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Let $S$ be a K3 surface and $\pi:S\rightarrow B$ be a SLAG $T^2$-fibration. I am struggling with a statement that

Fiberwise dualization does not change the topology of $S$.

Here by fiberwise dualization I mean dualizing smooth fibers and compactifying the dual fibration (I think this is the standard definition).

  1. My first question is, what do people mean by "compactification"?

Here is what I naively thought. Let $X$ be a Calabi--Yau n-fold and $\pi:X\rightarrow B$ a SLAG $T^n$-fibration. Let $B_0 \subset B$ be the set on which $\pi$ is smooth. Then $\pi^{-1}(B_0)$ and its dual $\pi^{-1}(B_0)^\vee$ is topologically the same, so we may compactify $\pi^{-1}(B_0)^\vee$ to get the original $X$. This should not happen in the SYZ picture. Of course this only holds at the topological level and ignores complex structure etc.

  1. My second question is, what makes difference between dimension two and higher?

The statement above is certainly not true in dimension three or higher. Otherwise SYZ mirror conjecture does not produce a mirror Calabi--Yau manifold, which in general has different topological type from the original one.

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2 Answers 2

The crucial point for the second question is the following. In arbitrary dimension, it is not true that $\pi^{-1}(B_0)$ and $\pi^{-1}(B_0)^{\vee}$ are homeomorphic as fibre bundles. This is essentially because there is no canonical isomorphism between a torus and its dual. If we think of an $n$-torus as $V/\Lambda$, with $V$ an $n$-dimensional vector space and $\Lambda$ a lattice in $V$, then the dual torus is $V^\vee/\Lambda^\vee\cong H^1(V/\Lambda,R/Z)$, while $V/\Lambda\cong H_1(V/\Lambda,R/Z)$. But in two dimensions, Poincare duality gives a canonical isomorphism between $H_1(V/\Lambda,R/Z)$ and $H^1(V/\Lambda,R/Z)$. This canonical isomorphism relativizes, so that if $X_0\rightarrow B_0$ is a two-torus bundle with section, there is a canonical fibrewise isomorphism $X_0^{\vee}\cong X_0$.

For the first question as to what "compactiifcation" means, we assume given a torus bundle $X_0\rightarrow B_0$ with $B_0\subseteq B$ and $B\setminus B_0=\Delta$ relatively nice (perhaps codimension two). One would then like to extend $X_0\rightarrow B_0$ to a proper map $X\rightarrow B$ which is sufficiently "nice". Typically our ability to do this depends on the monodromy of the torus bundle $X_0\rightarrow B_0$ near $\Delta$. For example, in the two-dimensional case, if one has monodromy around a point of $\Delta$ (monodromy acting on $H^1$ of a fibre) which takes the form $\pmatrix{1&0\cr 1&1\cr}$ in some basis, we can compactify by adding a pinched torus over the singular point. To do this, one needs to glue in a local model: see e.g., my paper http://arxiv.org/abs/math/9909015 for details on this kind of thing in dimensions two and three. In general, there is no canonical compactification, but if the monodromy is sufficiently nice, then there is a reasonable choice of compactification.

One can also ask for compactifications in various categories, e.g., not just topological compactifications, but symplectic or complex compactifications if $X_0$ is symplectic or complex. The symplectic case has been done in dimension three by Castano-Bernard and Matessi, while the complex case is far more subtle. Discussing this probably goes a bit too far afield for this question.

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I am not completely sure what is meant by this statement since the topology of the torus fibration certainly changes if the fibration doesn't have a section. If on the other hand we follow your prescription, dualize the smooth fibers, and compactify to a smooth space, then the total space of this new family of tori is a K3 again. And all K3 surfaces are diffeomorphic so the topology does not change.

Up to a hyper-Kaehler rotation the dualization is an algebraic process - it is the minimal resolution of the relative Jacobian of a holomorphic genus one fibration. It is unique and automatically a K3 hence diffeomorphic to the original K3.

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Thank you for the answer. I implicitly assume that there exists a section, but does it matter whether or not we have a section? As to the 3rd line, why is the new family again K3 surface? Why doesn't your argument work in dimension three? Thank you for your help. –  Tobias May 29 '13 at 20:11
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@Tobias: The compactified relative Picard automatically has a section. For a genus 1 fibration of a K3 surface over $\mathbb{P}^1$, there is a topological obstruction to the existence of even a continuous section (e.g., if there is a multiple fiber). Thus the topology of the fibration can change. The fact that the compactified relative Picard is again a K3 surface is a special case of a general theory of Mukai about moduli spaces of (semi)stable sheaves on hyper-Kaehler manifolds (although there may be a more direct proof for K3s). –  Jason Starr May 29 '13 at 21:59
    
The fact that the minimal resolution of the compactified relative Picard is again a K3 can be checked directly from Kodaira's classification of singular fibers. The type of the singular fiber is determined by the local monodromy and the local monodromy doesn't change under dualization because it is in $SL_2$. Once you know that singular fibers are the same as the original ones, the canonical class formula for an elliptic surface tells you that you have a K3. This doesn't work in higher dimensions already topologically as Mark explained above. –  Tony Pantev May 30 '13 at 12:37
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