# Topological conjugacy between dyadic map and tent map

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 Answer

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.

• 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. – Algebear Nov 30 '18 at 13:32
• 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"? – Algebear Nov 30 '18 at 13:35
• 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... – user120527 Nov 30 '18 at 13:58