2. 刚性定理

考虑二次多项式 f_{a}(x)=ax(1-x), a\in[0,4], f_{a}:[0,1]\rightarrow [0,1].

问题:

\{ a\in[0,4]: f_{a} \text{ satisfies Axiom A} \} 是否在 [0,4] 中稠密?

引理:f_{a} 满足 Axiom A \Leftrightarrow f_{a} 有双曲吸引周期轨。

定义 Kneading 序列: K(f_{a})=\{ i_{1}, i_{2},... \}, i_{k}=L \text{ if } f_{a}^{k}(\frac{1}{2})<\frac{1}{2};  i_{k}=c=\frac{1}{2} \text{ if } f_{a}^{k}(\frac{1}{2})=\frac{1}{2};  i_{k}=R \text{ if } f_{a}^{k}(\frac{1}{2})>\frac{1}{2}.

例子:

K(f_{4})=(R,L,L,L,...)=RLLL,

K(f_{1})=(L,L,L,L,...)=LLLL,

K(f_{2})=(c,c,c,c,...)=cccc,

K(f_{1.9})=(L,L,L,L,...)=LLLL,

在这里,f_{1}f_{1.9} 不是拓扑共轭的,即使它们的 Kneading 序列是一样的。

定义:f 和 g 称为拓扑共轭,如果存在同胚映射 h 使得 h\circ f= g \circ h.

性质1: 如果 f_{a_{1}} f_{a_{2}} 拓扑同胚,则有 K(f_{a_{1}})= K(f_{a_{2}}).

引理:如果 f_{a_{1}} f_{a_{2}} 没有双曲吸引或者双曲中性周期轨,则 K(f_{a_{1}})= K(f_{a_{2}}) \Rightarrow  f_{a_{1}} f_{a_{2}} 拓扑同胚 \Rightarrow a_{1}=a_{2}.

定义:拟共形映射的分析定义: \varphi: \Omega\rightarrow \tilde{\Omega}, 在这里 \Omega, \tilde{\Omega} 都是复平面上面的连通开集, \varphi 是保持定向的同胚映射,称 \varphi 是 K 拟共形映射 (K\geq 1), 如果

(1) \varphi 是 ACL 的,也就是线段上绝对连续,absolutely continuous on lines.

(2) | \frac{\partial \varphi}{\partial \overline{z}} | \leq \frac{K-1}{K+1} |\frac{\partial \varphi}{\partial z}| 几乎处处成立。

拟共形映射的一些性质:假设 \varphi 是 K-拟共形映射,K\geq 1.

(i) \varphi 几乎处处可微。对几乎所有的 z_{0}\in \Omega

\varphi(z) = \varphi(z_{0}) + \frac{\partial \varphi}{\partial z}(z_{0})(z-z_{0}) + \frac{\partial \varphi}{\partial \overline{z}}(z_{0})\overline{(z-z_{0})}+ o(|z-z_{0}|).

| \frac{\partial \varphi}{\partial z}|>0 几乎处处成立。

定义 \varphi 的复特征是 \mu_{\varphi}= \frac{\partial \varphi}{\partial \overline{z}} / \frac{\partial \varphi}{\partial z},||\mu_{\varphi}||_{\infty} \leq \frac{K-1}{K+1} <1.

(ii) Measurable Riemann Mapping Theorem ( Ahlfors-Bers )

 

Assume f_{a}(x)=ax(1-x), a_{0} \in (0,4]

Comb(a_{0})=\{ a\in(0,4]: K(f_{a})=K(f_{a_{0}}) \},

Top(a_{0})= \{ a\in (0,4]: f_{a} \text{ and } f_{a_{0}} \text{ are topological conjugate } \},

\Rightarrow Top(a_{0}) \subseteq Comb(a_{0}).

Qc(a_{0}) = \{ a\in (0,4]: f_{a} \text{ and } f_{a_{0}} \text{ are quasi-conformal conjugate} \},

Aff(a_{0}) = \{ a\in (0,4]: f_{a} \text{ and } f_{a_{0}} \text{ are linear conjugate} \},

\Rightarrow Aff(a_{0}) \subseteq Qc(a_{0}).

刚性问题:Comb(a_{0})=Qc(a_{0}) ? Comb(a_{0})=Aff(a_{0})?

 

定理:( Graczyk – Swiatek, Lyubich, 1997) 假设 f_{a_{0}} 没有双曲吸引或者中性周期轨,则 Comb(a_{0})=Qc(a_{0}).

推论:( Sullivan, 1988) Axiom A 系统在实系数二次多项式中稠密。

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s