# compute the Kähler moduli of an elliptic curve

Say given elliptic curve $\{ (x,y) | y^2 = (x^2-1)(x^2-k^2) \}$, what is the right form of the K$\ddot{a}$hler form and how to compute the K$\ddot{a}$hler moduli of this elliptic curve? Thank you.

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Put it into the more usual form of $y^2=f(x)$ with $f$ cubic (see Cassels book on elliptic curves for how to do this; Cassels gives lots of transformations for getting general genus 1 curves into this form, including the type you have here) and then it's just $dy/x$ (with the new coordinates). – Kevin Buzzard Sep 14 '10 at 6:38
Kevin, you have described an element of $H^0(\Omega^1)$. Isn't the Kahler form a 2-form, giving the hyperplane class in $H^2$ (so it should be a (1,1) form on the elliptic curve, not a (1,0) form). – Emerton Sep 14 '10 at 7:01
@Dan: $H^{1,1}$ is 1 dimensional. You just have to write down the Kähler form. Then the Kähler moduli is just a 1 dimensional space given by scalings of the Kähler form. – Kevin H. Lin Sep 14 '10 at 8:25
@Emerton: sounds like I misunderstood the question. I'd delete my comment were it not for the fact that it would make your comment look meaningless :-) Yes, I described a global holomorphic 1-form. I thought these were called Kaehler 1-forms by some people and assumed this was what the questioner was asking about. – Kevin Buzzard Sep 14 '10 at 9:51
Can't we make use of Kevin's comment? Take his (1,0) form \alpha and set \omega = i * \alpha \wedge \overline{\alpha}. Then \omega is a real (1,1)-form, and Kahler if it is non-degenarate (and positive, but take -\omega if it's negative). – Gunnar Þór Magnússon Sep 14 '10 at 11:20

I see what you mean, but I am still confused with the Fubini-Study like Kahler form here. We can choose a volume form (Kahler form) up to scaling, so the volume (Kahler moduli) will be a multiple of some area A. On the other side, the complex structure is normalised as $\tau$ ( say our lattice is $Z+Z\tau$). By the mirror map, $\tau \arrow \rho= b + iA$, my question is how can I fix $b$ and $A$? – Dan Sep 14 '10 at 21:55