A question on Ahlfors covering surface

Question

Given a transcendental entire function $f$, and three Jordan domains $D_1$, $D_2$, and $D_3$ such that the closures of the three Jordan domains do not intersect with each other. Then from Ahlfors covering surface (See in Walter Hayman's book), we can choose $i\in {1,2,3}$ such that there are INFINITELY MANY univalent island over $D_i$.

I wondered whether this interesting theorem can be proved by other method? Ahlfors original proof relies on his important inequality on covering surface, which seems highly non-trivial. Any comments and remarks will be appreciated.

Edit: In the original version of the question, I only assume three disjoint Jordan domains does not intersect with each other. In fact, this is not enough . Professor Alexandre Eremenko pointed out in fact it should be the closures of the three domains do not intersect with each other.

Answer 1

If you read French, the article "Sur la théorie d'Ahlfors des surfaces" by Duval may be interesting: http://arxiv.org/abs/1311.1589 .

It covers the full Ahlfors theory, not just the islands theorem. If you want only the latter, then Bergweiler's proof (which I think is actually very intuitive, well at least for the infinitesimal version) is probably your best bet.

Answer 2

In addition to the two papers mentioned by Malik Younsi and Lasse Rempe Gillen, there are these two papers:

Tôki, Yukinari, Proof of Ahlfors principal covering theorem. Rev. Math. Pures Appl. 2 1957 277–280.

de Thélin, Henry, Une démonstration du théorème de recouvrement de surfaces d'Ahlfors. Enseign. Math. (2) 51 (2005), no. 3-4, 203–209,

each containing a simplified proof of Ahlfors's two fundamental theorems. (The second fundamental theorem implies the island theorem). These four proofs make a complete list, I suppose.

Answer 3

Yes it can, see the paper A new proof of the Ahlfors five islands theorem by Walter Bergweiler. The proof is based on a result on Nevanlinna concerning perfectly branched values, Zalcman's rescaling lemma for non-normal families and the existence of solutions to the Beltrami equation (so-called Measurable Riemann mapping theorem).

You could also take a look at Bergweiler's other paper The role of the Ahlfors five islands theorem in complex dynamics for interesting applications.

Answer 4

Now I feel shy because there are some people watching that really know what they are talking about, unlike me. So a word of warning: anything I am going to write here may be too optimistic (or stupid). It is certainly extremely vague. With that said, the point of view I refer to is: consider the domain of f as a Riemann surface over the target. By this I mean that you can reconstruct dom(f) by patching pieces of the target using f to understand how to glue things. A the level of critical point, you have to remove the critical value and add several copies of the pieces that you may have to cut. This is already idealistic since there are also asymptotic values (points whose preimage is "at infinity", more rigorously, points $a$ in the range such that there is a path in the domain tending to the boundary of dom(f) and whose image by f tends to $a$). The prototype of the latter is the point 0 in the range of f=the exponential map, where a preimage of a neighborhood of 0 is an infinite cover over a punctured disk. But it may be less nice.

Some more heuristics: the infinity of C is small, the infinity of D is big. In fact, you can recompose D by patching many copies of a single compact patch, exponentially many as you wander off to the boundary of D, if you count the distance by how many patches you need to cross. Archetypical examples:

1. the universal cover of C minus two points,
2. Ahlfors' conjectured optimal function for the Bloch constant. This will not be possible for C.

In short: D is hyperbolic, C is parabolic (this is very vague, I know; there is at least one thing that makes me uneasy about all this presentation of things: that C nevertheless contains D.).

Now, anytime you have an island, you have a patch that is a copy of the domain D$_i$. If you start from a point in the domain, mapped in D$_i$, and try to lift D$_i$ by f and hit a problem, is may be because you need to put a ramification point, so this increases the number of patches you need... kind of... in fact you will probalbyhave to cut your patch into smaller pieces, because of more ramification points: but then the heuristics is that it only improves things (it makes dom(f) more hyperbolic). If the problem is not due to a ramification point, then it there is an asymptotic value, and again this leads to many patches: in some sense, it is even more hyperbolic than just having ramification points.

If $f$ is transcendental and some D$_i$ has only finitely many islands, then close enough to the boundary of dom(f), it is a bit like there is no islands. By the discussion above other preimages will be ramified, or worse. Now if there are enough D$_i$ with this property, then the game is to see that, on a purely combinatorial level, there must at least an exponential growth of the patches (this is idealistic but this gives an idea): this is what happens with the two examples above.

Let us look at an example that works not: cos(z). Its critical points are points of $\pi\mathbb Z$, half of which get mapped to -1 the other half to 1. There is no asymptotic value (in fact, infinity is). The number of patches (whatever that means) grows linearly with the combinatorial distance. More precisely if you take D$_1$ and D$_2$ two disks about -1 and 1, their preimages are all ramified of ramification number=2. Join them by a line. Then the preimage of this picture is many disks-like domains at all the points of $\pi\mathbb Z$ all joined to the next by a line. This graph is more "euclidean" than "hyperbolic".

Now if you have three domains D$_i$, take a point outside and link this point to each D$_i$ by disjoint curves, giving a Y-shaped picture with three disks attached at the tips. Ideally, the preimage of this graph contains (infinitely) many copies of this Y-shaped curve with disks at tips and to each of these disk (with finitely many exceptions) is attached at least two distinct copies of a Y. Such a graph is necessarily hyperbolic.

Again, I do not know the proof of Ahlfors' theorems, and I completely forgot the proof of Bloch's theorem, so all this may be far from the actual proofs. This is just my point of view on things.