## Archive for March, 2011

### The Carnot-Carathéodory metric

March 22, 2011

I remember looking at Gromov’s notes “Carnot-Carathéodory spaces seen from within” a long time ago and didn’t get anywhere. Recently I encountered it again through professor Guth. This time, with his effort in explaining, I did get some ideas. This thing is indeed pretty cool~ So I decided to write an elementary introduction about it here. We will construct a such metric in $\mathbb{R}^3$.

In general, if we have a Riemannian manifold, the Riemannian distance between two given points $p, q$ is defined as

$\inf_\Gamma(p,q) \int_0^1||\gamma'(t)||dt$

where $\Gamma$ is the collection of all differentiable curves $\gamma$ connecting the two points.

However, if we have a lower dimensional sub-bundle $E(M)$ of the tangent bundle (depending continuously on the base point). We may attempt to define the metric

$d(p,q) = \inf_{\Gamma'} \int_0^1||\gamma'(t)||dt$

where $\Gamma'$ is the collection of curves connecting $p, q$ with $\gamma'(t) \in E(M)$ for all $t$. (i.e. we are only allowed to go along directions in the sub-bundle.

Now if we attempt to do this in $\mathbb{R}^3$, the first thing we may try is let the sub-bundle be the say, $xy$-plane at all points. It’s easy to realize that now we are ‘stuck’ in the same height: any two points with different $z$ coordinate will have no curve connecting them (hence the distance is infinite). The resulting metric space is real number many discrete copies of $\mathbb{R}^2$. Of course that’s no longer homeomorphic to $\mathbb{R}^3$.

Hence for the metric to be finite, we have to require accessibility of the sub-bundle: Any point is connected to any other point by a curve with derivatives in the $E(M)$.

For the metric to be equivalent to our original Riemannian metric (meaning generate the same topology), we need $E(M)$ to be locally accessible: Any point less than $\delta$ away from the original point $p$ can be connected to $p$ by a curve of length $< \varepsilon$ going along $E(M)$.

At the first glance the existence of a (non-trivial) such metric may not seem obvious. Let’s construct one on $\mathbb{R}^3$ that generates the same topology:

To start, we first identify our $\mathbb{R}^3$ with the $3 \times 3$ real entry Heisenberg group $H^3$ (all $3 \times 3$ upper triangular matrices with “1”s on the diagonal). i.e. we have homeomorphism

$h(x,y,z) \mapsto \left( \begin{array}{ccc} 1 & x & z \\ 0 & 1 & y \\ 0 & 0 & 1 \end{array} \right)$

Let $g$ be a left-invariant metric on $H_3$.

In the Lie algebra $T_e(H_3)$ (tangent space of the identity element), the elements $X = \left( \begin{array}{ccc} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) , Y = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} \right)$ and $Z = \left( \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right)$ form a basis.

At each point, we take the two dimensional sub-bundle $E(H_3)$ of the tangent bundle generated by infinitesimal left translations by $X, Y$. Since the metric $g$ is left invariant, we are free to restrict the metric to $E(M)$ i.e. we have $||X_p|| = ||Y_p|| = 1$ for each $p \in M$.

The interesting thing about $H_3$ is that all points are accessible from the origin via curves everywhere tangent to $E(H_3)$. In other words, any points can be obtained by left translating any other point by multiples of elements $X$ and $Y$.

The “unit grid” in $\mathbb{R}^3$ under this sub-Riemannian metric looks something like:

Since we have

$\left( \begin{array}{ccc} 1 & x & z \\ 0 & 1 & y \\ 0 & 0 & 1 \end{array} \right) \left( \begin{array}{ccc} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) = \left( \begin{array}{ccc} 1 & x+1 & z \\ 0 & 1 & y \\ 0 & 0 & 1 \end{array} \right)$,

the original $x$-direction stay the same, i.e. a bunch of horizontal lines connecting the original $yz$ planes orthogonally.

However, if we look at a translation by $Y$, we have

$\left( \begin{array}{ccc} 1 & x & z \\ 0 & 1 & y \\ 0 & 0 & 1 \end{array} \right) \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} \right) = \left( \begin{array}{ccc} 1 & x & z+x \\ 0 & 1 & y+1 \\ 0 & 0 & 1 \end{array} \right)$

i.e. a unit length $Y$-vector not only add a $1$ to the $y$-direction but also adds a height $x$ to $z$, hence the grid of unit $Y$ vectors in the above three $yz$ planes look like:

We can now try to see the rough shape of balls by only allowing ourselves to go along the unit grid formed by $X$ and $Y$ lines constructed above. This corresponds to accessing all matrices with integer entry by words in $X$ and $Y$.

The first question to ask is perhaps how to go from $(0,0,0)$ to $(0,0,1)$. –since going along the $z$ axis is disabled. Observe that going through the following loop works:

We conclude that $d_C((0,0,0), (0,0,1)) \leq 4$ in fact up to a constant going along such loop gives the actual distance.

At this point one might feel that going along $z$ axis in the C-C metric is always takes longer than the ordinary distance. Giving it a bit more thought, we will find this is NOT the case: Imagine what happens if we want to go from $(0,0,0)$ to $(0,0,10000)$?

One way to do this is to go along $X$ for 100 steps, then along $Y$ for 100 steps (at this point each step in $Y$ will raise $100$ in $z$-coordinate, then $Y^{-100} X^{-100}$. This gives $d_C((0,0,0), (0,0,10000)) \leq 400$.

To illustrate, let’s first see the loop from $(0,0,0)$ to $(0,0,4)$:

The loop has length $8$. (A lot shorter compare to length $4$ for going $1$ unit in $z$-direction)

i.e. for large $Z$, it’s much more efficient to travel in the C-C metric. $d_C( (0,0,0), (0,0,N^2)) = 4N$

In fact, we can see the ball of radius $N$ is roughly an rectangle with dimension $R \times R \times R^2$ (meaning bounded from both inside and outside with a constant factor). Hence the volume of balls grow like $R^4$.

Balls are very “flat” when they are small and very “long” when they are large.

### Slides for my little Anosov talk

March 14, 2011

As promised~ have fun!

*Actually I’m a strong supporter of the idea that all talks should be done on blackboards…However, this time since the talk is only 25 minutes long and it takes me 5 minutes to draw a product Cantor set, I had to use slides…

Hence I made fake blackboard slides…

### 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)