Persistence of periodic points

55 Views Asked by At

I'm currently studying some of the more deeper results in complex dynamics theory and I've come across a statement that is sometimes used in proofs seemingly without justification.

Let $X$ be a complex manifold and let $(f_\lambda(z))$ be a holomorphic family of rational maps on the Riemann sphere $\hat{\mathbb{C}}$, that is, $(f_\lambda(z))$ is a family such that for every $\lambda \in X$, $f_\lambda: \hat{\mathbb{C}} \to \hat{\mathbb{C}}$ is a rational map, and for every $z \in \hat{\mathbb{C}}$, $f_\lambda(z)$ is holomorphic on $X$. Suppose $z_0 \in \hat{\mathbb{C}}$ is a repelling periodic point of $f_{\lambda_0}$ of period $\tau$. Then there exists an open neighbourhood $U$ of $\lambda_0$ in $X$ and a holomorphic map $w: U \to \hat{\mathbb{C}}$ satisfying $w(\lambda_0) = z_0$ such that for every $\alpha \in U$,

$$f_\alpha^\tau(w(\alpha)) = w(\alpha), \quad |(f_\alpha^\tau)'(w(\alpha))| > 1.$$

Here, $f_\alpha^\tau$ means the $\tau$-th composition of $f_\alpha$ and $(f_\alpha^\tau)'$ is the partial derivative of $f_\alpha^\tau$ with respect to $z$.

My attempt to prove this statement has so far involved employing the implicit function theorem to the map $g(\lambda, z)=f_\lambda^\tau(z)-z$ to obtain the desired open neighbourhood $U$ and holomorphic map $w$. (I think this can be done even though the spaces involved are a general complex manifold and the Riemann sphere, and not $\mathbb{C}^n$, but please feel free to correct me if this doesn't work.) My main problem is obtaining that derivative condition, so any suggestions would be much appreciated!

1

There are 1 best solutions below

1
On

To follow holomorphically a repelling periodic point $z$ of period $n$, you indeed want to apply the IFT to $F: (\lambda,z) \mapsto f_\lambda^n(z)-z$.

The condition required is $\frac{\partial F}{\partial z}(\lambda_0, z_0) \neq 0$, which is equivalent to $(f_{\lambda_0}^n)'(z_0) \neq 1$. This is satisfied if $z_0$ is a repelling periodic point for $f_{\lambda_0}$, because we then have $|(f_{\lambda_0}^n)'(z_0)|>1$.