[math-fun] Mandelbrot set question
Speaking of the Mandelbrot set, denoted by M: The functions z^2 + c, c in C, contain just one examplar of each linear equivalence class of quadratic maps. (I.e., in the set of maps Q = {f(z) = a z^2 + b z + c with a,b,c in C and a != 0} let f,g in Q be equivalent if there exists a linear map L(z) = d z + e with d,e in C and d != 0, such that Linv o f o L == g, where Linv is the inverse function of L.) But {z^2 + c} is only one of many ways to parametrize this set of equivalence classes, and the choice of parametrization will drastically affect the appearance of the Mandelbrot set. And so I'm wondering if it's a lot more symmetrical than it looks. Some ways to define its symmetry group might be 1) All maps f:M -> M that are restrictions of a conformal bijection f: U -> V between two open neighborhoods U, V of M, with f(M) = M. 2) All self-homeomorphisms h:M -> M, factored out by the equivalence relation of isotopy. (I.e., homeos h,k: M -> M are equivalent if there is a continuous mapping H: M x [0,1] -> M such that for all z in M and t in [0,1] we have * H(0,z) = h(z), * H(1,z) = k(z), and * the map H( ,t): M -> M is a self-homeo of M. ------------------------------------------------ Maybe 2) is more interesting, since it would show only the fundamentally different self-homeos of M. Does anyone know if there are results on this question? --Dan Sometimes the brain has a mind of its own.
My intuition here is that the "underlying" symmetry group of M is no bigger than what you see in the classic "projection", that is, the two-element group whose non-identical element is complex conjugation. Roughly, although some regions of M look similar to each other, they are all characterized by qualitatively different dynamics for the process z -> zz+c. For example, the point c = 1/4 is a cusp of the main cardioid, and has no stalk coming out of it; it's the only point in M that looks like that. On Thu, Jul 28, 2011 at 7:13 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Speaking of the Mandelbrot set, denoted by M:
The functions z^2 + c, c in C, contain just one examplar of each linear equivalence class of quadratic maps.
(I.e., in the set of maps
Q = {f(z) = a z^2 + b z + c with a,b,c in C and a != 0}
let f,g in Q be equivalent if there exists a linear map L(z) = d z + e with d,e in C and d != 0, such that Linv o f o L == g, where Linv is the inverse function of L.)
But {z^2 + c} is only one of many ways to parametrize this set of equivalence classes, and the choice of parametrization will drastically affect the appearance of the Mandelbrot set.
And so I'm wondering if it's a lot more symmetrical than it looks.
Some ways to define its symmetry group might be
1) All maps f:M -> M that are restrictions of a conformal bijection f: U -> V between two open neighborhoods U, V of M, with f(M) = M.
2) All self-homeomorphisms h:M -> M, factored out by the equivalence relation of isotopy. (I.e., homeos h,k: M -> M are equivalent if there is a continuous mapping
H: M x [0,1] -> M
such that for all z in M and t in [0,1] we have
* H(0,z) = h(z),
* H(1,z) = k(z), and
* the map H( ,t): M -> M is a self-homeo of M. ------------------------------------------------
Maybe 2) is more interesting, since it would show only the fundamentally different self-homeos of M.
Does anyone know if there are results on this question?
--Dan
Sometimes the brain has a mind of its own.
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The only "az^2+bz+c" variation I've seen is the "lambda map", f(z) = c z (1-z) = c z - c z^2. Here is an image: mrob.com/images/0-muency/lambda.jpg Notice the shape is different from the normal Mandelbrot shape, and they are related by "squaring" one to get the other. This version of the Mandelbrot set was actually published in Byte magazine[1] some months earlier than the Scientific American[2] that is normally thought to be the popular introduction of the Mandelbrot set. The Byte article was about fractals and simply showed a few pictures of the Lambda fractal, without a formula or other explanation. The "interior" was white, as in my image above, not black as the Mandelbrot set is more commonly drawn. I immediately recognized the importance of this new fractal and wished I could find out how it was calculated. I seem to remember reading somewhere that there aren't any other interesting shapes, but there must have been some qualification to that because you can get all sorts of shapes by applying a nonlinear transformation to the parameter, as in f(z)=z^2+1/c (see the "1/u plane" image at [3]) or f(z)=z^2+(K+exp(c)) (see my article on the exponential map, [4]) I guess the roots of f(z) are important, note that the lambda f(z) has roots at 0 and 1; the normal f(z)=z^2+c has a complex conjugate pair of roots +-sqrt(c) i. - Robert Munafo [1] Peter R. Sørensen. Fractals. Byte, Sep. 1984 p. 157. [2] A. K. Dewdney. Computer Recreations. Scientific American, Aug. 1985 Available online at: http://www.scientificamerican.com/media/inline/blog/File/Dewdney_Mandelbrot.... [3] http://www.matpack.de/Info/Mathematics/Fractals.html [4] http://mrob.com/pub/muency/exponentialmap.html On Thu, Jul 28, 2011 at 19:13, Dan Asimov <dasimov@earthlink.net> wrote:
Speaking of the Mandelbrot set, denoted by M:
The functions z^2 + c, c in C, contain just one examplar of each linear equivalence class of quadratic maps.
(I.e., in the set of maps
Q = {f(z) = a z^2 + b z + c with a,b,c in C and a != 0}
let f,g in Q be equivalent if there exists a linear map L(z) = d z + e with d,e in C and d != 0, such that Linv o f o L == g, where Linv is the inverse function of L.)
But {z^2 + c} is only one of many ways to parametrize this set of equivalence classes, and the choice of parametrization will drastically affect the appearance of the Mandelbrot set. [...]
-- Robert Munafo -- mrob.com Follow me at: gplus.to/mrob - fb.com/mrob27 - twitter.com/mrob_27 - mrob27.wordpress.com - youtube.com/user/mrob143 - rilybot.blogspot.com
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Robert Munafo