Let $X$ be a compact Riemann surface. Consider the operators $\partial$ and $\overline{\partial}$, where $\partial$ is considered as either an operator taking (complex-valued) functions to $(1, 0)$-forms or an operator taking $(0, 1)$-forms to $2$-forms; and where $\overline{\partial}$ is considered as either an operator taking functions to $(0, 1)$-forms or an operator taking $(1, 0)$-forms to $2$-forms.

Let $f$ be a smooth function on $X$; let $\alpha$ be a holomorphic $(1,0)$-form on $X$.

In Donaldson's book Riemann Surfaces, he says that the integral $$ \int_X \overline{\partial}(f \alpha) = 0 $$ vanishes by Stokes' theorem. Moreover, he frequently uses Stokes' theorem as justification for the vanishing of integrals of the form $\int \partial \beta$ or $\int \overline{\partial} \beta$ for appropriate $1$-forms $\beta$. But the only version of Stokes' theorem he writes down as a theorem is the usual one with the differential $\mathrm{d}$.

My question is whether there is indeed a more general version for Stokes' theorem, applicable to the complex differentials $\partial$ and $\overline{\partial}$, or if in each case I need to figure out why Stokes' theorem is applicable. And, if possible, can you help me understand why it is applicable in the special case mentioned above?

  • $\begingroup$ Is $f$ holomorphic? $\endgroup$ – Pratyush Sarkar Feb 17 '17 at 6:45
  • $\begingroup$ $f$ is not assumed to be holomorphic. $\endgroup$ – Doug Feb 17 '17 at 6:46

Since we have $d = \partial + \bar \partial$ and $\beta=f\alpha$ is a $(1,0)$-form, and note that

$$ d\beta= \partial \beta + \bar\partial \beta = \bar \partial \beta$$

as $\partial \beta = 0$ (there is no $(2,0)$-forms on a Riemann surface). Thus the equation follows from the usual Stokes theorem.

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    $\begingroup$ Oh, of course, that's great. I did have $\overline{\partial}(f \alpha) = d(f\alpha) - \partial f \wedge \alpha$ written down, but I missed the obvious vanishing of the last term...:) $\endgroup$ – Doug Feb 17 '17 at 6:49
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    $\begingroup$ So it appears that $\alpha$ being holomorphic had nothing to do with it, and you can use the Stokes' theorem for $\int \overline{\partial} \beta$ when $\beta$ is a $(1,0)$-form, and for $\int \partial \beta$ when $\beta$ is a $(0, 1)$-form. $\endgroup$ – Doug Feb 17 '17 at 7:35

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