Let $I^n:=[0,1]^n$ and $T$ be a homeomorphism on $I^n$.
If $T$ admits a decomposition of $I^n=A\cup B\cup C$ with $A,B,C$ Lebesgue measurable and mutually disjoint such that $$T(A)=B, T(B)=A \ \text{and}\ m(C)=0,$$ where $m(C)$ is the Lebesgue measure of $C$, then
What is the general solution of $T$?
I know some examples of $T$, such as $T$ with $T^{\circ 2k}=Id$ admits such a decomposition.
Another example, $T(x)=(t_1(x_1),t_2(x),...,t_n(x))$ with $t_1$ strictly monotone also admits a decomposition.

  • What do you mean by "solution"? – Ryan Budney Apr 4 '14 at 17:30
  • I mean other explicit way to describe $T$ or a general formula of $T$ perhaps. – Lucy Apr 4 '14 at 17:32
  • I think this question is interesting, but I don't quite understand the choice of tags. Maybe measure-theory and ds.dynamical-systems and more appropriate. – Pietro Majer Apr 4 '14 at 20:12
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    @plusepsilon.de: Functions with different numbers of fixed points can't be conjugates, and the number of fixed points could be any integer $\ge 1$. – Robert Israel Apr 6 '14 at 5:43
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    Homeomorphisms don't preserve null sets. For example, there is a homeomorphism of $[0,1]$ to itself that maps the usual Cantor set to a "fat" Cantor set of positive Lebesgue measure. Homeomorphisms that are bi-Lipschitz preserve null sets. – Robert Israel Apr 7 '14 at 17:06

Obvious necessary condition: the set of fixed points and periodic points of odd order must have measure $0$.

EDIT: And there's another class of examples that do not admit such a decomposition. Suppose there is a probability measure $\mu$ invariant under $T$ which is absolutely continuous with respect to Lebesgue measure. Then if $T$ admits a decomposition, it can't be strongly mixing with respect to $\mu$.

  • Yes, you are right. – Lucy Apr 7 '14 at 15:54

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