ISEGORIA / MATH ENCYCLOPEDIA
Ergodic theory: when time averages equal space averages
Ergodic theory asks what a single orbit of a dynamical system does in the long run. If a map preserves area or probability and cannot be split into two invariant pieces, then for almost every starting point the fraction of time the orbit spends in a region equals the size of that region. That one idea explains why an irrational rotation fills the circle evenly, why chaotic maps forget where they started, and why the powers of 2 begin with a 1 nearly a third of the time.
Before you begin: Measure theory and dynamical systems
Predict, manipulate, then check your reasoning against the example and question. Graphs illustrate the mathematics; they do not replace a proof.
1. Rotating a circle by an irrational angle
Turn a circle of circumference 1 by a fixed amount α again and again, and mark where the point 0 lands. If α is a fraction p/q the marks repeat after q turns. If α is irrational they never repeat, and something stronger happens: every arc of the circle receives a share of the marks equal to its length, a fact Weyl proved in 1916. This is the simplest ergodic system: the rotation preserves length, and no arc other than the whole circle and nothing is carried exactly onto itself. Add points one by one, watch the gaps between them, and move the arc to test the equidistribution.
Worked example. Take the golden mean \(\alpha=(\sqrt5-1)/2\approx0.618\). The first four points \(0,0.618,0.236,0.854\) cut the circle into gaps of lengths 0.146, 0.236 and 0.382, and \(0.382=0.146+0.236\); the fifth, 0.472, splits the longest gap and leaves only two lengths. After 100 points the arc \([0.1,0.35)\), of length 0.25, holds exactly 25 of them, and the star discrepancy is about 0.02, close to the best possible for any sequence.
Watch out. Equidistribution is a statement about the limit; how fast it happens depends on how well α can be approximated by fractions. The golden mean is the worst-approximable number and gives the most even spread, while \(\pi-3\) lies very close to 1/7, so its first fifty points bunch into seven clusters before spreading out. The points are computed in floating point, which is exact enough here.
Why are there never more than three different gap lengths?
Because the gaps are all made by the same rotation. Each gap starts at some point \(x_i\); shifting it by \(-\alpha\) gives the gap at \(x_{i-1}\), unless that step would leave the set, which can happen at only a few points (the newest point and those next to 0). So all gaps are copies of at most three, and the three-gap theorem, conjectured by Steinhaus and proved in the 1950s, shows the longest is the sum of the other two.
2. Mixing: the cat map and the baker’s map
Ergodicity says a single orbit visits everything fairly. Mixing is stronger: any blob of initial conditions spreads until it is evenly smeared over the whole space, so that knowing which blob you started in tells you nothing about where you are later. Arnold’s cat map does this on the torus with an integer matrix of determinant 1, which preserves area while stretching in one direction and squeezing in another. Move the disc, step the map, and compare it with a rotation of the torus, which is ergodic but never mixes.
Worked example. The matrix has eigenvalues \(2.618\) and \(0.382\), whose product is the determinant 1. A disc of radius 0.17 is stretched by 2.618 per step, so after 6 steps a filament of the disc is \(2.618^6\approx320\) times longer and wraps the torus dozens of times. With a test square of area 0.09, the fraction of the disc inside it is 0 at the start and stays within 0.01 of 0.09 from step 3 on.
Watch out. The picture is drawn by pulling each pixel back through the inverse map, so once the filaments are thinner than a pixel the image shows a fine grey-gold texture rather than the true filaments: the evenness you see is real, the texture is sampling. The curve counts 6000 sample points of the disc. On a finite grid of rational points the cat map is a permutation and pictures return after a fixed number of steps, which is not a contradiction: those points have measure zero.
Why does the rotation of the torus never mix, even though it is ergodic?
A rotation is an isometry: it moves the disc without changing its shape, so the disc stays a disc of the same radius forever. The fraction of it inside the square is then 0 whenever the disc is elsewhere and large when it passes through, and those values keep recurring. Ergodicity only needs the long-run average of that fraction to be \(\mu(A)\mu(B)\); mixing needs the fraction itself to converge, which requires stretching.
3. Birkhoff averages and Benford’s law
Birkhoff’s ergodic theorem says that for an ergodic map preserving a probability measure μ, the time average of any integrable quantity along almost every orbit converges to its average over space. The measure decides what “fair” means: for the doubling map it is length, for the logistic map at r = 4 it is the arcsine distribution, which crowds near the ends. The same theorem, applied to a rotation by log₁₀ 2, explains why the powers of 2 obey Benford’s law. Choose a system, drag the starting point and lengthen the orbit.
Worked example. For the logistic map, \(\int_0^1x^2\,\frac{dx}{\pi\sqrt{x(1-x)}}=\frac38\), while the uniform average would be \(\frac13\): an orbit of \(10^4\) steps from almost any start lands within about 0.005 of 0.375. Among the first \(10^4\) powers of 2, 3010 begin with a 1, against \(10^4\log_{10}2=3010.3\), and 458 begin with a 9, against 457.6.
Watch out. The theorem says almost every starting point, not every one: \(x_0=1/3\) under the doubling map cycles between 1/3 and 2/3 and averages 5/18 for \(x^2\). The doubling orbits here are generated as shifts of binary digits, with the first 30 digits from the chosen \(x_0\) and the rest random, because repeated doubling in floating point reaches 0 after 53 steps. The logistic orbit is iterated in floating point; it is not the exact orbit of \(x_0\) after a few dozen steps, but it stays close to some true orbit.
Why do the logistic map’s orbits pile up near 0 and 1?
Near \(x=1/2\) the map is flat, since its derivative \(4-8x\) vanishes there, so a whole interval of points near 1/2 is sent into a tiny interval near 1, and from there into a tiny interval near 0. Orbits therefore return to the ends far more often than an even spread would suggest. The substitution \(x=\sin^2(\pi\theta/2)\) turns the map into the tent map, which spreads \(\theta\) evenly, and the arcsine density is the image of that even spread.