For me a K3 surface will be a smooth complex projective variety of dimension 2 that is simply-connected and has trivial canonical bundle. Given a K3 surface $X$, an elliptic fibration $\pi \colon X \to \mathbb{C}P^1$ is a proper morphism with connected fibers such that all but finitely many fibers are smooth curves of genus 1. I've learned a little about these from here:

Generically a K3 surface admits no elliptic fibration, but among those that do, generically the fiber of $\pi$ is a smooth curve of genus 1 at all but 24 points, where the fiber is a rational curve with a single double point.

Huybrecht also catalogues the less generic cases where $\pi$ has fewer singular (i.e. non-smooth) fibers, but with correspondingly worse singularities. On MathOverflow there's a nice easy example: the Fermat quartic surface admits an elliptic fibration with 6 singular fibers, each of which has 4 double points.

But I'd like to see concrete easy examples of elliptic fibrations with 24 singular fibers.

By 'concrete easy examples' I mean that ideally I would like there to be simple explicit formulas for the K3 surface $X$, the hyper-Kähler structure on $X$, the elliptic fibration $\pi$, the 24 points on $\mathbb{C}\mathrm{P}^1$ with singular fibers, the double points on these fibers, and also the points of $X$ where $d\pi$ is not injective. But of course I'll settle for whatever I can get!

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    $\begingroup$ If you take homogeneous coordinates a,b,c on P^2 and x,y on P^1 then a surface cut out by a polynomial which is of bidegree 3,2 will give you a K3 (e.g. a sum of monomials like abcxy). The projection to the P^1 factor will have fibres that are cubics in P^2. You could try playing with these to get different combinations of singular fibres. $\endgroup$ Feb 2, 2022 at 22:57

3 Answers 3


Weierstrass equations are probably a good choice. You can try $$y^2 = x^3 - 3x +2 t^{12},$$ for example.

Here the singular fibers are when $t$ is a $24$th root of unity $\zeta$, and the double point is provided by $x = \zeta^{12}$ and $y=0$.

Aren't the points where $d\pi$ is not injective the same as the singular points?

To really describe this as a smooth surface over $\mathbb P^1$, first note that we can get a surface in $\mathbb P^2 \times \mathbb A^1$ by homogenizing, i.e. taking $(x:y:z)$ to be the coordinates of $\mathbb P^2$ and $t$ the coordinate of $\mathbb A^1$, look at the vanishing locus of the equation $$y^2 z= x^3 - 3xz^2 +2 t^{12}z^3$$ and then another chart is provided by changing variables to $\tilde{x} = x/ t^4$, $\tilde{y} = y/t^6$, $\tilde{z} = z$, $\tilde{t} =1/t$, where after the change of variables we are looking at the vanishing locus of the equation $$\tilde{y}^2 \tilde{z} = \tilde{x}^3 - 3 \tilde{t}^8 \tilde{x} \tilde{z}^2 +2 \tilde{z}^3$$ in $\mathbb P^2 \times \mathbb A^1$ with coordinates $(\tilde{x}:\tilde{y}:\tilde{z}),\tilde{t}$.

For me a hyper-Kähler structure is just a nowhere vanishing holomorphic 2-form. Such a 2-form is provided here by $$\frac{dx}{y} \wedge dt = \frac{d (t^4 \tilde{x} )}{t^6\tilde{y} } \wedge dt = \frac{t^4}{t^6} \frac{d\tilde{x}}{\tilde{y}} \wedge dt = - \frac{d\tilde{x}}{\tilde{y}}\wedge d \tilde{t}.$$

Let me explain how to make Jason Starr's construction explicit, modulo constructing a splitting. I'll construct two linearly independent complex-valued (but not holomorphic) vector fields, which give four linearly independent real vector fields.

Let $v_1$ be the (meromorphic) vector field on the elliptic surface, which is contained in the elliptic curve fibers, and which is inverse to the 1-form $dx/y$. In other words, if $u$ is any coordinate on the elliptic curve fiber, $v_1$ is given by $ y (dx/du)^{-1} \partial/\partial u$. A classical argument shows $dx/y$ is well-defined and non-degenerate everywhere, even on the points where $y$ or $dx/du$ vanishes or has a pole (since the zeroes and poles cancel).

Let $v_2$ be any smooth lift of the vector field $\partial/\partial t$ on $\mathbb A^1$, i.e. a vector field whose projection onto the $t$ line has size $1$ at each point. Such a $v_2$ exists by a partition of unity argument, but I don't know a formula for it. (Away from the singular fibers, a lift can be constructed using the natural flat torus structure on the fibers, flowing one into the next in the unique locally-affine zero-preserving way, but this vector field has poles at the singular fibers.)

Then $v_1$ has a pole of order $2$ at $t = \infty$, and no other poles or zeroes away from the $24$ singular points, while $v_2$ has a zero of order $2$ at $\infty$. Thus $v_2 / t^2 = v_2 \overline{t}^2/ |t|^4$ is well-defined and nonvanishing in a neighborhood of $t=\infty$, so $v_2 \overline{t}^2 / (1+ |t|^2)^2$ is well-defined and nonvanishing in a neighborhood of $t= \infty$.

Then $$ \frac{v_1}{ (1 + |t|^2)^2} + t^2 v_2 $$ and $$ \frac{v_1 \overline{t}^2 }{ (1 + |t|^2)^2} + i v_2$$

are well-defined complex-valued vector fields, and linearly independent at each nonsingular point:

Since $v_1$ and $v_2$ are well-defined and linearly independent away from $\infty$, these two linear combinations are linearly independent away from the vanishing locus of the determinant $\frac{i}{ (1+|t|^2)^2} + \frac{ t^2 \overline{t}^2}{(1+|t|^2)^2} = \frac{ |t|^4 +i}{ (1+ |t|^2)^2} $, which is never vanishing.)

At $t=\infty$, $ \frac{v_1 \overline{t}^2 }{ (1 + |t|^2)^2}$ and $t^2 v_2 $ are well-defined and linearly-independent, and $ \frac{v_1}{ (1 + |t|^2)^2} +$ and $i v_2$ vanish, so the two vector fields are well-defined at $t=\infty$ as well.

Then the real and imaginary parts of the two vector fields above are linearly independent and thus give a framing of the K3 surface minus the $24$ singular points. However, they can't be made explicit without an explicit choice of lift $v_2$.

  • $\begingroup$ @DavidRoberts Can you elaborate? I'm not sure exactly what you mean. $\endgroup$
    – Will Sawin
    Feb 3, 2022 at 13:18
  • $\begingroup$ I don't think of this as a lot of effort. I posted an n-Category Cafe article about a problem I was interested in and Jason Starr kindly explained that the key would be finding an elliptically fibered K3 with 24 singular points. I asked here for a concrete example and four hours later Will gave me one! However I still need to get an explicit action of the quaternions on the tangent bundle; I don't think a holomorphic symplectic structure instantly gives me that. $\endgroup$
    – John Baez
    Feb 3, 2022 at 18:41
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    $\begingroup$ @DavidRoberts No. The tangent bundle, restricted to the fiber, is an extension of two trivial line bundles. The extension class measures the rate of change of the elliptic curve in moduli space, so it is zero at a generic point. Thus the only holomorphic section is contained in the trivial subbundle, consisting of vector fields contained in the fiber. But now, no longer restricting to a fiber, this sub-bundle is the pullback of $\mathcal O(-2)$ on $\mathbb P^1$, which has nonvanishing Chern class and thus has no nowhere vanishing global sections, even non-analytic ones. $\endgroup$
    – Will Sawin
    Feb 4, 2022 at 3:22
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    $\begingroup$ @DavidRoberts Over the smooth fibers, there is a natural splitting, which uses the flat structure on the elliptic curve and the fact that there's a unique way to flow each fiber into the next by an identity-preserving affine transformation. However, the vector field provided by this splitting develops singularities on the whole singular fiber, not just on the singular point. $\endgroup$
    – Will Sawin
    Feb 4, 2022 at 3:45
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    $\begingroup$ @JohnBaez Edited to give an almost, but not quite, explicit formula for a framing, following Jason Starr's suggestion. $\endgroup$
    – Will Sawin
    Feb 4, 2022 at 19:59

As far as I know no one has “written down” a hyperkähler structure on a K3 surface. Indeed much of the work in Mirror symmetry revolves around trying to give an asymptotic expansion of such metrics (in this and other cases) in terms of curve counts and the like on the dual. Yau’s solution of the Calabi conjecture implies that every K3 surface admits a unique such structure. The holomorphic 2-form is then parallel. This together with the Kähler form (also parallel) implies that the bundle of self dual forms is trivial. For any self dual form $\omega$ on an oriented Riemannian four-manifold we have a map $$ J_\omega:T^*M \to T^*M $$ given by $$ J_\omega(\theta)=\star(\omega \wedge \theta) $$ where $\star$ is the Hodge star. One checks that $$ J_\omega^2=-2|\omega|^2. $$ The sphere of self-dual 2-forms of length $1/\sqrt{2}$ acts as almost complex structures. For our Yau metric this 2-sphere bundle admits a parallel trivializations and this gives the action of the imaginary quaternions.

  • $\begingroup$ Thanks! I didn't know that hyperkähler structure on K3 surface were so inexplicit. But actually for my purposes an explicit almost quaternionic structure would suffice. How about that? $\endgroup$
    – John Baez
    Feb 4, 2022 at 17:29
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    $\begingroup$ If you can write down the Kähler form and the non vanishing holomorphic two form the normalizing the two form gives you an orthogonal trivialization of $\Lambda^+$ which as above gives you the action of the imaginary quaternions on the cotangent bundle. $\endgroup$
    – Tom Mrowka
    Feb 4, 2022 at 17:37

One can get examples by taking a double cover of the Segre-Hirzebruch surface $\mathbb F_4$. Write $C^-$ for the negative section and $C^+$ for a zero section and $F$ for a fiber of the projection $\mathbb F_4\to \mathbb P^1$, so that $C^+$ is linearly equivalent to $C^-+4F$. Set $L=2C^-+6F$ and choose $D$ in the linear system $|3C^+|$. Then there is a double cover $f\colon X \to \mathbb F_4$ branched on $C^-+D$ and such that $f_*\mathcal O_X=\mathcal O_{\mathbb F_4}\oplus L^{-1}$; if $D$ is smooth, the surface $X$ is also smooth. Standard formulae for double covers give $h^1(\mathcal O_Y)=0$ and $K_Y=0$, so $Y$ is a K3.

The map $X\to \mathbb F_4\to \mathbb P^1$ is an elliptic fibration. The singular points of the elliptic fibers lie above the points where $D$ is tangent to the ruling of $\mathbb F_4$. If the tangency is simple the singularity of the elliptic fiber is a node. A dimension count shows that for a general choice of $D$ all the tangency points of $D$ to a ruling are simple. I think that using computer algebra it should not be hard to find explicit examples of $D$ where this condition holds.


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