Pugh-Shub Conjecture: Generic measure preserving partially hyperbolic diffeomorphism is ergodic. [PS (2000)]
The accessibility approach breaks this into two conjectures (and both are open):
Conjecture A: Generic partially hyperbolic diffeomorphism (measure preserving or not) is accessible.
Conjecture B: All measure preserving accessible partially hyperbolic diffeomorphisms are ergodic.
Note that if both conjecture A and conjecture B are true, then the Pugh-Shub Conjecture is true, but the failure of either won’t imply the conjecture being wrong.
Here we discuss the recent result of RHRHU (2008) which proves Pugh-Shub conjecture in the case where by using accessibility.
We are going to focus on the proof of Conjecture A, here’s a sketch of proof of Conjecture B when is assumed:
Theorem B: Let be compact manifold, be a measure preserving accessible partially hyperbolic diffeomorphism, , then is ergodic.
Proof: Let be invariant
Let , if then s.t. is a density point of .
At this point there are some technical details in the paper which we are going to skip, but the main idea is to the fact that (or in this case even the weaker hypothesis center brunching would work) to prove that in our case is a density point iff is a “leaf density point” in both its center-stable and center-unstable leaves. Hence by accessibility from to , we can “push” the point along the us-path that joins to and induce that is a “leaf density point” in hence a density point in .
all points are density points of hence .
Note that here if we replace accessibility by essential accessibility, we still get .
Hence , either or
is essentially constant. is ergodic.
This establishes theorem B.
Let be the set of measure preserving diffeomorphisms on that are of class
Theorem A: Accessibility is open dense in the space of diffeomorphisms in with .
For any , let denote the set of points that’s accessible from
Let is open
Fact: with is and
where is accessible and
and is integrable
Note that this actually requires some rather technical work which was done in the paper, here we skip the proof of this.
Let is open
It’s easy to see that is automatically open hence is compact.
Proposition: Let , the following are equivalent:
1) has non-empty interior
2) is open
3) has non-empty interior in
Proof: 1) 2) 3) 1)
Mainly by drawing pictures and standard topology.
Unweaving lemma: latex PH^r(M)$ with s.t. the distance between and is arbitrarily small, and is open.
The proof is left to the second part of the talk…