## Posts Tagged ‘Carnot-Carathéodory metric’

### 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.