# Generator determined by finitely many translates implies zero entropy

Let $T$ be a measure preserving transformation of a standard probability space $(X,\mathcal{B},\mu)$. A partition $\alpha$ of $X$ is said to be a generator for $T$ if the smallest $T$ invariant $\sigma$-algebra of measurable sets containing the pieces of $\alpha$ is all of $\mathcal{B}$.

Suppose there is a generator $\alpha$ for $T$ and a finite set $F \subseteq \mathbb{Z} \setminus \{0\}$ such that $\alpha$ is measurable with respect to $\bigvee_{n \in F} T^n \alpha$. Does this imply that $T$ has zero entropy?

## 1 Answer

No, this is not true. This is true if $F \subset \mathbb{N}$, but not in the general case. I am lost with the objects of ergodic theory so I start by rephrasing in the probabilistic language.

Let ${(Z_n)}_{n \geq 0}$ be the stationary (shift-invariant) process defined by $Z_n(x)=\alpha(T^n(x))$ where $\alpha(x)$ denotes the element of the partition $\alpha$ to which $x$ belongs. The (usual) definition of the entropy of $T$ with respect to $\alpha$ is $h(T,\alpha)=\lim \frac{H(Z_1,\ldots,Z_n)}{n}$, and it is well known that $h(T)=h(T,\alpha)$ when $\alpha$ is a generating partition. Moreover, introducing the decreasing sequence of $\sigma$-fields ${\cal F}_n=\sigma(Z_m; m \geq n)$, it is well known that $h(T,\alpha)=H({\cal F}_0 \mid {\cal F}_1)$.

Assume your $F$ is included in $\mathbb{N} \setminus \{0\}$. That means that $Z_0$ is measurable with respect to $\sigma(Z_k; k \in F) \subset {\cal F}_1$. Therefore ${\cal F}_0 \subset {\cal F}_1$ (then ${\cal F}_0={\cal F}_1)$ because ${\cal F}_0 = {\cal F}_1 \vee \sigma(Z_0)$. Hence $H({\cal F}_0 \mid {\cal F}_1)=0$.

If your $F$ contains some negative integers, then your assumption is equivalent to $\sigma(Z_{n_0}) \subset \sigma(Z_k; k \in F)$ where $F \subset \mathbb{N}\setminus\{n_0\}$. It is easy to get a counter-example for $n_0=1$ and $F=\{0,2\}$.

Take a stationary process ${(Z_n)}_{n \geq 0}$ with non-zero entropy, taking its values in a finite alphabet, and define a new stationary process ${(Y_n)}_{n \geq 0}$ by setting $Y_n = (Z_n, Z_{n+1})$. Then $Y_1 \subset \sigma(Y_0,Y_2)$. But it is clear from the formula defining $h(T,\alpha)$ that ${(Y_n)}_{n \geq 0}$ has the same entropy as ${(Z_n)}_{n \geq 0}$. This is also clear from the other formula because the process ${(Y_n)}_{n \geq 0}$ has the same decreasing sequence of $\sigma$-fields as ${(Z_n)}_{n \geq 0}$.

For example, if you take for ${(Z_n)}_{n \geq 0}$ a sequence of independent symmetric Bernoulli variables, then you can see ${(Y_n)}_{n \geq 0}$ as a stationary Markov chain on the vertices of a square $ABCD$, where $Y_n$ has the uniform distribution and it jumps from $A$ to $A$ or to the "next" point $B$ with equal probabilities, it jumps from $B$ to $B$ or to the "next" point $C$ with equal probabilities, etc. Its entropy is $\ln 2$ and it is easy to see that $Y_1 \subset \sigma(Y_0,Y_2)$.

So the answer is no, and it is still no with additional assumptions such as ergodicity, $K$, Bernoullicity.

I seize the opportunity to give a try to the markovchain R package:

library(markovchain)

Y <- new("markovchain", states = c("A", "B", "C", "D"),
transitionMatrix = 0.5*matrix(c(c(1, 1, 0, 0),
c(0, 1, 1, 0),
c(0, 0, 1, 1),
c(1, 0, 0, 1)),
byrow=TRUE, nrow = 4),
name = "Counter-example")

plot(Y) 