Let $X\subset\mathbb{P}^{3}$ be the Fermat quartic surface given by
$$x^4-y^4-z^4+w^4 = 0$$
and consider the involution 
$$i:X\rightarrow X,\: (x,y,z,w)\mapsto (y,x,w,z).$$
The surface $X$ can be seen as a narural elliptic fibration over $\mathbb{P}^{1}$ as explained here 


[construct the elliptic fibration of elliptic k3 surface](https://mathoverflow.net/questions/87633/construct-the-elliptic-fibration-of-elliptic-k3-surface)


The quotient $X/i$ inherits a fibration structure over $\mathbb{P}^{1}$ whose generic fiber is a smooth rational curve and with six special fibers which are union of two $\mathbb{P}^{1}$'s intersecting in a point. 


**Can one give an explicit description of this quotient?**



Involution of the Fermat quartic
Using the notation of the question [construct the elliptic fibration of elliptic k3 surface](https://mathoverflow.net/questions/87633/construct-the-elliptic-fibration-of-elliptic-k3-surface), one sees that the elliptic curve $C\_{[\lambda:\mu]}$ is sent to the curve $C\_{[-\lambda: \mu]}$ by the involution $i$. So the surface $S=X/i$ has an elliptic fibration over $\mathbb{P}^1$. 


The fixed locus of $i$ is given by the disjoint union of two $(-2)$-curves in $X$, namely the two lines $L\_1:\{x=y, \; z=w \}$ and $L\_2:\{x=-y, \; z=-w \}$, which are components of the the fibre $C\_{[1:0]}$.


Since $i$ has no isolated fixed points, the quotient $S$ is smooth, and the quotient map $\pi \colon X \to S$ is branched over two smooth rational curves, namely the images of $L\_1$ and $L\_2$.


Using the fact that the topological Euler number of $X$ is $24$ and that the branch locus of the double cover $\pi$ is homeomorphic to the disjoint union of two spheres, one finds that the topological Euler number of $S$ is $\frac{1}{2}(24-4)+4=14$. 


On the other hand, by Hurwitz formula one finds
$$K\_X=\pi^\*K\_S+L\_1+L\_2,$$
which yields $K\_S^2=\frac{1}{2}(K\_X-L\_1-L\_2)^2=-2$.


Using Noether formula we obtain $\chi(\mathscr{O}\_S)=(14-2)/12=1$, i.e. $p\_g(S)=q(S)$. In particular $S$ is not birational to a $K3$ surface, hence $i$ must be an anti-symplectic involution, namely $i^\* \omega = -\omega$ where $\omega$ is the holomorphic $2$-form on $S$.


By general results, if $i$ is an anti-simplectic involution on a $K3$ surface then $X/i$ is a rational surface or an Enriques surface, and the last case happens exactly when $|\textrm{Fix}(i)|=\emptyset$. Therefore in our case $S$ is a rational surface.


Summing up, the surface $S=X/i$ is a non-minimal rational surface with $K\_S^2=-2$ and an elliptic fibration over $\mathbb{P}^1$. Notice that such a fibration is not relatively minimal, since the fibre containing the branch locus also contains two $(-1)$-curves. Contracting those curves, one obtain a non-minimal rational surface $\widetilde{S}$ with $K\_{\widetilde{S}}^2=0$ and a relatively minimal elliptic fibration over $\mathbb{P}^1$.


By looking at the degenerate fibres on $X$, one checks that the degenerate fibres of $\widetilde{S}$ are two singular fibres of type $I\_2$ and two singular fibres of type $I\_4$ in Kodaira's classification; the existence of the last two fibres shows in particular that $\widetilde{S}$ is not isomorphic to $\mathbb{P}^2$ blown-up in nine points.


My guess is that $\widetilde{S}$ can be constructed in the following way: take a smooth quadric surface $Q$ and consider two reducible curves $T\_1$ and $T\_2$ of bidegree $(2,2)$, both composed by two lines in a ruling and two lines in the other ruling. Then $\widetilde{S}$ is obtained by blowing up the $8$ base points of the pencil of elliptic curves generated by $T\_1$ and $T\_2$. Notice that the $T\_i$ are precisely two degenerate fibres of type $I\_4$ in that pencil.


