Basic set of a smooth flow: Bowen’s trichotomy

This is a note taken from Bowen, Periodic Orbits for Hyperbolic Flows, American Journal of Mathematics (1972), 1–30.

Let $f:M\to M$ be a transitive Anosov flow. Then

1. either it is mixing: then strong stable and strong unstable manifold everywhere dense on $M$
2. or it is a suspension: choose the closure of a non-dense stable manifold as a cross-section and the induced roof-function is constant.

Anosov proved this in volume-preserving case and Plante proved it for the general case (Anosov flows, 1972).

Let $M$ be a closed manifold and $\phi_t:M\to M$ be a $C^1$ flow. Let $\Omega$ be a closed, invariant subset without fixed points, that is, the vector field of $\phi$ does not admits zeros on $\Omega$. Then $\Omega$ is said to be a basic set of $\phi$ if
$(\Omega,\phi_t)$ is (topologically) transitive and hyperbolic,
– close orbits are dense in $\Omega$
$(\Omega,\phi_t)$ is isolated: $\Omega=\bigcap_{\mathbb{R}}\phi_t(U)$ for some open neighborhood $U$ of $\Omega$.
The last one is equivalent to the local product structure.

Theorem 3.2 in [B]. There are three mutually exclusive types:
1. $\Omega$ consists of a single closed orbit of $\phi$;
2. the strong stable manifold $W^s(p)$ is dense in $\Omega$ for for each $p\in\Omega$;
3. $(\Omega,\phi_t)$ is the constant suspension of a Axiom A homeomorphism.

As remarked by Bowen, this is first proved by Anosov for volume-preserving Anosov flow, by Plante for general Anosov flow. We should view the following as a proof for Anosov flow case for first reading, and then take the induced topology on $\Omega$ for basic sets.

Proof. Clearly these three are disjoint. Let’s assume a basic set must be of type 3 if it fails 1 and 2.

Since the strong stable foliation is continuous, we can pick a periodic point $p\in\Omega$ such that $\Omega\backslash\overline{W^s(p)}\neq\emptyset$. Let $Y=\overline{W^s(p)}$ and $T>0$ be the period of $p$. So $\phi_T(p)=p$ and $\phi_T(Y)=Y$. Clearly $\phi_{t+kT}(Y)=\phi_t(Y)$ for all $t$ and $k$. The following is one key property:

Lemma 3.3. For any $\delta>0$, $D_\delta=\bigcup_{|t|\le \delta}\phi_t(Y)$ is a compact neighborhood of $Y$.

Proof. Let $\delta>0$ be given and choose $\epsilon>0$ accordingly with respect to the local product structure. It suffices to prove $B(W^u(p),\epsilon)\subset D_\delta$. Since close orbits are dense in $\Omega$, we reduce to show:

(a) $y\in D_\delta$ for every periodic point $y\in B(x,\epsilon)$ with some $x\in W^u(p),\epsilon)$.

Proof of (a). Consider the point $z=[x,y]=W^u_\delta(\phi_t x)\cap W^s_\delta(y)$ (for some $t$ with $|t|\le \delta$, which follows from the choice of $\epsilon$). Note that $z\in \phi_t(Y)$ and $\phi_{kT}z\in \phi_t(Y)$ for all $k$. Pick $k_n\to+\infty$ with $k_nT\to0(\!\mod r)$ where $r$ is the period of $y$. Firstly note that $d(y,\phi_{k_nT}y)\to 0$ since $\phi_r(y)=y$. So $d(y,\phi_t(Y))\le d(y,\phi_{k_nT}z)\le d(y,\phi_{k_nT}y)+d(\phi_{k_nT}y,\phi_{k_nT}z)\to 0$ since $z\in W^s(y)$. This implies $y\in \phi_t(Y)$ and finishes the proof of (a) (and hence Lemma 3.3).

Let $X=\bigcup_{0\le t\le T}\phi_t(Y)$. We see that $X$ is $\phi$-invariant (by periodicity), closed and also OPEN: $D_\delta(X)=X$ since $X$ is $\phi$-invariant. So $X=\Omega$.

(b) If $Y\cap \phi_t(Y)\neq\emptyset$, then $Y=\phi_t(Y)$.

Proof of (b). To derive a contradiction let’s assume $\phi_t(Y)\backslash Y\neq\emptyset$, say $y$. Since $Y$ is closed, we can assume $y\in\phi_t(W^u(p))=W^u(\phi_t(p))$ by an infinitesimal shifting of $y$. Pick $\delta>0$ small with $y\notin D_\delta$. Moreover we can pock $z\in W^u(\phi_t(p))\cap D_{\delta/2}$ since $D_{\delta/2}$ contains an open neighborhood of $Y$. Clearly $d(\phi_{-kT}y,\phi_{-kT}z)\to 0$ since both lie in $W^u(\phi_t(p))$. But $\phi_{-kT}z\in D_{\delta/2}$ for all $k$. Therefore $\phi_{-kT}y$ is getting closer and closer to $D_{\delta/2}$, and will lie in $D_\delta$ for $k$ large enough, which contradicts the choice of $\delta$. This finishes the proof of (b).

Consider the set $R=\{t\in\mathbb{R}:Y=\phi_t(Y)\}$ and $\tau=\inf\{t\in R:t>0\}$. We have

(c) $\tau>0$ and $R=\tau\mathbb{Z}$. Moreover $\Omega=\bigsqcup_{0\le t<\tau}\phi_t Y$.

Proof of (c). If there exists $t_n\in R\searrow 0$, then $t_n\mathbb{Z}\subset Z$ and hence $R$ is a dense subset of $\mathbb{R}$. Since $R$ is also closed, we see $R=\mathbb{R}$ and hence $Y=\Omega$. This is the (pre-excluded) second alternative. The rest follows automatically.

Proof of Theorem. Let $f=\phi_\tau|_Y$ and $(Y_\tau,f_t)$ be the constant suspension flow over $(Y,f)$. Set $H:Y\times [0,\tau]\to \Omega,(y,t)\mapsto \phi_t(y)$, which downgrades to a conjugate $h:Y_\tau\to\Omega$. This finishes the proof of Theorem.

In particular, the second alternative is mixing (the flow, and the time-1 map). But for the third alternative, invariant sets of time-1 map form a lamination over a circle via $Y_\tau\mapsto [0,\tau]/\{0,\tau\}$. Also there exist many non-constant suspensions with non transitive time-1 maps (suggested by mathoverflow user Alejandro at here). Namely let’s start with an arbitrary homeomorphism $f:X\to X$ and pick a ceiling function $\tau:X\to[\frac{2}{3},\frac{3}{2}]$. We want to find a layer $(x,c(x))$ that is covariant under $f_1:(x,c(x))\mapsto(fx,c(fx))$. This requires $c(x)+1=\tau(x)+c(fx)$, or equivalently, $\tau(x)=1+c(x)-c(fx)$, which can be viewed as the definition of our ceiling function.

Let $f:X\to X$ be a homeomorphism and $c$ be a continuous function on $X$ taking values in $[0,1/3]$. Then suspension of $(X,f)$ with respect to $\tau=1+c-c\circ f$ preserves the graph-layer of $c$.

Clearly $X_\tau=\bigsqcup_{0\le t<1} f_tY$ with $Y=\{(x,c(x)):x\in X\}$. So $X_\tau$ is just a re-labeling of the constant suspension with initial layer $Y$. Then we can start with a constant suspension, pick an arbitrary slowly-varying layer $c$, solve the ceiling $\tau$ and derive a non-constant suspension.