1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.