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!


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.

  • $\begingroup$ 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 '18 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 '18 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 '18 at 13:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.