 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 intervals. I want to go back to the paramedic representation that Francis showed you in the morning and actually Francis put forward a at the time very new approach to the computation of Feynman integrals which uses hyperlog rhythms throughout and is one attempt in one way to understand the prevalence of multiple poly logarithms and MZVs in in the calculation of many Feynman diagrams. So I want to talk about Feynman integrals, but specifically their connections with hyperlog rhythms and the motivation for this particular approach is or the goal is to understand why are so many Feynman integrals just abbreviated by Fi Expressible multiple polylog rhythms and Francis already showed you these functions in the morning. I just recall briefly So these are functions which you can define like like multiple series. I just write down now So they're indexed by a bunch of integers and want to indeed and they depend on several complex variables and one way to define them is via this summation over nested integers K1 up to Kd 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 stated by the index of the multiple polylog with them so You 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 of this form You know explicit count examples and we have many conjectures of graphs where we are pretty sure that they're 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 set of measure 0 in in whatever kind of measure 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 this form is very difficult and We will I will try to demonstrate to you one approach where which is decent Some cases makes makes it very clear why this is the case And I try to be very explicit and very down to earth So I will 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 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 view 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 There are three spokes Let's call these edges one two three Four five six There's you can think of it as a graph in five to the fourth here We spoke with four external legs, but because it's logarithmically divergent in four dimensions It has a 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 zero to infinity All the Schwinger parameters Well, not actually all of them Because remember in this form it's a projective integral 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 one over psi g squared That but I set alpha 6 to 1 sorry So this is the integral we want to compute and Recall that by definition This semantic polynomial psi g was the sum over all spanning trees And each spanning tree contributes a monomial which is the part of all edges not in the spanning tree so in particular this is linear with respect to Each individual variable and this means That that the first integrations are essentially elementary. So how does this work? You want to compute? Amplitude Let's let's just do the first interval over the first variable from zero to infinity of So over alpha 1 over psi g squared 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 zero infinity and We just get the one 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 the alpha 2 and I just continue right 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 and psi upper one lower to the constant part with respect to alpha 2 in the polynomial upper one So these are just shorthands 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 1 times the constant part So this is a quadratic polynomial, but it's factorized in two linear forms Yes, so these these are polynomials here So psi upper one Psi lower one are the coefficients with respect to alpha 1 so they are still polynomials in the remaining variables This is important No because these are linear polynomials in the variables They are linear in each variable I start with a point which is linear in each variable and I pick a coefficient So it's still linear in all the other variables That's the important point So I only have linear terms here, so I lower one upper to alpha 2 and the constant part So how do I do this integral while I do a partial fraction decomposition? so I get a Prefactor from the partial fraction decomposition Which you can work out very easily and then you're left with the integral of Over other 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 in the partial integral we have computed so far as a 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 edge one But which do contain edge two if you think about this actually means that this is the Ordinary graph polynomial of the graph where you delete edge one and where you contract edge two There's a so-called contraction deletion formula If you take the graph delete edge one, but contract edge two what happens is that six and four become parallel six and four 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? Well, it's again The spanning trees which do not contain alpha three so I can just delete edge three And then I'm essentially left with the 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 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 competition with respect to alpha three I look at the coefficient of alpha three and What happens when I said after three to zero so I contract after three and then everything becomes parallel Now we can just do the same game the other way around so I delete edge two and contract edge one then five and six become parallel and Three and four remain and Again, I can delete after 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 graph and I just told you that you just have to sum all the variables in the loop To get its polynomial And the last guy is the one where we set up a one and two to zero so we contract both edges after one and after two Which means that the edge three is now connect both endpoints of three are now connected because if you contracted 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 just summing over spending trees I mean no spending tree can contain edge three because it would have a loop and the spending tree is not allowed to have a loop So actually no spending tree contains alpha three which means that after three actually multiplies this polynomial It's a device of the polynomial And then the thing that remains is just again this sunrise polynomial again, so We'll 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 sub as you multiply this out the only thing that remains here is alpha six squared So what you find is I mean each time you write this sunrise polynomial it can take different variables It's always the same. I was just lazy. It's it's always four five six Yeah, I mean it's symmetric is it symmetric polynomial of the three variables I have only four variables remaining and I explicitly look what happens with alpha three and Four five six remain So we observe There's a miracle Happening here Name that this thing Actually is a complete square, which you can see over there. So it's just after three times alpha six plus the summarized 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 some 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 until 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 Yeah, 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 after three zero at infinity and we get other terms where you have to integrate still But the log became rational Then we do again separation Partial fraction decomposition then we again get locks, but the point is we don't get a dialogue over them yeah, 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 of the residue of this graph You get you get because of this integration by parts you get actually several terms, but you can You write them with a symmetry and you actually see that You can write the three times the same integral which in terms of alpha yeah for and then the sunrise polynomial and Logo with them four plus alpha five four plus alpha six over The sunrise polynomial Yeah, 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 dialogue with them at this stage Now I want to write this out explicitly So remember we have three variables left after four after five after six, but after six is set to one So we are actually only left with a two-dimensional integral and I would just rename the variable. So Alpha six is one Alpha five I call y And alpha four I call z And this is the integral zero to infinity D y over y and You can do the path refraction decomposition here Which I already did for you here in this expression So we get 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 one and Y plus one That's right over Y plus one So we boil down the computation of this integral to do this two-dimensional integral over logo with them and We again see mean that there were already Dodgson identities at place in the in the first Thing that I showed you in detail, but also in this integration by path procedure here We have to do new partial fractionary compositions and also there you actually make use of Dodgson identities So the the observation here is that again everything is linear everything factorizes into linear Functions in the next integration variables Yeah, so if you now integrate that everything is linear and then with respect to y and also the arguments here linear and z so Everything looks very linear But of course at this stage now We have to introduce more general special functions because you cannot compute this integral in terms of classical Logo rhythms and rational functions anymore So we do need a dialog with them But the question is how do you? Represent these poly logarithms 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 poly logarithms The multiple poly logarithms are not only sums, but you can also write them as iterate integrals and so for this Let us take a set of points in the complex plane should be a finite set and We think of we call these these elements of this set singularities or letters But I want to write them with a special symbol, so I introduced symbols omega sigma which should indicate that this is actually representing a differential form For all these letters So this is an abstract alphabet Then we can define special functions for each word in this alphabet If I'm sigma is no more the domain of integration Sigma was upstairs the domain of this is a capital sigma. Oh, no no no in the interval IG Yeah, but I mean the element. Oh, yeah, so sorry. Yeah, this is a difference sigma. Yeah, sorry so Then we define the hyper logarithm LW of Z associated to a word W and 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 normalization by n factorial Then we also want that If you take a hyper logarithm which has a word which begins with some letter then there come some other remaining letters That this first letter tells you the differential behavior so this should be one over Z minus sigma times The hyper logarithm associated to to the tail of the word This is just a 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 Limit When that goes to zero of these hyper logarithms zero Unless the word is Off the form Omega null to the n. So if you have a string of zeros We just fix them to be the logarithms, which of course diverge quite mildly at zero But for all the other hyper logarithms define in this way, you should think of them as iterated integrals from zero to Z So in fact Yes, sure, that's exactly what's going to happen next so example Well, what happens if you just take one letter And this is just the integral from zero to Z of Dt Over t minus sigma Because the integrant is given by the differential behavior, but it also has to vanish so sigma is now supposed to be non-zero Has to vanish at z equals zero. So it's this iterated integral and this is just the logarithm that minus sigma over minus sigma then There is L Omega zero omega one of Z What is this does this look like well we have to integrate from zero to Z D t1 or t1 minus Zero Yeah, this is the last integration and we differentiate with respect to that we have to get get down this Then we have the integral The nested integration Dt to the t2 minus one and if you recall Francis briefly mentioned the intro integral representation from li2. So this is actually minus li2 of Z and quite generally When you have You have a bunch of zeros. I'm sorry. I'm changing Orders now The 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 a 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 them 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 the benefit of writing them in this way is that it's trivially it is trivial to integrate in this representation because we just Define them as the iterate integrals So if you 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 The arguments of the logarithm so This integral zero to infinity d z one over z minus one over z plus y over one plus y This logarithm This is nothing but omega minus y and I have the logarithm With a single area z equals minus one And I have in the denominator The logarithm which has it's a single area with respect to z at minus y over one plus y Now I made sure that I have the right differential behavior I also have to think if I'm if I fix these constants Correctly which do not depend on that 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 us by definition just means that I prepend an omega zero and an omega minus y over one plus y To get a primitive of these and I evaluate it infinity So in this language, this just means The following so I introduce a short not 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 express the Penultimate 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 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 emma which you can prove suppose that we take 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 they are not fixed really complete the full total derivative And you can actually prove that there's 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 Some boundary terms of this formula so There's a sigma zero appearing here, which is defined to be Z and the sigma n plus one which is defined to be zero These correspond to the boundary terms of the integration. So actually there's a totally symmetric formula Now with this formula if you apply if you apply the total derivative now to such an expression You see this this grading by way this recursive structure which is so special about multiple logs 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 at the differential forms defining a hyper logarithm So this looks good and for this piece which in this case will still be a log over them 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 the 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 one over y integrate from zero to infinity So this just means that we have to put another omega zero in front so We finally arrive that The Residue of the beer with three spokes graph that we started with is three times This thing so we have now an iterate integral of weight three And as a hyper logarithm it has in this form evaluated at infinity Which still might look mysterious, but what is it? I mean this is an iterate integral C minus zero and minus one So we immediately know that it's a multiple zeta value you can also think of it As a period of m Zero six 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 one in some way So we can apply the mobius transformation which sends z to z over z plus one which means that Boundaries the upper boundary of infinity is now mapped to one Zero stays zero and you can compute the pullback of the differential forms Under this moogle transformation, which is a simple exercise And if you plug this all in You get that this is the same as Interval omega This Omega zero one this is now a Hyper logarithm evaluated at one which is essentially a multi-positor value There's still a little thing with the regularization you have to do because it starts with zero and you can use a shuffle product but if you use this You can show that this is the same as Yeah 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 And if you just take the definition of the multiple Polyloguism how they relate to these hyper logarithms namely this formula I can and see that this is essentially li three and this is li one comma two So this is three times li three of one Plus li one comma two of one Which is six times it all three so I spend an awful amount of time on explaining this first non-trivial example But I hope it was at least understandable and 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 in for example these single-scale master's fight of the four integrals that Prince was mentioning This is a case where we have I would think the the best knowledge Concerning the expansion in the loop number. How far can we get and still compute a lot of intervals? This is really Outstanding compared to other kinematical configurations. It can be much more complicated So we know we have already two loops where we get elliptic things these massive sunrise diagrams but in these master's diagrams following these lines one can get quite far and The the basic idea is exactly the same. So Let us recap what we did we computed Partial Feynman integrals meaning I call this I index K or friends called that I index K Which still depend on the remaining variables from K plus one up to the final variable So integrate from zero to infinity the first K variables and Let's 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 such as hyper logarithms, right? I was only telling you how to manipulate hyper logarithms But in many cases in the complicated ones the fine integral is not expressible as a hyper logarithm And at least not in the Schrodinger parameters and maybe it's 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 is a There are some devices. There's some variety 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 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 such given a such an implicit form Into a representation, which is explicit in y and then we can just find a primitive by Pre-pending letters or doing integration by parts and we can compute the fine 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. Yeah, what happens here? What are the precise poly logarithm which appear? This is not so important The only thing that is important is what kind of singularities are there and how do they depend on the variables? So I just want to briefly sketch the idea behind this so in the beginning Now we don't we didn't integrate any variable yet. So we just have one over psi squared the only Singularity is Psi can only have a singularity when Psi vanishes after one integration That's what happened after one integration I'm not sure if I still have it and we are still Almost do 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 If you reduce this then this was also linear polynomial, right? Which I discussed earlier. So this D D was squared So in this also in this case Paragraph is actually fulfilled that all the potential singularities are linear in the next integration variable Now if you want to understand Feynman diagram and 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's 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 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 mild but poly logarithms of a particular type so note The actual integrand Does not matter by which I mean we can take any Intergrant 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 There 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 Restricted to the one scale case. We can also look at general Feynman amplitudes We can also have such a polynomial But now allow for both graph polynomials Psi and the psi polynomial raised to arbitrary integer powers Subject only to the condition that it is a well-defined projected integral so in this case We start with a variety 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 exciting almost quadratic and you actually do not get very far But there are many applications for example without messes or just with few messes 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 what the outcome of this techniques is So we have reduced The study of the actual amplitudes and the integrals to algebraic or geometric Task of understanding how these Sets of polynomials which describe the singularities how they behave so we have The the amplitudes Which of course come from the graph But some of we take a detour if 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 finger parameters Yeah, so here we we have a chain how we get here via these integrals partial integrals okay, plus one 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 The we only need to know the single the 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 Yeah, 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's a theorem by him Let G have vertex with or actually able 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 is what would be called a triple box And you can have arbitrary and opal 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 poly logarithm and You can actually say what kind of multiple poly logarithm so you can write down an upper bound on Then on the differential forms which make up this iterate integral or in other words You can specify what the arguments of this multiple poly algorithm can be in the worst case And there's 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 and again You can generalize this to have three external momenta and Because these graphs are very special cometola structure. They're 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 poly logarithm 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 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 various tool 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 to scale 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 art part Okay, and that's exactly what most of people Involve not in exact computation, but in a factorized form or high energy do every day So there's an expansion in like yeah, yeah You try to guess 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 doing an asymptotic expansion like this on other cases I mean in physics, there's this whole business of expansion by regions for example You just take an individual Feynman diagram and look at the expandable some momentum gets big or some ask it's big Or you're doing this kind of thing You're always doing in this picture it has unified interpretation in some way because look at this integral Which is made up of this of this side polynomial and here in this polynomial You have your different masters and momentum now you're looking at what happens when one of those dominate and the others So essentially you you would hope that you can expand the integrand in this limit And then just compute the integrals the integrals themselves and the only problem is that if you do this expansion You might introduce in divergences Which you might have to take care of but if this is not the case then then you're right So you're using in this case Effectorization of this polynomial in this particular kinematic limit what I did what I did here is that I looked at a factorization of the psi polynomial Which does not involve any kinematic invariance But you also have factorization polynomials for the for the xy 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 you bet 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 So in the same spirit So if you have a gauge theory, okay, you have Because some of many diagrams at each individual diagram produces you some let's say power of Hard scale or whatever, okay, which is complete which has completely nothing to do with if you perform summation So do you have some clever way to treat numerators? Combine okay somehow I 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 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 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 right in a different form then they are manifest and then it's easy. So I Know that some people have tried this in other applications The two to try to combine different diagrams and see why certain cancellations happen engage theories But as far as I'm aware, this is still very much work in progress and we don't don't yet know The right approach for this the problem is really that the representation is not unique And you you would need a way to find the right representation in order to see it easily So and but probably finding that representation is as complicated as solving the problem, so What you have done here was not change of representation under steps of your computation You don't see it as a change of representation. You mean change of variables or yes I mean all those steps which you Yeah, but I only look at one integral at a time Yeah, and and I'm changing it that's that's right. I'm changing the representation of my actually I don't take the function I just rewrite in different way, which makes it essentially trivial to compute the interval in this new representation So this is 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 Where we know that an integral is a multiple poly logarithm 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 true-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 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 do some changes of variables and then it's Clear that it's a polylog over them or something related, but not in the original variables well For the formula with a general polynomial p. I think you in some cases that I know integration by part Enable to transform this into another form with different nb. It's more or less like we have seen in the previous talk Mm-hmm Yeah, so there are certainly tons of relations here, right, but the problem is I Mean that the question that came earlier was if you have different integrals and you want to combine them Yeah, and see that there are some cancellations If you bring them in a you if you said 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 the In this kind of canonical representation To master it might be easier and I'll remain to increase the integration by part in this case And I know as an example Question here you computed the residue in the example you can compute the residue But can you compute the the rest of the extension in the constant term? Well for this integral I mean it has been computed. I think two or three years ago In the in the on-tell massless case. I mean 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 Yeah, so you put one momentum to zero But take the other three momentous 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 poly algorithms to all orders in epsilon Close to all even dimensions with arbitrary powers on the 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 know what to do because it's just the common the kinematics is too complicated So the only case I know where this is computed with four external momenta is also cheating because it was all four external mental on-shell So it's only function of two variables And then it was computed by Hennens-Milnoff and was shown to be a multiple polylog with them 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 at some point after one integration you have a quadratic polynomial So it seems mysterious But then there's a change of variables which you can do which makes this thing factorize and then you can apply the algorithm You see you get this but it's one of the examples where I Would like to have a better reason to to see why it's a polylog with them 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 right in these exponents So if you do an expansion and epsilon the only thing you do is that you introduce logarithms of the polynomials Which are of course in the space of iterate intervals over these That I start with anyway So this is the statement that he had 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 finger 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 to the integral Reducability Yeah, so I'm I was quite sloppy actually with the way I how 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 Yeah, so 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 do not want to look at the actual integral Just take take whatever integrator which only has these singularities Do one integration what are the single areas that I could have here then by quite general arguments just by looking at the defibration 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 hyper logarithm form and The letter I get here this 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 decompositions, 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 hyper logarithm with these As an upper bound of the letters and denominators which appear so This is the point 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 reduction integration But on the other hand you 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