Kohn Laplacian

Remarks

The adjoint $\vartheta_b$ is often written $\bar\partial_b^*$, and $\square_b$ is also called the Kohn–Rossi Laplacian or the boundary Laplacian. On functions ($q = 0$) the first term is absent, and $\square_b u = \vartheta_b \bar\partial_b u$. Example: on the Heisenberg group $\mathbb{H} = \C \times \R$ with coordinates $(z, t)$, take $T^{0,1}\mathbb{H}$ spanned by $\bar Z = \partial/\partial \bar z - i z\, \partial/\partial t$, declare $\bar Z$ to have length $1$, and use Lebesgue measure. The vector field $\bar Z$ has divergence $0$, so integration by parts shows that the formal adjoint of $\bar Z$ is $-Z$, where $Z = \partial/\partial z + i \bar z\, \partial/\partial t$ is its complex conjugate. Hence $\square_b u = -Z \bar Z u$ for functions $u$. Non-example: the operator $-\bar Z Z$ is not the Kohn Laplacian on functions of $\mathbb{H}$. It differs from $-Z \bar Z$ by the commutator $[Z, \bar Z] = -2i\, \partial/\partial t$, which is not zero. The Kohn Laplacian is not elliptic, because its principal symbol vanishes on the nonzero real covectors that annihilate $T^{1,0}M \oplus T^{0,1}M$.