Volume of the subelliptic ball

Question

Let $\Omega \in \mathbb{R}^n$ a bounded open set when $n\geq 2$, and let $X_{1},X_{2},\cdots,X_{m}$ be real smooth vector fields that satisfy Hormander condition on $\Omega$. If we denote $Q(x)$ as the Hormander index of each $x\in \Omega$, then for each $1\leq i\leq Q(x)$ we denote $C_{i}(x)$ as the subspace of the tangent space $T_{x}(\Omega)$ which is spanned by the vector fields $\{X_{J}\}$ with $|J|\leq i$ ($X_{J}$ means $X_{1},X_{2},\cdots X_{m}$ together with all $|J|$ step repeated commutators).

Next we define $$ v(x)=\sum_{i=1}^{Q(x)}i(\dim C_{i}(x)-\dim C_{i-1}(x))\qquad \dim C_{0}(x)=0 $$ From Stein's article "Balls and metrics defined by vector fields I:Basic propoerties" We know that we can define a metrics via the vector fields by $$ d(x,y)=\inf\{\delta>0|\exists\varphi\in C(\delta)\qquad \varphi(0)=x,\varphi(1)=y \}.$$ Here $C(\delta)$ is a class of absolutely continuous mappings $\varphi:[0,1]\to \Omega$ which almost everywhere satisfy the differential equation $$\varphi'(t)=\sum_{j=1}^{q}a_{j}(t)Y_{j}(\varphi(t)) $$ with $|a_{j}(t)|<\delta^{d_{j}}$. Finally, we define the sub-elliptic Ball $ B(x,r)=\{y\in\Omega|d(x,y)<r\}$. I wonder if the volume of $B(x,r)$ has some relation with $r^{v(x)}$? in other words - can we find some positive constant $C_{1},C_{2}$ so that $$ C_{1}r^{v(x)}\leq |B(x,r)|\leq C_{2}r^{v(x)} ?$$

Here $|B(x,r)| $ means the Lebesgue measure of $B(x,r)$ in $\mathbb{R}^n$

Answer 1

What you defined is a special case of sub-Riemannian structure, and indeed there is such an estimate.

A basic (but non-trivial) fact in sub-Riemannian geometry is that, for any point $x$ there is a particular choice of coordinates $y_1,\ldots,y_n$, called privileged coordinates, which are in a sense adapted to the local geometry. Given such a set of coordinates, to each index $i \in 1,\ldots,n$ we define a weight $w_i$, in this way: the first $C_1(x)$ coordinates have weight $w_i(x)=1$, the next $C_2(x)-C_1(x)$ have weight $w_i(x) = 2$ and so on. Then, we define the weighted box as

$$\mathrm{Box}^w(\epsilon) := \{y \in \mathbb{R}^n \mid |y_i| \leq \epsilon^{w_i}, \quad i =1,\ldots,n \}.$$

Notice that this is a set with no intrinsic meaning, it makes sense only in this set of privileged coordinates. The ball-box theorem states that there exist positive constants $a < A$ and $\epsilon_0>0$ such that for all $\epsilon < \epsilon_0$

$$\mathrm{Box}(a \epsilon) \subset B(x,\epsilon) \subset \mathrm{Box}(A \epsilon). $$

Taking the Lebesgue volumes of the above inclusion gives the estimate:

$$(2a)^{v(x)} \epsilon^{v(x)} \leq |B(x,\epsilon)| \leq(2A)^{v(x)} \epsilon^{v(x)}.$$

This estimate holds only in this particular set of privileged coordinates, but at least for small $\epsilon_0$ (and with different constants) also in your original coordinates $x_1,\ldots,x_n$, since the transformation $(y_1,\ldots,y_n) \mapsto (x_1,\ldots,x_n)$ is a smooth change of coordinates, and small balls are compact.

However, the Lebesgue measure of balls has no intrinsic geometric meaning. A more interesting object is the Hausdorff measure, intrinsically associated with the sub-Riemannian distance.

If your structure is equiregular, that is $v(x)$ is constant, then the formula you used to define $v(x)$ is known as Mitchell's formula, and $Q = v(x)$ coincides with the Hausdorff dimension of the sub-Riemannian structure.

An excellent reference for this is Montgomery's book "A tour of sub riemannian geometries their geodesics and applications" (Chapters 2.3 and onward).

For the non-equiregular case (i.e. when $v(x)$ is not constant), the relation between the Lebesgue and Hausdorff measure is much more complicated, as the two are not absolutely continuous one with respect to the other. This interesting problem has been studied recently in this paper.