Tangential Cauchy–Riemann equations

Remarks

These are the Cauchy–Riemann equations restricted to the directions in which a function on $M$ can be differentiated. When $T^{0,1} M$ has constant rank, it is the conjugate of a CR structure on $M$, and the solutions are the CR functions; one writes the equations as $\bar\partial_b f = 0$.

For a hypersurface $M = \set{\rho = 0}$, with $\rho$ real and $d\rho \ne 0$ on $M$, the space $T^{0,1}_p M$ is spanned by the operators $$L_{jk} = \frac{\partial \rho}{\partial \bar z_j} \frac{\partial}{\partial \bar z_k} - \frac{\partial \rho}{\partial \bar z_k} \frac{\partial}{\partial \bar z_j}, \quad 1 \le j < k \le n,$$ so the equations are $L_{jk} f = 0$. Each $L_{jk}$ is tangent to $M$ because $L_{jk} \rho = 0$.

Example: on $S^3 \subset \C^2$, with $\rho = \abs{z_1}^2 + \abs{z_2}^2 - 1$, there is one equation $L_{12} f = 0$ with $L_{12} = z_1 \, \partial / \partial \bar z_2 - z_2 \, \partial / \partial \bar z_1$. The function $f = z_1 z_2$ is a solution, since it does not depend on $\bar z_1, \bar z_2$.

Non-example: $f = \abs{z_1}^2$ on $S^3$. Then $L_{12} f = -z_2 z_1$, which is nonzero at points where $z_1 z_2 \ne 0$, so $f$ does not satisfy the equations.

For a real curve $M$ in $\C$, $T^{0,1} M = 0$, so every function satisfies the tangential equations.