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I was just wondering, is there a general way to construct explicitly the jacobian of a curve, giving explicitly the vector space $\mathbb{C}^g$ and the lattice? (over the complex numbers is enough for what I'm studying) Or maybe for certain types of curves, for example, if $X$ is a hyperelliptic curve given by the equation $y^2=f(x)$ (nonsingular), can its jacobian be explicitly calculated?

This may be too broad of a question, but if anyone can point me possibly to a book, I'd appreciate it.

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1 Answer 1

up vote 6 down vote accepted

Integrating differential forms over homology classes gives an embedding $H^1(X,\mathbb Z) \hookrightarrow H^0(X,\Omega^1)^*,$ and the Jacobian is the quotient $H^0(X,\Omega^1)^*/H^1(X,\mathbb Z)$.

Note that $H^0(X,\Omega^1)$, and hence also its dual, is a complex vector space of dimension $g$, and so this expresses the Jacobian of the curve as $\mathbb C^g$ modulo the period lattice. If you want to actually realize it as a projective variety, the standard method is via $\theta$-functions.

How much more explicit do you want to get?

Added in response to the comment below:

For a hyperelliptic curve you can write down a basis of holomorphic differential (generalizing $dx/y$ in the elliptic curve case), but the lattice will (so to speak) be whatever it is. E.g. in the elliptic curve case, the lattice is (after rescaling) equal to the span of $1$ and $\tau$, where $\tau$ can be any element of the upper half-plane. In what sense do you want to identify $\tau$?

Perhaps one way is that in the elliptic curve case we can compute $\tau$ via certain hypergeometric functions in terms of the $\lambda$-invariant of the elliptic curve. Is this what you have in mind?

Added in response to the follow-up comment:

The curve $y^2 = x^6 - 1$ admits a surjection onto the curve $y^2 = x^3 - 1$, via $(x,y) \mapsto (x^2,y)$. Thus the latter curve (an elliptic curve with CM by $\sqrt{-3}$) is a factor (up to isogeny) of the Jacobian of $y^2 = x^6 - 1$. Thus this Jacobian is isogenous to the product of a pair of elliptic curves.

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I know the construction, I guess was just wondering if there's a way to explicitely calculate the homology basis (for example in the hyperelliptic case); knowing that you can integrate and find exactly what the lattice is in $\mathbb{C}^g$. That's what I was referring to. –  Robert Auffarth Sep 23 '11 at 18:40
    
@Robert: Dear Robert, I added an additional paragraph to my answer. Is it at all what you had in mind? Regards, –  Matt E Sep 23 '11 at 18:54
    
Thanks Matt, I guess my question is kind of vague... I was kind of hoping to calculate, for example, the Jacobian of the curve $X:y^2=x^6-1$ (I'm looking for genus $\geq 2$). My main problem is finding explicitly the homology basis for $H_1(X,\mathbb{Z})$, since I already know the basis for $\Omega_X^1$ ($\{dx/y,xdx/y\}$). I guess maybe pulling back curves that go between branch points of the projection $(x,y)\mapsto x$ would help? –  Robert Auffarth Sep 23 '11 at 20:53
    
@Robert: Dear Robert, Your example will be very special, because it has symmetries by $\mu_6$ (acting as multiplication on the $x$-variable), so the Jacobian will have extra automorphisms. It may well be a product of CM elliptic curves with $j$-invariant $0$ (up to isogeny). Regards, –  Matt E Sep 23 '11 at 21:09
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@MattE Even though I like "The Cure", I prefer curves though ;) –  Adrián Barquero Sep 24 '11 at 3:05

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