When I calculate the measure-theoretic entropy of different maps, I found that some of them is equal to zero, so I am curious about what is special about the map whose entropy is zero? Does the entropy equal to zeros mean the map is "regular", if yes what does this "regular" mean?

For example , in $X=[0,1)$, the map $\,T(x)=2x\;(\mod1)$'s entropy is $\log2$, but the entropy of $\,T(x)=x^2$ is zero. If I have a initial value, I can easily calculate what is the next value. So in this in this respect I don't understand what is the key difference between the two maps.

## Best Answer

Below I present two heuristics to think about the difference between zero entropy and positive entropy. I am conflating multiple ideas from entropy theory, but I also provide references where the ideas are presented more rigorously.

The first heuristic is purely ergodic theoretical. Fix a partition of the phase space; this corresponds to fixing a resolution with which one is observing the trajectories of the system. Entropy corresponds to how surprising it is to find out which cells of the partition a trajectory will hit in the future (i.e. the larger the entropy the more surprising the future is), and in which order, given complete past of the trajectory. Zero entropy means that if the whole past is known, there is no surprises in the future ever ("deterministic"; this is the heuristic Katok gives in the survey "Fifty Years of Entropy in Dynamics: 1958-2007" Oliver Díaz linked above (though this survey is not limited to Kolmogorov's work only); here is the link from Katok's own webpage: https://www.personal.psu.edu/axk29/pub/E50-aspublished.pdf), whereas in the case of positive entropy despite the full knowledge of the past, the future is still surprising.

To compare the examples in the OP, take the (a.e.) partition $\mathcal{P}=\{L=[0,1/2], R=[1/2,1]\}$ of $X=[0,1]$. Fix a point $x\in X$. First consider $x\mapsto 2x$. All possible past trajectories of $x$ w/r/t the partition $\mathcal{P}$ gives a tree which bifurcates into two at each time step. Knowing the full knowledge of the past of $x$ w/r/t $\mathcal{P}$ would be a choice of a branch of this tree. Say you are given such a branch. In which cell of $\mathcal{P}$ will $x$ end up? The answer is a coin flip; entropy is positive. (Your previous question Generator of doubling map when calculating its Measure-theoretic entropy was about a rigorous version of this argument.)

Now consider the trajectory of $x\in X$ under $x\mapsto x^2$ w/r/t the same partition $\mathcal{P}$. Note that this time either $x=0$ and for all time it stays fixed, or else it gets arbitrarily close to $1$, since the map is an increasing homeomorphism. Again we have a bifurcating tree of pasts. We claim that there is no surprises as to the demise of $x$. This is because either there is an $R$ in the past of $x$, in which case $x$ will be arbitrarily close to $1$; or else by going forward in time finitely many steps one can see an $R$, in which case the previous destiny applies. The only other option is all $L$'s, no matter how further (in finite steps) one goes in the future, in which case $x=0$. (See Direct proof for any homeomorphism on $[0,1]$ having zero entropy or Entropy of a homeomorphism on a unit circle is zero for rigorous versions of this argument, by way of the Variational Principle; see Relation between topological entropy and metric entropy)

The second heuristic is based on Katok's definition of measure theoretical entropy and it's intimately related to Lyapunov exponents (see his paper "Lyapunov exponents, entropy and periodic orbits for diffeomorphisms" (p.142), https://www.personal.psu.edu/axk29/pub/KatokIHES1980.pdf; for some discussion on the intuition for Lyapunov exponents see Lyapunov exponent for 2D map?). This heuristic is based on using the so-called Bowen metrics (see There is no expansive homeomorphism on $S^1$); which quantify the separation of trajectories in time. Then entropy is the exponential rate at which nearby orbits separate from each other.

To compare the two examples in the OP, for $x\mapsto 2x$, if $p$ and $q$ are two points very close to each other; their distance will grow like $n\mapsto 2^n$ initially, and then how distant they will be asymptotically will be a matter of a coin flip (if we have finite resolution) (a rigorous version of this is by way of the Pesin Entropy Formula: the entropy of a system is equal to the sum of its positive Lyapunov exponents). On the other hand, for $x\mapsto x^2$, if $p$ and $q$ are nearby points; we can distinguish in finite resolution if one of the points is $0$, in which case their distance will get arbitrarily close to $1$; else their distance will get arbitrarily close to $0$, so that the system is deterministic.

Positive entropy / indeterminism is also associated to "irreversibility". This could be misleading in our two examples, as $x\mapsto x^2$ is invertible and has zero entropy; while $x\mapsto 2x$ is non-invertible and has positive entropy. However from the perspective of entropy theory this non-invertibility is not a major distinguishing factor; indeed one can adjoin the pasts of the system to obtain an invertible system (see Uniqueness of a homomorphism in a projective limit dynamical system; showing that a projective limit system is compact - Is my proof correct? or Milnor's notes, p.2-5 at https://www.math.stonybrook.edu/~jack/DYNOTES/dn2.pdf). Likewise an invertible map may have positive entropy.

Finally, somewhat in jest, assuming that if there were anything special about zero entropy maps, zero entropy maps would be special, one can answer the question in the title negatively: on the contrary maps with non-zero entropy are special, by a classical theorem of Rohlin.

To make sense of this, consider the group $G$ of all bimeasurable, measure preserving self-maps of the $X=[0,1]$ endowed with the Borel $\sigma$-algebra and Lebesgue measure $\lambda$. Define two distances on $G$:

$$d_{\text{weak}}(T,S)= \sum_{n\in\mathbb{Z}_{\geq1}} \dfrac{1}{2^n}(\,\lambda (T(A_n)\,\triangle\, S(A_n))\,+\,\lambda (T^{-1}(A_n)\,\triangle\, S^{-1}(A_n))\,),$$

(where $A_\bullet:\mathbb{Z}_{\geq1}\to \mathcal{B}(X,\lambda)$ is a dense family of elements in the measure algebra w/r/t the distance $(B_1,B_2)\mapsto \lambda(B_1\triangle B_2)$; see Measurability of (random) set valued functions)

$$d_{\text{unif}}(T,S)=\lambda(\{x\in X\,|\, T(x)\neq S(x)\})$$

Both of these two distances make $G$ a topological group ($(G,d_{\text{weak}})$ is separable, $(G,d_{\text{unif}})$ is not). Then we have:

Theorem (Rohlin):The set of zero entropy elements in $G$ is a dense $G_\delta$ w/r/t $d_{\text{weak}}$ and $d_{\text{unif}}$.(See Parry's

Entropy and Generators in Ergodic Theory, p.103 for a discussion.)Thus generically a bimeasurable measure-preserving map has zero entropy. In this sense positive entropy is more special.