Complex Analysis in Several Variables, MTH 3108
Home Assignement 2. Solutions of selected problems.
Professor M.Shubin
Fall 1997


2. Let $\Omega\subset \hbox{{\bbb C}}^n$ be open and connected, $u\in A(\Omega)$.Prove that $\Omega\setminus u^{-1}(0)$ is connected.


Solution. Note first that if n=1 then the set u-1(0) is either discrete or coincides with $\Omega$. In particular, the desired statement is true for n=1. It follows that it is also true for any connected component of the intersection of arbitrary domain $\Omega\subset \hbox{{\bbb C}}^n$with any complex line.


Now let us take arbitrary points $z,w\in \Omega\setminus u^{-1}(0)$, and connect them by a continuous piecewise linear path $\gamma:[0,1]\to\Omega$, $\gamma(0)=z$, $\gamma(1)=w$.Let the values

\begin{displaymath} 0=t_0<t_1<\dots <t_N=1\end{displaymath}

be chosen so that $\gamma$ is linear on [tj,tj+1] for all $j=0,\dots,N-1$. Denote $z_j=\gamma(t_j)$, $j=0,\dots,N$, so z0=z, zN=w, and the path $\gamma$consists of the straight line segments [zj,zj+1] connecting zj and zj+1.


We can obviously assume that $u\not\equiv 0$ which implies that u-1(0) is nowhere dense. Therefore slightly perturbing the points $z_j,\ j=1,\dots, N-1,$ we can replace $\gamma$ by a path with the same properties and with additional requirement $z_j\not\in u^{-1}(0)$ for all j.


Now denote by Lj the complex line passing through zj and zj+1, i.e.

\begin{displaymath} L_j=\{\tau z_j+(1-\tau) z_{j+1}\vert\;\tau\in\hbox{{\bbb C}}\}\;.\end{displaymath}

The intersection $\Omega\cap L_j$ contains the segment [zj,zj+1]. Denote by Gj the connected component of zj in $\Omega\cap L_j$. Clearly $u\vert _{G_j}\not\equiv 0$, so we can connect zj and zj+1 by a path $\rho_j$ in $G_j\setminus u^{-1}(0)$.Then the union of all $\rho_j$, $j=0,\dots,N-1$, is a path connecting z and w in $\Omega\setminus u^{-1}(0)$.$\ecarre$


4. Let $\omega\in \Lambda^{p,q+1}(D)$ where D is an open polydisc in $\hbox{{\bbb C}}^n$, and $\bar{\partial}\omega=0$. Prove that there exists $\alpha\in\Lambda^{p,q}(D)$ such that $\bar{\partial} \alpha=\omega$ in D.


Solution. There are 2 cases here which should be considered separately: q=0 and q>0. The case q=0 is distinguished by the fact that the forms $\alpha\in \Lambda^{p,0}$with $\bar{\partial}\alpha=0$ are simply all forms of the type (p,0) which have holomorphic coefficients. So the proof in this case is a simple modification of the corresponding argument from the case n=1, which takes into account the fact that any holomorphic function on a polydisc can be approximated by polynomials (in z), e.g. by its Taylor polynomials.


In the case q>0 the coefficients of the form $\alpha\in\Lambda^{p,q}$ with $\bar{\partial}\alpha=0$ are not necessarily holomorphic. For example, consider the simplest case n=2 and take $\alpha=\alpha_1d\bar{z}_1+\alpha_2d\bar{z}_2\in\Lambda^{0,1}(\hbox{{\bbb C}}^2)$. The condition $\bar{\partial}\alpha=0$means that

\begin{displaymath} {\partial\alpha_1\over\partial\bar{z}_2}={\partial\alpha_2\over\partial\bar{z}_1}\;,\end{displaymath}

so for any $f\in C^\infty(\hbox{{\bbb C}}^2)$ we can take

\begin{displaymath} \alpha={\partial f\over \partial\bar{z}_1}d\bar{z}_1+{\partial f\over \partial\bar{z}_2}d\bar{z}_2\;.\end{displaymath}

The coefficients need not to be holomorphic (take e.g. $f(z,\bar{z})=\bar{z}_1^2$).


Now let us proceed to the proof of the statement in case q>0. Let $D_1,D_2,\dots$be a sequence of polydiscs such that $\overline{D_j}$ is a compact subset in Dj+1 and the union of all Dj is D. By the Dolbeault lemma we can construct $\alpha_1\in\Lambda^{p,q}(\hbox{{\bbb C}}^n)$ such that $\bar{\partial}\alpha_1=\omega$ in a neighbourhood of $\overline{D_1}$. Let us try to construct a form $\alpha_2\in\Lambda^{p,q}(\hbox{{\bbb C}}^n)$so that $\bar{\partial}\alpha_2=\omega$ in a neighbourhood of $\overline{D_2}$ and $\alpha_2=\alpha_1$in a neighbourhood of $\overline{D_1}$. Then repeating this procedure we will obtain the desired form $\alpha$ as the limit of the forms $\alpha_j$ as $j\to\infty$ in $\Lambda^{p,q}(D)$(it will stabilize on any compact subset of D).


First use the Dolbeault lemma to find a form $\beta\in\Lambda^{p,q}(\hbox{{\bbb C}}^n)$ such that $\bar{\partial}\beta=\omega$ in a neighbourhood of $\overline{D_2}$. Note that $\bar{\partial}(\beta-\alpha_1)=0$in a neighbourhood of $\overline{D_1}$. Therefore using again the Dolbeault lemma we can find a form $\gamma\in\Lambda^{p,q-1}(\hbox{{\bbb C}}^n)$ such that $\bar{\partial}\gamma=\beta-\alpha_1$or $\alpha_1=\beta-\bar{\partial}\gamma$ in a neighbourhood of $\overline{D_1}$.Now we can take $\alpha_2=\beta-\bar{\partial}\gamma$ in $\hbox{{\bbb C}}^n$. Obviously $\alpha_2=\alpha_1$ in a neighbourhood of $\overline{D_1}$, and besides $\bar{\partial}\alpha_2=\bar{\partial}\beta-\bar{\partial}^2\gamma=\bar{\partial}\beta=\omega$in a neibourhood of $\overline{D_2}$ as required. $\ecarre$

covering of D, and satisfy the relations on


 
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Sergey Bratus
11/8/1997