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Halving dynamical systems

Shaun Stevens    Tom Ward    and Stefanie Zegowitz

1 Introduction.

Halving (or doubling) a number is an innocuous operation. Unarguably it changes the size of a number, though for a rational number this change in the real size is exactly compensated by a change in its size as a 22-adic number. For a real number all the more interesting properties — being rational, algebraic, transcendent, well- or badly-approximable — are unchanged. Doubling (or halving) other objects also makes sense, but with potentially much greater impact on non-trivial properties.

For example, doubling a finite group GG may be thought of as forming another group HH for which the sequence of groups and homomorphisms

{1}C2HG{1}\{1\}\longrightarrow C_{2}\longrightarrow H\longrightarrow G\longrightarrow\{1\}

is exact, where C2C_{2} is the group with two elements. Once again this has a simple effect on the size, as |H|=2|G||H|=2|G|. However, the structure of the group clearly can change, and the property of being abelian or simple is not generally preserved. Halving only makes sense if |G||G| is even, in which case it might be thought of as follows. If |G||G| is even then GG contains at least one copy of C2C_{2}, and halving means forming the quotient space G/C2G/C_{2}. This operation does not even preserve the property of being a group, unless the copy of C2C_{2} happens to be a normal subgroup.

Our purpose here is to discuss doubling and halving in the context of topological dynamical systems, by which we mean pairs (X,T)(X,T), where XX is a compact metric space and T:XXT:X\to X is a homeomorphism with a fixed point and with finitely many points of period nn for each n1n\geqslant 1. Halving (and doubling) may be thought of as a relationship between pairs (X,T)(X,T) and (X¯,T¯)(\overline{X},\overline{T}) of the following form.

Suppose there is a continuous involution ı:XX\imath:X\to X that commutes with TT, and use this to define an equivalence relation on XX by saying that xyx\sim y if and only if x=ı(y)x=\imath(y). Writing [x][x]_{\mathord{\sim}} for the equivalence class {x,ı(x)}\{x,\imath(x)\} of xx we define X¯\overline{X} to be the quotient space X/X/\mathord{\sim} and T¯\overline{T} to be the map defined by T¯([x])=[T(x)]\overline{T}\left([x]_{\sim}\right)=[T(x)]_{\sim}. The process of passing from (X,T)(X,T) to (X¯,T¯)(\overline{X},\overline{T}) may be thought of as halving, and passing from (X¯,T¯)(\overline{X},\overline{T}) to (X,T)(X,T) as doubling.

The question we address is to ask which structures are preserved and which are not by doubling and halving in this sense. The most important quantity associated to a topological dynamical system is the topological entropy, which will not be of interest here as it is preserved by halving (see [3, Ch. 5] for the definition and details). We are particularly interested in the relationship between closed orbits in the two systems. It turns out that there are some constraints on the relationship between the numbers of closed orbits for TT and for T¯\overline{T} but that, within these constraints, essentially everything is possible. The constraints will be described in Lemma 5, and the freedom within those constraints in Corollary 10.

2 Closed orbits

We begin with some notational conventions for a map T:XXT:X\to X. For nn a natural number, we write

T(n)={xXTn(x)=x}\mathscr{F}_{T}(n)=\{x\in X\mid T^{n}(x)=x\}

for the set of points of period nn under iteration of TT, and 𝖥T(n)=|T(n)|\mathsf{F}_{T}(n)=|\mathscr{F}_{T}(n)| for the number of points of period nn. Similarly, write 𝒪T(n)\mathscr{O}_{T}(n) for the set of closed orbits of length nn under iteration of TT, and 𝖮T(n)=|𝒪T(n)|\mathsf{O}_{T}(n)=|\mathscr{O}_{T}(n)| for the number of closed orbits of length nn. The set of points of period nn comprises exactly the disjoint union of those points on a closed orbit of length dd for each dd dividing nn, and each such orbit consists of dd distinct points. Thus

𝖥T(n)=d|nd𝖮T(d),\mathsf{F}_{T}(n)=\sum_{d|n}d\mathsf{O}_{T}(d), (1)

and hence, by Möbius inversion,

𝖮T(n)=1nd|nμ(nd)𝖥T(d)\mathsf{O}_{T}(n)=\textstyle\frac{1}{n}\displaystyle\sum\limits_{d|n}\mu\left(\textstyle\frac{n}{d}\right)\mathsf{F}_{T}(d)

for any n1n\geqslant 1.

A convenient generating function for the periodic point data is the dynamical zeta function

ζT(z)=exp(n1𝖥T(n)znn)=n1(1zn)𝖮T(n),\zeta_{T}(z)=\exp\left(\sum_{n\geqslant 1}\mathsf{F}_{T}(n)\frac{z^{n}}{n}\right)=\prod_{n\geqslant 1}\left(1-z^{n}\right)^{-\mathsf{O}_{T}(n)},

(the second equality is equivalent to the identity (1)) which defines a function under the assumption that 𝖥T(n)<\mathsf{F}_{T}(n)<\infty for all n1n\geqslant 1, in which case it has radius of convergence given by

(lim supn(𝖥T(n))1/n)1.\left(\limsup_{n\to\infty}\left(\mathsf{F}_{T}(n)\right)^{1/n}\right)^{-1}.

3 Closed orbits and topological factors

The halving process of the introduction is a special case of forming a topological factor. If (X,T)(X,T) and (Y,S)(Y,S) are topological dynamical systems with a continuous surjective map π:XY\pi:X\to Y satisfying πT=Sπ\pi\circ T=S\circ\pi, then (Y,S)(Y,S) is called a topological factor of (X,T)(X,T). In general, closed orbits can behave very badly under a topological factor map, as the following examples illustrate.

Example 1.

Closed orbits in (X,T)(X,T) may be glued together in a topological factor, as illustrated in Figure 1(a). An extreme instance of this is to take Y={y}Y=\{y\} to be a singleton, set S(y)=yS(y)=y, and define the factor map π\pi by π(x)=y\pi(x)=y for all xXx\in X. Then, whatever the sequence (𝖮T(n))\left(\mathsf{O}_{T}(n)\right) of numbers of closed orbits under TT, the factor system has 𝖮S(1)=1\mathsf{O}_{S}(1)=1 and 𝖮S(n)=0\mathsf{O}_{S}(n)=0 for all n2n\geqslant 2.

Example 2.

Finite pieces of orbits in (X,T)(X,T) that are not closed may close up under the factor map, producing closed orbits in (Y,S)(Y,S), as illustrated in Figure 1(b). An extreme instance of this is once again to take any topological dynamical system (Y,S)(Y,S) and form the product system X=Y×𝕋X=Y\times\mathbb{T}, where 𝕋=/\mathbb{T}=\mathbb{R}/\mathbb{Z} is the additive circle, with map T(y,t)=(S(y),t+α(mod1))T(y,t)=(S(y),t+\alpha\pmod{1}) for some fixed α\alpha\notin\mathbb{Q}. Then (Y,S)(Y,S) is a topological factor of (X,T)(X,T) via the map

π:X\displaystyle\pi:X \displaystyle\longrightarrow Y\displaystyle Y
(y,t)\displaystyle(y,t) \displaystyle\longmapsto y.\displaystyle y.

Then, whatever the sequence (𝖮S(n))\left(\mathsf{O}_{S}(n)\right) of numbers of closed orbits under SS, the map TT has no orbits of finite length.

\psfrag{A}{$\pi$}\psfrag{B}{$\pi$}\psfrag{a}{(a)}\psfrag{b}{(b)}\includegraphics{factor.eps}
Figure 1: Closed orbits squashed to a point and created under a factor map.

Thus we cannot expect to be able to make statements about numbers of closed orbits in factor systems in general. For quotienting by an action of C2C_{2} (that is, halving), the situation is more restricted.

Example 3.

Let X=𝕋=/X=\mathbb{T}=\mathbb{R}/\mathbb{Z} be the additive circle, and define the map TT on XX by T(x)=2x(mod1)T(x)=2x\pmod{1}. This is not a homeomorphism, but is a convenient familiar map to use as an initial example. The involution ı(x)=1x\imath(x)=1-x commutes with TT, and so defines a halved system (X¯,T¯)(\overline{X},\overline{T}). A convenient way to visualize this system is to imagine looking sideways at the unit circle so that points identified by the map ı\imath are seen as a single point, as illustrated in Figure 2.

\psfrag{1}{$0$}\psfrag{m1}{$\textstyle\frac{1}{2}$}\psfrag{i}{$\textstyle\frac{3}{4}$}\psfrag{-i}{$\textstyle\frac{1}{4}$}\psfrag{A}{$[0]=\{0\}$}\psfrag{B}{$[\textstyle\frac{1}{2}]=\{\textstyle\frac{1}{2}\}$}\psfrag{C}{$[0]$}\psfrag{D}{$[\textstyle\frac{1}{2}]=\{\textstyle\frac{1}{2}\}$}\psfrag{II}{$[\textstyle\frac{1}{4}]=\{\textstyle\frac{1}{4},\textstyle\frac{3}{4}\}$}\includegraphics{figuretentmap.eps}
Figure 2: Halving the circle doubling map gives a tent map.

It is straightforward to check that the quotient system (X¯,T¯)(\overline{X},\overline{T}) is the tent map, with X¯=[0,12]\overline{X}=[0,\frac{1}{2}] and

T¯(x)={2x0x14,12x14x12.\overline{T}(x)=\begin{cases}2x&0\leqslant x\leqslant\frac{1}{4},\\ 1-2x&\frac{1}{4}\leqslant x\leqslant\frac{1}{2}.\end{cases}

For these two systems it is easy to calculate that

ζT(z)=1z12z\zeta_{T}(z)=\frac{1-z}{1-2z}

and

ζT¯(z)=112z.\zeta_{\overline{T}}(z)=\frac{1}{1-2z}.

In this example, a system with approximately 2n2^{n} points of period nn and a rational zeta function is halved to a system with the same properties.

The following example gives a natural way in which one could double a topological dynamical system, simply by putting together two copies of the system.

Example 4.

Given any dynamical system (Y,S)(Y,S) on a metric space (Y,d)(Y,d) we may form the doubled space X=Y×{0,1}X=Y\times\{0,1\} with the metric

dX((y,e),(y,e))=d(y,y)+{1if ee,0if e=e,d_{X}\left((y,e),(y^{\prime},e^{\prime})\right)=d(y,y^{\prime})+\begin{cases}1&\mbox{if }e\neq e^{\prime},\\ 0&\mbox{if }e=e^{\prime},\end{cases}

for y,yYy,y^{\prime}\in Y and e,e{0,1}e,e^{\prime}\in\{0,1\}, and define a map on the doubled space by

T:X\displaystyle T:X \displaystyle\longrightarrow X\displaystyle X
(y,e)\displaystyle(y,e) \displaystyle\longmapsto (S(y),e+1(mod2)).\displaystyle(S(y),e+1\pmod{2}).

The involution ı:(y,e)(y,e+1(mod2))\imath:(y,e)\longmapsto(y,e+1\pmod{2}) commutes with TT, giving a halved system (X¯,T¯)(\overline{X},\overline{T}) which can be identified with the original system (Y,S)(Y,S). Clearly

𝖥T(n)={0if n is odd;2𝖥S(n)if n is even,\mathsf{F}_{T}(n)=\begin{cases}0&\mbox{if $n$ is odd;}\\ 2\mathsf{F}_{S}(n)&\mbox{if $n$ is even,}\end{cases} (2)

so that

lim supn1nlog𝖥T(n)=lim supn1nlog𝖥S(n),\limsup_{n\to\infty}\tfrac{1}{n}\log\mathsf{F}_{T}(n)=\limsup_{n\to\infty}\tfrac{1}{n}\log\mathsf{F}_{S}(n),

meaning that the zeta functions ζT\zeta_{T} and ζT¯=ζS\zeta_{\overline{T}}=\zeta_{S} have the same radius of convergence. Moreover, the relation (2) may be written as

ζT(s)=ζS(s)ζS(s),\zeta_{T}(s)=\zeta_{S}(s)\zeta_{S}(-s), (3)

showing that if ζS\zeta_{S} is rational then ζT\zeta_{T} is also rational, though we will see later that the converse is not true (see Example 11).

Our purpose here is to show how extremely unrepresentative Examples 3 and 4 really are. In general, both the growth rate in the number of periodic points and the arithmetic nature of the zeta function do not survive under doubling or halving. However, in contrast to Examples 1 and 2 we will show that the change of the growth rate in the number of closed orbits is restricted.

4 Shortening, Surviving, and Gluing Orbits

We put ourselves back in the halving situation of the introduction. Thus (X,T)(X,T) is a topological dynamical system, which we recall means a pair with XX a compact metric space and T:XXT:X\to X a homeomorphism with a fixed point and with finitely many points of period nn for each n1n\geqslant 1, and ı\imath is a continuous involution on XX which commutes with TT. Then X¯\overline{X} is the quotient of XX under the equivalence relation induced by ı\imath, with quotient map π:XX¯\pi:X\mapsto\overline{X} given by π(x)={x,ı(x)}\pi(x)=\{x,\imath(x)\}. Note that X¯\overline{X} is a metric space: if dXd_{X} is the metric on XX, then the metric dX¯d_{\overline{X}} on X¯\overline{X} is given by

dX¯(x¯,y¯)=min{dX(x,y)xπ1(x¯),yπ1(y¯)}.d_{\overline{X}}(\overline{x},\overline{y})=\min\{d_{X}(x,y)\mid x\in\pi^{-1}(\overline{x}),y\in\pi^{-1}(\overline{y})\}.

The map T¯:X¯X¯\overline{T}:\overline{X}\to\overline{X} is (well-)defined by the relation πT=T¯π\pi\circ T=\overline{T}\circ\pi, and is a homeomorphism.

The factor map π\pi maps any closed orbit under TT to a closed orbit under T¯\overline{T}; conversely, since the fibres of π\pi are finite, the inverse image under π\pi of a closed orbit under T¯\overline{T} is a finite set closed under TT so is a finite union of closed orbits under TT. In particular, π\pi induces a surjective map

n=1𝒪T(n)n=1𝒪T¯(n).\bigsqcup_{n=1}^{\infty}\mathscr{O}_{T}(n)\to\bigsqcup_{n=1}^{\infty}\mathscr{O}_{\overline{T}}(n). (4)

In order to analyze this more closely, let τ={x,T(x),T2(x),,Tn(x)=x}\tau=\{x,T(x),T^{2}(x),\dots,T^{n}(x)=x\} be a closed orbit in (X,T)(X,T) of length nn. Then ı\imath and τ\tau can interact in just three ways.

  1. 1.

    It could fix the orbit τ\tau pointwise, (we say that τ\tau ‘survives’), that is x=ı(x)x=\imath(x) for all xτx\in\tau; we write 𝒪Ts(n)\mathscr{O}_{T}^{s}(n) for the set of closed orbits of length nn under TT that are fixed pointwise by ı\imath. Then the factor map π\pi induces an injective map 𝒪Ts(n)𝒪T¯(n)\mathscr{O}_{T}^{s}(n)\to\mathscr{O}_{\overline{T}}(n).

  2. 2.

    It could map τ\tau to another closed orbit τ\tau^{\prime} of the same length (we say that τ\tau is ‘glued’ to τ\tau^{\prime}), that is ı(x)τ\imath(x)\in\tau^{\prime} for all xτx\in\tau; we write 𝒪Tg(n)\mathscr{O}_{T}^{g}(n) for the set of closed orbits of length nn under TT that are glued together in pairs. The factor map π\pi induces a 22-to-11 map 𝒪Tg(n)𝒪T¯(n)\mathscr{O}_{T}^{g}(n)\to\mathscr{O}_{\overline{T}}(n).

  3. 3.

    It could preserve the orbit τ\tau but not fix it pointwise; in that case, we must have n=2kn=2k even and ı(x)=Tk(x)\imath(x)=T^{k}(x), for all xτx\in\tau, so that the orbit π(τ)\pi(\tau) of T¯\overline{T} has length kk (we say that τ\tau is ‘halved’). we write 𝒪Th(n)\mathscr{O}_{T}^{h}(n) for the set of closed orbits of length nn under TT that are halved in length, and π\pi induces an injective map 𝒪Th(2k)𝒪T¯(k)\mathscr{O}_{T}^{h}(2k)\to\mathscr{O}_{\overline{T}}(k).

Clearly

𝒪T(n)=𝒪Th(n)𝒪Tg(n)𝒪Ts(n)\mathscr{O}_{T}(n)=\mathscr{O}_{T}^{h}(n)\sqcup\mathscr{O}_{T}^{g}(n)\sqcup\mathscr{O}_{T}^{s}(n)

is a disjoint union, and so

𝖮T(n)=𝖮Th(n)+𝖮Tg(n)+𝖮Ts(n),\mathsf{O}_{T}(n)=\mathsf{O}_{T}^{h}(n)+\mathsf{O}_{T}^{g}(n)+\mathsf{O}_{T}^{s}(n), (5)

for any n1n\geqslant 1. Similarly, from the surjective map (4) and the three possible behaviours, we get

𝒪T¯(n)=π(𝒪Th(2n))π(𝒪Tg(n))π(𝒪Ts(n)),\mathscr{O}_{\overline{T}}(n)=\pi\left(\mathscr{O}_{T}^{h}(2n)\right)\sqcup\pi\left(\mathscr{O}_{T}^{g}(n)\right)\sqcup\pi\left(\mathscr{O}_{T}^{s}(n)\right),

and it follows that

𝖮T¯(n)=𝖮Th(2n)+12𝖮Tg(n)+𝖮Ts(n),\mathsf{O}_{\overline{T}}(n)=\mathsf{O}_{T}^{h}(2n)+\textstyle\frac{1}{2}\mathsf{O}_{T}^{g}(n)+\mathsf{O}_{T}^{s}(n), (6)

for any n1n\geqslant 1. Since these numbers are finite and

𝖮T¯(1)𝖮T(1)12𝖮Tg(1)12𝖮T(1)>0,\mathsf{O}_{\overline{T}}(1)\geqslant\mathsf{O}_{T}(1)-\textstyle\frac{1}{2}\mathsf{O}_{T}^{g}(1)\geqslant\textstyle\frac{1}{2}\mathsf{O}_{T}(1)>0,

we see that (X¯,T¯)(\overline{X},\overline{T}) is also a topological dynamical system.

The way in which the set of orbits of length nn under TT decomposes into those that halve in length, those that glue together, and those that survive, is not arbitrary. Whatever constraints on ı\imath arise from having to be a continuous involution on XX that commutes with TT, there are some purely combinatorial constraints as follows:

𝖮Th(n)=0 if n is odd;\mathsf{O}_{T}^{h}(n)=0\mbox{ if~$n$ is odd}; (7)
𝖮Tg(n) is even for all n1.\mathsf{O}_{T}^{g}(n)\mbox{ is even for all~$n\geqslant 1$}. (8)

These observations already constrain the effect that halving can have on the growth rate of closed orbits.

Lemma 5.

Suppose (X¯,T¯)(\overline{X},\overline{T}) is obtained from (X,T)(X,T) by halving. Then, for any n1n\geqslant 1,

  1. (a)

    12𝖥T(n)𝖥T¯(n)12(𝖥T(n)+𝖥T(2n))\frac{1}{2}\mathsf{F}_{T}(n)\leqslant\mathsf{F}_{\overline{T}}(n)\leqslant\frac{1}{2}\left(\mathsf{F}_{T}(n)+\mathsf{F}_{T}(2n)\right);

  2. (b)

    𝖮T¯(n)𝖮T(n)+𝖮T(2n)\mathsf{O}_{\overline{T}}(n)\leqslant\mathsf{O}_{T}(n)+\mathsf{O}_{T}(2n), and if nn is odd then 𝖮T¯(n)12𝖮T(n)\mathsf{O}_{\overline{T}}(n)\geqslant\frac{1}{2}\mathsf{O}_{T}(n).

Proof.

(a) The lower bound for 𝖥T¯(n)\mathsf{F}_{\overline{T}}(n) comes from the fact that the fibres of the factor map π\pi have cardinality at most 22; since π\pi maps T(n)\mathscr{F}_{T}(n) to T¯(n)\mathscr{F}_{\overline{T}}(n), we deduce that 𝖥T¯(n)12𝖥T(n)\mathsf{F}_{\overline{T}}(n)\geqslant\frac{1}{2}\mathsf{F}_{T}(n). The bound is achieved if all orbits of length dividing nn glue together in pairs.

The upper bound comes from the containment π1(T¯(n))T(2n)\pi^{-1}(\mathscr{F}_{\overline{T}}(n))\subseteq\mathscr{F}_{T}(2n). If xT(2n)T(n)x\in\mathscr{F}_{T}(2n)\setminus\mathscr{F}_{T}(n), then π(x)T¯(n)\pi(x)\in\mathscr{F}_{\overline{T}}(n) if and only if xx lies in an orbit which halves in length, in which case xx lies in a fibre of cardinality 22. Thus

𝖥T¯(n)\displaystyle\mathsf{F}_{\overline{T}}(n) =\displaystyle= |T¯(n)π(T(2n)T(n))|+|T¯(n)π(T(n))|\displaystyle\left|\mathscr{F}_{\overline{T}}(n)\cap\pi(\mathscr{F}_{T}(2n)\setminus\mathscr{F}_{T}(n))\right|+\left|\mathscr{F}_{\overline{T}}(n)\cap\pi(\mathscr{F}_{T}(n))\right|
\displaystyle\leqslant 12(𝖥T(2n)𝖥T(n))+𝖥T(n)\displaystyle\tfrac{1}{2}\left(\mathsf{F}_{T}(2n)-\mathsf{F}_{T}(n)\right)+\mathsf{F}_{T}(n)
=\displaystyle= 12(𝖥T(2n)+𝖥T(n)).\displaystyle\tfrac{1}{2}\left(\mathsf{F}_{T}(2n)+\mathsf{F}_{T}(n)\right).

The upper bound is achieved if all orbits of order dividing nn survive, while all other orbits of order dividing 2n2n halve.

(b) For the upper bound for 𝖮T¯(n)\mathsf{O}_{\overline{T}}(n), we have

𝖮T¯(n)=𝖮Th(2n)+12𝖮Tg(n)+𝖮Ts(n)𝖮T(2n)+𝖮T(n),\mathsf{O}_{\overline{T}}(n)=\mathsf{O}^{h}_{T}(2n)+\tfrac{1}{2}\mathsf{O}^{g}_{T}(n)+\mathsf{O}^{s}_{T}(n)\leqslant\mathsf{O}_{T}(2n)+\mathsf{O}_{T}(n),

by (5) and (6). Again this is achieved when all orbits of order nn survive, while all orbits of order 2n2n halve.

For the lower bound, when nn is odd we have 𝖮Th(n)=0\mathsf{O}_{T}^{h}(n)=0 so that

𝖮T(n)=𝖮Ts(n)+𝖮Tg(n)\mathsf{O}_{T}(n)=\mathsf{O}_{T}^{s}(n)+\mathsf{O}_{T}^{g}(n)

and, from (6),

𝖮T¯(n)12𝖮Tg(n)+𝖮Ts(n)12𝖮T(n).\mathsf{O}_{\overline{T}}(n)\geqslant\tfrac{1}{2}\mathsf{O}^{g}_{T}(n)+\mathsf{O}^{s}_{T}(n)\geqslant\tfrac{1}{2}\mathsf{O}_{T}(n).

This bound is achieved if all orbits of length nn are glued in pairs. ∎

Remark 6.

Note that Lemma 5 does not give a lower bound for 𝖮T¯(n)\mathsf{O}_{\overline{T}}(n) when nn is even. Specifically, when nn is even, all orbits of length nn might halve in length while orbits of length 2n2n retain their length. Thus the only possible lower bound is the trivial one 𝖮T¯(n)0\mathsf{O}_{\overline{T}}(n)\geqslant 0.

In the case that 𝖮T(n)\mathsf{O}_{T}(n) grow exponentially, Lemma 5 immediately gives bounds on the logarithmic growth rate of 𝖮T¯(n)\mathsf{O}_{\overline{T}}(n).

Corollary 7.

Let (X,T)(X,T) be a topological dynamical system, let ı\imath be a continuous involution on XX commuting with TT, and let (X¯,T¯)(\overline{X},\overline{T}) be the halved system. Suppose there is a real number λ>0\lambda>0 such that lim supn1nlog𝖥T(n)=λ\limsup\limits_{n\to\infty}\frac{1}{n}\log\mathsf{F}_{T}(n)=\lambda. Then

λlim supn1nlog𝖥T¯(n)2λ.\lambda\leqslant\limsup_{n\to\infty}\tfrac{1}{n}\log\mathsf{F}_{\overline{T}}(n)\leqslant 2\lambda.

We will see later (see Corollary 10) that every growth rate in the closed interval [λ,2λ][\lambda,2\lambda] is obtainable.

5 The basic lemma

Our main observation is that, if we are free to choose the topological dynamical systems, then (7) and (8) are the only constraints on the behaviour of closed orbits under halving. This is a simple extension of an elementary remark in [7]: for any sequence (an)(a_{n}) of non-negative integers, there is a topological dynamical system (X,T)(X,T) with 𝖮T(n)=an\mathsf{O}_{T}(n)=a_{n} for all n1n\geqslant 1.

Lemma 8.

Let (bns)(b_{n}^{s})(bng)(b_{n}^{g}), and (bnh)(b_{n}^{h}) be three sequences of non-negative integers, with b1s>0b_{1}^{s}>0. Define sequences (anh)(a_{n}^{h})(ang)(a_{n}^{g}), and (ans)(a_{n}^{s}) by

ans=bns,ang=2bng,andanh={bn/2hif n is even,0otherwise.a_{n}^{s}=b_{n}^{s},\quad a_{n}^{g}=2b_{n}^{g},\quad\mbox{and}\quad a_{n}^{h}=\begin{cases}b_{n/2}^{h}&\mbox{if~$n$ is even},\\ 0&\mbox{otherwise.}\end{cases}

Define an=ans+ang+anha_{n}=a_{n}^{s}+a_{n}^{g}+a_{n}^{h} and bn=bns+bng+bnhb_{n}=b_{n}^{s}+b_{n}^{g}+b_{n}^{h} for all n1n\geqslant 1. Then there are a topological dynamical system (X,T)(X,T) and a continuous involution ı:XX\imath:X\to X that commutes with TT, such that 𝖮T(n)=an\mathsf{O}_{T}(n)=a_{n} and 𝖮T¯(n)=bn\mathsf{O}_{\overline{T}}(n)=b_{n}, for all n1n\geqslant 1.

Notice in particular that taking bns=anb_{n}^{s}=a_{n} and bng=bnh=0b_{n}^{g}=b_{n}^{h}=0, for all n1n\geqslant 1 recovers the basic lemma of [7].

Before giving an algebraic proof, we give a more geometric sketch to give an idea of what is happening. We can construct XX as a closed (and hence compact) subset of the triangle {(x,y)0x1,0yx}2\{(x,y)\mid 0\leqslant x\leqslant 1,0\leqslant y\leqslant x\}\subset\mathbb{R}^{2}, with the metric inherited from 2\mathbb{R}^{2}. Above each point (1n,0)(\frac{1}{n},0) for n2n\geqslant 2 draw ana_{n} disjoint regular nn-gons in such a way that all of them are disjoint as subsets of the plane. By hypothesis a11a_{1}\geqslant 1, and we draw a11a_{1}-1 points (11-gons) above (1,0)(1,0). Finally locate a single 11-gon at (0,0)(0,0) (which may be thought of as a point ‘at infinity’). The space XX is now defined to be the union of all the vertices of the polygons. (See Figure 3.) It is closed because all but one point is isolated, and the accumulation point (0,0)(0,0) is, by construction, a member of XX. Number the vertices of each nn-gon with the numbers 11 to nn consecutively clockwise, so that we may speak of the ‘same’ point on two disjoint nn-gons as being the point with the same symbol. The homeomorphism TT is defined to be the map that takes each point on any nn-gon to the next point in a clockwise orientation around the same nn-gon (equivalently, adding one using the labels; the action of this map is illustrated by the lines joining vertices of the polygons in Figure 3). This defines a homeomorphism since all but one point is isolated in XX, and all the points close to the fixed point (0,0)(0,0) are moved by a very small distance. Then (X,T)(X,T) is a topological dynamical system, and by construction 𝖮T(n)=an\mathsf{O}_{T}(n)=a_{n} for all n1n\geqslant 1.

\psfrag{1}{$1$}\psfrag{2}{$\textstyle\frac{1}{2}$}\psfrag{3}{$\textstyle\frac{1}{3}$}\psfrag{4}{$\textstyle\frac{1}{4}$}\psfrag{I}{$\infty$}\includegraphics{triangle.eps}
Figure 3: Building the system (X,T)(X,T).

Now we define an action of C2C_{2} on XX using the numbers ans,anga_{n}^{s},a_{n}^{g}, and anha_{n}^{h} as follows.

  • For each n1n\geqslant 1 pick 12ang\frac{1}{2}a_{n}^{g} pairs of nn-gons above (1n,0)(\frac{1}{n},0) and define the action of ı\imath to send a point on any one of them to the same point on the paired nn-gon (these are the glued orbits).

  • For each n1n\geqslant 1 pick anha_{n}^{h} of the nn-gons above (1n,0)(\frac{1}{n},0), chosen from those that have not been chosen already for gluing, and on each polygon (which will by hypothesis have even length) define the action of ı\imath to be rotation by π\pi (these are the halved orbits).

  • On all the remaining points that are vertices of polygons that are neither glued nor halved, define ı\imath to be the identity map (these are the surviving orbits).

It is clear that ı\imath is continuous (since all points close to the fixed point are moved by a very small distance) and commutes with TT and so defines a halved system (X¯,T¯)(\overline{X},\overline{T}) which, by construction, has the required numbers of orbits.

Note that, in the proof below, we do not refer to the triangle in the plane and the metric on XX we give is not the same one as in this sketch, but it does give the same topology.

Proof of Lemma 8.

We write C2={e,ı}C_{2}=\{e,\imath\}, where ee is the identity element, for the cyclic group of order 22. We begin by describing XX as a set, before compactifying. It will take the form

X=n1Xn,X=\bigsqcup\limits_{n\geqslant 1}X_{n},

where each XnX_{n} will be the union of closed orbits of length nn. We set

Xn=XnsXngXnh,where{Xns={1,2,,ans}×/n,Xng={1,2,,bng}×C2×/n,Xnh={1,2,,anh}×/n.X_{n}=X_{n}^{s}\sqcup X_{n}^{g}\sqcup X_{n}^{h},\quad\mbox{where}\quad\begin{cases}X_{n}^{s}=\{1,2,\ldots,a_{n}^{s}\}\times\mathbb{Z}/n\mathbb{Z},\\ X_{n}^{g}=\{1,2,\ldots,b^{g}_{n}\}\times C_{2}\times\mathbb{Z}/n\mathbb{Z},\\ X_{n}^{h}=\{1,2,\ldots,a_{n}^{h}\}\times\mathbb{Z}/n\mathbb{Z}.\end{cases}

We define T:XXT:X\to X and ı:XX\imath:X\to X by describing their restrictions to each of the sets Xns,Xng,XnhX_{n}^{s},X_{n}^{g},X_{n}^{h}, which will be preserved:

  • for x=(i,k)Xnsx=(i,k)\in X_{n}^{s}, put T(x)=(i,k+1(modn))T(x)=(i,k+1\pmod{n}) and ı(x)=x\imath(x)=x;

  • for x=(i,γ,k)Xngx=(i,\gamma,k)\in X_{n}^{g}, put T(x)=(i,γ,k+1(modn))T(x)=(i,\gamma,k+1\pmod{n}) and ı(x)=(i,ıγ,k)\imath(x)=(i,\imath\gamma,k);

  • for x=(i,k)Xnhx=(i,k)\in X_{n}^{h}, so that nn is even, we put T(x)=(i,k+1(modn))T(x)=(i,k+1\pmod{n}) and ı(x)=(i,k+n2(modn))\imath(x)=(i,k+\frac{n}{2}\pmod{n}).

Then ı\imath commutes with TT and, by construction, the map TT and the induced map T¯\overline{T} on the quotient set X¯\overline{X} have the required numbers of orbits.

It remains to show that XX can be given the structure of a compact metric space, with respect to which TT and ı\imath are homeomorphisms. To do this, we pick a point in X1sX_{1}^{s} (which is non-empty by hypothesis) and call it \infty and define a metric as follows: if xXmx\in X_{m} and yXny\in X_{n}, with xy,x\not\in{y,\infty}, then

d(x,y)=d(y,x)={1mif y=,1min{m,n}otherwise,d(x,y)=d(y,x)=\begin{cases}\frac{1}{m}&\mbox{if }y=\infty,\\ \frac{1}{\min\{m,n\}}&\mbox{otherwise},\end{cases}

and d(x,x)=0d(x,x)=0. It is straightforward to check that this does indeed define a metric and, given any open set UU containing \infty, there exists N1N\geqslant 1 such that UU contains nNXn\bigsqcup_{n\geqslant N}X_{n} so that XUX\setminus U is finite; hence XX is compact. Moreover, since TT and ı\imath preserve the sets XnX_{n} and the point \infty, they are isometries, so homeomorphisms, and we are done. ∎

6 Growth in closed orbits

Lemma 8 shows that any pair of sequence (an),(bn)(a_{n}),(b_{n}) for which the combinatorial constraints (5)–(8) are satisfied does in fact arise as the orbit count of a pair of systems related by halving. However, it is not so easy to give conditions directly on the sequences (an),(bn)(a_{n}),(b_{n}) which guarantee that the combinatorial constraints are indeed satisfied. The following result gives some sufficient conditions.

Proposition 9.

Let (an)(a_{n}) be a sequence of non-negative integers with a11a_{1}\geqslant 1 such that there is an integer N1N\geqslant 1 with a2n12ana_{2n}\geqslant\frac{1}{2}a_{n}, for all nNn\geqslant N. Let (bn)(b_{n}) be any sequence of integers such that b1>12a1b_{1}>\frac{1}{2}a_{1} and

{12anbnanfor n<N,12anbna2nfor nN.\begin{cases}\tfrac{1}{2}a_{n}\leqslant b_{n}\leqslant a_{n}&\hbox{for }n<N,\\ \tfrac{1}{2}a_{n}\leqslant b_{n}\leqslant a_{2n}&\hbox{for }n\geqslant N.\end{cases}

Then there are a topological dynamical system (X,T)(X,T) and a continuous involution ı:XX\imath:X\to X that commutes with TT, such that 𝖮T(n)=an\mathsf{O}_{T}(n)=a_{n} and 𝖮T¯(n)=bn\mathsf{O}_{\overline{T}}(n)=b_{n}, for all n1n\geqslant 1.

The conditions on the pair of sequences (an),(bn)(a_{n}),(b_{n}) here are not necessary for the existence of a suitable halving system, but they are sufficient for our interests and are not so far from being necessary: for n>Nn>N odd, the condition

12anbna2n\tfrac{1}{2}a_{n}\leqslant b_{n}\leqslant a_{2n}

is necessary by Lemma 5.

Proof.

In order to use Lemma 8, we recursively define non-negative integers bns,bngb_{n}^{s},b_{n}^{g} and bnhb_{n}^{h} such that

bn=bns+bng+bnh,an=bns+2bng+bn/2h,andbnha2n,b_{n}=b_{n}^{s}+b_{n}^{g}+b_{n}^{h},\quad a_{n}=b_{n}^{s}+2b_{n}^{g}+b_{n/2}^{h},\quad\mbox{and}\quad b_{n}^{h}\leqslant a_{2n},

where we understand bn/2h=0b_{n/2}^{h}=0 when n/2n/2\not\in\mathbb{Z}. So suppose k1k\geqslant 1 and we have defined these for n<kn<k. Then there are two cases.

If bkakbk/2hb_{k}\leqslant a_{k}-b_{k/2}^{h}, which we note is always the case for k<Nk<N, then we put

bkg=akbkbk/2h,bks=bkbkg,bkh=0.b_{k}^{g}=a_{k}-b_{k}-b_{k/2}^{h},\quad b_{k}^{s}=b_{k}-b_{k}^{g},\quad b_{k}^{h}=0.

On the other hand, if bk>akbk/2hb_{k}>a_{k}-b_{k/2}^{h}, then we put

bkg=0,bks=akbk/2h,bkh=bkbks.b_{k}^{g}=0,\quad b_{k}^{s}=a_{k}-b_{k/2}^{h},\quad b_{k}^{h}=b_{k}-b_{k}^{s}.

These are non-negative and, in the latter case, we have bkhbka2kb_{k}^{h}\leqslant b_{k}\leqslant a_{2k}, since kNk\geqslant N. Note also that, in either case, b1s>0b_{1}^{s}>0. Now Lemma 8 implies the result. ∎

As a consequence, we get the following result when we consider exponential orbit growth rates.

Corollary 10.

Let λ,η,c\lambda,\eta,c be positive real numbers with λ>1\lambda>1 and

{η=λ and c12, orη(λ,λ2), orη=λ2 and 0<c1.\begin{cases}\mbox{$\eta=\lambda$ and~$c\geqslant\frac{1}{2}$, or}\\ \mbox{$\eta\in(\lambda,\lambda^{2})$, or}\\ \mbox{$\eta=\lambda^{2}$ and~$0<c\leqslant 1$.}\end{cases}

Then there exist a topological dynamical system (X,T)(X,T) and an involution ı\imath on XX commuting with TT, such that

𝖮T(n)λnand𝖮T¯(n)cηnas n.\mathsf{O}_{T}(n)\sim\lambda^{n}\quad\mbox{and}\quad\mathsf{O}_{\overline{T}}(n)\sim c\eta^{n}\quad\mbox{as $n\to\infty$}.
Proof.

Let N>1N>1 be any integer such that cηN<λ2Nc\eta^{N}<\lambda^{2N} and define sequences by

an=λn,bn={anif n<N,cηnif nN,a_{n}=\lceil\lambda^{n}\rceil,\quad b_{n}=\begin{cases}a_{n}&\mbox{if }n<N,\\ \lceil c\eta^{n}\rceil&\mbox{if }n\geqslant N,\end{cases}

for n1n\geqslant 1. This gives a pair of sequences satisfying the hypotheses of Proposition 9, from which the result follows. ∎

7 Dynamical zeta functions

Bowen and Lanford [1, Th. 2] showed that there are only countably many rational dynamical zeta functions, so Corollary 10 shows in particular that halving and doubling cannot preserve the property of having a rational zeta function. We now discuss some examples that give concrete instances of this phenomenon. The arguments all rely on the following facts: a power series with positive radius of convergence represents a rational function if and only if the coefficients satisfy a linear recurrence (see [4, Sec.1.1]). Moreover, the Skolem–Mahler–Lech Theorem says that, in any linear recurrence sequence (an)(a_{n}), the set of zeros, comprising those values of nn\in\mathbb{N} for which an=0a_{n}=0, is the union of a finite set of arithmetic progressions and a finite set (see [4, Ch. 1] for further details and a proof).

Example 11.

We revisit Example 4, so that (Y,S)(Y,S) is a topological dynamical system and X=Y×{0,1}X=Y\times\{0,1\} with the map T(y,e)=(S(y),e+1(mod2))T(y,e)=(S(y),e+1\pmod{2}). The involution ı:(y,e)=(y,e+1(mod2))\imath:(y,e)=(y,e+1\pmod{2}) commutes with TT, giving the halved system (X¯,T¯)=(Y,S)(\overline{X},\overline{T})=(Y,S). There is sufficient freedom in the choice of orbits of odd length under SS to allow us to find examples with ζT¯=ζS\zeta_{\overline{T}}=\zeta_{S} irrational but ζT\zeta_{T} rational. In particular, we may take

𝖥S(n)={2n+1if n is even;d|nd2(d1)/2if n is odd.\mathsf{F}_{S}(n)=\begin{cases}2^{n}+1&\mbox{if $n$ is even;}\\ \sum\limits_{d|n}d2^{(d-1)/2}&\mbox{if $n$ is odd.}\end{cases}

It is a pleasant exercise to verify that these really do arise from a topological dynamical system (that is, 1n𝖮S(n)\frac{1}{n}\mathsf{O}_{S}(n) is a non-negative integer for each nn). Then by (2)

ζT(z)=1(1z2)(14z2)\zeta_{T}(z)=\frac{1}{(1-z^{2})(1-4z^{2})}

is rational. On the other hand

zζS(z)ζS(z)=n=1𝖥S(n)zn=z1z+4z214z2+6z34z5(12z2)2+φ(z),z\frac{\zeta_{S}^{\prime}(z)}{\zeta_{S}(z)}=\sum_{n=1}^{\infty}\mathsf{F}_{S}(n)z^{n}=\frac{z}{1-z}+\frac{4z^{2}}{1-4z^{2}}+\frac{6z^{3}-4z^{5}}{(1-2z^{2})^{2}}+\varphi(z),

where

φ(z)=n=1z2n+1d|2n+1d1,2n+1d2(d1)/2.\varphi(z)=\sum_{n=1}^{\infty}z^{2n+1}\sum_{\begin{subarray}{c}d|2n+1\\ d\neq 1,2n+1\end{subarray}}d2^{(d-1)/2}.

We claim that φ\varphi is irrational, so that ζS\zeta_{S} is irrational. To see this, note that the coefficient of znz^{n} in φ(z)\varphi(z) vanishes precisely when nn is even or nn is an odd prime; since the set of primes is infinite, while any arithmetic progression contains composites, the Skolem–Mahler–Lech Theorem implies the sequence of coefficients cannot be a linear recurrence sequence and hence φ\varphi cannot be a rational function.

Our next examples use the sum of divisors function σ(n)=d|nd\sigma(n)=\sum_{d|n}d. There are sophisticated bounds for size of σ(n)\sigma(n), but for our purposes it is sufficient to note the trivial bounds

nσ(n)n2.n\leqslant\sigma(n)\leqslant n^{2}.

It follows that the complex power series

θ(z)=expn1σ(n)znn\theta(z)=\exp\sum_{n\geqslant 1}\sigma(n)\frac{z^{n}}{n}

has radius of convergence 11. It is known that

1θ(z)=1zz2+z5+z7,\frac{1}{\theta(z)}=1-z-z^{2}+z^{5}+z^{7}-\cdots,

where the powers of zz are those of the form (3k2±k)/2(3k^{2}\pm k)/2 (see, for example, Pólya and Szegö [6, Sec. VIII, Ex. 75]). This means that 1θ(z)\frac{1}{\theta(z)} is a power series with arbitrarily long consecutive sequences of zero coefficients. Thus, by the Skolem–Mahler–Lech Theorem, the coefficients of 1θ(z)\frac{1}{\theta(z)} are not a linear recurrence sequence and we deduce that θ(z)\theta(z) is not a rational function of zz.

Example 12.

In order to use the irrationality of θ(z)\theta(z), we define bng=1b_{n}^{g}=1 and bnh=0b_{n}^{h}=0, for all n1n\geqslant 1, and we choose bnsb_{n}^{s} later. Now we define bn,anb_{n},a_{n} as in Lemma 8 and denote by (X,T)(X,T)(X¯,T¯)(\overline{X},\overline{T}) the pair of systems given there. Thus TT has one extra orbit in each length, compared to TT, and the action of ı\imath on XX has the effect of gluing together exactly one pair of orbits of each length.

Now we first take bns=1nd|nμ(nd)2d1b_{n}^{s}=\frac{1}{n}\sum_{d|n}\mu\left(\frac{n}{d}\right)2^{d}-1, for n1n\geqslant 1, so that bnb_{n} is the number of orbits of length nn in any system with 2n2^{n} points of period nn. Then

ζT¯(z)=112z,\zeta_{\overline{T}}(z)=\frac{1}{1-2z},

while

𝖥T(n)=d|ndad=2n+σ(n),\mathsf{F}_{T}(n)=\sum_{d|n}da_{d}=2^{n}+\sigma(n),

so that

ζT(z)=112zθ(z).\zeta_{T}(z)=\frac{1}{1-2z}\theta(z).

By the remarks above, this is not a rational function.

In the reverse direction, we take b1s=1b_{1}^{s}=1 and bns=1nd|nμ(nd)2d2b_{n}^{s}=\frac{1}{n}\sum_{d|n}\mu\left(\frac{n}{d}\right)2^{d}-2, for n2n\geqslant 2, so that ana_{n} is the number of orbits of length nn in any system with 2n+12^{n}+1 points of period nn. Then

ζT(z)=1(1z)(12z),\zeta_{T}(z)=\frac{1}{(1-z)(1-2z)},

while

𝖥T¯(n)=d|ndbd=2n+1σ(n),\mathsf{F}_{\overline{T}}(n)=\sum_{d|n}db_{d}=2^{n}+1-\sigma(n),

so that

ζT¯(z)=1(1z)(12z)θ(z),\zeta_{\overline{T}}(z)=\frac{1}{(1-z)(1-2z)\theta(z)},

which is again irrational.

In fact, the zeta function in the previous example is worse than irrational. The function 1/θ(z)1/\theta(z) has integer coefficients and radius of convergence 11, but is not rational. Thus, by the Pólya–Carlson Theorem (see [2, 5]) it has the unit circle as natural boundary, and the function ζT¯\zeta_{\overline{T}} also has natural boundary here. Since the radius of convergence of ζT¯\zeta_{\overline{T}} is only 12\frac{1}{2} this is perhaps not so interesting, but our final example shows that it is possible for the circle of convergence and the natural boundary of ζT¯\zeta_{\overline{T}} to coincide, even when ζT\zeta_{T} is rational.

Example 13.

We begin by recursively defining an auxiliary sequence (cn)(c_{n}) of non-negative integers by the following conditions:

  • if n=1n=1 or nn is prime, then cn=0c_{n}=0;

  • if nn is composite then, for any prime pp dividing nn,

    cncn/p(modpordp(n)),c_{n}\equiv c_{n/p}\pmod{p^{\operatorname{ord}_{p}(n)}},

    and ncn<2nn\leqslant c_{n}<2n.

Note that, by the Chinese Remainder Theorem, these conditions determine (cn)(c_{n}) uniquely. Now set

an=1nd|nμ(nd)2d,bn=an+1nd|nμ(nd)cd2d.a_{n}=\tfrac{1}{n}\sum_{d|n}\mu\left(\tfrac{n}{d}\right)2^{d},\quad b_{n}=a_{n}+\tfrac{1}{n}\sum_{d|n}\mu\left(\tfrac{n}{d}\right)c_{d}2^{d}.

We assume for now that the sequences (an),(bn)(a_{n}),(b_{n}) are non-negative integers satisfying the conditions of Proposition 9 and denote by (X,T),(X¯,T¯)(X,T),(\overline{X},\overline{T}) the systems given there with these numbers of orbits. Then

ζT(z)=112z,\zeta_{T}(z)=\frac{1}{1-2z},

while

zζT¯(z)ζT¯(z)=2z12z+n=1cn2nzn.z\frac{\zeta_{\overline{T}}^{\prime}(z)}{\zeta_{\overline{T}}(z)}=\frac{2z}{1-2z}+\sum_{n=1}^{\infty}c_{n}2^{n}z^{n}.

Now the integer sequence (cn)(c_{n}) is not a linear recurrence sequence, since it is zero for all primes and non-zero for all composites, while the bound cn<2nc_{n}<2n implies that the power series

n1cnzn\sum_{n\geqslant 1}c_{n}z^{n}

has radius of convergence 11. Hence it has a natural boundary on the unit circle, and we deduce that zζT¯(z)ζT¯(z)z\frac{\zeta_{\overline{T}}^{\prime}(z)}{\zeta_{\overline{T}}(z)} has a natural boundary on the circle |z|=12|z|=\frac{1}{2}. Thus ζT¯\zeta_{\overline{T}} also has a natural boundary here, since it has radius of convergence 12\frac{1}{2}.

It remains to show that (an),(bn)(a_{n}),(b_{n}) are non-negative integers and satisfy the conditions of Proposition 9. First, note that ana_{n} is the number of closed orbits of period nn of the tent map so that (an)(a_{n}) is a sequence of non-negative integers. Now we prove that the sequence (bnan)(b_{n}-a_{n}) is also a sequence of non-negative integers. First we must check that

d|nμ(nd)cd2d\sum_{d|n}\mu\left(\tfrac{n}{d}\right)c_{d}2^{d}

is a non-negative integer divisible by nn, for all n1n\geqslant 1. To show that it is divisible by nn, we show that it is divisible by pordp(n)p^{\operatorname{ord}_{p}(n)}, for each prime pp dividing nn. For this, we use the following version of Euler’s generalization of Fermat’s Little Theorem: for any prime pp and integers r,mr,m, with p|mp|m, we have

rmrm/p(modpordp(m)).r^{m}\equiv r^{m/p}\pmod{p^{\operatorname{ord}_{p}(m)}}.

Thus, for any prime pp dividing nn,

d|nμ(nd)cd2d=d|npndμ(nd)(cd2dcd/p2d/p)0(modpordp(n)),\sum_{d|n}\mu\left(\tfrac{n}{d}\right)c_{d}2^{d}=\sum\limits_{\begin{subarray}{c}d|n\\ p\nmid\frac{n}{d}\end{subarray}}\mu\left(\tfrac{n}{d}\right)\left(c_{d}2^{d}-c_{d/p}2^{d/p}\right)\equiv 0\pmod{p^{\operatorname{ord}_{p}(n)}},

since, for any dd dividing nn with pndp\nmid\frac{n}{d}, we have ordp(d)=ordp(n)\operatorname{ord}_{p}(d)=\operatorname{ord}_{p}(n) and cdcd/p(modpordp(d))c_{d}\equiv c_{d/p}\pmod{p^{\operatorname{ord}_{p}(d)}} by construction.

For non-negativity, when nn is 11 or prime we have bnan=0b_{n}-a_{n}=0. On the other hand, for nn composite, the bounds cnnc_{n}\geqslant n and cd<2dnc_{d}<2d\leqslant n, for dd a divisor of nn, imply that

d|nμ(nd)cd2d>n2ndn2n2d>n(2n2n2+1)>0.\sum_{d|n}\mu\left(\tfrac{n}{d}\right)c_{d}2^{d}>n2^{n}-\sum_{d\leqslant\frac{n}{2}}n2^{d}>n(2^{n}-2^{\frac{n}{2}+1})>0.

Finally, for nn composite, the same bounds show that

bnan=1nd|nμ(nd)cd2d<(2n+2n2+1)<2n+1,b_{n}-a_{n}=\frac{1}{n}\sum_{d|n}\mu\left(\tfrac{n}{d}\right)c_{d}2^{d}<(2^{n}+2^{\frac{n}{2}+1})<2^{n+1},

and, similarly,

1n(2n2n2+1)<an<1n(2n+2n2+1)<1n2n+1.\frac{1}{n}(2^{n}-2^{\frac{n}{2}+1})<a_{n}<\frac{1}{n}(2^{n}+2^{\frac{n}{2}+1})<\frac{1}{n}2^{n+1}.

Thus, for n6n\geqslant 6, using that 22n>(2n+3)2n+12^{2n}>(2n+3)2^{n+1}, we have

a2n>12n(22n2n+1)>(n+1n)2n+1>(bnan)+an=bn.a_{2n}>\frac{1}{2n}(2^{2n}-2^{n+1})>\left(\frac{n+1}{n}\right)2^{n+1}>(b_{n}-a_{n})+a_{n}=b_{n}.

Finally, one checks that b4=19<30=a8b_{4}=19<30=a_{8} and, since bn=an<a2nb_{n}=a_{n}<a_{2n} for nn prime, the conditions of Proposition 9 are satisfied with N=2N=2.

8 Concluding remarks and questions

  1. 1.

    The simple observation in [7] that for any sequence (an)(a_{n}) of non-negative integers there is a topological dynamical system (X,T)(X,T) with 𝖮T(n)=an\mathsf{O}_{T}(n)=a_{n} for all n1n\geqslant 1 was extended by Windsor [8], who showed that the map may be required to be an infinitely differentiable map on the 22-torus. Does Lemma 8 also have a smooth version, in which both maps and the involution are differentiable maps on a manifold?

  2. 2.

    We have only considered quotients by an action of the group C2C_{2}. The same process makes sense if (X,T)(X,T) supports an action of some finite group GG commuting with TT, and similar questions arise. In this setting the structure of the group plays a larger role, and other complications arise; this is explored in [9], where generalizations of Corollaries 7 and 10 are obtained. A particularly interesting sample problem is to understand a version of the relation (3) for other groups of symmetries.

  3. 3.

    Achieving radius of convergence strictly smaller than 11 in Example 13 is important because with radius of convergence 11 the rational part of the Pólya–Carlson dichotomy is not particularly interesting: a rational Taylor series with integer coefficients and radius of convergence 11 has the form p(z)(1za)b\frac{p(z)}{(1-z^{a})^{b}} for some polynomial pp with integer coefficients and integers a,b0a,b\geqslant 0. In our settings, this would correspond to dynamical systems in which the number of closed orbits of length nn is constant for large nn.

References

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