2
$\begingroup$

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} \qquad\qquad\qquad(*) \end{equation} 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

$\endgroup$
  • 1
    $\begingroup$ why the downvote? $\endgroup$ – vidyarthi Mar 6 at 12:32
  • 3
    $\begingroup$ 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. $\endgroup$ – David Roberts Mar 6 at 12:44
  • 3
    $\begingroup$ Isn't this a special case of a partition of unity, namely the characteristic functions of the $V_n$? $\endgroup$ – Achim Krause Mar 6 at 12:45
  • $\begingroup$ 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. $\endgroup$ – Neil Strickland Mar 6 at 12:52
  • 1
    $\begingroup$ 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. $\endgroup$ – Achim Krause Mar 6 at 16:53

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.