# Example of integration over path on Riemann surface

Let $X$ be a Riemann surface $$X = \left\{ (z,w) \in \mathbb{C}^2 \mid z^3 + w^3 = 1 \right\}.$$ Then we have $z^2 dz + w^2 dw = 0$ and we can define a holomorphic form $\omega$ on $X$ by $$\omega = \left\{ \begin{array}{rl} \frac{dz}{w}, & w \neq 0 \\ -\frac{wdw}{z^2}, & z \neq 0 \end{array} \right.$$ Let $j = e^{\frac{2 i\pi}{3}}$ and $\gamma$ be a path on $\mathbb{C}_{z}$ given by the following image:

$\hskip4cm$ Let $\Gamma$ denote a closed path on $X$ obtained by lifting $\gamma$ and such that $\Gamma$ passes through $(0,1)$, and let $\Gamma_j$ denote a lifting of its part $\gamma_j$.

$\hskip3cm$ We will integrate $\omega$ over $\Gamma$. Let's start from $z = 0$ and move right. We have $$\int\limits_{\Gamma_1} \omega = \int\limits_{0}^{1-\varepsilon} \frac{dt}{(1-t^3)^{1/3}} =: I_{\varepsilon}\;.$$ Next we compute $\int_{\Gamma_2} \omega$. We can use the parametrisation $z = 1 + \varepsilon e^{i \varphi}$, $\phi \in [-\pi,\pi]$ and $$w = (1-(1+\varepsilon e^{i \varphi})^3)^{1/3} = (-\varepsilon e^{i \varphi}(3 + 3 \varepsilon e^{i \varphi} + \varepsilon^2 e^{2i \varphi}))^{1/3}\;.$$ For sufficiently small $\varepsilon$ we have $|w| \geqslant |\varepsilon|^{1/3} C_{1}$ on $\Gamma_2 = \Gamma_2(\varepsilon)$. Then $$\left| \int\limits_{\Gamma_2} \omega \right| \leqslant \int\limits_{-\pi}^{\pi} \frac{| \varepsilon i e^{i \varphi} | dt}{|w|} \leqslant C_{2} \varepsilon^{2/3} \to 0\text{ when }\varepsilon \to 0.$$ Using same techniques we obtain that $\int_{\Gamma_5} \omega$ and $\int_{\Gamma_8}\omega$ tend to $0$ when $\varepsilon \to 0$. Now we compute $\int_{\Gamma_3} \omega$. Let's remark that $w = \sqrt{1-z^3} = \sqrt{(1-z)(j-z)(j^2-z)} = \sqrt{1-z}\sqrt{(j-z)(j^2-z)}$. When $z$ goes around $1$ on the curve $\gamma_2$, the first factor's argument increases by $\frac{2 \pi}{3}$ because the argument of $(1-z)$ increases by $2 \pi$. The argument of the second factor in unchanged. Then the integral over $\Gamma_3$ is just $$\int\limits_{\Gamma_3} \omega = \int\limits_{1-\varepsilon}^{0} \frac{dt}{j(1-t^3)^{1/3}} = -j^2 I_{\varepsilon}$$ since $j^2 = \frac{1}{j}$. To take the integral over $\Gamma_4$ let's remark that $t^3 = (jt)^3$ and then $1-t^3 = 1-(jt)^3$. Then $$\int\limits_{\Gamma_4} \omega = \int\limits_{0}^{1-\varepsilon} \frac{ j dt}{ j (1-(jt)^3)^{1/3}} = I_{\varepsilon}\;.$$ Analogously to $\Gamma_3$ we have that $\int_{\Gamma_6} \omega = \int_{\Gamma_9} \omega = -j^2 I_{\varepsilon}$ and $\int_{\Gamma_7} \omega = I_{\varepsilon}$. Since the form $\omega$ is holomorphic on $X$ we must have $$\sum\limits_{k=1}^{9} \int\limits_{\Gamma_k} \omega = I$$ where $I$ is a so-called period, that doesn't depend on $\varepsilon$. Passing to the limit when $\varepsilon \to 0$ we obtain $$3(1-j^2) \int\limits_{0}^{1} \frac{dt}{(1-t^3)^{1/3}} = I\;.$$ My question is how to find such period $I$?

P.S. By the way, using other techniques (not related to Riemann surfaces) we can show that
$$\int\limits_{0}^{1} \frac{dt}{(1-t^3)^{1/3}} = \frac{2 \pi}{3 \sqrt{3}}\;.$$

• I haven't looked through the details yet, but why do you think that the integral of a holomorphic form over a closed path has to be $0$? This would be true if $X$ is simply connected, or if the path is homotopic to a point, but not in general. – Lukas Geyer Oct 31 '12 at 4:42
• @LukasGeyer ah, you're right! But in that case I have no ideas, how to find to what this integral is equal :( – Appliqué Oct 31 '12 at 4:47
• Your integration seems to be correct, and I am not sure that there is a much easier way to find the value of $I$. – Lukas Geyer Oct 31 '12 at 5:06
• @LukasGeyer I think there is. It is an exercise from Cartan's book. Author wants reader to find $\int_{0}^{1} \frac{dt}{(1-t^3)^{1/3}}$ using such integration – Appliqué Oct 31 '12 at 5:08
• Oh, I think I have an idea... – Lukas Geyer Oct 31 '12 at 5:14

The path is homotopic in the complement of the branch points $1,j,j^2$ to a large circle $|z|=R$, parametrized in the mathematically positive sense. Since the homotopy does not pass through branch points, it can be lifted to the surface $X$. Since the form is holomorphic, it is closed, so the integral only depends on the homotopy class, and it is equal to $$\int_{|z|=R} \frac{dz}{\sqrt{1-z^3}} = \int_{|z|=R} \frac{dz}{z\sqrt{z^{-3}-1}} = \frac{1}{e^{\frac{i \pi}{3}}} \int_{|z|=R} \frac{dz}{z \sqrt{1-z^{-3}}}$$ where I hope I got the branch of the third root right, and in the last expression it is the branch that maps $1$ to $1$. Expanding this out in a power series and using the fact that $\int\limits_{|z|=R} z^n \, dz = 0$ for $n \ne -1$ and $\int\limits_{|z|=R} z^{-1} \, dz =2\pi i$ (basically using the residue theorem at $\infty$) gives $$\frac{1}{e^{\frac{i \pi}{3}}} \int_{|z|=R} \frac{1}z\left(1-\frac13z^{-3}\pm\ldots\right) dz = 2\pi ie^{-\frac{i \pi}{3}}$$ and I think this is the same result you got.