Feynman integrals and hyperlogarithms
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Panzer, Eric
Many Feynman integrals evaluate to multiple polylogarithms and their special values like multiple zeta values. One particularly successful approach to understand this phenomenon is due to Francis Brown and uses iterated integrals called hyperlogarithms as a basis for the arising transcendental functions. I will go through the computation of a simple example in detail in order to explain this method, which combines calculus, algebra and combinatorics. The underlying geometric idea of linear reducibility will become clear as well. This hands-on talk aims to be accessible to students of both physics and mathematics.
 Thank you very much and thank you very much to you who stayed around in the afternoon to this last talk. I'm very grateful to the organizers to have this opportunity to speak here at the EHS. And I will try to connect with Francis's talk in the morning and after Ruth has showed you in the previous talk some other approaches and ideas on how to take a Feynman's integrals, I want to go back to the parametric representation that Francis showed you in the morning. And actually Francis put forward at the time a very new approach to the computation of Feynman's integrals which uses hyperlogisms throughout and is one attempt and one way to understand the prevalence of multiple polylogisms and MZVs in the calculation of many Feynman diagrams. So I want to talk about Feynman's integrals but specifically their connection with hyperlogisms and the motivation for this particular approach is, or the goal is to understand why are so many Feynman's integrals distributed by Fi expressible by multiple polylogisms. And Francis already showed you these functions in the morning, I just recall briefly. So these are functions which you can define like multiple series I just write down now. So they're indexed by a bunch of integers and 1 to Nd and they depend on several complex variables and one way to define them is via this summation over nested integers k1 up to kd and then the single sum is just the monomial index by this summation indices in the parameters divided by the summation indices raised to certain powers as dictated by the index of the multiple polylogism. So we have seen already some examples also in Ruth's talk where these functions show up in Feynman diagram calculations and it's still today sort of mysterious why there's so many of them where this is the case but you also know remember not all Feynman diagrams or Feynman integrals are of this form, we know explicit count examples and we have many conjectures of graphs where we are pretty sure that they're of a much more complicated form and if you think of the whole world of Feynman diagrams you might actually say probably this is a set of measures 0 in whatever kind of measures you put on the set of Feynman graphs but the surprise is that for most things which are relevant for physical calculation it's actually you get already very far with just these Feynman diagrams but still understanding why they are of this form is very difficult and I will try to demonstrate to you one approach where which is least in some cases makes it very clear why this is the case and I try to be very explicit and very down to earth so I decided to take an explicit example that we have seen already and try to compute it from beginning to end so if anything is unclear in the meantime please interrupt me because it should all be first or second year calculus I hope but it's still interesting because you see what's actually the important ingredients why this approach works so the idea that we want to follow is take the parametric representation and integrate out one Schwinger parameter at a time so recall that we had parameters associated to each edge of the graph called alpha e when we did the Schwinger trick and then we had the parametric representation and I will consider an explicit example the real with three spokes which was the first non-trivial example that Francis mentioned in his outlook so what was the wheel it is this graph so it is a wheel and three spokes it's called these edges 1 2 3 4 5 6 now as you can think of it as a graph in 5 to the 4th here we use 4 external legs but because it's logarithmically divergent in four dimensions it has this residue which I will denote by I this graph just the integral over omega g divided by the square of the graph polynomial the first semantic polynomial and what did this mean I will evaluate this integral finally so we just have to integrate from 0 to infinity all the Schwinger parameters well not actually all of them because remember in this form it's a projected integral so when you make it defined we can restrict an arbitrary hyperplane and I will choose the hyperplane and that form that I just set alpha 6 to 0 so I will integrate 1 over psi g squared but I set alpha 6 to 1 sorry 0 so this is the integral we want to compute and recall that by definition this semantic polynomial psi g was the sum over all spinning trees and each spinning tree contributes a monomial which is the product of all edges not in the spinning tree so in particular this is linear with respect to each individual variable and this means that the first integrations are essentially elementary how does this work we want to compute amplitude let's let's just do the first interval over the first variable from 0 to infinity of so over alpha 1 over psi g squared I just reminded you that this is a linear polynomial so let's just say that I call psi upper one the coefficient of alpha 1 this polynomial and psi lower one the constant part and this integral is clearly trivial to do so we just have the primitive of the integrand we have to evaluate this 0 at infinity and we just get the 1 over the product of psi upper one and psi lower one this was easy at the next stage we have to leave the rational functions so let us compute the next integral over alpha 2 psi upper one psi lower one now these are coefficients of the original polynomial which was linear in each individual variable so these polynomials are still linear in the next variables so we can do a similar decomposition d alpha 2 and I just continue with this labeling scheme so I call psi upper one to the coefficient of alpha 2 in psi upper one which is a polynomial depending on the other variables psi upper one lower two the constant part with respect to alpha 2 in the polynomial upper one so these are just short hands for particular spanning trees which contain h2 but not h1 or which contain neither of h1 and h2 and so on yeah so I started with the linear polynomial at a complete square so I just get the coefficient of alpha one times the constant part so this is a quadratic polynomial but it's factorized in two linear forms so these these are polynomials here so psi upper one to the lower one are the coefficients with respect to alpha one so they are still polynomials in the remaining variables this is important yeah because these are linear polynomials in the variables they are linear in each variable I start with a polynomial which is linear in each variable and I pick a coefficient so it's still linear in all the other variables this is the important point so I only have linear terms here so I lower one upper two alpha two and the constant part so how do I do this integral well I do a partial fraction decomposition so I get a pre-factor from the partial fraction decomposition which you can work out very easily and then you're left with the integral over alpha two so I hope you can all believe this formula and now we can just compute this integral in terms of logarithms that evaluated zero at infinity and then this thing here just becomes the logarithm of psi upper two lower one and psi upper one lower two and then the denominator we have the other two coefficients so upper one two and psi lower one two so what we've seen here is that we can very easily integrate the first two variables and at that stage the partial integral we have computed so far is the logarithm of some coefficients of the graph polynomial and it has a pre-factor which is given by this polynomial from the partial fraction decomposition now the idea is to continue with this and integrate out one variable after the other but in general now we're in a difficult situation because this is now genuinely a quadratic polynomial this is before because we multiply two linear things and if you want to integrate again we would like that this is actually continuous factorizing in these linear forms as we saw here so let us have a look what this actually is in this case so what is psi upper one lower two I want to integrate alpha three next so let's let's decompose this with respect to alpha three so first of all this is the coefficient of psi which of alpha one where alpha two is set to zero so it is the sum of all the spanning trees which do not contain h1 but which do contain h2 if you think about this actually means that this is the ordinary graph polynomial of the graph where you delete h1 and where you contract h2 this is so-called contraction deletion formula and if you take the graph delete h1 but contract h2 what happens is that six and four become parallel and you still have three and five in the business so this is actually this graph polynomial but I want to see how it depends on alpha three just for the fun of it so I know this is linear in each variable so what is the coefficient of alpha three but again the spanning trees which do not contain alpha three so I can just delete h3 and then I'm essentially left with a one loop graph and you know from one loop graph you just get the sum of all the edges in the loop and then we have a contribution the alpha three is zero so I have to contract alpha three so this gives me the graph polynomial of the sunrise which we have seen with the edges four five and six this is a reminder this was a symmetric polynomial in two variables which you have seen already in Francis's talk now we can yes so I do the completion with respect to alpha three I look at the coefficient of alpha three and what happens when I said alpha three to zero so I contract alpha three and then everything becomes parallel now we can just do the same game the other way around so I delete h2 and contract h1 then five and six become parallel and three and four remain and again I can delete alpha three also then I have only alpha five and alpha six remaining in the loop and I have the same when I contract three I still get the same sunrise polynomial I don't bother writing the indices here and we also have upper one two which means that I delete both edge one and edge two now what happens then is that you have this lonely six hanging around and the edges three four and five remaining this is just a one look rather than I just told you that you just have to sum all the variables in the loop to get its polynomial then the last guy is the one where we set alpha one and two to zero so we contract both edges alpha one and alpha two which means that the edge three is now connect both endpoints of three are now connected because if you contract it one and two this looks a little bit odd so we have this vertex when our edge three is like a self-loop and then we have the sunrise of four five and six here but what does the self-loop mean you're summing over spanning trees I mean no spanning tree can contain edge three because then it would have a loop and the spanning tree is not allowed to have a loop so actually no spanning tree contains alpha three which means that after three actually multiplies this polynomial it's a divisor of the polynomial and then the thing that remains is just again this sunrise polynomial again so you will soon see the reason why I wrote it this way so what is actually this denominator we want to understand this quadratic polynomial here so what is it well let's collect the terms quadratic in alpha three so we get terms quadratic in alpha three from the product of these two minus the product of these two so from this product we get just alpha four plus alpha six times alpha five plus alpha six and then we subtract the product of these two which just gives the sunrise then we have terms linear in alpha three but here the terms linear alpha three come from multiplying a psi with such a term and also here from the subtraction term the term linear alpha three is this times this so actually all these terms are divisible by the sunrise graph polynomial and we get alpha four plus alpha five and two alpha six from these and then we subtract alpha four and alpha five and then finally we have a constant part which is just the square of the sunrise polynomial and now you see that actually these cancel and what is this polynomial I wrote it down here so if you said as you multiply this out the only thing that remains here is alpha six squared so what you find is it's always the same I was just lazy it's always four five six yeah I mean it's symmetric it's a metric polynomial of the three variables four five six I have only four variables remaining and I explicitly look what happens after three and four five six remain so we observe there's a miracle happening here and I mean that this thing actually is a complete square which you can see over there so it's just alpha three times alpha six plus the sunrise polynomial squared and this thing in inside the square is again linear in each variable because we know that psi itself is linear in each variable so so this is a complete square of course there must be some explanation for this and this is an example of so-called Dodgson identities and these identities follow from the fact that the psi polynomial can actually be written as a determinant of a matrix and then there's a whole theory of identities between them in terms of so-called Dodgson polynomials which is something that also Francis and Karin Yates worked out in great detail you might know Mr. Dodgson is also known as Lewis Carroll so this is the one of Alice in Wonderland if you want the point is that these factorizations are extremely important for the fact that we see these multiple polylogisms in so many places because if this would not happen if this would just be a generic quadratic polynomial then order to continue the integration we would have to take the roots of this polynomial and introduce algebraic functions already but because we have a complete square we can just integrate by parts. So if you integrate by part we get one term but we just integrate the complete square and we keep the log but just evaluate it for 3 0 at infinity and we get other terms where you have to integrate still but the log became rational then we do again separation partial perfection decomposition then we again get logs but the point is we don't get a dialog with them. So in the next integration because of this complete square we remain in the world of logarithms and elementary functions and the result of this that the period or the residue of this graph you get because of this integration by part you get actually several terms but you can rewrite them with a symmetry and you actually see that you can write it three times the same integral which in terms of alpha 3 4 and then the sunrise polynomial and the logarithm 4 plus alpha 5 alpha 4 plus alpha 6 over the sunrise polynomial. This is an easy exercise to do this calculation that you get this expression. The point is that because of this complete square we didn't get a dialog with them at this stage. Now I want to write this out explicitly so remember we have three variables left alpha 4 alpha 5 alpha 6 but alpha 6 is set to 1 so we are actually only left with a two dimensional integral and I will just rename the variable so alpha 6 is 1 alpha 5 I call y and alpha 4 I call z and this is the integral 0 to infinity dy over y and you can do the path refraction decomposition here which I already did for you here in this expression. So this comes from the sunrise polynomial if you do these substitutions and do the path refraction decomposition and then here we just get this is now called z plus y z plus 1 and y plus 1 z plus y over y plus 1. So we boil down the computation of this integral to do this two dimensional integral over logarithm and we again see there were already dutch's identities at place in the first thing that I showed you in detail but also in this integration by path procedure here we have to do new path refraction decomposition and also there you actually make use of dutch's identities so the observation here is that again everything is linear everything factorizes into linear function in the next integration variables. So if you now integrate z everything is linear and then with respect to y and also the arguments here are linear and z so everything looks very linear and where is the… But of course at this stage now we have to introduce more general special functions because we cannot compute this integral in terms of classical logarithms and rational functions anymore so we do need a dialog with them but the question is how do you represent these polylogarithms? Just you raised the sum representation because you don't want to work with the sum here we want to have an iterated integral representation. So we want to exploit the iterated integral representation of multiple polylogarithms. The multiple polylogarithms are not only sums but you can also write them as iterated integrals and so for this let us take a set of points in the complex plane should be a finite set if you think of… we call these elements of the set singularities or letters but I want to write them with a special symbol so I introduce symbols omega sigma which should indicate that this is actually representing a differential form for all these letters. If this is an abstract alphabet then we can define special functions for each word in this alphabet. Sigma is no more the domain of integration. Sigma was upstairs the domain of integration. This is a capital sigma. No, no, no, it is integral i g but that means the element. Oh yeah, so sorry, this is a difference in… sorry. So then we define the hyperlogarithm Lw of z associated to a word w in a star so w is just a sequence of these letters by the following rules. First of all if you have a string of omega zeros this just should be the powers of the logarithm of z, the normalization by n factorial. Then we also want that if you take a hyperlogarithm which has a word which begins with some letter then there comes some other remaining letters that this first letter tells you the differential behavior so this should be 1 over z minus sigma times the hyperlogarithm associated to the tail of the word. This is just the reverse way of writing down an iterated integral but it does not fix the constant of integration of course which we require now such that the limit when z goes to zero of these hyperlogarithms is zero unless the word is of the form omega null to the n. So we have a string of zeros which just fix them to be the logarithms which of course diverge quite mildly at zero but for all the other hyperlogarithms defined in this way you should think of them as iterated integrals from zero to z. So in fact… Can we have an example now? Sorry? Can we have an example? Yes sure that's exactly what's going to happen next. So example… What happens if you just take one letter? This is just the integral from zero to z of dt over t minus sigma because the integral is given by the differential behavior but it also has to vanish so sigma is now supposed to be null zero has to vanish at z equals zero so it's this iterated integral and this is just the logarithm z minus sigma over minus sigma then there is l omega zero omega one of z. What does this look like? Well we have to integrate from zero to z dt one or dt one minus zero and this is the last integration and we differentiate with respect to z we have to get down this then we have the integral the nested integration dt two with t two minus one and if you recall Francis briefly mentioned the integral representation from Li two so this is actually minus Li two of z and quite generally when you have a bunch of zeros… Sorry I am changing orders now so people familiar with mzv will know this inside out so I take a word but I have to make distinctions whether a letter is zero or not because it's treated differently in the setup so let's suppose I have a word which ends in a non-zero letter then there comes a bunch of zeros then there comes a non-zero letter and bunch of zeros and so on so I have d non-zero letters and each of them might come with some zeros and this is actually the same as minus one to the d times the multiple polylog with these indices evaluated at the ratio of the non-zero letters. So we get all the multiple polylog with them just in a particular representation this is the point. Of course the benefit of writing them in this way is that it is trivial to integrate in this representation because we just define them as the iterate integrals so if we just continue our example over there let's just rewrite this logarithm here as an iterate integral in this form so we just have to look at the arguments of the logarithm so this integral zero to infinity dZ 1 over Z minus 1 over Z plus Y over 1 plus Y this logarithm this is nothing but L omega minus Y and I have the logarithm with singularity at Z equals minus 1 and I have in the denominator the logarithm which has its singularity with respect to Z at minus Y over 1 plus Y. Now I made sure that I have the right differential behavior I also have to think if I fix these constants correctly which do not depend on Z but I know if I take the logarithm here and set Z to zero and I just get Y in the numerator and also just get Y in the denominator so this vanishes at Z equals zero and these also define to vanish so this is the correct expression and now if I integrate this by definition just means that I prepend an omega zero and an omega minus Y over 1 plus Y to get a primitive of these and I evaluate it at infinity so in this language this just means the following so I introduce a short annotation I want to write linear combinations of words in the arguments and I just define this by linear extension now I just have to prepend this combination of letters so what I did now I expressed the ultimate integration in terms of an iterated integral evaluated at infinity but it still depends implicitly on the remaining integration variable Y which I still have to integrate so if you want to continue in this way you first have to understand this function as an iterate integral of Y so I hope this is clear the idea is that at each stage in this process we want to express the integral the integrand as an iterate integral namely a hyper logarithm in the next integration variable unfortunately I don't have enough time to do this in all glory detail so I have to take a little shortcut here but it's actually quite simple to how to do this so there is a little lemma which you can prove suppose that we take a hyper logarithm of some word at some argument that might be infinity like in this case and we compute its total derivative in the case where these sigmas are considered as functions so they are not fixed we really compute the full total derivative and you can actually prove that there is an simple explicit formula for this you have a sum over all the letters and you take the word where you delete one of those letters and then you have an explicit logarithmic total derivative of the consecutive differences of the letters so the proof is differentiate under the integral and integration by parts but it's really a very simple exercise to do this there are some boundary terms of this formula so there's a sigma 0 appearing here which is defined to be z and the sigma n plus 1 which is defined to be 0 these correspond to the boundary terms of the integration so actually there's a totally symmetric formula now with this formula if you apply the total derivative now to such an expression you see this grading by way this recursive structure which is so special about multiple polylogs coming into place here because you know if you apply the total derivative on the right hand side you only have lower weight multiple polylogs which have one letter less and explicit logarithmic derivatives now in our case these letters here are rational functions of y so if you compute the logarithmic derivative of the rational function you just have to factor the rational function into linear into the zeros and poles with respect to y and we get a hyper logarithm and the differential forms defining a hyper logarithm so this looks good and for this piece which in this case would still be a logarithm which remains we can just apply the machinery recursively so I just give you an example here but I won't work it out yeah I just tell you so this function here if you do all this it's just you can write it as in a hyper logarithm with respect to y so the point is that now all the letters are independent of y so you can convert this representation to this and this is algorithmic there's no magic in this process but now the final integration is trivial right because we just have to integrate we have to multiply this with 1 over y integrate from 0 to infinity so this just means that we have to put another omega 0 in front so we finally arrive that the residue of the B over 3 spokes graph that we started with is three times this thing so we have now an iterate integral of weight 3 and as a hyper logarithm it has this form evaluated at infinity which still might look mysterious but what is it I mean this is an iterate integral on c minus 0 and minus 1 so we immediately know that it is a multiple zeta value you can also think of it as a period of m 0 6 and if you want to write this multiple zeta value you just can you just use a Mobius transformations for example or associators which relate infinity to 1 in some way so we can apply the Mobius transformation which sends z to z over z plus 1 which means that the boundaries the upper boundary of infinity is now mapped to 1 0 stays 0 and you can compute the pullback of the differential forms under this Mobius transformation which is a simple exercise and if you plug this all in you get that this is the same as the integral omega that is omega 0 omega 1 this is now a hyper logarithm evaluated at 1 which is essentially a multi zeta value there's still a little thing with the regularization you have to do because it starts with 0 and you can use a shuffle product but if you use this you can show that this is the same as if you multiply this out and do this little thing which I don't have time to explain but it's really not that difficult you can rewrite it in this form if you just take the definition of the multiple poly logarithm how they relate to these hyper logarithms namely this formula I can see that this is essentially Li3 and this is Li1,2 so this is 3 times Li3 of 1 plus Li1,2 of 1 which is 6 times zeta of 3. So I spent an awful amount of time on explaining this first non-trivial example but I hope it was at least understandable to some degree the amazing aspect of this calculation is that it gets you ridiculously far in practice so at least in some families for example these single scale massless fight of the four integrals that Francis was mentioning this is a case where we have I would think the best knowledge concerning the expansion in the loop number how far can we get and still compute a lot of integrals this is really outstanding compared to other kinematical configurations which can be much more complicated so we know we already have two loops where we get elliptic things these massive sunrise diagrams but in these massless diagrams following these lines one can get quite far and the basic idea is exactly the same so let us recap what we did we computed partial assignment integrals meaning I call this i index k or Francis calls it i index k which still depend on the remaining variables from k plus 1 up to the final variable so integrate from 0 to infinity the first k variables and let us say we just compute such a residue so we compute these integrals one at a time so at each step we just integrate the next variable the prerequisite for doing this is that we can express these functions as hyper logarithms right I was only telling you how to manipulate hyper logarithms but in many cases in the complicated ones the five integral is not expressible as a hyper logarithm and at least not in the Schrodinger parameters and maybe it is not even a sensible iterate integral in any kind of form but in the case where this works we can just apply this procedure which I outlined to you so the prerequisite for this to work is that all singularities of this function which is a multi valued function of these variables so there are some devices there are some varieties which describes where this can pick up monodromes but we need that all of these singularities are linear in the next variable alpha k plus 1 because if this is the case then I just sketched you an algorithm based on this lemma and another lemma which tells you how you get the boundary constants there is an algorithm which in this situation can transform a representation given such an implicit form into a representation which is explicit in y and then we can just find a primitive by prepending letters or doing integration by parts and we can compute the five integral so the message is that all this will only work in very special situations but if it works the whole procedure is automated it is implemented in computer programs so we can try to focus our attention on the actual geometry which sits behind this computation so the actual integral itself what happens here what are the precise polylogon which appear this is not so important the only thing that is important is what kind of singularities are there and how would they depend on the variables so I just want briefly sketch the idea behind this so in the beginning we did not integrate any variable yet so we just have 1 over psi squared the only singularity is psi can only have a singularity when psi vanishes after one integration so what happened after one integration not sure if I still have it and we are still almost two so after one integration we did this we just got one over psi upper one psi lower one so apparently we now have two potentially different singularities so we have psi upper one and psi lower one and we also had i2 after two integrations which was this combination we had this denominator times logarithm psi upper one lower two psi upper two lower one the upper one two psi lower one two so here we have singularities well when either of these polynomials vanish but the important point is that the set of singularities if you take we reduce this then this was also linear polynomial right which I discussed earlier so this d was squared so in this also in this case the prerequisite is actually fulfilled that all this potential singularities are linear in the next integration variable now if you want to understand Feynman diagram and all the amplitudes that we might associate to this diagram it is enough to just look at these polynomials and how these singularities develop when you integrate out more and more variables so there is a name for this which is something that Francis introduced under the so-called polynomial reduction which is very interesting and it's very important but of course I only have five minutes left so I'll just say to you that there are algorithms under the name of polynomial reduction to compute upper bounds on these sets of singularities they actually have a name they're called Landau varieties and the bottom line is that once one has understood this polynomial reduction for a particular graph one knows that all the amplitudes one could associate with it will be via this algorithm express it on terms of multiple polylogisms of a particular type so note the actual integrand does not matter by which I mean we can take any integrand which is compatible with these singularities we started with so if you compute a polynomial reduction for this graph just with the first semantic polynomial then we can also make statements about intervals generalized intervals of this form we have omega g but we integrate but we raise psi to some higher power a and we put some polynomial in the numerator so that's that everything is homogeneous or more generally even though I stick to the one scale case we can also look at general Feynman amplitudes where we can also have such a polynomial but now allow for both graph polynomials the psi and the psi polynomial raised to arbitrary integer powers subject only to the condition that it is a well-defined projective integral so in this case we start with a variety which contains both polynomials which in general makes things more complicated if particularly if you have a graph where every edge is massive then you know that the psi polynomial is quadratic and you actually do not get very far but there are many applications for example without masses or just with few masses where you can still play the same game and get statements which are valid so I don't have time to explain this algorithm but I want to explain to you what the outcome of this techniques is so we have reduced the study of the actual amplitudes and the integrals to an algebraic or geometric task of understanding how these sets of polynomials which describe the singularities how they behave so we have the amplitudes ij which of course come from the graph but some of you take a detour we make things more complicated in order to understand the amplitude we understand something much more complicated we study how these partial integrals depend on all the on all these Schringer parameters and so here we we have a chain how we get here via these integrals partial integrals k plus 1 up to n and in case of kinematics they also depend on q and m but we do not even need to know any anything about these explicit functions we only need to know the single places where they could have singularities so actually we get information about these by looking at these sets of polynomials and if you want to understand this sequence of singularities you will have to take care of these resultants of these Dodgson identities which I mentioned earlier the point is that it do not appear always but they depend on the combinatorial structure of the graph so the question whether such a denominator you get from a partial fraction decomposition is linear factorized into linear factors is something which relates to combinatorial structures in the actual graph so here the graph feeds into as well and the bottom line is that you can get statements like the one Francis briefly mentioned so there is a theorem by him let G have vertex with or actually I will give another example because I have one minute left only consider the family of ladder boxes so these are the integrals with four external legs but you can make them arbitrarily big and you can make two legs massive on the side so you can have triple boxes what would be called a triple box and you can have arbitrary and upo boxes with two massive legs so we have an infinite family of Feynman graphs then if G is such a graph then the actual amplitude associated with this graph as a function of the four momenta with these side constraints is a multiple polylogism and you can actually say what kind of multiple polylogism so you can write down an upper bound on the differential forms which make up this integral or in other words you can specify what the arguments of this multiple polylogism can be in the worst case and there is another class of graphs which Francis studied which are vertex with three graphs so the idea is that those contain graphs for example with three external vertices like these things but again you can generalize this to have three external momenta and because these graphs are very special cometary structure they are very rigid you can use this structure to prove factorization identities of these Dodgson polynomials which then tells you that the actual amplitude is a multiple polylogism of a particular type and in practice you can also use this to do explicit computations so I want to end here and thank you for your attention and yeah thanks. Okay, any question? What happens if the dimension D is odd and I have square roots? Yeah I don't say anything about odd dimensions. I'm sorry for that. In the very story you have mentioned specifically in your example when you start with you use the contraction of line to have some vertices and to remove line and then you get also these lambda varieties and it's exactly what we use as physicists when we try to factorize some processes when you have two scales so for example you have two scales okay and you want to investigate how certain class of diagram can be factorized in order to describe the confinement part and the hard part okay and that's explicitly what most of people involved not in exact computation but in a factorized form or high energy do every day. So there's an expansion in like? Yeah, yeah to try to get asymptotic to even get the power expansion like what we call twist expansion in QCD so does all this machinery could say something interesting on this? Well I think so. Essentially what you're doing whenever you're doing an asymptotic expansion like this on other cases I mean in physics there's this whole business of expansion by regions for example where you just take an individual Feynman diagram and look at an expansion where some momentum gets big or some mass gets big or you know this kind of thing. You're always doing in this picture it has unified interpretation in some way because you look at this integral which is made up of this of this psi polynomial and here in this polynomial you have your different masses and momentum now you're looking at what happens when one of those dominate and the others. So essentially you would hope that you can expand the integrand in this limit and then just compute the integrals themselves and the only problem is that if you do this expansion you might introduce some divergences which you might have to take care of but if this is not the case then you're right so you're using in this case a factorization of this polynomial in this particular kinematic limit. What I did here is that I looked at the factorization of the psi polynomial which does not involve any kinematic invariance but you also have factorization polynomials for the psi polynomial and they are all very important also in the in the motivic approaches that were mentioned today so yeah I mean it's certainly very much related. What you also realize immediately when you do this is that when you do such an expansion you simplify the situation drastically because you replace this complicated bit by a product of too much simpler polynomials and then essentially the the polynomial reduction also breaks down into two independent pieces because the variables separate somehow. So yeah so we do have cases where we get for example in such an expansion everything is linearly reducible and you can compute the coefficients in terms of multiple polylogs whereas for the full actual function it's much more complicated and you do not know what happens so this is also one approach to try to get closer to something very complicated which you don't understand. Question? Nice. In the same spirit so if you have a gauge theory okay you have a constant sum of many diagrams at each individual diagram produces you some let's say power of hard scale or whatever okay which is completely which has completely nothing to do with if you perform summation so do you have some clever way to treat numerators in parametric to combine okay somehow? I must say no here of course the hope is that there should be something and you would hope to find but so far as I'm aware there have been some attempts but it is not yet clear how to do the problem is that you have all these integration by parts relations so in the ideal world you could just take the representations for different diagrams with the numerators and put them all in one numerator and look at this integral in one go and then you should you would hope that you see the cancellations of the leading degrees or whatever. The problem is that even though this is a sort of canonical representation for a diagram it's not the unique one because you have all these integration by parts identities so you can imagine situations where you take a sum of two terms and they don't look that there are no visible cancellations at all but when you do partial integration or write in different form then they are manifest and then it's easy so I know that some people have tried this in other applications to try to combine different diagrams and see why certain cancellations happen in gauge theories but as far as I'm aware this is still very much work in progress and we don't yet know the right approach for this. The problem is really that the representation is not unique and you would need a way to find the right representation in order to see it easily but probably finding that representation is as complicated as solving the problem. What you have done here was not change of representation, I mean the steps of your computation and you see it as a change of representation. You mean change of variables? Yes, I mean all those steps which you transform the integral of. But I only look at one integral at a time. And that's right, I'm changing the representation of my, I don't change the function I just rewrite it in different way which makes it essentially trivial to compute the integral in this new representation. So this is the point. What I should mention briefly is at least is that of course this is a very simple minded approach in the sense that I just take the Schwinger parameters which are 50 years old or 60 years old at least and try to do the integral in these variables. And it's sort of Alice's wonder that we actually get very far in many cases but we also have counter examples where we know that an integral is a multiple polylogism but we do not see it in this way because it's not linearly reducible. So you can have situations where something looks very complicated but it's just a sign that actually these variables are the bad variables. And of course all physicists know that it's extremely important to find the right variables. I mean also for tree level amplitudes if you want to be efficient in writing them down. And here you have a similar situation in the integration process. Of course absolutely unclear what the perfect representation in each case would be. So this is a completely open field I would say at the moment and there are many different ideas one could try out and try to follow but I can just say at the moment there are examples many examples where we have to twist this representation go to another representation to some changes of variables and then it's clear that it's a polylogism or something related but not in the original variables. Well for the formula with a general polynomial p I think in some cases that I know integration like part is enabled to transform this into another formula with different p and b it's more or less like we have seen in the previous talk. Yeah so there are certainly tons of relations here right but the problem is I mean the question that came earlier was if you have different integrals and you want to combine them and see that there are some cancellations if you bring them in if you select some master integrals and write everything in terms of these master integrals which are independent then you should see it on that form but it's not necessarily manifest in this kind of canonical representation. If you bring them in you can manage to get to master integrals by being easier and remain in integration by part in this case actually. And I know as an example integration might not give you a reason. Question here you computed the residue I mean in the example you gave you computed the residue but can you compute the rest of the external expansion at least the constant term? Well for this integral I mean it has been computed I think two or three years ago in the on-channel massless case. The problem is always what are the kinematics that you do? So if you look at the view of three spokes and you just look at it as a three point function and so you put one momentum to zero but take the other three moment as arbitrary then it's in this class of vertex with three graphs I didn't have time to describe these but this function you can compute explicitly in terms of multiple polylogisms to all orders in epsilon close to all even dimensions with arbitrary powers on the propagators. As soon as you go to four off-shell external momenta I don't even know what the right representation for the kinematics is. I mean even if you take the one loop box there with four off-shell momenta it is only known in four dimensions but if you go to four minus epsilon dimensions you don't really know what to do because it's just the kinematics is too complicated. So the only case I know where this is computed with four external momenta is also cheating because it puts all four external momenta on-shell so it's only function of two variables and then it was computed by Hennens-Milnoff and it was shown to be a multiple polylogism and this is actually one of these examples where when you do it with the Schringer parameters you do not get linear reducibility in the sense that I mentioned here so after one integration you have a quadratic polynomial so it seems mysterious but then there's a chain of variables which you can do which makes this thing factorize and then you can apply the algorithm if you see you get this. But it's one of the examples where I would like to have a better reason to see why it's a polylogism or not. But yes you can compute higher orders in the epsilon expansion. The reason is that when you go to, so the epsilon sits here in these exponents, so if you do an expansion in epsilon the only thing you do is that you introduce logarithms of the polynomials which are of course in the space of iterate integrals over these polynomials that I start with anyway. So this is the statement that here the actual integral doesn't really matter so I can allow for arbitrary powers but I can also expand in the powers and allow for logarithms. You can also have logarithms of individual Schringer parameters in the game and arbitrary powers of those. This does not play any role for the function theory. Of course it makes in practice a computation more complicated and takes more time but it does not change conceptually what happens to the integral. So if you have a linear reducibility without these logs, if you add the logs you are saying that you will stick with this property. So I was quite sloppy actually with the way I defined linear reducibility. So I didn't even mention the name did I? At least I didn't dare to write it down. Okay, so the idea is that it's really just something depending on the polynomials. So you take the polynomials then you see what is the worst case that could happen. If I take an arbitrary, so you do not exactly, exactly you do not want to look at the actual integral, just take whatever integral which only has these singularities. Do one integration, what are the singularities that I could have here? Then by quite general arguments just by looking at the defibrations of this variety you can prove these things, that these sets are upper bounds on the singularities. I mean I hope I made somehow clear how this actually arises in the computation because what happens if I want to compute such an integral? Well, this is a logarithm so I have this clear how to write it in this hyperlogarithm form and the letter I get here, this is psi upper one, alpha one plus psi lower one. So the letter will be something like, will be some L omega, the zero is at psi lower one divided by psi upper one. So after the integration I have of course singularities when these go to zero. I will have to do the partial fraction decomposition but these are all taken care of by D. So the changing the powers or anything changes the actual representation at the point but it does not change the fact that it is a hyperlogarithm with these as an upper bound of the letters and denominators which appear. So this is the point and in a way make it more complicated because you have to look at things depending on all these finger parameters which are completely unphysical of course, which only appear in the very last step when you did the last integration. But on the other hand you abstract from the actual integrand and you look at all amplitudes which you can assign to this graph at the same time because this is something which only depends on this geometry of the graph hyper surfaces. All right, one question? Okay, thank you. you
 Thank you very much and thank you very much to you who stayed around in the afternoon  to this last talk.  I'm very grateful to the organizers to have this opportunity to speak here at the EHS.  And I will try to connect with Francis's talk in the morning and after Ruth has showed  you in the previous talk some other approaches and ideas on how to take a Feynman's integrals,  I want to go back to the parametric representation that Francis showed you in the morning.  And actually Francis put forward at the time a very new approach to the computation of  Feynman's integrals which uses hyperlogisms throughout and is one attempt and one way  to understand the prevalence of multiple polylogisms and MZVs in the calculation of many Feynman  diagrams.  So I want to talk about Feynman's integrals but specifically their connection with hyperlogisms  and the motivation for this particular approach is, or the goal is to understand why are so  many Feynman's integrals distributed by Fi expressible by multiple polylogisms.  And Francis already showed you these functions in the morning, I just recall briefly.  So these are functions which you can define like multiple series I just write down now.  So they're indexed by a bunch of integers and 1 to Nd and they depend on several complex  variables and one way to define them is via this summation over nested integers k1 up  to kd and then the single sum is just the monomial index by this summation indices in  the parameters divided by the summation indices raised to certain powers as dictated by the  index of the multiple polylogism.  So we have seen already some examples also in Ruth's talk where these functions show  up in Feynman diagram calculations and it's still today sort of mysterious why there's  so many of them where this is the case but you also know remember not all Feynman diagrams  or Feynman integrals are of this form, we know explicit count examples and we have many  conjectures of graphs where we are pretty sure that they're of a much more complicated  form and if you think of the whole world of Feynman diagrams you might actually say probably  this is a set of measures 0 in whatever kind of measures you put on the set of Feynman  graphs but the surprise is that for most things which are relevant for physical calculation  it's actually you get already very far with just these Feynman diagrams but still understanding  why they are of this form is very difficult and I will try to demonstrate to you one approach  where which is least in some cases makes it very clear why this is the case and I try  to be very explicit and very down to earth so I decided to take an explicit example  that we have seen already and try to compute it from beginning to end so if anything is  unclear in the meantime please interrupt me because it should all be first or second  year calculus I hope but it's still interesting because you see what's actually the important  ingredients why this approach works so the idea that we want to follow is take the parametric  representation  and integrate out one Schwinger parameter at a time so recall that we had parameters  associated to each edge of the graph called alpha e when we did the Schwinger trick and  then we had the parametric representation and I will consider an explicit example the  real with three spokes which was the first non-trivial example that Francis mentioned  in his outlook so what was the wheel it is this graph so it is a wheel and three spokes  it's called these edges 1 2 3 4 5 6 now as you can think of it as a graph in 5 to the  4th here we use 4 external legs but because it's logarithmically divergent in four dimensions  it has this residue which I will denote by I this graph just the integral over omega  g divided by the square of the graph polynomial the first semantic polynomial and what did  this mean I will evaluate this integral finally so we just have to integrate from 0 to infinity  all the Schwinger parameters well not actually all of them because remember in this form  it's a projected integral so when you make it defined we can restrict an arbitrary hyperplane  and I will choose the hyperplane and that form that I just set alpha 6 to 0 so I will  integrate 1 over psi g squared but I set alpha 6 to 1 sorry 0 so this is the integral we  want to compute and  recall that by definition this semantic polynomial psi g was the sum over all spinning trees  and each spinning tree contributes a monomial which is the product of all edges not in the  spinning tree so in particular this is linear with respect to each individual variable and  this means that the first integrations are essentially elementary  how does this work we want to compute amplitude let's let's just do the first interval over  the first variable from 0 to infinity of so over alpha 1 over psi g squared I just reminded  you that this is a linear polynomial so let's just say that I call psi upper one the coefficient  of alpha 1 this polynomial and psi lower one the constant part and this integral is clearly  trivial to do so we just have the primitive of the integrand we have to evaluate this  0 at infinity and we just get the 1 over the product of psi upper one and psi lower one  this was easy at the next stage we have to leave the rational functions so let us compute  the next integral over alpha 2 psi upper one psi lower one now these are coefficients  of the original polynomial which was linear in each individual variable so these polynomials  are still linear in the next variables so we can do a similar decomposition d alpha 2  and I just continue with this labeling scheme so I call psi upper one to the coefficient  of alpha 2 in psi upper one which is a polynomial depending on the other variables psi upper  one lower two the constant part with respect to alpha 2 in the polynomial upper one so  these are just short hands for particular spanning trees which contain h2 but not h1  or which contain neither of h1 and h2 and so on yeah so I started with the linear polynomial  at a complete square so I just get the coefficient of alpha one times the constant part so this  is a quadratic polynomial but it's factorized in two linear forms so these these are polynomials  here so psi upper one to the lower one are the coefficients with respect to alpha one  so they are still polynomials in the remaining variables this is important yeah because these  are linear polynomials in the variables they are linear in each variable I start with a  polynomial which is linear in each variable and I pick a coefficient so it's still linear  in all the other variables this is the important point so I only have linear terms here so  I lower one upper two alpha two and the constant part so how do I do this integral well I do  a partial fraction decomposition so I get a pre-factor from the partial fraction decomposition  which you can work out very easily and then you're left with the integral over alpha two  so I hope you can all believe this formula and now we can just compute this integral  in terms of logarithms that evaluated zero at infinity and then this thing here just  becomes the logarithm of psi upper two lower one and psi upper one lower two and then the  denominator we have the other two coefficients so upper one two and psi lower one two so  what we've seen here is that we can very easily integrate the first two variables and at that  stage the partial integral we have computed so far is the logarithm of some coefficients  of the graph polynomial and it has a pre-factor which is given by this polynomial from the  partial fraction decomposition now the idea is to continue with this and integrate out  one variable after the other but in general now we're in a difficult situation because  this is now genuinely a quadratic polynomial this is before because we multiply two linear  things and if you want to integrate again we would like that this is actually continuous  factorizing in these linear forms as we saw here so let us have a look what this actually  is in this case so what is psi upper one lower two I want to integrate alpha three next so  let's let's decompose this with respect to alpha three so first of all this is the coefficient  of psi which of alpha one where alpha two is set to zero so it is the sum of all the  spanning trees which do not contain h1 but which do contain h2 if you think about this  actually means that this is the ordinary graph polynomial of the graph where you delete h1  and where you contract h2 this is so-called contraction deletion formula and if you take  the graph delete h1 but contract h2 what happens is that six and four become parallel  and you still have three and five in the business so this is actually this graph polynomial  but I want to see how it depends on alpha three just for the fun of it so I know this  is linear in each variable so what is the coefficient of alpha three but again the  spanning trees which do not contain alpha three so I can just delete h3 and then I'm  essentially left with a one loop graph and you know from one loop graph you just get  the sum of all the edges in the loop and then we have a contribution the alpha three is  zero so I have to contract alpha three so this gives me the graph polynomial of the  sunrise which we have seen with the edges four five and six this is a reminder this  was a symmetric polynomial in two variables which you have seen already in Francis's talk  now we can yes so I do the completion with respect to alpha three I look at the coefficient  of alpha three and what happens when I said alpha three to zero so I contract alpha three  and then everything becomes parallel now we can just do the same game the other way around  so I delete h2 and contract h1 then five and six become parallel and three and four remain  and again I can delete alpha three also then I have only alpha five and alpha six remaining  in the loop and I have the same when I contract three I still get the same sunrise polynomial  I don't bother writing the indices here and we also have upper one two which means that  I delete both edge one and edge two now what happens then is that you have this lonely six  hanging around and the edges three four and five remaining this is just a one look rather  than I just told you that you just have to sum all the variables in the loop to get its  polynomial then the last guy is the one where we set alpha one and two to zero so we contract  both edges alpha one and alpha two which means that the edge three is now connect both endpoints  of three are now connected because if you contract it one and two this looks a little  bit odd so we have this vertex when our edge three is like a self-loop and then we have  the sunrise of four five and six here but what does the self-loop mean you're summing  over spanning trees I mean no spanning tree can contain edge three because then it would  have a loop and the spanning tree is not allowed to have a loop so actually no spanning tree  contains alpha three which means that after three actually multiplies this polynomial  it's a divisor of the polynomial and then the thing that remains is just again this sunrise  polynomial again so you will soon see the reason why I wrote it this way so what is  actually this denominator we want to understand this quadratic polynomial here so what is  it well let's collect the terms quadratic in alpha three so we get terms quadratic in  alpha three from the product of these two minus the product of these two so from this  product we get just alpha four plus alpha six times alpha five plus alpha six and then  we subtract the product of these two which just gives the sunrise then we have terms  linear in alpha three but here the terms linear alpha three come from multiplying a psi with  such a term and also here from the subtraction term the term linear alpha three is this times  this so actually all these terms are divisible by the sunrise graph polynomial and we get  alpha four plus alpha five and two alpha six from these and then we subtract alpha four  and alpha five and then finally we have a constant part which is just the square of  the sunrise polynomial and now you see that actually these cancel and what is this polynomial  I wrote it down here so if you said as you multiply this out the only thing that remains  here is alpha six squared so what you find is it's always the same I was just lazy it's  always four five six yeah I mean it's symmetric it's a metric polynomial of the three variables  four five six I have only four variables remaining and I explicitly look what happens  after three and four five six remain so we observe there's a miracle happening here  and I mean that this thing actually is a complete square which you can see over there so it's  just alpha three times alpha six plus the sunrise polynomial squared and this thing  in inside the square is again linear in each variable because we know that psi itself is  linear in each variable so so this is a complete square of course there must be some explanation  for this and this is an example of so-called Dodgson identities  and these identities follow from the fact that the psi polynomial can actually be written  as a determinant of a matrix and then there's a whole theory of identities between them  in terms of so-called Dodgson polynomials which is something that also Francis and Karin  Yates worked out in great detail you might know Mr. Dodgson is also known as Lewis Carroll  so this is the one of Alice in Wonderland if you want the point is that these factorizations  are extremely important for the fact that we see these multiple polylogisms in so many  places because if this would not happen if this would just be a generic quadratic polynomial  then order to continue the integration we would have to take the roots of this polynomial  and introduce algebraic functions already but because we have a complete square we can  just integrate by parts.  So if you integrate by part we get one term but we just integrate the complete square  and we keep the log but just evaluate it for 3 0 at infinity and we get other terms where  you have to integrate still but the log became rational then we do again separation partial  perfection decomposition then we again get logs but the point is we don't get a dialog  with them.  So in the next integration because of this complete square we remain in the world of  logarithms and elementary functions and the result of this  that the period or the residue of this graph you get because of this integration by part  you get actually several terms but you can rewrite them with a symmetry and you actually  see that you can write it three times the same integral which in terms of alpha 3 4 and  then the sunrise polynomial and the logarithm 4 plus alpha 5 alpha 4 plus alpha 6 over the  sunrise polynomial.  This is an easy exercise to do this calculation that you get this expression.  The point is that because of this complete square we didn't get a dialog with them at  this stage.  Now I want to write this out explicitly so remember we have three variables left alpha  4 alpha 5 alpha 6 but alpha 6 is set to 1 so we are actually only left with a two dimensional  integral and I will just rename the variable so alpha 6 is 1 alpha 5 I call y and alpha  4 I call z and this is the integral 0 to infinity dy over y and you can do the path  refraction decomposition here which I already did for you here in this expression.  So this comes from the sunrise polynomial if you do these substitutions and do the path  refraction decomposition and then here we just get this is now called z plus y z plus  1 and y plus 1 z plus y over y plus 1.  So we boil down the computation of this integral to do this two dimensional integral over logarithm  and we again see there were already dutch's identities at place in the first thing that  I showed you in detail but also in this integration by path procedure here we have to do new path  refraction decomposition and also there you actually make use of dutch's identities so  the observation here is that again everything is linear everything factorizes into linear  function in the next integration variables.  So if you now integrate z everything is linear and then with respect to y and also the arguments  here are linear and z so everything looks very linear and where is the…  But of course at this stage now we have to introduce more general special functions because  we cannot compute this integral in terms of classical logarithms and rational functions  anymore so we do need a dialog with them but the question is how do you represent these  polylogarithms?  Just you raised the sum representation because you don't want to work with the sum here  we want to have an iterated integral representation.  So we want to exploit the iterated integral representation of multiple polylogarithms.  The multiple polylogarithms are not only sums but you can also write them as iterated integrals  and so for this let us take a set of points in the complex plane should be a finite set  if you think of… we call these elements of the set singularities or letters but I want  to write them with a special symbol so I introduce symbols omega sigma which should indicate  that this is actually representing a differential form for all these letters.  If this is an abstract alphabet then we can define special functions for each word in  this alphabet.  Sigma is no more the domain of integration.  Sigma was upstairs the domain of integration.  This is a capital sigma.  No, no, no, it is integral i g but that means the element.  Oh yeah, so sorry, this is a difference in… sorry.  So then we define the hyperlogarithm Lw of z associated to a word w in a star so w is  just a sequence of these letters by the following rules.  First of all if you have a string of omega zeros this just should be the powers of the  logarithm of z, the normalization by n factorial.  Then we also want that if you take a hyperlogarithm which has a word which begins with some letter  then there comes some other remaining letters that this first letter tells you the differential  behavior so this should be 1 over z minus sigma times the hyperlogarithm associated  to the tail of the word.  This is just the reverse way of writing down an iterated integral but it does not fix the  constant of integration of course which we require now such that the limit when z goes  to zero of these hyperlogarithms is zero unless the word is of the form omega null to the  n.  So we have a string of zeros which just fix them to be the logarithms which of course  diverge quite mildly at zero but for all the other hyperlogarithms defined in this way  you should think of them as iterated integrals from zero to z.  So in fact…  Can we have an example now?  Sorry?  Can we have an example?  Yes sure that's exactly what's going to happen next.  So example…  What happens if you just take one letter?  This is just the integral from zero to z of dt over t minus sigma because the integral  is given by the differential behavior but it also has to vanish so sigma is now supposed  to be null zero has to vanish at z equals zero so it's this iterated integral and this  is just the logarithm z minus sigma over minus sigma then there is l omega zero omega  one of z.  What does this look like?  Well we have to integrate from zero to z dt one or dt one minus zero and this is the  last integration and we differentiate with respect to z we have to get down this then  we have the integral the nested integration dt two with t two minus one and if you recall  Francis briefly mentioned the integral representation from Li two so this is actually minus Li two  of z and quite generally when you have a bunch of zeros…  Sorry I am changing orders now so people familiar with mzv will know this inside out so I take  a word but I have to make distinctions whether a letter is zero or not because it's treated  differently in the setup so let's suppose I have a word which ends in a non-zero letter  then there comes a bunch of zeros then there comes a non-zero letter and bunch of zeros  and so on so I have d non-zero letters and each of them might come with some zeros and  this is actually the same as minus one to the d times the multiple polylog with these  indices evaluated at the ratio of the non-zero letters.  So we get all the multiple polylog with them just in a particular representation this is  the point.  Of course the benefit of writing them in this way is that it is trivial to integrate in  this representation because we just define them as the iterate integrals so if we just  continue our example over there let's just rewrite this logarithm here as an iterate  integral in this form so we just have to look at the arguments of the logarithm so this  integral zero to infinity dZ 1 over Z minus 1 over Z plus Y over 1 plus Y this logarithm  this is nothing but L omega minus Y and I have the logarithm with singularity at Z  equals minus 1 and I have in the denominator the logarithm which has its singularity with  respect to Z at minus Y over 1 plus Y.  Now I made sure that I have the right differential behavior I also have to think if I fix these  constants correctly which do not depend on Z but I know if I take the logarithm here  and set Z to zero and I just get Y in the numerator and also just get Y in the denominator  so this vanishes at Z equals zero and these also define to vanish so this is the correct  expression and now if I integrate this by definition just means that I prepend an omega  zero and an omega minus Y over 1 plus Y to get a primitive of these and I evaluate it  at infinity so in this language this just means the following so I introduce a short  annotation I want to write linear combinations of words in the arguments and I just define  this by linear extension now I just have to prepend this combination of letters  so what I did now I expressed the ultimate integration in terms of an iterated integral  evaluated at infinity but it still depends implicitly on the remaining integration variable  Y which I still have to integrate so if you want to continue in this way you first have  to understand this function as an iterate integral of Y so I hope this is clear the  idea is that at each stage in this process we want to express the integral the integrand  as an iterate integral namely a hyper logarithm in the next integration variable unfortunately  I don't have enough time to do this in all glory detail so I have to take a little shortcut  here but it's actually quite simple to how to do this so there is a little lemma which  you can prove suppose that we take a hyper logarithm of some word at some argument that  might be infinity like in this case and we compute its total derivative in the case where  these sigmas are considered as functions so they are not fixed we really compute the  full total derivative and you can actually prove that there is an simple explicit formula  for this you have a sum over all the letters and you take the word where you delete one  of those letters and then you have an explicit logarithmic total derivative of the consecutive  differences of the letters so the proof is differentiate under the integral and integration  by parts but it's really a very simple exercise to do this there are some boundary terms of  this formula so there's a sigma 0 appearing here which is defined to be z and the sigma  n plus 1 which is defined to be 0 these correspond to the boundary terms of the integration so  actually there's a totally symmetric formula now with this formula if you apply the total  derivative now to such an expression you see this grading by way this recursive structure  which is so special about multiple polylogs coming into place here because you know if  you apply the total derivative on the right hand side you only have lower weight multiple  polylogs which have one letter less and explicit logarithmic derivatives now in our case these  letters here are rational functions of y so if you compute the logarithmic derivative  of the rational function you just have to factor the rational function into linear into  the zeros and poles with respect to y and we get a hyper logarithm and the differential  forms defining a hyper logarithm so this looks good and for this piece which in this  case would still be a logarithm which remains we can just apply the machinery recursively  so I just give you an example here but I won't work it out yeah I just tell you so this function  here if you do all this it's just you can write it as in a hyper logarithm with respect  to y so the point is that now all the letters are independent of y so you can convert this  representation to this and this is algorithmic there's no magic in this process but now the  final integration is trivial right because we just have to integrate we have to multiply  this with 1 over y integrate from 0 to infinity so this just means that we have to put another  omega 0 in front so we finally arrive that the residue of the B over 3 spokes graph that  we started with is three times this thing so we have now an iterate integral of weight  3 and as a hyper logarithm it has this form  evaluated at infinity which still might look mysterious but what is it I mean this is an  iterate integral on c minus 0 and minus 1 so we immediately know that it is a multiple  zeta value you can also think of it as a period of m 0 6 and if you want to write this multiple  zeta value you just can you just use a Mobius transformations for example or associators  which relate infinity to 1 in some way so we can apply the Mobius transformation which  sends z to z over z plus 1 which means that the boundaries the upper boundary of infinity  is now mapped to 1 0 stays 0 and you can compute the pullback of the differential forms under  this Mobius transformation which is a simple exercise and if you plug this all in you get  that this is the same as the integral omega that is  omega 0 omega 1 this is now a hyper logarithm evaluated at 1 which is essentially a multi  zeta value there's still a little thing with the regularization you have to do because  it starts with 0 and you can use a shuffle product but if you use this you can show  that this is the same as if you multiply this out and do this little thing which I don't  have time to explain but it's really not that difficult you can rewrite it in this form  if you just take the definition of the multiple poly logarithm how they relate to these hyper  logarithms namely this formula I can see that this is essentially Li3 and this is Li1,2  so this is 3 times Li3 of 1 plus Li1,2 of 1 which is 6 times zeta of 3.  So I spent an awful amount of time on explaining this first non-trivial example but I hope  it was at least understandable to some degree the amazing aspect of this calculation is  that it gets you ridiculously far in practice so at least in some families for example these  single scale massless fight of the four integrals that Francis was mentioning this is a case  where we have I would think the best knowledge concerning the expansion in the loop number  how far can we get and still compute a lot of integrals this is really outstanding compared  to other kinematical configurations which can be much more complicated so we know we  already have two loops where we get elliptic things these massive sunrise diagrams but  in these massless diagrams following these lines one can get quite far and the basic  idea is exactly the same so let us recap what we did we computed partial assignment  integrals meaning I call this i index k or Francis calls it i index k which still depend  on the remaining variables from k plus 1 up to the final variable so integrate from  0 to infinity the first k variables and let us say we just compute such a residue so we  compute these integrals one at a time so at each step we just integrate the next variable  the prerequisite for doing this  is that we can express these functions as hyper logarithms right I was only telling you  how to manipulate hyper logarithms but in many cases in the complicated ones the five  integral is not expressible as a hyper logarithm and at least not in the Schrodinger parameters  and maybe it is not even a sensible iterate integral in any kind of form but in the case  where this works we can just apply this procedure which I outlined to you so the prerequisite  for this to work is that all singularities of this function which is a multi valued function  of these variables so there are some devices there are some varieties which describes where  this can pick up monodromes but we need that all of these singularities are linear in the  next variable alpha k plus 1 because if this is the case then I just sketched you an algorithm  based on this lemma and another lemma which tells you how you get the boundary constants  there is an algorithm which in this situation can transform a representation given such  an implicit form into a representation which is explicit in y and then we can just find  a primitive by prepending letters or doing integration by parts and we can compute the  five integral so the message is that all this will only work in very special situations  but if it works the whole procedure is automated it is implemented in computer programs so  we can try to focus our attention on the actual geometry which sits behind this computation  so the actual integral itself what happens here what are the precise polylogon which  appear this is not so important the only thing that is important is what kind of singularities  are there and how would they depend on the variables so I just want briefly sketch the  idea behind this so in the beginning we did not integrate any variable yet so we just  have 1 over psi squared the only singularity is psi can only have a singularity when psi  vanishes after one integration so what happened after one integration not sure if I still  have it and we are still almost two so after one integration we did this we just got one  over psi upper one psi lower one so apparently we now have two potentially different singularities  so we have psi upper one and psi lower one and we also had i2 after two integrations  which was this combination we had this denominator times logarithm psi upper one lower two psi  upper two lower one the upper one two psi lower one two so here we have singularities  well when either of these polynomials vanish but the important point is that the set of  singularities if you take we reduce this then this was also linear polynomial right which  I discussed earlier so this d was squared so in this also in this case the prerequisite  is actually fulfilled that all this potential singularities are linear in the next integration  variable now if you want to understand Feynman diagram and all the amplitudes that we might  associate to this diagram it is enough to just look at these polynomials and how these  singularities develop when you integrate out more and more variables so there is a name  for this which is something that Francis introduced under the so-called polynomial reduction which  is very interesting and it's very important but of course I only have five minutes left  so I'll just say to you that there are algorithms under the name of polynomial reduction  to compute upper bounds  on these sets of singularities they actually have a name they're called Landau varieties  and the bottom line is that once one has understood this polynomial reduction for a particular  graph one knows that all the amplitudes one could associate with it will be via this algorithm  express it on terms of multiple polylogisms of a particular type so note the actual integrand  does not matter by which I mean we can take any integrand which is compatible with these  singularities we started with so if you compute a polynomial reduction for this graph just  with the first semantic polynomial then we can also make statements about intervals generalized  intervals of this form we have omega g but we integrate but we raise psi to some higher  power a and we put some polynomial in the numerator so that's that everything is homogeneous  or more generally even though I stick to the one scale case we can also look at general  Feynman amplitudes where we can also have such a polynomial but now allow for both graph  polynomials the psi and the psi polynomial raised to arbitrary integer powers subject  only to the condition that it is a well-defined projective integral so in this case we start  with a variety which contains both polynomials which in general makes things more complicated  if particularly if you have a graph where every edge is massive then you know that the  psi polynomial is quadratic and you actually do not get very far but there are many applications  for example without masses or just with few masses where you can still play the same game  and get statements which are valid so I don't have time to explain this algorithm but I  want to explain to you what the outcome of this techniques is so we have reduced the study  of the actual amplitudes and the integrals to an algebraic or geometric task of understanding  how these sets of polynomials which describe the singularities how they behave so we have  the amplitudes ij which of course come from the graph  but some of you take a detour we make things more complicated in order to understand the  amplitude we understand something much more complicated we study how these partial integrals  depend on all the on all these Schringer parameters and so here we we have a chain how we get  here via these integrals partial integrals k plus 1 up to n and in case of kinematics  they also depend on q and m but we do not even need to know any anything about these  explicit functions we only need to know the single places where they could have singularities  so actually we get information about these by looking at these sets of polynomials  and if you want to understand this sequence of singularities you will have to take care  of these resultants of these Dodgson identities which I mentioned earlier the point is that  it do not appear always but they depend on the combinatorial structure of the graph so  the question whether such a denominator you get from a partial fraction decomposition  is linear factorized into linear factors is something which relates to combinatorial  structures in the actual graph so here the graph feeds into as well and the bottom line  is that you can get statements like the one Francis briefly mentioned so there is a theorem  by him let G have vertex with or actually I will give another example because I have  one minute left only consider the family of ladder boxes so these are the integrals with  four external legs but you can make them arbitrarily big and you can make two legs massive on  the side so you can have triple boxes what would be called a triple box and you can have  arbitrary and upo boxes with two massive legs so we have an infinite family of Feynman  graphs then if G is such a graph then the actual amplitude associated with this graph  as a function of the four momenta with these side constraints is a multiple polylogism  and you can actually say what kind of multiple polylogism so you can write down an upper  bound on the differential forms which make up this integral or in other words you can  specify what the arguments of this multiple polylogism can be in the worst case and there  is another class of graphs which Francis studied which are vertex with three graphs so the  idea is that those contain graphs for example with three external vertices like these things  but again you can generalize this to have three external momenta and because these graphs  are very special cometary structure they are very rigid you can use this structure to prove  factorization identities of these Dodgson polynomials which then tells you that the  actual amplitude is a multiple polylogism of a particular type and in practice you can  also use this to do explicit computations so I want to end here and thank you for your  attention and yeah thanks.  Okay, any question?  What happens if the dimension D is odd and I have square roots?  Yeah I don't say anything about odd dimensions.  I'm sorry for that.  In the very story you have mentioned specifically in your example when you start with you use  the contraction of line to have some vertices and to remove line and then you get also these  lambda varieties and it's exactly what we use as physicists when we try to factorize  some processes when you have two scales so for example you have two scales okay and you  want to investigate how certain class of diagram can be factorized in order to describe the  confinement part and the hard part okay and that's explicitly what most of people involved  not in exact computation but in a factorized form or high energy do every day.  So there's an expansion in like?  Yeah, yeah to try to get asymptotic to even get the power expansion like what we call  twist expansion in QCD so does all this machinery could say something interesting on this?  Well I think so.  Essentially what you're doing whenever you're doing an asymptotic expansion like this on  other cases I mean in physics there's this whole business of expansion by regions for  example where you just take an individual Feynman diagram and look at an expansion where  some momentum gets big or some mass gets big or you know this kind of thing.  You're always doing in this picture it has unified interpretation in some way because  you look at this integral which is made up of this of this psi polynomial and here in  this polynomial you have your different masses and momentum now you're looking at what happens  when one of those dominate and the others.  So essentially you would hope that you can expand the integrand in this limit and then  just compute the integrals themselves and the only problem is that if you do this expansion  you might introduce some divergences which you might have to take care of but if this  is not the case then you're right so you're using in this case a factorization of this  polynomial in this particular kinematic limit.  What I did here is that I looked at the factorization of the psi polynomial which does not involve  any kinematic invariance but you also have factorization polynomials for the psi polynomial  and they are all very important also in the in the motivic approaches that were mentioned  today so yeah I mean it's certainly very much related.  What you also realize immediately when you do this is that when you do such an expansion  you simplify the situation drastically because you replace this complicated bit by a product  of too much simpler polynomials and then essentially the the polynomial reduction also breaks down  into two independent pieces because the variables separate somehow.  So yeah so we do have cases where we get for example in such an expansion everything is  linearly reducible and you can compute the coefficients in terms of multiple polylogs  whereas for the full actual function it's much more complicated and you do not know what  happens so this is also one approach to try to get closer to something very complicated  which you don't understand.  Question?  Nice.  In the same spirit so if you have a gauge theory okay you have a constant sum of many  diagrams at each individual diagram produces you some let's say power of hard scale or  whatever okay which is completely which has completely nothing to do with if you perform  summation so do you have some clever way to treat numerators in parametric to combine  okay somehow?  I must say no here of course the hope is that there should be something and you would hope  to find but so far as I'm aware there have been some attempts but it is not yet clear  how to do the problem is that you have all these integration by parts relations so in  the ideal world you could just take the representations for different diagrams with the numerators  and put them all in one numerator and look at this integral in one go and then you should  you would hope that you see the cancellations of the leading degrees or whatever.  The problem is that even though this is a sort of canonical representation for a diagram  it's not the unique one because you have all these integration by parts identities so  you can imagine situations where you take a sum of two terms and they don't look that  there are no visible cancellations at all but when you do partial integration or write  in different form then they are manifest and then it's easy so I know that some people  have tried this in other applications to try to combine different diagrams and see why  certain cancellations happen in gauge theories but as far as I'm aware this is still very  much work in progress and we don't yet know the right approach for this.  The problem is really that the representation is not unique and you would need a way to  find the right representation in order to see it easily but probably finding that representation  is as complicated as solving the problem.  What you have done here was not change of representation, I mean the steps of your computation  and you see it as a change of representation.  You mean change of variables?  Yes, I mean all those steps which you transform the integral of.  But I only look at one integral at a time.  And that's right, I'm changing the representation of my, I don't change the function I just  rewrite it in different way which makes it essentially trivial to compute the integral  in this new representation.  So this is the point.  What I should mention briefly is at least is that of course this is a very simple minded  approach in the sense that I just take the Schwinger parameters which are 50 years old  or 60 years old at least and try to do the integral in these variables.  And it's sort of Alice's wonder that we actually get very far in many cases but we also have  counter examples where we know that an integral is a multiple polylogism but we do not see  it in this way because it's not linearly reducible.  So you can have situations where something looks very complicated but it's just a sign  that actually these variables are the bad variables.  And of course all physicists know that it's extremely important to find the right variables.  I mean also for tree level amplitudes if you want to be efficient in writing them down.  And here you have a similar situation in the integration process.  Of course absolutely unclear what the perfect representation in each case would be.  So this is a completely open field I would say at the moment and there are many different  ideas one could try out and try to follow but I can just say at the moment there are  examples many examples where we have to twist this representation go to another representation  to some changes of variables and then it's clear that it's a polylogism or something  related but not in the original variables.  Well for the formula with a general polynomial p I think in some cases that I know integration  like part is enabled to transform this into another formula with different p and b it's  more or less like we have seen in the previous talk.  Yeah so there are certainly tons of relations here right but the problem is I mean the question  that came earlier was if you have different integrals and you want to combine them and  see that there are some cancellations if you bring them in if you select some master  integrals and write everything in terms of these master integrals which are independent  then you should see it on that form but it's not necessarily manifest in this kind of canonical  representation.  If you bring them in you can manage to get to master integrals by being easier and remain  in integration by part in this case actually.  And I know as an example integration might not give you a reason.  Question here you computed the residue I mean in the example you gave you computed the residue  but can you compute the rest of the external expansion at least the constant term?  Well for this integral I mean it has been computed I think two or three years ago in  the on-channel massless case.  The problem is always what are the kinematics that you do?  So if you look at the view of three spokes and you just look at it as a three point function  and so you put one momentum to zero but take the other three moment as arbitrary then it's  in this class of vertex with three graphs I didn't have time to describe these but this  function you can compute explicitly in terms of multiple polylogisms to all orders in epsilon  close to all even dimensions with arbitrary powers on the propagators.  As soon as you go to four off-shell external momenta I don't even know what the right  representation for the kinematics is.  I mean even if you take the one loop box there with four off-shell momenta it is only known  in four dimensions but if you go to four minus epsilon dimensions you don't really know what  to do because it's just the kinematics is too complicated.  So the only case I know where this is computed with four external momenta is also cheating  because it puts all four external momenta on-shell so it's only function of two variables and  then it was computed by Hennens-Milnoff and it was shown to be a multiple polylogism and  this is actually one of these examples where when you do it with the Schringer parameters  you do not get linear reducibility in the sense that I mentioned here so after one integration  you have a quadratic polynomial so it seems mysterious but then there's a chain of variables  which you can do which makes this thing factorize and then you can apply the algorithm if you  see you get this.  But it's one of the examples where I would like to have a better reason to see why it's  a polylogism or not.  But yes you can compute higher orders in the epsilon expansion.  The reason is that when you go to, so the epsilon sits here in these exponents, so if  you do an expansion in epsilon the only thing you do is that you introduce logarithms of  the polynomials which are of course in the space of iterate integrals over these polynomials  that I start with anyway.  So this is the statement that here the actual integral doesn't really matter so I can allow  for arbitrary powers but I can also expand in the powers and allow for logarithms.  You can also have logarithms of individual Schringer parameters in the game and arbitrary  powers of those.  This does not play any role for the function theory.  Of course it makes in practice a computation more complicated and takes more time but it  does not change conceptually what happens to the integral.  So if you have a linear reducibility without these logs, if you add the logs you are saying  that you will stick with this property.  So I was quite sloppy actually with the way I defined linear reducibility.  So I didn't even mention the name did I?  At least I didn't dare to write it down.  Okay, so the idea is that it's really just something depending on the polynomials.  So you take the polynomials then you see what is the worst case that could happen.  If I take an arbitrary, so you do not exactly, exactly you do not want to look at the actual  integral, just take whatever integral which only has these singularities.  Do one integration, what are the singularities that I could have here?  Then by quite general arguments just by looking at the defibrations of this variety you can  prove these things, that these sets are upper bounds on the singularities.  I mean I hope I made somehow clear how this actually arises in the computation because  what happens if I want to compute such an integral?  Well, this is a logarithm so I have this clear how to write it in this hyperlogarithm  form and the letter I get here, this is psi upper one, alpha one plus psi lower one.  So the letter will be something like, will be some L omega, the zero is at psi lower  one divided by psi upper one.  So after the integration I have of course singularities when these go to zero.  I will have to do the partial fraction decomposition but these are all taken care of by D.  So the changing the powers or anything changes the actual representation at the point but  it does not change the fact that it is a hyperlogarithm with these as an upper bound  of the letters and denominators which appear.  So this is the point and in a way make it more complicated because you have to look  at things depending on all these finger parameters which are completely unphysical of course,  which only appear in the very last step when you did the last integration.  But on the other hand you abstract from the actual integrand and you look at all amplitudes  which you can assign to this graph at the same time because this is something which  only depends on this geometry of the graph hyper surfaces.  All right, one question?  Okay, thank you.  you
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