Tag: volume preserving

A survey on ergodicity of Anosov diffeomorphisms

Conan3 Comments

This is in part a preparation for my 25-minutes talk in a workshop here at Princeton next week. (Never given a short talk before…I’m super nervous about this >.<) In this little survey post I wish to list some background and historical results which might appear in the talk.

Let me post the (tentative) abstract first:

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Title: Volume preserving extensions and ergodicity of Anosov diffeomorphisms

Abstract: Given a C^1 self-diffeomorphism of a compact subset in \mathbb{R}^n, from Whitney’s extension theorem we know exactly when does it C^1 extend to \mathbb{R}^n. How about volume preserving extensions?

It is a classical result that any volume preserving Anosov di ffeomorphism of regularity C^{1+\varepsilon} is ergodic. The question is open for C^1. In 1975 Rufus Bowen constructed an (non-volume-preserving) Anosov map on the 2-torus with an invariant positive measured Cantor set. Various attempts have been made to make the construction volume preserving.

By studying the above extension problem we conclude, in particular the Bowen-type mapping on positive measured Cantor sets can never be volume preservingly extended to the torus. This is joint work with Charles Pugh and Amie Wilkinson.

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A diffeomorphism f: M \rightarrow M is said to be Anosov if there is a splitting of the tangent space TM = E^u \oplus E^s that’s invariant under Df, vectors in E^u are uniformly expanding and vectors in E^s are uniformly contracting.

In his thesis, Anosov gave an argument that proves:

Theorem: (Anosov ’67) Any volume preserving Anosov diffeomorphism on compact manifolds with regularity C^2 or higher on is ergodic.

This result is later generalized to Anosov diffeo with regularity C^{1+\varepsilon}. i.e. C^1 with an \varepsilon-holder condition on the derivative.

It is a curious open question whether this is true for maps that’s strictly C^1.

The methods for proving ergodicity for maps with higher regularity, which relies on the stable and unstable foliation being absolutely continuous, certainly does not carry through to the C^1 case:

In 1975, Rufus Bowen gave the first example of an Anosov map that’s only C^1, with non-absolutely continuous stable and unstable foliations. In fact his example is a modification of the classical Smale’s horseshoe on the two-torus, non-volume-preserving but has an invariant Cantor set of positive Lebesgue measure.

A simple observation is that the Bowen map is in fact volume preserving on the Cantor set. Ever since then, it’s been of interest to extend Bowen’s example to the complement of the Cantor set in order to obtain an volume preserving Anosov diffeo that’s not ergodic.

In 1980, Robinson and Young extended the Bowen example to a C^1 Anosov diffeomorphism that preserves a measure that’s absolutely continuous with respect to the Lebesgue measure.

In a recent paper, Artur Avila showed:

Theorem: (Avila ’10) C^\infty volume preserving diffeomorphisms are C^1 dense in C^1 volume preserving diffeomorphisms.

Together with other fact about Anosov diffeomorphisms, this implies the generic C^1 volume preserving diffeomorphism is ergodic. Making the question of whether such example exists even more curious.

In light of this problem, we study the much more elementary question:

Question: Given a compact set K \subseteq \mathbb{R}^2 and a self-map f: K \rightarrow K, when can the map f be extended to an area-preserving C^1 diffeomorphism F: \mathbb{R}^2 \rightarrow \mathbb{R}^2?

Of course, a necessary condition for such extension to exist is that f extends to a C^1 diffeomorphism F (perhaps not volume preserving) and that DF has determent 1 on K. Whitney’s extension theorem gives a necessary and sufficient criteria for this.

Hence the unknown part of our question is just:

Question: Given K \subseteq \mathbb{R}^2, F \in \mbox{Diff}^1(\mathbb{R}^2) s.t. \det(DF_p) = 1 for all p \in K. When is there a G \in \mbox{Diff}^1_\omega(\mathbb{R}^2) with G|_K = F|_K?

There are trivial restrictions on K i.e. if K separates \mathbb{R}^2 and F switches complementary components with different volume, then F|_K can never have volume preserving extension.

A positive result along the line would be the following slight modification of Moser’s theorem:

Theorem: Any C^{r+1} diffeomorphism on S^1 can be extended to a C^r area-preserving diffeomorphism on the unit disc D.

For more details see this pervious post.

Applying methods of generating functions and Whitney’s extension theorem, as in this paper, in fact we can get rid of the loss of one derivative. i.e.

Theorem: (Bonatti, Crovisier, Wilkinson ’08) Any C^1 diffeo on the circle can be extended to a volume-preserving C^1 diffeo on the disc.

With the above theorem, shall we expect the condition of switching complementary components of same volume to be also sufficient?

No. As seen in the pervious post, restricting to the case that F only permute complementary components with the same volume is not enough. In the example, K does not separate the plane, f: K \rightarrow K can be C^1 extended, the extension preserves volume on K, and yet it’s impossible to find an extension preserving the volume on the complement of K.

The problem here is that there are ‘almost enclosed regions’ with different volume that are being switched. One might hope this is true at least for Cantor sets (such as in the Bowen case), however this is still not the case.

Theorem: For any positively measured product Cantor set C = C_1 \times C_2, the Horseshoe map h: C \rightarrow C does not extend to a Holder continuous map preserving area on the torus.

Hence in particular we get that no volume preserving extension of the Bowen map can be possible. (not even Holder continuous)

C^1 vs. C^1 volume preserving

Conan4 Comments

One of the things I’ve always been interested in is, for a given compact set say in \mathbb{R}^n, what maps defined on the set into \mathbb{R}^n can be extended to a volume preserving map (of certain regularity) on a larger set (for example, some open set containing the original set).

The analogues extension question without requiring the extended map to be volume preserving is answered by the famous Whitney’s extension theorem. It gives a beautiful necessary and sufficient condition on when the map has C^r extension – See this pervious post for more details.

A simple case of this type of question was discussed in my earlier Moser’s theorem post:

Question: Given a diffeomorphism on the circle, when can we extend it to a volume preserving diffeomorphism on the disc?

In the post, we showed that any C^r diffeomorphism on the circle can be extended to a C^{r-1} volume preserving diffeomorphism on the disc. Some time later Amie Wilkinson pointed out to me that, by using generating function methods, in fact one can avoid losing derivative and extend it to a C^r volume preserving.

Anyways, so we know the answer for the circle, what about for sets that looks very different from the circle? Is it true that whenever we can C^r extend the map, we can also so it volume-preserving? (Of course we need to rule out trivial case such as the map is already not volume-preserving on the original set or the map sends, say a larger circle to a smaller circle.)

Question: Is it true that for any compact set K \subseteq \mathbb{R}^n with connected complement, for any function f: K \rightarrow \mathbb{R}^n satisfying the Whitney condition with all candidate derivatives having determent 1, one can always extend f to a volume preserving F: \mathbb{R}^n \rightarrow \mathbb{R}^n.

Note: requiring the set to have connected complement is to avoid the ‘larger circle to small circle’ case and if some candidate derivative does not have determent 1 then the extended map cannot possibly be volume preserving near the point.

After thinking about this for a little bit, we (me, Charles and Amie) came up with the following simple example where the map can only be C^1 extended but not C^1 volume preserving.

Example: Let K \subset \mathbb{R}^2 be the countable union of segments:

K = \{0, 1, 1/2, 1/3, \cdots \} \times [0,1]

As shown below:

Define f: K \rightarrow K be the map that sends the vertical segment above 1/n to the vertical segment above 1/(n+1), preserves the y-coordinate and fixes the segment \{0\} \times [0,1]:

Claim: f can be extended to a C^1 map F: \mathbb{R}^2 \rightarrow \mathbb{R}^2.

Proof: Define g: \mathbb{R} \rightarrow \mathbb{R} s.t.

1) g is the identity on \mathbb{R}^{\leq 0}

2) g(x) = x-1/2 for x>1

3) g: 1/n \mapsto 1/(n+1)

4) g is increasing and differentiable on each [1/n, 1/(n-1)] with derivative no less than (1-1/n)(n^2-n)/(n^2+n) and the one sided derivative at the endpoints being 1.

It’s easy to check such g exists and is continuous:

Since \lim_{n \rightarrow \infty} (1-1/n)(n^2-n)/(n^2+n) = 1, we deduce g is continuously differentiable with derivative 1 at 0.

Let F = g \otimes \mbox{id}, F: \mathbb{R}^2 \rightarrow \mathbb{R}^2 is a C^1 extension of f.

Establishes the claim.

Hence the pair (K, f) satisfies the Whitney condition for extending to C^1 map. Furthermore, since the F as above has derivative being the identity matrix at all points of K, the determent of candidate derivatives are uniformly 1. In other words, this example satisfies all conditions in the question.

Claim: f cannot be extended to a C^1 volume preserving diffeomorphism of the plane.

Proof: The idea here is to look at rectangles with sides on the set K, if F preserves area, they have to go to regions enclosing the same area as the original rectangles, then apply the isoperimetric inequality to deduce that image of some edges of the rectangle would need to be very long, hence at some point on the edge the derivative of F would need to be large.

Suppose such extension F exists, consider rectangle R_n = [1/n, 1/(n-1)] \times [0,1]. We have

m_2(R_n) = 1/(n^2-n)

m_2(R_n) - m_2(R_{n+1})

=1/(n^2-n)-1/(n^2+n)=2/(n^3-n)

Hence in order for F(R_n) to have the same area as R_n, the image of the two segments

s_{n,0} = [1/n, 1/(n-1)] \times \{ 0\} and

s_{n,1}= [1/n, 1/(n-1)] \times \{ 1\}

would need to enclose an area of 2/(n^3-n) \sim n^{-3} outside of the rectangle R_{n+1}.

By isoparametric inequality, the sum of the length of the two curves must be at least \sim n^{-3/2}, while the length of the original segments is 2/(n^2-n) \sim n^{-2}.

Hence somewhere on the segments F needs to have derivative having norm at least

\ell(F(s_{n,0} \cup s_{n,1}) / \ell(s_{n,0} \cup s_{n,1})

\sim n^{-3/2}/n^{-2} = n^{1/2}

We deduce that there exists a sequence of points (p_n) converging to either (0,0) or (0,1) where

|| F'(p_n) || \sim n^{1/2} \rightarrow \infty.

Hence F cannot be C^1 at the limit point of (p_n).

Remark: In fact we have showed the stronger statement that no volume preserving Lipschitz extension could exist and gave an upper bound 1/2 on the best possible Holder exponent.

From this we know the answer to the above question is negative, i.e. not all C^1 extendable map can me extended in a volume preserving fashion. It would be very interesting to give criteria on what map on which sets can be extended. By applying same methods we are also able to produce an example where the set K is a Cantor set on the plane.

Remarks from the Oxtoby Centennial Conference

Conan1 Comment

A few weeks ago, I received this mysterious e-mail invitation to the ‘Oxtoby Centennial Conference’ in Philadelphia. I had no idea about how did they find me since I don’t seem to know any of the organizers, as someone who loves conference-going, of course I went. (Later I figured out it was due to Mike Hockman, thanks Mike~ ^^ ) The conference was fun! Here I want to sketch a few cool items I picked up in the past two days:

Definition:A Borel measure \mu on [0,1]^n is said to be an Oxtoby-Ulam measure (OU for shorthand) if it satisfies the following conditions:
i) \mu([0,1]^n) = 1
ii) \mu is positive on open sets
iii) \mu is non-atomic
iv) \mu(\partial [0,1]^n) = 0

Oxtoby-Ulam theorem:
Any Oxtoby-Ulam measure is the pull-back of the Lebesgue measure by some homeomorphism \phi: [0,1]^n \rightarrow [0,1]^n.

i.e. For any Borel set A \subseteq [0,1]^n, we have \mu(A) = \lambda(\phi(A)).

It’s surprising that I didn’t know this theorem before, one should note that the three conditions are clearly necessary: A homeo has to send open sets to open sets, points to points and boundary to boundary; we know that Lebesgue measure is positive on open sets, 0 at points and 0 on the boundary of the square, hence any pull-back of it must also has those properties.

Since I came across this at such a late time, my first reaction was: this is like Moser’s theorem in the continuous case! But much cooler! Because measures are a lot worse than differential forms: many weird stuff could happen in the continuous setting but not in the smooth setting.

For example, we can choose a countable dense set of smooth Jordan curves in the cube and assign each curve a positive measure (we are free to choose those values as long as they sum to 1. Now we can define a measure supported on the union of curves and satisfies the three conditions. (the measure restricted to each curve is just a multiple of the length) Apply the theorem, we get a homeomorphism that sends each Jordan curve to a Jordan curve with positive n dimensional measure and the n dimensional measure of each curve is equal to our assigned value! (Back in undergrad days, it took me a whole weekend to come up with one positive measured Jordan curve, and this way you get a dense set of them, occupying a full measure set in the cube, for free! Oh, well…>.<)

Question: (posed by Albert Fathi, 1970)
Does the homeomorphism \phi sending \mu to Lebesgue measure depend continuously on \mu?

My first thought was to use smooth volume forms to approximate the measure, for smooth volume forms, Moser’s theorem gives diffeomosphisms depending continuously w.r.t. the form (I think this is true just due to the nature of the construction of the Moser diffeos) the question is how large is the closure of smooth forms in the space of OU-measures. So I raised a little discussion immediately after the talk, as pointed out by Tim Austin, under the weak topology on measures, this should be the whole space, with some extra points where the diffeos converge to something that’s not a homeo. Hence perhaps one can get the homeo depending weakly continuously on \mu.

Lifted surface flows:

Nelson Markley gave a talk about studying flows on surfaces by lifting them to the universal cover. i.e. Let \phi_t be a flow on some orientable surface S, put the standard constant curveture metric on S and lift the flow to \bar{\phi}_t on the universal cover of S.

There is an early result:

Theorem: (Weil) Let \phi_t be a flow on \mathbb{T}^2, \bar{\phi}_t acts on the universal cover \mathbb{R}^2, then for any p \in \mathbb{R}^2, if \displaystyle \lim_{t\rightarrow \infty} ||\bar{\phi}_t(p)|| = \infty then \lim_{t\rightarrow \infty} \frac{\bar{\phi}_t(p)}{||\bar{\phi}_t(p)||} exists.

i.e. for lifted flows, if an orbit escapes to infinity, then it must escape along some direction. (No sprial-ish or wild oscillating behavior) This is due to the nature that the flow is the same on each unit square.

We can find its analogue for surfaces with genus larger than one:

Theorem: Let \phi_t be a flow on S with g \geq 2, \bar{\phi}_t: \mathbb{D} \rightarrow \mathbb{D}, then for any p \in \mathbb{D}, if \displaystyle \lim_{t\rightarrow \infty} ||\bar{\phi}_t(p)|| = \infty then \lim_{t\rightarrow \infty} \bar{\phi}_t(p) is a point on the boundary of \mathbb{D}.

Using those facts, they were able to prove results about the structure of \omega limiting set of such orbits (those that escapes to infinity in the universal cover) using the geometric structure of the cover.

I was curious about what kind of orbits (or just non-self intersecting curves) would ‘escape’, so here’s some very simple observations: On the torus, this essentially means that the curve does not wind around back and forth infinitely often with compatible magnitudes, along both generators. i.e. the curve ‘eventually’ winds mainly in one direction along each generating circle. Very loosely speaking, if a somewhat similar thing is true for higher genus surfaces, i.e. the curve eventually winds around generators in one direction (and non-self intersecting), then it would not be able to have very complicated \omega limiting set.

Measures on Cantor sets

In contrast to the Oxtoby-Ulam theorem, one could ask: Given two measures on the standard middle-third Cantor set, can we always find a self homeomorphism of the Cantor set, pushing one measure to the other?

Given there are so many homeomorphisms on the Cantor set, this sounds easy. But in fact it’s false! –There are countably many clopen subsets of the Cantor set (Note that all clopen subsets are FINITE union of triadic copies of Cantor sets, a countable union would necessarily have a limit point that’s not in the union), a homeo needs to send clopen sets to clopen sets, hence for there to exist a homeo the countably many values the measures take on clopen sets must agree.

So a class of ‘good measures’ on Cantor sets was defined in the talk and proved to be realizable by a pull back the standard Hausdorff measure via a homeo.

I was randomly thinking about this: Given a non-atomic measure \mu on the Cantor set, when can it be realized as the restriction of the Lebesgue measure to an embedding of the Cantor set? After a little bit of thinking, this can always be done. (One can simple start with an interval, break it into two pieces according to the measure \mu of the clopen sets before and after the largest gap, then slightly translate the two pieces so that there is a gap between them; iterate the process)

In any case, it’s been a fun weekend! ^^

Moser’s theorem with boundary

Conan2 Comments

In the process of constructing a diffeo with a uniformly hyperbolic set with intermediate measure, I came across the following problem which I find interesting in its own right:

Given a C^1 diffeo f:S^1 \rightarrow S^1, when does it extend to a volume preserving diffeo of the unit disc to itself?

Some people suggested me to look into Moser’s theorem for volume forms on compact manifolds, I found it pretty cool, so here it is (taken from Moser’s paper):

Theorem: Given two smooth volume forms \tau, \sigma with same total volume on a compact connected C^\infty manifold M, there exists diffeomorphism \phi: M \rightarrow M s.t. \sigma = \phi^* \tau.

However this does not directly apply to the boundary problem as we are dealing with manifolds (disc) with boundary rather than entire compact manifolds…Hence to make it applicable to the case, I would have to get some kind of ‘relative’ version of the theorem.

The expectation is that this can be done by plugging in the case into Moser’s proof and see if it can be modified.. I’ll update this pose when I got around to do that (hopefully in the next few days)

Okay…I think I have finally figured out how to do this. (with a huge amount of hints and directions from Keith Burns). But temporarily I have to lose one degree of regularity when making the extension (i.e. starting with a C^2 f, extend it to a C^1 volume preserving. But at this point I strongly believe that we can in fact do it with f being C^1. (the regularity is lost when I extended the diffeo locally)

Theorem: Given C^2 diffeomorphism f:S^1 \rightarrow S^1, we can find (Lebesgue) volume preserving diffeomorphism \hat{f}: \mathbb{D} \rightarrow \mathbb{D} that extends f. (where \mathbb{D} is the closed unit disc and S^1 is its boundary)

Proof:

First we extend the diffeo f:S^1 \rightarrow S^1 to a neignbourhood of S^1 in \mathbb{D}.

Let A = \{ \ r e^{i \theta} \ | \ 1/2 < r \leq 1 \} be the half-open annuals with radius 1/2 and 1.

Let C = \mathbb{R} / 2 \pi \mathbb{Z} \times [0, \infty) be the half-cylinder.

Define \phi: A \rightarrow C s.t. \phi (r e^{i \theta}) = (\theta, 1/2 - r^2/2)

\phi is a volume preserving diffeo from A to \mathbb{R} / 2 \pi \mathbb{Z} \times [0, 3/8)

Consider h = \phi \circ f \circ \phi^{-1}: \mathbb{R} / 2 \pi \mathbb{Z} \times \{0\} \rightarrow \mathbb{R} / 2 \pi \mathbb{Z} \times \{0\}

(h^{-1})' is continuous hence bounded.

Choose \epsilon < 1/(3 \max_\theta | (h^{-1})' (\theta) | )

Define \hat{h}: \mathbb{R} / 2 \pi \mathbb{Z} \times [0, \epsilon) \rightarrow C s.t.

\hat{h}(\theta, y) = ( \pi_1(h(\theta, 0) ), y |(h^{-1})'(\theta) | )

y |(h^{-1})'(\theta) | < \epsilon \max_\theta | (h^{-1})'(\theta) | 0 \ \mathbb{R} / 2 \pi \mathbb{Z} \times [0, \delta) \subseteq \hat{h}(\mathbb{R} / 2 \pi \mathbb{Z} \times [0, \epsilon/4)).

Let \epsilon'=1-\sqrt{1-2\epsilon}, \ \delta'=1-\sqrt{1-2\delta}. Let g:N_{\epsilon'}(S^1) \rightarrow A, \ g := \phi^{-1} \circ \hat{h} \circ \phi.

Hence g is volume preserving, g|_{S^1} = f and N_{\delta'}(S^1) \subseteq g(N_{\epsilon'/2}(S^1)).

Hence we have successfully extended f to a neighborhood of S^1 in a volume preserving way.

Now we further extend g|_{N_{\epsilon'/2}(S^1)} to a (non-volume-preserving) diffeo of \mathbb{D} to itself.

This can be done by first take f \times I on C, average it with g by a C^\infty bump function that is 1 on \mathbb{D} \backslash N_{\epsilon'}(S^1) and vanishes on N_{\epsilon'/2}(S^1).

Since both functions preserve vertical segments, taking the resulting map back to A will produce a diffeo except for at the point 0. We may smooth out the map at 0. Call the resulting diffeo \hat{g}: \mathbb{D} \rightarrow \mathbb{D}

Define measure \mu on \mathbb{D} by \mu(B) = \lambda(\hat{g}^{-1}(B)) where B is any Lebesgue measurable set and \lambda is the Lebesgue measure.

Since \hat{g}^{-1} is volume preserving on N_{\delta'}(S^1), hence \mu is equal to \lambda on N_{\delta'}(S^1).

Also we have \mu(\mathbb{D}) = \lambda(\hat{g}^{-1}(\mathbb{D}) = \lambda(\mathbb{D}).

Hence we can apply lemma 2 in Moser’s paper:

Lemma: If two C^r volume forms \mu_1, \ \mu_2 agree on an \epsilon neighbourhood of the boundary of the cube (disc in our case) with the same total volume, then there exists C^r diffeo \psi from the cube (disc) to itself s.t. \psi is identity on the \epsilon neighbourhood of the boundary and \mu_1(\psi(B)) = \mu_2(B) for all measurable set B.

We apply the lemma to \mu and \lambda, obtain \psi.

Hence \lambda(B) = \mu(\psi(B)) = \lambda(\hat{g}^{-1} \circ \psi(B))

\hat{g}^{-1} \circ \psi is a C^1 diffeo that preserved Lebesgue volume, so is its inverse \psi^{-1} \circ \hat{g}.

Let \hat{f} = \psi^{-1} \circ \hat{g}, \ \hat{f}|_{S^1} = \hat{g}|_{S^1} = f.

Hence \hat{f} is a volume preserving extension of f.

Generic Accessibility (part 1) – Andy Hammerlindl

Conan2 Comments

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 \dim (E^c)=1 by using accessibility.

We are going to focus on the proof of Conjecture A, here’s a sketch of proof of Conjecture B when \dim (E^c)=1 is assumed:

Theorem B: Let M be compact manifold, f: M \rightarrow M be a measure preserving accessible partially hyperbolic diffeomorphism, \dim (E^c)=1, then f is ergodic.

Proof: Let \phi: M \rightarrow \mathbb{R} be f invariant

Let A = \phi^{-1}(( - \infty, c]), if m(A)>0 then \exists p \in A s.t. p is a density point of A.

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 \dim (E^c)=1 (or in this case even the weaker hypothesis center brunching would work) to prove that in our case y \in M is a density point iff y is a “leaf density point” in both its center-stable and center-unstable leaves. Hence by accessibility from p to y, we can “push” the point p along the us-path that joins p to y and induce that y is a “leaf density point” in A hence a density point in A.

\therefore all points y \in M are density points of A hence m(A)=1.

Note that here if we replace accessibility by essential accessibility, we still get m(A)=1.

Hence \forall c \in \mathbb{R}, either m(\phi^{-1}(( - \infty, c]))=0 or m(\phi^{-1}(( - \infty, c])) = 1

\therefore \ \phi is essentially constant. \therefore f is ergodic.

This establishes theorem B.

Let PH^r(M) be the set of measure preserving diffeomorphisms on M that are of class C^r

Theorem A: Accessibility is open dense in the space of diffeomorphisms in PH^r(M) with \dim(E^c) = 1.

For any x \in M, let AC(x) denote the set of points that’s accessible from x

Let \mathcal{D} = \{ f \in {PH}^r (M) | \forall \ x \in Per(f), AC(x) is open \}

Fact: \mathcal{D} \subseteq PH^r(M) with \dim(E^c) = 1 is G_\delta and \mathcal{D} = \mathcal{A} \sqcup \mathcal{B}
where \mathcal{A} = \{ \ f \ | \ f is accessible \} and
\mathcal{B} = \{ \ f \ | \ per(f) = \phi and E^u \oplus E^s 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 U(f) = \{ \ x \in M \ | \ AC(x) is open \}

It’s easy to see that U(f) is automatically open hence V(f) = M \setminus U(f) is compact.

Proposition: Let x \in M, the following are equivalent:

1) AC(x) has non-empty interior

2) AC(x) is open

3) AC(x) \cap \mathcal{W}^c_{loc}(x) has non-empty interior in \mathcal{W}^c_{loc}(x)

Proof: 1) \Rightarrow 2) \Rightarrow 3) \Rightarrow 1)

Mainly by drawing pictures and standard topology.

Unweaving lemma: \forall x \in Per(f), \ \exists g \in latex PH^r(M)$ with \dim(E^c) = 1 s.t. the C^r distance between f and g is arbitrarily small, x \in Per(g) and AC_g(x) is open.

The proof is left to the second part of the talk…