Cantor spectrum for Schrödinger operators with potentials arising from generalized skew-shifts

Abstract

Abstract. We consider continuous SL(2,R)-cocycles over a strictly ergodic homeomorphism which fibers over an almost periodic dynamical system (generalized skew-shifts). We prove that any cocycle which is not uniformly hyperbolic can be approximated by one which is conjugate to an SO(2,R)-cocycle. Using this, we show that if a cocycle’s homotopy class does not display a certain obstruction to uniform hyperbolicity, then it can be C 0-perturbed to become uniformly hyperbolic. For cocycles arising from Schrödinger operators, the obstruction vanishes and we conclude that uniform hyperbolicity is dense, which implies that for a generic continuous potential, the spectrum of the corresponding Schrödinger operator is a Cantor set. 1. Statementofthe Results Throughout this paper we let X be a compact metric space. Furthermore, unless specified otherwise f: X→X will be a strictly ergodic homeomorphism (i.e., f is minimal and uniquely ergodic) that fibers over an almost periodic dynamical system. This means that there exists an infinite compact abelian groupGand an onto continuous map h: X→G such that h ( f (x))=h(x)+α for someα∈G. Examples of particular interest include: • minimal translations of the d-torusT d, for any d≥1; • the skew-shift (x, y)↦ → (x+α, y+x) onT 2, whereαis irrational. 1.1. Results for SL(2,R)-Cocycles. Given a continuous map A: X→SL(2,R), we consider the skew-product X×SL(2,R) → X×SL(2,R) given by (x,�)↦→ ( f (x), A(x)·�). This map is called the cocycle ( f, A). For n∈Z, A n is defined by ( f, A) n = ( f n, A n). We say a cocycle ( f, A) is uniformly hyperbolic 1 if there exist constants c>0,λ>1 such that‖A n (x)‖>cλ n for every x∈Xand n>0. 2 This is equivalent to the usual hyperbolic splitting condition: see [Y]. Recall that uniform hyperbolicity is an open condition in C 0 (X, SL(2,R)). We say that two cocycles ( f, A) and ( f, Ã) are conjugate if there exists a conjugacy B∈C 0 (X, SL(2,R)) such that Ã(x)=B ( f (x))A(x)B(x) −1. Our first result is: