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For trying to prove that the tent map $$T(x)= \begin{cases} 2x &\text{ if } x\in[0,\frac{1}{2}]\\ 2-2x &\text{ if } x\in[\frac{1}{2},1] \end{cases} $$ is ergodic, I have already shown that the dyadic map (period-doubling map) given by $E(x)=2x$ mod $1$ is ergodic using a Fourier transform of $f$ and comparing coefficients to show that any measurable $f:X\to\mathbb{R}$ with $\ f \circ E = f$ almost everywhere implies $f$ is constant almost everywhere.

Anyways, to show that the tent map is ergodic, I tried topological conjugation with the dyadic map (which is ergodic); I've done the following:

Lemma. The tent map $T$ is topologically semi-conjugate to the dyadic map $E(x)=2x$ mod 1.

Proof. Let $E: [0,1] \to [0,1]$ be the dyadic map $E(x) = 2x$ mod 1. Let $T$ be the tent map as before.

Let $\varphi: [0,1] \to [0,1]$ also be the tent map, the same as $T$; i.e. $\varphi\equiv T$. Since \begin{align*} \varphi\circ E(x)=T(2x\text{ mod 1}) &=\begin{cases} 2(2x\text{ mod 1})\ &\text{ if }0\leqslant2x\text{ mod } 1\leqslant\frac{1}{2}\\ 2-2(2x\text{ mod 1})\ &\text{ if }\frac{1}{2}\leqslant2x\text{ mod } 1\leqslant1 \end{cases}\\ &= \begin{cases} 4x\ &\text{ if }x\in[0,\frac{1}{4}]\cup[\frac{1}{2},\frac{3}{4}]\\ 2-4x \ &\text{ if }x\in[\frac{1}{4},\frac{1}{2}]\cup[\frac{3}{4},1] \end{cases}\\ &=T^2(x)=T\circ\varphi(x), %%Do not change, this is best way to write down, I noticed by trial and error \end{align*} we have that $\varphi\circ E = T\circ\varphi$; i.e. $T$ is a factor of $E$ (or $E$ is an extension of $T$). $\Box$.

I know that this is only semi-conjugation for $\varphi$ is not invertible. I think the argument for ergodicty does not go wrong only using semi-conjugation, but I would like to have "full" conjugation. This is the ergodicity argument assuming ergodicity of $E$:

Theorem. The tent map $T$ is ergodic.

Proof. Let $A$ be an invariant set in $[0,1]$ for $T$; i.e. $T^{-1}(A)=A$. Since $\varphi\circ E = T\circ\varphi$ with $\varphi,\ E$ and $T$ as in the lemma above, it follows that \begin{align*} (\varphi\circ E)^{-1}&=(T\circ\varphi)^{-1}\\ E^{-1}\circ\varphi^{-1}&=\varphi^{-1}\circ T^{-1} \end{align*} which, after plugging in $A$, gives \begin{equation*} E^{-1}(\varphi^{-1}(A))=\varphi^{-1}(T^{-1}(A))=\varphi^{-1}(A); \end{equation*} so $\varphi^{-1}(A)$ is invariant for $E$. Now since $E$ is ergodic, we have that $\varphi^{-1}(A)$ has zero or full Lebesgue measure. As $\varphi=T$ (and $\varphi^{-1}=T^{-1}$), we have that $\varphi^{-1}(A)=T^{-1}(A)=A$ and hence also $A$ has zero or full Lebesgue measure; i.e. $T$ is ergodic.$\Box$.

Question 1A: Is the proof of the theorem correct?

Question 1B: Does this last theorem proof that ergodicity is preserved under topological semi-conjugation?

Question 2: How does one prove topological conjugation between the tent map and the dyadic map?

Thanks in advance for time and help!

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1 Answer 1

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1.A. The proof is correct, although in the computation of $\varphi \circ E(x)$, the cases $1/2\leq x\leq 3/4$ and $x\geq 3/4$ were forgotten.

1.B. Yes, ergodicity of a system implies ergodicity of its factors. However, it must be pointed out that 'factor' must be understood as factor as for a measure preserving transformation (not topological semi-conjugacy): a system $(Y,\nu,S)$ is a factor of $(X,\mu,T)$ if there is a map $f:X\to Y$, measurable, such that $f_*\mu=\nu$, and $f\circ T=S \circ f$.

As a matter of fact, in your proof, you did not use the fact that $\varphi$ is continuous. However, at the end, you used the specific of the situation ($\varphi=T$) to conclude, but could easily have used the fact that $Leb(\varphi^{-1}A)=Leb(A)$, i.e. $\varphi_*(Leb)=Leb$, instead, and that would have been the general proof of the statement that a factor map is ergodic if the extension is.

  1. The answer depends on the precise topological model you choose.

model $E_1$ : $E_1:[0,1]\to [0,1]$,

model $E_2$ : $E_2:\mathbb{R}/\mathbb{Z} \to \mathbb{R}/\mathbb{Z}$,

model $T_1$ : $T_1:[0,1]\to [0,1]$,

model $T_2$ : $T_2:\mathbb{R}/\mathbb{Z}\to \mathbb{R}/\mathbb{Z}$.

$E_1$ and $T_2$ cannot be topologically conjugate as a circle is not homeomorphic to a closed interval. Same for $T_1$ and $E_2$. $E_1$ cannot be conjugate to $T_1$, because $T_1$ is continuous but $E_1$ is not. $E_2$ and $T_2$ cannot be conjugate, because the degree of $E_2$ is two, and the degree of $T_2$ is zero.

So unless one considers another topological model for these maps, they do not seem to be topologically conjugated.

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  • $\begingroup$ First of all, thank for considering the proof. 1A: I should have formulated it better, yes, but the mapping just repeats itself after [0,1/2]. It's a subtility well found by you. 1B: why is it not a semi-conjugation since $\varphi$ is a surjection? Also, $\lambda(\varphi^{-1}A)=\lambda(A)$ does need $\varphi$ to be measure preserving by definition; that may be needed in the proof that a factor map is ergodic. $\endgroup$
    – Algebear
    Nov 30, 2018 at 13:32
  • $\begingroup$ Moreover, on wikipedia it says (in part 1) that the tent map and the dyadic map are topologically conjugate (without proof). How is that different from your statement at 2 that one should use "another topological model"? $\endgroup$
    – Algebear
    Nov 30, 2018 at 13:35
  • $\begingroup$ My answer of 2 may be incomplete: I just wanted to say that one can call "dyadic map" several different map defined on several different spaces (e.g $[0,1]$, $[0,1)$ or $\mathbb{R}/\mathbb{Z}$ - all of which are measurably isomorphic, but not homeomorphic, what I called a topological model. And for the ones that spring to mind, dyadic map and tent map are not topologically conjugate, but perhaps I missed an obvious model that works... $\endgroup$
    – user120527
    Nov 30, 2018 at 13:58

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