# Homeomorphisms that admit a decomposition

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"? Apr 4, 2014 at 17:30
• I mean other explicit way to describe $T$ or a general formula of $T$ perhaps.
– Lucy
Apr 4, 2014 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. Apr 4, 2014 at 20:12
• @plusepsilon.de: Functions with different numbers of fixed points can't be conjugates, and the number of fixed points could be any integer $\ge 1$. Apr 6, 2014 at 5:43
• 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. Apr 7, 2014 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 $\mu$.