2001
DOI: 10.1112/s0025579300014376
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Geometry of the gauss map and lattice points in convex domains

Abstract: Let Q be a convex planar domain, with no curvature or regularity assumption on the boundary. Let N e (R) -card{RCl e r\Z 2 }, where Q e denotes the rotation of Q by 9. It is proved that, up to a small logarithmic transgression, N e (R) = \Cl\R 2 + O(R 2/i ), for almost every rotation. A refined result based on the fractal structure of the image of the boundary of Q under the Gauss map is also obtained.Introduction. Let Q be a bounded convex planar domain, and let = card{RQnZ 2 }. It was observed by Gauss that … Show more

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Cited by 11 publications
(7 citation statements)
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“…where ρB denotes the rotation of B by ρ ∈ S d−1 viewed as an element of SO(d). See, for example, [Ios2001], [BCIPT02], and [BIT01] for a detailed discussion of applications of average decay of the Fourier transform to lattice point problems. Also note that Theorem 1.2 below shows that convexity may be replaced by a C d , d > 1, is greater than d+1 2 , then ∆ B (ρB) has positive Lebesgue measure for almost every ρ ∈ S d−1 viewed as an element of SO(d).…”
Section: Introductionmentioning
confidence: 99%
“…where ρB denotes the rotation of B by ρ ∈ S d−1 viewed as an element of SO(d). See, for example, [Ios2001], [BCIPT02], and [BIT01] for a detailed discussion of applications of average decay of the Fourier transform to lattice point problems. Also note that Theorem 1.2 below shows that convexity may be replaced by a C d , d > 1, is greater than d+1 2 , then ∆ B (ρB) has positive Lebesgue measure for almost every ρ ∈ S d−1 viewed as an element of SO(d).…”
Section: Introductionmentioning
confidence: 99%
“…the proof of Lemma 2.2 below). We remark that it is well known that for almost all rotations A ∈ SO(d) the error terms E AΩ (t) improve, see [5], [31], [32], [23], [15], and [2].…”
Section: A Lattice Point Estimatementioning
confidence: 82%
“…We may assume that U is the unit cube, and that the support of χ has small diameter. We decompose 2]) so that ψ is equal to 1 on the support of ψ. Denote by T k the convolution operator with Fourier multiplier m δ,k and by R k the convolution operator with Fourier multiplier ψ(δ −1/2…”
Section: Proofmentioning
confidence: 99%
“…θ, with δ(p 0 ) > 1/p 0 for a given p 0 > 1, for certain class of convex planar domains whose curvature is allowed to vanish to infinite order. His result was extended, in the paper [1] by Brandolini, Colzani, Iosevich, Podkorytov, and Travaglini, to arbitrary convex planar domains with no curvature or regularity assumption on the boundary.…”
Section: Introductionmentioning
confidence: 90%