# Are $f(x)=2x+1$ and $g(x)=\frac{2x-1}3$ topologically conjugate on $\Bbb Z_2$?

Let $$f,g$$ be functions from $$\Bbb Z_2\to\Bbb Z_2^\times$$

$$f(x)=2x+1$$ and $$g(x)=\frac{2x-1}3$$

Then I have the notion that $$f,g$$ are topologically conjugate to each other in the 2-adic metric. I'd like to verify that but I find it offputting that the domains of the functions don't coincide with the range.

Anyway, proceding regardless, I can ask myself "is there a homeomorphism that conjugates one to the other?"

So I guess the homeomorphism would be $$x\mapsto\frac{x-2}3$$ or $$x\mapsto3x+2$$

So to my tiny brain, that homeomorphism looks good because it's one to one and onto (because the maps to and from zero fall outside of $$\Bbb Z_2^\times$$, preserves the 2-adic value $$\lvert x\rvert_2=1$$ and $$d(f(x),f(y))=d(x,y)$$. Is that correct?

I'm concerned I don't have any conjugation going on there, just a transformation from one to the other. But how could I, as the codomain is not within the range?

EDIT:

I've moved my thinking along. If I let $$h=\frac{x-2}3$$ and conjugate $$f$$ I seem to get $$hfh^{-1}=f$$ rather than $$g$$ so I'm definitely doing something wrong here. Note, I have no doubt they are topological conjugates, but I want to derive the homeomorphism.

The fact $$hfh^{-1}=f$$ suggests $$f,h$$ commute - so i checked whether $$ax+b$$ and $$cx+d$$ commute in general, and I got $$hfh^{-1}=ax+(1-a)d+bc$$, confirming they DON'T in general commute, but I substituted in this special case and got $$ax+(1-a)d+bc=2x+1$$. I found this which looks relevant but I seem to be going down a rabbit hole not relevant to the original question.

• As you've already computed, $x \mapsto \frac{x-2}{3}$ and $x \mapsto 3x + 2$ are not relevant to your question. This first map $h$ has been constructed to satisfy $hf = g$ which is not the conjugation condition, which is of course $hfh^{-1} = g$. Sep 15, 2022 at 20:15
• Thanks @QiaochuYuan I also had my domain and codomain the wrong way around. Makes more sense now I've swapped them. I'm gonna try substituting $h=ax+b$ into $hfh^{-1}=g$ and see if that gets me anywhere. Re $x \mapsto \frac{x-2}{3}$ satisfying $hfh^{-1}=f$ it's interesting they commute, we have $hf=fh=g$ so that means we have a little commutator group or something, right? math.stackexchange.com/q/4532615/334732 Sep 16, 2022 at 8:50
• Conjugating by a linear map won't change the coefficient of $x$; the best you can do that way is to move the location of the fixed point. Why do you restrict the codomain? Sep 16, 2022 at 17:55
• @QiaochuYuan I restricted the codomain because the action of the function restricts it. You can see this by the fact $\nu_2(2x+1)=0$ for all $x\in\Bbb Z_2$ and the same for $\frac{2x-1}3$. As for the substitution I was wondering if the substitution might give me a function in $x$ rather than just a coefficient. I do have a definition of (a) homeomorphism that topologically conjugates $f$ to $g$ but it acts on all of $\Bbb Z_2$ and I was hoping to see one or more derived which only act on this codomain, in order to understand how independent it is of $\Bbb Z_2\setminus\Bbb Z_2^\times$. Sep 16, 2022 at 19:15
• You're talking about the range there, not the codomain. It's fine for a function to not hit every point in the codomain. If you made the codomain $\mathbb{Z}_2$ then you could actually apply the definition of topological conjugacy here, because then the domain and codomain would be the same. Sep 16, 2022 at 19:19

It turns out that $$f$$ and $$g$$ actually are topologically conjugate.

Let's simplify the question first a bit by conjugating the two functions with the translation $$t(x)=x+1$$. We get $$\tilde{f}(x):=t(f(t^{-1}(x)))=(2(x-1)+1)+1=2x,$$ and $$\tilde{g}(x):=t(g(t^{-1}(x)))=([2(x-1)-1]/3)+1=2x/3.$$ Topological conjugacy is an equivalence relation, so $$f$$ and $$g$$ are conjugate if and only if $$\tilde{f}$$ and $$\tilde{g}$$ are.

Let us then define the function $$h:\Bbb{Z}_2\to\Bbb{Z}_2$$ by the recipe $$h(x)=\begin{cases}0,&\text{if x=0, and}\\ 3^{\nu_2(x)}x,&\text{otherwise.}\end{cases}$$ Here, for all $$x\neq0$$, $$\nu_2(x)=\ell$$, when $$x\in 2^\ell\Bbb{Z}_2\setminus2^{\ell+1}\Bbb{Z}_2$$. Then

• As $$3$$ is a $$2$$-adic unit, and $$\Bbb{Z}_2$$ is the disjoint union of the subsets $$U_\infty:=\{0\}$$ and $$U_\ell:=2^{\ell}\Bbb{Z}_2\setminus 2^{\ell+1} \Bbb{Z}_2$$, $$\ell=0,1,2,\ldots$$, it follows that $$h$$ maps each subset $$U_\ell, \ell=0,1,\ldots,\infty$$, bijectively onto itself. Therefore $$h$$ is itself also a bijection.
• If $$V=x+2^\ell\Bbb{Z}_2$$ is any basic open subset of $$\Bbb{Z}_2$$, then $$h^{-1}(V)=h^{-1}(x)+2^\ell\Bbb{Z}_2$$ is another basic open subset. Therefore $$h$$ is continuous.
• The argument of the previous bullet holds equally for the obvious inverse of $$h$$ (replace $$3$$ by $$1/3$$ everywhere), so we can conclude that $$h$$ is a homeomorphism.
• Finally, we also have, for all $$x\neq0$$ in $$\Bbb{Z}_2$$ $$h(\tilde{g}(x))=3^{\nu_2(2x/3)}(2x/3)=3^{1+\nu_2(x)}(2x/3)=3^{\nu_2(x)}\cdot 2x=\tilde{f}(h(x)).$$ So $$h$$ gives the topological conjugacy between $$\tilde{f}$$ and $$\tilde{g}$$ proving the main claim also.
• In the second bullet I had in mind the case $x\notin 2^\ell\Bbb{Z}_2$, when $\nu_2(y)=\nu_2(x)$ for all $y\in V$. We do need it in the case $x\in 2^\ell\Bbb{Z}_2$ as well. But then $V=2^\ell\Bbb{Z}_2$ and there is nothing to worry about in this case either. Nov 17, 2022 at 8:50
• Thanks for this. It will take me a while to digest. If we let $h$ cycle $(-\frac13,-1,1,\frac13)$ then apart from this cyclic set, it almost certainly maps all the positive ternary rationals into $X=\{\frac{n}3\in\Bbb N:3\nmid n\}$. This means it maps $h(X)\subset X$ (apart from the cyclic point $\frac13$). Do you think proving that is anywhere near within reach from the above? That would be equivalent to the Collatz Conjecture. Nov 17, 2022 at 9:10
• P.S. apologies, I just understood that you have given an explicit $h$ not compatible with that cyclic set. But there is a homeomorphism from your $h$ to a function that cycles those. Nov 17, 2022 at 9:33
• Another corrollary of your proof is that the Lyndon words classify cyclic orbits of the Collatz conjecture (over 2-adic numbers). There are two fixed points $(0),(-1)$, one cycle of order two $(1,2)$, two cycles of order three etc. This gives rise to a homeomorphism from the conjecture to standard models of chaos including the Dyadic Transformation in particular being of most interest, but also the Tent Map and logistic map. It's a shame I'm so slow at progressing this line of inquiry alone. Nov 17, 2022 at 10:13
• Yes. At least according to my understanding of topological conjugacy that's how the solution started. Apr 9 at 4:33