vector field on $b\Omega$ to a suitable tubular neighborhood of $b\Omega$. The differential condition arising in the description of $\dom\dbar^*$ is most easily expressed if we make the following choices.

Let $\varrho$ be the signed Euclidean distance from $b\Omega$ (negative inside, positive outside). In a suitable tubular neighborhood $U$ of $b\Omega$, this function $\varrho$ is well defined and in $C^\infty(\overline{U})$. We define the vector field $N$ on $U$ by setting $N=\text{grad\,}\varrho$. Then $N$ is the outward unit vector field, and if we set $N=\sum_{j=1}^{2n}\nu_j D_j$, then $N\nu_j =0$ on $U$. \[This last one is in fact the property that at several stages makes our computations easier, and the formulas appearing simplier.\]

\[domain\] *Let $\Omega$ be a smoothly bounded domain in ${{\Bbb C}^n}$. Let $\dbar^*$ be the ${W^s(\Omega)}$-Hilbert space adjoint of $\dbar$. Then $$\dom\dbar^* \cap C_{(0,q+1)}^\infty(\bar\Omega)
=\{ \psi: N^s \bigl(\psi \llcorner\dbar\varrho\bigr)_I 
=0 \text{ on }b\Omega, \text{ for all }I, 
|I|=q \} .$$*

Notice that $N^s$ denotes the $s$-fold composition of $N$ with itself.

Here the contraction of a $(0,q+1)$ form $\psi$ with a $(0,1)$ form $\omega=\sum_{k}\omega_k d\bar z_k$ is defined by the formula $$\psi\llcorner\omega:= \sum_I \sum_{kK}\e{K}{kI}\psi_K \bar\omega_k
d\bar z^I .$$

[ *Suppose that $r$ is a generic defining function for $\Omega$, 