Does this sequence of Blaschke Product have rescaling limit $z-1$?

Question

Background: The conformal conjugacy class of parabolic isometry of upper half plane $\mathbb{H}$ consists of $f(z) = z+1$ and $g(z)=z-1$.

Consider surjective proper holomorphic $F_n: \mathbb{H} \rightarrow \mathbb{H}$ of the form $$ F_n(z) = \frac{-1}{\cfrac{1}{n} - \cfrac{1}{z+1- \frac{1}{n^2 z}}} $$ It is clear algebraically $F_n \rightarrow z+1$ point wise. Observe $F_n$ is degree 2, and in the limit the degree drop by 1. $F_n$ is a representation of Blaschke product in Upper Half plane.

Some facts, $F_n$ is hyperbolic in a sense it has an attracting fixed point $p_n$ in upper half plane. Moreover the multiplier i.e. derivative $F_n'(p_n) \rightarrow 1$ leaving every horocycle. That is, $F_n'(p_n) = \delta_n + (1-\delta_n)e^{i \theta_n}$ for $\delta_n, \theta_n \rightarrow 0$. In particular, this sequence have no critical point that is quasi-fixed and diverge in moduli space of quadratic rational map.

Definition: We say $G:\mathbb{H} \rightarrow \mathbb{H}$ is a rescaling limit of $F_n$ if there is a sequence $M_n\in Aut(\mathbb{H})$ so $M_n \circ F_n \circ M_n^{-1} \rightarrow G$ converges algebraically (equivalently uniformly in compact set).

Question: Is $g(z) = z-1$ a rescaling limit of $F_n$?

Attempt:

My guess is no.

I want to reduce it to problem in degree one $\mathbb{H} \rightarrow \mathbb{H}$. I believe I can show if $M_n \in Aut(\mathbb{H})$ elliptic (i.e. has a fixed point $p_n$ in $\mathbb{H}$ and $M_n \rightarrow z+1$, then the multiplier $M_n'(p_n) \in \mathbb{D} \cap \mathbb{H}$ upper half disk.

Consider hyperbolic ball centered at $p_n$ fixed point of $F_n$ and containing $c_n$ critical point in the boundary. That is, if $d_\mathbb{H} (p_n, c_n) = r_n$, then I take $B_{\mathbb{H}}(p_n, r_n)$ and call it the critical disk.

$F_n$ behave like some automorphism $\phi_n$ inside the critical disk. Assume there exist rescaling $\tilde{F_n} \rightarrow z-1$. If can show the critical disk "Euclidean size" persist under rescaling of $F_n$, then I can reduce the problem to degree 1, and get a contradiction since the multiplier of $\tilde{F_n}$ must lies in lower half disk due to converging to $z-1$.

I do know the critical point of $F_n$, $c_n = \pm \frac{i}{n}$ while $p_n \rightarrow \infty$ along imaginary axis. Thus the Euclidean size of critical disk of $F_n$ increase to cover entire $\mathbb{H}$. However I do not have a good control of location of critical point of $\tilde{F_n}$.

Answer 1

The answer is no. Unfortunately the proof I came up with use heavy algebra. Would be nice to get a second pair of eyes to check if I miss some cases.

First we state definition of dependent moving frame, following Jan Kiwi

Definition: (Equivalent Moving Frame) We say that two moving frames $\{M_n\}, \{L_n\} \in Aut(\mathbb{H})$ are equivalent if there exists $M \in Aut(\mathbb{H})$ such that $M_n^{-1} \circ L_n \rightarrow M$ as $n \rightarrow \infty$. The equivalence class of moving frame $\{M_n\}$ will be denoted by $[M_n]$.

Given two equivalent moving frames $\{M_n\}$ and $\{L_n\}$, suppose that we have $ M_n^{-1} \circ G_n^+ \circ M_n \rightarrow z-1$. Then we have \begin{align*} L_n^{-1} \circ G_n^+ \circ L_n &= (L_n^{-1} \circ M_n) \circ (M_n^{-1} \circ G_n^+ \circ M_n) \circ (M_n^{-1} \circ L_n) \\ &\rightarrow M^{-1} \circ (z-1) \, \circ M. \end{align*} Notice the collection of affine transformation $\widetilde{Aut(\mathbb{H})}:= \{\varphi \in Aut(\mathbb{H}) : \varphi(\infty) = \infty \}$ acts transitively on $\mathcal{H}$. This mean given moving frame $\{L_n\} $, we can always find a moving frame $\{M_n\}$ so $M_n(z) \in \widetilde{Aut(\mathbb{H})}$ so that $M_n^{-1} \circ L_n (i) = i+1$ quasi fixed. Therefore after taking subsequence $M_n^{-1} \circ L_n$ will converge to some $M \in Aut(\mathbb{H})$.

Therefore it is enough to consider all possible rescaling of $G_n^{+}$ by affine map $M_n(z) = a_n z+ b_n$ for $a_n>0$ and $b\in \mathbb{R}$. However we need to consider the conjugacy class of $z-1$ since we do not have guarantee $M(\infty) = \infty$. Using the matrix form $PSL_2(\mathbb{R})$ in order to do simplification in algebra, we see functions in conjugacy class $[z-1]$ has to be of the form $$ z \mapsto \frac{(1-cd)z -d^2}{c^2 z + (1+cd)}, \,\, c,d \in \mathbb{R} \text{ not both 0.} $$ Now given $M_n(z) = a_n z + b_n$ for $a_n>0$ and $b_n \in \mathbb{R}$, we can write \begin{equation} M_n^{-1} \circ G_n^+ \circ M_n (z) = \cfrac{-1}{\cfrac{a_n}{k_n}-\cfrac{1}{z+\frac{b_n+1}{a_n}-\frac{1}{n^2(a_n^2z + a_n b_n)}}} -\cfrac{b_n}{a_n}. \end{equation} Call this our main equation.

Our task is to show the main equation cannot converge parabolic function algebraically. We consider two big cases

• Assume $\frac{a_n}{k_n} \rightarrow a \neq 0$. Then $a_n \rightarrow \infty$ so $\frac{1}{a_n} \rightarrow 0$. In this case the imaginary component $\Im(n^2(a_n^2z + a_n b_n))$ must go to $\infty$, thus $\frac{1}{n^2(a_n^2z + a_n b_n)} \rightarrow 0$. Denote $\frac{b_n}{a_n} \rightarrow b$. Then the expresion in the main equation converges algebraically to $$ \frac{-1}{a-\frac{1}{z+b}}-b = \frac{(1+ab)z+ab^2}{-az+1-ab}. $$ Suppose either $a=\infty$ or $b=\infty$, then we immediately get an expression that does not depend on $z$, so not in the conjugacy class $[z-1]$. Then we have equality $$ \frac{(1+ab)z+ab^2}{-az+1-ab} = \frac{(1-cd)z -d^2}{c^2 z + (1+cd)}. $$ Suppose $-a = c^2 > 0$, contradiction since $a_n >0$ so $a>0$ and $-a<0$. Therefore we have $$ \frac{-(1+ab)z-ab^2}{az-(1-ab)} = \frac{(1-cd)z -d^2}{c^2 z + (1+cd)}. $$ In this case we have $-1-ab = 1-cd$ and $-1+ab = 1+cd$. The first equation imply $cd-ab=2$ and the second equation imply $cd-ab=-2$, contradiction.

• Next assume $\lim \frac{a_n}{k_n} \rightarrow 0$. Denote $\lim_n \frac{b_n+1}{a_n} -\frac{1}{n^2(a_n^2z + a_n b_n)} = \widetilde{b}$ and $\lim_n \frac{b_n}{a_n} = b$. We cannot have $\widetilde{b}= \infty$ since then the limiting expression of the main equation does not have dependency in $z$. So $\widetilde{b}<\infty$. Assume $b = \infty$. Then the limiting expression of the main equation become $$ \cfrac{-1}{-\frac{1}{z+\widetilde{b}}}-\infty = \infty. $$ Again it does not have dependency with $z$, so $b<\infty$. This give identity $$ \widetilde{b} = b + \lim_n \left( \frac{1}{a_n} - \frac{1}{n^2(a_n^2z + a_n b_n)} \right). $$ Inputting everything to the main equation we have the following limit \begin{align*} \frac{-1}{-\frac{1}{z+\widetilde{b}}}-b &= z +\widetilde{b} - b \\ &= z + \lim_n \left( \frac{1}{a_n} - \frac{1}{n^2(a_n^2z + a_n b_n)} \right) = z -T, \text{ for } T>0. \end{align*} In order to get contradiction, it is enough to show to show the last equality is impossible. Suppose $a_n \rightarrow a_\infty \neq 0$, possibly infinity, then the imaginary component $\Im (n^2(a_n^2z + a_n b_n )) \rightarrow \infty$, giving $$ \lim_n \left( \frac{1}{a_n} - \frac{1}{n^2(a_n^2z + a_n b_n)} \right) = \lim \frac{1}{a_n} \geq 0, $$ contradiction since we need the limit to be negative number. Therefore $a_n \rightarrow 0$. Assume we have the following (I thank Lior Silberman for communicating me the argument from this point to the end of post) $$ \lim_n \left( \frac{1}{a_n} - \frac{1}{n^2(a_n^2z + a_n b_n)} \right) = -T, \text{ for }T>0. $$ Then for some sequence $e_n \rightarrow 0$, we can write $$ \frac{1}{a_n} - \frac{1}{n^2(a_n^2z + a_n b_n)} = -T +e_n. $$ Multiply both sides by $n^2 a_n^2$, we have the expression \begin{alignat}{2} && n^2 a_n - \frac{1}{z + \frac{b_n}{a_n}} &= n^2 a_n^2 (-T + e_n) \\ \iff&& n^2 a_n (1 + T a_n -a_n e_n) &= \frac{1}{z+ \frac{b_n}{a_n}} \\ \iff&& \lim_n n^2 a_n (1 + T a_n -a_n e_n) &= \lim_n \frac{1}{z+ \frac{b_n}{a_n}} = \frac{1}{z+c}. \end{alignat} However the left hand side of last equation does not depend on variable $z$, contradiction. We have exhausted all possibilities, finishing the proof