The Nielsen numbers of iterations of maps on infra-solvmanifolds of type $\mathrm {(R)}$ and periodic orbits

Abstract

We study the asymptotic behavior of the sequence of the Nielsen numbers $\{N(f^k)\}$, the essential periodic orbits of f and the homotopy minimal periods of f using the Nielsen theory of maps f on infra-solvmanifolds of type $\mathrm {(R)}$. We give a linear lower bound for the number of essential periodic orbits of such a map, which sharpens well-known results of Shub and Sullivan for periodic points and of Babenko and Bogatyǐ for periodic orbits. We also verify that a constant multiple of infinitely many prime numbers occur as homotopy minimal periods of such a map.

1 Introduction

Let $f:X\rightarrow X$ be a map on a finite complex X. A point $x\in X$ is a fixed point of f if $f(x)=x$; a periodic point of f with period n if $f^n(x)=x$. The smallest period of x is called the minimal period. We will use the following notations:

$$\begin{aligned}&\mathrm {Fix}(f)=\{x\in X\mid f(x)=x\},\\&\mathrm {Per}(f)=\text {the set of all minimal periods of }f,\\&P_n(f)=\text {the set of all periodic points of }f\text { with minimal period }n,\\&{\mathrm {HPer}(f)=\bigcap _{g\simeq f}\left\{ n\in \mathbb {N}\mid P_n(g)\ne \emptyset \right\} }\\&=\text {the set of all homotopy minimal periods of }f. \end{aligned}$$

Let $p:\tilde{X}\rightarrow X$ be the universal cover of X and $\tilde{f}:\tilde{X}\rightarrow \tilde{X}$ a lift of f, i.e., $p\circ \tilde{f}=f\circ p$. Two lifts $\tilde{f}$ and $\tilde{f}^\prime$ are called conjugate if there is a covering translation $\gamma : \tilde{X}\rightarrow \tilde{X}$ such that $\tilde{f}^\prime = \gamma \circ \tilde{f}\circ \gamma ^{-1}$. The subset $p(\mathrm {Fix}(\tilde{f}))\subset \mathrm {Fix}(f)$ is called the fixed point class of f determined by the lifting class $[\tilde{f}]$. A fixed point class is called essential if its index is nonzero. The number of essential fixed point classes is called the Nielsen number of f, denoted by N(f) [22].

The Nielsen number is always finite and is a homotopy invariant lower bound for the number of fixed points of f. In the category of compact, connected polyhedra the Nielsen number of a map is, apart from in certain exceptional cases, equal to the least number of fixed points of maps with the same homotopy type as f.

From the dynamical point of view, it is natural to consider the Nielsen numbers $N(f^k)$ of all iterations of f simultaneously. For example, N. Ivanov [17] introduced the notion of the asymptotic Nielsen number, measuring the growth of the sequence $N(f^k)$ and found the basic relation between the topological entropy of f and the asymptotic Nielsen number. Later on, it was suggested in [9,10,11, 36] to arrange the Nielsen numbers $N(f^k)$ of all iterations of f into the Nielsen zeta function

$$\begin{aligned} N_f(z)=\exp \left( \sum _{k=1}^\infty \frac{N(f^k)}{k}z^k\right) . \end{aligned}$$

The Nielsen zeta function $N_f(z)$ is a nonabelian analogue of the Lefschetz zeta function

$$\begin{aligned} L_f(z)=\exp \left( \sum _{k=1}^\infty \frac{L(f^k)}{k}z^k\right) , \end{aligned}$$

where

$$\begin{aligned} L(f^n) := \sum _{k=0}^{\dim X} (-1)^k\ \mathrm {tr\,}\!\left\{ f_{*k}^n:H_k(X;\mathbb {Q})\rightarrow H_k(X;\mathbb {Q})\right\} \end{aligned}$$

is the Lefschetz number of the iterate $f^n$ of f.

Nice analytical properties of $N_f(z)$ [11] indicate that the numbers $N(f^k)$ are closely interconnected. The manifestation of this is Gauss congruences

$$\begin{aligned} \sum _{d\mid k}\mu \!\left( \frac{k}{d}\right) N(f^d)\equiv 0\mod k, \end{aligned}$$

for any $k>0$, where f is a map on an infra-solvmanifold of type $\mathrm {(R)}$ [12].

The fundamental invariants of f used in the study of periodic points are the Lefschetz numbers $L(f^k)$, and their algebraic combinations, the Nielsen numbers $N(f^k)$ and the Nielsen–Jiang periodic numbers $N\!P_n(f)$ and $N\!\Phi _n(f)$.

The study of periodic points using the Lefschetz theory has been done extensively by many authors in the literatures such as [2, 8, 19, 22, 35]. A natural question is to know how much information we can get about the set of essential periodic points of f or about the set of (homotopy) minimal periods of f from the study of the sequence $\{N(f^k)\}$ of the Nielsen numbers of iterations of f. Even though the Lefschetz numbers $L(f^k)$ and the Nielsen numbers $N(f^k)$ are different from the nature and not equal for maps f on infra-solvmanifolds of type $\mathrm {(R)}$, we can utilize the arguments employed mainly in [2] and [19, Chap. III] for the Lefschetz numbers of iterations and thereby will study the asymptotic behavior of the sequence $\{N(f^k)\}$, the essential periodic orbits of f and the homotopy minimal periods of f by using the Nielsen theory of maps f on infra-solvmanifolds of type $\mathrm {(R)}$. We give a linear lower bound for the number of essential periodic orbits of such a map (Theorem 5.3), which sharpens well-known results of Shub and Sullivan for periodic points and of Babenko and Bogatyǐ for periodic orbits. We also verify that a constant multiple of infinitely many prime numbers occur as homotopy minimal periods of such a map (Theorem 6.1).

2 Nielsen numbers $N(f^k)$

Let S be a connected and simply connected solvable Lie group. A discrete subgroup $\Gamma$ of S is a lattice of S if $\Gamma \backslash {S}$ is compact and, in this case, we say that the quotient space $\Gamma \backslash {S}$ is a special solvmanifold. Let $\Pi \subset \mathrm {Aff}(S)$ be a torsion-free finite extension of the lattice $\Gamma =\Pi \cap S$ of S. That is, $\Pi$ fits the short exact sequence

Then, $\Pi$ acts freely on S and the manifold $\Pi \backslash {S}$ is called an infra-solvmanifold. The finite group $\Phi =\Pi /\Gamma$ is the holonomy group of $\Pi$ or $\Pi \backslash {S}$. It sits naturally in $\mathrm {Aut}(S)$. Thus, every infra-solvmanifold $\Pi \backslash {S}$ is finitely covered by the special solvmanifold $\Gamma \backslash {S}$. An infra-solvmanifold $M=\Pi \backslash {S}$ is of type $\mathrm {(R)}$ if S is of type $\mathrm {(R)}$ or completely solvable, i.e., if $\mathrm {ad}\, X:\mathfrak {S}\rightarrow \mathfrak {S}$ has only real eigenvalues for all X in the Lie algebra $\mathfrak {S}$ of S.

Recall that a connected solvable Lie group S contains a sequence of closed subgroups

$$\begin{aligned} 1=N_1\subset \cdots \subset N_k=S \end{aligned}$$

such that $N_i$ is normal in $N_{i+1}$ and $N_{i+1}/N_i\cong \mathbb {R}$ or $N_{i+1}/N_i\cong S^1$. If the groups $N_1,\cdots ,N_k$ are normal in S, the group S is called supersolvable. The supersolvable Lie groups are the Lie groups of type $\mathrm {(R)}$.

Lemma 2.1

([40, Lemma 4.1], [24, Proposition 2.1]). For a connected Lie group S, the following are equivalent:

• (1) S is supersolvable.

• (2) All elements of $\mathrm {Ad}(S)$ have only positive eigenvalues.

• (3) S is of type $\mathrm {(R)}$.

In this paper, we shall assume that $f:M\rightarrow M$ is a continuous map on an infra-solvmanifold $M=\Pi \backslash {S}$ of type $\mathrm {(R)}$ with holonomy group $\Phi$. Then, f has an affine homotopy lift $(d,D):S\rightarrow S$, and so $f^k$ has an affine homotopy lift $(d,D)^k=(d',D^k)$ where $d'=dD(d)\cdots D^{k-1}(d)$. By the averaging formulas for the Lefschetz and the Nielsen numbers $L(f^k)$ and $N(f^k)$ [29, Theorem 4.2] and [13], we have

$$\begin{aligned} L(f^k)&=\frac{1}{\#\Phi }\sum _{A\in \Phi }\det (I-A_*D_*^k),\\ N(f^k)&=\frac{1}{\#\Phi }\sum _{A\in \Phi }|\det (I-A_*D_*^k)|. \end{aligned}$$

(AV)

Concerning the Nielsen numbers $N(f^k)$ of all iterates of f, we recall the following results. These results about $N(f^k)$ are crucial in our discussion.

Theorem 2.2

([12, Theorem 11.4]). Let $f:M\rightarrow M$ be a map on an infra-solvmanifold M of type $\mathrm {(R)}$. Then

$$\begin{aligned} \sum _{d\mid k}\mu \!\left( \frac{k}{d}\right) N(f^d)\equiv 0\mod k \end{aligned}$$

(DN)

for all $k>0$.

Indeed, we have shown in [12, Theorem 11.4] that the left-hand side is non-negative because it is equal to the number of isolated periodic points of f with least period k. By [37, Lemma 2.1], the sequence $\{N_k(f)\}$ is exactly realizable.

Consider the sequences of algebraic multiplicities $\{A_k(f)\}$ and Dold multiplicities $\{I_k(f)\}$ associated with the sequence $\{N(f^k)\}$:

$$\begin{aligned} A_k(f)=\frac{1}{k}\sum _{d\mid k}\mu \!\left( \frac{k}{d}\right) N(f^d),\quad I_k(f)=\sum _{d\mid k}\mu \!\left( \frac{k}{d}\right) N(f^d). \end{aligned}$$

(2.1)

Then, $I_k(f)=kA_k(f)$ and all $A_k(f)$ are integers by Theorem 2.2. From the Möbius inversion formula, we immediately have

$$\begin{aligned} N(f^k)=\sum _{d\mid k}\ d ~A_d(f). \end{aligned}$$

(2.2)

Theorem 2.3

([7, Theorem 4.5]). Let $f:M\rightarrow M$ be a map on an infra-solvmanifold M of type $\mathrm {(R)}$. Then, the Nielsen zeta function of f

$$\begin{aligned} N_f(z)=\exp \left( \sum _{k=1}^\infty \frac{N(f^k)}{k}z^k\right) \end{aligned}$$

is a rational function.

In fact, it is well known that

$$\begin{aligned} L_f(z)=\prod _{k=0}^m\det (I-z\cdot {f_*}_k)^{(-1)^{k+1}} \end{aligned}$$

where ${f_*}_k:H_k(M;\mathbb {Q})\rightarrow H_k(M;\mathbb {Q})$. Hence, $L_f(z)$ is a rational function with coefficients in $\mathbb {Q}$. In [7, Theorem 4.5], it is shown that $N_f(z)$ is either

$$\begin{aligned} N_f(z)=L_f((-1)^nz)^{(-1)^{p+n}} \end{aligned}$$

or

$$\begin{aligned} N_f(z)=\left( \frac{L_{f_+}((-1)^nz)}{L_f((-1)^nz)}\right) ^{(-1)^{p+n}} \end{aligned}$$

where p is the number of real eigenvalues of $D_*$ which are $>1$ and n is the number of real eigenvalues of $D_*$ which are $<-1$. Here, $f_+$ is a lift of f to a certain twofold covering of M which has the same affine homotopy lift (d, D) as f. Consequently, $N_f(z)$ is a rational function with coefficients in $\mathbb {Q}$.

On the other hand, since $N_f(0)=1$ by definition, $z=0$ is not a zero nor a pole of the rational function $N_f(z)$. Thus, we can write

$$\begin{aligned} N_f(z)=\frac{u(z)}{v(z)}=\frac{\prod _i(1-\beta _iz)}{\prod _j(1-\gamma _jz)}=\prod _{i=1}^r(1-\lambda _iz)^{-\rho _i} \end{aligned}$$

(2.3)

with all $\lambda _i$ distinct nonzero algebraic integers (see for example [3] or [2, Theorem 2.1]) and $\rho _i$ nonzero integers. Taking log on both sides of the above identity, we obtain

$$\begin{aligned} \sum _{k=1}^\infty \frac{N(f^k)}{k}z^k&=\sum _{i=1}^r-\rho _i\log (1-\lambda _iz)\\&=\sum _{i=1}^r\rho _i\left( \sum _{k=1}^\infty \frac{(\lambda _iz)^k}{k}\right) \\&=\sum _{k=1}^\infty \frac{\sum _{i=1}^r\rho _i\lambda _i^k}{k}z^k. \end{aligned}$$

This induces

$$\begin{aligned} N(f^k)=\sum _{i=1}^{r(f)}\rho _i\lambda _i^k. \end{aligned}$$

(N1)

Note that r(f) is the number of zeros and poles of $N_f(z)$. Since $N_f(z)$ is a homotopy invariant, so is r(f). This argument tells us that whenever we have a rational expression of $N_f(z)$, we can write down all $N(f^k)$ directly from the expression. However, even though we can compute all $N(f^k)$ using the averaging formula, it can be rather complicated to write down the rational expression of $N_f(z)$.

We consider another generating function associated with the sequence $\{N(f^k)\}$:

$$\begin{aligned} S_f(z)=\frac{\mathrm{d}}{\mathrm{d}z}\log N_f(z). \end{aligned}$$

Then it is easy to see that

$$\begin{aligned} S_f(z)=\sum _{k=1}^\infty N(f^k)z^{k-1}=\sum _{k=1}^\infty \sum _{i=1}^{r(f)}\rho _i\lambda _i^kz^{k-1}=\sum _{i=1}^{r(f)}\frac{\rho _i\lambda _i}{1-\lambda _iz}, \end{aligned}$$

(2.4)

which is a rational function with simple poles and integral residues, and 0 at infinity. The rational function $S_f(z)$ can be written as $S_f(z)=u(z)/v(z)$ where the polynomials u(z) and v(z) are of the form

$$\begin{aligned} u(z)=N(f)+\sum _{i=1}^s a_iz^i,\quad v(z)=1+\sum _{j=1}^tb_jz^j \end{aligned}$$

with $a_i$ and $b_j$ integers, see $(3)\!\Rightarrow \!(5)$, Theorem 2.1 in [2] or [19, Lemma 3.1.31]. Let $\tilde{v}(z)$ be the conjugate polynomial of v(z), i.e., $\tilde{v}(z)=z^{t}v(1/z)$. Then, the numbers $\{\lambda _i\}$ are the roots of $\tilde{v}(z)$, and $r(f)=t$.

The following can be found in the proof of $(3)\Rightarrow (5)$, Theorem 2.1 in [2].

Lemma 2.4

If $\lambda _i$ and $\lambda _j$ are roots of the rational polynomial $\tilde{v}(z)$ which are algebraically conjugate (i.e., $\lambda _i$ and $\lambda _j$ are roots of the same irreducible polynomial), then $\rho _i=\rho _j$.

Proof

Let $\Sigma =\mathbb {Q}(\lambda _1,\cdots ,\lambda _r)\subset \mathbb {C}$ be the field of the rational polynomial $\tilde{v}(z)$ and let $\sigma$ be an automorphism of $\Sigma$ over $\mathbb {Q}$, i.e., $\sigma$ is the identity on $\mathbb {Q}$. The group of all such automorphisms is called the Galois group of $\Sigma$. Since the $\sigma (\lambda _i)$ are again the roots of $\tilde{v}(z)$, we have $\sigma (\lambda _i)=\lambda _{\sigma (i)}$. That is, $\sigma$ induces a permutation $\sigma$ on $\{1,\cdots ,r\}$. Applying $\sigma$ to the sequence $\{N(f^k)\}$, we obtain

$$\begin{aligned} \sigma \left( N(f^k)\right)&=\sigma \left( \sum _{i=1}^{r}\rho _i\lambda _i^k\right) =\sum _{i=1}^{r}\rho _i\sigma (\lambda _i)^k=\sum _{i=1}^{r}\rho _i\lambda _{\sigma (i)}^k =\sum _{i=1}^{r}\rho _{\sigma ^{{-}1}(i)}\lambda _i^k. \end{aligned}$$

Since the $N(f^k)$ are integers, $\sigma (N(f^k))=N(f^k)$ and consequently

$$\begin{aligned} \sum _{i=1}^{r}\rho _{i}\lambda _i^k=\sum _{i=1}^{r}\rho _{\sigma ^{{-}1}(i)}\lambda _i^k. \end{aligned}$$

As a matrix form, we can write

$$\begin{aligned} \left[ \begin{matrix} \lambda _1&{}\lambda _2&{}\cdots &{}\lambda _r\\ \lambda _1^2&{}\lambda _2^2&{}\cdots &{}\lambda _r^2\\ \vdots &{}\vdots &{}&{}\vdots \\ \lambda _1^r&{}\lambda _2^r&{}\cdots &{}\lambda _r^r\\ \end{matrix}\right] \left[ \begin{matrix}\rho _1\\ \rho _2\\ \vdots \\ \rho _r \end{matrix}\right] =\left[ \begin{matrix} \lambda _1&{}\lambda _2&{}\cdots &{}\lambda _r\\ \lambda _1^2&{}\lambda _2^2&{}\cdots &{}\lambda _r^2\\ \vdots &{}\vdots &{}&{}\vdots \\ \lambda _1^r&{}\lambda _2^r&{}\cdots &{}\lambda _r^r\\ \end{matrix}\right] \left[ \begin{matrix}\rho _{\sigma ^{-1}(1)}\\ \rho _{\sigma ^{-1}(2)}\\ \vdots \\ \rho _{\sigma ^{-1}(r)} \end{matrix}\right] . \end{aligned}$$

Since the $\lambda _i$ are distinct, the matrices in this equation are nonsingular (the Vandermonde determinant). Thus, $\rho _i=\rho _{\sigma ^{-1}(i)}$ for all $i=1,\dots ,r$. On the other hand, it is known that the Galois group acts transitively on the set of algebraically conjugate roots. Since $\lambda _i$ and $\lambda _j$ are conjugate roots of $\tilde{v}(z)$, we can choose $\sigma$ in the Galois group so that $\sigma (\lambda _i)=\lambda _j$. Hence, $\sigma (i)=j$ and so $\rho _i=\rho _j$. $\square$

Let $\tilde{v}(z)=\prod _{\alpha =1}^s\tilde{v}_\alpha (z)$ be the decomposition of the monic integral polynomial $\tilde{v}(z)$ into irreducible polynomials $\tilde{v}_\alpha (z)$ of degree $r_\alpha$. Of course, $r=r(f)=\sum _{\alpha =1}^sr_\alpha$ and

$$\begin{aligned} \tilde{v}(z)&=z^r+b_1z^{r-1}+b_2z^{r-2}+\cdots +b_{r-1}z+b_r\\&=\prod _{\alpha =1}^s\left( z^{r_\alpha }+b_1^\alpha z^{r_\alpha -1}+b_2^\alpha z^{r_\alpha -2}+\cdots +b_{r_\alpha -1}^\alpha z+b_{r_\alpha }^\alpha \right) =\prod _{\alpha =1}^s\tilde{v}_\alpha (z). \end{aligned}$$

If $\{\lambda _i^{(\alpha )}\}$ are the roots of $\tilde{v}_\alpha (z)$, then the associated $\rho$’s are the same $\rho _\alpha$. Consequently, we can rewrite (N1) as:

$$\begin{aligned} N(f^k)&=\sum _{\alpha =1}^s\rho _\alpha \left( \sum _{i=1}^{r_\alpha }(\lambda _i^{(\alpha )})^k\right) \\&=\sum _{\rho _\alpha >0}\rho _\alpha ^+\left( \sum _{i=1}^{r_\alpha }(\lambda _i^{(\alpha )})^k\right) -\sum _{\rho _\alpha <0}\rho _\alpha ^-\left( \sum _{i=1}^{r_\alpha }(\lambda _i^{(\alpha )})^k\right) . \end{aligned}$$

Consider the $r_\alpha \times r_\alpha$-integral square matrices

$$\begin{aligned} M_\alpha =\left[ \begin{matrix} 0&{}0&{}\cdots &{}0&{}-b^\alpha _{r_\alpha }\\ 1&{}0&{}\cdots &{}0&{}-b^\alpha _{r_\alpha -1}\\ \vdots &{}\vdots &{}&{}\vdots &{}\vdots \\ 0&{}0&{}\cdots &{}0&{}-b^\alpha _2\\ 0&{}0&{}\cdots &{}1&{}-b^\alpha _1 \end{matrix}\right] . \end{aligned}$$

The characteristic polynomial is $\det (zI-M_\alpha )=\tilde{v}_\alpha (z)$ and therefore $\{\lambda _i^{(\alpha )}\}$ are the eigenvalues of $M_\alpha$. This implies that $N(f^k)=\sum _{\alpha =1}^s\rho _\alpha \ \mathrm {tr\,}M_\alpha ^k$. Set

$$\begin{aligned} M_+=\bigoplus _{\rho _\alpha >0} \rho _\alpha ^+ M_\alpha ,\quad M_-=\bigoplus _{\rho _\alpha <0} \rho _\alpha ^- M_\alpha . \end{aligned}$$

Then

$$\begin{aligned} N(f^k)=\mathrm {tr\,}M_+^k-\mathrm {tr\,}M_-^k=\mathrm {tr\,}(M_+\oplus ( -M_-))^k. \end{aligned}$$

(N2)

Remark that from Theorem 2.3 (the rationality of the zeta functions on infra-solvmanifolds of type $\mathrm {(R)}$), it was possible to derive the identities (N2) above. Thus, we can reprove the Dold–Nielsen congruences (DN) of Theorem 2.2 using the following fact ([42, Theorem 1]): The Gauss congruence for the traces of powers of integer matrices

$$\begin{aligned} \sum _{d\mid k}\mu \left( \frac{k}{\mathrm{d}}\right) \mathrm {tr\,}(M^d)\equiv 0\mod k \end{aligned}$$

holds for all integer matrices M and all natural k. Recall also from [42] that the following Euler congruence

$$\begin{aligned} \mathrm {tr\,}(M^{p^r})\equiv \mathrm {tr\,}(M^{p^{r-1}})\mod p^r \end{aligned}$$

holds for all integer matrices M, all natural r, and all prime numbers p. Furthermore, it is shown in [42, Theorem 9] that the above two assertions are equivalent. Immediately we obtain the following Euler congruences for the Nielsen numbers, which are equivalent to the Gauss congruences for the Nielsen numbers in Theorem 2.2.

Theorem 2.5

Let $f:M\rightarrow M$ be a map on an infra-solvmanifold M of type $\mathrm {(R)}$. Then

$$\begin{aligned} N(f^{p^r})\equiv N(f^{p^{r-1}})\mod p^r \end{aligned}$$

(EN)

for all $r>0$ and all prime numbers p.

We remark also that the rationality of the Nielsen zeta function $N_f(z)$ on an infra-solvmanifold of type $\mathrm {(R)}$ is equivalent to the existence of a self-map g of some topological space X such that $N(f^k)=L(g^k)$ for all $k\ge 1$. In addition, due to the Gauss congruences (DN) in Theorem 2.2 we can choose X to be a compact polyhedron, see [2, Theorem 2.1 and Theorem(Dold)]. Thus, we can say that Nielsen theory on infra-solvmanifolds of type $\mathrm {(R)}$ is simpler than we have expected and is reduced to Lefschetz theory but on different spaces that are not necessarily closed nor aspherical manifolds.

We will show in Proposition 5.4 that if $A_k(f)\ne 0$ then $N(f^k)\ne 0$ and hence f has an essential periodic point of period k. In the following, we investigate some other necessary conditions under which $N(f^k)\ne 0$. Recall that

$$\begin{aligned} N(f^k)=\text { the number of essential fixed point classes of } f^k. \end{aligned}$$

If $\mathbb {F}$ is a fixed point class of $f^k$, then $f^k(\mathbb {F})=\mathbb {F}$ and the length of $\mathbb {F}$ is the smallest number p for which $f^p(\mathbb {F})=\mathbb {F}$, written $p(\mathbb {F})$. We denote by $\langle \mathbb {F}\rangle$ the f-orbit of $\mathbb {F}$, i.e., $\langle \mathbb {F}\rangle =\{\mathbb {F},f(\mathbb {F}),\dots , f^{p-1}(\mathbb {F})\}$ where $p=p(\mathbb {F})$. If $\mathbb {F}$ is essential, so is every $f^i(\mathbb {F})$ and $\langle \mathbb {F}\rangle$ is an essential periodic orbit of f with length $p(\mathbb {F})$ and $p(\mathbb {F})\mid k$. These are variations of Corollaries 2.3, 2.4 and 2.5 of [2].

Corollary 2.6

If $r(f)\ne 0$, then $N(f^i)\ne 0$ for some $1\le i\le r(f)$. In particular, f has at least $N(f^i)$ essential periodic points of period i and an essential periodic orbit with the length $p\mid i, i\le r(f)$.

Proof

Recall from (2.4) that

$$\begin{aligned} S_f(z)=\sum _{i=1}^{r(f)}\frac{\rho _i\lambda _i}{1-\lambda _iz}=\sum _{k=1}^\infty N(f^k)z^{k-1}. \end{aligned}$$

Assume that $N(f)=\cdots =N(f^{r(f)})=0$. For simplicity, write the above identity as $S_f(z)=u(z)/v(z)=s(z)$ where v(z) is a polynomial of degree r(f) and s(z) is the series of the right-hand side. Then, $u(z)=v(z)s(z)$. A simple calculation shows that the higher order derivative of v(z)s(z) up to order $r(f)-1$ at 0 are all zero. Since u(z) is a polynomial of degree $r(f)-1$, it shows that $u(z)=0$, a contradiction. $\square$

Recalling the identity $N(f^k)=\sum _{i=1}^{r(f)}\rho _i\lambda _i^k$, we define

$$\begin{aligned} \begin{aligned}&\rho (f)=\sum _{i=1}^{r(f)}\rho _i,\\&M(f)=\max \left\{ \sum _{\rho _i\ge 0}\rho _i,-\sum _{\rho _j<0}\rho _j\right\} ,\\&m(f)=\min \left\{ \sum _{\rho _i\ge 0}\rho _i,-\sum _{\rho _j<0}\rho _j\right\} . \end{aligned} \end{aligned}$$

(2.5)

Corollary 2.7

If $\rho (f)=0$ and $r(f)\ge 1$, then $r(f)\ge 2$ and $N(f^i)\ne 0$ for some $1\le i< r(f)$. In particular, f has at least $N(f^i)$ essential periodic points of period i and an essential periodic orbit with the length $p\mid i, i\le r(f)-1$.

Proof

The conditions $\rho (f)=0$ and $r(f)\ge 1$ immediately imply that $r(f)\ge 2$. Since $r(f)\ne 0$ by the previous corollary there exists $i\in \{1,\cdots ,r(f)\}$ such that $N(f^i)\ne 0$. Assume $N(f)=\cdots =N(f^{r(f)-1})=0$. So, $N(f^{r(f)})\ne 0$. As in the proof of the above corollary, we consider $u(z)/v(z)=s(z)$ where u(z) is a polynomial of degree $r(f)-1\ge 1$ and s(z) is a power series starting from the nonzero term $N(f^{r(f)})z^{r(f)-1}$. The derivative of order $r(f)-1$ on both sides of the identity $u(z)=v(z)s(z)$ yields that $N(f^{r(f)})=0$, a contradiction.

$\square$

Corollary 2.8

If $r(f)>0$, then $N(f^i)\ne 0$ for some $1\le i\le M(f)$. In particular, f has at least $N(f^i)$ essential periodic points of period i and an essential periodic orbit with the length $p\mid i, i\le M(f)$.

Proof

Assume that $N(f^k)=0$ for all $k=1,\cdots , M(f)$. From (N2), we have $J_k:=\mathrm {tr\,}M_+^k=\mathrm {tr\,}M_-^k$. For simplicity, suppose

$$\begin{aligned} m(f)=\sum _{\rho _i>0}\rho _i\le -\sum _{\rho _j<0}\rho _j=M(f). \end{aligned}$$

Then, the matrix $M_+$ has size m(f) and $M_-$ has size M(f). We write the eigenvalues of $M_+$ and $M_-$, respectively, as:

$$\begin{aligned} \mu _1,\cdots ,\mu _{m(f)};\ \tilde{\mu }_1,\cdots ,\tilde{\mu }_{M(f)}. \end{aligned}$$

Of course, $\{\mu _1,\cdots ,\mu _{m(f)},\tilde{\mu }_1,\cdots ,\tilde{\mu }_{M(f)}\}=\{\lambda _1,\cdots ,\lambda _r\}$ as a set. Now, the identities $\mathrm {tr\,}M_+^k=\mathrm {tr\,}M_-^k$ yield the M(f) equations

$$\begin{aligned} \mu _1^k+\cdots +\mu _{m(f)}^k+0+\cdots +0=\tilde{\mu }_1^k+\cdots +\tilde{\mu }_{M(f)}^k \end{aligned}$$

where $\mu _j=0$ when $m(f)<j\le M(f)$. By [4, p.72, Corollary], there exists a permutation $\sigma$ on $\{1,\cdots ,M(f)\}$ such that $\mu _i=\tilde{\mu }_{\sigma (i)}$. If $m(f)<M(f)$ then $0=\mu _{M(f)}=\tilde{\mu }_{\sigma (M(f))}=\lambda _j$ for some j, a contradiction. Hence, $m(f)=M(f)$ and the $\lambda _i$’s associated with $\rho _i>0$ and the $\lambda _j$’s associated with $\rho _j<0$ are the same. This implies that the rational function $N_f(z)$ has the same poles and zeros of equal multiplicity and hence $N_f(z)\equiv 1$, contradicting that $r(f)>0$.

$\square$

3 Radius of convergence of $N_f(z)$

From the Cauchy-Hadamard formula, we can see that the radii R of convergence of the infinite series $N_f(z)$ and $S_f(z)$ are the same and given by

$$\begin{aligned} \frac{1}{R}=\limsup _{k\rightarrow \infty }\left( \frac{N(f^k)}{k}\right) ^{1/k} =\limsup _{k\rightarrow \infty }N(f^k)^{1/k}. \end{aligned}$$

We will understand the radius R of convergence from the identity $N(f^k)$ $=\sum _{i=1}^{r(f)}\rho _i\lambda _i^k$. Recall that the $\lambda _i^{-1}$ are the poles or the zeros of the rational function $N_f(z)$.

Definition 3.1

We define

$$\begin{aligned} \lambda (f):=\max \{|\lambda _i|\mid i=1,\cdots ,r(f)\}. \end{aligned}$$

If $r(f)=0$, i.e., if $N(f^k)=0$ for all $k>0$, then $N_f(z)\equiv 1$ and $1/R=0$. In this case, we define customarily $\lambda (f)=0$. When $r(f)\ne 0$, we define

$$\begin{aligned} n(f) := \#\{i\mid |\lambda _i|=\lambda (f)\}. \end{aligned}$$

We shall assume now that $r(f)\ne 0$ or $\lambda (f)>0$. First, we can observe easily the following:

• 1. $\limsup z_k^{1/k}=\limsup (r_ke^{i\theta _k})^{1/k}=\limsup r_k^{1/k}e^{i\theta _k/k}=\limsup r_k^{1/k}$.

• 2. $\limsup (\lambda ^k)^{1/k}=|\lambda |$ by taking $z_k=\lambda ^k$ in (1).

• 3. When $\lim z_k=0$ in (1), $\limsup z_k^{1/k}=0$.

• 4. $\lim (z_k+\rho )^{1/k}=\lim z_k^{1/k}$ when $\lim z_k=\infty$. For, in this case 1 induces

  $$\begin{aligned} \lim \left( \frac{z_k+\rho }{z_k}\right) ^{1/k}=1. \end{aligned}$$

Assume $|\lambda _j|\ne \lambda (f)$ for some j; then we have

$$\begin{aligned} \frac{N(f^k)}{\lambda _j^k}=\sum _{i\ne j}\rho _i\left( \frac{\lambda _i}{\lambda _j}\right) ^k+\rho _j,\quad \lim \sum _{i\ne j}\rho _i\left( \frac{\lambda _i}{\lambda _j}\right) ^k=\infty . \end{aligned}$$

It follows from the above observations that $1/R=\limsup (\sum _{i\ne j}\rho _i\lambda _i^k)^{1/k}$. Consequently, we may assume that $N(f^k)=\sum _j\rho _j\lambda _j^k$ with all $|\lambda _j|=\lambda (f)$ and then we have

$$\begin{aligned} \frac{1}{R}=\limsup \left( \sum _{|\lambda _j|=\lambda (f)} \rho _j\lambda _j^k\right) ^{1/k}. \end{aligned}$$

Remark that if $\lambda (f)<1$ then $N(f^k)=\sum _{|\lambda _j|=\lambda (f)} \rho _j\lambda _j^k\rightarrow 0$ and so the sequence of integers are eventually zero, i.e., $N(f^k)=0$ for all k sufficiently large. This shows that $1/R=0$ and, furthermore, $N_f(z)$ is the exponential of a polynomial. Hence, the rational function $N_f(z)$ has no poles and zeros. This forces $N_f(z)\equiv 1$; hence, $\lambda (f)=0$. If $\lambda (f)>1$, then $N(f^k)\rightarrow \infty$ and by L’Hôpital’s rule we obtain

$$\begin{aligned} \limsup _{k\rightarrow \infty }\frac{\log N(f^k)}{k}&=\limsup _{k\rightarrow \infty }\frac{\log \left( \sum _j\rho _j\lambda _j^k\right) }{k} =\log \lambda (f)\ \Rightarrow \ \frac{1}{R}=\lambda (f). \end{aligned}$$

If $\lambda (f)=1$, then $N(f^k)\le \sum _j|\rho _j|<\infty$ is a bounded sequence and so it has a convergent subsequence. If $\limsup N(f^k)=0$, then $N(f^k)=0$ for all k sufficiently large and so by the same reason as above, $\lambda (f)=0$, a contradiction. Hence, $\limsup N(f^k)$ is a finite nonzero integer and so $1/R=1=\lambda (f)$.

Summing up, we have obtained that

Theorem 3.2

Let f be a map on an infra-solvmanifold of type $\mathrm {(R)}$. Let R denote the radius of convergence of the Nielsen zeta function $N_f(z)$ of f. Then, $\lambda (f)=0$ or $\lambda (f)\ge 1$, and

$$\begin{aligned} \frac{1}{R}=\lambda (f). \end{aligned}$$

(R1)

In particular,

• (1) $\lambda (f)=0$, which occurs if and only if $N_f(z)\equiv 1$;

• (2) if $\lambda (f)=1$, then $N(f^k)$ is a bounded sequence;

• (3) $\lambda (f)>1$ if and only if $N(f^k)$ is an unbounded sequence.

Remark 3.3

In this paper, the number $\lambda (f)$ will play a similar role as the “essential spectral radius” in [19] or the “reduced spectral radius” in [2]. Theorem 3.2 below shows that $1/\lambda (f)$ is the “radius” of the Nielsen zeta function $N_f(z)$. Note also that $\lambda (f)$ is a homotopy invariant.

By Theorem 3.2, we see that the sequence $N(f^k)$ is either bounded or exponentially unbounded.

Remark 3.4

Recall from (2.3) and (2.4) that

$$\begin{aligned}&S_f(z)=\sum _{i=1}^{r(f)}\frac{\rho _i\lambda _i}{1-\lambda _iz},\\&N_f(z)=\prod _{i=1}^{r(f)}(1-\lambda _iz)^{-\rho _i}=\frac{\prod _{\rho _j<0}(1-\lambda _jz)^{-\rho _j}}{\prod _{\rho _i>0}(1-\lambda _iz)^{\rho _i}}. \end{aligned}$$

These show that all the $1/\lambda _i$ are the poles of $S_f(z)$, whereas the $1/\lambda _i$ with corresponding $\rho _i>0$ are the poles of $N_f(z)$. The radius of convergence of a power series centered at a point a is equal to the distance from a to the nearest point where the power series cannot be defined in a way that makes it holomorphic. Hence, the radius of convergence of $S_f(z)$ is $1/\lambda (f)$ and the radius of convergence of $N_f(z)$ is $1/\max \{|\lambda _i|\mid \rho _i>0\}$. In particular, we have shown that

$$\begin{aligned} \lambda (f)=\max \{|\lambda _i|\mid i=1,\cdots ,r(f)\}=\max \{|\lambda _i|\mid \rho _i>0\}. \end{aligned}$$

Notice this identity in Example 3.7.

On the other hand, we can understand the radius R of convergence using the averaging formula. Compare our result with [12, Theorem 7.10]. Let $\{\mu _1,\cdots ,\mu _m\}$ be the eigenvalues of $D_*$, counted with multiplicities, where m is the dimension of the manifold M. We denote by $\mathrm {sp}(A)$ the spectral radius of the matrix A which is the largest modulus of an eigenvalue of A. From the definition, we have

$$\begin{aligned}&\mathrm {sp}(D_*)=\max \{|\mu _j|\mid j=1,\cdots ,m\},\\&\mathrm {sp}\left( \bigwedge D_*\right) ={\left\{ \begin{array}{ll} \prod \nolimits _{|\mu _j|>1}|\mu _j|&{}\text {when }\mathrm {sp}(D_*)>1;\\ 1&{}\text {when }\mathrm {sp}(D_*)\le 1. \end{array}\right. } \end{aligned}$$

Note $|\det (I-D_*^k)|=\prod _{j=1}^m|1-\mu _j^k|$. If some $\mu _j=1$, then $\det (I-D_*^k)=0$ for all $k>0$ and hence $\limsup |\det (I-D_*^k)|^{1/k}=0$. Assume now all $\mu _j\ne 1$; then, $\det (I-D_*^k)\ne 0$ for all $k>0$. Remark further from [12, Theorem 4.1] that

$$\begin{aligned} \log \left( \limsup _{k\rightarrow \infty }|\det (I-D_*^k)|^{1/k}\right)&=\limsup _{k\rightarrow \infty }\sum _{j=1}^m\frac{\log |1-\mu _j^k|}{k}\\&={\left\{ \begin{array}{ll} \sum \nolimits _{|\mu _j|>1}\log |\mu _j|&{}\text {when }\mathrm {sp}(D_*)>1\\ 0&{}\text {when }\mathrm {sp}(D_*)\le 1. \end{array}\right. } \end{aligned}$$

Now we ascertain that if $D_*$ has no eigenvalue 1, then

$$\begin{aligned} \frac{1}{R}=\limsup _{k\rightarrow \infty }|\det (I-D_*^k)|^{1/k}=\mathrm {sp}\left( \bigwedge D_*\right) . \end{aligned}$$

(R2)

From the averaging formula, we have $N(f^k)\ge |\det (I-D_*^k)|/\#\Phi$. This induces

$$\begin{aligned} \frac{1}{R}=\limsup _{k\rightarrow \infty }N(f^k)^{1/k}&\ge \limsup _{k\rightarrow \infty }\left( \frac{|\det (I-D_*^k)|}{\#\Phi }\right) ^{1/k}\\&=\limsup _{k\rightarrow \infty }|\det (I-D_*^k)|^{1/k}. \end{aligned}$$

Furthermore, for any $A\in \Phi$, we obtain (see the proof of [12, Theorem 4.3])

$$\begin{aligned} |\det (I-A_*D_*^k)|\le \prod _{j=1}^m(1+|\mu _j|^k) \end{aligned}$$

and hence from the averaging formula

$$\begin{aligned} N(f^k)\le \prod _{j=1}^m(1+|\mu _j|^k)\ \Rightarrow \frac{1}{R}\le \prod _{|\mu _j|>1}|\mu _j|. \end{aligned}$$

This finishes the proof of our assertion.

Following from (R1) and (R2), we immediately have

Theorem 3.5

Let f be a map on an infra-solvmanifold of type $\mathrm {(R)}$ with an affine homotopy lift (d, D). Let R denote the radius of convergence of the Nielsen zeta function of f. If $D_*$ has no eigenvalue 1, then

$$\begin{aligned} \frac{1}{R}=\mathrm {sp}\left( \bigwedge D_*\right) =\lambda (f). \end{aligned}$$

We recall that the asymptotic Nielsen number of f is defined to be

$$\begin{aligned} N^\infty (f):=\max \left\{ 1,\limsup _{k\rightarrow \infty }N(f^k)^{1/k}\right\} . \end{aligned}$$

We also recall that the most widely used measure for the complexity of a dynamical system is the topological entropy h(f). A basic relation between these two numbers is $h(f)\ge \log N^\infty (f)$, which was found by Ivanov in [17]. There is a conjectural inequality $h(f)\ge \log (\mathrm {sp}(f))$ raised by Shub [25, 26, 38]. This conjecture was proven for all maps on infra-solvmanifolds of type $\mathrm {(R)}$, see [12, 33, 34]. Now, we are able to state about relations between $N^\infty (f)$, $\lambda (f)$ and h(f).

Corollary 3.6

Let f be a map on an infra-solvmanifold of type $\mathrm {(R)}$ with an affine homotopy lift (d, D). If $D_*$ has no eigenvalue 1, then

$$\begin{aligned} N^\infty (f)=\lambda (f),\quad h(f)\ge \log \lambda (f). \end{aligned}$$

Proof

From [12, Theorem 4.3] and Theorem 3.5, we have that $N^\infty (f)=\mathrm {sp}(\bigwedge D_*)=\lambda (f)$. Hence by Ivanov’s inequality, we obtain that $h(f)\ge \log N^\infty (f)=\log \lambda (f)$. $\square$

The following example shows that the assumption in Theorem 3.5 and its Corollary that 1 is not in the spectrum of $D_*$ is essential.

Example 3.7

Let $f:M\rightarrow M$ be a map of type $(r,\ell ,q)$ on the Klein bottle M induced by an affine map $(d,D):\mathbb {R}^2\rightarrow \mathbb {R}^2$. Recall from [27, Theorem 2.3] and its proof that r is odd or $q=0$, and

$$\begin{aligned} (d,D)&={\left\{ \begin{array}{ll} \left( -\frac{1}{2}\left[ \begin{matrix}*\\ \ell \end{matrix}\right] , \left[ \begin{matrix}r&{}0\\ 0&{}q\end{matrix}\right] \right) &{}\text {when }r\text { is odd;}\\ \left( \left[ \begin{matrix}*\\ *\end{matrix}\right] , \left[ \begin{matrix}r&{}0\\ 2\ell &{}0\end{matrix}\right] \right)&\text {when }r\text { is even and }q=0, \end{array}\right. }\\ N(f^k)&={\left\{ \begin{array}{ll} |q^k(1-r^k)|={\left\{ \begin{array}{ll} q^k(r^k-1)&{}\text {when }r\text { is odd and }qr>0\\ (-1)^kq^k(r^k-1)&{}\text {when }r\text { is odd and }qr<0\\ \end{array}\right. }\\ |1-r^k|={\left\{ \begin{array}{ll}1&{}\text {when }q=r=0\\ r^k-1&{}\text {when }r>0\text { and }q=0\\ (-1)^k(r^k-1)&{}\text {when }r<0\text { and }q=0. \end{array}\right. } \end{array}\right. } \end{aligned}$$

A simple calculation shows that

q, r	$N(f^k)$	$N_f(z)$	$S_f(z)$	$(\lambda (f),\mathrm {sp}(\bigwedge D_*))$
$r=1$	0	1	0	$(0,\max \{1,|q|\})$
r odd, $qr>0$	$(qr)^k-q^k$	$\displaystyle {\frac{1-qz}{1-qrz}}$	$\displaystyle {\frac{qr}{1-qrz}-\frac{q}{1-qz}}$	(qr, qr)
r odd, $qr<0$	$(-qr)^k-(-q)^k$	$\displaystyle {\frac{1+qz}{1+qrz}}$	$\displaystyle {-\frac{qr}{1+qrz}+\frac{q}{1+qz}}$	$(-qr,-qr)$
$q=0$, $r=0$	1	$\displaystyle {\frac{1}{1-z}}$	$\displaystyle {\frac{1}{1-z}}$	(1, 1)
$q=0$, $r>0$	$r^k-1$	$\displaystyle {\frac{1+z}{1-rz}}$	$\displaystyle {\frac{r}{1-rz}-\frac{1}{1+z}}$	(r, r)
$q=0$, $r<0$	$(-r)^k-(-1)^k$	$\displaystyle {\frac{1+z}{1+rz}}$	$\displaystyle {-\frac{r}{1+rz}+\frac{1}{1+z}}$	$(-r,-r)$

These observations show that when one of the eigenvalues is 1, the invariants $N_f(z)$, $\mathrm {sp}(\bigwedge D_*)$ and $\lambda (f)$ still strongly depend on the other eigenvalue. Remark also in this example that the identity $\lambda (f)=\max \{|\lambda _i|\mid \rho _i>0\}$ holds.

4 Asymptotic behavior of the sequence $\{N(f^k)\}$

In this section, we study the asymptotic behavior of the Nielsen numbers of iterates of maps on infra-solvmanifolds of type $\mathrm {(R)}$.

We can write $N(f^k)=\sum _{i=1}^{r(f)}\rho _i\lambda _i^k$ as $N(f^k)=\Gamma (f^k)+\Omega (f^k)$, where

Since $|\lambda _i|<\lambda (f)$, it follows that

$$\begin{aligned} \frac{\Omega _k}{\lambda (f)^k} =\sum _{|\lambda _i|<\lambda (f)}\rho _i\left( \frac{\lambda _i}{\lambda (f)}\right) ^k \rightarrow 0. \end{aligned}$$

Theorem 4.1

For a map f of an infra-solvmanifold of type $\mathrm {(R)}$, one of the following three possibilities holds:

• (1) $\lambda (f)=0$, which occurs if and only if $N_f(z)\equiv 1$.

• (2) The sequence $\{N(f^k)/\lambda (f)^k\}$ has the same limit points as a periodic sequence $\{\sum _j\alpha _j\epsilon _j^k\}$ where $\alpha _j\in \mathbb {Z},\epsilon _j\in \mathbb {C}$ and $\epsilon _j^q=1$ for some integer $q>0$.

• (3) The set of limit points of the sequence $\{N(f^k)/\lambda (f)^k\}$ contains an interval.

Proof

For simplicity, we denote $\lambda (f)$ by $\lambda _0$. Recall that $\lambda _0=0$ if and only if all $N(f^k)=0$ and otherwise, $\lambda _0\ge 1$. Suppose that $\lambda _0\ge 1$. We may assume that

$$\begin{aligned} \lambda _1=\lambda _0e^{2i\pi \theta _1},\cdots , \lambda _{n(f)}=\lambda _0e^{2i\pi \theta _{n(f)}} \end{aligned}$$

are all the $\lambda _i$ of modulus $\lambda _0$ (see Definition 3.1 for n(f)). From ( $\Lambda$ ), we see that the sequence $\{N(f^k)/\lambda _0^k\}$ has the same asymptotic behavior as the sequence

$$\begin{aligned} \left\{ \frac{\Gamma _k}{\lambda _0^k}\right\} =\left\{ \sum _{j=1}^{n(f)}\rho _je^{2i\pi (k\theta _j)}\right\} . \end{aligned}$$

We consider the continuous function $T^{n(f)}\rightarrow [0,\infty )$ defined by

$$\begin{aligned} (\xi _1,\cdots ,\xi _{n(f)})\longmapsto \big |\sum _{j=1}^{n(f)}\rho _je^{2i\pi \xi _j}\big |. \end{aligned}$$

For any subset $\mathcal {S}$ of $\{1,\cdots ,n(f)\}$, we have a sub-torus

$$\begin{aligned} T^{|\mathcal {S}|}=\{(\xi _1,\cdots ,\xi _{n(f)})\in T^{n(f)}\mid \xi _j=0,\ \forall j\notin \mathcal {S}\}. \end{aligned}$$

The restriction of the above continuous function to the sub-torus $T^{|\mathcal {S}|}$ is continuous and has its maximum $m_\mathcal {S}$ because $T^{|\mathcal {S}|}$ is compact.

Now we show that either

• (i) $\{\sum _{j=1}^{n(f)} \rho _je^{2i\pi k\theta _j}\}$ is periodic

or

• (ii) there exists $\mathcal {S}\subset \{1,\cdots ,n(f)\}$ such that the sequence $\{\sum _{j=1}^{n(f)} \rho _je^{2i\pi k\theta _j}\}$ is dense in $[0,m_\mathcal {S}]$.

If $\dim _\mathbb {Z}\{\theta _1,\cdots ,\theta _{n(f)},1\}=1$, then all $\theta _j$ are rational $p_j/q_j$. Every $\lambda _j/\lambda _0=e^{2i\pi \theta _j}$ is a $q_j$th root of unity, and thus all $\lambda _j/\lambda _0=e^{2i\pi \theta _j}$ are roots of unity of degree $q=\mathrm {lcm}(q_1,\cdots ,q_{n(f)})$, and hence the sequence $\{\sum _{j=1}^{n(f)}\rho _je^{2i\pi (k\theta _j)}\}$ is periodic of period q. This proves (2).

Suppose $\dim _\mathbb {Z}\{\theta _1,\cdots ,\theta _{n(f)},1\}=s>1$. Then, there exists the smallest subset $\mathcal {S}=\{j_1,\cdots ,j_s\}\subset \{1,\cdots ,n(f)\}$ for which $\theta _{j_1},\cdots ,\theta _{j_s},1$ are linearly independent over the integers. This means that if $\sum _{i=1}^s\ell _i\theta _{j_i}=\ell$ with $\ell _i,\ell \in \mathbb {Z}$ then $\ell _{j_1}=\cdots =\ell _{j_s}=\ell =0$. Then, it follows from [6, Theorem 6, p.  91] that the sequence $(k\theta _{j_1},\cdots ,k\theta _{j_s})$ is dense in $T^{|\mathcal {S}|}$. This proves (3) with $[0,m_\mathcal {S}]$.

$\square$

Example 4.2

• (1) Let f be the identity on $S^1$. Then, $N(f^k)=0$ for all $k>0$ and so $\lambda (f)=0$ and $N_f(z)\equiv 1$.

• (2) Consider the map f on $T^2$ induced by the matrix

  $$\begin{aligned} D=\left[ \begin{matrix}0&{}1\\ -1&{}-1\end{matrix}\right] . \end{aligned}$$

The characteristic polynomial of D is $t^2+t+1$. The eigenvalues of D are $\omega =(-1+\sqrt{3}i)/2$ and $\bar{\omega }$. Hence

$$\begin{aligned} L(f^k)=\det (I-D^k)=(1-\omega ^k)(1-\bar{\omega }^k)=2-(\omega ^k+\bar{\omega }^k) ={\left\{ \begin{array}{ll} 0&{}\text {when }k=3\ell \\ 3&{}\text {when }k\ne 3\ell . \end{array}\right. } \end{aligned}$$

So, $N(f^k)=L(f^k)$ for all $k>0$. Therefore, we have

$$\begin{aligned} L_f(z)=N_f(z)&=\frac{(1-\omega z)(1-\bar{\omega }z)}{(1-z)^2}. \end{aligned}$$

Consequently, we have that

$$\begin{aligned}&\lambda (f)=\max \{|\omega |,|\bar{\omega }|,1\}=1,\\&\left\{ \frac{\Gamma (f^k)}{\lambda (f)^k}\right\} =\left\{ 2-(\omega ^k+\bar{\omega }^k)\right\} \ \text { is }3\text {-periodic.} \end{aligned}$$

• (3) Let f be the map on $T^4$ induced by the matrix

  $$\begin{aligned} D=\left[ \begin{matrix}0&{}1&{}0&{}0\\ 0&{}0&{}1&{}0\\ 0&{}0&{}0&{}1\\ -1&{}2&{}0&{}2\end{matrix}\right] . \end{aligned}$$

The characteristic polynomial of D is $t^4-2t^3-2t+1$. This polynomial has two complex roots $\omega ,\bar{\omega }$ of modulus 1 and two real roots $\alpha ,\beta$ with $0<\alpha <1$ and $\beta =1/\alpha$. Note that this is an example showing that there are algebraic integers $\omega ,\bar{\omega }$ of modulus 1 which are not roots of unity. Indeed, $\omega =\frac{1-\sqrt{3}}{2}+ yi\in S^1$ where $y^2=\frac{\sqrt{3}}{2}$.

Since

$$\begin{aligned} L(f^k)&=\det (I-D^k)=(1-\alpha ^k)(1-\beta ^k)(1-\omega ^k)(1-\bar{\omega }^k),\\ N(f^k)&=|\det (I-D^k)|=-(1-\alpha ^k)(1-\beta ^k)(1-\omega ^k)(1-\bar{\omega }^k)\\&=-2+\omega ^k+\bar{\omega }^k+2\alpha ^k-(\alpha \omega )^k-(\alpha \bar{\omega })^k\\&\quad +2\beta ^k-(\beta \omega )^k-(\beta \bar{\omega })^k-2(\alpha \beta )^k+(\alpha \beta \omega )^k+(\alpha \beta \bar{\omega })^k, \end{aligned}$$

we have

$$\begin{aligned} N_f(z)&=\frac{(1-z)^2(1-\alpha \omega z)(1-\alpha \bar{\omega }z)(1-\beta \omega z)(1-\beta \bar{\omega }z)(1-\alpha \beta z)^2}{(1-\omega z)(1-\bar{\omega }z)(1-\alpha z)^2(1-\beta z)^2(1-\alpha \beta \omega z)(1-\alpha \beta \bar{\omega }z)},\\ \lambda (f)&=\max \{1,\alpha ,\beta ,\alpha \beta \}=\beta ,\quad \frac{\Gamma (f^k)}{\lambda (f)^k}=2-\omega ^k-\bar{\omega }^k. \end{aligned}$$

This is an example of Case (3) with [0, 2] in Theorem 4.1.

In Theorem 3.5, we showed that if $D_*$ has no eigenvalue 1 then $\lambda (f)=\mathrm {sp}(\bigwedge D_*)$. In Example 3.7, we have seen that when $D_*$ has an eigenvalue 1, there are maps f for which $\lambda (f)\ne \mathrm {sp}(\bigwedge D_*)$ with $\lambda (f)=0$, and $\lambda (f)=\mathrm {sp}(\bigwedge D_*)$ with $\lambda (f)\ge 1$. In fact, we prove in the following that the latter case is always true.

Lemma 4.3

Let f be a map on an infra-solvmanifold of type $\mathrm {(R)}$ with an affine homotopy lift (d, D). If $\lambda (f)\ge 1$, then $\lambda (f)=\mathrm {sp}(\bigwedge D_*)$.

Proof

Since $\lambda (f)\ge 1$, by Corollary 2.6, $N(f^k)\ne 0$ for some $k\ge 1$ and then by the averaging formula, there is $B\in \Phi$ such that $\det (I-B_*D_*^k)\ne 0$. Choose $\beta \in \Pi$ of the form $\beta =(b,B)$. Then $\beta (d,D)^k$ is another homotopy lift of $f^k$. We have observed above that there are numbers $\nu _1,\cdots ,\nu _m$ such that $\det (I-B_*D_*^k)=\prod _{i=1}^m(1-\nu _i)$ and $\{(\mu _i^k)^\ell \}=\{\nu _i^\ell \}$ for some $\ell >0$. Since $\det (I-B_*D_*^k)\ne 0$, $B_*D_*^k$ has no eigenvalue 1. Hence by Theorem 3.5, we have $\lambda (f^k)=\mathrm {sp}(\bigwedge B_*D_*^k)$. Recall that $N(f^k)=\sum _{i=1}^{r(f)}\rho _i\lambda _i^k$ and $\lambda (f)=\max \{|\lambda _i|\}$. Since $\lambda (f)\ge 1$, it follows that $\lambda (f^k)=\lambda (f)^k$. Observe further that $\mathrm {sp}(\bigwedge B_*D_*^k)=\prod _{|\nu _i|\ge 1}|\nu _i|=\prod _{|\mu _i|\ge 1}|\mu _i^k|=\mathrm {sp}(\bigwedge D_*)^k$. Consequently, we obtain the required identity $\lambda (f)=\mathrm {sp}(\bigwedge D_*)$. $\square$

Example 4.4

Consider the 3-dimensional orientable flat manifold with fundamental group $\mathfrak {G}_2$ generated by $\{t_1,t_2,t_3,\alpha \}$ where

Thus,

$$\begin{aligned} \mathfrak {G}_2=\left\langle t_1,t_2,t_3,\alpha \mid [t_i,t_j]=1, \alpha ^2=t_1, \alpha t_2\alpha ^{-1}=t_2^{-1}, \alpha t_3\alpha ^{-1}=t_3^{-1}\right\rangle . \end{aligned}$$

Let $\varphi :\mathfrak {G}_2\rightarrow \mathfrak {G}_2$ be any homomorphism. Every element of $\mathfrak {G}_2$ is of the form $\alpha ^pt_2^mt_3^n$. Thus, $\varphi$ has the form

$$\begin{aligned} \varphi (t_2)=\alpha ^{p_2}t_2^{m_2}t_3^{n_2},\quad \varphi (t_3)=\alpha ^{p_3}t_2^{m_3}t_3^{n_3},\quad \varphi (\alpha )=\alpha ^{p}t_2^{m}t_3^{n}. \end{aligned}$$

The relations $\alpha t_2\alpha ^{-1}=t_2^{-1}$ and $\alpha t_3\alpha ^{-1}=t_3^{-1}$ yield that $p_2=p_3=0$, $(-1)^pm_i=-m_i$ and $(-1)^pn_i=-n_i$. Hence, when p is even, we have $m_i=n_i=0$. Further, $\varphi (t_1)=t_1^p$.

Now we shall determine an affine map (d, D) satisfying $\varphi (\beta )(d,D)=(d,D)\beta$ for all $\beta \in \mathfrak {G}_2$.

Case $p=2\ell$.

In this case, we have $\varphi (t_2)=\varphi (t_3)=1$ and $\varphi (\alpha )=\alpha ^pt_2^mt_3^n=t_1^\ell t_2^mt_3^n$ and, hence, we need to determine (d, D) satisfying

$$\begin{aligned}&(d,D)=(d,D)(e_2,I)\Rightarrow D(e_2)=\mathbf{0},\\&(d,D)=(d,D)(e_3,I)\Rightarrow D(e_3)=\mathbf{0},\\&(\ell e_1+m e_2+n e_3,I)(d,D)=(d,D)(a,A)\\&\qquad \Rightarrow \ell e_1+me_2+ne_3+d=d+D(a),\ D=DA. \end{aligned}$$

Hence, the second and the third columns of D must be $\mathbf{0}$ and so $D=DA$ is automatically satisfied and the first column of D is $2\left[ \begin{matrix}\ell&m&n\end{matrix}\right] ^t$. That is,

$$\begin{aligned} (d,D)=\left( \left[ \begin{matrix}*\\ {*}\\ {*}\end{matrix}\right] , \left[ \begin{matrix}2\ell &{}0&{}0\\ 2m&{}0&{}0\\ 2n&{}0&{}0\end{matrix}\right] \right) . \end{aligned}$$

The eigenvalues of D are 0 (multiple) and $2\ell$, and $N(f^k)=|(2\ell )^k-1|$ and

$$\begin{aligned} N_f(z)={\left\{ \begin{array}{ll} \frac{1}{1-z}&{}\text {when }\ell =0;\\ \frac{1-z}{1-2\ell z}&{}\text {when }\ell \ge 1;\\ \frac{1+z}{1+2\ell z}&{}\text {when }\ell \le -1. \end{array}\right. } \end{aligned}$$

It follows that $\lambda (f)=\max \{1,2|\ell |\}=\mathrm {sp}(\bigwedge D_*)$. Moreover, the sequence $\{N(f^k)/\lambda (f)^k\}$ is asymptotically the constant sequence $\{1\}$. In fact, if for example $\ell \ge 1$ then $N(f^k)=(-1)\cdot 1^k+1\cdot (2\ell )^k$, $\lambda (f)=2\ell$ and $\Gamma (f^k)=1\cdot (2\ell )^k$, hence $\Gamma (f^k)/\lambda (f)^k\equiv 1$. We then have Case (2) of Theorem 4.1.

Case $p=2\ell +1$.

In this case, we have $\varphi (t_2)=t_2^{m_2}t_3^{n_2},\ \varphi (t_3)=t_2^{m_3}t_3^{n_3}$ and $\varphi (\alpha )=\alpha ^pt_2^mt_3^n=\alpha t_1^\ell t_2^mt_3^n$ and, hence, we need to determine (d, D) satisfying

$$\begin{aligned}&\varphi (t_2)(d,D)=(d,D)t_2\Rightarrow D(e_2)=m_2e_2+n_2e_3,\\&\varphi (t_3)(d,D)=(d,D)t_3\Rightarrow D(e_3)=m_3e_2+n_3e_3,\\&\varphi (\alpha )(d,D)=(d,D)\alpha \\&\qquad \Rightarrow a+A(\ell e_1+me_2+ne_3)+A(d)=d+D(a),\ AD=DA. \end{aligned}$$

These yield

$$\begin{aligned} (d,D)=\left( \left[ \begin{matrix}*\\ -\frac{m}{2}\\ -\frac{n}{2}\end{matrix}\right] , \left[ \begin{matrix}2\ell +1&{}0&{}0\\ 0&{}m_2&{}m_3\\ 0&{}n_2&{}n_3\end{matrix}\right] \right) . \end{aligned}$$

Now, we consider some explicit examples of such D. First we take D to be

$$\begin{aligned} D=\left[ \begin{matrix}-1&{}0&{}0\\ 0&{}d&{}e\\ 0&{}-e&{}d\end{matrix}\right] . \end{aligned}$$

Then, D has eigenvalues $-1$ and $\mu =d\pm ei$. Clearly, $N(f^k)=0$ for all even integers $k>0$. For odd k, $D^k$ and $AD^k$ are, respectively, of the form

$$\begin{aligned} D^k=\left[ \begin{matrix}-1&{}0&{}0\\ 0&{}x&{}y\\ 0&{}-y&{}x\end{matrix}\right] \ \text { and }\ AD^k=\left[ \begin{matrix}-1&{}0&{}0\\ 0&{}-x&{}-y\\ 0&{}y&{}-x\end{matrix}\right] . \end{aligned}$$

Then

$$\begin{aligned} N(f^k)&=\frac{1}{2}\left\{ |\det (I-D^k)|+|\det (I-AD^k)|\right\} \\&=\frac{1}{2}\left\{ 2\left( (1-x)^2+y^2\right) +2\left( (1+x)^2+y^2\right) \right\} =2(1+x^2+y^2). \end{aligned}$$

Here, $x^2+y^2=\mu ^k\bar{\mu }^k=|\mu |^{2k}$. Consequently, $N(f^k)=2(1+|\mu |^{2k})$ for odd k. This yields that

$$\begin{aligned} N_f(z)&=\exp \left( \sum _{k=1}^\infty \frac{2(1+|\mu |^{2(2k-1)})}{2k-1}z^{2k-1}\right) \\&=\exp \left( \sum _{k=1}^\infty \frac{2}{2k-1}z^{2k-1}+\sum _{k=1}^\infty \frac{2}{2k-1}(|\mu |^2z)^{2k-1}\right) \\&=\exp \left( \log \frac{1+z}{1-z}+\log \frac{1+|\mu |^2z}{1-|\mu |^2z}\right) \\&=\frac{(1+z)(1+|\mu |^2z)}{(1-z)(1-|\mu |^2z)}. \end{aligned}$$

Moreover, $\lambda (f)=\max \{1,|\mu |^2\}=\mathrm {sp}(\bigwedge D_*)$. We can see also that the sequence $\Gamma (f^k)/\lambda (f)^k$ is $(-1)^{k+1}+1$ if $|\mu |\ne 1$ and $2((-1)^{k+1}+1)$ if $|\mu |=1$, hence the sequence $\Gamma (f^k)/\lambda (f)^k$ is 2-periodic. We thus have Case (2) of Theorem 4.1.

Secondly, we take D to be

$$\begin{aligned} D=\left[ \begin{matrix}-1&{}0&{}0\\ 0&{}d&{}e\\ 0&{}e&{}f\end{matrix}\right] . \end{aligned}$$

Let D have eigenvalues $-1$ and $\mu _1$ and $\mu _2$. For odd k, $D^k$ and $AD^k$ are, respectively, of the form

$$\begin{aligned} D^k=\left[ \begin{matrix}-1&{}0&{}0\\ 0&{}x&{}y\\ 0&{}y&{}z\end{matrix}\right] \ \text { and }\ AD^k=\left[ \begin{matrix}-1&{}0&{}0\\ 0&{}-x&{}-y\\ 0&{}-y&{}-z\end{matrix}\right] . \end{aligned}$$

Then

$$\begin{aligned} N(f^k)&=\frac{1}{2}\left\{ |\det (I-D^k)|+|\det (I-AD^k)|\right\} \\&=\frac{1}{2}\left\{ 2\left( (1-x)(1-z)-y^2\right) +2\left( (1+x)(1+z)-y^2\right) \right\} \\&=2(1+xz-y^2). \end{aligned}$$

Here, $xz-y^2=\mu _1^k\mu _2^k$. Hence, $N(f^k)=(1+(-1)^{k+1})(1+\mu _1^k\mu _2^k)$. This yields that

$$\begin{aligned} N_f(z)&=\exp \left( \sum _{k=1}^\infty \frac{2(1+(\mu _1\mu _2)^{2k-1})}{2k-1}z^{2k-1}\right) \\&=\exp \left( \sum _{k=1}^\infty \frac{2}{2k-1}z^{2k-1}+\sum _{k=1}^\infty \frac{2}{2k-1}(\mu _1\mu _2 z)^{2k-1}\right) \\&=\exp \left( \log \frac{1+z}{1-z}+\log \frac{1+\mu _1\mu _2 z}{1-\mu _1\mu _2 z}\right) \\&=\frac{(1+z)(1+\mu _1\mu _2 z)}{(1-z)(1-\mu _1\mu _2 z)}. \end{aligned}$$

Observe also that $\lambda (f)=\max \{1, |\mu _1\mu _2|\}=\mathrm {sp}(\bigwedge D_*)$. Whether $\mu _i$ are real or complex, we can see that the sequence $N(f^k)/\lambda (f)^k$ is asymptotically periodic, and so we have Case (2) of Theorem 4.1.

It is important to know not only the rate of growth of the sequence $\{N(f^k)\}$ but also the frequency with which the largest Nielsen number is encountered. The following theorem shows that this sequence grows relatively dense. The following are variations of Theorem 2.7, Proposition 2.8 and Corollary 2.9 of [2].

Theorem 4.5

Let $f:M\rightarrow M$ be a map on an infra-solvmanifold of type $\mathrm {(R)}$. If $\lambda (f)\ge 1$, then there exist $\gamma >0$ and a natural number N such that for any $m> N$ there is an $\ell \in \{0,1,\cdots ,n(f)-1\}$ such that $N(f^{m+\ell })/\lambda (f)^{m+\ell }>\gamma$.

Proof

As in the proof of Theorem 4.1, for any $k>0$, we can write $N(f^k)=\Gamma _k+\Omega _k$ so that $\Gamma _k/\lambda (f)^k=\sum _{j=1}^{n(f)}\rho _je^{2i\pi (k\theta _j)}$. Consider the following n(f) consecutive identities

$$\begin{aligned} \frac{\Gamma _{k+\ell }}{\lambda (f)^{k+\ell }}=\sum _{j=1}^{n(f)}\left( \rho _je^{2i\pi (k\theta _j)}\right) e^{2i\pi (\ell \theta _j)}, \quad {\ell =0,\cdots ,n(f)-1.} \end{aligned}$$

Let $W=W(\theta _1,\cdots ,\theta _{n(f)})$ be the Vandermonde operator on $\mathbb {C}^{n(f)}$

$$\begin{aligned} W(\theta _1,\cdots ,\theta _{n(f)}) =\left[ \begin{matrix} 1&{}1&{}\cdots &{}1\\ e^{2i\pi \theta _1}&{}e^{2i\pi \theta _2}&{}\cdots &{}e^{2i\pi \theta _{n(f)}}\\ e^{2i\pi (2\theta _1)}&{}e^{2i\pi (2\theta _2)}&{}\cdots &{}e^{2i\pi (2\theta _{n(f)})}\\ \vdots &{}\vdots &{}&{}\vdots \\ e^{2i\pi (n(f)-1)\theta _1}&{}e^{2i\pi (n(f)-1)\theta _2}&{}\cdots &{}e^{2i\pi (n(f)-1)\theta _{n(f)}} \end{matrix}\right] \end{aligned}$$

and let $2\gamma =||W^{-1}||^{-1}$. Then, the vector $\mathbf {\rho }= \left( \rho _1e^{2i\pi (k\theta _1)},\cdots ,\rho _{n(f)}e^{2i\pi (k\theta _{n(f)})}\right)$ satisfies $||W\mathbf {\rho }||\ge ||W^{-1}||^{-1}||\mathbf {\rho }||=2\gamma ||\mathbf {\rho }||\ge 2\gamma \sqrt{n(f)}$. Thus, there is at least one of the coordinates of the vector $W\mathbf {\rho }$ whose modulus is $\ge 2\gamma$. That is, there is an $\ell \in \{0,1,\cdots ,n(f)-1\}$ such that ${|\Gamma _{k+\ell }|}/{\lambda (f)^{k+\ell }}\ge 2\gamma$. On the other hand, since $\Omega _k/\lambda (f)^k\rightarrow 0$, we can choose N so large that $m>N\Rightarrow |\Omega _m|/\lambda (f)^m<\gamma$.

In all, whenever $m>N$ there is an $\ell \in \{0,1,\cdots ,n(f)-1\}$ such that

$$\begin{aligned} \frac{N(f^{m+\ell })}{\lambda (f)^{m+\ell }}&\ge \frac{|\Gamma _{m+\ell }|}{\lambda (f)^{m+\ell }} -\frac{|\Omega _{m+\ell }|}{\lambda (f)^{m+\ell }} >2\gamma -\gamma =\gamma . \end{aligned}$$

This finishes the proof. $\square$

Corollary 4.6

If $\lambda (f)\ge 1$, then we have

$$\begin{aligned} \limsup _{k\rightarrow \infty }\frac{N(f^k)}{\lambda (f)^k}=\limsup _{k\rightarrow \infty }\frac{\Gamma (f^k)}{\lambda (f)^k}>0,\quad \lim _{k\rightarrow \infty }\frac{\Omega (f^k)}{\lambda (f)^k}=0. \end{aligned}$$

Proposition 4.7

([2, Proposition 2.8]) Let $f:M\rightarrow M$ be a map on an infra-solvmanifold of type $\mathrm {(R)}$ such that $\lambda (f)>1$. Then, for any $\epsilon >0$, there exists N such that if $N(f^m)/\lambda (f)^m\ge \epsilon$ for $m>N$, then the Dold multiplicity $I_m(f)$ satisfies

$$\begin{aligned} |I_m(f)|\ge \frac{\epsilon }{2}\lambda (f)^m. \end{aligned}$$

Proof

From the definition of Dold multiplicity $I_k(f)$, we have

$$\begin{aligned} |I_k(f)|=\Big |\sum _{d\mid k}\mu \left( \frac{k}{d}\right) N(f^d)\Big | \ge N(f^k)-\Big |\sum _{d\mid k, d\ne k}\mu \left( \frac{k}{d}\right) N(f^d)\Big |. \end{aligned}$$

Let C be any number such that $2M(f)\le C$. Then, for any $d>0$

$$\begin{aligned} N(f^d)\le \sum _{i=1}^{r(f)}|\rho _i|\lambda (f)^d\le 2M(f)\lambda (f)^d\le C\lambda (f)^d. \end{aligned}$$

Thus, we have

$$\begin{aligned} |I_k(f)|&\ge N(f^k)-C\sum _{d\mid k, d\ne k}\lambda (f)^d \ge N(f^k)-C\tau (k)\lambda (f)^{k/2}\\&=N(f^k)-C\frac{\tau (k)}{\lambda (f)^{k/2}}\lambda (f)^k \end{aligned}$$

where $\tau (k)$ is the number of divisors of k. Since $\tau (k)\le 2\sqrt{k}$, see [19, Ex 3.2.17], and since $\lambda (f)>1$, we have $\lim _{k\rightarrow \infty }\tau (k)/\lambda (f)^{k/2}=0$, and so there exists an integer N such that $C\tau (k)/\lambda (f)^{k/2}<\epsilon /2$ for all $k>N$. Let $m>N$ such that $N(f^m)/\lambda (f)^m\ge \epsilon$. The above inequality induces the required inequality

$$\begin{aligned} |I_m(f)|&\ge \left( \frac{N(f^m)}{\lambda (f)^m}-C\frac{\tau (m)}{\lambda (f)^{m/2}}\right) \lambda (f)^m \ge \frac{\epsilon }{2}\lambda (f)^m. \end{aligned}$$

$\square$

Theorem 4.5 and Proposition 4.7 imply immediately the following:

Corollary 4.8

Let $f:M\rightarrow M$ be a map on an infra-solvmanifold of type $\mathrm {(R)}$ such that $\lambda (f)>1$. Then, there exist $\gamma >0$ and a natural number N such that if $m\ge N$ then there exists $\ell$ with $0\le \ell \le n(f)-1$ such that $|I_{m+\ell }(f)|/\lambda (f)^{m+\ell }\ge \gamma /2$. In particular $I_{m+\ell }(f)\ne 0$ and so $A_{m+\ell }(f)\ne 0$.

5 Essential periodic orbits

In this section, we shall give an estimate from below the number of essential periodic orbits of maps on infra-solvmanifolds of type $\mathrm {(R)}$.

First of all, we recall the following:

Theorem 5.1

([39], see also [26]) If $f:M\rightarrow M$ is a $C^1$-map on a smooth compact manifold M and $\{L(f^k)\}$ is unbounded, then the set of periodic points of f, $\bigcup _k\mathrm {Fix}(f^k)$, is infinite.

This theorem is not true for continuous maps. Consider the one-point compactification of the map of the complex plane $f(z)=2z^2/||z||$. This is a continuous degree two map of $S^2$ with only two periodic points. But $L(f^k)=2^{k+1}$.

However, when M is an infra-solvmanifold of type $\mathrm {(R)}$, the theorem is true for all continuous maps f on M. In fact, using the averaging formula ([29, Theorem 4.3], [13]), we obtain

$$\begin{aligned} |L(f^k)|\le \frac{1}{\#\Phi }\sum _{A\in \Phi }|\det (I-A_*D_*^k)|=N(f^k). \end{aligned}$$

If $L(f^k)$ is unbounded, then so is $N(f^k)$ and hence the number of essential fixed point classes of all $f^k$ is infinite. In fact, the inequality $|L(f)|\le N(f)$ for any map f on an infra-solvmanifold was proved in [41].

Corollary 5.2

Let f be a map on an infra-solvmanifold of type $\mathrm {(R)}$. Suppose $\{N(f^k)\}$ is unbounded. If every periodic point of f is isolated, then the set of minimal periods of f is infinite.

Proof

By assumption, each $\mathrm {Fix}(f^m)$ consists of isolated points, so $\mathrm {Fix}(f^m)$ and hence $P_m(f)$ are finite. If, in addition, f has finitely many minimal periods then f must have finitely many periodic points. This implies that $\{N(f^k)\}$ is bounded, a contradiction. $\square$

Recall that any map f on an infra-solvmanifold of type $\mathrm {(R)}$ is homotopic to a map $\bar{f}$ induced by an affine map (d, D). By [12, Proposition 9.3], every essential fixed point class of $\bar{f}$ consists of a single element x with index $\mathrm {sign~\!}\det (I-df_x)$. Hence, $N(f)=N(\bar{f})$ is the number of essential fixed point classes of $\bar{f}$. It is a classical fact that a homotopy between f and $\bar{f}$ induces a one-one correspondence between the fixed point classes of f and those of $\bar{f}$, which is index preserving. Consequently, we obtain

$$\begin{aligned} |L(f^k)|\le N(f^k)\le \#\mathrm {Fix}(f^k). \end{aligned}$$

This induces the following conjectural inequality (see [38, 39]) for infra-solvmanifolds of type $\mathrm {(R)}$:

$$\begin{aligned} \limsup _{k\rightarrow \infty }\frac{1}{k}\log |L(f^k)|\le \limsup _{k\rightarrow \infty }\frac{1}{k}\log \#\mathrm {Fix}(f^k). \end{aligned}$$

We denote by $\mathcal {O}(f,k)$ the set of all essential periodic orbits of f with length $\le k$. Thus

$$\begin{aligned} \mathcal {O}(f,k)=\{\langle \mathbb {F}\rangle \mid \mathbb {F}\text { is an essential fixed point class of }f^m\text { with }m\le k\}. \end{aligned}$$

Theorem 5.3

Let f be a map on an infra-solvmanifold of type $\mathrm {(R)}$. Suppose that the sequence $N(f^k)$ is unbounded. Then, there exists a natural number $N_0$ such that

$$\begin{aligned} k\ge N_0\Longrightarrow \#\mathcal {O}(f,k)\ge \frac{k-N_0}{r(f)}. \end{aligned}$$

Proof

As mentioned earlier, we may assume that every essential fixed point class $\mathbb {F}$ of any $f^k$ consists of a single element $\mathbb {F}=\{x\}$. Denote by $\mathrm {Fix}_e(f^k)$ the set of essential fixed point (class) of $f^k$. Thus, $N(f^k)=\#\mathrm {Fix}_e(f^k)$. Recalling also that f acts on the set $\mathrm {Fix}_e(f^k)$ from the proof of [12, Theorem 11.4], we have

$$\begin{aligned} \mathcal {O}(f,k)=\{\langle x\rangle \mid x \text { is a essential periodic point of }f\text { with length }\le k\}. \end{aligned}$$

Observe further that if x is an essential periodic point of f with minimal period p, then $x\in \mathrm {Fix}_e(f^{q})$ if and only if $p\mid q$. The length of the orbit $\langle x\rangle$ of x is p, and

$$\begin{aligned}&\mathrm {Fix}_e(f^k)=\bigcup _{d\mid k}\ \mathrm {Fix}_e(f^d),\\&\mathrm {Fix}_e(f^d)\bigcap \mathrm {Fix}_e(f^{d'})=\mathrm {Fix}_e(f^{\gcd (d,d')}). \end{aligned}$$

Recalling that

$$\begin{aligned} A_m(f)&=\frac{1}{m}\sum _{k\mid m}\mu \left( \frac{m}{k}\right) N(f^k) =\frac{1}{m}\sum _{k\mid m}\mu \left( \frac{m}{k}\right) \#\mathrm {Fix}_e(f^k), \end{aligned}$$

we define $A_m(f,\langle x\rangle )$ for any $x\in \bigcup _i\mathrm {Fix}_e(f^i)$ to be

$$\begin{aligned} A_m(f,\langle x\rangle )&=\frac{1}{m}\sum _{k\mid m}\mu \left( \frac{m}{k}\right) \# \left( \langle x\rangle \cap \mathrm {Fix}_e(f^k)\right) . \end{aligned}$$

Then, we have

$$\begin{aligned} A_m(f)=\sum _{\begin{array}{c} \langle x\rangle \\ x\in \mathrm {Fix}_e(f^m) \end{array}} A_m(f,\langle x \rangle ). \end{aligned}$$

We begin with new notation. For a given integer $k>0$ and $x\in \bigcup _m\mathrm {Fix}_e(f^m)$, let

$$\begin{aligned} \mathcal {A}(f,k)= & {} \left\{ m\le k\mid A_m(f)\ne 0\right\} ,\\ \mathcal {A}(f,\langle x\rangle )= & {} \left\{ m\mid A_m(f,\langle x\rangle )\ne 0\right\} . \end{aligned}$$

Remark that if $A_m(f)\ne 0$ then there exists an essential periodic point x of f with period m such that $A_m(f,\langle x\rangle )\ne 0$. Consequently, we have

$$\begin{aligned} \mathcal {A}(f,k)\subset \bigcup _{\langle x\rangle \in \mathcal {O}(f,k)} \mathcal {A}(f,\langle x\rangle ) \end{aligned}$$

Since $N(f^k)$ is unbounded, we have that $\lambda (f)>1$, see the observation just above Theorem 3.2. By Corollary 4.8, there is $N_0$ such that if $n\ge N_0$ then there is i with $n\le i\le n+n(f)-1$ such that $A_i(f)\ne 0$. This leads to the estimate

$$\begin{aligned} \#\mathcal {A}(f,k)\ge \frac{k-N_0}{n(f)}\quad \forall k\ge N_0. \end{aligned}$$

Assume that x has minimal period p. Then, we have

$$\begin{aligned} A_m(f,\langle x\rangle )&=\frac{1}{m}\sum _{p\mid n\mid m}\mu \left( \frac{m}{n}\right) \#\langle x\rangle =\frac{p}{m}\sum _{p\mid n\mid m}\mu \left( \frac{m}{n}\right) . \end{aligned}$$

Thus, if m is not a multiple of p then by definition $A_m(f,\langle x\rangle )=0$. It is clear that $A_p(f,\langle x\rangle )=\mu (1)=1$, i.e., $p\in \mathcal {A}(f,\langle x\rangle )$. Because $p\mid n\mid rp\Leftrightarrow n=r'p$ with $r'\mid r$, we have

$$\begin{aligned} A_{rp}(f,\langle x\rangle )=\frac{1}{r} \sum _{p\mid n\mid rp}\mu \left( \frac{rp}{n}\right) =\frac{1}{r}\sum _{r'\mid r}\mu \left( \frac{r}{r'}\right) \end{aligned}$$

which is 0 when and only when $r>1$. Consequently, $\mathcal {A}(f,\langle x\rangle )=\{p\}$.

In all, we obtain the required inequality

$$\begin{aligned} \frac{k-N_0}{r(f)}&\le \#\mathcal {A}(f,k)\le \#\mathcal {O}(f,k). \end{aligned}$$

$\square$

We consider the set of periodic points of f with minimal period k

$$\begin{aligned} P_k(f)=\mathrm {Fix}(f^k)-\bigcup _{d\mid k, d<k}\mathrm {Fix}(f^d). \end{aligned}$$

It is clear that $\mathrm {Fix}(f)\subset \mathrm {Fix}(f^2)$, i.e., any fixed point class of f is naturally contained in a unique fixed point class of $f^2$. It is also known that $\mathrm {Fix}_e(f)\subset \mathrm {Fix}_e(f^2)$. We define

$$\begin{aligned} \mathrm {E}P_k(f)=\mathrm {Fix}_e(f^k)-\bigcup _{d\mid k, d<k}\mathrm {Fix}_e(f^d), \end{aligned}$$

the set of essential periodic points of f with minimal period k. Because

$$\begin{aligned} \mathrm {Fix}_e(f^k)=\coprod _{d\mid k}\mathrm {E}P_d(f), \end{aligned}$$

we have

$$\begin{aligned} N(f^k)=\#\mathrm {Fix}_e(f^k)=\sum _{d\mid k}\# \mathrm {E}P_d(f). \end{aligned}$$

Proposition 5.4

For every $k>0$, we have

$$\begin{aligned} \# \mathrm {E}P_k(f)=\sum _{d\mid k}\mu \left( \frac{k}{d}\right) N(f^d)=I_k(f). \end{aligned}$$

In particular, if $I_k(f)\ne 0$ then $N(f^k)\ne 0$.

Proof

We apply the Möbius inversion formula to the above identity and then we obtain $\# \mathrm {E}P_k(f)=\sum _{d\mid k}\mu \left( \frac{k}{d}\right) N(f^d)$, which is exactly $I_k(f)$ by its definition. $\square$

Definition 5.5

We consider the mod 2 reduction of the Nielsen number $N(f^k)$ of $f^k$, written $N^{(2)}(f^k)$. A positive integer k is a $N^{(2)}$-period of f if $N^{(2)}(f^{k+i})=N^{(2)}(f^i)$ for all $i\ge 1$. We denote the minimal $N^{(2)}$-period of f by $\alpha ^{(2)}(f)$.

Proposition 5.6

([35, Proposition 1]). Let p be a prime number and let A be a square matrix with entries in the field $\mathbb {F}_p$. Then, there exists k with $(p,k)=1$ such that

$$\begin{aligned} \mathrm {tr\,}A^{k+i}=\mathrm {tr\,}A^i \end{aligned}$$

for all $i\ge 1$.

Recalling (N2): $N(f^k)=\mathrm {tr\,}M_+^k-\mathrm {tr\,}M_-^k=\mathrm {tr\,}(M_+\oplus (-M_-))^k$, we can see easily that the minimal $N^{(2)}$-period $\alpha ^{(2)}(f)$ always exists and must be an odd number.

Now, we obtain a result which resembles [35, Theorem 2].

Theorem 5.7

Let f be a map on an infra-solvmanifold of type $\mathrm {(R)}$. Let $k>0$ be an odd number. Suppose that $\alpha ^{(2)}(f)^2\mid k$ or $p\mid k$ where p is a prime such that $p\equiv 2^i\mod \alpha ^{(2)}(f)$ for some $i\ge 0$. Then

$$\begin{aligned} \#\{\langle x\rangle \mid x\in \mathrm {E}P_k(f)\}=\# \mathrm {E}P_k(f)/k \end{aligned}$$

is even.

Proof

By Proposition 5.4, $\# \mathrm {E}P_k(f)=I_k(f)$. Hence, it is sufficient to show that $I_k(f)$ is even.

Let $\alpha =\alpha ^{(2)}(f)$. Consider the case where $\alpha ^2\mid k$. If $d\mid k$ and $\mu (k/d)\ne 0$, then it follows that $\alpha \mid d$. By the definition of $\alpha$, $N^{(2)}(f^d)=N^{(2)}(f^\alpha )$. This induces that

$$\begin{aligned} I_k(f)&\equiv \sum _{d\mid k}\mu \left( \frac{k}{d}\right) N^{(2)}(f^d) =N^{(2)}(f^\alpha )\sum _{d\mid k}\mu \left( \frac{k}{d}\right) =0 \mod 2. \end{aligned}$$

Assume p is a prime such that $p\mid k$ and $p\equiv 2^i \!\mod \alpha$ for some $i\ge 0$. Write $k=p^jr$ where $(p,r)=1$. Then

$$\begin{aligned} I_k(f)&=\sum _{d\mid r}\mu \left( \frac{r}{d}\right) \left( \sum _{e\mid p^j}\mu \left( \frac{p^j}{e}\right) N((f^d)^e)\right) \\&=\sum _{d\mid r}\mu \left( \frac{r}{d}\right) \left( \mu \left( 1\right) N((f^d)^{p^j})+\mu \left( p\right) N((f^d)^{p^{j-1}})\right) \\&=I_r(f^{p^j})-I_r(f^{p^{j-1}}). \end{aligned}$$

Since $\alpha$ is a $N^{(2)}$-period of f, it follows that the sequence $\{I_r(f^i)\}_i$ is $\alpha$-periodic in its $\mod 2$ reduction, i.e., $I_r(f^{j+\alpha })\equiv I_r(f^i)\mod 2$ for all $j\ge 1$. Since $p\equiv 2^i\mod \alpha$, we have $I_r(f^{p^s})\equiv I_r(f^{2^{is}})\mod 2$ for all $s\ge 0$. Recall [5, Proposition 5]: For any square matrix B with entries in the field $\mathbb {F}_p$ and for any $j\ge 0$, we have $\mathrm {tr\,}B^{p^j}=\mathrm {tr\,}B$. Due to this result, we obtain

$$\begin{aligned} N^{(2)}(f^{2^j})\equiv \mathrm {tr\,}(M_+\oplus (-M_-))^{2^j}\equiv \mathrm {tr\,}(M_+\oplus (-M_-))\equiv N^{(2)}(f) \mod 2 \end{aligned}$$

and it follows that $I_r(f^{2^{is}})\equiv I_r(f)\mod 2$. Consequently, we have

$$\begin{aligned} I_k(f)=I_r(f^{p^j})-I_r(f^{p^{j-1}}) \equiv I_r(f^{2^{ij}})-I_r(f^{2^{i(j-1)}}) \equiv I_r(f)-I_r(f)=0 \mod 2. \end{aligned}$$

This finishes the proof. $\square$

6 Homotopy minimal periods

In this section, we study (homotopy) minimal periods of maps f on infra-solvmanifolds of type $\mathrm {(R)}$. We want to determine $\mathrm {HPer}(f)$ only from the knowledge of the sequence $\{N(f^k)\}$. This approach was used in [1, 14, 23] for maps on tori, in [18,19,20,21, 31, 32] for maps on nilmanifolds and some solvmanifolds, and in [28, 30] for expanding maps on infra-nilmanifolds.

Theorem 6.1

Let f be a map on an infra-solvmanifold of type $\mathrm {(R)}$ with $\lambda (f)>1$. If the sequence $\{N(f^k)/\lambda (f)^k\}$ is asymptotically periodic, then there exist an integer $m>0$ and an infinite sequence $\{p_i\}$ of primes such that $\{mp_i\}\subset \mathrm {Per}(f)$. Furthermore, $\{mp_i\}\subset \mathrm {HPer}(f)$.

Proof

For simplicity, we denote $\tilde{N}(f^k)=\Gamma (f^k)/\lambda (f)^k$. By Corollary 4.6, $\lambda (f)\ge 1$ implies that $\limsup N(f^k)/\lambda (f)^k=\limsup \tilde{N}(f^k)>0$. We consider the condition that the sequence $\{\tilde{N}(f^k)\}$ is periodic. By Theorem 4.1, we can choose q such that the sequence $\{\tilde{N}(f^k)\}$ is q-periodic and nonzero. Consequently, there exists m with $1\le m\le q$ such that $\tilde{N}(f^m)\ne 0$.

Let $h=f^m$. Then, $\lambda (h)=\lambda (f^m)=\lambda (f)^m\ge 1$. The periodicity $\tilde{N}(f^{m+\ell q})=\tilde{N}(f^m)$ induces $\tilde{N}(h^{1+\ell q})=\tilde{N}(h)$ for all $\ell >0$. By Corollary 4.6 again, we can see that there exists $\gamma >0$ such that $N(h^{1+\ell q})\ge \gamma \lambda (h)^{1+\ell q}>0$ for all $\ell$ sufficiently large. Since $\lambda (f)>1$, we have $\lambda (h)>1$ and it follows from Proposition 4.7 that the Dold multiplicity $I_{1+\ell q}(h)$ satisfies $|I_{1+\ell q}(h)|\ge (\gamma /2)\lambda (h)^{1+\ell q}$ when $\ell$ is sufficiently large.

According to Dirichlet prime number theorem, since $(1,q)=1$, there are infinitely many primes p of the form $1+\ell q$. Consider all primes $p_i$ satisfying $|I_{p_i}(h)|\ge (\gamma /2)\lambda (h)^{p_i}$. Remark that for any prime number p,

$$\begin{aligned} I_p(h)&=\sum _{d\mid p}\mu \left( \frac{p}{d}\right) N(h^d)=\mu (p)N(h)+\mu (1)N(h^p)=N(h^p)-N(h)\\&=\#\mathrm {Fix}_e(h^p)-\#\mathrm {Fix}_e(h)=\# \left( \mathrm {Fix}_e(h^p)-\mathrm {Fix}_e(h)\right) \end{aligned}$$

where the last identity follows from that fact that $\mathrm {Fix}_e(h)\subset \mathrm {Fix}_e(h^p)$. Since p is a prime, the set $\mathrm {Fix}_e(h^p)-\mathrm {Fix}_e(h)$ consists of essential periodic points of h with minimal period p.

Because $|I_{p_i}(h)|>0$, each $p_i$ is the minimal period of some essential periodic point of h. Thus, $mp_i$ is a period of f. This means that $m_ip_i$ is the minimal period of f for some $m_i$ with $m_i\mid m$. Choose a subsequence $\{m_{i_k}\}$ of the sequence $\{m_i\}$ bounded by m which is constant, say $m_0$. Consequently, the infinite sequence $\{m_0p_{i_k}\}$ consists of minimal periods of f, or $\{m_0p_i\}\subset \mathrm {Per}(f)$.

These arguments also work for all maps homotopic to f. Hence $\{m_0p_i\}\subset \mathrm {HPer}(f)$, which completes the proof.

We next consider the condition that the sequence $\{\tilde{N}(f^k)\}$ contains an interval. This means by Theorem 4.1 that the set of limit points of the sequence $\{N(f^k)/\lambda (f)^k\}$ contains an interval. $\square$

The following example shows that the condition $\lambda (f)>1$ in Theorem 6.1 is essential. This condition is equivalent to the unboundedness of the sequence $\{N(f^k)\}$ by Theorem 3.2.

Example 6.2

Consider the map f on $T^2$ induced by the matrix

$$\begin{aligned} D=\left[ \begin{matrix}0&{}1\\ -1&{}-1\end{matrix}\right] . \end{aligned}$$

We have observed in Example 4.2 that

$$\begin{aligned} \{N(f^k)\}= & {} \{2-(\omega ^k+\bar{\omega }^k)\}\ \text { is bounded where } \omega =(-1+\sqrt{3}i)/2,\\ \lambda (f)= & {} \max \{|\omega |,|\bar{\omega }|,1\}=1,\\ \{\tilde{N}(f^k)\}= & {} \{2-(\omega ^k+\bar{\omega }^k)\}\ \text { is } 3\text {-periodic.} \end{aligned}$$

Observe also that since $f^3=\mathrm {id}$ we have $\mathrm {Per}(f)\subset \{1,2,3\}$. In fact, we can see that $\mathrm {Per}(f)=\{1,3\}$.

In the proof of Theorem 6.1, we have shown the following, which proves that the algebraic period is a homotopy minimal period when it is a prime number.

Corollary 6.3

Let f be a map on an infra-solvmanifold of type $\mathrm {(R)}$. For any prime p, if $A_p(f)\ne 0$ then $p\in \mathrm {HPer}(f)$.

Corollary 6.4

Let f be a map on an infra-solvmanifold of type $\mathrm {(R)}$ with $\lambda (f)>1$. If the sequence $\{N(f^k)\}$ is eventually monotone increasing, then there exists N such that the set $\mathrm {HPer}(f)$ contains all primes larger than N.

Proof

Since $\lambda (f)>1$, by Theorem 4.5 there exist $\gamma >0$ and N such that if $k>N$ then there exists $\ell =\ell (k)<r(f)$ such that $N(f^{k-\ell })/\lambda (f)^{k-\ell }>\gamma$. Then for all sufficiently large k, the monotonicity induces

$$\begin{aligned} \frac{N(f^k)}{\lambda (f)^k}\ge \frac{N(f^{k-\ell })}{\lambda (f)^k}=\frac{N(f^{k-\ell })}{\lambda (f)^{k-\ell }\lambda (f)^\ell } \ge \frac{\gamma }{\lambda (f)^\ell }\ge \frac{\gamma }{\lambda (f)^{r(f)}}. \end{aligned}$$

Applying Proposition 4.7 with $\epsilon =\gamma /\lambda (f)^{r(f)}$, we see that $I_k(f)\ne 0$ and so $A_k(f)\ne 0$ for all k sufficiently large. Now our assertion follows from Corollary 6.3. $\square$

We next recall the following:

Definition 6.5

A map $f:M\rightarrow M$ is essentially reducible if any fixed point class of $f^k$ being contained in an essential fixed point class of $f^{kn}$ is essential, for any positive integers k and n. The space M is essentially reducible if every map on M is essentially reducible.

Lemma 6.6

([1, Proposition 2.2]). Let $f:M\rightarrow M$ be an essentially reducible map. If

$$\begin{aligned} \sum _{\frac{m}{k}:{ \mathrm{prime}}} N(f^k)< N(f^m), \end{aligned}$$

then any map which is homotopic to f has a periodic point with minimal period m, i.e., $m\in \mathrm {HPer}(f)$.

Lemma 6.7

Every infra-solvmanifold of type $\mathrm {(R)}$ is essentially reducible.

Proof

Let $f:M\rightarrow M$ be a map on an infra-solvmanifold $M=\Pi \backslash {S}$ of type $\mathrm {(R)}$. Then, $\Pi$ fits a short exact sequence

where $\Gamma =\Pi \cap S$ and the holonomy group $\Phi$ of $\Pi$ naturally sits in $\mathrm {Aut}(S)$. By [29, Lemma 2.1], we know that $\Pi$ has a fully invariant subgroup $\Lambda$ of finite index and $\Lambda \subset \Gamma$. Therefore, $\Lambda \subset \Gamma \subset S$ and $\bar{M}=\Lambda \backslash {S}$ is a special solvmanifold which covers M. Since $\Lambda$ is a fully invariant subgroup of $\Pi$, it follows that any map $f:M\rightarrow M$ has a lifting $\bar{f}:\bar{M}\rightarrow \bar{M}$, and $\bar{M}$ is a regular covering of M. By [15, Corollary 4.5], $\bar{f}$ is essentially reducible and then by [30, Proposition 2.4], f is essentially reducible. $\square$

We can not only extend but also strengthen Corollary 6.4 as follows:

Proposition 6.8

Let f be a map on an infra-solvmanifold of type $\mathrm {(R)}$. Suppose that the sequence $\{N(f^k)\}$ is strictly monotone increasing. Then:

• (1) All primes belong to $\mathrm {HPer}(f)$.

• (2) There exists N such that if p is a prime $>N$ then $\{p^n\mid n\in \mathbb {N}\}\subset \mathrm {HPer}(f)$.

Proof

Observe that for any prime p

$$\begin{aligned} N(f^p)-\sum _{\frac{p}{k}:{ \mathrm{prime}}} N(f^k)=N(f^p)-N(f)=I_p(f). \end{aligned}$$

The strict monotonicity implies $A_p(f)=pI_p(f)>0$ and, hence, $p\in \mathrm {HPer}(f)$ by Corollary 6.3. This proves (1).

The strict monotonicity of $\{N(f^k)\}$ implies that $\lambda (f)>1$. Under this assumption, we have shown in the proof of Corollary 6.4 that there exists N such that $k>N\Rightarrow I_k(f)>0$. Let p be a prime $>N$ and $n\in \mathbb {N}$. Then

$$\begin{aligned}&N(f^{p^n})-\sum _{\frac{p^n}{k}:{ \mathrm{prime}}} N(f^k) =\sum _{i=0}^n I_{p^{i}}(f)-N(f^{p^{n-1}})=I_{p^n}(f)>0. \end{aligned}$$

By Lemma 6.6, we have $p^n\in \mathrm {HPer}(f)$, which proves (2). $\square$

For the set of algebraic periods $A(f)=\{m\in \mathbb {N}\mid A_m(f)\ne 0\}$, its lower density $\mathrm {DA}(f)$ was introduced in [19, Remark 3.1.60]:

$$\begin{aligned} \mathrm {DA}(f)=\liminf _{k\rightarrow \infty }\frac{\#(A(f)\cap [1,k])}{k}. \end{aligned}$$

We can consider as well the lower densities of $\mathrm {Per}(f)$ and $\mathrm {HPer}(f)$, see also [16]:

$$\begin{aligned}&\mathrm {DP}(f)=\liminf _{k\rightarrow \infty }\frac{\#(\mathrm {Per}(f)\cap [1,k])}{k},\\&\mathrm {DH}(f)=\liminf _{k\rightarrow \infty }\frac{\#(\mathrm {HPer}(f)\cap [1,k])}{k}. \end{aligned}$$

Since $I_k(f)=\#\mathrm {E}P_k(f)$ by Proposition 5.4, it follows that $\mathcal {A}(f)\subset \mathrm {HPer}(f)\subset \mathrm {Per}(f)$. Hence, we have $\mathrm {DA}(f)\le \mathrm {DH}(f)\le \mathrm {DP}(f)$.

By Corollary 4.8, when $\lambda (f)>1$, we have a natural number N such that if $m\ge N$ then there is $\ell$ with $0\le \ell <n(f)$ such that $A_{m+\ell }(f)\ne 0$. This shows that $\mathrm {DA}(f)\ge 1/n(f)$.

On the other hand, by Theorem 4.1, we can obtain the following: If $\lambda (f)=0$ then $N(f^k)=0$ and $A_k(f)=0$ for all k, which shows that $\mathrm {DA}(f)=0$. Consider first Case (2), i.e., the sequence $\{N(f^k)/\lambda (f)^k\}$ is asymptotically a periodic and nonzero sequence $\{\sum _{j=1}^{n(f)}\rho _je^{2i\pi (k\theta _j)}\}$ of some period q. Now from the identity (2.2), it follows that $\mathrm {DA}(f)\ge 1/q$. Finally consider Case (3). Then, the sequence $\{N(f^k)/\lambda (f)^k\}$ asymptotically has a subsequence $\{\sum _{j\in \mathcal {S}}\rho _je^{2i\pi (k\theta _j)}\}$ where $\mathcal {S}=\{j_1,\cdots ,j_s\}$ and $\{\theta _{j_1},\cdots ,\theta _{j_s},1\}$ is linearly independent over the integers. Therefore by [6, Theorem 6, p.  91], the sequence $(k\theta _{j_1},\cdots ,k\theta _{j_s})$ is uniformly distributed in $T^{|\mathcal {S}|}$. From the identity (2.2), it follows that $\mathrm {DA}(f)=1$ (see [19, Remark 3.1.60]).

Theorem 6.1 studies the homotopy minimal periods for maps of Case (2) in Theorem 4.1. Now we can state immediately the following result for maps of Case (3) in Theorem 4.1.

Corollary 6.9

Let f be a map on an infra-solvmanifold of type $\mathrm {(R)}$. Suppose that the sequence $\{N(f^k)/\lambda (f)^k\}$ asymptotically contains an interval. Then, $\mathrm {DA}(f)=\mathrm {DH}(f)=\mathrm {DP}(f)=1$.

Corollary 6.10

Let f be a map on an infra-solvmanifold of type $\mathrm {(R)}$. Suppose that the sequence $\{N(f^k)\}$ is unbounded and eventually monotone increasing. Then, $\mathrm {HPer}(f)$ is cofinite and $\mathrm {DA}(f)=\mathrm {DH}(f)=\mathrm {DP}(f)=1$.

Proof

Under the same assumption, we have shown in the proof of Corollary 6.4 that there exists N such that if $k>N$ then $I_k(f)>0$. This means $\mathrm {E}P_k(f)$ is nonempty by Proposition 5.4 and hence $k\in \mathrm {HPer}(f)$. $\square$

Now we can prove the main result of [30].

Corollary 6.11

([28, Theorem 4.6], [30, Theorem 3.2]). Let f be an expanding map on an infra-nilmanifold. Then, $\mathrm {HPer}(f)$ is cofinite.

Proof

Since f is expanding, we have that $\lambda (f)=\mathrm {sp}(\bigwedge D_*)>1$. For any $k>0$, we can write as before $N(f^k)=\Gamma (f^k)+\Omega (f^k)$ so that $\Omega (f^k)\rightarrow 0$ and $\Gamma (f^k)\rightarrow \infty$ as $k\rightarrow \infty$. This implies that $N(f^k)$ is eventually monotone increasing. The assertion follows from Corollary 6.10. $\square$