Cointeracting Bialgebras
Foissy, Loïc
Pairs of cointeracting bialgebras recently appears in the literature of combinatorial Hopf algebras, with examples based on formal series, on trees (Calaque, Ebrahimi-Fard, Manchon), graphs (Manchon), posets... We will give several results obtained on pairs of cointeracting bialgebras: actions on the group of characters, antipode, morphisms to quasi-symmetric functions... and we will give applications to Ehrhart poylnomials and chromatic polynomials.
 First of all, happy birthday to Dirk. So I'm going to start with a monoid or a group, which is proto-algebraic, so it does not really matter what it means, but it just has a good ring or algebra of polynomial functions. So this algebra just reflects the topology of the group, and you have more, you have a co-product, which sends one polynomial function, two pairs of polynomial functions, which is the co-product data, which you can see here. We just reflect the composition of the group. So this is what is called a bi-algebra if you have a monoid, and if it is a group, it is a hop-algebra. And you can recover, it's quite well known, the group of a monoid from the bi-algebra just by taking its characters, the characters are just bi-algebra morphisms from your algebra to the field, which is for me the complex field. And there is a product on character, which is the convolution, which is in some sense this dual of the co-product. So I'm not satisfied with only one group or one monoid, I need two groups, and in fact I want to do some semi-direct product or thing like this. So for this, I take two groups or two monoids with a good bi-algebras, and I suppose that the second one, g-prime, acts on the first one by monoid andomorphisms, which is exactly what I need if I want to do a semi-direct product. So what does this mean? I have first a monoid g, so it has a hop-algebra, which I call or bi-algebra, which I call a. I've got a second monoid g-prime, which has also bi-algebra, which is b. g-primes acts on g, so b co-acts on a. So there is a correction, which is a map, an algebra map from a to a tensor b. So this is just to reflect the action of g-prime on j, and if I want to translate the fact that the action of g-primes is by monoid andomorphisms, this means that a is a bi-algebra in the category of b-commodules. So what does this mean exactly, this fourth point? This means this axiom. So the first one just means that this is a correction. This is the axiom of a right correction. The second one means that the product of a is a morphism in the category of co-module over b, and it also means that the correction is an algebra morphism. This is the same. The fourth one means that the co-unit of a is a co-module morphism from a to the base field. And the last one, which is the Mohr-Schwisting, means that the co-product of a, the big delta, is a co-module morphism from a to a tensor a, which is also a b-commodule. So this means that if you do first the correction and then the co-product on the left, this is equal to do the first the co-product, then the correction on both sides of the co-product, and then to regroup together the two terms, which belongs to b. So this is this product m1324, which case four elements, and we group the second and the fourth one at the end. So for example, let's just take a very simple example. You can consider the group c with the addition, which is a group, a Boolean group. And on this, the group c star with the multiplication naturally acts by group automorphisms. So this is exactly what you have before. So the first group g is c plus. So it's algebra, it's hop-algebra is the polynomial algebras on c, which is the polynomial ring with one in the terminate. With the co-product, which is additive, the co-product of x is x tensor one plus one tensor x, this is hop-algebra. The second group g prime is c star with the multiplication. So the polynomial algebras on c star is the Laurent polynomial algebra, c of x and x minus one. I have to add an inverse to x because there is no zero in c star. With another co-product, which is multiplicative, which sends x to x tensor x. So for the first one, the big delta x is a primitive element. For the second one, it's the group line element. So this is also hop-algebra because there is an inverse. And b co-acts on a because c star acts on c with this correction, rho, which sends also x and x tensor x. In fact, more or less the co-action on the co-product, the second co-product, are more or less the same. So I don't act very much this x minus one. I would prefer to have cx. So I just forget it and I obtain a hop, not a hop-algebra, but a set-algebra, which is b, which sees the same algebra as a with another co-product. So it's no more hop-algebra because x has no more inverse. It's just a bi-algebra. Then it co-acts in the same way on a with the same correction. And for this, the correction on the co-product is the same. So this is the frame I will use now. So what I have is an algebra a with a single product and two co-products, big delta on small delta such that a m delta is a bi-algebra, a m small delta is a bi-algebra. And the second one co-acts on the first one with the correction rho, which is also the second co-product. So what I have is an object with one product and two co-products. For both co-products, it's a bi-algebra. And moreover, there is a compatibility between the two co-products, which is something like this. Replace just rho by delta and you obtain the second, the compatibility between big delta and small delta. So if you want to be complete, this should be called bi-algebras in a category of co-modules of another bi-algebra, which is quite long. So I just now called them double bi-algebras or co-interacting bi-algebras and things like this. And the first example we have is just the polynomial ring, c of x, with these two co-products, big delta which is additive and small delta which is multiplicative. We know more examples. So for example, the well-known cone-cramer hop-algebra of trees, which is based on rooted forest, and I won't belong on the remainder on it. So it's just based on rooted trees. For me, the trees, the roots of the trees are at the bottom. So if you have the trees with two vertices, three vertices, four vertices, the product is the Dijon union, so it has the basis of forests. And the first co-product is the cone-cramer one, just given by admissible cuts. So you take your tree or your forest, you just cut some branches, you put the branches on the right and the trunk on the left. So for example, for this tree, you can cut nothing and the tree turns to one, or you can cut everything, one turns to the tree, or you can cut a leaf in two possible ways. You obtain two turns to the trunk, turns to the leaf, or you can cut the two leaves which go on the right and three minutes on either roots. And the same for this, you can cut nothing or everything, or the trunk after the root, or the trunk just before the leaf, or you obtain these four turns. There is a primitive part, the first two turns, which means that the cone is very simple. The cone unit of a forest is one, is the forest has no vertex and zero otherwise. And you can observe that it is graded, obviously, by the number of vertices. If you cut a forest, you don't lose any vertex. Some goes on the left, the other on the right, but you don't lose any vertex, so this is graded by the number of vertices. So this is the most famous co-product on it, but there is a second one, which was first described by Damien Calac, Koro Shebrahini-Fart and Dominique Monson in 2008, I think, which is given by the process of construction on extraction. So what does this mean? For example, for this tree, you can separate your trees into disjoint sub-trees. So for example, you can disjoint it into three sub-trees, which has only the vertex on the left to contract these sub-trees, but nothing appears, and on the right to put these sub-trees. So here you don't do anything. You can contract the edge on the left, so this sub-tree, so this gives this tree in two possible ways. You can contract the whole tree, it only remains one vertex, and put the other tree on the right. So this is another co-product, which is also co-associative. It's not co-commutative, you can see it. And it's not a hop-file algebra because you have a group like, which is the tree with only one vertex with no inverse, so you only obtain a bi-algebra for this co-product. And they prove that this is a double bi-algebra, so this means that this co-product really co-acts in a good way on the first cone-crimer co-product. And there are similar constructions on post-sets, finite post-sets, you can see trees as post-sets just by taking the order, the partial order to be higher in the tree. So if you have a tree, you have a post-sets. And such a construction also exists on finite post-sets, or more generally on finite topologies. So this is the first example of, not the first, but second examples of double bi-algebra. I'm going to give another one based on graphs. So for this, the basis of my hop-file algebra of graphs is this whole set of graphs. So these are just simple graphs. So here you have all graphs connected or not with one, two, three, or four vertices. There is a simple product on it, which is the Dijon union. The unit is the empty graph, which is here. So for example, this graph is the product of this by itself, something like this. There is another simple product. There is a very simple co-product, which is just given by separating two graphs into two parts. So take your set of vertices, you put some of the vertices on the left with the edges between them, the other vertices on the right, also with the edges between them. And you obtain a nice co-product, which was first defined, I think, in a paper of Schmitt on incidence hop-file zebras. This is an example of incidence hop-file zebras based on a set of graphs. This is just a set of graphs with a given set of vertices with expression of edges. So this is a co-product. This is a co-associative. It's really not difficult to see. And it's co-commutative. This is very different from the co-crime or co-product. This is co-commutative. There is a second one, which is also can be found in the paper of Schmitt with another incidence by a zebra, and which was also described in a paper of Dominique in 2011, with various examples on graphs, oriented graphs, acyclic oriented graphs, something like this. So this is the same idea as for trees. So for trees, the first co-product is just cutting the trees into two parts. So this is the same for graphs. The second co-product was given by extraction and contraction. So this is the same for graphs. Just take a graph. You can take some equivalences on the set of vertices. So this means that you just do a partition of your vertices. On the right, you just contract your equivalent cases. So this means that you contract some sub-subgraphs of your graphs to vertices. And on the left, you just forget the edges, which are not between, which are in vertices, which are not equivalent. So I'm just going to give an example. So for this one, you can contract everything, just remain a vertex. And on the right, you obtain the world graph. You can contract only one edge, for example, this one. If you contract, you just obtain a graph with two vertices and one wedge between them, this. And the extraction is given by this edge and the other vertex, so something like this. You have three possible ways to do this. You can just extract the vertices, just go on the right, and if you contract the vertices, nothing happens. So the graph stays itself. So this is also a second core product. This is also by algebra. It is not a hop-adjibra because there is a group like, the graph is one vertex, and it has no inverse, so you don't have any anti-pod. It's not a big problem, but well, it's not a hop-adjibra. And the code is very simple. In fact, this is because of these parts. For if graph, you obtain a vertex and some of the graph plus a graph tensor vertices and more terms. So the code unit is just given by sending any graph to one. If your graph is totally disconnected with no edge or zero otherwise, and you can prove that this is really a double by algebra. So the second core product, this extraction contraction core product really coax on the first one. And the last example, which is the hop-adjibra of quasi-symmetric functions. So as a algebra, it's based on the set of compositions. The composition is just the finite sequences of positive integers. With these products, which was used just before by Dominique, this is the quasi-triple products on composition. The core product, the first core product is given by decon-catination. You just cut yours into two parts between two letters. And there is a second one, which is given by extraction and contraction. So for example, for this, you will cut your words into two, into several parts. One part, two parts, two parts, and so three parts, two parts, two parts, and one part. You just contract the parts. So contracting just means that you sum all the letters of your word. Because the letters are integers, you can sum them. So this gives the terms on the left and on the right, you just quasi-suffer your parts. So this is another core product. You can find it into the papers of Thibault, Noveli, and the other on this subject. I'm not totally sure they proved this is really a double by algebra. I'm not sure they proved the co-interaction, but well, it's true. And they prove it with a trick, which is based on manipulation of alphabets. So it's not really obvious, but you can do it without too many combinatorics, just with algebraic tricks. So well, there is a co-unit, I forgot it, which is also, she's given by this. You take a word, a composition, if it's of 0 or 1, the co-unit is 1 and 0 otherwise. And it turns out that this is a character of Cucine, which appears in another paper by Agar, Bergeron, and Sotile. In this paper, they define the category of combinatorial hopp-vagebras, which are graded, connected hopp-vagebras with a character, which is their pairs. And they prove that in this category, Cucine with this character, epsilon prime, they don't mention that this is the co-unit of a certain co-product, by the way, but this pair is a terminal object. So this means that if you take a combinatorial hopp-vagebras, so a connected, graded hopp-vagebras with a character, you automatically obtain a morphism to Cucine, compatible with this co-unit. Okay, so that's nice. These are nice objects, but the question is, what you can do with this, what you can deduce on this construction, and well, what will it give on your, on these examples of graphs and trees and things like this? So first of all, take a double bi-algebra, A, with one product and two co-products, another bi-algebra, B, and you're looking at morphism of hopp-vagebras or bi-algebra from A to B. So it turns out that the monoid of characters of A for the second co-product acts on the set of bi-algebra morphisms with the help of the co-action of the second co-products. So this means that if you obtain a bi-algebra morphism from A to B, in fact, you have a lot more bi-algebra morphisms from A to B. You can deform any bi-algebra morphisms with the help of characters of A. If you have this one, you will have everyone's just by using the actions. So let's try to do this for forests. So forests form a double bi-algebra. So this means that there should be a unique morphism from a forest to with both co-products. You can prove that you can compute it uniquely. For example, let's just start with the first tree, the tree with only one vertex. Well, it's primitive for trees. So its image should be primitive for polynomials. The set of primitive elements of k of x is one-dimensional, generated by x. So this means that phi 1 of this tree should be a multiple of x. Moreover, phi 1 is compatible with a second co-product and with its co-unit epsilon prime. Epsilon prime of this tree is one. So epsilon prime of its image should be one. So epsilon prime of lambda x is lambda. So lambda should be one. So you entirely determine phi 1 of this tree, this should be x. For the second one, you do the same. Let's first compute the co-product of this tree. This is this. This is only one admissible cost, non-trivial. Let us apply phi 1 to this. Phi 1 is compatible with back delta. So you should find something like this. Phi 1 of this is this. So this means that phi 1 of this tree should be x2 over 2 plus a primitive element. So lambda of x. This morphism phi 1 is compatible with the co-unit epsilon prime. So epsilon prime of this polynomial should be epsilon prime of this tree. So this should be equal to 0. And you obtain that lambda is equal to minus 1 over 2. And you obtain that phi 1 of this tree. This is exactly this. So what you are doing now is to put that this morphism is unique. What is not clear is that it is really compatible with the second co-product. I only use that it's compatible with the co-unit. But you obtain for free, but it is in fact compatible with the second co-product. Just for free. So you can continue like this. For this tree, you obtain something like this, which is a real bad polynomial. Quite famous. For this tree, this is normal, real bad polynomial, but something like this. Maybe you recognize it. This polynomial counts the sum of the square. This evaluate to n is 1 square plus 2 square plus, etc. plus n square. So these are quite special polynomial. You can do more in fact. Well, this is a nice way to compute the variant phi 1, but it's quite long. And in fact, you can do better. You can put some formula like this. If you take an element of your double by algebra a, you can compute phi 1 of a in this way. First, you can compute all its reduced co-products. So the reduced co-products is just by forgetting the primitive parts. So this means that for trees, you just forget the trivial cuts and the cuts of everything. You can compute it and iterate and iterate and iterate and something like this. And you know that at certain point, it will stop. The iterated co-products should be zero after a certain point. So you compute all of them. You obtain tons of trees and things like this. You just apply the co-unit on the left and then you multiply the tons by your Hilbert polynomial. And this means that your invariance really counts something. If you evaluate x into an integer, this is a binomial coefficient. So really phi 1 of a just counts something for forests. And this counts something like this. This is quite a well-known construction. If you take a forest with k vertices, well, just choose an index session of the forest. It does not really matter. Then you can associate with a polytope of dimension k. In fact, this polytope is defined by some inequality. If you take one vertex, which is another one, x1, xi is below the vertex j in your tree. Then you associate to it an inequality xi times smaller than xj. So this defines a polytope. And you want first to delete it by an integer, i minus 1 times the polytope. So just do an amortesy. And you want to count the number of integral points inside. So it's quite a famous result that this is, in fact, given by a polynomial sequence. So this defines a polynomial, which is called the Erard polynomial. The strict Erard polynomial is the same. But just count the number of vertices inside your polytopes. So this is quite known that this defines two polynomials, which are related, I will say later. So just to mention a problem, usually in the literature on this, the Erard polynomial in n, just count the number of integral points of the dilated of the polytope by n. Here, I have a problem. If I do this, it does not work. So I have to do some translation by one. So for example, for this, so you have three vertices, which I did an index by one, two three from bottom to left. One is smaller than two, two is smaller than three. So your polytope is defined by x smaller than one, smaller than z, between zero and one. So the polytope associated to f is just a simplex. So if you want to count the number of integral points into the dilated of f, what you count is the number of points x, y, z, which are integers such that x is smaller than one, is smaller than z, is smaller than n minus one. And it's not very difficult to count them on two pools, but this is n, n plus one, n plus two over six. So this is for the Erard polynomial. For the strict Erard polynomial, you count the point inside. So this means that you replace your smaller or equal by just strictly smaller, and n minus one by n plus one. So these counts things like this, and it's not difficult to prove that this is this polynomial, which by the way is in fact phi one of f evaluated to n. You can do the same thing for this tree. So here you can try to draw the polytope. This is a pyramid with a square basis. You can count the number of points inside, which is this. And so this is the Erard polynomial, the strict Erard polynomial, which is again phi one of f evaluated to n. And in fact, this is exactly this. The unique morphism compatible with the product and the two co-products is in fact the strict Erard polynomial. OK. So if you look at this, you can observe that the strict Erard and the Erard polynomial are very similar. More or less they are the same coefficient, and things like this. And in fact, you can prove that there is another morphism from the conch primer to polynomials, which is compatible with the product and two co-products. This is not the Erard polynomial, but more or less itself. You just replace x by minus x, and you have to correct the same by multiplying by the power of minus one. And it's not very difficult to prove combinatorially, but this is compatible also with the product and both co-products. But there is only one morphism compatible with the product and the two co-products. So this means that this morphism phi is the same as phi one. And what you obtain is an algebraic proof and the 180 principle for Erard polynomial. In fact, the strict Erard polynomial and the Erard polynomial are really closely rated. This is just by evaluating x into minus one, minus x, and replace the same. So here, as an application, you have a proof of the 180 principle for Erard polynomial without, more or less without any real combinatorial stuff. The usual proof uses some Mobius inversion in some posets. So these are really combinatorial. Can I ask a quick question here? Sorry. So this, you exemplified this as in the case of these polytops you get from trees via this post-order. Can you also use a similar kind of argument to get this Erard polynomial duality result for arbitrary polytops? I did not manage to do it. In fact, I would need a hopper algebra structure on polytops. A good algebra structure on polytops. I have a product, just the usual product of polytops, but I don't find the core products. In fact, those coming from forest are very special. You see they are defined by some equations which are nice. If you take any polytops, I don't know how to quote any polytops. So. Yeah. Thank you. So let's do it for graphs now. So for graphs, I can apply the formula for phi 1, which I gave before. So what does this mean? I have to do all the iterated core products of graphs. So this means that I have to cut the graphs into any number of parts I want. So the iterated core product just means that I really cut the graphs into a lot of parts. And then on each part, I will apply the co-unit for the second core product. So the co-unit is one if the part has no edge. And otherwise, it's zero. So this means that in phi 1, which appear, only decomposition of your graphs into parts with no edges. The decomposition of the graph is equivalent to a coloration of a graph. So a coloration is just associated to any vertex of the graph, a color, which usually is a number. So a partition of the graph is just the same as a coloration. And the coloration happening in my phi 1 of A are just the coloration which are called valid. So this means that if two vertices have the same colors, they should not be neighbors in the graph with no edges between them. So this means that in phi 1 of A, I just take in account valid colorations. So this is a polynomial with constant like this, and which is called the chromatic polynomial. So in fact, what I find with graphs, the unit morphism compatible with both core products and the core product is the chromatic polynomial, which is perhaps not a big surprise. Because if you work with graphs, you know that chromatic polynomials is an essential tool for such an event. So this is perhaps an explanation why it's so important. In fact, it's a unique polynomial invariance on graphs, which will be compatible with all these structures of extraction or contraction and extraction of subclass. OK, so something else, no, another application. So I'm going back to a theoretical result. In fact, I'm looking for the antipode. So in all my examples for the big delta, this is a hop algebra. So it has an antipode. And for the second component, it's just a bi-algebra, so no antipode. In fact, I can prove that if I want to compute the antipode of A, I just have to compute the inverse of a special character, which is the coordinate of the second coproduct. So the coordinate of the second coproduct is the unit for the convolution of the second coproduct. But for the first coproduct, it's just a character with no spatial property. Maybe it's invertible. If it is, well, you know that A is a hop-algebra, and you have a nice formula for your hop-algebra. Just apply the second coproduct, and then the character and the first components. So for double bi-algebras, if you want to compute the algebra, you just have to compute a special character. There is something more. In fact, maybe you observe that all my examples of double bi-algebras are commutative. In fact, what you obtain here, like this, is that S is a composition of algebra morphisms. So this means that in double bi-algebras, the antipode of A is an algebra morphism. But usually it's an anti-algebra morphism. So double bi-algebras are special bi-algebras, so that the antipode is both an algebra and an anti-algebra morphisms. So this means that essentially, if you have a double bi-algebra, it should be commutative. So this is the reason why all my examples are commutative. In fact, you cannot obtain any double bi-algebra, which is not commutative, because, this is one of the reasons the antipode should be an algebra morphism. You can do more. In fact, you have to compute the inverse of a character, which is not so obvious, in fact. You can do it inductively, but it's not so obvious. But if you know how to compute phi 1, well, it's very easy to find the inverse of a character. Just take an element of your algebra, a small a, just compute phi 1 of a, this is a polynomial, and then evaluate it into minus 1. And this is really the character alpha of a. So for example, for rooted forests, if you want to compute the antipode, you need to compute the error polynomials of any forest evaluated to minus 1. And it's very easy, just a problem of minus 1. So alpha of s is just a problem of minus 1. And you obtain this formula for the antipode, which was proved by Conan Kremer, not in this way at all. It's just an indicative proof. I can obtain this, something like this, with no induction. It's more interesting for graphs. For graphs, for a long time, the antipode was not known, not really. You can compute it inductively, of course, but it was not so clear. Last year, in fact, formula was proved by Ben-Yaditi Bergeron on match-ad-sec, I'm not totally sure of the pronunciation, with a combinatorial method, which was quite complicated. There is a Mobius inversion, and curiously, the number of acyclic orientation appears. And in fact, this is obtained with my method by this. If you want to do this, you have to compute phi 1 of the graph evaluated in minus 1. Phi 1 of the graph is chromatic polynomial, and it's quite a famous result in graph theory that the graph polynomial evaluated in minus 1 counts the number of acyclic orientations. And to prove this, it's not so difficult, it's just a combinatorial proof by induction of a number of vertices. So what you obtain is the formula by Ben-Yaditi Bergeron on match-ad-sec, with no more combinatorics, more or less. It just applies the chromatic polynomial evaluated to minus 1. OK. You can do better for the chromatic character. In fact, there is a very simple half-algebra morphism from graphs to polynomials, which is just sending a G to a monomial, thanks to the power of the number of vertices of G. This is really easy to show that this is half-algebra morphism. Of course, it's just compatible with the first core product, not with the second one. The unique morphism compatible with the second one is phi 1, the chromatic polynomial. And I mentioned before that any half-algebra morphism from graphs to polynomials should be obtained from the chromatic polynomial by the action of the character. So we can write that phi 0, this very simple polynomial, very simple invariant, should be obtained from chromatic polynomial by action of the character, which I denote by lambda, which is very easy to compute. You just send any graph to 1. So lambda is a very simple character. But which is more interesting is that it is invertible for the second convolution. So this means that you can obtain the chromatic polynomial from this very simple morphism just by acting a certain character, which is not so easy to find. So this means that you obtain a formula for the chromatic polynomial. In fact, the chromatic polynomial is the sum of all possible contractions of your graph thanks to the power of the number of classes of your contraction and then a scalar, which is, which can be objectively computed. So lambda, this is a chromatic character. You can compute its values just by induction. And you can observe on this example that it's never 0. The chromatic polynomial never goes to 0. And it only depends on the number of vertices. With one vertex, it's positive. Two vertices, it's negative. Three vertices, it's positive. Four vertices, negative. And you can prove it just by induction. So just by something like this. I don't have any more time, so just cut it a little bit. So just by this, and with this formula, you can prove that in the chromatic polynomial, the things of it are alternating, which is a result proof by Rota in the 70s, I think, with complicated combinatorial methods. Here you obtain each just by a small combinatorial tool, which is the construction extraction of edges. And then this formula for the chromatic polynomial related to the chromatic character. Okay. I think I still have five minutes. So for the moment, I talk about morphism with values to polynomials. Now I'm going to talk about morphism with values in the quasi-symmetric algebra. So I mentioned before that by Agar, Bergeron, and Sotile, I know that there are a lot of morphism to it because it's a terminal object. If I want a morphism to Q-sym, I just have to choose a character. And then I will obtain homogenous morphism compatible with the product and the first co-product. And I've got a formula for this, which is similar than the formula for the polynomial invariant. If I want to construct a morphism from A to Q-sym, I just take all the iterative co-products. I apply some projection on it. So I project on the homogenous parts. Then I applied the co-unit on all the parts. I take in account the degrees of the parts with a composition, something like this. So what can I prove is if I'm looking for a morphism from A to Q-sym compatible with both bi-algebraic structures, so compatible with the products and the two co-products, this is the only possibilities. If I'm looking for something compatible with the products and the co-products, this is the only one. We should work. And happily, it does not work any time. I need another condition. I need another condition of the gradation. In fact, I need that the gradation more or less just respect the first components. If this condition, this technical condition is not satisfied, well, phi 1 is not compatible with the second co-product. So this means that I won't have any morphism compatible with both structures. And this is what happens for flow rates. In fact, my condition means that on the second co-products, I should obtain only things with three vertices on the left. And this is not the case. There are some red trees. We should have three vertices if I want to be compatible with the second co-products. And unhappily, well, it has only two vertices. So in this case, I won't have any morphism compatible with both structures from trees to cussing. So I have to cheat a little bit. When I just replace forests by decorated forests, so this means that on any vertex on my forest, I had a decoration, which is an integer. So if I do some contraction, for example, for this, I'm contracting the edge between A and B. Well, I don't forget A and B, and I just replace the vertex, the decoration of the vertex, by A plus B. So for this, now the co-product is homogeneous. Here, the weight is A plus B plus C. And on the left, and also on the right, I only obtain trees with weight A plus B plus C. So I can just put them into black. And now, with this trick, the technical condition on the second co-product, which I changed, is satisfied. So this means that I obtained for free a hop-edge-abramorphism from forests to cussing, which is homogeneous for this godvation by the weight, incompatible with all structures. And this is a generalization of the Erhard polynomial, which I call the Erhard-casisimetric function. It is something like this. And the same for trees, so for graphs, sorry. And what I obtain is a generalization of the chromatic polynomial, which is called the chromatic-casisimetric function. But in fact, it's not quasi-symmetric. It's symmetric for a reason of co-computativity. And this is an object which was already known by combinatoricists on graphed-heresists. Well, they know it, but they didn't know that it was compatible with a zero structure on the quasi-symmetric functions. And the last word, I said before that double bi-Agebra does not run very well with non-computative bi-Agebras. In fact, you can do some non-computative bi-Agebra, you can replace by index tree or planar tree, something like this. And you can also define two co-products. Okay, so two co-products. They are no more compatible as before, but they are still existing, so why not? And you can generalize your chromatic series or Erhard series. All the morphisms I mentioned before still exist in a non-computative way, but you cannot choose anymore the formalism of double bi-Agebra. There are no more double bi-Agebra. So if you want to do this, if you want to explain it, you have to work in another category, which is in some sense bigger, which is the category of species. In fact, all my objects are traces, this means that they are imaged by the functions of something which exists in a category of species. In the category of species, with two functions, which will give, in part, the co-productive objects, which are double bi-Agebras, and non-computative objects, which are not double bi-Agebras, but which are traces, non-computative traces, or double bi-Agebras in the category of species. So we stop here and thank you very much for your attention. Thank you very much, Loic. Thank you. Are there questions? Okay. Yes. Did you actually... Yeah. Did you actually upgrade all your formalisms in the species setting? Yes. In fact, in the species settings, you can do the same. So the QC on KX is replaced by the same object, which is species of composition. Yeah. Which is also double bi-Agebras, something like this. And it turns out that you have two functions, the fog function, the full fog function. The one will send you to commutative objects, and the other one to non-computative objects. So in fact, you have bi-Agebras, you obtain... The double bi-Agebras in the setting of species are commutative in the species settings. Yeah. It's sort of strong commutativity, so with the second function, there are no more commutative in the user space in the setting. So you can do it all this in the set of species. More or less the same proof, something like this. More or less, there are some technicalities in some sense, but these are the same ideas. If I may, I'd like to ask or make a comment or ask a question. Thank you, Loïc, for your nice talk. You were wondering about co-product on polytopes. There is one on cones, and polytopes can be seen as intersections of cones, and that's the way, formally, on polytopes of the kind you were talking about and be derived. So I think it should be possible, maybe, going that way, following the path of Bavinoc, Berlin, then I wonder whether it's possible. So I can say that I found some co-products on polytopes, but more or less they are useless. There are stupid co-products, and you don't obtain the era of polynomers as an invariant. You obtain stupid things just by sending, for example, a polynomers to the number of vertices or something like this. I just found stupid co-products. Not interesting one. Okay. The one I'm thinking of is very geometric, and it serves similar purposes. It's four counting into two points on cones, so it's made for that. And it's implicit in people's work in toric geometry. So it could be helpful, maybe. It's a geometric one. Okay. So it should be better than I wanted. I will look at it. There's a question from Janek Vargas in the Q&A, and I have said him so that he can unmute, but he asks, is there a polytope associated to double-post it the same way you define the polytope associated to a forest? Yes. In fact, everything I did for forests can be done for post-sets or topologies. And you also obtain some era polynomials with the same properties. I just speak myself to forest because it was easy to describe and post-sets or topologies. David, do you want to ask your question? Yes. I wanted to make a comment because it's directly related to the title of this conference, an algebraic structures and quantum field theory. Now, look, you know very well, Dex, work on the hot algebra of renormalization, but I hope you're also aware of his recent work on the co-action associated with the monodromy of the functions we get from the phymer diagrams, and that's associated with cutting lines. It was an interesting question as to how that relates to co-action and multiple polar logarithms that we obtain and functions beyond that. So there is, facing us at present in quantum field theory, a compatibility question. It might not be directly related to your talk. And that is what about calculations in which we both cut lines to discover analytic structure, but we also have sub-divergences that we have to renormalize. Yeah, yeah, that's precisely the right question, David, but I think there will be an answer pretty soon. And it has a lot to do with Schloich's co-interacting biologmas. Excellent. Thank you. Okay. Thank you. So I think there's also some simpler connection that just when we were looking at the general tree-phyman rolls, then having the co-structure on trees, the co-interacting structure on trees, it picks out a particular one, a particular choice for the lower order terms. So that's a more trivial observation than what David and Dirk are getting at, but it's another connection to the quantum field theory situation. I see Mischi has his hand up. Yes. Thanks, and thanks, Loik, for the nice talk. I also have a question regarding the co-products on polytopes. And it's about, I mean, if there's a relation or if you're aware of this stuff, this newer stuff by Agia together with Agia on the hopped mono structure of generalized parameter hydra, which are polytopes, and on them you can define this co-product. Is this too specific for your purposes? At least they asked some questions about erhard polynomials on the polytopes. Yes, but the parameter hydra are very specific polytopes. All right. But in this case, it's easier to define some co-products on these sort of objects, which have a strong, they have really a strong combinatorial structure than them. If you take any polytopes, not the case, just the problem. The problem is really the generalization then, too. So this only works for these specific polytopes. Yes. I think that for a special family of polytopes with strong structure, you can define a co-product which should give you the erhard polynomial or things like this. All right. So any polytopes, in fact, this co-product is quite, you can see it easily on the post set, you just cut or things like this. On the polytop, geometrically, it's not so clear for me. This is the section of faces or things like this, which is really something more complicated. Okay. So I don't know exactly what it is geometrically. Okay. But, okay. I see. Thank you. So the problem for me is that I can understand polytopes in three-dimensional, but no more. But this is only a few, three or four examples, which I can manage. So it's not enough for me to understand what happens for bigger polytopes. So I don't know exactly what the co-product is for polytopes. All right. In the interest of time, why don't we thank Loic again now? Thank you. Thank you.
 First of all, happy birthday to Dirk.  So I'm going to start with a monoid or a group, which is proto-algebraic, so it does not really  matter what it means, but it just has a good ring or algebra of polynomial functions.  So this algebra just reflects the topology of the group, and you have more, you have  a co-product, which sends one polynomial function, two pairs of polynomial functions,  which is the co-product data, which you can see here.  We just reflect the composition of the group.  So this is what is called a bi-algebra if you have a monoid, and if it is a group, it  is a hop-algebra.  And you can recover, it's quite well known, the group of a monoid from the bi-algebra  just by taking its characters, the characters are just bi-algebra morphisms from your algebra  to the field, which is for me the complex field.  And there is a product on character, which is the convolution, which is in some sense  this dual of the co-product.  So I'm not satisfied with only one group or one monoid, I need two groups, and in fact  I want to do some semi-direct product or thing like this.  So for this, I take two groups or two monoids with a good bi-algebras, and I suppose that  the second one, g-prime, acts on the first one by monoid andomorphisms, which is exactly  what I need if I want to do a semi-direct product.  So what does this mean?  I have first a monoid g, so it has a hop-algebra, which I call or bi-algebra, which I call  a.  I've got a second monoid g-prime, which has also bi-algebra, which is b.  g-primes acts on g, so b co-acts on a.  So there is a correction, which is a map, an algebra map from a to a tensor b.  So this is just to reflect the action of g-prime on j, and if I want to translate the fact that  the action of g-primes is by monoid andomorphisms, this means that a is a bi-algebra in the category  of b-commodules.  So what does this mean exactly, this fourth point?  This means this axiom.  So the first one just means that this is a correction.  This is the axiom of a right correction.  The second one means that the product of a is a morphism in the category of co-module  over b, and it also means that the correction is an algebra morphism.  This is the same.  The fourth one means that the co-unit of a is a co-module morphism from a to the base  field.  And the last one, which is the Mohr-Schwisting, means that the co-product of a, the big delta,  is a co-module morphism from a to a tensor a, which is also a b-commodule.  So this means that if you do first the correction and then the co-product on the left, this  is equal to do the first the co-product, then the correction on both sides of the co-product,  and then to regroup together the two terms, which belongs to b.  So this is this product m1324, which case four elements, and we group the second and  the fourth one at the end.  So for example, let's just take a very simple example.  You can consider the group c with the addition, which is a group, a Boolean group.  And on this, the group c star with the multiplication naturally acts by group automorphisms.  So this is exactly what you have before.  So the first group g is c plus.  So it's algebra, it's hop-algebra is the polynomial algebras on c, which is the polynomial ring  with one in the terminate.  With the co-product, which is additive, the co-product of x is x tensor one plus one tensor  x, this is hop-algebra.  The second group g prime is c star with the multiplication.  So the polynomial algebras on c star is the Laurent polynomial algebra, c of x and x minus  one.  I have to add an inverse to x because there is no zero in c star.  With another co-product, which is multiplicative, which sends x to x tensor x.  So for the first one, the big delta x is a primitive element.  For the second one, it's the group line element.  So this is also hop-algebra because there is an inverse.  And b co-acts on a because c star acts on c with this correction, rho, which sends also  x and x tensor x.  In fact, more or less the co-action on the co-product, the second co-product, are more  or less the same.  So I don't act very much this x minus one.  I would prefer to have cx.  So I just forget it and I obtain a hop, not a hop-algebra, but a set-algebra, which is  b, which sees the same algebra as a with another co-product.  So it's no more hop-algebra because x has no more inverse.  It's just a bi-algebra.  Then it co-acts in the same way on a with the same correction.  And for this, the correction on the co-product is the same.  So this is the frame I will use now.  So what I have is an algebra a with a single product and two co-products, big delta on  small delta such that a m delta is a bi-algebra, a m small delta is a bi-algebra.  And the second one co-acts on the first one with the correction rho, which is also the  second co-product.  So what I have is an object with one product and two co-products.  For both co-products, it's a bi-algebra.  And moreover, there is a compatibility between the two co-products, which is something like  this.  Replace just rho by delta and you obtain the second, the compatibility between big delta  and small delta.  So if you want to be complete, this should be called bi-algebras in a category of co-modules  of another bi-algebra, which is quite long.  So I just now called them double bi-algebras or co-interacting bi-algebras and things like  this.  And the first example we have is just the polynomial ring, c of x, with these two co-products,  big delta which is additive and small delta which is multiplicative.  We know more examples.  So for example, the well-known cone-cramer hop-algebra of trees, which is based on rooted  forest, and I won't belong on the remainder on it.  So it's just based on rooted trees.  For me, the trees, the roots of the trees are at the bottom.  So if you have the trees with two vertices, three vertices, four vertices, the product  is the Dijon union, so it has the basis of forests.  And the first co-product is the cone-cramer one, just given by admissible cuts.  So you take your tree or your forest, you just cut some branches, you put the branches  on the right and the trunk on the left.  So for example, for this tree, you can cut nothing and the tree turns to one, or you  can cut everything, one turns to the tree, or you can cut a leaf in two possible ways.  You obtain two turns to the trunk, turns to the leaf, or you can cut the two leaves which  go on the right and three minutes on either roots.  And the same for this, you can cut nothing or everything, or the trunk after the root,  or the trunk just before the leaf, or you obtain these four turns.  There is a primitive part, the first two turns, which means that the cone is very simple.  The cone unit of a forest is one, is the forest has no vertex and zero otherwise.  And you can observe that it is graded, obviously, by the number of vertices.  If you cut a forest, you don't lose any vertex.  Some goes on the left, the other on the right, but you don't lose any vertex, so this is  graded by the number of vertices.  So this is the most famous co-product on it, but there is a second one, which was first  described by Damien Calac, Koro Shebrahini-Fart and Dominique Monson in 2008, I think, which  is given by the process of construction on extraction.  So what does this mean?  For example, for this tree, you can separate your trees into disjoint sub-trees.  So for example, you can disjoint it into three sub-trees, which has only the vertex on the  left to contract these sub-trees, but nothing appears, and on the right to put these sub-trees.  So here you don't do anything.  You can contract the edge on the left, so this sub-tree, so this gives this tree in two  possible ways.  You can contract the whole tree, it only remains one vertex, and put the other tree on the  right.  So this is another co-product, which is also co-associative.  It's not co-commutative, you can see it.  And it's not a hop-file algebra because you have a group like, which is the tree with  only one vertex with no inverse, so you only obtain a bi-algebra for this co-product.  And they prove that this is a double bi-algebra, so this means that this co-product really  co-acts in a good way on the first cone-crimer co-product.  And there are similar constructions on post-sets, finite post-sets, you can see trees as post-sets  just by taking the order, the partial order to be higher in the tree.  So if you have a tree, you have a post-sets.  And such a construction also exists on finite post-sets, or more generally on finite topologies.  So this is the first example of, not the first, but second examples of double bi-algebra.  I'm going to give another one based on graphs.  So for this, the basis of my hop-file algebra of graphs is this whole set of graphs.  So these are just simple graphs.  So here you have all graphs connected or not with one, two, three, or four vertices.  There is a simple product on it, which is the Dijon union.  The unit is the empty graph, which is here.  So for example, this graph is the product of this by itself, something like this.  There is another simple product.  There is a very simple co-product, which is just given by separating two graphs into two  parts.  So take your set of vertices, you put some of the vertices on the left with the edges  between them, the other vertices on the right, also with the edges between them.  And you obtain a nice co-product, which was first defined, I think, in a paper of Schmitt  on incidence hop-file zebras.  This is an example of incidence hop-file zebras based on a set of graphs.  This is just a set of graphs with a given set of vertices with expression of edges.  So this is a co-product.  This is a co-associative.  It's really not difficult to see.  And it's co-commutative.  This is very different from the co-crime or co-product.  This is co-commutative.  There is a second one, which is also can be found in the paper of Schmitt with another  incidence by a zebra, and which was also described in a paper of Dominique in 2011, with various  examples on graphs, oriented graphs, acyclic oriented graphs, something like this.  So this is the same idea as for trees.  So for trees, the first co-product is just cutting the trees into two parts.  So this is the same for graphs.  The second co-product was given by extraction and contraction.  So this is the same for graphs.  Just take a graph.  You can take some equivalences on the set of vertices.  So this means that you just do a partition of your vertices.  On the right, you just contract your equivalent cases.  So this means that you contract some sub-subgraphs of your graphs to vertices.  And on the left, you just forget the edges, which are not between, which are in vertices,  which are not equivalent.  So I'm just going to give an example.  So for this one, you can contract everything, just remain a vertex.  And on the right, you obtain the world graph.  You can contract only one edge, for example, this one.  If you contract, you just obtain a graph with two vertices and one wedge between them,  this.  And the extraction is given by this edge and the other vertex, so something like this.  You have three possible ways to do this.  You can just extract the vertices, just go on the right, and if you contract the vertices,  nothing happens.  So the graph stays itself.  So this is also a second core product.  This is also by algebra.  It is not a hop-adjibra because there is a group like, the graph is one vertex, and  it has no inverse, so you don't have any anti-pod.  It's not a big problem, but well, it's not a hop-adjibra.  And the code is very simple.  In fact, this is because of these parts.  For if graph, you obtain a vertex and some of the graph plus a graph tensor vertices  and more terms.  So the code unit is just given by sending any graph to one.  If your graph is totally disconnected with no edge or zero otherwise, and you can prove  that this is really a double by algebra.  So the second core product, this extraction contraction core product really coax on the  first one.  And the last example, which is the hop-adjibra of quasi-symmetric functions.  So as a algebra, it's based on the set of compositions.  The composition is just the finite sequences of positive integers.  With these products, which was used just before by Dominique, this is the quasi-triple products  on composition.  The core product, the first core product is given by decon-catination.  You just cut yours into two parts between two letters.  And there is a second one, which is given by extraction and contraction.  So for example, for this, you will cut your words into two, into several parts.  One part, two parts, two parts, and so three parts, two parts, two parts, and one part.  You just contract the parts.  So contracting just means that you sum all the letters of your word.  Because the letters are integers, you can sum them.  So this gives the terms on the left and on the right, you just quasi-suffer your parts.  So this is another core product.  You can find it into the papers of Thibault, Noveli, and the other on this subject.  I'm not totally sure they proved this is really a double by algebra.  I'm not sure they proved the co-interaction, but well, it's true.  And they prove it with a trick, which is based on manipulation of alphabets.  So it's not really obvious, but you can do it without too many combinatorics, just with  algebraic tricks.  So well, there is a co-unit, I forgot it, which is also, she's given by this.  You take a word, a composition, if it's of 0 or 1, the co-unit is 1 and 0 otherwise.  And it turns out that this is a character of Cucine, which appears in another paper  by Agar, Bergeron, and Sotile.  In this paper, they define the category of combinatorial hopp-vagebras, which are graded,  connected hopp-vagebras with a character, which is their pairs.  And they prove that in this category, Cucine with this character, epsilon prime, they don't  mention that this is the co-unit of a certain co-product, by the way, but this pair is a  terminal object.  So this means that if you take a combinatorial hopp-vagebras, so a connected, graded hopp-vagebras  with a character, you automatically obtain a morphism to Cucine, compatible with this  co-unit.  Okay, so that's nice.  These are nice objects, but the question is, what you can do with this, what you can deduce  on this construction, and well, what will it give on your, on these examples of graphs  and trees and things like this?  So first of all, take a double bi-algebra, A, with one product and two co-products, another  bi-algebra, B, and you're looking at morphism of hopp-vagebras or bi-algebra from A to B.  So it turns out that the monoid of characters of A for the second co-product acts on the  set of bi-algebra morphisms with the help of the co-action of the second co-products.  So this means that if you obtain a bi-algebra morphism from A to B, in fact, you have a  lot more bi-algebra morphisms from A to B. You can deform any bi-algebra morphisms with  the help of characters of A.  If you have this one, you will have everyone's just by using the actions.  So let's try to do this for forests.  So forests form a double bi-algebra.  So this means that there should be a unique morphism from a forest to with both co-products.  You can prove that you can compute it uniquely.  For example, let's just start with the first tree, the tree with only one vertex.  Well, it's primitive for trees.  So its image should be primitive for polynomials.  The set of primitive elements of k of x is one-dimensional, generated by x.  So this means that phi 1 of this tree should be a multiple of x.  Moreover, phi 1 is compatible with a second co-product and with its co-unit epsilon prime.  Epsilon prime of this tree is one.  So epsilon prime of its image should be one.  So epsilon prime of lambda x is lambda.  So lambda should be one.  So you entirely determine phi 1 of this tree, this should be x.  For the second one, you do the same.  Let's first compute the co-product of this tree.  This is this.  This is only one admissible cost, non-trivial.  Let us apply phi 1 to this.  Phi 1 is compatible with back delta.  So you should find something like this.  Phi 1 of this is this.  So this means that phi 1 of this tree should be x2 over 2 plus a primitive element.  So lambda of x.  This morphism phi 1 is compatible with the co-unit epsilon prime.  So epsilon prime of this polynomial should be epsilon prime of this tree.  So this should be equal to 0.  And you obtain that lambda is equal to minus 1 over 2.  And you obtain that phi 1 of this tree.  This is exactly this.  So what you are doing now is to put that this morphism is unique.  What is not clear is that it is really compatible with the second co-product.  I only use that it's compatible with the co-unit.  But you obtain for free, but it is in fact compatible with the second co-product.  Just for free.  So you can continue like this.  For this tree, you obtain something like this, which is a real bad polynomial.  Quite famous.  For this tree, this is normal, real bad polynomial, but something like this.  Maybe you recognize it.  This polynomial counts the sum of the square.  This evaluate to n is 1 square plus 2 square plus, etc. plus n square.  So these are quite special polynomial.  You can do more in fact.  Well, this is a nice way to compute the variant phi 1, but it's quite long.  And in fact, you can do better.  You can put some formula like this.  If you take an element of your double by algebra a, you can compute phi 1 of a in this way.  First, you can compute all its reduced co-products.  So the reduced co-products is just by forgetting the primitive parts.  So this means that for trees, you just forget the trivial cuts and the cuts of everything.  You can compute it and iterate and iterate and iterate and something like this.  And you know that at certain point, it will stop.  The iterated co-products should be zero after a certain point.  So you compute all of them.  You obtain tons of trees and things like this.  You just apply the co-unit on the left and then you multiply the tons by your Hilbert polynomial.  And this means that your invariance really counts something.  If you evaluate x into an integer, this is a binomial coefficient.  So really phi 1 of a just counts something for forests.  And this counts something like this.  This is quite a well-known construction.  If you take a forest with k vertices, well, just choose an index session of the forest.  It does not really matter.  Then you can associate with a polytope of dimension k.  In fact, this polytope is defined by some inequality.  If you take one vertex, which is another one, x1, xi is below the vertex j in your tree.  Then you associate to it an inequality xi times smaller than xj.  So this defines a polytope.  And you want first to delete it by an integer, i minus 1 times the polytope.  So just do an amortesy.  And you want to count the number of integral points inside.  So it's quite a famous result that this is, in fact, given by a polynomial sequence.  So this defines a polynomial, which is called the Erard polynomial.  The strict Erard polynomial is the same.  But just count the number of vertices inside your polytopes.  So this is quite known that this defines two polynomials, which are related, I will say  later.  So just to mention a problem, usually in the literature on this, the Erard polynomial in  n, just count the number of integral points of the dilated of the polytope by n.  Here, I have a problem.  If I do this, it does not work.  So I have to do some translation by one.  So for example, for this, so you have three vertices, which I did an index by one, two  three from bottom to left.  One is smaller than two, two is smaller than three.  So your polytope is defined by x smaller than one, smaller than z, between zero and one.  So the polytope associated to f is just a simplex.  So if you want to count the number of integral points into the dilated of f, what you count  is the number of points x, y, z, which are integers such that x is smaller than one,  is smaller than z, is smaller than n minus one.  And it's not very difficult to count them on two pools, but this is n, n plus one, n plus  two over six.  So this is for the Erard polynomial.  For the strict Erard polynomial, you count the point inside.  So this means that you replace your smaller or equal by just strictly smaller, and n minus  one by n plus one.  So these counts things like this, and it's not difficult to prove that this is this polynomial,  which by the way is in fact phi one of f evaluated to n.  You can do the same thing for this tree.  So here you can try to draw the polytope.  This is a pyramid with a square basis.  You can count the number of points inside, which is this.  And so this is the Erard polynomial, the strict Erard polynomial, which is again phi  one of f evaluated to n.  And in fact, this is exactly this.  The unique morphism compatible with the product and the two co-products is in fact the strict  Erard polynomial.  OK.  So if you look at this, you can observe that the strict Erard and the Erard polynomial  are very similar.  More or less they are the same coefficient, and things like this.  And in fact, you can prove that there is another morphism from the conch primer to polynomials,  which is compatible with the product and two co-products.  This is not the Erard polynomial, but more or less itself.  You just replace x by minus x, and you have to correct the same by multiplying by the  power of minus one.  And it's not very difficult to prove combinatorially, but this is compatible also with the product  and both co-products.  But there is only one morphism compatible with the product and the two co-products.  So this means that this morphism phi is the same as phi one.  And what you obtain is an algebraic proof and the 180 principle for Erard polynomial.  In fact, the strict Erard polynomial and the Erard polynomial are really closely rated.  This is just by evaluating x into minus one, minus x, and replace the same.  So here, as an application, you have a proof of the 180 principle for Erard polynomial  without, more or less without any real combinatorial stuff.  The usual proof uses some Mobius inversion in some posets.  So these are really combinatorial.  Can I ask a quick question here?  Sorry.  So this, you exemplified this as in the case of these polytops you get from trees via this  post-order.  Can you also use a similar kind of argument to get this Erard polynomial duality result  for arbitrary polytops?  I did not manage to do it.  In fact, I would need a hopper algebra structure on polytops.  A good algebra structure on polytops.  I have a product, just the usual product of polytops, but I don't find the core products.  In fact, those coming from forest are very special.  You see they are defined by some equations which are nice.  If you take any polytops, I don't know how to quote any polytops.  So.  Yeah.  Thank you.  So let's do it for graphs now.  So for graphs, I can apply the formula for phi 1, which I gave before.  So what does this mean?  I have to do all the iterated core products of graphs.  So this means that I have to cut the graphs into any number of parts I want.  So the iterated core product just means that I really cut the graphs into a lot of parts.  And then on each part, I will apply the co-unit for the second core product.  So the co-unit is one if the part has no edge.  And otherwise, it's zero.  So this means that in phi 1, which appear, only decomposition of your graphs into parts  with no edges.  The decomposition of the graph is equivalent to a coloration of a graph.  So a coloration is just associated to any vertex of the graph, a color, which usually  is a number.  So a partition of the graph is just the same as a coloration.  And the coloration happening in my phi 1 of A are just the coloration which are called  valid.  So this means that if two vertices have the same colors, they should not be neighbors in  the graph with no edges between them.  So this means that in phi 1 of A, I just take in account valid colorations.  So this is a polynomial with constant like this, and which is called the chromatic polynomial.  So in fact, what I find with graphs, the unit morphism compatible with both core products  and the core product is the chromatic polynomial, which is perhaps not a big surprise.  Because if you work with graphs, you know that chromatic polynomials is an essential  tool for such an event.  So this is perhaps an explanation why it's so important.  In fact, it's a unique polynomial invariance on graphs, which will be compatible with all  these structures of extraction or contraction and extraction of subclass.  OK, so something else, no, another application.  So I'm going back to a theoretical result.  In fact, I'm looking for the antipode.  So in all my examples for the big delta, this is a hop algebra.  So it has an antipode.  And for the second component, it's just a bi-algebra, so no antipode.  In fact, I can prove that if I want to compute the antipode of A, I just have to compute  the inverse of a special character, which is the coordinate of the second coproduct.  So the coordinate of the second coproduct is the unit for the convolution of the second  coproduct.  But for the first coproduct, it's just a character with no spatial property.  Maybe it's invertible.  If it is, well, you know that A is a hop-algebra, and you have a nice formula for your hop-algebra.  Just apply the second coproduct, and then the character and the first components.  So for double bi-algebras, if you want to compute the algebra, you just have to compute  a special character.  There is something more.  In fact, maybe you observe that all my examples of double bi-algebras are commutative.  In fact, what you obtain here, like this, is that S is a composition of algebra morphisms.  So this means that in double bi-algebras, the antipode of A is an algebra morphism.  But usually it's an anti-algebra morphism.  So double bi-algebras are special bi-algebras, so that the antipode is both an algebra and  an anti-algebra morphisms.  So this means that essentially, if you have a double bi-algebra, it should be commutative.  So this is the reason why all my examples are commutative.  In fact, you cannot obtain any double bi-algebra, which is not commutative, because, this is  one of the reasons the antipode should be an algebra morphism.  You can do more.  In fact, you have to compute the inverse of a character, which is not so obvious, in  fact.  You can do it inductively, but it's not so obvious.  But if you know how to compute phi 1, well, it's very easy to find the inverse of a character.  Just take an element of your algebra, a small a, just compute phi 1 of a, this is a polynomial,  and then evaluate it into minus 1.  And this is really the character alpha of a.  So for example, for rooted forests, if you want to compute the antipode, you need to  compute the error polynomials of any forest evaluated to minus 1.  And it's very easy, just a problem of minus 1.  So alpha of s is just a problem of minus 1.  And you obtain this formula for the antipode, which was proved by Conan Kremer, not in this  way at all.  It's just an indicative proof.  I can obtain this, something like this, with no induction.  It's more interesting for graphs.  For graphs, for a long time, the antipode was not known, not really.  You can compute it inductively, of course, but it was not so clear.  Last year, in fact, formula was proved by Ben-Yaditi Bergeron on match-ad-sec, I'm not  totally sure of the pronunciation, with a combinatorial method, which was quite complicated.  There is a Mobius inversion, and curiously, the number of acyclic orientation appears.  And in fact, this is obtained with my method by this.  If you want to do this, you have to compute phi 1 of the graph evaluated in minus 1.  Phi 1 of the graph is chromatic polynomial, and it's quite a famous result in graph theory  that the graph polynomial evaluated in minus 1 counts the number of acyclic orientations.  And to prove this, it's not so difficult, it's just a combinatorial proof by induction  of a number of vertices.  So what you obtain is the formula by Ben-Yaditi Bergeron on match-ad-sec, with no more combinatorics,  more or less.  It just applies the chromatic polynomial evaluated to minus 1.  OK.  You can do better for the chromatic character.  In fact, there is a very simple half-algebra morphism from graphs to polynomials, which  is just sending a G to a monomial, thanks to the power of the number of vertices of G.  This is really easy to show that this is half-algebra morphism.  Of course, it's just compatible with the first core product, not with the second one.  The unique morphism compatible with the second one is phi 1, the chromatic polynomial.  And I mentioned before that any half-algebra morphism from graphs to polynomials should  be obtained from the chromatic polynomial by the action of the character.  So we can write that phi 0, this very simple polynomial, very simple invariant, should be  obtained from chromatic polynomial by action of the character, which I denote by lambda,  which is very easy to compute.  You just send any graph to 1.  So lambda is a very simple character.  But which is more interesting is that it is invertible for the second convolution.  So this means that you can obtain the chromatic polynomial from this very simple morphism  just by acting a certain character, which is not so easy to find.  So this means that you obtain a formula for the chromatic polynomial.  In fact, the chromatic polynomial is the sum of all possible contractions of your graph  thanks to the power of the number of classes of your contraction and then a scalar, which  is, which can be objectively computed.  So lambda, this is a chromatic character.  You can compute its values just by induction.  And you can observe on this example that it's never 0.  The chromatic polynomial never goes to 0.  And it only depends on the number of vertices.  With one vertex, it's positive.  Two vertices, it's negative.  Three vertices, it's positive.  Four vertices, negative.  And you can prove it just by induction.  So just by something like this.  I don't have any more time, so just cut it a little bit.  So just by this, and with this formula, you can prove that in the chromatic polynomial,  the things of it are alternating, which is a result proof by Rota in the 70s, I think,  with complicated combinatorial methods.  Here you obtain each just by a small combinatorial tool, which is the construction extraction  of edges.  And then this formula for the chromatic polynomial related to the chromatic character.  Okay.  I think I still have five minutes.  So for the moment, I talk about morphism with values to polynomials.  Now I'm going to talk about morphism with values in the quasi-symmetric algebra.  So I mentioned before that by Agar, Bergeron, and Sotile, I know that there are a lot of  morphism to it because it's a terminal object.  If I want a morphism to Q-sym, I just have to choose a character.  And then I will obtain homogenous morphism compatible with the product and the first  co-product.  And I've got a formula for this, which is similar than the formula for the polynomial  invariant.  If I want to construct a morphism from A to Q-sym, I just take all the iterative co-products.  I apply some projection on it.  So I project on the homogenous parts.  Then I applied the co-unit on all the parts.  I take in account the degrees of the parts with a composition, something like this.  So what can I prove is if I'm looking for a morphism from A to Q-sym compatible with  both bi-algebraic structures, so compatible with the products and the two co-products,  this is the only possibilities.  If I'm looking for something compatible with the products and the co-products, this is  the only one.  We should work.  And happily, it does not work any time.  I need another condition.  I need another condition of the gradation.  In fact, I need that the gradation more or less just respect the first components.  If this condition, this technical condition is not satisfied, well, phi 1 is not compatible  with the second co-product.  So this means that I won't have any morphism compatible with both structures.  And this is what happens for flow rates.  In fact, my condition means that on the second co-products, I should obtain only things with  three vertices on the left.  And this is not the case.  There are some red trees.  We should have three vertices if I want to be compatible with the second co-products.  And unhappily, well, it has only two vertices.  So in this case, I won't have any morphism compatible with both structures from trees  to cussing.  So I have to cheat a little bit.  When I just replace forests by decorated forests, so this means that on any vertex on my forest,  I had a decoration, which is an integer.  So if I do some contraction, for example, for this, I'm contracting the edge between  A and B.  Well, I don't forget A and B, and I just replace the vertex, the decoration of the vertex,  by A plus B.  So for this, now the co-product is homogeneous.  Here, the weight is A plus B plus C. And on the left, and also on the right, I only obtain  trees with weight A plus B plus C.  So I can just put them into black.  And now, with this trick, the technical condition on the second co-product, which I changed,  is satisfied.  So this means that I obtained for free a hop-edge-abramorphism from forests to cussing, which is homogeneous  for this godvation by the weight, incompatible with all structures.  And this is a generalization of the Erhard polynomial, which I call the Erhard-casisimetric  function.  It is something like this.  And the same for trees, so for graphs, sorry.  And what I obtain is a generalization of the chromatic polynomial, which is called the  chromatic-casisimetric function.  But in fact, it's not quasi-symmetric.  It's symmetric for a reason of co-computativity.  And this is an object which was already known by combinatoricists on graphed-heresists.  Well, they know it, but they didn't know that it was compatible with a zero structure on  the quasi-symmetric functions.  And the last word, I said before that double bi-Agebra does not run very well with non-computative  bi-Agebras.  In fact, you can do some non-computative bi-Agebra, you can replace by index tree or planar tree,  something like this.  And you can also define two co-products.  Okay, so two co-products.  They are no more compatible as before, but they are still existing, so why not?  And you can generalize your chromatic series or Erhard series.  All the morphisms I mentioned before still exist in a non-computative way, but you cannot  choose anymore the formalism of double bi-Agebra.  There are no more double bi-Agebra.  So if you want to do this, if you want to explain it, you have to work in another category,  which is in some sense bigger, which is the category of species.  In fact, all my objects are traces, this means that they are imaged by the functions of something  which exists in a category of species.  In the category of species, with two functions, which will give, in part, the co-productive  objects, which are double bi-Agebras, and non-computative objects, which are not double  bi-Agebras, but which are traces, non-computative traces, or double bi-Agebras in the category  of species.  So we stop here and thank you very much for your attention.  Thank you very much, Loic.  Thank you.  Are there questions?  Okay.  Yes.  Did you actually...  Yeah.  Did you actually upgrade all your formalisms in the species setting?  Yes.  In fact, in the species settings, you can do the same.  So the QC on KX is replaced by the same object, which is species of composition.  Yeah.  Which is also double bi-Agebras, something like this.  And it turns out that you have two functions, the fog function, the full fog function.  The one will send you to commutative objects, and the other one to non-computative objects.  So in fact, you have bi-Agebras, you obtain...  The double bi-Agebras in the setting of species are commutative in the species settings.  Yeah.  It's sort of strong commutativity, so with the second function, there are no more commutative  in the user space in the setting.  So you can do it all this in the set of species.  More or less the same proof, something like this.  More or less, there are some technicalities in some sense, but these are the same ideas.  If I may, I'd like to ask or make a comment or ask a question.  Thank you, Loïc, for your nice talk.  You were wondering about co-product on polytopes.  There is one on cones, and polytopes can be seen as intersections of cones, and that's  the way, formally, on polytopes of the kind you were talking about and be derived.  So I think it should be possible, maybe, going that way, following the path of Bavinoc, Berlin,  then I wonder whether it's possible.  So I can say that I found some co-products on polytopes, but more or less they are useless.  There are stupid co-products, and you don't obtain the era of polynomers as an invariant.  You obtain stupid things just by sending, for example, a polynomers to the number of  vertices or something like this.  I just found stupid co-products.  Not interesting one.  Okay.  The one I'm thinking of is very geometric, and it serves similar purposes.  It's four counting into two points on cones, so it's made for that.  And it's implicit in people's work in toric geometry.  So it could be helpful, maybe.  It's a geometric one.  Okay.  So it should be better than I wanted.  I will look at it.  There's a question from Janek Vargas in the Q&A, and I have said him so that he can unmute,  but he asks, is there a polytope associated to double-post it the same way you define  the polytope associated to a forest?  Yes.  In fact, everything I did for forests can be done for post-sets or topologies.  And you also obtain some era polynomials with the same properties.  I just speak myself to forest because it was easy to describe and post-sets or topologies.  David, do you want to ask your question?  Yes.  I wanted to make a comment because it's directly related to the title of this conference, an  algebraic structures and quantum field theory.  Now, look, you know very well, Dex, work on the hot algebra of renormalization, but I  hope you're also aware of his recent work on the co-action associated with the monodromy  of the functions we get from the phymer diagrams, and that's associated with cutting lines.  It was an interesting question as to how that relates to co-action and multiple polar logarithms  that we obtain and functions beyond that.  So there is, facing us at present in quantum field theory, a compatibility question.  It might not be directly related to your talk.  And that is what about calculations in which we both cut lines to discover analytic structure,  but we also have sub-divergences that we have to renormalize.  Yeah, yeah, that's precisely the right question, David, but I think there will be an answer  pretty soon.  And it has a lot to do with Schloich's co-interacting biologmas.  Excellent.  Thank you.  Okay.  Thank you.  So I think there's also some simpler connection that just when we were looking at the general  tree-phyman rolls, then having the co-structure on trees, the co-interacting structure on  trees, it picks out a particular one, a particular choice for the lower order terms.  So that's a more trivial observation than what David and Dirk are getting at, but it's  another connection to the quantum field theory situation.  I see Mischi has his hand up.  Yes.  Thanks, and thanks, Loik, for the nice talk.  I also have a question regarding the co-products on polytopes.  And it's about, I mean, if there's a relation or if you're aware of this stuff, this newer  stuff by Agia together with Agia on the hopped mono structure of generalized parameter hydra,  which are polytopes, and on them you can define this co-product.  Is this too specific for your purposes?  At least they asked some questions about erhard polynomials on the polytopes.  Yes, but the parameter hydra are very specific polytopes.  All right.  But in this case, it's easier to define some co-products on these sort of objects, which  have a strong, they have really a strong combinatorial structure than them.  If you take any polytopes, not the case, just the problem.  The problem is really the generalization then, too.  So this only works for these specific polytopes.  Yes.  I think that for a special family of polytopes with strong structure, you can define a co-product  which should give you the erhard polynomial or things like this.  All right.  So any polytopes, in fact, this co-product is quite, you can see it easily on the post  set, you just cut or things like this.  On the polytop, geometrically, it's not so clear for me.  This is the section of faces or things like this, which is really something more complicated.  Okay.  So I don't know exactly what it is geometrically.  Okay.  But, okay.  I see.  Thank you.  So the problem for me is that I can understand polytopes in three-dimensional, but no more.  But this is only a few, three or four examples, which I can manage.  So it's not enough for me to understand what happens for bigger polytopes.  So I don't know exactly what the co-product is for polytopes.  All right.  In the interest of time, why don't we thank Loic again now?  Thank you.  Thank you.
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First of all, happy birthday to Dirk. So I'm going to start with a monoid or a group, which is proto-algebraic, so it does not really matter what it means, but it just has a good ring or algebra of polynomial functions. So this algebra just reflects the topology of the group, and you have more, you have a co-product, which sends one polynomial function, two pairs of polynomial functions, which is the co-product data, which you can see here. We just reflect the composition of the group. So this is what is called a bi-algebra if you have a monoid, and if it is a group, it is a hop-algebra. And you can recover, it's quite well known, the group of a monoid from the bi-algebra just by taking its characters, the characters are just bi-algebra morphisms from your algebra to the field, which is for me the complex field. And there is a product on character, which is the convolution, which is in some sense this dual of the co-product. So I'm not satisfied with only one group or one monoid, I need two groups, and in fact I want to do some semi-direct product or thing like this. So for this, I take two groups or two monoids with a good bi-algebras, and I suppose that the second one, g-prime, acts on the first one by monoid andomorphisms, which is exactly what I need if I want to do a semi-direct product. So what does this mean? I have first a monoid g, so it has a hop-algebra, which I call or bi-algebra, which I call a. I've got a second monoid g-prime, which has also bi-algebra, which is b. g-primes acts on g, so b co-acts on a. So there is a correction, which is a map, an algebra map from a to a tensor b. So this is just to reflect the action of g-prime on j, and if I want to translate the fact that the action of g-primes is by monoid andomorphisms, this means that a is a bi-algebra in the category of b-commodules. So what does this mean exactly, this fourth point? This means this axiom. So the first one just means that this is a correction. This is the axiom of a right correction. The second one means that the product of a is a morphism in the category of co-module over b, and it also means that the correction is an algebra morphism. This is the same. The fourth one means that the co-unit of a is a co-module morphism from a to the base field. And the last one, which is the Mohr-Schwisting, means that the co-product of a, the big delta, is a co-module morphism from a to a tensor a, which is also a b-commodule. So this means that if you do first the correction and then the co-product on the left, this is equal to do the first the co-product, then the correction on both sides of the co-product, and then to regroup together the two terms, which belongs to b. So this is this product m1324, which case four elements, and we group the second and the fourth one at the end. So for example, let's just take a very simple example. You can consider the group c with the addition, which is a group, a Boolean group. And on this, the group c star with the multiplication naturally acts by group automorphisms. So this is exactly what you have before. So the first group g is c plus. So it's algebra, it's hop-algebra is the polynomial algebras on c, which is the polynomial ring with one in the terminate. With the co-product, which is additive, the co-product of x is x tensor one plus one tensor x, this is hop-algebra. The second group g prime is c star with the multiplication. So the polynomial algebras on c star is the Laurent polynomial algebra, c of x and x minus one. I have to add an inverse to x because there is no zero in c star. With another co-product, which is multiplicative, which sends x to x tensor x. So for the first one, the big delta x is a primitive element. For the second one, it's the group line element. So this is also hop-algebra because there is an inverse. And b co-acts on a because c star acts on c with this correction, rho, which sends also x and x tensor x. In fact, more or less the co-action on the co-product, the second co-product, are more or less the same. So I don't act very much this x minus one. I would prefer to have cx. So I just forget it and I obtain a hop, not a hop-algebra, but a set-algebra, which is b, which sees the same algebra as a with another co-product. So it's no more hop-algebra because x has no more inverse. It's just a bi-algebra. Then it co-acts in the same way on a with the same correction. And for this, the correction on the co-product is the same. So this is the frame I will use now. So what I have is an algebra a with a single product and two co-products, big delta on small delta such that a m delta is a bi-algebra, a m small delta is a bi-algebra. And the second one co-acts on the first one with the correction rho, which is also the second co-product. So what I have is an object with one product and two co-products. For both co-products, it's a bi-algebra. And moreover, there is a compatibility between the two co-products, which is something like this. Replace just rho by delta and you obtain the second, the compatibility between big delta and small delta. So if you want to be complete, this should be called bi-algebras in a category of co-modules of another bi-algebra, which is quite long. So I just now called them double bi-algebras or co-interacting bi-algebras and things like this. And the first example we have is just the polynomial ring, c of x, with these two co-products, big delta which is additive and small delta which is multiplicative. We know more examples. So for example, the well-known cone-cramer hop-algebra of trees, which is based on rooted forest, and I won't belong on the remainder on it. So it's just based on rooted trees. For me, the trees, the roots of the trees are at the bottom. So if you have the trees with two vertices, three vertices, four vertices, the product is the Dijon union, so it has the basis of forests. And the first co-product is the cone-cramer one, just given by admissible cuts. So you take your tree or your forest, you just cut some branches, you put the branches on the right and the trunk on the left. So for example, for this tree, you can cut nothing and the tree turns to one, or you can cut everything, one turns to the tree, or you can cut a leaf in two possible ways. You obtain two turns to the trunk, turns to the leaf, or you can cut the two leaves which go on the right and three minutes on either roots. And the same for this, you can cut nothing or everything, or the trunk after the root, or the trunk just before the leaf, or you obtain these four turns. There is a primitive part, the first two turns, which means that the cone is very simple. The cone unit of a forest is one, is the forest has no vertex and zero otherwise. And you can observe that it is graded, obviously, by the number of vertices. If you cut a forest, you don't lose any vertex. Some goes on the left, the other on the right, but you don't lose any vertex, so this is graded by the number of vertices. So this is the most famous co-product on it, but there is a second one, which was first described by Damien Calac, Koro Shebrahini-Fart and Dominique Monson in 2008, I think, which is given by the process of construction on extraction. So what does this mean? For example, for this tree, you can separate your trees into disjoint sub-trees. So for example, you can disjoint it into three sub-trees, which has only the vertex on the left to contract these sub-trees, but nothing appears, and on the right to put these sub-trees. So here you don't do anything. You can contract the edge on the left, so this sub-tree, so this gives this tree in two possible ways. You can contract the whole tree, it only remains one vertex, and put the other tree on the right. So this is another co-product, which is also co-associative. It's not co-commutative, you can see it. And it's not a hop-file algebra because you have a group like, which is the tree with only one vertex with no inverse, so you only obtain a bi-algebra for this co-product. And they prove that this is a double bi-algebra, so this means that this co-product really co-acts in a good way on the first cone-crimer co-product. And there are similar constructions on post-sets, finite post-sets, you can see trees as post-sets just by taking the order, the partial order to be higher in the tree. So if you have a tree, you have a post-sets. And such a construction also exists on finite post-sets, or more generally on finite topologies. So this is the first example of, not the first, but second examples of double bi-algebra. I'm going to give another one based on graphs. So for this, the basis of my hop-file algebra of graphs is this whole set of graphs. So these are just simple graphs. So here you have all graphs connected or not with one, two, three, or four vertices. There is a simple product on it, which is the Dijon union. The unit is the empty graph, which is here. So for example, this graph is the product of this by itself, something like this. There is another simple product. There is a very simple co-product, which is just given by separating two graphs into two parts. So take your set of vertices, you put some of the vertices on the left with the edges between them, the other vertices on the right, also with the edges between them. And you obtain a nice co-product, which was first defined, I think, in a paper of Schmitt on incidence hop-file zebras. This is an example of incidence hop-file zebras based on a set of graphs. This is just a set of graphs with a given set of vertices with expression of edges. So this is a co-product. This is a co-associative. It's really not difficult to see. And it's co-commutative. This is very different from the co-crime or co-product. This is co-commutative. There is a second one, which is also can be found in the paper of Schmitt with another incidence by a zebra, and which was also described in a paper of Dominique in 2011, with various examples on graphs, oriented graphs, acyclic oriented graphs, something like this. So this is the same idea as for trees. So for trees, the first co-product is just cutting the trees into two parts. So this is the same for graphs. The second co-product was given by extraction and contraction. So this is the same for graphs. Just take a graph. You can take some equivalences on the set of vertices. So this means that you just do a partition of your vertices. On the right, you just contract your equivalent cases. So this means that you contract some sub-subgraphs of your graphs to vertices. And on the left, you just forget the edges, which are not between, which are in vertices, which are not equivalent. So I'm just going to give an example. So for this one, you can contract everything, just remain a vertex. And on the right, you obtain the world graph. You can contract only one edge, for example, this one. If you contract, you just obtain a graph with two vertices and one wedge between them, this. And the extraction is given by this edge and the other vertex, so something like this. You have three possible ways to do this. You can just extract the vertices, just go on the right, and if you contract the vertices, nothing happens. So the graph stays itself. So this is also a second core product. This is also by algebra. It is not a hop-adjibra because there is a group like, the graph is one vertex, and it has no inverse, so you don't have any anti-pod. It's not a big problem, but well, it's not a hop-adjibra. And the code is very simple. In fact, this is because of these parts. For if graph, you obtain a vertex and some of the graph plus a graph tensor vertices and more terms. So the code unit is just given by sending any graph to one. If your graph is totally disconnected with no edge or zero otherwise, and you can prove that this is really a double by algebra. So the second core product, this extraction contraction core product really coax on the first one. And the last example, which is the hop-adjibra of quasi-symmetric functions. So as a algebra, it's based on the set of compositions. The composition is just the finite sequences of positive integers. With these products, which was used just before by Dominique, this is the quasi-triple products on composition. The core product, the first core product is given by decon-catination. You just cut yours into two parts between two letters. And there is a second one, which is given by extraction and contraction. So for example, for this, you will cut your words into two, into several parts. One part, two parts, two parts, and so three parts, two parts, two parts, and one part. You just contract the parts. So contracting just means that you sum all the letters of your word. Because the letters are integers, you can sum them. So this gives the terms on the left and on the right, you just quasi-suffer your parts. So this is another core product. You can find it into the papers of Thibault, Noveli, and the other on this subject. I'm not totally sure they proved this is really a double by algebra. I'm not sure they proved the co-interaction, but well, it's true. And they prove it with a trick, which is based on manipulation of alphabets. So it's not really obvious, but you can do it without too many combinatorics, just with algebraic tricks. So well, there is a co-unit, I forgot it, which is also, she's given by this. You take a word, a composition, if it's of 0 or 1, the co-unit is 1 and 0 otherwise. And it turns out that this is a character of Cucine, which appears in another paper by Agar, Bergeron, and Sotile. In this paper, they define the category of combinatorial hopp-vagebras, which are graded, connected hopp-vagebras with a character, which is their pairs. And they prove that in this category, Cucine with this character, epsilon prime, they don't mention that this is the co-unit of a certain co-product, by the way, but this pair is a terminal object. So this means that if you take a combinatorial hopp-vagebras, so a connected, graded hopp-vagebras with a character, you automatically obtain a morphism to Cucine, compatible with this co-unit. Okay, so that's nice. These are nice objects, but the question is, what you can do with this, what you can deduce on this construction, and well, what will it give on your, on these examples of graphs and trees and things like this? So first of all, take a double bi-algebra, A, with one product and two co-products, another bi-algebra, B, and you're looking at morphism of hopp-vagebras or bi-algebra from A to B. So it turns out that the monoid of characters of A for the second co-product acts on the set of bi-algebra morphisms with the help of the co-action of the second co-products. So this means that if you obtain a bi-algebra morphism from A to B, in fact, you have a lot more bi-algebra morphisms from A to B. You can deform any bi-algebra morphisms with the help of characters of A. If you have this one, you will have everyone's just by using the actions. So let's try to do this for forests. So forests form a double bi-algebra. So this means that there should be a unique morphism from a forest to with both co-products. You can prove that you can compute it uniquely. For example, let's just start with the first tree, the tree with only one vertex. Well, it's primitive for trees. So its image should be primitive for polynomials. The set of primitive elements of k of x is one-dimensional, generated by x. So this means that phi 1 of this tree should be a multiple of x. Moreover, phi 1 is compatible with a second co-product and with its co-unit epsilon prime. Epsilon prime of this tree is one. So epsilon prime of its image should be one. So epsilon prime of lambda x is lambda. So lambda should be one. So you entirely determine phi 1 of this tree, this should be x. For the second one, you do the same. Let's first compute the co-product of this tree. This is this. This is only one admissible cost, non-trivial. Let us apply phi 1 to this. Phi 1 is compatible with back delta. So you should find something like this. Phi 1 of this is this. So this means that phi 1 of this tree should be x2 over 2 plus a primitive element. So lambda of x. This morphism phi 1 is compatible with the co-unit epsilon prime. So epsilon prime of this polynomial should be epsilon prime of this tree. So this should be equal to 0. And you obtain that lambda is equal to minus 1 over 2. And you obtain that phi 1 of this tree. This is exactly this. So what you are doing now is to put that this morphism is unique. What is not clear is that it is really compatible with the second co-product. I only use that it's compatible with the co-unit. But you obtain for free, but it is in fact compatible with the second co-product. Just for free. So you can continue like this. For this tree, you obtain something like this, which is a real bad polynomial. Quite famous. For this tree, this is normal, real bad polynomial, but something like this. Maybe you recognize it. This polynomial counts the sum of the square. This evaluate to n is 1 square plus 2 square plus, etc. plus n square. So these are quite special polynomial. You can do more in fact. Well, this is a nice way to compute the variant phi 1, but it's quite long. And in fact, you can do better. You can put some formula like this. If you take an element of your double by algebra a, you can compute phi 1 of a in this way. First, you can compute all its reduced co-products. So the reduced co-products is just by forgetting the primitive parts. So this means that for trees, you just forget the trivial cuts and the cuts of everything. You can compute it and iterate and iterate and iterate and something like this. And you know that at certain point, it will stop. The iterated co-products should be zero after a certain point. So you compute all of them. You obtain tons of trees and things like this. You just apply the co-unit on the left and then you multiply the tons by your Hilbert polynomial. And this means that your invariance really counts something. If you evaluate x into an integer, this is a binomial coefficient. So really phi 1 of a just counts something for forests. And this counts something like this. This is quite a well-known construction. If you take a forest with k vertices, well, just choose an index session of the forest. It does not really matter. Then you can associate with a polytope of dimension k. In fact, this polytope is defined by some inequality. If you take one vertex, which is another one, x1, xi is below the vertex j in your tree. Then you associate to it an inequality xi times smaller than xj. So this defines a polytope. And you want first to delete it by an integer, i minus 1 times the polytope. So just do an amortesy. And you want to count the number of integral points inside. So it's quite a famous result that this is, in fact, given by a polynomial sequence. So this defines a polynomial, which is called the Erard polynomial. The strict Erard polynomial is the same. But just count the number of vertices inside your polytopes. So this is quite known that this defines two polynomials, which are related, I will say later. So just to mention a problem, usually in the literature on this, the Erard polynomial in n, just count the number of integral points of the dilated of the polytope by n. Here, I have a problem. If I do this, it does not work. So I have to do some translation by one. So for example, for this, so you have three vertices, which I did an index by one, two three from bottom to left. One is smaller than two, two is smaller than three. So your polytope is defined by x smaller than one, smaller than z, between zero and one. So the polytope associated to f is just a simplex. So if you want to count the number of integral points into the dilated of f, what you count is the number of points x, y, z, which are integers such that x is smaller than one, is smaller than z, is smaller than n minus one. And it's not very difficult to count them on two pools, but this is n, n plus one, n plus two over six. So this is for the Erard polynomial. For the strict Erard polynomial, you count the point inside. So this means that you replace your smaller or equal by just strictly smaller, and n minus one by n plus one. So these counts things like this, and it's not difficult to prove that this is this polynomial, which by the way is in fact phi one of f evaluated to n. You can do the same thing for this tree. So here you can try to draw the polytope. This is a pyramid with a square basis. You can count the number of points inside, which is this. And so this is the Erard polynomial, the strict Erard polynomial, which is again phi one of f evaluated to n. And in fact, this is exactly this. The unique morphism compatible with the product and the two co-products is in fact the strict Erard polynomial. OK. So if you look at this, you can observe that the strict Erard and the Erard polynomial are very similar. More or less they are the same coefficient, and things like this. And in fact, you can prove that there is another morphism from the conch primer to polynomials, which is compatible with the product and two co-products. This is not the Erard polynomial, but more or less itself. You just replace x by minus x, and you have to correct the same by multiplying by the power of minus one. And it's not very difficult to prove combinatorially, but this is compatible also with the product and both co-products. But there is only one morphism compatible with the product and the two co-products. So this means that this morphism phi is the same as phi one. And what you obtain is an algebraic proof and the 180 principle for Erard polynomial. In fact, the strict Erard polynomial and the Erard polynomial are really closely rated. This is just by evaluating x into minus one, minus x, and replace the same. So here, as an application, you have a proof of the 180 principle for Erard polynomial without, more or less without any real combinatorial stuff. The usual proof uses some Mobius inversion in some posets. So these are really combinatorial. Can I ask a quick question here? Sorry. So this, you exemplified this as in the case of these polytops you get from trees via this post-order. Can you also use a similar kind of argument to get this Erard polynomial duality result for arbitrary polytops? I did not manage to do it. In fact, I would need a hopper algebra structure on polytops. A good algebra structure on polytops. I have a product, just the usual product of polytops, but I don't find the core products. In fact, those coming from forest are very special. You see they are defined by some equations which are nice. If you take any polytops, I don't know how to quote any polytops. So. Yeah. Thank you. So let's do it for graphs now. So for graphs, I can apply the formula for phi 1, which I gave before. So what does this mean? I have to do all the iterated core products of graphs. So this means that I have to cut the graphs into any number of parts I want. So the iterated core product just means that I really cut the graphs into a lot of parts. And then on each part, I will apply the co-unit for the second core product. So the co-unit is one if the part has no edge. And otherwise, it's zero. So this means that in phi 1, which appear, only decomposition of your graphs into parts with no edges. The decomposition of the graph is equivalent to a coloration of a graph. So a coloration is just associated to any vertex of the graph, a color, which usually is a number. So a partition of the graph is just the same as a coloration. And the coloration happening in my phi 1 of A are just the coloration which are called valid. So this means that if two vertices have the same colors, they should not be neighbors in the graph with no edges between them. So this means that in phi 1 of A, I just take in account valid colorations. So this is a polynomial with constant like this, and which is called the chromatic polynomial. So in fact, what I find with graphs, the unit morphism compatible with both core products and the core product is the chromatic polynomial, which is perhaps not a big surprise. Because if you work with graphs, you know that chromatic polynomials is an essential tool for such an event. So this is perhaps an explanation why it's so important. In fact, it's a unique polynomial invariance on graphs, which will be compatible with all these structures of extraction or contraction and extraction of subclass. OK, so something else, no, another application. So I'm going back to a theoretical result. In fact, I'm looking for the antipode. So in all my examples for the big delta, this is a hop algebra. So it has an antipode. And for the second component, it's just a bi-algebra, so no antipode. In fact, I can prove that if I want to compute the antipode of A, I just have to compute the inverse of a special character, which is the coordinate of the second coproduct. So the coordinate of the second coproduct is the unit for the convolution of the second coproduct. But for the first coproduct, it's just a character with no spatial property. Maybe it's invertible. If it is, well, you know that A is a hop-algebra, and you have a nice formula for your hop-algebra. Just apply the second coproduct, and then the character and the first components. So for double bi-algebras, if you want to compute the algebra, you just have to compute a special character. There is something more. In fact, maybe you observe that all my examples of double bi-algebras are commutative. In fact, what you obtain here, like this, is that S is a composition of algebra morphisms. So this means that in double bi-algebras, the antipode of A is an algebra morphism. But usually it's an anti-algebra morphism. So double bi-algebras are special bi-algebras, so that the antipode is both an algebra and an anti-algebra morphisms. So this means that essentially, if you have a double bi-algebra, it should be commutative. So this is the reason why all my examples are commutative. In fact, you cannot obtain any double bi-algebra, which is not commutative, because, this is one of the reasons the antipode should be an algebra morphism. You can do more. In fact, you have to compute the inverse of a character, which is not so obvious, in fact. You can do it inductively, but it's not so obvious. But if you know how to compute phi 1, well, it's very easy to find the inverse of a character. Just take an element of your algebra, a small a, just compute phi 1 of a, this is a polynomial, and then evaluate it into minus 1. And this is really the character alpha of a. So for example, for rooted forests, if you want to compute the antipode, you need to compute the error polynomials of any forest evaluated to minus 1. And it's very easy, just a problem of minus 1. So alpha of s is just a problem of minus 1. And you obtain this formula for the antipode, which was proved by Conan Kremer, not in this way at all. It's just an indicative proof. I can obtain this, something like this, with no induction. It's more interesting for graphs. For graphs, for a long time, the antipode was not known, not really. You can compute it inductively, of course, but it was not so clear. Last year, in fact, formula was proved by Ben-Yaditi Bergeron on match-ad-sec, I'm not totally sure of the pronunciation, with a combinatorial method, which was quite complicated. There is a Mobius inversion, and curiously, the number of acyclic orientation appears. And in fact, this is obtained with my method by this. If you want to do this, you have to compute phi 1 of the graph evaluated in minus 1. Phi 1 of the graph is chromatic polynomial, and it's quite a famous result in graph theory that the graph polynomial evaluated in minus 1 counts the number of acyclic orientations. And to prove this, it's not so difficult, it's just a combinatorial proof by induction of a number of vertices. So what you obtain is the formula by Ben-Yaditi Bergeron on match-ad-sec, with no more combinatorics, more or less. It just applies the chromatic polynomial evaluated to minus 1. OK. You can do better for the chromatic character. In fact, there is a very simple half-algebra morphism from graphs to polynomials, which is just sending a G to a monomial, thanks to the power of the number of vertices of G. This is really easy to show that this is half-algebra morphism. Of course, it's just compatible with the first core product, not with the second one. The unique morphism compatible with the second one is phi 1, the chromatic polynomial. And I mentioned before that any half-algebra morphism from graphs to polynomials should be obtained from the chromatic polynomial by the action of the character. So we can write that phi 0, this very simple polynomial, very simple invariant, should be obtained from chromatic polynomial by action of the character, which I denote by lambda, which is very easy to compute. You just send any graph to 1. So lambda is a very simple character. But which is more interesting is that it is invertible for the second convolution. So this means that you can obtain the chromatic polynomial from this very simple morphism just by acting a certain character, which is not so easy to find. So this means that you obtain a formula for the chromatic polynomial. In fact, the chromatic polynomial is the sum of all possible contractions of your graph thanks to the power of the number of classes of your contraction and then a scalar, which is, which can be objectively computed. So lambda, this is a chromatic character. You can compute its values just by induction. And you can observe on this example that it's never 0. The chromatic polynomial never goes to 0. And it only depends on the number of vertices. With one vertex, it's positive. Two vertices, it's negative. Three vertices, it's positive. Four vertices, negative. And you can prove it just by induction. So just by something like this. I don't have any more time, so just cut it a little bit. So just by this, and with this formula, you can prove that in the chromatic polynomial, the things of it are alternating, which is a result proof by Rota in the 70s, I think, with complicated combinatorial methods. Here you obtain each just by a small combinatorial tool, which is the construction extraction of edges. And then this formula for the chromatic polynomial related to the chromatic character. Okay. I think I still have five minutes. So for the moment, I talk about morphism with values to polynomials. Now I'm going to talk about morphism with values in the quasi-symmetric algebra. So I mentioned before that by Agar, Bergeron, and Sotile, I know that there are a lot of morphism to it because it's a terminal object. If I want a morphism to Q-sym, I just have to choose a character. And then I will obtain homogenous morphism compatible with the product and the first co-product. And I've got a formula for this, which is similar than the formula for the polynomial invariant. If I want to construct a morphism from A to Q-sym, I just take all the iterative co-products. I apply some projection on it. So I project on the homogenous parts. Then I applied the co-unit on all the parts. I take in account the degrees of the parts with a composition, something like this. So what can I prove is if I'm looking for a morphism from A to Q-sym compatible with both bi-algebraic structures, so compatible with the products and the two co-products, this is the only possibilities. If I'm looking for something compatible with the products and the co-products, this is the only one. We should work. And happily, it does not work any time. I need another condition. I need another condition of the gradation. In fact, I need that the gradation more or less just respect the first components. If this condition, this technical condition is not satisfied, well, phi 1 is not compatible with the second co-product. So this means that I won't have any morphism compatible with both structures. And this is what happens for flow rates. In fact, my condition means that on the second co-products, I should obtain only things with three vertices on the left. And this is not the case. There are some red trees. We should have three vertices if I want to be compatible with the second co-products. And unhappily, well, it has only two vertices. So in this case, I won't have any morphism compatible with both structures from trees to cussing. So I have to cheat a little bit. When I just replace forests by decorated forests, so this means that on any vertex on my forest, I had a decoration, which is an integer. So if I do some contraction, for example, for this, I'm contracting the edge between A and B. Well, I don't forget A and B, and I just replace the vertex, the decoration of the vertex, by A plus B. So for this, now the co-product is homogeneous. Here, the weight is A plus B plus C. And on the left, and also on the right, I only obtain trees with weight A plus B plus C. So I can just put them into black. And now, with this trick, the technical condition on the second co-product, which I changed, is satisfied. So this means that I obtained for free a hop-edge-abramorphism from forests to cussing, which is homogeneous for this godvation by the weight, incompatible with all structures. And this is a generalization of the Erhard polynomial, which I call the Erhard-casisimetric function. It is something like this. And the same for trees, so for graphs, sorry. And what I obtain is a generalization of the chromatic polynomial, which is called the chromatic-casisimetric function. But in fact, it's not quasi-symmetric. It's symmetric for a reason of co-computativity. And this is an object which was already known by combinatoricists on graphed-heresists. Well, they know it, but they didn't know that it was compatible with a zero structure on the quasi-symmetric functions. And the last word, I said before that double bi-Agebra does not run very well with non-computative bi-Agebras. In fact, you can do some non-computative bi-Agebra, you can replace by index tree or planar tree, something like this. And you can also define two co-products. Okay, so two co-products. They are no more compatible as before, but they are still existing, so why not? And you can generalize your chromatic series or Erhard series. All the morphisms I mentioned before still exist in a non-computative way, but you cannot choose anymore the formalism of double bi-Agebra. There are no more double bi-Agebra. So if you want to do this, if you want to explain it, you have to work in another category, which is in some sense bigger, which is the category of species. In fact, all my objects are traces, this means that they are imaged by the functions of something which exists in a category of species. In the category of species, with two functions, which will give, in part, the co-productive objects, which are double bi-Agebras, and non-computative objects, which are not double bi-Agebras, but which are traces, non-computative traces, or double bi-Agebras in the category of species. So we stop here and thank you very much for your attention. Thank you very much, Loic. Thank you. Are there questions? Okay. Yes. Did you actually... Yeah. Did you actually upgrade all your formalisms in the species setting? Yes. In fact, in the species settings, you can do the same. So the QC on KX is replaced by the same object, which is species of composition. Yeah. Which is also double bi-Agebras, something like this. And it turns out that you have two functions, the fog function, the full fog function. The one will send you to commutative objects, and the other one to non-computative objects. So in fact, you have bi-Agebras, you obtain... The double bi-Agebras in the setting of species are commutative in the species settings. Yeah. It's sort of strong commutativity, so with the second function, there are no more commutative in the user space in the setting. So you can do it all this in the set of species. More or less the same proof, something like this. More or less, there are some technicalities in some sense, but these are the same ideas. If I may, I'd like to ask or make a comment or ask a question. Thank you, Loïc, for your nice talk. You were wondering about co-product on polytopes. There is one on cones, and polytopes can be seen as intersections of cones, and that's the way, formally, on polytopes of the kind you were talking about and be derived. So I think it should be possible, maybe, going that way, following the path of Bavinoc, Berlin, then I wonder whether it's possible. So I can say that I found some co-products on polytopes, but more or less they are useless. There are stupid co-products, and you don't obtain the era of polynomers as an invariant. You obtain stupid things just by sending, for example, a polynomers to the number of vertices or something like this. I just found stupid co-products. Not interesting one. Okay. The one I'm thinking of is very geometric, and it serves similar purposes. It's four counting into two points on cones, so it's made for that. And it's implicit in people's work in toric geometry. So it could be helpful, maybe. It's a geometric one. Okay. So it should be better than I wanted. I will look at it. There's a question from Janek Vargas in the Q&A, and I have said him so that he can unmute, but he asks, is there a polytope associated to double-post it the same way you define the polytope associated to a forest? Yes. In fact, everything I did for forests can be done for post-sets or topologies. And you also obtain some era polynomials with the same properties. I just speak myself to forest because it was easy to describe and post-sets or topologies. David, do you want to ask your question? Yes. I wanted to make a comment because it's directly related to the title of this conference, an algebraic structures and quantum field theory. Now, look, you know very well, Dex, work on the hot algebra of renormalization, but I hope you're also aware of his recent work on the co-action associated with the monodromy of the functions we get from the phymer diagrams, and that's associated with cutting lines. It was an interesting question as to how that relates to co-action and multiple polar logarithms that we obtain and functions beyond that. So there is, facing us at present in quantum field theory, a compatibility question. It might not be directly related to your talk. And that is what about calculations in which we both cut lines to discover analytic structure, but we also have sub-divergences that we have to renormalize. Yeah, yeah, that's precisely the right question, David, but I think there will be an answer pretty soon. And it has a lot to do with Schloich's co-interacting biologmas. Excellent. Thank you. Okay. Thank you. So I think there's also some simpler connection that just when we were looking at the general tree-phyman rolls, then having the co-structure on trees, the co-interacting structure on trees, it picks out a particular one, a particular choice for the lower order terms. So that's a more trivial observation than what David and Dirk are getting at, but it's another connection to the quantum field theory situation. I see Mischi has his hand up. Yes. Thanks, and thanks, Loik, for the nice talk. I also have a question regarding the co-products on polytopes. And it's about, I mean, if there's a relation or if you're aware of this stuff, this newer stuff by Agia together with Agia on the hopped mono structure of generalized parameter hydra, which are polytopes, and on them you can define this co-product. Is this too specific for your purposes? At least they asked some questions about erhard polynomials on the polytopes. Yes, but the parameter hydra are very specific polytopes. All right. But in this case, it's easier to define some co-products on these sort of objects, which have a strong, they have really a strong combinatorial structure than them. If you take any polytopes, not the case, just the problem. The problem is really the generalization then, too. So this only works for these specific polytopes. Yes. I think that for a special family of polytopes with strong structure, you can define a co-product which should give you the erhard polynomial or things like this. All right. So any polytopes, in fact, this co-product is quite, you can see it easily on the post set, you just cut or things like this. On the polytop, geometrically, it's not so clear for me. This is the section of faces or things like this, which is really something more complicated. Okay. So I don't know exactly what it is geometrically. Okay. But, okay. I see. Thank you. So the problem for me is that I can understand polytopes in three-dimensional, but no more. But this is only a few, three or four examples, which I can manage. So it's not enough for me to understand what happens for bigger polytopes. So I don't know exactly what the co-product is for polytopes. All right. In the interest of time, why don't we thank Loic again now? Thank you. Thank you.
polynomial function, two pairs of polynomial on this, the group c star with the multiplication bi-algebras or co-interacting bi-algebras and morphisms with the help of characters of A. If usually in the literature on this, the Erard from graphs to polynomials should be obtained structure on the quasi-symmetric functions. And chromatic series or Erhard series. All the will give, in part, the co-productive objects, co-structure on trees, the co-interacting