I am interested in the topological entropy of the map: $T_s(x) = min\{ sx, s(1-x)\}$, the tent map with slopes $\pm s$ and peak at $x = \frac{1}{2}$. For $s \in [1, 2]$.

When $s = 2$, this is the standard tent map and the topological entropy is log(2).

I do not need an exact answer, I am just wondering if either

a) The topological entropy is log(s)

b) The topological entropy is NOT log(s).

I have found one source that claims (without clear source) that the answer is log(s). I am inclined to believe it is probably NOT log(s), but I cannot find a clear argument for why this is the case. If you could provide any direction or intuition for either argument, that would be greatly appreciated.

Thank you in advance.


Kind of a duplicate of toplogical entropy of general tent map, but OK.

As pointed out in the answer of the other question, up to a countable set (the backward orbits of $0$ and $1$), these tent maps are homeomorphically conjugate to a full shift on 2 symbols, therefore the topological entropy is $\log 2$.

The conjugation comes from the Markov partition given by $[0,1/s] \cup [1-1/s, 1]$.

Edit: this was for $s>2$.

Heuristic for why $h(T)=\log s$ if $1<s<2$:

Let's work with Bowen-Dinabourg's definition of the topological entropy in terms of the maximal number $N(n,\epsilon)$ of $(n, \epsilon)$ separated points.

Clearly, $N(0,\epsilon) \sim 1/\epsilon$ by taking regularly spaced points. Now notice that if you take points that are regularly spaced at intervals $\epsilon/s$, almost every pair is going to be $(1,\epsilon)$ separated: indeed, the distance between points on the same side of $1/2$ is going to be stretched by a factor $s$. For points on different sides of $1/2$, this is not necessarily true, but most of those points are starting far away from each other (ie at distance more than $\epsilon$) so they are also $(1,\epsilon)$ sepearated.

Similarly, you could convince yourself that $N(n,\epsilon) \sim s^n/\epsilon$. From there it follows that the entropy is $\log s$.

  • $\begingroup$ I understand and see that your solution is correct in the other case. It doesn't seem to me that $[0, 1/s] \cup [1-1/s, 1]$ is a Markov partition in this case because $T_s$ is not monotone on $[1-1/s, 1]$. Wouldn't the point $x = 1/2$ also have to be included in the partition? $\endgroup$ – curiousaboutthings Sep 13 '16 at 13:16
  • $\begingroup$ i guess it depends on the value of $s$. i was assuming $s>2$ $\endgroup$ – Glougloubarbaki Sep 13 '16 at 13:18
  • $\begingroup$ Okay, I made a little note that I am primarily interested in $s \in [1, 2]$. $\endgroup$ – curiousaboutthings Sep 13 '16 at 13:23
  • $\begingroup$ @katherine all right, I completed my answer by an heuristic $\endgroup$ – Glougloubarbaki Sep 13 '16 at 13:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.