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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.

Thank you in adavance

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    $\begingroup$ Non-ellipticity is easy, just check that the principal symbol is degenerate (compute determinant or whatever you like). For subellipticity, are you happy with the "big gun" of Hormander's theorem, or do you want a direct argument? $\endgroup$
    – Nate Eldredge
    Commented Feb 2, 2016 at 17:30
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    $\begingroup$ 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? $\endgroup$
    – Nate Eldredge
    Commented Feb 2, 2016 at 17:39
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    $\begingroup$ I would just use the definition of sub-elliptic of a operator $\endgroup$
    – Z. Alfata
    Commented Feb 2, 2016 at 17:44
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    $\begingroup$ 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. $\endgroup$
    – Nate Eldredge
    Commented Feb 2, 2016 at 20:02
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    $\begingroup$ Subellipticity is mostly useful as a substitute for elliptic regularity, and those definitions don't imply anything of the kind. Even if we strengthen them by adding a bracket-generating type condition, so that they do imply subellipticity in the sense of Folland, the proof of this implication is not trivial. $\endgroup$
    – Nate Eldredge
    Commented Feb 2, 2016 at 20:03

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