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 be a transitive Anosov flow. Then

1. either it is mixing: then strong stable and strong unstable manifold everywhere dense on

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 be a closed manifold and be a flow. Let be a closed, invariant subset without fixed points, that is, the vector field of does not admits zeros on . Then is said to be a basic set of if

– is (topologically) transitive and hyperbolic,

– close orbits are dense in

– is isolated: for some open neighborhood of .

The last one is equivalent to the local product structure.

Theorem 3.2 in [B]. There are three mutually exclusive types:

1. consists of a single closed orbit of ;

2. the strong stable manifold is dense in for for each ;

3. 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 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 such that . Let and be the period of . So and . Clearly for all and . The following is one key property:

Lemma 3.3. For any , is a compact neighborhood of .

Proof. Let be given and choose accordingly with respect to the local product structure. It suffices to prove . Since close orbits are dense in , we reduce to show:

(a) for every periodic point with some .

Proof of (a). Consider the point (for some with , which follows from the choice of ). Note that and for all . Pick with where is the period of . Firstly note that since . So since . This implies and finishes the proof of (a) (and hence Lemma 3.3).

Let . We see that is -invariant (by periodicity), closed and also OPEN: since is -invariant. So .

(b) If , then .

Proof of (b). To derive a contradiction let’s assume , say . Since is closed, we can assume by an infinitesimal shifting of . Pick small with . Moreover we can pock since contains an open neighborhood of . Clearly since both lie in . But for all . Therefore is getting closer and closer to , and will lie in for large enough, which contradicts the choice of . This finishes the proof of (b).

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

(c) and . Moreover .

Proof of (c). If there exists , then and hence is a dense subset of . Since is also closed, we see and hence . This is the (pre-excluded) second alternative. The rest follows automatically.

Proof of Theorem. Let and be the constant suspension flow over . Set , which downgrades to a conjugate . 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 . 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 and pick a ceiling function . We want to find a layer that is covariant under . This requires , or equivalently, , which can be viewed as the definition of our ceiling function.

Let be a homeomorphism and be a continuous function on taking values in . Then suspension of with respect to preserves the graph-layer of .

Clearly with . So is just a re-labeling of the constant suspension with initial layer . Then we can start with a constant suspension, pick an arbitrary slowly-varying layer , solve the ceiling and derive a non-constant suspension.