4
$\begingroup$

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?

$\endgroup$
3
  • 2
    $\begingroup$ 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. $\endgroup$
    – rita
    Oct 9, 2012 at 20:23
  • $\begingroup$ 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. $\endgroup$
    – Puzzled
    Oct 10, 2012 at 12:29
  • $\begingroup$ I had tought of this possibility. If I have time I can try to work out a description. Or maybe Francesco will do it... $\endgroup$
    – rita
    Oct 10, 2012 at 16:46

1 Answer 1

4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.