 Solution

For $V={\mathbb{R}}^2$ with basis $e_1,e_2$, the step-two Lie algebra has basis $e_1,e_2,[e_1,e_2]$ and its last element is central. WriteIts first level is $v=(x,y)$ and its second level isThe step-two multiplication givesConsequently the mapis an isomorphism onto the real Heisenberg group. The shift by $xy/2$ converts exponential coordinates into the matrix coordinates used for that group. In exponential coordinates the left-invariant horizontal vector fields areTaking $X,Y$ orthonormal gives exactly the Carnot-Carathéodory distance inherited from $G^2({\mathbb{R}}^2)$.