Exercise from Geometry of algebraic curves by Arbarello, Cornalba, Griffiths, Harris

Question

Let $\pi:C^{'} \rightarrow C$ an unramified double cover of a complex Riemann surface $C$ of genus $g$. With the symbol $Nm_{\pi}$ we mean the norm application that takes a meromorphic function on $C^{'}$ named $h$ and produce a meromorphic function on $C$ named $Nm_{\pi}( h)$ defined as $$Nm_{\pi}(h)(p)=\prod_{q \in \pi^{-1}(p)} h(q)^{\nu(q)}$$
where $\nu(q)$ is the multiplicity of $q$ in the fiber of $p$ respect to $\pi$. Write $(f)=(Nm_{\pi}(h))$ the principal divisor associated to the function $Nm_{\pi}(h)$.
I would like to prove that:
1) if $\tau: C^{'} \rightarrow C^{'}$ is the sheet interchange involution for the double cover $\pi$. Show that there exist a function $s$ meromorphic on $C^{'}$ such that $$ s(q)=-s(\tau(q))$$ 2)Let $\tau$ as in the point 1) and let $f$ a meromorphic function on $C^{'}$ such that $f=-f \circ \tau$. Show that $(Nm_{\pi}(f))$ is divisible by 2 i.e. we can write $(Nm_{\pi}(f))=2D$.

Answer 1

By Riemann existence theorem the meromorphic functions on a compact Riemann surface separate the points, so consider two different points $p,p'$ on $C'$ on a fiber of $\pi$ and a meromorphic function $f$ on $C'$ which is holomorphic at these points and $f(p)\neq f(p')$. Define $$s = f - f\circ \tau.$$ Then we have $$s\circ \tau = f\circ \tau - f\circ \tau\circ \tau = f\circ \tau - f = -s,$$ on the other hand $s(p) = f(p)-f(p')\neq 0$ so $s$ is a nontrivial solution for 1).

2) If $f$ satisfies $f=-f\circ \tau$ then for every $q\in C'$, $\mathrm{ord}_f(q) = \mathrm{ord}_f(\tau(q))$ and $$(Nm_\pi(f))(\pi(q)) = \mathrm{ord}_f(q)+\mathrm{ord}_f(\tau(q)) = 2\mathrm{ord}_f(q).$$

Note also that the condition of unramified-ness is not necessary for above statements, although we have used it.