Consider injective homolomorphic functions $f:\mathbb D\to \mathbb C$ on the unit disk $|z|\leq 1$, normalized by the conditions $f(0)=0$ and $f'(0)=1$.

Thus for $|z|\leq 1$ we have $ f(z)=\sum_{k=0}^{\infty} a_k z^k $ with $a_0=0$ and $a_1=1$. Ludwig Bieberbach conjectured in 1916 and Louis de Branges proved in 1984 that for all $k \in \mathbb N$ the inequality $|a_k|\leq k$ holds.

Question Is there an analogue to de Branges's result for functions defined on a suitable domain (polydisk, ball,...?) in $\mathbb C^n$ for $n\geq 2$ ?

  • $\begingroup$ Did you mean to say for all $k \in N$? As stated, I don't see how $n$ enters into your statement of Bieberbach's result. $\endgroup$ Mar 18 '11 at 16:17
  • $\begingroup$ Dear drbobmeister, of course I meant "for all $k\in \mathbb N$" and I have just corrected this typo: thanks a lot for catching it. $\endgroup$ Mar 18 '11 at 18:28
  • $\begingroup$ My pleasure, Georges. $\endgroup$ Mar 18 '11 at 19:39

Such a result would have to be quite different in several variables, because the holomorphic automorphism group of $\mathbb{C}^n$ is very big when $n \geq 2$. For injectivity, we need to look at equidimensional mappings $F$ from the domain (whatever it may be), and into $\mathbb{C}^n$ say. For simplicity, choose $n = 2$, which already contains the main features of the general case. Any mapping $\psi$ of the form $(z,w) \mapsto (z,w + h(z))$ (a "shear") with $h$ an arbitrary entire function is holomorphic everywhere on $\mathbb{C}^2$ and injective. Thus $\psi \circ F$ is also an injective holomorphic mapping from the domain into $\mathbb{C}^2$. Assuming the domain contains the origin and that $F$ is normalized (taking the origin to the origin and having the identity as derivative at the origin), we can ensure that $\psi \circ F$ is also normalized by choosing the entire function $h$ to have a double zero at the origin. The freedom of choosing $h$ implies that the power series coefficients of the mappings in the normalized family of injective holomorphic mappings have no universal bounds over the family; unlike in the one-dimensional case, the normalized family is not compact.

The key difficulty is this: In one variable the automorphism group of the target $\mathbb{C}^1$ consists of mappings of the form $z \mapsto az + b$, and we can mod out this automorphism group simply by fixing $f(0)$ and $f^{\prime}(0)$, and then it turns out that injectivity implies compactness. When $n \geq 2$ the automorphism group is enormous (a dense subgroup of it was determined explicitly by E. Andersén and L. Lempert, so it is known in a sense), thus a linear normalization is very inadequate, and we should normalize by the whole automorphism group. But it is not clear what it means in practice to mod out by the automorphism group, and whether injectivity would imply compactness after moding out when $n \geq 2$. If we obtain a compact family, there will exist sharp bounds on all coefficients, of course, though we might not be able to establish them.

A precise question (though in an unexplicit form) would be: When can we write the family of all injective holomorphic mappings from a domain $\Omega$ into $\mathbb{C}^n$ by composing $\mathrm{Aut}(\mathbb{C}^n)$ with a compact family? Trivially we can do it when $\Omega = \mathbb{C}^n$ by choosing the compact family to consist of the identity map alone.

If one is willing to impose strong geometric conditions (e. g. starlikeness) on the injective mappings, one can get compactness. See the book Geometric function theory in one and several variables by I. Graham and G. Kohr for example.

ADDED: Having thought about this a little more, I believe it unlikely that for $n \geq 2$ compactness can hold for $F : \mathbb{B}_n \rightarrow \mathbb{C}^n$ even after moding out by the automorphism group of $\mathbb{C}^n$. For generic domains have only the identity automorphism, and so we could replace $\mathbb{C}^n$ by a slightly smaller domain $D$ with only the identity automorphism and look at injective holomorphic mappings $F : \mathbb{B}_n \rightarrow D$. Since $D$ is nearly as "roomy" as $\mathbb{C}^n$ the family should be noncompact, but obviously cannot be made compact by moding out by automorphisms. So moding out by automorphisms seems to be an "accidental" device, so to speak, and unlikely to work when $n \geq 2$. I would support this conclusion as follows:

One way to see that the family of injective holomorphic functions $f : \mathbb{D} \rightarrow \mathbb{C}$ with $f(0) = 0$ and $f'(0) = 1$ is compact is to apply the Schwarz Lemma to the inverse $f^{-1}$ to see that $f$ omits a point on the circle $|w| = 1$. Since the omitted set of $f$ on the unit sphere is a continuum containing $\infty$, $f$ also omits a point on the circle $|w| = 2$. A well known theorem states that if a family of holomorphic functions omits three points on the Riemann sphere, then it is precompact ("normal" in complex analysis parlance). Here we have three points, but two are "movable" (on the circles) and the third is fixed (at infinity) and a routine extension of the theorem mentioned shows that the family is still precompact, because the three omitted points stay uniformly away from each other. Since the normalized family of injective holomorphic functions $f$ is closed by a theorem of Hurwitz, the family is compact.

Now let us see what this yields for injective holomorphic mappings $F : \mathbb{B}_n \rightarrow D$ with $F(0) = 0$ and $F'(0) = I$ (we assume $0 \in D$) for $n \geq 2$. We can still apply the Schwarz Lemma (in several variables) to the inverse $F^{-1}$ to conclude that $F$ omits a point on the sphere $|w| = 1$, so that the omitted set of $F$ is an unbounded continuum with a point on that sphere. But this information is hardly strong enough to conclude that the family is precompact. The higher-dimensional analogue of the theorem on three omitted points is that if we remove $2n+1$ hyperplanes in general position from $\mathbb{C}^n$ then any family of holomorphic mappings into what remains is precompact. So it seems that we have removed much too little when $n \geq 2$.

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    $\begingroup$ +1 for this very convincing and well-written argument that even formulating a reasonable analogue of Bieberbach's conjecture is a difficult problem. Thanks a lot, Marius! $\endgroup$ Mar 20 '11 at 22:22

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