TITLE: construct the elliptic fibration of elliptic k3 surface
QUESTION [20 upvotes]: 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!

REPLY [7 votes]: 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. 

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

Since each fiber is a plane cubic, this construction limits the possible singularities of the fibers quite a bit. 
The line we started with is a triple section. 
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 $C\cdot E=1$.
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. 
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.