Counting Orbits in a Chaotic Attractor: Periodic Points, Entropy and Dimension — Epoche B2
Challenging the Simplicity of Attractors in Dynamical Systems For a long time the attractors anyone could name were a fixed point and a limit cycle, and both are simple in a precise sense: they are single orbits, and the long-run behaviour of the system is described by naming one of them. Chaotic attractors are not simple in that sense, but saying so is easy and worth very little on its own. What is worth something is being able to say how much is inside one. This essay takes the geometric mechanism that produces chaos — a square of state space stretched, folded and put back where it came from — and counts what survives the operation: how many periodic orbits of each length, how fast that number grows, and what dimension the surviving set has. Every one of those quantities can be worked out exactly for the idealised case, and the answers are the reason the word simple has to go. From a flow to a map Continuous systems are awkward to count in, so we replace the flow by a map, in the standard way set out in Strogatz's textbook [1] . Choose a surface $\Sigma$ cutting across the trajectories of a three-dimensional flow, and record only the points where a trajectory pierces $\Sigma$ in a fixed direction. Successive piercings define the Poincaré map $$ \mathbf{x}_{k+1} = P(\mathbf{x}_k), \qquad \mathbf{x}_k\in\Sigma\subset\mathbb{R}^{2}, $$ where $\mathbf{x}_k$ is the $k$-th crossing point. The map loses nothing that matters: a periodic orbit of the flow becomes a periodic point of $P$, a bounded aperiodic trajectory becomes a bounded aperiodic sequence of crossings, and the dimension drops from three to two. A fixed point $\mathbf{q}=P(\mathbf{q})$ is a saddle when the derivative matrix $DP(\mathbf{q})$ has one eigenvalue with $|\mu|\gt 1$ and one with $|\lambda|\lt 1$: iterating pushes points away along one direction and pulls them in along the other. Its stable manifold $W^{s}(\mathbf{q})$ is the set of points whose forward iterates converge on $\mathbf{q}$, its unstable manifold $W^{u}(\mathbf{q})$ the set whose backward iterates do. What the map does to a square Stephen Smale's construction [2] , expounded in his 1967 survey of differentiable dynamical systems, takes a square $S$ in $\Sigma$ and applies one map with two ingredients. First stretch $S$ by a factor $\mu \gt 2$ horizontally and squeeze it by $\lambda \lt 1/2$ vertically, producing a long thin ribbon of the same area if $\mu\lambda = 1$ and less if $\mu\lambda \lt 1$. Then bend the ribbon into a horseshoe and lay it back across $S$. The bend is essential: it is what keeps everything inside a bounded region, and it is the only nonlinear thing in the whole construction. The intersection $f(S)\cap S$ consists of two vertical strips, call them $V_0$ and $V_1$, each of width $\lambda$, since that is the thickness of the ribbon that was laid across the square; the bend itself falls outside $S$. Pulling them back, $f^{-1}(V_0)$ and $f^{-1}(V_1)$ are two horizontal strips $H_0$ and $H_1$ inside $S$, each of height $1/\mu$ — a strip has to be that thin for the stretch by $\mu$ to carry it the whole way across. So a point stays in $S$ for one step exactly when it starts in $H_0$ or $H_1$, and everything in between is thrown out for ever. On the left, a square marked S contains two horizontal amber strips labelled H-zero and H-one. An arrow labelled f points to the right. On the right, the same square is shown with the image of the map drawn over it: a horseshoe-shaped band whose two arms cross the square as vertical blue strips labelled V-zero and V-one, joined by a bend that lies outside the square above it. Only the points lying in the two horizontal strips on the left have images inside the square on the right. H 0 H 1 S f V 0 V 1 S with f(S) drawn over it Fig. 1 — One step of the horseshoe map. The square $S$ is stretched horizontally by $\mu\gt2$, squeezed vertically by $\lambda\lt1/2$, bent, and laid back across itself. Only the two amber horizontal strips $H_0$ and $H_1$ have images inside $S$; they land as the two blue vertical arms $V_0$ and $V_1$, while the bend falls outside the square and those points never return. Repeating the operation forwards and backwards leaves the invariant set $\Lambda$, the product of two Cantor sets. Counting what survives Ask which points remain in $S$ for ever, forwards and backwards. Define $$ \Lambda = \bigcap_{n=-\infty}^{\infty} f^{\,n}(S). $$ Staying in for two forward steps requires being in one of the two horizontal strips and having an image in one of them, which selects two sub-strips inside each strip, of height $1/\mu^{2}$: four in all. After $n$ steps there are $2^{n}$ horizontal sub-strips of height $\mu^{-n}$. In the limit the forward-invariant set is a Cantor set in the vertical direction, with ratio $1/\mu$; running the argument backwards gives a Cantor set in the horizontal direction, with ratio $\lambda$, since that is the rate at which the vertical strips thin; and $\Lambda$ is the product of the two. The bookkeeping now becomes exact. Label each point of $\Lambda$ by which of the two strips it occupies at every step, past and future. This assigns to each point a doubly infinite string of $0$s and $1$s, and the labelling is a bijection: distinct points get distinct strings and every string occurs. Applying the map advances the string by one place, so on $\Lambda$ the dynamics is the shift $\sigma$ on the space $\{0,1\}^{\mathbb{Z}}$, $$ f\big|_{\Lambda} \;\cong\; \sigma, \qquad \sigma\bigl((s_k)\bigr)_k = s_{k+1}. $$ Every question about the dynamics on $\Lambda$ is now a question about strings, and the answers are arithmetic. A point has period $n$ exactly when its string repeats every $n$ places, so it is determined by one block of $n$ symbols: $$ \#\{\mathbf{x}\in\Lambda : f^{\,n}(\mathbf{x})=\mathbf{x}\} = 2^{\,n}. $$ Removing the points whose true period divides $n$ turns that into the count of genuine orbits, which for small $n$ runs $2, 1, 2, 3, 6, 9, 18, 30$ for $n = 1