in the last five minutes should be rather open-ended so I want to talk about this notion of weight that has been creeping in that has been creeping into the picture so the majority of the known amplitudes which are for example the polylogs and the MZVs which is 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 z values and we can ask where on earth this weight comes from if it's got anything to do with physics so recall that a V a vector space of a Q is graded if and only if it admits the action of a group where so a lambda in Q star a 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 a possible action of a group on amplitudes which a priori doesn't make much sense so here's an important comment in general 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 so we should be careful of this and it will be a filtration will be an increasing filtration and if we stick with this the normalisation of the weights we've used up to now which is the physicist's one then we must expect half-introduction half-integer weights so this is a fact well known to algebraic geometries but I don't think it's been fully assimilated into the physics community so what 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 generalised amplitudes which is the space which certainly contains let's call it F or something so this is the vector space over which field I won't say of regularised because these integrals are possibly divergent residualised integrals of the form of the general shape some numerator or A and B at integers and pure polynomial which is homogeneous such that the whole integrand is homogeneous at 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 let me fix fix D even I think this there should be something similar in odd dimensions as well 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 preserve by the action of this group there should be a huge group of symmetries acting on all possible amplitudes so C is would be called a cosmic Galois group 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 if you have to replace not on actual numbers it's a transcendence sure sure sure sure this is 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 there's no reason why the the Galois conjugates of some some crazy number like this should still be an amplitude there seems to be the case so in some sense there's somehow the periods of some operate 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 formulae for the action and here there's some sort of operate structure that's material and so this this is a useful concept it really tells you something about amplitudes so it says that the action of this action of this group on an amplitude will be expressible in terms of smaller amplitudes in some controllable way as I've said it 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 to because there were the abstract of one of the talks there was mentions of a co-product so let me explain where this comes from so so we should have this group acting on the space of 5 integrals of 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 integrans 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 is equivalent to a co-action F goes to F tensor 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 one 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 so this doesn't make sense on numbers because you run into transcendence conjectures we don't know that z to 5 could be rational so we don't know that it makes sense but indeed in these in this family you the conjecture due to panzernschnets is that this space is closed under the space of under this co-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 5 and z to 3 which are two smaller diagrams so the fact that you're allowed to have z to 5 3 you know that you're allowed to have z to 5 3 is allowed to appear at this place and only in this place because you've previously had a z to 3 and a z to 5 in the right places before you're not allowed to have z to 3 times z to 2 because that would give a z to 2 and there's no graph giving z to 2 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-committee graph because you said that when the graph is non-committee the residue the integral giving the residue is infinite so you need to it diverges and you have to re-normalise it so I didn't have time to explain no so we did an overall regularisation for the overall divergence because the superficial degree of divergence was 0 so that's a regularisation but all the subgraphs which have superficial degree of divergence are a need to be regularising some sense as well and that needs to be done consistently so that's re-normalisation so there's a way to do that I didn't have time to explain that unfortunately