`Clean' Partitions

Figure 13: The four points in the orbit of (0111) provide a `clean' partition of the interval into three subintervals. The return map on these subintervals is shown. This map is used to construct a transition matrix, from which the topological entropy is computed.

The points in a periodic orbit provide good points for a partition of the interval into subintervals. This partition can be used to estimate the topological entropy for that periodic orbit. This is an estimate of the number of periodic orbits (by period) which the existence of the original periodic orbit implies (forces). We illustrate the idea first for the period four orbit $(0111)$ whose permutation is $(1324)$. The construction is illustrated in Fig. 4.13. The four points on this orbit are used to partition the interval between the extremal points 1 and 4 into three subintervals $I_{12}, ~I_{23}$, and $I_{34}$. Then a return map is drawn. This shows that 1 is mapped to 3, 2 to 4, 3 to 2, and 4 to 1. Since $f(1)=3$ and $f(2)=4$, $f(I_{12})=I_{34}$. We also easily see that $f(I_{23})=I_{23} \cup I_{34}$ and $f(I_{34})=I_{12}$. From this, or directly by inspection of the return map for the partition using this orbit, we find the following Markov transition matrix

$$\left[ \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 1 & 1\\ 1 & 0 & 0\end{array}\right]$$

The largest eigenvalue of this matrix is +1, so the topological entropy of this orbit is 0. This makes sense, since this period four orbit does not imply the existence of anything but its mother period two orbit (01) and its grandmother (1).

Figure 14: Three, four, and five points on the orbits $3_1$, $4_2$, and $5_3$ divide the interval into two, three, and four subintervals (left to right). The return maps for these `clean' partitions are shown.

Three more interesting cases are shown in Fig. 4.14. These are the return maps for the three orbits $3_1$, $4_2$, and $5_3$. For these three orbits we have

$$\begin{array}{cccc} {\rm Orbit}&3_1 & 4_2 & 5_3 \\ {\rm Per... ...975 \\ h_T &0.481211825 &0.609377863 &0.656255979 \end{array}$$ (17)