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?


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.

  • $\begingroup$ 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 $hfh^{-1} = g$. $\endgroup$ Sep 15, 2022 at 20:15
  • $\begingroup$ 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 $\endgroup$ Sep 16, 2022 at 8:50
  • $\begingroup$ 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? $\endgroup$ Sep 16, 2022 at 17:55
  • $\begingroup$ @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 $\Bbb Z_2\setminus\Bbb Z_2^\times$. $\endgroup$ Sep 16, 2022 at 19:15
  • $\begingroup$ 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. $\endgroup$ Sep 16, 2022 at 19:19

1 Answer 1


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.
  • $\begingroup$ 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. $\endgroup$ Nov 17, 2022 at 8:50
  • $\begingroup$ 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. $\endgroup$ Nov 17, 2022 at 9:10
  • $\begingroup$ 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. $\endgroup$ Nov 17, 2022 at 9:33
  • $\begingroup$ 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. $\endgroup$ Nov 17, 2022 at 10:13
  • 1
    $\begingroup$ Yes. At least according to my understanding of topological conjugacy that's how the solution started. $\endgroup$ Apr 9 at 4:33

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