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For integrate on a Riemann Surface $X$ we need a holomorphic $1$-form $\omega$ and a piecewise smooth path $\gamma : [a,b] \to X$ (This means $\phi \circ \gamma(t) : [a,b] \to \mathbb{C}$ is smooth for $\phi$ a chart on $X$). Then we have this tiny version of Stoke's theorem: If $T\subset X$ is a triangle on the domain of a chart then $$\int_{\partial T} \omega = 0 $$ if we consider the triangle $T$ as a homeomorphic image of a triangle in $\mathbb{R}^2$. But, Why can I say that $\partial T$ is a piecewise smooth path?

ÝTAN
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  • First I think you probably want $d\omega = 0$ for your theorem to holds. I would say that this only make sense for $T$ diffeomorphic to a usual triangle, unless you have a definition for integrate a differential form along a continuous path ? –  Jul 04 '17 at 09:37
  • @N.H. Oh sorry, I edited... I mean for holomorphic 1-forms. I'm trying to copy the theory of differential forms to holomorphic forms whithout to pass by 2-forms. – ÝTAN Jul 04 '17 at 09:41
  • Ok, this make sense, but I still feel a bit confused how you can integrate over something just homeomorphic to a triangle, maybe you need to ask $T$ to be diffeomorphic to a triangle ? –  Jul 04 '17 at 09:47
  • @N.H. Yes, I think that. But in Miranda's Riemann Surfaces book (page 123) the author says that we can give to $\partial T$ the form of a piecewise smooth path without a convincing reason. – ÝTAN Jul 04 '17 at 09:52
  • Ok, I read this passage, and indeed he's taking only homeomorphic image of triangle, so I don't understand how is defined integration. I think you can safely assume that we are taking triangle with piecewise smooth boundary. –  Jul 04 '17 at 09:59
  • Ok, his definition of path is a piecewise $C^{\infty}$ path, and he integrates $1$-form only on such path, so definitely you can assume that the boundary of the triangle is a piecewise smooth path, i.e that the map $f : T \to X$ is a diffeomorphism and not simply an homeomorphism, this is surely a typo. –  Jul 04 '17 at 10:03
  • Then, how can I connect this with the homology group, in the way to define the jacobian of an arbitrary compact riemman surface $X$ ? Because we need to integrate over triangulable closed sets. If you want to read the next pages about integrate over the homology group – ÝTAN Jul 04 '17 at 10:12
  • For any continuous path $\delta$ there is an homologous path $\gamma$ which is smooth and homologous/homotopic to $\delta$. (For this, you can use that on a compact segment, any continuous function is limits of polynomial) and then define $\int_{\delta} \omega := \int_{\gamma} \omega$. –  Jul 04 '17 at 10:21
  • @N.H. Where can I find a reference to this result? Thanks! That is useful – ÝTAN Jul 04 '17 at 10:23
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    here is the statement of the Weierstrass approximation theorem. Details might be a bit boring to write but at least this shows you that really, you can assume that you are integrating over smooth path. –  Jul 04 '17 at 10:26
  • Possibly of interest: free homotopy class of closed paths in a compact Riemannian manifold. – Andrew D. Hwang Jul 04 '17 at 10:39

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