1.4: Inequivalence of Ball and Polydisc

Definition: Biholomorphic

Two domains $U \subset \mathbb{C}^n$ and $V \subset \mathbb{C}^n$ are said to be biholomorphic or biholomorphically equivalent if there exists a one-to-one and onto holomorphic map $f \colon U \to V$ such that the inverse $f^{-1} \colon V \to U$ is holomorphic. The mapping $f$ is said to be a biholomorphic map or a biholomorphism.

As function theory on two biholomorphic domains is the same, one of the main questions in complex analysis is to classify domains up to biholomorphic transformations. In one variable, there is the rather striking theorem due to Riemann:

Theorem $\PageIndex{1}$: Riemann Mapping Theorem

If $U \subset \mathbb{C}$ is a nonempty simply connected domain such that $U \neq \mathbb{C}$, then $U$ is biholomorphic to $\mathbb{D}$.

In one variable, a topological property on $U$ is enough to classify a whole class of domains. It is one of the reasons why studying the disc is so important in one variable, and why many theorems are stated for the disc only. There is simply no such theorem in several variables. We will show momentarily that the unit ball and the polydisc, $$\mathbb{B}_{n} = \bigl\{ z \in \mathbb{C}^n : ||z|| < 1 \bigr\} \qquad \text{and} \qquad \mathbb{D}^n = \bigl\{ z \in \mathbb{C}^n : |z_j| < 1 ~\text{for $j=1,\ldots,n$} \bigr\} ,$$ are not biholomorphically equivalent. Both are simply connected (have no holes), and they are the two most obvious generalizations of the disc to several variables. They are homeomorphic, that is, topology does not see any difference.

Exercise $\PageIndex{1}$

Prove that there exists a homeomorphism $f \colon \mathbb{B}_{n} \to \mathbb{D}^n$, that is, $f$ is a bijection, and both $f$ and $f^{-1}$ are continuous.

Let us stick with $n=2$. Instead of proving that $\mathbb{B}_{2}$ and $\mathbb{D}^2$ are biholomorphically inequivalent we will prove a stronger theorem. First a definition.

Definition: Proper Map

Suppose $f \colon X \to Y$ is a continuous map between two topological spaces. Then $f$ is a proper map if for every compact $K \subset \subset Y$, the set $f^{-1}(K)$ is compact.

The notation “$\subset \subset$” is a common notation for relatively compact subset, that is, the closure is compact in the relative (subspace) topology. Often the distinction between compact and relatively compact is not important. For instance, in the definition above we can replace compact with relatively compact. So the notation is sometimes used if “compact” is meant.

Vaguely, “proper” means that “boundary goes to the boundary.” As a continuous map, $f$ pushes compacts to compacts; a proper map is one where the inverse does so too. If the inverse is a continuous function, then clearly $f$ is proper, but not every proper map is invertible. For example, the map $f \colon \mathbb{D} \to \mathbb{D}$ given by $f(z) = z^2$ is proper, but not invertible. The codomain of $f$ is important. If we replace $f$ by $g \colon \mathbb{D} \to \mathbb{C}$, still given by $g(z)=z^2$, then the map is no longer proper. Let us state the main result of this section.

Theorem $\PageIndex{2}$: Rothstein 1935

There exists no proper holomorphic mapping of the unit bidisc $\mathbb{D}^2 = \mathbb{D} \times \mathbb{D} \subset \mathbb{C}^2$ to the unit ball $\mathbb{B}_{2} \subset \mathbb{C}^2$.

As a biholomorphic mapping is proper, the unit bidisc is not biholomorphically equivalent to the unit ball in $\mathbb{C}^2$. This fact was first proved by Poincaré by computing the automorphism groups of $\mathbb{D}^2$ and $\mathbb{B}_{2}$, although his proof assumed the maps extended past the boundary. The first complete proof was by Henri Cartan in 1931, though popularly the theorem is attributed to Poincaré. It seems standard practice that any general audience talk about several complex variables contains a mention of Poincaré, and often the reference is to this exact theorem.

We need some lemmas before we get to the proof of the result. First, a certain one-dimensional object plays an important role in the geometry of several complex variables. It allows us to apply one-variable results in several variables. It is especially important in understanding the boundary behavior of holomorphic functions. It also prominently appears in complex geometry.

Definition: Analytic Disc

A nonconstant holomorphic mapping $\varphi \colon \mathbb{D} \to \mathbb{C}^n$ is called an analytic disc. If the mapping $\varphi$ extends continuously to the closed unit disc $\overline{\mathbb{D}}$, then the mapping $\varphi \colon \overline{\mathbb{D}} \to \mathbb{C}^n$ is called a closed analytic disc.

Often we call the image $\Delta = \varphi(\mathbb{D})$ the analytic disc rather than the mapping. For a closed analytic disc we write $\partial \Delta = \varphi( \partial \mathbb{D})$ and call it the boundary of the analytic disc.

In some sense, analytic discs play the role of line segments in $\mathbb{C}^n$. It is important to always have in mind that there is a mapping defining the disc, even if we are more interested in the set. Obviously for a given image, the mapping $\varphi$ is not unique.

Let us consider the boundaries of the unit bidisc $\mathbb{D} \times \mathbb{D} \subset \mathbb{C}^2$ and the unit ball $\mathbb{B}_{2} \subset \mathbb{C}^2$. We notice the boundary of the unit bidisc contains analytic discs $\{p\} \times \mathbb{D}$ and $\mathbb{D} \times \{p\}$ for $p \in \partial \mathbb{D}$. That is, through every point in the boundary, except for the distinguished boundary $\partial \mathbb{D} \times \partial \mathbb{D}$, there exists an analytic disc lying entirely inside the boundary. On the other hand, the ball contains no analytic discs in its boundary.

Proposition $\PageIndex{1}$

The unit sphere $S^{2n-1} = \partial \mathbb{B}_{n} \subset \mathbb{C}^n$ contains no analytic discs.

Proof

Suppose there is a holomorphic function $g \colon \mathbb{D} \to \mathbb{C}^n$ such that the image $g(\mathbb{D})$ is inside the unit sphere. In other words, $$||g(z)||^2 = |g_1(z)|^2 + |g_2(z)|^2 + \cdots + |g_n(z)|^2 = 1$$ for all $z \in \mathbb{D}$. Without loss of generality (after composing with a unitary matrix) assume that $g(0) = (1,0,0,\ldots,0)$. Consider the first component and notice that $g_1(0) = 1$. If a sum of positive numbers is less than or equal to 1, then they all are, and hence $|g_1(z)| \leq 1$. Maximum principle says that $g_1(z) = 1$ for all $z \in \mathbb{D}$. But then $g_j(z) = 0$ for all $j=2,\ldots,n$ and all $z \in \mathbb{D}$. Therefore, $g$ is constant and thus not an analytic disc.

The fact that the sphere contains no analytic discs is the most important geometric distinction between the boundary of the polydisc and the sphere.

Exercise $\PageIndex{2}$

Modify the proof to show some stronger results.

1. Let $\Delta$ be an analytic disc and $\Delta \cap \partial \mathbb{B}_{n} \not= \emptyset$. Prove $\Delta$ contains points not in $\overline{\mathbb{B}_{n}}$.
2. Let $\Delta$ be an analytic disc. Prove that $\Delta \cap \partial \mathbb{B}_{n}$ is nowhere dense in $\Delta$.
3. Find an analytic disc in $\mathbb{C}^2$, such that $(1,0) \in \Delta$, $\Delta \cap \mathbb{B}_{2} = \emptyset$, and locally near $(1,0) \in \partial \mathbb{B}_{2}$, the set $\Delta \cap \partial \mathbb{B}_{2}$ is the curve defined by $\Im z_1=0$, $\Im z_2=0$, ${(\Re z_1)}^2+ {(\Re z_2)}^2 = 1$.

Before we prove the theorem, let us prove a lemma making the statement about proper maps taking boundary to boundary precise.

Lemma $\PageIndex{1}$

Let $U \subset \mathbb{R}^n$ and $V \subset \mathbb{R}^m$ be bounded domains and let $f \colon U \to V$ be continuous. Then $f$ is proper if and only if for every sequence $\{ p_k \}$ in $U$ such that $p_k \to p \in \partial U$, the set of limit points of $\bigl\{ f(p_k) \bigr\}$ lies in $\partial V$.

Proof

Suppose $f$ is proper. Take a sequence $\{ p_k \}$ in $U$ such that $p_k \to p \in \partial U$. Then take any convergent subsequence $\bigl\{ f(p_{k_j}) \bigr\}$ of $\bigl\{ f(p_k) \bigr\}$ converging to some $q \in \overline{V}$. Consider $E = \bigl\{ f(p_{k_j}) \bigr\}$ as a set. Let $\overline{E}$ be the closure of $E$ in $V$ (subspace topology). If $q \in V$, then $\overline{E} = E \cup \{ q \}$ and $\overline{E}$ is compact. Otherwise, if $q \notin V$, then $\overline{E} = E$. The inverse image $f^{-1}(\overline{E})$ is not compact (it contains a sequence going to $p \in \partial U$) and hence $\overline{E}$ is not compact either as $f$ is proper. Thus $q \notin V$, and hence $q \in \partial V$. As we took an arbitrary subsequence of $\bigl\{ f(p_k) \bigr\}$, $q$ was an arbitrary limit point. Therefore, all limit points are in $\partial V$.

Let us prove the converse. Suppose that for every sequence $\{ p_k \}$ in $U$ such that $p_k \to p \in \partial U$, the set of limit points of $\bigl\{ f(p_k) \bigr\}$ lies in $\partial V$. Take a closed set $E \subset V$ (subspace topology) and look at $f^{-1}(E)$. If $f^{-1}(E)$ is not compact, then there exists a sequence $\{ p_k \}$ in $f^{-1}(E)$ such that $p_k \to p \in \partial U$. That is because $f^{-1}(E)$ is closed (in $U$) but not compact. The hypothesis then says that the limit points of $\bigl\{ f(p_k) \bigr\}$ are in $\partial V$. Hence $E$ has limit points in $\partial V$ and is not compact.

Exercise $\PageIndex{3}$

Let $U \subset \mathbb{R}^n$ and $V \subset \mathbb{R}^m$ be bounded domains and let $f \colon \overline{U} \to \overline{U}$ be continuous. Suppose $f(U) \subset V$, and $g \colon U \to V$ is defined by $g(x) = f(x)$ for all $x \in U$. Prove that $g$ is proper if and only if $f(\partial U) \subset \partial V$.

Exercise $\PageIndex{4}$

Let $f \colon X \to Y$ be a continuous function of locally compact Hausdorff topological spaces. Let $X_\infty$ and $Y_\infty$ be the one-point compactifications of $X$ and $Y$. Then $f$ is a proper map if and only if it extends as a continuous map $f_\infty \colon X_\infty \to Y_\infty$ by letting $f_\infty |_X = f$ and $f_\infty(\infty) = \infty$.

We now have all the lemmas needed to prove the theorem of Rothstein.

Proof of Theorem $\PageIndex{2}$. Suppose there is a proper holomorphic map $f \colon \mathbb{D}^2 \to \mathbb{B}_{2}$. Fix some $e^{i\theta}$ in the boundary of the disc $\mathbb{D}$. Take a sequence $w_k \in \mathbb{D}$ such that $w_k \to e^{i\theta}$. The functions $g_k(\zeta) = f(\zeta,w_k)$ map the unit disc into $\mathbb{B}_{2}$. By the standard Montel’s theorem, by passing to a subsequence we assume that the sequence of functions converges (uniformly on compact subsets) to a limit $g \colon \mathbb{D} \to \overline{\mathbb{B}}_2$. As $(\zeta,w_k) \to (\zeta,e^{i\theta}) \in \partial \mathbb{D}^2$, then by Lemma $\PageIndex{1}$ we have that $g(\mathbb{D}) \subset \partial \mathbb{B}_{2}$ and hence $g$ must be constant.

Let $g_k'$ denote the derivative (we differentiate each component). The functions $g_k'$ converge to $g' = 0$. So for an arbitrary fixed $\zeta \in \mathbb{D}$, $\frac{\partial f}{\partial z_1} (\zeta, w_k) \to 0$. This limit holds for all $e^{i\theta}$ and some subsequence of an arbitrary sequence $\{ w_k \}$ where $w_k \to e^{i\theta}$. The holomorphic mapping $w \mapsto \frac{\partial f}{\partial z_1} (\zeta, w)$, therefore, extends continuously to the closure $\overline{\mathbb{D}}$ and is zero on $\partial \mathbb{D}$. We apply the maximum principle or the Cauchy formula and the fact that $\zeta$ was arbitrary to find $\frac{\partial f}{\partial z_1} \equiv 0$. By symmetry $\frac{\partial f}{\partial z_2} \equiv 0$. Therefore, $f$ is constant, which is a contradiction as $f$ was proper.

The proof is illustrated in Figure $\PageIndex{1}$.

Figure $\PageIndex{1}$

In the picture, on the left-hand side is the bidisc, and we restrict $f$ to the horizontal gray lines (where the second component is fixed to be $w_k$) and take a limit to produce an analytic disc in the boundary of $\mathbb{B}_{2}$. We then show that $\frac{\partial f}{\partial z_1} = 0$ on the vertical gray line (where the first component is fixed to be $\zeta$). The right-hand side shows the disc where $z_1 = \zeta$ is fixed, which corresponds to the vertical gray line on the left.

The proof says that the reason why there is not even a proper mapping is the fact that the boundary of the polydisc contains analytic discs, while the sphere does not. The proof extends easily to higher dimensions as well, and the proof of the generalization is left as an exercise.

Theorem $\PageIndex{3}$

Let $U = U' \times U'' \subset \mathbb{C}^n \times \mathbb{C}^k$, $n,k \geq 1$, and $V \subset \mathbb{C}^m$, $m \geq 1$, be bounded domains such that $\partial V$ contains no analytic discs. Then there exist no proper holomorphic mapping $f \colon U \to V$.

Exercise $\PageIndex{5}$

Prove Theorem $\PageIndex{3}$

The key takeaway from this section is that in several variables, to see if two domains are equivalent, the geometry of the boundaries makes a difference, not just the topology of the domains.

The following is a fun exercise in one dimension about proper maps of discs:

Exercise $\PageIndex{6}$

Let $f \colon \mathbb{D} \to \mathbb{D}$ be a proper holomorphic map. Then $$f(z) = e^{i\theta} \prod_{k=1}^m \frac{z-a_k}{1-\bar{a}_k z} ,$$ for some real $\theta$ and some $a_k \in \mathbb{D}$ (that is, $f$ is a finite Blaschke product). Hint: Consider $f^{-1}(0)$.

In several variables, when $\mathbb{D}$ is replaced by a ball, this question (what are the proper maps) becomes far more involved, and if the dimensions of the balls are different, it is not solved in general.

Exercise $\PageIndex{7}$

Suppose $f \colon U \to \mathbb{D}$ be a proper holomorphic map where $U \subset \mathbb{C}^n$ is a nonempty domain. Prove that $n=1$. Hint: Consider the same idea as in Exercise 1.2.19.

Exercise $\PageIndex{8}$

Suppose $f \colon \overline{\mathbb{B}_{n}} \to \mathbb{C}^m$ is a nonconstant continuous map such that $f|_{\mathbb{B}_{n}}$ is holomorphic and $||f(z)|| = 1$ whenever $||z||=1$. Prove that $f|_{\mathbb{B}_{n}}$ maps into $\mathbb{B}_{m}$ and furthermore that this map is proper.