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\author{\sc G\'erard H. E. Duchamp, }

\title{\Large\bf Combinatoire, Informatique et Physique :\\
Alg\`ebre des Diagrammes}

\date{18.06.05 07:23}

\begin{document}

\maketitle

\tableofcontents


\setcounter{section}{-1}
\section{Tentative plan}
\subsection*{1) Second quantized partition function $Tr(exp(-\beta H))$}
\subsection*{2) Reordering problem for $H=H(a^+,a)$ (with examples)}
\subsection*{3) The $exp(G(x))$ and $exp(F(xd/dy))G(y)\big|_{y=0}$ problems}
\subsection*{4) Diagrammatic expansion for the double exponential form}
\subsection*{5) The Hopf algebra structures $\Delta_{WS}$ and $\Delta_{BS}$}
\subsection*{6) Relation of the Hopf structures to the physical problems}

\newpage
\setcounter{section}{2}
\section{Les probl\'emes \`a une et deux exponentielles :\\
\centerline{ $\mathbf{exp(g(x))}$ et $\mathbf{exp(f(xd/dy))exp(g(y))\big|_{y=0}}$}}

\subsection{Matrice de Stirling generalis\'ee}

Commen\c cons par d\'efinir l'alg\`ebre de Wey, $W$. Ce sera ici la $\C$-alg\`ebre associative unif\`ere 
(c'est \`a dire avec \'el\'ement unit\'e \cite{Bo2} III. 1 apr\`es la d\'efinition 1), d\'efinie par la pr\'esentation suivante 
\begin{equation}
\langle a^+,a;\ [a^+,a]=1\rangle_{\C \textbf{ -alg-ass-unit}}
\end{equation}
c'est \`a dire par les deux g\'en\'erateurs $\{a,a^+\}$ et l'unique relation $[a,a^+]=1$. Cette alg\`ebre est de dimension de 
Gelfand-Kirillov $2$ et a une base constitu\'ee par les mon\^omes $\left\{(a^+)^ka^l\right\}_{k,l\geq 0}$.\\
Lorsqu'un \'el\'ement $\Omega$ 
\footnote{Souvent appel\'e op\'erateur \`a cause de diff\'erentes repr\'esentations comme celle de Bargmann-Fock.} 
de $W$ est exprim\'e selon cette base,\\ 
on dit qu'il est en ordre normal (ou forme normale) soit
\begin{equation}
\Omega=\sum_{k,l}\al(k,l) (a^+)^k a^l.
\end{equation}

On sait qu'il est impossible de trouver une repr\'esentation $\rho : W\mapsto \mathcal{L}(B)$ o\`u $B$ est un espace de 
Banach et $\rho(a),\rho(a^+)$ sont des op\'erateurs continus, mais on peut trouver des repr\'esentations fid\`eles, par des 
op\'erateurs continus dans des espaces de Fr\'echet. La plus c\'el\`ebre de ces repr\'esentations est la repr\'esentation de 
{\sc Bargmann-Fock} qui agit sur l'espace des polyn\^omes $\C[x]$ muni de la topologie induite du produit (\cite{T1}).
Les cr\'eations et annihilations sont les suivantes
\begin{equation}
a\mapsto \frac{\textrm{d}}{\textrm{d}x};\ a^+\mapsto x
\end{equation}
o\`u $x$ d\'esigne ici l'endomorphisme $P\mapsto xP$.\\
Ceci rend \'evident le fait d'adopter la fonction de poids $weight(a)=-1;\ weight(a^+)=1$.\\
Soit donc $\Omega\in W$ un operateur homog\`ene par rapport \`a cette fonction de poids. 
Il est de la forme
\begin{equation}
\Omega=\sum_{k-l=e\geq 0}c(k,l)(a^+)^ka^l
\end{equation}
o\`u $e$ est le degr\'e, qui est appel\'e {\it exc\'edent} par les physiciens car il repr\'esente l'exc\'edent (peut-\^etre n\'egatif) 
des cr\'eations sur les annihilations.\\
Selon que celui-ci est positif ou n\'egatif, la forme normale de $\Omega^n$ est 

\begin{equation}\label{normal_form1}
{\mathcal N}(\Omega^n)=(a^+)^{ne}\left(\sum_{k=0}^\infty S_\Omega (n,k) (a^+)^ka^k\right)\ \textrm{or}\
\left(\sum_{k=0}^\infty S_\Omega (n,k) (a^+)^ka^k\right)(a)^{n|e|}
\end{equation}

Nous obtenons ainsi des quantit\'es combinatoires \footnote{Pour l'instant cela qualifiera des \'el\'ements de $\N$, mais nous 
verrons des cas o\`u elles comptent quelque chose.} \`a deux indices $\left\{S_\Omega (n,k)\right\}_{n,k\geq 0}$ c'est \`a dire 
une matrice doublement infinie de format $\N\times\N$ qui sera appel\'ee {\it matrice de Stirling g\'en\'eralis\'ee} de $\Omega$.

*** 
In fact, it is easily checked that, if the coefficients
 $c(k,l)$ of $\Omega$ are non-negative integers, so are the entries ($S_\Omega (n,k)$) of this matrix.   

\smallskip
For $\Omega=a^+a$, one gets the usual matrix of Stirling numbers of the second kind.
\begin{equation}
\left\lceil
{\begin{array}{rrrrrrrr}
1 & 0 & 0 & 0 & 0 & 0 & 0 &\cdots\\
0 & 1 & 0 & 0 & 0 & 0 & 0 &\cdots\\
0 & 1 & 1 & 0 & 0 & 0 & 0 &\cdots\\
0 & 1 & 3 & 1 & 0 & 0 & 0 &\cdots\\
0 & 1 & 7 & 6 & 1 & 0 & 0 &\cdots\\
0 & 1 & 15 & 25 & 10 & 1 & 0 &\cdots\\
0 & 1 & 31 & 90 & 65 & 15 & 1&\cdots\\
\vdots & \vdots & \vdots  & \vdots  & \vdots  & \vdots  & \vdots &\ddots\\
\end{array}}
 \right.
\end{equation}

\smallskip
For $\Omega=a^+a+ aa^+$, we have
\begin{equation}
\left\lceil
{\begin{array}{rrrrrrrr}
1 & 0 & 0 & 0 & 0 & 0 & 0 &\cdots\\
1 & 2 & 0 & 0 & 0 & 0 & 0 &\cdots\\
1 & 8 & 4 & 0 & 0 & 0 & 0 &\cdots\\
1 & 26 & 36 & 8 & 0 & 0 & 0 &\cdots\\
1 & 80 & 232 & 128 & 16 & 0 & 0 &\cdots\\
1 & 242 & 1320 & 1360 & 400 & 32 & 0 &\cdots\\
1 & 728 & 7084 & 12160 & 6320 & 1152 & 64 &\cdots\\
\vdots & \vdots & \vdots  & \vdots  & \vdots  & \vdots  & \vdots &\ddots\\
\end{array}}
\right.
\end{equation}

\smallskip
and for $\Omega=a^+aa^+ + a^+$, we gets
\begin{equation}
\left\lceil
{\begin{array}{rrrrrrrr}
1 	& 0 	& 0 	& 0 	& 0 	& 0 	& 0 	&\cdots\\
2 	& 1 	& 0 	& 0 	& 0 	& 0 	& 0 	&\cdots\\
6 	& 6 	& 1 	& 0 	& 0 	& 0 	& 0 	&\cdots\\
24 	& 36 	& 12 	& 1 	& 0 	& 0 	& 0 	&\cdots\\
120 	&  240 	& 120 	& 20 	& 1 	& 0 	& 0	&\cdots\\
720 	& 1800 & 1200 & 300 	& 30 	& 1 	& 0 	&\cdots\\
5040 & 15120 	&12600 &4200 &630 	& 42 	& 1 	&\cdots\\
\vdots & \vdots & \vdots  & \vdots  & \vdots  & \vdots  & \vdots &\ddots\\
\end{array}}
\right.
\end{equation}

but, for $w=a^+aaa^+a^+$, one gets
\begin{equation}
\left\lceil
{\begin{array}{rrrrrrrrrr}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots\\
2 & 4 & 1 & 0 & 0 & 0 & 0 & 0 & 0 &\cdots\\
12 & 60 & 54 & 14 & 1 & 0 & 0 & 0 & 0 &\cdots\\
144 & 1296 & 2232 & 1296 & 306 & 30 & 1 & 0 & 0 &\cdots\\
2880 & 40320 & 109440 & 105120 & 45000 & 9504 & 1016 & 52 & 1 &\cdots\\
\vdots & \vdots & \vdots  & \vdots  & \vdots  & \vdots  & \vdots & \vdots & \vdots &\ddots\\
\end{array}}
\right.
\end{equation}

Let $\Omega=\sum_{k,l\geq 0\geq 0}c(k,l)(a^+)^ka^l$ (finite supported sum) be a general term 
of $W_n$ in normal form and call {\it dominant term} the sum of monomials with maximum 
length $k+l$. It is not difficult to prove that, if $\Omega$ is homogeneous, the dominant term 
consists in a single monomial $c(k_0,l_0) (a^+)^{k_0}a^{l_0}$. Thus, the dominant term of 
$\Omega^n$ is $c(k_0,l_0)^n (a^+)^{nk_0}a^{nl_0}$
in the generalized Stirling matrix of $\Omega$, the rightmost non-zero coefficient of the line 
$n$ has address $(n,n(k_0-l_0))$ and bears the coefficient $c(k_0,l_0)^n$.    
Thus all the matrices are row-finite and triangular iff $r=1$, which case will be 
of special interest for us in the following.

\begin{rem} There is a beautiful expression of the normal form of $w^n$ in case $w$ is a single word 
in $\{a,a^+\}$. The normal form of $w$ is 
\begin{equation}
{\mathcal N}(w)= \sum_{k\geq 0}r(B,k) (a^+)^{r-k}a^{s-k}
\end{equation}
where $r(B,k)$ is the $k$th rook number of a certain board $B$ constructed after $w$
(see  \cite{GOF4,Varvak} {\tt put a more detailed reference}), and $r=|w|_{a^+};\ s=|w|_{a}$ 
are the number of occurences of $a^+$ and $a$ in $w$. 
\end{rem}

\subsection{One parameter groups and the connected graph theorem}

In case $\Omega$ is homogeneous and sum of monomials with only one ``$a$'', one has 
\begin{equation} 
\Omega=\sum_{0\leq p\leq r}c_{1}(r,p)(a^+)^{r-p}a(a^+)^{p}
\end{equation}
and the normal form of $\Omega$ writes $q(a^+)a+v(a^+)$ and the image of $\Omega$ under the 
Bargmann-Fock correspondence is 
\begin{equation}
\omega=q(x)\frac{d}{dx}+v(x)
\end{equation}


{\tt
\begin{itemize}
\item cas des op\'erateurs avec un seul ``$a$'' :  $\Omega=\sum_{0\leq p\leq r}c_{1}(r,p)(a^+)^{r-p}a(a^+)^{p}$
\item lien avec les champs de vecteurs, les groupes � un param\`etre, conjugacy trick
\item lien entre l'int\'egration et les formes normales, formule (avec $e$ l'ex\'edent)
\begin{eqnarray}\label{corresp}
\sum_{n,k\geq 0} S_\Omega(n,k)\frac{x^n}{n!}y^k=g(x)e^{y\phi(x)}\\
U_\lambda[f](x)=g(\lambda x^e)f\left(x\left(1+\phi(\lambda x^e)\right) \right)
\end{eqnarray}

\item exemple de $\Omega=x^2\frac{d}{dx}x + x\frac{d}{dx}x^2$
\begin{itemize}
\item on a d'abord la forme conjugu\'ee
\begin{equation}
\Omega=x^2\frac{d}{dx}x + x\frac{d}{dx}x^2=x^{-\frac{3}{2}}(2x^3\frac{d}{dx})x^{\frac{3}{2}}
\end{equation}
\begin{equation}\label{conjexp}
e^{\la\Omega}[f]=x^{-\frac{3}{2}}e^{\la(2x^3\frac{d}{dx})}x^{\frac{3}{2}}
\end{equation}
\begin{equation}\label{expfield}
e^{\la(2x^3\frac{d}{dx})}[f](x)=f(u^{-1}(u(x)+\la))\ \textrm{for some straightening factor}\ u
\end{equation}
\item differentiating with respect to $\la$ one obtains $\frac{1}{u'(x)}=2x^3$ and then with \mref{expfield}, one has the one-parameter group 
of transformations 
\begin{equation}
e^{\la (2x^3\frac{d}{dx})}[f](x)=f(s_\la(x)=f\left( \sqrt{\frac{x^2}{1-4\la x^2}}\right)
\end{equation}
thus, in view of \mref{expfield} one gets
\begin{equation}
e^{\la\Omega}[f](x)=x^{-\frac{3}{2}}f(s_\la(x))s_\la(x)^{\frac{3}{2}}=\sqrt[4]{\frac{1}{(1-4\la x^2)^3}}f\left( \sqrt{\frac{x^2}{1-4\la x^2}}\right)
\end{equation}
\item and, with \mref{corresp}, we get 
\begin{equation}
\sum_{n,k\geq 0}S_\Om(n,k) \frac{x^n}{n}y^k=(1-4x)^{-\frac{3}{4}}exp\left(y\sum_{n>0}\binomial{2n}{n}x^n\right)
\end{equation}
{\tt peut-\^etre montrer que les coefficients de la s\'erie exponentielle $(1-4x)^{-\frac{3}{4}}$ sont entiers et appliquer le produit \'etiquet\'e.}
\end{itemize}
\end{itemize}
}

A great, powerful and celebrated result:
(For certain classes of graphs)

If $C(x)$ is the EGF of {\it CONNECTED} graphs (of a certain class), then
$exp(C(x))$ is the EGF of {\it ALL} graphs of the same class. 
(Uhlenbeck, Mayer, Touchard,�)

This implies that the matrix 
\begin{equation}
M(n,k)=\textrm{number of graphs with $n$ vertices and 
having $k$ connected components} 
\end{equation}

is the matrix of a substitution (like $S_\omega (n,k)$ previously
but without prefactor). 
One can prove, using a Zariski-like argument, that, if 
$M$ is such a matrix (with identity diagonal) then, all its powers (positive, negative and fractional) are 
substitution matrices and form a one-parameter group of substitutions, thus coming from a vector field on the line which could (in theory) 
be computed.\\ 
But no nice combinatorial principle seems to emerge.
{\tt expliquer cela}
For example, to begin with the Stirling substitution 
$z\mapsto e^z-1$. We know that there is a unique one-parameter 
group of substitutions $s_\la(z)$ such that, for $\la$ integer, 
one has the value ($s_2(z) \lra \textrm{partition of partitions}$)

\begin{equation}
s_2(z)=e^{e^{z}-1}-1\ ;\ s_3(z)=e^{e^{e^{z}-1}-1}-1\ ;\ s_{-1}(z)=ln(1+z)
\end{equation}
In fact, we have no nice description of this group nor of the vector field generating it.

{\tt Description de l'expression $g(x)e^{y\phi(x)}$ avec le produit \'etiquet\'e}


\subsection{Product formula}
The Hadamard product of two sequences $(a_n)_{n\geq 0};\ (b_n)_{n\geq 0}$ is given by the pointwise product $(a_nb_n)_{n\geq 0}$. 
We can at once transfer this law on EGFs by

\begin{equation}
\Big(\sum_{n\geq 0}a_n \frac{x^n}{n!}\Big)\ \odot\ 
\Big(\sum_{n\geq 0}b_n \frac{x^n}{n!}\Big)
:=\sum_{n\geq 0}a_nb_n \frac{x^n}{n!}
\end{equation}

But, it is not difficult to check that the family 
\begin{equation}
\left(\frac{(y\frac{\partial}{\partial x})^n}{n!}\ \frac{x^m}{m!}\right)_{n,m\in \N}
\end{equation}
is summable in $C[[x,y]]$ as we have 
\begin{equation}
\frac{(y\frac{\partial}{\partial x})^n}{n!}\ \frac{x^m}{m!}=
\left\{ 
\begin{array}{cl}
0 				&\mathrm{if}\ n>m\\
\frac{y^nx^{m-n}}{n!(m-n)!}	&\mathrm{otherwise}
\end{array}
\right.
\end{equation}
therefore, for $F(x)=\sum_{n\geq 0}a_n\frac{x^n}{n!}$ and $G(x)=\sum_{n\geq 0}b_n\frac{x^n}{n!}$ yields the {\it product formula}

\begin{equation}
(F\odot G)(x):=F(y\frac{\partial}{\partial x})G(x)\big|_{x=0}=\sum_{n\geq 0}a_nb_n \frac{y^n}{n!}
\end{equation}

for this reason, we will call here $\odot$ the Hadamard exponential product. With this product, the set of series form a commutative 
associative algebra with unit, which is actually the product algebra $\C^\N$ and we will return later on to the combinatorial significance 
of some of its elements (see section \mref{hadalg}).

\subsection{The double exponential formula}

The case $F(0)=G(0)=1$ will be of special interest in the following study. Every series with constant term $1$ can be represented by 
an exponential $e^{\sum_{n\geq 1}L_n \frac{x^n}{n!}}$ which can be developped using Bell polynomials and Fa\`a di Bruno coefficients. 
Let us now recall some fact about these combinatorial features.\\
We still consider the alphabet $\L=\{L_1,L_2,\cdots\}=\{L_i\}_{i\geq 1}$, then the complete Bell polynomials \cite{Co1} are defined by
\begin{equation}\label{bellpol}
exp(\sum_{m\geq 1} L_m \frac{x^m}{m!})=\sum_{n\geq 0} Y_n(\L) \frac{x^n}{n!}
\end{equation}
We will denote alternatively $Y_n(L_1,\cdots L_n)$ for $Y_n(\L)$ as this polynomial is independant from the subalphabet $(L_m)_{m>n}$. We 
know \cite{Co1} that 
\begin{equation}\label{bellpoldev}
Y_n(\L)=Y_n(L_1,\cdots L_n)=\sum_{||\al||=n}((\al)) \L^\al=\sum_{||\al||=n}((\al)) x_1^{\al_1}x_2^{\al_2}\cdots x_n^{\al_n}
\end{equation}
where $\al=(\al_1,\al_2,\cdots \al_n)$ is an integral vector, $\L^\al=L_1^{\al_1}L_2^{\al_2}\cdots L_n^{\al_n}$ is the multiindex 
standard notation and 
\begin{equation}
((\al))=\frac{||\al||!}{(1!)^{\al_1}(2!)^{\al_2}\cdots (n!)^{\al_n}(\al_1)!\cdots (\al_n)!}
\end{equation}
is the Fa\`a di Bruno coefficient \cite{Co2,JR} which will be interpreted, in the next section, as enumerating structures called {\it set partitions}.\\
Combining \mref{bellpol} and \mref{bellpoldev} one gets

\begin{equation}\label{def}
exp\Bigg(\sum_{m\geq 1} L_m \frac{(y\frac{\partial}{\partial x})^m}{m!}\Bigg)exp\Bigg(\sum_{n\geq 1} V_n \frac{x^n}{n!}\Bigg)\bigg|_{x=0}=
\sum_{k\geq 0} \frac{y^k}{k!}\Bigg(\sum_{|\al|=|\be|} ((\al))((\be))\L^\al\V^\be\Bigg)
\end{equation}
Formula \mref{def} will be called in the sequel the {\it double exponential formula}. 

\section{Monomial expansion of the double exponential formula}

In this paragraph, we will use unordered and ordered set partitions. By unordered partition $P$ of the set $X$ we mean a finite subset 
$P\subset(\frak{P}(X)-\{\emptyset\})$ ($\frak{P}(X)$ is the set of all subsets of $X$ \cite{Co1}) such that 
\begin{equation}
\bigcup_{Y\in P}Y=X \textrm{ and }(Y_1,Y_2\in P,\ Y_1\not=Y_2 \Longrightarrow Y_1\cup Y_2=\emptyset)
\end{equation}
this explains why {\it without any convention} the classical Stirling number of second kind $S(0,0)$ equals $1$. The elements of $P$ are called 
{\it blocks}.\\ 
Following Comtet (\cite{Co2} p 39), we will say that a partition $P$ is of type $\al=(\al_1,\al_2,\cdots ,\al_m)$ iff there is no $j$-block 
for $j>m$ and $\al_j$ $j$-block(s) for each $j\leq m$. This implies in particular that the set $X$ is of cardinality $||\al||:=\sum_{j=1}^m j\al_j$. 
The number of partitions of type $\al$ of a given set is (see \cite{Co2,JR})
\begin{equation}
((\al))=\frac{||\al||!}{\prod_{j=1}^m (j!)^{\al_j}\al_j !}.
\end{equation}
This number is called the Fa\`a di Bruno coefficient of type $\al$ \cite{JR}.\\
Here one can see easily that the number of blocks of a partition of type $\al$ is $|\al|=\sum_{j=1}^m \al_j$.
An ordered partition of type $\al$ of the set $X$ is just a partition in which the blocks are labelled from $1$ to $|\al|$ ($|\al|$ is the number 
of blocks). \\
In other words, one could say that an ordered partition is a {\it list} of subsets and an unordered partition is a {\it set} of subsets.\\
%the type of an ordered partition $P=(B_1,B_2,\cdots ,B_k)$ is just the type of the corresponding unordered partition 
To every ordered partition $P=(B_1,B_2,\cdots ,B_{|\al|})$ corresponds an unordered one\\ 
$\Phi_p(P)=\{B_1,B_2,\cdots ,B_{|\al|}\}$  
where $\Phi_p$ is the ``forgetful'' function which forgets the order. Now to a pair $(P^{(1)},P^{(2)})$ of ordered partitions of the same set (call it $X$)
\begin{equation}
P^{(1)}=(B^{(1)}_1,B^{(1)}_2,\cdots ,B^{(1)}_{k_1});\ P^{(2)}=(B^{(2)}_1,B^{(2)}_2,\cdots ,B^{(2)}_{k_2})
\end{equation}
one can associate the intersection matrix $IM_o(P^{(1)},P^{(2)})$ such that the entry of address $(i,j)$ is the number of elements of the intersection of the block $i$ of the 
first partition and the block $j$ of the second. For example with partitions of $X=\{1,2,3,4,5,6\}$, and specifying  
$$P^{(1)}=(\{1,2,5\},\{3,4,6\}), P^{(2)}=(\{1,2\},\{3,4\},\{5,6\}),$$ 
one gets 
\begin{center}
\begin{tabular}{c|c|c|c|}
   		& $\{1,2\}$ 	& $\{3,4\}$ 	& $\{5,6\}$\\
\hline
$\{1,2,5\}$	&  2		& 0		& 1\\	
\hline
$\{3,4,6\}$	&  0		& 2		& 1\\	
\hline
\end{tabular}
\end{center}
and hence the matrix 
\begin{equation}
\begin{pmatrix}
2 & 0 & 1\\
0 & 2 & 1
\end{pmatrix}
\end{equation}
Formally, $IM_o(P^{(1)},P^{(2)})$ is the matrix of size $k_1\times k_2$ such that 
\begin{equation}
IM_o(P^{(1)},P^{(2)})[i,j]=\sharp\  \left(B^{(1)}_i\bigcap B^{(2)}_j\right)
\end{equation}
The matrices obtained this way are identical to the so-called packed matrices defined in  \cite{DHT}. All packed matrix can be obtained through the 
matching procedure illustrated above.\\
If we consider now a pair of unordered partitions $(Q^{(1)},Q^{(2)})$, we cannot associate to them a single matrix but, rather a class of matrices obtained 
from the preimages of $(Q^{(1)},Q^{(2)})$ under $\Phi_p\times \Phi_p$. In a compact formulation, the set of matrices so obtained is
\begin{equation}
\Big\{IM_0(P^{(1)},P^{(2)})\Big\}_{\Phi_p(P^{(1)})=Q^{(1)}\atop \Phi_p(P^{(2)})=Q^{(2)}  }
\end{equation}

For example, with 
\begin{equation}
(Q^{(1)},Q^{(2)})=\left(\left\{\{1,2,5\},\{3,4,6\}\right\},\left\{\{1,2\},\{3,4\},\{5,6\}\right\}\right)
\end{equation}
one gets the 12 preimages $(P^{(1)}_i,P^{(2)}_j)$, where $P^{(1)}_i$ are among the two preimages of $Q^{(1)}$ and 
$P^{(2)}_j$ are among the 6 preimages of $Q^{(2)}$. Explicitely 
\begin{eqnarray*}
P^{(1)}_1=(\{1,2,5\},\{3,4,6\}) && P^{(1)}_2=(\{3,4,6\},\{1,2,5\})
\end{eqnarray*}
are the preimages of $Q^{(1)}$ and
\begin{eqnarray*}
P^{(2)}_1=(\{1,2\},\{3,4\},\{5,6\}) && P^{(2)}_2=(\{1,2\},\{5,6\},\{3,4\})\\
P^{(2)}_3=(\{3,4\},\{1,2\},\{5,6\}) &&  P^{(2)}_4=(\{3,4\},\{5,6\},\{1,2\})\\
P^{(2)}_5=(\{5,6\},\{1,2\},\{3,4\}) && P^{(2)}_6=(\{5,6\},\{3,4\},\{1,2\})
\end{eqnarray*}
are the preimages of $Q^{(2)}$.\\
The set of matrices so obtained reads
\begin{equation*}
IM_u(M)=\bigg\{
\begin{pmatrix}2 & 0 & 1\\0 & 2 & 1\end{pmatrix},
\begin{pmatrix}2 & 1 & 0\\0 & 1& 2\end{pmatrix},
\begin{pmatrix}1 & 2 & 0\\1 & 0 & 2\end{pmatrix},
\begin{pmatrix}0 & 2 & 1\\2 & 0 & 1\end{pmatrix},
\begin{pmatrix}0 & 1 & 2\\2 & 1 & 0\end{pmatrix},
\begin{pmatrix}1 & 0 & 2\\1 & 2 & 0\end{pmatrix}
\bigg\}
\end{equation*}

This is the orbit of one of them under permutation of lines and columns. The acting group is the direct product $\SG_r\times \SG_c$ where $r$ 
(resp. $c$) is the number of raws (resp. columns) of the given matrix. The actions of the elementary generators of this group are, for each 
$i\in [1,r]$ (resp. $j\in [1,c]$), the $\sigma^r_i$ (resp. $\sigma^c_j$) i.e. the exchange of raw $i$ with raw $i+1$ 
(resp. the exchange of column $j$ with column $j+1$).\\
Let us go back to the example and fix one of them, say $M=\begin{pmatrix}2 & 0 & 1\\0 & 2 & 1\end{pmatrix}$, 
the number of elements in the orbit is 
\begin{equation}
\frac{12}{\sharp \textrm{ symmetry group of $M$}}=6
\end{equation}
the isotropy group of $M$ being here the pair $\{Id,\sigma^c_1\sigma^l_1\}$.\\
The correspondence which, to a pair of unordered partitions associates a class of matrices (under permutations of lines and columns) will be 
denoted $IM_u$ as in the example.

\ss
Thus, one gets a commutative diagram of mappings
\begin{equation}
\begin{CD}
\textrm{Pairs of ordered partitions} @> \Phi_p\times \Phi_p >> \textrm{Pairs of unordered partitions} \\
@V{IM_o}VV						@VV{IM_u}V\\
\textrm{Packed matrices}                 @> Class >> \textrm{Classes of packed matrices}\\
@V{Dg_o}VV						@VV{Dg_u}V\\
\textrm{Labelled diagrams}                 @> \Phi_d>> \textrm{Diagrams}\\
\end{CD}
\end{equation}
The scheme presented above shows how to associate to a pair of unordered (resp. ordered) set partitions a packed matrix (resp. a class of packed 
matrices). The packed matrices can be alternatively represented by {\it labelled diagrams} which are unoriented bipartite multigraphs. The two sets of 
vertices being a column of white spots and column of  black spots as shown below

\vspace{5mm}
\begin{center}
\setlength{\unitlength}{0.7mm}
\begin{picture}(40,40)
\put(5,10){\circle{5}}
\put(5,30){\circle{5}}

\put(35,5){\circle*{5}}
\put(35,20){\circle*{5}}
\put(35,35){\circle*{5}}

\put(5,32){\line(6,1){30.41}}
\put(5,28){\line(6,1){30.41}}
\put(5,12){\line(3,1){30.41}}
\put(5,08){\line(3,1){30.41}}

\put(6,29){\line(6,-5){30}}
\put(6,9){\line(6,-1){27}}

\end{picture}

\bs
Labelled diagram of the matrix  $\begin{pmatrix}2 & 0 & 1\\0 & 2 & 1\end{pmatrix}$.
\end{center}
\vspace{1cm}


In the sequel we will set $Diag_u:=Dg_u\circ IM_u$ and $Diag_o:=Dg_o\circ IM_o$ for the mapping which associates diagrams to pairs of partitions.
Now, the multiplicity of a diagram $\D$ is the number of pairs $(P^{(1)},P^{(2)})$ of unordered partitions such that $Dr_u(IM_u(P^{(1)},P^{(2)}))=\D$. 
We remark that the types of the left (resp. right) components $\al(P^{(1)})$ (resp. $\al(P^{(2)})$ ) are the same. These common types are therefore 
characteristic of the diagram itself and we will refer to it as the left (resp. the right) type of $\D$, $\al(\D)$ (resp. $\be(\D)$).\\
The product formula now reads
\begin{eqnarray}\label{prodform}
&&exp\Big(\sum_{m\geq 1}\frac{L_m}{m!}\left(y \frac{\partial}{\partial x}\right)^m\Big) exp\Big(\sum_{n\geq 1}\frac{V_n}{n!}x^n\Big)\Bigg|_{x=0}=
\sum_{n\geq 0} \frac{y^n}{n!} \Big(\sum_{\D\textrm{ diagram}\atop |\D|=n}mult(\D) \mathbb{L}^{\al(\D)}\mathbb{V}^{\be(\D)}\Big)\cr
&& =\sum_{\D\ \textrm{diagram}}\frac{mult(\D)}{|d|!} m(\D,\L,\V,y)
\end{eqnarray}
with 
\begin{equation}\label{m}
m(\D,\L,\V,y):=\L^{\al(\D)}\V^{\be(\D)} y^{|d|}
\end{equation}

\section{Diagrammatic expansion of the double exponential formula}

The main interest of the expansion \mref{def} is that we can impose (at least) two types of rules on the diagrams 
\begin{itemize}
\item on the diagrams themselves (selection rules) : on the outgoing degrees, ingoing degrees, total or partial weights (the graph is 
supposed oriented from white to black spots)
\item {\tt d\'ecrire ce que l'on sait d\'ej\`a faire sur les mon\^�mes : produit et coproduit. Donner l'exemple d'un produit et du coprodt de 
$x^3y^5$} 
\item on the set of diagrams (composition and decomposition rules) : product and coproduct on the space of diagrams.
\end{itemize}
The last item suggests us to (try to) structure the (specs of) diagrams into a Hopf algebra. To do so, we have to proceed in three steps. 

\begin{itemize}
\item First Step : Define the space
\item Second Step : Define a product
\item Third Step : Define a coproduct
\end{itemize}

{\sc First step}\pointir Let $\Diag_\C$ (resp. $LDiag_\C$) be the $\C$-vector space freely generated by the diagrams (resp. labelled diagrams) i.e. 

\begin{eqnarray}
\Diag_\C:=\bigoplus_{d\ \textrm{diagram}}\C d &;& \LDiag_\C:=\bigoplus_{d\ \textrm{labelled diagram}} \C d 
\end{eqnarray}
at this stage, we have a linear arrow $\LDiag_\C\mapsto \Diag_\C$ provided by the linear extension of $\Phi_d$. 

{\sc Second step}\pointir The product on the set of diagrams is just the is just the superposition of diagrams 

\begin{equation}
d_1\star d_2=\fbox{$\begin{matrix} d_1\cr d_2 \end{matrix}$}.
\end{equation}

This amounts to the blockdiagonal product of packed matrices. Indeed, for $M_1,\ M_2$ packed one has 
\begin{equation}
Dg_o\left(\begin{pmatrix}M_1 & 0\cr 0 & M_2\end{pmatrix}\right)=Dg_o(M_1)\star Dg_o(M_2).
\end{equation}
This product yields the product of monomials in the following way. For $\D$ a diagram via the ``$m$'' correspondence as we have, 
with the setting of \mref{m} 

\begin{equation}\label{prodmon}
m(d_1\star d_2,\L,\V,z)=m(d_1,\L,\V,z)m(d_2,\L,\V,z)
\end{equation}

This product is associative with unit (the empty diagram) and thus endows the set of labelled diagrams with a structure of monoid. 
It is compatible with the arrow $\Phi_d$ and so defines the product on $\Diag$ which, in turn is compatible with the product of monomials.

\begin{equation}
\begin{CD}
\textrm{Labelled diagrams}^2 @> \Phi_d\times \Phi_d >> \textrm{Diagrams}^2 @> m(?,\L,\V,z)\times m(?,\L,\V,z) >> \textrm{Monomials}^2\\
@ V\textrm{product}VV				 @ V\textrm{product}VV		@ V\textrm{product}VV\\
\textrm{Labelled diagrams} @> \Phi_d >> \textrm{Diagrams} @> m(?,\L,\V,z) >> \textrm{Monomials}\\
\end{CD}
\end{equation}

{\it Third step}\pointir For the coproduct on $\LDiag$, we have several possibilities :

\begin{enumerate}
\item split with respect to the white spots (two ways : by intervals and by subsets) 
\item split with respect to the black spots (two ways : by intervals and by subsets) 
\item split with respect to the edges
\end{enumerate}

\begin{comment} : i) (3) does not give a nice identity with the monomials (when applying $d\mapsto m(d,?,?,?))$) nor do (2) and (3) by intervals.\\ 
ii) (2) and (3) are essentially the same (because of the WS $\lra$ BS symmetry).\\
In fact (2) and (3) by subsets give a good representation and, moreover, they are appropriate  for several physics models. 
\end{comment} 

In section \mref{hopf}, we will develop the possibility (1) and (2) by subsets.


\subsection{FBD, bitypes and packed matrices}

\subsection{Counting the multiplicities}
\subsubsection{For ordered bi-partitions}

We here indicate a way to compute the factor $mult(\D)$ od a given diagram $\D$. Let $M$ be a packed matrix such that 
\begin{equation}
\Phi_d(Dr_o(M))=\D
\end{equation}
we can formalize the notion of {\it labelling the edges} of $\D$ in the following way. We will say that a mapping 
\begin{equation}
\lambda : [1..|\D|]\mapsto [1..|\al(\D)|]\times [1..|\be(\D)|]
\end{equation}
is $M$-compatible iff, for each address $(i,j)\in [1..|\al(\D)|]\times [1..|\be(\D)|]$, one has \\
$|\lambda^{-1}(i,j)|=M[i,j]$. Every pair $(P^{(1)},P^{(2)})$ of ordered partitions
\begin{equation}
P^{(1)}=(B^{(1)}_1,B^{(1)}_2,\cdots ,B^{(1)}_{k_1});\ P^{(2)}=(B^{(2)}_1,B^{(2)}_2,\cdots ,B^{(2)}_{k_2})
\end{equation}
such that $IM_o(P^{(1)},P^{(2)})=M$ (hence $k_1=\al(\D);\ k_2=\be(\D)$) defines  labelling $M$-compatible by 
\begin{equation}
\lambda(s)=(i,j)\Longleftrightarrow s\in B^{(1)}_i\cap B^{(2)}_j
\end{equation}
conversely, to every $M$-compatible labelling, one can associate the pair of ordered partitions 
\begin{equation}
OParts(\lambda)=\left(\left(pr_1\circ\lambda)^{-1}(i)\right)_{1\leq i\leq k_1},\left((pr_2\circ\lambda)^{-1}(j)\right)_{1\leq j\leq k_2}\right)
\end{equation}
and, by construction, these two correspondences are reciprocal and then one-to-one. Let similarily define
\begin{equation}
UParts(\lambda)=\left(\left\{(pr_1\circ\lambda)^{-1}(i)\right\}_{1\leq i\leq k_1},\left\{(pr_2\circ\lambda)^{-1}(j)\right\}_{1\leq j\leq k_2}\right)
\end{equation}
(this time not one-to-one). By construction, one has, for  $\Phi_d(Dr_o(M))=\D$,
\begin{equation}
mult_o(\D)=\sharp\{UParts(\lambda)\}_{\lambda\textrm{ is $M$-compatible}}
\end{equation}

\subsubsection{For unordered bi-partitions}

The multiplicity used in the formula \mref{prodform} is, for a given diagram $\D$, the number of pairs $(Q_1,Q_2)$ such that $IM_u(Q_1,Q_2)=\D$. 
One can give a direct formula for the computation of these quantities. First, one can remark that $mult(\D)$ can be divided by $((\al(\D)))$, in the 
sense that the set of solutions of $DG_u\circ IM_u(Q_1,Q_2)=\D$ can be partitionned into disjoint slices with respect to the first coordinate. Call 
\begin{equation}
\mathrm{Slice}_Q(\D)=\{Q_2 | DG_u\circ IM_u(Q,Q_2)=\D\}
\end{equation}
it is not difficult to check that these slices are non void iff $\mathrm{type}(Q)=\al(\D)$, and that, thank to the relabelling, they have the same 
number of elements.\\ 
Let now $M\in \N^{k\times l}$ be a packed matrix such that $Dg_u(class(M))=\D$ and $\la_1,\la_2,\cdots \la_k$ the types corresponding to the lines of $M$ (i.e. 
$\la_j$ is the set of non-zero entries, with their multiplicities, rearranged in decreasing order. Set 
\begin{equation}
(((\al))):=\frac{||\al||!}{\prod_{j=1}^m (j!)^{\al_j}}.
\end{equation}
(it is the number of {\it ordered partitions} of type $\al$). Now, let $C_j$ denote the $j$-th column of $M$. The equivalence on $[1..l]$ defined by 
$i\sim j \Longleftrightarrow C_i=C_j$ is a set partition. Let $\be=(\be_1, \be_2,\cdots \be_l)$ be its type. Then

\begin{equation}
mult(\D)=((\al(\D)) \frac{(((\la_1)))(((\la_2)))\cdots (((\la_k)))}{\be_1 ! \be_2 !\cdots \be_l !}
\end{equation}

\section{The Hopf algebra structures $\Delta_{WS}$ and $\Delta_{BS}$}\label{hopf}

The space spanned by the packed matrices has already received a structure of Hopf algebra, the algebra $\MQSym$ \cite{DHT}. We briefly review 
the structure of this Hopf algebra.\\
{\tt review NCSFVI paragraph 5 and explain as well how the coproducts are linked with the addition of alphabets}\\
Here, doubling the variables of $\L$ or $\V$ suggests two new coproducts which will be denoted $\DWS$ (for ``white spots'') 
and $\DBS$ (for ``black spots'').\\
We describe in details $\DWS$ as the other coproduct is actually got by the same process but applied on the columns instead of the lines. 
Let $M$ be a packed matrix of dimensions $k_1\times k_2$ for every subset $X\in [1..k_1]$ we consider the matrix $\pi_X(M):=pack(M[X,[1..k_2]])$, 
the restriction to the lines of $X$ and then packed (with this restriction to the lines, we only need to perform a horizontal packing). Thus, the 
coproduct $\DWS$ reads
\begin{equation}
\DWS(M)=\sum_{X+Y=[1..k_1]}\pi_X(M)\otimes \pi_Y(M)
\end{equation}
To avoid confusion we will call the supporting space $\HWS$ ($=\MQSym$). We keep the (total) grading of $\MQSym$ by the total weight (i.e. the 
sum of the coefficients) of the matrices. The packed matrices are a linear basis of $\HWS=\MQSym$, thus every element expresses uniquely
\begin{equation}
x=\sum_{M \textrm{ packed}} \lambda_M(x) M
\end{equation}
The coproduct above is cocommutative and with counit $\ep:=\lambda_{\emptyset}$ where $\emptyset$ is the void matrix corresponding to 
the void diagram this particular matrix will be denoted $\1H$.\\
For example, with the packed matrix above one has 

\begin{eqnarray*}
\DWS(\begin{pmatrix} 2 & 0\\0 & 2\\1 & 1\end{pmatrix})=\begin{pmatrix} 2 & 0\\0 & 2\\1 & 1\end{pmatrix}\otimes \1H + 
\begin{pmatrix} 2\end{pmatrix}\otimes \begin{pmatrix} 0 & 2\\1 & 1\end{pmatrix} + 
\begin{pmatrix} 2\end{pmatrix}\otimes \begin{pmatrix} 2 & 0\\1 & 1\end{pmatrix} + 
\begin{pmatrix} 1 & 1\end{pmatrix}\otimes \begin{pmatrix} 2 & 0 \\ 0 & 2\end{pmatrix} +\\
\begin{pmatrix} 2 & 0 \\ 0 & 2\end{pmatrix}\otimes \begin{pmatrix} 1 & 1\end{pmatrix} + 
\begin{pmatrix} 2 & 0 \\ 1 & 1\end{pmatrix}\otimes \begin{pmatrix} 2\end{pmatrix} + 
\begin{pmatrix} 0 & 2 \\ 1 & 1\end{pmatrix}\otimes \begin{pmatrix} 2\end{pmatrix} + 
\1H \otimes \begin{pmatrix} 2 & 0\\0 & 2\\1 & 1\end{pmatrix}
\end{eqnarray*}

This coproduct is compatible with the usual coproduct on the monomials for an infinite family of alphabets $\V$. More precisely set, for all 
$N\geq 0$, 
\begin{equation}
\mathbf{1}_N:=\{\underbrace{1,1,\cdots 1}_{N\ \textrm{times}},0,0,0,\cdots\}\ 
\textrm{i.e. }V_n=1\ \textrm{for}\ n\leq N\ \textrm{and}\ V_n=0\ \textrm{otherwise.}
\end{equation}
Then, using Sweeler's denotation, for theses $\V$s, if $\DWS(d)=\sum d_{(1)}\otimes d_{(2)}$, one has
\begin{equation}
m(d,\L'+\L'',\mathbf{1}_N)=\sum m(d_{(1)},\L',\mathbf{1}_N)m(d_{(2)},\L'',\mathbf{1}_N)
\end{equation}

Thus, it can be shown that, with this structure (product with unit, coproduct 
and the counit), $\LDiag_\C$ is a graded bialgebra infinite dimensions and then a Hopf algebra. The arrow 
$\LDiag_\C\mapsto \Diag_\C$ endows $\Diag_\C$ with a structure of Hopf algebra. 


\subsection{Structure results}

By a theorem of Cartier, Milnor and Moore \cite{Ca, MM}, it must be the envelopping algebra of its primitive elements (which we will described 
in a forthcoming paper), $\LDiag_\C$ is a Hopf algebra.

 
\section*{Acknowledgments}
We thank ...

\bigskip



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\end{document}