# Piecewise smooth? triangles on a Compact Riemann Surface

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?

• 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 ? – user171326 Jul 4 '17 at 9:37
• @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. – user6565190 Jul 4 '17 at 9:41
• 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 ? – user171326 Jul 4 '17 at 9:47
• @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. – user6565190 Jul 4 '17 at 9:52
• 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. – user171326 Jul 4 '17 at 10:26