# Division by Zero

A blog about math, puzzles, teaching, and academic technology
Posted by: Dave Richeson | January 12, 2011

## Beautiful theorems about dynamical systems on the plane

I was reading through some papers written by my Ph.D. advisor (John Franks) from the early 1990′s and was reminded of a few beautiful results about the dynamics of planar homeomorphisms. So I thought I’d share them here.

For those of you who are not familiar with the terminology, a planar homeomorphism is a bijective function ${f:\mathbb{R}^{2}\rightarrow \mathbb{R}^{2}}$ for which ${f}$ and ${f^{-1}}$ are continuous. A simple example of a planar homeomorphism is a translation, such as ${f(x,y)=(x+1,y)}$.

We will look at these homeomorphisms as discrete dynamical systems. That is, we are interested in orbits of points: ${x,f(x),f(f(x)),\ldots}$ For simplicity we write ${f^{k}(x)}$ for ${f(f(f(\cdots(x))))}$ (${k}$ compositions of ${f}$). Intuitively you can think of the point ${x}$ hopping around the plane as we repeatedly apply the function ${f}$.

A point ${x}$ is a fixed point of ${f}$ if ${x=f(x)=f^{2}(x)=\cdots}$ and ${x}$ is a periodic point if ${x=f^{p}(x)=f^{2p}(x)=\cdots}$

From here onward I will assume (without saying it explicitly) that the homeomorphisms are orientation-preserving. This means that the image of a circle oriented clockwise is a closed curve oriented clockwise. A translation is always orientation preserving, but the reflection ${f(x,y)=(x,-y)}$ is orientation reversing.

To warm up, let’s give a theorem of Brouwer’s.

Theorem [Brouwer]. If ${f}$ has a periodic point, then it has a fixed point.

In fact, this theorem can be strengthened considerably. Roughly speaking, if ${f}$ has just about any type of recurrent behavior, then it must have a fixed point.

Here are two examples:

Theorem [Barge, Franks (1993)]. If there are disjoint arcs (or disjoint disks) ${A_{1},\ldots,A_{n}}$ such that ${f(A_{i})\cap A_{i}=\emptyset}$ for all ${i}$, and some iterate of ${A_{1}}$ intersects ${A_{2}}$, some iterate of ${A_{2}}$ intersects ${A_{3}}$, etc., and some iterate of ${A_{n}}$ intersects ${A_{1}}$, then ${f}$ has a fixed point.

In 2002 Jim Wiseman and I gave a short proof of the following result:

Theorem. If the orbit of every point intersects the unit disk ${x^{2}+y^{2}\le1}$, then there is a fixed point in the disk. (Actually, the hypotheses of this theorem are so strong that it holds when $f$ is not invertible and when the space is ${\mathbb{R}^{n}}$.)

The meta-contrapositive of this collection of theorems is that if there is no fixed point, then there is no recurrent behavior. In fact, as Brouwer discovered, if ${f}$ has no fixed point then ${f}$ behaves like a translation.

An open connected set ${L}$ is a domain of translation for ${f}$ if its boundary is ${B\cup f(B)}$, where ${B}$ is a proper embedding of ${\mathbb{R}}$ that separates ${L}$ and ${f^{-1}(L)}$ (as in the image below).

Theorem [Brouwer's plane translation theorem] If ${f}$ has no fixed points, then every point is contained in some domain of translation.

See Franks (1992) for a short proof of the theorem. Apparently Brouwer wrote several papers on the plane translation theorem (1909-1919), and since 1920 others have had to go back and clean up the statement and the proof of his theorem. As Brouwer discovered, one has to be very careful with the topology of the plane. For instance, one pathological example that sent Brower back to the drawing board was the Lakes of Wada (isn’t that a great name?). In 1917 Takeo Wada discovered that it is possible to find three disjoint connected open sets in the plane that all have the same boundary! Here’s a picture of three such sets.

Now, consider the iterates of a set ${S\subset \mathbb{R}^{2}}$, ${f(S),f^{2}(S),\ldots}$, and keep track of which sets ${f^{k}(S)}$ are disjoint from ${S}$. Call this collection of integers ${k>0}$, ${E(S)}$; that is,

${E(S)=\{k>0:f^{k}(S)\cap S=\emptyset\}}$.

Theorem [Barge, Franks]. If ${f}$ has no fixed points and ${S}$ is an open or closed connected set, then ${E(S)}$ is closed under addition.

For example, if ${f^{2}(S)\cap S=\emptyset}$ and ${f^{5}(S)\cap S=\emptyset}$, then ${f^{k}(S)\cap S=\emptyset}$ for ${k=2+2=4}$, ${k=2+4=6}$, ${k=2+5=7}$, etc.

An immediate consequence of this theorem is that if ${S}$ and ${f(S)}$ are disjoint (i.e., ${1\in E(S)}$), then so are ${S}$ and ${f^{k}(S)}$ for all ${k>0}$ (i.e., ${E(S)=\mathbb{Z}^{+}}$). In particular it follows that:

Corollary. If ${S}$ and ${f(S)}$ are disjoint, then ${f^{k}(S)}$ and ${f^{j}(S)}$ are disjoint for all ${k\ne j}$.

In particular, if we added ${f^{2}(S), f^{3}(S), f^{4}(S), \ldots}$ to the the image below, then they would all be disjoint.

Franks and Barge also prove a converse to this theorem.

Theorem [Barge, Franks]. Suppose ${E}$ is a set of positive integers that is closed under addition. Then there is a translation ${T:\mathbb{R}^{2}\rightarrow\mathbb{R}^{2}}$ and an open topological disk ${D}$ such that ${E(D)=E}$.

Two-dimensional dynamical systems is a fascinating area of mathematics. These are only a few of the many beautiful theorems.

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Posted in Math | Tags: Brouwer, dynamical systems, John Franks, Lakes of Wada, Marcy Barge, plane, surface, topology

## Responses

1. Isn’t it “such that T(D)=E” instead of “such that E(D)=D” in the last theorem.

By: Kévina on January 13, 2011
at 7:49 am

• No, that’s correct. I could have been more precise by writing $E(D,T)$ rather than $E(D)$.

By: Dave Richeson on January 13, 2011
at 9:52 am

2. [...] blogs: MathBlogging.org Dynamical systems on a plane Exotic spheres ESP and statistics Sums of [...]

By: Weekend miscellany — The Endeavour on January 15, 2011
at 9:58 am

3. So, the open disk D in the last theorem should be unbounded in general, something like an infinite centipede, isn’t it?
For closed disks the theorem is failing (at least for translations), but do you know some results about the set E(D) in this case, refining the fact that it is closed under adition? For example, in the case of translation the complement of E(D) should be finite, as D is bounded.

si-top

By: simba-top on March 29, 2011
at 1:11 am
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