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I have a simple question regarding complex geometry: is there an analog for the Stokes Theorem for the Dolbeault Operator $\bar{\partial}$? For instance, suppose that $M$ is a closed complex manifold and I am looking for some identity like

$$\int_M \bar{\partial}(\cdots)=\cdots$$

For it to make sense, I suppose that $(\cdots)$ should be a $(0,n-1)$-form, and that we have a complex volume form... But anyway, which is the best identity one may get?

In case there is no such analog, could someone find a 'deep' reason ensuring this non-existence (a very crucial difference...)?

EDIT (just for enlarging the scope of the question)

Is there a possibility that something may be said for $(0,n-1)$-forms provided that there is some complex (or 'anticomplex') volume form, that is, a $(0,n)$-form which is locally $\theta=dz_1\wedge\cdots\wedge dz_n$ coming from (say) a $SU(n)$ structure? I know that this point of view is far away of the initial scope of the question, but... it is a soft question indeed.

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    $\begingroup$ You will first have to define the quantity you hope to evaluate. How do you plan to integrate a $(0,n)$-form over a manifold of real dimension $2n$? $\endgroup$
    – Robert Bryant
    Commented Aug 6, 2015 at 18:02
  • $\begingroup$ @RobertBryant Yes, you're right!! Please see the comment I have made for the answer below. Perhaps the question is not very accurate. $\endgroup$
    – Jjm
    Commented Aug 6, 2015 at 18:15

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If $M$ is an $n$-dimensional complex manifold, then $M$ is a $2n$-dimensional smooth manifold, so you should integrate a $2n$-form (note that $\bar{\partial}$ of a $(0, n-1)$-form is an $n$-form).

Consider the expression $\int_M\bar{\partial}\alpha$ where $\alpha$ is a complex $(2n-1)$-form. As $\mathcal{E}^{2n-1}(X, \mathbb{C}) = \mathcal{E}^{n,n-1}(X)\oplus\mathcal{E}^{n-1,n}(X)$, we can write $\alpha = \alpha^{n,n-1} + \alpha^{n-1,n}$ where the superscripts denote the bidegree. Note that

$$\bar{\partial}\alpha = \bar{\partial}\alpha^{n,n-1} + \bar{\partial}\alpha^{n-1, n} = \bar{\partial}\alpha^{n, n-1}.$$

As $\partial\alpha^{n,n-1} = 0$, $\bar{\partial}\alpha^{n,n-1} = d\alpha^{n,n-1}$ so

$$\int_M\bar{\partial}\alpha = \int_M d\alpha^{n,n-1} = \int_{\partial M}\alpha^{n,n-1} = 0.$$

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  • $\begingroup$ Is there a possibility that something may be said for $(0,n-1)$-forms provided that there is some complex (or 'anticomplex') volume form, that is, a $(0,n)$-form which is locally $\theta=dz_1\wedge\cdots\wedge dz_n$ coming from (say) a $SU(n)$ structure? I know that this point of view is far away of the initial scope of the question, but... it is a soft question indeed. $\endgroup$
    – Jjm
    Commented Aug 6, 2015 at 18:13
  • $\begingroup$ I'm not sure what can be said. First of all, it's not clear to me why you would call a $(0, n)$-form a volume form. Second of all, you can only integrate an $n$-form on an real $n$-dimensional manifold. If $n = 1$, $M$ is a compact Riemann surface. In this scenario, what would you integrate a $(0, 1)$-form over? $\endgroup$
    – Michael Albanese
    Commented Aug 6, 2015 at 18:21
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    $\begingroup$ Well, that such $\theta$ is called a complex volume form come from some articles... I think it is no standard notation. Now, if we have $\theta$ $(0,n)$-form, any other $(0,n)$-form may be seen as a function (and we may integrate it). $\endgroup$
    – Jjm
    Commented Aug 6, 2015 at 18:26
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    $\begingroup$ I should have pointed this out before, if $\theta = dz_1\wedge\dots\wedge dz_n$, then $\theta$ is a $(n, 0)$-form, not a $(0, n)$-form. What is the volume form you are using to integrate the function you extract? $\endgroup$
    – Michael Albanese
    Commented Aug 6, 2015 at 18:35

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