Let $\mathcal{M}$ the Mandelbrot set,
$\mathcal{M}=\{c \in \mathbb{C}: \{Q_c^n(0) \}_{n \in \mathbb{N}} \text{ is bounded, where } Q_c(z)=z^2+c \}$
And let the hyperbolic or stable component, $H_n=\{ c \in \mathcal{M} \text{ so that } Q_c \text{ has an attracting cycle of period n} \}$.
It is known that stable components are given by only one model dynamics, meaning any two maps $Q_c,Q_d$ are quasiconformally conjugate if $c,d \in H_n$ for some $n$. I am recently looking at some Piecewise linearization of the double standard family studied by Misiurewicz and Rodrigues in their 2007 paper and was wondering if the stable components in this family also are given by only one model dynamics.
Now by a result from combinatorial equivalence it is known that for piecewise linear $l$-modal interval maps, under certain circumstances an order preserving conjugacy between the grand orbit of turning points can be extended as a conjugacy on the whole interval (De Melo- Van strien, Theorem 3.1, chapter 2). But it will soon become clear as I ask my question that this result will not be applicable. For instance, the circle maps I am considering are orientation preserving so there can't be turning points. Let me now ask my question:
- Question: We may use $\mathbb{R/Z},\mathbb{T},\mathbb{S^1}$ interchangeably for the circle and we will identify $\mathbb{T}$ with $[0,1]$ ( $0$ and $1$ identified). Let $f,g: \mathbb{T} \to \mathbb{T}$ be two continuous circle maps satisfying the following.
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- $f,g$ are both degree $2$, continuous, piecewise linear circle maps with a change of slope only at $1/2$. Both $f,g$ have slope greater than $1$ on $[0,1/2]$ and positive slope less than $1$ on $[1/2,1]$. Having positive slopes makes $f,g$ orientation preserving or increasing.
(It is clear that for both $f,g$, the sum of the slopes of the linear pieces on $[0,1/2]$ and $[1/2,1]$ must be equal to $4$ to maintain that they have degree $2$)
- Both $f$ and $g$ have exactly one attracting cycle of same period $n\geq1$ and same type. This means that if $F,G$ are the respective lifts with $F(0),G(0) \in [0,1)$ and $x_f,x_g$ be some attracting periodic points then there exists some common $1\leq k\leq2^n-1$ so that $F^n(x_f)=x_f+k$ and $G^n(x_g)=x_g+k$.
(We will not consider $x_f$ if $x_f$ is preperiodic to $1/2$, same for $g$.)
Then is true that $f$ and $g$ are topologically conjugate?
Remark: Now for the Mandelbrot set the multiplier of the attracting cycle is actually the Riemann map and the operation of establishing conjugacy is packaged in the proof which is then showed to vary holomorphically. The way to establish conjugacy is by a quasiconformal surgery: by taking advantage of the simple connectedness of the basin component, one imports the the complex structure from say $Q_c$ and pastes it on the distinguished basin component of $Q_d$ via changing the multiplier using a degree $2$ Blaschke product. Then by Measurable Riemann mapping theorem one chooses a correctly normalized Straightening Quasiconformal map which turns out to be the exact candidate for the conjugacy between $Q_c$ and $Q_d$ via the Weyl's lemma. So this operations function because of the rigidity of holomorphic maps. These piecewise linear maps $f,g$ are also somehow rigid so that is why I was wondering if an analogous procedure can be performed. But this is all speculation!
Any help will be appreciated. (I would love to know about any relevant literature!)
