TITLE: A K3 over $P^1$  with six singular $A_1$- fibers?
QUESTION [11 upvotes]: Hirzebruch, in the paper 'Arrangements of Lines and Algebraic Surfaces'
constructs a special  $K3$ surface out of a 'complete quadrilateral' in 
 $CP^2$.  A complete quadritlateral consists of 
 4 points in general position and the  $6$ lines joining them.
Over each line Hirzebruch  forms the local 2:1 fold cover to get
a   new surface which comes  as  a branched
covering of  $CP^2$,  branched over the $6$ lines.
Away from the lines the covering has degree  $2^{6-1}$.
This surface has singularities of conical type at the original $4$ points
(At these points 3 lines are coincident. )
Blow up the singularities coming from these 4 points..  The resulting smooth surface  is Hirzebruch's $K3$.
Viewed from a different perspective, I believe  that I can get this same
$K3$ has an elliptic surface over $CP^1$ with $6$ singular fibers.
I also believe that each of the  singular  fibers are of $A_1$ type, meaning  two $CP^1$'s intersecting transversally
(as in $xy = 0$), but am less sure of this. 
The corresponding singular points on $CP^1$ can be taken to be   the vertices of the octahedron.
And I believe that the  manifest symmetry group of order $4! = 24$ seen in  Hirzebruch's construction (permute the 
original 4 points) agrees with the symmetry group of the  octahedron. 
Questions.  Do you know this second K3?  If so, could you give me a reference for it?
Have you seen a place which shows that the second $K3$  is the same  as Hirzebruch's?
More generally, what are the first few 'simplest'  elliptic $K3$'s? 
By   'elliptic' I mean   expressed as elliptic surface $f: X \to  CP^1$, over
$CP^1$. By  'simplest' I mean a small number of singular fibers whose singularities
are as  'simple' as possible. For example, if all  singular fibers are of $A_1$-type,
what is the fewest number of fibers? Must this number be $6$? 
(I have  looked in Barth-Hulek-Peters-van de Ven's 'Compact Complex Surfaces',
 esp.  ch. V, sec. 2 and   suppose this information is buried there somehow
or other, but  is rather beyond me  to untangle it from there. Neither did I
find this 2nd $K3$ in Gompf and Stipsicz's book )

REPLY [3 votes]: As explained by Noam Elkies, the fibers you want are of Kodaira type:

$I_2$ : two rational curves intersecting at two distinct points 
$III$ : two rational curves meeting at a double point 

Since these are the only fibers you want, I will recommand not using a Weierstrass equation but rather a Jacobi quartic form for  your elliptic fibration. 
Indeed, a smooth Weierstrass model admits only singular fibers of type $I_1$  (nodal curves) and type $II$ (cuspidial curve). So if you want to have singular fibers of type $I_2$ and $III$, you will have to introduce singularities in the total space of your fibration and resolve them following  Tate's algorithm. 
You can avoid dealing with the resolution of singularities if you start with the Jacobi quartic form. Consider a weighted projective plane $\mathbb{P}^2_{1,2,1}$ bundle, a smooth quartic equation will define an elliptic curve as you can see by using the adjunction formula. The canonical form of such a curve is 
$$
\text{Jacobi form}: \quad y^2=x^4+ e x^2 z^2+ f x z^3+ g z^4
$$
where $[x:y:z]$ are the projective coordinates of weight $1$, $2$ and $1$ respectively. 
For a fibration $Y\rightarrow B$, the weighted projective plane is replaced by a  weighted  projective bundle $\mathbb{P}_{1,2,1}[\mathscr{O}\oplus\mathscr{L}\oplus\mathscr{L}^2]$ with weight  $1,2,1$,  where $\mathscr{L}\  $    is a line bundle over the base $B$ of your fibration. In that equation $e,f,g$ are now  sections of $\mathscr{L}^2$, $\mathscr{L}^3$ and $\mathscr{L}^4$ respectively. The projective coordinates  $x$, $z$, $y$ are respectively sections  of $\mathscr{L}\otimes \mathscr{O}(1)$, $\mathscr{O}(1)$ and $\mathscr{L}^2\otimes\mathscr{O}(2)$, where $\mathscr{O}(1)$ is the tautological line bundle of the weighted projective bundle. 

Using the adjuction, you can see that you have a $K3$ elliptic fibration iff  $c_1(\mathscr{L})=c_1(B)$.

In case you actually want a Weierstrass model, the Jacobian fibration will give you a birationally equivalent Weierstrass model: $y^2 =x^3+ Fx + G$ with $F=-4 g +e^2/3$ and $G=f^2-8/3 eg +2e^3/27$. You can compute the birational transformation using Maple.
Going back to the Jacobi form, it is not difficult to see that a smooth elliptic fibration will have fibers of type $I_1$, $II$, $I_2$ and $III$.
In particular, the fibers are:

type $I_1$ at a general point of $4  
   F^3+27 G^2=0$   
type $II$ at $F=G=0$
and $e\neq 0$ 
type  $I_2$ iff $f= 
   e^2-4 g=0$ and $g\neq 0$ 
type  $III$    iff $f=e=g=0$.
With a $\mathbb{P}^1$ base and
$\mathscr{L}=O(2H)$, you have a $K3$
surface with fibers $I_1,II, I_2,
 III$.  

You can avoid the fiber $I_1$ and $II$ by manipulating the equation. For example if you take the equation with a $\mathbb{P}^1$  base  and $\mathscr{L}=O(2H)$, you can write the following K3 elliptic surface

$$y^2=x^4+ g z^4\Longrightarrow \   \text{8 fibers of type}\ III $$

You need  $g$ to be a smooth polynomial of degree 8 in $P^1$.  For example $g=x_1^8+x_0^8$ with $[x_0:x_1]$ the projective coordinates of $\mathbb{P}^1$. You will have $8$ fibers of type $III$ at the zero of $g$.Note that with that choice, the singular fibers are at the vertices of an octagon on $\mathbb{P}^1$ defined by $x_1^8+x_0^8=0$.
These types of fibrations based on the Jacobi quartic form are known as $E_7$ elliptic fibration in the string theory literature. They have some nice topological properties and transitions. You can read more about them   in this paper.