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

  • $\begingroup$ In general, I don't think there are useful sufficient conditions. $\endgroup$ Oct 21, 2015 at 14:55
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    $\begingroup$ For commuting circle shifts, your question boils down to asking for sufficient criteria for the sum of two irrational numbers to be irrational. It seems difficult to come up with any usable criteria other than the tautological "the sum is not rational". $\endgroup$
    – Terry Tao
    Oct 21, 2015 at 16:08
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    $\begingroup$ @TerryTao this is actually a nice condition IMO. One can well imagine that extending the question to a larger family of maps including non-commuting maps, maps on higher dimensional tori etc, might lead to some very nice number-theoretic conditions analogous to your "the sum is irrational" condition. I agree however that at the level of generality asked for by the OP there is unlikely to be a useful, non-tautologically obvious condition. $\endgroup$
    – Dan Romik
    Oct 21, 2015 at 19:01
  • $\begingroup$ Sorry for pinging you on an unrelated question - I just wanted to let you know that there is a discussion on meta related to one of your questions. (That question is now deleted, so I was not able to ping you there.) $\endgroup$ Sep 19, 2017 at 4:04

1 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 $\dim(\text{fix}(L_{h}(L_{f}+L_{g})/2)$.


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 $(L_{f}^{*}+L_{g})L_{f}v/2=v$.


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