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.

  • $\begingroup$ What do you mean by "solution"? $\endgroup$ – Ryan Budney Apr 4 '14 at 17:30
  • $\begingroup$ I mean other explicit way to describe $T$ or a general formula of $T$ perhaps. $\endgroup$ – Lucy Apr 4 '14 at 17:32
  • $\begingroup$ 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. $\endgroup$ – Pietro Majer Apr 4 '14 at 20:12
  • 1
    $\begingroup$ @plusepsilon.de: Functions with different numbers of fixed points can't be conjugates, and the number of fixed points could be any integer $\ge 1$. $\endgroup$ – Robert Israel Apr 6 '14 at 5:43
  • 1
    $\begingroup$ 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. $\endgroup$ – 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 $\mu$.

| cite | improve this answer | |
  • $\begingroup$ Yes, you are right. $\endgroup$ – Lucy Apr 7 '14 at 15:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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