mathcal{Z}\Big(\displaystyle\sum_iz_i\Big)=\displaystyle\sum_i\mathcal{Z}(z_i)=\Delta\displaystyle\sum_iw_i\in\mathcal{J}_1\\
& \mathcal{Z}\Big(\displaystyle\sum_iw_i\Big)=\displaystyle\sum_i\mathcal{Z}(w_i)=\displaystyle\sum_i P_i=0\in\mathcal{J}_1 \text{ by the first assertion of Lemma \ref{lem:teccom}}
\end{align*}
With a similar computation, using the other assertions of Lemma \ref{lem:teccom}
it is possible to prove that $\Sigma_2$, $\Sigma_3$ 
and~$\Sigma_4$ are $\chi$-invariant.
\end{proof}

\begin{cor}
{\sl Let $\mathrm{Ch}(X_1,X_2,X_3,X_4)$ be a $4$-chambar
on an open set $\mathcal{U}\subset\mathbb{C}$, with 
$X_j=y_j\frac{\partial}{\partial x}$, $y_j\in\mathcal{O}(\mathcal{U})$, 
$1\leq j\leq 4$.

Suppose that $y_k(x_0)=y_{\ell}(x_0)$ for some $x_0\in\mathcal{U}$ and $k\not=\ell$.
Then $y_k(x)=y_\ell(x)$ for all $x\in\mathcal{U}$.
Moreover, if $k=1$ and $\ell=2$, for instance,
then either the chambar is constant and $2a_1+a_3+a_4=0$ or 
$y_j(x)=a_j\sqrt{\lambda x+\mu}$ with $\lambda\not=0$, 
$a_1=a_2=-\frac{1}{3}$ and $a_3$ and $a_4$ the roots of $3z^2+2z+3=0$.}
\end{cor}

\begin{proof}[{\sl Proof}]
The first assertion is consequence of Theorem \ref{thm:4chamonC2} 
and the other of Theorem \ref{thm:loc3bfcweight}.
\end{proof}

Let us denote by $\mathrm{Ch}(4,1)$ the set of $4$-tuples 
$(X_1,X_2,X_3,X_4)$ of germs at $0\in\mathbb{C}$ of
holomorphic vector fields whose flows satisfy the 
barycentric conditions.

\begin{cor}
{\sl The set $\mathrm{Ch}(4,1)$ is isomorphic to an algebraic
submanifold of $\mathbb{C}^{12}$ whose irreducible 
components have dimension at most six.}
\end{cor}

\begin{proof}[{\sl Proof}]
According to Theorems \ref{thm:4chamonC1} and 
\ref{thm:4chamonC2} 
any $4$-chambar on $\mathbb{C}$ gives origin to a 
trajectory $(y,z,w)\colon(\mathbb{C},0)\to\mathbb{C}^{12}$
tangent to the $\chi$-invariant submanifold 
$\Sigma=\Sigma_1\cap\Sigma_2\cap\Sigma_3$ of 
$\mathbb{C}^{12}$. This defines an embedding of 
$\mathrm{Ch}(4,1)$ on $\Sigma$. 
\end{proof}


\section{Linear chambars}\label{sec:linnil}

\begin{thm}\label{thm:linnil}
{\sl Let $X_1$, $X_2$, $\ldots$, $X_p$ be some 
linear vector fields on 
$\mathbb{R}^n$ 
$($resp. $\mathbb{C}^n)$. 

If they satisfy the barycentric property, then 
they are nilpotent.}
\end{thm}

\begin{proof}[{\sl Proof}]
The flow $\varphi_t^k$ of $X_k$ can be written
\[
\varphi_t^k(x)=(\exp tA_k)(x)
\]
where the $A_k$ belong to $\mathrm{End}(\mathbb{R}^n)$
or $\mathrm{End}(\mathbb{C}^n)$. We identify 
the $A_k$ to some matrices. The barycentric property
is equivalent to 
\[
\displaystyle\sum_{k=1}^p\sum_{\ell=0}^\infty\frac{t^\ell}{\ell!}A_k^\ell=p\mathrm{Id}
\]
that implies $\displaystyle\sum_{k=1}^p A_k^\ell=0$ for 
any $\ell\geq 1$. Let $\lambda_{k,j}$ be the eigenvalues
of $A_k$, $1\leq j\leq n$. We get for all $\ell\geq 1$
\[
0=\mathrm{Tr}\Big(\displaystyle\sum_{k=1}^pA_k^\ell\Big)=\displaystyle\sum_{k=1}^p\sum_{j=1}^n\lambda_{k,j}^n.
\]
As a result all the $\lambda_{k,j}$ are equal to zero. 
\end{proof}

\begin{rem}
The $\varphi_t^k$ are polynomial in $x$ and $t$.
\end{rem}

\begin{rem}
If $p=2$, then the indices of nilpotence are $2$ 
({\it i.e.} $A^2=0$) and we recover the fact that
the trajectories are straight lines. Note also 
that if $X$ is a nilpotent vector field of 
index $2$, then the pair $(X,-X)$ is a $2$-chambar.
\end{rem}

\begin{eg}
Let $X$ be a nilpotent linear vector field of 
order $p$. Let $\sigma=\exp\left(\frac{2\mathbf{i}\pi}{p}\right)$
be a primitive $p$-th root of unity. Then 
the vector fields $X$, $\sigma X$, $\sigma^2X$, 
$\ldots$, $\sigma^{p-1}X$ satisfy the barycentric 
property.
\end{eg}


\begin{rem}
Let $\mathrm{Ch}(X_1,X_2,\ldots,X_p)$ be a linear
$p$-chambar. Denote by $k$ the maximal order of 
nilpotence of the $X_i$'s. Take $\ell<k$ an 
integer. Then 
$\mathrm{Ch}(X_1^\ell,X_2^\ell,\ldots,X_p^\ell)$
is a $q$-chambar for some $q\leq p$. The 
inequality comes from the fact that two 
$X_k^\ell$ can be equal or $X_k^\ell$ can be zero. 
The fact that $q<p$ measures some degeneration and
if $q=p$ for any $\ell<k$ it gives some
condition of transversality.
\end{rem}


\begin{rem}
Let $\mathrm{Ch}(X_1,X_2,\ldots,X_p)$ be a singular
$p$-chambar such that $X_k(0)=0$. Denote by 
$A_i$ the linear part of $X_i$ for $1\leq i\leq p$.

Assume that the $A_i$'s generate a linear $p$-chambar
$\mathrm{Ch}(A_1,A_2,\ldots,A_p)$.

Consider the homothety $h_s\colon x\mapsto sx$, 
$s\in\mathbb{C}^*$ and 
\[
X_k^s=h_{s*}X_k=A_k+s(\ldots)
\]
We construct in this way a family  
$\mathrm{Ch}^s=\mathrm{Ch}(X_1^s,X_2^s,\ldots,X_p^s)$
of $p$-chambars, all conjugate for $s\not=0$, 
and that joins the initial chambar 
$\mathrm{Ch}^1=\mathrm{Ch}(X_1,X_2,\ldots,X_p)$
to the linear chambar 
$\mathrm{Ch}^0=\mathrm{Ch}(A_1,A_2,\ldots,A_p)$.
\end{rem}

\subsection{Linear $p$-chambars in dimension $2$}



\begin{lem}
{\sl Let $B$ be $(2\times 2)$-matrix with complex coefficients.
If $\mathrm{Tr}(B)=0$, then $B$ is the sum of two 
nilpotent matrices.}
\