[This post](https://mathoverflow.net/questions/194683/are-carnot-groups-as-carnot-caratheodory-metric-spaces-doubling) shows that every Carnot group is a [doubling metric space](https://en.wikipedia.org/wiki/Doubling_space). However, what is its doubling constant?



Doubling constant of Carnot group
If the homogeneous dimension of the Carnot group is $s$, then the $s$-dimensional Hausdorff measure satisfies $\mathcal{H}^s(B(x,r))=Cr^s$ with a fixed constant $C$ independent of the center and the radius of the ball. Therefore
$\mathcal{H}^s(B(x,2r))=C(2r)^s=2^sCr^s=2^s\mathcal{H}^s(B(x,r))$ and the doubling constant is $2^s$.


For example, the Hesinberg group $\mathbb{H}^1$ is $\mathbb{R}^3$, but $\mathbb{H}^1$ is equipped with the Carnot-Caratheodory metric which is non-euclidean. With respect to this metric, the Hausdorff dimension of $\mathbb{H}^1$ is $4$ and hence $\mathcal{H}^4(B(x,r))=Cr^4$. It is rather surprising, but the $4$-dimensional measure on $\mathbb{H}^1$ coincides with the $3$-dimensional Lebesgue measure on $\mathbb{R}^3$ (up to a multiplicative factor). How is that possible? This is because balls with respect to the Carnot-Caratheodory metric are roughly speaking close in the shape to ellipsoids with two semi-axes of length $r$ and one semiaxis of length $r^2$. Therefore the Euclidean volume of such a ball is $Cr^4$.


The same is true in any Carnot group of homogeneous dimension $s$. Any such Carnot group is an euclidean space $\mathbb{R}^n$ for some $n\leq s$, but it is equipped with a non-euclidean Carnot-Caratheodory metric and the $s$-Hausdorff measure equals to the $n$-dimensional Lebesgue measure up to a multiplicative factor. One should also note that $s$ is always an integer greater than or equal $n$.


