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SYZ mirror symmetry for K3 surfaces

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My question is essentially related to this post, but let me formulate it again. Let $f:S \rightarrow \mathbb{P}^1$ be an elliptic fibration, then this can be a SLAG fibration with respect to another complex structure on $S$, say $S_K$. Since the compactified dual fibration $f^\vee$ is naturally identified with $f$ (see the above post, especially Gross's answer), it seems the mirror manifold of $S_K$ is again $S_K$. However, this does not seem compatible with mirror symmetry of K3 surfaces (in the sense of Dolgachev for example).

Can anyone clarify the problem? A possible mistake is that the dual fibration $f^\vee$ cannot be identified with $f$...


This post imported from StackExchange MathOverflow at 2015-04-02 13:13 (UTC), posted by SE-user Vladhagen

asked Jun 10, 2014 in Mathematics by Vladhagen (10 points) [ revision history ]
edited Apr 2, 2015 by Dilaton
as far as I know, Dolgachev's version of mirror symmetry is not compatible with the physicists' one

This post imported from StackExchange MathOverflow at 2015-04-02 13:13 (UTC), posted by SE-user Misha Verbitsky

1 Answer

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My answer in the link given above is purely at a topological level, saying that if we have a $T^2$-fibration, the dual is canonically homeomorphic. However, $T$-duality should also be viewed as exchanging complex and symplectic structure . For K3 surfaces, this can be described in terms of forms, and I sketched this in an answer to a different question, Mirror symmetry for hyperkahler manifold.

Dolgachev's mirror symmetry can be viewed as a subset of physicist's mirror symmetry. The key paper explaining mirror symmetry for K3 surfaces from a physics point of view is a paper of Aspinwall and Morrison, http://arxiv.org/abs/hep-th/9404151. There is a Teichmuller space of SCFTs on a K3 surface, essentially the space of space-like four-planes in $H^{even}(X,{\mathbb R})$, equipped with the Mukai pairing and lattice $H^{even}(X,{\mathbb Z})$, which has signature $(4,20)$. To first approximation, one can view one of these four-planes as the subspace spanned by the real and imaginary parts of a holomorphic two-form, the Kaehler form, and the exponential of the $B$-field (although the actual description in terms of this data is a bit more complicated). The actual moduli space of SCFTs is obtained by dividing out by the group of automorphisms of the lattice $H^{even}(X,{\mathbb Z})$. This group is generated by the "classical" identifications, coming from automorphisms of $H^2(X,{\mathbb Z})$, and additional automorphisms coming from integral shifts in the $B$-field and finally a choice of "mirror involution". This comes from a choice of a hyperbolic plane $H\subset H^2(X,{\mathbb Z})$, and the mirror involution exchanges the hyperbolic plane $H^0(X,{\mathbb Z})\oplus H^4(X,{\mathbb Z})$ with $H$ and leaves everything else fixed.

The choice of hyperbolic plane $H$ can be viewed as the choice of $H$ in the Dolgachev construction. This involution acts on the full Teichmuller space of SCFTs, but after making some choices, one sees that it restricts to Dolgachev's description of mirror K3 families. I can provide more details if needed.

This post imported from StackExchange MathOverflow at 2015-04-02 13:13 (UTC), posted by SE-user Mark Gross
answered Jul 22, 2014 by Mark Gross (20 points) [ no revision ]

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