## Posts Tagged ‘Whitney’s extension theorem’

### 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.

### Whitney’s extension theorem revisited

March 9, 2010

I (very surprisingly) bumped into Charles Fefferman at Northwestern this afternoon…Hence we talked math for a little bit. Among other things I mentioned that I’ve been trying to extend $C^1$ functions to the disc volume-preservingly. After trying on the board for a while, he laughs out loud when he saw that this may be obtained applying his favorite Whitney’s extension theorem. (I’ll discuss what he did later in the poster)

Mean while, it’s a pity that I’ve never written a post on Whitney’s extension theorem, hence here it is~

Given a compact subset $K$ in $\mathbb{R}^n$ and a function $f: K \rightarrow \mathbb{R}$, when can we extend it to a $C^r$ function on the whole $\mathbb{R}^n$?

First we note that there are obvious cases for which this can’t be done: for example, if we take $E$ to be a segment in $\mathbb{R}^2$ and $f$ a one-variable function of lower regularity than $r$, then of course there are no way to find a $C^r$ extension.

Hence it’s only reasonable to restrict our attention to those $f$ that has ‘candidate derivatives’ of all orders no larger than $r$ at all points in $E$.

i.e. For any $k$-fold subscript $d= (d_1, d_2, \cdots, d_k)$ with $d_1+d_2+ \cdots +d_k \leq r$ (we will denote $d_1+d_2+ \cdots +d_k = |d|$, there is a continuous function $f_d: K \rightarrow \mathbb{R}$ with the following property:

For all $x_o \in K$, $\displaystyle f_d(x) = \sum_{|l| \leq r-|d|} \frac{f_{l+d}(x)}{l!}(x-x_0)^l+R_d(x, x_0)$  where $R_d(x, x_0) \sim o(|x-x_0|^{r-|d|})$ as $x \rightarrow x_0$ and is uniform in $x_0$.

i.e. The functions $f_\alpha$ are compatible as Taylor coefficients of some $C^r$ function on $\mathbb{R}^n$, which is absolutely necessary for a $C^r$ extension to exist.

Whitney’s extension theorem: (classical version)

Suppose a set of functions $f_\alpha$ with all multi-index $| \alpha | \leq r$ satisfying the above Taylor condition at all points in $K$. Then there is a $C^r$ function $\hat{f}: \mathbb{R}^n \rightarrow \mathbb{R}$ s.t. $\hat{f}|_K = f_{\bar{0}}$ and for all $\alpha \leq r$, $(D^\alpha \hat{f})|_K = f_\alpha$. Furthermore, $\hat{f}$ can be taken real analytic on $\mathbb{R}^n \backslash K$.

This is indeed the best one could hope for. i.e. there is a $C^r$ extension whenever possible, furthermore the extension is at worst $C^r$ at the points which it is given to be only $C^r$ and much better (analytic) everywhere else.

However, sometimes we would like to control the $C^r$ norm of the resulting function in terms of the $C^r$ norm of the function on $K$.

Theorem: (Fefferman)

For any $n, \ r$, there exists $C$ such that the extension $||\hat{f}|| \leq C \cdot ||f||$ where the norm is the $C^r$ norm.