1

Let $Z$ be the the set of dyadic and ternary rationals in the interval $\left[\frac12,1\right)$ whose 3-adic valuation is either $-1$ or $0$, with the standard absolute value topology inherited from the real line.

Let $X=\{z\in Z:\nu_3(x)=0\}$.

Let $Y=\{z\in Z:\nu_3(x)=-1\}$.

Now define an equivalence relation $\sim$ which partitions $Z$ into equivalence classes $\{x, y\}$ of cardinality $2$ with $x \in X$, $y \in Y$, and $f(x) = y$, where $f:X\to Y$ is given by $$f(x)=\begin{cases}\frac{4x}3 &\text{if}& x<\frac34\\ \frac{2x}3& \text{if}& x>\frac34.\end{cases}$$

So for example $\frac5{8}\sim\frac56$ and $\frac{7}{8}\sim\frac{7}{12}$.

The quotient (pseudo)metric $d_\sim$ on $Z/{\sim}$ is the infimum distance by which one can traverse from any equivalence class to another, by stepping on up to infinitely many equivalence classes in-between and summing only the distance between classes and not the distance travelled within classes. More formally this is defined at @EricWofsey's answer to "Is there a conceptual reason why topological spaces have quotient structures while metric spaces don't?".

Question

I seek an explicit definition of $d_\sim$ in this instance. Of course, in a sense Eric's answer does give that, but it leaves me with a requirement to somehow iterate over all possible sequences of equivalence classes and determine the shortest path, something well beyond my capabilities.

Also Note

While this question stands alone without reference to the Collatz conjecture, I feel more comfortable declaring that identifying this metric is a component of my study of the conjecture, partly in the spirit of full-disclosure, but also becase it may be material to the answer, to be mindful of the following observations:

  • $g(x)=x+\frac132^{\nu_2(x)}$ is both a surjection $X\to Y$ and a surjection $Z/{\sim}\to Z/{\sim}$ and seen as a map $g:Z/{\sim}\to Z/{\sim}$ it is essentially the Collatz graph and its graph is connected if and only if the Collatz conjecture is true. One should not be surprised therefore, if some proof that $d_\sim$ is the trivial metric, $\forall [z_0],[z_1]:d_\sim([z_0],[z_1])=0$, were related to the claim that the graph of the orbit of $g$ through $Z/{\sim}$ is connected.

  • The $n$-indexed sequences of the form $s_n(x)=x+(1-2^{-6n})\cdot2^{\nu_2(x)}\cdot3^{\nu_3(x)-1}$ form an exact cover of $X$ (up to subsequences) and $g$ is their infinite limit. Moreover, for every $y\in Y$ there are precisely two $s_n$ (up to subsequences) whose union is the level set by $g$ of $y$.

  • If $d_\sim$ is not the trivial (pseudo)metric, it seems likely a proof that some sequence gives the infimum for the metric may use the sequences given in the two bullets above.

  • This question has a "dual", if you like, which is to enquire into the quotient pseudometric obtained when setting $x\sim' g(x)$ instead of $x\sim f(x)$ and moreover the conjecture is equivalent to the claim that $Z/({\sim}\cdot{\sim'})$ is a singleton having $Z$ as its only element.

LSpice
  • 11,423

1 Answers1

1

One gets the trivial semi-distance. Let $d$ denote the semi-distance on $Z$ that defines $d_\sim$ on $Z/\sim$, that is $d_\sim([z],[z']):=d(z,z')$ for all $z$ and $z'$ in $Z$. Thus $d(z,z')\le |z-z'|$ for all $z$ and $z'$ in $Z$ and $d(x,f(x))=0$ for all $x\in X$.

It is convenient to extend the definition of $f$ to a self-map $[1/2,1)\to[1/2,1)$, still denoted $f$: $$f(x)=\begin{cases}\frac{4x}3 &\text{if}& 0\le x<\frac34\\ \frac{2x}3& \text{if}& \frac34\le x <1.\end{cases}$$

The key point is that all orbits of $f$ are dense. So for every $z$ and $z'$ in $Z$, and for every $\epsilon>0$, there is $m\in\mathbb N$ such that $\lvert f^m(z)-z'\rvert< \epsilon$. Since $X$ is also dense, and since $f$ is everywhere (right) continuous, for all indices $ 0\le j\le m$ there are $x_j\in X$ such that \begin{align*} \bigl\lvert f^j(z)-x_j\bigr\rvert<\frac\epsilon m, & \qquad (0\le j\le m) \\ \bigl\lvert f(x_j)-f^{j+1}(z)\bigr\rvert<\frac\epsilon m, & \qquad (0\le j\le m-1). \end{align*} Therefore for $0\le j\le m-1$ $$\bigl\lvert f(x_j)-x_{j+1}\bigr\rvert \le \bigl\lvert f(x_j)-f^{j+1}(z)\bigr\rvert+\bigl\lvert f^{j+1}(z)-x_{j+1}\bigr\rvert\le \frac{2\epsilon}m$$ and
$$d(z,z')\le\bigl\lvert z-x_0\bigr\rvert+ \bigg(\sum_{j=0}^{m-1} \bigl\lvert f(x_j)-x_{j+1}\bigr\rvert \bigg) +\big|x_m-z'\big|\le 4\epsilon,$$ which proves that $d$ vanishes identically.

All orbits of $f$ are dense: Indeed, having all orbits dense is a property invariant by conjugation, and $f$ is conjugate to the irrational translation (modulo $1$) on $[0,1)\sim\mathbb{R/Z }$ given by $T_c:x\mapsto {x -c}$, where $c:=\frac {\log 3}{\log 2}-1$ (all orbits of a translation $T_c$ are dense iff $c$ is irrational).

Consider the homeomorphism $h:[0,1)\to[1/2,1)$ defined by $h(x)=2^{x-1}$. Then it is easy to check that $h\circ T_c=f \circ h$.

Pietro Majer
  • 56,550
  • 4
  • 116
  • 260
  • Thank-you most kindly. This will take some time for me to digest properly to my lack of skill but I think the gist is that the pseudometric is $d_\sim=0$. Does this necessarily push through to the quotient space not being Hausdorff? I assume not because the quotient space is richer than the quotient metric. – it's a hire car baby Jan 04 '21 at 22:32
  • P.S. was this fairly instantly obvious to a better mathematician than I, or did it take some degree of thought? It seems fairly obvious the way you put it about every orbit being dense - must imply the distances between them are zero. – it's a hire car baby Jan 04 '21 at 22:33
  • 1
    I think so (quotient being Hausdorff) but I do not see immediately. – Pietro Majer Jan 04 '21 at 23:02
  • 1
    On the PS: To me, it was not obvious, and I wasn't even sure that the distance was trivial. How I proceeded, it's nothing special, but here it is. 1) It seemed that $X$ and $Y$ do not play a special role in studying these objects, and that it is healthier to have $f$ as a self-map, to make iterations. 2) one works better on the interval, than on the quotient. – Pietro Majer Jan 04 '21 at 23:22
  • 1
  • in this semi-distance, $1/2$ is close to the right-end, for $f(3/4-\epsilon)= 1-4\epsilon/3$ and $f(3/4+\epsilon)= 1/2-2\epsilon/3$. So nothing changes if one starts from a circle, identifying the endpoints of $[1/2,1]$, instead using $[1/2,1)$, which makes $f$ a homeomorphism. This allows to use the tools of homeomorphims of the circle. I wrote $f$ as a time-1 flow to compute its rotation number and find a conjugation with a rotation: only at the end I realized the conjugation was so easy.
  • – Pietro Majer Jan 04 '21 at 23:28
  • 1
    PS: The quotient (semi)distance $\tilde d$ on $\tilde X$ for a quotient $\pi:X\to\tilde X$ is the maximum (semi)distance on $\tilde X$ that makes $\pi$ a $1$-Lipschitz map. This leads to the construction of $\tilde d$ via chains of jumps from a class to anothe (which I would call the construction rather than the definition of $\tilde d$ ). – Pietro Majer Jan 05 '21 at 10:30
  • 1
    Yes, it was meant to be a proof (sorry if it wasn't clear). Having every orbit dense is a conjugation invariant (recall that $f=hgh^{-1}\Rightarrow f^m=hg^mh^{-1}$ for all $m$), and the translation $x\mapsto x+c\mod 1$ on $\mathbb{ T:=R/Z}$ does have every orbit dense if (and only if) $c$ is irrational (for the orbit of $x$ is $x+c \mathbb Z\mod 1 $), so every orbit of $f$ is dense too. – Pietro Majer Jan 05 '21 at 11:39
  • Don't worry, it is perfectly normal that we sometimes miss something, and we have the right to ask any question --even because sometimes "trivialities" hide not-at-all-trivial facts. – Pietro Majer Jan 05 '21 at 12:16
  • It seems LSpice overlooked the condition in the first sentence: whose $3$-adic valuation is either $−1$ or $0$. This constrains the equivalence classes to have cardinality two. I'm just checking you didn't overlook the same, and whether your answer still stands notwithstanding this? – it's a hire car baby Jan 05 '21 at 15:54
  • 1
    You don't really need to quotient, as your quotient set $Z/\sim$ can be identified with $X$. Alternatively, you may even consider $f$ on all rationals, and quotient on the $\mathbb Z$ action, so that equivelence classes are the whole orbits of $f$. Any class then contains exactly one element of $X$. – Pietro Majer Jan 05 '21 at 16:21
  • I'm sorry to push for more - please feel free to ignore. This is to do with the motivation behind this question: Consider the $n$-indexed sequence $\dfrac{2^{6n+2}-1}6=\frac12,\frac{85}{128},\frac{5461}{8192}\ldots$. This converges in $\Bbb R$ to $\frac23$ and in $\Bbb Q_2$ to $-\frac16$. In the trivial pseudometric here it trivially converges as it's the identity sequence. But it also converges to $[\frac12\sim\frac23]$ in the quotient topology (non-pseudometric). Any pointers how I might attempt to derive/construct this quotient topology and show it is Hausdorff or nontrivial? – it's a hire car baby Jan 07 '21 at 15:54
  • ok but which quotient topology? – Pietro Majer Jan 07 '21 at 16:44
  • Simply $Z/{\sim}$ in the quotient topology as per https://math.stackexchange.com/a/3810857/334732 which may be stronger than the quotient pseudometric. I.e. the standard absolute value metric from $\Bbb R$ reduced by the equivalence relation $\sim$. Or maybe it's no stronger and it coincides with the pseudometric topology in this particular case? – it's a hire car baby Jan 07 '21 at 17:23
  • So you mean the quotient topology on $Z/!\sim$ for the topological space $Z$ (as a subspace of $\mathbb R$) and the equivalence relation, as defined in your post here? – Pietro Majer Jan 07 '21 at 17:25
  • Yes, that's it. That is the ultimate motivation of this question... I hoped this question would either be it, or be a stepping stone towards it. The convergent sequences or open sets I'm interested have the recurrence relation $x_{n+1}=x_n+21\cdot2^{\nu_2(x_n)-6}\cdot3^{\nu_3(x_n)}$ – it's a hire car baby Jan 07 '21 at 17:44
  • Thank-you. This will provide a good month or two of food for thought for me. I am much obliged to you. – it's a hire car baby Jan 07 '21 at 20:10
  • Topologic conjugacy, as it is written in the last line. – Pietro Majer Jan 09 '21 at 20:34
  • 1
    @samerivertwice, I agree: $(f \circ h)(x) = \frac{2^{x + 1}}3 = 2^{x - \log_2(3) + 1} = h(x - c + 1)$ for $x < c$, and $(f \circ h)(x) = \frac{2^x}3 = 2^{(x - \log_2(3) + 1) - 1} = h(x - c)$ for $x \ge c$. – LSpice Jan 11 '21 at 15:58
  • You say I don't really need to quotient as the quotient can be identified with $X$. I want to show $g=x+\frac{21}{64}\cdot2^{\nu_2(x)}$ converges on transfinite iteration.

    For some $x:\lvert x\rvert_3=1$, the infinite limit of $g^n(x)$ satisfies $\lvert x\rvert_3=3$. Fine, so apply $f^{-1}$ and I'm back in $X$. Therefore I can just use $j(x)=f^{-1}\circ \lim_{n\to\infty}g^n:X\to X$. Does your $h$ help me with the proof that $j$ converges on iteration? The claim that it does converge is equivalent to the Collatz conjecture.

    – it's a hire car baby Jan 20 '21 at 13:59