when the composition of two ergodic maps is ergodic?

Question

I would like to know if there are sufficient criteria for the composition of two ergodic maps to be still ergodic.

My context is piecewise affine transformations of the torus in arbitrary dimensions

Answer 1

If $f:X\rightarrow X$ is a measure-preserving transformation, then let $L_{f}:L^{p}(X)\rightarrow L^{p}(X)$ be the mapping defined by letting $L_{f}(h)=h\circ f$. If $f$ is invertible, then the ergodicity of $g\circ f^{-1}$ is equivalent to some properties of the averaging transformation $(L_{f}+L_{g})/2$ whenever $1<p<\infty$ regardless of whether one chooses the real or complex $L_{p}(X)$ spaces. Furthermore, if $p=2$, then the ergodicity of $g\circ f$ is equivalent to some properties of the average $(L_{f}^{*}+L_{g})/2$. These characterizations are analogous to the fact that $f$ is ergodic if and only if the only fixed points of $L_{f}$ are the constant functions. The space of fixed points of a linear operator $L$ shall be denoted by $\text{fix}(L)$.

Observe that if $f,g$ are measure preserving transformations, then $f\circ g$ is ergodic if and only if $g\circ f$ is also ergodic.

Proposition: Let $1<p<\infty$. Suppose that $f,g:X\rightarrow X$ are measure preserving transformations with $f$ bijective. Then the following are equivalent:

1. $g\circ f^{-1}$ is ergodic.

2. Whenever $h\in L^{p}(X)$, if $\|h\|_{p}=\|(L_{f}+L_{g})(h)/2\|_{p}$, then $h$ is a constant function.

3. Whenever $A$ is a Banach space and $L:A\rightarrow L^{p}(X),M:L^{p}(X)\rightarrow A$ are linear operators with $\|L\|\leq 1,\|M\|\leq 1$ and $L$ injective, we have $\dim(\text{fix}(M(L_{f}+L_{g})L/2))\leq 1$.

4. Whenever $h:X\rightarrow X$ is a bijective measure preserving transformation, we have $\dim(\text{fix}(L_{h}(L_{f}+L_{g})/2)\leq 1$.

Proof: $3\rightarrow 4$. $4$ is a special case of $3$.

$2\rightarrow 3$. Suppose that $h$ is a non-constant function. Then $$\|h\|_{p}>\|(L_{f}+L_{g})(h)/2\|_{p}\geq\|M(L_{f}+L_{g})(h)/2\|_{p}.$$ Therefore, if $a,b\in A$, and $a,b$ are linearly independent, then $L(a),L(b)$ are also linearly independent. Therefore, at least one of $L(a),L(b)$ is non-constant. Suppose therefore that $L(a)$ is non-constant. Then $$\|a\|\geq\|L(a)\|_{p}>\|M(L_{f}+L_{g})L(a)/2\|,$$ so $$a\neq M(L_{f}+L_{g})L(a)/2.$$

$1\rightarrow 2$. Suppose that $g\circ f^{-1}$ is ergodic. Suppose now that $\|h\|_{p}=\|(L_{f}+L_{g})(h)/2\|_{p}$. Then observe that $\|(L_{f}+L_{g})(h)/2\|_{p}=\|L_{f}(h)/2\|_{p}+\|L_{g}(h)/2\|_{p}$. Therefore, there are some positive $\alpha,\beta$ where $\alpha L_{f}(h)=\beta L_{g}(h)$, and we clearly have $\alpha=\beta=1$. Therefore, $h\circ f=L_{f}(h)=L_{g}(h)=h\circ g$, so $h=h\circ g\circ f^{-1}$. Therefore, since $g\circ f^{-1}$ is ergodic, we conclude that $h$ is a constant function.

$1\rightarrow 4.$ Suppose that $g\circ f^{-1}$ is not ergodic. Then there is some $A$ with $A=(g\circ f^{-1})^{-1}[A]$ and $0<\mu(A)<1$. Let $h=f^{-1}$. Then $$L_{h}(L_{f}+L_{g})\chi_{A}=L_{h}(\chi_{f^{-1}[A]}+\chi_{g^{-1}[A]})$$ $$=\chi_{A}+\chi_{h^{-1}[g^{-1}[A]]}=\chi_{A}+\chi_{(g\circ h)^{-1}[A]} =\chi_{A}+\chi_{A}.$$ Thus, $\chi_{A},1$ are two linearly independent elements of $\dim(\text{fix}(L_{h}(L_{f}+L_{g})/2)$.

Q.E.D.

Observe that in the above result, the measure preserving transformations $f,g$ need not be ergodic.

In the complex case, let $L^{p}_{-}(X)$ be the subspace of $L^{p}(X)$ consisting of all functions $f$ with $\int fd\mu=0$. Then whenever $A$ is a Banach space and $L:A\rightarrow L^{p}_{-}(X),M:L^{p}_{-}(X)\rightarrow A$ are linear operators where $L$ is injective, and $\lambda$ is an eigenvalue of $M(L_{f}+L_{g})L/2$, then $|\lambda|<\|L\|\cdot\|M\|$.

What if $f$ is not invertible?

Analogous results hold when $f$ is not invertible. We simply need to set $p=2$ so we have access to adjoints and replace the role of $L_{f^{-1}}$ with $L_{f}^{*}$. The adjoint $L_{f}^{*}$ is commonly known as the transfer operator of the map $f$.

Proposition: Let $p=2$. Suppose that $f,g:X\rightarrow X$ are measure preserving transformations. Then the following are equivalent.

1. $g\circ f$ is ergodic

2. If $h\in L^{2}(X)$ and $\|h\|_{2}=\|(L_{f}^{*}+L_{g})(h)/2\|_{2}$, then $h$ is a constant function.

3. If $A$ is a Banach space, and $L:A\rightarrow L^{2}(X),M:L^{2}(X)\rightarrow A$ are bounded linear operators where $L$ is injective and $\|L\|\leq 1,\|M\|\leq 1$, then $\dim(\text{fix}(M(L_{f}^{*}+L_{g})L/2))\leq 1$.

Proof: The directions $1\rightarrow 2,2\rightarrow 3$ are nearly the same as they were in the original problem, so these directions are omitted.

$3\rightarrow 1$. Suppose that $g\circ f$ is not ergodic. Then one can show that $f\circ g$ is not ergodic either. Then there is some $A$ with $(f\circ g)^{-1}[A]=A$ but where $0<\mu(A)<1$. Then for each $u\in L^{2}(X)$, we have $$\langle(L_{f}^{*}+L_{g})L_{f}\chi_{A},u\rangle =\langle L_{g}L_{f}\chi_{A},u\rangle+\langle L_{f}^{*}L_{f}\chi_{A},u\rangle$$ $$=\langle L_{f\circ g}\chi_{A},u\rangle+\langle L_{f}\chi_{A},L_{f}u\rangle=\langle\chi_{A},u\rangle+\langle\chi_{A},u\rangle=\langle 2\chi_{A},u\rangle.$$

Therefore, $(L_{f}^{*}+L_{g})L_{f}\chi_{A}/2=\chi_{A}$, so $\chi_{A},1$ are two linearly independent elements $v$ where $(L_{f}^{*}+L_{g})L_{f}v/2=v$.