If I have a specific quadratic function

$$f(z) = \frac{az^2 + bz + c}{z^2 + dz + e}$$

on the Riemann sphere, then can one describe its Julia set completely? That is to say, by possibly applying a Mobius transformation, can it be reduced to some sort of a 'canonical form' where the answer is known?

  • 2
    $\begingroup$ Short answer: No. First off, that's not a quadratic function, it's a rational function of degree 2. Even if we restrict to just quadratics, there are quadratics whose Julia sets are not computable. $\endgroup$ – Mark McClure Mar 9 '17 at 3:12

As Mark said, there's no easy way to figure out what is in the Julia set, and indeed there are cases where the Julia set is not computable. Also these functions are quadratic rational, or rational functions of degree 2, not quadratics.

For a more canonical form, consider looking at Remarks on Quadratic Rational Maps by John Milnor, Appendix C. It describes various normal forms for these maps

$$\frac{Az^2 + B}{Cz^2 + D}$$

And also see that the moduli space of them is of dimension 2 (you only need two parameters rather than 4, or 6).

You can also find more normal forms there, and some derivations of them in this paper by Adam Epstein.

Milnor also has a paper on rational maps with two critical points. Which you can check that quadratic rational maps do.

HINT: By Riemann Hurwitz, it must have at most $2$, and so it has $0$, $1$ or $2$. Use mobius transforms to assume that infinity is not a possible critical point. Call the function $F(z)/G(z)$Then, differentiate this form using quotient rule. $$\frac{F'(z)G(z) - G'(z)F(z)}{G(z)^2}$$ You will have a fraction with the top being a quadratic and the bottom being a quartic. Show they can't share any roots. Now the critical points would be solutions to this quadratic over $\mathbb{C}$, ruling out having no critical points. The discriminant of the polynomial is $\mathrm{res}(F,G) $ which by definition is non zero for $F,G$ chosen. Thus you have $2$ solutions to this equation. All three of these are interesting if you are looking for normal forms.


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