In this post, we discuss the density of closed orbits of a prescribed length for an SFT and a suspension flow over an SFT.

Main motivation

Let $\sigma : X_A \rightarrow X_A$ be an SFT (see the next section for the precise definition). It is well-known that the collection of periodic points $P$ for $\sigma$ is dense in the space. In this post, we are interested in quantifying that rate of density. For a compact metric space $(X,d)$, we will say that a subset $U \subseteq X$ is $\eta$-dense if for each $x \in X$, we have $d(x, U) \leq \eta$ (in other words, the set $U$ approximates all of $X$ at scale $\eta$). The goal is to show that there are constants $C, \delta > 0$ such that if we look at the collection of periodic points with period at most $n$, then it is $\eta$-dense in the entire space, where $\eta \leq C e^{-n\delta}$. Thus, not only is $P$ dense, but it becomes exponentially dense as we look at more and more closed orbits. I’ll refer to this as the effective density of closed orbits. Furthermore, we will also show that the same holds true for suspension flows.

Preliminaries

Perron-Frobenius theorem

We briefly recall the famous Perron-Frobenius theorem.

Theorem: If $A$ is a $k \times k$ real-matrix with entries $A(i,j) > 0$, then

1. there is a unique $1$-dimensional eigenspace $E$ such that $E \cap {(x_1, \ldots, x_n) \in \mathbb{R}^n : x_i > 0} \neq \varnothing$, and
2. the corresponding eigenvalue $\lambda_E$ is positive and satisfies the condition that $\vert \lambda_E \vert > \vert \lambda \vert$ for all other eigenvalues $\lambda$.

There is a nice proof by Kohlberg and Pratt which we sketch without details below. First, we recall the contracting mapping principle.

Theorem: Let $0 < C < 1$ and let $f : X \rightarrow X$ be a mapping of a compact metric space such that $d(f(x), f(y)) \leq C d(x,y)$ for all $x,y \in X$. There exists a unique $x \in X$ such that $f(x) = x$.

Remark: If a map $f$ satisfies the assumption of this theorem, then we call it a contracting map.

The aim is to construct this eigenspace using the contracting mapping principle. To that end, let

\[\Delta_1^+ := \left\{(x_1, \ldots, x_n) \in \mathbb{R}^n \mid \sum_{i=1}^n x_i = 1 \text{ and } x_i \geq 0\right\},\]

and

\[B := \{ (x_1, \ldots, x_n) \in \mathbb{R}^n \mid x_i \geq 0 \text{ and } x_i = 0 \text{ for some } 1 \leq i \leq n\}.\]

Observation: Given $v, w$ in the interior of $\Delta_1^+$, there is a unique $v’, w’ \in B$ such that $v, v’, w, w’$ are all collinear, $v’$ is the closest to $v$, and $w’$ is the closest to $w$.

We can use this observation to define the Hilbert metric on $\Delta_1^+$ by

\[d(v,w) := \log \frac{\|w - v'\| \cdot \|v - w'\|}{\|v - v'\| \cdot \|w - w'\|}.\]

Observation: The Hilbert metric is actually an honest metric.

Finally, we observe the following (which is essentially due to compactness).

Lemma: If $f : \Delta_1^+ \rightarrow \Delta_1^+$ is a continuous map such that $\overline{f(\Delta_1^+)} \subseteq \text{Int}(\Delta_1^+)$, then there is a uniform $C < 1$ such that $d(f(v), f(w)) \leq C d(v,w)$ for all $v,w \in \Delta_1^+$.

With this, we now have the tools to prove the Ruelle-Perron theorem.

Sketch of Proof: We will prove the result assuming $A$ is symmetric. Consider the map

\[\hat{A} : \Delta_1^+ \rightarrow \Delta_1^+, \quad \hat{A}v := \frac{Av}{\|Av\|_1}.\]

Since $A$ has all positive entries, we have $\overline{\hat{A}(\Delta_1^+)} \subset \text{Int}(\Delta_1^+)$, so by the previous lemma it is a contracting map. By the contracting mapping principle, there must be a unique fixed point. This unique fixed point corresponds to a subspace $E$, and so this gives us an eigenspace for $A$. Let $\lambda_E$ be the corresponding eigenvalue. We now finish the proof in four steps. Let $v \in E$ be an eigenvector with all positive entries.

Step 1. We need to show that if $w$ is also an eigenvector for $A$ corresponding to $\lambda_E$, then $w \in E$. If $w$ is such an eigenvector and not in $E$, then there must be (at least) one entry of $w$ which is not strictly positive. Consider now the vector $z = v + tw$ for $t \in \mathbb{R}$. Since these are both eigenvectors, we have that $z$ is an eigenvector. Furthermore, if we choose $t$ small enough, then $z$ will have all positive entries, and hence must lie in $E$ by the contracting mapping principle. Since $E$ is a subspace, this forces $w$ to lie in $E$, a contradiction. Thus, $\lambda_E$ is a simple eigenvalue.

Step 2. Let’s first show that for any other eigenvalue $\lambda$, we have $\vert \lambda_E \vert \geq \vert \lambda \vert$. Let $w$ be such that $Aw = \lambda w$. Notice that

\[\lambda w_i = (Aw)_i = \sum_{j=1}^k A(i,j) w_j.\]

Now taking the absolute value of both sides, we have

\[\vert \lambda \vert \vert w_i \vert \leq \sum_{j=1}^k A(i,j) \vert w_j \vert.\]

Let $z = (\vert w_1 \vert, \ldots, \vert w_k \vert)^T$. This shows us that for each $i$, we have $\vert \lambda \vert z_i \leq (A z)_i$. Let $v$ be the left eigenvector corresponding to $\lambda_E$ with all positive entries. We see that for each $j$ we have

\[\vert \lambda \vert z_i \cdot v_i \leq \sum_{j=1}^k A(i,j) z_j \cdot v_i.\]

Summing over $i$, we get

\[\vert \lambda \vert v z \leq vAz = \lambda_E vz.\]

Step 3. We will show that if $\vert \lambda \vert = \lambda_E$, then $z$ (which is the vector whose components are the absolute value of the components of $w$) must lie in $E$. Notice that

\(0 = vAz - \vert \lambda \vert vz = v \left(Az - \vert \lambda \vert z \right).\) Since the entries of $v$ are all positive, we must have $Az = \vert \lambda \vert z$. If $\vert \lambda \vert = \lambda_E$, then this implies that $z \in E$. Since $A$ is symmetric, this shows us that $z = C v^T$ for some $C > 0$.

Step 4. We now need to show that if $\vert \lambda \vert = \lambda_E$, then $\lambda = \lambda_E$. In the previous step, we had shown that $z \in E$. Using Cauchy-Schwarz, we have

\[\left\vert \sum_{j=1}^k w_j v_j\right\vert^2 \leq \left(\sum_{j=1}^k v_j^2\right) \left( \sum_{j=1}^k z_j^2 \right),\]

with equality if and only if $v$ and $w$ are linearly dependent. Thus, we aim to show that

\[\left\vert \sum_{j=1}^k w_j v_j\right\vert = C\left(\sum_{j=1}^k v_j^2\right),\]

Observe that

\[\sum_{j=1}^k w_j v_j = v w = \lambda_E^{-1} vAw = \lambda_E^{-1} \sum_{i=1}^k \sum_{j=1}^k v_iA(i,j) w_j.\]

Taking the absolute value,

\[\left\vert \sum_{j=1}^k w_j v_j \right\vert = \lambda_E^{-1} \left\vert \sum_{i=1}^k v_i \sum_{j=1}^k A(i,j)w_j \right\vert.\]

Now, using previous step, we know that

\[\sum_{j=1}^k A(i,j) \vert w_j \vert = \sum_{j=1}^k A(i,j) z_j = \lambda_E z_i =\vert \lambda \vert z_i = \vert (Aw)_i \vert = \left \vert \sum_{j=1}^k A(i,j) w_j \right\vert,\]

so multiplying by $v_i,$ using the $v_i$ are positive, and adding yields

\[\left\vert \sum_{j=1}^k w_j v_j \right\vert = \lambda_E^{-1} \left\vert \sum_{i=1}^k v_i \sum_{j=1}^k A(i,j)w_j \right\vert = \lambda_E^{-1} \sum_{i=1}^k v_i \sum_{j=1}^k A(i,j) z_j = \lambda_E^{-1} v A z = vz = C \sum_{i=1}^k v_i^2,\]

as desired. This shows that $w$ must lie in $E$, and so $\lambda = \lambda_E$. $\square$

Remark: The steps are mostly the same for the non-symmetric case, but one has to take more care in Steps 2, 3, and 4.

Shifts of finite type

We briefly recall the set up for symbolic dynamics and what an SFT is. Many of the details follow the first chapter of Parry and Pollicott’s book as well as the beautiful notes by Lalley.

We will use the notation $\mathbb{N} := {1, \ldots}$, and we let $X := {1, \ldots, k}^\mathbb{N}$. Given $x \in X$, we will write $x = (x_1, \ldots)$ for the projections onto the respective spaces. The shift map is given by

\[\sigma : X \rightarrow X, \quad \sigma(x) = y, \text{ where } y_i = x_{i+1}.\]

Recall that if we give ${1, \ldots, k}$ the discrete topology and $X$ the product topology, then the product topology on $X$ is generated by the collection of all sets of the form

\[[x_1, \ldots, x_n] := \{y \in X : y_i = x_i \text{ for } 1 \leq i \leq n \}.\]

We call these sets the cylinders, and the length of the cylinder is the number of coordinates that are being fixed.

We now recall how this can be connected to a metric. For $\theta \in (0,1)$, we adopt the convention $\theta^\infty = 0$. Given $x, y \in X$, we let $n(x,y) := \min{n \in \mathbb{N} \cup {\infty} \mid x_n \neq y_n}$, and for $\theta \in (0,1)$ we can define a metric by $d_\theta(x,y) = \theta^{n(x,y)}$. It is easy to check that this is a metric, and furthermore the topology generated by this metric coincides with the topology on $X$.

Let $A$ be a $k \times k$ matrix whose $(i,j$)-entry is given by $A(i,j) \in {0,1}$. We say that the matrix $A$ is irreducible if for each $(i,j)$, there is an $m \geq 1$ such that $A^{m}(i,j) > 0$. As an example, the matrix

\[A = \begin{pmatrix}1 & 0 \\ 1 & 1 \end{pmatrix}\]

will be irreducible, while the matrix

\[A = \begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix}\]

is clearly not irreducible.

Standing assumption 1: All matrices $A$ will be irreducible.

Let

\[X_A := \{x \in X : A(x_i, x_{i+1}) = 1 \text{ for } i \geq 0\}.\]

It is easy to see that $X_A$ is invariant for the shift map, and so from now on we will work with the shift on the level of $X_A$, say, $\sigma : X_A \rightarrow X_A$.

One may also be interested in mixing properties of the shift map, which leads us to the notion of period. The period map for $A$ is the map \(p : \{1, \ldots, k\} \rightarrow \mathbb{N}, \quad p(i) := \text{GCD}(\{n \in \mathbb{N} : A^n(i,i) > 0\}).\) We will say that $A$ is aperiodic if the greatest common divisor of ${p(1), \ldots, p(k)}$ is $1$, and one can show that this is equivalent to checking that $\text{GCD}({n \in \mathbb{N} : A^n(i,i) > 0 \text{ and } i \in {1,\ldots, k}}) = 1$.

While not every matrix is aperiodic, one can decompose $X_A$ into a finite union of invariant subsets along which the (corresponding) matrix is aperiodic, and so the next standing assumption is not so outrageous.

Standing assumption 2: All matrices $A$ will be aperiodic.

From now on, given $A$ satisfying our standing assumptions, we say that $\sigma : X_A \rightarrow X_A$ is a shift of finite type, or SFT for short. Note that not all cylinders are non-empty anymore; for example, if $A(x_0, x_1) = 0$, then the cylinder $[x_0, x_1] = \varnothing$. We write $\mathcal{C}(n)$ for all cylinders of length $n$ which are non-trivial. The topological entropy of the SFT is then given by

\[h := \lim_{n \rightarrow \infty} \frac{1}{n} \log(\# \mathcal{C}(n)).\]

Remark: Notice that

\[\# \mathcal{C}(n) = \sum_{1 \leq i,j \leq k} A^{n-1}(i,j) = 1^T A^{n-1} 1,\]

where $1$ is the vector of all ones. Using the Perron-Frobenius theorem, we have that

\[\# \mathcal{C}(n) \sim c \lambda^{n-1},\]

where $\lambda$ is the dominant eigenvalue. This shows that this limit exists, and, in fact, the entropy corresponds to the dominant eigenvalue coming from the Perron-Frobenius theorem.

Moreover, we will say that a finite sequence $(x_1, \ldots, x_n)$ is admissible if $[x_1, \ldots, x_n] \in \mathcal{C}(n)$, or, in other words, $A(x_i, x_{i+1}) = 1$ for all $1 \leq i < n$.

One key property that we will need is the specification property, which we describe in the following lemma.

Lemma: There is a uniform constant $M > 0$ such that $A^M(i,j) > 0$ for all $i,j \in {1, \ldots, k}$.

In particular, this has the following (important) consequence.

Observation: Given finite sequences $(x_1, \ldots, x_n)$ and any symbol $z \in {1, \ldots, k}$, there is a finite sequence $(x_{n+1}, \ldots, x_{n+R})$, where $R \leq M$, such that $(x_1, \ldots, x_{n+R}, z)$ is admissible.

Finally, we set

\[P(n) := \bigcup_{k \leq n} \text{Fix}(\sigma^k), \text{ and } P := \bigcup_{n \geq 0} P(n).\]

Remark: We will not take the time to write it in detail, but essentially all of the same details and definitions hold if we replace $X$ with the set ${1, \ldots, k}^{\mathbb{Z}}$. It will be clear from context which set we decide to work with.

Suspension flows

A function $r : X_A \rightarrow \mathbb{R}$ is $\theta$-Lipschitz if there is a constant $C > 0$ such that we have $d_\theta(r(x), r(y)) \leq C d_\theta(x,y)$ for all $x,y \in X_A$. We also adopt the notation

\[r^n(x) := \sum_{k=0}^{n-1} r(\sigma^k(x)).\]

Given a $\theta$-Lipschitz map $r$, which we will call the roof function, we can define the suspension space corresponding to the roof function as

\[X_A^r := \{(x,t) : x \in X_A, 0 \leq t \leq r(x)\}/\sim\]

where $(x, r(x)) \sim (\sigma(x), 0)$. We equip this space with the metric

\[d_\theta((x,t),(y,s)) := d_\theta(x,t) + \vert t-s \vert.\]

The suspension flow is then the flow on $X_A^r$ with corresponding vector field $\partial/\partial t$. However, we can state it more precisely – it is the flow $\sigma^t : X_A^r \rightarrow X_A^r$ given by $\sigma^t(x,s) := (x, s+t)$ for small $t$. A closed orbit for the suspension flow is then a curve $\gamma(t) := \sigma^t(x,s)$ such that $\gamma(0) = \gamma(T)$ for some $T > 0$. The smallest such $T$ is the period of the flow.

We observe that every closed orbit must pass through the roof function at some point (since the vector field is only pointing up). As a consequence, given a closed orbit $\gamma$, we can associate to it a point $(x,0) \in X_A^r$. Observe that $\sigma^T(x,0) = (x,0)$ happens if and only if $r^n(x) = T$ and $\sigma^n(x) = x$. Thus, the collection of points on closed orbits with length up to $T$, $P(T)$, has some correspondence to the set

\[\Gamma(T) := \{ x \in X_A^r : r^n(x) \leq T \text{ and } \sigma^n(x) = x\}.\]

The correspondence is that $\Gamma(T)$ is the projection of $P(T)$ onto the base $X_A \times {0}$ along orbits. Furthermore, the flow is isometric in the second coordinate (by construction), and so we have the following.

Observation: The set $P(T)$ is $\eta$-dense in $X_A^r$ with respect to $d_\theta$ if and only if $\Gamma(T)$ is $\eta$-dense in $X_A$ with respect to $d_\theta$.

Markov partitions

We now turn to the smooth setting, following the notes by Walkden. Let $M$ be a closed manifold equipped with some Riemannian metric, and let $f : M \rightarrow M$ be a $C^\infty$-diffeomorphism (note that lower regularity is allowed in other texts, but for todays purposes we’ll just ignore this). We say that $f$ is Anosov if there is a $df$-invariant splitting of the tangent bundle $TM = E^s \oplus E^u$ and constants $C, \lambda > 0$ such that for all $n \geq 0$, if $v \in E^s(x)$, then $|d_xf^n(v)| \leq C e^{-n\lambda}$, and if $v \in E^u(x)$ then $|d_xf^n(v)| \geq C e^{n\lambda}$.

Remark: While we need to equip $M$ with some metric so that this is well-defined, the property of being Anosov is independent of the metric.

The classical example of an Anosov diffeomorphism is Arnol’d’s cat map, which is the diffeomorphism on $\mathbb{T}^2$ corresponding to the matrix

\[A = \begin{pmatrix} 2 & 1 \\ 1 & 1\end{pmatrix}.\]

It turns out that, if we are only interested in Holder properties of SFT’s, then we can translate these to statements about Anosov diffeomorphisms through a nice covering of our manifold $M$. We make this a little more precise, although the details are technical.

Let $\epsilon > 0$ be small. The local stable manifold of radius $\epsilon > 0$ centered at $x \in M$ is the set

\[W^{s}_\epsilon(x) := \{y \in M : d(f^n(x), f^n(y)) \leq \epsilon \text{ for all } n \geq 0\}.\]

Similarly, the local unstable manifold of radius $\epsilon > 0$ centered at $x \in M$ is the set

\[W^{u}_\epsilon(x) := \{y \in M : d(f^{-n}(x), f^{-n}(y)) \leq \epsilon \text{ for all } n \geq 0\}.\]

The local stable/unstable manifold theorem says there is a uniform $\epsilon > 0$ such that the local stable and unstable manifolds are actually smoothly embedded disks for all $x\in M$ (and hence the “manifold” part of the name is not arbitrary). Also important is the following: There is a uniform $\eta > 0$ such that if $d(x,y) < \eta$, then

\[\#( W^u_\epsilon(x) \cap W^s_\epsilon(y)) = 1.\]

We denote this unique intersection point by $[x,y]$, and we call this the Bowen bracket of the two points.

A subset $R \subseteq M$ is a rectangle if $x,y \in R$ implies that $[x,y] \in R$. We will be dealing with proper rectangles, which just means that $R$ is proper in the topological sense (i.e., it is equal to the closure of its interior). If $R$ is a proper rectangle and $x \in R^\mathrm{o}$, then we write $W^{s/u}(x,R) := W^{s/u}_\epsilon(x) \cap R$. Finally, a Markov partition of $M$ is a family of proper rectangles $\mathcal{R} = {R_1, \ldots, R_k}$ which satisfies the following:

1. we have $\bigcup_{i=1}^k R_i = M$,
2. $R_i^{\mathrm{o}} \cap R_j^{\mathrm{o}} = \varnothing$ if $i \neq j$,
3. if $x \in R_i^{\mathrm{o}}$ and $f(x) \in R_j^{\mathrm{o}}$, then $f(W^s(x,R_i)) \subseteq W^s(f(x), R_j)$, and
4. if $x \in R_i^{\mathrm{o}}$ and $f^{-1}(x) \in R_j^{\mathrm{o}}$, then $f^{-1}(W^u(x,R_i)) \subseteq W^u(f(x), R_j)$. The diameter of the Markov partition will be $\sup_{1\leq i \leq k} \text{Diam}(R_i)$.

The famous theorem (which is due to Sinai and later expanded on by Bowen) says that these partitions always exist.

Theorem: If $f : M \rightarrow M$ is an Anosov diffeomorphism, then there exists a Markov partition $\mathcal{R}$ with arbitrarily small diameter.

With this Markov partition, we define the $k \times k$ matrix by $A(i,j) = 1$ if and only if $f(R_i^{\mathrm{o}}) \cap R_j^\mathrm{o} \neq \varnothing$. Let $X_A$ be the associated sequence space to this matrix (which are bi-infinite sequences), and define

\[\pi : X_A \rightarrow M, \quad \pi(x) = \bigcap_{i=-\infty}^\infty f^{-i}(R_{x_i}).\]

Theorem: The map $\pi : X_A \rightarrow M$ is

1. Holder continuous (for some $\alpha \in (0,1)$) and surjective,
2. injective on a set of full measure and on a dense residual set,
3. bounded-to-one, and
4. $f \circ \pi = \pi \circ \sigma$.

Remark: The same thing can be said for Anosov flows, which we will not get into for time’s sake.

Main result

With the preliminaries out of the way, we now dive in to analyzing the density of closed orbits with a prescribed length.

Shifts of finite type

The goal of this section is to prove the following.

Theorem: Let $\theta \in (0,1)$ and $\sigma : X_A \rightarrow X_A$ be a shift of finite type. There exists constants $C, \delta > 0$ such that for every $n$ large enough, we have $P(n)$ is $\eta$-dense, where $\eta \leq C e^{-n \delta}$. Furthermore, $\delta$ depends only on $\theta$, and $C$ depends on $\sigma$ and $\theta$.

Proof: Let $M$ be our specification constant. Supposing that $n$ is large enough, we can use the specification property to show that for any finite admissible sequence $(x_1, \ldots, x_{n-M-1})$, we can add on more symbols to get a sequence $(x_1, \ldots, x_{n-M}, x_{n-M+1}, \ldots, x_1)$ which is admissible. This shows that for every cylinder in $C \in \mathcal{C}(n-M-1)$, we have $C \cap P(n) \neq \varnothing$. From this, we deduce that for every $x \in X_A$, we have $d_\theta(P(n), x) \leq \theta^{n-M-1} = \theta^{-M-1} e^{n\log(\theta)}.$ Let $\delta := - \log(\theta)$ and $C := \theta^{-M-1}$. $\square$

An easy corollary is the following (well-known) observation.

Corollary: If $\sigma : X_A \rightarrow X_A$ is a shift of finite type, then $P$ is dense.

We claim that this also extends to Anosov diffeomorphisms through Markov partitions. Let

\[P(n) := \bigcup_{k \leq n} \text{Fix}(f^k).\]

Corollary: Let $f : M \rightarrow M$ be an Anosov diffeomorphism. There are constants $C, \delta > 0$ such that for every $n$ large enough, we have $P(n)$ is $\eta$-dense, where $\eta \leq C e^{-n\delta}$.

Proof: Let $\pi : X_A \rightarrow M$ be the coding map coming from our Markov partition construction. Let $p \in M$ be arbitrary. We can write $p = \pi(x)$ where $x \in X_A$. Using the theorem, we have that there is a $z \in P(n)$ such that $d_\theta(x, z) \leq C e^{-n \delta}$, so $d(\pi(x), \pi(z)) \leq C’ d_\theta(x,z)^\alpha \leq C’ \cdot C^\alpha e^{-n \alpha \delta}.$ The result now follows if we can show that $\pi(z) \in P(n)$. Since $z \in P(n)$, we have $\sigma^k(z) = z$ for some $k \leq n$, so $f^k(\pi(z)) = \pi(\sigma^k(z)) = \pi(z)$, and hence $\pi(z) \in P(n)$. $\square$

Another interesting corollary is the following.

Corollary: Let $r : X_A \rightarrow \mathbb{R}$ be Lipschitz continuous, and let $\sigma^t : X_A^r \rightarrow X_A^r$ be a suspension over a shift of finite type. There exists constants $C, \delta > 0$ such that for every $n$ large enough, we have $P(T)$ is $\eta$-dense, where $\eta \leq C e^{-n \delta}$.

Proof: We use the fundamental observation from that section, which is that we can reduce the study of $P(T)$ to the study of $\Gamma(T)$. Now let $M := |r|_\infty$, and notice that if $x \in P(\lfloor T/M \rfloor)$, then for $k$ such that $\sigma^k(x) = x$, we have $r^k(x) \leq k M \leq T$. Thus, $P(\lfloor T/M \rfloor) \subseteq \Gamma(T)$, and from this we can deduce that $\Gamma(T)$ is $\eta$-dense, where $\eta \leq C e^{-(T/M - 1)\delta}$ and $C$ and $\delta$ are from the theorem. The result now follows. $\square$

Finally, nearly the same argument extends this from the suspension case to the case of Anosov flows.

Follow up remarks

• Since we are using Markov partitions, one could directly jump to Axiom A diffeomorphisms and flows, but for the sake of presentation we don’t get into that.
• It would be interesting to see what an optimal choice of $\delta$ is. For example, using more geometric means, Butt was able to get a similar result for Anosov geodesic flows in Lemma 2.9.
• It might be interesting to see how this would follow from an effective equidistribution result (in the vein of Kadyrov in here, and later improved by O’Hare in here).