## construct the elliptic fibration of elliptic k3 surface

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!

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One elliptic fibration of the Fermat quartic that was known (in all but name) to Euler is described on pages 12-13 of the lecture notes at math.harvard.edu/~elkies/euler_11c.pdf, which also describe a few other ways to work with elliptic fibrations of K3's. – Noam D. Elkies Feb 6 2012 at 5:41
Certainly the topological fibrations of elliptically fibered K3 surfaces are not all topologically equivalent. The number of singular fibers, as well as the types of singularities, can vary. The "weighted sum" of the singularities of fibers always equals 24, but that leaves a lot of room for variation. – Jason Starr Feb 6 2012 at 11:20
Some more questions in this direction: -- @ Jason Starr: What is the weight you are mentioning? --So the number of singular fibers (which I think is the degree of j-map to $\mathbb{P}^1$ can be different in different examples? --Are there examples where the singular fiber is "not" normal-crossing" (or semi-stable)? – Mohammad F.Tehrani Mar 27 at 2:07
When talking about j-map above, I am assuming there is a section. – Mohammad F.Tehrani Mar 27 at 2:52
This looks fun: grdb.lboro.ac.uk/search/ellk3 – Mohammad F.Tehrani Mar 27 at 2:57

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 ${}$. 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.

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 Thanks a lot! It is quite helpful! – Jay Feb 7 2012 at 5:35 Good! So is this answer satisfactory or you wanted to know more specific things? – diverietti Feb 7 2012 at 9:40 that is great~ thanks! by the way, do you know anything about the existence of sections? maybe just a topological section... – Jay Feb 8 2012 at 5:11