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.

We start with Anosov dichotomy:

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.

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