Posts Tagged ‘Amie Wilkinson’

A train track on twice punctured torus

April 22, 2012

This is a non-technical post about how I started off trying to prove a lemma and ended up painting this:

One of my favorite books of all time is Thurston‘s ‘Geometry and Topology of 3-manifolds‘ (and I just can’t resist to add here, Thurston, who happen to be my academic grandfather, is in my taste simply the coolest mathematician on earth!) Anyways, for those of you who aren’t topologists, the book is online and I have also blogged about bits and parts of it in some old posts such as this one.

I still vividly remember the time I got my hands on that book for the first time (in fact I had the rare privilege of reading it from an original physical copy of this never-actually-published book, it was a copy on Amie‘s bookshelf, which she ‘robbed’ from Benson Farb, who got it from being a student of Thurston’s here at Princeton years ago). Anyways, the book was darn exciting and inspiring; not only in its wonderful rich mathematical content but also in its humorous, unserious attitude — the book is, in my opinion, not an general-audience expository book, but yet it reads as if one is playing around just to find out how things work, much like what kids do.

To give a taste of what I’m talking about, one of the tiny details which totally caught my heart is this page (I can’t help smiling each time when flipping through the book and seeing the page, and oh it still haunts me >.<):

This was from the chapter about Kleinian groups, when the term ‘train-track’ was first defined, he drew this image of a train(!) on moving on the train tracks, even have smoke steaming out of the engine:

To me such things are simply hilarious (in the most delightful way).

Many years passed and I actually got a bit more into this lamination and train track business. When Dave asked me to ‘draw your favorite maximal train track and test your tube lemma for non-uniquely ergodic laminations’ last week, I ended up drawing:

Here it is, a picture of my favorite maximal train track, on the twice punctured torus~! (Click for larger image)

Indeed, the train is coming with steam~

Since we are at it, let me say a few words about what train tracks are and what they are good for:

A train track (on a surface) is, just as one might expect, a bunch of branches (line segments) with ‘switches’, i.e. whenever multiple branches meet, they must all be tangent at the intersecting point, with at least one branch in each of the two directions. By slightly moving the switches along the track it’s easy to see that generic train track has only switches with one branch on one side and two branches on the other.

On a hyperbolic surface $S_{g,p}$, a train track is maximal if its completementry region is a disjoint union of triangles and once punctured monogons. i.e. if we try to add more branches to a maximal track, the new branch will be redundant in the sense that it’s merely a translate of some existing branch.

As briefly mentioned in this post, train tracks give natural coordinate system for laminations just like counting how many times a closed geodesic intersect a pair of pants decomposition. To be slightly more precise, any lamination can be pushed into some maximal train track (although not unique), once it’s in the track, any laminations that’s Hausdorff close to it can be pushed into the same track. Hence given a maximal train track, the set of all measured laminations carried by the train track form an open set in the lamination space, (with some work) we can see that as measured lamination they are uniquely determined by the transversal measure at each branch of the track. Hence giving a coordinate system on $\mathcal{ML})(S)$.

Different maximal tracks are of course them pasted together along non-maximal tracks which parametrize a subspace of $\mathcal{ML}(S)$ of lower dimension.

To know more about train tracks and laminations, I highly recommend going through the second part of Chapter 8 of Thurston’s book. I also mentioned them for giving coordinate system on the measured lamination space in the last post.

In any case I shall stop getting into the topology now, otherwise it may seem like the post is here to give exposition to the subject while it’s actually here to remind myself of never losing the Thurston type childlike wonder and imagination (which I found strikingly larking in contemporary practice of mathematics).

A survey on ergodicity of Anosov diffeomorphisms

March 7, 2011

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

January 23, 2011

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.

Hausdorff dimension of projections

March 30, 2010

A few days ago, professor Wilkinson asked me the following question on google talk (while I was in Toronto):

Say that a set in $\mathbb{R}^n$ is a $k$-zero set for some integer $k if for every $k$-dimensional subspace $P$, saturating the set $X$ by planes parallel to $P$ yields a set of $n$-dimensional Lebesgue measure zero. How big can a $k$-zero set be?”

On the spot my guess was that the Hausdorff dimension of the set is at most $n-k$. In deed this is the case:

First let’s note that $n$-dimensional Lebesgue measure of the $P$-saturated set is $0$ iff the $n-k$ dimensional Lebesgue measure of the projection of our set to the $n-k$ subspace orthogonal to $P$ is $0$.

Hence the question can be reformulated as: If a set $E \subseteq \mathbb{R}^n$ has all $n-k$ dimensional projection being $n-k$ zero sets, how big can the set be?

Looking this up in the book ‘The Geometry of Fractal Sets’ by Falconer, indeed it’s a theorem:

Theorem: Let $E \subseteq \mathbb{R}^n$ compact, $\dim(E) = s$ (Hausdorff dimension), let $G_{n,k}$ be the Garssmann manifold consisting of all $k$-dimensional subspaces of $\mathbb{R}^n$, then
a) If $s \leq k$, $\dim(\mbox{Proj}_\Pi E) = s$ for almost all $\Pi \in G_{n,k}$

b) If $s > k$, $\mbox{Proj}_\Pi E$ has positive $k$-dimensional Lebesgue measure for almost all $\Pi \in G_{n,k}$.

In our case, we have some set with all $n-k$-dimensional projection having measure $0$, hence the set definitely does not satisfy b), i.e. it has dimensional at most $n-k$. Furthermore, a) also gives that if we have a uniform bound on the dimension of almost all projections, this is also a bound on the dimension of our original set.

This is strict as we can easily find sets that’s $n-k$ dimensional and have all such projections measure $0$. For example, take an $n-k$ subspace and take a full-dimension measure $0$ Cantor set on the subspace, the set will have all projections having measure $0$.

Also, since the Hausdorff dimension of any projection can’t exceed the Hausdorff dimension of the original set, a set with one projection having positive $n-k$ measure implies the dimension of the original set is $\geq n-k$.

Question 2: If one saturate a $k$-zero set by any smooth foliations with $k$-dimensional leaves, do we still get a set of Lebesgue measure $0$?

We answer the question in the affective.

Given foliation $\mathcal{F}$ of $\mathbb{R}^n$ and $k$-zero set $E$. For any point $p \in E$, there exists a small neighborhood in which the foliation is diffeomorphic to the subspace foliation of the Euclidean space. i.e. there exists $f$ from a neighborhood $U$ of $p$ to $(-\epsilon, \epsilon)^n$ where the leaves of $\mathcal{F}$ are sent to $\{\bar{q}\} \times (-\epsilon, \epsilon)^k$, $\bar{q} \in (-\epsilon, \epsilon)^{n-k}$.

By restricting $f$ to a small neighborhood (for example, by taking $\epsilon$ to be half of the origional $\epsilon$), we may assume that $f$ is bi-Lipschitz. Hence the measure of the $\mathcal{F}$-saturated set inside $U$ of $U \cap E$ is the same as $f(U \cap E)$ saturated by parallel $k$-subspaces inside $(-\epsilon, \epsilon)^n$. Dimension of $f(E)$ is the same as dimension of $E$ which is $\leq n-k$, if the inequality is strict, then all projections of $f(E)$ onto $n-k$ dimensional subspaces has measure $0$ i.e. the saturated set by $k$-planes has $n$ dimensional measure $0$.

…to be continued

Billiards

September 24, 2009

Amie Wilkinson asked me the following question some time ago:

Given a smooth convex Jordan curve $J \subseteq \mathbb{R}^2$, consider the billiard map $\varphi: J \times (0, \pi) \rightarrow J \times (0, \pi)$, let $\pi_\theta: J \times (0, \pi) \rightarrow (0, \pi)$ be the projection.

a) If $\forall (p, \theta) \in J \times (0, \pi), \pi_\theta \circ \varphi (p, \theta) = \theta$, does this imply $J$ is a circle?

(Yes, in fact we only need $\varphi$ to fix the second component of points in $\{p, q \} \times (0, \pi)$ for a chosen pair of points $p, q$. Classical geometry)

b) If $\forall p \in J, \pi_\theta \circ \varphi (p, \pi/2) = \pi/2$, does this imply $J$ is a circle?

(No, my example was a cute construction that attaches six circular arcs together)

c) What’s the smallest set $S \subseteq (0, \pi)$ s.t. if $\varphi$ fixes the second component on $J \times S$ then $J$ has to be a circle?

I am still thinking about c)…My guess is that any sub interval would work, and of course any dense subset inside a given set works equally well as the whole set…

But is it possible to have only finitely many angles? Maybe even two angles?