Heisenberg group

Question

I want to determine the board of the of the disk $D=\{z=(z_1,z_2)\in \mathbb{C^2}:|z|^2=|z_1|^2+|z_2|^2 <1 \}$.

Is it the Heisenberg group i,e., $\delta D=\mathbb{H^2}$?

Answer 1

The boundary of the unit ball cannot be isomorphic to the Heisenberg Lie group. For the correct definition of the Heisenberg group in this context see section 1.4 here. We obtain the Heisenberg group as the group of affine holomorphic bijections of $U^n$.

Answer 2

The highlighted sentence is misleading, if not outright false. The boundary of the unit ball in $\mathbb C^{n+1}$ is the sphere $\mathbb S^{2n+1}$, which is compact; while the Heisenberg group $H^n$ is not, so you cannot identify the Heisenberg group with the boundary of the ball, either as Lie groups or as manifolds.

I suspect that what the author had in mind was one or both of the following:

1. Both the $(2n+1)$-sphere and the Heisenberg group are CR manifolds, which are the abstract models of boundaries of domains in $\mathbb C^{n+1}$. The sphere obtains its CR structure as the boundary of the ball $\mathbb B^{n+1}$, while the Heisenberg group inherits its as the boundary of the Siegel domain $\mathscr D^{n+1} = \{(z^1,\dots,z^n,w): \operatorname{Im} w > |z|^2\}$. The ball and the Siegel domain are biholomorphic via the $(n+1)$-dimensional Cayley transform, which is the map $K\colon \mathbb B^{n+1}\to \mathscr D^{n+1}$ given by $$ K(\zeta^1,\dots,\zeta^{n+1}) = \left( \frac{\zeta^1}{1+\zeta^{n+1}},\dots, \frac{\zeta^n}{1+\zeta^{n+1}}, i\frac{1-\zeta^{n+1}}{1+\zeta^{n+1}}\right). $$ This map extends smoothly to a diffeomorphism from $\overline{\mathbb B^{n+1}}\smallsetminus \{(0,\dots,0,-1)\}$ to $\overline{\mathscr D^{n+1}}$, and then restricts to a CR diffeomorphism from the sphere minus a point to the Heisenberg group. For more about this, see sections 3 and 4 in my paper with David Jerison on the CR Yamabe problem.
2. As Dietrich Burde noted, the group structure on $H^n$ can be realized by identifying $H^n$ with the subgroup $G$ of the biholomorphism group of $\mathscr D^{n+1}$ consisting of affine maps. These biholomorphisms extend smoothly to $\overline {\mathscr D^{n+1}}$, and $G$ acts simply transitively on $\partial \mathscr D^{n+1} \approx H^n$, so the map from $G$ to $H^n$ given by $g\mapsto g\cdot(0,\dots,0)$ is a bijection.