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Let $A$ and $B$ be two abelian varieties with dual Abelian varieties $\widehat A$, $\widehat B$. An isomorphism of Abelian varieties $f\colon A\times\widehat A\to B\times\widehat B$ represented by the matrix $$\left(\begin{array}{cc} a_{11} & a_{12}\\ a_{21} & a_{22}\end{array}\right)$$ (where $a_{11}\in Hom(A,B)$, $a_{12}\in Hom(\widehat A,B)$, etc.) is called symplectic if one has $f^{-1}=f^\dagger$ where $f^\dagger\colon B\times\widehat B\to A\times\widehat A$ is the morphism represented by the matrix $$\left(\begin{array}{cc} \hat a_{22} & -\hat a_{12}\\ -\hat a_{21} & \hat a_{11}\end{array}\right)$$ where $\hat a_{ij}$ denotes the transposed homomorphism between the dual Abelian varieties.

In the paper: D. Orlov, Derived categories of coherent sheaves on Abelian varieties and equivalences between them. Izv. Math. 66 569 (2002) (available also on the arxiv: http://arxiv.org/abs/alg-geom/9712017), it is claimed just after Proposition 4.12 that every symplectic isomorphism $f\colon A\times\widehat A\to B\times\widehat B$ can be factored as $f=f_1\circ f_2$ where $f_1\colon A\times \widehat A\to B\times \widehat B$, $f_2\colon A\times \widehat A\to A\times\widehat A$ are symplectic isomorphisms such that the $12$ entries of their corresponding matrices are isogenies. It is easy to prove this if $A$ and $B$ are simple Abelian varieties.


Any ideas on how can it be proved in the general case?

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I don't know how to answer this question but I'm curious now. Do you think you could present your solution in the case of simple Abelian varieties? –  solbap May 27 '10 at 22:42

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