0
$\begingroup$

Edit: So, my original question (stated below) was to find an error in my "proof" that immediate parabolic basins for rational maps are always simply connected. Since I have not received any answers as of yet I would ask alternatively if someone could point out an explicit example of a rational map with a parabolic fixed point which has a non-simply connected immediate basin, so that I could hopefully check by examining this example where my argument goes wrong. Kind regards an idiot

Hello! Sorry if this seems stupid. I know there must be an error in my thinking.
Let f be a rational map on the Riemann sphere with a parabolic fixed point $f(z_0)=z_0$, $f'(z_0)=e^{2\pi t}$ with $t\in \mathbb{Q}$.
I will try to demonstrate that each parabolic immediate basin is simply connected. I know this is wrong. But I don't find the mistake in my "proof". So please help me out.
By Leau-Fatou Flower theorem, in each immediate parabolic basin of $z_0$ there is an attractive petal $V$ such that each point in that immediate basin tends to $z_0$ via $V$.
Let $V_0$ be such a petal, and for simplicity's sake let's say that $f(\overline{V_0})\subset V_0\cup{z_0}$ (i.e. no periodic jumping between different petals, can be achieved by simply taking an iterate $f^n$ instead of $f$ for suitable n).
So we have:
- $f(\overline{V_0})\subset V_0\cup{z_0}$
- $V_0\subset A^*(z_0)$ open and simply connected
- For every $z\in A^*(z_0)$ there is some $n\in\mathbb{N}$ with $f^n(z)\in V_0$

We may slightly shrink $V_0$ if necessary such that $\partial V_0$ does not contain any postcritical points and $\overline{V_0}$ is homeomorphic to a closed disk.

Now for $k\in\mathbb{N}$ let $V_k$ be the component of $f^{-1}(V_0)$ that contains $V_0$. It's easy to see that $A^*(z_0)=\cup_{k=0}^{\infty}V_k$.

If $A^*(z_0)$ is not simply connected then there must be a minimal $m\in\mathbb{N}$ such that $V_m$ is not simply connected.
In that case let $B$ be a component of $ \hat{\mathbb{C}}-\overline{V_m}$, such that $\partial B$ does not contain $z_0$.
Then $\partial B\subset \partial V_m$ and so $f(\partial B)\subset\partial V_{m-1}$.
Since $\partial V_0$ contains no postcritical points, $\cup_{k=0}^m \partial V_m$ contains no critical points.
Thus $f^m$ is locally injective on $\partial B\subset\partial V_m$ and $f^m(\partial B)$ is a full component of $\partial V_0$ (proper covering), hence $f^m(\partial B)=\partial V_0$, since $\partial V_0$ has only one component.
But then there is $z\in\partial B\subset F(f)$ with $f^n(z)=z_0\in J(f)$. That's a contradiction.

Can someone help me see my mistake? I hope it's a simple one.

$\endgroup$
1
  • $\begingroup$ Just to bring this question to the top once more (hope this is allowed). I just edited the question and asked an alternative question which might help me. Kind regards, an idiot $\endgroup$
    – idiot_1337
    Aug 12, 2012 at 12:11

1 Answer 1

3
$\begingroup$

The mistake is in the statement that $\partial B\subset F(f)$. There can be points on $\partial B$ and $\partial V_m$ which are in $J(f)$, namely preimages of $z_0$ :-)

An example is $f(z)=z+1-1/z$. There is one petal for the neutral point at infinity. Let $A$ be the dmain of attraction of $\infty$. Critical points are $\pm i$. Everything is symmetric with respect to the real line, because the function is real. One critical point is in $A$, so by symmetry the other one is also in $A$. The map $f:A\to A$ is 2-to-1 (because $f$ is of degree $2$), so Riemann and Hurwitz tell us that $A$ is infinitely connected.

$\endgroup$
1
  • $\begingroup$ Thank you. And sorry, I was very sloppy. Where I wrote: "In that case let $B$ be a component of $ \hat{\mathbb{C}}-\overline{V_m}$", I now added: "such that $\partial B$ does not contain $z_0$."<br> But I see now that this also does not help.<br> Thank you very much. I truly am an idiot.<br> And thanks for the example. May the force be with you. $\endgroup$
    – idiot_1337
    Aug 12, 2012 at 14:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.