Someone recently gave me the following way to view the Heisenberg group, but I have forgotten some of the details.
Here is what I remember:
Take $S^3 \subset \mathbb R^4$. Every point $x \in S^3$ has a normal vector $n_x \in \mathbb R^4$, and we can identify the hyperplane tangent to $n_x$ with the tangent space $T_x S^3$. (Everything is over ℝ so far.)
Next, we view $n_x$ as an element of $\mathbb C^2$. Then we can consider $H_x S^3 := \{v \in \mathbb C^2 ~|~ \langle n_x, v \rangle = 0 \} \subset T_xS^3$ (where we take the complex inner product).
My question is:
What do we do from here? Since $S^3$ is three dimensional, for any point $x \in S^3$, there is a chart $(U, \psi)$ with $x \in U$ and $\psi : U \to \mathbb R^3$ a homeomorphism onto its image. Is there a proper/natural way to choose $\psi$, so that the image of $HS^3$ under $d\psi$ coincides with the horizontal subbundle of the Heisenberg group?