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 )


3 Answers 3


[EDITED to give likely identification of the OP's surface at the end]

An elliptic K3 surface has discriminant $\Delta$ of degree $24$, and $\Delta$ has valuation 2 or 3 at an $A_1$ fiber depending on whether the Kodaira type is I$_2$ or III, so if that's the only kind of singular fiber there must be at least $24/3=8$ such fibers (and that is possible, e.g. $y^2 = x^3 + ax$ over the $t$-line, with $a=a(t)$ a polynomial of degree $8$ with distinct roots).

Your description of an $A_1$ fiber is not quite right: it's two lines meeting at two points, counted with multiplicity, so either transversely at two different points (type I$_2$) or at the same point with multiplicity 2 (type III). The intersection graph is the extended Dynkin diagram $\tilde A_1$. Likewise for the other A-D-E fiber types.

For the K3 surface you're studying: in general a K3 surface can have many (albeit finitely many) non-isomorphic elliptic fibrations, so without further information I can't compute the configuration of singular fibers and other invariants of your fibration. How did you obtain it? Do you have an explicit formula?

[added the next day] On further thought, there's a natural elliptic fibration on this surface that's probably what the OP had in mind, with octahedral symmetry respecting the $S_4$ symmetry of Hirzebruch's construction, and it is indeed a well-known fibration: the universal elliptic curve over the modular curve ${\rm X}(4)$, i.e. the universal curve with full level-4 structure. This fibration in fact respects all $2^5 4!$ symmetries of the Hirzebruch construction ($4!$ permuting the four vertices, and $2^5$ deck transformations): the octahedral symmetries give ${\rm PSL}_2({\bf Z}/4{\bf Z}) \cong S_4$, and translation by the $4^2$ torsion points together with inversion in the group law gives the factor of $2^5$. The six reducible fibers, coming from the six cusps of ${\rm X}(4)$, are of type I$_4$, so contribute $A_3$ each to the Néron-Severi lattice. I don't know whether this identification is in the literature already; if anybody would know whether it's been published already, Dolgachev would.

To get this fibration, start with the pencil of conics through the four special points $P_1,\ldots,P_4$. A generic such conic is a rational curve on which Hirzebruch's $2^5:1$ cover restricts to a $2^3:1$ cover branched at four points, giving an elliptic curve isomorphic with the double cover branched at the same points. That's not quite a fibration of our K3, because the missing factor of $2^2$ comes from extracting square roots of functions that are constant but not necessarily square on the generic conic; we need a $2^2:1$ cover of the base curve of the pencil. This base change is branched at three points, corresponding to the three reducible conics in the pencil (unions of lines {$P_1 P_2$ and $P_3 P_4$}, {$P_1 P_3$ and $P_4 P_2$}, or {$P_1 P_4$ and $P_2 P_3$}). Thus the actual base curve has six cusps, with $S_4$ symmetry coming from the $S_3$ permutations of the reducible conics together with the $2^2:1$ cover.

To do this explicitly it is convenient to put the $P_m$ at the four points with projective coordinates $(\pm 1 : \pm 1 : \pm 1)$. The action of $S_4$ is then by signed permutation matrices. The six lines $P_m P_n$ are $x = \pm y$, $y = \pm z$, and $z = \pm x$. The pencil of conics is $Ax^2 + By^2 + Cz^2 = 0$ with $(A:B:C)$ on the line $A+B+C=0$. The $2^2:1$ base change is $(A:B:C) = (a^2:b^2:c^2)$.

  • $\begingroup$ ...and the base is then the conic $a^2+b^2+c^2=0$, with the octahedral symmetry given by the signed permutation group, and the six vertices of the octahedron at $abc=0$. $\endgroup$ Sep 25, 2011 at 19:52
  • $\begingroup$ Yes! That is our base! We, Rick Moeckel and I, arrive at it by starting with the the planar 4 body problem of celestial mechanics. Quotient by translations and rotations. Levi-Civita'' regularize the binary collisions. Result:the phase space is the cotangent bundle of the cone of Hirzebruch's surface BEFORE regularizing the 4 points. Do the same for 3 bodies. Result: the cotangent bundle of the cone over Noam's $a^2 + b^2 + c^2 =0$ conic. Fibration: forget one of the bodies''. $\endgroup$ Sep 26, 2011 at 3:56

In Ch. III, section 11 of Barth-Peters-Van de Ven (I only have the old edition), Proposition 11.4, there is a formula that relates the topological Euler characteristic of a fibered surface with the characteristic of the fiberse and of the base. In the case of a K3 with an elliptic fibration it gives: $24=\sum_{s\in T} e(F_s)$, where $T\subset {\mathbb P}^1$ is the set of critical values of the fibration, and $e(F_s)$ is the Euler characteristic of the fiber $F_s$ over $s\in T$. So if all the singular fibers are of type $I_2$, you have 12 singular fibers.

The Hirzebruch K3 (call it $S$) that you describe has at least 5 elliptic fibrations. If $P_1,P_2,P_3,P_4$ are the 4 vertices of the quadrilateral, these pencils are obtained by pulling back the pencil of lines through any of the $P_i$ and the pencil of conics through the points $P_1,\dots P_4$. I suspect that they can all be transformed one into the other: besides the $S_4$ symmetry that you point out, one can perfom a standard Cremona transformation centered, say, at $P_1,P_2,P_3$ and fixing $P_4$. This exchanges the pencil of lines through $P_4$ with the pencil conics through the 4 points.

Consider the pencil $|F_1|$ of lines through $P_1$: one can check that the preimage in $S$ of a general element of $|F_1|$ is the disjoint union of $4$ smooth elliptic curves. The singular fibers correspond to the lines $P_1P_i$, $i=2,3,4$. It seems to me that the preimage of each of these is the disjoint union of 4 double fibers of type $2I_2$.

EDIT: in fact I was wrong, the preimage of $P_1P_i$ is the disjoint union of two fibers of type $I_4$.

As I remarked above, the configuration of singular fibers should be the same for all the 5 fibrations by symmetry.

  • $\begingroup$ If $f: X \to C$ is an elliptic fibration over a curve $C$, then this formula: $\chi(X) = \sum_{s\in T} e(F_s)$, from Barth et al that Rita referenced has been wonderfully helpful. If the symmetry group of $X$ is fiber-preserving and permutes the singular fibers, as in my case, that formula becomes $24 = n e(F)$ where $n$ is the number of these singular fibers. For me $n = 6$ so $e(F) = 4$. This $4$' is the same 4' of the 4 spheres ($P^1$'s) making Noam Elkie's $I_4$. $\endgroup$ Oct 1, 2011 at 0:48
  • $\begingroup$ In fact, I had made a mistake: the preimage of the line $P_1P_i$ is the disjoint union of 2 $I_4$ fibers. Thanks for pointing this out. I'm going to edit my answer. $\endgroup$
    – rita
    Oct 1, 2011 at 6:49

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.

  • $\begingroup$ Wait, $x_0^8 + x_1^8 = 0$ gives a regular octagon, not an octahedron (which has six vertices, not eight). If you want 8 points with octahedral symmetry, you're looking for vertices of a cube, e.g. the roots of $x_0 x_1 (x_0^3-x_1^3) (x_0^3 + 8 x_0^3)$. $\endgroup$ Sep 25, 2011 at 17:35
  • $\begingroup$ You're welcome. Of course I in turned committed a typo: the last factor should be $x_0^3 + 8 x_1^3$, not $x_0^3 + 8 x_0^3$... $\endgroup$ Sep 25, 2011 at 19:40

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