Archive for March 22nd, 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.