Complexity of a fixed point - MathOverflow

Complexity of a fixed point

2011-05-16T15:15:46Z

Let $\varphi:\mathbb{R}^{2}\rightarrow\mathbb{R}^{2}$ be a homeomorphism of the plane with fixed point $p$, i.e. $\varphi(p)=p$, and no other periodic points. Let $r$ be a fixed natural number. My question is:

Is it possible to partition the plane into a finite number of closed sets $A_{i}$, $i=1,...,k$ ($\bigcup_{i=1}^{k}A_{i}=\mathbb{R}^{2}$), such that $\varphi^{j}(A_{i})\cap A_{i}\subset\{p\}$ for any $j=1,...,r$, $i=1,...,k$. (This condition means that the intersection $\varphi^{j}(A_{i})\cap A_{i}$ is either empty, or the point $p$). The problem here is the finiteness of the family ${A_{i}}$, as the answer is clearly affirmative for a countable family of $A_{i}$'s.

[I came across this problem while considering some concrete systems in the plane with a finite number of periodic points. Then it is possible to formulate an analoguous question, but I am asking the most simple variant here, since I cannot imagine neither a counterexample, nor a proof even in this case...]

s::l

Answer by rpotrie for Complexity of a fixed point

2011-05-17T04:12:39Z

You can extend your homeomorphism to the sphere with two fixed points and no other periodic points, and if it preserves a probability measure with total support it is called an irrational pseudo-rotation in this paper (see proposition 0.2 and recall Oxtoby-Ulam's theorem stating that if a homeomorphism preserves a probability measure with total support then it is conjugated to a conservative one).

There, it is proved that an irrational pseudo-rotation has, for every $n \geq 0$ a curve joining the fixed points such that it is disjoint from its first $n$ iterates and ordered exactly as in the irrational rotation. This allows to construct the desired $A_i$ if $n$ is sufficiently large compared with $r$.

When it does not preserve a measure with total support, the result does not apply, but many of the tools there may be useful, in particular, the Brower translation theorem.

Answer by Rodrigo A. Pérez for Complexity of a fixed point

2012-12-09T21:45:12Z

Let $A_i$ be the sector $\theta \in \left[ \frac{i}{2r}2\pi, \frac{(i+1)}{2r}2\pi \right]$ with $i = 0, \ldots, (2r-1)$.

Think of $\mathbb{R}^2$ as $\mathbb{C}$, and let $\varphi: z \mapsto \frac{1}{2}{\rm e}^{2\pi{\rm i}/r}z$. This rotates sectors two places counterclockwise, thus ensuring that $\varphi^j(A_i) \cap A_i = \{ 0 \}$ (the only fixed point of $\varphi$) for all $j$ from $1$ to $r$, while the factor $\frac{1}{2}$ ensures no other periodic points.