YC
1. Introduction
A natural approach to determine the existence or non-existence of rational points on a variety
$X/\mathbb{Q}$
is to first check the existence of
$\mathbb{Q}_p$
-rational points for all
$p \leq \infty$
, where with
$\mathbb{Q}_{\infty}$
we mean
$\mathbb{R}$
. Consider a family of varieties
$\{X_i \}_{i \in I}$
. If, for all
$X_i$
, we have
$X_i(\mathbb{Q}) \neq \emptyset$
if and only if
$X_i(\mathbb{Q}_p) \neq \emptyset$
for all
$p \leq \infty$
, then we say that the family
$\{X_i \}$
satisfies the Hasse principle. This is indeed the case for quadrics and Severi–Brauer varieties, for instance, but in general the Hasse principle may fail. The failure is often explained by the Brauer–Manin obstruction; see [
Reference Colliot-Thélène and Skorobogatov4
, Chapter 13] for a precise formulation. The idea is that inside the set of adelic points
$X(\mathbb{A}_{\mathbb{Q}})$
lies the so-called Brauer–Manin set
$X(\mathbb{A}_{\mathbb{Q}})^{\operatorname{Br}} \subset X(\mathbb{A}_{\mathbb{Q}})$
, consisting of those adelic points that satisfy certain conditions imposed by the Brauer group. Class field theory shows that these conditions are always satisfied by the rational points
$X(\mathbb{Q})$
. The key point is that emptiness of the Brauer–Manin set can give an obstruction to the existence of rational points, even when
$X(\mathbb{Q}_p) \neq \emptyset$
for all
$p\leq \infty$
.
This approach to rational points justifies the interest in studying the Brauer groups of varieties, with the goal of computing them as abstract groups and describing their elements in a way amenable to the obstruction computation. The Brauer group
$\operatorname{Br}\!(X)$
is endowed with a natural filtration
$\operatorname{Br}_0(X) \subset \operatorname{Br}_1(X) \subset \operatorname{Br}\!(X)$
, defined as
\[\operatorname{Br}_0(X)=\mathrm{Im}(\!\operatorname{Br}\!(\mathbb{Q}) \to \operatorname{Br}\!(X)) \subset \operatorname{Br}_1(X)=\mathrm{ker}(\!\operatorname{Br}\!(X) \to \operatorname{Br}\!(\overline{X}) ), \]
where
$\overline{X}=X \times_{\mathbb{Q}} \overline{\mathbb{Q}}$
. The elements in
$\operatorname{Br}_0(X)$
are called constant elements and these will be discarded since they never yield obstructions to rational points. The algebraic Brauer group
$\operatorname{Br}_1(X)$
can play a role in the obstruction and it is usually computed using the Hochschild–Serre spectral sequence, which provides the isomorphism
$\operatorname{Br}_1(X)/\operatorname{Br}_0(X) \cong \operatorname{H}^1(\mathbb{Q}, \operatorname{Pic}(\overline{X}))$
. Finally, the elements of
$\operatorname{Br}\!(X)/\operatorname{Br}_1(X)$
are called transcendental elements and they are troublesome to study, since knowledge of
$\operatorname{Br}\!(\overline{X})^{\Gamma_{\mathbb{Q}}}$
is not in general sufficient to determine which elements descend to the base field
$\mathbb{Q}$
.
The Brauer–Manin obstruction is not the only obstruction to the Hasse principle, as first shown by Skorobogatov in his famous paper [ Reference Skorobogatov21 ]. However, it is conjectured to be the only obstruction for some important families of varieties, such as curves ([ Reference Skorobogatov22 ]), geometrically rationally connected varieties ([ Reference Colliot-Thélène5 ]), and K3 surfaces ([ Reference Skorobogatov23 ]).
The connection between certain K3 surfaces and abelian surfaces or elliptic curves has been exploited to study the Brauer group and the corresponding obstruction; see the results on the transcendental Brauer groups of Kummer surfaces in [ Reference Skorobogatov and Zarhin24 ] and [ Reference Skorobogatov and Zarhin25 ] or the results on diagonal quartic surfaces in [ Reference Ieronymou, Skorobogatov and Zarhin10 ] and [ Reference Ieronymou and Skorobogatov9 ], for example.
This paper extends work of Van Luijk on the Brauer group of a certain family of generalised Kummer surfaces, constructed as follows. Let k be a number field and let C be a diagonal cubic curve over k defined by
$C\;:\; ax^3+by^3+cz^3=0$
. The product
$C\times C$
admits an automorphism
$\rho$
of order 3; see Section 3.1 for the definition of
$\rho$
. The generalised Kummer surface
$Y_C$
is defined to be the minimal desingularisation of the quotient of
$C \times C$
by
$\rho$
. By [
Reference van Luijk26
, Proposition 4·2], the algebraic Brauer groups of these surfaces contain only constant elements, except potentially when abc is a cube in k.
Our first main theorem investigates the relationship between the transcendental Brauer groups of
$Y_C$
and
$C \times C$
, obtaining a result analogous to [
Reference Skorobogatov and Zarhin24
, Theorem 2·4].
For an abelian group G and a positive integer n, we denote by
$G_n$
the subgroup of elements of order dividing n.
Theorem 1·1 (Theorem 3·11). Let C be a smooth projective cubic curve defined over a number field. There is an embedding
\[ \frac{\operatorname{Br}\!(Y_C)_n}{\operatorname{Br}_1(Y_C)_n} \hookrightarrow \frac{\operatorname{Br}\!(C \times C)_n}{\operatorname{Br}_1(C \times C)_n}, \]
which is an isomorphism if n is coprime to 3. This induces an isomorphism
\[ \left(\frac{\operatorname{Br}\!(Y_C)}{\operatorname{Br}_1(Y_C)}\right )_n \xrightarrow{\sim} \left(\frac{\operatorname{Br}\!(C \times C)}{\operatorname{Br}_1(C \times C)}\right)_n \]
for all n coprime to 3.
This result allows us to explicitly compute the Brauer group of
$Y_C$
over
$\mathbb{Q}$
, using the description of the transcendental Brauer group of products of CM elliptic curves given by Newton in [
Reference Newton16
].
Theorem 1·2 (Corollary 4·7, Theorem 5·3). Let
$C/\mathbb{Q}$
be a diagonal cubic curve
$ax^3+by^3+cz^3=0$
such that abc is not a cube in
$\mathbb{Q}$
. Then,
\[\frac{\operatorname{Br}\!(Y_C)}{\operatorname{Br}_0(Y_C)}= \begin{cases} \mathbb{Z}/2\mathbb{Z} &\text{ if } 4abc \text{ is a cube in } \mathbb{Q}; \\ 0 &\text{ otherwise.}\end{cases} \]
Moreover, there is no Brauer–Manin obstruction to the Hasse principle for
$Y_C$
.
Theorem 1·2 builds upon [
Reference van Luijk26
, Proposition 4·2], wherein Van Luijk showed that
$\operatorname{Br}_1(Y_C)=\operatorname{Br}_0(Y_C)$
. Skorobogatov has conjectured that the Brauer–Manin obstruction is the only obstruction to the Hasse principle for K3 surfaces. Assuming this conjecture and everywhere local solubility of
$C/\mathbb{Q}$
, Theorem 1·2 proves the existence of Galois cubic points on C, i.e. L-rational points on C for a Galois extension
$L/\mathbb{Q}$
of degree 3.
Theorem 1·3 (Theorem 5·2). Let
$C/\mathbb{Q}$
be an everywhere locally soluble smooth curve with equation
$ax^3+by^3+cz^3=0$
. Then, Skorobogatov’s conjecture implies the existence of Galois cubic points on C.
The proof of Theorem 1·3 gives a strengthening of [
Reference van Luijk26
, Theorem 1·1] in the case
$k = \mathbb{Q}$
; it shows that one can replace the algebraic Brauer group in the statement of [
Reference van Luijk26
, Theorem 1·1] with the full Brauer group of
$Y_C$
. Thus, it shows that Skorobogatov’s conjecture implies a negative answer to [
Reference van Luijk26
, Question 2] over
$\mathbb{Q}$
. The author would be very interested in unconditionally proving the existence of Galois cubic points on C as in Theorem 1·3, or relating this problem to the Brauer–Manin obstruction for curves instead.
1·1. Outline of the paper
After recalling some technicalities in Section 2, Section 3 is devoted to the proof of Theorem 1·1. In Section 4 we prove Theorem 1·2. Section 5 briefly discusses the Hasse principle for the surface
$Y_C$
and the implications for Galois cubic points on C.
1·2. Notation
We fix some notation.
-
(1) k is a number field and
$\Gamma_k$
is the absolute Galois group
$\operatorname{Gal}(\overline{k}/k)$
; -
(2) for a k-variety X, we denote by
$\overline{X}$
the base change to an algebraic closure
$X \times_k \overline{k}$
; for
$x \in \operatorname{Br}\!(X)$
, we denote by
$\overline{x}$
the image of x via the natural map
$\operatorname{Br}\!(X) \to \operatorname{Br}\!(\overline{X})$
; -
(3) for an abelian group G and a positive integer n, we denote by
$G_n$
the subgroup of elements of order dividing n; for a prime number l, we define
$G\{l\}=\bigcup_{i \geq 1} G_{l^i}$
; for a G-module M, we denote by
$M^G$
the submodule of G-invariants; -
(4) for an abelian variety A, we denote by
$A^{\vee}\;:\!=\;\operatorname{Pic}^0 A$
the dual abelian variety and, if
$f\;:\;A \to B$
is an isogeny of abelian varieties, we denote by
$f^{\vee}\;:\; B^{\vee} \to A^{\vee}$
the dual isogeny, which is induced by the pullback map
$f^*\;:\; \operatorname{Pic} B \to \operatorname{Pic} A$
.
2. Preliminaries
In this section we gather the most important technical tools (well-known in the literature) that are required for the proofs in Section 3. We begin with the statement, and sketch of the proof, of a very general lemma. In this section we assume X to be a smooth, projective, geometrically integral variety over a field k.
Lemma 2·1. Let U be an open subset whose complement in X has codimension
$\geq 2$
. Then the following hold:
\[ k[U]=k[X], \qquad \operatorname{Pic}(U)=\operatorname{Pic}(X), \qquad \operatorname{Br}\!(U)=\operatorname{Br}\!(X). \]
Proof. The first equality comes from the fact that U is dense in X because X is irreducible. Since codimension of
$X \setminus U$
is at least two, the restriction gives a correspondence between (Weil-)divisors of X and U. Finally, the result about the Brauer groups derives directly from Grothendieck’s purity theorem.
Let
$\rho_X$
be the rank of the Néron–Severi group
$\operatorname{NS}(\bar{X})$
and let
$b_2$
be the second Betti number of X. Then for every prime number l, we have the following exact sequence ([
Reference Colliot-Thélène and Skorobogatov4
, Proposition 5·2·9])
\[0 \to (\mathbb{Q}_l/\mathbb{Z}_l)^{b_2-\rho_X} \to \operatorname{Br}\!(\bar{X})\{ l\} \to \operatorname{H}^3_{\acute{\mathrm{e}\mathrm{t}}}(\bar{X},\mathbb{Z}_l(1))_{\mathrm{tors}}\to 0.\]
Moreover, since
$k \subset \mathbb{C}$
, we have that
$\bigoplus_{l} \operatorname{H}^3_{\acute{\mathrm{e}\mathrm{t}}}(\bar{X},\mathbb{Z}_l(1))_{\mathrm{tors}}$
is isomorphic to the torsion subgroup of
$\operatorname{H}^3(X(\mathbb{C}),\mathbb{Z})$
by the comparison theorem between étale cohomology and Betti cohomology, see [
Reference Milne13
, Theorem III·3·12]. If X is an abelian surface or a K3 surface, then
$\operatorname{H}^1(X(\mathbb{C}),\mathbb{Z})$
is torsion-free, thus by Poincaré duality
$\operatorname{H}^3(X(\mathbb{C}),\mathbb{Z})$
is also torsion-free, which implies
\[\operatorname{Br}\!(\bar{X}) \cong (\mathbb{Q}/\mathbb{Z})^{b_2-\rho_X}.\]
Finally, a crucial proposition relating the Brauer groups for a Galois covering of smooth varieties.
Proposition 2·2 ([
Reference Colliot-Thélène and Skorobogatov4
, Theorem 3·8·5]). Let X and Y be smooth varieties and let
$f \;:\; Y \to X$
be a finite flat morphism of degree d such that k(Y) is a Galois extension of k(X) with Galois group G. Then, for any n coprime to
$d=\lvert G \rvert$
, the natural map
$f^*\;:\; \operatorname{Br}\!(X)_n \to \operatorname{Br}\!(Y)^G_n$
is an isomorphism.
2·1. Néron–Severi groups of split abelian surfaces
In this section we present some facts about the Néron–Severi groups of products of abelian varieties; in particular we are interested in the case of split abelian surfaces. We recommend [ Reference Mumford15 , pp. 189-190] and the appendix in [ Reference Kani11 ] for more details on this matter.
Let A be an abelian variety over an algebraically closed field and recall that the elements of the Néron–Severi group
$\operatorname{NS}(A) = \operatorname{Pic}(A)/\operatorname{Pic}^0(A)$
are classes of line bundles on A up to algebraic equivalence. For any line bundle
$\mathcal{L}$
, denote with
$\lambda_{\mathcal{L}}$
the polarisation given by
$\mathcal{L}$
, i.e. the map
\begin{align*} \lambda_{\mathcal{L}}\;:\; A & \to \operatorname{Pic}^0(A) \\ a & \mapsto t^*_a \mathcal{L} \otimes \mathcal{L}^{-1}, \end{align*}
where
$t_a \;:\; A \to A$
is the translation by
$a \in A$
. We assume that A is principally polarised with polarisation
$\lambda \;:\; A \xrightarrow{\sim} A^{\vee}$
. Using the Rosati involution
$r_{\lambda}$
on
$\operatorname{End}(A)$
defined by
$r_{\lambda} (\alpha)= \lambda^{-1} \alpha^{\vee} \lambda$
, the group
$\operatorname{NS}(A)$
can be described as a subgroup of
$\operatorname{End}(A)$
in the following way.
Proposition 2·3 ([
Reference Mumford15
]). The map
$D \mapsto \lambda^{-1} \lambda_D$
defines a functorial isomorphism
\[ \Phi_{\lambda}\;: \operatorname{NS}(A) \xrightarrow{ \sim} \operatorname{End}_{\lambda}(A) \;:\!=\;\{ \alpha \in \operatorname{End}(A)\;:\; r_{\lambda}(\alpha)=\alpha \}. \]
It is possible to be more explicit in the case of a product of two elliptic curves. Let
$A=E_1 \times E_2$
and let
$p_1\;:\; A \to E_1$
,
$p_2\;:\;A \to E_2$
be the projections and let
$e_1\;:\; E_1 \to A$
,
$e_2\;:\;E_2 \to A$
be the natural embeddings. Then,
$p\;:\!=\;p_1^{\vee}+p_2^{\vee} \;:\; E_1^{\vee} \times E_2^{\vee} \to A^{\vee}$
is an isomorphism and given two principal polarisations
$\lambda_i$
of
$E_i$
we can define a principal polarisation on A as
$\lambda_1 \times \lambda_2 \;:\!=\; p \circ (\lambda_1 \times \lambda_2)\; :\; A \xrightarrow{\sim} A^{\vee}$
, called the product polarisation. Recall that if
$\alpha \in \operatorname{End}$
$(E_1 \times E_2)$
we can describe
$\alpha$
as a
$2 \times 2$
matrix
$(\alpha_{ij})$
by putting
$\alpha_{ij}=p_i \alpha e_j \in \operatorname{Hom}\!(E_j,E_i)$
, i.e.
\[\operatorname{End}(E_1 \times E_2) = \Bigl\{ \begin{pmatrix} \alpha_{11} & \alpha_{12} \\ \alpha_{21} & \alpha_{22}\end{pmatrix} :\; \alpha_{ij} \in \operatorname{Hom}\!(E_j,E_i) \Bigr\}. \]
Proposition 2·4 ([ Reference Kani11 , Proposition 60]). We have that
\[\operatorname{End}_{\lambda_1 \times \lambda_2}(E_1 \times E_2) = \Bigl\{ \begin{pmatrix} \alpha_{11} & \alpha'_{21} \\ \alpha_{21} & \alpha_{22}\end{pmatrix} :\; \alpha_{ii} \in \operatorname{End}_{\lambda_i}(E_i), \alpha_{21} \in \operatorname{Hom}\!(E_1,E_2) \Bigr\}, \]
where
$\alpha'_{21}=r_{\lambda_1, \lambda_2}(\alpha_{21})=\lambda_1^{-1} \alpha_{21}^{\vee}\lambda_2$
. Thus, the map
$(\alpha_1, \alpha_2, \beta) \mapsto$
$\begin{pmatrix} \alpha_1 & \beta' \\ \beta & \alpha_2 \end{pmatrix}$
defines an isomorphism:
\[ \phi \;:\; \operatorname{End}_{\lambda_1}(E_1) \oplus \operatorname{End}_{\lambda_2}(E_2) \oplus \operatorname{Hom}\!(E_1,E_2) \xrightarrow{\sim} \operatorname{End}_{\lambda_1 \times \lambda_2}(A), \]
equivalently written as
\[ \phi \;:\; \operatorname{NS}(E_1) \oplus \operatorname{NS}(E_2) \oplus \operatorname{Hom}\!(E_1,E_2) \xrightarrow{\sim} \operatorname{NS}(A).\]
Finally, since
$\operatorname{NS}(E) \cong \mathbb{Z}$
by the degree map, we can be more precise.
Corollary 2·5 ([ Reference Kani11 , Proposition 62]). The above isomorphism
\[ \phi\;:\; \mathbb{Z} \oplus \mathbb{Z} \oplus \operatorname{Hom}\!(E_1,E_2) \xrightarrow{\sim} \operatorname{NS}(A)\]
is explicitly given by
\[(a, b, f) \mapsto (a-1) \mathrm{Im}(e_1) + (b- \deg f) \mathrm{Im}(e_2) + \Gamma_{f},\]
where
$\Gamma_f$
is the divisor associated to the graph
$\{ (P,f(P))\}_{P \in E_1}$
.
Lemma 2·6. Given an isogeny
$f\;:\;E_2 \to E_1$
, denote with
$\Gamma^{-1}_f$
the divisor
$\{(f(P),P) \}_{P \in E_2} \in \operatorname{NS}(E_1 \times E_2)$
. Then the following relation holds
\[ \Gamma^{-1}_f= \Gamma_{f^{\vee}} + (\!\deg f -1) (\mathrm{Im}(\mathrm{e_1})-\mathrm{Im}(\mathrm{e_2})) \]
as divisors in
$\operatorname{NS}(E_1 \times E_2)$
.
Proof. We map the divisor
$\Gamma_{f}^{-1}\in \operatorname{NS}(E_1 \times E_2)$
to the divisor
$\Gamma_f=\{P, f(P) \}_{P \in E_2} \in \operatorname{NS}(E_2 \times E_1)$
, via the coordinates switching map. By Corollary 2·5,
$\phi^{-1}(\Gamma_f)=(1, \deg f, f) \in \operatorname{NS}(E_2) \oplus \operatorname{NS}(E_1) \oplus \operatorname{Hom}\!(E_2,E_1)$
. Now, we use Proposition 2·4 and we identify
$E_1$
and
$E_2$
with their duals, i.e. we identify
$\lambda_1$
and
$\lambda_2$
with the identities so that
$\alpha'_{21}=\alpha^{\vee}_{21}$
. Therefore, we have
\[ \Phi(1, \deg f, f)= \begin{pmatrix} 1 & f^{\vee} \\ f & \deg f\end{pmatrix} \in \operatorname{End}_{\lambda_2 \times \lambda_1}(E_2 \times E_1),\]
and, switching back the coordinates, this corresponds to the element
$ \begin{pmatrix} \deg f & f \\ f^{\vee} & 1\end{pmatrix} \in \operatorname{End}_{\lambda_1 \times \lambda_2}(E_1 \times E_2)$
. Finally Proposition 2·4 and Corollary 2·5 imply the statement.
3. Transcendental Brauer group of
$Y_C$
As mentioned in the introduction, this paper applies a strategy similar to the one followed by Skorobogatov and Zarhin in [ Reference Skorobogatov and Zarhin24 ] in order to compute the transcendental Brauer group of the family of K3 surfaces studied by Van Luijk in [ Reference van Luijk26 ] and which we now define again.
3·1. Construction of K3 surfaces
We will follow very closely [ Reference van Luijk26 , Section 2], trying to adopt the same notation as well.
Let k be a number field and let
$C \subset \mathbb{P}^2_k$
be a smooth projective cubic curve. Let P, Q, R be three (geometric) points of C; we say P, Q, R are collinear if the divisor
$(P)+(Q)+(R)$
is linearly equivalent to a linear section of C. This simply extends the usual collinearity property in order to count tangency points and inflexion points with the suitable multiplicity. By Bezout’s theorem, for every two points P, Q of C, there exists a unique third point R collinear with P and Q (possibly equal to P or Q). This fact yields the following natural isomorphism
\[ S\;:\!=\; C \times C \cong \{ (P,Q,R) \in C^3 \;:\; P,Q,R \text{ are collinear} \}.\]
Let
$\rho$
be the automorphism of
$C^3$
that permutes cyclically the three coordinates; at the level of
$C \times C $
,
$\rho$
maps (P, Q) to (Q, R) with P, Q, R collinear. This defines an automorphism of S of order 3. Let
$X_C$
be the quotient
$S/\rho$
and let
$X=X_C$
when the curve C is understood.
The argument on page 3 in [
Reference van Luijk26
] shows that
$X_C$
has nine singular points, all double points, which correspond to the locus of
$\rho$
-fixed points
$S_3 \subset S$
given by (P, P) for each inflexion point P of C. Let
$Y_C$
(or Y when the dependency on C is clear) denote the blow-up of
$X_C$
at its singular points; as the singular locus of
$X_C$
is defined over k, so is
$Y_C$
. Moreover, defining
$Y_0\;:\!=\;(S \setminus S_3) / \rho$
, we have that
$Y \setminus Y_0$
splits over
$\bar{k}$
into nine pairs of smooth (-2)-curves. The two curves within a pair intersect in one point and each pair corresponds to a geometric point of
$S_3$
.
Proposition 3·1 ([
Reference van Luijk26
, Proposition 2·2 and 3·12]). The surface Y is a smooth K3 surface with
$\mathrm{rk}\operatorname{NS}(Y)=18+ \operatorname{rk} \operatorname{End}(\!\operatorname{Jac} C)$
.
3·2. Equivariant resolutions of singularities
Let
$S_0\;:\!=\;S \setminus S_3$
. The aim of this section is to find a surface S
′, birationally equivalent to S (in particular containing an open subset isomorphic to
$S_0$
), endowed with a proper surjective morphism
$S' \to Y$
such that the diagram
commutes. Indeed, we want to extend the action of
$\rho$
to S
′ and require
$S'/\rho$
to be a resolution of singularities of
$S/\rho$
, so that we obtain a birational morphism
$S'/\rho \to Y$
.
Let us first consider the situation at the geometric level, i.e. over
$\bar{k}$
. We recall the considerations from [
Reference van Luijk26
]. Every singularity of
$S/\rho$
comes from a fixed point of
$\rho$
on S. These take the form (P, P), where P is any of the nine inflexion points of C. To work out the local behaviour around these points, we choose two copies r, s of a uniformizer for C at P. These are then local parameters for S at (P, P). The local description of
$\rho$
modulo squares is
$(r,s) \mapsto (s,-r-s)$
because it corresponds to the description of the formal group law modulo squares with respect to the point P; this describes the action of
$\rho$
on the tangent space of S at (P, P). The subring of
$\rho$
-invariants in k[r, s] is generated by
$a=r^2+rs+s^2$
,
$b=-3rs(r+s)$
,
$c=r^3+3r^2s-s^3$
, which satisfy the equation
$a^3=b^2+bc+c^2$
. This shows that
$S/\rho$
has 9 (geometric) singularities of type
$A_2$
.
Up to a linear change over
$\bar{k}$
, we can describe the tangent space of S at a fixed point of
$\rho$
as
$\mathbb{A}^2=\operatorname{Spec} \overline{k}[x,y]$
with
$\rho$
acting by the matrix
$\begin{pmatrix} \zeta_3 & 0 \\ 0 & \zeta_3^2\end{pmatrix}$
, where
$\zeta_3$
is a primitive cube root of unity. Let S
′′ be the blow-up of S at
$S_3$
. The exceptional divisor above every point in
$S_3$
is isomorphic to
$\mathbb{P}^1_{\overline{k}}$
with the action of
$\rho$
given as a projective transformation by the same matrix. Hence each exceptional divisor contains two fixed points. Consider a chart
$U_0$
containing a fixed point
$P_0$
. Up to an étale map, S
′′ can be described by
$\{xT_1=yT_0\}$
in
$\mathbb{A}^2_{x,y} \times \mathbb{P}^1_{[T_0:T_1]}$
. The open set
$U_0$
is given by
$\{ xT_1=y \}$
in
$\mathbb{A}^3_{x,y,T_1}= S'' \cap \{T_0=1 \}$
, where
$P_0$
is
$\{x=y=T_1=0\}$
. Clearly
$U_0 \cong \mathbb{A}^2_{x,T_1}$
and
$\rho$
acts on
$U_0$
by the matrix
$\begin{pmatrix} \zeta_3 & 0 \\ 0 & \zeta_3\end{pmatrix}$
. Therefore, the blow-up of S
′′ at
$P_0$
has an exceptional divisor that is pointwise fixed by
$\rho$
. The reasoning is analogous for an open chart around the point
$P_1=\{x=y=T_0=0 \}$
.
In summary, let S
′′ be the blow-up of S at
$S_3$
and let S
′ be the blow-up of S
′′ at the locus of fixed points of the induced action of
$\rho$
. The computation above shows that the locus of
$\rho$
-fixed points in S
′ is a divisor, so the quotient
$S'/\rho$
is smooth by the Chevalley–Shepard–Todd theorem, see for instance [
Reference Shephard and Todd19
, Theorem 5·1]. Since Y is the minimal resolution of
$S/\rho$
, we obtain a birational morphism
$S'/\rho \to Y$
, which fits into the diagram (3·1). Finally, the construction of S
′ descends to k, since the blow-up locus is Galois stable.
Remark 3·2. Note that the morphism
$S'/\rho \to Y$
is not an isomorphism, because in
$S'/\rho$
there are three geometric curves above each singularity of X. Using the projection formula it is possible to show that those curves consist of a
$(\!-\!1)$
-curve intersecting two disjoint
$(\!-\!3)$
-curves. This implies that the morphism
$S'/\rho \to Y$
is the blow-down of these nine geometric
$(\!-\!1)$
-curves.
The additional geometry of S
′ and
$S'/\rho$
does not affect the Brauer groups.
Lemma 3·3. The morphism
$S'/\rho \to Y$
induces an isomorphism
$\operatorname{Br}\!(Y) \xrightarrow{\sim} \operatorname{Br}\!(S'/\rho)$
and the inclusion
$S_0 \hookrightarrow S'$
induces an isomorphism
$ \operatorname{Br}\!(S') \xrightarrow{\sim} \operatorname{Br}\!(S_0)$
.
Proof. Since the morphism
$S'/\rho \to Y$
is a birational morphism of smooth surfaces, the induced map
$\operatorname{Br}\!(Y) \to \operatorname{Br}\!(S'/\rho)$
is an isomorphism ([
Reference Grothendieck6
, Corollary 7·2]). Similarly, the morphism
$S' \to S$
induces an isomorphism
$ \operatorname{Br}\!(S) \to \operatorname{Br}\!(S')$
. The composition of the injective maps
\[ \operatorname{Br}\!(S) \xrightarrow{\sim} \operatorname{Br}\!(S') \xrightarrow{} \operatorname{Br}\!(S_0) \]
is an isomorphism by Lemma 2·1, so the restriction map
$\operatorname{Br}\!(S') \to \operatorname{Br}\!(S_0)$
is an isomorphism.
Lemma 3·4. The map
$\pi \;:\; S_0 \to Y_0\;:\!=\;S_0/\rho$
is a finite flat covering of regular surfaces such that
$k(S_0)/k(Y_0)$
is Galois of degree 3. The same holds for the map
$\pi\;:\; S' \to S'/\rho$
.
Proof. Every finite surjective map of regular varieties is flat, so
$\pi \;:\; S_0 \to Y_0$
is flat (from [
Reference Liu and Erne12
, Remark 4·3·11]). Moreover,
$\rho \in \operatorname{Aut}_{Y_0}(S_0)$
induces a nontrivial automorphism of
$k(S_0)$
over
$k(Y_0)$
, because we can find a coordinate
$f \in k(S_0)$
such that
$f(x)=0$
and
$f(\rho(x)) \neq 0$
for some
$x \in S_0$
. For instance, we choose
$g,h \in k(C)$
such that
$\mathrm{div}(g)$
and
$\mathrm{div}(h)$
have disjoint support and such that
$g(P)=0$
for some point
$P \in C$
which is not an inflexion point of C. Moreover, we can choose
$Q, R \in C$
collinear with P such that Q, R are not in the supports of g and h. Consider
$f=g \otimes h \in k(C) \otimes k(C) = k(C \times C)=k(S_0)$
, and we have that
$f((P,Q))=0$
but
$f(\rho(P,Q))=f((Q,R)) \neq 0$
.
Now
$k(S_0) / k(S_0)^{\langle \rho \rangle}$
is a Galois extension of degree 3 and, since
$k(Y_0) \subset k(S_0)^{\langle \rho \rangle} \subset k(S_0)$
with
$k(S_0)/k(Y_0)$
degree 3, we have that
$k(Y_0) = k(S_0)^{\langle \rho \rangle}$
. Since the function field is a birational invariant, the same conclusions hold for the map of regular surfaces
$S' \to S'/\rho$
.
The surface S
′ allows us to relate the Brauer groups of
$Y_0=S_0/\rho$
and Y.
Lemma 3·5. For every n, the inclusion
$Y_0 \subset Y$
induces a natural injection of groups
\[ \operatorname{Br}\!(Y)_n \hookrightarrow \operatorname{Br}\!(Y_0)_n, \]
which is an isomorphism for n coprime to 3.
Proof. The inclusion
$Y_0 \subset Y$
induces an injection
$\operatorname{Br}\!(Y)\hookrightarrow \operatorname{Br}\!(Y_0)$
. Moreover, by Proposition 2·2 and Lemmas 3·3, 3·4 we have
$\operatorname{Br}\!(Y)_n \cong \operatorname{Br}\!(S'/\rho)_n \cong \operatorname{Br}\!(S')^{\rho}_n$
for n coprime to 3. We also have
$\operatorname{Br}\!(S')^{\rho}_n \cong \operatorname{Br}\!(S_0)^{\rho}_n$
by Lemma 3·3 and, finally,
$\operatorname{Br}\!(S_0)^{\rho}_n \cong \operatorname{Br}\!(Y_0)_n$
by Proposition 2·2 again.
3·3. Geometric Brauer groups
In order to relate the transcendental Brauer groups of S and Y, we first need to study the relationship between their geometric Brauer groups.
Let A be the abelian variety
$E \times E$
, where
$E= \operatorname{Jac}(C)$
, so that
$\bar{S} \cong \bar{A}$
. We start by explicitly computing
$\operatorname{H}^1(\rho, \operatorname{NS}(\bar{A}))$
using the following description of
$\operatorname{NS}(\overline{E} \times \overline{E})$
:
\[ \operatorname{NS}(\overline{E} \times \overline{E}) \cong \mathbb{Z} \cdot e \oplus \operatorname{End}(\overline{E}) \oplus \mathbb{Z} \cdot e', \]
where e and e
′ are the divisors
$\overline{E} \times \{ 0_E\}$
and
$\{ 0_E \} \times \overline{E}$
, respectively.
Lemma 3·6. We have
$\operatorname{H}^1(\rho, \operatorname{NS}(\bar{A}))=0$
.
Proof. The group
$\operatorname{H}^1(\rho, \operatorname{NS}(\bar{A}))$
can be described as
\[\{ x \in \operatorname{NS}(\bar{A}) \;:\; (1+\rho+\rho^2)(x)=0 \}/(\rho -1)\operatorname{NS}(\bar{A}).\]
Our computational strategy will be to show that
$(\rho -1)\operatorname{NS}(\bar{A})$
is a primitive sublattice of
$\operatorname{NS}(\bar{A})$
and that
$(1+\rho+\rho^2)(x) \neq 0$
for all
$x \in \operatorname{NS}(\bar{A}) \setminus (\rho -1)\operatorname{NS}(\bar{A})$
.
Let
$\phi\;:\; \mathbb{Z} \cdot e \oplus \mathbb{Z} \cdot e' \oplus \operatorname{End}(\overline{E}) \to \operatorname{NS}(\overline{A})$
be the isomorphism given by Corollary 2·5
\[(a, b, f) \mapsto (a-1) e + (b- \deg f) e' + \Gamma_{f},\]
where
$\Gamma_f$
is the divisor associated to the graph
$\{ (P,f(P))\}_{P \in E}$
. We study the induced action of
$\rho$
on the left-hand side using
$\phi$
and the action of
$\rho^*$
on
$\operatorname{NS}(\overline{A})$
.
First, suppose that E has CM by an order
$\mathcal{O}$
, which is generated by
$\alpha$
. Let
$x^2+cx+d \in \mathbb{Z}[x]$
be the minimal polynomial of
$\alpha$
. Note that
$\Gamma_0=e$
:
-
(i)
$\rho(1,0,0)=\rho(\phi^{-1}(\Gamma_0)) = \phi^{-1}(\rho^*(\Gamma_0))=\phi^{-1} (\Gamma_{-1})=(1, 1, -1) $
; -
(ii)
$\rho(0,1,0)= \phi^{-1}(\rho^*(e'))=\phi^{-1}(e)=(1,0,0)$
; -
(iii)
$\rho(0,0, 1)=\phi^{-1}(\rho^*(\!-e + \Gamma_{1}-e' )) = (\!-\!1,-1, 1) + (\!-\!1,0,0) + \phi^{-1}(\rho^*(\Gamma_{1}))$
; we have that
$\rho^*(\Gamma_{1})$
is the divisor supported on
$\Gamma^{-1}_{-2}=\{(\!-\!2P,P)\}_{P \in E}$
. Using Lemma 2·6, we have
$\phi^{-1}(\rho^*(\Gamma_{1}))=(4,1,-2)$
, thus
$\rho(0,0,1)= (\!-\!2, -1, 1) + (4,1,-2)= (2,0,-1)$
; -
(iv)
$\rho(0,0, \alpha)=\phi^{-1}(\rho^*(\!-e+\Gamma_{\alpha}-d\cdot e'))= (\!-\!1,-1,1) + (\!-d,0,0) + \phi^{-1}(\rho^*(\Gamma_{\alpha}))$
; as before, Lemma 2·6 implies
$\rho^*(\Gamma_{\alpha}) = \Gamma^{-1}_{-1-\alpha}=\phi^{-1}(1-c+d, 1, (\!-\!1-\alpha)^{\vee})$
. Using that
$(\!-\!1-\alpha)^{\vee}=-1+ \alpha +c$
, we obtain
$\rho(0,0,\alpha)=(\!-d-1,-1,1)+$
$(1-c+d,1,-1+\alpha+c)=(\!-c,0,\alpha+c)$
.
The sublattice
$(\rho -1)\operatorname{NS}(\bar{A})$
is the primitive sublattice generated by
$(0, 1, -1)$
and
$(1,0,-1)$
; a direct complement of it in
$\operatorname{NS}(\bar{A})$
is the sublattice generated by (0, 1, 0) and
$(0,0,\alpha)$
. Calculations lead to
$(1+\rho+\rho^2)(0,1,0)=(2,2,-1)$
and
$(1+\rho+\rho^2)(0,0,\alpha)=(\!-c,-c, 3\alpha+2c)$
, hence any nontrivial
$\mathbb{Z}$
-linear combination of these cannot vanish.
In the case of a non-CM elliptic curve E, the same computation shows that the sublattice
$(\rho -1)\operatorname{NS}(\bar{A})$
is the primitive sublattice generated by
$(0, 1, -1)$
and
$(1,0,-1)$
, and that
$(1+\rho+\rho^2)(0,1,0)=(2,2,-1)$
, yielding the same result.
Proposition 3·7. The natural map
$\pi^*\;:\; \operatorname{Br}\!(\bar{Y}_0) \to \operatorname{Br}\!(\bar{S}_0)$
is an isomorphism, which induces an isomorphism of
$\Gamma$
-modules
$\operatorname{Br}\!(\bar{Y}) \to \operatorname{Br}\!(\bar{S})$
; in particular, the action of
$\rho$
on
$\operatorname{Br}\!(\bar{S})$
is trivial.
Proof. Consider the short exact sequence
\[ 0 \to A^{\vee}(\bar{k}) \to \operatorname{Pic}(\bar{A}) \to \operatorname{NS}(\bar{A}) \to 0 \]
on which
$\rho$
acts via pullbacks. Thus, we have a long exact sequence of group cohomology involving the terms
\[ \cdots \to \operatorname{H}^1(\rho, A^{\vee}(\bar{k})) \to \operatorname{H}^1(\rho, \operatorname{Pic}(\bar{A})) \to \operatorname{H}^1(\rho, \operatorname{NS}(\bar{A})) \to \cdots. \]
Notice that
$\rho$
acts on
$E \times E$
by mapping
$(P,Q) \mapsto (Q, -P-Q)$
, so on
$(E \times E)^{\vee}$
by mapping
$(P,Q) \mapsto (\!-P-Q,P)$
. Then,
$(\rho -1)(P,Q)=(\!-\!2P-Q, P-Q)$
, which is surjective on
$A^{\vee}(\bar{k})$
, hence
$\operatorname{H}^1(\rho, A^{\vee}(\bar{k}))=0$
. Using Lemma 3·6, the exact sequence in group cohomology shows that
$\operatorname{H}^1(\rho, \operatorname{Pic}(\bar{A}))=0$
.
By Lemma 3·4
$\pi\;:\; S_0 \to Y_0$
is a flat Galois covering with Galois group
$\mathbb{Z} /3$
, therefore there is the associated Hochschild–Serre spectral sequence ([
Reference Milne13
, Theorem III·2·20])
\[ \operatorname{H}^p\big(\mathbb{Z} /3, \operatorname{H}^q_{\acute{\mathrm{e}\mathrm{t}}} (\bar{S}_0, \mathbb{G}_m)\big) \Longrightarrow \operatorname{H}^{p+q}_{\acute{\mathrm{e}\mathrm{t}}}(\bar{Y}_0, \mathbb{G}_m),\]
which yields the following exact sequence:
\begin{align*} \cdots \to \operatorname{H}^2\big(\rho, \operatorname{H}^0_{\acute{\mathrm{e}\mathrm{t}}}(\bar{S}_0,\mathbb{G}_m)\big) &\to \ker\big(\!\operatorname{H}^2_{\acute{\mathrm{e}\mathrm{t}}}(\bar{Y}_0, \mathbb{G}_m) \to \operatorname{H}^0(\rho, \operatorname{H}^2_{\acute{\mathrm{e}\mathrm{t}}}(\bar{S}_0,\mathbb{G}_m)) \big)\\ &\to \operatorname{H}^1\big(\rho, \operatorname{H}^1_{\acute{\mathrm{e}\mathrm{t}}}(\bar{S}_0,\mathbb{G}_m)\big) \to \cdots. \end{align*}
Moreover,
\[\operatorname{H}^2\big(\rho, \operatorname{H}^0_{\acute{\mathrm{e}\mathrm{t}}}(\bar{S}_0,\mathbb{G}_m)\big)= \operatorname{H}^2\big(\rho, \bar{k}^{\times}\big)= \bar{k}^\times/\bar{k}^{\times 3} =0,\]
\[\operatorname{H}^1\big(\rho, \operatorname{H}^1_{\acute{\mathrm{e}\mathrm{t}}}(\bar{S}_0,\mathbb{G}_m)\big)=\operatorname{H}^1(\rho, \operatorname{Pic}(\bar{S}))=0,\]
so we deduce an injection
$\operatorname{Br}\!(\bar{Y}) \hookrightarrow \operatorname{Br}\!(\bar{Y}_0) \hookrightarrow \operatorname{H}^0(\rho, \operatorname{Br}\!(\bar{S}_0)) \hookrightarrow \operatorname{Br}\!(\bar{S}_0) \cong \operatorname{Br}\!(\bar{S})$
. The discussion in Section 2 shows that, for an abelian surface or a K3 surface V, we have
\[\operatorname{Br}\!(\bar{V})=(\mathbb{Q}/\mathbb{Z})^{b_2(V) - \rho(V)}.\]
Now,
$b_2(\bar{Y})=22$
and
$b_2(\bar{S})=6$
; see for instance [
Reference Huybrechts8
, Chapter 1, Section 2] and [
Reference Milne14
, Section 15]. By [
Reference van Luijk26
, Proposition 3·11, 3·12] we have
$\rho(\bar{Y})=16+\rho(\bar{S})$
, which implies that
\[b_2(\bar{Y}) - \rho(\bar{Y})=b_2(\bar{S}) - \rho(\bar{S}).\]
This shows that
$\operatorname{Br}\!(\bar{Y})$
and
$\operatorname{Br}\!(\bar{S})$
are both abstractly isomorphic to
$(\mathbb{Q}/\mathbb{Z})^2$
if
$E=\operatorname{Jac} C$
has CM or
$(\mathbb{Q}/\mathbb{Z})^3$
in the non-CM case.
To conclude the proof, it is sufficient to notice that every injective homomorphism
$(\mathbb{Q}/\mathbb{Z})^n \hookrightarrow (\mathbb{Q}/\mathbb{Z})^n$
is an isomorphism, because, when restricted to the m-torsion, it induces an isomorphism
$(\mathbb{Z}/m\mathbb{Z})^n \hookrightarrow (\mathbb{Z}/m\mathbb{Z})^n$
for every m.
Remark 3·8. Consider the exact sequence coming from the Kummer sequence
\[ 0 \to \operatorname{NS}(\bar{S}) \otimes \mathbb{Z}/l^n \to \operatorname{H}^2_{\acute{\mathrm{e}\mathrm{t}}}(\bar{S},\mu_{l^n}) \to \operatorname{Br}\!(\bar{S})_{l^n} \to 0. \]
In the Kummer surface case, the action of the involution
$[\!-\!1]_{\bar{S}}$
induces a trivial action on
$\operatorname{H}^2_{\acute{\mathrm{e}\mathrm{t}}}(\bar{S},\mu_{l^n}) = \bigwedge^{2} \operatorname{H}^1_{\acute{\mathrm{e}\mathrm{t}}}(\bar{S},\mu_{l^n})$
, whereas the action of
$\rho$
on
$\operatorname{H}^2_{\acute{\mathrm{e}\mathrm{t}}}(\bar{S},\mu_{l^n}) $
is not trivial because it is nontrivial on
$\operatorname{NS}(\bar{S})$
. Nevertheless, we are lucky since the nontriviality of the action of
$\rho$
comes only from the Néron–Severi group, i.e. the action on the quotient
$\operatorname{Br}\!(\bar{S})_{l^n}$
is still trivial and this allows a result analogous to [
Reference Skorobogatov and Zarhin24
, Theorem 2·4].
Remark 3·9. The proof of Proposition 3·7 also shows that the injection
$\operatorname{Br}\!(\bar{Y}) \hookrightarrow \operatorname{Br}\!(\bar{Y_0})$
is an isomorphism. This can also be shown directly using the sequence
\[ 0 \to \operatorname{Br}\!(\bar{Y}) \to \operatorname{Br}\!(\bar{Y_0}) \to \bigoplus \operatorname{H}^1_{\acute{\mathrm{e}\mathrm{t}}} (Z_i, \mathbb{Q} / \mathbb{Z}), \]
where the sum ranges over all connected components of
$\overline{Y}\setminus \overline{Y}_0$
. Indeed, every
$Z_i$
is isomorphic to
$\mathbb{P}^1_{\overline{k}} \cup_{pt} \mathbb{P}^1_{\overline{k}}$
, i.e. two genus 0 curves intersecting in one point, since
$Z_i$
arises from the resolution of a
$A_2$
-singularity. We have
$\operatorname{H}^1_{\acute{\mathrm{e}\mathrm{t}}} (\mathbb{P}^1_{\overline{k}} \cup_{pt} \mathbb{P}^1_{\overline{k}} , \mathbb{Q} / \mathbb{Z})=0$
, since
$\pi_1^{top}(\mathbb{P}^1_{\mathbb{C}} \cup_{pt} \mathbb{P}^1_{\mathbb{C}}, \bar{x})=0$
by the Seifert–Van Kampen Theorem.
3·4. Transcendental Brauer groups
A final lemma before proving the main result of this section:
Lemma 3·10. For every n, we have a natural injection
\[ \frac{\operatorname{Br}\!(Y)_n}{\operatorname{Br}_1(Y)_n} \hookrightarrow \frac{\operatorname{Br}\!(Y_0)_n}{\operatorname{Br}_1(Y_0)_n}, \]
which is an isomorphism for n coprime to 3.
Proof. For every n, we have
where
$i^*\;:\; \operatorname{Br}\!(Y)_n \to \operatorname{Br}\!(Y_0)_n$
is the injective map induced by the inclusion. The isomorphism in the diagram follows from Remark 3·9. In order to quotient by the algebraic Brauer groups, we need
\[i^* \operatorname{Br}\!(Y)_n \cap \operatorname{Br}_1(Y_0)_n= i^* \operatorname{Br}_1(Y)_n. \]
From one side we have
$i^* \operatorname{Br}_1(Y)_n \subset i^* \operatorname{Br}\!(Y)_n$
and
$i^* \operatorname{Br}_1(Y)_n \subset \operatorname{Br}_1(Y_0)_n$
. On the other hand, if
$i^*x \in i^* \operatorname{Br}\!(Y)_n \cap \operatorname{Br}_1(Y_0)_n $
then,
$i^*\bar{x}=0$
hence
$\bar{x}=0$
by the isomorphism, implying
$x \in \operatorname{Br}_1(Y)_n$
. The statement for n coprime to 3 follows by Lemma 3·5.
Theorem 3·11. The map
$\pi^*\;:\; \operatorname{Br}\!(Y_0) \to \operatorname{Br}\!(S_0)$
induces an embedding
\[ \frac{\operatorname{Br}\!(Y)_n}{\operatorname{Br}_1(Y)_n} \hookrightarrow \frac{\operatorname{Br}\!(S)_n}{\operatorname{Br}_1(S)_n}, \]
which is an isomorphism if n is coprime to 3. This induces an isomorphism
\[ \left(\frac{\operatorname{Br}\!(Y)}{\operatorname{Br}_1(Y)}\right )_n \xrightarrow{\sim} \left(\frac{\operatorname{Br}\!(S)}{\operatorname{Br}_1(S)}\right)_n \]
for all n coprime to 3.
Proof. By Proposition 3·7, the commutative diagram
gives the embedding
$\operatorname{Br}\!(Y_0)_n/\operatorname{Br}_1(Y_0)_n \hookrightarrow \operatorname{Br}\!(S_0)_n/\operatorname{Br}_1(S_0)_n \cong \operatorname{Br}\!(S)_n/\operatorname{Br}_1(S)_n$
. Hence Lemma 3·10 implies the existence of the required embedding for all n.
Consider now n coprime to 3. The natural map
$\operatorname{Br}\!(S) \to \operatorname{Br}\!(\bar{S})$
naturally commutes with
$\rho$
. Since
$\rho$
acts trivially on
$\operatorname{Br}\!(\bar{S})$
by Proposition 3·7, we have that
$\rho(\bar{x})-\bar{x}=0$
for any
$x \in \operatorname{Br}\!(S)_n $
. Therefore
$\rho(x)-x \in \operatorname{Br}_1(S)_n$
and, similarly,
$\rho^2(x)-x \in \operatorname{Br}_1(S)_n$
. Then,
\[ 3x = (x+ \rho(x) +\rho^2(x)) + (x - \rho(x))+(x-\rho^2(x)) \in \operatorname{Br}\!(S)_n^{\rho} + \operatorname{Br}_1(S)_n\]
and the assumption that n is coprime to 3 implies that
$\operatorname{Br}\!(S)_n=\operatorname{Br}\!(S)_n^{\rho} + \operatorname{Br}_1(S)_n$
, hence
\[ \operatorname{Br}\!(S)_n/\operatorname{Br}_1(S)_n = \operatorname{Br}\!(S)_n^{\rho}/\operatorname{Br}_1(S)_n^{\rho}. \]
Moreover, we obtain the diagram
by applying Proposition 2·2 to the triple covering
$\pi\;:\;S_0 \to Y_0$
together with
$\operatorname{Br}\!(S_0)=\operatorname{Br}\!(S)$
by Lemma 3·3. Combining the diagram with Lemma 3·10 proves the first statement.
Finally, the second statement follows from the claim that an element of order coprime to 3 in
$\operatorname{Br}\!(Y)/\operatorname{Br}_1(Y)$
comes from
$\operatorname{Br}\!(Y)_n$
for some n coprime to 3. The proof of this claim only uses that the Brauer group is torsion.
Indeed, let us fix
$x \in [x] \in (\!\operatorname{Br}\!(Y)/\operatorname{Br}_1(Y))_n$
with n coprime to 3; we have
$n \cdot x \in \operatorname{Br}_1(Y)$
, so there exists
$m\gt0$
such that
$mn\cdot x=0$
, so
$x \in \operatorname{Br}\!(Y)_{mn}$
. Let
$m=3^am'$
with m
′ coprime to 3; there exists
$a'\gt a$
such that
\[ x'\;:\!=\; 3^{a'} \cdot x \in [x] \in (\!\operatorname{Br}\!(Y)/\operatorname{Br}_1(Y))_n\]
by considering any a
′ such that
$3^{a'} \equiv 1 \pmod{n}$
. The proof is concluded, since
$x' \in [x]$
and
$ x' \in \operatorname{Br}\!(Y)_{m'n}$
with m
′
n coprime to 3.
4. Transcendental Brauer group of S
In the previous section we have related the transcendental Brauer groups of
$S=C \times C$
and of the K3 surface Y. Let us now study the former, exploiting the fact that S is a torsor of an abelian surface.
In this section, we assume C to be a diagonal cubic curve:
$ax^3+by^3+cz^3=0$
, defined over
$\mathbb{Q}$
, so we can assume a, b, c to be a primitive triple of integers. Let L be a cubic extension of
$\mathbb{Q}$
such that
$C_L$
has a rational point, e.g. the number field
$\mathbb{Q}(\sqrt[3]{{a}/{b}})$
. We have
$C_L \cong E_L$
, where
$E=\operatorname{Jac}(C)$
is described by
$y^2=x^3-2^43^3a^2b^2c^2$
. The equation of the Jacobian derives from classical results like [
Reference An, Kim, Marshall, Marshall, McCallum and Perlis1
]. Notice that this elliptic curve has CM by
$\mathcal{O}_K$
with
$K=\mathbb{Q}(\sqrt{-3})$
, as expected from the diagonal form of the curve C.
The compatibility of the restriction maps
implies the injection
$\operatorname{Br}\!(C \times C)/\operatorname{Br}_1(C \times C) \hookrightarrow \operatorname{Br}\!(E_L \times E_L)/\operatorname{Br}_1(E_L \times E_L)$
. In full generality, we would need to study the right-hand side, but, away from 3, we can reduce this to
$\operatorname{Br}\!(E \times E)$
thanks to the following proposition.
Proposition 4·1 ([
Reference Várilly-Alvarado and Viray28
, 3·3]). Let N be a positive integer, let X and X’ be principal homogeneous spaces of an abelian variety A, defined over the same field, and let
$f \;:\; X' \to X$
be an N-covering. Then, for any integer n coprime to N, we have the isomorphism
\[f^*\;:\; \frac{\operatorname{Br}\!(X)_n}{\operatorname{Br}_1(X)_n} \xrightarrow{\sim} \frac{\operatorname{Br}\!(X')_n}{\operatorname{Br}_1(X')_n}. \]
Fortunately, the transcendental Brauer groups of products of CM elliptic curves have been extensively studied in [ Reference Newton16 , Reference Newton17 ]. We now record the required results from that paper, stated directly under our assumptions on L and K.
Theorem 4·2 ([
Reference Newton16
, 2·9 and 2·2]). Let
$E/L$
be an elliptic curve with CM by
$\mathcal{O}_K$
; let l be a prime number, then
\[ \bigg(\frac{\operatorname{Br}\!(E \times E )}{\operatorname{Br}_1(E \times E )} \bigg)_{l^{\infty}}= \bigg(\frac{\operatorname{Br}\!(E \times E )_{l^m}}{\operatorname{Br}_1(E \times E )_{l^m}} \bigg) \cong \mathbb{Z}/l^m, \]
where
$m=m(l)$
is the largest integer k such that every prime
$\mathfrak{q}$
of KL of good reduction for
$E_{KL}$
and coprime to l satisfies
$\psi_{E/KL}(\mathfrak{q}) \in \mathcal{O}_{l^k}=\mathbb{Z} + l^k \mathcal{O}_K$
for the Hecke character
$\psi_{E/KL}$
. Let n(l) be the largest integer k such that the ring class field
$K_{l^k}$
of the order
$\mathcal{O}_{l^k}$
embeds into KL; then,
$m(l) \leq n(l)$
.
Moreover, Newton explicitly computes the transcendental Brauer group of
$E \times E$
for a CM elliptic curve
$E\;:\; y^2=x^3+D$
defined over
$\mathbb{Q}$
.
Proposition 4·3 ([
Reference Newton16
, Example 3]). Let E be the elliptic curve over
$\mathbb{Q}$
with equation
$y^2=x^3+D$
with
$D \in \mathbb{Z} \setminus \{ 0\}$
. Then
\[ \frac{\operatorname{Br}\!(E \times E)}{\operatorname{Br}_1(E \times E)} = \begin{cases} \mathbb{Z}/2\mathbb{Z} &\text{ if } D \text{ is a cube in } \mathbb{Z}; \\ \mathbb{Z}/3\mathbb{Z} &\text{ if } 4D \text{ is a cube in } \mathbb{Z}; \\ 0 &\text{ otherwise.}\end{cases} \]
We recall the following property of Hecke characters:
Lemma 4·4. Let
$E/K$
be an elliptic curve with CM by
$\mathcal{O}_K$
and let L be a number field. Let
$\mathfrak{q}$
be a prime in KL lying over the prime
$\mathfrak{p}$
in K. Then,
\[\psi_{E/KL}(\mathfrak{q})=\psi_{E/K}(\mathfrak{p})^{f_{\mathfrak{q}/\mathfrak{p}}},\]
where
$f_{\mathfrak{q}/\mathfrak{p}}$
is the relative inertia degree.
Proof. This derives from the definition of the Hecke character
$\psi_{E}$
and the compatibility of the Artin map with the norm map, see [
Reference Silverman20
, II·9·1, II·3·5].
Finally, we need an elementary lemma before computing the transcendental Brauer group of
$C \times C$
in our case of interest.
Lemma 4·5. Let a, b, c be positive integers such that abc is not a cube in
$\mathbb{Z}$
. Then, there exists a choice of
$\lambda \in \{a/b,b/c, c/a\}$
such that
$\mathbb{Q}(\sqrt[3]{abc}) \neq \mathbb{Q}(\sqrt[3]{\lambda})$
.
Proof. First recall that
$\mathbb{Q}(\sqrt[3]{abc}) \neq \mathbb{Q}(\sqrt[3]{\lambda})$
if and only if
$abc \not\sim \lambda$
and
$abc \not\sim \lambda^2$
in
$\mathbb{Q}^{\times}/\mathbb{Q}^{\times 3}$
. In particular, if
$\lambda=a/b$
, the conditions become
$b^2c \not\sim 1$
and
$a^2c \not\sim 1$
. If
$\lambda=b/c$
, then the conditions become
$ac^2\not\sim 1$
and
$ab^2 \not\sim 1$
. Finally, if
$\lambda=c/a$
, they are
$a^2b \not\sim 1$
and
$bc^2 \not\sim 1$
.
Notice that at least one between
$b^2c \not\sim 1$
and
$a^2c \not\sim 1$
is satisfied, since
$abc \not \sim 1$
. If they are both satisfied, choose
$\lambda=a/b$
. Otherwise let us start by assuming
$b^2c \sim 1$
and therefore
$a^2c \not\sim 1$
; using again that
$abc \not\sim 1$
, we have
$ab^2 \not\sim 1$
, so
$\lambda=b/c$
works. On the other hand, if we assume
$a^2c \sim 1$
and therefore
$b^2c \not\sim 1$
, then
$a^2b \not \sim 1$
and
$\lambda= c/a$
works.
Theorem 4·6. Let
$C/\mathbb{Q}$
be a diagonal cubic curve
$ax^3+by^3+cz^3=0$
such that abc is not a cube in
$\mathbb{Q}$
. Then,
\[\frac{\operatorname{Br}\!(C \times C)}{\operatorname{Br}_1(C \times C)}= \begin{cases} \mathbb{Z}/2\mathbb{Z} &\text{ if } 4abc \text{ is a cube in } \mathbb{Q}; \\ 0 &\text{ otherwise.}\end{cases} \]
Proof. If
$C(\mathbb{Q}) \neq \emptyset$
, then
$C \cong E= \operatorname{Jac}(C)$
and therefore Proposition 4·3 applies directly with
$D=-2^43^3a^2b^2c^2$
. We can therefore assume that
$C(\mathbb{Q}) = \emptyset$
; the proof is still a modification of the reasoning in [
Reference Newton16
, Example 3].
The statement for the l-primary part of the transcendental Brauer group for
$l \neq 3$
is a direct implication of Proposition 4·1 and Proposition 4·3. Again let
$D=-2^43^3a^2b^2c^2$
and let
$\mathfrak{q}$
be a prime in KL coprime to D; denote by
$\pi_{\mathfrak{p}}$
the unique generator of
$\mathfrak{p}=\mathfrak{q} \cap \mathcal{O}_K$
such that
$\pi_{\mathfrak{p}} \equiv 1 \pmod{3}$
. Then, Lemma 4·4 and [
Reference Silverman20
, Example II·10·6] show that
\[ \psi_{E/KL}(\mathfrak{q})= \bigg[ {\left( \frac{4D}{\pi_{\mathfrak{p}}} \right) }^{-1}_6 \cdot \pi_{\mathfrak{p}} \bigg]^{f_{\mathfrak{q}/\mathfrak{p}}} , \]
where
${\left( \frac{\cdot}{\cdot} \right) }_6$
denotes the sextic residue symbol on
$\mathbb{Z}[\zeta_3]$
.
Let
$l=3$
. We have assumed that 4D is not a cube in
$\mathbb{Z}$
. Moreover, choose
$L=\mathbb{Q}(\sqrt[3]{\lambda})$
given by Lemma 4·5, so that
$KL \neq K(\sqrt[3]{4D})$
(indeed fixed an embedding of KL into
$\mathbb{C}$
, L is the only real subfield of degree 3 over
$\mathbb{Q}$
). Since 4D is not a cube in K, then by [
Reference Cassels and Fröhlich3
, Exercise 6·1] there are infinitely many primes
$\mathfrak{p}$
of K such that 4D is not a cube mod
$\mathfrak{p}$
, which is equivalent to
$\mathfrak{p}$
being inert in
$K(\sqrt[3]{4D})/K$
. Again by [
Reference Cassels and Fröhlich3
, Exercise 6·1], we can also assume that
$\mathfrak{p}$
is split in
$KL/K$
, since
$KL \neq K(\sqrt[3]{4D})$
. Given all these choices, we have found a prime
$\mathfrak{q}$
in KL above
$\mathfrak{p}$
such that
$f_{\mathfrak{q}/\mathfrak{p}}=1$
and
${({4D}/{\pi_{\mathfrak{p}}}) }_6 \neq \pm 1$
, therefore
$\psi_{E/KL}(\mathfrak{q}) \not\in \mathcal{O}_3=\mathbb{Z}+3\mathcal{O}_K$
, giving
$m(3)=0$
in Theorem 4·2.
Corollary 4·7. Let
$C/\mathbb{Q}$
be a diagonal cubic curve
$ax^3+by^3+cz^3=0$
such that abc is not a cube in
$\mathbb{Q}$
. Then,
\[\frac{\operatorname{Br}\!(Y_C)}{\operatorname{Br}_0(Y_C)}= \begin{cases} \mathbb{Z}/2\mathbb{Z} &\text{ if } 4abc \text{ is a cube in } \mathbb{Q}; \\ 0 &\text{ otherwise.}\end{cases} \]
Proof. By comparing Theorems 3·11 and 4·6, we obtain
$\operatorname{Br}\!(Y_C) / \operatorname{Br}_1(Y_C)$
and by [
Reference van Luijk26
, Proposition 4·2], we have that
$\operatorname{Br}\!(\mathbb{Q}) \to \operatorname{Br}_1(Y_C)$
is surjective.
Remark 4·8. This result agrees with Várilly-Alvarado’s conjecture on the uniform boundedness of the Brauer groups of K3 surfaces, see [
Reference Várilly-Alvarado27
, 5·5]. Moreover, the cardinality of the transcendental Brauer group of
$Y_C$
can be bounded depending only on the base field k, thanks to the bound of n(l), given by class field theory as discussed in [
Reference Newton16
, Remark 1]. Indeed, Várilly-Alvarado’s conjecture has been proven for CM K3 surfaces (such as our case) in [
Reference Orr and Skorobogatov18
].
5. On the Hasse principle
We can now discuss the implications of the computation of the Brauer group of
$Y_C$
assuming Skorobogatov’s conjecture for these K3 surfaces.
Conjecture 5·1 (Skorobogatov, [ Reference Skorobogatov23 ]). The Brauer–Manin obstruction is the only obstruction to the Hasse principle for algebraic K3 surfaces over number fields.
Theorem 5·2. Let
$C/\mathbb{Q}$
be an everywhere locally soluble smooth curve with equation
$ax^3+by^3+cz^3=0$
. Then, Skorobogatov’s conjecture implies the existence of Galois cubic points on C.
Proof. Section 2 in [
Reference van Luijk26
] describes how to relate k-rational points on
$Y_C$
and C, for a field k of characteristic zero. First, the surface
$Y_C$
has a k-rational point if and only if
$S/\rho$
does. Secondly, a
$\overline{k}$
-point of
$S/\rho$
, described by the orbit
$\{(P,Q), (Q,R), (R,P) \}$
, is k-rational if either
-
(1) P, Q, R are k-rational collinear points on C, or
-
(2) P, Q, R are collinear l-rational points on C, where
$l/k$
is a Galois cubic extension and P, Q, R are permuted by
$\operatorname{Gal}(l/k)$
.
If neither abc nor 4abc is a cube in
$\mathbb{Q}$
, then Corollary 4·7 and Conjecture 5·1 imply the Hasse principle for
$Y_C$
, which implies the existence of Galois cubic points on C. If abc or 4abc is a cube in
$\mathbb{Q}$
, we can show that C must have a rational point, as in [
Reference van Luijk26
, Remark 4·3]. We first reduce to the case of
$a,b,c \in \mathbb{Z}$
with
$\gcd(a,b,c)=1$
. If abc is a cube in
$\mathbb{Q}$
, then for any prime
$p \mid abc$
, we can assume that
$p \nmid a$
,
$p\mid\mid b$
and
$p^2 \mid\mid c$
. By an infinite descent argument, these conditions imply
$C(\mathbb{Q}_p)=\emptyset$
, contradicting the hypothesis of C being everywhere locally soluble. This shows that a, b, c are units in
$\mathbb{Z}$
and therefore C has a rational point. If 4abc is a cube in
$\mathbb{Q}$
, the analogous argument shows that we can reduce to the case
$a=\pm1$
,
$b=\pm 1$
,
$c=\pm 2$
and all such C have a rational point.
Let us finally consider the case where
$Y_C$
is everywhere locally soluble, which, as explained in the proof of Theorem 5·2, is a weaker assumption than C being everywhere locally soluble. We are able to obtain the same result as in Theorem 5·2, assuming that abc is not a rational cube.
Theorem 5·3. Let
$C/\mathbb{Q}$
be a smooth curve with equation
$ax^3+by^3+cz^3=0$
, with abc not a rational cube, such that the surface
$Y_C$
is everywhere locally soluble. Then, there is no Brauer–Manin obstruction to the Hasse principle on
$Y_C$
. Consequently, Skorobogatov’s conjecture implies the existence of Galois cubic points on C.
Proof. If 4abc is not a rational cube, then the result follows from Corollary 4·7. Assume now that 4abc is a rational cube. Then Corollary 4·7 shows that
$\operatorname{Br}\!(E \times E)/\operatorname{Br}_1(E \times E) \cong \operatorname{Br}\!(C \times C)/\operatorname{Br}_1(C \times C) \cong \operatorname{Br}\!(Y_C)/\operatorname{Br}_0(Y_C) \cong \mathbb{Z}/2\mathbb{Z}$
. Let
$\beta=(x-3,u-3) \in \operatorname{Br}\!(k(E \times E))$
, where the two components of
$E \times E$
are given by
$y^2=x^3-27$
and
$v^2=u^3-27$
. Now, [
Reference Skorobogatov and Zarhin24
, Proposition 3·3] states that
\[ \frac{\operatorname{Br}\!(E \times E)_2}{\operatorname{Br}_1(E \times E)_2} \cong \frac{\operatorname{End}_{\Gamma_\mathbb{Q}}(E_2)}{(\!\operatorname{End}(E)/2)^{\Gamma_\mathbb{Q}}}. \]
In our case, the only nontrivial class is represented by the endomorphism given by complex conjugation on
$E_2$
. The discussion in [
Reference Skorobogatov and Zarhin24
, Section 3] shows that
$ \bar{\beta} \in \operatorname{Br}\!(\overline{E} \times \overline{E})_{2}^{\Gamma_{\mathbb{Q}}}$
and that
$\bar{\beta}$
corresponds, via the isomorphism above, to the complex conjugation endomorphism of
$E_2$
. Alternatively, we can prove the second claim by proving that
$\bar{\beta}$
is nontrivial, similarly to [
Reference Harari and Skorobogatov7
, Proposition 3·8], in the following way.
Let
$K=\bar{\mathbb{Q}}(E)$
and let
$\pi_1 \;:\; \bar{E} \times \bar{E} \to \bar{E}$
be the projection onto the first component. Let
$E_K$
be the generic fibre of
$\pi_1$
, so we view
$E_K$
as an elliptic curve defined over
$K=\bar{\mathbb{Q}}(x)[y]/(y^2-x^3+27)$
by the equation
$v^2=u^3-27$
. Let
$\beta_K$
be the restriction of
$\bar{\beta}$
to
$E_K$
. It is sufficient to show that
$\beta_K$
is a nontrivial element of
$\operatorname{Br} E_K$
. From [
Reference Skorobogatov22
, Theorem 4·1·1, example, p. 63 and exercise 2, p. 91], every element of
$\operatorname{Br} E_K$
which vanishes at the origin is of the form
$(A, u-3) + (B, u-3\zeta_3)$
for some
$A,B \in K^{\times}$
. This element is trivial if and only if the class of
$(A,B) \in (K^{\times}/K^{\times 2})^2$
is in the image of the Kummer map
$E_K(K)/2E_K(K) \to \operatorname{H}^1(K, \mu_2^2)$
. In our case,
$\beta_K$
vanishes at the origin because
$\operatorname{Br}\!(K)=0$
by Tsen’s theorem, so
$\beta_K$
corresponds to the pair
$(A,B)=(x-3,1)$
. Notice that
$E_K(K)= \mathrm{Mor}(\bar{E},\bar{E})=\operatorname{End}(\bar{E}) \oplus E(\bar{\mathbb{Q}})$
, since K-rational points of
$E_K$
are in bijection with sections of
$\pi_1$
. Now,
$E(\bar{\mathbb{Q}})$
is 2-divisible, so
$E_K(K)/2E_K(K) \cong \operatorname{End}(\bar{E})/2\operatorname{End}(\bar{E})$
is generated by the points
$(u,v)=(x,y)$
and
$(u,v)=(\zeta_3 x, y)$
. We conclude by observing that the image of the Kummer map is
$\langle (x-3, x-3\zeta_3), (x-3\zeta_3^2, x-3) \rangle$
, which does not contain
$(x-3,1)$
.
We now discuss the local
$\mathbb{Q}_2$
-evaluation map of
$\beta$
. First notice that the evaluation of
$\beta$
at the origin of
$E \times E$
is trivial. Consider the point
$P=(3,0)$
, which is the only nontrivial rational 2-torsion point of E. The evaluation of
$\beta=(x-3, u-3)=(x^2+3x+9, u^2+3u+9)$
at (P, P) is the quaternion algebra (3, 3) which ramifies at the primes
$p=2,3$
. Thus, we have shown that
$ev_{\beta}\;:\; (E \times E) (\mathbb{Q}_2) \to \mathbb{Z}/2$
is surjective. Moreover, let
$\beta_E=\beta+\rho^*(\beta)+(\rho^{2})^*(\beta)$
. This is a
$\rho$
-invariant element, which represents the class of
$\beta$
in
$\operatorname{Br}\!(E \times E)/\operatorname{Br}_1(E \times E)$
because
$\bar{\beta_E}=3\bar{\beta}=\bar{\beta}$
by Proposition 3·7. We also have that
$ev_{\beta_E}\;:\; (E \times E) (\mathbb{Q}_2) \to \mathbb{Z}/2$
is surjective, because
$ev_{\beta_E}(0_E, 0_E)=3 \cdot ev_{\beta}(0_E, 0_E)=0$
and
$ev_{\beta_E}(P, P)= ev_{\beta}(P, P)+ev_{\beta}(P, 0_E)+ev_{\beta}(0_E, P)=ev_{\beta}(P, P)$
. The strategy now is to chase the points via the natural maps to prove that the surjectivity holds for
$Y_C$
as well.
Note that
$C(\mathbb{Q}_2) \neq \emptyset$
; this is a consequence of the fact that every
$t \in \mathbb{Z}_2^{\times}$
is a cube in
$\mathbb{Z}_2$
and the assumption that 4abc is a rational cube. Let
$\iota \;:\; C \to E$
be a 3-covering generated by a rational linear section D of C, i.e. the map
$\iota(Q)=3[Q]-D \in \operatorname{Pic}^0(C)=E$
. Let
$\tau$
be a
$\mathbb{Q}_2$
-isomorphism
$\tau\;:\; C_{\mathbb{Q}_2} \to E_{\mathbb{Q}_2}$
that makes
$\iota$
into the multiplication
$[3]\;:\; E_{\mathbb{Q}_2} \to E_{\mathbb{Q}_2}$
as in the following diagram.
Let
$\beta_C=(\iota \times \iota)^*(\beta_E) \in \operatorname{Br}\!(C \times C)_2$
; by Proposition 4·1,
$\beta_C$
represents the nontrivial class in
$\operatorname{Br}\!(C \times C)/\operatorname{Br}_1(C \times C)$
. Let
$\beta_{E,2}$
(resp.
$\beta_{C,2}$
) be the base change of
$\beta_E$
(resp.
$\beta_C$
) to
$\mathbb{Q}_2$
. By [
Reference Milne14
, Theorem 15·1], we have
$([3] \times [3])^* \beta_{E,2}=9\beta_{E,2}=\beta_{E,2}$
, thus
$(\tau \times \tau)^*\beta_{E,2}=\beta_{C,2}$
. Since
$\tau$
is an isomorphism, this implies that
$ev_{\beta_{C,2}}\;:\; (C \times C) (\mathbb{Q}_2) \to \mathbb{Z}/2$
is surjective and therefore the same holds for
$\beta_C$
. Notice that our choice of D implies that
$\iota \times \iota$
is a
$\rho$
-equivariant morphism, therefore
$\beta_C$
is
$\rho$
-invariant.
The blow-up map
$S' \to C \times C$
induces an isomorphism of Brauer groups, see Lemma 3·3. Let
$\beta_{S'} \in \operatorname{Br}\!(S')$
be the image of
$\beta_C$
via the blow-up map; then
$\beta_{S'}$
represents the only nontrivial class in
$\operatorname{Br}\!(S')/\operatorname{Br}_1(S')$
and
$ev_{\beta_{S'}}\;:\; S'(\mathbb{Q}_2) \to \mathbb{Z}/2$
is also surjective. Indeed, in the case where the locus
$S_3$
of the blow-up has a
$\mathbb{Q}_2$
-rational point R, the exceptional divisor above R is a pair of genus 0 curves intersecting in a
$\mathbb{Q}_2$
-rational point R
′ and by functoriality we have
$ev_{\beta_{S'}}(R')=ev_{\beta_{C}}(R)$
. Given that the blow-up map is
$\rho$
-equivariant, we also have that
$\beta_{S'}$
is
$\rho$
-invariant.
Finally, consider the map
$\pi\;:\; S' \to Y_C$
. By Proposition 2·2 and Lemma 3·3, we can define
$\beta_Y=(\pi^*)^{-1}\beta_{S'}$
. Theorem 3·11 yields that
$\beta_Y$
represents the only nontrivial class in
$\operatorname{Br}\!(Y_C)/\operatorname{Br}_1(Y_C)=\operatorname{Br}\!(Y_C)/\operatorname{Br}_0(Y_C)$
. By functoriality,
$ev_{\beta_Y}\;:\; Y_C(\mathbb{Q}_2) \to \mathbb{Z}/2$
is surjective. The surjectivity of the evaluation map at a single prime is sufficient to show the non-emptiness of the Brauer–Manin set, see for instance [
Reference Bright2
, Proposition 7·3]. The discussion of rational points on
$Y_C$
in the proof of Theorem 5·2 implies the statement on the existence of Galois cubic points on C.
Acknowledgements
I would like to express my gratitude to my supervisor Rachel Newton for introducing me to this topic and project, giving me careful supervision and proposing several helpful comments on the first draft of the paper. I would also like to thank Alexei Skorobogatov for helpful suggestions, as well as Dario Beraldo, Federico Bongiorno and Riccardo Calini for their patience in our algebraic geometry discussions. Special thanks to Eric Zhu for carefully checking the first draft of the preprint and thanks to the anonymous referee for the many good suggestions on how to improve the exposition. Last but not least, I really appreciated the stimulating and relaxed research atmosphere in Rachel’s group. This work was supported by the Engineering and Physical Sciences Research Council [EP/S021590/1], the EPSRC Centre for Doctoral Training in Geometry and Number Theory (The London School of Geometry and Number Theory), University College London.
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