Finiteness results for K3 surfaces over arbitrary fields
Abstract.
Over an algebraically closed field, various finiteness results are known regarding the automorphism group of a K3 surface and the action of the automorphisms on the Picard lattice. We formulate and prove versions of these results over arbitrary base fields, and give examples illustrating how behaviour can differ from the algebraically closed case.
Key words and phrases:
K3 surfaces; automorphism groups2010 Mathematics Subject Classification:
Primary 14J28; Secondary 14J50, 14G271. Introduction
The geometry of K3 surfaces over the complex numbers has a long history, with many results known about the cohomology, the Picard group, and the automorphism group of an algebraic K3 surface, and how these objects interact. Such results over the complex numbers carry over to other algebraically closed fields of characteristic zero, and similar results are also known over algebraically closed fields of other characteristics. For a comprehensive treatment of the geometry of K3 surfaces, we refer the reader to the lecture notes of Huybrechts [12]. Much of the theory we use was originally developed by Nikulin [15].
K3 surfaces are also interesting from an arithmetic point of view, with much recent work on understanding the rational points, curves, Brauer groups and other invariants of K3 surfaces over number fields. In this article, we investigate the extent to which some standard finiteness results for K3 surfaces over algebraically closed fields remain true over more general base fields. In particular, we show how to define the correct analogue of the Weyl group, and give an explicit description of it. This allows us to formulate and prove finiteness theorems over arbitrary fields, modelled on those already known over algebraically closed fields. The tools we use include representability of the Picard and automorphism schemes, classification of transitive group actions on Coxeter–Dynkin diagrams, and an explicit description of the walls of the ample cone. We follow these theoretical results with several detailed examples, showing how the relationship between the Picard group and the automorphism group can be different from the geometric case. We end by proving that a surface over of the form has finite automorphism group for that is not in the subgroup generated by squares together with .
The specific finiteness results we address go back to Sterk [20]. To state them, we need some definitions; we follow the notation of [12]. In this article, by a K3 surface we will always mean an algebraic K3 surface, which is therefore projective.
Let be a field, and let be a K3 surface over . Denote the group of isometries of by . Reflection in any class in defines an isometry of , and we define the Weyl group to be the subgroup generated by these reflections.
We recall the definitions of the positive, ample and nef cones associated to the K3 surface ; see also [12, Chapter 8]. Let denote the real vector space . By the Hodge index theorem, the intersection product on has signature ; so the set consists of two connected components. The positive cone is the connected component containing all the ample classes.
The ample cone is the cone in generated by all classes of ample line bundles. The nef cone is defined as
Finally, we define to be the real convex hull of . An application of the criterion of Nakai–Moishezon–Kleiman shows that is the interior of and is the closure of : see [12, Corollary 8.1.4].
The following finiteness theorems are due to Sterk [20] for and to Lieblich and Maulik [13] when has positive characteristic not equal to . As in Huybrechts [12, Chapter 8], a fundamental domain for the action of a discrete group acting continuously on a topological manifold is defined as the closure of an open subset such that and such that for the intersection does not contain interior points of or .
Theorem 1.1.
Let be an algebraically closed field of characteristic not equal to , and let be a K3 surface over .

[12, Corollary 8.2.11] The cone is a fundamental domain for the action of on the positive cone .

[12, Theorem 8.4.2] The action of on admits a rational polyhedral fundamental domain.

[12, Corollary 8.4.6] The set of orbits under of curves on is finite. More generally, for any there are only finitely many orbits under of classes of irreducible curves of selfintersection .
In Section 3 we will prove analogues of the various statements of Theorem 1.1 when is replaced by an arbitrary base field of characteristic different from .
A consequence of Theorem 1.1(2) is that the finiteness of depends only on the lattice . Those possible Picard lattices for which is finite have been classified [14]. Over an arbitrary base field we will see that, instead of using the Weyl group , we must use the Galoisinvariant part of the geometric Weyl group. This means that the finiteness of is no longer determined purely by the Picard lattice ; rather, it depends on the geometric Picard lattice together with the Galois action. In Section 4, we give several examples that illustrate this difference to the classical case.
We thank the referee, whose thorough report helped us improve the exposition and correct a number of errors in the paper. The second author would like to thank the Tutte Institute for Mathematics and Computation for its partial support for a visit to the University of Leiden during which much of this research was done.
2. Lemmas on lattices
In this section we will study lattices with the action of a group. Given a lattice with the action of a finite group , we consider the group of automorphisms that commute with , and the group of automorphisms that preserve the sublattice fixed by .
Definition 2.1.
A lattice is a free abelian group of finite rank with a nondegenerate integervalued symmetric bilinear form. If the form is (positive or negative) definite, we likewise refer to as definite. The group of automorphisms of preserving the form is denoted . A sublattice of is a subgroup on which the restriction of the form is nondegenerate. Given a sublattice , we use for the subgroup of fixing as a set. For a subgroup , let be the subgroup of consisting of the elements fixed by every element of (note that according to our conventions may not be a lattice, because the quadratic form on may be degenerate when restricted to ). The vector space will be denoted .
Our goal is to prove the following proposition.
Proposition 2.2.
Let be a lattice and a subgroup such that is a lattice. Then the following hold:

the natural map has image of finite index;

suppose that is definite. Then has finite kernel, and the centralizer is a finiteindex subgroup of .
Our interest in this situation arises from the geometry of K3 surfaces. Let be a projective K3 surface defined over a field , and let be a Galois extension. Let with the intersection pairing, and let be the image of in . If has a rational point over , we have . (See Section 3 for more details. This statement holds slightly more generally: for example, if is a number field and has points everywhere locally over .) The Hodge index theorem states that has signature and has signature : therefore is definite.
Before giving the proof we first collect a few helpful statements, which are probably well known.
Lemma 2.3.
Let be a lattice and a sublattice. Let be a subgroup of such that is a lattice. For groups , let denote the centralizer of in .

is a sublattice of .

There is a natural injection with image of finite index.

is contained in .

The kernel of the map is contained in .
Proof.
To prove (1), we just have to prove that the pairing on the subspace is nondegenerate; this is [7, Satz 1.2].
Let . By definition restricts to an endomorphism of . Let be the saturation of in . Clearly . Now, has the same rank as and contains , so it is equal to . So if , then the image of does not contain , contradicting the hypothesis that is an automorphism of . Since is a subgroup of finite index of , this implies that . But , so it follows that . Thus there is a map . Now, if and , then for all , so , and we get a map in the same way. Combining these two maps gives a map .
If , then is the identity on , which is a subgroup of of finite index. Because is torsionfree, this forces to be the identity. To show that has finite image in , et be the smallest positive integer such that , and let . Every element of that fixes as a set is in the image of , because the induced automorphism of extends to an automorphism of with the same action on . Since there are only finitely many subgroups of index in , the stabilizer of is of finite index, and we have proved (2).
To prove (3), let , and let and . Then and . So , establishing that .
Finally we prove (4). Choose in the kernel and , and let . We will view and as automorphisms of . In we may write , where and . Then However, , so , and is in the kernel of the map to , so . It follows that . Similarly, , establishing that commutes with . ∎
Proof of Proposition 2.2.
The map is a composition . In part (2) of the lemma just proved we showed that the first map has image of finite index. The second map is surjective, so the composition has image of finite index as well.
We now suppose that is definite to prove the second statement. Then is finite, so is a composition of an injective map with a map with finite kernel and so its kernel is finite. Let . Then has finite index in , because both inject into the finite group . Therefore has finite index in too. ∎
3. Finiteness results for K3 surfaces
In this section we formulate and prove analogues of the statements of Theorem 1.1 when is an arbitrary field. We first look at the case of separably closed, which is straightforward.
Lemma 3.1.
Let be a separably closed field, and let be an algebraic closure of . Let be a K3 surface over , and let be the base change of to . Then the natural maps and are isomorphisms.
Proof.
As is projective, the Picard scheme exists, is separated and locally of finite type over , and represents the sheaf , which is defined to be the sheafification on the big étale site over of the presheaf
(see [9, Theorem 9.4.8]). In particular, because and are both separably closed, we have and . From it follows from [9, Theorem 9.5.11] that is étale over . So every point of is defined over , and is an isomorphism.
The functor taking a scheme to the group is represented by a scheme : see [9, Theorem 5.23]. A standard argument in deformation theory shows that the tangent space at the identity element is isomorphic to , where denotes the tangent sheaf on . Indeed, an element of the tangent space is given by a morphism extending the morphism sending to the identity automorphism. Such a morphism corresponds to an automorphism of restricting to the identity on the central fibre. By [9, Theorem 8.5.9], the set of these morphisms forms an affine space under . In our case, the group is zero [12, Theorem 9.5.1], so the scheme is étale over , and is an isomorphism. ∎
Corollary 3.2.
In the situation of Lemma 3.1, every curve on is defined over .
Proof.
Let be a curve on . Then Lemma 3.1 shows that there is a line bundle on whose base change to is isomorphic to . The Riemann–Roch theorem and flat base change give . So all nonzero sections of cut out the same divisor and the base change of to must coincide with . In other words, is defined over . ∎
We now pass to the case of a general field. Let be a field; fix an algebraic closure of , and let be the separable closure of in . Let be a K3 surface over , and let and denote the base changes of to and , respectively. Write .
The group acts on preserving intersection numbers, giving a representation . Let act on by conjugation, that is, such that for all . For a class , denote the reflection in by ; then we have . So the action of on restricts to an action on .
Definition 3.3.
Define to be the group .
Recall that is contained in, but not necessarily equal to, the fixed subgroup . The Hochschild–Serre spectral sequence gives rise to an exact sequence
If has a point, then evaluation at that point gives a left inverse to , showing that is an isomorphism. In general this does not have to be true. However, because is torsion and is finitely generated, the above sequence shows that is of finite index in .
It is easy to see that the action of on preserves , but it is not immediately obvious that this action preserves . To show that this is the case, we use an explicit description of provided by a theorem of Hée and Lusztig, for which Geck and Iancu gave a simple proof. Before stating their theorem, we establish some notation and conventions for Coxeter systems.
Definition 3.4.
Let be a group generated by a set of elements of order . For , let be the order of if has finite order and otherwise. Suppose that the relations for with are a presentation of . Then is a Coxeter system.
Let be a graph with vertices and such that are adjacent in if and only if does not commute with ; in this case, label the edge joining to with for and otherwise. We refer to as the Coxeter–Dynkin diagram of . The Coxeter system is said to be irreducible if its Coxeter–Dynkin diagram is connected.
Let the length of an element be the length of a shortest word in the that represents it. If is finite, there is such that for all (see [2, Proposition 2.3.1]). We refer to as the longest element of .
Let be a permutation of . Then there is at most one way to extend to a homomorphism , because generates . If there is such an extension, it is an automorphism, because extends to its inverse, and we speak of it as the automorphism induced by .
If is a Coxeter system, and is a subset of , let denote the subgroup of generated by the elements of . Then is a Coxeter system: see [2, Proposition 2.4.1 (i)].
Theorem 3.5 ([11], Theorem 1).
Let be a Coxeter system. Let be a group of permutations of that induce automorphisms of . Let be the set of orbits for which is finite, and for let be the longest element of . Then is a Coxeter system.
We will apply this theorem with and being the set of reflections in curves on .
Proposition 3.6.
Let be the set of Galois orbits of curves on of the following two types:

consists of disjoint curves;

consists of disjoint pairs of curves, each pair having intersection number 1.
Then the following statements hold.

For each , let be the subgroup of generated by reflections in the classes of curves in , and let be the longest element of the Coxeter system . Then is a Coxeter system.

For each , let be the sum of the classes in . Then acts on as reflection in the class .

The action of on preserves .
Proof.
Let be a Galois orbit of curves, and suppose that the subgroup is finite. We will show that is of one of the two types described. Firstly, no two curves in have intersection number greater than , for then the corresponding reflections would generate an infinite dihedral subgroup of . Since is finite, its Coxeter–Dynkin diagram is a finite union of trees [2, Exercise 1.4]. In particular, it contains a vertex of degree . However, the Galois group acts transitively on the diagram, so we conclude that either every vertex has degree , or every vertex has degree . These two possibilities correspond to the two types of orbits described. Now (1) follows from Theorem 3.5.
We prove (2) separately for the two types of orbits. In the first case we have . The reflections in the all commute, so is isomorphic to the Coxeter group . The longest element is . For , the intersection numbers are all equal, and one calculates
that is, coincides on with reflection in the class , of selfintersection .
In the second case, write , where and the other intersection numbers are all zero. The two reflections and generate a subgroup isomorphic to the Coxeter group , in which the longest element is
Thus we have and the longest element in is the product of the longest elements in the factors , that is, it equals . For , all the and are equal, and we have
which coincides with reflection of in the class , of selfintersection .
Finally, each class is, by construction, the class of a Galoisfixed divisor on , so lies in . So in both cases the formula for given above shows that reflection in preserves , and therefore the action of preserves , proving (3). ∎
We now turn to the ample and nef cones. As ampleness is a geometric property, and the nef cone is the closure of the ample cone over any base field, it follows that
the intersections taking place inside . The following result is well known when is algebraically closed (see [12, Corollary 8.2.11]), and descends easily to arbitrary .
Proposition 3.7.
Let be a K3 surface over . The cone is a fundamental domain for the action of on the positive cone , and this action is faithful.
Proof.
We will prove two things: first, that every class in is equivalent to an element of ; and second, that the translates of by two distinct elements of meet only along their boundaries. The second of these shows in particular that the action is faithful. (When we refer to the boundary of or one of its translates, we mean the boundary within . The boundary of in is just the boundary of intersected with , so that distinction is not so important.)
To prove the first statement, let . Suppose first that has trivial stabilizer in . Then, by [12, Corollary 8.2.11], there exists a unique such that lies in the interior of . We claim that lies in . For any , we have
since the Galois action preserves the properties of being nef and positive. By uniqueness of , we conclude that , that is, lies in . It then follows that lies in and therefore in .
Now suppose that has nontrivial stabilizer. Then lies on at least one of the walls defined by the action of on ; see [12, Section 8.2]. By [12, Proposition 8.2.4], the chamber structure of this group action is locally polyhedral within , so a small enough neighbourhood of meets only finitely many chambers. Also note that is not contained in any of the walls, because admits an ample divisor. This allows us to construct a sequence of elements of , tending to and all lying in the interior of the same chamber of . As in the previous paragraph, there is a unique satisfying for all . By continuity, also lies in .
We now prove the second statement, that intersects the translate by any nontrivial element of only in its boundary. Suppose that lies in the intersection , for some nontrivial . By [12, Corollary 8.2.11], we see that lies in the boundary of . The following lemma shows that lies in the boundary of . ∎
Lemma 3.8.
Let be a real vector space, let be a closed convex cone, and let be a subspace having nonempty intersection with the interior of . Then we have
Proof.
The dual cone is defined by
By the supporting hyperplane theorem, has the property that a point lies in if and only if there exists a nonzero satisfying . (The hyperplane is called a supporting hyperplane of at .) Let be the natural restriction map; then the dual is equal to (see [16, Corollary 16.3.2]).
Let be a point of . Then there is a supporting hyperplane to at , that is, there exists a nonzero satisfying . The condition that meet the interior of implies that does not vanish identically on , so is nonzero. Thus is a nonzero element of vanishing at , so lies in .
Conversely, suppose that lies in . Then there is a supporting hyperplane to at , that is, there exists a nonzero satisfying . Let satisfy ; then we have and so . ∎
Remark 3.9.
We can also give an explicit description of the walls of . According to [12, Corollary 8.1.6], a class in lies in if and only if it has nonnegative intersection number with every curve on , or, equivalently, with every Galois orbit of curves. The question is to determine which Galois orbits are superfluous, and which actually define walls of .
Let be a Galois orbit of curves on , and suppose that the subgroup generated by reflections in the elements of is finite, that is, is as described in Proposition 3.6. The longest element of acts on by reflection in the class , which has negative selfintersection. The hyperplane orthogonal to is a wall of : by the same argument as in [20], given a class , the class
is orthogonal to but has positive intersection number with all curves outside .
On the other hand, let be a Galois orbit of curves such that the subgroup is infinite. Then, for a curve , the hyperplane orthogonal to in does not meet , as the following argument shows. Let be a class in orthogonal to ; then is also orthogonal to all the other curves in , and so is fixed by . Therefore is also orthogonal to the infinitely many images of under the action of , contradicting the fact that the chamber structure induced by on is locally polyhedral.
Next we would like to prove an analogue of Theorem 1.1 (2). However, while the authors do not have an explicit counterexample, there seems to be no reason for the image of in to be normal. Instead, we will see that there is a natural homomorphism from a semidirect product to having finite kernel and image of finite index.
Note that the natural action of on by conjugation fixes and commutes with the Galois action, so fixes . This gives an action of on , and a homomorphism from the associated semidirect product to . Since and both fix , we also obtain a homomorphism .
Proposition 3.10.
Let be a K3 surface over a field of characteristic different from . Then the natural map
has finite kernel and image of finite index.
Remark 3.11.
In the literature, this statement appears in various different but equivalent forms.

Because the action of fixes the ample cone, the image of in meets only in the identity element. Therefore the finiteness of the kernel in Proposition 3.10 is equivalent to the finiteness of the kernel of .

Let be the subgroup of symplectic automorphisms [12, Definition 15.1.1]. Over an algebraically closed field of characteristic zero, the induced map from to is injective, so that can be viewed as a subgroup of . Instead of our Proposition 3.10, Huybrechts [12, Theorem 15.2.6] makes the statement that has finite index in . This is equivalent to our formulation, because is of finite index in . However, when is not algebraically closed there is no reason for to be injective, so there is no advantage to stating the result in terms of .

Lieblich and Maulik [13, Theorem 6.1] define to be the subgroup of elements preserving the ample cone, and then show that has finite kernel and cokernel. Over an algebraically closed base field , the subgroups and generate : any element of permutes the classes in and therefore takes the ample cone to one of its translates under ; so composing with an element of gives an element of . Therefore that condition is also equivalent to the condition of Proposition 3.10.
Before proving Proposition 3.10, we first state two lemmas which are well known in the case of abelian groups.
Lemma 3.12.
Let be a finite group, and a homomorphism of (possibly noncommutative) modules having finite kernel and image of finite index. Then the induced homomorphism also has finite kernel and image of finite index.
Proof.
The kernel of is contained in the kernel of , so is finite. For the statement about the image, consider the short exact sequence
The associated exact sequence of cohomology gives
and is a finite set, showing that is of finite index in . On the other hand, we claim that is of finite index in . Indeed, by hypothesis is of finite index in , and this property is preserved on intersecting with the subgroup . Thus is of finite index in . ∎
The following lemma is standard and easy to prove; we state it for reference.
Lemma 3.13.
Let and be two homomorphisms of groups.

If and both have finite kernel and image of finite index, then so does the composition .

If has finite kernel and image of finite index, and has finite kernel, then has finite kernel and image of finite index.

If has finite kernel and image of finite index, and has image of finite index, then has finite kernel and image of finite index.
Proof.
We leave this as an exercise for the reader. ∎
Proof of Proposition 3.10.
By [12, Theorem 15.2.6] in characteristic zero, or [13] in characteristic , the natural map
has finite kernel and image of finite index. Lemma 3.1 shows that the same is true for . The action of on all of these groups factors through a finite quotient, so Lemma 3.12 shows that the induced homomorphism
(1) 
also has finite kernel and image of finite index. Because the automorphism group functor is representable, we have .
There is an exact sequence
and the homomorphism (1) factors through , and hence through the injective map
Therefore, by Lemma 3.13, the map
has finite kernel and image of finite index. Since the image of in meets only in the identity element, this shows that the natural map
(2) 
also has finite kernel and image of finite index.
We apply Proposition 2.2 with and being the image of in . The centralizer is then , so part (2) of Proposition 2.2 shows that is of finite index in . Parts (1) and (2) of Proposition 2.2 combined show that the map
has finite kernel and image of finite index; by Lemma 3.13 so does the map .
Composing with (2) and applying Lemma 3.13 shows that
has finite kernel and image of finite index. Since the actions of both and on preserve , this last map factors as
(3) 
As the second map in this composition is clearly injective, Lemma 3.13 shows that the first one has finite kernel and image of finite index.
Finally, is of finite index in , so its orthogonal complement in the latter is trivial. Lemma 2.3(2) shows that the natural map
is injective, and its image has finite index. Composing with the first map of (3) and applying Lemma 3.13 again, we deduce that has finite kernel and image of finite index. ∎
Remark 3.14.
The finiteness of the kernel can also be proved directly, by the same proof as over an algebraically closed field.
Having proved Proposition 3.10, we can deduce the remaining results exactly as in the classical case. Define the cone to be the real convex hull of .
Corollary 3.15.
The action of on admits a rational polyhedral fundamental domain.
Proof.
This is as in the case of ; we briefly recall the argument of Sterk [20], making the necessary adjustments.
Let be a lattice of signature and let be an arithmetic subgroup (for example, a subgroup of finite index in ). Let be one of the two components of ; this is a selfadjoint homogeneous cone [1, Remark 1.11]. Let be the convex hull of . Pick any ; the argument of [20, p. 511] shows that the set
is rational polyhedral (for this, the reference to Proposition 11 of the first edition of [1] has become Proposition 5.22 in the second edition) and is a fundamental domain for the action of on . (Sterk does not explicitly prove that two translates of intersect only in their boundaries, but this is easy to show from the description above.) In our case, applying this with and being the image of gives a rational polyhedral fundamental domain for the action of on .
If we choose to be an ample class in , then the resulting is contained in , as we now show. Let be a Galois orbit of curves on such that the corresponding group is finite. Proposition 3.6 states that the longest element of acts on as reflection in the class , and that these elements generate . Taking in the definition of shows that is contained in the halfspace . As this holds for all such , Remark 3.9 shows that is contained in .
We conclude as in [20]. If is a class in then, since is a fundamental domain for the action of on , we can find and such that lies in . But now and both lie in , so they are equal and lie in . This shows that is a fundamental domain for the action of on . ∎
Corollary 3.16.

There are only finitely many orbits of rational curves on .

There are only finitely many orbits of primitive Picard classes of irreducible curves on of arithmetic genus .

For , there are only finitely many orbits of Picard classes of irreducible curves on of arithmetic genus .
Proof.
Let be a rational polyhedral fundamental domain for the action of on , as in Corollary 3.15. Every rational curve on defines a wall of , by Remark 3.9. Since meets only finitely many walls of , this proves (1).
Gordan’s Lemma states that the integral points of the dual cone of a rational polyhedral convex cone form a finitely generated monoid. Applying this to the dual cone of , let be a minimal set of generators for . Since these all lie in , we have for all , and for . As observed in [20], this implies that, for any , there are only finitely many classes in of selfintersection ; and there are only finitely many primitive classes in of selfintersection zero. The class of an irreducible curve of arithmetic genus has selfintersection and therefore lies in , so this proves (2) and (3). ∎
Remark 3.17.
It is not true that every irreducible curve on of arithmetic genus is a rational curve. However, there are not many possibilities, as we now show. Let be such a curve, and let be the geometric components of ; the Galois group acts transitively on them. In order to achieve , we must have for all , so each is a curve. Consider the intersection matrix : the sum of the entries in the matrix is , and the Galois action shows that every row sum is the same. Therefore the number of rows divides , and there are only two options: , so that is a rational curve; or