Several Complex Variables (III): Hartogs's Phenomenon

06 Sep 2018

• several complex variables

One of the most dramatic differences between theory of single variable and several variable holomorphic functions is Hartogs’s phenomenon: for $n>1$, there are domains (nonempty connected open sets) $U\subsetneq V\subseteq\bb C^n$ such that every holomorphic function on $U$ extends to a holomorphic function on $V$.

This is in stark contrast to the $n=1$ case: if $p\in V\cap\del U$, then $f(z)=\frac1{z-p}$ is holomorphic in $U$ but does not extend to $V$.

A class of examples

Consider the polydisc $\Delta(0,r)\subseteq\bb C^n$ ($n\geq2$), with polyradius $r=(r_1,\ldots,r_n)$. We can view the first $n-1$ coordinates and the $n$-th coordinate separately:

$$\Delta(0,r)=\Delta(0,r')\times\Delta(0,r_n),\qquad z=(z',z_n).$$

where $r’=(r_1,\ldots,r_{n-1})$, $z’=(z_1,\ldots,z_{n-1})$. We can picture this as if the $n$-th coordinate $z_n$ were drawn vertically, with the other $n-1$ coordinates $z’$ drawn horizontally.

Take any region $U\subseteq\Delta(0,r)$. For each $z’\in\Delta(0,r’)$, we can consider the vertical slice

$$U_{z'}\coloneqq\{z_n\in\bb C\,:\,(z',z_n)\in U\}\subseteq\Delta(0,r_n).$$

(picture)

Intuitively, to guarantee that every holomorphic function on $U$ extends to $\Delta(0,r)$, $U$ has to be ‘large’ in some sense. The technical conditions we need are as follows.

Theorem: Assume that $f$ is a holomorphic function on $U$ such that:

1. For some $s< r_n$, we have $\Delta(0,r)\backslash U\subseteq\Delta(0,r’)\times\overline\Delta(0,s)$, ie. any vertical slice of $U$ contains the annulus $\Delta(0,r_n)\backslash\overline\Delta(0,s)$;
2. $U_{z’}=\Delta(0,r_n)$ for all $z’$ in some open subset of $\Delta(0,r’)$, ie. over some horizontal open set, the vertical slice is all of the disc $\Delta(0,r_n)$.

Then every holomorphic function on $U$ has a holomorphic extension to $\Delta(0,r)$.

Proof: For $s<\tilde s< r_n$, consider the function $f_{\tilde s}:\Delta(0,r’)\times\Delta(0,\tilde s)\to\bb C$ given by

$$f_{\tilde s}(z',z_n)\coloneqq\frac1{2\pi i}\int_{\del\Delta(0,\tilde s)}\!\frac{f(z',\zeta)}{\zeta-z_n}d\zeta.$$

By condition 1, $\del\Delta(0,\tilde s)$ is contained in every vertical slice, so this integral is well-defined.

Note that $f_{\tilde s}$ is:

• continuous: the integrand is continuous in $(z’,z_n)$;
• analytic in $z_n$: for fixed $z’$, $f_{\tilde s}$ is a Cauchy integral in $z_n$;
• analytic in $z’$: for fixed $z_n$, the integrand is an absolutely convergent power series about $z’$, so by exchanging summation and integral we get a power series for $f_{\tilde s}$.

Thus by Osgood’s lemma, $f_{\tilde s}$ is analytic in its domain of definition. (See the previous post for results on equivalent definitions of holomorphic functions.)

Moreover, by the Cauchy integral formula, the integral above is independent of the contour, as long as $z_n$ lies in the region bounded by the contour. Hence for any other $s<\tilde s_0< r_n$, the functions $f_{\tilde s}$ and $f_{\tilde s_0}$ agree on their common domain of definition. Now gluing together all the $f_{\tilde s}$ gives a function $\hat f$ holomorphic on $\Delta(0,r)$ (since it is holomorphic on each $\Delta(0,r’)\times\Delta(0,\tilde s)$).

It remains to show that $\hat f=f$ in $U$. By condition 2 and the Cauchy integral formula, this holds in the vertical slices over some horizontal open set, so $\hat f=f$ in some open subset of $U$. Now we finish by applying the following theorem to $g=\hat f-f$. $\qed$

Identity theorem: Let $g$ be holomorphic in some domain $U\subseteq\bb C^n$, such that $g$ and its partial derivatives of all orders vanish at some point in $U$. Then $g\equiv0$ on $U$.

The proof is the same as in the one variable case:

Proof: Consider the set $S$ of all points in $U$ at which $g$ and its partial derivatives of all orders vanish. Note that this is the intersection of the zero sets of each partial derivative, which are all closed sets, so $S$ is closed. On the other hand, if $\zeta\in S$ then the power series for $g$ is identically zero in some neighbourhood $W$ of $\zeta$, so $W\subseteq S$; hence $S$ is open. Therefore $S$ is a nonempty clopen subset of the connected set $U$, so $S=U$. In particular, $g$ vanishes on $U$. $\qed$

Examples

We now explicitly construct domains $U$ on which Hartogs’s phenomenon holds.

Take any compact $K\subseteq\Delta(0,r)$ such that $U=\Delta(0,r)\backslash K$ is connected. Then $U$ satisfies conditions 1 and 2, so any holomorphic function on $U$ has a holomorphic extension to $\Delta(0,r)$.

Corollary: A holomorphic function in $n\geq2$ variables cannot have isolated zeroes.

Proof: Otherwise, assume without loss of generality that $f$ has isolated zero at $0$, so it is the only zero of $f$ in a sufficiently small $\Delta(0,r)$. Take $K=\overline\Delta(0,\frac r2)$, $U=\Delta(0,r)\backslash K$, so $\frac1f$ has a holomorphic extension from $U$ to $\Delta(0,r)$, agreeing with $\frac1f$ everywhere on $\Delta(0,r)\backslash\{0\}$ by identity theorem. In particular, $\frac1f$ has finite limit as $z\to0$, contradiction. $\qed$

The above result corresponds to the idea that the zero set of a $n$-variable holomorphic function should have dimension $n-1$ instead of 0. This will only be made rigorous later in the course, when we develop the concept of dimension for analytic varieties.

For our next example, consider the Hartogs figure

$$U=\{(z_1,z_2)\in\Delta(0,(1,1))\,:\,|z_1|<\frac12\text{ or }|z_2|>\frac12\}.$$

The cross-section of $U$ along the two real axes $x_1,x_2$ (ie. $y_1=y_2=0$) is shown below. As the picture suggests, $U$ is homeomorphic to $\Delta(0,(1,1))$.

(diagram)

Again, $U$ satisfies conditions 1 and 2, so any holomorphic function on $U$ has a holomorphic extension to $\Delta(0,(1,1))$.