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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

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?

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It seems to me that the fibers of the fibration described in mathoverflow.net/questions/87633/… are permuted by $i$, so the quotient surface has an elliptic fibration. –  rita Oct 9 '12 at 20:23
    
You are right. The fibration of mathoverflow.net/questions/87633/… is not exactly the one I had in mind, sorry. It is just because I was considering a different fibration on $\mathbb{P}^{1}$, namely the one given by $$ C_{[\mu,\lambda]} = \left\{\begin{array}{l} \lambda (x^{2}-y^{2}) = \mu (z^{2}-t^{2})\\ \lambda (x^{2}+y^{2}) = \mu (z^{2}+t^{2}) \end{array} \right. $$ In this way $X/i$ is a fibration over $\mathbb{P}^{1}$ with $\mathbb{P}^{1}$ as general fiber. –  MorFel1921 Oct 10 '12 at 12:29
    
I had tought of this possibility. If I have time I can try to work out a description. Or maybe Francesco will do it... –  rita Oct 10 '12 at 16:46

1 Answer 1

up vote 4 down vote accepted

Using the notation of the question 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.

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