Permute Wada Lakes keeping the coastline intact? (still open in dim >2)

Question

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.

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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 $h$

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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.

Answer 1

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.

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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).

Answer 2

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.

Answer 3

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 $U_i$.

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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 $Q$.