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Elementary partial and dbar calculations

Example

On $C^{2}$ , let $f (z) = z_{1}^{2} {\overset{ˉ}{z}}_{2} + {\overset{ˉ}{z}}_{1}^{2}$ and $η = f d z_{2}$ . Then $\partial f = 2 z_{1} {\overset{ˉ}{z}}_{2} d z_{1}, \overset{ˉ}{\partial} f = 2 {\overset{ˉ}{z}}_{1} d {\overset{ˉ}{z}}_{1} + z_{1}^{2} d {\overset{ˉ}{z}}_{2},$ and $\partial η = 2 z_{1} {\overset{ˉ}{z}}_{2} d z_{1} \land d z_{2}, \overset{ˉ}{\partial} η = 2 {\overset{ˉ}{z}}_{1} d {\overset{ˉ}{z}}_{1} \land d z_{2} + z_{1}^{2} d {\overset{ˉ}{z}}_{2} \land d z_{2} .$ In the last term, $d {\overset{ˉ}{z}}_{2} \land d z_{2} = - d z_{2} \land d {\overset{ˉ}{z}}_{2}$ .

Facts & Assumptions

Given: The polynomial $f = z_{1}^{2} {\overset{ˉ}{z}}_{2} + {\overset{ˉ}{z}}_{1}^{2}$ and the smooth form $η = f d z_{2}$ on $C^{2}$ .

[F1]

On a form $a_{I, J} d z^{I} \land d {\overset{ˉ}{z}}^{J}$ , the $\partial$ coefficient formula is $(\partial_{z_{j}} a_{I, J}) d z_{j} \land d z^{I} \land d {\overset{ˉ}{z}}^{J}$ (Bigraded complex forms and the Dolbeault operators).

[F2]

The Wirtinger operator is $\partial_{z_{k}} f = \frac{1}{2} (\partial_{x_{k}} f - i \partial_{y_{k}} f)$ (Wirtinger operators in $C^{m}$ ).

[F3]

The Wirtinger operator is $\partial_{{\overset{ˉ}{z}}_{k}} f = \frac{1}{2} (\partial_{x_{k}} f + i \partial_{y_{k}} f)$ (Wirtinger operators in $C^{m}$ ).

[F4]

The $\overset{ˉ}{\partial}$ coefficient formula is $(\partial_{{\overset{ˉ}{z}}_{j}} a_{I, J}) d {\overset{ˉ}{z}}_{j} \land d z^{I} \land d {\overset{ˉ}{z}}^{J}$ (Bigraded complex forms and the Dolbeault operators).

[F5]

The two type operators obey the graded product rule; on functions the sign is positive (The d, partial and dbar identities).

[F6]

The exterior derivative is the sum $d = \partial + \overset{ˉ}{\partial}$ (The d, partial and dbar identities).

Proof

technique · direct

1.1F2F3F5givenalgebra

The coordinate Wirtinger derivatives give the displayed derivatives of $f$ . [F2, F3, F5, given, algebra] From [F2]–[F3] and $z_{k} = x_{k} + i y_{k}$ , one has $\partial_{z_{j}} z_{k} = δ_{j k}$ , $\partial_{z_{j}} {\overset{ˉ}{z}}_{k} = 0$ , $\partial_{{\overset{ˉ}{z}}_{j}} z_{k} = 0$ , and $\partial_{{\overset{ˉ}{z}}_{j}} {\overset{ˉ}{z}}_{k} = δ_{j k}$ . The scalar product rule [F5] therefore gives $\partial_{z_{1}} f = 2 z_{1} {\overset{ˉ}{z}}_{2}$ , $\partial_{z_{2}} f = 0$ , $\partial_{{\overset{ˉ}{z}}_{1}} f = 2 {\overset{ˉ}{z}}_{1}$ , and $\partial_{{\overset{ˉ}{z}}_{2}} f = z_{1}^{2}$ . Wedge these coefficients with their corresponding one-forms to obtain the displayed $\partial f$ and $\overset{ˉ}{\partial} f$ .

2.1F1F4step 1.1givenalgebra

Applying the type coefficient formulas to $η = f d z_{2}$ yields the two displayed form derivatives. [F1, F4, step 1.1, given, algebra] The only type factor in $η$ is $d z_{2}$ . Thus [F1]–[F4] give $\partial η = (2 z_{1} {\overset{ˉ}{z}}_{2} d z_{1}) \land d z_{2}$ and $\overset{ˉ}{\partial} η = (2 {\overset{ˉ}{z}}_{1} d {\overset{ˉ}{z}}_{1} + z_{1}^{2} d {\overset{ˉ}{z}}_{2}) \land d z_{2}$ . Anticommutativity gives $d {\overset{ˉ}{z}}_{2} \land d z_{2} = - d z_{2} \land d {\overset{ˉ}{z}}_{2}$ , so the sign in the last summand is as stated.

3.1F6step 2.1givenalgebra∎

Their sum is the exterior derivative of this example. [F6, step 2.1, given, algebra] By [F6], $d η = \partial η + \overset{ˉ}{\partial} η$ , so the two explicitly computed type components in step 2.1 add to $d η$ with the same wedge sign.

Depends on

Used by

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Sources

Lebl, Tasty Bits of Several Complex Variables, v4.4, Chapter 4 §4.4 (standard reference, not scraped)