Dbar operator

Remarks

The operator $\partial$ is defined in the same way with $\partial / \partial z_k$ and $dz_k$; it sends $(p, q)$-forms to $(p+1, q)$-forms. On forms on $U$, the exterior derivative is $d = \partial + \bar\partial$, and $\bar\partial \circ \bar\partial = 0$. It obeys the Leibniz rule $\bar\partial(\alpha \wedge \beta) = \bar\partial \alpha \wedge \beta + (-1)^{k} \alpha \wedge \bar\partial \beta$ for $\alpha$ of degree $k$. Example: for $f(z_1, z_2) = \abs{z_1}^2 + z_2$, $\partial f / \partial \bar z_1 = z_1$ and $\partial f / \partial \bar z_2 = 0$, so $\bar\partial f = z_1 \, d\bar z_1$. Non-example: $\bar\partial$ is not the exterior derivative, since $\bar\partial z_1 = 0$ while $d z_1 = dz_1 \ne 0$; the $(1, 0)$ part of $d$ is dropped. A $C^1$ function $f$ satisfies $\bar\partial f = 0$ on $U$ if and only if it is holomorphic on $U$. Because holomorphic changes of coordinates preserve types, $\bar\partial$ is well defined on complex manifolds.