Let $P_{c}(z)=z^2+c$. It seems from the software that the map between the parameter $c$ and the Julia set $J(P_c)$ is an injective map. Is there some reference about it? Any comments and reference will be appreciated.
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3 Answers
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I expect that you will find the answer (and a lot more) in the following paper, since it's easy to figure out which quadratic polynomials (if any) commute with one another:
Commuting polynomials and polynomials with same Julia set, Pau Atela, Jun Hu, Int. J. Bifurcation Chaos 06, 2427 (1996). DOI: http://dx.doi.org/10.1142/S0218127496001570
Abstract: It has been known since Julia that polynomials commuting under composition have the same Julia set. More recently in the works of Baker and Eremenko, Fern\'andez, and Beardon, results were given on the converse question: When do two polynomials have the same Julia set? We give a complete answer to this question and show the exact relation between the two problems of polynomials with the same Julia set and commuting pairs.
The paper is also available on the ArXiv
answered Sep 2, 2016 at 14:50
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$\begingroup$ Thank you, professor. Very nice paper. This is first time I know Fernandez's result, which claims that if f and g has the same degree and the same leading coefficient, then $J(f)=J(g)$ implies $f=g$. $\endgroup$– yaoxiaoCommented Sep 2, 2016 at 15:05
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Joe's answer and the cited paper are, indeed, quite nice. I think the basic idea is fairly self-contained and probably easier than the reference.
Two complex polynomials $f$ and $g$ of the same degree have geometrically similar Julia sets precisely when there is an affine function $\varphi(z)=mz+b$ with $m\neq0$ such that $$\varphi(f(z)) = g(\varphi(z)).$$ The function $\varphi$ is called a conjugacy and it's easy to see why this should work. By induction, $$\varphi(f^n(z)) = g^n(\varphi(z)).$$ As a result, $\varphi$ maps an orbit of $f$ containing the point $z_0$ onto an orbit of $g$ containing the point $\varphi(z_0)$ so we expect the dynamics of $f$ and $g$ to be closely related; the nicer the function $\varphi$. The nicer the relationship and our $\varphi$ is about as nice as it could be.
Now the question is, when is there an affine conjugacy between $$f(z)=z^2+c_1\: \text{ and } \: g(z)=z^2+c_2?$$ If you simply expand the conjugation equation or $$m(z^2+c_1)+b = (mz+b)^2+c_2$$ and collect terms you find the system \begin{align} m^2-m &= 0 \\ 2bm &= 0 \\ mc_1-c_2-b^2+b &= 0. \end{align} Recalling that $m$ can't be zero for $\varphi$ to be an affine function, we see that the first equation implies that $m=1$ so that the second equation implies $b=0$. Plugging those into the last equation we see that $c_1=c_2$.
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3$\begingroup$ But isn't the hard part proving that if $J(f)=J(g)$, then there is an affine conjugacy? Once you know that, then rest, as you've indicated, is straightforward. $\endgroup$ Commented Sep 2, 2016 at 19:28
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For $z^2+c$ it is really easy: the answer is yes. In general, if Julia sets of polynomials coincide, these polynomials are equal unless their common Julia set has a rotation symmetry, or polynomials are permutable. (Baker, Eremenko, Ann. Acad. Sci. Fenn., 12 (1987) 229-236.)
http://www.acadsci.fi/mathematica/Vol12/vol12pp229-236.pdf
For $z^2+c$ we never have rotational symmetry and never have permutable polynomials. The result cited above was reproved and generalized by many other people. For rational functions, the question in full generality is still open.
answered Sep 5, 2016 at 13:02
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$\begingroup$ Thanks for your comment and reference, professor. Would you think it may be too optimistic to guess that $\lambda\rightarrow J(\lambda*\exp)$ be injective map for the paramters $\lambda$ for which $J(\lambda*exp)\neq \mathbb{C}$. $\endgroup$– yaoxiaoCommented Sep 7, 2016 at 14:33
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$\begingroup$ @yaoxiao: I don't know how to prove or disprove this. $\endgroup$ Commented Sep 9, 2016 at 4:21