Irrational skew shift 2026-10-06
For irrational , the irrational skew shift is the skew product on the torus . It preserves normalized Lebesgue measure and has iterates modulo one. Its Koopman operator sends the Fourier basis character to . The resulting infinite chains of Fourier coefficients show that it is an ergodic transformation. Moreover, uniform equidistribution of an irrational skew shift proves unique ergodicity.
For an irrational rotation of the circle, let on , and let be any invariant Borel probability measure. For each integer , set . Invariance gives
For , irrationality forces , so the corresponding Fourier coefficient is zero. For it is one. These are exactly the Fourier coefficients of normalized Lebesgue measure . By the Stone-Weierstrass theorem, trigonometric polynomials are uniformly dense in the continuous functions on the circle group; hence and integrate every continuous function equally and are the same Borel probability measure. Since is invariant, the irrational rotation of the circle is uniquely ergodic, with unique measure .
For the irrational skew shift on the two-dimensional torus, write
The map is invertible, with modulo one, and it preserves as allowed in the question. We first prove the ergodic transformation property by the invariant-function characterization of ergodicity.
Let satisfy . Its expansion in the Fourier basis is in . Direct calculation of the Koopman operator gives
Uniqueness of the Fourier coefficients therefore implies
For , the magnitudes of the Fourier coefficients along all distinct indices , , are equal. By the Bessel inequality they are square summable, so every such coefficient must be zero. When , the relation becomes , which forces for . Only remains. Thus every invariant function is constant, and
To prove unique ergodicity, we will establish uniform averages for every continuous function directly. No theorem on unique ergodicity of skew products is needed. Induction on , using the old first coordinate in the second coordinate of , yields
In particular, for a Fourier basis element,
If and , this is a geometric series in , and
uniformly in .
For , we give the finite Van der Corput inequality for finite scalar sequences and its proof. For , extend by zero outside . Fix an integer and set . Every original summand appears times in the identity
The Cauchy-Schwarz inequality, followed by expansion of the squared window sums, gives
Consequently the Van der Corput inequality for finite scalar sequences is
The finite prefactor is important; the order of limits will be with fixed, followed by .
For the irrational skew shift character sequence above, differencing cancels the quadratic term:
For every fixed , is irrational, so another geometric series estimate gives
uniformly in . With fixed , the Van der Corput inequality for finite scalar sequences therefore implies
Letting proves that the averages of every nonconstant Fourier basis element converge uniformly to zero. The constant character has average one. This establishes uniform equidistribution of an irrational skew shift on all trigonometric polynomials.
The Stone-Weierstrass theorem makes these trigonometric polynomials uniformly dense in . If approximates a continuous with , then
Taking and then proves uniform convergence to for every continuous .
Finally, if is any invariant Borel probability measure for the irrational skew shift, invariance and this uniform convergence give
Thus , since continuous functions determine Borel probability measures on a compact metric space. We conclude
The uniform-average proof also shows that every starting point has the same limiting continuous-function averages, a stronger conclusion than the almost-everywhere assertion provided by the pointwise ergodic theorem.