Let $X$ an abelian variety /$k$, char($k$)=0, $k=\overline k$, and be $\widehat{X}$ its dual. With $P$ I will denote the (normalized) Poincaré bundle over $X\times_k\widehat X$. We have an action of $Z/2Z$ over the abelian variety $X\times_k\widehat X$, given by the product $i\times i$ of the two inversions. Since the $P$ is symmetric there is a unique isomorphism $\rho:P\longrightarrow(i\times i)^*P$ such that $\rho(0,\widehat 0)=$ identity over $P(0,\widehat 0)$. Given a point of order 2 $(x,\alpha)\in X\times_k\widehat X$ one can define $e(x,\alpha)$ as the scalar quantity, $a$ such that $\rho(x,\alpha)$ is given by multiplication by $a$. This is either 1 or -1. My question is: Is there a quick way to find the quantity $e(x,\alpha)$ for the points of the form (x,\widheat 0) with $x$ a point of order two? Could it be that it is always 1?
I know from Mumford (On the equation defining abelian varieties I, Proposition 2 pg 307) that if $D$ is a symmetric divisor such that $P=O_X(D)$ then $$e(x,\alpha)=(-1)^{m(x,\alpha)-m(0,\widehat 0)}$$ where $m(x,\alpha)$ denote the multiplicity on $D$ at $(x,\alpha)$. What I do not know is how a symmetric divisor $D$ such that $P=O_X(D)$ looks like and, moreover how to compute its multiplicity in the point of order two.
Thank you very much for all the answer I will receive! Stgemain
