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Let $S$ be a K3 surface and $f:S\rightarrow \mathbb{P}^1$ a $T^2$-fibration (not necessarily holomorphic, I have a special Langrangian fibration in mind). Assume there is a $k$-section, then a fiber and the $k$-section generate a sublattice $L\subset H^{2}(S,\mathbb{Z})$, which is isomorphic to $U(k)$ (hyperbolic lattice multiplied by $k \in \mathbb{N}$).

Assume also that there is an holomorphic involution $\sigma$ of $S$ such that induced action $\sigma^{*}$ acts as $-id$ on $L$ (especially preserves $L\subset H^{2}(S,\mathbb{Z})$). Is it true that $\sigma$ preserves the fibration? If so, could one tell how $\sigma$ acts on each smooth fiber of $f$? If not so, what additional condition is required?

Edit The original question does not mach much sense. The fibration is NOT holomorphic. the following is the motivation of my question. I have a K3 surface $S$ with an anti-symplectic involution $\sigma$. Assume that, by using another complex structure, we can construct an elliptic fibration $f:S\rightarrow \mathbb{P}^{1}$ (possibly with no section). I hope this map to be a special Lagrangian $T^2$-fibration with respect to the original complex structure. I now want to understnad how $\sigma$ and the map $f$ are related.

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  • $\begingroup$ Are you saying that $\sigma$ takes an effective class to a negative effective class? If not, what does it mean that $\sigma^*$ acts as $-id$? $\endgroup$
    – Sándor Kovács
    Commented Sep 5, 2012 at 23:39
  • $\begingroup$ Indeed, $-id$ on an elliptic surface, if it exists and preserves $L$, acts as $id$ on $L$, since it preserves degree and preserves the class of a fiber. If $\sigma$ acts as $id$, then it preserves the fibration, since it preserves the class of the fiber, and the fibration is just the map to $\mathbb P^1$ coming from the global sections of the line bundle whose divisor class is the class of the fiber. $\endgroup$
    – Will Sawin
    Commented Sep 6, 2012 at 0:11
  • $\begingroup$ My first question does not make much sense. I made a major change in my question, also adding some motivation of my question. I am sorry for the confusion. $\endgroup$
    – Carmen
    Commented Sep 6, 2012 at 0:31

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