Classification of Newton maps: realization

August 8, 2009

On the way to realize combinatorial models as real Newton maps the plan was to use Thurston’s theorem. In order to do so, one needs to prove that the branched covering, specified by the combinatorial data, doesn’t have Thurston obstructions.
This is the point where theorem of K.Pilgrim and T.Lei comes in use. It basically says that obstructions don’t intersect the preimages of arc systems.
The main difficulty in our case was to eliminate the case when the obstruction does intersect the arc system itself. Due to the theorem mentioned above this would imply that it cannot intersect the preimages of the arc systems – that is exactly the point where we aim at a contradiction.
Suppose it happens. Then as follows from the theorem the obstruction must be a Levy cycle. It is also known (BFH) that the complement components of the curves in the Levy cycles can only contain periodic mark points which in our case would be periodic points on extended Hubbard trees lying in the Julia set.
Such points being repelling for corresponding polynomial-like maps, have at least one external ray landing on it, i.e. there exists a preimage of the arc system which intersects the Thurston obstruction.
Contradiction.

Hubbard Trees

August 7, 2009

Hubbard Tree of z^2+i

It is known that Hubbard trees serve as a model for Julia sets of post-critically finite polynomials. What does it mean precisely? In fact, there is an inverse limit space of sequences (x_1,x_2,\ldots,x_n,\ldots) together with the shift map, such that the obtained dynamical system is topologically conjugate to the dynamical system p:J \to J, where $J$ is the Julia set of the polynomial $p$.

One of the explanations of the fact comes from the theory of IMG (iterated monodromy groups). Perhaps, the fact itself could be proven much easier way without involving the notion of IMG, but this is just one point of view on it.
IMG itself could be defined of partial self-coverings. A bit more general situuation arises when one defines IMG on topological automaton.

Definition. Topological automaton is a quadruple (M,M_1,f,\iota), where M,M_1 – topological spaces, f:M_1 \to M is a finite covering map, \iota:M_1 \to M is a continuous map.

In the case of partial self-coverings, when M_1 \subset M, \iota is just an embedding. But in this general setting we have in fact two maps between two topological spaces. One can iterate this topological automaton and associate an inverse limit system to it.
For example M_2 is defined as a pullback of the map f:M_1 \to M_0=M under the map \iota. M_2 = \{ (x,y) \in M^2_1: f(y)=\iota(x) \}.
In the same way M_n = \{ (x_1,x_2,\ldots,x_n) \in M^n_1 | f(x_{i+1})=\iota(x_i) \}

Topological Entropy

August 3, 2009

There are several qualitative invariants of dynamical systems, such as density of periodic points, topological mixing, hyperbolicity. The answer whether a concrete dynamical system has this property or not would be “Yes” or “No”. There do exists quantitative invariants, for example the growth of period points. But perhaps the most important quantitative invariant is a topological entropy.

Definition. In some sense, the topological entropy describes the growth of orbits all at the same time. To be precise, for two points x,y \in X one can define the metric

\displaystyle d_n^f(x)=max_{0 \leq i \leq n-1} d(f^i(x),f^i(y))

for a given positive integer n. Let B(x,r,n)=\{y \in X: d_n^f(x,y)<r \}, the ball of radius r. Let S_d(f,r,n) be the minimal number of such ball, sufficient to cover the whole of X.

Then

\displaystyle h_{top}(f)=\lim_{r \to 0} \lim_{n \to \infty} \frac{ \log S_d(f,r,n)}{n}  .

There are several other definitions for the topological entropy. Sometimes it is not quite useful to use covering sets for computation reasons. For example, one can use the so-called separating sets, i.e. the sets such that the pairwise distance between any two points in the metric d_n^f > r  . The cardinality of the maximal separating set is denoted by N_d(f,r,n) and can be used similarly for the definition of the topological entropy, substituting S_d(f,r,n) by N_d(f,r,n) in the formula above.

For example, for the expanding map E_m(x)=mx \quad mod (1), the topological entropy h_{top}(E_m)=\log|m|.

Proof. In general, for expanding maps the distance between two points growth until it becomes larger than some constant (1/2m for E_m). Let us choose two points x,y with d(x,y)<\frac{1}{2m^n}.Then

d_n^{E_m}(x,y)=d(E_m^{n-1}(x),E_m^{n-1}(x)) = m^{n-1}d(x,y).

If we want to have

d_n^{E_m}(x,y)> m^{-k}, then we must have d(x,y)>m^{-k-n}.

Hence if we choose S=\{i\cdot m^{-k-n}: 0\leq i \leq m^{k+n}-1 \} all the points from S will have pairwise distances between each other at least m^{-k}. And the set S can serve as a separating set with |S|=m^{k+n}, we obtain that

h_{top}(E_m)=\log |m|.

Bowen has asked to find a topological entropy of the complex polynomial of degree d. More generally M.Lyubich gave the answer for any rational function in the complex plane. The lower bound h_{top}(f) \geq \log (deg f) was known before due to Misiurewisz-Przytycki theorem.

The answer is as follows:

Theorem. Let f: S^2 \to S^2 be a rational function, that is not a constant. Then the topological entropy

h_{top}(f)=\log (deg f).

Self maps of hyperbolic surfaces with two fixed points

July 23, 2009

Problem. Let f be a self map of a hyperbolic surface X such that f has two fixed points. Show that f is a finite order automorphism, i.e. there exists n such that f^{\circ n} = id_X.

Solution. Let \pi be a covering map from the unit disk D to X. First of all, notice that due to Schwarz-Pick theorem such an f is either

  • A global isometry, i.e. a conformal isomorphisms of X
  • A covering map, but is not one-to-one, i.e. a local isometry
  • Strictly decreases the Poincare metric

It is easy to see that in the last case f cannot have two fixed points. Therefore let’s assume that f is a covering map. Since the covering is universal, one can lift the map f to the map \tilde{f}: D \to D.

Also note that \pi \circ \tilde{f} = f \circ \pi. Assume that p,q are the fixed points of f and \pi(0)=p, perhaps composing \tilde{f} with an automorphism of D we can also assume that \tilde{f}(0)=0.  Remember that f is not strictly decreasing Poincare metric, the same applies to its lift \tilde{f}, hence \tilde{f} must be a rotation around the origin (it is not difficult to classify all automorpisms of the unit disk), \tilde{f}(z)=e^{2\pi i \alpha} for some real \alpha.

Note that \pi \circ \tilde{f}^{\circ n} = f^{\circ n} \circ \pi, hence \tilde{f}^{\circ n}(\tilde{q}) is in the fiber of q and therefore the set \{ \tilde{f}^{\circ n}(\tilde{q}) \in N \} is a discrete set which implies that \alpha is rational and we are done.

P.S. Actually, if we were talking about self-coverings of compact surfaces, the situation would be easier: for compact Riemann surfaces we do have Riemann-Hurwitz formula:

2g(x)-2=n(2g(Y)-2)+\sum_{j=1}^n (e_j-1),

where f:X \to Y is a ramified covering of degree n. If $latex  X=Y$ then we would obtain that either g=1 or n=1. Inother words, for compact Riemann surfaces of genus >1 the only self-covering is a biholomorphism.

Rational Maps don’t have Levy cycles

July 13, 2009

Here is a sketch of the following:

Theorem. The rational functions doesn’t have Levy cycles.

Proof. Suppose it does and denote by \Gamma the corresponding Levy cycle. Choose \gamma_1,\gamma_2,\ldots, \gamma_n the geodesics representatives in \Gamma. Then the map f: \bar{C}-f^{-1}(P_f) \to \bar{C}-P_f is the covering map, hence a local isometry and preserves the distances locally. Therefore

l_{\bar{C}-f^{-1}(P_f)} (\gamma'_i) = l_{\bar{C}-P_f} (\gamma_{i+1}) where \gamma'_i is the component of f^{-1}(\gamma_{i+1}) which is in the same homotpy class with \gamma_i. From the other hand \bar{C}-f^{-1}(P_f) \subset \bar{C}-P_f and

l_{\bar{C}-P_f} (\gamma_{i+1}) =  l_{\bar{C}-f^{-1}(P_f)} (\gamma'_i)  >  l_{\bar{C}-P_f} (\gamma_i) and we get that each \gamma_i is strictly shorter than \gamma_{i+1}, which is impossible.

Linearization and circle diffeomorphisms

July 11, 2009

Today, i have learn that there is quite interesing interplay between the theory of linearization of maps f(z)=\lambda z + O(z^2) and analytic circle diffeomorphism theory.

In fact, according to results of P.Marco, for every such f(z) there is a fully invariant set K, which is called a Siegel continua, or Hedgehog (in the Cremer case) such that if we consider the conformal representation of the complement \bar{C} - K to the complement of the unit disk, then it conjugates the map f(z) to an anylitc circle diffeomorphism, which actually even has the same roattion number.And the last property is crucial!

The idea of the proof of the existence of such K is basically the following: in case of the rational multiplier \lambda there are quite a lot of things known about the topology in the neighborhood of the fixed point 0:  Leau-Fatou flower theorem. Which states that one can construct attracting and repelling petals, all have 0 on the boundary, which interchange and form the whole neighborhood of 0.  As a candidate for K in the rational case one can take the union of intersections of attracting and repelling petals: they are clearly fully invariant and the other properties can also be established. Then it left to use the density argument and show that the density of rational translates to our case in terms of existence of these continua K.

It follows that we have a dictionary between these two theories.  Even though a lot of things are the same in this dictionary, there are differences. For example (the only one i know so far), the set of lineariziable quadratic maps with irrationally indifferent fixed point – is the set of Bruno numbers (one way – Bruno, Siegel,  the necessity – Yoccoz). However, for circle diffeomorphisms, the linearizability condition is for the rotation number to be Herman!

Hello world!

July 11, 2009

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