Let $X$ be a K3 surface (algebraic, complex). An involution $\sigma:X\rightarrow X$ is called non-symplectic if it acts trivially on $H^{2,0}(X)=\Bbb{C}\omega\_X$ $\ $ (where $\omega\_X$ is any nowhere vanishing 2-form), i.e. if $\sigma^\ast\omega\_X=-\omega\_X$.


Now, the Picard group can be identified with the lattice $S\_X=\{x\in H^2(X,\Bbb{Z}): \ x\cdot\omega\_X = 0 \}$.


Denote by $S(\sigma)=\{x\in H^2(X,\Bbb{Z}): \ \sigma^\ast x = x \}$ the invariant lattice. 


It is clear that $S(\sigma)\subset S\_X$ (recall that $\sigma^\ast$ is an isometry). My question is whether the equality holds, and if it is an easy fact to prove.



K3 surface with a non-symplectic involution: a basic question
A typical example is given by K3 surfaces of genus 2, that is, double coverings of $\Bbb{P}^2$ branched along a sextic plane curve $C$. Here $S(\sigma )$ is 1-dimensional (it is the pull back of $\mathrm{Pic}(\Bbb{P}^2)$), while $\mathrm{rank}(S\_X)$ can take all values from 1 to 20 -- for instance it is 20 when $C$ is the Fermat sextic.


