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Given an explicit description (as an intersection) of an abelian surface $A$ is there an algorithm for computing the period lattice of the surface? For the specific examples that I am interested in, the ideal of $A$ has been obtained by Weil restriction from the affine model of an elliptic curve.

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I don't think that an abelian surface can be a complete intersection. – Piotr Achinger Sep 27 '12 at 23:49
This is explicit: The Weil restriction of an elliptic curve over a quadratic extension is an abelian surface. Restriction of scalars of the ideal of the curve gives two equations in four variables. I'm not an algebraic geometer, but I think that makes it a complete intersection. – Johnson-Leung Sep 28 '12 at 0:33
Any complete intersection surface in projective space has $q=0$, so it can not be an Abelian surface. – Mohan Sep 28 '12 at 0:44
OK. I edited it to not imply that I have a projective model. Formally applying Weil restriction, I get the surface as the intersection of two affine varieties. – Johnson-Leung Sep 28 '12 at 0:54
up vote 5 down vote accepted

It seems to me that it's best to go back up to the quadratic extension and view it as a product of two elliptic curves $E \times E^\sigma$. If you can compute the Weierstrass equations for these elliptic curves, you can compute their $j$ invariants, and then you just need to find $\tau_1,\tau_2$ such that $j(\tau_1)=j(E)$, $j(\tau_2)=j(E^\sigma)$. Then the lattice is $\mathbb C^2 /((1,0),(\tau_1,0),(0,1),(0,\tau_2))$.

So if you have a Weierstrass equation for the curve you're Weil restricting, you're done, modulo the analytic process of inverting $j$, which should be computable to an arbitrary degree of accuracy, and exactly computable if your $j$ is a special value.

A question I'm not sure about is, if you have the equation for the Weil-restricted surface in some other form, how you can get it into product-of-Weierstrass form.

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