My question is in some sense a less ambitious version of the following MO question where the answer was inconclusive.
A Carnot group of step $N$ can be identified within the tensor algebra, modulo tensors order $N$. The fact that this is a Lie group, and the fact that the Carnot groups are studied in sub-Riemannian geometry makes me curious if there should be a nice characterisation of what it "does" geometrically in the same way $SO(n)$ performs rotations of $n$ space.
The specific case of the Carnot group on 2 dimensional vector spaces gives the Heisenberg group. This has two representations, one being matrices acting on an extended plane $(x,y,1)^T$ and the other acting on $L^2(\mathbb{R})$ functions relevant to quantum mechanics. Neither of these seem generalisable to other Carnot groups and seem to crucially depend on the fact that the "step 2" component is scalar.
My other hunch was inspired by the construction from Bourbaki's Algebra 1, (credit to this MSE this question I found) and to generalise the central extension structure of the Heisenberg group to exact sequence like:
$$G^N/ \, V \subset G^N\rightarrow G^N \rightarrow V$$
and then say that elements of $V$, possibly identified as $(1,V,0 \ldots)$ act on "higher order" elements of the form $(1,0, [V,V], \ldots )$ which have $0$ in the $V$ slot. I'm sure this is certainly not correct and this isn't a proper exact sequence unless $G^N / V = \mathbb{R}$ like the Heisenberg group, but I actually got somewhere following this line of thought, until I ran into the trouble that the group action ends up being trivial, at least in the most easy/interesting step 2 case.
Is such an action possible? Or is the Heisenberg group truly uniquely "more geometric" than another other Carnot group?
