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I am trying to define complex integration over a surface. I have found some notes and books here and there but nothing that defines it rigorously.

So suppose we have the complex integral of a function $f(z)$ over a surface $A$ of $\mathbb{C}$. My thought is that we can define:

$ \int \int_A f(z) dz \wedge d\bar{z}$ as $\int \int_B f(\gamma(t,s)) | J(\gamma) | dt \wedge ds $, where $\gamma: B \subset \mathbb{R}^2 \to A \subset \mathbb{R}^2 $ is a parametrization of the surface and $J(\gamma)$ is its Jacobian matrix.

Now I have also seen that $dz \wedge d\bar{z} = -2i dx\wedge dy$ where $z=x+iy$.

So, suppose we want to calculate the area of a given square $A=[a,b]\times [ai,bi]$ in $\mathbb{C}$. Then:

$\int \int_A dz \wedge d\bar{z} = (-2i) \int_{ai}^{bi} \int_{a}^b dxdy = 2(b-a)^2,$

and we get this annoying $2$ on the integral, where is my mistake ? (My point is to understand complex integration over surfaces not just this concrete example). Thanks!

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You cannot define the integral of a function over a (Riemann) surface without having a measure, or, more conveniently, a $2$-form. If this is a complex $1$-manifold, i.e., a Riemann surface, you want to integrate a form of type $(1,1)$, as in your example.

The standard area $2$-form in the plane is $\omega=\dfrac i2 dz\wedge d\bar z = dx\wedge dy$, using coordinates $z=x+iy$. Your square should be $[a,b]\times [a,b]$ (no $i$, as $y$ is real here). So your integral is in fact $-2i$ times the area, and, as I said, you correct by multiplying by $i/2$.

A general Riemann surface will need a Kähler form $\omega$ analogous to the form $\omega$ we had on $\Bbb C$. This is equivalent to having a hermitian metric (just as we have the flat Euclidean metric on the plane). One integrates a complex $2$-form no differently from the integration of real $2$-forms on smooth $2$-manifolds.

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