# Genus 2 Elliptic curves & their periods

The first part of my question is just a check of my knowledge on elliptic curves. I'm fairly happy with the number theory side of things (torsions, rank, whatever) but is my understanding of the more geometric/topoological side correct?

It is well known that an elliptic curve (smooth plane cubic) is equivalent, via the Weierstrass function, to a complex torus, that is, $\mathbb{C}/\Lambda$ for some non-degenerate lattice $\Lambda=\omega_1\mathbb{Z}+\omega_2\mathbb{Z}$ such that $\omega_1/\omega_2\notin \mathbb{R}$. In this case, $\omega_1,\omega_2$ are the periods of the elliptic curve and can be computed as

$$\omega_1=\int_\alpha \omega=\int_\alpha \frac{dx}y, \quad \omega_2=\int_\beta \omega=\int_\beta \frac{dx}y,$$

$\omega$ being the invariant differential 1-form. The paths here are the two path independent ways in the torus -- i.e. one loops 'around' the middle hole and one loops 'through' the hole. The integrals can also be expressed as elliptic integrals integrating between the roots of the RHS of $y^2=x^3+ax+b$.

Now my question is how does this all generalise to curves of genus 2?

My understanding is that we get hyperelliptic curves (plane quintic or sextic with distinct roots) and you can think of it as some complex torus of the form $$\mathbb{C}^2/(\omega_1\mathbb{Z}+\omega_2\mathbb{Z}+\omega_3\mathbb{Z}+\omega_4\mathbb{Z})$$ or some two-hole doughnut. How would the periods genearlise, now that we have 2 invariant differentials and 4 independent paths on the torus?

• Just to be clear: a complex torus of the form you wrote is not a curve: it's a complex surface. Conversely, no curve of genus $\geq 2$ is a complex torus. – user64687 Jan 23 '15 at 13:03
• (To clarify my clarifying comment: I meant to say that a complex torus of the second form you wrote isn't a curve.) – user64687 Jan 23 '15 at 13:12
• On the other hand the Jacobian of a genus 2 curve is exactly the kind of object you are talking about, and that is where the kind of action you are talking about takes place. Maybe you could revise the question to clear up that point. – user64687 Jan 23 '15 at 14:25

Let say you have a degree $d$ smooth curve, $P(x,y)$ over $\mathbb{C}$. Then $g = \frac{(d-1)(d-2)}{2}$.
$$M_{ij} := \int_{\gamma_j} \omega_i$$
Where, as you said, $\gamma_1, ..., \gamma_{2g} \in H_1(C)$, and $\omega_1, ..., \omega_g \in H^1(C)$. Here, $\omega_1, ..., \omega_g$ is a basis of holomorphic 1 forms for $C$. This basis comes to you as $g \cdot \frac{dx}{\frac{\partial P}{\partial y}}$ for $g \in \mathbb{C}[x,y]$ such that $\text{deg}(g) \leq d-3$ (see Harris, Corollary 6.2).
Thus, for fixed $k$, the period is $M_{kj}$, that is, the $k$th column vector of the period matrix.