Background: Let $V_\lambda$ be the elliptic curve $x^3+y^3+z^3 - \lambda xyz=0$. Then, when we consider $\omega_\lambda \in H^1_{dR}(V_\lambda)$, since $H^1_{dR}(V_\lambda)$ is only 2-dimensional, this implies that $\omega_\lambda$, $\frac{\partial \omega_\lambda}{\partial \lambda}$, and $\frac{\partial^2 \omega_\lambda}{\partial \lambda^2}$ must be linearly related.

The coefficients of this relationship can be derived by brute force finding $B$ and $C$ such that $\frac{d^2\omega}{d\lambda^2} + B\frac{d\omega}{d\lambda} + C \omega$ is a closed 1-form.

I am particularly curious about the comparison between the lattice model of (the Jacobian of) a variety and its integral presentation, and how to get between to two. I am in love with the elliptic curve story -- solving the Picard-Fuchs equation to get a lattice model of an elliptic curve as $\mathbb{C}/\Lambda$.

From reading about the Gauss-Manin connection, I have hope that there is a method of going from an integral model of parameterized families of a higher dimensional variety to a lattice model of its Jacobian as $\mathbb{C}^g/\Lambda$.

I have a particular example in mind: Let $A$ over $\mathbb{C}$ be the variety cut out by the zero set of $P (x,y) := y^3 - x(x- \zeta_0) (x- \zeta_1) (x - \zeta_ 2)$. This is a genus 3 curve, so, $H^1_{dR} (A) \simeq \mathbb{Z}^{6}$. What is the lattice model of the Jacobian of this curve; i.e., how can we present $Jac(A)$ as $\mathbb{C}^3/\Lambda$?

We can divide through $P$ to get rid of one of the parameters but we are still left with two parameters, say we are down to $P(x, y) = y^3 - x(x- 1) (x- \zeta_1) (x - \zeta_ 2)$

Q. What are the analogous 7 basis vectors in this case that yield an analogous differential equation? How do I solve for multiple periods to form a basis for the higher dimensional (abelian) variety?

  • $\begingroup$ Don't you confuse a curve with its Jacobian? $\endgroup$ – Sasha Jun 16 '17 at 12:03
  • $\begingroup$ @Sasha, yes, thank you, I have edited the question to reflect this. $\endgroup$ – Catherine Ray Jun 16 '17 at 13:20

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