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.

## 1 Answer

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 $\mu$.

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