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What's the image of the function $f_r(x)=\left(\dfrac{3x}{2^r}+2^{v_2\left(\tfrac{x}{2^r}\right)}\right)$ on the Prufer 2-group $\mathbb{Z}_2(2^{\infty})$?

For each case of $r=0, 1,\text{ or }2$?

To well-define the function for $r=1,2$, choose $x$ in the range $0\leq x<1$ and pre-divide by $2^r$

For the sake of argument let's say $2^{v_2(0)}=1$ (which is the logical thing to do, since $1\cong0$ in this group).

Taking $r=0$ for starters. If we ignore values around the $\mod1$ region we can easily show that the image is a superset of:

$\mathbb{Z}_2(2^{\infty})\setminus\left\{\dfrac{3p}{2^q}:p\in\mathbb{N},q\in\mathbb{N}\right\}$

This is because a) the action of this function on odd numerators is essentially the Collatz function from one odd number to the next, and b) every odd number bar multiples of $3$ has a predecessor in the Collatz function (odd to odd), while no multiple of $3$ has an odd predecessor.

What I'm unclear on is in the upper region in which $f(x)$ would be greater than $1$ if we ignored the group's modulo $1$ rule, what do these values map to once reduced modulo $1$, and to what degree do they replace the missing multiples of $3$ in the set?

It's clear they replace some, if not all of the multiples of $3$. For example, there is $f(\frac{7}{16})\equiv\frac{11}{8}\mod 1=\frac{3}{8}$ and $f(\frac{7}{8})=\frac{3}{4}$.

I think if we set $r=1$ then we only get one of these values $f(\frac{7}{8})\equiv\frac{11}{4}\mod 1=\frac{3}{4}$ and if we set $r=2$ we get none because $x<1\implies (3x+2^{v_2(x)})/4<1,$ irrespective of reduction modulo $1$.

Choosing $r=1$ looks promising because we have on the face of it:

$f_1(\frac{1}{2}+\frac{1}{4})=f_1(\frac{3}{4})=\frac{1}{2}=0+\frac{1}{2}=f_1(\frac{1}{2})+f_1(\frac{1}{4})$

If all of $\left\{\dfrac{3p}{2^q}:p\in\mathbb{N},q\in\mathbb{N}\right\}$ are replaced in this way then it would seem we might (with a bit of luck and the wind behind us) be able to define a group morphism $f_1:\{\mathbb{Z}_2(2^{\infty})\}\to\{\mathbb{Z}_2(2^{\infty})\}$ albeit with a (possibly) modified group function $f_1:+\to *$.

It seems there's still promise here for various proofs of the Collatz conjecture:

Given that we know one element converges to the identity under composition of $f_1$, all values do.

The fixed point $f(0)=0$ precludes cycles of other orders within the group.

Under Sharkovskii's theorem; some continuous bijective map from $\mathbb{Z}_2(2^{\infty})\times(\mathbb{Z}_2^{\times}\setminus\mathbb{Z})$ to the real line exists, which would preclude cycles of other orders (since cycles of any order other than $1$ would imply the existence of a 2-cycle, contradicting known results).

EDIT

I have found counterexamples to the possibility this is a group automorphism for $r=0,1$.

Robert Frost
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1 Answers1

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$f$ is a surjective function on the Prüfer $2$-group $\Bbb Z(2^\infty) =\Bbb Z[\frac{1}{2}]/\Bbb Z$.

Proof: $f(0) = 0$. Let $\displaystyle \frac{y}{2^n}$ with $y$ odd represent an arbitrary nonzero element. Set

$$x := \begin{cases} \displaystyle \frac{4y-1}{3} & \text{if }y \equiv 1 \text{ mod } 6\\ \displaystyle \frac{4y-1-2^{n+2}}{3} & \text{if }y \equiv 3 \text{ mod } 6, n \text{ odd}\\ \displaystyle \frac{2y-1-2^{n+1}}{3} & \text{if }y \equiv 3 \text{ mod } 6, n \text{ even}\\ \displaystyle \frac{2y-1}{3} & \text{if }y \equiv 5 \text{ mod } 6.\\ \end{cases}$$

Check that in all cases, $x$ is odd. Then in the first two cases, $f(\displaystyle \frac{x}{2^{n+2}}) =\frac{y}{2^n}$, whereas in the last two cases, $f(\displaystyle \frac{x}{2^{n+1}}) =\frac{y}{2^n}$. QED.

About $r\ge 1$ in your new definition (as of 4-II-2018): If I'm not mistaken, you are basically asking what is the image, under the original function $f$, of the set $A_r$ which consists of those elements in $\Bbb{Z}(2^\infty)$ whose representative $\frac{x}{2^n}$ with $0\le x < 2^n$ satisfies $x\le 2^{n-r}$. For $r=1$ one can easily see from the above that all $\frac{y}{2^n}$ with $y \equiv \pm 1$ mod $6$ are still in the image. So are those with $y \equiv 3$ mod $6$ as long as $y\le 2^{n-1}$. I think I can show that the others are not in the image. In other words, for $r=1$ the image misses exactly the cosets of

$$\frac{3}{4}, \frac{9}{16}, \frac{15}{16}, \frac{21}{32}, \frac{27}{32}, \frac{33}{64}, \frac{39}{64}, \frac{45}{64}, \frac{51}{64}, \frac{57}{64}, \frac{63}{64}, \frac{69}{128}, ...$$

For $r=2$ it seems to me that the image consists just of the cosets of $\frac{y}{2^n}$ where $y \equiv \pm 1$ mod $6$ and $0\le\frac{y}{2^n} <3/4$.

To be honest I have not proven these claims for $r=1,2$ in every detail, so beware.

Further, let me point out again that the group endomorphisms of $\Bbb{Z}(2^\infty)$ are precisely the multiplications $x\mapsto kx$ with a fixed $2$-adic number $k \in \Bbb Z_2$, they are all surjective (except $k=0$), and your $f_r$'s are far from being any of them.

Another misconception in your post I want to point out is

some continuous bijective map from $\Bbb Z(2^{\infty})\times(\Bbb Z_2^\times \setminus \Bbb Z)$ to the real line exists, which would preclude cycles of other orders

Namely , if $X$ is any space with cardinality of the real numbers, endowed with the discrete topology (say), then there is a continuous bijective map from $X$ to the real line. And I can't even begin to think how many (continuous) maps $f: X\rightarrow X$ exist, and how many cycles of arbitrary lengths they could have.

Torsten Schoeneberg
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  • Re what you say about cycles. I was referring to some continuous bijection $g$ between $\Bbb Z(2^{\infty})\times(\Bbb Z_2^\times \setminus \Bbb Z)$ and $\mathbb{R}$ such that a cyclic point $f^m(x)=x$ in $f:\Bbb Z(2^{\infty})\times(\Bbb Z_2^\times \setminus \Bbb Z)\to\Bbb Z(2^{\infty})\times(\Bbb Z_2^\times \setminus \Bbb Z)$ in the 2-adic metric space is a cyclic point $f_g^m(x)=x:\mathbb{R}\to\mathbb{R}$ where $f_g$ is the image of $f$ in $\mathbb{R}$ and $\mathbb{R}$ has the standard metric. Bearing in mind continuity in the two spaces is measured by different metrics, if such a map $g$... – Robert Frost Jan 24 '18 at 11:43
  • ...existed, this would mean the existence of any cycle of length $m\notin{0,1}$ would imply (by Sharkovskii's theorem) existence of a cycle of length $m=2$, which is known not to be the case and therefore the trivial cycle is the only cycle. – Robert Frost Jan 24 '18 at 11:44
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    About the cycles, the point is -- if I understand your notation correctly -- that what you call $f_g$ (which I assume would be $g\circ f\circ g^{-1}$) has no reason to be continuous w.r.t. the standard metric on $\Bbb R$. So you cannot apply Sharkovskii to it. (That $g$ is continuous is not sufficient for $f_g$ to be continuous. Look at the counterexample I gave.) – Torsten Schoeneberg Jan 25 '18 at 09:55
  • ok thanks for that. Nevertheless; where i say there is hope, I mean I believe some function $g$ does exist such that $f_g$ is a continuous function on $\mathbb{R}$ and to find it proves the weakened collatz conjecture (no nontrivial loops). What we have here is the Collatz conjecture translated onto an uncountable space, which is a step closer to finding $g$ than having a countable space. – Robert Frost Jan 25 '18 at 13:22
  • The reason I haven't accepted this excellent answer yet is that for $f_1$ I think we lose at least some of the $\dfrac{3p}{2^q}$ and all of them for $f_2$ – Robert Frost Jan 30 '18 at 16:27
  • As I said, $f_r$ is not well-defined for $r \ge 1$. When you tell me what it's supposed to mean, I could think about what its image is. – Torsten Schoeneberg Jan 31 '18 at 01:36
  • Sorry, my bad. I thought I'd rewritten $f$ such that $0\leq x<1$ for all $r>0$ but I hadn't. I've done it now so please let me know if still a problem. – Robert Frost Jan 31 '18 at 09:06
  • I just noticed you've added to your answer for $f_n$ and wanted to say thank-you. $f_1$ is the crucial function. Where you say it misses precisely $\equiv3\mod 6$ I think this is a coincidence at the heart of proving the Collatz conjecture. It turns out these numbers are the precisely the leaves of the Collatz tree ($\Bbb{N}\setminus2\Bbb{N}\to\Bbb{N}\setminus2\Bbb{N}$) since no integer multiple of $3$ is in the range of $(3x+1)\lvert3x+1\rvert_2$ – Robert Frost Feb 10 '18 at 18:38
  • For the record, nothing "misses precisely" classes of representatives $y/2^n$ with ($0\le y<2^n$, $y$ odd and) $y\equiv 3$ mod $6$, and I did not claim that. For $r=1$, $f(7/16) = 3/8$ (or in your definition, $f_1(7/8) = 3/8$), so some $y\equiv 3$ are included. For $r\ge 2$, e.g. $7/8$ is not in the image, and $7\equiv 1$ mod $6$. – Torsten Schoeneberg Feb 11 '18 at 06:01
  • Yes I get that; I think you mean it misses the cosets of every largest element whose numerator in lowest terms is $\equiv3 \mod 6$. So this for $3/4$ is $3/16, 3/32,\ldots$ – Robert Frost Feb 11 '18 at 20:05