Realizations of Hopf algebras of graphs by alphabets

TL;DR

This paper constructs polynomial realizations of Hopf algebras related to graphs and posets using totally quasi-ordered alphabets, providing explicit algebraic embeddings and coproduct structures.

Contribution

It introduces new polynomial realizations of Hopf algebras on graphs and posets via totally quasi-ordered alphabets, with explicit coproduct definitions.

Findings

Polynomial embeddings of graph-related Hopf algebras into polynomial algebras.

Explicit coproduct structures via alphabet doubling and squaring.

Identification of cointeracting bialgebras in the commutative case.

Abstract

We here give polynomial realizations of various Hopf algebras or bialgebras on Feynman graphs, graphs, posets or quasi-posets, that it to say injections of these objects into polynomial algebras generated by an alphabet. The alphabet here considered are totally quasi-ordered. The coproducts are given by doubling the alphabets; a second coproduct is defined by squaring the alphabets, and we obtain cointeracting bialgebras in the commutative case.

Equations

$$M_{(a_1,\dots ,a_k)}(X)=\sum _{x_1<\dots <x_k\text{mbox}inX}{x}_1^{a_1}\dots x_n^{a_n}.$$

$$M_{(a)}(X)M_{(b)}(X)$$

$$=\sum _{x<y}{x}^a{y}^b+\sum _{y<x}{x}^a{y}^b+\sum _x{x}^{a+b}$$

$$=M_{(a,b)}(X)+M_{(b,a)}(X)+M_{(a+b)}(X),$$

$$M_{(a)}{M}_{(b)}=M_{(a,b)}+M_{(b,a)}+M_{(a+b)}.$$

$$\Delta (M_{(a_1,\dots ,a_k)})(X,Y)=M_{(a_1,\dots ,a_k)}(X\bigsqcup Y).$$

$$\Delta (M_{(a,b)})(X,Y)$$

$$=\sum _{x<y\text{mbox}inX}{x}^a{y}^b+\sum _{x<y\text{mbox}inY}{x}^a{y}^b+\sum _{(x,y)\in X\times Y}{x}^a{y}^b$$

$$=M_{(a,b)}(X)+M_{(a,b)}(Y)+M_{(a)}(X)M_{(b)}(Y),$$

$$\Delta (M_{(a,b)})=M_{(a,b)}\otimes 1+1\otimes M_{(a,b)}+M_{(a)}\otimes M_{(b)}.$$

$$\delta (M_{(a_1,\dots ,a_k)})(X,Y)=M_{(a_1,\dots ,a_k)}(XY).$$

$$\delta (M_{(a,b)})(X,Y)$$

$$=\sum _{\begin{array}{c}x<x^{\prime }\text{mbox}inX,\\ y,y^{\prime }\in Y\end{array}}{x}^a{y}^a{x}^{\prime b}{y}^{\prime b}+\sum _{\begin{array}{c}x\in X,y<y^{\prime }\text{mbox}inX,\end{array}}{x}^a{y}^a{x}^b{y}^{\prime b}$$

$$=M_{(a,b)}(X)(M_{(a,b)}(Y)+M_{(b,a)}(Y)+M_{(a+b)}(Y))+M_{(a+b)}(X)M_{(a,b)}(Y),$$

$$\delta (M_{(a,b)})=M_{(a,b)}\otimes (M_{(a,b)}+M_{(b,a)}+M_{(a+b)})+M_{(a+b)}\otimes M_{(a,b)}.$$

$$(X\bigsqcup Y)Z=(XZ)\bigsqcup (YZ).$$

$$(\Delta \otimes Id)\circ \delta =m_{1,3,24}\circ (\delta \otimes \delta )\circ \Delta ,$$

$$G=\mbox{\parbox{8.53581pt}{\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern 7.11108pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern 0.0pt\offinterlineskip\halign{\entry@#!@&&\entry@@#!@\cr\\\\\\\crcr}}}\ignorespaces{\hbox{\kern-3.0pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise 0.0pt\hbox{$\textstyle{}$}}}}}}}{{{\hbox{\ellipsed@{7.11108pt}{6.0pt}}}}\hbox{\kern-7.11108pt\raise-33.0pt\hbox{\hbox{\kern 6.0pt\raise 0.0pt\hbox{$\textstyle{\>}$}}}}}\ignorespaces\ignorespaces\ignorespaces{}\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern-4.30554pt\raise-3.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{}\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern 4.30554pt\raise-3.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{{{\hbox{\ellipsed@{7.11108pt}{6.0pt}}}}\hbox{\kern-7.11108pt\raise-69.0pt\hbox{\hbox{\kern 6.0pt\raise 0.0pt\hbox{$\textstyle{\>}$}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{}{}{{}{}{}{}{}{}{}{}{}{}{}{}{}}{}\ignorespaces\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{{}{}{{}}{{}{}{}}{}}}}\ignorespaces{}\ignorespaces{}{}{}{{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}}{\hbox{\kern-3.33269pt\raise-38.99854pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}\ignorespaces\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{{}{}{}{{}}{{}{}{}\lx@xy@spline@}{}}}}\ignorespaces{}\ignorespaces\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{{}{}{{}}{{}{}{}}{}}}}\ignorespaces{}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{}{}{{}{}{}{}{}{}{}{}{}{}{}{}{}}{}\ignorespaces\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{{}{}{{}}{{}{}{}}{}}}}\ignorespaces{}\ignorespaces{}{}{}{{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}}{\hbox{\kern 3.33269pt\raise-38.99854pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}\ignorespaces\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{{}{}{}{{}}{{}{}{}\lx@xy@spline@}{}}}}\ignorespaces{}\ignorespaces\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{{}{}{{}}{{}{}{}}{}}}}\ignorespaces{}{\hbox{\kern-3.0pt\raise-102.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise 0.0pt\hbox{$\textstyle{\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces{}\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern 0.0pt\raise-74.99997pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}\ignorespaces}}}}\ignorespaces}},$$

$$\Delta (G)$$

$$\Delta_{CK}(G)=G\otimes\mbox{\parbox{8.53581pt}{\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern 7.11108pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern 0.0pt\offinterlineskip\halign{\entry@#!@&&\entry@@#!@\cr&\\&\\&\crcr}}}\ignorespaces{\hbox{\kern-3.0pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise 0.0pt\hbox{$\textstyle{}$}}}}}}}{\hbox{\kern 35.22217pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise 0.0pt\hbox{$\textstyle{}$}}}}}}}{{{\hbox{\ellipsed@{7.11108pt}{6.0pt}}}}\hbox{\kern-7.11108pt\raise-33.0pt\hbox{\hbox{\kern 6.0pt\raise 0.0pt\hbox{$\textstyle{\>}$}}}}}\ignorespaces\ignorespaces\ignorespaces{}\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern-4.30554pt\raise-3.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{}\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern 4.30554pt\raise-3.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{{{\hbox{\ellipsed@{7.11108pt}{6.0pt}}}}\hbox{\kern 31.11108pt\raise-33.0pt\hbox{\hbox{\kern 6.0pt\raise 0.0pt\hbox{$\textstyle{\>}$}}}}}\ignorespaces\ignorespaces\ignorespaces{}\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern 33.91663pt\raise-3.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{}\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern 42.52771pt\raise-3.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern-3.0pt\raise-66.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise 0.0pt\hbox{$\textstyle{\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces{}\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern 0.0pt\raise-38.99997pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern 35.22217pt\raise-66.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise 0.0pt\hbox{$\textstyle{\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces{}\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern 33.91663pt\raise-38.99997pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{}\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern 42.52771pt\raise-38.99997pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}\ignorespaces}}}}\ignorespaces}}+\mbox{\parbox{8.53581pt}{\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern 7.11108pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern 0.0pt\offinterlineskip\halign{\entry@#!@&&\entry@@#!@\cr\\\\\crcr}}}\ignorespaces{\hbox{\kern-3.0pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise 0.0pt\hbox{$\textstyle{}$}}}}}}}{{{\hbox{\ellipsed@{7.11108pt}{6.0pt}}}}\hbox{\kern-7.11108pt\raise-33.0pt\hbox{\hbox{\kern 6.0pt\raise 0.0pt\hbox{$\textstyle{\>}$}}}}}\ignorespaces\ignorespaces\ignorespaces{}\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern-4.30554pt\raise-3.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{}\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern 4.30554pt\raise-3.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern-3.0pt\raise-66.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise 0.0pt\hbox{$\textstyle{\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces{}\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern 0.0pt\raise-38.99997pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}\ignorespaces}}}}\ignorespaces}}\otimes G.$$

$$(H_{\mathcal{FG}},{.}_q,\Delta )\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$\forall i\in X,$$

$$\forall i,j,k\in X,$$

$$\forall i,j\in X,$$

$$\forall i,j\in X\bigsqcup Y,$$

$$\text{mbox}or(i,j\in Y\text{mbox}andi{\le }_{y}j)$$

$$\text{mbox}or(i\in X,j\in Y).$$

$$\forall i,j\in X\bigsqcup Y,$$

$$(X\bigsqcup Y)\bigsqcup Z=X\bigsqcup (Y\bigsqcup Z).$$

$$\forall i,i^{\prime }\in X,j,j^{\prime }\in Y,$$

$$(XY)Z$$

$$A_q(X)$$

$${\mathbf{A}}_q(X)$$

$$M=\prod _{i\in X}{x}_i^{{ϵ}_i}\prod _{i{\le }_{X}j}{x}_{i,j}^{{\alpha }_{i,j}}\prod _{j\in X}{x}_{-\infty ,j}^{{\beta }_j}\prod _{i\in X}{x}_{i,+\infty }^{{\gamma }_i},$$

Full text

Keywords. Combinatorial Hopf algebras; Feynman graphs; posets

AMS classification. 16T05, 05C25, 06A11

1. 1 Operations on alphabets

2. 1.1 Quasi-ordered alphabets

3. 1.2 Disjoint union

4. 1.3 Product

5. 2 Algebras attached to alphabets

6. 2.1 Definition

7. 2.2 Doubling the alphabets

8. 2.3 Squaring the alphabets

9. 3 Feynman graphs

10. 3.1 Definition

11. 3.2 Monomials and Feynman graphs

12. 3.3 The first coproduct

13. 3.4 The second coproduct

14. 3.5 Restriction to ordered alphabets

15. 4 Quotients of Feynman graphs

16. 4.1 Simple oriented graphs

17. 4.2 Simple graphs with no cycle

18. 4.3 quasiposets

19. 4.4 Posets

20. 4.5 Dual product

Introduction

Some combinatorial Hopf algebras admit a polynomial realization, which gives an efficient way to prove the existence of the coproduct and more structures, see [8, 19, 15, 7, 17]. Let us explicit a well-known example. The algebra $\mathbf{QSym}$ of quasi-symmetric functions has a basis $(M_{a_{1},\ldots,a_{k}})$ indexed by compositions, that is to say finite sequences of positive integers.

For any totally ordered alphabet $X$, let us consider the following elements of the ring of formal series $\mathbb{K}[[X]]$ generated by $X$:

[TABLE]

This defines a map from $\mathbf{QSym}$ to $\mathbb{K}[[X]]$, injective if, and only if, $X$ is infinite. This map is an algebra morphism. For example, if $a,b\geq 1$, for any totally ordered alphabet $X$:

[TABLE]

and, in $\mathbf{QSym}$:

[TABLE] 2. 2.

If $X$ and $Y$ are totally ordered alphabets, then $X\sqcup Y$ is too, the elements of $X$ being smaller than the elements of $Y$. Identifying $\mathbb{K}[[X\sqcup Y]]$ with a subalgebra of $\mathbb{K}[[X]]\otimes\mathbb{K}[[Y]]$, we define a coproduct on $\mathbf{QSym}$ by:

[TABLE]

For example:

[TABLE]

and, in $\mathbf{QSym}$:

[TABLE]

The coassociativity of $\Delta$ is easily obtained from the equality $(X\sqcup Y)\sqcup Z=X\sqcup(Y\sqcup Z)$. 3. 3.

If $X$ and $Y$ are totally ordered alphabets, then $XY=X\times Y$ is too, with the lexicographic order. We consider $\mathbb{K}[[XY]]$ as a subalgebra of $\mathbb{K}[[X]]\otimes\mathbb{K}[[Y]]$, identifying $(x,y)$ with $x\otimes y$. We can define a second coproduct on $\mathbf{QSym}$ by:

[TABLE]

For example:

[TABLE]

and, in $\mathbf{QSym}$:

[TABLE]

The coassociativity of $\delta$ is easily obtained from the equality $(XY)Z=X(YZ)$. Moreover:

[TABLE]

This implies that in $\mathbf{QSym}$:

[TABLE]

where $m_{1,3,24}:\mathbf{QSym}^{\otimes 4}\longrightarrow\mathbf{QSym}^{\otimes 3}$ sends $x\otimes y\otimes z\otimes t$ to $x\otimes z\otimes yt$. This means that the Hopf algebra $(\mathbf{QSym},m,\Delta)$ is a Hopf algebra in the category of right comodules over the bialgebra $(\mathbf{QSym},m,\delta)$, the coaction being $\delta$ itself: we call this a pair of bialgebras in cointeraction.

For other examples of such objects and applications, see [16, 18, 10, 9, 11].

We here give other examples of cointeracting bialgebras coming from the manipulation of alphabets and polynomial realizations. We use here totally quasi-ordered alphabets, that is to say sets with a total transitive reflexive (but not necessarily antisymmetric) relation. The associated algebras $A_{q}(X)$ are slightly more complicated, see Definition 6. Their different sets of generators allows to polynomially realize Feynman graphs (one set for vertices, one set for internal edges, one set for incoming half-edges and a last one for outgoing half-edges); this gives a family of products $\cdot_{q}$ on the space $H_{\mathcal{FG}}$ generated by isoclasses of Feynman graphs, indexed by a scalar $q$ (Theorem 17). If $F$ and $G$ are two Feynman graphs, $F\cdot_{q}G$ is a sum of graphs obtained by gluing together vertices of $F$ and $G$; in particular, if $q=0$, this is reduced to the disjoint union of $F$ and $G$. The trick of doubling the alphabet gives $\mathcal{H}_{\mathcal{FG}}$ a coproduct $\Delta$, given by ideals (Theorem 19), and the trick of squaring the alphabet gives it a second coproduct $\delta$; we obtain in this way a pair of cointeracting bialgebras (Corollary 20). For example, for the following graph:

[TABLE]

we obtain:

[TABLE]

The coproduct $\delta$ is similar to the Connes-Kreimer’s one $\Delta_{CK}$ [2, 3, 4, 6, 5, 1, 16], but slightly different. For example:

[TABLE]

We shall then consider several quotients of $A_{q}(X)$, leading to quotient bialgebras of $\mathcal{H}_{\mathcal{FG}}$. We obtain in this way a polynomial realization of a Hopf algebra of simple oriented graphs $\mathcal{H}_{\mathcal{SG}}$, and then a polynomial realization of the Hopf algebra on quasi-posets $\mathcal{H}_{\mathcal{QP}}$ of [12, 13, 14, 10]. Restricting to ordered alphabets, instead of quasi-ordered alphabets, we obtain quotients bialgebras, namely $H_{\mathcal{NCFG}}$ based on Feynman graphs with no cycle in Theorem 25 , $H_{\mathcal{NCSG}}$ on simple oriented graphs with no cycle in Theorem 33, and $H_{\mathcal{P}}$ on posets in Theorem 43, obtaining diagrams of Hopf algebras:

[TABLE]

We also show that these Hopf algebra admit noncommutative versions, replacing the algebras $A_{q}(X)$ by a noncommutative analogue. The last paragraph is devoted to the description of the dual Hopf algebra of posets, using the notion of system of edge between two posets.

1 Operations on alphabets

All the proofs of this section are elementary and left to the reader.

1.1 Quasi-ordered alphabets

Definition 1**.**

A quasi-ordered alphabet is a pair $(X,\leq_{X})$, where $X$ is an alphabet and $\leq_{X}$ is a total quasi-order on $X$, that is to say a relation on $X$ such that:

[TABLE]

If $\leq_{X}$ is an order, we shall say that $(X,\leq_{X})$ is an ordered alphabet.

Notations 1*.*

Let $(X,\leq_{X})$ be a quasi-ordered alphabet. We define an equivalence $\sim_{X}$ on X by:

[TABLE]

For all $i,j\in X$, we shall denote $i<_{X}j$ if $i\leq_{X}j$ and not $j\leq_{X}i$.

1.2 Disjoint union

Proposition 2**.**

Let $(X,\leq_{X})$ and $(Y,\leq_{Y})$ be two quasi-ordered alphabets. The set $X\sqcup Y$ is given a relation $\leq_{X\sqcup Y}$:

[TABLE]

Then $(X\sqcup Y,\leq_{X\sqcup Y})$ is a quasi-ordered alphabet.

Remark 1*.*

If $X$ and $Y$ are ordered alphabets, then $X\sqcup Y$ is also ordered.

Lemma 3**.**

Let $X,Y$ be quasi-ordered alphabets.

[TABLE] 2. 2.

Let $X$, $Y$ and $Z$ be quasi-ordered alphabets. Then:

[TABLE]

1.3 Product

Proposition 4**.**

Let $(X,\leq_{X})$ and $(Y,\leq_{Y})$ be two quasi-ordered alphabets. The set $XY=X\times Y$ is given a relation $\leq_{XY}$ in the following way:

[TABLE]

Then $(XY,\leq_{XY})$ is a quasi-ordered alphabet, which we denote by $XY$.

Remark 2*.*

If $X$ and $Y$ are ordered alphabets, then $XY$ is also ordered, and $\leq_{XY}$ is the lexicographic order.

Lemma 5**.**

Let $X$ and $Y$ be quasi-ordered alphabets.

[TABLE] 2. 2.

Let $X$, $Y$ and $Z$ be quasi-ordered alphabets. Then:

[TABLE]

2 Algebras attached to alphabets

2.1 Definition

Definition 6**.**

Let $X$ be a quasi-ordered alphabet and let $q\in\mathbb{K}$. We put:

[TABLE]

Both of them are given their usual topology of rings of formal series.

Elements of $A_{q}(X)$ are formal infinite spans of monomials

[TABLE]

where $\epsilon_{i}\in\{0,1\}$, $\alpha_{i,j},\beta_{j},\gamma_{i}\in\mathbb{N}$, with only a finite number of them non-zero. Elements of $\mathbf{A}_{q}(X)$ are formal infinite spans of monomials

[TABLE]

where $i_{1},\ldots,i_{k}$ are elements of $X$, all distinct, $\alpha_{i,j},\beta_{j},\gamma_{i}\in\mathbb{N}$, with only a finite number of them non-zero.

2.2 Doubling the alphabets

Proposition 7**.**

Let $X,Y$ be two quasi-ordered alphabets. We define a continuous algebra morphism $\Delta_{X,Y}$ from $A_{q}(X\sqcup Y)$ to $A_{q}(X)\otimes A_{q}(Y)$ or from $\mathbf{A}_{q}(X\sqcup Y)$ to $\mathbf{A}_{q}(X)\otimes\mathbf{A}_{q}(Y)$ by:

[TABLE]

Proof.

In the commutative case, we have to check that for any $i\in X\sqcup Y$, $\Delta_{X,Y}(x_{i}^{2}-qx_{i})=0$. Indeed:

[TABLE]

So $\Delta_{X,Y}$ is well-defined. The proof in the noncommutative case is similar. ∎

Proposition 8**.**

Let $X$, $Y$ and $Z$ be quasi-ordered alphabets. Then:

[TABLE]

seen as morphisms from $A_{q}(X\sqcup Y\sqcup Z)$ to $A_{q}(X)\otimes A_{q}(Y)\otimes A_{q}(Z)$, or from $\mathbf{A}_{q}(X\sqcup Y\sqcup Z)$ to $\mathbf{A}_{q}(X)\otimes\mathbf{A}_{q}(Y)\otimes\mathbf{A}_{q}(Z)$.

Proof.

It is enough to apply these two algebra morphisms on generators. We find:

[TABLE]

When applied to $x_{i,j}$, we find for both of them:

[TABLE]

So these morphisms are equal. ∎

2.3 Squaring the alphabets

Proposition 9**.**

Let $X,Y$ be two nonempty quasi-ordered alphabets. There exists a unique continuous algebra morphism $\delta_{X,Y}$ from $A_{q}(XY)$ to $A_{q_{1}}(X)\otimes A_{q_{2}}(Y)$, or from $\mathbf{A}_{q}(XY)$ to $\mathbf{A}_{q_{1}}(X)\otimes\mathbf{A}_{q_{2}}(Y)$ such that:

[TABLE]

if, and only if, $q=q_{1}q_{2}$.

Proof.

In the commutative case, we have to check that for any $(i,i^{\prime})\in XY$, $\delta_{X,Y}(x_{(i,i^{\prime})}^{2}-qx_{(i,i^{\prime})})=0$. We compute:

[TABLE]

So this holds if, and only if, $q_{1}q_{2}=q$. The noncommutative case is proved in the same way. ∎

Remark 3*.*

In particular, if $q=q_{1}=q_{2}$, $\delta_{X,Y}$ exists if, and only if, $q=1$ or $q=0$.

Lemma 10**.**

Let $q_{1},q_{2},q_{3}\in\mathbb{K}$ and let $X$, $Y$ and $Z$ be quasi-ordered alphabets. The following diagrams commute:

[TABLE]

Proof.

As these two maps are algebra morphisms, it is enough to prove that they coincide on generators of $A_{q_{1}q_{2}q_{3}}(XYZ)$ or $\mathbf{A}_{q_{1}q_{2}q_{3}}(XYZ)$. Let $i,i^{\prime}\in X$, $j,j^{\prime}\in Y$, $k,k^{\prime}\in Z$.

• •

Both send $x_{(i,j,k)}$ to $x_{i}\otimes x_{j}\otimes x_{k}$.

• •

Both send $x_{(i,j,k),\infty}$ to $x_{i,\infty}\otimes x_{j,\infty}\otimes x_{k,\infty}$.

• •

Both send $x_{-\infty,(i^{\prime},j^{\prime},k^{\prime})}$ to $x_{-\infty,i^{\prime}}\otimes x_{-\infty,j^{\prime}}\otimes x_{-\infty,k^{\prime}}$.

• •

Both send $x_{(i,j,k),(i^{\prime},j^{\prime}k^{\prime})}$ to $\begin{cases}x_{i,i^{\prime}}\otimes x_{j,\infty}x_{-\infty,j^{\prime}}\otimes x_{k,\infty}x_{-\infty,k^{\prime}}\mbox{ if }i<_{X}i^{\prime},\\ 1\otimes x_{j,j^{\prime}}\otimes x_{k,\infty}x_{-\infty,k^{\prime}}\mbox{ if }i\sim_{X}i^{\prime}\mbox{ and }j<_{Y}j^{\prime},\\ 1\otimes 1\otimes x_{k,k^{\prime}}\mbox{ if }i\sim_{X}i^{\prime}\mbox{ and }j\sim_{Y}j^{\prime}.\end{cases}$

So $(\delta_{X,Y}\otimes Id)\circ\delta_{XY,Z}=(Id\otimes\delta_{Y,Z})\circ\delta_{X,YZ}$. ∎

Lemma 11**.**

Let $q_{1},q_{2}\in\mathbb{K}$ and let $X$, $Y$, $Z$ be quasi-ordered alphabets. The following diagram commutes:

[TABLE]

where:

[TABLE]

Proof.

First step. Let us prove that $m_{1,3,24}$ is an algebra morphism. Let $X=x\otimes z_{1}\otimes y\otimes z_{2}$ and $X^{\prime}=x^{\prime}\otimes z_{1}^{\prime}\otimes y^{\prime}\otimes z_{2}^{\prime}$ in $A_{q_{1}}(X)\otimes A_{q_{2}}(Z)\otimes A_{q_{1}}(Y)\otimes A_{q_{2}}(Z)$.

[TABLE]

As $A_{q_{1}q_{2}}(Z)$ is commutative, $m_{1,3,24}(XX^{\prime})=m_{1,3,24}(X)m_{1,3,24}(X^{\prime})$.

Second step. By composition, both $(\delta_{X,Y}\otimes Id)\circ\delta_{X\sqcup Y,Z}$ and $(Id\otimes\delta_{Y,Z})\circ\delta_{X,Y\sqcup Z}$ are algebra morphisms: it is enough to prove that they coincide on the generators of $A_{q_{1}q_{2}}((X\sqcup Y)Z)$ or $\mathbf{A}_{q_{1}q_{2}}((X\sqcup Y)Z)$. Let $i,i^{\prime}\in X$, $j,j^{\prime}\in Y$ and $k,k^{\prime}\in Z$.

• •

Both send $x_{(i,k)}$ to $x_{i}\otimes 1\otimes x_{k}$ and $x_{j,k}$ to $1\otimes x_{j}\otimes x_{k}$.

• •

Both send $x_{(i,k),\infty}$ to $x_{i,\infty}\otimes 1\otimes x_{k,\infty}$ and $x_{(j,k),\infty}$ to $1\otimes x_{j,\infty}\otimes x_{k,\infty}$.

• •

Both send $x_{-\infty,(i^{\prime},k^{\prime})}$ to $x_{-\infty,i^{\prime}}\otimes 1\otimes x_{-\infty,k^{\prime}}$ and $x_{-\infty,(j^{\prime},k^{\prime})}$ to $1\otimes x_{-\infty,j^{\prime}}\otimes x_{-\infty,k^{\prime}}$.

• •

Both send $x_{(i,k),(i^{\prime},k^{\prime})}$ to $\begin{cases}x_{i,i^{\prime}}\otimes 1\otimes x_{k,\infty}x_{-\infty,k^{\prime}}\mbox{ if }i<_{X}i^{\prime},\\ 1\otimes 1\otimes x_{k,k^{\prime}}\mbox{ if }i\sim_{X}i^{\prime}.\end{cases}$

• •

Both send $x_{(j,k),(j^{\prime},k^{\prime})}$ to $\begin{cases}1\otimes x_{j,j^{\prime}}\otimes x_{k,\infty}x_{-\infty,k^{\prime}}\mbox{ if }j<_{Y}j^{\prime},\\ 1\otimes 1\otimes x_{k,k^{\prime}}\mbox{ if }j\sim_{Y}j^{\prime}.\end{cases}$

• •

Both send $x_{(i,k),(j^{\prime},k^{\prime})}$ to $x_{i,\infty}\otimes x_{-\infty,j}\otimes x_{k,\infty}x_{-\infty,k^{\prime}}$.

Therefore, they are equal. ∎

Remark 4*.*

This does not work for morphisms

[TABLE]

But, if we put $\rho_{X,Y}=(Id\otimes p_{Y})\circ\delta_{X,Y}:\mathbf{A}_{q_{1}q_{2}}(X\sqcup Y)\longrightarrow\mathbf{A}_{q_{1}}(X)\otimes A_{q_{2}}(Y)$, where $p_{Y}$ is the canonical surjection from $\mathbf{A}_{q_{2}}(Y)$ to $A_{q_{2}}(Y)$, then

[TABLE]

seen as morphisms

[TABLE]

The proof is identical to the one of Lemma 11.

3 Feynman graphs

3.1 Definition

Definition 12**.**

A Feynman graph $G$ is given by:

• •

A non-empty, finite set $HE(G)$ of half-edges, with a map $type_{G}:HE(G)\longrightarrow\{out,in\}$.

• •

A non-empty, finite set $V(G)$ of vertices.

• •

An incidence map for half-edges, that is to say an involution $i_{G}:HE(G)\longrightarrow HE(G)$.

• •

A source map for half-edges, that is to say a map $s_{G}:HE(G)\longrightarrow V(G)$.

The incidence rule must be respected:

[TABLE]

The set of external half-edges of $G$ is:

[TABLE]

The set of Feynman graphs is denoted by $\mathcal{FG}$.

We shall use graphical representations of Feynman graphs: vertices are represented by , half-edges of type $out$ are represented by , half-edges of type $in$ by ; the incidence map glues two such half-edges to obtain an oriented internal edge of the Feynman graph . In the sequel, we shall write Feynman graph instead of isoclass of Feynman graphs. We shall also consider ordered Feynman graphs, that is to say Feynman graphs such that the set of vertices is given a total order.

Remark 5*.*

We restraint ourselves to Feynman graphs with a unique type of edges. It is possible to do the same for several type of edges, adding generators to the algebras associated to totally quasi-ordered alphabets.

Example 1*.*

Here are examples of Feynman graphs:

[TABLE]

Definition 13**.**

Let $G_{1},\ldots,G_{k}$ be Feynman graphs and let $\sigma:V(G_{1})\sqcup\ldots\sqcup V(G_{k})\twoheadrightarrow C$ be a surjective map. We define a Feynman graph $G=\sigma(G_{1},\ldots,G_{k})$ in the following way:

• •

$V(\sigma(G_{1},\ldots,G_{k}))=C$.

• •

The set of half-edges of $\sigma(G_{1},\ldots,G_{k})$ is:

[TABLE]

• •

If $e\in HE(G_{i})\cap HE(\sigma(G_{1},\ldots,G_{k}))$, then:

[TABLE]

Roughly speaking, $G\sqcup_{\sigma}H$ is obtained by identifying the vertices of the disjoint union of Feynman graphs $G_{1}\ldots G_{k}$ with the same image by $\sigma$, and deleting the loops created in this process. In particular, if $\sigma=Id_{V(G_{1})\sqcup\ldots\sqcup V(G_{k})}$, or more generally if $\sigma$ is bijective, then $\sigma(G_{1},\ldots,G_{k})$ is (isomorphic to) the disjoint union $G_{1}\ldots G_{k}$.

Remark 6*.*

If $G_{1},\ldots,G_{k}$ are ordered Feynman graphs, then $G_{1}\ldots G_{k}$ is also ordered: if $x,y\in V(G_{1})\sqcup\ldots\sqcup V(G_{k})$,

[TABLE]

We deduce a total order on $\sigma(G_{1},\ldots,G_{k})$ in this way: for any $x,y\in C$,

[TABLE]

Example 2*.*

If $G$ and $H$ are the Feynman graphs of Example 1 and:

[TABLE]

then:

[TABLE]

We shall use the following particular case:

Definition 14**.**

Let $G$ and $H$ be Feynman graphs, $A\subseteq V(G)$ and $\sigma:A\longrightarrow V(H)$ an injection. We define an equivalence $\sim$ on $V(G)\sqcup V(H)$ by:

[TABLE]

Let $\pi$ be the canonical surjection from $V(G)\sqcup V(H)$ to $(V(G)\sqcup V(H))/\sim$. The Feynman graph $\pi(G,H)$ is denoted by $G\sqcup_{\sigma}H$.

In particular, if $A=\emptyset$, $G\sqcup_{\sigma}H$ is the disjoint union $GH$.

3.2 Monomials and Feynman graphs

Definition 15**.**

Let $M$ be a monomial of $A_{q}(X)$:

[TABLE]

1. (a)

We shall say that $M$ is admissible if:

[TABLE] 2. (b)

If $M$ is admissible, we attach to $M$ a Feynman graph $fg(M)$, defined in this way:

• •

The set of vertices of $fg(M)$ is the set of elements $i\in X$, such that $\epsilon_{i}=1$.

• •

*The number of internal edges * \textstyle{i}$$\textstyle{j}

• in $fg(M)$ is $\alpha_{i,j}$.*

• •

*The number of external edges * $\textstyle{j}$

• in $fg(M)$ is $\beta_{j}$.*

• •

*The number of external edges * $\textstyle{i}$

• in $fg(M)$ is $\gamma_{i}$.*

Let $M$ be a monomial of $\mathbf{A}_{q}(X)$:

[TABLE]

We shall say that $M$ is admissible if its image $\overline{M}$ in $A_{q}(X)$ (which is a monomial) is admissible. The Feynman graph $fg(\overline{M})$ attached to the image of $M$ in the quotient $A_{q}(X)$ of $\mathbf{A}_{q}(M)$ is ordered: the set of its vertices is $\{i_{1},\ldots,i_{k}\}$, totally ordered by $i_{1}<\ldots<i_{k}$. This ordered Feynman graph is denoted by $fg(M)$.

Example 3*.*

Let $a,b\in X$, $a\leq_{X}b$, $a\neq b$. In $\mathbf{A}_{q}(X)$:

[TABLE]

In $A_{q}(X)$:

[TABLE]

Lemma 16**.**

Let $M$ and $N$ be two admissible monomials in $A_{q}(X)$ or in $\mathbf{A}_{q}(X)$. We put $A=V(fg(M))\cap V(fg(N))$ and $\sigma:A\longrightarrow V(fg(N))$ be the canonical injection. There exists an admissible monomial $P$, such that in $A_{q}(X)$:

[TABLE]

Moreover:

[TABLE]

Proof.

We work in $A_{q}(X)$; the proof for $\mathbf{A}_{q}(X)$ is similar. We put:

[TABLE]

then, as $x_{i}^{2}=qx_{i}$ in $A_{q}(X)$:

[TABLE]

and $MN=q^{|A|}P$. ∎

Theorem 17**.**

Let $G$ be a Feynman graph and let $X$ a quasi-ordered alphabet. We put:

[TABLE]

Then $H_{\mathcal{FG}}(X)$ is a subalgebra of $A_{q}(X)$. For any Feynman graphs $G$ and $H$:

[TABLE]

Moreover, there exists a quasi-ordered alphabet $X$, such that the elements $M_{G}(X)$ are linearly independent in $A_{q}(X)$. 2. 2.

1. (a)

Let $G$ be an ordered Feynman graph and let $X$ a quasi-ordered alphabet. We put:

[TABLE]

Then $\mathbf{H}_{\mathcal{FG}}(X)$ is a subalgebra of $\mathbf{A}_{q}(X)$. For any ordered Feynman graphs $G$ and $H$:

[TABLE]

There exists a quasi-ordered alphabet $X$, such that the elements $\mathbf{M}_{G}(X)$ are linearly independent in $\mathbf{A}_{q}(X)$.

Proof.

1. By Lemma 16, $M_{G}(X)M_{H}(X)$ is a linear sums of terms $q^{|A|}P$, with $fg(P)=\pi_{\sigma}(F,G)$. It remains to show that all such terms are obtained, which is immediate. So $H_{\mathcal{FG}}(X)$ is a subalgebra of $A_{q}(X)$.

Let $X$ be an infinite set. We give it a relation $\leq_{X}$ by:

[TABLE]

making it a quasi-ordered alphabet. Let $G$ be a Feynman graph. For any $i,j\in V(G)$, we define:

• •

$\alpha_{i,j}$ is the number of internal edges \textstyle{i}$$\textstyle{j} in $G$.

• •

$\beta_{j}$ is the number of external edges $\textstyle{j}$ in $G$.

• •

$\gamma_{i}$ is the number of external edges $\textstyle{i}$ in $G$.

Let $\tau:V(G)\longrightarrow X$ be an injection: this exists, as $V(G)$ is finite and $X$ is infinite. We consider:

[TABLE]

Obviously, $M$ is admissible and $fg(M)\approx G$, so $M_{G}(X)$ is non-zero. As the elements of $M_{G}(X)$ are all non-zero and their support are disjoint, they are linearly independent.

2. Similar proof. ∎

Remark 7*.*

In particular, if $q=0$:

[TABLE]

3.3 The first coproduct

Definition 18**.**

Let $G$ be a Feynman graph, and let $A\subseteq V(G)$.

We define a Feynman graph $G_{\mid A}$ by the following:

• •

The set of vertices of $G_{\mid A}$ is $A$.

• •

The set of half-edges of $G_{\mid A}$ is the set of half-edges $e\in HE(G)$ such that $s_{G}(e)\in A$.

• •

For all half-edge $e$ of $G_{\mid A}$:

[TABLE] 2. 2.

We shall say that $A$ is an ideal of $G$ if for all $i,j\in V(G)$, if $i\in A$ and if there is an edge from $i$ to $j$ in $G$, then $j\in A$. The set of ideals of $G$ is denoted by $I(G)$.

Note that if $G$ is an ordered Feynman graph, then for any $A$, $G_{\mid A}$ is also ordered.

Theorem 19**.**

Let $G$ be a Feynman graph. For any quasi-ordered alphabets $X$ and $Y$:

[TABLE]

Proof.

We consider the two following sets:

• •

$\mathcal{A}$ is the set of triples $(M,M_{1},M_{2})$, where $M$, $M_{1}$ and $M_{2}$ are monomials of respectively $A_{q}(X\sqcup Y)$, $A_{q}(X)$ and $A_{q}(Y)$, such that $\Delta_{X,Y}(M)=M_{1}\otimes M_{2}$ and $fg(M)=G$.

• •

$\mathcal{B}$ is the set of triples $(A,M_{1},M_{2})$, where $A$ is an ideal of $G$, $M_{1}$ is an admissible monomial of $A_{q}(X)$ such that $fg(M_{1})=G_{\mid V(G)\setminus A}$, $M_{2}$ is an admissible monomial of $A_{q}(Y)$ such that $fg(M_{2})=G_{\mid A}$.

Let $(M,M_{1},M_{2})\in\mathcal{A}$. Let $A$ be the set of vertices of $fg(M)$ belonging to $Y$. By Definition of $\Delta_{X,Y}$, as $\Delta_{X,Y}(M)=M_{1}\otimes M_{2}$, $fg(M_{1})=G_{\mid V(G)\setminus A}$ and $fg(M_{2})=G_{\mid A}$. Moreover, if $i\in A$ and if there is an edge from $i$ to $j$ in $G$, this means that $x_{i,j}$ appears in $M$, so $i\leq_{X\sqcup Y}j$. As $i\in Y$, $j\in Y$, so $j\in A$: $A\in I(G)$. We define in this way a map:

[TABLE]

It is not difficult to prove that it is a bijection. Hence:

[TABLE]

The proof in $\mathbf{A}_{q}(X\sqcup Y)$ is similar. ∎

Corollary 20**.**

Let $H_{\mathcal{FG}}$ be the vector space generated by the set of Feynman graphs and let $q\in\mathbb{K}$. It is given a Hopf algebra structure: for any Feynman graphs $G$ and $H$,

[TABLE] 2. 2.

Let $\mathbf{H}_{\mathcal{FG}}$ be the vector space generated by the set of ordered Feynman graphs and let $q\in\mathbb{K}$. It is given a Hopf algebra structure by (3)-(4).

Proof.

1. For any quasi-ordered alphabet $X$, there is linear map:

[TABLE]

By Theorems 17 and 19, for any quasi-ordered alphabets $X$ and $Y$, for any $a,b\in H_{\mathcal{FG}}$:

[TABLE]

Choosing quasi-ordered alphabets $X$, $Y$ and $Z$, such that $\Theta_{X}$, $\Theta_{Y}$ and $\Theta_{Z}$ (Proposition 17), we obtain that $H_{\mathcal{FG}}$ is an algebra. For all $a,b\in H_{\mathcal{FG}}$:

[TABLE]

As $\Theta_{X}$ and $\Theta_{Y}$ are injective, $\Theta_{X}\otimes\Theta_{Y}$ is too, so $\Delta(ab)=\Delta(a)\Delta(b)$. Moreover:

[TABLE]

By the injectivity of $\Theta_{X}\otimes\Theta_{Y}\otimes\Theta_{Z}$, $\Delta$ is coassociative, so $H_{\mathcal{FG}}$ is a bialgebra.

Observe that $(H_{\mathcal{FG}},m,\Delta)$ is filtered by the cardinality of the set of vertices of Feynman graphs, even graded if $q=0$; as its components of degree [math] is reduced to $\mathbb{K}$, it is connected. Hence, it is a Hopf algebra.

2. Similar proof. ∎

3.4 The second coproduct

Definition 21**.**

Let $G$ be a Feynman graph, and let $\sim$ be an equivalence on $V(G)$.

We shall say that $\sim$ is $G$-compatible if the following assertions hold:

• •

If $A\subseteq V(G)$ is an equivalence class of $\sim$, then $G_{\mid A}$ is a connected Feynman graph.

• •

If there is a path $x_{1}\rightarrow x_{2}\rightarrow\ldots\rightarrow x_{k}$ in $G$, with $x_{1}\sim x_{k}$, then $x_{1}\sim x_{2}\sim\ldots\sim x_{k}$.

The set of $G$-compatible equivalences will be denoted by $CE(G)$. 2. 2.

If $A_{1},\ldots,A_{k}$ are the equivalence classes of $\sim$, we denote by $G_{\mid\sim}$ the disjoint union of the Feynman graphs $G_{\mid A_{i}}$, $1\leq i\leq k$. 3. 3.

We denote by $G/\sim$ the following Feynman graph:

• •

The set of vertices of $G/\sim$ is $V(G)$.

• •

The half-edges of $G/\sim$ are the half-edges of $G$ such that $i(e)=e$ or not $s(e)\sim s(i(e))$.

• •

For any half-edge $e$ of $G/\sim$:

[TABLE]

Roughly speaking, $G_{\mid\sim}$ is obtained by the deletion in $G$ of all the internal edges whose two extremities are not $\sim$-equivalent, whereas $G/\sim$ is obtained the deletion in $G$ of all the internal edges whose two extremities are $\sim$-equivalent.

Remark 8*.*

If $G$ is ordered, then, as $V(G_{\mid\sim})=V(G/\sim)=V(G)$, the Feynman graphs $G_{\mid\sim}$ and $G/\sim$ are also ordered. 2. 2.

The Feynman graph $G/\sim$ is not the usual contraction of $G$ according to a subgraph used by Connes and Kreimer to define a coproduct on Feynman graphs [2, 3, 4, 6, 5, 1, 16], as here all the vertices are conserved.

Notations 2*.*

Let $q\in\mathbb{K}$. For any Feynman graph $G$, if $G_{1},\ldots,G_{k}$ are its connected components, we put:

[TABLE]

In particular, $\psi_{0}(G)=G$.

Proposition 22**.**

Let $X$, $Y$ be two quasi-ordered alphabets.

We assume that $q=q_{1}q_{2}$. We consider $\delta_{X,Y}:A_{q}(XY)\longrightarrow A_{q_{1}}(X)\otimes A_{q_{2}}(Y)$. For any Feynman graph $G$:

[TABLE] 2. 2.

We consider $\delta_{X,Y}:\mathbf{A}_{0}(XY)\longrightarrow\mathbf{A}_{0}(X)\otimes\mathbf{A}_{0}(Y)$. For any ordered Feynman graph $G$, in $\mathbf{A}_{0}(X)$:

[TABLE]

Proof.

1. Let $M$ be a monomial of $A_{q}(XY)$, such that $fg(M)=G$. The set of vertices of $fg(M)$ is denoted by $V(G)=\{(i_{1},j_{1}),\ldots,(i_{k},j_{k})\}$. We define an equivalence on $V(G)$ by:

[TABLE]

Let us assume that there is a path $(i_{p_{1}},j_{p_{1}})\rightarrow\ldots\rightarrow(i_{p_{k}},j_{p_{k}})$ in $G$, with $i_{p_{1}}\sim_{X}j_{p_{1}}$. Then, as $G=fg(M)$, in $XY$:

[TABLE]

so, in $X$:

[TABLE]

As $i_{p_{1}}\sim_{X}i_{p_{k}}$, we obtain that $i_{p_{1}}\sim_{X}\ldots\sim_{X}i_{p_{k}}$, and finally:

[TABLE]

Moreover, $\delta_{X,Y}(M)=M_{1}\otimes M_{2}$, with, by Definition of $\delta_{X,Y}$, $fg(M_{1})=G/\sim$ and $fg(M_{2})=G_{\mid\sim}$. Let $\sim^{\prime}$ be the equivalence which equivalent classes are the connected components of $G_{\mid\sim}$. Then $\sim^{\prime}\in CE(G)$ and:

[TABLE]

Moreover, $\sim^{\prime}$ is the unique element of $CE(G)$ such that (5) holds. Finally, we proved that $\delta_{X,Y}$ is a sum of terms $M_{1}\otimes M_{2}$, such that there exists $\sim^{\prime}\in CE(G)$, such that $fg(M_{1})$ is a monomial of $\psi_{q_{1}}(G/\sim^{\prime})$ and $fg(M_{2})$ is a monomial of $\psi_{q_{2}}(G_{\mid\sim^{\prime}})$. It is not difficult to see that all these terms are obtained.

2. Similar proof. ∎

Corollary 23**.**

We define two coproducts on $H_{\mathcal{FG}}$ for any Feynman graph $G$ by:

[TABLE]

Then $(H_{\mathcal{FG}},._{0},\delta)$ and $(H_{\mathcal{FG}},._{1},\delta_{1})$ are bialgebras. For any $q_{1},q_{2}\in\mathbb{K}$, we define a coaction by:

[TABLE]

If $(q_{1},q_{2})=(q,1)$ with $q\in\mathbb{K}$, or $(0,0)$, then $(H_{\mathcal{FG}},._{q_{1}},\Delta)$ is a bialgebra in the category of right $(H_{\mathcal{FG}},._{q_{2}},\delta_{q_{2}})$-comodules by the coaction $\rho_{q_{1},q_{2}}$, that is to say:

• •

For all $x,y\in H_{\mathcal{FG}}$, $\rho_{q_{1},q_{2}}(x._{q_{1}}y)=\rho_{q_{1},q_{2}}(x)._{q_{1},q_{2}}\rho_{q_{1},q_{2}}(y)$, where:

[TABLE]

• •

$\rho_{q_{1},q_{2}}(1)=1\otimes 1$.

• •

$(\Delta\otimes Id)\circ\rho_{q_{1},q_{2}}=m_{1,3,24}\circ(\rho_{q_{1},q_{2}}\otimes\rho_{q_{1},q_{2}})\circ\Delta$, where:

[TABLE]

• •

For all $x\in H_{\mathcal{FG}}$, $(\varepsilon_{\Delta}\otimes Id)\circ\rho_{q_{1},q_{2}}(x)=\varepsilon_{\Delta}(x)1$.

Note that $\rho_{0,0}=\delta$. 2. 2.

Similarly, (6) defines a coproduct $\delta$ on $\mathbf{A}_{q}(0)$, making it a bialgebra on the category of right $(H_{\mathcal{FG}},._{0},\delta)$-comodules.

makes of $(H_{\mathcal{FG}},._{0},\delta)$-comodules,

Proof.

We take $(q_{1},q_{2})=(q,1)$ or $(0,0)$. Hence, $q_{1}q_{2}=q_{1}$. For any quasi-ordered alphabets $X$, $Y$, $(\Theta_{X}\otimes\Theta_{Y})\circ\delta=\delta_{X,Y}\circ\Theta_{XY}$. The proof that $(H_{\mathcal{FG}},._{0},\delta)$, $(\mathbf{H}_{\mathcal{FG}},._{0},\delta)$, $(H_{\mathcal{FG}},._{1},\delta_{1})$ and $(\mathbf{H}_{\mathcal{FG}},._{1},\delta_{1})$ are bialgebras is similar to the proof of corollary 20. The only non-trivial remaining assertion to prove is point 3. Let us take $X,Y,Z$ alphabets such that $\Theta_{X}$, $\Theta_{Y}$ and $\Theta_{Z}$ are injective.

[TABLE]

We conclude by the injectivity of $\Theta_{X}\otimes\Theta_{Y}\otimes\Theta_{Z}$. ∎

3.5 Restriction to ordered alphabets

Let $G$ be a Feynman graph. A cycle in $G$ is a sequence of vertices $(i_{1},\ldots,i_{k})$, with $k\geq 2$, all distinct, such that there exist internal edges $e_{1},\ldots,e_{k}$, with:

[TABLE]

Lemma 24**.**

Let $G$ be a Feynman graph. The following conditions are equivalent:

For any ordered alphabet, $M_{G}(X)=0$. 2. 2.

$G$* has a cycle.*

Moreover, there exists an ordered alphabet $X$, such that for any Feynman graph with no cycle, $M_{G}(X)\neq 0$.

Proof.

$2.\Longrightarrow 1$. Let us assume that $G$ has a cycle $(i_{1},\ldots,i_{k})$, with $M_{G}(X)\neq 0$. There exists $j_{1},\ldots,k_{k}\in X$, all distinct, such that $x_{j_{1},j_{2}}\ldots x_{j_{k-1},j_{k}}x_{j_{k},j_{1}}$ appears in $M_{G}(X)$, so $j_{1}\leq_{X}\ldots\leq_{X}j_{k-1}\leq_{X}j_{k}\leq_{X}j_{1}$. As $\leq_{X}$ is an order, $j_{1}=\ldots=j_{k}$, which is a contradiction. So $M_{G}(X)=0$.

$1.\Longrightarrow 2$. Let us prove that there exists a total order $\leq$ on $V(G)$, such that if there is an edge from $i$ to $j$ in $G$, then $i\leq j$. We proceed by induction on $|V(G)|$. If $|V(G)|=1$, this is obvious. If $|V(G)|\geq 2$, as $G$ has no cycle, it has a source $v_{1}$, that is to say a vertex with no internal incoming edge. The Feynman graph $G^{\prime}$ obtained from $G$ by deleting $v_{0}$ and all the attached half-edges has no cycle, so, by the induction hypothesis, the set of its vertices inherits a total order $v_{2}\leq\ldots\leq v_{k}$, compatible with the internal edges of $G^{\prime}$. We give $V(G)$ a total order by $v_{1}\leq v_{2}\leq\ldots\leq v_{k}$; as $v_{1}$ is a source, this order is compatible with the internal edges of $G$.

We take $X=\mathbb{N}$, with its usual order. There exists a monomial of $A_{q}(X)$ of the form:

[TABLE]

such that $fg(M)=G$: $\alpha_{i,j}$ is the number of internal edges between between $v_{i}$ and $v_{j}$, $\beta_{i}$ is the number of incoming external edges in $v_{i}$, and $\gamma_{j}$ is the number of external half-edges outgoing from $v_{j}$, if $v_{1}\leq\ldots\leq v_{k}$ is the previously defined order on the set of vertices of $G$. So $M_{G}(X)\neq 0$. ∎

Observe that if $G$ has no cycle and $\sim\in CE(G)$, then both $G_{\mid\sim}$ and $G/\sim$ has no cycle. Hence, considering only ordered alphabets, we obtain a Hopf algebra and a bialgebra of Feynman graphs with no cycle:

Theorem 25**.**

Let $H_{\mathcal{NCFG}}$ be the vector space generated by the set of Feynman graphs with no cycle and let $q\in\mathbb{K}$.

It is given a Hopf algebra structure: for any Feynman graphs $G$ and $H$ with no cycle,

[TABLE] 2. 2.

We define a second coproduct by:

[TABLE]

Then $(H_{\mathcal{NCFG}},._{0},\delta)$ is a bialgebra. Moreover, $(H_{\mathcal{NCFG}},._{0},\Delta)$ is a bialgebra in the category of $(H_{\mathcal{NCFG}},._{0},\delta)$-comodules. 3. 3.

Let us consider the map:

[TABLE]

It is a Hopf algebra morphism from $(H_{\mathcal{FG}},._{q},\Delta)$ to $(H_{\mathcal{NCFG}},._{q},\Delta)$ and from $(H_{\mathcal{FG}},._{0},\delta)$ to $(H_{\mathcal{NCFG}},._{0},\delta)$.

Here is its non-commutative version:

Theorem 26**.**

Let $\mathbf{H}_{\mathcal{NCFG}}$ be the vector space generated by the set of ordered Feynman graphs with no cycle and let $q\in\mathbb{K}$.

It is given a Hopf algebra structure by (7)-(8). 2. 2.

We define a second coproduct by (9). Then $(\mathbf{H}_{\mathcal{NCFG}},._{0},\delta)$ is a bialgebra. Moreover, $(\mathbf{H}_{\mathcal{NCFG}},._{0},\Delta)$ is a bialgebra in the category of $(H_{\mathcal{NCFG}},._{0},\delta)$-comodules. 3. 3.

Let us consider the map:

[TABLE]

It is a Hopf algebra morphism from $(\mathbf{H}_{\mathcal{FG}},._{q},\Delta)$ to $(\mathbf{H}_{\mathcal{NCFG}},._{q},\Delta)$ and from $(\mathbf{H}_{\mathcal{FG}},._{0},\delta)$ to $(\mathbf{H}_{\mathcal{NCFG}},._{0},\delta)$.

Remark 9*.*

There is a canonical injection $\iota:H_{\mathcal{NCFG}}\longrightarrow H_{\mathcal{FG}}$. This injection $\iota$ is compatible with $._{q}$ if, and only if, $q=0$: indeed, as $._{0}$ is the disjoint union, it is compatible with $\iota$. If $q\neq 0$, taking G=H=\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern 6.0pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern 0.0pt\offinterlineskip\halign{\entry@#!@&&\entry@@#!@\cr&\crcr}}}\ignorespaces{{{\hbox{\ellipsed@{6.0pt}{6.0pt}}}}\hbox{\kern-6.0pt\raise 0.0pt\hbox{\hbox{\kern 6.0pt\raise 0.0pt\hbox{\textstyle{}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern 36.24512pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{{{\hbox{\ellipsed@{6.0pt}{6.0pt}}}}\hbox{\kern 36.24512pt\raise 0.0pt\hbox{\hbox{\kern 6.0pt\raise 0.0pt\hbox{\textstyle{}}}}}}\ignorespaces}}}}\ignorespaces:

• •

In $H_{\mathcal{FG}}$:

[TABLE]

• •

In $H_{\mathcal{NCFG}}$:

[TABLE]

However, $\iota$ is compatible with $\Delta$ and $\delta$.

4 Quotients of Feynman graphs

4.1 Simple oriented graphs

Let $X$ be a quasi-ordered alphabet. We consider the following quotients of $A_{q}(X)$ and $\mathbf{A}_{q}(X)$:

[TABLE]

Elements of $A^{\prime}_{q}[X]$ are formal spans of monomials

[TABLE]

where the $\epsilon_{i},\alpha_{i,j}\in\{0,1\}$, with only a finite number of them non-zero. Elements of $\mathbf{A}^{\prime}_{q}[X]$ are formal spans of monomials

[TABLE]

where $i_{1},\ldots,i_{k}$ are elements of $X$, all distinct, $\alpha_{i,j}\in\{0,1\}$, with only a finite number of them non-zero. The canonical surjections from $A_{q}(X)$ to $A^{\prime}_{q}(X)$ or form $\mathbf{A}_{q}(X)$ to $\mathbf{A}^{\prime}_{q}(X)$ are both denoted by $\varpi^{\prime}_{X}$.

Proposition 27**.**

Let $X,Y$ be two quasi-ordered alphabets, and $q,q_{1},q_{2}\in\mathbb{K}$, such that $q=q_{1}q_{2}$.

There exist unique algebra morphisms $\Delta_{X,Y}:A^{\prime}_{q}(X\sqcup Y)\longrightarrow A^{\prime}_{q}(X)\otimes A^{\prime}_{q}(Y)$ and $\delta_{X,Y}:A^{\prime}_{q}(XY)\longrightarrow A^{\prime}_{q_{1}}(X)\otimes A^{\prime}_{q_{2}}(Y)$, such that the following diagrams commute:

[TABLE]

They are given in the following way: if $i,i^{\prime}\in X$, $j,j^{\prime}\in Y$,

[TABLE] 2. 2.

The same assertions hold if one replaces $A_{q}$ and $A^{\prime}_{q}$ by $\mathbf{A}_{q}$ and $\mathbf{A}^{\prime}_{q}$ everywhere.

Proof.

Immediate verifications. ∎

Definition 28**.**

Let $G$ be a Feynman graph. We denote by $S(G)$ the simple oriented graph obtained by the following procedure:

Delete all the external edges of $G$. 2. 2.

Delete the loops, that is to say internal edges with two identical extremities. 3. 3.

If $G$ has several edges from $i$ to $j$, where $i,j\in V(G)$, keep only one edge from $i$ to $j$ in $S(G)$.

Lemma 29**.**

Let $G$ and $H$ be two Feynman graphs. The following conditions are equivalent:

$\varpi^{\prime}_{X}(M_{G}(X))=\varpi^{\prime}_{X}(M_{H}(X))$* for any quasi-ordered alphabet $X$.* 2. 2.

$S(G)=S(H)$.

Proof.

$2.\Longrightarrow 1$. Let $X$ be a quasi-ordered alphabet. As $S(G)=S(H)$, we can assume that $V(G)=V(H)$. We denote by $E$ the set of pairs $(i,j)$ such that there is an edge from $i$ to $j$ in $G$ and in $H$, with $i\neq j$. There exist non-zero scalars $\alpha_{i,j}$, $\alpha^{\prime}_{i,j}$, for $(i,j)\in E$, and scalars $\alpha_{i},$, $\alpha^{\prime}_{i}$, $\beta_{i}$, $\beta^{\prime}_{i}$, $\gamma_{j}$, $\gamma^{\prime}_{j}$, and a set of injections $\Lambda$ from $V(G)$ to $X$ such that:

[TABLE]

Their image under $\varpi^{\prime}_{X}$ are both equal to:

[TABLE]

$1.\Longrightarrow 2$. Let us choose an alphabet $X$, such that $M_{G}(X)$ and $M_{H}(X)$ are non-zero. Let $M$ be a monomial of $M_{G}(X)$:

[TABLE]

Then:

[TABLE]

where $\tilde{\alpha}^{\prime}_{i,j}=1$ if $\alpha_{i,j}\neq 0$ and [math] otherwise. This monomial appears in $M_{H}(X)$, so there is a monomial $M^{\prime}$ in $M_{H}(X)$, of the form:

[TABLE]

with, if $i\neq j$, $\alpha^{\prime}_{i,j}\neq 0$ if, and only if, $\tilde{\alpha}_{i,j}=1$ if, and only if, $\alpha_{i,j}\neq 0$. This implies that $S(G)=S(H)$. ∎

For any simple graph $G$, we denote $M_{G}(X)=\varpi^{\prime}_{X}(M_{H}(X))$ and $\mathbf{M}_{G}(X)=\varpi^{\prime}_{X}(\mathbf{M}_{H}(X))$, where $H$ is any Feynman graph such that $S(H)=G$ (for example, $H=G$). These elements, if they are all non-zero, form a basis of a subalgebra of $A^{\prime}_{q}(X)$ or $\mathbf{A}^{\prime}_{q}(X)$. We obtain a quotient of $H_{\mathcal{FG}}$ and $\mathbf{H}_{\mathcal{FG}}$ based on simple graphs. To sum up:

Theorem 30**.**

Let $H_{\mathcal{SG}}$ be the vector space generated by the set of simple graphs and let $q\in\mathbb{K}$.

It is given a Hopf algebra structure: for any simple graphs $G$ and $H$,

[TABLE] 2. 2.

We define a second coproduct by:

[TABLE]

Then $(H_{\mathcal{SG}},._{0},\delta)$ is a bialgebra. Moreover, $(H_{\mathcal{SG}},._{0},\Delta)$ is a bialgebra in the category of $(H_{\mathcal{SG}},._{0},\delta)$-comodules. 3. 3.

Let us consider the map:

[TABLE]

It is a Hopf algebra morphism from $(H_{\mathcal{FG}},._{q},\Delta)$ to $(H_{\mathcal{SG}},._{q},\Delta)$ and from $(H_{\mathcal{FG}},._{0},\delta)$ to $(H_{\mathcal{SG}},._{0},\delta)$.

It has a non-commutative version:

Theorem 31**.**

Let $\mathbf{H}_{\mathcal{SG}}$ be the vector space generated by the set of ordered simple graphs and let $q\in\mathbb{K}$.

It is given a Hopf algebra structure by (10)-(11). 2. 2.

We define a second coproduct by (12). Then $(\mathbf{H}_{\mathcal{SG}},._{0},\delta)$ is a bialgebra. Moreover, $(\mathbf{H}_{\mathcal{SG}},._{0},\Delta)$ is a bialgebra in the category of $(H_{\mathcal{SG}},._{0},\delta)$-comodules. 3. 3.

Let us consider the map:

[TABLE]

It is a Hopf algebra morphism from $(\mathbf{H}_{\mathcal{FG}},._{q},\Delta)$ to $(\mathbf{H}_{\mathcal{SG}},._{q},\Delta)$ and from $(\mathbf{H}_{\mathcal{FG}},._{0},\delta)$ to $(\mathbf{H}_{\mathcal{SG}},._{0},\delta)$.

Remark 10*.*

As simple graphs are Feynman graphs, there is a canonical injection $\kappa:H_{\mathcal{SG}}\longrightarrow\mathcal{FG}$. It is compatible with $._{q}$ if, and only if, $q=0$. Indeed, the disjoint union product $._{0}$ is compatible with $\kappa$. If $q\neq 0$, taking G=H=\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern 6.0pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern 0.0pt\offinterlineskip\halign{\entry@#!@&&\entry@@#!@\cr&\crcr}}}\ignorespaces{{{\hbox{\ellipsed@{6.0pt}{6.0pt}}}}\hbox{\kern-6.0pt\raise 0.0pt\hbox{\hbox{\kern 6.0pt\raise 0.0pt\hbox{\textstyle{}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern 75.77625pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{{{\hbox{\ellipsed@{6.0pt}{6.0pt}}}}\hbox{\kern 75.77625pt\raise 0.0pt\hbox{\hbox{\kern 6.0pt\raise 0.0pt\hbox{\textstyle{}}}}}}\ignorespaces}}}}\ignorespaces:

[TABLE]

Hence, $\kappa$ is not compatible with $._{q}$. Moreover, $\kappa$ is not compatible with $\Delta$ and $\delta$. For example, if G=\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern 6.0pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern 0.0pt\offinterlineskip\halign{\entry@#!@&&\entry@@#!@\cr&\crcr}}}\ignorespaces{{{\hbox{\ellipsed@{6.0pt}{6.0pt}}}}\hbox{\kern-6.0pt\raise 0.0pt\hbox{\hbox{\kern 6.0pt\raise 0.0pt\hbox{\textstyle{}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern 75.77625pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{{{\hbox{\ellipsed@{6.0pt}{6.0pt}}}}\hbox{\kern 75.77625pt\raise 0.0pt\hbox{\hbox{\kern 6.0pt\raise 0.0pt\hbox{\textstyle{}}}}}}\ignorespaces}}}}\ignorespaces:

[TABLE]

4.2 Simple graphs with no cycle

Lemma 32**.**

Let $G$ be a simple graph. The following conditions are equivalent:

For any ordered alphabet, $M_{G}(X)=0$. 2. 2.

$G$* has a cycle.*

Moreover, there exists an ordered alphabet $X$, such that for any simple graph with no cycle, $M_{G}(X)\neq 0$.

Proof.

Similar to the proof of Lemma 24. ∎

Consequently, we obtain Hopf algebras structures on simple graphs with no cycle:

Theorem 33**.**

Let $H_{\mathcal{NCSG}}$ be the vector space generated by the set of simple graphs with no cycle and let $q\in\mathbb{K}$. It is given a Hopf algebra structure: for any simple graphs with no cycle $G$ and $H$,

[TABLE]

Let us consider the map:

[TABLE]

It is a Hopf algebra morphism from $(H_{\mathcal{SG}},._{q},\Delta)$ to $(H_{\mathcal{NCSG}},._{q},\Delta)$.

Here is its non-commutative version:

Theorem 34**.**

Let $\mathbf{H}_{\mathcal{NCSG}}$ be the vector space generated by the set of ordered simple graphs with no cycle and let $q\in\mathbb{K}$. It is given a Hopf algebra structure by (13)-(14). Let us consider the map:

[TABLE]

It is a Hopf algebra morphism from $(\mathbf{H}_{\mathcal{SG}},._{q},\Delta)$ to $(\mathbf{H}_{\mathcal{NCSG}},._{q},\Delta)$ and from $(\mathbf{H}_{\mathcal{SG}},._{0},\delta)$ to $(\mathbf{H}_{\mathcal{NCSG}},._{0},\delta)$.

4.3 quasiposets

Let $X$ be a quasi-ordered alphabet. We consider the following quotients of $A^{\prime}_{q}(X)$ and $\mathbf{A}^{\prime}_{q}(X)$:

[TABLE]

The canonical surjection from $A^{\prime}_{q}(X)$ to $A^{\prime\prime}_{q}(X)$ and from $\mathbf{A}^{\prime}_{q}(X)$ to $\mathbf{A}^{\prime\prime}_{q}(X)$ are both denoted by $\varpi^{\prime\prime}_{X}$.

Definition 35**.**

Let $G$ be a simple oriented graph, which set of vertices is denoted by $V(G)$. We define a quasi-order on $V(G)$, by:

[TABLE]

Note that any quasi-poset, that is to say any pair $P=(V(P),\leq_{P})$, where $V(P)$ is a finite set and $\leq_{P}$ is a quasi-order on $V(P)$, can be obtained in this way: consider the arrow diagram $G$ of $P$, which is a simple graph such that $\leq_{G}=\leq_{P}$.

Lemma 36**.**

Let $X$ be a quasi-ordered alphabet. For any simple graphs $G$, $H$, the following conditions are equivalent:

$\varpi^{\prime\prime}_{X}(M_{G}(X))=\varpi^{\prime\prime}_{X}(M_{H}(X))$* for any quasi-ordered alphabet $X$.* 2. 2.

The quasi-posets $(V(G),\leq_{G})$ and $(V(H),\leq_{H})$ are isomorphic.

Proof.

We define a congruence on the set of monomials of $A^{\prime}_{q}(X)$ by $x_{i,j}x_{j,k}x_{i,k}\equiv x_{i,j}x_{j,k}$. Then two monomials of $A^{\prime}_{q}(X)$ have the same image under $\varpi^{\prime\prime}_{X}$ if, and only if, they are congruent. At the level of graphs, this gives that for any quasi-ordered alphabet $X$, $\varpi^{\prime\prime}_{X}(M_{G}(X))=\varpi^{\prime\prime}_{X}(M_{H}(X))$ if, and only if, one can go from $G$ to $H$ by a sequence of transformations:

[TABLE]

that is to say if, and only if, $\leq_{G}$ and $\leq_{H}$ are isomorphic. ∎

Proposition 37**.**

Let $X,Y$ be two quasi-ordered alphabets. There exists a unique algebra morphism $\Delta_{X,Y}:A^{\prime\prime}_{q}(X\sqcup Y)\longrightarrow A^{\prime\prime}_{q}(X)\otimes A^{\prime\prime}_{q}(Y)$ such that the following diagram commutes:

[TABLE] 2. 2.

The same assertion hold, after replacing $A_{q}$ and $A^{\prime}_{q}$ by $\mathbf{A}_{q}$ and $\mathbf{A}^{\prime}_{q}$ everywhere.

Proof.

1. If $i\leq_{X\sqcup Y}j\leq_{X\sqcup Y}k$, in $A^{\prime}_{q}(X\sqcup Y)$:

[TABLE]

So $\Delta_{X,Y}$ is defined from $A^{\prime\prime}_{q}(X\sqcup Y)$ to $A^{\prime\prime}_{q}(X)\otimes A^{\prime\prime}_{q}(Y)$.

2. Similar proof. ∎

Remark 11*.*

Unfortunately, this does not work for $\delta_{X,Y}$, except if $X$ is a totally ordered alphabet.

For any quasi-poset $P$, we denote $M_{P}(X)=\varpi^{\prime\prime}_{X}(M_{G}(X))$ and $\mathbf{M}_{G}(X)=\varpi^{\prime\prime}_{X}(\mathbf{M}_{G}(X))$, where $G$ is any simple graph such that $\leq_{G}=\leq_{P}$, for example the arrow graph or the Hasse graph of $\leq_{P}$. These elements, if all non-zero, are a basis of a subalgebra of $A^{\prime\prime}_{q}(X)$ or $\mathbf{A}^{\prime\prime}_{q}(X)$. We obtain a quotient of $H_{\mathcal{SG}}$ and $\mathbf{H}_{\mathcal{SG}}$ based on quasi-posets. We shall need the following definitions to describe it:

Definition 38**.**

Let $P,Q$ be two quasi-posets, $A\subseteq V(P)$ and $\sigma:A\hookrightarrow V(Q)$ be an injective map. We consider the quotient, already used in Definition 14:

[TABLE]

We define a relation $\mathcal{R}$ on $V(P)\sqcup V(Q)$ by:

[TABLE]

The transitive closure of $\mathcal{R}$ is denoted by $\leq_{P\sqcup_{\sigma}Q}$, and $P\sqcup_{\sigma}Q=(V(P)\sqcup_{\sigma}V(Q),\leq_{P\sqcup_{\sigma}Q})$ is a quasi-poset.

Note that if $P$ and $Q$ are ordered quasi-posets, then $P\sqcup_{\sigma}Q$ is also an ordered quasi-poset.

Definition 39**.**

Let $P$ be a quasi-poset and let $A\subseteq V(P)$.

We denote by $P_{\mid A}$ the quasi-poset $(A,(\leq_{P})_{\mid A})$. Note that if $P$ is a poset, $P_{\mid A}$ is too. 2. 2.

We shall say that $A$ is an ideal (or an open set) of $P$ if:

[TABLE]

The set of ideals of $P$ is denoted by $I(P)$.

Theorem 40**.**

Let $H_{\mathcal{QP}}$ be the vector space generated by the set of quasi-posets and let $q\in\mathbb{K}$.

It is given a Hopf algebra structure: for any quasi-posets $P,Q$,

[TABLE] 2. 2.

Let us consider the map:

[TABLE]

It is a Hopf algebra morphism from $(H_{\mathcal{SG}},._{q},\Delta)$ to $(H_{\mathcal{QP}},._{q},\Delta)$.

Here is its non-commutative version:

Theorem 41**.**

Let $\mathbf{H}_{\mathcal{QP}}$ be the vector space generated by the set of ordered quasi-posets and let $q\in\mathbb{K}$.

It is given a Hopf algebra structure by (15)-(16). 2. 2.

Let us consider the map:

[TABLE]

It is a Hopf algebra morphism from $(\mathbf{H}_{\mathcal{SG}},._{q},\Delta)$ to $(\mathbf{H}_{\mathcal{QP}},._{q},\Delta)$.

The Hopf algebras $(H_{\mathcal{QP}},._{0},\Delta)$ and $\mathbf{H}_{\mathcal{QP}},._{0},\Delta)$ are introduced and studied in [12, 13, 14, 10].

4.4 Posets

Lemma 42**.**

Let $P$ be a quasi-poset. The following conditions are equivalent:

For any ordered alphabet $X$, $M_{P}(X)=0$. 2. 2.

$P$* is a poset.*

Moreover, there exists an ordered alphabet $X$, such that for any poset $P$, $M_{P}(X)\neq 0$.

Proof.

Similar to the proof of Lemma 24. ∎

Consequently, we obtain a Hopf algebra structure on posets.

Theorem 43**.**

Let $H_{\mathcal{P}}$ be the vector space generated by the set of posets and let $q\in\mathbb{K}$.

It is given a Hopf algebra structure: for any posets $P$ and $Q$,

[TABLE] 2. 2.

Let us consider the map:

[TABLE]

It is a Hopf algebra morphism from $(H_{\mathcal{NCSG}},._{q},\Delta)$ to $(H_{\mathcal{P}},._{q},\Delta)$.

Here is its non-commutative version:

Theorem 44**.**

Let $\mathbf{H}_{\mathcal{P}}$ be the vector space generated by the set of ordered posets and let $q\in\mathbb{K}$.

It is given a Hopf algebra structure by (17)-(18). 2. 2.

Let us consider the map:

[TABLE]

It is a Hopf algebra morphism from $(\mathbf{H}_{\mathcal{NCSG}},._{q},\Delta)$ to $(\mathbf{H}_{\mathcal{P}},._{q},\Delta)$.

Remark 12*.*

Using Remark 11, we could use totally ordered alphabets to define a second coproduct on $H_{\mathcal{P}}$. We would obtain the coproduct given for any poset $P$ of order $n$ by:

[TABLE]

which is coassocative, with a right counit but no left counit.

4.5 Dual product

We now describe the dual product of $\Delta$ on $H_{\mathcal{P}}$. We identify $H_{\mathcal{P}}$ and its graded dual through the symmetric pairing defined for any pair $(P,Q)$ of posets by:

[TABLE]

where $s_{P}$ is the number of automorphisms of $P$.

Proposition 45**.**

Let us define a coproduct $\blacktriangle$ on $\mathcal{H}_{\mathcal{P}}$ in the following way: If $P$ is a poset, denoting by $P_{1},\ldots,P_{k}$ its connected components,

[TABLE]

Then $\blacktriangle$ is coassociative and counitary, and for any $x,y,z\in H_{\mathcal{P}}$:

[TABLE]

Proof.

The coassociativity of $\blacktriangle$ is immediate. Its counit is the map defined by $\varepsilon(P)=\delta_{P,1}$ for any poset $P$. Let $P$, $Q$, $R$ be three posets. Let us denote by $P_{1},\ldots,P_{k}$ the different isoclasses of connected components of $P$, $Q$ and $R$. There exist $\alpha=(\alpha_{i})_{i\in[k]}$, $\beta=(\beta_{i})_{i\in[k]}$, $\gamma=(\gamma_{i})_{i\in[k]}$, such that:

[TABLE]

Moreover:

[TABLE]

Hence:

[TABLE]

Moreover:

[TABLE]

Finally, $\langle\blacktriangle(P),Q\otimes R\rangle=\langle P,QR\rangle$. ∎

Definition 46**.**

Let $P=(V(P),\leq_{P})$ and $Q=(V(Q),\leq_{Q})$. We denote by $\mathcal{P}(V(Q))$ the set of subsets of $V(Q)$. A system of edges from $P$ to $Q$ is a map $\Theta:V(P)\longrightarrow\mathcal{P}(V(Q))$ such that:

For any $x,y\in V(P)$, such that $x<_{P}y$, then for any $x^{\prime}\in\Theta(x)$, $y^{\prime}\in\Theta(y)$, we do not have $y^{\prime}\leq_{Q}x^{\prime}$. 2. 2.

For any $x\in V(P)$, for any $x^{\prime},x^{\prime\prime}\in\Theta(x)$, then $x^{\prime}\leq_{Q}x^{\prime\prime}$ if, and only if, $x^{\prime}=x^{\prime\prime}$.

Proposition 47**.**

Let $P$, $Q$ be two posets and $\Theta$ be a system of edges from $P$ to $Q$. We define an order on $V(P)\sqcup V(Q)$ in the following way:

• •

For any $x,y\in V(P)$, $x\leq y$ if $x\leq_{P}y$.

• •

For any $x,y\in V(Q)$, $x\leq y$ if $x\leq_{Q}y$.

• •

For any $x\in V(P)$, $y\in V(Q)$, $x\leq y$ if there exists $x^{\prime}\in V(P)$, $y^{\prime}\in\Theta(x^{\prime})$, such that $x\leq_{P}x^{\prime}$ and $y^{\prime}\leq_{Q}y$.

This poset is denoted by $P\sqcup_{\Theta}Q$. Moreover, $V(Q)$ is an ideal of $P\sqcup_{\Theta}Q$ and the edges of the Hasse graph of $P\sqcup_{\Theta}Q$ are:

• •

The edges of the Hasse graph of $P$,

• •

The edges of the Hasse graph of $Q$,

• •

The edges $(x,x^{\prime})$, with $x\in V(P)$ and $x^{\prime}\in\Theta(x)$.

Proof.

Note that if $x\leq y$, we cannot have $x\in V(Q)$ and $y\in V(P)$.

$\leq$ is obviously reflexive. If $x\leq y$ and $y\leq x$, then both $x$ and $y$ belong to $V(P)$, or both belong to $V(Q)$. Hence, $x\leq_{P}y$ and $y\leq_{P}x$, or $x\leq_{Q}y$ and $y\leq_{Q}x$. In both cases, $x=y$, so $\leq$ is antisymmetric. Let us assume that $x\leq y$ and $y\leq z$. Four cases hold.

$(x,y,z)\in V(P)^{3}$. Then $x\leq_{P}y\leq_{P}z$, so $x\leq z$. 2. 2.

$(x,y,z)\in V(P)^{2}\times V(Q)$. There exist $y^{\prime}\in V(P)$, $z^{\prime}\in\Theta(x^{\prime})$, such that $x\leq_{P}y\leq_{P}y^{\prime}$ and $z^{\prime}\leq_{Q}z$. So $x\leq z$. 3. 3.

$(x,y,z)\in V(P)\times V(Q)^{2}$. There exist $x^{\prime}\in V(P)$, $y^{\prime}\in\Theta(x^{\prime})$, such that $x\leq_{P}x^{\prime}$ and $y^{\prime}\leq_{Q}y\leq_{Q}z$. So $x\leq z$. 4. 4.

$(x,y,z)\in V(Q)^{3}$. Then $x\leq_{Q}y\leq_{Q}z$, so $x\leq z$.

Hence, $\leq$ is an order.

Let $(x,y)$ be an edge of the Hasse graph of $P\sqcup_{\Theta}Q$, that is to say:

• •

$x<y$.

• •

If $x\leq z\leq y$, then $z\in\{x,y\}$.

Three cases are possible.

$(x,y)\in V(P)^{2}$. As $(P\sqcup Q)_{\mid V(P)}=P$, $(x,y)$ is an edge of the Hasse graph of $P$. 2. 2.

$(x,y)\in V(Q)^{2}$. As $(P\sqcup Q)_{\mid V(Q)}=Q$, $(x,y)$ is an edge of the Hasse graph of $Q$. 3. 3.

$(x,y)\in V(P)\times V(Q)$. Then there exists $x^{\prime}\in V(P)$, $y^{\prime}\in\Theta(x^{\prime})$, such that $x\leq_{P}x^{\prime}$ and $y^{\prime}\leq_{Q}y$. By definition, $x^{\prime}<y^{\prime},$ so $x\leq x^{\prime}<y^{\prime}\leq y$. As $(x,y)$ is an edge, $x=x^{\prime}$ and $y=y^{\prime}$, so $y\in\Theta(x)$.

Conversely:

• •

Let $(x,y)$ be an edge of the Hasse graph of $P$. If $x\leq z\leq y$, necessarily $z\in V(P)$ as $y\in V(P)$, so $x\leq_{P}z\leq_{P}y$, which implies $z\in\{x,y\}$.

• •

Similarly, if $(x,y)$ is an edge of the Hasse graph of $Q$, then it is an edge of the Hasse graph of $P\sqcup_{\Theta}Q$.

• •

If $x\in V(P)$ and $y\in\Theta(x)$, then $x<y$. If $x\leq z\leq y$, two cases are possible.

If $z\in V(P)$, there exists $z^{\prime}\in V(P)$, $y^{\prime}\in\Theta(z^{\prime})$, such that $x\leq_{P}z\leq_{P}z^{\prime}$, $y^{\prime}\leq_{Q}y$. So $x\leq z\leq z^{\prime}<y^{\prime}\leq y$. As $(x,y)$ is an edge, $x=z=z^{\prime}$ and $y^{\prime}=y$. 2. 2.

Similarly, if $z\in V(Q)$, $z=y$.

Obviously, $V(Q)$ is an ideal of $P\sqcup_{\Theta}Q$. ∎

Lemma 48**.**

Let $P_{1},P_{2},P$ be three posets. We consider the two following sets:

• •

$C(P_{1},P_{2},P)$* is the set of triple $(I,\phi_{1},\phi_{2})$, where $I$ is an ideal of $P$, $\phi_{1}$ is an isomorphism from $P_{1}$ to $P_{\mid V(P)\setminus I}$ and $\phi_{2}$ is an isomorphism from $P_{2}$ to $P_{\mid I}$.*

• •

$D(P_{1},P_{2},P)$* is the set of pairs $(\Theta,\phi)$, where $\Theta$ is a system of edges from $P_{1}$ to $P_{2}$ and $\phi$ is an isomorphism from $P_{1}\sqcup_{\Theta}P_{2}$ to $P$.*

Then $C(P_{1},P_{2},P)$ and $D(P_{1},P_{2},P)$ are in bijection.

Proof.

We shortly denote $C=C(P_{1},P_{2},P)$ and $D=D(P_{1},P_{2},P)$.

First step. Let $F:C\longrightarrow D$ defined by $F(I,\phi_{1},\phi_{2})=(\Theta,\phi)$, with:

• •

$\phi$ is defined by $\phi_{\mid V(P_{1})}=\phi_{1}$ and $\phi_{\mid V(P_{2})}=\phi_{2}$.

• •

For any $x\in V(P_{1})$, $\Theta(x)$ is the set of $y\in V(P_{2})$ such that $(\phi_{1}(x),\phi_{2}(y))$ is an edge of the Hasse graph of $P$.

Let us prove that $F$ is well-defined.

Let $x,y\in V(P_{1})$, $x^{\prime}\in\Theta(x)$, $y^{\prime}\in\Theta(y)$, such that $x<_{P_{1}}y$ and $y^{\prime}\leq_{P_{2}}x^{\prime}$. Then $\phi_{1}(x)<_{P}\phi_{1}(y)$, $\phi_{2}(y^{\prime})\leq_{P}\phi_{2}(x^{\prime})$, $(\phi_{1}(x),\phi_{2}(x^{\prime}))$ and $(\phi_{1}(y),\phi_{1}(y^{\prime}))$ are edges of the Hasse graph of $P$. We obtain:

[TABLE]

This contradicts that $(\phi_{1}(x),\phi_{2}(x^{\prime}))$ is an edge of the Hasse graph of $P$.

Let $x\in V(P_{1})$, $x^{\prime},x^{\prime\prime}\in\Theta(x)$, with $x^{\prime}\leq_{P_{2}}x^{\prime\prime}$. Then $\phi_{1}(x)<_{P}\phi_{2}(x^{\prime})\leq\phi_{2}(x^{\prime\prime})$. As $(\phi_{1}(x),\phi_{2}(x^{\prime\prime})$ is an edge, $\phi_{2}(x^{\prime})=\phi_{2}(x^{\prime\prime})$, so $x^{\prime}=x^{\prime\prime}$. We proved that $\Theta$ is a system of edges from $P_{1}$ to $P_{2}$.

By the preceding lemma, the image by $\phi$ of the edges of $P_{1}\sqcup_{\Theta}P_{2}$ are the edges of $P$, so $\phi$ is an isomorphism. We proved that $F$ is well-defined.

Second step. Let $G:D\longrightarrow C$ defined by $G(\Theta,\phi)=(I,\phi_{1},\phi_{2})$, where:

• •

$I=\phi(V(P_{2}))$.

• •

$\phi_{1}=\phi_{\mid V(P_{1})}$ and $\phi_{2}=\phi_{\mid V(P_{2})}$.

By the preceding lemma, $G$ is well-defined.

Last step. Let $(I,\phi_{1},\phi_{2})\in C$. We put $F(I,\phi_{1},\phi_{2})=(\Theta,\phi)$ and $G(\Theta,\phi)=(I^{\prime},\phi_{1}^{\prime},\phi_{2}^{\prime})$. Then $I^{\prime}=\phi(V(P_{2}))=I$, $\phi^{\prime}_{1}=\phi_{\mid V(P_{1})}=\phi_{1}$ and similarly, $\phi^{\prime}_{2}=\phi_{2}$. So $G\circ F=Id_{C}$.

Let $(\Theta,\phi)\in D$. We put $G(\Theta,\phi)=(I,\phi_{1},\phi_{2})$ and $F(I,\phi_{1},\phi_{2})=(\Theta^{\prime},\phi^{\prime})$. For any $x\in V(P_{1})$, $y\in V(P_{2})$:

[TABLE]

Therefore, $\Theta^{\prime}=\Theta$. Moreover, for $i\in\{1,2\}$, $\phi^{\prime}_{\mid V(P_{i})}=\phi_{i}=\phi_{\mid V(P_{i})}$, so $\phi^{\prime}=\phi$. Hence, $F\circ G=Id_{D}$. ∎

Theorem 49**.**

We define a product $\star$ on $H_{\mathcal{P}}$ in the following way: for any posets $P,Q$,

[TABLE]

Then $(\mathcal{H}_{\mathcal{P}},\star,\blacktriangle)$ is a Hopf algebra. Moreover, for any $x,y,z\in H_{\mathcal{P}}$:

[TABLE]

Proof.

Let $P_{1},P_{2},P$ be posets. By the preceding lemma:

[TABLE]

As $\langle-,-\rangle$ is non degenerate and $(H_{\mathcal{P}},m,\Delta)$ is a Hopf algebra, dually $(\mathcal{H}_{\mathcal{P}},\star,\blacktriangle)$ is a Hopf algebra.

Let $x,y,z\in H_{\mathcal{P}}$. For any $t\in H_{\mathcal{P}}$:

[TABLE]

We conclude by the non degeneracy of $\langle-,-\rangle$. ∎