 Thank you Lowekand, thank you Karen, Eric for organizing the event. So I will be talking about more combinatorial results. So at the end, we should be seeing an application for code diagrams in interpreting Feynman diagrams in Yukawa theory and quench QED and in a way that will help us to evaluate or estimate the asymptotic behavior of some power series. So let's see. So the first part of the talk will be just recovering what is code diagrams and some properties of two connected code diagrams and then we'll go into the applications there. So probably you have seen this earlier in the conference but let's just start by defining what is a code diagram. Imagine you have a circle, you have an even number of vertices on it and you have some connections between the pairs of vertices. We will be considering rooted code diagrams. So imagine again the same circle, but this time you have a distinguished vertex. So this will give you a way to represent code diagrams linearly like that. These spacings are going to be called intervals. The generating series for rooted code diagrams will be denoted by d of x and it's clearly seen that the number of rooted code diagrams on n-codes is 2n minus 1 double factorial just like matching. If we are going to talk about connected code diagrams, this means something like that where you can't isolate pieces away from each other. So for example this one is not connected because we can isolate this piece. It's not hooked with any other code diagram. One of the classic relations between connected code diagrams and general rooted code diagrams is this one. You can see where this comes from. So a rooted code diagram is either empty so you have the one or you can somehow pull up whatever the largest connected piece you can pull and next to each vertex in it you will have from the right you will have a code diagram and from the left you will have another code diagram. That's why you get the xd of x squared. Another useful identity which is also easily obtained is this one. These two equations are useful when you want to study the asymptotic behavior of connected code diagrams and that's what was done in the work of Michi especially this equation if we apply the suitable alien derivative to both sides we eventually get the desired asymptotic information for c of x and that's what we will be interested in doing for two connected code diagrams. So what should be a two connected code diagram? You probably can guess so it will be a code diagram which in order to get disconnected you have to remove at least two codes so if you if you wish to define it formally it will be you have no set of no set s of consecutive end points with proper size which is paired with the actually this definition is for k connected which is paired with less than k end points out of s. So for example the diagram here is is two connected but it's not three connected because we have we can disconnect the the diagram by removing these two bold codes but we can't disconnect it with less than that. The generating function for two connected code diagrams will be denoted by c greater than or equal to 2 and next we will be studying the asymptotic behavior of this power series. In an older result by Kreetman he gave an argument that there is a proportion for k connected code diagrams with respect to the old code the connected code diagrams is is approaching e to the minus k as n equals large. We'll be able to extend this result for the case of for the case of two connected code diagrams this means k is two and we'll see that this this value here is which is going to correspond only to the first term in an infinite expansion. So again in the work of Michi he obtained this value for the alien derivative with parameters two and half applied to x applied to c of x and used that to get the asymptotic information. So the first question do we have an alternative for this relation here? I mean we can dream of something like c of x equaling some expression involving c greater than or equal to 2 of x and then apply it to something else so that we can do the same trick we apply alien derivatives and get the the asymptotic information. So that would be the first thing we do we want something like that involving two connected code diagrams and that's how it's going to look like. Connected code diagrams can be expressed like this pairs of connected code diagrams by the by x minus two connected code diagrams applied to pairs of connected code diagrams again. I will not be giving the proof of this result here but you can find it in the phd feed. So now what we are going to do is as I said we are going to apply alien derivatives to this equation to get the desired information for c greater than or equal to 2 and our way for doing that is through a factorial division power series. I guess you already know the definition of factorial division power series in previous talks but let's see it again. So a power series is going to be said to be a factorially divergent power series with four parameters alpha and beta if we can express the coefficients of it like that. So we have you can consider this modified gamma functions as expressing the divergence of the factorial that's why it's called factorial division power series and these coefficients are just real numbers so and the modified gamma functions are for these parameters are just defined through the original gamma functions in this way. So our passage from such a coefficient to an ordinary power series is going to be through extracting these coefficients and that's what the alien derivatives does. An alien derivative is going to be applied for a factorial division power series to give you an ordinary power series whose coefficients are these real numbers that were besides the gamma functions. You can think of that as we have just executed the factorial divergence and we are concerned with the remaining information about the series. It turns out this is a derivation as Michi proves in his work and the ring of such series is a subring of the ring of power series. We'll call these things alien derivatives and again we have this. So let's apply the suitable alien derivative to the equation involving two connected core diagrams. We eventually get this expression after some work and after using Lagrangian version in the middle. Eventually we get that and if we simplify it we don't have to go into the details here but at the end we are going to express the nth coefficient of two connected core diagrams like that and here you can see that treatment's result is standing for the first term just the first term here but now we can get better approximations as much as we can by truncating this series. So let's see how useful this will become in working in QFT. Just pay attention that we didn't need any similarity analysis to obtain this expression, this expansion and we'll see that the same happens for many series from eukaryotic theory and quench QED. So I will start by eukaryotic theory. So again in the work of Michi he has this table. Let me go quickly through the notation here. So these are the the proper green functions of the zero-dimensional eukaryotic theory and if we have so for example the first line is the proper green function for graphs with no external legs at all. These are the vacuum diagrams. The second one here is we have one we have one boson edge one external boson edge and zero fermion edges and so on. So these are the proper green functions and we can readily notice here that the coefficients for this one for example which is the most apparent one are exactly the number of connected chord diagrams. These are the tadpoles of eukaryotic theory. So let's see how to prove a bijection between tadpoles and connected chord diagrams and that's the main result I want to talk about here and from which all of these other lines will follow. So again for the sake of an example the coefficient of h bar to the 4 in this power series which in its original setting counts tadpole graphs with loop number 4 is 27. This coefficient is 27 and that's exactly the number of of these tadpoles but in the same time it's the number of connected chord diagrams with four codes. So why is this relation happening? How can we jump from such a tadpole into a connected chord diagram and for example for these more complicated ones how can we express them in terms of connected chord diagrams and would that be useful at all? So that's what we are going to prove that the nth coefficient here which stands for 1pi tadpole graphs with loop number n is exactly the number of connected chord diagrams on n codes. Let's see so what we need to do is because we don't see the relation yet we don't see how to move from here to from tadpoles to connected chord diagrams so at least we are going to try proving that tadpoles satisfy the same recurrence as connected chord diagrams and that's this equation which we have given in the beginning for connected chord diagrams. We want to see that t of x is going to satisfy that. The first step is to make a standard way for drawing tadpoles so like that so remember in this theory these two tadpoles are considered different because we have different directions for these for these loops but in our notation let's agree that we are going to forget about the directions but we are going to pay the price we are going to pay the price so for example this one becomes that one but this one just for the sake of fixing a direction counterclockwise we are going to like flip it like that and we'll have we consider we are having this overlapping here. By this way we can ignore talking about the directions because they have been encoded in another way. Also for the sake of more notation we have so for example this vertex which is the root vertex for the tadpole is going to be denoted by vt where t is the tadpole this is the rt edge this is boson edge boson edges are always going to be denoted by a light line and the fermion edges are more bold than that. If we have a vertex b the the the fermion loop it stands on is going to be denoted by loop of b and the next edge the next fermion edge counterclockwise to a vertex is the fermion and similarly the boson of it because it's unique. So let's see how to move on big u10 is the is just the notation for the class of the tadpoles and our first lemma is to prove that the number of independent cycles in such a tadpole or the loop number is the same as the number of all boson edges so that's our first step the number of the loop number is exactly the number of the boson edges so we now have to get the bijection between the tadpole and connected chord diagrams taking into account this fact so that's our algorithm to move from so again let me let me just remind you we want to prove this relation and this combinatorially this means we want to take we want to take a tadpole and another tadpole with a with a distinguished vertex and we prove that from this one we can uniquely get either either for the class we can either get two tadpoles in a unique way or one tadpole and also in a unique way so let's see how we can do so we start with a tadpole t1 and the tadpole t2 with a with a special vertex or a special a fermion edge because they are the same they come the same thing and we have we have this one so if the if the special vertex is exactly the is exactly the root we'll just return the two tadpoles by forgetting this extra information here if it is not we are going to do the following let me explain that on an example it will be more clear so let's see here we have these two tadpoles and we wish to get and this one have an extra information which is this d over here now let's see how we can merge them to get a unique tadpole so first of all we notice that d is not u2 so we are in the second case so we are going to do the following we we take yeah so we take this whole diagram here and we insert it in here in a special way like that since we don't have any internal edges it will follow the following scheme we are going to just put v1 next to d over here and we and next we are going to place u2 next to v1 which we have just placed and the root is going to be at v1 it sounds strange but that's a you this way is unique because that's the only case where you get a two connected graph so if reversely you are going to move backwards let's see this later sorry for that but this example will be more clear in representing the idea so here we have two bigger tadpoles our algorithm is going to be like that first of all we we move from v1 we determine what is the next vertex to it it's w we detach w and then we insert w next to d and we take v2 we take u2 and we place it just next to v1 now if we wish to see how this produces a unique tadpole let's try to move backwards what i'm going to do is i will start by the root that's the only information i know in a tadpole i will check the first vertex next to it counterclockwise and i will detach it i will i i will imagine i cut this edge now the the resulting graph is either one connected or it is two connected if it is two connected this this happens in the only case where we have the first tadpole empty from inside like this one because in the second case we always have a bridge somewhere and that's how we mark our way back so if i'm going to start from this tadpole again i will start by the root the only information i know i move counterclockwise the count is the first vertex i meet i detach then i check if the graph has bridges this one for example has bridges it has this bridge it has this bridge so i will determine one bridge and that's possible in polynomial time yeah i will determine one bridge and then again i will check the the remaining graph which contains the the root i will again check if it has more bridges and i will keep repeating this process searching for bridges till i find the last bridge and this will mark the end of the algorithm i'll just detach there and i will put the w again next to the root of the second tadpole and i will recover the two tadpoles if the other case if it's two connected i will know i have this situation where there is a simple tadpole and a more complicated tadpole and to recover them i just need to remove uh and i just need to detach the vertex next to the root like that and then i will i will remove the root forever and create a simple like yeah i actually i have the drawings here so like here i start by this one i detach the first vertex like that i search for bridges imagine i determine this bridge first then i will keep searching for more bridges here i find b1 and if i search again for bridges i will not find any so that's the last bridge i detach it like that and now i have the the t2 is it's the extra information in here that's where i detach w and then all i have to do is to place w next to the root which is the only information i i ever know in a tadpole so that's how we prove that the recurrence is satisfied by tadpoles but can we have some means of jumping directly from connected chord diagrams to tadpoles and vice versa yes and that's that's what we do in this theory um so it it only takes the advantage of the root shared decomposition and the map psi which we have which is some kind of an order defined of the tadpoles so so let me let me explain this on this diagram here so if i start with a tadpole like this one please ignore these red numbers because they should be information you get at the end if i start by such a tadpole i just apply the the function big psi we have defined in the algorithm i split it into two with a distinguished verdict in the second tadpole and then i keep doing that all the time till the end and these are just corresponding to simple chord diagrams one chord one chord one chord then i will move backwards again using the chord diagrams and the extra information i had here and my way in inserting the chord diagrams is by reversing the the work of the ratio of the root shape decomposition and eventually i get that that's the uh uh that's the corresponding chord diagram in the middle of doing that we need an order on the vertices of the tadpole and that's what i that's what i have defined uh using this map psi but i i will i don't have the time to uh to write it down here but it is simple it's just imitating the root shape decomposition on tadpoles making use of the internal structures that we have discovered here so that's it we have tadpoles corresponding to connected chord diagrams it follows from that that's vacuum vacuum diagrams can be expressed like this pairs of connected chord diagrams of 2x if we have two external boson edges it will be expressed like that and here we see that two connected chord diagrams come into the picture it's you can see it here if we have two boson edges like that first of all i will move from this fixed vertex here i will search for the last bridge uh actually i will first detach this vertex and i will search for the last bridge which in this case is this one similarly from starting from here i will detach this one and i will search for the first bridge which is this one in that case and these will mark where i should cut i will get that thing and this thing which again is going to be isolated like that so you get one of these and two tadpoles and we can express this one also in terms of connected chord diagrams that's that's how we can express it it turns out that it is expressible in terms of two connected chord diagrams i don't think i have the time for the other for the other parts but just to say all the lines or all the the graphs with the different external structure turned out to be expressible in terms of connected and two connected chord diagrams and the old follow more or less they follow from the the result of connecting chord diagrams and tadpoles what is the advantage of doing that we can get the asymptotic behavior of such a green function without singularity analysis because we already had we already know everything asymptotically about connected and two connected chord diagrams a similar result is for quenched qed i could prove that the counter terms or these special counter terms of quenched qed are counted by two connected chord diagrams as well and that's it we see advantage of all of that is to evaluate the asymptotic behavior more easy thank you thank you all right are there questions for ali David has his hand up yes congratulations on this wonderful bijection so i have an obvious question in unquenched qed what happens if you restrict your tadpoles by forest theorem to those in which fermion loops contain only a odd number of vertices what can you say about the subset of chord diagrams that arises from your bijection so you mean you are allowing fermion loops to existing in that setting you mean no i impose forest theorem you went from all tadpoles to all chord diagrams but your some of your tadpoles would not exist in qed because on your fermion loops you had both odd and even number of vertices so if i restrict it by forest theorem that says that we only have an even number of vertices on a fermion loop then that identifies by your bijection a subset of chord diagrams so what is the kind how on the chord diagram side do you describe that subset of tadpoles which satisfies forest theorem um yeah but it's not necessarily that we have an even number of vertices on on these tadpoles it can be arbitrary if i if i understand the question properly so for example this simple one is having just this vertex also this one have like three vertices it's not that we are just taking taking this and putting it as a chord it's it's it's different it's more or less different but again we can have odd number of vertices but the thing ali is if you impose if you looked at just the subset of diagrams that had this oddness restriction then you would get a subset of the chord diagrams and david's asking which ones you would get okay i see um i i'm not sure david um i i didn't think about that but i think it would be interesting to it's a very interesting question because implicit in your work is that subset of chord diagrams that corresponds by your explicit bijection to forest theorem so you call them for a third diagram you know write a paper about about for his chord diagrams yeah anyway great work thank you david yep great thank you are there any other questions for ali i just have an obvious question maybe briefly um what about the higher order cases like the three four five connected i have thought about that michi uh but i didn't go anywhere i don't know if i have told you earlier but yeah i i didn't know uh what happens with k connected chord diagrams generally um it doesn't look uh easy and i think just as the case of k connected graphs we we were not able to uh to move forward i mean even for graphs not not chord diagrams we we only can express the generating series for two connected graphs and the same happened here but i hope that the result by cleatman is an evidence for uh that we can't do something uh i mean it's it's giving you a fixed pattern for the proportion of k connected chord diagrams so maybe there is a way to express k connected chord diagrams in terms of the previous ones but i don't see it now happy to see you right yeah thanks for the great talk ali good to see you too thank you all right let's thank ali again