# How to prove that the Laplace oparator of the Heisenberg group is sub-elliptic?

The Heisenberg group $H^3$ is the set $\mathbb C\times \mathbb R$ endowed with the group law $$(z,t)\cdot(w,s) =\left (z+w, \,t+s+\tfrac{1}{2}\Im m(z \bar{w})\right).$$ For $z=x+ i y \in \mathbb C$ and $t\in \mathbb R$, the Laplacian operator on the Heisenberg group $H^3$ is given by $$\Delta= \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + (x^2+y^2 ) \frac{\partial^2}{\partial t^2} + 2(x\frac{\partial}{\partial y} -y \frac{\partial}{\partial x} ) \frac{\partial}{\partial t} .$$ I want to show that the Laplacian operator $\Delta$ is a sub-elliptic but not elliptic.

• Also, for subellipticity, do you just want the "qualitative" statement "if $\Delta u \in C^\infty$ then $u \in C^\infty$", or do you want the "quantitative" subelliptic estimates on Sobolev norms? – Nate Eldredge Feb 2 '16 at 17:39
• That talk shows two (inequivalent) definitions of "subelliptic", both of which are satisfied by the Heisenberg Laplacian, which is very easy to check. If you're using one of those definitions, then I don't understand what your question is. But those are strange definitions, since they don't lead directly to any useful analytic properties; and they're satisfied by truly degenerate operators like $\partial_x^2 + \partial_y^2$ on $\mathbb{R}^3$, or the 0 operator. – Nate Eldredge Feb 2 '16 at 20:02