$\DeclareMathOperator\Pic{Pic}$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 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) ).

$\DeclareMathOperator\Pic{Pic}$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) ).

$\DeclareMathOperator\Pic{Pic}$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) ).

$\DeclareMathOperator\Pic{Pic}$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 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) ).

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 calssified 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 not either.

So my qeustion 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 surface with the property b) only ( not satisfying the property a)  )

$\DeclareMathOperator\Pic{Pic}$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) ).