# Are there periodic orbits of two-adic isometries not having period a power of two?

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.

UPDATE

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 $$EA and that implies $$EA but by isometry $$EA=AB$$, a contradiction so no pentagons.

## 1 Answer

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 $$2$$.

• 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? Commented Nov 18, 2022 at 15:44
• @samerivertwice yes Commented Nov 18, 2022 at 17:15
• I can't parse $\zeta_{2^k}$ here. Is it to do with this? en.wikipedia.org/wiki/P-adic_L-function Commented Nov 18, 2022 at 23:10
• @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 $\Bbb{C}=\Bbb{R}(i)\cong \Bbb{R}[t]/(t^2+1)$) Commented Nov 18, 2022 at 23:19
• ok thanks will try again. Also I added a proof of no pentagons to the question. Commented Nov 18, 2022 at 23:45