Kohn–Rossi cohomology

Remarks

The formula makes sense because $T^{0,1}M$ is closed under the Lie bracket, which is the integrability condition in the definition of a CR manifold; this also gives $\bar\partial_b^2 = 0$. It is the formula for the exterior derivative of a differential form, applied only to vectors in $T^{0,1}M$. For $q = 0$, $\bar\partial_b u$ is the restriction of $du$ to $T^{0,1}M$, so $H^{0,0}_b(M)$ is the space of smooth CR functions on $M$. Example: on $M = \C^n$ with $T^{1,0}M$ the usual holomorphic tangent bundle, $\bar\partial_b = \bar\partial$, and $H^{0,q}_b(\C^n)$ is the Dolbeault cohomology $H^{0,q}(\C^n)$, which is $0$ for $q \ge 1$. Non-example: on the unit sphere $S^3 \subset \C^2$, with $\bar L = z_2 \partial / \partial \bar z_1 - z_1 \partial / \partial \bar z_2$, the function $u = \bar z_1$ satisfies $\bar\partial_b u (\bar L) = \bar L \bar z_1 = z_2 \ne 0$, so it is not a cocycle and gives no element of $H^{0,0}_b(S^3)$. More generally one defines groups $H^{p,q}_b(M)$ using forms of type $(p,q)$. For a compact strongly pseudoconvex CR manifold of hypersurface type and real dimension $2n + 1$, Kohn proved that $H^{0,q}_b(M)$ is finite-dimensional for $1 \le q \le n - 1$.