Mirror symmetry for hyperkahler manifold

2013-01-26T00:10:59Z

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!

Answer by Mark Gross for Mirror symmetry for hyperkahler manifold

2013-01-26T15:29:10Z

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.

Mirror symmetry for hyperkahler manifold - MathOverflow

Mirror symmetry for hyperkahler manifold

Answer by Mark Gross for Mirror symmetry for hyperkahler manifold