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1.4: Inequivalence of Ball and Polydisc

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Definition: Biholomorphic

Two domains UCn and VCn are said to be biholomorphic or biholomorphically equivalent if there exists a one-to-one and onto holomorphic map f:UV such that the inverse f1:VU 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 1.4.1: Riemann Mapping Theorem

If UC is a nonempty simply connected domain such that UC, then U is biholomorphic to 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, Bn={zCn:||z||<1}andDn={zCn:|zj|<1 for j=1,,n}, 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 1.4.1

Prove that there exists a homeomorphism f:BnDn, that is, f is a bijection, and both f and f1 are continuous.

Let us stick with n=2. Instead of proving that B2 and D2 are biholomorphically inequivalent we will prove a stronger theorem. First a definition.

Definition: Proper Map

Suppose f:XY is a continuous map between two topological spaces. Then f is a proper map if for every compact K⊂⊂Y, the set f1(K) is compact.

The notation “⊂⊂” 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:DD given by f(z)=z2 is proper, but not invertible. The codomain of f is important. If we replace f by g:DC, still given by g(z)=z2, then the map is no longer proper. Let us state the main result of this section.

Theorem 1.4.2: Rothstein 1935

There exists no proper holomorphic mapping of the unit bidisc D2=D×DC2 to the unit ball B2C2.

As a biholomorphic mapping is proper, the unit bidisc is not biholomorphically equivalent to the unit ball in C2. This fact was first proved by Poincaré by computing the automorphism groups of D2 and B2, 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 φ:DCn is called an analytic disc. If the mapping φ extends continuously to the closed unit disc ¯D, then the mapping φ:¯DCn is called a closed analytic disc.

Often we call the image Δ=φ(D) the analytic disc rather than the mapping. For a closed analytic disc we write Δ=φ(D) and call it the boundary of the analytic disc.

In some sense, analytic discs play the role of line segments in Cn. 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 φ is not unique.

Let us consider the boundaries of the unit bidisc D×DC2 and the unit ball B2C2. We notice the boundary of the unit bidisc contains analytic discs {p}×D and D×{p} for pD. That is, through every point in the boundary, except for the distinguished boundary D×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 1.4.1

The unit sphere S2n1=BnCn contains no analytic discs.

Proof

Suppose there is a holomorphic function g:DCn such that the image g(D) is inside the unit sphere. In other words, ||g(z)||2=|g1(z)|2+|g2(z)|2++|gn(z)|2=1 for all zD. Without loss of generality (after composing with a unitary matrix) assume that g(0)=(1,0,0,,0). Consider the first component and notice that g1(0)=1. If a sum of positive numbers is less than or equal to 1, then they all are, and hence |g1(z)|1. Maximum principle says that g1(z)=1 for all zD. But then gj(z)=0 for all j=2,,n and all zD. 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 1.4.2

Modify the proof to show some stronger results.

  1. Let Δ be an analytic disc and ΔBn. Prove Δ contains points not in ¯Bn.
  2. Let Δ be an analytic disc. Prove that ΔBn is nowhere dense in Δ.
  3. Find an analytic disc in C2, such that (1,0)Δ, ΔB2=, and locally near (1,0)B2, the set ΔB2 is the curve defined by z1=0, z2=0, (z1)2+(z2)2=1.

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

Lemma 1.4.1

Let URn and VRm be bounded domains and let f:UV be continuous. Then f is proper if and only if for every sequence {pk} in U such that pkpU, the set of limit points of {f(pk)} lies in V.

Proof

Suppose f is proper. Take a sequence {pk} in U such that pkpU. Then take any convergent subsequence {f(pkj)} of {f(pk)} converging to some q¯V. Consider E={f(pkj)} as a set. Let ¯E be the closure of E in V (subspace topology). If qV, then ¯E=E{q} and ¯E is compact. Otherwise, if qV, then ¯E=E. The inverse image f1(¯E) is not compact (it contains a sequence going to pU) and hence ¯E is not compact either as f is proper. Thus qV, and hence qV. As we took an arbitrary subsequence of {f(pk)}, q was an arbitrary limit point. Therefore, all limit points are in V.

Let us prove the converse. Suppose that for every sequence {pk} in U such that pkpU, the set of limit points of {f(pk)} lies in V. Take a closed set EV (subspace topology) and look at f1(E). If f1(E) is not compact, then there exists a sequence {pk} in f1(E) such that pkpU. That is because f1(E) is closed (in U) but not compact. The hypothesis then says that the limit points of {f(pk)} are in V. Hence E has limit points in V and is not compact.

Exercise 1.4.3

Let URn and VRm be bounded domains and let f:¯U¯U be continuous. Suppose f(U)V, and g:UV is defined by g(x)=f(x) for all xU. Prove that g is proper if and only if f(U)V.

Exercise 1.4.4

Let f:XY be a continuous function of locally compact Hausdorff topological spaces. Let X and Y 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:XY by letting f|X=f and f()=.

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

Proof of Theorem 1.4.2. Suppose there is a proper holomorphic map f:D2B2. Fix some eiθ in the boundary of the disc D. Take a sequence wkD such that wkeiθ. The functions gk(ζ)=f(ζ,wk) map the unit disc into B2. 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:D¯B2. As (ζ,wk)(ζ,eiθ)D2, then by Lemma 1.4.1 we have that g(D)B2 and hence g must be constant.

Let gk denote the derivative (we differentiate each component). The functions gk converge to g=0. So for an arbitrary fixed ζD, fz1(ζ,wk)0. This limit holds for all eiθ and some subsequence of an arbitrary sequence {wk} where wkeiθ. The holomorphic mapping wfz1(ζ,w), therefore, extends continuously to the closure ¯D and is zero on D. We apply the maximum principle or the Cauchy formula and the fact that ζ was arbitrary to find fz10. By symmetry fz20. Therefore, f is constant, which is a contradiction as f was proper.

The proof is illustrated in Figure 1.4.1.

clipboard_e59e81390b42aea4792ad6d9033b0b8a5.png

Figure 1.4.1

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 wk) and take a limit to produce an analytic disc in the boundary of B2. We then show that fz1=0 on the vertical gray line (where the first component is fixed to be ζ). The right-hand side shows the disc where z1=ζ 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 1.4.3

Let U=U×UCn×Ck, n,k1, and VCm, m1, be bounded domains such that V contains no analytic discs. Then there exist no proper holomorphic mapping f:UV.

Exercise 1.4.5

Prove Theorem 1.4.3

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 1.4.6

Let f:DD be a proper holomorphic map. Then f(z)=eiθmk=1zak1ˉakz, for some real θ and some akD (that is, f is a finite Blaschke product). Hint: Consider f1(0).

In several variables, when 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 1.4.7

Suppose f:UD be a proper holomorphic map where UCn is a nonempty domain. Prove that n=1. Hint: Consider the same idea as in Exercise 1.2.19.

Exercise 1.4.8

Suppose f:¯BnCm is a nonconstant continuous map such that f|Bn is holomorphic and ||f(z)||=1 whenever ||z||=1. Prove that f|Bn maps into Bm and furthermore that this map is proper.


This page titled 1.4: Inequivalence of Ball and Polydisc is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Jiří Lebl via source content that was edited to the style and standards of the LibreTexts platform.

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