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.

  • $\begingroup$ Sorry but in my comment i said that the coefficients of $\alpha_{*}(D)$ are even. Why do you say in your edit that it's always the case.....? $\endgroup$ – dario Feb 10 '15 at 16:48
  • $\begingroup$ I don't understand why does the coefficent of $\alpha_{*}(E)$ is always even? $\endgroup$ – dario Feb 13 '15 at 17:26
  • $\begingroup$ can you post a new answer with the complete proof? $\endgroup$ – dario Feb 13 '15 at 18:09

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.

  • $\begingroup$ In your proof you have to use the coefficent $\frac{a}{2}$ but you are not sure that is even. $\endgroup$ – dario Feb 4 '15 at 9:31
  • $\begingroup$ Soory. It is just to be clear. You write $E-\alpha^{*} \left( \frac{1}{2} \alpha_{*} E \right) \equiv D- \tau D$. Does this relation implies $E \equiv D-\tau D$? $\endgroup$ – dario Feb 4 '15 at 9:52
  • $\begingroup$ i've found this relations: 1) if a is the coefficent of the point $p \in supp(\alpha_{*}(E))$ then we get $a=a_1 + a_2$ where $a_i$ is the order of the point $q_i \in \pi^{-1}(p)$ of the funtion $g$. 2) if we put $D=E+(g)$ then using the fact that $\alpha_{*}(D)=\alpha_{*}(E)+\alpha_{*}(g)$ , with $\alpha_{*}(E)=(Nm_{\pi}(g))$, where $Nm_{\pi}$ is the norm function, and the fact that $\alpha_{*}(g)=(Nm_{\pi}(g))$ we get $\alpha_{*}(D)=2(Nm_{\pi}(g))$ $\endgroup$ – dario Feb 4 '15 at 14:53
  • $\begingroup$ dario: 1) Yes, I said that this works if $a$ is even. 2) You assumed that $\alpha_*E$ is trivial and hence $\equiv 0$ 3) I'm not convinced by your norm calculation. $\endgroup$ – Sándor Kovács Feb 4 '15 at 22:15
  • $\begingroup$ Can you expain the relation at the beginning of the third line of your answer? I don't understand Why at the left part of the equation there is $a_{2}$. $\endgroup$ – dario Feb 4 '15 at 23:25

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.

  • $\begingroup$ OK, I think this indeed does it. $\endgroup$ – Sándor Kovács Feb 9 '15 at 19:26

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