# Discriminant locus of elliptic K3 surfaces

Given a complex elliptic K3 surface $$\pi\colon X\rightarrow \mathbb P^1$$, its discriminant locus is the divisor $$D = \sum_{i = 1}^s n_i P_i$$ on $$\mathbb P^1$$ such that $$n_i$$ is equal to the Euler-Poincaré characteristic of the fiber $$\pi^{-1}(P_i)$$, where the sum runs over the points $$P_i \in \mathbb P^1$$ such that $$\pi^{-1}(P_i)$$ is singular. It is well known that $$\deg D = 24$$.

Conversely, given an effective divisor $$D$$ of degree $$24$$ on $$\mathbb P^1$$, when is it the discriminant locus of a complex elliptic K3 surface?

I am particularly curious about the minimal possible $$s$$. The maximal Euler-Poincaré characteristic of a singular fiber is $$20$$, so $$s \geq 2$$. But in case, say, $$n_1 = 20$$, then the fibration is of type $$I_{14}^*,I_1,I_1,I_1,I_1$$ (see Schütt-Schweizer), so indeed $$s = 5$$. Are smaller $$s$$ possible?

• Since the moduli space of polarized K3 surfaces is hyperbolic, the inequality $s\geq 3$ holds. An example of an elliptic fibration with $s=3$ is provided by the Legendre elliptic curve $y^2= x(x-1)(x-\lambda)$ over $\mathbb{C}-\{0,1\}$. The total space of this fibration is not K3 though. Oct 8, 2018 at 12:51
• Sorry, I actually miswrote something in my comment. I meant to say that the moduli "space" of elliptic curves is hyperbolic. (Although the moduli space of polarized K3 surfaces is hyperbolic as well, it is the moduli space of elliptic curves which plays a role here.) Indeed, if $D$ is the support of the discriminant divisor of the elliptic surface $f:X\to \mathbb{P}^1$, then there is a non-constant morphism $\mathbb{P}^1 \setminus D \to \mathcal{M}$ induced by the Jacobian of $X\setminus f^{-1}D\to \mathbb{P}^1\setminus D$, where $\mathcal{M}$ is the moduli of elliptic curves. Oct 8, 2018 at 14:46

The minimal $$s$$ is $$3$$.
It is attained by several elliptic K3's, including $$y^2 = x^3 + (t^2-t)^4$$ which has IV* fibers at $$t = 0, 1, \infty$$ and no other singular fibers. The comment by Ariyan Javanpeykar gives one argument that $$s$$ can be no smaller. (See postscript. This uses characteristic zero; in small positive characteristic $$s$$ can be as small as $$1$$, e.g. in characteristic 2 the elliptic K3 surface $$y^2 + y = x^3 + t^9$$ has only one reducible fiber, at $$t = \infty$$.)
P.S. There are other ways to prove $$s>2$$; for example it follows from Szpiro's inequality, which has an elementary proof via the Mason-Stothers theorem (polynomial ABC). See MO 190530, Are there nonisotrivial elliptic curves over $$\mathbb{G}_m$$?, for this and some related ideas. (That question is related because $${\bf G}_m({\bf C}) = {\bf CP}^1 - \{0, \infty\}$$ and if an elliptic surface $$\pi: X \to {\bf P}^1$$ has $$s \leq 2$$ then one can choose the coordinate on $${\bf P}^1$$ so that each bad fiber maps to $$0$$ or $$\infty$$.)
• Another "easy" way of showing that $s>2$ is to use that the modular curve $Y(n)$ is of genus at least two when $n>>0$. Indeed, suppose that $s\leq 2$ for $X\to \mathbb{P}^1$. Now, choose $n >0$ such that $X(n)$ has genus at least two. Note that adding level $n$ structure induces a new family $X'\to \mathbb{P}^1$ (because $s<3$), and thus a morphism $\mathbb{P}^1\to X(n)$. Oct 9, 2018 at 21:48
• Yes, that kind of argument was given in both Daniel Litt's answer to MO 190530 and in mine. (It's related to your "hyperbolic" idea: ${\bf G}_m$ is not hyperbolic, so neither are its unramified covers.) Oct 9, 2018 at 21:53
• @NoamD.Elkies Hi Noam, how do you verify that the singular fibers of $y^2 = x^3 + (t^2-t)^4$ are IV* at $t=0,1, \infty$ and that this fibration has no other singular fibers? Aug 18, 2020 at 2:44
• @AmorFati In general the singular fibers are the values of $t$ where the discriminant $\Delta$ vanishes. Here $\Delta$ is a multiple of $(t^2-t)^8$ [really $(t_1 t_0 (t_1-t_0))^8$ where $(t_1:t_0)$ are projective coordinates, so the $t_0=0$ point "$\infty$" is also a zero of $\Delta$; clearly the only others are $t_1=0$ and $t_1=t_0$, i.e. $t=0$ and $t=1$]. The reduction type can be determined by Tate's algorithm. Since we are not in characteristic 2 or 3 here, we can recognize Type IV* from the valuations $v(c_4) \geq 3$, $v(c_6) = 4$. Aug 18, 2020 at 22:52