How to write a divisor: Exercise n 20 pag 285 from Geometry of algebraic curves by Arbarello, Cornalba, Griffiths, Harris

Question

Let $C$ a complex Riemann surface (compact) and $\alpha:C^{'} \rightarrow C$ an unramified double cover of $C$.
Define the application $\alpha_{*} :Div(C^{'}) \rightarrow Div(C)$ as follow $$\forall E\in Div(C^{'}), E=\sum_{i=1}^{n}q_i \Rightarrow \alpha_{*}(E)=\sum_{i=1}^{n}\alpha(q_i) .$$
(then we can extend the application to all divisor using linearity)
Now suppose that $E \in Div(C^{'})$ is a divisor such that $\alpha_{*}(E)$ is equivalent to the zero divisor, i.e $\alpha_{*}(E) \equiv 0$ where the relation of equivalence is the classical relation defined on divisor (i.e. $D,D^{'} \in Div(C^{'})$ are equivamente if and only if $D-D^{'}=(f)$ with $f$ a meromorphic function on $C^{'}$).
Put $\tau: C^{'} \rightarrow C^{'}$ the involution sheet exchange.
Then find a divisor $D\in Div(C^{'})$ such that $E \equiv D- \tau D$.
Such divisor is involved in the characterization of the Weil pairing.
Hint from book: due to theory we know that the norm map $Nm_{\alpha}:M^{*}(C^{'})\rightarrow M^{*}(C)$ is surjective. So using the hypothesis, $\alpha_{*}(E)=(f)$ where $f$ is a meromorphic funtion on $C$, we get: $\alpha_{*}(E)=Nm_{\alpha}(g)$ with $g$ a meromorphic funtion on $C^{'}$.
Put $D=E+(g)$. Why is this the divisor that satisfy $E\equiv D- \tau D$?
Thanks in advance for any suggestions or comments.

Answer 1

Do this: Let $P\in C$ be a point that appears in the support of $\alpha_*E$ with coefficient $a$ and let $P_1,P_2=\tau(P_1)\in C'$ be the points that map to $P$ with coefficients $a_1$ and $a_2$ in $E$. Observe that since $a_1+a_2=a$, $-\tau\left((a_1-\frac a2)P_1\right)=(a_2-\frac a2)P_2$, which means that $E-\alpha^*\left(\frac a2P\right)$ can be written as desired near $P_1$ and $P_2$. Doing this for all $P\in C$ means that there is a $D$ such that $$ E-\alpha^*\left(\frac 12\alpha_*E\right) \equiv D -\tau D $$ In other words, this gives you what you want if $\alpha_*E$ is linearly equivalent to a divisor with only even coefficients.

EDIT: According to dario's comment+answer this is always the case.

Answer 2

Too long for comment. WHAT I SAID ABOUT THE NORM.

Let $D=E+(g)$ where $g$ is the meromorphic funcion determined in the question. The application $\alpha_{*}$ is linear so if we compute:

$$\alpha_{*}(D)=\alpha_{*}(E)+ \alpha_{*}(g),$$
using the hypothesis we know that $\alpha{*}(E)= \left( Nm_{\pi}(g) \right)$ where in the map $Nm_{\pi}$ is defined as: $$ \forall p \in C :Nm_{\pi}(g)(p)= \prod_{q \in\pi^{-1}(p)}g(q)^{\nu(q)},$$ and $\nu(q)$ is the moltiplicity of $q$ in the fiber of $\pi$.

Due to the fact that $\pi$ is unramified double cover we get: $$ \forall p \in C: Nm_{\pi}(g)(p)=g(q_1)g(q_2)$$, setting ${q_1,q_2}\in \pi^{-1}(p)$.
We can check that $\alpha_{*}(g)= \left( Nm_{\pi}(g) \right)$. So using this fact we get: $$\alpha_{*}(D)=\alpha_{*}(E)+ \alpha_{*}(g)=\alpha_{*}(Nm_{\pi}(g))+\alpha_{*}(Nm_{\pi}(g))=2\alpha_{*}(Nm_{\pi}(g)).$$
I don't know if this is usefull for our problem.