 Thank you very much, Min. I would like to start by thanking the organizers of this conference. Yes, Gerard. So I'm very delighted to be here and I think to keep up on organizing such meetings is extremely important during these times. So I would like very much to praise this opportunity and to keep up doing research so that life can continue. Anyway, today I would like to present to you this work. It's a joint work with Sanjay of Queen Marie. Can I draw a little? Why can't I see my yes, Sanjay? Yes, from Queen Marie University of London. You can find the main things here on this archive paper that appears recently, but you made it also with two paper to understand better if you want the detail and some of the theorem that I will present you today. So as I have many to say, I will just move to the next slide. Isn't it just my arrows that make it moving? I cannot move the screen. Isn't it just moving with the arrows that make it moving? Yes, yes, we see the arrow moving. Okay, good. So here. So good. Now I have some, can I, I think, raise everything here. So the outline of my talk is the following. I will try of course to motivate why, what are we doing here and where does it come from? What is this question about the chronicle? And I will try to have an overview in the introduction. Then I will start by setting up our tools, which is a particular algebra here, which is also nilbert space. And this algebra has a reason from counting graphs. So everything has started from here. And using permutation in fact, so counting graph using permutation methods. That was a starting point of this story. And then we will move through three steps for proving our main results, step one, step two, and step three. The first thing is to introduce operators, amiltonians on this algebra. And we will show that these amiltonians have some integrality structure due to the fact that even the structure constant of this algebra is by itself integral. So this is an important feature that I need to discuss. Oops. Oh, it just jumped to, let me go back. It just jumped to the right thing. So I want to erase this small thing, but I think if I just cannot, so I will move on. And then having linear operators that are amiltonians on my illbert space here, we could discuss something like a quantum mechanical model. So these two things shows you that there is a quantum mechanical quantum mechanics on that illbert space. And that is already interesting. Dear Joseph, we see your drawings. Can you clean them? Yes. So this is what I try to do. But I think as this slide is kind of is produced from from the latex, I cannot remove them. I try to by using the eraser here. And but they Oh, yes. Okay, it works now. Previously, it didn't work, but it's okay. Sorry for the disturbance. So, yes. So I was saying the step two was to introduce this amiltonian, the step three is to conquer, you know, to to to to discuss. First of all, how do you extract the square of the conical coefficient out of this algebra and then the conical by itself. And the thing here is here is to find an interpretation or combinatorial interpretation for for the conical. And that is a longstanding problem that I will just describe before drawing some conclusion here. So let me erase this before it's going to propagate along the slides and move on to the introduction. I think it's taking a little lag before it's moving. It doesn't move. So I think it's making a slight moment before is moving from one slide to the next. I don't understand why it's not moving for the moment. Do you see the next slide or do you still on the outline? Sorry, outline, outline. So I doesn't move. I don't think I don't know if it is a default of my connection or can I can I escape from here and come back? Maybe it is the key you use, the key. Well, you just have to push one arrow, the arrow for moving from one slide to the other. So let me escape this. I think I'm kind of stuck. Okay. And I don't have any more control in my device. So let me first off. Okay, good. It's moving now. Okay. So I think when the annotation here makes me some trouble, I won't be able to use them anyway. So let me push this around and start. Okay. So the conical coefficient, what is it? It's counting just the multiplicity of an irreducible representation in the tensor product representation of the symmetric group. Here Sn is the symmetric group of permutation of m objects and taking two spurt module, v mu tensor v nu, you can expand them back in irreducibles up to, you know, multiplicity, which is your conical here. That's my conical coefficient. Mu, nu and lambda are just partitions of n's or young diagrams, they are equivalent. And so here you have this direct somewhere up here v's c, which is the conical coefficients. So that's the multiplicity. In a more symmetric way, you can rewrite the same object as that average sum of a convoluted product of characters. So chi here, chi mu is my character of my symmetric group, given, you know, a partition mu. So it is a longstanding problem to characterize, to give a combinatorial rule to understand this conical coefficients. And as they are integers, as we see, we see them here, see here, nothing but multiplicities, a dimension of multiplicities. So they are just integers, positive integer, in fact, so, and as they are integers, positive integer, are they counting some object? So this is a very old problem stated by Murnahan. We are in 1938. So this problem has been treated along, you know, along years, for instance, Stanley has stated it in his 10th problem. But here he's just using symmetric functions for, you know, for, for, for seeing the c appearing. So and he has this comment, often these coefficients meant by this C in particular, we'll have a representation theoretic interpretation such as the multiplicity of an irreducible representation within some larger representation. This is exactly what I described before. Sometimes the only known proof of positivity will be such an interpretation. And the problem will be to find a combinatorial, combinatorial proof of that. So how do we recharacterize combinatorial and combinatorially the conical? And in fact, the conical coefficient is an object which is very well studied in theoretical computer science, theoretical computational complexity theory, as it is the main object that people in the program of geometric complexity theory program are studying. Geometric complexity theory has been introduced by Murnahan and Sohoni. And one, they try to answer one of the most famous problem in computer science, which is how do you distinguish how do you separate the classes P and NP? Along the years, many people have contributed to that. And still recently, with understanding how you can separate the classes still open, of course, but let's have a look about this comment by Bakken Panova saying the following. More importantly, the conical coefficient actually are famously sharply hard to compute and NP hard to decide if they are non-zero. So one should not expect the closed formula. What makes matters worse? What makes the matter worse? It is a longstanding problem, citing Stanley, to find a combinatorial interpretation of the conical. So it is not even clear what we are counting. In this paper of Sanjay and I, we provide a combinatorial answer to that problem. And in fact, the conical coefficient is counting the dimension, which means vectors of an integer sublattice in the lattice generated by bipartite-ribbon graphs, with some specific feature that I will, of course, make precise, regarded as vector, spanning a given space. So that's what they are counting these conical coefficients. And along my talk, I would like to prove this statement. So let's review our main tool, which is this KN algebra. And actually, all are started in theoretical physics and leading us to combinatorics. And as we were interested in counting graphs at the initial level. So I have to say that there is permutations methods for counting graph have encountered a very success in theoretical physics. In particular, they have been applied to compute multi-matrix correlators in Super Young Mills theory with UN gauge symmetry. A lot of people have worked on that. And also for exploring half BBS sector and ADS-CFT correspondence. So the recent paper by Rangulam and Kemp actually in 2020 would make me give you perhaps a kind of another view of what's going on there. So it's a well established method for counting graphs, counting observables. As I was, and Sanjay asking some questions about enumerations of graphs and observables of unitary tensor model. So this is where we are starting. And we have very simple question about how do you enumerate observables of unitary tensor model. So that was a question that we have back then. And we were interested, you can, you may be interesting of tensor of rank three. And how do you count all observables of these types? And we found that actually these observables were in one-to-one correspondence with bipartite ribbon graphs. Counting bipartite ribbon graphs actually I have to be to make things clear here. I'm restricted with ribbon graph with N edges. And as they are bipartite, I will focus only on, I will say at most N, vertex of a given type say white and N vertex of a given white of a given of another type say black. So that's the type of object whenever I say in this talk bipartite ribbon graph, it's at fixed N, at most fixed, at fixed N edges, at most fixed N vertices and N black vertices. I hope it's clear. So we were interested to count these type of objects. So what is a bipartite ribbon graph? You can consider it as an equivalence class or an orbit. So take the symmetric group of N objects, take a pair in SN cross SN, right? And now you let act on that couple of permutation of these pairs, you know, the simultaneous adjoint action on both of the slots. All right. So having the entire orbit here is should be seen as, you know, a ribbon graph. How does it work? So let me just show you. So the construction of the ribbon graph works like this. The cycle of the first element here Sigma one will be, will define vertices of a given type, say black type. The labels in the cyclic order, you know, where you draw the label in the cyclic order, they appear in the cycle you want to consider. And then you have to give a unique orientation for all these vertices. Play the same game with Sigma two. So you have a second type of vertices say white, you draw the label in the cyclic order and et cetera. Then you just have to glue all labels, you know, label by one from one side to the other one, the labels given by two from one side to the other one. So this may be a little bit too theoretical. So let me, let me give you an example. For instance, this is, if you want the identity I make here, this parenthesis empty for, for, for denoting the identity. So you have here two cycle of length one, perhaps here I need my annotation tools. And perhaps this is useful for giving you explicit example, maybe I can take here another risk to use the pencil. See the mouse anymore here. So let me take here the pencil and draw you. So here it's my identity. It is made with two cycles of length one. Hope you are able to see. So that's my Sigma one on one side and the same for Sigma two. So I'm with that example here. So the same for Sigma two, right? So it's two identities that I'm trying to draw the ribbon graph associated with that. So as you see, this is a cycle of length one. So it's a vertex with only one label. So that's that vertex for instance. So and I put the label one as in this cycle you have the label one. So this is what I'm putting here. Second here, you have the second cycle, which is the cycle of which contains only two. So you draw here the cycle of length two here, which we have just one label, sorry, which is the two. So here we go, you have this. Great. So you play the same game with this on the other side. And you do have for Sigma two, now the same vertices here. Okay, so again, and then you join the labels, you may join labels one with one, two with two, et cetera. I hope this is clear. And so that you see. So let's make something more complicated. Let's have a look about on this side. We can look on this side. So I have your one cycle, Sigma one, made with two labels one, two. So this is the way that you draw them in cyclic ordering one and two labels hooked to that vertex, that black vertex. On the other side here, you have the second, the second permutation, Sigma two, where you see here again, two labels one and two. If you don't have too many labels, it's becoming a little bit trivial, of course. But when you have more labels, it's not obvious that you will have such a simple figure. And so let's have a look what's happening if you go in higher rank. Okay, so let's let's me take three vertices. Okay, I have to clean this. Perhaps. Can I just move on? Doesn't want to. Okay, sorry. Okay, good. So here is, for instance, the vertices with more edges, sorry, n equals 3. So I need to clean a little bit the board. There is a small lag between the moment that I am annotating and the moment is diffusing. I don't see any more my mouse. Anyway, so let me keep up. Look at the figure, for instance, the 11 figure below. Okay, I don't see even my mouse. Okay, so here we go. So let's focus on this figure number 11, which is right below. It is made with two cycles, one, two, three, and one, three, two. So you just list the labels one, two, three in that order in that cyclic ordering and one, three, two in the other cyclic ordering. Okay, here. And then you join one label one with label one, label two and label two, label three and label three, label two and label two. So in this situation, you generate a graph and the cyclic ordering here defines for you the embedding of the graph in the surface. Sometimes, of course, you may generate graphish genus. So for instance, this 10th graph, I hope you see my mouse because now I don't, I'm not able to draw easily on the slide. It would be more convenient to do, but I think there is an issue with that. So the 10th graph, as you see, it's one, two, three, one, two, three, and there is now a crossing of edges that makes this 10th graph non-planar. So these are certainly a genus. This is a graph with genus. I hope this is clear. We can move on and I need to clean here a bit the board. Dear Joseph, you have a dustbin. You have also trash. Yes. So, and if I'm trashing, he's doing it all along. Clean or drawing. Okay, great. That's fantastic. Thank you very much. This I think will be useful. All right. So counting, counting graphs and counting orbits. So as I was explaining to you, so a reborn graph is an equivalence class of SN course SN under some diagonal action, diagonal adjoint action on both slots. So if you are interested in counting how many reborn graph do you have, you have to count these cosets. And it is straightforward by using bandside lemma that counting orbits is counting fixed points. So you can implement the following formula where the delta function here is the delta function on the symmetry group telling you delta of the identity is one otherwise that's zero. And that is the formula for counting those reborn graphs. And you can certainly play with your favorite program to compute how many of these number you have. So this is your sequence of number. The fourth, the fourth number that you see here, the fourth is because you have four guys here. And here you see you have 11 pictures. That's the 11 that you see here and is going for much complicated objects in higher end. All right. So this is the counting. You may ask yourself, what if you revisit the same counting under a different light, that of representation theory? Okay. So this is if you want direct space computation with only permutation and groups, what if you go to the representation theory? So the delta function in representations actually develops, expands in terms of the character. Okay. So this is my character R where that DR is the dimension of your presentation. So it's that's an expansion of the delta functions. R is a partition of N or a young diagram. And that's a definition of your delta function. So here I'm just playing a trick. I'm adding more variable in order to quickly expand them all and recognize that I'm in fact discounting is nothing but the sum of squares of the Conecker coefficient. So with this small game, you are able to say the following. In fact, the sum of the Conecker squares are nothing but the number of bipartite rebone graph with N edges. All right. So this is already something that is interesting because you do have a combinatorial interpretation of the sum of squares. So but not yet the C by itself, but the sum of squares are indeed constructible because you just need to draw all these type of rebone graph to know what is the sum of squares. All right. So but this formula looks looks interesting by itself. It's a sum of square. Does it have an interpretation? Does it have a meaning? And the answer is yes, but you need to define an algebra constructed around these rebone graph and this is what I would like to explain you. We call it the graph algebra because it's an algebra made out of the graph seen as vectors. All right. So consider the group algebra C of Sn and this is just the vector space generated by linear combination of that sort, the coefficient being taken in C. And looking at your orbits of this equivalence relation, consider the following subspace of the tensor product of C of Sn times C of Sn. So the tensor product of two copies of the group algebra C of Sn. Well, inside here, you can sum over all these elements, all the orbits. All right. So this is given a sigma one and a sigma two. You take the sum over the orbits of these tensor product of the two slots here and you generate, sorry, you generate here an algebra, a vector space first, sorry, a vector space. And it's a fact that the dimension of that vector space is nothing but the number of rebone that you have before. Why? Because I'm just having here all for each sigma one and sigma two, I'm generating the orbit of it. So you don't have more than these numbers of vectors. You don't have more than that. Okay. So the span of it gives you the already dimension of K of N, which is this Z that we have already counted. So the second fact that you have to understand here is, so you do have the rebone graph as a geometric surface. It's in one-to-one correspondence with the orbit, an orbit, that orbit, but it's now also a vector. Okay. So you have these triple correspondence between a rebone graph, orbits, in C, in SN cross SN, but also a vector, a vector of, in K of N, a vector spanning that K of N. And this, in fact, this K of N is more than that. It's an algebra. You can check it in the following way. Take that vector, that particular vector, spanning your space, multiply it by another one, and you will see that it's some, is giving you back a sum of element of that sort, up to some composition here. So you are, the multiplication is stable. So you are in an algebra, and you can even prove that this algebra is associative. This is just because you, it's just inheriting, it's inherited from the fact that the permutation are, in fact, associative. So you're, you do have a associative multiplication, your K N, which was a vector space becomes an algebra. And as I said, there is more than that. There is a pairing on that algebra, which is just given by this formula 14. I'm just taking, first of all, the delta on each element of your, the product of delta on your groups, that you extend by linearity to K N, and you can show that indeed this pairing is non-degenerate. This makes K of N a semi-simple algebra, and we do have the following result. So K N is a unital, associative, semi-simple algebra. The unique is coming from the identity times the identity thing. So K N is a unital, semi-simple algebra, and there is a representation theoretic base, Q, which that makes the Verdemba-Arting decomposition manifest. So the Verdemba-Arting decomposition, it's the one which tells you that every semi-simple algebra decomposes as a direct sum of matrix of algebras. So this is the fact by Verdemba-Arting. And now you would like to know what is the basis? What makes, how do you see the basis that make manifest that decomposition in matrix blocks? So that's the Q base that you see here, and it is labeled by three young diagrams, and A B below here are exactly the matrix if you want the entries of your matrix, the indices of the entries of your matrix. So how do you produce this base? First of all, introduce the representation base of C of Sn. So it looks like this formula 15. I'm summing over all groups element. I'm having here the so-called Wigner matrix, if you want associated, that's a representation matrix in dimension R. You sum over all these matrix elements and this produce you an element here, okay, Q R, R J, at fix R, at fix level here, R, and I J, which are also fixed. So this element, you can show that actually it's a base, it's a basis element of that, of that algebra. The K, the Kappa R that you see here is just a normalization factor that you use to make it autonormal, okay, so it's just a pre-factor that plays that role. How do you produce now the basis of K of N? So here is the generic way of doing that. So take the ordinary base of C of Sn cross C of Sn, the tensor product, so I'm taking two of these guys, okay, and take the tensor product so that they belong to C of Sn, or times twice. I make it invariant by acting left and right, you know, on each slot by the adjunct action on each slot. So this is already invariant, but still you do have these indices, I J, I J, and I'm using two Glebsch-Gordon, two Glebsch-Gordon to contract those indices. This I J, I don't want to see them anymore, so I'm using here some Glebsch-Gordon to neutralize this, to contract that. So, but you have to pay attention that on the Glebsch-Gordon of the symmetry group, you do have here a tau index. That tau index is the multiplicity. It counts the multiplicity of the tensor product of R3, of R3 inside the tensor product of R1 cross R2. So you do have this multiplicity, and that's the multiplicity ranging from one to the conical coefficient by itself. So here is your, actually this is a, you can show that this is invariant, this block here with Q is at fixed R1, R2, R3. It's a matrix in tau 1, tau 2, and it's a matrix of size C, the conical by itself. So that's a block decomposition. That's your block decomposition of your entire space, the KN space. Kappa here that I'm putting here is again a normalization factor to make this, you know, orthogonal, to make it or to normal if you wish. Another important thing about KN, it's in fact an Hilbert space. It supports a sesquiliniform that is non-degenerate. This is your sesquiliniform, nothing really fancy, but something really simple. You just take linear combination of indistensor product, another element indistensor product, you just bar on one side, take the complex conjugate on one side, take the coefficient on the other side and you know delta function of the symmetry group to pair the permutation that you have left. So you can show this is again non-degenerate and you do have an Hilbert space here. There is another operator that it turns out to be important in our setting which is the conjugation. There is a conjugation or an evolution in fact that we call conjugation if you wish and it runs like this, take a linear combination of that sort, you just need to invert the permutation here and of course you can check that playing this game twice, applying S on this element twice, you get back to the initial element so you do have indeed an evolution here. So now you are ready you do have everything in place, KN as an algebra, KN as an Hilbert space, we can push now and define operators acting on that space and for doing this I need a priori, first of all to introduce some particular elements inside this space and I will introduce this notation, I'm using a representative remember that each pair tau 1, tau 2, you know if you let act, if you act by adjoint diagonal adjoint action on this pair, you generate an orbit that is also Ribbon Graph, that's important, so let me fix here a particular label of one of my Ribbon Graph, they're running from 1 to Rib N this is the cardinality, if you want the Z of N that I have before, now I just write it like Rib N. Alright so let me take a particular representative inside this orbit and let act again adjoint action on both slots so you are again on the same space okay so this is the same formula but I'm just referring here with R, I'm just reporting that I'm in Ribbon label by R I hope it's clear, you can see that you do have an automorphism subgroup of SN leaving tau 1 and tau 2 that pair invariant, so you can recast this summation in terms of some of the all distinct element in the orbit you factorize now the automorphism group outside, so this is simple, so you extract the automorphism group, but this pre-factor here is by the orbit stabilizer theorem and this is done over the orbit okay something more interesting also that this basis in fact the previous one is orthogonal respect to the billionaire pairing that I have already introduced and they are orthogonal and this is your pre-factor your normalization factor so let's have a look how do you multiply two of those elements, these two elements they are the same basis and the coefficient here, the structure constants TRST, CRST here, if this is what I would like to understand how does it work, okay what is this number okay I will play a little bit with some notations here rather than having two tensor product here, I'm just writing it as just a sigma R, rather than having two elements two sigma 1 R cross sigma 1 I just write it in that way, sigma of R and multiplying this means I am acting on both slots, okay so in that sort the previous formula looks simpler and can be written in that way or in the other way here so now you are ready for the computation so you just compute this product ERES and what do you obtain is the following, you just need to do a small algebra, the small steps are here you end up with this formula 3023 and it looks like this, so multiplying these two elements, you sum over all elements of the same kind, one over orbit of R and a sum of a delta function acting on orbits, okay so what is it that, what do you have in here so the algebra is here, I will let you know the slide to Gerard to upload it if people are interested in the detail of the steps, you can have it there so let's have a look on that, what's that so this is the orbit of all elements in the orbit of R acting on the representative of ES here and you check if this orbit has an intersection with that orbit T so that summation that you have here over this delta function counts the number of time that the multiplication of elements from the orbit R with a fixed element in the orbit S gives you back an element in the orbit T, so it's a counting procedure here so it's an integer and now here you see that this number here is just an integer in the orbit by this orb R that the size of the orbit now let's introduce our Hamiltonian in the linear operator acting on Kn that we will call Hamiltonian so take N, K in N okay, K could be anything from 2 to N and I will write CK for the conjugacy class of some elements of C of SN made a single cycle of length K and all remaining cycles with length 1 for instance, if I'm sticking, if I'm just using N equals 3, K equals 2 C2 will have a permutation of the following form, a length, a single length of size 2 and 1 and everything else which means here one element left of size 1, so these are there so a single cycle of length 2, cycle of length 1 cycle of length 2, cycle of length 1 so this is C2 for instance, you may play the same game for any CK, K wrenching from 2 to N now these are interesting elements I'm summing for all elements of this class of this conjugacy class, I'm summing all these elements on this class, you can show that these elements are central elements in C of SN so the first thing important is a lemma from Camp and Rangolam and they're showing the following the set T2 up to K star and TK star N, so this is a subset among all the T2 up to TN that are able to generate the center of the algebra so that's a thing that you must understand and this KN needs not to be the N it may be even smaller, there is a sequence an interesting sequence of what can be KN but I won't be able to discuss it here so you do have a subset of TKs that generate the center that's an important fact, the second fact is the basis, the basis Q that I have introduced you before is the eigenvector of this TK operator and what are the eigenvalue normalized character, so you have the character in the representation R, if you have QR, this character will be in that representation as well divided by the dimension I will call this ratio here the normalized character, so TK as for eigenvector, the Q base, the representation theoretic base of the group algebra and the eigenvalue are normalized character to prove this you just need sure lemma and you're gonna be over it so it's not complicated to show now let's introduce our operator on KFN, define the following element in the tensor product C of SN tensor C of SN and they are defined in that way, I'm taking T1K which is the TK acting on the first slot T2K is the TK that you I introduced before acting on the second slot this time and T3K is the same sum over the conjugacy class but acting on both slots it cannot be represented as a TK if you want but it's almost that in that sense that's the way of you can write it so these three operators are interesting, they are linear they show that in fact they are in that algebra by just multiplying on the left and on the right by proper element you will see that you can show this but what is more important here is the following, TK here acts as linear operator on KFN this can be simply seen and there are matrices they do have matrices that I'm denoting like MIKTS so when the TK act on ES you do have a matrix here telling you MIK for TIK and then TS for the element of the matrices what is important here is these matrix elements are non-negative integers how does it work, in fact it's coming from the fact that the TK by themselves are proportional to the ER that I've introduced you before and what is the coefficient of proportionality is nothing but the size of an orbit so the TK is proportional to a certain R, KI, a certain element a certain basis element that is called ERKI and the coefficient here is nothing but the size of the orbit therefore the action of an ER on an ES gives you one of a orb, the sum of the E and the delta function so now you immediately see that the orbit here cancels away and you do have a nice coefficient which is an integer so that's your matrix element here so the matrix element that you look for which is the matrix element of this operator acting on KN counts a number of times that the matrix element in that sum acting on a fixed orbit produce you an element in the orbit T so this lesson we already know now what is also interesting and important is that the TKI by themselves are emission operator on K we can play with this formula, you want to prove this formula and the steps are here so they are integral and the matrix that they are their matrix are integer, are integer coefficients but they are also emission operators and that is great because you can discuss now you are in an algebra, you are also in a nilbert space you do have an emission operator you can discuss about quantum evolution so you are right in a setting of quantum mechanics and this is why the talk is called to this setting as a quantum mechanics of of ribbon graphs so your ribbon graph are vectors and you can let them evolve and think about all things that important for quantum mechanical system I won't be able to discuss more about this because I'm kind of already short in time but I'm anyone want to discuss more about it is more than welcome there are consequences of that and it's maybe important the proposition that you have to see that you have to understand is the following so T1, T1K, T2K, T3K do have eigenvector which is exactly the basis, the Weidenberg-Arting basis so you remember that was my Weidenberg-Arting basis and the TK acting on those actually produce you the characters so how this is working so each TK so the T1K acting on the Q gives you the character of R1 the first slot first slot when the T2 acts on the Q it gives you the character of the representation of the second slot here and then the T3 gives you the R3 which is the third third slots here so the T do have eigenvectors and which are known how this is working remember the TIK are formed by the TK and the Q are formed by the QI but you know that this one has for eigenvectors these guys so you understand quickly that this is going to act like that on this in the same way that this is acting on that and the proof is follows rather easily so what is also important we will start by this as an input in our framework and this is certainly an input from representation theory so this is just to mention that for instance if you want your combinatorial proof your combinatorial interpretation to be totally free from representation theory well in our setting here we do have an input but at the end of the day we will have a fully combinatorial interpretation of our result okay so we can move on and another important result is that of Kemp and Rangulam and they are telling you the following in fact the list of T1 up to 2 through to K a certain K star T2 to a certain K star T3 to a certain K star here uniquely determines the triple R1 R2 R3 so you can you are able with this with the eigenvalue of this operator to uniquely speak your triple R1 R2 R3 this is important and now because you know you will immediately see that this operator actually have an eigenspace that is exactly spanned by your variable acting block so this is the block that they generate their as exactly of the dimension of this block which is C squared and that is important because you want to generate the following matrix where the MIK here are exactly the TIK matrix if I'm subtracting the eigenvalue be careful here you should have an identity of the same size of this matrix I omitted it but you should have here an identity all for the rest here and what is happening here you have this block thing acting on a vector if you the null space of this problem is exactly the dimension of the variable acting block which is of dimension C squared it is important to know that this character this normalized character are known combinatorially by the Murnahan Nakayama rule so this is normalized character we know how to extract them by a rational but we know how to construct them using combinatorial rules second thing they might be even rational but we can choose the least common divisor multiply by the entire setting to remove the denominator if you wish and now you end up with a fully integral operator which mean an operator with only a positive integer entries multiplying by some vector and collector zero so you know that this operator as well will have a null space of this dimension of C squared and you've done the step one so you have produced your operators this operator do have a kernel and that kernel is of dimension C squared so how do we extract the conica we go quickly as I am already almost out of time so here we go so the null space of this operator as dimension C squared in fact for matrices with integer coefficient there is procedure there are procedures for determining their null space and it is known that their null space actually is formed by a factor generating lattices sub lattices in fact of the entire space so how does it work let me just write X for my operator and in fact you do have the decomposition what we call hermit normal form is a decomposition of your matrix into two matrices one which is unibodular that's my U and the H which is upper triangular decomposition the way that you proceed is just linear integer linear combination among the rows swaps of rows multiplying rows by minus one these are allowed operations and all these operation actually are concatenate are the one that that generate the U what is important is the dimension of the null space of X counts the number of null row of H oh I didn't say perhaps I did H is upper triangular and is unique so the number of null rows that are listed below you know in H are exactly the null space of X the null space of X by itself is spanned by the the rows of U corresponding to the indices of the null rows of H so this is your null space and the entire procedure only involve you know discrete steps something which is constructable so you may say that your construction here is integer so you are ready for the interpretation of C square by itself so for every triple of young diagram with n boxes this is your entire lattice z to the power of the number of ribbon graphs so that's your space that lattice of linear integer coefficient of geometric ribbon graphs remember geometric ribbon vector E of R in this space contained a sublative of dimension C squared spanned by integer null vectors of the linear operator L so that's already an interpretation and a constructable way of getting C squared but we want the space itself so we need another refound counting if you want by using now the S squared so S you know divide this space so for the moment this V is nothing but the vector space now associated with K of n so I'm forgetting everything and just show you if I'm looking at this space just as a vector space V to the power the dimension of your space which is a rib of n so if this space decomposes in the block having S as eigenvalue S equals 1 as eigenvalue or S equal minus 1 as I'm going to do here and look at the verdemba art in the composition you do have this decomposition in this V rib R1 R2 R3 where this space is of dimension C squared now you can project this space on S1 and just the way that S acts on the Q states here you can you can count the number of degrees of freedom and you say that the dimension of this space is C C plus 1 divided by 2 for minus 1 space you do have you know the dimension of the space projected on S equals minus 1 is C C minus 1 divided by 2 so if you want the projection from this entire space to that vector space you need just to stack the two matrices below the previous matrices and play the same game of the H and F construction for extracting the new null space from this block here now you do have C C plus 1 C C plus 1 divided by 2 you do have C C minus 1 divided by 2 how do you get C the difference of these two numbers or you can it's just amount to to make precise an injection from this space to that space and this will give you a constructive interpretation of C and so this is a second theorem that you achieve for every triple of young diagram R1 R2 R3 with n boxes there are three constructable sub lattice in the entire of respective dimension C C plus 1 divided by 2 C C minus 1 divided by 2 and C by itself and this answers the question so in conclusion let me quickly say that the coefficient is a dimension of a constructable sub lattice of in the lattice of ribbon graph here Z ribbon the proof relies on Hermitian operator acting with integral eigenvalues acting on the Lillebert space which is also an algebra built over ribbon graphs so this entire could be seen as also quantum mechanical system the HNF form the HNF algorithm offers the lattice interpretation directly the method that we have discussed can be generalized to other type of things which are much more generalized general than the chronicle coefficients for instance if you take the multiplicity of Rd in the tensor product of R1 up to Rd minus 1 so this will be the generalized way of of extending the chronicle coefficients you can play the same game and you will have also answers about this and it can be also generalized to other coefficients such that the literary chat zone but for other group theoretical framework our open problems of course there are people working on combinatorial interpretation of the chronicle we must be making contact with those studies because we know what counts the chronicle for all three rectangular shapes or hoop shapes or a mixture of those but only on this specific case you are able to define the combinatorial interpretation of the chronicle so we must be trying to make contact with those studies so this must be done but also I would like to say part of our proof relied on representation theory perhaps in the spirit of Stanley he doesn't like this we don't know but if you want to remove this representation theoretical input in your framework we need to look at again this eigenvalue problem this eigenvalue thing where the tk acts on the q and you collect the normalized character times q so if we are able to provide an interpretation of the eigenvalue of tk you know that was the sum of a sigma belonging to a class without relying on this q we will be done so why the question is why the tk the eigenvalue of the tk satisfied if we could answer this well the entire stating will be fully combinatorial and we will be able to give a full combinatorial proof for the chronicle equation so thank you for your attention I will stop there thank you Joseph are there questions maybe Darij had questions maybe no actually this was resolved in the chat okay so there are a few questions on the chat maybe I can address this somebody okay is the multiplication of C of Senn iterated it is a tensor product C of Senn tensor C of Senn yes yes yes our Sanjay is there and already answers it okay I have a question go ahead it is about the the conjugacy classes of type n-1 1 so hold on I think my presentation has disappeared yes it has it has okay sorry so I need to reboot it again sorry so let me go back to that so maybe I can yes hold on a second let me try to reboot it I have experienced some trouble here so let me put full screen first yes your t1 your t2 to tk star n conjugacy classes of type n-1 1 because you have sub cycle of length n-1 and fix point okay so hold on Gerard so you yes so come again your question please yes your elements t2 until tk star n yes yes keep up yes I have not my question is that Antonio Maki maybe you know of him he made the work many many times ago not proving that it is I think not proving that it yes exactly I think not proving that it generates a center but it is prep-preparating this proposition because these these classes tk are multiplicative if you multiply two of them you find a sum of of other if you consider the Z module generated by the ck up to my remembrance it is closed by multiplication do you know this work it is work of Maki you say yes Antonio Maki Antonio Maki I didn't know personally I didn't know about it but I think it's right as the center is stable so what you say is entirely right as the center is stable so if you multiply this you're going to end up there as well yes of course now you know more that you can say that his work of 20 years ago can be deduced from this work now I am just pointing you yes yes maybe not so complete but important to cite in the bibliography so Antonio Maki and I can try to recover the paper because it was an idea of Professor Schutzenberger can I make a quick comment yes go ahead so in general the linear basis for the center is given by T sub p where p is a partition of n here this set sort of selects out those partitions which have one cycle of length 2 and remaining cycles of length 1 that's the T2 or T3 which is one cycle of length 3 remaining cycles of length 1 so it just there's only at most n minus 1 of these guys it's a subset of all the partitions but if you take these guys and take products of them you generate the whole thing which is linearly spanned by all partitions so it's a rather small subset in a bigger center of SN so but we'll be happy to have a look at that paper certainly okay yes but it was just a pointer of your merit you know oh yes okay so thank you very much for this nice talk