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\begin{document}
\noindent

\vspace{1cm}
\hfill{\it Paris, le 12 Septembre 2003, 16h09}

\medskip

\begin{center}
{\large\bf Title (to be defined)}
\end{center}

\begin{abstract}
In this paper, we study, by means of a statistical version of the exponential formula, various sequences transforms 
associated to partitions with constraints and idempotent functions statistics. The link with Bender's diagrams with 
arbitrary constraints is given and we completely solve the problem of monomial separability in free zero-dimensional 
QFT.
\end{abstract}

\section{Sequence transformations}
\subsection{Framework}
\subsubsection{General setting}
The main tool is here to consider the sequences as ``measuring'' polynomials (i.e. the space of formal power series is 
dual to polynomials \cite{BR}). The linear fonctional defined by $S=\sum_{k=0}^\infty a_kz^k$ acts on a polynomial 
$P=\sum_{k=0}^n c_kz^k$ by (we set $c_k=0$ for $k>n$)
\begin{equation}
\scal{S}{P}=\sum_{k\geq 0}a_kc_k
\end{equation}
It is now possible to seek when some power series have a meaning for classical analysis whenever they do not 
represent a Taylor expansion.\\ 
Let's give an example.\\
It is possible to write the OGF of the Bell numbers
\begin{equation}
B(z)=\sum_{n} B_nz^n
\end{equation}
and compute formally (give some identities, prove obstructions) \cite{Kl}, but one can also address the problem of 
the meaning of $B(z)$ as a functional even if it's convergence radius is $0$.\\
The answer is that $B(z)$ {\it as a fonctional over the polynomials} coincide with (and therefore is) a distribution. The  
expression is given by Dobi\`nski's formula
$$
B_n=e^{-1}\sum_{k=0}^\infty \frac{k^n}{k!}
$$
so, the functional $\phi$ such that $\phi(z^n)=B_n$ is $\frac{1}{e}\sum_{k=0}^\infty \frac{\epsilon_k}{k!}$
(where $\epsilon_k$ denote the Dirac measure with support $\{k\}$). And we have, as distributions \cite{Ze}
\begin{equation}
B(z)\equiv \frac{1}{e}\sum_{k=0}^\infty \frac{\epsilon_k}{k!}.
\end{equation}
More general expressions can be found in \cite{BPS}.

\bs
Now it is natural to consider sequences of such functionals. A functional as above can be written as a raw 
$(a_0,a_1,\cdots)$ and then, sequences of such functionals can be seen as infinite matrices.\\
We therefore will be concerned with ``kernel-type'' transformations of sequences such that, to every integer sequence 
$(a_k)_{k\geq 0}$, the transform is $(b_n)_{n\geq 0}$ such that 
\begin{equation}\label{raw_finite_kernel}
b_n=\sum_{k=0}^\infty K(n,k) a_k
\end{equation}
where $K(n,k)$ is a complex-valued double sequence and which are able to transform {\it any} integer sequence. 
The following statement gives us the framework\footnote{An abundant litterature is devoted to such transformations, 
the interested reader is refered to \cite{?1}.}.

\begin{prop} Let $K(n,k)_{n,k\geq 0}$ be a complex-valued double sequence. In order that, for every integer sequence 
$(a_k)_{k\geq 0}$, the series 
$$
\sum_{k=0}^\infty K(n,k) a_k
$$
be convergent, it is necessary and sufficient that every row $(K(n,k))_{k\geq 0}$ be with finite support.
\end{prop}

\Proof The sufficiency is straightforward.\\
Then write $K=K_1+iK_2$ (real and imaginary parts of the kernel). We know that 
$\sum_{k=0}^\infty K(n,k) a_k$ converges iff $\sum_{k=0}^\infty K_1(n,k) a_k$ and $\sum_{k=0}^\infty K_2(n,k) a_k$ do 
(since the ``test'' sequences are real). So the criterium can be proven for $K$ real. Now suppose that there is a row  
(with index $n_0$) such that the support of the sequence $(K(n_0,k))_{k\geq 0}$ be infinite. Let $P$ (resp. $M$) be the set of 
indices $k$ such that $K(n_0,k)>0$ (resp. $K(n_0,k)<0$). One of these subsets (at least) must be infinite. Suppose it is 
$P$ (the reasonning is similar in case it is $M$) and define 
$$
a_k =
\left\{
\begin{array}{cl}
0 & \ \hbox{if} \ k \notin P \\
\lceil   \frac{1}{K(n_0,k)}\rceil& \ \hbox{if} \ k\in P
\end{array}
\right.
$$
a simple computation then gives $\sum_{k=0}^\infty K(n_0,k) a_k=+\infty$.\cqfd

\bs
In the sequel, we will be specially interested by sequence transformations induced by substitutions.\\ 
Let us give a first example of such a transformation, induced by the subtitution $z\ra \frac{z}{1-z}$ .
If we attach to $(a_k)_{k\geq 0}$ the OGF $f(z)=\sum_{k\geq 0}a_kz^k$, then $f(\frac{z}{1-z})=\sum_{n\geq 0}b_nz^n$ with 
$b_n=\sum_{k\geq 0}K(n,k)a_k$ for some kernel $K$. An easy computation gives

\begin{enumerate}
\item $K(0,0)=1,\ K(0,k)=K(n,0)=0,\ {\rm for}\ k,n\not=0$
\item for $0<k\leq n,\ K(n,k)=(-1)^{n-k}\frac{(n-1)!}{(n-k)!(k-1)!}=(-1)^{n-k}\binomial{n-1}{k-1}$
\item\label{lt} if $k>n,\ K(n,k)=0$.
\end{enumerate} 

\begin{note}
i) Remark that \mref{lt} above means that the matrix $K(n,k)$ is lower triangular (this phenomenon will be seen as caracteristic of 
substitutions of type $z\ra \phi(z)=\al z +\ \sum {\rm higher\ terms}$ - see section \mref{subs} for more details).\\
ii) In case $\alpha=1,\ \phi(z)=z +\ \sum {\rm higher\ terms})$ the corresponding matrix is unipotent, so we 
will all these substitutions {\it unipotent} (see \mref{unipot}).
\end{note}

\subsection{A statistical (categorical) version of the exponential formula}\label{expf}

Throughout the paper, we will be interested to compute various examples of EGF for combinatorial objects having 
(a finite set of) nodes (their set-theoretical support) so we use as central concept the mapping $\sigma$ which 
associates to every structure, its set of nodes.\\
We need to draw what could be called ``square-free decomposable objects'' (SFD). This version is suited to our needs 
for the ``exponential formula''. It is sufficiently general to contain, as a particular case, the case of multivariate series. 
For other points of view, see \cite{Jo, St, Vi}\\
Let $\Cl$ be a class of (combinatorial) objects and $\fst$ the catgory of finite sets, $\Cl$ will be called (SFD) if it fulfills the 
two following conditions.

\bs
(DS) {\it Direct sum}\pointir There is a (partial) binary law $\oplus$ on $\Cl$, defined for couples of objects 
$(\om_1,\om_2)$ such that $\si(\om_1)\cap \si(\om_2)=\emptyset$, which is associative, commutative and such that  
\begin{equation}\label{plus}
\Cl_{F_1}\times \Cl_{F_2}\stackrel{\oplus}{\ra} \Cl_{F_1\cup F_2}
\end{equation}
is into.\\
Moreover, $\Cl_\emptyset$ consists in a single element $\{\ep\}$ which is neutral in the sense that, identically 
\begin{equation}
\ep\oplus \om=\om \oplus \ep=\om
\end{equation}
(LP) {\it Levi's property}\pointir Let $\om=\om_1\oplus \om_2=\om^1\oplus \om^2$ be two decompositions. 
Then it can be found a four terms decomposition $\om=\oplus_{i,j=1,2}\om_j^i$ refining the original data in the sense 
that the maginal sums give the factors of the decompositions i.e. 
\begin{equation}
\om_j=\om_j^1\oplus \om_j^2\  {\rm and}\ \om^i=\om_1^i\oplus \om_2^i;\ i,j=1,2
\end{equation}

\begin{note}
Condition \mref{plus} implies that $\si(\om_1\oplus \om_2)=\si(\om_1)\sqcup \si(\om_2)$.
\end{note}

Now, an {\it atom} is any object $\om\not=\ep$ which cannot be split, formally 
\begin{equation}
\om=\om_1\oplus \om_2\Longrightarrow \ep\in \{\om_1,\om_2\}
\end{equation}

\begin{expl}
As example of this setting we have:\\
\begin{enumerate}\label{ex1}
\item\label{intsf} the positive square-free integers $\si(n)$ being the set of primes which divide $n$, the atoms being the 
primes.
\item\label{int} the positive integers $\si(n)$ being the set of primes which divide $n$, the atoms being the 
primes.
\item\label{graphs} all graphs, hypergraphs and weighted graphs, $\si(G)$ being the set of nodes and $\oplus$ the 
juxtaposition, here the atoms are connected graphs.
\item\label{endof} the class of endofunctions $f$ with $\si(f)=dom(f)$
\item\label{ex13} the (multivariate) polynomials in $\N[X]$ with $\si=Alph$ and $\oplus=+$.
\item\label{ex14} the square-free monic (for a given order on the variables) polynomials ; $\si(P)$ being the set of irreducible 
monic divisors of $P$ and $\oplus$ being the multiplication.
\item\label{ex15} complex algebraic curves ; $\si(V)$ being the set of monic irreducible bivariate polynomials vanishing 
on an infinite subset of $V$.
\end{enumerate}
\end{expl}

The prescriptions (DS,LP) imply that decomposition of objects into atoms always exists and is unique. 
\begin{prop}
Let $\om\in \Cl$ then $\om=\om_1\oplus\om_2\oplus\cdots\oplus\om_l$ the $\om_i$ being (dinstinct) atoms and the set 
$\{\om_1,\om_2\cdots\om_l\}$ depends only on $\om$.
\end{prop}

In the class $\Cl$, objects are conceived to be  ``measured'' by different parameters (data in statistical language). So, to get 
a general purpose tool, we suppose that the statistics takes it's values iin a ring $K$ which contains $\Q$ (as, to write 
EGFs it is convenient to have no trouble with the fractions $\frac{1}{n!}$). Let then $c: \Cl\ra K$ be the 
given statistic. In order to write generating series, we need 
\begin{enumerate}
\item that the sum $c_F=\sum_{\om \in \Cl_F}c(w)$ exists 
\item that $F\ra c_F$ should depend only of the cardinality of $F$.
\item $c(\om_1\oplus\om_2)=c(\om_1).c(\om_2)$
\end{enumerate}
We formalize it in 

\bs
(LF) {\it Local finiteness}\pointir For each finite set $F$, the subclass 
\begin{equation}
\Cl_F=\{\om\in \Cl| \si(\om)=F\}
\end{equation}
is a finite set.\\
(Eq) {\it Equivariance}\pointir 
\begin{equation}
|F_1|=|F_2|\Longrightarrow c_{F_1}=c_{F_2}
\end{equation}
(Mu) {\it Multiplicativity}\pointir 
\begin{equation}
c(\om_1\oplus\om_2)=c(\om_1).c(\om_2)
\end{equation}

\begin{note}
a) In fact, (LF) is a property of the class $\Cl$, while (Eq) is a property of the statistics. 
In practice, we choose $\Cl$ which is locally finite and choose equivariant statistics for instance
$$
c(\om)=x^{\rm (number\ of\ cycles)}y^{\rm (number\ of\ fixed\ points)}
$$ 
for some variables $x,y$.

\bs
b) More generally, it is typical to take integer-valued partial (additive) statistics $c_1, \cdots c_i,\cdots ,c_r$ 
(for every $\omega\in \Cl$, $c_i(\omega)\in \N$) and set
$c(\om)=x_1^{c_1(\omega)}x_2^{c_2(\omega)}\cdots x_r^{c_r(\omega)}$.

\bs 
c) The class of examples \ref{ex1}.\ref{int} is not locally finite, but other examples satisfy (LF): \ref{ex1}.\ref{graphs} 
if one asks that the number of arrows and weight is finite, \ref{ex1}.\ref{intsf} and  \ref{ex1}.\ref{ex13} to 
\ref{ex1}.\ref{ex15} in any case.
\end{note}

Now, we are in position to state the exponential formula as it will be used throughout the paper.

\begin{prop} Let $\Cl$ be a locally finite (SFD) and $c: \Cl\ra K$ an equivariant statistics on $\Cl$.  
For every subclass $\F$ one sets the following exponential generating series
\begin{equation}
E(\F)=\sum_{n=0}^\infty c(\F_{[1..n]}) \frac{z^n}{n!}
\end{equation}
Let $\Cl^a$ be the set of atoms of $\Cl$. Then, one has 
\begin{equation}
E(\Cl)=e^{E(\Cl^a)}
\end{equation}
\end{prop} 
\Proof  (First Step)\pointir We consider the subclasses of objects the atoms of which have a support of cardinality $n$ 
i.e.
\begin{eqnarray}
\Cl[n]=\hfill\cr
\{\om\in \Cl| \om=\om_1\oplus\om_2\oplus\cdots \oplus\om_s=\om\ {\rm with}\ \om_i\in\Cl^a,\ i=1..s\ {\rm then}\ 
|\si(\om)|=n, i=1..s\}
\end{eqnarray}

These subclasses are closed under compositions (i.e. under $\oplus$) and decompositions and their 
atoms $\Cl^a[n]=\{\om\in  \Cl[n] \cap \Cl^a \}$. 
Now, one has, thanks to the partitions of $[1..n]$
\begin{eqnarray}
\Cl_{[1..n]}=\bigsqcup_{k\geq 0,\ 0<n_1<n_2<\cdots <n_k\atop n_1+n_2+\cdots n_k=n}
\bigsqcup_{|P_j|=n_j\atop  P_1\sqcup P_2\sqcup \cdots P_k=[1..n]} 
\Cl_{P_1}\oplus \Cl_{P_2}\oplus\cdots\oplus \Cl_{P_k}\\
c(\Cl_{[1..n]})=\sum_{k\geq 0}\sum_{0<n_1<n_2<\cdots <n_k\atop n_1+n_2+\cdots n_k=n}
\sum_{|P_j|=n_j\atop  P_1\sqcup P_2\sqcup \cdots P_k=[1..n]} 
c(\Cl_{P_1}) c(\Cl_{P_2})\cdots c(\Cl_{P_k})
\end{eqnarray}
as, for disjoint sets, it is easy to check that $c(\Cl_X\oplus \Cl_Y)=c(\Cl_X)c(\Cl_Y)$. Now, due to the equivariance of $c$ and to the 
fact that partitions $(P_1,P_2, \cdots ,P_k)$ such that $P_j=n_j$ and  $P_1\sqcup P_2\sqcup \cdots P_k=[1..n]$ are in number 
$$
\frac{n!}{n_1!n_2!\cdots n_k!}
$$
we get
\begin{equation}
c(\Cl_{[1..n]})=\sum_{k\geq 0}\sum_{0<n_1<n_2<\cdots <n_k\atop n_1+n_2+\cdots n_k=n}
\frac{n!}{n_1!n_2!\cdots n_k!}
c(\Cl_{[1..n_1]}) c(\Cl_{[1..n_2]})\cdots c(\Cl_{[1..n_k]})
\end{equation}
and then 
\begin{equation}
E(\Cl)=\prod_{n>0}E(\Cl[n])
\end{equation}
We then compute the factors. 
\begin{equation}
E(\Cl[n])=\sum_{k\geq 0} c(\Cl[n]_{[1..nk]}) \frac{z^{nk}}{(nk)!}
\end{equation}
but 
\begin{equation}
E(\Cl^a[n])=c(\Cl^a_{[1..n]}) \frac{z^{n}}{n!}
\end{equation}
(one monomial) and
\begin{equation}
e^{E(\Cl^a[n])}=\sum_{k\geq 0}c(\Cl^a_{[1..n]})^k \frac{z^{nk}}{(n!)^kk!}=
\sum_{k\geq 0}c(\Cl^a_{[1..n]})^k \frac{z^{nk}}{(nk!)}\frac{(nk)!}{(n!)^kk!}=
\sum_{k\geq 0} c(\Cl[n]_{[1..nk]}) \frac{z^{nk}}{(nk)!}=E(\Cl[n])
\end{equation}
due to the fact that the number of (unordered) partitions of $[1..nk]$ into $k$ blocs of cardinality $n$ is $\frac{(nk)!}{(n!)^kk!}$. 
To end the proof, it suffices to remark that $\Cl^a=\sqcap_{n>0}\Cl^a[n]$ and then 
\begin{equation}
E(\Cl)=\prod_{n>0}E(\Cl[n])=\prod_{n>0}e^{E(\Cl^a[n])}=e^{\sum_{n>0}E(\Cl^a[n])}=e^{E(\Cl^a)}
\end{equation}
\cqfd

\begin{note}
The proof suggests us that it is fruitful to factor a class $\Cl$ into 
(full) subclasses i.e. that are generated by a partition of the atoms 
(see below section \mref{msection}).
\end{note}

\subsection{Transformations induced by substitutions}\label{tsubs}

We consider substitutions of the form $f(z)\ra f(\phi(z))$ where 
\begin{equation}\label{noconst}
\phi(z)=c_1z+\sum_{n\geq 2}c_kz^k.
\end{equation}
Then, setting $f(z)=\sum_{n=0}^\infty a_n\frac{z^n}{n!}$ and $f(g(z))=\sum_{n=0}^\infty b_n \frac{z^n}{n!}$ one can see that the 
transformation $(a_k)_{k}\ra (b_n)_{n}$ is of the type \mref{raw_finite_kernel}.\\
Then, one can address the question: when a transformation of type \mref{raw_finite_kernel} ?

\bs
First, remark that in  the case of substitutions of type \mref{noconst}, the matrix of coefficients $K(l,c)$ is lower triangular.\\
One considers conversely a transformation defined by $(K(l,c))_{l,c\geq 0}$ which is lower triangular.\\
The transformed sequence is
\begin{equation}
b_l=\sum_{c}K(l,c)a_c
\end{equation}
Now, if the transformation is the substitution by a function $\phi$ (as in \mref{noconst}), the transform of $\frac{z^c}{c!}$ reads 
\begin{equation}
\frac{(\phi(z))^c}{c!}=\sum_{l=0}^\infty K(l,c)\frac{z^l}{l!}
\end{equation}
giving the answer. 

\begin{prop}\label{subs}
Let $K(l,c)$ be a lower-triangular kernel.\\ 
i) The following conditions are equivalent 
\begin{enumerate}
\item the transformation defined by $K$ on exponential sequences is induced by a substitution by a function $\phi$ without 
constant term ($\phi(z)=c_1z+\sum_{k\geq 2}c_kz^k)$)
\item $K(0,0)=1,\ K(0,k)=K(n,0)=0,\ {\rm for}\ k,n\not=0$ and if one sets $\phi(z)=\sum_{l}K(l,1)\frac{z^l}{l!}$, one has, for every 
other $c$ (i.e. $c\geq 2$) 
\begin{equation}
\frac{(\phi(z))^c}{c!}=\sum_{l=0}^\infty K(l,c)\frac{z^l}{l!}
\end{equation}
\item $\sum_{n,k\geq 0}K(n,k)x^k\frac{z^n}{n!}=e^{x\phi(z)}$
\end{enumerate}
ii) In the preceding conditions, $K(n,n)=c_1^n$. This implies that the matrix $K$ is unitriangular (unipotent, i.e. with every 
diagonal coefficient equal to $1$) iff $c_1=1$.
\end{prop}


\section{Applications}
\subsection{Generalized Stirling Numbers}

In their papers, \cite{BPS,BPS2,BPS3}, P. B\l asiak, K. Penson, A. I. Solomon, defined generalized Stirling numbers by 
\begin{equation}
(u^rd^s)^n=u^{n(r-s)}\sum_{k=s}^{ns}S_{r,s}(n,k)u^kd^k
\end{equation}

\begin{note} These numbers have a clear interpretation in terms of coefficients of {\it rook polynomials} of {\it certain Ferrers boards} 
(see \mref{??}).
\end{note}

From one of their formula (\cite{BPS3} {\it form. 16}), one can prove that the double sequence $S_{r,1}(n,k)$ defines a substitution (for $s>1$, 
the $S_{r,s}(n,k)$ do not define a lower triangular matrix). 
This formula is the following
\begin{equation}
\prod_{j=1}^n (x+(j-1)(r-s))^{\ul{s}}=\sum_{k=s}^{ns}S_{rs}(n,k)x^{\ul{k}}
\end{equation}
where $X^{\ul{m}}=X(X-1)\cdots (X-m+1)$ is the falling factorial.\\
For $s=1,\ r>1$ one has
\begin{eqnarray}
\sum_{k=1}^{n}S_{r1}(n,k)x^{\ul{k}}=\prod_{j=1}^n (x+(j-1)(r-1))=(r-1)^n \prod_{j=1}^n \left( \frac{x}{r-1}+j-1 \right)=\cr
(r-1)^n \left( \frac{x}{r-1}\right)^{\ol{n}}=(r-1)^n \sum_{l=1}^n |St1(n,l)| \left( \frac{x^l}{(r-1)^l}\right)=\cr
(r-1)^n \sum_{l=1}^n |St1(n,l)| \frac{1}{(r-1)^l}\sum_{k=1}^l St2(l,k) x^{\ul{k}}
\end{eqnarray}
with $X^{\ul{m}}=X(X+1)\cdots (X+m-1)$ the ascending factorial.\\
Thus, since $(x^{\ul{k}})_{k\geq 0}$ is a basis of the space of univariate polynomials, one has
\begin{equation}
S_{r1}(n,k)=(r-1)^n \sum_{l=1}^n |St1(n,l)| \frac{1}{(r-1)^l}\sum_{k=1}^l St2(l,k)
\end{equation}
where $Sti$ are the Stirling numbers of the $i eth$ kind.\\
In view of proposition \mref{subs} let us compute the double series $\sum_{n,k\geq 0}S_{r1}(n,k) \frac{x^n}{n!}y^k$.
\begin{eqnarray}
\sum_{n,k\geq 0}S_{r1}(n,k) \frac{x^n}{n!}y^k=1+\sum_{n,k\geq 1}S_{r1}(n,k) \frac{x^n}{n!}y^k=\cr
1+\sum_{n,k\geq 1}(r-1)^n \sum_{l=1}^n |St1(n,l)| \frac{1}{(r-1)^l}\sum_{k=1}^l St2(l,k)\frac{x^n}{n!}y^k
\end{eqnarray}
which is, due to the boundary properties of the Stirling numbers,
\begin{eqnarray}
\sum_{n,l,k\geq 0}(r-1)^n |St1(n,l)| \frac{1}{(r-1)^l}St2(l,k)\frac{x^n}{n!}y^k=\cr
\sum_{n,l,k\geq 0}\frac{1}{(r-1)^l}St2(l,k)y^k(r-1)^n |St1(n,l)| \frac{x^n}{n!}=\cr
\sum_{l,k\geq 0}\frac{1}{(r-1)^l}St2(l,k)y^k\sum_{n\geq 0} |St1(n,l)| \frac{(r-1)^nx^n}{n!}=\cr
\sum_{l,k\geq 0}\frac{1}{(r-1)^l}St2(l,k)y^k\frac{\left(-log(1-(r-1)x)\right)^l}{l!}=
\sum_{l,k\geq 0}St2(l,k)y^k\frac{\left(\frac{-log(1-(r-1)x)}{r-1}\right)^l}{l!}=\cr 
e^{\left({y(e^{\frac{-log(1-(r-1)x)}{r-1}}-1)}\right)}=e^{\left({y(\frac{1}{1-(r-1)x})^{\frac{1}{r-1}}-1)}\right)}\cr
\end{eqnarray}
which proves that the matrix $S_{r1}(n,k)$ defines the substitution $\{z\ra (-1+(\frac{1}{1-(r-1)z})^{\frac{1}{r-1}})\}$ which can 
be checked to be unipotent. For $r=2$, we have the substitution\\
 $\{z\ra (-1+\frac{1}{1-z})=\frac{z}{1-z}\}$. This will be checked in \mref{??} by other means.



\subsection{Idempotent transformations and the like}
\subsubsection{Idempotent transformations}
Let $I(n,k)$ be the number of idempotent endofunctions (i.e. 
such that $f^2=f$) on a (fixed) set with $n$ elements and $k$ 
fixed points. In the setting \mref{expf}, the class $\Cl$ is the class of idempotent endofunctions, the composition is 
just the juxtaposition of endofunctions (that we will denote by $\E$ 
in the next section) defined on disjoint domains, the atoms being the (idempotent) 
endofunctions having only one fixed point and the statistic is $c(f)=x^{\rm (number\ of\ fixed\ points)}$. It is clear then that 
$c(f\oplus g)=c(f)c(g)$. One has, by definition
\begin{equation}
E(\Cl)=\sum_{n,k\geq 0}I(n,k)x^k\frac{z^n}{n!}=e^{E(\Cl^a)}=e^{\sum_{n\geq 0}\frac{nxz^n}{n!}}=e^{xze^z}
\end{equation}
using the criterium \mref{subs}, one gets that $I(n,k)$ defines the substitution $z\ra ze^z$.

\subsection{Higher order idempotent transformations}

The idempotent endofunctions are exactly the endofunctions 
satisfying $f^2=f$.\\
 Can we pursue the investigation of such equations ? \\
The answer is positive, but we have to analyse carefully 
arithmetic phenomena that are not apparent in the case 
of idempotent endofunctions.\\ 
Call $\E_m$ the class of endofunctions satisfying the equation 
$f^m=f^{m+1}$. It is a full subclass of $\E$. We first remark that, 
if $k|m$, one has  $\E_k\subset \E_m$. So, we will have a 
finer classification if we consider the {\it strict} classes.\\
We denote $\ES_m$ the class of endofunctions that have only 
$m$-cycles and satisfy $f^m=f^{m+1}$. Then it is not difficult 
to check that 
\begin{enumerate}
\item the $\ES_m$ are full subclasses of $\E$
\item\label{unat} $(\E_m)^a=\sqcup_{k|m} (\ES_m)^a$
\item $((\ES_m)^a)$ consists in the endofunctions having only 
a $m$-cycle (and satisfying $f^m=f^{m+1}$, which is equivalent 
to say that the image of every point is cyclic)
\item\label{fesm} $E((\ES_m)^a)=\sum_{n\geq m}\binomial{n}{m}(m-1)!m^{n-m}\frac{z^n}{n!}
=\frac{z^m}{m}e^{mz}$
\end{enumerate}
we deduce from \mref{unat} and \mref{fesm} above that 
\begin{equation}
E(\E_m)=e^{\sum_{k|m\atop k\geq 1}\frac{z^k}{k}e^{kz}}
\end{equation}
and if one counts the number of cycles so that $I_m(n,c)$ is 
the number of endofunctions in $[1..n]$ satisfying $f^m=f^{m+1}$
and having $c$ cycles, one has the double statistics
\begin{equation}
\sum_{n,c\geq 0}I_m(n,c)(x^c)\frac{z^n}{n!}=e^{x(\sum_{k|m\atop k\geq 1}\frac{z^k}{k}e^{kz})}
\end{equation}
which implies that $I_m(n,c)$ is the matrix associated to the substitution 
$z\ra \sum_{k|m\atop k\geq 1}\frac{z^k}{k}e^{kz}$.

\bs
Let $\E_\infty$ be the class of all endofunctions satisfying some equation $f^m=f^{m+1}$ 
(Burnside condition with $k=1$,see below), and denote $I(n,c)$ the number of such endofunctions 
on $[1..n]$ with $c$ cycles. We have (directly or by limiting process)
\begin{equation}
\sum_{n,c\geq 0}I(n,c)(x^c)\frac{z^n}{n!}=e^{x(\sum_{k\geq 1}\frac{z^k}{k}e^{kz})}=e^{xlog(\frac{1}{1-ze^z})}
\end{equation}
so, again, $I(n,c)$ is the matrix associated to the substitution $z\ra log(\frac{1}{1-ze^z})$. In particular
\begin{equation}
E(\E_{\infty})=e^{log(\frac{1}{1-ze^z})}=\frac{1}{1-ze^z}
\end{equation}
this rather simple equation, which seems to belong to the realm of OGFs and is, for us, a mystery.

\subsection{The Lie group of unipotent substitutions}\label{unipot}

The substitutions $z\ra \phi(z)=z +\ \sum {\rm higher\ terms}$ form a group under composition. in the next section, 
we see that, in a certain sense, this group is an (infinite dimensional) Lie group. In particular it is possible to compute 
one parameter subgroups of substitutions.\\
The aim of his section is then to prove that for any subsitution  $z\ra \phi(z)=z +\ \sum {\rm higher\ terms}$ it is 
possible to compute a continuous family $\phi_t;\ t\in \R$ such that 
\begin{equation}
\phi_0(z)=z;\ \phi_1=\phi;\ \phi_r(\phi_s(z))=\phi_{r+s}(z)
\end{equation}

\begin{expl}
For example, with $\phi(z)=\frac{z}{1-z}$, one has $\phi_t(z)=\frac{z}{1-tz}$.
\end{expl}

Hence the analytic substitutions form a Lie group $\sbs$. 

\begin{expl} We consider the generalized Stirling numbers of \cite{BPS3} $S_{21}(n,k)$. One has 
\begin{equation}
\sum_{n,k\geq 0} S_{21}(n,k)\frac{x^n}{n!}y^k=e^{y\frac{x}{1-x}}
\end{equation}
\end{expl}


Before claiming that $\sbs$ is a Lie group, we need to put some topology on it and study an algebra in which it is 
embedded.

\bs
{\bf The (topological) algebra $\T$}\pointir


Let $\T$ be the ($\C$-) algebra of infinite $\N\times\N$ lower triangular matrices with complex coefficients 
\begin{equation}
\T=\{K\in \C^{\N\times\N}|\ c>l\Longrightarrow K(l,c)=0\}
\end{equation}
the following properties are not difficult to check
\begin{enumerate}
\item $\T$ is a $\C$-algebra of infinite matrices
\item let $\tau_n :\ \T\ra\T_n$ be the {\it truncation} mapping such that $\tau_n(K)$ is the $[0..n]\times [0..n]$ submatrix 
of $K$ (we denote $\T_n$ the algebra of $[0..n]\times [0..n]$ triangular matrices).
\item For $n\leq m$ let $\tau_{nm}:\ \T_m\ra\T_n$ be the truncation mapping. One has $\tau_{nm}\tau_m=\tau_n$
\end{enumerate}
We put on $\T$ the topology of local convergence i.e. determined by the morphisms $\tau_n$ (more precisely, 
it is the coarsest topology for which the $\tau_n$ are continuous, the algebras $\T_n$ being endowed with the standard
topology namely a filter, a sequence converges iff all it's tuncations do.

In view of proposition \mref{subs}, to be the matrix of a substitution is a local condition as the equality 
$\sum_{n,k\geq 0}K(n,k)x^k\frac{z^n}{n!}=e^{x\phi(z)}$ is equivalent to the fact that, for all $k$
\begin{equation}
\sum_{n\geq 0}K(n,k)\frac{z^n}{n!}=\frac{1}{k!}(\sum_{n\geq 0}K(n,1)\frac{z^n}{n!})^k
\end{equation}
which, in turn expresses that, for all  $k,N\geq 0$
\begin{equation}
\sum_{0\leq n\leq N}K(n,k)\frac{z^n}{n!}=\frac{1}{k!}\left( (\sum_{0\leq n\leq N}K(n,1)\frac{z^n}{n!})^k\right)_{\leq N}
\end{equation}
where $(T(z))_{\leq N}$ denotes the truncation of the series after at order $N$.

\subsection{(Internal) More on Burnside matrices}

For $m,k\geq 1$ and $n,c\geq 0$, let $B_{m,k}(n,c)$ be the number of endofunctions of $[1..n]$ 
such that $f^m=f^{m+k}$ and having $k$ cycles. It is not difficult to prove, as there is 
one and only one cycle in each connected component, that 
\begin{equation}
\sum_{n,c\geq 0}B_{m,k}(n,c)(x^c)\frac{z^n}{n!}=e^{x(\sum_n B_{m,k}(n,1)\frac{z^n}{n!})}
\end{equation}
but the computation of $B_{m,k}(n,1)$ seems difficult.

\subsection{Multisections and full subclasses}\label{msection}

Bi- and multisection is the art of selecting terms in a given power series. The first procedures are:
\begin{enumerate}
\item take one term over 2 (i.e. the even and odd part of a given function)
\item generalization: the first, the second ... the $m$-ieth in each $m$ terms
\item take the first $m$ terms, withdraw the first $m$ terms
\end{enumerate}
Applying one of these precedures to the exponent of an exponential formula amounts to select some 
among the atoms of the class under consideration. Then the formula counts the objects having only 
the selected atoms in their maximal decomposition.

\section{Separability in zero-dimensional QFT}

\subsection{Hadamard product of Taylor series}
In this section, the central r\^ole is played by an interesting expression of the {\it Hadamard product} of two 
sequences. Recall that the {\it Hadamard product} \cite{BR,?} or {\it pointwise product} of two sequences 
$A=(a_k)_{k\geq 0},\ B=(b_k)_{k\geq 0}$ is 
\begin{equation}
A\odot B=(a_kb_k)_{k\geq 0}
\end{equation}
this can, of course be transposed to the corresponding Taylor series \\
$E_A(z)=\sum_{k\geq 0}a_k\frac{z^k}{k!},\ E_B(z)=\sum_{k\geq 0}b_k\frac{z^k}{k!}$ by 
\begin{equation}
E_{A\odot B}(z)=\sum_{k\geq 0}a_kb_k\frac{z^k}{k!}
\end{equation}
Many interesting (and non-trivial) features are known about Hadamard product such as the fact that the Hadamard 
product of the Taylor expansions of two rational functions is of the same type (the computation of the coefficients of the 
result is however non-trivial).\\
A formula, classical in QFT, says that the Hadamard product of $A(z),\ B(z)$ can be obtained by 
\begin{equation}
E_{A\odot B}(y)=E_A(y\frac{d}{dz})E_B(z)|_{z=0}
\end{equation}
[motivation, explanation, interest].
... as I could understand the following formula is of utmost importance in QFT ??
\begin{equation}\label{F_m}
e^{\sum_{m\geq 1}L_m \frac{(y\frac{d}{dz})^m}{m!}}e^{\sum_{m\geq 1}V_m \frac{z^m}{m!} }|_{z=0}=
\sum_{n\geq 0}F_n((L_m)_{m\geq 1},(V_m)_{m\geq 1})\frac{y ^n}{n!}
\end{equation}
it is not difficult to prove that the functions $F_n$ are, in fact, polynomials in the double set of variables 
$(L_m)_{m\geq 1},(V_m)_{m\geq 1}$. So, the separation problem by these quantity (potential, field) as they vary 
reads
\begin{center}
What are the coefficients and the monomials of the polynomials $F_n$ ?
\end{center}


\subsection{Separation theorem}

As we know that $\sum_{n\geq 0}F_n((L_m)_{m\geq 1},(V_m)_{m\geq 1})\frac{z^n}{n!}$ is the Hadamard product 
of $e^{\sum_{m\geq 1}L_m \frac{z^m}{m!}}$ and  $e^{\sum_{m\geq 1}V_m \frac{z^m}{m!}}$ we must begin in 
computing the coefficients of the expansion
\begin{equation}
e^{\sum_{m\geq 1}L_m \frac{z^m}{m!}}=\sum_{n\geq 0}P_n((L_m)_{m\geq 1}) \frac{z^n}{n!}
\end{equation}
We have 
\begin{equation}\label{pot}
e^{\sum_{m\geq 1}L_m \frac{z^m}{m!}}=\sum_{k\geq 0}\frac{(\sum_{m\geq 1}L_m \frac{z^m}{m!})^k}{k!}
\end{equation}
with the following conventions
\begin{enumerate}
\item $\N^{(\N^*)}$ is the set of (integer) sequences $(\al(1),\al(2),\cdots \al(j)\cdots)$ with finite support
\item for $\al\in\N^{(\N^*)}$, $|\al(i)|=\sum_{i\geq 1}\al(i)$ and $||\al||=\sum_{i\geq 1}i\al(i)$
\item for $\al\in\N^{(\N^*)}$, $\al!=\prod_{i\geq 1}\al(i)!$ and $\L^\al=\prod_{i\geq 1}L_i^{\al(i)}$
\end{enumerate}
one has
\begin{equation}
\frac{1}{k!}(\sum_{m\geq 1}L_m \frac{z^m}{m!})^k=
\frac{1}{k!}\sum_{\al\in \N^{(\N^*)}\atop |\al|=k}\frac{k!}{\al!}\L^\al \frac{z^{||\al||}}{\prod_{i\geq 1}(i!)^{\al(i)}}
\end{equation}
from what we derive
\begin{equation}
P_n((L_m)_{m\geq 1})=\sum_{\al\in \N^{(\N^*)}\atop ||\al||=n}\frac{n!}{\al!\prod_{i\geq 1}(i!)^{\al(i)}}\L^\al 
\end{equation}
but now the coefficient of $\L^\al$ has a clear combinatorial interpretation. It is the number of (unordered) 
partitions of $[1..n]$ into (nonempty) subsets: $\al(1)$ with one element (singletons), $\al(2)$ with two elements 
(unordered pairs), $\al(3)$ with three elements,$\cdots$ $\al(j)$ with $j$ elements, $\cdots$ hence the condition 
$||\al||=n$. Such partitions will be called {\it of type $\al$} and their number will be denoted by $numpart(\al)$. Then
\begin{equation}
numpart(\al)=\frac{n!}{\al!\prod_{i\geq 1}(i!)^{\al(i)}}.
\end{equation}
From \mref{F_m}, we see that 
\begin{equation}
F_n((L_m)_{m\geq 1},(V_m)_{m\geq 1})=\sum_{||\al||=||\be||=n}numpart(\al)numpart(\be)\L^\al\V^\be
\end{equation}

We then, have proved the following separation property which solves the question of the preceding 
section.
\begin{theorem}
Let $\L=(L_m)_{m\geq 1};\ \V=(V_m)_{m\geq 1})$ two (disjoint) alphabets (sets of free variables), and 
$F_n$, the functions defined by 
\begin{equation}
\sum_{n\geq 0}F_n((L_m)_{m\geq 1},(V_m)_{m\geq 1})\frac{y ^n}{n!}=
e^{\sum_{m\geq 1}L_m \frac{(y\frac{d}{dz})^m}{m!}}e^{\sum_{m\geq 1}V_m \frac{z^m}{m!} }|_{z=0}
\end{equation}
then
\begin{enumerate}
\item the functions $F_n$ are polynomials
\item we have, with the usual mlultiindex notations 
\begin{equation}
F_n(\L,\V)=\sum_{||\al||=||\be||=n}numpart(\al)numpart(\be)\L^\al\V^\be
\end{equation}
\end{enumerate}
\end{theorem}

\subsection{Drawn diagrams, diagrams and their separation power}

In this section, we deal with the combinatorial interpretation of the products $numpart(\al)numpart(\be)$ 
linked with some special bipartite graphs.\\
Remark that $numpart(\al)numpart(\be)$ is the number of couples of partitions with bitype $(\al,\be)$.\\
First, let us review some basic knowledge about partitions and types. 
\begin{definition}\label{unionref}
Let $X$ be a set and $P\subset {\goth P}(X)-\{\emptyset\}$.

\ss
i) The collection $P$ is called a partition (unordered) of $X$ iff 
\begin{enumerate}
\item $\cup_{Y\in P}Y=X$
\item\label{unionpart} $Y,Z\in P\Longrightarrow (Y=Z)\ {\rm or}\ (Y\cap Z=\emptyset)$
\end{enumerate}
in another denotation $X=\sqcup_{Y\in P}Y$.\\
ii) If $X$ is finite, the type of $X$ is $\al\in \N^{\N^*}$ 
\begin{equation}
\al(i)=\#\{Y\in P | \#Y =i\}
\end{equation}
\end{definition}

\begin{note} i) The condition of definition \mref{unionref} i)\ref{unionpart}) implies that, if the set $P$ is empty 
then so is $X$. The converse is obvious.\\
ii) The number of partitions of a set of $n$ elements is called the Bell number of order $n,\ B(n)$. Below 
the first Bell numbers.

\begin{center}
\begin{tabular}{l|| c|c|c|c|c|c|c|c|c|c}
$n$ 	& 0 	&1 	&2 	&3 	&4 	&5 	&6 	&7 	&8 	&9 	\\
\hline
$B(n)$ & 1	&1	&2	&5	&15	&52	&203	&877	&4140	&21147
\end{tabular}
\end{center}

iii) The ``partition type'' of a partition of $[1..n]$ is a partition in the Diophantine sense. Below the partitons of 
$[1..4]$ and their types (if there are several choices, the multiplicity is indicated between brackets).
\begin{center}
{\fns
\begin{tabular}{l|| c| c| c |c |c}
types  		& $[1111]$			& $[112]$			& $[2,2]$		     & $[13]$		         & $[4]$\\
\hline
partitions	&$\{\{1\},\{2\},\{3\},\{4\}\}$	& $\{\{a,b\},\{c\},\{d\}\}$ (6)	& $\{\{a,b\},\{c,d\}\}$ (3)  & $\{\{a\},\{b,c,d\}\}$ (4) & $\{1,2,3,4\}$
\end{tabular}
}
\end{center}
\end{note}

From what has been said it is clear that, for two (Diophantine) partitions $\al,\be$ of $n$, the number $numpart(\al)numpart(\be)$ 
(which is the coefficient of the monomial $\L^\al\V\be$ in the function $F_n$) is the number of (ordered) pairs 
$P_1,P_2$, the $P_i;\ i=1,2$ being (set-theoretical) partitions of $[1..n]$.

\bs
{\bf Bi-partitions}\pointir\\
These couples are representable by bipartite (multi-)graphs between the sets $P_1,\ P_2$, the node $Y_1\in P_1$ and 
$Y_2\in P_2$ being joined by as many arrows as they have elements in common. So, for every couple of partitions of 
the same set, we get an incidence matrix $\M(P_1,P_2)$ (the incidence matrix of the bipartite graph described previously. 
The table below shows the incidence matrices for $P_1=\{\{1\},\{2\},\{3,4\}\}$ and $P_2$ of type $[2,2]$

\bs
\begin{center}
\begin{tabular}{r|| c|c|c}
$P_1\backslash P_2$ & $\{\{1,2\},\{3,4\}\}$	& $\{\{1,3\},\{2,4\}\}$	& $\{\{1,4\},\{2,3\}\}$\\
\hline
$\{\{1\},\{2\},\{3,4\}\}$	&  $\pmatrix{1 & 0 \cr 1 & 0\cr 0 & 2}$ &  $\pmatrix{1 & 0 \cr 0 & 1\cr 1 & 1}$&  $\pmatrix{1 & 0 \cr 0 & 1\cr 1 & 1}$
\end{tabular}
\end{center}

for example, the graph associated to the first matrix is

\bs
\begin{center}
\begin{tabular}{ccc}
$\{1\}$ & --- & \{1,2\}\\
&$\slash$ &\\
$\{2\}$ & 	  & \\
$\{3,4\}$ & $\stackrel{-}{-}$   & $\{3,4\}$
\end{tabular}
\end{center}

\begin{note} It is equivalent also to ``unlabel'' the vertives and label the edges from $1$ to $n$, since every vertex contains 
the set of labels which are connected to it.
\end{note}

\bs
{\bf First oblivion: drawn diagrams}\pointir\\

We can forget some of the information in ``unlabelling'' the parts of the partitions involved, replacing the subsets of the left column 
by white spots ($\circ$) and the others (right column) by black spots ($\bullet$)

\bs
\begin{center}
\begin{tabular}{ccc}
$\circ$ & --- & $\bullet$\\
&$\slash$ &\\
$\circ$ & 	  & \\
$\circ$ & $\stackrel{-}{-}$   & $\bullet$
\end{tabular}
\end{center}

of course, the vertices remain labelled in some way because as far a a choice is made to draw the spots, we can consider 
that they are labelled from top to bottom. This is what we call ``drawn diagrams'' their data is equivalent to the (integer) 
matrices with no zero raw and no zero column. The information contained in such a diagram is not sufficient to recover 
the bi-partition (see above the second and third matrix) they come from but it is to recover the bi-type as the first type is 
given by the sums of the raws and the second type is given by the sums of the columns. In this sense, we can say that 
the coefficient $numpart(\al)numpart(\be)$ ``contains'' all the drawn diagrams of the given bi-type and enumerate them 
with their multiplicities (which is the number of bi-partitions giving the drawn diagram under consideration).

\bs
{\bf Second oblivion: diagrams}\pointir\\

When manipulating the diagrams, one says that two diagrams are ``the same'' if they differ by permutations 
of the white spots and black spots (between themselves). This amounts to consider the orbit of the matrix 
representing the diagram under permutations of the raws and th columns (between themselves). Again 
these items contain the information of the bi-type, so the coefficients of the generic formula \mref{F_m} can 
be seen to be enumerations of diagrams with their multiplicities.
 

\subsection{Restricted Bender diagrams}

\section{Concluding remarks}

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\end{document} 