Dendrites and measures with discrete spectrum

It is clear from Proposition 3.1 that each set $J_{\theta}$ is closed, connected, and has non-empty intersection with $M$. It is also clear that $f(J_{\theta})=J_{\tau(\theta)}$. However, since the sets $f^{i}(D_{k})\cap f^{j}(D_{k})$ are allowed to intersect at a point, it is not clear if the sets $J_{\theta}$ are pairwise disjoint. We prove this fact now. Suppose there are $\theta\neq\theta^{\prime}$ with $J_{\theta}\cap J_{\theta^{\prime}}\neq\emptyset$. Find $k$ minimal such that $\theta_{k}\neq\theta^{\prime}_{k}$. Then clearly

for some single point $a\in X$. Taking the image by $f^{\alpha_{k}}$ we have

since $D_{k}$ is periodic with period $\alpha_{k}$. This shows that $a$ is periodic with period $\alpha_{k}$. In particular, it does not belong to the infinite minimal set $M$. Now for $n\in\mathbb{N}$ let $J_{n}=J_{\tau^{n\alpha_{k}}(\theta)}$. Then $a\in J_{n}$ and we can choose an additional point $m_{n}\in M\cap J_{n}$ for all $n$. Thus the $J_{n}$’s are nondegenerate subdendrites and intersect pairwise only at $a$. In particular, the sets $(a,m_{n}]$ are pairwise disjoint connected subsets of $X$, so their diameters must converge to zero (see eg. [17, Lemma 2.3]). But since $M$ is closed, this shows that $a\in M$, a contradiction.

Now that the sets $J_{\theta}$, $\theta\in\Omega$ have been shown to be pairwise disjoint, we see immediately that $\pi$ is well-defined. It is also easy to see that $\pi$ is continuous and $\tau\circ\pi=\pi\circ f$.

Again, since the sets $J_{\theta}$, $\theta\in\Omega$ are pairwise disjoint connected sets in $X$, only countably many of them can have positive diameter. It follows that $\pi^{-1}(\theta)$ is a singleton except for countably many $\theta$. It remains to show that $M\cap J_{\theta}$ is countable when $J_{\theta}$ is non-degenerate. Since $M$ is minimal and $J_{\theta}$ never returns to itself, we must have $M\cap J_{\theta}\subset\operatorname{Bd}(J_{\theta})$, where $\operatorname{Bd}(J_{\theta})$ stands for the boundary of $J_{\theta}$. But the boundary in $X$ of the subdendrite $J_{\theta}$ is a subset of $E(J_{\theta})\cup B(X)\cup\overline{E(X)}$, which is countable by the assumption that $E(X)$ has countable closure. Here we use the well-known facts that $B(X)$ is countable in any dendrite, and the cardinality of the endpoint set of a dendrite cannot increase when we pass to a subdendrite, see e.g. [19, 20]. ∎

Using Lemma 3.2 we know that there are uncountably many singleton sets $J_{\theta}$. Now a dendrite whose endpoint set has countable closure is always the union of a countable sequence of free arcs and a countable set, see [6, Theorem 2.2]. It follows that we can find $\theta$ with the singleton $J_{\theta}$ in the interior of a free arc $A$ in $X$. Since $J_{\theta}$ is the nested intersection $\bigcap_{N}f^{\theta_{N}}(D_{N})$ we can find $N$ large enough that $f^{\theta_{N}}(D_{N})$ is contained in $A$. Then $f^{\theta_{k}}(D_{k})$ is a free arc for all $k\geq N$. ∎

Throughout the proof we will use freely the following well-known property of $\omega$-limit sets (e.g. [7]): if for fixed $n\geq 2$ we write $W_{i}=\omega_{f^{n}}(f^{i}(x))$ for $0\leq i<n$, then $\omega_{f}(x)=\bigcup_{i=0}^{n-1}W_{i}$ and $f(W_{i})=W_{i+1\;(\text{mod }n)}$. In particular, if $\omega_{f}(x)$ is uncountable, then so is each $W_{i}$, and if $\omega_{f}(x)$ contains a given fixed point, then each $W_{i}$ contains it as well. We continue to use the notation $[x,y]$ for the unique arc in $X$ with endpoints $x,y\in X$, and if $z\in(x,y)=[x,y]\setminus\{x,y\}$ we will say for simplicity that $z$ lies between $x$ and $y$.

Now let $L=\omega_{f}(x)$ where $x$ is recurrent but not periodic. Then $L$ is the closure of the orbit of $x$, hence it is a perfect uncountable set.

Step 1: $L$ does not contain a periodic point with a free arc neighborhood in $X$.

For suppose to the contrary that $a\in L$, $f^{N}(a)=a$, and some free arc $C$ is a neighborhood of $a$ in $X$. Then by the standard properties mentioned above $a\in\omega_{f^{N}}(f^{i}(x))$ for some $0\leq i<N$. Replacing $f$ with its iterate and $x$ with its image we may safely assume that $N=1$ and $i=0$, i.e. $a$ is a fixed point in $L=\omega_{f}(x)$.

Since periodic points are never isolated in infinite $\omega$-limit sets we know that $L$ accumulates on $a$ from at least one side in the free arc $C$. So choose an endpoint $b$ of $C$ such that $L\cap[a,b]$ accumulates on $a$. For convenience we let $C$ carry its natural order as an arc, oriented in such a way that $a<b$. Choose five points $y_{i}\in L\cap[a,b]$ with $a<y_{1}<y_{2}<y_{3}<y_{4}<y_{5}<b$. Choose three small arc neighborhoods $I_{2},I_{3},I_{4}$ containing $y_{2},y_{3},y_{4}$ respectively and let them be pairwise disjoint and lie between $y_{1}$ and $y_{5}$. Since the orbit of $x$ visits each of these neighborhoods $I_{i}$ infinitely often, there must be points in $I_{2}$ and $I_{4}$ which visit $I_{3}$, so by [17, Theorem 2.13] $f$ has a periodic point $c$ between $y_{1}$ and $y_{5}$. Replacing $x$ with a point from its orbit near $y_{1}$ we may assume that $a<x<c<y_{5}<b$. Let $r$ be the period of $c$ and put $g=f^{r}$. Since $a$ was already fixed for $f$ we have $a\in\omega_{g}(x)$ as well. Note that since $x$ is recurrent for $f$ it is also recurrent for $g$.

Claim: There is an arc $I$ invariant for $g$ with $[a,x]\subseteq I\subseteq[a,c]$.

To prove the claim put $I=\overline{\bigcup_{n=0}^{\infty}g^{n}([a,x])}$. Since $a$ is fixed and $x$ is recurrent, it suffices to show that $g^{n}([a,x])\subseteq[a,c]$ for all $n$. If this is not true, then there is $z\in[a,x]$ and $n_{0}\geq 1$ such that $a$ is between $g^{n_{0}}(z)$ and $c$ or $c$ is between $g^{n_{0}}(z)$ and $a$. We treat these two cases separately.

Suppose first that $a$ is between $g^{n_{0}}(z)$ and $c$. Then $a\in g^{n_{0}}([z,c])$, so there is $a_{-1}$ between $z$ and $c$ with $g^{n_{0}}(a_{-1})=a$. Then $f^{n}(a_{-1})=a$ for all $n\geq n_{0}\cdot r$. Since $L\cap[a,b]$ accumulates on $a$ we can find a point $x^{\prime}\in\mathrm{Orb}_{f}(x)$ between $a$ and $a_{-1}$. Since $y_{5}\in\omega_{f}(x)$ we can find $n\geq n_{0}\cdot r$ such that $f^{n}(x^{\prime})$ is close to $y_{5}$ and $a<x^{\prime}<a_{-1}<c<f^{n}(x^{\prime})$. Put $J=[a,x^{\prime}]$ and $K=[x^{\prime},a_{-1}]$. Then $f^{n}(J)\cap f^{n}(K)\supseteq J\cup K$, so $f$ possess an arc horseshoe and thus has positive topological entropy, a contradiction.

Suppose instead that $c$ is between $g^{n_{0}}(z)$ and $a$. Then $c\in g^{n_{0}}([a,x])$, so there must be $c_{-1}$ between $a$ and $x$ with $g^{n_{0}}(c_{-1})=c$. Since $a\in\omega_{g}(x)$ we can choose $n>n_{0}$ with $g^{n}(x)$ close to $a$ so that $g^{n}(x)<c_{-1}<x<c$. Put $J=[c_{-1},x]$ and $K=[x,c]$. Then again $g^{n}(J)\cap g^{n}(K)\supseteq J\cup K$, so $g$ has positive topological entropy and so does $f$, a contradiction. This completes the proof of the claim.

Now we may use the claim to finish Step 1. Since $x$ belongs to the closed invariant set $I$ we have $\omega_{g}(x)=\omega_{g|_{I}}(x)$. But $g|_{I}$ is an interval map, and when an infinite $\omega$-limit set for an interval map contains a periodic point, the topological entropy must be positive (see [18]), a contradiction.

Step 2: $L$ does not contain any periodic points.

Suppose to the contrary that $a\in L$ is a periodic point. As in Step 1 we may assume that $a$ is fixed. By [6, Theorem 2.2] the dendrite $X$ is the union of a countable sequence of free arcs together with a countable set. In particular, we can find a free arc $C$ not containing $a$ with $L\cap C$ uncountable. Write $C=[u,v]$ with $v$ between $u$ and $a$ and let $<$ denote the order in $C$ with $u<v$. Since $L\cap C$ is infinite we may choose four points $x_{i}\in\mathrm{Orb}_{f}(x)$ with $u<x_{1}<x_{2}<x_{3}<x_{4}<v$. As in Step 1 we can use small arc neighborhoods of $x_{2},x_{3},x_{4}$ to find a periodic point $c$ with $u<x_{1}<c<v$, and since $x_{1}$ is in the orbit of $x$ we may redefine $x=x_{1}$ without changing $\omega_{f}(x)$. Let $r$ denote the period of $c$ and put $g=f^{r}$. Since $x$ is recurrent also for $g$ we have $\mathrm{Orb}_{g}(x)\cap[u,c]$ infinite, so we can find two points $x_{5},x_{6}\in\mathrm{Orb}_{g}(x)$ with $u<x_{5}<x_{6}<c$ and passing forward along the orbit we can redefine $x=x_{6}$ without changing $\omega_{g}(x)$. In particular, $x_{5}\in\omega_{g}(x)=\overline{\mathrm{Orb}_{g}(x)}$, so we can choose $p\geq 1$ with $g^{p}(x)$ close to $x_{5}$ so that $u<g^{p}(x)<x<c$.

Let $l=\omega_{g^{p}}(x)$. We have $x\in l$ because $x$ is recurrent and $a\in l$ because $a$ is a fixed point in $\omega_{f}(x)$. Moreover $c\not\in l$ as a result of Step 1. So let $X_{0},$ $X_{1}$ denote the connected components of $X\setminus\{c\}$ containing $x$ and $a$, respectively, and put $l_{i}=l\cap X_{i}$. Then $l=l_{0}\cup l_{1}$ expresses $l$ as the disjoint union of two nonempty open subsets (in the topology induced from $X$ to $l$). Recall that every $\omega$-limit set $\omega_{f}(x)$ is weakly incompressible, i.e. $f(\overline{U})\not\subset U$ for any set $U\subsetneq\omega_{f}(x)$ open in $\omega_{f}(x)$ (see, e.g., [23]). Thus we have $g^{p}(l_{0})\cap l_{1}\neq\emptyset$. Therefore we may choose $y\in l_{0}$ with $g^{p}(y)\in l_{1}$, and since $\mathrm{Orb}_{g^{p}}(x)$ is dense in $l_{0}$ we may choose $y$ from the orbit of $x$. We finish the proof in two cases, depending on the location of $y$.

Suppose first that $y$ is between $x$ and $c$. In the ordering of the arc $[g^{p}(x),g^{p}(y)]$ we have $g^{p}(x)<x<y<c<g^{p}(y)$. Put $I=[x,y]$ and $J=[y,c]$. Clearly $g^{p}(I)\supseteq I\cup J$. Since $y\in\mathrm{Orb}_{g}(x)$ we have $\omega_{g}(y)=\omega_{g}(x)\supset\mathrm{Orb}_{g}(x)\ni g^{p}(x)$. In particular, we may choose $n>p$ to make $g^{n}(y)$ as close to $g^{p}(x)$ as we like, so that $x,y\in[g^{n}(y),c]$. But then $g^{n}(J)\supseteq I\cup J$. We conclude that $g$ possess an arc horseshoe and thus $g$ has positive topological entropy, which is a contradiction with $h_{top}(f)=0$.

Suppose instead that $x$ is between $y$ and $c$. Then $c\in[g^{p}(x),g^{p}(y)]$, so there is $c_{-1}\in(x,y)$ with $g^{p}(c_{-1})=c$. In the ordering of the arc $[y,g^{p}(y)]$ we have $y<c_{-1}<x<c$ and we also have $x\in(g^{p}(x),c)$. Put $I=[c_{-1},x]$ and $J=[x,c]$. Since $y\in\mathrm{Orb}_{g}(x)\subset\omega_{g}(x)$ we can find $n>p$ with $g^{n}(x)$ as close to $y$ as we like. In particular we can get $x,c_{-1}\in[g^{n}(x),c]$. But then $g^{n}(I)\cap g^{n}(J)\supseteq I\cup J$. Again we conclude that $g$ has positive topological entropy, which is a contradiction. This ends the proof. ∎

Let $x\in\mathrm{Rec}(f)$. If $x$ is periodic then it is minimal, so assume $x$ is not periodic. Let $L=\omega(x)$. Let $M\subset L$ be a minimal set. By Lemma 3.4 $L$ contains no periodic orbits, so $M$ is an infinite minimal set. Then Proposition 3.1 applies and we get a sequence of $f$-periodic subdendrites $(D_{k})_{k\geq 1}$ and periods $(\alpha_{k})$ satisfying properties (1)–(5) of that Proposition. By Lemma 3.3 for all sufficiently large $k$ we have that $f^{i}(D_{k})$ is a free arc for suitable $i$. Since $M$ is infinite and $D_{k}$ is periodic, we have $M\cap\operatorname{int}f^{i}(D_{k})\neq\emptyset$ and as a consequence $Orb_{f}(x)\cap D_{k}\neq\emptyset$, for every sufficiently large $k$. Hence $\bigcap_{k\geq 1}Orb_{f}(D_{k})$ contains $L$, that is, property (4) still holds with $L$ in the place of $M$.

We claim that property (5) also holds with $L$ in the place of $M$. Fix $k$ and denote $L_{i}=f^{i}(D_{k})\cap L$ for $0\leq i<\alpha_{k}-1$. Observe that $L$ does not contain periodic points, and the set $\mathrm{Orb}(f^{i}(D_{k})\cap f^{j}(D_{k}))$ is always finite and invariant for any $i\neq j$ (can be empty) and hence $L_{i}\cap f^{j}(D_{k})=\emptyset$ for $i\neq j$. This shows that the sets $L_{i}\cap L_{j}=\emptyset$ for $i\neq j$. Clearly $f(L_{i})\subseteq L_{i+1(\text{mod }\alpha_{k})}$, and $f(L)=L$ since $\omega$-limit sets are always mapped onto themselves. This shows that $f(L_{i})=L_{i+1(\text{mod }\alpha_{k})}$. In particular, we conclude that $L_{i}$ is uncountable for each $i$.

Again using Lemma 3.3 choose $k$ large enough that $f^{i}(D_{k})$ is a free arc for some $0\leq i<\alpha_{k}-1$ and let $A=f^{i}(D_{k})$ denote that free arc. We have just shown that $L_{i}=A\cap L$ is uncountable, so since $A$ is a free arc there are points from $\omega(x)$ in its interior. Thus we can find a point $y=f^{l}(x)$ from the forward orbit of $x$ in $A$. Then $\omega_{f}(x)=\omega_{f}(y)$ and $y$ is also recurrent for $f$. Since $\mathrm{Rec}(f^{\alpha_{k}})=\mathrm{Rec}(f)$, $y$ is also recurrent for $f^{\alpha_{k}}$. But the restriction of $f^{\alpha_{k}}$ to $A$ is an interval map with zero topological entropy. For such a map, all recurrent points are minimal points, see e.g. [7, Chapter VI. Proposition 7]. Thus $y$ is a minimal point for $f^{\alpha_{k}}$, and hence also for $f$. This shows that $\omega_{f}(y)=\omega_{f}(x)$ is a minimal set, and hence $x$ itself is minimal. ∎

Let $\mu$ be any finite invariant measure concentrated on $\bigcup_{i}A_{i}$. Since each $A_{i}$ is invariant, i.e. $f(A_{i})\subset A_{i}$, and $f$ preserves $\mu$, we may assume by throwing away a set in $X$ of $\mu$-measure zero that $f^{-1}(A_{i})=A_{i}$ for each $i$.

We may take the index set for the variable $i$ to be $\{1,\ldots,n\}$ in the finite case or $\mathbb{N}$ in the countable case. Then putting $B_{i}=A_{i}\setminus\bigcup_{j<i}A_{j}$ for each $i$, we get a collection $\{B_{i}\}$ of pairwise disjoint invariant Borel sets. Now let $I=\{i~{}:~{}\mu(B_{i})>0\}$ and write $\mu_{i}=\mu|_{B_{i}}$ for the (unnormalized) restriction of $\mu$ to $B_{i}$. Then we get a direct sum decomposition of Hilbert spaces $L^{2}_{\mu}(X)=\bigoplus_{i\in I}L^{2}_{\mu_{i}}(B_{i}).$ We may extend each function $\phi\in L^{2}_{\mu_{i}}(B_{i})$ to an element of $L^{2}_{\mu_{i}}(X)$ by letting $\phi$ vanish outside of $B_{i}$. Since $f^{-1}(B_{i})=B_{i}$, we see that if $\phi\circ f=\lambda\phi$ holds $\mu_{i}$ almost-everywhere in $B_{i}$, then by letting $\phi$ vanish outside $B_{i}$ it continues to hold $\mu$-almost everywhere in $X$. Thus we have the equivalent direct sum decomposition

and an eigenfunction in a coordinate space is still an eigenfunction in the whole space. For each $i\in I$, the normalized measure $\mu_{i}/\mu(B_{i})$ is an invariant probability measure for $f$ concentrated on $B_{i}\subset A_{i}$, so by hypothesis the eigenfunctions of the Koopman operator on the space $L^{2}_{\mu_{i}/\mu(B_{i})}(X)$ have dense linear span. Dropping the normalizing constant, the same holds for $L^{2}_{\mu_{i}}(X)$. Passing through the direct sum decomposition, it follows that the eigenfunctions of the Koopman operator on the space $L^{2}_{\mu}(X)$ have dense linear span, that is, $\mu$ has discrete spectrum.

The last statement of the lemma follows by the Poincaré recurrence theorem, whereby if $\mathrm{Rec}(f)\subseteq\bigcup A_{i}$, then every measure $\mu\in M_{f}(X)$ is concentrated on $\bigcup A_{i}$. ∎

Suppose that $F$ has positive entropy. Then by [22] there exists an arc horseshoe $I_{1},I_{2}$ with $F^{n}(I_{1}\cap I_{2})\supset I_{1}\cup I_{2}$ for some $n\in\mathbb{N}$. Then $F^{i}(I_{j})$ is not a single point for any $i=1,\ldots,n$ and $j=1,2$. But if $F(J)$ is nondegenerate for an arc $J$ then $f(J)\supset F(J)$ which implies that $f^{n}(I_{1})\cap f^{n}(I_{2})\supset I_{1}\cup I_{2}$ which implies that $f$ has positive topological entropy. A contradiction. ∎

Let $Z=\{z\in\overline{E(X)}~{}:~{}\omega_{f}(z)\text{ is an infinite minimal set}\}.$ Following arguments in [19, Theorem 10.27], let $(T_{n})_{n\in\mathbb{N}}\subset X$ be an increasing sequence of topological trees with endpoints in $E(X)$ defined as follows. We inductively construct the sequence $(T_{n})_{n\in\mathbb{N}}$ starting with $T_{1}=\{e_{1}\}$ for some $e_{1}\in E(X)$. Then for $n\geq 1$, we attach to $T_{n}$ an arc $[e,e_{n+1}]$ whose one endpoint $e_{n+1}$ belongs to $E(X)\setminus T_{n}$ and $e\in T_{n}$. Since $E(X)$ is countable we can put every endpoint into one of the trees, that is, we let the sequence $(e_{n})_{n\in\mathbb{N}}$ be an enumeration of $E(X)$, and then $\bigcup_{n\geq 1}T_{n}$ being a connected set must coincide with the whole dendrite $X$.

Let $\hat{T}_{n}=\bigcap_{i=0}^{\infty}f^{-i}(T_{n})$ be the maximal invariant set completely contained in $T_{n}$. Let $\mathrm{Per}(f)$ be the set of periodic points of $f$. We claim that:

To see this, let $x$ be a non-periodic recurrent point whose orbit is not contained in any of the trees $T_{n}$. This means that there are points $f^{n_{i}}(x)$ which belong to $T_{m_{i}}\setminus T_{m_{i}-1}$ for some strictly increasing sequences $m_{i}$, $n_{i}\to\infty$. Then the arcs $[f^{n_{i}}(x),e_{m_{i}}]$ in $X$ are pairwise disjoint, so by [17, Lemma 2.3] their diameters tend to zero. This shows that $\liminf_{n\to\infty}d(f^{n}(x),E(X))=0$. Therefore $\omega_{f}(x)\cap\overline{E(X)}\neq\emptyset$. By Theorem 3.5, $\omega_{f}(x)$ is a minimal set, so choosing $z\in\omega_{f}(x)\cap\overline{E(X)}$ we have $\omega_{f}(x)=\omega_{f}(z)$. This establishes (4.2).

Now observe that any finite invariant measure concentrated on $\mathrm{Per}(f)$ has discrete spectrum, see eg. [14, Theorem 2.3]. As for the sets $\hat{T}_{n}$, note that for each $n\in\mathbb{N}$ the map $F=R\circ f$, where $R\colon X\to T_{n}$ is a retraction, satisfies $F|_{\hat{T}_{n}}=f|_{\hat{T}_{n}}$ by the definition and therefore each $f$-invariant measure concentrated on $\hat{T}_{n}$ (a subset of a tree) has discrete spectrum, as, by [14], all invariant measures of $F$ have discrete spectrum.

Finally, we claim that any invariant measure concentrated on $\omega_{f}(z)$, $z\in Z$, has discrete spectrum. Let $(D_{k})$ be the periodic subdendrites with periods $(\alpha_{k})$ described in Proposition 3.1 and let $\pi:(\omega_{f}(z),f)\to(\Omega,\tau)$ be the factor map onto the odometer described in Lemma 3.2. Let $\mu\in M_{f}(X)$ be any invariant measure concentrated on $\omega_{f}(z)$. Then the pushforward measure $\pi_{*}(\mu)$ is invariant for the odometer, so by unique ergodicity it is the Haar measure on $\Omega$ and by well-known properties of odometers it has discrete spectrum. Now since $\omega(z)$ contains no periodic points, we know that $\mu$ is non-atomic and therefore countable sets have measure zero. Then in the category of measure preserving transformations, the factor map $\pi:(\omega_{f}(z),\mu,f)\to(\Omega,\pi_{*}(\mu),\tau)$ is in fact an isomorphism, since by Lemma 3.2 it is invertible except on a set of $\mu$-measure zero. This implies that $\mu$ has discrete spectrum.

We have shown that an invariant measure concentrated on any of the countably many invariant sets in (4.2) has discrete spectrum. By Lemma 4.1 this completes the proof. ∎