Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Wada Lakes are three disjoint open subsets of $\mathbb R^2$ with common boundary. Originally they were constructed by hand, but they also arise naturally in the real life, that is, theory of dynamical systems. (See, for example, this http://www.math.cornell.edu/~hubbard/pendulum.pdf paper of John Hubbard.)

From dynamical construction it is clear that one can have a homeomorphism that permutes the lakes. For example, this expository paper http://www.ams.org/notices/200601/fea-coudene.pdf gives such a homeomorphism (in fact, a diffeomorphism) for lakes on the sphere $S^2$. The dynamics on the boundary of the lakes is chaotic.

==

Question 1: Let $U_1$, $U_2$ and $U_3$ be open subset of $\mathbb R^d$ with common boundary $C$. Assume that each $U_i$ is an image of an injective continuous map $\mathbb R^d\to\mathbb R^d$. Is there a homeomorphism of $\mathbb R^d$ that permutes open sets $U_i$ and is identity on $C$?

Question 2: Let $U_i\subset \mathbb R^d$, $i\in \mathbb Z$, be a disjoint collection of open sets each of which is homeomorphic to $\mathbb R^d$. Is there a homeomorphism of $\mathbb R^d$ that satisfies the following

  • $h$ maps $U_i$ onto $U_{i+1}$
  • every point in $\partial U_i$ is periodic under $h$

==

Edit: Below André Henriques has produced two examples when such a homeomorphism does not exist. Both examples make use of some local structure that cannot be permuted. It is still not clear what happens for "abstact Wada Lakes": are these examples exceptions or arbitrary Wada Lakes cannot be permuted as well?

Edit2 (Jan. 9th 2012)

  • In is clear now that there is no such homeomorphism if the dimension $d=2$. (See the comments below).
  • The answer to Q1 is also negative in higher dimension as explained in the reference given by Andres Koropecki.
  • For $d\ge 3$ there are examples by André Henriques of open sets for which one cannot find such a homeomorphism. In general, Question 2 is still open.
share|improve this question
2  
As far as I could tell, this bounty thing rarely works: if no one knows a good answer, it's unlikely to change upon offering them a bonus! A few recent featured questions ended up with no satisfactory answers (and some with no answers at all). –  Victor Protsak Aug 8 '10 at 4:51
    
I agree, but this question is important to me and what else can I do with points anyway. –  Andrey Gogolev Aug 8 '10 at 5:27
1  
Here's a variant: does there EXIST a pair U_1, U_2, U_3 as above and a homeomorphism as above? Do you know the answer? –  André Henriques Aug 8 '10 at 18:20
    
@Andrey: Do I understand you correctly that in you edit "what happens in general is still not clear", you want to know what happens when d=2? Or do you also want existence results? –  André Henriques Aug 9 '10 at 10:32
6  
I enjoyed seeing a question by Andrey answered by André and Andres. –  Andreas Blass Jan 10 '12 at 0:51
show 7 more comments

3 Answers 3

up vote 5 down vote accepted

In dimension 2, the answer is also "no".

Recall the classical construction of the Wada lakes. In the linked picture, one sees little "straits" connecting the red/blue/green regions at stage $n$ with the extra windy strip that is added at stage $n+3$. For example, one sees a blue strait roughly in the middle of the picture, and a red strait (barely visible) on the upper left part. The location of these straights are free parameters in the construction of the Wada lakes.


Now I proceed to construct the desired counterexample. Let $P$ be a point on the boundary of the yet-to-be-constructed Wada lakes. And let $U$ be a fixed ball around $P$. The straights can be picked so that:

  • The blue straights forms a converging sequence with limit point $P$.
  • The red and green straights all lie outside $U$.

In that case, there is no homeomorphism of $\mathbb{R}^2$ fixing $P$, and exchanging the blue lake with a lake of another color. The reason is the following:

  • There exist arbitrarily small neighborhoods $V$ of $P$ such that all connected components of Red-Lake $\cap V$ and Green-Lake $\cap V$ intersect $\partial V$ in exactly two intervals (taking $V$ be a metric ball will do).
  • For every sufficiently small neighborhood $V$ of $P$, there exist connected components of Blue-Lake $\cap V$ that intersect $\partial V$ in at least three intervals (the components containing the straights).
share|improve this answer
    
Thank you, I like this one even better. After your answers I imagine one can come up with various local obstructions near a fixed point P. –  Andrey Gogolev Aug 9 '10 at 20:12
    
I can't follow your procedure. In the standard construction of the lakes of Wada, at stage $n$ one of the lakes ($n$ mod 3) is enlarged by digging channels so that it comes within $a_n$ of each dry point on the island ($a_n\to 0$). If, as you say, the red and green straights all lie outside of U, then P will not be on the boundary of red and green lakes, which contradicts the Wada property. Are you somehow separating the lakes into "straights" and "channels" at every stage? –  Victor Protsak Aug 9 '10 at 20:45
    
Victor, staits are not the long channels that are responsible for the "common boundary feature". Straits are short channels that are being dug to make sure that lakes are connected. –  Andrey Gogolev Aug 10 '10 at 0:39
add comment

I'm late by a year, but just in case, the result from this one-page paper seems to answer your question (but it has nothing to do with the Wada property):

M. Brown and J. M. Kister, Invariance of complementary domains of a fixed point set, Proc. Amer. Math. Soc. 91 (1984), no. 3, 503–504. MR 744656

The theorem is as follows: Let $f$ be a homeomorphism of a connected topological manifold $M$ with fixed point set $F$. Then either (1) each connected component of $M-F$ is invariant, or (2) there are exactly two components and $f$ interchanges them.

share|improve this answer
    
Thank you very much for your reference. The 2d problem is indeed not too hard, it actually was resolved in the comments to the question some time ago. –  Andrey Gogolev Dec 12 '11 at 22:13
    
There is no restriction on the dimension of M in the result of Brown and Kister, so it should answer your general question (unless I'm missing something). –  Andres Koropecki Jan 2 '12 at 22:00
    
Yes indeed. I'm sorry, somehow I have looked up a wrong paper. –  Andrey Gogolev Jan 10 '12 at 0:16
add comment

The answer to your question is "no".

I'm working in $\mathbb{R}^3$. Let $C_1, C_2$ be open balls whose boundaries touch along an interval $J := \partial C_1 \cap \partial C_2$, and let $P \in J$ be a point on that interval. For concreteness:

$C_1 := ]0,1[ \times ]0,1[ \times ]0,1[ = (0,1)^3$

$C_2:= ]1,2[ \times ]1,2[ \times ]0,1[$

$J := \lbrace 1\rbrace \times \lbrace 1\rbrace \times ]0,1[$,

and $P := (1, 1, 0.5).$

It is possible to construct $U_1, U_2, U_3 \subset \mathbb{R}^3$ with common boundary, such that

  • $C_1\subset U_1$
  • $C_2 \subset U_2$
  • No $2$-disk in $U_3$ bounds the interval $J$.

The above properties exclude the possibility of a homeomorphism that fixes $P$ and exchanges $U_3$ with another $U_i$.


I now describe the open sets $U_1, U_2, U_3$ in a neighborhood of $P$.

Let $Q := ]1,2[ \times ]0,1[ \times ]0,1[$ and $R := ]0,1[ \times ]1,2[ \times ]0,1[$.

I'll describe what $U_1, U_2, U_3$ look like inside $Q$ and $R$, and leave the rest of the construction unspecified. Actually, the construction will be identical in $Q$ and in $R$, so I'll just describe the construction in $Q$.

Let $J' \subset J$ be a countable dense subset, $J' = \lbrace j_1, j_2, j_3, j_4, \dots \rbrace$, and let $Q' \subset Q$ be a countable dense subset $Q' = \lbrace q_1, q_2, q_3, q_4, \dots \rbrace$. Let $F:= \partial Q$ \ $(\partial C_1 \cup \partial C_2)$ be the part of the boundary of $Q$ that doesn't touch the other cubes. We inductively pick "cones" $V_n \subset Q$ with the following properties:

  • $V_n$ is homeomorphic to a 3-ball
  • $j_n \in \partial V_n$, and it is the unique point of $J$ in the boundary of $V_n$
  • $V_n \cap F$ is homeomorphic to a $2$-disc.
  • The closure of $V_n$ doesn't intersect the closures of the other $V_m$'s.
  • Unless $q_n$ is already in the closure of some other $V_m, m < n$, we make sure that $q_n \in V_n$.

One can then partition the $V_n$'s into three classes: $V_{3k+1}, V_{3k+2}, V_{3k+3}$. The open sets $U_i$ will be such that

$$ U _ i \cap Q = \bigcup_{k\ge 1} V _ {3k+i}. $$

Unless one does something very stupid, the subsets $U_i \cap Q$ will all have the same boundary within $Q$.

share|improve this answer
    
Thank you very much for your example! It gives me some hope that the answer in general might me "no" as well. One minor remark is that some effort is required to show that U_1, U_2 and U_3 are balls, but I think this can be arranged. –  Andrey Gogolev Aug 9 '10 at 0:30
    
To make sure that U_i is a ball, one needs to construct it by iterating the following method. At stage n, one has a 3-ball. Stage n+1 is constructed from stage n by gluing another 3-ball along an open 2-disc of its boundary. The infinite union is then homeomorphic to a 3-ball. It's not too difficult (but certainly messy to explain) to see that the above construction can be made in such a way that one gets the local picture described in my answer. –  André Henriques Aug 9 '10 at 10:08
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.