Finiteness results for K3 surfaces over arbitrary fields
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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.
Keywords
K3 surfaces Automorphism groups Picard groups Nonalgebraically closed fieldsMathematics Subject Classification
14J28 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 \(\mathbb {Q}\) of the form \(x^4y^4 = c(z^4w^4)\) has finite automorphism group for \(c \in \mathbb {Q}^*\) that is not in the subgroup generated by squares together with \(1, 2\).
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 k be a field, and let X be a K3 surface over k. Denote the group of isometries of \(\mathrm{Pic\,}X\) by \({\mathrm {O}}(\mathrm{Pic\,}X)\). Reflection in any \((2)\)class in \(\mathrm{Pic\,}X\) defines an isometry of \(\mathrm{Pic\,}X\), and we define the Weyl group\(\mathrm{W}(\mathrm{Pic\,}X) \subset {\mathrm {O}}(\mathrm{Pic\,}X)\) to be the subgroup generated by these reflections.
We recall the definitions of the positive, ample and nef cones associated to the K3 surface X; see also [12, Chapter 8]. Let \((\mathrm{Pic\,}X)_\mathbb {R}\) denote the real vector space Open image in new window . By the Hodge index theorem, the intersection product on \((\mathrm{Pic\,}X)_\mathbb {R}\) has signature \((1, \rho 1)\); so the set Open image in new window consists of two connected components. The positive cone Open image in new window is the connected component containing all the ample classes.
The following finiteness theorems are due to Sterk [20] for \(k=\mathbb {C}\) and to Lieblich and Maulik [13] when k has positive characteristic not equal to 2. As in Huybrechts [12, Chapter 8], a fundamental domain for the action of a discrete group G acting continuously on a topological manifold M is defined as the closure \(\overline{U}\) of an open subset \(U\subset M\) such that \(M = \bigcup _{g \in G} g \overline{U}\) and such that for \(g \ne h \in G\) the intersection \(g\overline{U} \cap h\overline{U}\) does not contain interior points of gU or hU.
Theorem 1.1
 (1)
([12, Corollary 8.2.11]) The cone \(\mathrm{Nef}(X) \cap {\mathscr {C}}_X\) is a fundamental domain for the action of \(\mathrm{W}(\mathrm{Pic\,}X) \subset {\mathrm {O}}(\mathrm{Pic\,}X)\) on the positive cone \({\mathscr {C}}_X\).
 (2)
([12, Theorem 15.2.6], [13, Proposition 5.2]) The subgroup \(\mathrm{W}(\mathrm{Pic\,}X)\) is normal in \({\mathrm {O}}(\mathrm{Pic\,}X)\); the natural map \(\mathrm{Aut}\,X \rightarrow {\mathrm {O}}(\mathrm{Pic\,}X)/\mathrm{W}(\mathrm{Pic\,}X)\) has finite kernel and image of finite index.
 (3)
([12, Theorem 8.4.2]) The action of \(\mathrm{Aut}\,X\) on \(\mathrm{Nef}^{\,\mathrm e} X\) admits a rational polyhedral fundamental domain.
 (4)
([12, Corollary 8.4.6]) The set of orbits under \(\mathrm{Aut}\,X\) of \((2)\)curves on X is finite. More generally, for any d there are only finitely many orbits under \(\mathrm{Aut}\,X\) of classes of irreducible curves of selfintersection 2d.
In Sect. 3 we will prove analogues of the various statements of Theorem 1.1 when k is replaced by an arbitrary base field of characteristic different from 2.
A consequence of Theorem 1.1 (2) is that the finiteness of \(\mathrm{Aut}\,X\) depends only on \(\mathrm{Pic\,}X\). Those possible Picard lattices for which the quotient \({\mathrm {O}}(\mathrm{Pic\,}X)/\mathrm{W}(\mathrm{Pic\,}X)\) is finite have been classified in [14]. Over an arbitrary base field we will see that, instead of using the Weyl group \(\mathrm{W}(\mathrm{Pic\,}X)\), we must use the Galoisinvariant part of the geometric Weyl group. This means that the finiteness of \(\mathrm{Aut}\,X\) is no longer determined purely by the Picard lattice \(\mathrm{Pic\,}X\); rather, it depends on the geometric Picard lattice together with the Galois action. In Sect. 4, we give several examples that illustrate this difference to the classical case.
2 Lemmas on lattices
In this section we will study lattices with the action of a group. Given a lattice \(\Lambda \) with the action of a finite group H, we consider the group of automorphisms that commute with H, and the group of automorphisms that preserve the sublattice fixed by H.
Definition 2.1
A lattice\(\Lambda \) 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 \(\Lambda \) as definite. The group of automorphisms of \(\Lambda \) preserving the form is denoted \({\mathrm {O}}(\Lambda )\). A sublattice of \(\Lambda \) is a subgroup on which the restriction of the form is nondegenerate. Given a sublattice \(M \subseteq \Lambda \), we use \({\mathrm {O}}(\Lambda ,M)\) for the subgroup of \({\mathrm {O}}(\Lambda )\) fixing M as a set. For a subgroup \(H \subseteq {\mathrm {O}}(\Lambda )\), let Open image in new window be the subgroup of \(\Lambda \) consisting of the elements fixed by every element of H (note that according to our conventions Open image in new window may not be a lattice, because the quadratic form on \(\Lambda \) may be degenerate when restricted to Open image in new window ). The vector space Open image in new window will be denoted as \(\Lambda _\mathbb {Q}\).
Our goal is to prove the following proposition.
Proposition 2.2
 (1)
the natural map \({\mathrm {O}}(\Lambda ,M) \rightarrow {\mathrm {O}}(M)\) has image of finite index;
 (2)
suppose that \(M^{\perp }\) is definite. Then \({\mathrm {O}}(\Lambda ,M) \rightarrow {\mathrm {O}}(M)\) has finite kernel, and the centralizer \(Z_{{\mathrm {O}}(\Lambda )} H\) is a finiteindex subgroup of \({\mathrm {O}}(\Lambda ,M)\).
Our interest in this situation arises from the geometry of K3 surfaces. Let X be a projective K3 surface defined over a field F, and let K / F be a Galois extension. Let \(\Lambda = \mathrm{Pic\,}X_K\) with the intersection pairing, and let H be the image of \(\mathrm{Gal}(K/F)\) in \({\mathrm {O}}(\Lambda )\). If X has a rational point over F, we have Open image in new window (See Sect. 3 for more details. This statement holds slightly more generally: for example, if F is a number field and X has points everywhere locally over F.) The Hodge index theorem states that \(\Lambda \) has signature (1, n) and Open image in new window has signature (1, m): therefore \(M^{\perp }\) is definite.
Before giving the proof we first collect a few helpful statements, which are probably well known.
Lemma 2.3
 (1)
\(M^{\perp }\) is a sublattice of \(\Lambda \).
 (2)
There is a natural injection Open image in new window with image of finite index.
 (3)
\(Z_{{\mathrm {O}}(\Lambda )} H\) is contained in Open image in new window .
 (4)
The kernel of the map Open image in new window is contained in \(Z_{{\mathrm {O}}(\Lambda )} H\).
Proof
To prove (1), we just have to prove that the pairing on the subspace Open image in new window is nondegenerate; this is [7, Satz 1.2].
Let \(\phi \in {\mathrm {O}}(\Lambda ,M)\). By definition \(\phi \) restricts to an endomorphism of M. Let \(S_M\) be the saturation of M in \(\Lambda \). Clearly \(\phi (S_M) \subseteq S_M\). Now, \(\phi ^{1}(S_M)\) has the same rank as \(S_M\) and contains \(S_M\), so it is equal to \(S_M\). So if \(\phi (S_M) \ne S_M\), then the image of \(\phi \) does not contain \(S_M\), contradicting the hypothesis that \(\phi \) is an automorphism of \(\Lambda \). Since M is a subgroup of finite index of \(S_M\), this implies that Open image in new window . But \(\phi (M) \subseteq M\), so it follows that \(\phi (M) = M\). Thus there is a map \({\mathrm {O}}(\Lambda ,M) \rightarrow {\mathrm {O}}(M)\). Now, if \(\phi \in {\mathrm {O}}(\Lambda ,M)\) and Open image in new window then Open image in new window for all \(m \in M\), so Open image in new window and we get a map \({\mathrm {O}}(\Lambda ,M) \rightarrow {\mathrm {O}}(M^{\perp })\) in the same way. Combining these two maps gives a map Open image in new window .
If \(d(\phi ) = 1\), then d is the identity on Open image in new window which is a subgroup of \(\Lambda \) of finite index. Because \(\Lambda \) is torsionfree, this forces \(\phi \) to be the identity. To show that \(\mathrm{im\,}d\) has finite image in Open image in new window , let n be the smallest positive integer such that Open image in new window and let Open image in new window . Every element of Open image in new window that fixes \(n\Lambda \) as a set is in the image of d, because the induced automorphism of \(n\Lambda \) extends to an automorphism of \(\Lambda \) with the same action on Open image in new window Since there are only finitely many subgroups of index k in Open image in new window the stabilizer of \(n\Lambda \) is of finite index, and we have proved (2).
To prove (3), let \(\phi \in Z_{{\mathrm {O}}(\Lambda )} H\), and let Open image in new window and \(h \in H\). Then \(h(m) = m\) and Open image in new window . So \(h(\phi (m)) = \phi (h(m)) = \phi (m)\), establishing that Open image in new window
Finally we prove (4). Choose \(\phi \) in the kernel and \(h \in H\), and let \(x \in \Lambda \). We will view \(\phi \) and h as automorphisms of \(\Lambda _\mathbb {Q}\). In \(\Lambda _\mathbb {Q}\) we may write \(x = x_1 + x_2\), where Open image in new window and Open image in new window Then \(h(\phi (x)) = h(\phi (x_1 + x_2)) = h(\phi (x_1)) + h(\phi (x_2))\). However, Open image in new window , so \(h(\phi (x_1)) = \phi (x_1)\), and \(\phi \) is in the kernel of the map to Open image in new window , so \(\phi (x_2) = x_2\). It follows that \(h(\phi (x)) = \phi (x_1) + h(x_2)\). Similarly, \(\phi (h(x)) = \phi (h(x_1+x_2)) = \phi (h(x_1)) + \phi (h(x_2)) = \phi (x_1) + h(x_2) = h(\phi (x))\), establishing that \(\phi \) commutes with h. \(\square \)
Proof of Proposition 2.2
The map \({\mathrm {O}}(\Lambda ,M) \rightarrow {\mathrm {O}}(M)\) is a composition Open image in new window . 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 \(M^{\perp }\) is definite to prove the second statement. Then \({\mathrm {O}}(M^{\perp })\) is finite, so \({\mathrm {O}}(\Lambda ,M) \rightarrow {\mathrm {O}}(M)\) is a composition of an injective map with a map with finite kernel and so its kernel is finite. Let Open image in new window . Then \(Z_{{\mathrm {O}}(\Lambda )} H/K\) has finite index in \({\mathrm {O}}(\Lambda ,M)/K\), because both inject into the finite group \({\mathrm {O}}(M^{\perp })\). Therefore \(Z_{{\mathrm {O}}(\Lambda )} H\) has finite index in \({\mathrm {O}}(\Lambda ,M)\) too. \(\square \)
3 Finiteness results for K3 surfaces
In this section we formulate and prove analogues of the statements of Theorem 1.1 when k is an arbitrary field. We first look at the case of k separably closed, which is straightforward.
Lemma 3.1
Let \(k = k^\mathrm{s}\) be a separably closed field, and let \(\overline{k}\) be an algebraic closure of k. Let X be a K3 surface over k, and let \({\overline{X}}\) be the base change of X to \(\overline{k}\). Then the natural maps \(\mathrm{Pic\,}X \rightarrow \mathrm{Pic\,}{\overline{X}}\) and \(\mathrm{Aut}\,X \rightarrow \mathrm{Aut}\,{\overline{X}}\) are isomorphisms.
Proof
The functor taking a kscheme S to the group Open image in new window is represented by a scheme \(\mathbf {Aut}_{X/k}\): see [9, Theorem 5.23]. A standard argument in deformation theory shows that the tangent space at the identity element is isomorphic to \(\mathrm {H}^0(X, T_X)\), where \(T_X\) denotes the tangent sheaf on X. Indeed, an element of the tangent space is given by a morphism \(S=\mathrm{Spec\,}k[\varepsilon ]/(\varepsilon ^2) \rightarrow \mathbf {Aut}_{X/k}\) extending the morphism sending \(\mathrm{Spec\,}k\) to the identity automorphism. Such a morphism corresponds to an automorphism of Open image in new window restricting to the identity on the central fibre. By [9, Theorem 8.5.9], the set of these morphisms forms an affine space under \(\mathrm {H}^0(X,T_X)\). In our case, the group \(\mathrm {H}^0(X, T_X)\) is zero [12, Theorem 9.5.1], so the scheme \(\mathbf {Aut}_{X/k}\) is étale over k, and \(\mathrm{Aut}\,X \rightarrow \mathrm{Aut}\,{\overline{X}}\) is an isomorphism. \(\square \)
Corollary 3.2
In the situation of Lemma 3.1, every \((2)\)curve on \({\overline{X}}\) is defined over k.
Proof
Let \(\overline{C}\) be a \((2)\)curve on \({\overline{X}}\). Then Lemma 3.1 shows that there is a line bundle L on X whose base change to \({\overline{X}}\) is isomorphic to \({\mathscr {O}}_{{\overline{X}}}(\overline{C})\). The Riemann–Roch theorem and flat base change give \(h^0(X,L) = h^0({\overline{X}},{\mathscr {O}}_{{\overline{X}}}(\overline{C})) = 1\). So all nonzero sections of L cut out the same divisor \(C \subset X\) and the base change of C to \({\overline{X}}\) must coincide with \(\overline{C}\). In other words, \(\overline{C}\) is defined over k. \(\square \)
We now pass to the case of a general field. Let k be a field; fix an algebraic closure \(\overline{k}\) of k, and let \(k^\mathrm{s}\) be the separable closure of k in \(\overline{k}\). Let X be a K3 surface over k, and let \({X^\mathrm{s}}\) and \({\overline{X}}\) denote the base changes of X to \(k^\mathrm{s}\) and \(\overline{k}\), respectively. Write \({\Gamma _k}= \mathrm{Gal}(k^\mathrm{s}/k)\).
The group \({\Gamma _k}\) acts on \(\mathrm{Pic\,}{X^\mathrm{s}}\) preserving intersection numbers, giving a representation \({\Gamma _k}\rightarrow {\mathrm {O}}(\mathrm{Pic\,}{X^\mathrm{s}})\). Let \({\Gamma _k}\) act on \({\mathrm {O}}(\mathrm{Pic\,}{X^\mathrm{s}})\) by conjugation, that is, so that Open image in new window for all \(x \in \mathrm{Pic\,}{X^\mathrm{s}}\). For a \((2)\)class Open image in new window denote the reflection in \(\alpha \) by \(r_\alpha \); then we have \(({{}^{\sigma } r_\alpha }) = r_{\sigma \alpha }\). So the action of \({\Gamma _k}\) on \({\mathrm {O}}(\mathrm{Pic\,}{X^\mathrm{s}})\) restricts to an action on \(\mathrm{W}(\mathrm{Pic\,}{X^\mathrm{s}})\).
Definition 3.3
Define \(R_X\) to be the group Open image in new window
It is easy to see that the action of \(R_X\) on \(\mathrm{Pic\,}{X^\mathrm{s}}\) preserves Open image in new window but it is not immediately obvious that this action preserves \(\mathrm{Pic\,}X\). To show that this is the case, we use an explicit description of \(R_X\) 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 W be a group generated by a set \(T \subset W\) of elements of order 2. For Open image in new window , let Open image in new window be the order of Open image in new window if Open image in new window has finite order and 0 otherwise. Suppose that the relations \(t_i^2 = 1\), Open image in new window for i, j with Open image in new window are a presentation of W. Then (W, T) is a Coxeter system.
Let G be a graph with vertices T and such that Open image in new window are adjacent in G if and only if \(t_i\) does not commute with Open image in new window ; in this case, label the edge joining \(t_i\) to Open image in new window with \(n_{i,j}  2\) for Open image in new window and 0 otherwise. We refer to G as the Coxeter–Dynkin diagram of (W, T). The Coxeter system (W, T) is said to be irreducible if its Coxeter–Dynkin diagram is connected.
Let the length\(\ell (w)\) of an element \(w \in W\) be the length of a shortest word in the \(t_i\) that represents it. If W is finite, there is \(w_0 \in W\) such that \(\ell (w_0) > \ell (w)\) for all \(w \ne w_0 \in W\) (see [2, Proposition 2.3.1]). We refer to \(w_0\) as the longest element of W.
Let \(\sigma \) be a permutation of T. Then there is at most one way to extend \(\sigma \) to a homomorphism \(W \rightarrow W\), because T generates W. If there is such an extension, it is an automorphism, because \(\sigma ^{1}\) extends to its inverse, and we speak of it as the automorphism induced by\(\sigma \).
If (W, T) is a Coxeter system, and I is a subset of T, let \(W_I\) denote the subgroup of W generated by the elements of I. Then \((W_I,I)\) is a Coxeter system: see [2, Proposition 2.4.1 (i)].
Theorem 3.5
([11, Theorem 1]) Let (W, T) be a Coxeter system. Let G be a group of permutations of T that induce automorphisms of W. Let F be the set of orbits \(I \subset T\) for which \(W_I\) is finite, and for \(I \in F\) let \(w_{I,0}\) be the longest element of \((W_I,I)\). Then Open image in new window is a Coxeter system.
We will apply this theorem with \(W=\mathrm{W}(\mathrm{Pic\,}{X^\mathrm{s}})\) and T being the set of reflections in \((2)\)curves on Open image in new window
Proposition 3.6
 (i)
I consists of disjoint \((2)\)curves;
 (ii)
I consists of disjoint pairs of \((2)\)curves, each pair having intersection number 1.
 (1)
For each \(I \in F\), let \(W_I\) be the subgroup of \(\mathrm{W}(\mathrm{Pic\,}{X^\mathrm{s}})\) generated by reflections in the classes of curves in I, and let \(r_I\) be the longest element of the Coxeter system \((W_I,I)\). Then Open image in new window is a Coxeter system.
 (2)
For each \(I \in F\), let \(C_I \in (\mathrm{Pic\,}{X^\mathrm{s}})^{{\Gamma _k}}\) be the sum of the classes in I. Then \(r_I\) acts on \((\mathrm{Pic\,}{X^\mathrm{s}})^{{\Gamma _k}}\) as a reflection in the class \(C_I\).
 (3)
The action of \(R_X\) on \(\mathrm{Pic\,}{X^\mathrm{s}}\) preserves \(\mathrm{Pic\,}X\).
Proof
Let I be a Galois orbit of \((2)\)curves, and suppose that the subgroup \(W_I\) is finite. We will show that I is of one of the two types described. Firstly, no two \((2)\)curves in I have intersection number greater than 1, for then the corresponding reflections would generate an infinite dihedral subgroup of \(W_I\). Since \(W_I\) is finite, its Coxeter–Dynkin diagram is a finite union of trees [2, Exercise 1.4]. In particular, it contains a vertex of degree \(\leqslant 1\). However, the Galois group \({\Gamma _k}\) acts transitively on the diagram, so we conclude that either every vertex has degree 0, or every vertex has degree 1. These two possibilities correspond to the two types of orbits described. Now (1) follows from Theorem 3.5.
Finally, each class \(C_I\) is, by construction, the class of a Galoisfixed divisor on Open image in new window so lies in \(\mathrm{Pic\,}X\). So in both cases the formula for \(r_I\) given above shows that reflection in \(C_I\) preserves \(\mathrm{Pic\,}X\), and therefore the action of \(R_X\) preserves \(\mathrm{Pic\,}X\), proving (3). \(\square \)
Proposition 3.7
Let X be a K3 surface over k. The cone \(\mathrm{Nef}(X) \cap {\mathscr {C}}_X\) is a fundamental domain for the action of \(R_X\) on the positive cone \({\mathscr {C}}_X\), and this action is faithful.
Proof
We will prove two things: first, that every class in \({\mathscr {C}}_X\) is \(R_X\)equivalent to an element of \(\mathrm{Nef}(X) \cap {\mathscr {C}}_X\); and second, that the translates of \(\mathrm{Nef}(X) \cap {\mathscr {C}}_X\) by two distinct elements of \(R_X\) meet only along their boundaries. The second of these shows in particular that the action is faithful. (When we refer to the boundary of \(\mathrm{Nef}(X)\) or one of its translates, we mean the boundary within \((\mathrm{Pic\,}X)_\mathbb {R}\). The boundary of \(\mathrm{Nef}(X) \cap {\mathscr {C}}_X\) in \({\mathscr {C}}_X\) is just the boundary of \(\mathrm{Nef}(X)\) intersected with \({\mathscr {C}}_X\), so that distinction is not so important.)
Now suppose that \(D \in {\mathscr {C}}_X\) has nontrivial stabilizer. Then D lies on at least one of the walls defined by the action of \(\mathrm{W}(\mathrm{Pic\,}{\overline{X}})\) on \((\mathrm{Pic\,}{\overline{X}})_\mathbb {R}\); see [12, Section 8.2]. By [12, Proposition 8.2.4], the chamber structure of this group action is locally polyhedral within \({\mathscr {C}}_{\overline{X}}\), so a small enough neighbourhood of D meets only finitely many chambers. Also note that \(\mathrm{Pic\,}X\) is not contained in any of the walls, because X admits an ample divisor. This allows us to construct a sequence \((D_i)_{i=1}^{\infty }\) of elements of \({\mathscr {C}}_X\), tending to D and all lying in the interior of the same chamber of \({\mathscr {C}}_{\overline{X}}\). As in the previous paragraph, there is a unique \(g \in R_X\) satisfying \(g D_i \in \mathrm{Nef}(X) \cap {\mathscr {C}}_X\) for all i. By continuity, gD also lies in \(\mathrm{Nef}(X) \cap {\mathscr {C}}_X\).
We now prove the second statement, that \(\mathrm{Nef}(X) \cap {\mathscr {C}}_X\) intersects the translate by any nontrivial element of \(R_X\) only in its boundary. Suppose that \(x \in {\mathscr {C}}_X\) lies in the intersection \(\mathrm{Nef}(X) \cap g \mathrm{Nef}(X)\), for some nontrivial \(g \in R_X\). By [12, Corollary 8.2.11], we see that x lies in the boundary of \(\mathrm{Nef}({\overline{X}})\). The following lemma shows that x lies in the boundary of \(\mathrm{Nef}(X)\). \(\square \)
Lemma 3.8
Proof
Let x be a point of \(\partial _V(C) \cap S\). Then there is a supporting hyperplane to C at x, that is, there exists a nonzero \(\phi \in C^*\) satisfying \(\phi (x)=0\). The condition that S meet the interior of C implies that \(\phi \) does not vanish identically on S, so \(f(\phi )\) is nonzero. Thus \(f(\phi )\) is a nonzero element of Open image in new window vanishing at x, so x lies in Open image in new window .
Conversely, suppose that x lies in Open image in new window . Then there is a supporting hyperplane to \(C \cap S\) at x, that is, there exists a nonzero Open image in new window satisfying \(\psi (x)=0\). Let \(\phi \in C^*\) satisfy \(f(\phi )=\psi \); then we have \(\phi (x)=0\) and so \(x \in \partial _V(C)\). \(\square \)
Remark 3.9
We can also give an explicit description of the walls of \(\mathrm{Nef}(X) \cap {\mathscr {C}}_X\). According to [12, Corollary 8.1.6], a class in \({\mathscr {C}}_X\) lies in \(\mathrm{Nef}(X)\) if and only if it has nonnegative intersection number with every \((2)\)curve on Open image in new window or, equivalently, with every Galois orbit of \((2)\)curves. The question is to determine which Galois orbits are superfluous, and which actually define walls of \(\mathrm{Nef}(X) \cap {\mathscr {C}}_X\).
On the other hand, let I be a Galois orbit of \((2)\)curves such that the subgroup \(W_I\) is infinite. Then, for a \((2)\)curve \(C \in I\), the hyperplane orthogonal to C in \((\mathrm{Pic\,}{X^\mathrm{s}})_\mathbb {R}\) does not meet \(\mathrm{Nef}(X) \cap {\mathscr {C}}_X\), as the following argument shows. Let x be a class in \({\mathscr {C}}_X\) orthogonal to C; then x is also orthogonal to all the other curves in I, and so is fixed by \(W_I\). Therefore x is also orthogonal to the infinitely many images of C under the action of \(W_I\), contradicting the fact that the chamber structure induced by \(\mathrm{W}(\mathrm{Pic\,}{X^\mathrm{s}})\) on \({\mathscr {C}}_{X^\mathrm{s}}\) 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 \(R_X\) in \({\mathrm {O}}(\mathrm{Pic\,}X)\) to be normal. Instead, we will see that there is a natural homomorphism from a semidirect product Open image in new window to \({\mathrm {O}}(\mathrm{Pic\,}X)\) having finite kernel and image of finite index.
Note that the natural action of \(\mathrm{Aut}\,X \subset \mathrm{Aut}\,{\overline{X}}\) on \({\mathrm {O}}(\mathrm{Pic\,}{\overline{X}})\) by conjugation fixes \(\mathrm{W}(\mathrm{Pic\,}{\overline{X}})\) and commutes with the Galois action, so fixes \(R_X\). This gives an action of \(\mathrm{Aut}\,X\) on \(R_X\), and a homomorphism from the associated semidirect product Open image in new window to \({\mathrm {O}}(\mathrm{Pic\,}{\overline{X}})\). Since \(\mathrm{Aut}\,X\) and \(R_X\) both fix \(\mathrm{Pic\,}X\), we also obtain a homomorphism Open image in new window .
Proposition 3.10
Remark 3.11

Because the action of \(\mathrm{Aut}\,X\) fixes the ample cone, the image of \(\mathrm{Aut}\,X\) in \({\mathrm {O}}(\mathrm{Pic\,}X)\) meets \(R_X\) only in the identity element. Therefore the finiteness of the kernel in Proposition 3.10 is equivalent to the finiteness of the kernel of \(\mathrm{Aut}\,X \rightarrow {\mathrm {O}}(\mathrm{Pic\,}X)\).

Let Open image in new window be the subgroup of symplectic automorphisms [12, Definition 15.1.1]. Over an algebraically closed field k of characteristic zero, the induced map from Open image in new window to \({\mathrm {O}}(\mathrm{Pic\,}X)\) is injective, so that Open image in new window can be viewed as a subgroup of \({\mathrm {O}}(\mathrm{Pic\,}X)\). Instead of our Proposition 3.10, Huybrechts [12, Theorem 15.2.6] makes the statement that Open image in new window has finite index in \({\mathrm {O}}(\mathrm{Pic\,}X)\). This is equivalent to our formulation, because \(\mathrm {Aut_s} X\) is of finite index in \(\mathrm{Aut}\,X\). However, when k is not algebraically closed there is no reason for Open image in new window to be injective, so there is no advantage to stating the result in terms of Open image in new window .

Lieblich and Maulik [13, Theorem 6.1] define \(\Gamma \subset {\mathrm {O}}(\mathrm{Pic\,}X)\) to be the subgroup of elements preserving the ample cone, and then show that \(\mathrm{Aut}\,X \rightarrow \Gamma \) has finite kernel and cokernel. Over an algebraically closed base field k, the subgroups \(\Gamma \) and \(\mathrm{W}(\mathrm{Pic\,}X)\) generate \({\mathrm {O}}(\mathrm{Pic\,}X)\): any element of \({\mathrm {O}}(\mathrm{Pic\,}X)\) permutes the \((2)\)classes in \(\mathrm{Pic\,}X\) and therefore takes the ample cone to one of its translates under \(\mathrm{W}(\mathrm{Pic\,}X)\); so composing with an element of \(\mathrm{W}(\mathrm{Pic\,}X)\) gives an element of \(\Gamma \). 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 G be a finite group, and \(f :A \rightarrow B\) a homomorphism of (possibly noncommutative) Gmodules having finite kernel and image of finite index. Then the induced homomorphism Open image in new window also has finite kernel and image of finite index.
Proof
The following lemma is standard and easy to prove; we state it for reference.
Lemma 3.13
 (1)
If f and g both have finite kernel and image of finite index, then so does the composition Open image in new window .
 (2)
If Open image in new window has finite kernel and image of finite index, and g has finite kernel, then f has finite kernel and image of finite index.
 (3)
If Open image in new window has finite kernel and image of finite index, and f has image of finite index, then g has finite kernel and image of finite index.
Proof
We leave this as an exercise for the reader. \(\square \)
Proof of Proposition 3.10
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 \(\mathrm{Nef}^{\,\mathrm e}(X)\) to be the real convex hull of \(\mathrm{Nef}(X) \cap \mathrm{Pic\,}X\).
Corollary 3.15
The action of \(\mathrm{Aut}\,X\) on \(\mathrm{Nef}^{\,\mathrm e}(X)\) admits a rational polyhedral fundamental domain.
Proof
This is as in the case of \(k=\mathbb {C}\); we briefly recall the argument of Sterk [20], making the necessary adjustments.
If we choose y to be an ample class in \(\mathrm{Pic\,}X\), then the resulting \(\Pi \) is contained in \(\mathrm{Nef}(X)\), as we now show. Let I be a Galois orbit of \((2)\)curves on \({X^\mathrm{s}}\) such that the corresponding group \(W_I \subset \mathrm{W}(\mathrm{Pic\,}{X^\mathrm{s}})\) is finite. Proposition 3.6 states that the longest element w of \(W_I\) acts on \(\mathrm{Pic\,}X\) as reflection in the class \(C_I = \sum _{E \in I} E\), and that these elements generate \(R_X\). Taking \(\gamma =w\) in the definition of \(\Pi \) shows that \(\Pi \) is contained in the halfspace Open image in new window . As this holds for all such I, Remark 3.9 shows that \(\Pi \) is contained in \(\mathrm{Nef}(X)\).
We conclude as in [20]. If x is a class in \(\mathrm{Nef}^{\,\mathrm e}(X)\) then, since \(\Pi \) is a fundamental domain for the action of Open image in new window on \(({\mathscr {C}}_X)_+\), we can find \(\phi \in \mathrm{Aut}\,X\) and \(r \in R_X\) such that \(r \phi (x)\) lies in \(\Pi \). But now \(\phi (x)\) and \(r \phi (x)\) both lie in \(\mathrm{Nef}^{\,\mathrm e}(X)\), so they are equal and lie in \(\Pi \). This shows that \(\Pi \) is a fundamental domain for the action of \(\mathrm{Aut}\,X\) on \(\mathrm{Nef}^{\,\mathrm e}(X)\). \(\square \)
Corollary 3.16
 (1)
There are only finitely many \(\mathrm{Aut}\,X\)orbits of krational \((2)\)curves on X.
 (2)
There are only finitely many \(\mathrm{Aut}\,X\)orbits of primitive Picard classes of irreducible curves on X of arithmetic genus 1.
 (3)
For \(g \geqslant 2\), there are only finitely many \(\mathrm{Aut}\,X\)orbits of Picard classes of irreducible curves on X of arithmetic genus g.
Proof
Let \(\Pi \) be a rational polyhedral fundamental domain for the action of \(\mathrm{Aut}\,X\) on \(\mathrm{Nef}^{\,\mathrm e}(X)\), as in Corollary 3.15. Every krational \((2)\)curve on X defines a wall of \(\mathrm{Nef}(X)\), by Remark 3.9. Since \(\Pi \) meets only finitely many walls of \(\mathrm{Nef}(X)\), 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 \(\Pi \), let \(D_1, \cdots , D_r\) be a minimal set of generators for \(\Pi \cap \mathrm{Pic\,}X\). Since these all lie in \(\mathrm{Nef}(X)\), we have Open image in new window for all i, and Open image in new window for \(i \ne j\). As observed in [20], this implies that, for any \(n > 0\), there are only finitely many classes in \(\Pi \cap \mathrm{Pic\,}X\) of selfintersection n; and there are only finitely many primitive classes in \(\Pi \cap \mathrm{Pic\,}X\) of selfintersection zero. The class of an irreducible curve of arithmetic genus \(g \geqslant 1\) has selfintersection \(2g2\) and therefore lies in \(\mathrm{Nef}(X)\), so this proves (2) and (3). \(\square \)
Remark 3.17
It is not true that every irreducible curve on X of arithmetic genus 0 is a krational \((2)\)curve. However, there are not many possibilities, as we now show. Let C be such a curve, and let \(C_1, \cdots , C_r\) be the geometric components of C; the Galois group \({\Gamma _k}\) acts transitively on them. In order to achieve Open image in new window , we must have \(C_i^2<0\) for all i, so each \(C_i\) is a \((2)\)curve. Consider the intersection matrix Open image in new window : the sum of the entries in the matrix is , and the Galois action shows that every row sum is the same. Therefore the number r of rows divides 2, and there are only two options: \(r=1\), so that C is a rational \((2)\)curve; or \(r=2\), and \(C=C_1 \cup C_2\) is the union of two conjugate \((2)\)curves meeting transversely in a single point. By Corollary 3.15, both types of curves define walls of \(\mathrm{Nef}(X)\), so in fact both types fall into finitely many orbits under the action of \(\mathrm{Aut}\,(X)\).
Both types occur on the surfaces considered in Sect. 4.3. Indeed, the line \(x = y\), \(z = w\) on the surface \(x^4  y^4 = c(z^4  w^4)\) is a rational \((2)\)curve, while the line \(x = y\), \(z = iw\) meets its conjugate transversely in a single point.
Remark 3.18
In the case \(g=1\), the condition that the class be primitive cannot be omitted, for the following reason. Take for example \(k=\mathbb {Q}\), and suppose that X admits an elliptic fibration A general fibre of such a fibration is a smooth, geometrically irreducible curve E of genus 1, whose class in \(\mathrm{Pic\,}X\) has selfintersection 0 and is primitive. (In fact, the class is even primitive in \(\mathrm{Pic\,}{\overline{X}}\).) If \(s \in \mathbb {P}^1\) is a point of degree \(m>1\), then in general the fibre \(\pi ^{1}(s)\) will be an irreducible curve on X of arithmetic genus 1, linearly equivalent to mE. As m varies, this construction gives infinitely many such classes that are clearly not \(\mathrm{Aut}\,X\)equivalent.
4 Examples
In this section we give three examples that illustrate some of the theory developed up to this point. First, we will give a K3 surface X with finite automorphism group, even though all K3 surfaces V over \(\overline{\mathbb {Q}}\) with \(\mathrm{Pic\,}V \cong \mathrm{Pic\,}X\) have infinite automorphism group. Second, we will construct a surface Y with finite automorphism group, even though, for all extensions \(k/\mathbb {Q}\), all K3 surfaces V over \(\overline{\mathbb {Q}}\) with \(\mathrm{Pic\,}V \cong \mathrm{Pic\,}Y_k\) have infinite automorphism group. For a third example, we will prove that the quartic surface Z in \(\mathbb {P}^3\) defined by \(x^4  y^4 = c(z^4  w^4)\) over a field k of characteristic 0 has finite automorphism group when \(c \in k\) is such that the Galois group of the field of definition of the Picard group has degree 16, the largest possible.
4.1 First example
We construct a surface X such that \(\mathrm{Aut}\,\overline{X}\) is finite, and thus so is \(\mathrm{Aut}\,X\), while any K3 surface over \(\overline{\mathbb {Q}}\) having the same Picard lattice as X has infinite automorphism group. In contrast, over an algebraically closed field of characteristic not equal to 2, finiteness of the automorphism group depends only on the isomorphism type of the Picard lattice. This is an immediate consequence of statement (2) of Theorem 1.1.
Proposition 4.1
Let V be a K3 surface over \(\overline{\mathbb {Q}}\) with \(\mathrm{Pic\,}V \cong L_M\). Then \(\mathrm{Aut}\,V\) is infinite.
Proof
There is an obvious embedding of the hyperbolic lattice U with Gram matrix \(\left( {\begin{matrix}0&{}1\\ 1&{}0\end{matrix}} \right) \) into \(L_M \cong \mathrm{Pic\,}V\). By [12, Remark 11.1.4], this implies that there is an elliptic fibration \(\pi :V \rightarrow \mathbb {P}^1\) with a section. Let E be the class of a fibre of \(\pi \) and O the class of a section; then E and O generate a sublattice of \(\mathrm{Pic\,}V\) isomorphic to U (though not necessarily the obvious one). If \(\pi \) were to have a reducible fibre, then any component of that fibre other than the one meeting O would lie in \({\langle E,O \rangle }^{\perp }\) and have selfintersection \(2\). However, \({\langle E,O \rangle }\) is a lattice of rank 2 and determinant \(1\), so the determinant of \({\langle E,O \rangle }^{\perp }\) is the negative of that of \(\mathrm{Pic\,}V\) and its rank is 2 less. Thus it is generated by a single vector of norm \(8\) and so there are no reducible fibres. It follows by the Shioda–Tate formula [12, Corollary 11.3.4] that the Mordell–Weil group of the fibration has rank 1. Translation by a nontorsion section gives an automorphism of V of infinite order. \(\square \)
Remark 4.2
In fact, Shimada [18, Remark 9.3] has proved that this translation, and negation in the Mordell–Weil group, generate the whole of \(\mathrm{Aut}\,V\).
In contrast with Proposition 4.1, we will see that there exist K3 surfaces over \(\mathbb {Q}\) with Picard lattice isomorphic to \(L_M\) for which the automorphism group is finite. We use the standard Kodaira symbols for reducible fibres of elliptic fibrations, as in [19, IV.9, Table 4.1].
Proposition 4.3
 (1)
The rank of \(\mathrm{Pic\,}V\) is at most 6.
 (2)
The Picard lattice \(\mathrm{Pic\,}V\) is isomorphic to \(L_N\).
 (3)
The Mordell–Weil group associated to \(\pi \) is trivial, and there are only four reducible fibres.
Before proving the proposition, we state and prove a lemma.
Lemma 4.4
Let V be a K3 surface containing four pairwise disjoint smooth rational curves \(C_1, C_2, C_3, C_4\). Then the class Open image in new window cannot be divided by 2 in \(\mathrm{Pic\,}V\).
Proof
Suppose otherwise, and let Open image in new window . Then Open image in new window , so either D or \(D\) is effective; since 2D is effective, we deduce that D is effective. Now Open image in new window for \(1 \leqslant i \leqslant 4\), so the \(C_i\) are base components of the linear system D and Open image in new window is also effective, giving a contradiction. \(\square \)
Proof of Proposition 4.3
The implication (2) \(\Rightarrow \) (1) is trivial; we now prove (1) \(\Rightarrow \) (2). As above, we have an embedding \(L_N \hookrightarrow \mathrm{Pic\,}V\), so \(\mathrm{Pic\,}V\) must have rank exactly 6. Since \(\mathrm{Pic\,}V\) has rank 6 this embedding has finite index, and we must prove it to be an isomorphism. The determinant of N is \(16\); the square of the index Open image in new window must divide this, so the index is 1, 2, or 4. If it is not 1, there is some element of \(L_N\) that can be divided by 2 in \(\mathrm{Pic\,}V\) but not in \(L_N\). We take this element to be of the form Open image in new window , where all the coefficients are 0 or 1 and not all are 0. Then \(a = b = 0\), for otherwise the intersection number with [O] or [E] would be odd; and all the \(c_i\) must be equal, because the selfintersection of any divisor on a K3 surface is even and hence that of any divisor that can be divided by 2 is a multiple of 8. Thus all \(c_i\) are equal to 1. However, Lemma 4.4 shows that Open image in new window is not divisible by 2. This proves \(\mathrm{Pic\,}V \cong L_N\).
Finally, Nikulin [14, Theorem 3.1] has listed the finitely many possibilities for \(\mathrm{Pic\,}V\) of rank \(\geqslant 6\) that give rise to finite automorphism groups. The lattice Open image in new window is in the list, showing that \(\mathrm{Aut}\,V\) is finite. \(\square \)
Proposition 4.5
Let X be a K3 surface over a field k with an elliptic fibration \(\pi :X \rightarrow \mathbb {P}^1\) that has a section. Suppose that \(\pi \) has four conjugate fibres of type \(\textit{I}_2\) or \(\textit{III}\) and that the rank of \(\mathrm{Pic\,}\overline{X}\) is at most 6. Then there are compatible isomorphisms \(\mathrm{Pic\,}\overline{X} \cong L_N\) and \(\mathrm{Pic\,}X \cong L_M\), and \(\mathrm{Aut}\,X\) is finite.
Proof
Applying Proposition 4.3 to the surface \(\overline{X}\) shows that \(\mathrm{Pic\,}\overline{X}\) is isomorphic to \(L_N\). More precisely, the proof shows that there is an isomorphism \(\mathrm{Pic\,}\overline{X} \cong L_N\) that identifies the basis \((e_1, \cdots , e_6)\) of \(L_N\) with the basis \(([E], [E]+[O], [W_1], \cdots , [W_4])\) of \(\mathrm{Pic\,}\overline{X}\), where E is a fibre of \(\pi \) and O is a section, and \(W_1, \cdots , W_4\) are the components of the four reducible fibres that do not meet O. Since by Proposition 4.3 there are no other reducible fibres, the Galois action permutes \(W_1, \cdots , W_4\) transitively and so the Galoisinvariant subgroup is identified with \(L_M\). As X contains a krational curve O of genus 0, it has rational points over k and hence Open image in new window .
Proposition 4.3 also shows that \(\mathrm{Aut}\,\overline{X}\) is finite, and a fortiori that \(\mathrm{Aut}\,X\) is finite. \(\square \)
We now construct a K3 surface X over \(\mathbb {Q}\) satisfying the conditions of Proposition 4.5, as the Jacobian of a genus1 fibration on a quartic U in Open image in new window See [12, Sect. 11.4] for the properties of the Jacobian of a genus1 fibration on a K3 surface.
Let U be a smooth quartic surface in \(\mathbb {P}^3\) over \(\mathbb {Q}\) containing a line L. Projection away from L induces a morphism \(\pi _L :U \rightarrow \mathbb {P}^1\) whose fibres are the residual intersections with U of planes containing L. The generic fibre is a smooth curve of genus 1, and the induced morphism \(L \rightarrow \mathbb {P}^1\) has degree 3.
Proposition 4.6
Let U be a smooth quartic surface in \(\mathbb {P}^3_\mathbb {Q}\) containing a rational line L and a Galois orbit \(\mathscr {L}\) of four lines that meet L. Suppose in addition that each of the four planes containing L and a line in \(\mathscr {L}\) meets U in one further component, which is a smooth conic. Then \(\pi _L\) has four conjugate fibres of type \(\textit{I}_2\) or \(\textit{III}\).
Let \(X \rightarrow \mathbb {P}^1\) be the relative Jacobian of \(\pi _L\), and suppose in addition that \(\mathrm{Pic\,}\overline{U}\) has rank at most 6. Then there are compatible isomorphisms \(\mathrm{Pic\,}\overline{X} \cong L_N\) and \(\mathrm{Pic\,}X \cong L_M\), and \(\mathrm{Aut}\,X\) is finite.
Proof
Let H be one of the four conjugate planes containing L and a line Open image in new window , and let C denote the residual smooth conic. The union Open image in new window is a fibre of \(\pi _L\). We have Open image in new window , because Open image in new window are a line and a conic in the plane H. Either \(L'\) is tangent to C, in which case we have a fibre of type \(\textit{III}\), or they intersect in two distinct points, and the fibre is of type \(\textit{I}_2\). The same description holds for the other three planes that are Galois conjugates of H.
The relative Jacobian X is a K3 surface [12, Proposition 11.4.5] that has the same geometric Picard number as U [12, Corollary 11.4.7 and the discussion following it], and \(X \rightarrow \mathbb {P}^1\) has the same geometric fibres as \(\pi _L\) [12, Chapter 11, Equation (4.1)]. Now apply Proposition 4.5. \(\square \)
A very general surface U constructed according to this proposition has Picard group generated by the classes of L, the lines in \(\mathscr {L}\) and a fibre of \(\pi _L\). This does not imply that such a surface exists over \(\mathbb {Q}\), but we will see that it is not difficult to find an example.
Example 4.7
Point counts on the surface \(U_3\)
n  1  2  3  4  5  6  7  8 

\(\#\,U_3(\mathbb {F}_{3^n})\)  16  94  676  7066  60, 076  533, 818  4, 785, 076  43, 101, 802 
Remark 4.8
We found U by first fixing the lines L and Open image in new window The space of rational quartic polynomials vanishing on L and on the conjugates of \(L'\) is easily checked to have dimension 14. (This is the expected dimension. The space of quartic polynomials has dimension 35 and vanishing on a line is equivalent to vanishing on five points of the line and so imposes five conditions. However, if two lines meet in a point, that point gives the same condition for both lines. With four pairs of intersecting lines, we therefore expect the dimension to be Open image in new window .) We randomly chose elements of this space until we obtained one defining a smooth surface with good reduction at 3 and suitable characteristic polynomial of Frobenius.
Lemma 4.9
The surface \(\overline{X}\) has exactly nine smooth rational curves.
Proof
Remark 4.10
The surface \(\overline{U}\), on the other hand, has infinitely many smooth rational curves. Indeed, by [12, Chapter 11, Equation (4.5)], the Picard lattice of \(\overline{U}\) has discriminant \(144\). However, the list of Picard lattices of rank \(\geqslant 6\) giving finite automorphism group in [14, Theorem 3.1] does not include any lattices of rank 6 and discriminant \(144\). Hence \(\mathrm{Aut}\,\overline{U}\) is infinite. Let \(\mathscr {C}\) be the union of \(\{L\}\) and the set of components of the reducible fibres of \(\pi _L\). By construction \(\mathscr {C}\) spans a subgroup of \(\mathrm{Pic\,}\overline{U}\) of finite index (in fact, by computing the discriminant one checks that \(\mathscr {C}\) generates \(\mathrm{Pic\,}\overline{U}\)). It follows that the stabilizer of \(\mathscr {C}\) in \(\mathrm{Aut}\,\overline{U}\) is finite and hence that the orbit is infinite.
Remark 4.11
We now explain why we do not construct the Jacobian of U directly as a smooth surface in a projective space.
Thus an ample divisor class cannot give an embedding into Open image in new window for \(n < 9\).
4.2 Second example
Now we give an example that is perhaps more surprising: a K3 surface \(Y/\mathbb {Q}\) for which \(\mathrm{Aut}\,Y\) is finite even though, for all field extensions \(L/\mathbb {Q}\), a K3 surface over \(\overline{\mathbb {Q}}\) whose Picard lattice is isomorphic to \(\mathrm{Pic\,}Y_L\) would have infinite automorphism group.
Definition 4.12
We will choose Y to be a K3 surface whose Picard lattice over \(\mathbb {Q}\) is isomorphic to \(L_M\), while over \(\overline{\mathbb {Q}}\), and indeed over a certain quadratic extension \(K_Y\), the Picard lattice is isomorphic to \(L_N\). The Galois group will act through the quotient \(\mathbb {Z}/2\mathbb {Z}\) by exchanging the second and third generators.
Proposition 4.13
Let Y be a K3 surface in Open image in new window given as the intersection of a quadric and a cubic, containing two disjoint Galoisconjugate conics \(C_1, C_2\) defined over a quadratic field \(K_Y\), and having Picard number 3 over \(\overline{\mathbb {Q}}\). Then \(\mathrm{Pic\,}Y \cong L_M\) and \(\mathrm{Pic\,}Y_{\overline{\mathbb {Q}}} \cong L_N\).
Proof
Let H be a hyperplane section. Then the divisors \(H + C_1 + C_2, C_1, C_2\) have the intersection matrix N, whose determinant is Open image in new window . Thus we have an embedding \(L_N \hookrightarrow \mathrm{Pic\,}Y_{\overline{\mathbb {Q}}}\) whose image has index 1 or 2. If it is 2, then either \([H] + [C_1]\) or \([H] + [C_2]\) can be divided by 2. If \([H] + [C_1] \sim 2D\), then, letting \(\sigma \) be an extension of the nontrivial automorphism of the field of definition of \(C_1\) to that of D, we have Open image in new window and so both classes can be divided by 2 and the index is a multiple of 4, a contradiction. Similarly, \([H] + [C_2]\) cannot be divided by 2. We conclude that the index is 1: that is, \(\mathrm{Pic\,}Y_{\overline{\mathbb {Q}}} \cong L_N\). There are Galoisinvariant divisors of the classes \([H] + [C_1] + [C_2], [C_1] + [C_2]\), and these span the invariant subspace, so they generate \(\mathrm{Pic\,}Y\). Their intersection matrix is M. \(\square \)
Remark 4.14
The divisor class [H] is very ample, because it is the hyperplane class on the smooth projective surface Y. The divisor class \(D = [H]+[C_1]+[C_2]\) is not ample, but we will show that it is nef. Indeed, for any irreducible curve C other than \(C_1\) and \(C_2\), the intersection numbers Open image in new window , Open image in new window , Open image in new window are all nonnegative, while Open image in new window and Open image in new window are both zero. In fact, one can show that D is the hyperplane class for a model of Y as a surface of degree 10 in \(\mathbb {P}^6\) with two ordinary double points (\(A_1\) singularities).
Proposition 4.15
 (1)
The group \({\mathrm {O}}(L_M)\) is infinite. If \(\alpha \in {\mathrm {O}}(L_M)\) has infinite order and A is a normal subgroup of \({\mathrm {O}}(L_M)\) containing \(\alpha \) then \({\mathrm {O}}(L_M)/A\) is finite.
 (2)
Let V be a K3 surface over an algebraically closed field having Picard lattice isomorphic to \(L_M\). Then \(\mathrm{Aut}\,V\) is infinite.
 (3)
Let W be a K3 surface over an algebraically closed field having Picard lattice isomorphic to \(L_N\). Then \(\mathrm{Aut}\,W\) is infinite.
Proof
We first prove (1). To find \({\mathrm {O}}(L_M)\), we consider the quadratic form Open image in new window associated to M. Its automorphism group is generated by the sign changes Open image in new window and Open image in new window and by multiplication by a generator of the group of totally positive units of \({\mathrm {O}}_{\mathbb {Q}(\sqrt{10})}\). (This generator is \(19 + 6 \sqrt{10}\) and it takes (x, y) to \((19x+12y,30x+19y)\).) Thus \({\mathrm {O}}(L_M)\) has a subgroup of finite index isomorphic to \(\mathbb {Z}\), so the quotient by any infinite normal subgroup is finite.
We continue with (2). Working mod 5 we see that \(\mathrm{Pic\,}V\) has no vectors of norm \(2\), whence V has no rational curves. Thus \(\mathrm{Aut}\,V\) coincides up to finite index with the infinite group \({\mathrm {O}}(\mathrm{Pic\,}V) = {\mathrm {O}}(L_M)\): this follows from Theorem 1.1 (2), because \(\mathrm{W}(\mathrm{Pic\,}V)\) is trivial.
We now turn to (3). Let \(D_1, D_2, D_3\) be divisors on W whose intersection matrix is N, and let \(D = D_12D_2D_3\). Since \(D^2 = 0\) and D is primitive, for some \(\alpha \in {\mathrm {O}}(L_N)\) there is a genus1 fibration \(\pi \) with fibres of class \(\alpha (D)\). There is no section of \(\pi \) (indeed, no two curves on W have odd intersection), but there is a 2section. The Jacobian J of \(\pi \) is a K3 surface of Picard number 3 [12, Corollary 11.4.7] on which the determinant of the intersection pairing is Open image in new window [12, Equation (11.4.5)]. As in Proposition 4.1, this shows that J has no reducible fibres: the nonidentity component of a reducible fibre would be orthogonal to the classes of both a fibre and the zerosection, and have selfintersection \(2\), which is incompatible with the required determinant. The Shioda–Tate formula then shows that the Mordell–Weil group of J has rank 1. This Mordell–Weil group acts faithfully on W, so it follows as in Proposition 4.1 that \(\mathrm{Aut}\,W\) is infinite. \(\square \)
Proposition 4.15 states that any K3 surface over an algebraically closed field having Picard lattice isomorphic to either \(L_M\) or \(L_N\) has infinite automorphism group. In contrast, the following proposition shows that a K3 surface over \(\mathbb {Q}\) having Picard lattice isomorphic to \(L_M\) or \(L_N\) over any algebraic extension of \(\mathbb {Q}\) may have finite automorphism group. Indeed, the last part of this section will be devoted to finding an example of such a surface.
Proposition 4.16
Let Y be a K3 surface as in Proposition 4.13. Then \(\mathrm{Aut}\,Y\) is finite.
Proof
Let E be an effective divisor with \([E] = aD_1  b_1 [C_1]  b_2 [C_2]\). By Remark 4.14, \(D_1\) is nef, so Open image in new window , and therefore \(a \geqslant 0\). Moreover, if E is irreducible, then the inequality Open image in new window yields Open image in new window , and if furthermore we have \(a>0\), then the inequality Open image in new window gives \(b_i \geqslant 0\).
Remark 4.17
Given equations for Y, it is quite practical to make the curve R used in the proof explicit. For any variety Open image in new window let I(V) be the ideal in the homogeneous coordinate ring of Open image in new window of polynomials that vanish on V. Then the dimension of the degree6 part of \(((I(Y)+I(C_1)^3) \cap (I(Y)+I(C_2)^4))/I(Y)\) is 1; let \(f_6\) be a generator, so that the divisor cut out on Y by \(f_6\) is of the form \(C + 3C_1 + 4C_2\). Using Magma one checks that C is indeed an irreducible curve of arithmetic genus 0.
Remark 4.18
The numerical properties of F follow from the fact that (19, 6) is a solution to the Pell equation \(x^2  10y^2 = 1\). More generally, let k be even and suppose that \(D_1, [C_1], [C_2]\) are classes of pairwise disjoint divisors on a surface with \(D_1^2 = k\) and \(C_1^2 = C_2^2\), where \(C_1\) and \(C_2\) constitute a Galois orbit. Suppose further that \(m^2  kn^2 = 1\): clearly m is odd. Let D be the divisor class \(nD_1  \bigl \lfloor \frac{m}{2} \bigr \rfloor [C_1]  \bigl \lceil \frac{m}{2} \bigr \rceil [C_2]\) and let \(D^{\sigma }\) be its Galois conjugate. Then Open image in new window and Open image in new window .
The remainder of this section will be devoted to constructing a K3 surface Y satisfying the conditions of Proposition 4.13. The verification will be more complicated than that of Example 4.7, because the Picard number over \(\overline{\mathbb {Q}}\) in this example is odd, so that the reduction map cannot induce an isomorphism of \(\mathrm{Pic\,}Y_{\overline{\mathbb {Q}}}\) to Open image in new window . We use the following proposition.
Proposition 4.19
 (1)
Y has good reduction at p and q;
 (2)
the reduction Open image in new window contains a line L disjoint from the reductions of \(C_1, C_2\), and its Picard group over Open image in new window has rank 4;
 (3)
the reduction \(Y_{\mathbb {F}_q}\) base changed to \(\overline{\mathbb {F}}_q\) contains no lines disjoint from the reductions of \(C_1\) and \(C_2\).
Remark 4.20
To speak of Open image in new window we first need to choose a model of Y over \(\mathbb {Z}\). When we give equations with integral coefficients for Y over \(\mathbb {Q}\), it is understood that the model over \(\mathbb {Z}\) is defined by the same equations. In each case it may be checked that the ideal defining this model is generated by the intersection of the ideal defining Y over \(\mathbb {Q}\) with the ring of polynomials in the same variables with integral coefficients.
Proof
Let Open image in new window be the specialization homomorphism Open image in new window , which is injective by [21, Proposition 6.2]. Then by [8, Theorem 1.4] the cokernel of Open image in new window is torsionfree. Note that specialization preserves intersection numbers [10, Section 20.3]. If Open image in new window is surjective, the class of L is in the image. The class \(L_0\) with has positive intersection with the hyperplane class and selfintersection \(2\), so it is effective; its intersection with the hyperplane class is 1, so it is the class of a line on \(Y_{\overline{\mathbb {Q}}}\). We have Open image in new window , so Open image in new window and Open image in new window as well. This contradicts hypothesis (3). It follows that Open image in new window . On the other hand, the hyperplane class in Open image in new window and the classes of \(C_1, C_2\) are independent, so \(\mathrm{rk\,}\mathrm{Pic\,}Y_{\overline{\mathbb {Q}}} \geqslant 3\). \(\square \)
Proposition 4.21
Point counts on the surface \(Y_3\)
n  \(\#\,Y_3(\mathbb {F}_{3^n})\) 

1  18 
2  104 
3  846 
4  6776 
5  59,658 
6  532,694 
7  4,790,811 
8  43,068,056 
9  387,398,079 
Proof
The verification that \(Y_3\) is smooth and contains the given curves is routine. To see that \(Y_3\) has Picard number 4, we proceed as in Example 4.7. In order to count the number of points on \(Y_3\) over \(\mathbb {F}_{3^n}\) for \(1\leqslant n \leqslant 9\), we write \(Y_3\) as an elliptic surface using the fibration of Picard class \(H  C_{3,1}\), for which L is a section, and we sum the number of points of the fibres. We obtain point counts as in Table 2.
Now let \(C_{5,1}, C_{5,2}\) be the reductions of \(C_1\) and \(C_2\), respectively, at one of the two primes above 5.
Proposition 4.22
Proof
Again, it is easy to check that \(Y_5\) is smooth and that the \(C_{5,i}\) are on \(Y_5\). It is also easy to verify directly that \(Y_5\) contains no lines defined over \(\mathbb {F}_5\) by checking that no line through two distinct \(\mathbb {F}_5\)points of \(Y_5\) is contained in \(Y_5\); however, this is not sufficient. Suppose that L is a line contained in \(Y_5\) and disjoint from \(C_{5,1}\). We embed \(Y_5\) in \(\mathbb {P}^5_{\mathbb {F}_5}\) by the divisor class \(H + [C_{5,1}]\), where H is the hyperplane class in Open image in new window . In this embedding, L is still a line, because Open image in new window .
The image is a surface \(T_5\) defined by three quadrics \(Q_1, Q_2, Q_3\). For each \(Q_i\), let \(Z_i\) be the subscheme of Open image in new window consisting of pairs of points \(P, P'\) for which \(P = P'\) or the line joining P to \(P'\) is on the variety defined by \(Q_i = 0\). The image of \(Z_i\) in \(\mathbb {P}^{14}\) under the Plücker embedding is isomorphic to the Fano scheme of lines on \(Q_i\). It is easy to check in Magma that the intersection of the images of the \(Z_i\) in \(\mathbb {P}^{14}\) is empty, which implies that there are no lines on \(T_5\) even over \(\overline{\mathbb {F}}_5\). \(\square \)
Remark 4.23
In fact, we claim that \(Y_5\) contains no lines at all. Indeed, a calculation similar to the one in the proof of Proposition 4.22 shows that there are also no lines on \(Y_5\) disjoint from \(C_{5,2}\). To finish the proof of the claim, it suffices to show that there are no lines that meet both \(C_{5,1}\) and \(C_{5,2}\). Let \(Z_2\) be the closure of the subset of Open image in new window consisting of all pairs of points (x, y) such that \(x \ne y\) and the line through x and y lies on the hypersurface defined by \(F_5\). Define \(Z_3\) similarly for \(G_5\). Since \(C_{5,1}\) and \(C_{5,2}\) are disjoint, the intersection Open image in new window consists of all pairs (x, y) with \(x \in C_{5,1}\) and \(y\in C_{5,2}\) for which there is a line on \(Y_5\) through x and y. We then check in Magma that this intersection is empty. This is done in our Magma scripts [5].
In principle one can prove directly that \(Y_5\) has no lines by checking that the images of \(Z_2\) and \(Z_3\) under the Plücker embedding in Open image in new window are disjoint, but this calculation is very slow.
It remains to show that there exists a surface Open image in new window containing the conics \(C_1,C_2\) and lifting the surfaces \(Y_3,Y_5\) described above. Denote by Open image in new window the vector space of homogeneous forms of degree 2, with coefficients in \(\mathbb {Q}\), vanishing on \(C_1 \cup C_2\). One easily computes that Open image in new window has dimension 5; let \(q_1, \cdots , q_5\) be a \(\mathbb {Z}\)basis for the free abelian group Open image in new window of forms having integer coefficients. One also easily computes that the analogous spaces Open image in new window and Open image in new window both have dimension 5 over \(\mathbb {F}_3\) and \(\mathbb {F}_5\) respectively, which implies that the images of \(q_1, \cdots , q_5\) span these spaces. Now the Chinese remainder theorem guarantees the existence of a quadratic form Open image in new window satisfying both \(F \equiv F_3 \,(\mathrm{mod}\, 3)\) and \(F \equiv F_5 \,(\mathrm{mod}\, 5)\). Similarly, the analogous spaces of cubic forms vanishing on \(C_1 \cup C_2\) are all of dimension 21, which implies the existence of a cubic form G satisfying \(G \equiv G_3 \,(\mathrm{mod}\, 3)\) and \(G \equiv G_5\, (\mathrm{mod}\, 5)\). Remark 4.24 below shows that one can avoid explicit computations to calculate these dimensions. The surface Y given by \(F=G=0\) reduces modulo 3 and 5 to \(Y_3\) and \(Y_5\), respectively.
Remark 4.24
4.3 Finite automorphism group for a family of diagonal surfaces
 1.
Determination of \(\mathrm{Pic\,}Z\), and comparison to the Picard lattice of a K3 surface V with finite automorphism group over \(\overline{k}\).
 2.
Determination of \({\mathrm {O}}(\mathrm{Pic\,}V)\) and \({\mathrm {O}}(\mathrm{Pic\,}Z)\) up to finite index.
 3.
Construction of elements of \(R_Z\) and verification that \({\mathrm {O}}(\mathrm{Pic\,}Z)/R_Z\) is finite.
Remark 4.25
If the characteristic of k is congruent to \(1 \ \mathrm{mod} \ 4\), then k contains the \(4\hbox {th}\) roots of 1 and this Galois action is not possible. If it is \(3 \ \mathrm{mod} \ 4\), then Z is a supersingular surface and \(Z_{\overline{k}}\) has Picard number 22, so the argument given here does not apply. If k has characteristic 2, then of course Z is not a smooth surface.
Lemma 4.26
There is a sublattice of \(L_N\) of index 4 that is isomorphic to \(\mathrm{Pic\,}Z\).
Proof
The Picard group of \(\overline{Z}\) is generated by the classes of the 48 lines. (See [17, Sect. 6.1] for a history of proofs of this fact.) One easily verifies that the subspace fixed by the action of Galois is of rank 6 and that the intersection form has discriminant Open image in new window . Since the surface has rational points such as (1, 1, 0, 0), the fixed subspace of the Picard group of \(\overline{Z}\) is in fact the Picard group of Z.
We now proceed to step 2, determining \({\mathrm {O}}(L_N)\). Let V be a K3 surface over \(\overline{k}\) with Picard lattice \(L_N\), such as those constructed in Proposition 4.6. We choose a basis \(e_1, \ldots , e_6\) for \(L_N \cong \mathrm{Pic\,}V\) to consist of the classes \(E, E+O, C_1, C_2, C_3, C_4\), where E is the class of an elliptic fibration with zero section O and the \(C_i\) are the components of the reducible fibres that do not meet O. In this basis, the Gram matrix of \(\mathrm{Pic\,}V\) is as above. Let \(G \subset {\mathrm {O}}(L_N)\) be the image of the symmetric group \({\mathscr {S}}_4\) by the homomorphism taking a permutation \(\pi \) to the automorphism fixing E and \(E+O\) and taking \(C_i\) to \(C_{\pi (i)}\). Let \(A_{L_N}\) be the discriminant group of \(L_N\) with its discriminant form, a quadratic form with values in \(\mathbb {Q}/2\mathbb {Z}\) [12, Chapter 14].
Lemma 4.27
\(A_{L_N}\) is generated by the classes of \(e_i/2\) for \(3 \leqslant i \leqslant 6\). The discriminant form takes Open image in new window to Open image in new window , and the natural map \(G \rightarrow \mathrm{Aut}\,A_{L_N}\) is an isomorphism.
Proof
The \(e_i/2\) for \(3 \leqslant i \leqslant 6\) belong to \(L_N^*\) and they generate a group that contains \(L_N\) with index 16. Since \(L_N\) has discriminant 16, they generate \(L_N^*\) and their classes generate \(A_{L_N}\). The quadratic form takes Open image in new window to Open image in new window . It follows that the only elements of \(A_{L_N}\) taken to Open image in new window by the quadratic form are the basis vectors, so every automorphism of \(A_{L_N}\) must permute these. Conversely it is clear that every permutation of the basis vectors extends to an automorphism of \(A_{L_N}\) and that this automorphism is the image of a unique element of G. \(\square \)
Recall [12, Definition 15.1.1] that an automorphism of V is symplectic if it acts trivially on \(\mathrm {H}^0(V,\Omega ^2_V)\).
Proposition 4.28
The group \(\mathrm{Aut}\,V\) is finite cyclic and the subgroup Open image in new window of symplectic automorphisms of V is trivial.
Proof
By Lemma 4.9, there are exactly nine smooth rational curves on V. Every automorphism of V must preserve the configuration of these curves. Exactly one of them, the section of the elliptic fibration, intersects four of the others. Hence, the only automorphisms of the configuration are those that permute the reducible fibres. Since the nine curves generate \(\mathrm{Pic\,}V\), we conclude that every automorphism of V acts on \(\mathrm{Pic\,}V\) by elements of G. Now, a symplectic automorphism \(\alpha \) of a K3 surface V in characteristic 0 acts as the identity on the transcendental lattice in \(\mathrm {H}^2(V,\mathbb {Z})\) [12, Remark 15.1.2] and hence on its discriminant group, so in addition it acts trivially on the discriminant group of \(\mathrm{Pic\,}V\) by [12, Lemma 14.2.5]. Since \(\alpha \) acts on \(\mathrm{Pic\,}V\) through G, combining this with the last lemma shows that \(\alpha \) acts trivially on \(\mathrm{Pic\,}V\). It follows that \(\alpha \) acts trivially on \(\mathrm {H}^2(V,\mathbb {Z})\); it is therefore the identity by [12, Corollary 15.1.6].
Remark 4.29
There is always an automorphism of order 2 of V given by negation of the elliptic fibration with respect to the unique section. For very general V the order of \(\mathrm{Aut}\,V\) is 2, but it might be possible for \(\mathrm{Aut}\,V\) to be of order \(m = 4, 6\), or 8 if there is an automorphism that permutes the reducible fibres of the fibration nontrivially. In this case the field of definition of \(\mathrm{Pic\,}V\) would contain \(\mu _m\), which does not happen generically.
By [6, Proposition 5.10], we know that \({\mathrm {O}}(L_N)\) is generated by the automorphisms of the ample cone and the reflections in the classes of rational curves, together with \(1\); the automorphisms of the ample cone are just the G of Lemma 4.27. Since G is generated by two elements and has three orbits on the set of rational curves, we have a set of six generators for \({\mathrm {O}}(L_N)\). To pass to \({\mathrm {O}}(\mathrm{Pic\,}Z)\), we use a simple lemma.
Lemma 4.30
Let \(\Lambda \) be a lattice with sublattices \(\Lambda _1, \Lambda _2\) with Open image in new window finite. Then the stabilizer of \(\Lambda _1 \cap \Lambda _2\) in \({\mathrm {O}}(\Lambda _1)\) has finite index. If in addition \(\Lambda _1, \Lambda _2\) have equal rank, then \({\mathrm {O}}(\Lambda _1) \cap {\mathrm {O}}(\Lambda _2)\) has finite index in \({\mathrm {O}}(\Lambda _1)\) and \({\mathrm {O}}(\Lambda _2)\), where the \({\mathrm {O}}(\Lambda _i)\) are regarded as subgroups of Open image in new window .
Proof
There are only finitely many sublattices of \(\Lambda _1\) of index Open image in new window , so the stabilizer is of finite index. If Open image in new window is finite and \(\Lambda _1, \Lambda _2\) have equal rank then Open image in new window is finite as well and \({\mathrm {O}}(\Lambda _1) \cap {\mathrm {O}}(\Lambda _2)\) is the stabilizer of \(\Lambda _1 \cap \Lambda _2\) in \({\mathrm {O}}(\Lambda _i)\) for \(i \in \{1,2\}\). \(\square \)
Remark 4.31
The group generated by \({\mathrm {O}}(\Lambda _1)\) and \({\mathrm {O}}(\Lambda _2)\) usually does not contain \({\mathrm {O}}(\Lambda _1)\) with finite index.
Using Magma [3] to compute the permutation representation of \({\mathrm {O}}(L_N)\) on sublattices with quotient Open image in new window we find a set of 30 generators for \({\mathrm {O}}(L_N) \cap {\mathrm {O}}(\mathrm{Pic\,}Z)\). However, some of these are products of others, so we easily reduce to a set of nine generators. We now perform step 3, the construction of elements of \(R_Z\).
Proposition 4.32
There are 14 Galois orbits of lines and eight of conics on \(\overline{Z}\) that give finite Coxeter groups.
Proof
We list the conics by observing that their Picard classes are of the form \(H/2 + M\), where Open image in new window is such that Open image in new window and Open image in new window . Since Open image in new window is a definite lattice, the set of M with these properties is finite. Such a class is represented by a conic if and only if it is integral and not the sum of two classes of lines, both of which are easily checked. We finish the proof by referring to Proposition 3.6 for the configurations of curves that give finite Coxeter groups. \(\square \)
Remark 4.33
It appears that there are no further generators of \(R_Z\).
Theorem 4.34
\(\mathrm{Aut}\,Z\) is finite.
Proof
We use Proposition 3.10. To do so, we search for relations among the generators of \({\mathrm {O}}(L_N) \cap {\mathrm {O}}(\mathrm{Pic\,}Z)\) and the 22 elements of \(R_Z\) that we have found, simply by checking which products of up to five generators of \({\mathrm {O}}(L_N) \cap {\mathrm {O}}(\mathrm{Pic\,}Z)\) and up to four of these together with the generators of \(R_Z\) give the identity matrix. Magma quickly verifies that these relations are sufficient to show that Open image in new window is finite, from which we conclude by Proposition 3.10 that \(\mathrm{Aut}\,Z\) is finite. \(\square \)
Corollary 4.35
There are finitely many classes of geometrically irreducible curves on Z of each genus.
Proof
This follows by combining the theorem with Corollary 3.16. \(\square \)
Remark 4.36
We conclude that in order to prove that the rational points of Z are Zariskidense, we do not have an infinite automorphism group \(\mathrm{Aut}\,Z\) to our disposal. In particular, we will not be able to find an elliptic fibration on Z with a section of infinite order, as translation by such a section would be an automorphism of infinite order.
Notes
Acknowledgements
We thank the referee, whose thorough report helped us improve the exposition and correct a number of errors in the paper.
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