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$\DeclareMathOperator\Pic{Pic}$Let $X$ be a K3 surface (algebraic, complex). An involution $\sigma:X\rightarrow X$ is called non-symplectic if it acts as multiplication by $-1$ on $H^{2,0}(X)=\Bbb{C}\omega_X$ (where $\omega_X$ is any nowhere vanishing 2-form), i.e. $\sigma^\ast\omega_X=-\omega_X$, or equivalently the quotient $X/\langle \sigma \rangle$ is a rational surface or an Enriques surface.

The invarant sublattice $\Pic(X)^\sigma:= \{ l \in \Pic(X): \sigma^*(l)= l\}$ of the Picard lattice was completely classified by Nikulin.

I have been looking for an example of K3 surface $X$ with two non-symplectic involutions $\sigma_1$, $\sigma_2$ such that

a) The lattice $L:=\Pic(X)^{\sigma_1} \cap \Pic(X)^{\sigma_2} $ contains an ample divisor class of $X$ and

b) $L$ is a proper subset of $\Pic(X)^{\sigma_i}$ for both $i=1, 2$.

I have kept trying to find such an example both by concrete geometric construction, by using the Torelli theorem on K3 surfaces but I could not.

I have also tried to see any obstruction in constructing such an example but I couldn't either.

So my question is:

Does there exist such an example of a K3 surface?

By the way, there are many examples of K3 surfaces such that $L=\Pic(X)^{\sigma_1}\subsetneq \Pic(X)^{\sigma_2}$ with the property a).

And there are also many examples of K3 surfaces with the property b) only (not satisfying the property a) ).

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  • $\begingroup$ Seems that property (a) always holds for the following reason: taking the quotient of $X$ by the subgroup of $H$ of $Aut(X)$ generated by $\sigma_1$ and $\sigma_2$, the resulting (singular) surface $X/H$ will have an ample divisor, whose pullback to X will still be ample? $\endgroup$
    – Evgeny Shinder
    Commented May 24, 2022 at 11:58
  • $\begingroup$ You seem to be essentially asking about two involutions $\sigma_1$, $\sigma_2$ in $O(NS(X))$ which both act by $-1$ on the discriminant lattice, and whose $+$-eigenspaces on $NS(X)$ are distinct? $\endgroup$
    – Evgeny Shinder
    Commented May 24, 2022 at 12:07
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    $\begingroup$ @Evgeny Shinder, the subgroup $H$ does not need to be finite and so you don't necessarily have a surface $X/H$. In fact, one can find many examples of K3 surfaces with $\sigma_1, \sigma_2$ that satisfy the property b) only. I just added a comment in the main post about this. $\endgroup$
    – Basics
    Commented May 24, 2022 at 12:17
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    $\begingroup$ Ah, I see. Maybe one can still hope for commuting non-symplectic involutions? Another idea is to look at multiplication by (-1) for two different elliptic structures on X. $\endgroup$
    – Evgeny Shinder
    Commented May 24, 2022 at 13:27
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    $\begingroup$ @Evgeny Shinder, for your first idea: Yes. In fact, if the composite $\sigma_1 \circ \sigma_2$ is of finite order, then the property a) is satisfied. $\endgroup$
    – Basics
    Commented May 24, 2022 at 13:35

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