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

Hi there,

I have some questions about the mirror symmetry of hyperkahler manifold and K3 surface.

The well-known result said: the mirror symmetry for K3 surface is just given by its hyperkahler rotation.

1) In what sense, the rotation gives the mirror map?

2) Does this means:

if we start from $(M,\omega_I,I)$, here the k3 surface $M$ has a special lagrangian fibration structure with respect to $\omega_I$ and $I$, and also a special lagrangian section. Denote its SYZ mirror by $(M^{mirror},J_{mirror})$. Then exist a (fiberwised) diffeomorphism $\phi: M \rightarrow M^{mirror}$, s.t. $\phi^{*} (J_{mirror}) = K$?

(Here $I,K$ are the standard base of the $S^2$-family of compatible complex structure of the hyperkahler metric on $M$.)

Thanks!

share|improve this question
    
I recommend you read section 7 of this paper of Gross arxiv.org/pdf/math/9809072.pdf, section 1 of Gross-Wilson arxiv.org/pdf/math/0008018v3.pdf, and this paper of Dolgachev arxiv.org/pdf/alg-geom/9502005v2.pdf –  YangMills Jan 26 '13 at 2:10
    
arxiv.org/abs/hep-th/9512195 Mirror Symmetry for hyperkaehler manifolds Misha Verbitsky We prove the Mirror Conjecture for Calabi-Yau manifolds equipped with a holomorphic symplectic form. Such manifolda are also known as complex manifolds of hyperkaehler type. We obtain that a complex manifold of hyperkaehler type is Mirror dual to itself. The Mirror Conjecture is stated (following Kontsevich, ICM talk) as the equivalence of certain algebraic structures related to variations of Hodge structures. We compute th –  Alexander Chervov Jan 26 '13 at 13:11
    
thank you so much! i have not expect such direct approach.. the kahler form gives the deformation of the complex structure directly through Tian-Todorov coordinates. inspiring! –  Jay Feb 6 '13 at 18:48

1 Answer 1

Thanks, YangMills, for the references to my papers. I want to elaborate, because I disagree with the statement that mirror symmetry is given by hyperkahler rotation. It may be the case for certain choices of K3, but I think this happens by accident and that it's not a useful principle. Here is how I view mirror symmetry for K3 surfaces. Choose a rank 2 sublattice of the K3 lattice generated by $E$ and $F$ with $E^2=F^2=0, E.F=1$. Consider a K3 surface $X$ with a holomorphic $2$-form with $E.\Omega\not=0$. We can assume after rescaling $\Omega$ that $E.\Omega=1$, and then write $\Omega=F+\check B+i\check\omega \mod E$ for some classes $\check B,\check\omega$ in $E^{\perp}/E$. The K3 surface will be equipped also with a Kaehler form $\omega$ and a B-field $B$, which we write as $B+i\omega$. We choose this data in $E^{\perp}/E \otimes {\mathbb C}$, although the class of $\omega$ is determined in $E^{\perp}$ by its image in $E^{\perp}/E$ by the fact that $\omega\wedge \Omega$ must be zero. Then the mirror $\check X$ is taken to have holomorphic form $\check\Omega=F+B+i\omega\mod E$ and complexified Kaehler class $\check B+i\check\omega$.

Note that there is no particular reason to expect this new K3 surface to be a hyperkaehler rotation, as the mirror complex structure depends on $B$, which gives far too many parameters worth of choices: there is only a two-dimensional family of hyperkaehler rotation of $X$.

Note that we can hyperkaehler rotate $X$ so that special Lagrangians become holomorphic. The new holomorphic form is $\check\omega + i \omega \mod E$. If we multiply this form by $i$, we get $-\omega +i\check\omega\mod E$. A change of Kaehler form followed by another hyperkaehler rotation will give the mirror for certain choices of $B$-field, but note this involves two hyperkaehler rotations with respect to different metrics.

share|improve this answer

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.