Christopher J. Bishop — Page 2

arXiv1995

The dimension of the Brownian frontier is greater than 1

Consider a planar Brownian motion run for finite time. The frontier or ``outer boundary'' of the path is the boundary of the unbounded component of the complement. Burdzy (1989) showed that the frontier has infinite length. We improve this by showing that the Hausdorff dimension of the frontier is strictly greater than 1. (It has been conjectured that the Brownian frontier has dimension $4/3$, but this is still open.) The proof uses Jones's Traveling Salesman Theorem and a self-similar tiling of the plane by fractal tiles known as Gosper Islands.

math.PR↗

arXiv1994

Hausdorff dimension and Kleinian groups

Let G be a non-elementary, finitely generated Kleinian group, Lambda(G) its limit set and Omega(G) = S \ Lambda(G) (S = the sphere) its set of discontinuity. Let delta(G) be the critical exponent for the Poincaré series and let Lambda_c be the conical limit set of G. Suppose Omega_0 is a simply connected component of Omega(G). We prove that (1) delta(G) = dim(Lambda_c). (2) A simply connected component Omega is either a disk or dim(Omega)>1$. (3) Lambda(G) is either totally disconnected, a circle or has dimension > 1, (4) G is geometrically infinite iff dim(Lambda)=2. (5) If G_n \to G algebraically then dim(Lambda) <= \liminf dim(Lambda_n). (6) The Minkowski dimension of Lambda equals the Hausdorff dimension. (7) If Area(Lambda)=0 then delta(G) = dim(Lambda(G)). The proof also shows that \dim(Lambda(G)) > 1 iff the conical limit set has dimension > 1 iff the Poincaré exponent of the group is > 1. Furthermore, a simply connected component of Omega(G) either is a disk or has non-differentiable boundary in the the sense that the (inner) tangent points of \partial Omega have zero 1-dimensional measure. Almost every point (with respect to harmonic measure) is a twist point.

math.DS↗