Complexity of a fixed point - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T10:04:02Z http://mathoverflow.net/feeds/question/65146 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/65146/complexity-of-a-fixed-point Complexity of a fixed point symboleon 2011-05-16T15:15:46Z 2012-12-09T21:45:12Z <p>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:</p> <p>Is it possible to partition the plane into a <em>finite</em> 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.</p> <p>[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...]</p> <p>s::l</p> http://mathoverflow.net/questions/65146/complexity-of-a-fixed-point/65202#65202 Answer by rpotrie for Complexity of a fixed point rpotrie 2011-05-17T04:12:39Z 2011-05-17T04:12:39Z <p>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 <a href="http://arxiv.org/PS_cache/math/pdf/0506/0506041v1.pdf" rel="nofollow">this paper</a> (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). </p> <p>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 $r$. </p> <p>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. </p> http://mathoverflow.net/questions/65146/complexity-of-a-fixed-point/115926#115926 Answer by Rodrigo A. Pérez for Complexity of a fixed point Rodrigo A. Pérez 2012-12-09T21:45:12Z 2012-12-09T21:45:12Z <p>Let $A_i$ be the sector $\theta \in \left[ \frac{i}{2r}2\pi, \frac{(i+1)}{2r}2\pi \right]$ with $i = 0, \ldots, (2r-1)$.</p> <p>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 <code>$\varphi^j(A_i) \cap A_i = \{ 0 \}$</code> (the only fixed point of $\varphi$) for all $j$ from $1$ to $r$, while the factor $\frac{1}{2}$ ensures no other periodic points.</p>