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.