## Archive for July, 2009

### 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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