My teacher handed out an excerpt from a book by Robinson on chaotic dynamical systems. The excerpt is from a chapter on Markov partitions, and the following part has me confused:
Let $$f(x)= \frac{4}{3}x \quad (\text{mod }1).$$ This map has a discontinuity at $x= 3/4$. It does not have a Markov partition because $(4/3)^n$ is never an integer, so $f^n(1)$ never returns to $0$.
I'm not entirely surely why this contradicts the existence of a Markov partition. Unfortunately the excerpt doesn't actually contain a definition of Markov partition, but I think I have a vague understanding of the concept from earlier reading.
Could someone please explain why $f^n(1)$ not returning to $0$ prevents the existence of a Markov partition?
EDIT: I'll elaborate a bit on what I understand Markov partitions to be. Assuming the function above is defined on $[0,1]$, we can partition the domain into intervals $I_1,I_2,\ldots I_n$. Then, to each $x \in [0,1]$ we assign anitinerary, a sequence $s_0 s_1 s_2 s_3 \ldots$ such that $s_k = i$ if $f^k(x) \in I_i$. The idea is now that we can define a shift map on the space of such sequences, and try to show that this shift map has the same dynamics as $f$. This practice is called symbolic dynamics.
I understand that not all partitions are Markov partitions, but what exactly are the properties that a partition must have in order to qualify as a Markov partition? Specifically, why does $f^n(1) \not= 0$ prevent a Markov partition from existing?
- $\begingroup$Not having a definition of Markov partitions at hand, any answer will be shooting in the dark, because there are several definitions that work at several levels of abstraction: the subsets of the Markov partition can be as special as intervals and as general as measurable sets. Can you give some hints as to your vague understanding of the definition?$\endgroup$Lee Mosher– Lee Mosher2014-12-20 15:23:51 +00:00CommentedDec 20, 2014 at 15:23
- $\begingroup$@LeeMosher I've elaborated in an edit now.$\endgroup$Andrea– Andrea2014-12-20 20:31:32 +00:00CommentedDec 20, 2014 at 20:31
1 Answer1
In the context of a partition into subintervals, one way to turn this into a Markov partition with symbolic dynamics as you describe is to require that the restriction of $f$ to the interior of each subinterval $I_m$ is a homeomorphism onto the interior of a finite union of one or more consecutive subintervals $I_j \cup I_{j+1} \cup \cdots \cup I_k$. This is the definition used, for example, in the paper of Milnor and Thurston "On iterated maps of the interval"; their context requires the map to be continuous, but having discontinuity at the endpoints is still okay.
The best I can do is to point out that in order for this definition to be satisfied, the sequence $f^n(1) = (4/3)^n$ would have to take on only finitely many values, namely the endpoints of the intervals $I_1,I_2,…,I_n$. That is clearly impossible for the sequence $(4/3)^n$.
- $\begingroup$Okay. So if I understand correctly, since $f$ must map homeomorphically between the interiors of the $I_i$, then the endpoints must map only to other endpoints?$\endgroup$Andrea– Andrea2014-12-21 11:22:23 +00:00CommentedDec 21, 2014 at 11:22
- $\begingroup$@Andreas: According to the definition I gave, that's correct.$\endgroup$Lee Mosher– Lee Mosher2014-12-21 13:37:16 +00:00CommentedDec 21, 2014 at 13:37
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