Hyperbolic manifolds, discrete groups and ergodic theory

Math 277 – Fall 1996 – Berkeley C. McMullen

Homework Solutions 1

4. Let T : S^{1} → S^{1} be the irrational rotation T (z) = exp(2πiγ)z. Let I ⊂ S^{1} be a fixed interval, 0 < m(I) < 1, and let

a_{n}(z) = \frac{1}{n + 1} ∑_{0}^{n} χ_{I}(T^{i}z).

Let E_{n} = sup_{z} |a_{n}(z) − m(I)| denote the maximum error obtained if you estimate the measure of I by the frequency of visits of the orbit of z.

(c*) Estimate the size of E_{n} when γ = (1 + \sqrt{5})/2 is the golden ratio. Solution. (D. Thurston) We have E_{n} = O(log n/n). To see this, let

⟨q_{1}, q_{2}, q_{3}, …⟩ = ⟨1, 2, 3, 5, 8, 13, …⟩

be the Fibonacci numbers; and let p_{i} = q_{i−1}; then

|γ − \frac{p_{i}}{q_{i}} | ≤ \frac{C}{q_{i}^{2}}

for some constant C. Therefore

|kγ − \frac{kp_{i}}{q_{i}} | ≤ \frac{C}{q_{i}}

for k = 1, 2, …, q_{i}. This shows that by moving each point in the orbit of length q_{i} a distance less than C/q_{i}, we obtain equally spaced points. Since the error for q_{i} equally spaced points is O(1/q_{i}), we find that E_{q_{i}} = O(1/q_{i}).

A general n can be written as a sum of about log n Fibonacci numbers, so we obtain E_{n} = O(log n/n).