Calculus integral on Riemann Surface (Stokes' theorem)

Question

I made a question:

Calculus itegral on Riemann surfaces

So I thought a bit about the solution and would like some help on what I did:

Following the hint,

$\int_{\partial \Omega} i \space \partial\Phi =\int_{\partial\Omega} $.

Therefore, by Stokes' Theorem,

$\int_{\partial\Omega} = \int_\Omega div\space(\Phi_x,\Phi_y,\Phi_z)$.

Now I need to use that $\Phi$ is a real-valued function which is positive on and vanishes on the boundary of $\Omega$ to justify that $\int_\Omega div\space(\Phi_x,\Phi_y,\Phi_z)\ge 0$. And that would solve the problem. So if anyone can help me put the pieces together, I appreciate it.

Answer 1

There's no Stokes's Theorem involved in this. For one thing, you don't know anything about the Laplacian of $\Phi$ on the interior of $\Omega$. However, if you write everything out carefully in $\Bbb C$, you have \begin{align*} i\partial\Phi &= \frac i2\big(\frac{\partial\Phi}{\partial x}-i\frac{\partial\Phi}{\partial y}\big)\big(dx+i\,dy\big) \\ &= \frac12\left(i\big(\frac{\partial\Phi}{\partial x}\,dx + \frac{\partial\Phi}{\partial y}\,dy\big) - \big(\frac{\partial\Phi}{\partial x}dy - \frac{\partial\Phi}{\partial y}\,dx\big)\right) = \tfrac12\big(i\,d\Phi - \star(d\Phi)\big). \end{align*} The integral of the first term around the closed curve $\partial\Omega$ is $0$. The integral of the second term is the negative of the flux of $\text{grad}\,\Phi$ across $\partial\Omega$. Since $\Phi$ is positive on $\Omega$ and zero on $\partial\Omega$, $\text{grad}\,\Phi$ points inward everywhere (or is zero). Thus, the flux is negative, and we have our answer. (This can be extended to a Riemann surface by working with local (holomorphic) parametrizations. The geometry of the flux works out just fine.)