GALOIS COVERINGS OF THE PLANE BY K3 SURFACES

A. MUHAMMED ULUDAĞ

Abstract. We give a complete list of branched Galois coverings of the projective plane by smooth K3 surfaces, with the branching curve of degree d < 6, and some examples of such coverings with d = 6. These coverings are not assumed to be abelian.

1. Introduction

The aim of the present article is to study the problem: “classify the branched Galois coverings of P^{2} by smooth K3 surfaces”. A basic example of such a covering is the following: Let C ⊂ P^{2} be a smooth quartic given by the homogeneous polynomial f(x, y, z). Then the surface X_{1} ⊂ P^{3} with equation w^{4} = f(x, y, z) is a smooth K3 surface. The projection ψ : [x : y : z : w] ∈ X_{1} → [x : y : z] ∈ P^{2} is a covering with ℤ/(4) as the Galois group, branched along C with branching index 4. Viewing the pair O := (P^{2}, 4C) as an orbifold, the projection ψ : X → O is a uniformization of this orbifold. The curve C is called the locus of O. Our approach to the problem consists in the study of orbifolds over P^{2} uniformized by smooth K3 surfaces. As a result, we obtain a complete list of orbifolds over P^{2} uniformized by smooth K3 surfaces and with an abelian Galois group (Theorem 1) and a complete list of orbifolds over P^{2} with a locus of degree ≤ 5, uniformized by smooth K3 surfaces and with non-abelian Galois groups (Theorem 2). We also present some K3 orbifolds over P^{2} with a sextic locus. There are no K3 orbifolds with a locus of degree > 6.

An alternative approach to this problem is to study the K3 surfaces X with the action of a group G ⊂ Aut(X) such that X/G ≃ P^{2}. A triple (G, X, ρ) is called a K3 group if X is a K3 surface and G is a group with a faithful action ρ : G × X → X. A K3 group (G, X, ρ) is called symplectic if G preserves a nowhere vanishing 2-form ω_{X} on X. Symplectic K3 groups were classified by Nikulin [19] for the abelian case and by Mukai [18] in the general case: they are 80 subgroups of a certain set of 11 subgroups of the Mathieu group M_{23}. Mukai also shows that if (G, X, ρ) is symplectic, then the minimal desingularization of X/G is also a K3 surface, see also [29] and [13] . The case where X/G is a Kummer surface has been studied in detail, see [25],[17], [23].

Since the form ω_{X} is unique up to a constant multiple, any K3 group (G, X, ρ) has a 1-dimensional representation g → α(g) ∈ ℂ, where α is defined by g^{∗}ω_{X} = α(g)ω_{X}. The kernel of G_{N} of this representation preserves ω_{X}, and hence is a symplectic K3 group. Hence, the classification of K3 groups can be reduced to the study of “geometrically realizable” extensions of symplectic groups. In a recent preprint [21], Oguiso announced that this classifification will be carried out in [11]. In contrast with the extensive literature on K3 groups (X, G, ρ), there are not many results concerning the orbifold structure of X/G (see however [30]

A classical construction in the theory of algebraic surfaces is that of a double plane, which has been applied extensively to plane sextics to obtain concrete examples of K3 surfaces [26], [1] [24]. Arguing in the opposite direction, Yang [32] used the classification theory of K3 surfaces to classify all irreducible plane sextics with only simple singularities. Yang’s results can be used to extend our results to orbifolds with sextic locus.

Note that the double plane construction yields a smooth surface X only when C is a smooth sextic. Excluding this case, we shall not discuss double covers in this article. We are exlusively concerned with those coverings X → P^{2} where X is a “honestly” smooth K3 surface.

2. Orbifolds

2.1. Definitions. We shall mostly follow the terminology settled in [31]. One may also consult [6] and [33]. Let M be a connected smooth complex manifold, G ⊂ Aut(M) a properly discontinuous subgroup and put X := M/G. Then the projection ϕ : M → X is a branched Galois covering endowing X with a map β_{ϕ} : X → ℕ defined by β_{ϕ}(p) := |G_{q}| where q is a point in ϕ^{−1}(p) and G_{q} is the isotropy subgroup of G at q. In this setting, the pair (X, β_{ϕ}) is said to be uniformized by ϕ : M → (X, β_{ϕ}). An orbifold is a pair (X, β) of an irreducible normal analytic space X with a function β : X → ℕ such that the pair (X, β) is locally finitely uniformizable. In case γ|β, the orbifold (X, γ) is said to be a suborbifold of (X, β). Let ϕ : (Y, 1) → (X, γ) be a uniformization of (X, γ), e.g. β_{ϕ} = γ. Then ϕ : (Y, β′) → (X, β) is called an orbifold covering, where β′ := β o ϕ/γ o ϕ. The orbifold (Y, β′) is called the lifting of (X, β) to the uniformization Y of (X, γ) and will be denoted by (X, β)/(X, γ).

Let (X, β) be an orbifold, B_{β} := supp(β − 1) be its locus and let B_{1}, …, B_{n} be the irreducible components of B_{β}. Then β is constant on B_{i}\sing(B_{β}); so let b_{i} be this number. The orbifold fundamental group π_{1}^{orb}(X, β) of (X, β) is the group defined by π_{1}^{orb}(X, β) := π_{1}(X\B_{β})/⟨⟨μ_{1}^{b_{1}}, …, μ_{n}^{b_{n}}⟩⟩ where μ_{i}^{b_{i}} is a meridian of B_{i} and ⟨⟨⟩⟩ denotes the normal closure. The local orbifold fundamental group π_{1}^{orb}(X, β)_{p} at a point p ∈ X is the orbifold fundamental group π_{1}^{orb}(B, p) of the restriction of (X, β) to a sufficiently small neighbourhood of B of p.

2.2. Orbifold singularities in dimension 2. Let (X, β) be an orbifold with X being a smooth surface. Then the map β is determined by the numbers b_{i}; in fact β(p) is the order of the local orbifold fundamental group at p. In case dim X = 2 the orbifold condition (i.e. locally finitely uniformizability) is equivalent to the finiteness of the local fundamental groups. This implies that the locus B_{β} of (X, β) may have ADE singularities only, and the admissible values for b_{i} are very restricted. These singularities and the branching numbers can either be characterized as the branching loci of the quotients of ℂ^{2} by finite reflection groups or as the germs with a finite orbifold fundamental group (see [33]). Below is a list of these singularities.

1  2  3  4  5  6  7

\begin{matrix}No & Equation & Condition & Order \\ 1 & xy & −− & pq \\ 2 & xy(x + y) & 0 < ρ := \frac{1}{p} + \frac{1}{q} + \frac{1}{r} − 1 & 4ρ^{−2} \\ 3 & x^{n} − y^{m} (gcd(n, m) = 1) & 0 < ρ := \frac{1}{n} + \frac{1}{m} + \frac{1}{p} − 1 & \frac{4}{nm}ρ^{−2} \\ 4 & x^{2} − y^{2n} (n ≥ 2) & 0 < ρ := \frac{1}{p} + \frac{1}{q} + \frac{1}{n} − 1 & \frac{4}{n}ρ^{−2} \\ 5 & y(x^{2} − y^{2n}) & 0 < ρ := \frac{1}{p} + \frac{1}{q} + \frac{1}{nr} − 1 & \frac{4}{n}ρ^{−2} \\ 6 & y(x^{2} − y^{n}) (n odd) & −− & 2nq^{2} \\ 7 & x(x^{2} − y^{3}) & −− & 96 \end{matrix}

In the last column of the table above, order of the local orbifold fundamental group of the corresponding singularity is given. For the node (1), this order is evidently pq. For the case (2), i.e. triple point, the orders of the corresponding reflection groups are well known, see e.g. [33]. For the remaining cases, the orders can be computed by using Yoshida’s result that any of the remaining singular loci is an orbifold covering of one of the triple points in (2) ([33], pp 167-169).

2.3. Finite uniformizations. Let X be a smooth manifold, (X, β) an orbifold, and let f : π_{1}^{orb}(X, β) → G be a surjection onto a finite group G. Then there is a corresponding branched covering ϕ : M → (X, β) with G as the Galois group. This is a finite smooth uniformization of X if and only if ϕ coincides locally with the local finite uniformizations of (X, β). In terms of G, this condition is expressed as follows:

Lemma 1. Let ι_{p} : π_{1}^{orb}(X, β)_{p} → π_{1}^{orb}(X, β) be the homomorphism induced by the inclusion. Then M is a finite smooth uniformization of (X, β) if and only if for any p ∈ X, the map f ∘ ι_{p} : π_{1}^{orb}(X, β)_{p} → G is an injection.

Let (X, β) be an orbifold, (X, γ) a suborbifold, (Y, 1) → (X, γ) a universal uniformization, and (Y, β′) ≃ (X, β)/(X, γ). Then one has the exact sequence of fundamental groups

0 → π_{1}^{orb}(Y, β′) → π_{1}^{orb}(X, β) → π_{1}^{orb}(X, γ) → 0

so that if π_{1}^{orb}(X, β) is finite, then |π_{1}^{orb}(X, β)| = |π_{1}^{orb}(X, γ)||π_{1}^{orb}(Y, β′)|. The following lemma will be useful.

Lemma 2. Let O = (P^{2}, B) be an orbifold where B = ∑_{i=1}^{n} b_{i}B_{i} with B_{i} being irreducible plane curve of degree d_{i}. If B is a nodal curve then

π_{1}^{orb}(O) ≃ ⟨μ_{1}, μ_{2}, …, μ_{n} | μ_{1}^{b_{1}} = μ_{2}^{b_{2}} = ⋯ = μ_{n}^{b_{n}} = μ_{1}^{d_{1}}μ_{2}^{d_{2}} … μ_{n}^{d_{n}} = 1⟩

Proof. By the Zariski conjecture proved by Deligne and Fulton (see [5]) the fundamental group of a nodal curve is abelian. Therefore

π_{1}(P^{2}\B) ≃ H_{1}(P^{2}\B, ℤ) ≃ ⟨μ_{1}, μ_{2}, …, μ_{n} | μ_{1}^{d_{1}}μ_{2}^{d_{2}} … μ_{n}^{d_{n}} = 1⟩

the classes μ_{i} being representatives of meridians around B_{i} for i ∈ [1, n]. The lemma follows by adding the orbifold relations to the above presentation. □.

2.4. Orbifold Chern numbers. Hereafter, we shall exclusively be concerned with orbifolds (X, β) with X being a smooth algebraic surface. Since in this case β is determined by its values b_{i} on B_{i}\sing(B_{β}), an orbifold can alternatively be defined as a pair (X, B) where B := b_{1}B_{1} + … b_{n}B_{n} is a divisor on X with b_{i} ≥ 1. For an orbifold (X, B) the corresponding map X → ℕ will be denoted by β_{B}.

Let O := (P^{2}, B) be an orbifold where B = b_{1}B_{1} + ⋯ + b_{n}B_{n} with B_{i} being an irreducible curve of degree d_{i}.

Definition. The orbifold Chern numbers of O are defined as

c_{1}^{2}(O) := [−3 + ∑_{1≤i≤n} d_{i} (1 − \frac{1}{b_{i}} )]^{2}

e(O) := 3 − ∑_{1≤i≤n} (1 − \frac{1}{b_{i}} ) e(B_{i}\sing(B)) − ∑_{p∈sing(B)} (1 − \frac{1}{β_{B}(p)})

Let O := (P^{2}, B) be an orbifold where B = b_{1}B_{1}+⋯+b_{n}B_{n} with B_{i} being an irreducible curve of degree d_{i}. If (X, 1) → (P^{2}, β) is a finite uniformization of degree d then the Chern numbers of X are given by e(X) = de(O) and c_{1}^{2}(X) = dc_{1}^{2}(O) (see [12]).

2.5. K3 orbifolds. We call O := (P^{2}, B) a K3 orbifold if there is a finite uniformization ϕ : M → O by a smooth K3 surface M. Since M is simply connected, ϕ must be the universal uniformization of O, which means that the corresponding Galois group is the full group π_{1}^{orb}(O), and the conditions of Lemma 1 must hold for G = π_{1}^{orb}(O). Moreover, since c_{1}^{2}(M) = 0 and e(M) = 24, one has the following lemma:

Lemma 3. Let O := (P^{2}, B) be a K3 orbifold where B = b_{1}B_{1} + ⋯ + b_{n}B_{n} with B_{i} being an irreducible curve of degree d_{i}. Then

(i) One has c_{1}^{2}(O) = 0. In particular, 4 ≤ deg(B) ≤ 6, and if deg(B)=6, then b_{1} = b_{2} = ⋯ = b_{n} = 2. (ii) π_{1}^{orb}(O) is of order 24/e(O). In particular, 24/e(O) is an integer.

3. Abelian K3 orbifolds

In 3.1-3.5, we shall give a complete list of K3 orbifolds with abelian uniformizing Galois group. The completeness of the list will be proved in 3.6. The locus of any orbifold considered in this section is assumed to be a nodal curve.

3.1. The K3 orbifolds C_{1}, C_{2} and C_{3}. These orbifolds are defined as follows: C_{1} is the orbifold (P^{2}, 4K) where K is a smooth quartic; C_{2} is the orbifold (P^{2}, 2L_{1} + 2L_{1} + 2L_{3} + 4Q) where L_{1}, L_{2} and L_{3} are three distinct lines and Q is a smooth quadric; C_{3} is the orbifold (P^{2}, 4L_{1} + 4L_{1} + 4L_{3} + 4L_{4}) where L_{1}, L_{2}, L_{3} and L_{4} are four distinct lines.

C_{1} (4)  C_{2} (16)  C_{3} (64)

3.1.1. The orbifold C_{1}. This is the orbifold mentioned in the beginning of the introduction. Obviously, one has |π_{1}^{orb}(C_{1})| = 4.

3.1.2. The orbifold C_{2}. Suppose that L_{1}, L_{2}, L_{3} and Q are given respectively by the equations x = 0, y = 0, z = 0 and P [x : y : z] = 0. The suborbifold A_{2} := (P^{2}, 2L_{1} +2L_{2} +2L_{3}) of C_{2} is uniformized by P^{2} via the covering ϕ_{2} : [x : y : z] → [x^{2} : y^{2} : z^{2}]. Lifting the orbifold C_{2} to this uniformization gives the orbifold C_{1} = (P^{2}, 4K), where K is the smooth quartic P (x^{2}, y^{2}, z^{2}) = 0. In other words, one has C_{1} ≃ C_{2}/A_{2}, and there is a covering ϕ_{2} : C_{1} → C_{2} of degree 4. The orbifold C_{1} in turn is uniformized by the K3 surface X_{2} defined by the equation w^{4} = P (x^{2}, y^{2}, z^{2}). One has

|π_{1}^{orb}(C_{2})| = |π_{1}^{orb}(A_{2})||π_{1}^{orb}(C_{1})| = 4 × 4 = 16

3.1.3. The orbifold C_{3}. Since any two arrangement of four lines in general position in P^{2} are projectively equivalent, one can assume that the lines L_{1}, L_{2}, L_{3} and L_{4} are given by the equations x = 0, y = 0, z = 0 and x + y + z = 0 respectively. Consider the suborbifold A_{4} := (P^{2}, 4L_{1} + 4L_{2} + 4L_{3}) of C_{3}, which is uniformized by P^{2} via the covering ϕ_{4} : [x : y : z] → [x^{4} : y^{4} : z^{4}]. Lifting the orbifold C_{3} to this uniformization gives the orbifold C_{1} = (P^{2}, 4C), where C is the Fermat quartic given by the equation x^{4} + y^{4} + z^{4} = 0. In other words, one has C_{1} ≃ C_{3}/A_{4}, and there is a covering ϕ_{4} : C_{1} → C_{3} of degree 16. The orbifold C_{1} in turn is uniformized by the Fermat K3 surface X_{3} defined by the equation w^{4} = x^{4} + y^{4} + z^{4}. One has

|π_{1}^{orb}(C_{3})| = |π_{1}^{orb}(A_{4})||π_{1}^{orb}(C_{1})| = 16 × 4 = 64

Note that A′_{2} := (P^{2}, 2L_{1} + 2L_{2} + 2L_{3}) is also a suborbifold of C_{3}, which is uniformized by P^{2} via the covering ϕ′_{2} : [x : y : z] → [x^{2} : y^{2} : z^{2}]. Lifting the orbifold C_{3} to this uniformization gives the orbifold C_{2} = (2L_{1} + 2L_{2} + 2L_{3} + 4Q), where L_{1}, L_{2} and L_{3} are respectively given by the equations x = 0, y = 0, z = 0 and Q is the smooth quadric given by the equation x^{2} + y^{2} + z^{2} = 0. In other words, one has C_{2} ≃ C_{3}/A′_{2}, and there is a covering ϕ′_{2} : C_{2} → C_{3} of degree 4.

3.2. The K3 orbifolds D_{1}, D_{2}, D_{3}, D_{4} and D_{5}. Let L_{1}, L_{2}, L_{3}, L_{4} be four distinct lines. D_{1} is the orbifold (P^{2}, S) where S is a smooth sextic; D_{2} is the orbifold (P^{2}, 3L_{1} + 3L_{2} + 3L_{3} + 2Q) where Q is a smooth quadric; D_{3} is the orbifold (P^{2}, 6L_{1} + 6L_{2} + 6L_{3} + 2L_{4}); D_{4} is the orbifold (P^{2}, 3L_{1} + 3L_{2} + 6Q) where Q is a smooth quadric. D_{5} is the orbifold (P^{2}, 2L_{1} + 2L_{2} + 2L_{3} + 2C) where C is a smooth cubic.

D_{1} (2)  D_{2} (18) D_{3} (72) D_{4} (18)  D_{5} (8)

3.2.1. The orbifold D_{1}. The double covering Y_{1} of the plane along the sextic is the uniformization in this case. One has |π_{1}^{orb}(D_{1})| = 2.

3.2.2. The orbifold D_{2}. Consider the suborbifold A_{3} := (P^{2}, 3L_{1} + 3L_{2} + 3L_{3}) of D_{3}, which is uniformized by P^{2} via ϕ_{3}. Assume that L_{1}, L_{2}, L_{3} and Q are defined respectively by x = 0, y = 0, z = 0 and P [x : y : z] = 0. Lifting D_{2} by ϕ_{3} gives the orbifold D_{1} = (P^{2}, 2S), where S is the smooth sextic P (x^{3}, y^{3}, z^{3}) = 0. In other words, one has D_{1} ≃ D_{2}/A_{3}, and there is a covering ϕ_{3} : D_{1} → D_{2} of degree 9. The orbifold D_{1} in turn is uniformized by the K3 surface Y_{2} which is the double cover of the plane along S. One has

|π_{1}^{orb}(D_{2})| = |π_{1}^{orb}(A_{3})||π_{1}^{orb}(D_{1})| = 9 × 2 = 18

3.2.3. The orbifold D_{3}. Assume that L_{1}, L_{2}, L_{3} and L_{4} are defined respectively by x = 0, y = 0, z = 0 and x + y + z = 0. Consider the suborbifold A_{6} := (P^{2}, 6L_{1} + 6L_{2} + 6L_{3}) of D_{3}, which is uniformized by P^{2} via ϕ_{6}. Lifting D_{3} by ϕ_{6} gives D_{1} = (P^{2}, 2S), where S is the Fermat sextic x^{6} + y^{6} + z^{6} = 0. In other words, one has D_{1} ≃ D_{3}/A_{6}, and there is a covering ϕ_{6} : D_{1} → D_{3} of degree 36. The orbifold D_{1} in turn is uniformized by the K3 surface Y_{3} which is the double cover of the plane along S. One has

|π_{1}^{orb}(D_{3})| = |π_{1}^{orb}(A_{6})||π_{1}^{orb}(D_{1})| = 36 × 2 = 72

3.2.4. The orbifold D_{4}. One has D_{4} ≃ D_{3}/A_{2}, where A_{2} := (P^{2}, 2L_{2} + 2L_{3} + 2L_{4}). Hence |π_{1}^{orb}(D_{4})| = |π_{1}^{orb}(D_{3})|/|π_{1}^{orb}(A_{2})| = 72/4 = 18.

3.2.5. The orbifold D_{5}. One has D_{5} ≃ D_{3}/A_{3}, where A_{3} := (P^{2}, 3L_{1} + 3L_{2} + 3L_{3}). Hence |π_{1}^{orb}(D_{5})| = |π_{1}^{orb}(D_{3})|/|π_{1}^{orb}(A_{3})| = 72/9 = 8.

3.3. The K3 orbifolds E_{1} and E_{2}. These are defined as follows: E_{1} is the orbifold (P^{2}, 2Q_{1} + 2Q_{2} + 2Q_{3}) where Q_{1}, Q_{2}, Q_{3} are three smooth quadrics; E_{2} is the orbifold (P^{2}, ∑_{i=1}^{6} 2L_{i}) where L_{i} (i ∈ [1, 6]) are six distinct lines.

E_{1} (8) E_{2} (32)

3.3.1. The orbifold E_{2}. By Lemma 2 the orbifold fundamental group is the abelian group with the presentation

π_{1}^{orb}(E_{2}) ≃ ⟨λ_{1}, …, λ_{6} | λ_{1}^{2} = … λ_{6}^{2} = λ_{1} … λ_{6} = 1⟩,

which is isomorphic to (ℤ/(2))^{5} and is of order 32. In order to show that E_{1} is uniformizable, one must verify that the subgroup ⟨λ_{i}⟩ is the group ℤ/(2) and the subgroup ⟨λ_{i}, λ_{j}⟩ is the group ℤ/(2) ⊕ ℤ/(2) for any i ≠ j ∈ [1, 6]. This is easily verified. Moreover, the euler number of this orbifold is easily seen to be 3/4, so that the euler number of the uniformization is 3|π_{1}^{orb}(E_{2})|/4 = 24. Since the first Chern number of E_{2} is 0, it follows that this is a K3 orbifold.

3.3.2. The orbifold E_{1}. One has E_{1} ≃ E_{2}/A_{2}, where A_{2} := (P^{2}, 2L_{1} + 2L_{2} + 2L_{3}). Hence |π_{1}^{orb}(E_{1})| = |π_{1}^{orb}(E_{2})|/|π_{1}^{orb}(A_{2})| = 32/4 = 8.

3.4. The K3 orbifolds F_{1} and F_{2}. These are defined as follows: F_{1} is the orbifold (P^{2}, 2Q + 2K) where Q is a smooth quadric and K is a smooth quartic; F_{2} is the orbifold (P^{2}, 2Q + ∑_{i=1}^{4} 2L_{i}) where L_{i} (i ∈ [1, 4]) are four distinct lines and Q is a smooth quadric.

F_{1} (4)  F_{2} (16)

3.4.1. The orbifold F_{2}. By Lemma 2 the orbifold fundamental group is the abelian group with the presentation

π_{1}^{orb}(F_{2}) ≃ ⟨κ, λ_{1}, …, λ_{4} | λ_{1}^{2} = … λ_{4}^{2} = λ_{2}λ_{3}λ_{4}λ_{5} = 1⟩,

which is isomorphic to (ℤ/(2))^{4} and is of order 16. In order to show that F_{1} is uniformizable, one must verify that the subgroups ⟨κ⟩ and ⟨λ_{i}⟩ is isomorphic to ℤ/(2) and the subgroups ⟨κ, λ_{j}⟩ and ⟨λ_{i}, λ_{j}⟩ are isomorphic to ℤ/(2) ⊕ ℤ/(2) for any i ≠ j ∈ [1, 4]. These are easily verified. Moreover, the euler number of this orbifold is easily seen to be 3/4, so that the euler number of the uniformization is 3|π_{1}^{orb}(F_{2})|/4 = 24. Since the first Chern number of F_{2} is 0, it follows that this is a K3 orbifold.

3.4.2. The orbifold F_{1}. One has F_{1} ≃ F_{2}/A_{2}, where A_{2} := (P^{2}, 2L_{1} + 2L_{2} + 2L_{3}). Hence |π_{1}^{orb}(F_{1})| = |π_{1}^{orb}(F_{2})|/|π_{1}^{orb}(A_{2})| = 16/4 = 4.

3.5. The K3 orbifold G. This is the orbifold (P^{2}, 2Q + 3C) where Q is a smooth quadric and C is a smooth cubic.

G (6)

By Lemma 2 one has

π_{1}^{orb}(O) ≃ ⟨λ, κ | λ^{2} = κ^{3} = 1⟩ ≃ ℤ/(6),

In order to show that G is uniformizable, one must verify that the subgroups ⟨κ⟩ and ⟨λ⟩ are isomorphic respectively to ℤ/(3) and ℤ/(2), and the subgroup ⟨κ, λ⟩ is isomorphic to ℤ/(2) ⊕ ℤ/(3) since the local orbifold fundamental group at any node of Q ∪ C is generated by λ and κ. These are easily verified. Moreover, the euler number of G is easily seen to be 4, so that the euler number of the uniformization is 4|π_{1}^{orb}(F_{2})| = 24. Since the first Chern number of G is 0, it follows that this is a K3 orbifold.

3.6. Completeness of the list.

Theorem 1. The list C_{1}, C_{2}, C_{3}, D_{1}, D_{2}, D_{3}, D_{4}, D_{5}, E_{1}, E_{2}, F_{1}, F_{2}, G is the complete list of K3 orbifolds with an abelian fundamental group.

Suppose that O = (P^{2}, B) is a K3 orbifold over P^{2}, the locus of B = ∑_{i=1}^{n} b_{i}B_{i} being a plane curve of degree d. By Lemma 3 one has 4 ≤ d ≤ 6 and ∑_{i=1}^{n} deg(B_{i})/b_{i} = 1 if d = 4. In case d = 5 one has ∑_{i=1}^{n} deg(B_{i})/b_{i} = 2. If d = 6, one has b_{1} = ⋯ = b_{n} = 2. Hereafter it will be assumed that these conditions holds.

On the other hand, the local fundamental group of an orbifold singularity is abelian if and only if the singularity is nodal. By the condition of smoothness on the uniformization, the local orbifold fundamental group of each singular point should inject into the global group π_{1}^{orb}(O). Since this latter group is assumed to be abelian, and since the only orbifold singularities with abelian local fundamental group are nodes, B must be a nodal curve. We claim moreover that irreducible components of B must be smooth: Assume the contrary; e.g. suppose that an irreducible component B_{i} of B has a node p.

The local orbifold fundamental group of this node has the presentation ⟨μ_{1}, μ_{2} | [μ_{1}, μ_{2}] = μ_{1}^{b_{i}} = μ_{2}^{b_{i}}⟩ ≃ ℤ/(b_{i}) ⊕ ℤ/(b_{i}), where μ_{1} and μ_{2} are meridians of the branches of B_{i} meeting at p. However, since B_{i} is irreducible, the meridians μ_{1} and μ_{2} of B_{i} are conjugate elements of the group π_{1}^{orb}(O). Since π_{1}^{orb}(O) is abelian, one actually has μ_{1} = μ_{2}. Hence, the subgroup of π_{1}^{orb}(O) generated by μ_{1} and μ_{2} is at most ℤ/(b) and cannot be isomorphic to the local orbifold fundamental group ℤ/(b_{i}) ⊕ ℤ/(b_{i}) at p.

The proof of Theorem 1 will be given in 3.6.1-3.6.3.

3.6.1. Quartics. Suppose that B = b_{1}L_{1} + b_{2}L_{2} + b_{3}L_{3} + b_{4}L_{4} with L_{1}, L_{2}, L_{3}, L_{4} being four lines in general position. There are 14 four-tuples (b_{1}, b_{2}, b_{3}, b_{4}) with 1/b_{1} + 1/b_{2} + 1/b_{3} + 1/b_{4} = 1. The euler number of O reads

e(O) = 1 − (\frac{1}{b_{1}} + \frac{1}{b_{2}} + \frac{1}{b_{3}} + \frac{1}{b_{4}} ) + ∑_{1≤i≠j≤4} \frac{1}{b_{i}b_{j}}

By Lemma 3, the number 24/e(O) should be an integer. The only 4-tuples (b_{1}, b_{2}, b_{3}, b_{4}) satisfiying this are (4, 4, 4, 4) and (2, 6, 6, 6), corresponding respectively to the orbifolds C_{3} and D_{3}.

Suppose that B = b_{1}L_{1} +b_{2}L_{2} +bQ where L_{1}, L_{2} are two lines and Q is a smooth quadric. The only triples (b_{1}, b_{2}, b) satisfying 1/b_{1} + 1/b_{2} + 2/b = 1 are (12, 4, 3), (6, 6, 3), (4, 4, 4) and (3, 3, 6). One has

e(O) = 2 − (\frac{1}{b_{1}} + \frac{1}{b_{2}} + \frac{2}{b}) + \frac{1}{b_{1}b_{2}} + \frac{2}{b_{1}b} + \frac{2}{b_{2}b}

One has 24/e(O) ∈ ℕ only for the triple (3, 3, 6), corresponding to the orbifold D_{4}.

Suppose that B = b_{1}Q_{1}+b_{2}Q_{2} where Q_{1}, Q_{2} are two smooth quadrics. The only pairs (b_{1}, b_{2}) satisfying 2/b_{1} + 2/b_{2} = 1 are (4, 4) and (6, 3). One has e(O) = 3 − (2/b_{1} + 2/b_{2}) + 4/(b_{1}b_{2}). The number 24/e(O) is never an integer.

Suppose that B = b_{1}L + b_{2}C where L is a line and C is a smooth cubic. The only pairs (b_{1}, b_{2}) satisfying 1/b_{1} + 3/b_{2} = 1 are (4, 4) and (2, 6). One has e(O) = 3 − (1/b_{1} + 3/b_{2}) + 3/(b_{1}b_{2}). The number 24/e(O) is never an integer.

Suppose that B = b_{1}K, where K is a smooth quartic. Then b_{1} = 4 and e(O) = 6, corresponding to the orbifold C_{1}.

3.6.2. Quintics. Suppose that B = ∑_{i=1}^{5} b_{i}L_{i}, where L_{1}, …, L_{5} are five lines in general position. There are only 3 five-tuples (2, 2, 2, 3, 6), (2, 2, 2, 4, 4), (2, 2, 3, 3, 3) with reciprocals summing up to 2. The euler number of O reads

e(O) = 3 − ∑_{1≤i≤5} \frac{2}{b_{i}} + ∑_{1≤i≠j≤5} \frac{1}{b_{i}b_{j}}

None of the above 5-tuples satisfies 24/e(O) ∈ ℕ.

Suppose that B = b_{1}L_{1} + b_{2}L_{2} + b_{3}L_{3} + bQ with L_{1}, L_{2} and L_{3} being three lines and Q a smooth quadric. The only four-tuples (b_{1}, b_{2}, b_{3}, b) satisfying 1/b_{1}+1/b_{2}+1/b_{3}+2/b = 2 are (2, 3, 6, 2), (2, 2, 2, 4), (2, 4, 4, 2), (3, 3, 3, 2) and (2, 2, 3, 3). One has

e(O) = 4 − (\frac{2}{b_{1}} + \frac{2}{b_{2}} + \frac{2}{b_{3}} + \frac{4}{b}) + \frac{1}{b_{1}b_{2}} + \frac{1}{b_{2}b_{3}} + \frac{1}{b_{3}b_{1}} + \frac{2}{b_{1}b} + \frac{2}{b_{2}b} + \frac{2}{b_{3}b}

The number 24/e(O) is integer only for the four-tuples (2, 2, 2, 4) and (3, 3, 3, 2) which corresponds to the orbifolds C_{2} and D_{2} respectively.

Suppose that B = bL_{1} + b_{1}Q_{1} + b_{2}Q_{2} with Q_{1}, Q_{2} being two smooth quadrics, and L a line. The only triples (b, b_{1}, b_{2}) satisfying 1/b_{,}+2/b_{1}+ 2/b_{2} = 2 are (2, 2, 4), and (3, 3, 2). One has

e(O) = 5 − (\frac{4}{b_{1}} + \frac{4}{b_{2}} + \frac{2}{b}) + \frac{2}{bb_{1}} + \frac{2}{b_{2}b} + \frac{4}{b_{1}b_{2}}

The number 24/e(O) is never an integer.

Suppose that B = b_{1}L_{1} + b_{2}L_{2} + bC where C is a smooth cubic. The only triples (b_{1}, b_{2}, b) satisfying 1/b_{1} + 1/b_{2} + 3/b = 2 are (4, 4, 2), (3, 6, 2) and (2, 2, 3). One has

e(O) = 6 − (\frac{2}{b_{1}} + \frac{2}{b_{2}} + \frac{6}{b}) + \frac{1}{b_{1}b_{2}} + \frac{3}{b_{3}b_{1}} + \frac{3}{b_{2}b_{3}}

The number 24/e(O) is never an integer.

Suppose that B = b_{1}Q_{+}b_{2}C where Q is a smooth quadric and C is a smooth cubic. The only pairs (b_{1}, b_{2}) satisfying 2/b_{1} + 3/b_{2} = 2 are (4, 2) and (2, 3). One has e(O) = 7 − (4/b_{1} + 6/b_{2}) + 6/(b_{1}b_{2}). One has 24/e(O) ∈ ℕ only for the pair (2, 3), corresponding to the orbifold G.

In case B = b_{1}L_{1} + b_{2}K with K being a smooth quartic, there are no pairs (b_{1}, b_{2}) satisfying 1/b_{1} + 4/b_{2} = 2. Similarly, in case B = bP with P being a smooth quintic, one never has 5/b = 2.

3.6.3. Sextics. It suffices to show that if deg(B) = 6 and O is not one of the orbifolds D_{1}, D_{5}, E_{1}, E_{2}, F_{1}, F_{2} then O is not a K3 orbifold.

If B = 2(L + P) where P is a smooth quintic, then by Lemma 2 one has π_{1}^{orb}(O) ≃ ⟨λ, κ | λ^{2} = κ^{2} = λκ^{5} = 1⟩ ≃ ℤ/(2), whereas the local orbifold fundamental group of a node of B is ℤ/(2)⊕ℤ/(2), which cannot be a subgroup of π_{1}^{orb}(O). Hence O is not uniformizable.

If B = 2(C_{1} + C_{2}) where C_{1} and C_{2} are smooth cubics, then one has π_{1}^{orb}(O) ≃ ⟨λ, κ | λ^{2} = κ^{2} = λ^{3}κ^{3} = 1⟩ ≃ ℤ/(2), so that O is not uniformizable.

If B = 2(L_{1} + L_{2} + K) with K being a smooth quartic then one has π_{1}^{orb}(O) ≃ ⟨λ_{1}, λ_{2}, κ | λ_{1}^{2} = λ_{2}^{2} = κ^{2} = λ_{1}λ_{2} = 1⟩. Consider the intersection point p of L_{1} and L_{2}. The local orbifold fundamental group at the node p is generated by λ_{1} and λ_{2} and is isomorphic to ℤ/(2) ⊕ ℤ/(2), whereas the subgroup of π_{1}^{orb}(O) generated by λ_{1} and λ_{2} is isomorphic to ℤ/(2). Hence O is not uniformizable.

If B = 2(L + Q + C) with Q being a quadric and C a cubic, then one has π_{1}^{orb}(O) ≃ ⟨λ, κ, μ | λ^{2} = κ^{2} = μ^{2} = λμ = 1⟩. Consider the intersection point p of L and C. The local orbifold fundamental group at the node p is generated by λ and μ and is isomorphic to ℤ/(2)⊕ℤ/(2), whereas the subgroup of π_{1}^{orb}(O) generated by λ and μ is isomorphic to ℤ/(2). Hence O is not uniformizable.

If B = 2(L_{1} + L_{2} + Q_{1} + Q_{2}) with Q_{1} and Q_{2} being smooth quadrics then π_{1}^{orb}(O) ≃ ⟨λ_{1}, λ_{2}, κ_{1}, κ_{2} | λ_{1}^{2} = λ_{2}^{2} = κ_{1}^{2} = κ_{2}^{2} = λ_{1}λ_{2} = 1⟩. Consider the intersection point p of L_{1} and L_{2}. The local orbifold fundamental group at the node p is generated by λ_{1} and λ_{2} and is isomorphic to ℤ/(2)⊕ℤ/(2), whereas the subgroup of π_{1}^{orb}(O) generated by λ_{1} and λ_{2} is isomorphic to ℤ/(2). Hence O is not uniformizable.

4. Non-Abelian K3 orbifolds

4.1. The K3 orbifolds H_{1} and H_{2}. These are defined as follows: H_{1} is the orbifold (P^{2}, 2L_{1} + 2L_{2} + 2Q_{1} + 2Q_{2}), where Q_{1}, Q_{2} are two smooth quadrics tangent to each other at two distinct points, and L_{1}, L_{2} are two lines in general position with respect to Q_{1} ∪ Q_{2}; H_{2} is the orbifold (P^{2}, 2L_{1} + 2L_{2} + 2L_{3} + 4L_{4} + 4L_{5}), where L_{1}, L_{2} and L_{3} are three lines with a common point and L_{4}, L_{5} are two lines in general position with respect to L_{1} ∪ L_{2} ∪ L_{3}.

H_{1} (64)  H_{2} (16)

4.1.1. The orbifold H_{2}. One has H_{2}/A_{2} ≃ C_{2}, where A_{2} := (P^{2}, 2L_{1} + 2L_{2} + 2L_{4}). Since C_{2} is a K3 orbifold, so is H_{2}. One has

|π_{1}^{orb}(H_{2})| = |π_{1}^{orb}(A_{2})||π_{1}^{orb}(C_{2})| = 4 × 16 = 64.

4.1.2. The orbifold H_{1}. One has H_{1} ≃ H_{2}/A′_{2}, where A′_{2} := (P^{2}, 2L_{1} + 2L_{4} + 2L_{5}). Hence |π_{1}^{orb}(H_{1})| = |π_{1}^{orb}(H_{2})|/|π_{1}^{orb}(A_{2})| = 64/4 = 16.

4.2. The K3 orbifolds I_{1} and I_{2}. These are defined as follows: I_{1} is the orbifold (P^{2}, 2L_{1} + 2Q_{1} + 2Q_{2}), where Q_{1}, Q_{2} are two smooth quadrics tangent to each other at two distinct points, and L_{1} is a line in general position with respect to Q_{1}∪Q_{2}; I_{2} is the orbifold (P^{2}, 2L_{1}+ 2L_{2} +3L_{3} +2L_{4} +6L_{5}) where L_{1}, L_{2}, L_{3} are three lines with a common point and L_{4} ∪ L_{5} is in general position with respect to L_{1} ∪ L_{2} ∪ L_{3}.

I_{1} (18)  I_{2} (72)

4.2.1. The orbifold I_{2}. One has I_{2}/A_{2} ≃ D_{4}, where A_{2} := (P^{2}, 2L_{1} + 2L_{2} + 2L_{4}). Hence |π_{1}^{orb}(I_{2})| = |π_{1}^{orb}(A_{2})||π_{1}^{orb}(D_{4})| = 4 × 18 = 72. Moreover, one has I_{2}/A′_{2} ≃ D_{2}, where A′_{2} := (P^{2}, 2L_{1} + 2L_{2} + 2L_{5}).

4.2.2. The orbifold I_{1}. One has I_{1} ≃ I_{2}/A′′_{2}, where A′′_{2} := (P^{2}, 2L_{1} + 2L_{4} + 2L_{5}). Hence |π_{1}^{orb}(I_{1})| = |π_{1}^{orb}(I_{2})|/|π_{1}^{orb}(A′′_{2})| = 72/4 = 18.

4.3. The K3 orbifolds J_{1}, J_{2}, J_{3}, J_{4} and J_{5}. These are defined as follows: J_{1} is the orbifold (P^{2}, ∑_{i=1}^{6} 2L_{i}), where the lines L_{1}, L_{2}, L_{3} meet at a point, L_{3}, L_{4}, L_{5} meet at another point, and L_{1} ∪ L_{2} ∪ L_{2} is in general position with respect to L_{4} ∪ L_{5} ∪ L_{6}; J_{2} is the orbifold (P^{2}, 2L_{1}+2L_{2}+2L_{3}+4Q) where L_{1}, L_{2}, L_{3} meet at a point not lying on the smooth quadric Q and L_{1}, L_{3} are tangents to Q; J_{3} is the orbifold (P^{2}, 2L_{1} + 2L_{2} + 2L_{3} + 2L_{4} + 2Q) where L_{1}, L_{2}, L_{3} meet at a point not lying on Q, the lines L_{1}, L_{3} are tangents to Q and L_{4} pass through these tangence points; J_{4} is the orbifold (P^{2}, 4L_{1} + 2L_{2} + 2L_{3} + 2L_{4} + 4L_{5}) where L_{3} pass through the points L_{1}∩L_{2} and L_{4}∩L_{5}; J_{5} is the orbifold (P^{2}, 2L_{1} + 2L_{2} + 2K) where K is a quartic with a tacnode and L_{1}, L_{2} are two flexes (with multiplicity 4) of K.

J_{1} (64)  J_{2} (64)  J_{3} (64)  J_{4} (256)  J_{5} (16)

One has J_{4}/A_{2} ≃ C_{3}, where A_{2} := (P^{2}, 2L_{2} + 2L_{3} + 2L_{4}) is a suborbifold of J_{4}. Hence |π_{1}^{orb}(J_{4})| = |π_{1}^{orb}(A_{2})||π_{1}^{orb}(C_{3})| = 4 × 64 = 256. Moreover, let A′_{2} := (P^{2}, 2L_{1} +2L_{3} +2L_{5}), A′′_{2} := (P^{2}, 2L_{1} +2L_{2} +2L_{4}) and A′′′_{2} := (P^{2}, 2L_{1} + 2L_{2} + 2L_{5}) be suborbifolds of J_{4}. Then one has J_{4}/A′_{2} ≃ J_{1}, J_{4}/A′′_{2} ≃ J_{2} and J_{4}/A′′′_{2} ≃ J_{3}, showing that J_{1}, J_{2} and J_{3} are K3 orbifolds.

On the other hand, one has J_{1}/A_{2} ≃ H_{1} where A_{2} := (P^{2}, 2L_{1} + 2L_{2} + 2L_{4}) is a suborbifold of J_{1}, J_{3}/A_{2} ≃ H_{1} where A_{2} := (P^{2}, 2L_{1} + 2L_{3} + 2L_{4}) is a suborbifold of J_{3}, and J_{3}/A′_{2} ≃ J_{5} where A_{2} := (P^{2}, 2L_{1} + 2L_{2} + 2L_{4}) is a suborbifold of J_{3}.

4.4. The K3 orbifold K. This is the orbifold (P^{2}, 3L_{1} + 3L_{2} + 3L_{3} + 2L_{4} + 4L_{5}) where L_{3} pass through the points L_{1} ∩ L_{2} and L_{4} ∩ L_{5}. One has K/A_{3} ≃ J_{1}, showing that K is a K3 orbifold.

K (576)

4.5. The K3 orbifolds L_{1}, L_{2} and L_{3}. The orbifolds L_{1}, L_{2} are defined respectively as (P^{2}, 3L_{1} +3L_{2} +3L_{3} +2Q) and (P^{2}, 4L_{1} +4L_{2} + 2L_{3} + 2Q) where L_{1}, L_{2} are tangent to Q and L_{3} is in general position with respect to L_{1} ∪ L_{2} ∪ Q. By definition, L_{3} is the orbifold L_{1}/A_{3}, where A_{3} := (P^{2}, 3L_{1} + 3L_{2} + 3L_{3}) is a suborbifold of L_{1}. The locus of L_{3} is an irreducible sextic with 6 cusps and no other singularities. For other examples of non-abelian branched coverings along sextic see [27].

L_{1} (54)  L_{2} (64)  L_{3} (6)

4.5.1. The orbifold L_{2}. One has L_{2}/A_{2} ≃ P_{2}, which is a K3 orbifold (see 4.8).

4.5.2. The orbifold L_{1}. An application of Zariski-Van Kampen (see [7]) gives the (redundant) presentation

π_{1}(P^{2}\(L_{1} ∪ L_{2} ∪ Q)) ≃ ⟨κ, τ_{1}, τ_{2} | (κτ_{1})^{2} = (τ_{1}κ)^{2}, κ^{2}τ_{1}τ_{2} = 1⟩,

where κ is a meridian of Q and τ_{i} is a meridian of L_{i}. Since L_{3} is in general position with respect to L_{1} ∪ L_{2} ∪ Q, the fundamental group π_{1}(P^{2}\(L_{1} ∪ L_{2} ∪ L_{3} ∪ Q)) admits the presentation

⟨κ, τ_{1}, τ_{2}, τ_{3} | (κτ_{1})^{2} = (τ_{1}κ)^{2}, [τ_{3}, κ] = [τ_{3}, τ_{1}] = [τ_{3}, τ_{2}] = κ^{2}τ_{3}τ_{2}τ_{1} = 1⟩, so that the group π_{1}^{orb}(L_{1}) admits the presentation (see [9])

(1) π_{1}^{orb}(L_{1}) ≃ ⟨κ, τ_{1}, τ_{2}, τ_{3} | \begin{matrix}(κτ_{1})^{2} = (τ_{1}κ)^{2} \\ [τ_{3}, κ] = [τ_{3}, τ_{1}] = [τ_{3}, τ_{2}] = 1 \\ τ_{3}τ_{2}τ_{1} = 1 \\ τ_{1}^{3} = τ_{2}^{3} = τ_{3}^{3} = κ^{2} = 1 \end{matrix} ⟩

The group theory software GAP [8] gives |π_{1}^{orb}(L_{1})| = 54. In order to show that L_{1} is a K3 orbifold, one must verify (i) |⟨τ_{i}⟩| = 3, (ii) |⟨τ_{i}, τ_{j}⟩| = 9 (i ≠ j), (iii) |⟨κ⟩| = 2, (iv) |⟨κ, τ_{3}⟩| = 6 and (v) |⟨κ, τ_{1}⟩| = |⟨κ, τ_{2}⟩| = 18 in the group π_{1}^{orb}(L_{1}). The conditions (i)-(iv) already holds in the abelianized orbifold fundamental group and are easily verified. By symmetry, it suffices to verify only |⟨κ, τ_{1}⟩| = 18 in (v). This latter is a routine computer check by GAP. Hence, L_{1} is uniformized by a smooth surface Y with c_{1}^{2}(Y) = 0. Since e(L_{1}) = 3/2, one has e(Y) = |π_{1}^{orb}(L_{1})|e(L_{1}) = 24, and Y is a K3 surface.

4.6. The orbifolds M_{1} and M_{2}. The orbifolds M_{1}, M_{2} are defined respectively as (P^{2}, 2L_{1}+2L_{2}+2L_{3}+4Q) and (P^{2}, 2L_{1}+2L_{2}+3L_{3}+3Q) where Q is a smooth quadric and L_{1}, L_{2}, L_{3} are three distinct tangents. Note that any two such arrangements are projectively equivalent.

M_{1} (256) M_{2} (576)

4.6.1. The orbifold M_{1}. One has M_{1}/A_{2} ≃ C_{3}, where A_{2} := (P^{2}, 2L_{1}+ 2L_{2} + 2L_{3}). Note that M_{1} is invariant under the S_{3} action on P^{2}.

4.6.2. The orbifold M_{2}. The group π_{1}(P^{2}\(Q ∪ L_{1} ∪ L_{2} ∪ L_{3})) is the braid group on two strands of the thrice punctured sphere, and its presentation is known, see [2] and see [3], [15] for generalizations. Adding the orbifold relations to this presentation gives

(2) π_{1}^{orb}(M_{2}) ≃ ⟨ \begin{matrix}τ_{1}, τ_{2}, τ_{3}, \\ κ_{1}, κ_{2}, κ_{3} \end{matrix} | \begin{matrix}κ_{i} = τ_{i−1}κ_{i−1}τ_{i−1}^{−1}, 2 ≤ i ≤ 3 \\ (κ_{i}τ_{i})^{2} = (τ_{i}κ_{i})^{2}, 1 ≤ i ≤ 3 \\ [κ_{i}^{−1}τ_{i}κ_{i}, τ_{j}] = 1, 1 ≤ i < j ≤ 3 \\ τ_{3}τ_{2}τ_{1}κ_{1}^{2} = τ_{1}^{2} = τ_{2}^{2} = τ_{3}^{3} = κ^{3} = 1 \end{matrix} ⟩

where κ_{i} are meridians of Q and τ_{i} is a meridian of T_{i} for 1 ≤ i ≤ 3. GAP gives |π_{1}^{orb}(M_{2})| = 576.

As a by-product of the method used in the computation of the group π_{1}(P^{2}\(Q ∪ L_{1} ∪ L_{2} ∪ L_{3})) in [2], the generators of the local fundamental groups around the singular points of Q ∪ L_{1} ∪ L_{2} ∪ L_{3} can be read easily from the above presentation. For the nodes the local groups are ⟨κ_{i}^{−1}τ_{i}κ_{i}, τ_{j}⟩, and for the tangency points they are ⟨κ_{i}, τ_{i}⟩. In order to show that M_{2} is uniformizable, one needs to verify the followings: (i) |⟨κ_{1}⟩| = |⟨τ_{3}⟩| = 3 and |⟨τ_{1}⟩| = |⟨τ_{2}⟩| = 2, (ii) |⟨κ_{1}^{−1}τ_{1}κ_{1}, τ_{2}j⟩| = 4 and |⟨κ_{1}^{−1}τ_{1}κ_{1}, τ_{3}⟩| = |⟨κ_{2}^{−1}τ_{2}κ_{2}, τ_{3}⟩| = 6, (iii) |⟨κ_{1}, τ_{1}⟩| = |⟨κ_{2}, τ_{2}⟩| = 18 and |⟨κ_{3}, τ_{3}⟩| = 72. This is a routine check by GAP. Hence, the orbifold M_{2} is uniformizable by a smooth surface Z. Since e(M_{2}) = 1/24, one has e(Z) = e(M_{2})|π_{1}^{orb}(M_{2})| = 24, so that Z is a K3 surface, showing that M_{2} is a K3 orbifold.

4.7. The sextic orbifolds N_{1}, N_{2}, N_{3}, N_{4} and N_{5}. The orbifolds N_{1}, N_{2} and N_{3} are defined as follows: Let B = 2L_{1} +2L_{2} +2L_{3} +2L_{4} +2Q, where L_{1}, L_{2}, L_{3} are three tangents of Q. The case where L_{4} is in general position with respect to Q ∪ L_{1} ∪ L_{2} ∪ L_{2} gives N_{1}. The case where L_{4} passes through the point L_{1} ∩ L_{2} and not through the point L_{3} ∩ Q gives N_{2}. Finally, the case where L_{4} is the line through the points L_{1} ∩ L_{2} and L_{3} ∩ Q gives N_{3}. The orbifolds N_{4} and N_{5} are defined as follows: Let B = 2L_{1} + 2L_{2} + 2Q_{1} + 2Q_{2}, where L_{1}, L_{2} are two common tangents of the quadrics Q_{1} and Q_{2}. Assume that the lines do not pass through any point in Q_{1} ∩Q_{2}. The case where Q_{1} ∩Q_{2} consists of four distinct points gives N_{4}, and the case where Q_{1} ∩ Q_{2} consists of three distinct points gives N_{5}.

N_{1} (64)  N_{2} (128)  N_{3} (256)  N_{4} (32)  N_{5} (64)

Let A_{2} := (P^{2}, 2L_{1} + 2L_{2} + 2L_{3}) and A′_{2} := (P^{2}, 2L_{1} + 2L_{2} + 2L_{4}) be suborbifolds of N_{1}. One has N_{1}/A_{2} ≃ F_{2}. Since F_{2} is a K3 orbifold, so is N_{1} and |π_{1}^{orb}(N_{1})| = |π_{1}^{orb}(A_{2})||π_{1}^{orb}(F_{2})| = 4 × 16 = 64. On the other hand, one has N_{1}/A′_{2} = N_{3}, N_{3}/A_{2} = J_{2} and N_{3}/A′_{2} = N_{5}.

One has N_{2}/A_{2} ≃ E_{2}. Since E_{2} is a K3 orbifold, so is N_{2} and |π_{1}^{orb}(N_{1})| = |π_{1}^{orb}(A_{2})||π_{1}^{orb}(F_{2})| = 4 × 16 = 64. On the other hand, one has N_{2}/A′_{2} ≃ N_{4}.

4.8. The sextic K3 orbifolds P_{1}, P_{2}, P_{3}, P_{4} and P_{5}. Let P_{1} be the orbifold (P^{2}, B), where B = ∑_{i=1}^{∞} 2L_{i} such that L_{1}, L_{2} and L_{3} meet at a point, L_{3}, L_{4} and L_{5} meet at another point, and L_{6} is in general position with respect to ∪_{i=1}^{5}L_{i}. One has P_{1}/A_{2} ≃ F_{2}, where A_{2} := (P^{2}, 2L_{2} + 2L_{3} + 2L_{4}). Since F_{2} is a K3 orbifold, so is P_{1}. On the other hand, one has one has P_{1}/A′_{2} ≃ P_{2} where A′_{2} := (P^{2}, 2L_{1} + 2L_{2} + 2L_{6}); P_{1}/A′′_{2} ≃ P_{3} where A′′_{2} := (P^{2}, 2L_{2} + 2L_{4} + 2L_{6}); P_{1}/A′′′_{2} ≃ P_{4} where A′′′_{2} := (P^{2}, 2L_{2} + 2L_{3} + 2L_{6}); P_{1}/A′′′′_{2} ≃ P_{5} where A′′′′_{2} := (P^{2}, 2L_{2} + 2L_{3} + 2L_{4}).

P_{1} (64)  P_{2} (16)  P_{3} (16)  P_{4} (16)  P_{5} (16)

4.9. The sextic K3 orbifolds Q_{1} and Q_{2}. Let Q_{1} be the orbifold (P^{2}, B), where B = ∑_{i=1}^{∞} 2L_{i} such that L_{1}, L_{2}, L_{3} meet at a point, L_{3}, L_{4}, L_{5} meet at another point, and L_{5}, L_{6}, L_{1} meet at another point. The orbifold Q_{2} is defined to be the lifting Q_{1}/A_{2}, where A_{2} := (P^{2}, 2L_{2} + 2L_{4} + 2L_{6}); its locus consists of three quadrics Q_{1}, Q_{2}, Q_{3} with six tacnodes. On the other hand, one has Q_{1}/A′_{2} ≃ E_{2}, where A′_{2} := (P^{2}, 2L_{1} + 2L_{3} + 2L_{5}), showing that Q_{1} and Q_{2} are both K3 orbifolds.

Q_{1} (128)  Q_{2} (32)

4.10. Completeness of the list.

Theorem 2. The list H_{2}, I_{1}, I_{2}, J_{2}, J_{4}, K, L_{1}, L_{2}, M_{1}, M_{2} is the complete list of K3 orbifolds over P^{2}, with a locus of degree ≤ 5 and with a non-abelian fundamental group.

It is easily shown that there are no quartic K3 orbifolds with a nonabelian fundamental group, see 4.10.1. Suppose that O = (P^{2}, B) is a quintic K3 orbifold with a non-abelian orbifold fundamental group. In particular, the group π_{1}(P^{2}\B) should be non-abelian. All quintic curves with a non-abelian fundamental group has been listed and computed by Degtyarev [4], so it suffices to consider the orbifolds with loci in Degtyarev’s list. Evidently, those quintics with non-simple singularities are immediately eliminated. The proof of the theorem for quintic orbifolds will be given in 4.10.2- 4.10.6.

4.10.1. Quartics. Suppose that O = (P^{2}, B) is a quartic K3 orbifold with a non-abelian orbifold fundamental group, where B = b_{1}B_{1} + ⋯ + b_{n}B_{n}. The condition c_{1}^{2}(O) = 0 forces b_{1} = ⋯ = b_{n} = 4. Since π_{1}^{orb}(O) is non-abelian, B must have a non-nodal singular point. This point cannot be a reducible singularity of B, since two branches of a reducible non-nodal orbifold point cannot have branching 4 simultaneously. Hence, B may have irreducible singularities only, in particular B itself is irreducible. The only irreducible quartic with a non-abelian fundamental group is the 3-cuspidal quartic K (see [34], [7]). The group π_{1}(P^{2}\K) is non-abelian of order 12. The euler number of the orbifold O = (P^{2}, 4K) is easily computed: Since K is rational, one has e(E) = 2 and e(P^{2}\K) = 1. Since e(C\sing(K) = 2 − 3 = −1, one has e(O) = 1 − 1/4 + 3/96 = 25/32, so that 24/e(O) ∉ ℕ.

4.10.2. Linear quintics. For the Sections 4.10.2- 4.10.6 set β_{i} := 1/b_{i}. Let O = (P^{2}, B), where B = ∑_{i=1}^{5} b_{i}L_{i}. One has c_{1}^{2}(O) = 0 if and only if ∑_{i=1}^{5} β_{i} = 2. There are only 3 five-tuples (2, 2, 2, 3, 6), (2, 2, 2, 4, 4), (2, 2, 3, 3, 3) satisfying this condition.

Suppose that L_{1}, L_{2}, L_{3} meet at a point, and that L_{4} ∪ L_{5} is in general position with respect to L_{1} ∪ L_{2} ∪ L_{3}. Then e(O) is

2−(β_{1}+β_{2}+β_{3}+2β_{4}+2β_{5})+(β_{4}+β_{5})(β_{1}+β_{2}+β_{3})+ \frac{1}{4}[β_{1}+β_{2}+β_{3}−1]^{2}

For the triple point to be an orbifold point one must have: β_{1}+β_{2}+β_{3} > 1. The only five-tuples satisfiying this are (2, 2, 2, 4, 4) and (2, 2, 3, 6, 2), corresponding to the orbifolds H_{2} and I_{2} respectively.

Suppose that L_{1}, L_{2}, L_{3} meet at a triple point, and L_{1}, L_{4}, L_{5} meet at another point. Then e(O) is

1 − (β_{2} + β_{3} + β_{4} + β_{5}) + (β_{4} + β_{5})(β_{2} + β_{3})+

\frac{1}{4} [β_{1} + β_{2} + β_{3} − 1]^{2}+\frac{1}{4} [β_{1} + β_{5} + β_{4} − 1]^{2}

For the triple points to be orbifold points one must have: β_{1}+β_{2}+β_{3} > 1 and β_{1} + β_{4} + β_{5} > 1. The only five-tuples satisfiying these conditions are (2, 4, 2, 2, 4) and (3, 3, 2, 3, 2), corresponding to the orbifolds J_{4} and K respectively.

4.10.3. A quadric with three lines. Let B = b_{1}L_{1} + b_{2}L_{2} + b_{3}L_{3} + bQ. One has c_{1}^{2}(O) = 0 if and only if β_{1} + β_{2} + β_{3} + 2β = 2. The possible values of (b_{1}, b_{2}, b_{3}, b) are (i) (2, 3, 6, 2), (ii) (2, 4, 4, 2), (iii) (2, 2, 2, 4), (iv) (3, 3, 3, 2) and (v) (2, 2, 3, 3).

Suppose that L_{1}, L_{2}, L_{3} have a common point and L_{1} ∪ L_{2} ∪ L_{3} is in general position with respect to Q. Then

e(O) = 3−(β_{1} +β_{2} +β_{3} +4β)+2β_{4}(β_{1} +β_{2} +β_{3})+ \frac{1}{4} [β_{1} + β_{2} + β_{3} − 1]^{2}

For the triple point to be an orbifold point one must have: β_{1}+β_{2}+β_{3} > 1. This occurs only in (iii), which forces b_{1} = b_{2} = b_{3} = 2 and b = 4. But then e(O) = 21/16, so that 24/e(O) ∉ ℕ.

Suppose that L_{1}, L_{2}, L_{3} have a common point, L_{1} is tangent to Q, and L_{2} ∪ L_{4} is in general position with respect to Q. Then e(O) is

2−(β_{2}+β_{3}+3β)+2β(β_{2}+β_{3})+ \frac{1}{4} [β_{1} + β_{2} + β_{3} − 1]^{2}+ \frac{1}{2} [β_{1} + β − \frac{1}{2}]^{2}

For the triple point to be an orbifold point one must have: β_{1}+β_{2}+β_{3} > 1. This occurs only in (iii), which forces b_{1} = b_{2} = b_{3} = 2 and b = 4. But then e(O) = 31/16, so that 24/e(O) ∉ ℕ.

Suppose that L_{1}, L_{2}, L_{3} have a common point and L_{1}, L_{2} are tangent to Q. For the triple point to be orbifold point one must have: β_{1} + β_{2} + β_{3} > 1. This occurs only in (iii), which forces b_{1} = b_{2} = b_{3} = 2 and b = 4. But then O is the orbifold J_{2}.

Suppose that L_{1}, L_{2}, L_{3} have no point in common, L_{1}, L_{2} are tangent to Q and L_{3} is in general position with respect to Q. Then

e(O) = 2 − (β_{1} + β_{2} + 2β_{3} + 2β) + β_{1}β_{2} + β_{2}β_{3} + β_{3}β_{1} + 2ββ_{3}+

\frac{1}{2} [β + β_{1} − \frac{1}{2}]^{2}+\frac{1}{2} [β + β_{2} − \frac{1}{2}]^{2}

For the tangence points to be orbifold singularities one must have: β +β_{1} > 1/2 and β +β_{2} > 1/2. The number 24/e(O) is an integer only for 4 four-tuples (b_{1}, b_{2}, b_{3}, b): The four-tuples (3, 3, 3, 2) and (4, 4, 2, 2) yield the K3 orbifolds L_{1} and L_{2}, respectively. The remaining two orbifolds are not uniformizable: Indeed, for (3, 6, 2, 2), one should have |π_{1}^{orb}(O)| = 24/e(O) = 64, whereas a meridian of L_{1} in this group should be of order 3. For (2, 6, 3, 2) one has the presentation

(3) π_{1}^{orb}(L_{1}) ≃ ⟨κ, τ_{1}, τ_{2}, τ_{3} | \begin{matrix}(κτ_{1})^{2} = (τ_{1}κ)^{2} \\ [τ_{3}, κ] = [τ_{3}, τ_{1}] = [τ_{3}, τ_{2}] = 1 \\ τ_{3}τ_{2}τ_{1}κ^{2} = 1 \\ τ_{1}^{2} = τ_{2}^{6} = τ_{3}^{3} = κ^{2} = 1 \end{matrix} ⟩

This group is of order 24. But then the corresponding orbifold is not uniformizable since 24/e(O) = 54 ≠ |π_{1}^{orb}(O)|.

Suppose that L_{1}, L_{2}, L_{3} have no point in common, L_{1}, L_{2} are tangent to Q, the line L_{3} pass through L_{1} ∩ Q and in general position with respect to L_{2} ∪ Q. Then e(O) is

1−(β_{2}+β_{3}+β)+β_{1}β_{2}+β_{2}β_{3}+ββ_{3}+\frac{1}{2} [β + β_{1} + \frac{β_{3}}{2} − 1]^{2}+\frac{1}{2} [β + β_{2} − \frac{1}{2}]^{2}

For the tangence points to be orbifold points one must have: β + β_{1} + β_{3}/2 > 1 and β + β_{2} > 1/2. The number 24/e(O) is never an integer.

Let B = b_{1}L_{1} + b_{2}L_{2} + b_{3}L_{3} + bQ, where L_{1}, L_{2}, L_{3} have no point in common, L_{1}, L_{2} are tangent to Q and L_{3} pass through the tangence points. Then

e(O) = β_{1}β_{2} + \frac{1}{2} [β + β_{1} + \frac{β_{3}}{2} − 1]^{2} + \frac{1}{2} [β + β_{2} + \frac{β_{3}}{2} − 1]^{2}

For the tangence points to be orbifold singularities one must have: β + β_{1} + β_{3}/2 > 1 and β + β_{2} + β_{3}/2 > 1. The number 24/e(O) is never an integer.

Let B = b_{1}L_{1} + b_{2}L_{2} + b_{3}L_{3} + bQ, where L_{1}, L_{2}, L_{3} are tangent to Q. This case was settled in [28]. It yields the orbifolds M_{1} and M_{2}.

4.10.4. Two quadrics with a line. Let B = bL + b_{1}Q_{1} + b_{2}Q_{2}. Since β + 2β_{1} + 2β_{2} = 2, one has (b, b_{1}, b_{2}) ∈ {(2, 2, 4), (3, 2, 3)}.

Suppose that Q_{1}, Q_{2} have an intersection point of multiplicity 4. The point Q_{1} ∩ Q_{2} is an orbifold point only when Q_{1} ∩ Q_{2} ∉ L and (b_{1}, b_{2}) is one of (2, 2), (3, 2) or (2, 3). Since β + 2β_{1} + 2β_{2} = 2, the case (b_{1}, b_{2}) = (2, 2) cannot occur. This leaves the possibilities (3, 2, 3) and (3, 3, 2) for (b, b_{1}, b_{2}). If L is in general position with respect to Q_{1}∪Q_{2}, then for (b, b_{1}, b_{2}) = (3, 3, 2) one has 24/e(O) = 384/17 ∉ ℕ. On the other hand, if L is tangent to Q_{1}, then for the triple (3, 2, 3) one has e(O) = −7/144, whereas for the triple (3, 3, 2) one has e(O) = 25/48.

Suppose that Q_{1} and Q_{2} are tangent at two distinct points. If L is in general position with respect to Q_{1} ∪ Q_{2}, then

e(O) = 3 − 2(β + β_{1} + β_{2}) + 2β(β_{1} + β_{2}) + [β_{1} + β_{2} − \frac{1}{2}]^{2}

Since β + 2(β_{1} + β_{2}) = 2, this becomes e(O) = (5 + 2β − 3β^{2})/4. Recall that b can only take the values 2 and 3. For b = 2, one has 24/e(O) = 384/21 ∉ ℕ. For b = 3 one has 24/e(O) = 18. The corresponding triple is (b, b_{1}, b_{2}) = (3, 3, 2), which gives the orbifold I_{1}.

If L is tangent to Q_{1} and in general position with respect to Q_{2}, then

e(O) = 2 − (β + β_{1} + 2β_{2}) + 2ββ_{2} + \frac{1}{2} [β + β_{1} − \frac{1}{2}]^{2} + [β_{1} + β_{2} − \frac{1}{2}]^{2} ,

and 24/e(O) is never an integer. If L intersects Q_{1}∪Q_{2} transversally at one or both of the points of Q_{1} ∩Q_{2}, then this point is an orbifold point only if β/2+β_{1}+β_{2} > 1, which contradicts the condition β+2β_{1}+2β_{2} = 2. If L is tangent to Q_{1} ∪ Q_{2} at one of the points Q_{1} ∩ Q_{2}, then this point is an orbifold point only if b = b_{1} = b_{2} = 2. But then one can not have β + 2β_{1} + 2β_{2} = 2.

4.10.5. A cubic with two lines. Let B := bC + b_{1}L_{1} + b_{2}L_{2}, where C is a cubic curve. The condition 3β + β_{1} + β_{2} = 2 implies that (b, b_{1}, b_{2}) is one of (4, 4, 2), (2, 2, 3) or (3, 6, 2). In this subsection, we shall adopt Degtyarev’s notation for quintic curves [4] and give only the euler numbers of the orbifolds. Non of these orbifolds are uniformizable.

The quintic C_{3}(A_{2}) ⊔ {×3} ⊔ {×2, ×1} : The euler number e(O) is

1−(2β+β_{2})+β_{2}(β+β_{1})+\frac{3}{2} [β − \frac{1}{6}]^{2}+\frac{1}{2} [β_{1} + β − \frac{1}{2}]^{2}+\frac{3}{4} [β_{2} + β − \frac{2}{3}]^{2}

The quintic C_{3}(A_{2}) ⊔ {A_{2}^{∗}} ⊔ {×3} : The line L_{1} is tangent to C_{3}(A_{2}) at the cusp A_{2}. This is an orbifold point only if b = b_{1} = 2. But then one cannot have 3β + β_{1} + β_{2} = 2.

The quintics C_{3}(A_{2})⊔\{\underline{×3}\}⊔\{A_{2}, \underline{×1}\} and C_{3}(A_{2})⊔\{\underline{×3}\}⊔\{\underline{×1}, ×1, ×1\}: The flex point of C_{3}(A_{2}) is an orbifold point only if β + β_{1} + β_{2}/3 > 1. But then one cannot have 3β + β_{1} + β_{2} = 2.

The quintic C_{3}(A_{2}) ⊔ {×3} ⊔ {A_{2}, ×1} : Euler number e(O) is

e(O) = 1−(β_{2}+β)+β_{2}(β_{1}+β)+ \frac{3}{4} [β + β_{1} − \frac{2}{3}]^{2}+ \frac{3}{2} [β + β_{2}/3 − \frac{1}{2}]^{2}

The quintic C_{3}(A_{2}) ⊔ {×3} ⊔ {×1, ×1, ×1} : Euler number e(O) is

e(O) = 2 − (2β_{2} + 3β) + β_{2}(3β + β_{1}) + \frac{3}{2} [β − \frac{1}{6}]^{2} + \frac{3}{4} [β + β_{1} − \frac{2}{3}]^{2}

The quintic C_{3}(A_{2}) ⊔ \{\underline{×2}, ×1\} ⊔ \{×2, \underline{×1}\} : Euler number e(O) is

e(O) = 1−2β+β_{1}β+\frac{3}{2} [β − \frac{1}{6}]^{2}+\frac{1}{2} [β + β_{2} − \frac{1}{2}]^{2}+\frac{1}{2} [β + β_{2}/2 + β_{1} − 1]^{2}

The quintic C_{3}(A_{2}) ⊔ \{A_{2}, \underline{×1}\} ⊔ \{×2, \underline{×1}\} : Euler number e(O) is

e(O) = 1−β+\frac{1}{2} [β + β_{2} − \frac{1}{2}]^{2}+\frac{1}{4} [β + β_{2} + β_{1} − 1]^{2}+\frac{3}{2} [β + β_{1}/3 − \frac{1}{2}]^{2}

The cusp of C is an orbifold point only if β + β_{1}/3 > \frac{1}{2}, which forces b = 2. But then β_{1} + β_{2} + 3β = 2 implies that β + β_{1} + β_{2} = 1. On the other hand, the triple point is an orbifold point only if β + β_{1} + β_{2} > 1.

The quintic C_{3}(A_{1}) ⊔ {×3} ⊔ {×3} : Euler number e(O) is

e(O) = 1 − 2β + β^{2} + β_{1}β_{2} + \frac{3}{4} [β + β_{1} − \frac{2}{3}]^{2} + \frac{3}{4} [β + β_{2} − \frac{2}{3}]^{2}

The quintic C_{3}(A_{1}) ⊔ \{\underline{×3}\} ⊔ \{×2, \underline{×1}\} : The point of intersection of the lines is an orbifold point only if β = β_{1} = 2. But then β_{1} + 3β = 2, so that one cannot have β_{1} + β_{2} + 3β = 2.

The quintic C_{3}(A_{1}) ⊔ \{×2, \underline{×1}\} ⊔ \{×2, \underline{×1}\} : Euler number e(O) is

2−3β +β^{2} + \frac{1}{2} [β + β_{2} − \frac{1}{2}]^{2} + \frac{1}{2} [β + β_{1} − \frac{1}{2}]^{2} + \frac{1}{4} [β + β_{2} + β_{1} − 1]^{2}

4.10.6. A cubic with a quadric. Let B = b_{1}Q + b_{2}C. Then (b_{1}, b_{2}) is one of (4, 2) or (2, 3). Suppose that Q intersects C at two points of multiplicity 3 at each. Then

e(O) = 1 − β_{2} + \frac{3}{2} [β_{2} − \frac{1}{6}]^{2} + \frac{3}{2} [β_{1} + β_{2} − \frac{2}{3}]^{2}

For (β_{1}, β_{2}) = (4, 2) one has e(O) = 65/96. For (b_{1}, b_{2}) = (2, 3) one has 24/e(O) = 32. But if |π_{1}^{orb}(O)| = 32 then this group cannot contain elements of order 3. Hence, this orbifold is not uniformizable neither.

Acknowledgements. This work was partially realized at Max Planck Institute für Mathematik in Bonn in 2003. The author is grateful to the institute for providing the opportunity to study in its stimulating atmosphere.

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Galatasaray University, Department of Mathematics, Ortaköy/İstanbul, Turkey e-mail adress: muludag@gsu.edu.tr