The *Julia set* is not the same as the *boundedness set* for a polynomial map. The Julia set is the set of points such that every neighborhood eventually spreads out to cover a dense subset of C. It does not include points that are attracted to a periodic cycle. Hausdorff dimension of the Julia set is not continuous in general: in particular, it fails to be continuous at any parameter where there is a rationally indifferent periodic point, that is, a periodic point where the derivative of the return map is a root of unity. These are points where mini-Mandelbrot sets are attached. For a sequence of points converging to such a point, the Julia set of the limit is always contained in the limit of Julia sets, but the limit of Julia sets can be much bigger. You can see this by playing around with a Julia set visualizer, (if you use a mac with recent operating system, there are good programs such as fractalworks available from the App Store). However, Hausdorff dimension is continuous as long as the critical point tends toward an attracting periodic cycle, infinity allowed. So, I think you'd see the Mandelbrot set as basically the discontinuity set for the graph of Hausdorff dimension, since rationally indifferent periodic cycles are presumed to be dense in the Mandelbrot set. Bill Thurston On Aug 1, 2011, at 5:50 PM, Huddleston, Scott wrote:
... I especially wonder whether the Mandelbrot set would be in any way "visible" in this dimension plot.
Every Julia set with nonempty interior has Hausdorff dimension 2. These are indexed by attracting Mandelbrot cycles, so the interior of the Mandelbrot set has value 2 in this map.
I think every repelling Mandelbrot cycle indexes a Julia set with 1 <= Hausdorff dimension < 2, and disconnected Julia sets (indexed by the Mandelbrot complement) have Hausdorff dimension < 1.
-- Scott
-----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun- bounces@mailman.xmission.com] On Behalf Of Allan Wechsler Sent: Monday, August 01, 2011 1:46 PM To: Dan Asimov; math-fun Subject: Re: [math-fun] Hausdorff dimension of Julia sets
That's an excellent question, and I haven't seen it posed before. First of all, one must know that a given Julia set in fact has a well-defined dimension (the boundary of the Mandelbrot set doesn't), but I think this is well-established. I don't know what technology there is for computing dimensions of Julia sets, and of course I especially wonder whether the Mandelbrot set would be in any way "visible" in this dimension plot.
On Mon, Aug 1, 2011 at 4:26 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Has anyone seen a graph {(c, H(c)) in R^3 | c in C=R^2} of the Hausdorff dimension H(c) of the Julia set of f(z) = z^2 + c, c in C ?
--Dan
Sometimes the brain has a mind of its own.
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