Hi all,


As we know, every elliptic k3 surface admits an elliptic fibration over $P^1$, but generally how do we construct this fibration? For example, how to get such a fibration for Fermat quartic?


Moreover, as we know all (elliptic) k3 surfaces are differential equivalent to each other, does this mean: topologically the elliptic fibration we get for each elliptic fibraion is the same, which is just the torus fibration over $S^2$ with 24 node singularities? Or, the totally space is the same, but different complex data(structure) provides different way or "direction" of projection onto $S^2$, thus induces different type of fibrations?


Thanks!



construct the elliptic fibration of elliptic k3 surface
Let $S$ be a smooth projective $K3$ surface, say over the complex numbers, and suppose that $S$ admits a (non-constant) fibration $\pi\colon S\to C$ over a curve $C$. 


By the universal property of the normalization, we can suppose that $C$ is normal, hence smooth. Now, this curve $C$ must be $\mathbb P^1$, since otherwise you would have (by pulling-back) some non-trivial global holomorphic $1$-form on $S$, contradicting $h^{1,0}(S)=0$.


Next, this fibration is clearly given by the linear system $|\pi^\*H^0(\mathbb P^1,\mathcal O(1))|\subset |L|$ of the pull-back $L:=\pi^\*\mathcal O(1)$, which is spanned by two independent sections, say $\sigma$ and $\tau$. Now, take a general fiber: it is a smooth curve $F\subset S$ which is a divisor in the above-mentioned linear system, of the form $\{\lambda s+\mu t=0\}$. In particular $\mathcal O\_S(F)\simeq L$. Moreover, by definition, $\mathcal O\_F(F)\simeq L|\_F=\pi^\*\mathcal O(1)|\_F$ which is trivial.


By adjunction, $K\_F\simeq (K\_S\otimes\mathcal O\_S(F))|\_F\simeq\mathcal O\_F$ is trivial, so that $F$ is an elliptic curve. 


Thus, any fibration of a smooth projective $K3$ surface is an elliptic fibration over $\mathbb P^1$, obtained as above.


Now, let's consider the more specific case of the Fermat's quartic $S:\{x^4-y^4-z^4+t^4=0\}$ in $\mathbb P^3$ (it is the standard Fermat's quartic up to multiplying $y$ and $z$ by a $4$th root of $-1$). Then, we can factorize it in the following way
$$
(x^2+y^2)(x^2-y^2)-(z^2+t^2)(z^2-t^2)=0.
$$
This shows that, for $[\lambda:\mu]\in\mathbb P^1$, the complete intersection given by
$$
C\_{[\lambda:\mu]}:=\begin{cases} 
\lambda(x^2-y^2)=\mu(z^2+t^2) \\
\mu(x^2+y^2)=\lambda(z^2-t^2)
\end{cases} 
$$
is contained in $S$. For generic $[\lambda:\mu]\in\mathbb P^1$, this is a smooth elliptic curve, since its tangent bundle fits in the following short exact sequence
$$
0\to T\_{C\_{[\lambda:\mu]}}\to T\_{\mathbb P^3}|\_{C\_{[\lambda:\mu]}}\to\mathcal O\_{C\_{[\lambda:\mu]}}(2)\oplus\mathcal O\_{C\_{[\lambda:\mu]}}(2)\to 0.
$$
The function $[\lambda:\mu]$ defines a map from $S$ onto $\mathbb P^1$, which is the elliptic pencil on $S$ you were looking for.


Note that, for $\lambda/\mu=0,\pm 1,\pm i,\infty$, $C\_{[\lambda:\mu]}$ degenerates into a cycle of four lines. This gives you the 24 singularities. 


