# Integration over a Surface without using Partition of Unity

Suppose we are given a compact Riemann surface $$M$$, an open cover $$\mathscr{U}=\{U_1,U_2,\dots\}$$ of $$M$$, charts $$\{(U_1,\phi_1),(U_2,\phi_2),\dots\}$$, holomorphic coordinates, $$\phi_m:p\in U_m\mapsto z_m$$, holomorphic transition functions $$z_m=f_{mn}(z_n)$$ on patch overlaps $$U_m\cap U_n$$ (with appropriate cocycle relations, $$f_{mn}\circ f_{n\ell}\circ f_{\ell m}=1$$ on triple overlaps $$U_m\cap U_n\cap U_\ell$$), and a globally defined top form, $$\omega=\omega(z,\bar{z})dz\wedge d\bar{z}$$.

I want to integrate $$\omega$$ over $$M$$, $$I= \int_M\omega,$$ using the above data (without using a partition of unity), and I'm wondering whether my reasoning is correct. In particular, suppose we construct non-overlapping sets $$\{V_1,V_2,\dots\}$$ such as those shown in the figure: My question is whether I can compute $$I$$ via a sum of integrals over the $$\{V_1,V_2,\dots\}$$ (to guarantee no overcounting) without using a partition of unity; i.e., is it true that: \begin{aligned} I &= \sum_m\int_{V_m}\omega\\ &=\sum_m\int_{V_m}\omega_m(z_m,\bar{z}_m)dz_m\wedge d\bar{z}_m \end{aligned} \qquad\qquad\qquad(*) In the examples I have considered (such as $$\omega=\frac{i}{2\pi}\mathcal{R}_{z\bar{z}}dz\wedge d\bar{z}$$ the curvature tensor or $$\omega=ig_{z\bar{z}}dz\wedge d\bar{z}$$ the volume form) this gives the right answer (e.g. the Euler characteristic in the first example and area in the second) but I would like to know if it is true in general. Thanks!

Update I: It should presumably be possible to derive (*) using a partition of unity subordinate to $$\mathscr{U}$$. If so, how might one derive the last equality in: $$I=\int_M\omega=\int_{M} \omega\big(\sum_m\lambda_m\big)=\sum_m\int_{U_m}\omega\lambda_m = \sum_m\int_{V_m}\omega$$

Update II: This question now has a clear and explicit answer at here

• why the downvote? – vidyarthi Mar 6 at 12:32
• You can indeed do such a computation. Any top form defines a measure on the surface, making is a measure space, and the boundaries of the $V_m$s have measure zero. – David Roberts Mar 6 at 12:44
• Isn't this a special case of a partition of unity, namely the characteristic functions of the $V_n$? – Achim Krause Mar 6 at 12:45
• Are you trying to integrate numerically over a specific Riemann surface, or are you doing something more theoretical? I have some possibly relevant experience for numerical calculations. – Neil Strickland Mar 6 at 12:52
• By the characteristic function of $V_n$ I just mean the function that is constantly 1 on $V_n$ and 0 outside of it. That's a partition of 1 if the $V_n$ cover the whole space and don't overlap. – Achim Krause Mar 6 at 16:53