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: enter image description here 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{equation} \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} \end{equation} In the examples I have considered this gives the right answer but I would like to know if it is true in general. Thanks!


1 Answer 1


Yes, from a computational perspective that works fine, assuming that one can describe the decomposition $\{V_1,V_2,\ldots\}$ with sufficient rigor (in particular, their boundaries should have measure zero).

The proof would go something like this.

Define a new partition $\{W_k\}$ whose elements are obtained by taking all possible finite intersections of elements of the $\{U_i\}$ covering and elements of the $\{V_j\}$ decomposition. For example, in your picture one of the $W$'s would be $U_n \cap U_l \cap V_m$ which is the tiny quadrilateral in your $V$ diagram having two straight sides on the boundary of $V_m$, one curved side on the boundary of $U_n$, and one curves side on the boundary of $U_l$.

First one proves that $I = \int_M \omega = \sum_k \int_{W_k} \omega$. One has $\int_M \omega = \sum_i \int_{U_i} \phi_i \omega$ by definition. Then one decomposes $U_i$ into a disjoint union of $W_k$'s, and rewrites $\int_{U_i} \phi_i \omega$ as a sum over the terms of that disjoint union. Then for each $k$ one collects the $W_k$ terms and takes their sum, using the partition of unity property to get a sum equal to $\int_{W_k} \omega$.

Added: By request, here are more details in the case of $W_k = U_n \cap U_l \cap V_m$. The terms that one collects to get $\int_{W_k} \omega$ are: $$\int_{W_k} \phi_n \omega + \int_{W_k} \phi_l \omega + \int_{W_k} \phi_m \omega = \int_{W_k} (\phi_n+\phi_l+\phi_m) \omega = \int_{W_k} 1 \cdot \omega = \int_{W_k} \omega $$

Next one simply collects the $\int_{W_k} \omega$ terms in a different way to get $\sum_j \int_{V_j} \omega$.

  • $\begingroup$ excellent!! thank you $\endgroup$ Mar 6, 2019 at 15:25
  • $\begingroup$ it would be helpful (also for future readers) if you could add a bit more detail on constructing the integral, $\int_{W_k}\omega$, over the specific quadrilateral you mentioned in the example, in particular how the partition of unity enters and what precise terms are contained in the sum for the given $k$ $\endgroup$ Mar 6, 2019 at 15:44
  • 1
    $\begingroup$ I added a few more details. $\endgroup$
    – Lee Mosher
    Mar 6, 2019 at 16:31
  • $\begingroup$ that's now crystal clear and very helpful -- many thanks again $\endgroup$ Mar 6, 2019 at 16:33

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