Show that the Laplacian operator on the Heisenberg group is negative

Question

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 $$ \Delta= \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + (1+x^2+y^2 ) \frac{\partial^2}{\partial t^2} + 2(x\frac{\partial}{\partial y} -y \frac{\partial}{\partial x} ) \frac{\partial}{\partial t} .$$ I want to show that the Laplacian operator $ \Delta$ is negative on $\left( L^{2}(\mathbb R^3); \left<.,.\right>_{2}\right)$, where $$ \left<f,g\right>_{2} =\int_{\mathbb R^3} f(x) \, \bar{g}(x) \, dx .$$ that's what I did: we have

$\begin{align} \Delta &= \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + (1+x^2+y^2 ) \frac{\partial^2}{\partial t^2} + 2(x\frac{\partial}{\partial y} -y \frac{\partial}{\partial x} ) \frac{\partial}{\partial t} \\ &= \Delta_{\mathbb R^2} + (1+x^2+y^2 ) \frac{\partial^2}{\partial t^2} + 2(x\frac{\partial}{\partial y} -y \frac{\partial}{\partial x} ) \frac{\partial}{\partial t} \end{align}$

we know that $\Delta_{\mathbb R^2} $ and $ (1+x^2+y^2 ) \frac{\partial^2}{\partial t^2} $ are negative operator.

My question, how to show that $A:= (x\frac{\partial}{\partial y} -y \frac{\partial}{\partial x} ) \frac{\partial}{\partial t}$ is a negative operator. For it, let $u\in C^{\infty}_{0}(\mathbb R^3) \subset L^{2}(\mathbb R^3) $, we have \begin{align} \left<A u,u\right>_{2} &= \int_{\mathbb R^3} \left[(x\frac{\partial}{\partial y} -y \frac{\partial}{\partial x} ) \frac{\partial}{\partial t} u(x,y,t) \right]\, u(x,y,t) \, dxdydt \end{align}

I tried the integration by parts, but I can not show that $\left<A u,u\right>_{2} \leq 0 $.

Thanks you in advance

Answer 1

I think the simplest way to prove that $\Delta$ is negative is to write $$\Delta = X^2 + Y^2 + T^2$$ where $$\begin{align*} X &= \frac{\partial}{\partial x} - y \frac{\partial}{\partial t} \\ Y &= \frac{\partial}{\partial y} + x \frac{\partial}{\partial t} \\ T &= \frac{\partial}{\partial t}. \end{align*}$$ For any test function $u$, integration by parts shows that $$\int_{\mathbb{R}^3} X^2u \cdot u = -\int_{\mathbb{R}^3} (Xu)^2 \le 0$$ and similarly for $Y$ and $T$. Thus you have $$\int_{\mathbb{R}^3} \Delta u \cdot u = - \int_{\mathbb{R}^3} ((Xu)^2 + (Yu)^2 + (Tu)^2) \le 0.$$

I am not sure whether your operator $A$ is actually negative, so I don't know whether that approach will work.

Answer 2

The factorization of the Heisenberg Laplacian is well known, see

http://www.scirp.org/journal/PaperDownload.aspx?paperID=22807

Answer 3

Using the following

Proposition: If we consider a left invariant vector field $X$ as a first-order differential operator on a group $G$, then it is skew-symmetric with respect to the inner product defined by a Haar measure.

So, the the left invariant vector fields $X, Y$ and $T$ are skew-symmetrics with respect to the inner product $\left<.,.\right>_{2} \leq 0$,then $\left<X^2 u,u\right>_{2} = - \left<X u,Xu\right>_{2} \leq 0$, and hence $\Delta= X^2 + Y^2+T^2$ is a negative operator on $(L^{2}(\mathbb R^3),\left<.,.\right>_{2})$