1999
DOI: 10.1112/s0024610799006961
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Piecewise Absolutely Continuous Cocycles Over Irrational Rotations

Abstract: A. IWANIK, M. LEMAN ! CZYK  C. MAUDUIT A For an irrational rotation α of the circle group T l R\Z and a piecewise absolutely continuous function f : T ,-R, the unitary operator Vh(x) l e# π if(x) h(xjα) on L#(T) is studied. It is shown that if f has a single discontinuity with non-integer jump then V is κ-weakly mixing for some κ with 0 QκQ 1. In particular V has continuous singular spectrum. The property of κ-weak mixing (with possible change of the value of κ, 0 QκQ 1) holds for all irrational rota… Show more

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Cited by 24 publications
(17 citation statements)
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“…For a class of cocycles inspired by [15,17], we show that the corresponding unitary operators have purely Lebesgue spectrum. The result is not new [14] but our proof is completely new and does not rely on the study of the Fourier coe cients of the spectral measure. In addition, our approach leads naturally to a limiting absorption principle and to the obtention of a large class of locally smooth operators.…”
Section: Introductionmentioning
confidence: 95%
See 2 more Smart Citations
“…For a class of cocycles inspired by [15,17], we show that the corresponding unitary operators have purely Lebesgue spectrum. The result is not new [14] but our proof is completely new and does not rely on the study of the Fourier coe cients of the spectral measure. In addition, our approach leads naturally to a limiting absorption principle and to the obtention of a large class of locally smooth operators.…”
Section: Introductionmentioning
confidence: 95%
“…The rst part of Proposition 3.2 is not new; the nature of the spectrum of U was already determined in [14,15] under a slightly weaker assumption (h absolutely continuous with h 0 of bounded variation). On the other hand, we have not been able to nd in the literature any information about globally U-smooth operators.…”
Section: Cocycles Over Irrational Rotationsmentioning
confidence: 99%
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“…For example, S(ϕ) = 0 implies continuous spectrum on H ⊥ 0 (see [9]). Moreover, if ϕ has a single discontinuity with S(ϕ) ∈ R \ Z, then T ϕ has continuous singular spectrum on H ⊥ 0 .…”
Section: (T × T λ ⊗ λ) Moreover the Operator U T ϕ : H M → H M Is Umentioning
confidence: 99%
“…Veech proved that µ θ (E) exists for every interval E if and only if θ has bounded partial quotients in its continued fraction expansion, and in this case if t ∈ 2Z · θ + Z then µ θ (E) = 1/2. For closely related results, see [8], [9], [11], [13], [15].…”
mentioning
confidence: 99%