Several Complex Variables (IV): Domains of Holomorphy

06 Sep 2018

• several complex variables

In this post, we will look at domains of holomorphy, which are (roughly speaking) domains which do not exhibit Hartogs’s phenomenon anywhere. Historically, characterising domains of holomorphy was a central problem in several complex variables theory; here we will only present some basic results.

Domains of holomorphy

An open set $U\subseteq\bb C^n$ is called a domain of holomorphy if there is a holomorphic function $f$ on $U$ such that for any $w\in\del U$ and any polyradius $r$, there is no holomorphic function on $\Delta(w,r)$ that is equal to $f$ on any component of $\Delta(w,r)\cap U$. Intuitively, this means that $f$ has no local holomorphic extension across any boundary point of $U$.

Note that this is not the same as saying that $f$ has no global extension to a larger domain. For instance, the principal branch of $\log$ on $\bb C\backslash(-\infty,0)$ has no holomorphic (or even continuous) extension to a larger domain. However, any point on $(-\infty,0)$ has a neighbourhood where $\log$ has a branch agreeing with the principal branch on, say, the upper half plane.

This definition is only useful for more than one variable, because of the following:

Proposition: Any domain $U\subseteq\bb C$ is a domain of holomorphy.

Proof: For any $p\in\del U$, the function $f(z)=\frac1{z-p}$ does not have a holomorphic extension to any neighbourhood of $p$. $\qed$

Now we give some simple constructions of domains of holomorphy.

Proposition: If $U\subseteq\bb C^n$ and $V\subseteq\bb C^m$ are domains of holomorphy, then so is $U\times V$.

Proof: Take any $(p,q)\in\del(U\times V)=(\del U\times V)\cup(U\times\del V)$; without loss of generality assume $(p,q)\in\del U\times V$. Since $U$ is a domain of holomorphy, there exists $f$ holomorphic on $U$ which cannot be extended on any polydisc around $p$. Hence the function $F(u,v)\coloneqq f(u)$ has no holomorphic extension on any polydisc around $(p,q)$. $\qed$

In particular, any polydisc is a domain of holomorphy. Compare this with the Hartogs figure (constructed in the previous post), which is homeomorphic to a polydisc but not a domain of holomorphy.

Proposition: Any convex domain $U\subseteq\bb C^n$ is a domain of holomorphy.

Proof: At any point $p\in\del U$, there is a supporting hyperplane of $U$, ie. a real-valued $\bb R$-linear functional $\ell:\bb C^n\to\bb R$ with $\ell(u)>\ell(p)$ for all $u\in U$. Now we have

$$\ell(z)=\sum_{k=1}^n\alpha_kz_k+\beta_k\overline{z_k},$$

and since $\ell$ is real-valued we have $\beta_k=\overline{\alpha_k}$. Hence $\ell(z)=\Re h(z)$, where $h(z)=2\sum\alpha_kz_k$ is a $\bb C$-linear functional. Hence $f(z)=\frac1{h(z)-h(p)}$ is holomorphic on $U$ and has no holomorphic extension on any polydisc around $p$. $\qed$

The Cartan-Thuller theorem

If a domain $U$ is not a domain of holomorphy, ie. $U$ exhibits Hartogs’s phenomenon, then there exists a point $p\in\del U$ such that every holomorphic function on $U$ can be extended to a neighbourhood of $p$; in particular, $f$ cannot blow up near $p$. It is natural to ask if this property characterises domains of holomorphy.

The above notion is formalised as follows. Given $K\subseteq U$, define the holomorphically convex hull of $K$ to be

$$\hat K\coloneqq\{w\in U\,:\,|f(w)|\leq\|f\|_K\text{ for all }f\text{ holomorphic on }U\},$$

where $\lVert f\rVert_K=\sup_{z\in K}\lvert f(z)\rvert$. We say that $U$ is holomorphically convex if $\hat K$ is compact for all compact $K\subseteq U$.

Why does this capture the property of blowing up at a boundary point? Note that for $K$ compact,

• $\hat K$ is bounded: take $f$ to be the coordinate functions $z_k$;
• $\hat K$ is relatively closed in $U$: $\hat K$ is the intersection of closed sets in $U$, one for each $f$.

Hence $\hat K$ fails to be compact exactly when it accumulates at some point $p\in\del U$, in which case no holomorphic function on $U$ blows up near $p$.

Cartan-Thuller theorem: A domain $U\subseteq\bb C^n$ is a domain of holomorphy if and only if it is holomorphically convex.

Proof: $(\Rightarrow)$: Let $K\subseteq U$ be compact. Then there exists $s>0$ such that $\Delta(a,\hat s)\coloneqq\Delta(a,(s,\ldots,s))\subseteq U$ for all $a\in K$. By estimates from the Cauchy integral formula, we have

$$\left|\frac{\del^{i_1+\cdots+i_n}}{\del z_1^{i_1}\cdots\del z_n^{i_n}}f(a)\right|\leq\frac{i_1!\cdots i_n!}{r^{i_1+\cdots+i_n}}\|f\|_{\bigcup_{a\in K}\Delta(a,\hat s)},$$

for all $a\in K$, and thus for all $a\in\hat K$. Hence the power series for $f$ about $a$ converges in $\Delta(a,\hat s)$ for all $a\in\hat K$.

As noted before, if $\hat K$ is not compact then it shares some boundary point $b$ with $U$. Take $a\in\hat K\cap\Delta(b,\hat s)$, so $b\in\Delta(a,\hat s)$. But every holomorphic function $f$ has an extension to $\Delta(a,\hat s)$, which is a neighbourhood of $b$, contradicting the fact that $U$ is a domain of holomorphy. Hence $\hat K$ is compact for all compact $K\subseteq U$, so $U$ is holomorphically convex.

$(\Leftarrow)$: Let $\{K_j\}$ be an increasing sequence of compact subsets of $U$ such that every compact subset of $U$ is contained in some $K_j$; for instance, we can take

$$K_j\coloneqq\{z\in\Delta(0,(j,\ldots,j))\,:\,d(z,\bb C^n\backslash U)\leq\frac1j\}.$$

Since $U$ is holomorphically convex, $\hat K_j$ is compact.

Let $\{w_j\}$ be an enumeration of a countable dense subset of $U$ (for instance, the points in $U$ with all real and imaginary coordinates rational), and let $S_j$ be the sphere of maximal radius contained in $U$ centred at $w_j$. Since $S_j$ is either all of $\bb C^n$ or contains points close to $\del U$, there exists $z_j\in S_j\backslash\hat K_j$.

By definition of $\hat K_j$, there exists a holomorphic function $f_j$ with $f_j(z_j)=1$ and $\lVert f_j\rVert_{K_j}<1$. By taking $f_j$ to a large power, we may assume that $\lVert f_j\rVert_{K_j}<2^{-j}$. Now on each $K_j$, the sum $\sum_{j=1}^\infty jf_j$ converges uniformly, so the product $f\coloneqq\prod_{j=1}^\infty(1-f_j)^j$ converges uniformly as well. Note that $f$ is:

• holomorphic on $U$;
• not identically zero: the infinite product converges, hence is nonzero, on $K_1$;
• has a zero at $z_j$ of total order at least $j$ (ie. all terms of order less than $j$ in the power series for $f$ at $z_j$ vanish).

Let $\Delta$ be a ball of radius $\delta$ centred at some $p\in\del U$. Then any component $\mc C$ of $U\cap\Delta$ contains a subsequence $w_{j_k}\to p$, say with $\lvert w_{j_k}-p\rvert<\frac\delta2$. Then the radius of $S_{j_k}$ is less than $\frac\delta2$, so $\mc C$ contains every $S_{j_k}$, and in particular every $z_{j_k}$. Hence $\mc C$ contains a sequence of zeroes of $f$ of arbitrarily high total order with limit $p$.

Now if $\hat f$ is a holomorphic continuation of $f$ from $\mc C$ to $\Delta$, then $\hat f$ and its partial derivatives of all orders must vanish at $p$, so $\hat f\equiv0$ on $\Delta$, contradicting the fact that $f$ is not identically zero on $U$. Hence $U$ is a domain of holomorphy. $\qed$