on $\ti \n$. However, to keep the notation as simple as possible we will forget the tilde and keep the $\n$.

Resolution of a Dirichlet Problem
---------------------------------

Consider the generic manifold $\n \times
\R^n \subset \C^{k+m}\times \C^n$.

\[l2\] A basis $\L$ of the CR vector fields $\n \times
\R^n$ consists of vector field $ L_j$ of the form

$$\label{e17}
  L_j={{\partial} \over {\partial \overline z_j}}+\sum_{l=1}^mf_{j,l}
{{\partial} \over {\partial s_l}},$$

where the $  L_j$ can be chosen to verify the following property.

If $s \in \R^{m} \setminus B_{2\epsilon}$ we have $$L_j={{\partial} \over {\partial \overline z_j}}.$$

The proof of the first part of lemma \[l2\] can be found in any textbook on CR geometry. The second part is just a consequence of the construction of $  \n$.

### Construction of the operator $\Delta_{\n\times\R^n}$

\[l3\] There exists $m$ vector fields $R_j$ of the form $$R_j=\sum_{l=1}^m a_{jl}(z,s){{\partial} \over {\partial  s_l}}$$ where the $a_{jl}$ are smooth functions, such that

\(i) $R_{j}(  w'_l)=\delta_{jl}$, $j,l \in \{1,...,m\}$,

\(ii) $[R_j,R_l]=0$,

\(iii) $[  L_j,R_l]=0$,

\(iv) the set $\{  L_1,...,  L_k, \overline{   L_1},...,\overline
{  L_k},R_1,...,R_{m}\}$ span the complex tangent plane to $  \n $ near the origin.

The proof of the lemma is classic. (ii) (iii) and (iv) are a consequence 