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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

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Let me see if I understand the question. You want to compute e(x,0)? (That line seems to have a typo.) If so, this is 1 by definition: after all, the restriction of P to A\times{0} is trivial, and \rho is normalized to be the trivial map. More generally, e(x,y) is a bilinear pairing. –  t3suji Jun 22 '10 at 2:40
    
Thank you very much for your comment/answer. I really had not understood that the canonical bilinear pairng $e_2:X\times\widehat X$ was the same of what Mumford dentoes as $e_*^P(\cdot,\cdot)$ with $P$ the Poincarè bundle. Thank you again for the time you dedicated to me Best Regards, –  Rurik Jun 22 '10 at 8:32
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It seems to me this isn't really a "soft question" and I'd bet there are more specific tags that one could replace that with, but as I don't know much about algebraic geometry, I'll let someone else change that. –  j.c. Jun 22 '10 at 10:19
    
Yes, you are right! A friend of mine told me the same thing, yesterday, thus I removed the tag! Itagget it a s a soft question because after t3suji answered me I understood it was really easy (and I should have known it form the beginning :'(). Thank you for your comment! –  Rurik Jun 23 '10 at 8:45
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