| Posted on Saturday, 27 December, 2003 - 07:50 pm: |
I was reading recently about some of the ideas behind the theory of Julia sets, and, in particular, came across the fact that in order to prove some of the fundamental properties of Julia sets (such as the fact that they form the boundary of any basin of attraction, or that the inverse iterates of any point in a Julia set are dense in the Julia set), an alternative definition to the usual one is required. The definition seen most frequently is that the Julia set is the closure of the set of repelling fixed and periodic points of a function f. However the alternative is the set of all points z Î C such that S={fn:n Î N} is not a normal family at z, i.e. there is a sequence in S that does not have a subsequence that converges uniformly in every closed disc contained in a neighbourhood of z. Proving that this is equivalent to the usual definition requires Montel's theorem, a theorem from complex analysis. One text states the theorem as ''Let {gk} be a family of complex analytic functions on an open domain U. If {gk} is not a normal family, then for all w Î C with at most one exception we have gk(z)=w for some z Î U and some k.'' and refers the reader to 'the literature' for a proof. However, the only other statement I have found (with proof) is of the form ''If A is an open subset of C and S is a set of functions analytic on A that is uniformly bounded on closed discs in A, then ... S is a normal family.'' It is not immediately obvious to me why these two statements should be equivalent. The first statement can be derived from the contrapositive of the second if the fact that S is not uniformly bounded on closed discs in A implies that for all w Î C (except possibly one value) we can find some f Î S such that z Î A: f(z)=w. Is this a property true in general of families of analytic functions? I would be very grateful if anyone could clarify this point, or point me in the direction of any useful references on this subject.
| Posted on Monday, 29 December, 2003 - 09:56 am: |
Here are rough details of linking second quote to first quote. See references at end maybe for more.
Suppose that the gk in your first quote omit 2 values in the complex plane. That is, let E be the complex plane minus two points: we assume all gk map into E.
E is a hyperbolic domain, it has universal covering space the unit disc D, covering map p. Thus we can construct a sequence of maps fk from U to D satisfying pfk =gk . The fk s are normal by your second quote, therefore the gk s are likewise.
There's a fair amount going on here. I'm pretty sure there are details in my supervisor's book
--> Iteration of Rational Functions (A.F. Beardon)
also try
--> Complex Dynamics (Carleson & Gamelin)
((titles here might be slightly wrong)).
The first book, I THINK, explains and constructs explicitly D as a universal covering surface of E.
Please post if problems (maybe with the covering surface part),
Ian
| Posted on Tuesday, 30 December, 2003 - 06:12 pm: |
Ian,
Many thanks for your reply. I am quite unfamiliar with a lot of this stuff, in particular the definitions of a 'hyperbolic domain' and 'universal covering space'. My understanding of the latter term is that there is a mapping p from D to E which satisfies certain criteria. However, one point which seems counter-intuitive is that such a mapping exists if E is
with two points
removed, but not with one or no points removed! Is there
a way of explaining this fact?Many thanks,
Alex.
| Posted on Tuesday, 30 December, 2003 - 06:35 pm: |
The exponential map certainly misses 0 and nothing else in C . So if we have the disc as the universal covering of C (or the punctured complex plane), then we have a bounded analytic function on the whole of complex plane. By Liouville's theorem, such a function must be constant. Hence such a map cannot possibly exist.
The fact that we have the unit disc as the covering space of twice punctured complex plane is a matter of construction (or if you like, a corollary of the uniformisation lemma).
Kerwin
| Posted on Wednesday, 31 December, 2003 - 07:29 pm: |
Hi Alex,
To reiterate, possibly, on what Kerwin has said, it is possible to construct a map f from D to E that is a covering map (my notation). It is not entirely straightforward to state what the map is, it's nothing simple like ez .
Regarding what a 'covering map' is, it suffices to know that this map has the property that if another continuous function g maps a connected space F to E then there exists a further map h from F to D such that g=fh. It'll probably take a moment to digest this and a diagram helps.
This is the only property of universal covering spaces used in the proof (and hyperbolic means covering space is D). There are two different definitions of covering map that you'll meet in the literature.
Ian