Glendinning chaos

A continuous interval map is chaotic in Glendinning's sense when some positive iterate has a horseshoe for an interval map.

An interval map $F:I\to I$ has a horseshoe when there are an open interval $J\subseteq I$ and disjoint open intervals $K_0,K_1\subseteq J$ such thatIterating inverse branches produces full two-symbol itinerary dynamics.

Suppose a interval map $F:A\to A$ maps every nondegenerate subinterval onto $A$ after finitely many iterates. Choose two separated closed intervals inside the interior of $A$ and a common iterate mapping each onto $A$. Between first-passage endpoint levels in each interval, take an open interval that maps exactly onto the interior of $A$. These two disjoint intervals are the branches of a horseshoe for an interval map for that iterate.

// Target: dynamical-systems.bigb

For a continuous interval map $F$ with a three-cycle $a<b<c$, after reflecting the coordinate if needed its ordering is $F(a)=b$, $F(b)=c$, $F(c)=a$. Then $F^2(a)=c$, $F^2(b)=a$. Also the intermediate value theorem gives $d\in (b,c)$ with $F(d)=b$, hence $F^2(d)=c$. Each of $[a,b]$ and $[b,d]$ therefore maps across $[a,c]$ under $F^2$. Choose first-passage subintervals between endpoint levels to avoid overshoots; their interiors are disjoint and both map exactly onto $(a,c)$. This is a horseshoe for an interval map, and establishes Glendinning chaos.

If a directed covering graph of an interval map has two distinct closed walks of the same length $n$, based at the same interval and with different first edges, successive use of the interval covering relation produces two subintervals with disjoint interiors that $F^n$ maps across the base interval. Hence $F^n$ has a horseshoe for an interval map.

Order the positive integers by odd numbers from $3$ upward, then twice the odds, then four times the odds, and so on, followed by descending powers of two and finally $1$. If a continuous interval map has a cycle of one period, it has cycles of every period later in this order. In particular, period three forces every positive period.

A period-three orbit forces an interval-covering graph with adjacency matrixIts number of closed pointed length-$n$ itineraries is $\mathrm{tr}(A^n)=L_n$. Möbius removal of lower periods and division by $n$ give four primitive cyclic classes at $n=7$ and five at $n=8$.