construct the elliptic fibration of elliptic k3 surface

Question

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!

Answer 1

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.

Answer 2

Here is an alternative way to construct elliptic fibrations of quartic surfaces containing a line.

Let $X\subset \mathbb P^3$ be a smooth quartic surface and assume that it contains a line $L\subset X$. Let $\mathfrak d$ denote the $1$-dimensional linear system of hyperplanes in $\mathbb P^3$ containing $L$ restricted to $X$. Further let $H\in \mathfrak d$ and $E=H-L$.

Observe that $E^2=H^2-2H\cdot L L^2= 4 - 2\times 1 -2=0$ and that by construction $\dim |E|\geq 1$. It follows that for any $E_1,E_2\in |E|$, $E_1\cap E_2=\emptyset$ and hence $|E|$ is basepoint-free and $\dim |E|=1$.

Therefore $|E|$ defines a morphism $f: X\to \mathbb P^1$ and it follows from the construction that the fibers of $f$ are cubic plane curves, so this is indeed an elliptic fibration.

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Remark 1: One can get more data out of this construction.

1. Since each fiber is a plane cubic, this construction limits the possible singularities of the fibers quite a bit.
2. The line we started with is a triple section.
3. If $X$ contains another line, say $C$, such that $C\cap L=\emptyset$ then $C$ gives a section of $f$: Indeed if $C$ is a line, then $C\cdot H=1$ and if $C\cap L=\emptyset$, then $C\cdot E=1$.
4. If $X$ contains another line, say $C$, such that $C\cap L\neq\emptyset$, then $C$ is contained in a fiber intersecting $L$ exactly once.
5. You can keep playing with this to get more.

Remark 2: For an elliptic K3 the Picard number is at least $2$ and if it is exactly $2$, then it admits at most two different elliptic fibrations. If the Picard number is at least $3$, then an elliptic K3 may admit infinitely many elliptic fibrations, so obviously in many cases you don't get all the elliptic fibrations this way.

Remark 3: Finally, note that the Fermat quartic contains lines: Using diverietti's notation the 16 lines $$\ell_{\xi,\xi'}=\big\{[a:\xi a:b:\xi' b] \ \vert\ [a:b]\in \mathbb P^1\big\}$$ where $\xi,\xi'$ are 4$^\text{th}$ roots of unity all lie on that Fermat quartic, so one obtains 16 different elliptic fibrations this way.

Further note that if $\xi\neq \xi''$ and $\xi'\neq \xi'''$, then $\ell_{\xi,\xi'}\cap \ell_{\xi'',\xi'''}=\emptyset$, so this way we get a couple of sections for each of the elliptic fibrations as described above.