This is the problem IV.3.F I found at page 127 of the book "Algebraic Curves and Riemann Surfaces" of Rick Miranda.
Let $\tau \in \mathbb{C}$ such that $Im(\tau)>0$ and define the lattice $L=\mathbb{Z}+\mathbb{Z}\tau$. Let $h$ be a meromorphic function on $\mathbb{C}$ which is $L$-periodic, i.e. $h(z+1)=h(z+\tau)=h(z)$ for all $z$. For $p\in \mathbb{C}$, let $\gamma_p$ be the path which is the anticlockwise boundary of the parallelogram with vertices $p, p+1, p+\tau+1, p+\tau,p$. Assume that $p$ is chosen so that there aren't any zeroes or poles of $h$ on $\gamma_p$. Show that $$\frac{1}{2\pi i}\int_{\gamma_p}z\frac{h'(z)}{h(z)}dz\in L$$
Remark
The solution should involve Riemann Surfaces.
My solution
Let X be the torus defined as the quotient group $\mathbb{C}/L$. We know that the natural projection map $\pi:\mathbb{C}\rightarrow X$ is a holomorphic map between Riemann surfaces. The following paths
$\gamma_1:[0,1]\rightarrow \mathbb{C}$ defined as $\gamma_1(t)=p+t$
$\gamma_2:[0,1]\rightarrow \mathbb{C}$ defined as $\gamma_2(t)=p+1+\tau t$
$\gamma_3:[0,1]\rightarrow \mathbb{C}$ defined as $\gamma_3(t)=p+\tau +1-t$
$\gamma_4:[0,1]\rightarrow \mathbb{C}$ defined as $\gamma_4(t)=p+\tau (1-t)$
run along the sides of the parallelogram considered in an anticlockwise direction, therefore $\gamma_p$ is their concatenation. Observe that
$$\int_{\gamma_3}z \frac{h'(z)}{h(z)}dz=\int_{0}^{1}(p+\tau +1-t)\frac{h'(p+\tau +1-t)}{h(p+\tau +1-t)}(-1)dt=$$
$$=\int_{0}^{1}(p +1-t)\frac{h'(p +1-t)}{h(p +1-t)}(-1)dt+\tau\int_{0}^{1}\frac{h'(p +1-t)}{h(p +1-t)}(-1)dt=$$
$$=\int_{-\gamma_1}z \frac{h'(z)}{h(z)}dz+\tau\int_{-\gamma_1}\frac{h'(z)}{h(z)}dz $$ where $-\gamma_1$ is the reverse path of $\gamma_1$. Similarly, we have that $$\int_{\gamma_4}z \frac{h'(z)}{h(z)}dz=\int_{-\gamma_2}z \frac{h'(z)}{h(z)}dz-\int_{-\gamma_2}\frac{h'(z)}{h(z)}dz, $$ therefore
$$\int_{\gamma_p}z\frac{h'(z)}{h(z)}dz=\sum_{j=1}^{4}\int_{\gamma_j}z\frac{h'(z)}{h(z)}dz=-\tau\int_{\gamma_1}\frac{h'(z)}{h(z)}dz+\int_{\gamma_2}\frac{h'(z)}{h(z)}dz. $$
Since $h$ is $L$-periodic, we can "transfer" it on X defining $\tilde{h}:X\rightarrow \mathbb{C}$ as $\tilde{h}(z+L)=h(z)$, so it results $h=\tilde{h}\circ \pi$. Let's define now $H:X\rightarrow\mathbb{C}_\infty$ as
$$H(x) = \left\{\begin{matrix}
\tilde{h}(x) & \text{if $x$ is not a pole of } \tilde{h} \\
\infty & \text{if $x$ is a pole of } \tilde{h} \\
\end{matrix}\right.$$
$H$ is a holomorphic map between the two Riemann surfaces X and $\mathbb{C}_\infty$. Note that $\pi \gamma_j$ and $\eta_j=H\pi \gamma_j$ are closed paths respectively on X and $\mathbb{C}_\infty$, moreover, since $h$ hasn't zeroes or poles on $\gamma_p$, $\eta_j$ doesn't pass through the points $0$ and $\infty$. Let $z$ be a local coordinate of X and define the meromorphic 1-form $\omega=\frac{h'(z)}{h(z)}dz$. It's well defined on all X and a generic change of local coordinate is given by $z=T(w)=w+\ell$ for some $\ell \in L$, so $T'(w)=1$ and then $\omega=\frac{h'(w)}{h(w)}dw$. It results
$$\frac{1}{2\pi i}\int_{\gamma_1}\frac{h'(z)}{h(z)}dz=\frac{1}{2\pi i}\int_{\gamma_1}\pi^*\omega=\frac{1}{2\pi i}\int_{\pi \gamma_1}\omega=\frac{1}{2\pi i}\int_{\pi \gamma_1}H^*\frac{d\zeta}{\zeta}=\frac{1}{2\pi i}\int_{\eta_1}\frac{d\zeta}{\zeta}=$$
$$=Ind_{\eta_1}(0)=m \in \mathbb{Z},$$ the winding number of $\eta_1$ about $0$. Similarly, we have that $$\frac{1}{2\pi i}\int_{\gamma_2}\frac{h'(z)}{h(z)}dz=Ind_{\eta_2}(0)=n \in \mathbb{Z},$$ therefore $$\frac{1}{2\pi i}\int_{\gamma_p}z\frac{h'(z)}{h(z)}dz=n-m\tau\in L.$$
Note $\pi^*\omega$ is the pull-back of $\omega$ via $\pi.$
Questions
- Is my solution correct?
- Is it possible to simplify it?
It's easy to see that $\pi \gamma_2=-\pi \gamma_4$ and $\pi \gamma_3=-\pi \gamma_1$ therefore $$\int_{\gamma_p}\pi^*\mu=\int_{\pi \gamma_p}\mu=0.$$
where $$\mu=z\frac{h'(z)}{h(z)}dz$$ - Is there another solution, involving Riemann surfaces, for example, by using the Residue Theorem?
Thanks for any comments or workarounds.