Involution of the Fermat quartic

2012-10-09T17:35:33Z

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

http://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?

Answer by Francesco Polizzi for Involution of the Fermat quartic

2012-10-10T00:05:29Z

Using the notation of the question http://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.

Involution of the Fermat quartic - MathOverflow

Involution of the Fermat quartic

Answer by Francesco Polizzi for Involution of the Fermat quartic