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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.

it's a hire car baby
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  • 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 $hfh^{-1} = g$. – Qiaochu Yuan Sep 15 '22 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? https://math.stackexchange.com/q/4532615/334732 – it's a hire car baby Sep 16 '22 at 08:50
  • 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? – Qiaochu Yuan Sep 16 '22 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 $\Bbb Z_2\setminus\Bbb Z_2^\times$. – it's a hire car baby Sep 16 '22 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. – Qiaochu Yuan Sep 16 '22 at 19:19
  • @QiaochuYuan Happy to have codomain $\Bbb Z_2$. I guess having $h:\Bbb Z_2\to \Bbb Z_2$ might be sufficient. Either way, does that help me show they're homeomorphic? I wasn't sure if this might be a hint (and the answer the the question itself should be obvious to me) or purely a pointer re the range / codomain to be taken at face value. – it's a hire car baby Sep 18 '22 at 10:55
  • It's not a hint, it just allows you to actually apply the definition of topological conjugacy. I don't actually know the answer to your question either way; some messing around with the $p$-adic logarithm might tell you something. – Qiaochu Yuan Sep 18 '22 at 21:25
  • @JyrkiLahtonen Thanks, that's helpful. Qiaochu alluded to; functions don't need to surject in order to be topologically conjugate, so had I defined $f,g$ to be $\Bbb Z_2\to\Bbb Z_2$ they would have both gone $X\to X$ as you say with $X=Y$ and met the definition of topologically conjugate. I'm working up to is a question about extending the conjugacy between these two functions to $\Bbb Q_2$, which I have intuitively straight in my mind but I need to straighten out my language and definitions first. – it's a hire car baby Nov 16 '22 at 17:09

1 Answers1

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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.
Jyrki Lahtonen
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  • 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. – Jyrki Lahtonen Nov 17 '22 at 08:50
  • 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. – it's a hire car baby Nov 17 '22 at 09:10
  • 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. – it's a hire car baby Nov 17 '22 at 09:33
  • 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. – it's a hire car baby Nov 17 '22 at 10:13
  • Am I right in thinking $\tilde{f}(x)$ is topologically conjugate to $f(x)$? – it's a hire car baby Apr 09 '23 at 03:57
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    Yes. At least according to my understanding of topological conjugacy that's how the solution started. – Jyrki Lahtonen Apr 09 '23 at 04:33
  • Oh hang on ignore that, it does compute because it's just exchanging $(0,1)$ in the radix. It just looked wrong / counterintuitive at first. – it's a hire car baby Apr 09 '23 at 07:15
  • I can't parse why the set in bullet two is some arbitrary open set, but this is due to my own ineptitude at topology rather than any shortcoming of the question. But I accept $h$ is a homeomorphism due to the metric space axioms. – it's a hire car baby Apr 10 '23 at 10:44
  • Related: https://math.stackexchange.com/questions/4784358/ – it's a hire car baby Oct 10 '23 at 14:51
  • In case you are interested in a particularly elegant proof of this, I have now found this in a paper, Lagarias and Bernstein 2019 https://www.semanticscholar.org/paper/The-3x-%2B-1-Conjugacy-Map-Bernstein-Lagarias/56dfdf65e1d03de836f6cf5c92b5c5876abe8145 . They show that the function is solenoidal, i.e. it induces a permutation mod $2^n$ and this is equivalent to isometry. It's a pretty neat paper and very clear throughout. – it's a hire car baby Nov 27 '23 at 22:49