Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
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
1  
@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
2  
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

1 Answer 1

up vote 5 down vote accepted

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.

share|improve this answer
    
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
1  
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. –  Eric Zaslow Sep 15 '10 at 0:32
    
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 –  Dan Sep 16 '10 at 1:35
    
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 –  Eric Zaslow Sep 16 '10 at 20:38
    
Thanks a lot. It is very nice of you. –  Dan Sep 16 '10 at 21:57

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.