2

Question

What functions on 2-adic space are in the conjugacy class of $f(x)=x+2^{\nu_2(x)}$?

A conjugacy class is the set of functions whose action conjugates to any other element of the class, in this case via a homeomorphism. Conjugacy, being reflexive, symmetric and transitive is an equivalence relation.

I'm looking for a classification which is at least exhaustive over functions of the form:

$f(x)=ax+b\cdot2^{\nu_2(x)}$

I can give some simple and more complex examples.

Example 1

$a=1,b=-1$ is an easy trivial example via the homeomorphism $x\mapsto -x$

Example 2

Moreover, examples varying $b$ are easy. 2-adic numbers can be represented with any radix $(0,d)$ where $\nu_2(d)=0$ so the case $b=d$ gives us all $\{(a,b):a=1,\nu_2(b)=0\}$

Example $3$

I know of examples for $a\neq1$. A corollary of Jyrki Lahtonen's answer here is that $3x+2^{\nu_2(x)}$ is conjugate to $x+2^{\nu_2(x)}$

If we take as a given that $\nu_2(b)=0$ then can we pick arbitrary $a$ such that $\nu_2(a)=0$ ?

Observation

Relevant to the above question is the fact that $f(x)$ commutes with $g(x)=2x$, i.e. $f(2x)=2f(x)$. This means for any $a,b$ which are conjugate, $2^m(a,b)$ is also conjugate. So one can pick either $a,b$ as satisfying $\nu_2=0$ and have a canonical representation of the class with $a,b$ drawn from $\Bbb Z_2^\times$. Essentially here I am asking a very similar question as what functions of the form $ax+b$ are conjugate to $x+1$.

Hypothesis

Is it the case that $f$ is conjugate for all $\{(a,b):a\in\Bbb Z_2^\times, b\in\Bbb Z_2^\times\}$?

it's a hire car baby
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    The argument in my answer seems to work for any $2$-adic unit in place of $3$, doesn't it? – Jyrki Lahtonen Apr 07 '23 at 10:18
  • I don't understand how example 3 follows from Jyrki's proof in the link. – Torsten Schoeneberg Apr 10 '23 at 04:46
  • @TorstenSchoeneberg the inverse functions of what Jyrki conjugates there, are $(3x+1)/2$ and $(2x+1)/2$ (or possibly $2x-1$ I can't remember.) But $x\mapsto-x$ is a homeomorphism so it doesn't matter which. Is that sufficient or do you need more? – it's a hire car baby Apr 10 '23 at 10:36
  • @JyrkiLahtonen yes I agree, in my view your linked example also works for arbitrary choices of $a\in\Bbb Z_2^\times$ in $f(x)=ax+1$ and I also think arbitrary choices of $b\in\Bbb Z_2^\times$ can be substituted for $1$ in $f(x)=ax+b$. So we have that all ${f(x)=ax+b:a,b\in\Bbb Z_2^\times}$ are topologically conjugate, correct? And it follows that all ${f(x)=ax+b\cdot2^{\nu_2(x)}:a,b\in\Bbb Z_2^\times}$ are topologically conjugate to each other too. – it's a hire car baby Apr 10 '23 at 10:47
  • @TorstenSchoeneberg I was careless in my last comment, the inverse functions are $f(x)=(x-1)/2$ and $f(x)=(3x+1)/2$. We are free to exchange $x\mapsto -x$ in one function as doing so is a reflection which survives conjugation. That gives us $3x+1\sim x+1$ and $3x+2^{\nu_2(x)}\sim x+2^{\nu_2(x)}$ follows immediately from that as either can be composed with $x\mapsto 2x$ at will. In this question I assume we are in $\Bbb Q_2$ which makes it a little more certain, but I think it holds in $\Bbb Z_2$ as well. – it's a hire car baby Apr 10 '23 at 11:51
  • I see how Jyrki's proof implies that $(x-1)/2$ is conjugate to $(3x+1)/2$ but I do not see how to conclude from there, not even how to remove the $2$'s in the denominator. I.e. I doubt that $(x-1)$ is conjugate to $(3x+1)$. Certainly not via Jyrki's functions because at least one of them does not commute with powers of $2$. – Torsten Schoeneberg Apr 11 '23 at 22:49
  • @TorstenSchoeneberg ok. Sketch the complete directed binary graph over $\Bbb Z_2$ of ${x}\mapsto {2x,2x+1}$, rooted at $\overline{01}_2=-\frac13$. Next replicate the graph but this time sketch ${x}\mapsto{(3x+1)/2,2x}$ rooted at $1$. You should see that you have a homeomorphism from one graph to the other, which is isometric over both $\Bbb Z_2^\times$ and $\Bbb Z_2$. One might say the homeomorphism in $\Bbb Z_2^\times$ respects the rays of the quotient map $\Bbb Z_2\to\Bbb Z_2/\langle2\rangle$ and it can take representatives from $\Bbb Z_2^\times$. – it's a hire car baby Apr 12 '23 at 15:12
  • This is of course only the homeomorphism between one pair of connected graphs, but one only need pick all the other $u$ from the following in order to see the collection of all connected subgraphs which together cover $\Bbb Q_2$: ${2^tu+v:t\in\Bbb Z,u\in\Bbb Z_2^\times,v\in\Bbb Z[\frac12]}$. I'm pretty sure isometry of the homeomorphism is a sufficient condition, because it means convergence on the left corresponds with convergence on the right. – it's a hire car baby Apr 12 '23 at 15:14
  • @TorstenSchoeneberg Sorry in case that was ambiguous, where I wrote "pick all the other $u$" it is probably clearer if I say "replace $-\frac13$ on the left hand graph with any other $u\in\Bbb Z_2^\times$, and relabel the graph on the right by assigning each vertex the binary string corresponding to its parity sequence." Although it is possibly worth mentioning that any choice of all possible $u$ to exactly cover $\Bbb Q_2$ constitutes a Vitali set. – it's a hire car baby Apr 12 '23 at 16:03
  • @TorstenSchoeneberg sorry I made an error in this comment https://math.stackexchange.com/questions/4674472/what-functions-on-2-adic-completion-of-the-rationals-are-in-the-conjugacy-class?noredirect=1#comment9889845_4674472 , where I wrote ${x}\mapsto{(3x+1)/2,2x}$ I should have written ${x}\mapsto{(2x-1)/3,2x}$. I gave the inverse function the left vertex. – it's a hire car baby Apr 12 '23 at 19:19
  • I have no idea what you are talking about, but I have pretty much convinced myself that for the maps $f(x) = 3x+1$ and $g(x) = x-1$, there is no homeomorphism either $c: \mathbb Q_2 \rightarrow \mathbb Q_2$ or $c:\mathbb Z_2 \rightarrow \mathbb Z_2$ such that $c\circ f\circ c^{-1} = g$. If I have time next week or so I might double check my proofs and, if I find no mistake, share them. – Torsten Schoeneberg Apr 13 '23 at 02:48
  • @TorstenSchoeneberg did you draw the graphs? The isometry is definitely correct and if you draw the graphs you will probably see it within 5 minutes. One possible error you may be making, is rooting the graphs incorrectly. I designed the illustration to dispel this possibility. – it's a hire car baby Apr 13 '23 at 03:31
  • No I did not and will not draw graphs. My doubt has little to do with graphs. If you think there is a map $c$ as in my previous comment, just tell me what map it is and I will check it. – Torsten Schoeneberg Apr 13 '23 at 03:58
  • @TorstenSchoeneberg sure no problem. The map is as follows. Let $t(x)=3x+2^{\nu_2(x)}$. Let $t^n$ indicate $n$ compositions of $t$. Then $c=\sum_{k=0}^\infty 2^{\nu_2(t^n(x))}$. It's a $2$-adic isometry over $\Bbb Q_2$. But I still think drawing the graphs I suggested is clearer if you are willing to give that a go. Moreover following the discussion with Jyrki I'm satisfied one can substitute any $f(x)$ in this question for $t$ to get a new homeomorphism. – it's a hire car baby Apr 13 '23 at 08:02
  • @TorstenSchoeneberg another matter drawing the graph would have clarified is that one must be careful about the domain and range, and the graph shows how the homeomorphism behaves with respect to extending the map from $\Bbb Z_2^\times\to\Bbb Z_2$ to one covering all of $\Bbb Q_2$ which was a matter of discussion of my previous question. – it's a hire car baby Apr 13 '23 at 08:12
  • @JyrkiLahtonen I added an answer but really it just says that your previous answer generalises. – it's a hire car baby Apr 13 '23 at 14:26
  • No $c: \mathbb Q_2 \rightarrow \mathbb Q_2$ as in my comment exists, for the simple reason that for $x=-1/2$, we have $f(x) =x$ and hence $c \circ f \circ c^{-1} = g$ implies $c(x) = c(x)-1$. You seem to be talking about $3x+1$ and $3x+2^{v_2(x)}$ interchangeably, those are different functions, I asked about the first. – Torsten Schoeneberg Apr 14 '23 at 05:13
  • @TorstenSchoeneberg I agree, you're right. $3x+1$ and $x+1$ are only topologically conjugate as functions $\Bbb Z_2\times\to2\Bbb Z_2$. My apologies, I thought you were offering that as a counterexample to the proposition in the question but I realise you were not. I only meant to say there is no counterexample to the proposition in the question. The matter of handling the domain and range which was explained to me in the comments here: https://math.stackexchange.com/questions/4531981/ . – it's a hire car baby Apr 14 '23 at 10:04
  • I am not sure if you understand what people explained to you there about domain and codomain. As said I am also pretty sure that there cannot be a $c:\mathbb Z_2\rightarrow \mathbb Z_2$ conjugating $f$ and $g$ as in my comment. Before I put the effort into writing up a proof for that, may I ask if such proof would convince you that $f$ and $g$ are not topologically conjugate? If not, you would have to say what you mean by them being topologicallly conjugate, because I would have disproved the only two meanings of that I can think of. – Torsten Schoeneberg Apr 15 '23 at 21:40
  • @TorstenSchoeneberg It is unfortunate that Jyrki has pruned his comments on that linked question, where he claimed in error that I had not understood the matter of the domain and range, because it was clearer when his comments were still present. Since $c$ is an isometry, $\lvert c(1/2)=\rvert_2=2$. Find $c(-\frac12)$ by finding $c(-1)$ and halving it. $-1$ is a fixed point of the isometry and therefore $c(-\frac12)=-\frac12$. I have given you the isometry and elucidated in my own answer. I refer here to the $f$ in my question and $g(x)=3x+2^{\nu_2(x)}$. – it's a hire car baby Apr 17 '23 at 07:26
  • @TorstenSchoeneberg sorry I meant your $f,g$ are topologically conjugate $\Bbb Z^\times_2\to\Bbb Z_2$ and this extends to an isometry $c$ over $\Bbb Q_2$ via the extension $c(2x)=2c(x)$ – it's a hire car baby Apr 17 '23 at 10:13
  • @TorstenSchoeneberg I just wondered if you have no counterexamples to my answer below, which I would like to accept as correct, but I would not like to do so if you were of the view that it is wrong. – it's a hire car baby Apr 27 '23 at 09:28
  • I do not have time to check that now or presumably in the next few weeks. I have raised my objections to some claims above. Whether those objections invalidate your answer I cannot assess. I remain skeptical. – Torsten Schoeneberg Apr 29 '23 at 00:01
  • @TorstenSchoeneberg it's fairly easily shown that the conjugacy map I define here is solenoidal i.e. it induces a bijection mod $2^n$. This is equivalent to it being an isometry for any pair of units $a,b$. I have now found this in Lagarias and Bernstein 2019 https://www.semanticscholar.org/paper/The-3x-%2B-1-Conjugacy-Map-Bernstein-Lagarias/56dfdf65e1d03de836f6cf5c92b5c5876abe8145 – it's a hire car baby Nov 27 '23 at 22:44

1 Answers1

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Jyrki's answer to my previous linked question generalises to $f_{a,b}(x)=ax-b\cdot2^{\nu_2(x)}$ with $a,b$ any pair of $2$-adic units and therefore all these are topologically conjugate.

(Here $2^{\nu_2(x)}$ is the highest power of $2$ that divides $x$.)

Then if we let $f^n(x)$ indicate the $n^{th}$ composition of $f(x)$ we can define a 2-adic isometry $T$ as follows:

Let $\displaystyle T=\sum_{n=0}^\infty 2^{\nu_2(f^n(x))}$

And $T:\Bbb Q_2\to\Bbb Q_2$ is the isometry which topologically conjugates $f(x)$ to $g(x)=x-2^{\nu_2(x)}$ by satisfying the equation $T(f(T^{-1}(x)))=g(x)$

I will call $T$ the parity vector of $f$.

And there is a little more. Let $T_{a,b}$ be the parity vector associated with $f_{a,b}$ by the above method.

We can say $T_{1,1}(x)=g(x)$ is the identity of a group sitting as a subgroup of the homeomorphisms of $\Bbb Q_2$ because $T_{1,1}(x)=x$.

And for every $f_{1,b}(x)$ it is self-evident that $T$ simply exchanges the radix $(0,1)$ for $(0,b)$ which implies $f_{1,b}(x)=bx$ and therefore we have that the group inverse of $T_{1,b}(x)$ is $T_{1,b^{-1}}(x)$ and the group operation for $a=1$ is $T_{1,b_1\times b_2}(x)$.

Then as a minimum scope of the group we have $T_{1,b}():b\in\Bbb Z_2^\times$ and the group is the "scalings" of $\Bbb Q_2$ by $b\in\Bbb Z_2^\times$ although I'm aware scalings is probably the wrong word given that $b$ are units.

The perhaps less trivial case is the operation when $a\neq 1$ which may have implications for a range of Collatz-like problems. It looks likely $f_{a_1,b_1}(x)\circ f_{a_2,b_2}(x)=f_{a_1\times a_2,b_1\times b_2}(x)$.

I will move on to investigate whether $b$ has an inverse, and what it is, next.

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