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\title{\Large\bf Independance of hyperlogarithms\\ 
over function fields via\\
 algebraic combinatorics.}

\author{{\sc M. Deneufch\^atel, G. H. E. Duchamp},\\ 
{\sc Hoang Ngoc Minh}    
\thanks{\rm LIPN - UMR 7030
CNRS - Universit\'e Paris 13
F-93430 Villetaneuse, France}\\
{\rm and} {\sc  Allan I. Solomon}
\thanks{\rm Open University}
}
\date{}

\begin{document}

\maketitle
%\isodayandtime

\tableofcontents

\begin{abstract}
Group-like series ...
\end{abstract}

\section{Introduction}

.../...\\
the method of Weshung consists in a recurrence based on
the total degree. However this method cannot be used with
variable coefficients. Another proof was given in
[Joris and al. 1998] based on monodromy.
Here, we give a general theorem on differential algebra
and show that, at the cost of using variable domains (which
is the realm of germ spaces) and replace the recurrence on total
degree by a recursion on the words (with the graded lexicographic ordering), 
one can encompass the previous results mentionned above and get much larger 
rings of coefficients
 
\section{Non commutative differential equations (abstract setting).}

The ground field $k$ is supposed commutative and of characteristic zero. We suppose given a commutative differential $k$-algebra $(\A,d)$ that is a $k$-algebra (associative and commutative with unit) $\A$ endowed with an element 
$d\in \frak{Der}(\A)$. We will suppose that the ring of constants $ker(d)$ is exactly $k$.     

\ss
An alphabet $X$ being given, one can at once extend the derivation $d$ to a derivation of the algebra $\ncs{\A}{X}$ by
\begin{equation}
	\bd(S)=\sum_{w\in X^*}d(\scal{S}{w})w\ .
\end{equation}

\begin{theorem} Let $(\A,d)$ be a $k$-commutative associative differential algebra with unit ($ch(k)=0$) and $\calC$ be a differential subfield of $\A$ (i.e. $d(\calC)\subset \calC$). 
We suppose that $S\in \ncs{\A}{X}$ is a solution of the differential equation 
\begin{equation}
	\bd(S)=MS\ ;\ \scal{S}{1}=1
\end{equation}
where the multiplier $M=\sum_{x\in X}u_x x\in \ncs{\calC}{X}$ is an homogeneus series (a polynomial in case $X$ is finite) of degree $1$.\\
The following condition are equivalent :\\
i) The family $(\scal{S}{w})_{w\in X^*}$ of coefficients of $S$ is free over $\calC$.\\
ii) The family of coefficients $(\scal{S}{y})_{y\in X\cup \{1_{X^*}\}}$ is free over $\calC$.\\
iii) The family $(u_x)_{x\in X}$ is such that, for $f\in \calC$ and $\al_x\in k$   
\begin{equation}\label{prim_ind}
d(f)=\sum_{x\in X} \al_x u_x \Longrightarrow (\forall x\in X)(\al_x=0)\ .
\end{equation}
iv) The family $(u_x)_{x\in X}$ is free over $k$ and 
\begin{equation}
	d(\calC)\cap span_k\Big((u_x)_{x\in X}\Big)=\{0\}\ .
\end{equation}
\end{theorem}

\Proof (i)$\Longrightarrow$(ii) Obvious.\\
(ii)$\Longrightarrow$(iii)\\
Suppose that the family $(\scal{S}{y})_{y\in X\cup \{1_{X^*}\}}$ (coefficients taken at letters and the emty word) of coefficients of $S$ is free over $\calC$ and let us consider a relation as eq. \mref{prim_ind}
\begin{equation}
d(f)=\sum_{x\in X} \al_x u_x\ .
\end{equation}
We form the polynomial $P=-f1_{X^*}+\sum_{x\in X} \al_x x$. One has $\bd(P)=-d(f)1_{X^*}$ and
\begin{equation}
	d(\scal{S}{P})=\scal{\bd(S)}{P}+\scal{S}{\bd(P)}=\scal{MS}{P}-d(f)\scal{S}{1_{X^*}}=
	(\sum_{x\in X} \al_x u_x)-d(f)=0
\end{equation}
and, then $\scal{S}{P}$ must be a constant, say $\lambda\in k$. For $Q=P-\lambda.1_{X^*}$, we have 
$$
supp(Q)\subset X\cup \{1_{X^*}\}\ \textrm{and}\ \scal{S}{Q}=\scal{S}{P}-\lambda\scal{S}{1_{X^*}}=\scal{S}{P}-\lambda=0
\ .
$$
This implies that $Q=0$ and, as $Q=-(f+\lambda)1_{X^*}+\sum_{x\in X} \al_x x$, one has, in particular, all the $\al_x=0$.\\
(iii)$\Longleftrightarrow$(iv)\\
Obvious, (iv) being a geometric reformulation of (iii).\\   
(iii)$\Longleftrightarrow$(i)\\
Let $\mathcal{K}$ be the kernel of $P\mapsto \scal{S}{P}$ (a linear form 
$\ncp{\calC}{X}\ra \calC$) i.e.  
\begin{equation}
	\mathcal{K}=\{P\in \ncp{\calC}{X}| \scal{S}{P}=0\}\ .
\end{equation}
If $\mathcal{K}=\{0\}$, we are done. Otherwise, let us adop the following strategy.\\ 
First, we order $X$ by some well-ordering $<$ (\cite{B_Set} III.2.1) and $X^*$ by the graded lexicographic ordering 
$\prec$ defined by
\begin{equation}
	u\prec v \Longleftrightarrow |u|<|v|\ \textrm{or}\ (u=pxs_1\ ,\ v=pys_2\ \textrm{and}\ x<y)
\end{equation}
it is easy to check that $\prec$ is also a well-ordering relation. For each nonzero polynomial $P$, we note $lead(P)$ its leading monomial i.e. the greatest element of its support $supp(P)$ (for $\prec$).\\
Now, as $\mathcal{R}=\mathcal{K}-\{0\}$ is not empty, let $w_0$ be the minimmal element of $lead(\mathcal{R})$ and choose a $P\in \mathcal{R}$ such that $lead(P)=w_0$. We write 
\begin{equation}
	P=fw_0+\sum_{u\prec w_0}\scal{P}{u}u\ ;\ f\in \calC-\{0\}\ .
\end{equation}
the polynomial $Q=\frac{1}{f}P$ is also in $\mathcal{R}$ with the same leading monomial, but the leading coefficient is now $1$ and $Q$ reads
\begin{equation}
	Q=w_0+\sum_{u\prec w_0}\scal{Q}{u}u\ .	
\end{equation}

Differentiating $\scal{S}{Q}=0$, one gets
\begin{eqnarray}
	0=\scal{\bd(S)}{Q}+\scal{S}{\bd(Q)}=\scal{MS}{Q}+\scal{S}{\bd(Q)}=\cr
	\scal{S}{M^\dagger Q}+\scal{S}{\bd(Q)}=\scal{S}{M^\dagger Q+\bd(Q)}
\end{eqnarray}
with 
\begin{equation}
M^\dagger Q+\bd(Q)=\sum_{x\in X} u_x (x^\dagger Q)+\sum_{u\prec w_0}d(\scal{Q}{u})u\in \ncp{\calC}{X}	\ .
\end{equation}
It is impossible that $M^\dagger Q+\bd(Q)\in \mathcal{R}$ because it would be of leading monomial strictly less than $w_0$, hence $M^\dagger Q+\bd(Q)=0$. This is equivalent to the recursion  
\begin{equation}
	d(\scal{Q}{u})=-\sum_{x\in X} u_x \scal{Q}{xu}\ ;\ \textrm{for}\ x\in X\ ,\ v\in X^*
\end{equation}
From this last relation, we derive that $\scal{Q}{w}\in k$ for every $w$ of length $deg(Q)$ and,   
because $\scal{S}{1}=1$, one must have $deg(Q)>0$. Then, write $w_0=x_0v$ and compute the coefficient at $v$
\begin{equation}
d(\scal{Q}{v})=-\sum_{x\in X} u_x \scal{Q}{xv}=\sum_{x\in X} \al_x u_x 
\end{equation}
with coefficients $\al_x=-\scal{Q}{xv}\in k$ as $|xv|=\deg(Q)$ for all $x\in X$.
Condition \textbf{PI} implies that all coefficients $\scal{Q}{xu}$ are zero, in particular, as 
$\scal{Q}{x_0u}=1$, we get a contradiction. This proves that $\mathcal{K}=\{0\}$.\\ 
\cqfd   

\section{Series with variable coefficients.}

\subsection{Motivations.}

In this section, we implement an abstract setting which is intended to apply on function spaces. As a motivation, let us illustrate this by an example.\\ 
Let $V$ be a connected and simply connected analytic variety of dimension one (for example, the doubly cut plane $\C-(]-\infty, 0[\cup ]1,+\infty[)$, or the universal covering of $\C-\{0,1\}$), $\H=C^\omega(V;\C)$ be the space of all analytic fonctions on $V$. This space is a differential algebra with the derivative $\frac{d}{dz}$. One extends at once this derivative (then denoted by $\bd$) to $\ncs{\H}{X}$ by 
\begin{equation}\label{ext_d_1}
	\bd(S)=\sum_{w\in X^*} \frac{d}{dz}(\scal{S}{w}) w
\end{equation}
it is easy to check (proof in the general case below) that $\bd$ is a derivation of the algebra $\ncs{\H}{X}$. Differential equations of the type 
\begin{equation}\label{ED_ex}
	\bd(S)=MS
\end{equation}
where $M=\sum_{x\in X}u_x(z)x$ were widely considered in the domains of (\todo{automatique, Drinfel'd, Weshung,  etc...}) and provide, through integrators build by iterated integrals, spaces of special functions. An immediate application of theorem (\todo{??}) below provides the result  that, for any solution of \mref{ED_ex} (with $\scal{S}{1}=1$), the family of functions $(\scal{S}{w})_{w\in X^*}$ is free over the field of rational functions on $V$ . 

\subsection{General setting}

Let $\M$, be a locally finite monoid \cite{Eil} and $\ncs{\H}{\M}$, be the large algebra \cite{B_Alg_III} of $\M$ with coefficients in $\H$). 
Let $\M$ be a locally finite monoid \cite{Eil} and $V$ be a set (where the variable $z$ stands) and $\H\subset C^V$, an algebra of fonctions. Every series $S\in \ncs{\H}{\M}$ can be written  
\begin{equation}
	S:=\sum_{m\in \M}\scal{S}{m}m
\end{equation}
(as the family $(\scal{S}{m}m)_{m\in \M}$ is summable). Thus, one can consider the series 
of $\ncs{\H}{\M}$ as functions on $V$ (with values in $\ncs{\C}{\M}$) and specialize them by 
%
\begin{equation}
	S(z_0):=\sum_{m\in \M}\Big(\scal{S}{m}\Big)\Big|_{z=z_0}m\ .
\end{equation}
%
Moreover, if $d$ is a derivation in $\H$, its extension to $\ncs{\H}{\M}$ ``coefficient by coefficient'' given as 
%
\begin{equation}
	\bd(S):=\sum_{m\in \M}d(\scal{S}{m})m
\end{equation}
%
is a derivation of $\ncs{\H}{\M}$.
%
Let us show, on an example, how the proof of theorem \todo{??} below works on an example.\\
The data are
\begin{enumerate}
\item $X=\{x_0,x_1\}$
\item $V$ is a connected and simply connected subset of $\C$  
\item $\H=C^\omega(V,\C)$ which does not contain $\{0,1\}$, eendowed with the derivation $\frac{d}{dz}$
\item $M=\frac{x_0}{z}+\frac{x_1}{1-z}$
\item \textbf{Statement} : If $S$ is any solution of 
\begin{equation}
	\bd(S)=MS\ ;\ \scal{S}{1}=1
\end{equation}
 then the functions $(\scal{S}{w})_{w\in X^*}$ are linearly independant over the field of rational functions i.e. if 
\begin{equation}
f_i=\frac{p_i}{q_i},\ i=1\cdots N\ ;\ p_i,q_i\in \C[z]
\end{equation}
and $w_i\in X^*$ are such that 
\begin{equation}
	\sum_{i=1}^N f_i(z)\scal{S}{w_i}=0
\end{equation}
on some open (non void) set (which does not contain the opes of the $f_i$) then 
\begin{equation}
	(\forall i\in [1\cdots N])(f_i\equiv 0)\ .
\end{equation}
\item \textbf{An example for the recursion} Order $X$ by $x_0<x_1$ and use $\prec_{glex}$, the graded lexicographic ordering on $X^*$. For each non-trivial relation (\textsc{Rel}) (if there are such) 
\begin{equation}
	\sum_{i=1}^N f_i(z)\scal{S}{w_i}=0
\end{equation}
we consider the leading monomial $lead(\textsc{Rel})=\sup\{w_i|f_i\not\equiv 0\}$\ .
If there were nontrivial relatons, we could take one with the least possible leading monomial and the relation itself.\\ 
\textbf{Assumption 1} Suppose that the set of its monomials be $\{x_0,x_1,x_1x_0,x_0^2x_1\}$ with $lead(\textsc{Rel})=x_0^2x_1$. 
One has 
\begin{equation}\label{rel1}
	f_{x_0}(z)\scal{S}{x_0}+f_{x_1}(z)\scal{S}{x_1}+f_{x_1x_0}(z)\scal{S}{x_1x_0}+f_{x_0^2x_1}(z)\scal{S}{x_0^2x_1}=0\ .
\end{equation}
which is defined on $U$ intersection of the domains $U_0=dom(f_{w_i})$.\\
At the cost of restricting the relation to $U_1=U_0\setminus \O_{f_{x_0^2x_1}}$ one has also 
\begin{equation}\label{rel2}
	g_{x_0}(z)\scal{S}{x_0}+g_{x_1}(z)\scal{S}{x_1}+g_{x_1x_0}(z)\scal{S}{x_1x_0}+\scal{S}{x_0^2x_1}=0
\end{equation}
with $g_i=\frac{f_i}{f_{x_0^2x_1}}$. Differentiating \mref{rel2}, we get 
\begin{eqnarray}\label{diff_rel2}	g'_{x_0}(z)\scal{S}{x_0}+g_{x_0}(z)\scal{S'}{x_0}+g'_{x_1}(z)\scal{S}{x_1}+g_{x_1}(z)\scal{S'}{x_1}+\cr g'_{x_1x_0}(z)\scal{S}{x_1x_0}++g_{x_1x_0}(z)\scal{S'}{x_1x_0}+\scal{S'}{x_0^2x_1}=0\ .
\end{eqnarray}
As $S'=MS$, we have 
\begin{eqnarray}	
\scal{S'}{x_0u}=\scal{MS}{x_0u}=\frac{1}{z}\scal{S}{u}\ \textrm{and}\cr \scal{S'}{x_1u}=\scal{MS}{x_1u}=\frac{1}{1-z}\scal{S}{u}\ .  
\end{eqnarray} 
With this in hand \mref{diff_rel2} becomes 
\begin{eqnarray}\label{diff_rel2}	g'_{x_0}(z)\scal{S}{x_0}+g_{x_0}(z)\frac{1}{z}\scal{S}{1}+g'_{x_1}(z)\scal{S}{x_1}+
g_{x_1}(z)\frac{1}{1-z}\scal{S}{1}+\cr g'_{x_1x_0}(z)\scal{S}{x_1x_0}++g_{x_1x_0}(z)\frac{1}{1-z}\scal{S}{x_0}+\frac{1}{z}\scal{S}{x_0x_1}=0\ .
\end{eqnarray}
which is of rank strictly less than \mref{rel1} and then should be trivial. A contradiction as $\frac{1}{z}$ is zero in no non-void open subset ; ``\textbf{Assumption 1}'' must be false and we are done.\\  
\cqfd
\end{enumerate}
 
\subsection{Non commutative differential equations (abstract setting).}

The ground field $k$ is supposed commutative and of characteristic zero. We suppose given a commutative differential $k$-algebra $(\A,d)$ that is a $k$-algebra (associative and commutative with unit) $\A$ endowed with an element 
$d\in \frak{Der}(\A)$. We will suppose that the ring of constants $ker(d)$ is exactly $k$.     

\ss
An alphabet $X$ being given, one can at once extend the derivation $d$ to the algebra $\ncs{\A}{X}$, as in \mref{ext_d_1} by
\begin{equation}
	\bd(S)=\sum_{w\in X^*}d(\scal{S}{w})w\ .
\end{equation}

\begin{theorem} Let $(\A,d)$ be a $k$-commutative associative differential algebra with unit ($ch(k)=0$) and $\calC$ be a differential subfield of $\A$ (i.e. $d(\calC)\subset \calC$). 
We suppose that $S\in \ncs{\A}{X}$ is a solution of the differential equation 
\begin{equation}
	\bd(S)=MS\ ;\ \scal{S}{1}=1
\end{equation}
where the multiplier $M=\sum_{x\in X}u_x x\in \ncs{\calC}{X}$ is an homogeneus series (a polynomial in case $X$ is finite) of degree $1$.\\
The following condition are equivalent :\\
i) The family $(\scal{S}{w})_{w\in X^*}$ of coefficients of $S$ is free over $\calC$.\\
ii) The family of coefficients $(\scal{S}{y})_{y\in X\cup \{1_{X^*}\}}$ is free over $\calC$.\\
iii) The family $(u_x)_{x\in X}$ is such that, for $f\in \calC$ and $\al_x\in k$   
\begin{equation}\label{prim_ind}
d(f)=\sum_{x\in X} \al_x u_x \Longrightarrow (\forall x\in X)(\al_x=0)\ .
\end{equation}
iv) The family $(u_x)_{x\in X}$ is free over $k$ and 
\begin{equation}
	d(\calC)\cap span_k\Big((u_x)_{x\in X}\Big)=\{0\}\ .
\end{equation}
\end{theorem}

\Proof (i)$\Longrightarrow$(ii) Obvious.\\
(ii)$\Longrightarrow$(iii)\\
Suppose that the family $(\scal{S}{y})_{y\in X\cup \{1_{X^*}\}}$ (coefficients taken at letters and the emty word) of coefficients of $S$ is free over $\calC$ and let us consider a relation as eq. \mref{prim_ind}
\begin{equation}
d(f)=\sum_{x\in X} \al_x u_x\ .
\end{equation}
We form the polynomial $P=-f1_{X^*}+\sum_{x\in X} \al_x x$. One has $\bd(P)=-d(f)1_{X^*}$ and
\begin{equation}
	d(\scal{S}{P})=\scal{\bd(S)}{P}+\scal{S}{\bd(P)}=\scal{MS}{P}-d(f)\scal{S}{1_{X^*}}=
	(\sum_{x\in X} \al_x u_x)-d(f)=0
\end{equation}
and, then $\scal{S}{P}$ must be a constant, say $\lambda\in k$. For $Q=P-\lambda.1_{X^*}$, we have 
$$
supp(Q)\subset X\cup \{1_{X^*}\}\ \textrm{and}\ \scal{S}{Q}=\scal{S}{P}-\lambda\scal{S}{1_{X^*}}=\scal{S}{P}-\lambda=0
\ .
$$
This implies that $Q=0$ and, as $Q=-(f+\lambda)1_{X^*}+\sum_{x\in X} \al_x x$, one has, in particular, all the $\al_x=0$.\\
(iii)$\Longleftrightarrow$(iv)\\
Obvious, (iv) being a geometric reformulation of (iii).\\   
(iii)$\Longleftrightarrow$(i)\\
Let $\mathcal{K}$ be the kernel of $P\mapsto \scal{S}{P}$ (a linear form 
$\ncp{\calC}{X}\ra \calC$) i.e.  
\begin{equation}
	\mathcal{K}=\{P\in \ncp{\calC}{X}| \scal{S}{P}=0\}\ .
\end{equation}
If $\mathcal{K}=\{0\}$, we are done. Otherwise, let us adop the following strategy.\\ 
First, we order $X$ by some well-ordering $<$ (\cite{B_Set} III.2.1) and $X^*$ by the graded lexicographic ordering 
$\prec$ defined by
\begin{equation}
	u\prec v \Longleftrightarrow |u|<|v|\ \textrm{or}\ (u=pxs_1\ ,\ v=pys_2\ \textrm{and}\ x<y)
\end{equation}
it is easy to check that $\prec$ is also a well-ordering relation. For each nonzero polynomial $P$, we note $lead(P)$ its leading monomial i.e. the greatest element of its support $supp(P)$ (for $\prec$).\\
Now, as $\mathcal{R}=\mathcal{K}-\{0\}$ is not empty, let $w_0$ be the minimmal element of $lead(\mathcal{R})$ and choose a $P\in \mathcal{R}$ such that $lead(P)=w_0$. We write 
\begin{equation}
	P=fw_0+\sum_{u\prec w_0}\scal{P}{u}u\ ;\ f\in \calC-\{0\}\ .
\end{equation}
the polynomial $Q=\frac{1}{f}P$ is also in $\mathcal{R}$ with the same leading monomial, but the leading coefficient is now $1$ and $Q$ reads
\begin{equation}
	Q=w_0+\sum_{u\prec w_0}\scal{Q}{u}u\ .	
\end{equation}

Differentiating $\scal{S}{Q}=0$, one gets
\begin{eqnarray}
	0=\scal{\bd(S)}{Q}+\scal{S}{\bd(Q)}=\scal{MS}{Q}+\scal{S}{\bd(Q)}=\cr
	\scal{S}{M^\dagger Q}+\scal{S}{\bd(Q)}=\scal{S}{M^\dagger Q+\bd(Q)}
\end{eqnarray}
with 
\begin{equation}
M^\dagger Q+\bd(Q)=\sum_{x\in X} u_x (x^\dagger Q)+\sum_{u\prec w_0}d(\scal{Q}{u})u\in \ncp{\calC}{X}	\ .
\end{equation}
It is impossible that $M^\dagger Q+\bd(Q)\in \mathcal{R}$ because it would be of leading monomial strictly less than $w_0$, hence $M^\dagger Q+\bd(Q)=0$. This is equivalent to the recursion  
\begin{equation}
	d(\scal{Q}{u})=-\sum_{x\in X} u_x \scal{Q}{xu}\ ;\ \textrm{for}\ x\in X\ ,\ v\in X^*
\end{equation}
From this last relation, we derive that $\scal{Q}{w}\in k$ for every $w$ of length $deg(Q)$ and,   
because $\scal{S}{1}=1$, one must have $deg(Q)>0$. Then, write $w_0=x_0v$ and compute the coefficient at $v$
\begin{equation}
d(\scal{Q}{v})=-\sum_{x\in X} u_x \scal{Q}{xv}=\sum_{x\in X} \al_x u_x 
\end{equation}
with coefficients $\al_x=-\scal{Q}{xv}\in k$ as $|xv|=\deg(Q)$ for all $x\in X$.
Condition \textbf{PI} implies that all coefficients $\scal{Q}{xu}$ are zero, in particular, as 
$\scal{Q}{x_0u}=1$, we get a contradiction. This proves that $\mathcal{K}=\{0\}$.\\ 
\cqfd   

\section{Applications of the main theorem.}

\subsection{Independance of the polylogarithms.}

Let $V$ be a connected and simply connected analytic variety of dimension $1$ (for example, the doubly cut plane $\C-(]-\infty, 0]\cup [1,+\infty[)$, or the universal covering of $\C-\{0,1\}$), $\A=C^\omega(V;\C)$ be the space of analytic fonctions on $V$ endowed with the derivative $d=\frac{d}{dz}$. Let $X=\{x_0,x_1\}$ and $X^*$ be the free monoid on $X$. It is locally finite \cite{Eil} and we note $\ncs{\A}{X}$ its large algebra \cite{B_Alg_III} (with coefficients in $\A$). 



\section{R\'eserve \`a ordonner}

We will use three types of differential equations.\\  
% 
a) {\bf Left-sided equation}
%
\begin{equation}\label{LSE}
\frac{d}{dz} S(z)=M(z)S(z)	
\end{equation}
%
b) {\bf Right-sided equation}
%
\begin{equation}\label{RSE}
\frac{d}{dz} S(z)=S(z)M(z)	
\end{equation}
%
c) {\bf Two-sided equation}
%
\begin{equation}\label{TSE}
\frac{d}{dz} S(z)=M_1(z)S(z)+S(z)M_2(z)	
\end{equation}
%
with $M,M_i\in \ncs{\H_{\geq 1}}{\M}$. One first give the resolution of equations of type \mref{TSE} as their properties specialize, with $M_2=0$ (resp. $M_1=0$) to the type \mref{LSE} (resp. \mref{RSE}).

\begin{theorem}\label{thmA} With the preceding assumptions.\\ 
i) Equation \mref{TSE} has solutions all of the form
\begin{equation}
	S=(H_{z_0}^z)^*S_0
\end{equation}
where $H_{z_0}^z$ is the operator 
\begin{equation}\label{Chen alpha}
G\mapsto \int_{z_0}^z \Big(M_1(s)G(s)+G(s)M_2(s)\Big) ds  
\end{equation}
and $S_0=S(z_0)$ is a constant series.\\
ii) Two solutions which coincide at a point do coincide everywhere.\\ 
iii) Let $\Delta$ be a closable comultiplication with constant coefficients and suppose that $M_1,M_2$ in \mref{TSE} are primitive elements for $\Delta$ i.e. 
$$
\Delta(M_i)=M_i\otimes 1+1\otimes M_i\ ;\ i=1,2\ .
$$ 
Then if $S$, a solution of \mref{TSE}, is group-like at a point of $V$, it is group-like everywhere in $V$.\\
iv) The constant term of $S$ is contant on $V$ (and is that of $S_0$), in particular, if a solution is invertible at a point, it is so everywhere (these solutions will be called {\it regular}).\\ 
v) Let $S$ be a regular solution of an equation of type \mref{LSE} with primitive multiplier. Let $\F$ be a filter on $V$ (neighbourhoods of $0$, of $1$, of infinity etc.) and one supposes that $S$ is asymtotically equivalent to $G$ (w.r.t. $\F$ i.e. $lim_\F(G^{-1}S)=1$). Then $S$ is group-like. 
\end{theorem}

\subsubsection{Proof of theorem \mref{thmA}}

i) The integrator $H=H_{z_0}^z\in\End(\ncs{\H}{\M})$ satisfies  
$$
H^n(\ncs{\H}{\M})\subset \ncs{\H_{\geq n}}{\M}
$$ for all $n\in \N$. This implies that $(H^n)_{n\geq 0}$ is summable for the $\frak{M}$-adic topology of $\End^{filtr}(\ncs{\H_{\geq 1}}{\M})$ given by the ideal $\frak{M}$ of operators $\phi$ such that $\phi\Big(\ncs{\H}{\M}\Big)\subset \ncs{\H_{\geq 1}}{\M}$. Let $H^*=\sum_{n\geq 0}H^n$ be its sum. For $S_0$ a constant series, one has 
\begin{eqnarray}
\frac{d}{dz}(H^*[S_0])=\frac{d}{dz}(I+HH^*)[S_0]=\cr
\frac{d}{dz}(S_0)+\frac{d}{dz}(HH^*)[S_0]=M_1H^*[S_0]+H^*[S_0]M_2
\end{eqnarray}
Conversely, if $S$ is a solution of \mref{LSE}, then $S_0=(I-H)[S]$ is a constant series 
as 
$$
\frac{d}{dz}(S_0)=\frac{d}{dz}(I-H)[S]=\frac{d}{dz}S-\frac{d}{dz}(H)[S]=
M_1S+SM_2-(M_1S+SM_2)=0\ .
$$ 
Moreover $S=(I-H)^{-1}[S_0]=H^*[S_0]$ and then $S_0=S(\al)$.\\
ii) If two solutions $S_1,S_2$ coincide at $z_1\in V$, one can construct the operator $H$ with formula \mref{Chen alpha} and $\al=z_1$. One has then $S_1=H^*(S_1(z_1))=H^*(S_2(z_1))=S_2$.\\
iii) A comultiplication with constant coefficients permute with the derivation operator hence, 
if $S$ is a solution of \mref{TSE}, one has 
\begin{eqnarray}
\frac{d}{dz}(\Delta(S))=\Delta(\frac{d}{dz}(S))=\Delta(M_1S+SM_2)=\cr
\Delta(M_1)\Delta(S)+\Delta(S)\Delta(M_2)
\end{eqnarray}
this proves that $\Delta(S)$ satifies a two sided differential equation. 
On the other hand, $S\otimes S$ satisfies
\begin{eqnarray}
\frac{d}{dz}(S\otimes S)=\frac{d}{dz}(S)\otimes S+S\otimes \frac{d}{dz}(S)=\cr
(M_1S+SM_2)\otimes S+S\otimes (M_1S+SM_2)=\cr
(M_1\otimes 1+1\otimes M_1)(S\otimes S)+(S\otimes S)(M_1\otimes 1+1\otimes M_1)
\end{eqnarray}
which proves that, if $M_1$ and $M_2$ are primitive, $\Delta(S)$ and $S\otimes S$ 
satisfy the same differential equation. By virtue of (ii), if $S$ is group-like at a point, it is so everywhere. In particular (and this will be used in (v)), if $S$ is an invertible solution of an equation of type \mref{TSE}, with $M_1$ and $M_2$ primitive, then, for $z_0\in V$, $S(z)S(z_0)^{-1}$ is group-like. 

iv) Indeed 
\begin{eqnarray}
\scal{S}{1_{X^*}}=\scal{H^*[S_0]}{1_{X^*}}=\scal{(I+HH^*)[S_0]}{1_{X^*}}=\cr
\scal{S_0}{1_{X^*}}+\scal{HH^*[S_0]}{1_{X^*}}=\scal{S_0}{1_{X^*}}\ .
\end{eqnarray}
%
v) For $z_0\in V$, one defines
$$
R(z,z_0)=S(z)S(z_0)^{-1}G(z_0)G(z)^{-1}=S(z)\Big(G(z_0)^{-1}S(z_0)\Big)^{-1}G(z)^{-1}\ .
$$
$R=R(z,z_0)$ is the product of two group-like series ($S(z)S(z_0)^{-1}$ and $G(z_0)G(z)^{-1}$). Thus, $z$ being fixed, one has 
%\begin{eqnarray}\label{limit1}
%lim_{z_0:\F} \Delta(R(z,z_0))&=_{(1)}& \Delta(lim_{z_0:\F}(R(z,z_0))=\cr
%lim_{z_0:\F} R(z,z_0)\otimes R(z,z_0)&=_{(2)}& \Delta(S(z)G(z)^{-1})=\cr
%(S(z)G(z)^{-1})\otimes (S(z)G(z)^{-1})&=_{(3)}& \Delta(S(z))\Delta(G(z)^{-1})
%\end{eqnarray}
\begin{eqnarray}\label{limit1}
\Big(S(z)\otimes S(z)\Big)\Big(G(z)^{-1}\otimes G(z)^{-1}\Big)=
(S(z)G(z)^{-1})\otimes (S(z)G(z)^{-1})=\cr
lim_{z_0:\F} R(z,z_0)\otimes R(z,z_0)=
lim_{z_0:\F} \Delta(R(z,z_0))=\cr
\Delta(lim_{z_0:\F}(R(z,z_0))=\Delta(S(z)G(z)^{-1})=\Delta(S(z))\Delta(G(z)^{-1})=\cr
\Delta(S(z))\Big(G(z)^{-1}\otimes G(z)^{-1}\Big)
\end{eqnarray}
%
and, finally $\Delta(S)=S\otimes S$.\cqfd

\begin{remark} The proof of the theorem provides an integrator 
$$
H(G)=\sum_{w\in\M}\Big(\sum_{uv=w}\int_{z_0}^z\Big(
\scal{M_1}{u}(s)\scal{G}{v}(s)+\scal{G}{u}(s)\scal{M_2}{v}(s)\Big)ds\Big)w
$$
but any similar operator $H$ such that $\frac{d}{dz}(H(G))=M_1G+GM_2$ would do. In particular, one can construct operators with varied lower integration bounds. For example, the operator 
\begin{equation}\label{pieds_variables}
H(G)=\sum_{w\in\M}\Big(\sum_{uv=w}\Big(\int_{a(u)}^z
\scal{M_1}{u}(s)\scal{G}{v}(s)+\int_{b(v)}^z\scal{G}{u}(s)\scal{M_2}{v}(s)\Big)ds\Big)w
\end{equation}
is fairly admissible. We will see in Paragraph \mref{LPH} an application of such a principle.
\end{remark}

\section{Coordinates of group-like elements.}

\subsection{Through the looking glass: passing from right to left.}\label{fonctions_S}

Let $S\in\ncs{\H}{X}$, we call $\F(S)$ the $\C$-vector space generated by the coefficients of $S$, one has 
\begin{equation}
	\F(S)=\{\scal{S}{P}\}_{P\in \ncp{\C}{X}}\ .
\end{equation}
We will use the following increasing filtrations 
\begin{equation}\label{fine_filtr_coeff}
	\F_{\leq \al}(S)=\{\scal{S}{P}\}_{P\in \ncp{\C_{\leq \al}}{X}}\ .
\end{equation}
or
\begin{equation}\label{filtr_coeff}
	\F_{\leq n}(S)=\{\scal{S}{P}\}_{P\in \ncp{\C_{\leq n}}{X}}\ .
\end{equation}

\begin{proposition}\label{kernull} We have the following properties :\\ 
i) If $T\in\ncs{\C}{X}$ then $\F(ST)\subset \F(S)$ and one has equality if $T$ is invertible.\\
ii) If $S$ is group-like, then $\F(S)$ is a unital sub-algebra of $\H$, which is filtered w.r.t. \mref{fine_filtr_coeff} and \mref{filtr_coeff} i.e.
\begin{equation}
\F_{\leq \al}(S)\F_{\leq \be}(S)\subset \F_{\leq \al+\be}(S)
\end{equation}
\end{proposition}

\Proof (i) The space $\F(ST)$ is spanned by the 
$$
\scal{ST}{w}=\sum_{uv=w}\scal{S}{u}\scal{T}{v}\in \F(S)
$$
if $T$ is invertible one has $\F(S)=\F(STT^{-1})\subset \F(ST)$ which proves the equality.\\
ii) If $S$ is group-like, one has 
\begin{equation}
	\scal{S}{u}\scal{S}{v}=\scal{S\otimes S}{u\otimes v}=\scal{\Delta(S)}{u\otimes v}=\scal{S}{u\shuffle v}
\end{equation}

In the case when all functions $\scal{S}{w}$ are $\C$-linearly independant, one has a correspondence between the Differential Galois group of a differential equation of type \mref{LSE} (acting on the right) and the group of automorphisms of $\F(S)$ compatible with the preceding filtration (they turn out to be unipotent). 

\begin{proposition}\label{ind_lin}
Let $S$ be a group-like series. The following conditions are equivalent:\\
i) For every $x\in X$, $ker_\C(S)\subset ker_\C(Sx)$.\\
ii) For every $x\in X$, there is a derivation $\delta_x\in \frak{Der}(\F(S))$ such that 
\begin{equation}
	\delta_x(S)=Sx
\end{equation}
iii) For every $x\in X$, there is a one-paramater group of automorphisms $\phi_{x}^t\in Aut(\F(S));\ t\in \R$ such that 
\begin{equation}
	\phi_{x}^t(S)=Se^{tx}
\end{equation} 
iv) For every $C\in \ls{C}{X}$, there is $\delta\in \frak{Der}(\F(S))$ such that 
\begin{equation}
	\delta(S)=SC
\end{equation} 
v) For every $C\in \ls{C}{X}$, there is $\phi\in Aut(\F(S))$ such that 
\begin{equation}
	\phi(S)=Se^C
\end{equation} 
vi) The functions $(\scal{S}{w})_{w\in X^*}$ are $\C$-linearly independant.
\end{proposition}

\Proof $i)\Longrightarrow ii)$ From the inclusion, we derive that, for all $x\in X$ there exists a $\C$-linear mapping $\phi\in \End(\F(S))$ such that for all $w\in \M$,  $\phi(\scal{S}{w})=\scal{Sx}{w}$. It must be a derivation of $\F(S)$ as 
\begin{eqnarray}
\phi(\scal{S}{u}\scal{S}{v})=\phi(\scal{S}{u\shuffle v})=\scal{Sx}{u\shuffle v}=
\scal{S}{(u\shuffle v)x^{-1}}=\cr
\scal{S}{(ux^{-1}\shuffle v)+(u\shuffle vx^{-1})}=\scal{S}{(ux^{-1}\shuffle v)}\scal{S}{(u\shuffle vx^{-1})}=\cr 
\scal{Sx}{u}\scal{S}{v}+\scal{S}{u}\scal{Sx}{v}=
\phi(\scal{S}{u})\scal{S}{v}+\scal{S}{u}\phi(\scal{S}{v})
\end{eqnarray}
from the fact that $(\scal{S}{w})_{w\in X^*}$ spans $\F(S)$.\\
$ii)\Longrightarrow iv)$ As $(\scal{S}{w})_{w\in X^*}$ spans $\F(S)$, the derivation $\phi$ is uniquely defined. Let us note it $\delta_x$ and notice that, doing so, we have constructed a mapping $\Phi : X\ra \frak{Der}(\F(S))$ (which is Lie algebra. Therefore, there is a unique extension of this mapping as a morphism $\lp{\C}{X}\ra \frak{Der}(\F(S))$. This correspondence, which we will note $P\ra \delta(P)$ is (uniquely) recursively defined by 
\begin{equation}
	\delta(x)=\delta_x\ ;\ \delta([P,Q])=[\delta(P),\delta(Q)]\ .
\end{equation}
For $C=\sum_{n\geq 0}C_n\in \ls{\C}{X}$ with $C_n\in \lp{\C}{X}_{n}$, we remark that the sequence $\scal{S\sum_{0\leq n\leq N}C_n}{w}$ is stable (for large $N$). Set 
$\delta_{\leq N}:=\delta(\sum_{0\leq n\leq N}C_n)$. We see that $\delta_{\leq N}$ is stable (for large $N$) on every $\F_\al$ and we note $\delta(C)$ its limit. It is clear that this limit is a derivation and that it corresponds to $C$.\\ 
$iv)\Longrightarrow v)$ For evey $C=\sum_{n\geq 0}C_n\in \ls{\C}{X}$, the exponential $e^C$ defines a mapping $\phi\in \End(\F(S))$ as indeed $e^{\delta_{\leq N}}$ is stationnary. It is easily checked that this mapping is an automorphism of algebra of $\F(S)$.\\
$v)\Longrightarrow iii)$ For $C_i\in \ls{\C}{X};\i=1,2$ which commute we have 
\begin{equation}
Se^{C_1}e^{C_2}=\phi_{C_1}(S)e^{C_2}=\phi_{C_1}(Se^{C_2})=\phi_{C_1}\phi_{C_2}(S)
\end{equation}
this proves the existence, for a $C\in \ls{\C}{X}$ of a one-parameter (rational) group $\phi_C^{t}$ in $\Aut(\F(S))$ such that $Se^{tC}=\phi_C^{t}(S)$. This one-parameter (rational) group can be extended to $\R$ as continuity is easily checked by taking the scalar products 
$\scal{\phi_C^{t}(S)}{w}=\scal{Se^{tC}}{w}$ and it suffices to specialize the result to $C=x$.\\
$iii)\Longrightarrow ii)$ By stationary limits one has 
\begin{equation}
	\scal{Sx}{w}=\lim_{t\ra 0}\frac{1}{t}(\scal{Se^{tx}}{w}-\scal{S}{w})=
	\lim_{t\ra 0}\frac{1}{t}(\scal{\phi_x^{t}(S)}{w}-\scal{S}{w})
\end{equation}
$v)\Longrightarrow i)$ Let $x\in X, t\in \R$, we take $C=tx$ and $\phi_t\in \Aut(\F(S))$  s.t. $\phi_t(S)=Se^{tx}$. It there is $P\in \ncp{\C}{X}$ such that $\scal{S}{P}=0$ one has
\begin{equation}
0=\scal{S}{P}=\phi_t(\scal{S}{P})=\scal{\phi_t(S)}{P}=\scal{Se^{tx}}{P}=\sum_{n=0}^{\deg(P)}\frac{t^n}{n!}\scal{Sx^n}{P}
\end{equation}
and then, for all $z\in V$, the polynomial
\begin{equation}
\sum_{n=0}^{\deg(P)}\frac{t^n}{n!}\scal{S(z)x^n}{P}
\end{equation}
is identically zero over $\R$ hence so are all of its coeefficients in particular $\scal{S(z)x}{P}$ for all $z\in V$. This proves the claim.\\  
%
$i)\Longrightarrow vi)$ Let $P\in ker_\C(S)$ if $P\not=0$ take it of minimal degree with this property. For all $x\in X$, one has $P\in ker_\C(Sx)$ which means $\scal{Sx}{P}=0$ and then $Px^\dagger=0$ as $\deg(Px^\dagger)=\deg(P)-1$. The reconstruction lemma implies that
\begin{equation}
	P=\scal{P}{1}+\sum_{x\in X} (Px^\dagger)x=\scal{P}{1}
\end{equation}
Then, one has $0=\scal{S}{P}=\scal{S}{1}\scal{P}{1}=\scal{P}{1}$ which shows that $ker_\C(S)=\{0\}$. This is equivalent to the statement (vi). \\
$vi)\Longrightarrow i)$ Is obvious as $ker_\C(S)=\{0\}$.\\
\cqfd

It is possible to enlarge somehow the range of proposition \mref{kernull} to coefficients that are analytic functions $f:\ dom(f)\ra\C$. 

\begin{definition} We call here \textrm{differential field of germs} w.r.t. a filter basis $\B$ of open connected subsets of $V$, a map $\calC$ defined on $\B$ such that for every $U\in \B$, $\calC[U]$ is a subring of $C^\omega(U,\C)$ and 

%\begin{definition} We here call \textrm{function field} a set $\calC$ of analytic functions\\ 
%$f:\ dom(f)\ra\C$ such that 
\begin{enumerate}
	\item $\calC$ is compatible with restrictions i.e. if $U,V\in\B$ and  $V\subset U$, one has 
	$$
	res_{VU}(\calC[U])\subset \calC[V]
	$$ 
	\item if $f\in \calC[U]\setminus \{0\}$ then there exists $V\in\B$ s.t. 
	$V\subset U-\mathcal{O}_f$ and $f^{-1}$ (defined on $V$) is in $\calC[V]$\ .
\end{enumerate}

For any $U\in \B$, we note $\calC[U]$ the ring of functions in $\calC$ defined on $U$ and restricted to this set.
\end{definition}


There are important cases when the conditions \mref{kernull} are satified as shows the following theorem.

\begin{theorem}\label{lin_indep} Let $V$ be a simply connected non-void open subset of $\C-\{a_0,\cdots a_n\}$ ($\{a_0,\cdots a_n\}$ are distinct points), $M=\sum_{i=0}^n \frac{\lambda_i x_i}{z-a_i}$ be a multiplier on $X=\{x_0,\cdots x_n\}$ with all $\lambda_i\not=0$ and $S$ be any regular solution of
\begin{equation}
	\frac{d}{dz}S=MS\ .
\end{equation}
Then, let $\calC$ be a differential field of functions defined on $V$ which do not contain linear combinations of logarithms on any domain but which contains $z$ and the constants (as, for example the rational functions).\\
If $U$ is a non-void domain of $\calC$ and $P\in \ncp{\calC[U]}{X}$, one has 
\begin{equation}
	\scal{S}{P}=0\Longrightarrow P=0
\end{equation}
\end{theorem}

\Proof Let $U \in\B$. For every non-zero $Q\in \ncp{\calC[U]}{X}$, we note $lead(Q)$ the greatest word in the support of $Q$ for the graded lexicographic ordering $\prec$ (we have endowed $X$ with any linear ordering) and call $Q$ monic if the leading coefficient $\scal{Q}{lead(Q)}$ is $1$. A monic polynomial then reads
\begin{equation}
	Q=w+\sum_{u\prec w}\scal{Q}{u}u\ .
\end{equation}
Suppose now that it is possible to find $U$ and $P\in\ncp{\calC[U]}{X}$ (not necessarily monic) such that $\scal{S}{P}=0$, we choose $P$ with $lead(P)$ minimal for $\prec$.

\ss 
Then 
\begin{equation}
	P=f(z)w+\sum_{u\prec w}\scal{P}{u}u
\end{equation}
with $f\not\equiv 0$. Thus $U_1=U\setminus\mathcal{O}_f\in \B$
and $Q=\frac{1}{f(z)}P\in \ncp{\calC[U_1]}{X}$ is monic and satisfies 
\begin{equation}\label{linear_rel}
\scal{S}{Q}=0\ .	
\end{equation}
Differentiating eq. \mref{linear_rel}, we get 
\begin{equation}
0=\scal{S'}{Q}+\scal{S}{Q'}=\scal{MS}{Q}+\scal{S}{Q'}=\scal{S}{Q'+M^\dagger Q}\ .	
\end{equation}
Remark that one has 
\begin{equation}
	Q'+M^\dagger Q\in \ncp{\calC[U_1]}{X}
\end{equation}
If $Q'+M^\dagger Q\not= 0$, one has $lead(Q'+M^\dagger Q)\prec lead(Q)$ and this is not 
possible because of the minimality hypothesis of $lead(Q)=lead(P)$. Hence, one must have 
$R=Q'+M^\dagger Q= 0$. With $|w|=n$, write now
\begin{equation}
	Q=Q_n+\sum_{|u|< n}\scal{Q}{u}u\ .
\end{equation}
where $Q_n=\sum_{|u|=n}\scal{Q}{u}u$ is the dominant homogeneous component of $Q$. 
For every $|u|=n$ we have 
\begin{equation}
	(\scal{Q}{u})'=-\scal{M^\dagger Q}{u}=-\scal{Q}{Mu}=0
\end{equation}
thus all the coefficients of $Q_n$ are constant.

\ss
If $n=0$, $Q\not=0$ is constant which is impossible by eq. \mref{linear_rel} and because $S$ is regular. If $n>0$, for any word $|v|=n-1$, we have 
\begin{equation}
	(\scal{Q}{v})'=-\scal{M^\dagger Q}{v}=-\scal{Q}{Mv}=
	-\sum_{i=0}^n \frac{\lambda_i}{z-a_i} \scal{Q}{x_iv}=
	-\sum_{i=0}^n \frac{\lambda_i}{z-a_i} \scal{Q_n}{x_iv}
\end{equation}
Because all $x_iv$ are of length $n$.\\ 
Then 
\begin{equation}
	\scal{Q}{v}=-\sum_{i=0}^n \scal{Q_n}{x_iv} \int_{\al}^z \frac{\lambda_i}{s-a_i} ds + const
\end{equation}
But all the functions $\int_{\al}^z \frac{\lambda_i}{s-a_i} ds$ are linearly independant over $\C$ and not all the scalars $\scal{Q_n}{x_iv}$ are zero (write $w=x_kv$ and choose $v$ such). This contradicts the fact that $Q\in \ncp{\calC[U_1]}{X}$ as $\calC$ does contain no linear combination of logarithms. 
\cqfd

\begin{remark} i) If a series satifies the equivalent conditions of the theorem 
\mref{ind_lin}, then every $Se^C$ so does.\\
ii) Series as the one of polylogarithms and all the exponential solutions of equation 
\begin{equation}
	\frac{d}{dz}(S)=(\frac{x_0}{z}+\frac{x_1}{1-z})S
\end{equation}
satisfy conditions of the theorem \mref{ind_lin} as shows theorem \mref{lin_indep}.\\ 
iii) One could ask oneself what happens when these conditions are not satisfied. In fact the set of Lie series $C\in \ls{\C}{\X}$ such that it exists a $\phi\in\End(\F(S))$ (then a derivation) s.t. $SC=\phi(S)$ is a closed Lie subalgebra of $\ls{\C}{\X}$ which we will note $Lie_S$. For example 
\begin{itemize}
	\item for $X=\{x_0,x_1\}$ and $S=e^{zx_0}$ one has $x_0\in Lie_S\ ;\ x_1\notin Lie_S$ 
	\item for $X=\{x_0,x_1\}$ and $S=e^{z(x_0+x_1)}$, one has $x_0,x_1\notin Lie_S$ but $(x_0+x_1)\in Lie_S$. 
\end{itemize}
\end{remark}
 
\subsection{Polylogarithms and related functions}\label{LPH}

Here $X$ is still the finite alphabet $\{x_0,x_1\}$ equipped with the order $x_0<x_1$
and let $\calC$ be the ring $\C[z,z^{-1},(1-z)^{-1}]$.

The iterated integral over $\omega_0,\omega_1$ associated to
$w=x_{i_1}\cdots x_{i_{k}}$ over $X$ and along the integration path
$z_0\path z$ is the following multiple integral defined by
\begin{eqnarray}\label{intint}
\int\limits_{z_0\path z}\omega_{i_1}\cdots\omega_{i_k}=
\int_{z_0}^{z}\omega_{i_1}(t_1)\int_{z_0}^{t_1}\omega_{i_2}(t_2)\ldots
\int_{z_0}^{t_{r-2}}\omega_{i_r}(t_{r-1})\int_{z_0}^{t_{r-1}}\omega_{i_r}(t_{r}),
\end{eqnarray}
where $t_1\cdots t_{r-1}$ is a subdivision of the path $z_0\path z$.
In a shortened notation, we denote this integral by
$\alpha_{z_0}^z(w)$ and\footnote{Here,
$1_{X^*}$ stands for the empty word over $X$.} $\alpha_{z_0}^z(1_{X^*})=1$.
One can check that the polylogarithm $\Li_{s_1,\ldots,s_r}$
is also the value of the iterated integral over
$\omega_0,\omega_1$ and along the integration path $0\path z$ \cite{FPSAC95,FPSAC96}~:
\begin{eqnarray}
\Li_w(z)=\alpha_0^z(x_0^{s_1-1}x_1\ldots x_0^{s_r-1}x_1).
\end{eqnarray}
The definition of polylogarithms is extended over the words $w\in X^*$ by putting
$\Li_{x_0}(z):=\log z$.
The $\{\Li_w\}_{w\in X^*}$ are $\calC$-linearly independent \cite{FPSAC98,SLC43}. In order to, define $L=\sum_{w\in X^*}\Li_w w$, one also can use an integrator with variable lower integration bounds as one described by \mref{pieds_variables} with $M_2=0$, $a(u)=1$ for $u\in x_0^*$ and $a(u)=0$ for $u\in X^*x_1X^*$.\\
Indeed, $L$ is group-like but, to show this one cannot use Thm \ref{thmA} (iii) because the lower bounds of the integrals are different. So one first shows that 

\begin{eqnarray}\label{boundarycondition}
lim_{z\ra 0}\exp(-x_0\log z)\mathrm{L}(z)=lim_{z\ra 0}\mathrm{L}(z)\exp(-x_0\log z)=1
\end{eqnarray}
which can be done as follows. One first remarks that, in case $w$ contains at least ons $x_1$ (i.e. $|w|_{x_1}\geq 1$) and for every $k$
%
\begin{equation}
lim_{z\ra 0}\log(z)^k\scal{\mathrm{L(z)}}{w}=0
\end{equation}
then, setting $\mathrm{L^+(z)}=\sum_{|w|_{x_1}\geq 1}\scal{\mathrm{L(z)}}{w}w$, one has 
\begin{equation}
lim_{z\ra 0}\exp(-x_0\log z)\mathrm{L^+(z)}=\mathrm{L^+(z)}\exp(-x_0\log z)=0
\end{equation}
and as 
\begin{equation}
L(z)=\mathrm{L^+(z)}+\sum_{w\in (x_0)^*}\scal{\mathrm{L(z)}}{w}w=
\mathrm{L^+(z)}+\exp(x_0\log z)
\end{equation}
the result follows. 

\ss 
The following functions
\begin{eqnarray}
\forall w\in X^*,&& \fP_w(z)=(1-z)^{-1}\Li_w(z),
\end{eqnarray}
are also $\calC$-linearly independent, as $\calC$ is an integral domain, by the following lemma easy to check
\begin{lemma} Let $\calA$ be an integral domain and $M$ an $\calA$-module. If $(x_i)_{i\in I}$ is a linearly independant family and $b\not= 0$ in $\calA$, then $(bx_i)_{i\in I}$ is linearly independant. 
\end{lemma}
Since, for any $w\in Y^*,\fP_w$ is the ordinary generating function of
the sequence $\{\H_w(N)\}_{N\ge0}$~:
\begin{eqnarray}
\fP_w(z)=\sum_{N\ge0}\H_w(N)\;z^N
\end{eqnarray}
then, as a consequence of the classical isomorphism between convergent Taylor series
and their associated sums, the harmonic sums $\{\H_w\}_{w\in Y^*}$ are 
$\C$-linearly independent. Firstly, $\ker\fP=\{0\}$ and $\ker\H=\{0\}$,
and secondly, $\fP$ is a morphism transporting the stuffle to the Hadamard product~:
\begin{eqnarray}
\fP_u(z)\odot\fP_v(z)
=\sum_{N\ge0}\H_u(N)\H_v(N)z^N
=\sum_{N\ge0}\H_{u\ministuffle v}(N)z^N
=\fP_{u\ministuffle v}(z).
\end{eqnarray}
%
\section{Conclusion}
%
To sum up what has been done in this paper? we can state that the deformed algebra $\LDIAG(q_c,q_s)$, which originates from a special quantum field theory \cite{BBM}, is free and its law can be constructed from very general procedures: it is a shifted twisted law. Before shifting, one can observe that the law is, in fact, dual to a comultiplication on a free algebra. This comultiplication is a perturbation, with $q_s$ (the superposition parameter) of the shuffle comultiplication on this free algebra. The parameter $q_s$ is obtained by addition of a perturbating factor which is just dual to a (diagonally) deformed law of a semigroup whereas the crossing parameter $q_c$ is obtained by extending to the tensor structure (i.e. to words) a colour factor of an algebra.


\newpage

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\end{document}
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