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.

  • $\begingroup$ 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). $\endgroup$ – Kevin Buzzard Sep 14 '10 at 6:38
  • $\begingroup$ 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). $\endgroup$ – Emerton Sep 14 '10 at 7:01
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    $\begingroup$ @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. $\endgroup$ – Kevin H. Lin Sep 14 '10 at 8:25
  • $\begingroup$ @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. $\endgroup$ – Kevin Buzzard Sep 14 '10 at 9:51
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    $\begingroup$ 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). $\endgroup$ – Gunnar Þór Magnússon Sep 14 '10 at 11:20

The curve you wrote in equations lies in C^2, while the "elliptic curve" of your text is presumably a compact projective variety -- meaning you imagine making your equations homogeneous (or even quasi-homogeneous) and considering the closure of the set of points described by your equation in a (quasi-)projective plane. Not every "homogenization" will lead to an elliptic curve (Calabi-Yau) upon compactification, so you have to do this correctly (as noted by Kevin Buzzard above).

Having said that, the answer is that every projective variety is also Kahler: just restrict e.g. the Fubini-Study(-like) Kahler form. In plain English, since a Kahler form on a complex curve is just a volume form, the volume of the compact curve inside projective space gives you your answer.

  • $\begingroup$ 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 $A$? $\endgroup$ – Dan Sep 14 '10 at 21:55
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    $\begingroup$ You're thinking of mirror symmetry now, not pure geometry. In mirror symmetry, the Kahler form is "complexified," referring to a sum of a general (1,1)-form (with no positivity conditions) plus "i" times a geometric Kahler form (which lives in the [real] Kahler cone). More generally, when you integrate the exponential of this form over the surface (i.e. take the pairing with a class in H_2) you get a complex number of modulus less than one, which is the weighting factor for the Gromov-Witten invariant in that homology class. Ideally, the sum over all such numbers will converge. $\endgroup$ – Eric Zaslow Sep 15 '10 at 0:32
  • $\begingroup$ Thank you for your comments. Let me put the question in another way, how can one verify some phenomenon is mirror symmetry of dimension one? I am reading "motives from diffraction" by Jan Stienstra arxiv.org/abs/math/0511485 $\endgroup$ – Dan Sep 16 '10 at 1:35
  • $\begingroup$ That question ("some phenomenon") is a bit vague. For more you can read Kontsevich's 1994 ICM paper or this paper by Polishchuk et al: front.math.ucdavis.edu/9801.5119 $\endgroup$ – Eric Zaslow Sep 16 '10 at 20:38

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