What orders of cycles are impossible in a two-adic isometry? Are there orders other than powers of two?

Sharkovskii's theorem says that if a continuous function on an interval of the real line has periodic points of order $n$ then it has periodic points of every order $m$ such that $m$ follows $n$ in the following ordering:

  • The odd numbers greater than one in increasing order

  • Two times the odd numbers greater than one in increasing order

  • Four times the odd numbers greater than one in increasing order

  • etc...

  • The powers of two in decreasing order.

So for example, if a function has periodic points of order six, it has periodic points of every order other than possibly three. Three coming first in this ordering is seen by many as a symptom (or sign) of the well-known "period three implies chaos" adage.

Now consider a 2-adic isometry $T$ having an orbit of periodic points $x_0,x_1,\ldots$ such that $x_{n+1}=T(x_n)$. Isometry guarantees $d(x_n,x_{n+1})=d(x_{n+1},x_{n+2})$ and since there are no equilateral triangles in $\Bbb Z_2$, there can be no $3$-cycles. This is ever so slightly tantalising because of three implies chaos. It got me wondering, what periodic points are possible in a 2-adic isometry? Does the set of orbits comply with Sharkovskii's ordering for example?

Well since for any isometry $T$ we also have $T^n$ is an isometry, we can count out periods of every multiple of $3n$ too. What about other periods having some odd number as a factor? Any 2-adic isometry having an odd number greater than one as a factor of the period of any point, would be a counterexample to the conjecture that 2-adic isometries have periods compatible with Sharkovskii's ordering. Can you give one?

I have examples of 2-adic isometries with points of period $1, 2$ and $4$, weakly suggesting any period $2^n$ is possible. From the above factorisation rule, we have that the final line of Sharkovskii's ordering is obeyed - i.e. if an isometry exists of order $2^n$ then isometries exist of every order $2^m:m<n$.


I think I've made some little progress by excluding $5$ as a factor, perhaps the principle can be extended...

Consider edge lengths of a pentagon $ABCDE$ then by the same no equilateral triangles rule we have $AB=BC>AC$ and by isometry / topological conjugacy I suspect we must also have $AC=BD=CE=DA=EB$. If so, then by the same rule we have $EA<AC=CE$ and that implies $EA<AC<AB$ but by isometry $EA=AB$, a contradiction so no pentagons.


1 Answer 1


The map $$\Bbb{Q}_2 \to \Bbb{Q}_2(\zeta_{2^k}), \qquad \sum_{n\ge -N} a_n 2^n \mapsto \sum_{n\ge -N} a_n (\zeta_{2^k}-1)^n, \qquad a_n \in \{0,1\}$$ is an isometry with suitable normalization of the absolute values ($|2|_{\Bbb{Q}_2}=1/2=|\zeta_{2^k}-1|_{\Bbb{Q}_2(\zeta_{2^k})}$)

The multiplication by $\zeta_{2^k}$ is an isometry of order $2^k$ on $\Bbb{Q}_2(\zeta_{2^k})$, this gives an isometry of order $2^k$ on $\Bbb{Q}_2$.

Next, every orbit under an isometry $f:\Bbb{Q}_2\to \Bbb{Q}_2$ is either infinite or of size a power of $2$.

We assume that $a$ is a periodic point and that $v(a-f(a))=r$.

$f(a+2^r\Bbb{Z}_2)=f(a)+2^r\Bbb{Z}_2 = a+2^r \Bbb{Z}_2$

$f(a+2^{r+1}\Bbb{Z}_2) = f(a)+2^{r+1}\Bbb{Z}_2 = a+2^r+2^{r+1}\Bbb{Z}_2$

$f(a+2^r+2^{r+1}\Bbb{Z}_2)\cap f(a+2^{r+1}\Bbb{Z}_2)=\emptyset$

Whence $f(a+2^r+2^{r+1}\Bbb{Z}_2)=a+2^{r+1}\Bbb{Z}_2$

ie. $f$ is swapping $a+2^r+2^{r+1}\Bbb{Z}_2$ and $a+2^{r+1}\Bbb{Z}_2$.

This implies that the size of the orbit of $a$ is even.

Considering the orbits under $f^{2^k}$ for each $k$, all of which are either even or trivial, we get that the size of the orbit is a power of $2$.

  • $\begingroup$ Thanks. Do I understand correctly this generalises my last comment that I expect to see isometries with preiodic points of every order $2^n$? So it's a partial answer, but not confirming or refuting the conjecture that we have no prime factors other than two among periods of 2-adic isometries? $\endgroup$ Commented Nov 18, 2022 at 15:44
  • 1
    $\begingroup$ @samerivertwice yes $\endgroup$
    – reuns
    Commented Nov 18, 2022 at 17:15
  • $\begingroup$ I can't parse $\zeta_{2^k}$ here. Is it to do with this? en.wikipedia.org/wiki/P-adic_L-function $\endgroup$ Commented Nov 18, 2022 at 23:10
  • $\begingroup$ @samerivertwice No it is a $2^k$-th root of unity, so $\Bbb{Q}_2(\zeta_{2^k})$ is an algebraic extension of $\Bbb{Q}_2$, the splitting field of $x^{2^k}-1\in \Bbb{Q}_2[x]$. Concretely $\Bbb{Q}_2(\zeta_{2^k})$ is isomorphic to the quotient ring $\Bbb{Q}_2[t]/(t^{2^{k-1}}+1)$ (try first with $\Bbb{C}=\Bbb{R}(i)\cong \Bbb{R}[t]/(t^2+1)$) $\endgroup$
    – reuns
    Commented Nov 18, 2022 at 23:19
  • $\begingroup$ ok thanks will try again. Also I added a proof of no pentagons to the question. $\endgroup$ Commented Nov 18, 2022 at 23:45

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