# Do singular fibers determine the elliptic K3 surface, generically?

General elliptic K3 surfaces. Consider K3 surfaces of Picard rank two with Neron-Severi lattice isomorphic to $$\left[\begin{array}{cc} 2d & t \\ t & 0 \end{array}\right]$$ for some positive integers $$d, t$$. Such K3 surfaces are necessarily elliptic (because they have class of square zero), and conversely any projective elliptic K3 surface of Picard rank two has such a Neron-Severi lattice. Here $$t$$ is the minimal degree of a multisection (which is uniquely defined) and $$2d$$ is the degree of polarization (which is only well-defined mod. $$t$$). Elliptic surfaces with a section correspond to the case $$t = 1$$: for any $$d$$ such lattice is isomorphic to the hyperbolic plane $$U$$.

The moduli space. By the Torelli theorem for K3 surfaces, for fixed $$d, t > 0$$ such K3 surfaces exist and they form an $$18$$-dimensional moduli space $$\mathcal{M}_{d,t}$$. (These elliptic surfaces have either one or two elliptic fibrations depending on $$d, t$$.)

Singular fibers. Assume that singular fibers of the elliptic fibrations have a single node (type $$I_1$$) so that by the Euler characteristic count there are $$e(X) = 24$$ singular fibers. These singular fibers are parametrized by a $$21$$-dimensional variety $$W_{24}$$ as we consider $$24$$ distinct unordered points on $$\mathbb{P}^1$$ with three of them can be fixed to be $$0, 1, \infty$$.

Question: are there finitely many elliptic K3 from $$\mathcal{M}_{d,t}$$ with fixed positions of singular fibers? If not, what can be said about the dimension of the moduli space of those elliptic K3 surfaces with a prescribed set of $$24$$ singular fibers? Equivalently I am asking about the fibers of the rational map $$\mathcal{M}_{d,t} \to W_{24}$$ which takes elliptic fibration to its set of branch points.

Possibly if one deals with the $$\mathcal{M}_{0,1}$$ (elliptic surfaces with a section) the general case will follow by considering the Jacobian fibration (as suggested by user25309 in the comments); this will need the argument that fibers of taking the Jacobian fibration map $$\mathcal{M}_{d,t} \to \mathcal{M}_{0,1}$$ are also finite.

Any suggestions or references welcome.

• What is the discriminant of your Picard lattice? At least for K3 surfaces that are obtained as deformations of double covers of elliptic rational surfaces, this follows from the corresponding statement for the elliptic rational surfaces (by deforming away from the locus of double covers): the orbit of a line in the $\mathbb{P}^9$ of plane cubics is determined up to finite indeterminacy by its intersection with the discriminant (degree $12$) hypersurface. Jun 5, 2021 at 12:47
• Here is a reference for the result about rational elliptic surfaces: arxiv.org/abs/math/9910077 Jun 5, 2021 at 12:48
• A maybe obvious remark: given an elliptic K3 surface with 24 nodal singular fibers and without section, the corresponding Jacobian surface is a different elliptic K3 surface with section and with 24 nodal singularities at the same 24 points. Jun 5, 2021 at 13:35
• Finiteness holds for (non-isotrivial) elliptic surfaces with a section (and therefore for all $\mathcal{M}_{d,t}$) much more generally: given any finite set $S$ of points on any smooth projective curve $C$ there are only finitely many non-isotrivial families of elliptic curves on $C - S$. I am not sure who first proved this (maybe Shafarevich) but for more general results see the paper of Faltings "Arakelov's theorem for abelian varieties" and Deligne "Un theoreme de finitude pout la monodromie"
– naf
Jun 7, 2021 at 7:42
• I haven't been able to find the original reference (for elliptic curves) but the proof is actually quite easy: after replacing $C$ by a finite cover of bounded degree, unramified outside $S$, we may assume that the family of elliptic cuvrves has level $n$ structure for some large integer $n$. This gives a map to the modular curve of level $n$ (which determines the family) and by the de Franchis theorem there are only finitely many such maps (since $n$ is big).
– naf
Jun 8, 2021 at 4:10

I am expanding naf's comments to make a self-contained community wiki answer. By an elliptic fibration we mean a smooth projective relatively minimal surface $$f: X \to C$$ with general fiber given by an elliptic curve.

Main Proposition. If $$C$$ is a smooth projective curve and $$S \subset C$$ a finite subset, then the number of nonisotrivial elliptic surfaces with a section $$f: X \to C$$ and having singular fibers only over points of $$S$$ is finite.

The proof consists in making a base change of finite degree, ramified only at $$S$$ such that the new elliptic surface is rigidified by $$n$$-level structure and using that moduli spaces of elliptic curves with $$n$$-level structure is representable, together with some standard finiteness results. We go through the arguments in detail. To deal with elliptic fibrations without a section, one needs to use the Jacobian fibration trick.

Warm up. Let us first give a simple proof of a related but easier statement: if $$C$$ is a smooth projective curve and $$f: X \to C$$ is an elliptic surface with no singular fibers, then $$f$$ is isotrivial (all fibers isomorphic).

Proof: the $$j$$ invariant of each fiber gives an algebraic map $$C \to \mathbb{A}^1$$ (the $$j$$-line). Since $$C$$ is projective, such a map is constant, and so all fibers have the same $$j$$-invariant, hence are isomorphic.

Representing moduli spaces of elliptic curves. The $$j$$-line $$\mathbb{A}^1$$ is the coarse moduli space of the stack $$\mathcal{M}_{1,1}$$ of elliptic curves. The standard way to get a representing variety, not a stack, is to get rid of automorphisms by parametrizing pairs $$(E, \phi)$$ where $$E$$ is an elliptic curve and $$\phi: E[n] \simeq (\mathbb{Z}/n)^2$$; in other words we choose a basis of $$n$$-torsion points on $$E$$. The corresponding stack is the modular curve $$X(n)$$. It is a usual smooth curve for $$n \ge 4$$ and has positive genus for $$n$$ large enough. As we have defined it $$X(n)$$ is not projective; one can compactify and then points at infinity parametrize certain singular curves with level structure. There are other variants of modular curves such as $$X_1(n)$$ and $$X_0(n)$$; any of them will work.

Level structure and monodromy. Of course, our elliptic surface $$f: X \to C$$ admits level structure at each smooth fiber, but not necessarily in a family. This is controlled by the monodromy representation: if $$U = C \setminus S$$ is the locus of smooth fibers, and $$0 \in U$$ is any point, $$E = f^{-1}(0)$$, then $$\pi_1(U, 0)$$ acts on $$E[n] \simeq (\mathbb{Z}/n)^2$$; we get a homomorphism $$\gamma: \pi_1(U) \to \mathrm{GL}_2(\mathbb{Z}/n).$$ By definition, a family over $$U$$ admits a level $$n$$ structure if $$\gamma$$ is trivial, that is you can canonically identify $$n$$-torsion points at all fibers.

Lemma 1. Fix $$n > 0$$. For any elliptic surface $$f: X \to C$$ with singular fibers only at $$S \subset C$$ we can make a base change $$C' \to C$$ of bounded degree and ramified only at $$S$$ such that the pullback of $$f$$ to $$C'$$ admits $$n$$-level structure on the smooth part.

Proof. Let $$G = \gamma^{-1}(e)$$. This is a subgroup in $$\pi_1(U)$$ of index bounded by $$|\mathrm{GL}_2(\mathbb{Z}/n)|$$ and by the covering theory there exists $$U' \to U$$ of degree equal to $$[\pi_1(U):G]$$ such that $$\pi_1(U) = G$$ so that the new mondoromy representation is $$\gamma|_G$$ which is trivial. Hence the pullback of $$f$$ to $$U$$ admits the $$n$$-level structure. We may extend our unramified cover to a finite map $$C' \to C$$ between smooth projective curves ramified only at $$S$$.

In the proof of the Proposition we will use finiteness of such covers:

Lemma 2. Given $$S \subset C$$ there are only finitely many covers $$C' \to C$$ of bounded degree ramified only at $$S$$.

This Lemma is a starting point for various beautiful theories such as Hurwitz numbers counting covers of $$\mathbb{P}^1$$ and Grothendieck's dessin's d'enfant dealing with covers of $$\mathbb{P}^1$$ branched at $$3$$ points. The proof is standard and also uses representations of the fundamental group of $$C \setminus S$$.

Proof of the Proposition

Take $$n$$ large enough so that $$X(n)$$ is a curve of genus $$g \ge 2$$. By Lemma 1 and Lemma 2 we may make a base change and assume that $$f$$ admits an $$n$$-level structure over $$U$$. In this case due to representability of the modular curves above to give such an elliptic fibration is the same as to give a map $$U \to X(n)$$. Such maps are in bijection with maps $$C \to \overline{X}(n)$$ (the smooth projective model). Finally result follows from:

Lemma 3. Given smooth projective curves $$C_1$$, $$C_2$$ with $$g(C_1) \ge 2$$, $$g(C_2) \ge 2$$ there are at most finitely many nonconstant maps $$C_1 \to C_2$$. (This is of course not true if genus is $$0$$ or $$1$$; it suffices to ask for $$g(C_2) \ge 2$$.) This is the De Franchis Theorem.

Exit quiz. Use the same proof to show that any elliptic fibration $$f: X \to \mathbb{P}^1$$ with only two singular fibers is isotrivial!

Historical notes. Results of this kind in much larger generality go back to Shafarevich 1962 (hyperelliptic curves), Parshin 1968, Arakelov 1971, Faltings 1983 (abelian varieties with some restrictions) and Deligne 1987 (Hodge structures). The form discussed here was presumably known to Shafarevich in 1962.

Edit notice: As Evgeny Shinder pointed out in the comment, it is unclear why we have $$S = J^{-1}(\{0,1,\infty\})$$ where $$J : C \to \mathbf{P}^1$$ is the $$j$$-invariant map. The problem is that $$X \to S$$ might contain smooth fibers of period $$0$$ or $$1$$. After revision, the conclusion is now weaker than the original answer.

This is meant to be a modest complementary of naf and Evgeny Shinder's answer to this question, showing that a weaker version of the main proposition (over $$\mathbf{C}$$) also follows from Kodaira's work of elliptic surfaces. We will prove it under the stronger assumption that $$S$$ parameterizes exactly fibers of periods $$0$$ or $$1$$ and singular fibers and that $$X \to S$$ is relatively minimal.

We use the notations set up by Evgeny Shinder. As a consequence of the classification result summarized in BHPV's compact complex surface (Theorem V.11.1), the $$j$$-invariant map $$J : C \to \mathbf{P}^1$$ determines the elliptic surface $$f: X \to C$$ with a section up to finite ambiguity. It remains to show that there exist at most finitely many surjective maps $$J : C \to \mathbf{P}^1$$ such that $$S = J^{-1}(S')$$ where $$S' := \{ 0,1,\infty \}$$.

We first show that the degree $$d$$ of such a map $$J : C \to \mathbf{P}^1$$ is bounded. Indeed, if $$R$$ denotes the ramification degree of $$J$$, then since $$S = J^{-1}(S')$$, we have $$3d- |S| \le R$$ and Riemann-Hurwitz shows that $$2g(C) - 2 = -2d + R \ge d - |S|.$$ Hence $$d \le 2g(C) - 2+ |S|$$.

If we fix $$g : S \to S'$$, then the moduli space $$\mathrm{Hom}(C,\mathbf{P}^1;g)$$ of such maps $$J : C \to \mathbf{P}^1$$ satisfying the additional condition that $$J_{|S} = g$$ has tangent space at $$J$$ isomorphic to $$H^0(C,J^*T_{\mathbf{P}^1} \otimes I_{S})$$. As $$S = J^{-1}(S')$$ and $$J$$ is surjective, we have $$\deg (J^*T_{\mathbf{P}^1} \otimes I_{S}) < 0$$, and it follows from the boundedness of the degree that $$\mathrm{Hom}(C,\mathbf{P}^1;g)$$ is finite. Since there are only finitely many choices of $$g$$, the finiteness of the elliptic surfaces in question follows.

• Why is $S = J^{-1}(\{0,1,\infty\})$? I would think that $J(S) = \{\infty\}$? Jun 12, 2021 at 7:23
• You're right, this is so embarrassing. So the argument only proves an assertion weaker than the main proposition, in which we need to strengthen the assumption so that $S$ parameterizes exactly fibers of periods $0$ or $1$ and singular fibers. Let me edit the answer.
– HYL
Jun 12, 2021 at 8:58
• From Kodaira's table (e.g. Table 6 in BHPV), in general it's possible that $J(S)$ contains $0$,$1$. But I agree that $S=J^{-1}(\{0,1,\infty \})$ seems unclear, contrary to what I believed and what BHPV asserted in the paragraph before Theorem V.11.1. Thanks for pointing this out. Another tiny remark: we should assume that the elliptic surfaces are relatively minimal in the main proposition to avoid infinitely many elliptic surfaces constructed by blowups.
– HYL
Jun 12, 2021 at 13:27
• I actually wonder if the J-map alone can prove the easier "Exit quiz" claim (only isotrivial fibrations over $\mathbb{P}^1 \setminus \{0,\infty\})$. Clearly there are infinitely many (families of) nonconstant maps $\mathbb{P}^1 \setminus \{0,\infty\}) \to \mathbb{A}^1$ but somehow they are not realized by elliptic fibrations. Jun 12, 2021 at 14:34