# Neighbors and arithmetic of isogenous K3 surfaces

Domenico Valloni

###### Abstract.

We use lattice theory to study the isogeny class of a K3 surface. Starting
from isotropic Brauer classes, we construct isogenies via Kneser method of
neighboring lattices. We also determine the fields of definition of isogenous
K3 surfaces, and study Kneser construction over number fields. We then apply
our results to relate conjectures about the finiteness of Brauer groups and
Néron-Severi lattices of K3 surfaces.

##  1\. Introduction

Two complex K3 surfaces $X$ and $Y$ are said to be isogenous if there exists a
Hodge isometry
$f\colon\mathrm{T}(Y,\mathbf{Q})\xrightarrow{\sim}\mathrm{T}(X,\mathbf{Q})$
between their rational transcendental lattices. Isogenies between K3 surfaces
have been studied since Mukai’s seminal work [Muk87] on the moduli spaces of
stable sheaves on K3 surfaces, and his ideas eventually led to a proof that
every rational Hodge isometry is algebraic (see also [Nik07] and [Bus19] for
the cases not considered by Mukai, and [Huy19] where isogenies are studied via
twisted derived equivalences). Our first aim in this paper, is to apply a
lattice-theoretical method due to Kneser to construct isogenies. In fact, it
is difficult in general to produce K3 surfaces that are isogenous to a given
$X$: Mukai’s construction of isogenous K3 surfaces as moduli spaces can only
yield finitely many new K3 surfaces, corresponding to transcendental lattices
that contain $\mathrm{T}(X)$ with finite index and cyclic cokernel. Given an
even lattice $L$, Kneser method of neighboring lattices allows one to
construct new lattices $L^{\prime}\subset L\otimes\mathbf{Q}$ that belong to
the same genus of $L$ starting from isotropic lines in
$L\otimes\mathbf{Z}/d\mathbf{Z}$, which we interpret as cyclic subgroups of
the relevant Brauer group. To explain his construction, consider
$\mathrm{T}\subset\mathrm{T}(X)$ a sublattice such that
$\mathrm{T}(X)/\mathrm{T}\cong\mathbf{Z}/d\mathbf{Z}$ and such that $d$ and
$\mathrm{discr}(\mathrm{T}(X))$ coprime to each other. The sublattice
$\mathrm{T}$ is said to be split if its discriminant form
$A_{\mathrm{T}}\coloneqq\mathrm{T}^{\vee}/\mathrm{T}$ splits as
$A_{\mathrm{T}}=A_{X}\oplus\mathbf{Z}/d\mathbf{Z}\oplus\mathbf{Z}/d\mathbf{Z}$,
where $A_{X}\cong\mathrm{T}(X)^{\vee}/\mathrm{T}(X)$ and $\mathrm{T}^{\vee}$
denotes the dual of $\mathrm{T}$. The results of Kneser can be formulated as a
natural one-to-one correspondence between

  1. (1)

Isotropic lines in
$\mathrm{T}(X)\otimes\mathbf{Z}/d\mathbf{Z}\cong\operatorname{Br}(X)[d]$;

  2. (2)

Split sublattices of index $d$ in $\mathrm{T}(X)$;

  3. (3)

And $d$-neighbors of $\mathrm{T}(X)$, that is, even lattices
$\mathrm{T}\subset\mathrm{T}(X,\mathbf{Q})$ such that

$\mathrm{T}(X)/\mathrm{T}(X)\cap\mathrm{T}\cong\mathrm{T}/\mathrm{T}(X)\cap\mathrm{T}\cong\mathbf{Z}/d\mathbf{Z}.$

An isotropic line is by definition a cyclic subgroup
$\mathbf{Z}/d\mathbf{Z}\subset\mathrm{T}(X)\otimes\mathbf{Z}/d\mathbf{Z}$
which is isotropic with respect to the natural perfect pairing on
$\mathrm{T}(X)\otimes\mathbf{Z}/d\mathbf{Z}$. It is a non-trivial result of
Kneser that $d$-neighbors always belong to the same genus (see Theorem 3.8).

In 3.4 we generalize this correspondence to the case when $d$ and
$\mathrm{discr}(\operatorname{NS}(X))$ are not coprime. Thanks to the global
Torelli theorem for K3 surfaces [Pvv71], to any complex K3 surface $X$ and any
isotropic line $\ell\subset\operatorname{Br}(X)[d]$ we associate a well-
defined K3 surface $X_{\ell}$. This comes with an isogeny
$f_{\ell}\colon\mathrm{T}(X_{\ell},\mathbf{Q})\rightarrow\mathrm{T}(X,\mathbf{Q})$
such that the image of $\mathrm{T}(X_{\ell})$ is the corresponding neighbor of
$\mathrm{T}(X)$, and such that
$\operatorname{NS}(X_{\ell})\cong\operatorname{NS}(X)$. We call this K3
surface the $d$-neighbor of $X$ associated to $\ell$. When
$\operatorname{End}_{\operatorname{Hdg}}(\mathrm{T}(X)_{\mathbf{Q}})=\mathbf{Q}$
or a totally real field, then $X_{\ell}\cong X_{\ell^{\prime}}$ if and only if
$\ell=\ell^{\prime}$. When $\rho(X)\geq 3$, we also construct another K3
surface $Y_{\ell}$, this time well-defined up to derived equivalence, endowed
with a Hodge isometry
$g_{\ell}\colon\mathrm{T}(Y_{\ell},\mathbf{Q})\rightarrow\mathrm{T}(X,\mathbf{Q})$
such that
$g_{\ell}(\mathrm{T}(Y_{\ell}))=f_{\ell}(\mathrm{T}(X_{\ell}))\cap\mathrm{T}(X)$
is the associated $d$-split sublattice. In particular
$\mathrm{discr}(\operatorname{NS}(Y_{\ell}))=d^{2}\cdot\mathrm{discr}(\operatorname{NS}(X))$.
When the K3 surface $Y_{\ell}$ exists, both the original $X$ and its neighbor
$X_{\ell}$ can be interpreted as moduli spaces of sheaves on some Fourier-
Mukai partner of $Y_{\ell}$ (3.19).

Although we do not have a precise geometric interpretation of this
construction, we believe that ‘taking neighbors’ is a very natural operation
for K3 surfaces, which in some sense is the analogue of taking quotients of
abelian varieties. For instance, we show that for a Kummer surface
$X=\mathrm{Km}(E_{1}\times E_{2})$, where $E_{1},E_{2}$ are two non-isogenous,
non-CM elliptic curves, and for any odd $d$, the $d$-neighbors of $X$ are in
one-to-one correspondence to $\mathrm{Km}(E_{1}/C_{1}\times E_{2}/C_{2})$,
where $C_{1}\cong C_{2}\cong\mathbf{Z}/d\mathbf{Z}$.

Next, we study the surfaces $X_{\ell}$ and $Y_{\ell}$ over number fields. In
fact, we prove a more general criterion that allows us to understand whether
an isogenous K3 surface is defined over a given number field:

######  Theorem 1.1.

Let $K\subset\mathbf{C}$ be a number field and let $X/K$ be a K3 surface.
Denote by $\rho_{X}$ the Galois representation associated to
$\mathrm{T}(\overline{X},\mathbf{A}_{f})=\mathrm{T}(\overline{X},\mathbf{Q})\otimes\widehat{\mathbf{Z}}$.
Let $Y$ be a complex K3 surface and let
$f\colon\mathrm{T}(Y,\mathbf{Q})\rightarrow\mathrm{T}(X_{\mathbf{C}},\mathbf{Q})$
be a Hodge isometry. Then, the K3 surface $Y$, the Hodge isometry $f$, and
$\operatorname{NS}(Y)$ can defined over $K$ if and only if:

  1. (1)

The $\widehat{\mathbf{Z}}$-lattice
$f(\mathrm{T}(Y))\otimes\widehat{\mathbf{Z}}$ is stable under $\rho_{X}$;

  2. (2)

The induced $G_{K}$ action on
$A_{Y}\coloneqq\mathrm{T}(Y)^{\vee}/\mathrm{T}(Y)$ is trivial.

By (2) we mean the following: since
$f(\mathrm{T}(Y))\otimes\widehat{\mathbf{Z}}$ is stable under
$G_{K}=\operatorname{Gal}(\overline{K}/K)$ by point (1), there is an induced
Galois representation on

$\frac{(f(\mathrm{T}(Y))\otimes\widehat{\mathbf{Z}})^{\vee}}{f(\mathrm{T}(Y))\otimes\widehat{\mathbf{Z}}}\cong
A_{Y},$

and we require this to be trivial. We remark that $\operatorname{NS}(Y)$ and
$f$ can be defined after a field extension of degree bounded independently
from the K3 surfaces involved ( 2.3 and 2.7). This means that the difference
between a smaller field of definition of $Y$ and a field where $Y$, $f$, and
$\operatorname{NS}(Y)$ are simultaneously defined can be universally bounded
(4.3). We also remark that point (2) is the only difference between the
analogue result for abelian varieties, and it is also the most difficult thing
to verify in practice. Regarding neighboring K3 surfaces, we have the
following corollary:

######  Corollary 1.2.

Let $K$ be a number field and let $X/K$ be a K3 surface such that
$\operatorname{NS}(\overline{X})^{G_{K}}=\operatorname{NS}(\overline{X})$. Let
$\ell\subset\operatorname{Br}(\overline{X})[d]$ be an isotropic line.

  1. (1)

If $\ell$ is $G_{K}$-stable, then $(X_{\ell},f,\operatorname{NS}(X_{\ell}))$
can be defined over $K$;

  2. (2)

It $G_{K}$ acts trivially on $\ell$ and $\rho_{X}\geq 3$, then
$(Y_{\ell},g_{\ell},\operatorname{NS}(Y_{\ell}))$ can be defined over $K$.

We also apply Theorem 1.1 to understand wheter a K3 surface $X/K$ that is
geometrically a moduli space of sheaves on some other K3 surface, is also a
moduli space over $K$, see Corollary 4.6 and the discussion that precedes it.

In the last part of the paper, we give some applications of our results.

######  Conjecture 1.3 (Shafarevich).

Only finitely many lattices can appear as $\operatorname{NS}(\overline{X})$
where $X$ is any K3 surface over any number field of given degree.

Coleman conjectured an analogue of 1.3 for abelian varieties, where the Néron-
Severi lattice is replaced by the endomorphism ring of the variety. It is
known that Colemann conjecture implies a uniform bound of
$|A(K)_{\mathrm{tors}}|$ for abelian varieties of bounded dimension over
fields of bounded degree; as well as of the size of their isogeny classes
[R1́8, Theorem 1.1.]. In [OSZ19], the authors show that Colemann conjecture
also bounds $|\operatorname{Br}(\overline{X})^{G_{K}}|$ uniformily for abelian
varieties and K3 surfaces. On the other hand, there are no known relations
between Shafarevich conjecture and the boundeness of the Brauer groups of K3
surfaces.

######  Conjecture 1.4 (Folklore).

Let $N$ be an integer and let $X/\overline{\mathbf{Q}}$ be a K3 surface. Then,
the number of isomorphism classes of K3 surfaces isogenous to $X$ that can be
defined over extensions of degree at most $N$ can be bounded only in terms on
$N$.

In the following, $\ell(A_{X})$ denotes the minimal number of generators of
the finite abelian group $A_{X}$. We prove:

######  Theorem 1.5.

  1.   2. (1)

Assume that Shafarevich Conjecture is true and let $N>0$ be an integer. Then,
there is a integer $M$ such that, for any K3 surface $X$ defined over a number
field $K$ with $[K\colon\mathbf{Q}]\leq N$, which satisfies both
$\rho_{\bar{X}}\geq 3+\ell(A_{X})$ and $\ell(A_{X})\leq 19-\rho(X)$, if
$\operatorname{Br}(\overline{X})[d]\subset\operatorname{Br}(\overline{X})^{G_{K}}$
then $d\leq M$.

  3. (2)

Assume that Conjecture 1.4 is true and let $N>0$ be an integer. Then, there is
a integer $M$ such that, for any K3 surface $X$ defined over a number field
$K$ with $[K\colon\mathbf{Q}]\leq N$, such that
$\operatorname{End}(\mathrm{T}(X_{\mathbf{C}}))=\mathbf{Q}$ or a totally real
field and $\ell(A_{X})\leq 19-\rho(X)$, if
$\operatorname{Br}(\overline{X})[d]\subset\operatorname{Br}(\overline{X})^{G_{K}}$
then $d\leq M$.

If one considers only level structures with $d$ and
$\mathrm{discr}(\operatorname{NS}(X))$ coprime, one can remove the condition
$\ell(A_{X})>19-\rho(X)$ from both statements, and one can replace the
condition $\rho_{\bar{X}}\geq 3+\ell(A_{X})$ with $\rho_{\bar{X}}\geq 3$ in
the first statement.

Note that the condition $\ell(A_{X})\leq 19-\rho(X)$ holds automatically when
$\rho(X)\leq 9$. The result can be strengthened if instead of level structures
one considers the finiteness of the isotropic Brauer lines or classes.

For sake of completeness, we include in the end of the paper an alternative
proof of the following finiteness:

######  Theorem 1.6.

Let $X/\overline{\mathbf{Q}}$ be a K3 surface and suppose that
$\operatorname{End}_{\operatorname{Hdg}}(T(X_{\mathbf{C}}))=\mathbf{Z}$. Then,
for any integer $N>0$, only finitely many isomorphism classes of K3 surfaces
$Y/\overline{\mathbf{Q}}$ that are isogenous to $X$ can be defined over field
extensions $K/\mathbf{Q}$ of degree $[K\colon\mathbf{Q}]\leq N$.

This is well-known to the experts, and it can be proved, for instance, using
the Masser-Wüstholz’s isogeny estimates for abelian varieties [MW93] and the
Zarhin trick for K3 surfaces invented by Charles (see [Cha16]) in complete
analogy to the strategy employed in [OSZ19] to prove the finiteness of CM K3
surfaces. We give an elementary proof more in line with the techniques
expressed in the paper, and which only relies on the open-adelic image proved
by Cadoret and Moonen in [CM20] and some classical finiteness properties of
adelic algebraic groups.

###  1.1. Acknowledgment

This work was funded by the Europen Research Council (ERC) under the European
Union’s Horizon 2020 research and innovation programme - grant agreement No
948066.

I wish to thank Salvatore Floccari, Giacomo Mezzedimi, Nebojsa Pavic, and
Stefan Schreieder for the many enjoyable discussions about mathematics and
more.

###  1.2. Notation and convention

All the fields in this paper are considered inside a fixed field $\mathbf{C}$
of the complex numbers. If $X/K$ is a K3 surface over a number field, we
denote by $\overline{X}$ its base-change to $\overline{K}$, and by
$X_{\mathbf{C}}$ its base-change to $\mathbf{C}$. By a lattice we shall always
mean a free, finitely generated $\mathbf{Z}$-module $L$ endowed with a non-
degenerate bilinear form $L\times L\rightarrow\mathbf{Z}$. We denote by
$L^{\vee}$ its dual and we say that $L$ is even if $(x,x)\in 2\mathbf{Z}$ for
every $x\in L$. For a finite abelian group $A$, we denote by $\ell(A)$ its
length, that is, the minimal number of generators.

##  2\. Preliminaries on Hodge structures of K3 surfaces and Galois
representations

In this section we set-up the notation and we recall some well-known theory of
K3 surfaces. Our only contribution is Proposition 2.7.

###  2.1. Singular cohomology of K3 surfaces

Let $X/\mathbf{C}$ be an algebraic K3 surface and let
$\mathrm{H}^{2}_{\text{sing}}(X,\mathbf{Z})$ be its second singular cohomology
group. This is a free $\mathbf{Z}$ module of rank $22$. The intersection
pairing

$\mathrm{H}^{2}_{\text{sing}}(X,\mathbf{Z})(1)\times\mathrm{H}^{2}_{\text{sing}}(X,\mathbf{Z})(1)\rightarrow\mathrm{H}_{\text{sing}}^{4}(X,\mathbf{Z})(2)\cong\mathbf{Z}$

endows $\mathrm{H}^{2}_{\text{sing}}(X,\mathbf{Z})(1)$ with a non-degenerate
quadratic form, and from Poincare duality one deduces that
$\mathrm{H}^{2}_{\text{sing}}(X,\mathbf{Z})(1)$ is unimodular, i.e., that
$|\mathrm{discr}(\mathrm{H}^{2}_{\text{sing}}(X,\mathbf{Z})(1))|=1$. The
isomorphism class of this lattice does not depend on the chosen $X$, and one
can also prove that $\mathrm{H}^{2}_{\text{sing}}(X,\mathbf{Z})(1))\cong
U^{3}\oplus E_{8}(-1)^{2},$ where $U$ denotes the hyperbolic lattice and
$E_{8}(-1)$ the unique even, negative definite, unimodular lattice of rank
$8$. There is a natural identification between $\operatorname{Pic}(X)$ and the
Néron-Severi lattice
$\operatorname{NS}(X)\subset\mathrm{H}^{2}_{\text{sing}}(X,\mathbf{Z})(1)$.
The orthogonal complement of $\operatorname{NS}(X)$ is the transcendental
lattice $\mathrm{T}(X)\subset\mathrm{H}^{2}_{\text{sing}}(X,\mathbf{Z})(1).$
This is an integral Hodge structure of weight $0$ with weights
$(1,-1),(0,0),(-1,1)$ and with $\dim_{\mathbf{C}}(T^{1,-1}_{X})=1$. The Hodge
structure
$\mathrm{T}(X,\mathbf{Q})=\mathrm{T}(X)\otimes_{\mathbf{Z}}\mathbf{Q}$ is
irreducible whenever $X$ is projective, and $\mathrm{T}(X)$ can also be
characterized as the biggest, primitive sub-Hodge structure of
$\mathrm{H}^{2}(X,\mathbf{Z})(1)$ whose complexification contains the period
of $X$.

######  Definition 2.1.

The finite group $A_{X}=\operatorname{NS}(X)^{\vee}/\operatorname{NS}(X)$ is
called the discriminant form of $X$.

Here $L^{\vee}$ denotes the dual of a lattice $L$. The group $A_{X}$ carries a
natural quadratic form $q\colon A_{X}\rightarrow\mathbf{Q}/2\mathbf{Z}$ given
by $q(\ell+\operatorname{NS}(X)):=(\ell,\ell)+2\mathbf{Z}$ (this is well
defined since $\operatorname{NS}(X)$ is even). Similarly, one has a quadratic
form $q^{\prime}$ on $\mathrm{T}(X)^{\vee}/\mathrm{T}(X)$, and a result of
Nikulin [Nik79, Proposition 1.6.1] shows that there is a natural isomorphism

(2.1) $(A_{X},q)\cong(\mathrm{T}(X)^{\vee}/\mathrm{T}(X),-q^{\prime}),$

due to the fact that $\operatorname{NS}(X)$ and $\mathrm{T}(X)$ are orthogonal
complements in the unimodular lattice
$\mathrm{H}^{2}_{\text{sing}}(X,\mathbf{Z})$. Note that
$|A_{X}|=|\mathrm{disc}(\mathrm{T}(X))|=|\mathrm{disc}(\operatorname{NS}(X))|.$
For the next proposition, see [Nik79, Corollary 1.5.2].

######  Proposition 2.2.

Let $X,Y$ be complex K3 surfaces, and let
$f_{t}\colon\mathrm{T}(X)\rightarrow\mathrm{T}(Y)$ and
$f_{n}\colon\operatorname{NS}(X)\rightarrow\operatorname{NS}(Y)$ be isometries
such that the two induced maps $A_{X}\rightarrow A_{Y}$ coincide. Then there
exists a unique isometry
$\mathrm{H}^{2}_{\text{sing}}(X,\mathbf{Z})(1)\rightarrow\mathrm{H}^{2}_{\text{sing}}(Y,\mathbf{Z})(1)$
that restrict to $f_{t}$ and $f_{n}$ on $\mathrm{T}(X)$ and
$\operatorname{NS}(X)$ respectively.

We recall that the Grothendieck-Brauer group
$\operatorname{Br}(X)=\mathrm{H}^{2}_{\textrm{\'{e}t}}(X,\mathcal{O}_{X}^{\times})$
of $X$ can be written via the exponential sequence as
$\operatorname{Br}(X)\cong\mathrm{T}^{\vee}(X)_{X}\otimes_{\mathbf{Z}}\mathbf{Q}/\mathbf{Z}\cong\operatorname{Hom}(\mathrm{T}(X),\mathbf{Q}/\mathbf{Z}).$
In particular, $\operatorname{Br}(X)$ is a torsion abelian group of the form
$(\mathbf{Q}/\mathbf{Z})^{22-\rho_{X}}.$

###  2.2. Ètale cohomology and Galois action

Suppose now that $X$ is defined over a number field $K\subset\mathbf{C}$ and
let $\bar{K}$ be the algebraic closure of $K$ in $\mathbf{C}.$ Let
$\mathrm{H}^{2}_{\textrm{\'{e}t}}(\bar{X},\widehat{\mathbf{Z}})(1)$ be the
Tate twist of the second étale cohomology group of
$\bar{X}:=X\times_{K}\operatorname{Spec}(\bar{K})$. The inclusion
$\bar{K}\subset\mathbf{C}$ induces a comparison isomorphism

(2.2)
$\mathrm{H}^{2}_{\textrm{\'{e}t}}(\bar{X},\widehat{\mathbf{Z}})(1)\cong\mathrm{H}^{2}_{\text{sing}}(X_{\mathbf{C}},\mathbf{Z})(1)\otimes\widehat{\mathbf{Z}},$

so that we can view
$\mathrm{H}^{2}_{\text{sing}}(X,\mathbf{Z})(1)\subset\mathrm{H}^{2}_{\textrm{\'{e}t}}(\bar{X},\widehat{\mathbf{Z}})(1)$
via the map $v\mapsto v\otimes 1.$ We denote the orthogonal complement of
$\operatorname{NS}(\bar{X})$ in
$\mathrm{H}^{2}_{\textrm{\'{e}t}}(\bar{X},\widehat{\mathbf{Z}})(1)$ by
$T(\overline{X},\widehat{\mathbf{Z}})$. As the notation suggests, via the
isomorphism 2.2, one has

$T(\overline{X},\widehat{\mathbf{Z}})\cong\mathrm{T}(X_{\mathbf{C}})\otimes\widehat{\mathbf{Z}}$

and a natural isomorphism

$\operatorname{Br}(\bar{X})\cong\mathrm{T}(\overline{X},\widehat{\mathbf{Z}})^{\vee}\otimes\mathbf{Q}/\mathbf{Z}.$

To $X/K$ there is associated a (continuous) Galois representation

$\rho_{X}\colon
G_{K}\rightarrow\operatorname{GL}(\mathrm{H}_{\text{sing}}^{2}(X,\mathbf{Z}))(\widehat{\mathbf{Z}}).$

Since the Galois action respects Poincaré duality, one sees that $\rho_{X}$
acts by $\widehat{\mathbf{Z}}$-isometries. The action on the Néron-Severi
lattice is not very interesting due to the following well-known phenomenon:

######  Lemma 2.3.

There exists a constant $C_{n}$ with the following property: for any K3
surface $X/K$ as above, there is a field extension $K^{\prime}/K$ of degree
$[K\colon K^{\prime}]\leq C_{n}$ such that $G_{K^{\prime}}$ acts trivially on
$\operatorname{NS}(\overline{X})$.

###### Proof.

Since $\operatorname{NS}(\overline{X})$ is a free $\mathbf{Z}$-module and
$\rho_{X}$ is continuous, the image of $\rho_{X}$ must be a finite subgroup of
$\operatorname{GL}(\operatorname{NS}(\overline{X}))(\mathbf{Z})$. The
statement follows from the classical fact that any such subgroup embedds into
$\operatorname{GL}(\operatorname{NS}(\overline{X}))(\mathbf{Z}/3)$, and that
for K3 surfaces the rank of the Néron-Severi group is always smaller than
$20$. ∎

Let $\operatorname{MT}_{X}\subset\mathrm{GL}(\mathrm{T}(X,\mathbf{Q}))$ be the
Mumford-Tate group of $X$; Mumford-Tate groups of K3 surfaces are completely
classified by Zarhin, and they only depends on the field
$E_{X}\coloneqq\operatorname{End}_{\operatorname{Hdg}}(\mathrm{T}(X,\mathbf{Q}))$
and on its action on the quadratic space $\mathrm{T}(X,\mathbf{Q})$. When
$E_{X}=\mathbf{Q}$, one has
$\operatorname{MT}_{X}=\mathrm{SO}(\mathrm{T}(X,\mathbf{Q}))$. Since the
Mumford conjecture holds for K3 surfaces, see [Tan95] and [And96], one knows
due to a result of Serre (see [LP92, Proposition 6.14]) that that there is a
finite extension $K^{\prime}/K$ such that
$\rho_{X}(G_{K})\subset\operatorname{MT}_{X}(\mathbf{A}_{f})$. The following
important theorem is due to Cadoret and Moonen:

######  Theorem 2.4 ([CM20]).

Let $X$ be a K3 surface over a number field $K$ and assume that
$\rho_{X}(G_{K})\subset\operatorname{MT}_{X}(\mathbf{A}_{f})$. Then,
$\rho_{X}(G_{K})\subset\operatorname{MT}_{X}(\mathbf{A}_{f})$ is a compact
open subgroup for the adelic topology.

Note that the same result does not hold in general for abelian varieties for
which the Mumford-Tate conjecture is known, because their Mumford-Tate groups
are not always Hodge-maximal.

###  2.3. Galois action on Hodge classes

Let
$E_{X}\coloneqq\operatorname{End}_{\operatorname{Hdg}}(\mathrm{T}(X,\mathbf{Q}))$
be the algebra of Hodge endomorphisms of $\mathrm{T}(X,\mathbf{Q})$ and let
$U_{X}\subset E_{X}$ be the subgroup of Hodge isometries. One knows [Zar83]
that $E_{X}$ is always a field, either totally real, or CM. One also knows
that $E_{X}\subset\mathrm{H}^{4}(X\times X,\mathbf{Q}(2))$ consists of
motivated classes in the sense of André or absolute Hodge classes in the sense
of Deligne, and this is also true for
$\mathrm{Hom}_{\text{Hdg}}(\mathrm{T}(X,\mathbf{Q}),\mathrm{T}(Y,\mathbf{Q}))\subset\mathrm{H}_{\mathrm{sing}}^{4}(X\times
Y,\mathbf{Q}(2))$ for another K3 surface $Y$. In both situations, for
$\sigma\in\operatorname{Aut}(\mathbf{C})$ and any
$f\in\mathrm{Hom}_{\text{Hdg}}(\mathrm{T}(X,\mathbf{Q}),\mathrm{T}(Y,\mathbf{Q}))$
there corresponds
$f^{\sigma}\in\mathrm{Hom}_{\text{Hdg}}(\mathrm{T}(X^{\sigma},\mathbf{Q}),\mathrm{T}(Y^{\sigma},\mathbf{Q}))$,
in such a way that the following diagram commutes

(2.3)
${\mathrm{T}(X^{\sigma},\mathbf{A}_{f})}$${\mathrm{T}(Y^{\sigma},\mathbf{A}_{f})}$${\mathrm{T}(X,\mathbf{A}_{f})}$${\mathrm{T}(Y,\mathbf{A}_{f}).}$$\scriptstyle{f^{\sigma}}$$\scriptstyle{\sigma^{*}_{X}}$$\scriptstyle{\sigma_{Y}^{*}}$$\scriptstyle{f}$

Here, $\sigma^{*}_{X}$ denotes the natural pullback map in étale cohomology
induced by the morphism of schemes $\sigma_{X}\colon X\rightarrow X^{\sigma}$,
and similarly for $Y$.

######  Definition 2.5.

Assume that $X$ and $Y$ can be defined over a number field $K$ and let
$f\in\mathrm{Hom}_{\text{Hdg}}(\mathrm{T}(X,\mathbf{Q}),\mathrm{T}(Y,\mathbf{Q})).$
We say that $f$ is defined over $K$ if $f^{\sigma}=f$ for every $\sigma\in
G_{K}$.

Note that by (2.3), if $X=Y$, this happens if and only if $f\in E_{X}$
commutes with the action of $G_{K}$. There is a straightforward way to
understand whether an element of $E_{X}$ is defined over $K.$ One first notes
that if $X/\mathbf{C}$ is a complex K3 surface, there is a natural map

(2.4)
$E_{X}\rightarrow\operatorname{End}(T^{1,-1}(X_{\mathbf{C}}))=\mathbf{C}$

which can be easily proved to be an embedding of fields, due to the fact that
$\mathrm{T}(X,\mathbf{Q})$ is irreducible whenever $X$ is projective. It
follows from [Val21] that $E_{X}$ is defined over $K$ if and only if, for one
embedding $\sigma\colon K\rightarrow\mathbf{C}$, and hence any, the image of
(2.4) is contained in $\sigma(K)$. Since $[E_{X}\colon\mathbf{Q}]\leq 20$
always, we obtain the following corollary:

######  Corollary 2.6.

For any K3 surface $X/K$ there exists a finite extension $K^{\prime}/K$ of
degree $[K^{\prime}\colon K]\leq 20$ over which all the elements of $E_{X}$
are defined.

We conclude the section with a uniform bound similar to Lemma 2.3 and to the
Corollary above.

######  Proposition 2.7.

There exists a universal constant $C_{h}>0$ with the following property.
Suppose that $X$ and $Y$ are K3 surfaces, and assume that they are defined
over a number field $K$. Let
$f\colon\mathrm{T}(Y_{\mathbf{C}},\mathbf{Q})\rightarrow\mathrm{T}(X_{\mathbf{C}},\mathbf{Q})$
be a Hogde isometry. Then, there exists a field extension $K^{\prime}/K$ of
degree $[K^{\prime}\colon K]\leq C_{h}$ such that $f$ is defined over
$K^{\prime}$.

###### Proof.

Up to an extension of degree bounded by $20$, we can assume that $E_{X}$ is
defined over $K$. Since both $Y$ and $X$ are defined over $K$, there is a
natural $G_{K}$ action on the set
$\text{Isom}_{\operatorname{Hdg}}(\mathrm{T}(Y_{\mathbf{C}},\mathbf{Q}),\mathrm{T}(X_{\mathbf{C}},\mathbf{Q})).$
The map $G_{K}\ni\sigma\mapsto f^{\sigma}\circ f^{-1}\in U_{X}$ is a
continuous $1-$cocycle of $G_{K}$ with values in $U_{X}$ (where $G_{K}$ has
its natural profinite topology and $U_{X}$ the discrete one). Since the Galois
action on $U_{X}$ is trivial, it follows that the map $\sigma\mapsto
f^{\sigma}\circ f^{-1}$ is a continuous morphism of groups, and hence it must
factorize as $G_{K}\rightarrow(U_{X})_{\text{tors}}=\mu(E_{X})\subset U_{X},$
where $\mu(E_{X})$ is the group of the roots of unity in $E_{X}$. The order
$|\mu(E_{X})|$ is universally bounded for K3 surfaces, since
$\mathrm{T}(X_{\mathbf{C}},\mathbf{Q})$ is always a vector space over
$\mathbf{Q}(\mu(E_{X}))$, and
$20\geq\dim_{\mathbf{Q}}\mathrm{T}(X_{\mathbf{C}},\mathbf{Q})\geq\phi(|\mu(E_{X})|),$
where $\phi$ denotes the Euler totient function. ∎

As the proof shows, in the general case where the only roots of unity in
$E_{X}$ are $\\{\pm 1\\}$, the extension can be chosen to be quadratic.

##  3\. Isogeny class and lattices

In this section we study the isogeny class of a complex projective K3 surface
$X$ in terms of lattices inside $T(X)_{\mathbf{Q}}$.

###  3.1. Basics

Let $X$ be a complex projective K3 surface. The marked isogeny class of $X$,
denoted by $\mathcal{H}^{m}_{X}$, consists of all the pairs $(Y,f)$ where
$Y\in\mathcal{H}_{X}$ and
$f\colon\mathrm{T}(Y,\mathbf{Q})\rightarrow\mathrm{T}(X,\mathbf{Q})$ is a
Hodge isometry, up to the natural notion of isomorphism.

Consider the set $\mathcal{L}$ of all the free $\mathbf{Z}$-modules
$L\subset\mathrm{T}(X,\mathbf{Q})$ of maximal rank such that the restriction
of the quadratic form to $L$ is even, and hence also integral. In this way, we
obtain a map $\mathcal{H}^{m}_{X}\rightarrow\mathcal{L}$ given by
$(Y,f)\mapsto f(\mathrm{T}(Y))$ and we denote by
$\mathcal{L}_{K3}\subset\mathcal{L}$ its image. The map
$\mathcal{H}^{m}_{X}\rightarrow\mathcal{L}_{K3}$ descends to a map
$\mathcal{T}\colon\mathcal{H}_{X}\rightarrow U_{X}\backslash\mathcal{L}_{K3}$
where $U_{X}\backslash\mathcal{L}_{K3}$ denotes the set of orbits in
$\mathcal{L}_{K3}$ under the natural left action of $U_{X}$, that is, the set
of $U_{X}$-classes of lattices. Elements in the same fibre of $\mathcal{T}$
have isometric integral transcendental lattices and it follows that
$\mathcal{T}$ has finite fibres. Note that if $E_{X}$ is totally real one has
$U_{X}=\\{\pm 1\\}$, so that up to Fourier-Mukai partners, the correspondence
between $\mathcal{H}_{X}$ and $\mathcal{L}_{K3}$ is one-to-one.

######  Proposition 3.1.

$\mathcal{L}_{K3}$ consists of all the lattices in $\mathrm{T}(X,\mathbf{Q})$
that can be embedded primitively into the K3 lattice. Moreover, if
$L\in\mathcal{L}$, and if there exists $L^{\prime}\in\mathcal{L}_{K3}$ such
that $L^{\prime}\subset L$, then $L\in\mathcal{L}_{K3}$.

###### Proof.

To prove the first statement, let $L\subset\mathrm{T}(X,\mathbf{Q})$ and fix a
primitive embedding of $L$ into $\Lambda_{K3}$. The lattice $L$ is endowed
with a natural Hodge structure, determined by the fact that the inclusion
$L\subset\mathrm{T}(X,\mathbf{Q})$ respects the Hodge decomposition. Endowing
$\Lambda_{K3}$ with such Hodge structure, by the surjectivity of the period
map, we obtain a K3 surface $Y$ endowed with a Hodge isometry
$f\colon\mathrm{T}(Y,\mathbf{Q})\rightarrow\mathrm{T}(X,\mathbf{Q})$ such that
$f(\mathrm{T}(Y))=L$.

For the other statement, we need to show that if a lattice $L$ embedds
primitively into $\Lambda_{K3}$, then every even overlattice $L\subset
L^{\prime}$ of finite index embedds primitively into $\Lambda_{K3}$ as well.
This follows from Nikulin’s theory: a primitive embedding of $L$ into
$\Lambda_{K3}$ with orthogonal complement isomorphic to $N$ is determined by
an isomorphism of finite quadratic forms $A_{L}\cong-A_{N}$. The overlattice
$L^{\prime}$ is determined uniquely by an isotropic subgroup $H\subset A_{L}$
and there is a natural isomorphism of finite quadratic forms
$A_{L^{\prime}}\cong H^{\perp}/H$ ([Nik79, Proposition 1.4.1]). Under the
identification $A_{L}\cong-A_{N}$, we see that $H$ also determines an
overlattice $N^{\prime}$ of $N$, and that there is an isomorphism of quadratic
forms $A_{L^{\prime}}\cong-A_{N^{\prime}}$. This shows that there is a
primitive embedding of $L^{\prime}$ into a unimodular, even lattice of
signature type $(3,19)$ with orthogonal complement isomorphic to $N^{\prime}$.
Since the K3 lattice is the unique unimodular even lattice of signature
$(3,19)$, we conclude. ∎

In particular, when $\rho(X)\geq 12$, the map
$\mathcal{T}\colon\mathcal{H}_{X}\rightarrow U_{X}\backslash\mathcal{L}$ is a
bijection because every even lattice of rank smaller then $10$ embedds
primitively into the K3 lattice, and because, in this case, $\mathrm{T}(X)$
determines $X$ uniquely.

In the next, we make some considerations that often appear in the theory of
Shimura varieties and moduli spaces of K3 surfaces. An adelic point
$g\in\mathrm{SO}_{X}(\mathbf{A}_{f})$ of $\mathrm{SO}_{X}$ is given by a
collection of elements $g_{p}\in\mathrm{SO}_{X}(\mathbf{Q}_{p})$ indexed by
each prime $p$ and such that, for almost all $p$, one has

$g_{p}(\mathrm{T}(X)\otimes\mathbf{Z}_{p})=\mathrm{T}(X)\otimes\mathbf{Z}_{p},$

with the equality occurring in $\mathrm{T}(X)\otimes\mathbf{Q}_{p}$. It is
well-known that for any $g\in\mathrm{SO}_{X}(\mathbf{A}_{f})$ and any lattice
$L\in\mathcal{L}$, there exists a unique lattice $gL\in\mathcal{L}$ such that,
for each prime $p$, one has
$gL\otimes\mathbf{Z}_{p}=g_{p}(L\otimes\mathbf{Z}_{p})$. In this way one lets
$\mathrm{SO}_{X}(\mathbf{A}_{f})$ act on $\mathcal{L}$.

######  Lemma 3.2.

The action of $\mathrm{SO}_{X}(\mathbf{A}_{f})$ on $\mathcal{L}$ respects
$\mathcal{L}_{K3}\subset\mathcal{L}$.

###### Proof.

Let $g\in\mathrm{SO}_{X}(\mathbf{A}_{f})$, let $L\in\mathcal{L}_{K3}$ and
consider $gL\in\mathcal{L}$. The lattices $L$ and $gL$ are locally isometric.
By [Nik79, Theorem 1.12.2] of Nikulin, an even lattice $L$ embedds primitively
into the lattice $\Lambda_{K3}$ if and only if some local conditions are
satisfied at every prime of $\mathbf{Z}$; from this it follows that also $gL$
embedds primitively in $\Lambda_{K3}$. ∎

Thus, if $L\in\mathcal{L}_{K3}$, the whole orbit
$\mathrm{SO}_{X}(\mathbf{A}_{f})\cdot L$ lives inside $\mathcal{L}_{K3}$. The
set $\mathrm{SO}_{X}(\mathbf{A}_{f})\cdot L$ is called the genus of $L$, and
it is a classical result that $\mathrm{SO}_{X}(\mathbf{A}_{f})\cdot L$ can be
written as a finite union of classes, sets of the form
$\mathrm{SO}_{X}(\mathbf{Q})\cdot L^{\prime}$ for some lattices
$L^{\prime}\in\mathcal{L}_{K3}$. If $L$ is an indefinite lattice, then in most
of the situations one has that its genus consists only of one class. This for
example happens when $\ell(A_{L})\leq\mathrm{rank}(L)-2$ and
$\mathrm{rank}(L)\geq 3$, see [Nik79, Theorem 1.13.2]. Nevertheless, the genus
may be in general non-trivial. The following notion is a particular case of
one introduced by Shimura, see [Shi10, Chapter 29].

######  Definition 3.3.

A lattice $L\in\mathcal{L}$ is said to be maximal if there is no other lattice
$L^{\prime}\in\mathcal{L}$ such that $L^{\prime}\subsetneq L$. We denote the
set of maximal lattices as $\mathcal{L}^{\mathrm{max}}\subset\mathcal{L}$ and
we denote by $\mathcal{H}_{X}^{\mathrm{max}}$ the subset of the isogeny class
consisting of K3 surfaces $Y$ such that $\mathrm{T}(Y)$ is maximal.

Clearly, the action of $\mathrm{SO}_{X}(\mathbf{A}_{f})$ on the set
$\mathcal{L}$ also respects $\mathcal{L}^{\mathrm{max}}$. The following is
[Shi10, Lemma 29.2, Lemma 29.9].

######  Proposition 3.4.

Any lattice $L\in\mathcal{L}$ is contained in some maximal lattice. Moreover,
if $L,L^{\prime}\in\mathcal{L}^{\mathrm{max}}$, then there exists
$g\in\mathrm{SO}_{X}(\mathbf{A}_{f})$ such that $L^{\prime}=gL$.

######  Remark 3.5.

The result above says that $\mathcal{L}^{\mathrm{max}}$ consists of a single
genus with respect to $\mathrm{SO}_{X}$.

As a corollary to the previous proposition and of Lemma 3.2, we obtain

######  Corollary 3.6.

One has that $\mathcal{L}^{\mathrm{max}}\subset\mathcal{L}_{K3}$. In
particular, there are always maximal K3 surfaces in each isogeny class.

###  3.2. Constructing elements of $\mathcal{H}_{X}$

If $A/\mathbf{C}$ is an abelian variety, one can construct all of its isogeny
class by taking quotients of $A$ by its finite subgroups. For K3 surfaces, one
does not have similar straightforward ways to build isogenies. On the other
hand, due to Proposition 3.1, in order to construct isogenous K3 surfaces one
only needs a way to construct elements in $\mathcal{L}_{K3}$. It is reasonable
to believe that the role of torsion points on abelian varieties should be
replaced by elements of $\mathrm{T}(X)\otimes\mathbf{Q}/\mathbf{Z}$ or,
similarly, by Brauer classes. One natural way to construct elements of
$\mathcal{L}$ via Brauer classes on $X$ is as follows. Let
$B\subset\operatorname{Br}(X)$ be a finite subgroup. Using that
$\operatorname{Br}(X)\cong\operatorname{Hom}(\mathrm{T}(X),\mathbf{Q}/\mathbf{Z})$
one defines $\ker(B)\subset\mathrm{T}(X)$ as the obvious sublattice. If
$\ker(B)$ embedds primitively into the K3 lattice (this is automatically true
when $\mathrm{rank}(\mathrm{T}(X))\leq 10$) then there is at least a K3
surface $Y$ whose transcendental lattice is Hodge isometric to $\ker(B)$,
i.e., the K3 surface $Y$ has a Hodge isometry
$f\colon\mathrm{T}(Y,\mathbf{Q})\rightarrow\mathrm{T}(X,\mathbf{Q})$ such that
$f(\mathrm{T}(Y))=\ker(B)$. The problem in this situation is twofold: first,
this will not work generically for K3 surfaces of small Picard rank, and
second, even if $\ker(B)$ does embedd primitively into the K3 lattice, the
associated K3 surface $Y$ will satisfy
$|\mathrm{discr}(\operatorname{NS}(Y))|=|B|^{2}\cdot|\mathrm{discr}(\operatorname{NS}(X))|$,
and in particular one never obtains transcendental lattices that are in the
same genus of $\mathrm{T}(X)$.

Some examples with $B=\mathbf{Z}/2\mathbf{Z}$ have a geometric interpretation,
see [van05] and [MSTVA17]. Most examples of this situation come from the
theory of moduli spaces of sheaves on K3 surfaces developed by Mukai [Muk87],
as we shall now briefly explain. Let $X$ be a complex K3 surface and let
$v\in\mathrm{H}^{*}(X,\mathbf{Z})$ be a primitive and isotropic Mukai vector.
Let $X=M(v)$ be the moduli space of stable sheaves on $X$ whose Mukai vector
is $v$, where ‘stable’ is intended with respect to a generic polarization.
Then, if $M(v)$ is non-empty, it is a K3 surface. Mukai shows that on $X\times
Y$ there is a quasi-universal family, and a slight modification of its Chern
class induces an isogeny $f\colon\mathrm{T}(Y)\rightarrow\mathrm{T}(X)$ such
that $f(\mathrm{T}(Y))\subset\mathrm{T}(X)$. In fact, there is a Brauer class
$\alpha\in\operatorname{Br}(X)\cong\operatorname{Hom}(\mathrm{T}(X),\mathbf{Q}/\mathbf{Z})$
such that $\mathrm{ker}(\alpha)=f(\mathrm{T}(Y))$, and in particular the
quotient $\mathrm{T}(X)/f(\mathrm{T}(Y))$ is cyclic. This Brauer class is
computed in [C0̆2] and can be seen as an obstruction to the existence of a
universal family on $X\times Y$, that is, to $X$ being a fine moduli space: if
$\alpha=0$ then $\mathrm{T}(X)\cong\mathrm{T}(Y)$, the moduli space $X$ is
fine, and $X$ and $Y$ are Fourier-Mukai partners. In particular, if $Y$ is
maximal, then every such $X$ is a fine moduli space. One sees that via this
construction only finitely many lattices in $\mathrm{T}(Y,\mathbf{Q})$ can be
obtained. [Nik79, Proposition 1.4.1]

We recall some basic lattice theory concerning this situation. For an even
lattice $L$ we denote by $A_{L}\cong L^{\vee}/L$ its discriminant form, and by
$q\colon A_{L}\rightarrow\mathbf{Q}/2\mathbf{Z}$ the associated quadratic
form. Note that $A_{L}$ is also endowed with a non-denerate pairing
$A_{L}\times A_{L}\rightarrow\mathbf{Q}/\mathbf{Z}$. Given an inclusion of
lattices $N\subset L$ of finite cokernel, there is a flag

(3.1) $N\subset L\subset L^{\vee}\subset N^{\vee},$

and $L/N\subset A_{N}$ is an isotropic subgroup of $A_{N}$, i.e., the
quadratic form $q\colon A_{N}\rightarrow\mathbf{Q}/2\mathbf{Z}$ is zero when
restricted to $L/N$. In fact, even overlattices of $N$ corresponds one-to-one
to isotropic subgroups of $A_{N}$, see [Nik79, Proposition 1.4.1]. One also
has a natural isomorphism

(3.2) $N^{\vee}/L^{\vee}\cong\operatorname{Hom}(L/N,\mathbf{Q}/\mathbf{Z}),$

and that $(L/N)^{\perp}=L^{\vee}/N$, where the orthogonal is taken with
respect to the natural pairing $A_{N}\times
A_{N}\rightarrow\mathbf{Q}/\mathbf{Z}$, and in particular
$(L/N)^{\perp}/(L/N)\cong A_{L}$. In case of $N=\ker(B)$ and
$L=\mathrm{T}(X)$, one readily sees that
$B\cong\operatorname{Hom}(\mathrm{T}(X)/\ker(B),\mathbf{Q}/\mathbf{Z})$.

######  Definition 3.7.

The lower isogeny class $\mathcal{H}^{\mathrm{low}}_{X}$ of $X$ consists of
all the $Y\in\mathcal{H}_{X}$ for which there exists a Hodge isometry
$f\colon\mathrm{T}(Y,\mathbf{Q})\rightarrow\mathrm{T}(X,\mathbf{Q})$ such that
$f(\mathrm{T}(Y))=\mathrm{T}(X)$.

###  3.3. Kneser neighbors

Kneser method of neighboring lattices has two important advantages with
respect to the situation before: it works for every K3 surface, and it allows
one to construct lattices in the same genus of $\mathrm{T}(X)$. The original
reference is [Kne57], for the following treatment we follow Chapter 3 of
[DF21]. We give a self contained proof of most of the results using Nikulin’s
language. Let $L$ be an even lattice, and let $d>1$ be an integer that is
coprime to $\mathrm{discr}(L)$. Denote by $\tilde{q}\colon
L\rightarrow\mathbf{Z}$ the quadratic form given by $2\tilde{q}(x)=(x,x)$
where the brackets denote the bilinear form of $L$. Another even lattice
$L^{\prime}\subset L\otimes\mathbf{Q}$ is called a $d$-neighbor of $L$ if

$L/L\cap L^{\prime}\cong L^{\prime}/L\cap
L^{\prime}\cong\mathbf{Z}/d\mathbf{Z}.$

The original definition of Kneser only contemplated the case $d=2$, but his
results can be generalized to any integer $d$.

######  Theorem 3.8 (Kneser).

If $L$ and $L^{\prime}$ are $d$-neighbors, then $L$ and $L^{\prime}$ belong to
the same genus.

###### Proof.

Let $M\coloneqq L\cap L^{\prime}$, and let $p$ be a prime that divides $d$.
Then, the lattice $L_{p}=L\otimes\mathbf{Z}_{p}$ is unimodular (and even, if
$p=2$) and the lattice $L^{\prime}_{p}=L^{\prime}\otimes\mathbf{Z}_{p}$ is
integral (and even, if $p=2$). We have a short exact sequence $0\rightarrow
L_{p}/M_{p}\rightarrow A_{M,p}\rightarrow M_{p}^{\vee}/L_{p}\rightarrow 0$
where we denote
$A_{M,p}=A_{M}\otimes\mathbf{Z}_{p}=A_{M\otimes\mathbf{Z}_{p}}$. The
overlattice $L^{\prime}_{p}$ of $M_{p}$ is given by an isotropic subgroup
$L^{\prime}_{p}/M_{p}\subset A_{M,p}$ which is transverse to $L_{p}/M_{p}$. In
particular $A_{M,p}\cong L_{p}/M_{p}\oplus L^{\prime}_{p}/M_{p}$, with both
summands isomorphic to $\mathbf{Z}/p^{n}\mathbf{Z}$ where $p^{n}$ is the
maximal power of $p$ that divides $d$. An element $[v]\in
L^{\prime}_{p}/M_{p}$ orthogonal to $L_{p}/M_{p}$ is represented by an element
$v\in L^{\prime}_{p}$ such that $(v,L_{p})\subset\mathbf{Z}_{p}$; since the
lattice $L_{p}$ is unimodular, it is also maximal, and we conclude that $v\in
L_{p}$. By Nikulin [Nik79, Proposition 1.4.1], the discriminant form of
$L^{\prime}_{p}$ is given by
$(L^{\prime}_{p}/M_{p})^{\perp}/(L^{\prime}_{p}/M_{p})=0$. In particular,
$L_{p}$ and $L_{p}^{\prime}$ are two $p$-adic lattices (even, if $p=2$) with
same rank and discriminant forms. Moreover, their discriminants satisfy
$\mathrm{discr}(L_{p})=p^{2n}\mathrm{discr}(M_{p})=\mathrm{discr}(L^{\prime}_{p})$,
and one concludes that $L_{p}$ and $L^{\prime}_{p}$ are isometric due to
[Nik79, Corollary 1.9.3]. For primes $p$ that do not divide $d$, one has
$L_{p}=L^{\prime}_{p}$, and we conclude the proof. ∎

One constructs $d$-neighbors of a lattice $L$ using isotropic lines in $L/dL$.
Since $d$ is coprime to $\mathrm{discr}(L)$, the induced quadratic form
$\tilde{q}\colon L/dL\rightarrow\mathbf{Z}/d\mathbf{Z}$ is non-degenerate. A
$d$-isotropic line $\ell$ is by definition a cyclic subgroup
$\mathbf{Z}/d\mathbf{Z}\subset L/dL$ which is isotropic with respect to this
quadratic form. One has the line map

$\\{d\text{-neighbors of}\,\,L\\}\rightarrow\\{d\text{-isotropic lines in
}L/dL\\}$

that associated to any $d$-neighbor $L^{\prime}$ of $L$ the line
$\ell=\operatorname{Im}(dL^{\prime}\rightarrow L\rightarrow L/dL)$.

######  Proposition 3.9.

The line map is well defined, and it is a bijection.

###### Proof.

Let $\ell\subset L/dL$ be an isotropic line generated by $v\in L$. We begin by
showing that we can choose $v$ such that $v^{2}\in 2d^{2}\mathbf{Z}$. For any
$w\in L$, we compute
$(v+dw)^{2}=2\tilde{q}(v+dw)=2\tilde{q}(v)+2d^{2}\tilde{q}(w)+2d(v,w)$. Write
$\tilde{q}(v)=dk$ for $k\in\mathbf{Z}$ and consider the equation
$k+(v,w)=0\mod d.$ Since the bilinear pairing on $L/dL$ is perfect, we deduce
that we can always solve this equation, and hence that we can always find a
$w$ such that $(v+dw)^{2}\in p^{2}\mathbf{Z}$. Choose such $v$ and consider
the map $h\colon L\rightarrow\mathbf{Z}/d\mathbf{Z}$ given by pairing with it,
and let $M=\ker(h)$. Note that $M^{\vee}$ is generated by $1/dv$ and $L$, and
we want to show that the lattice $L^{\prime}$ generated by $1/dv$ and $M$ is a
$d$-neighbor of $L$. By the fact that $v$ generates a group isomorphic to
$\mathbf{Z}/d\mathbf{Z}$ one deduces that $L\cap L^{\prime}=M$, and since
$v^{2}\in 2d^{2}\mathbf{Z}$ one easily sees that $L^{\prime}$ is even. It is
straightforward to check that this is the inverse of the line map. ∎

Note that since $d$ and $\mathrm{discr}(L)$ are coprime then
$L\otimes\mathbf{Z}/d\mathbf{Z}\cong\operatorname{Hom}(L,\mathbf{Z}/d)\cong(d^{-1}L)/L.$

######  Lemma 3.10.

Consider an isotropic line $\ell\subset L\otimes\mathbf{Z}/d\mathbf{Z}$ as a
subgroup $\ell\subset\operatorname{Hom}(L,\mathbf{Z}/d)$. Then,
$\ker(\ell)=L\cap L^{\prime}$ where $L^{\prime}$ is the $d$-neighbor
associated to $\ell$. Moreover, one has a natural identification
$\operatorname{Hom}(\ell,\mathbf{Q}/\mathbf{Z})\cong L/M$.

###### Proof.

The first part follows from the fact that one can choose a generator of $\ell$
to satisfy $v^{2}\in 2d^{2}\mathbf{Z}$ as explained in the previous proof. The
second part is (3.2). ∎

######  Lemma 3.11.

Let $L$ and $L^{\prime}$ be $d$-neighbors, and let $M=L\cap L^{\prime}$. Let
$H\cong\mathbf{Z}/d\mathbf{Z}\subset A$ be the isotropic subgroup associated
to the overlattice $L$ and define $H^{\prime}$ similarly. Then, $H^{\prime}$
is the unique isotropic subgroup $H^{\prime}\cong\mathbf{Z}/d\mathbf{Z}\subset
A_{M}$ with $H\cap H^{\prime}=0$.

###### Proof.

Due to cardinality reason, if another group $H^{\prime\prime}$ that satisfies
the properties in the statement existed, then $H^{\prime\prime}\subset H\oplus
H^{\prime}$ necessarily. So one can assume that $H^{\prime\prime}$ is
generated by $a+b$ with $a\in H$ and $b\in H^{\prime}$. One has
$q(a+b)=2(a,b)\in\mathbf{Q}/2\mathbf{Z}$ and one notes that this is never $0$
due to the perfect pairing from the lemma above, unless $a=0$ or $b=0$. ∎

We remark that the unicity above fails in general when $L/M$ is not cyclic,
that is, the fact that $L/M$ is cyclic is fundamental for the notion of Kneser
neighbor and this situation cannot be generalized.

######  Definition 3.12.

Let $L$ be a lattice. A sublattice $M\subset L$ is said to be $d$-split if $d$
is corpime to $\mathrm{discr}(L)$, the quotient
$L/M\cong\mathbf{Z}/d\mathbf{Z}$ and the isotropic subgroup $L/M\subset A_{M}$
has a unique transversal as in Lemma 3.11.

In this way, one obtains a one-to-one correspondence between $d$-split
sublattices, $d$-neighbors, and $d$-isotropic lines: to any $d$-isotropic line
$\ell$, one associates the $d$-split sublattice $M\coloneqq\ker(\ell)$. Let
$H^{\prime}\subset A_{M}$ be the unique isotropic subgroup from Lemma 3.11.
This determines a unique overlattice $L^{\prime}$ of $M$, and one readily
shows that $L^{\prime}$ is a $d$-neighbor of $L$. Finally, the isotropic
$\ell=\operatorname{Im}(dL^{\prime}\rightarrow L\rightarrow L/dL)$ is the one
we have started with. As explained in the proof of 3.9, one can also construct
$L^{\prime}$ directly from $\ell$.

######  Definition 3.13.

A K3 surface $Y\in\mathcal{H}_{X}$ is a $d$-neighbor of $X$ if $d$ is coprime
to $\mathrm{discr}(\operatorname{NS}(X))$ and if there exists a Hodge isometry
$f\colon\mathrm{T}(Y,\mathbf{Q})\rightarrow\mathrm{T}(X,\mathbf{Q})$ such that
$f(\mathrm{T}(Y))$ is a $d$-neighbor of $\mathrm{T}(X)$.

Due to 3.1 and 3.8 we conclude that neighboring K3 surfaces can be constructed
from isotropic lines in
$\mathrm{T}(X)/d\mathrm{T}(X)\cong\operatorname{Br}(X)[d]$, and if $E_{X}$ is
totally real, then the isomorphism classes of transcendental lattices of K3
surfaces that are $d$-neighbors of $X$ correspond one-to-one to isotropic
lines in $\mathrm{T}(X)/d\mathrm{T}(X)$.

######  Remark 3.14.

As a consequence of the Lang-Weil estimates, one sees that if
$\mathrm{rank}(L)\geq 3$ then $d\rightarrow\infty$ implies that
$|\\{d\text{-isotropic lines}\\}|\rightarrow\infty$ too.

We need the following lemma:

######  Lemma 3.15.

Let $L$ be an even lattice and let $L^{\prime}$ be a lattice in
$L\otimes\mathbf{Q}$ with discriminant forms $A$ and $A^{\prime}$
respectively. There is a natural isomorphism $A\cong A^{\prime}$ whenever one
of the following condition is satisfied:

  1. (1)

$L,L^{\prime}\in\mathcal{L}^{\mathrm{max}}$;

  2. (2)

$L,L^{\prime}$ belong to the same genus and $[L\colon L\cap L^{\prime}]$ and
$\mathrm{discr}(L)$ are coprime.

###### Proof.

Let $L\coloneqq L_{1}\cap L_{2}$. Since
$\mathrm{discr}(L_{1})=\mathrm{discr}(L_{2})$ because $L_{1}$ and $L_{2}$ are
in the same genus, we see that

$[L_{1}\colon
L]^{2}=\frac{\mathrm{discr}(L)}{\mathrm{discr}(L_{1})}=\frac{\mathrm{discr}(L)}{\mathrm{discr}(L_{2})}=[L_{2}\colon
L]^{2},$

i.e., that $[L_{1}\colon L]=[L_{2}\colon L]$. Let $A$ be the finite quadratic
form associated to $L$, and let $H_{1},H_{2}\subset A$ be the isotropic
subgroups that determine $L_{1}$ and $L_{2}$ respectively. Since $L=L_{1}\cap
L_{2}$ by definition, we see that $H_{1}\cap H_{2}=0$, and hence that
$H_{1}\oplus H_{2}\subset A$. Moreover, we also have natural isomorphisms
$A_{1}\cong H_{1}^{\perp}/H_{1}$ and $A_{2}\cong H_{2}^{\perp}/H_{2}$.

  1. (1)

To prove point (1), we want to show that the map
$H_{1}^{\perp}/H_{1}\rightarrow A/(H_{1}\oplus H_{2})$ is injective, which is
equivalent to $H_{1}^{\perp}\cap H_{2}=0$. This is in turn equivalent to
$L_{1}^{\vee}\cap L_{2}=L$. To show this last equality, assume that $v\in
L_{1}^{\vee}\cap L_{2}$, and consider the lattice $\tilde{L}_{1}$ generated by
$v$ and $L_{1}$ (note that this might not be primitive). Since any element of
$\tilde{L}_{1}$ can be written as $av+bw$ with $a,b\in\mathbf{Z}$ and $w\in
L_{1}$, we compute

$(av+bw,av+bw)=a^{2}(v,v)+2ab(v,w)+b^{2}(w,w),$

and since $(w,w)\in 2\mathbf{Z}$, $(w,v)\in\mathbf{Z}$ (because $v\in
L_{1}^{\vee}$) and $(v,v)\in 2\mathbf{Z}$ (because $v\in L_{2}$) we conclude
that $\tilde{L}_{1}$ must be even, and hence integral. But since
$L_{1}\subset\tilde{L}_{1}$ and $L_{1}$ is maximal by assumption, we must have
that $L_{1}=\tilde{L}_{1}$, i.e., that $v\in L_{1}$. It follows that $L\subset
L_{1}^{\vee}\cap L_{2}\subset L$. Thus we prove the injectivity of
$H_{1}^{\perp}/H_{1}\rightarrow A/(H_{1}\oplus H_{2})$. But by cardinality
reasons, this map must also be surjective, and hence it is an isomorphism.
Carrying out the specular argument for $L_{2}$, we have constructed natural
isomorphisms

$A_{1}\cong H_{1}^{\perp}/H_{1}\cong A/(H_{1}\oplus H_{2})\cong
H_{2}^{\perp}/H_{2}\cong A_{2},$

and point (1) of the Lemma is proved.

  2. (2)

To prove point (2), let $D=\mathrm{discr}(L_{1})$. Then, since
$|H_{1}|=|H_{2}|$ is coprime to $D$, we must have a natural decomposition
$A=H_{1}\oplus H_{2}\oplus A^{\prime}$, where $A^{\prime}$ corresponds to the
$D$-torsion of $A$. From this one easily shows that $H_{1}^{\perp}=H_{1}\oplus
A^{\prime}$ and $H_{2}^{\perp}=H_{2}\oplus A^{\prime}$, and proceeds as
before.

∎

######  Proposition 3.16.

Let $X/\mathbf{C}$ be a K3 surface. For any isotropic line
$\ell\subset\operatorname{Br}(X)[d]$ one can constuct a well-defined K3
surface $X_{\ell}$ endowed with a Hodge isometry $f_{\ell}$ such that
$f_{\ell}(\mathrm{T}(X_{\ell}))$ is the $d$-neighbor of $\mathrm{T}(X)$
associated to $\ell$, and such that
$\operatorname{NS}(X)\cong\operatorname{NS}(X_{\ell})$.

###### Proof.

Denote by $\mathrm{T}_{\ell}(X)\subset\mathrm{T}(X)_{\mathbf{Q}}$ the
$d$-neighbor associated to $\ell$. By the lemma above, there is a natural
isomorphism between $A_{X}$ and the discriminant form of
$\mathrm{T}_{\ell}(X)$ \- which we denote by $A$. Consider now the direct sum
$\mathrm{T}_{\ell}(X)\oplus\operatorname{NS}(X)$ endowed with its natural
Hodge structure. We get natural identifications $A\cong-
A_{X}\cong-\operatorname{NS}(X)^{\vee}/\operatorname{NS}(X)$, which we use to
glue the direct sum into the K3 lattice. By the global Torelli, we have
constructed a unique projective K3 surface $X_{\ell}$. ∎

In order to construct the auxiliary K3 surface, we need to ensure that
$\ker(\ell)$ embedds primitively into the K3 lattice.

######  Proposition 3.17.

Let $X/\mathbf{C}$ be a K3 surface and assume that $\rho(X)\geq 3$. For any
isotropic line $\ell\subset\operatorname{Br}(X)[d]$ one can construct a K3
surface $Y_{\ell}$ endowed with a Hodge isometry $g_{\ell}$ such that
$g_{\ell}(\mathrm{T}(Y_{\ell}))=\ker(\ell)$.

###### Proof.

Let $A$ be the discriminant form of $\ker(\ell)$ and let $p$ be a prime
number. If $p$ divides $\mathrm{discr}(\operatorname{NS}(X))$ then
$A\otimes\mathbf{Z}_{p}\cong A_{X}\otimes\mathbf{Z}_{p}$ and
$A\otimes\mathbf{Z}_{p}\cong\mathbf{Z}/p\oplus\mathbf{Z}/p\mathbf{Z}$ if $p$
divides $d$ and $A\otimes\mathbf{Z}_{p}=0$ otherwise. If $\rho(X)\geq 3$ we
know from [Nik79, Corollary 1.12.3] that $\ker(\ell)$ embedds primitively into
the K3 lattice, and hence that there is a K3 surface $Y_{\ell}$ with a Hodge
isometry that maps $\mathrm{T}(Y_{\ell})$ to $\ker(\ell)$. This is well-
defined up to Fourieri-Mukai partners. ∎

Also the next is a consequence of Nikulin results.

######  Lemma 3.18.

In order for $Y_{\ell}$ to be unique, it is enough that $\rho(X)\geq 4$ and
that $\ell(A_{X})\leq\rho(X)-2$.

###### Proof.

We only need to show that in this case $\ker(\ell)$ embedds uniquely into the
K3 lattice. This follows from [Nik79, Corollary 1.12.3]. ∎

The next follows directly from [Muk87, Proposition 6.6].

######  Proposition 3.19.

Assume that $Y_{\ell}$ exists. Then, both $X$ and $X_{\ell}$ are moduli spaces
of sheaves over some Fourier-Mukai partner of $Y_{\ell}$.

When $Y_{\ell}$ does not exists, then the relation between $X$ and its
neighbors can be understood in terms of twisted derived equivalences as
explained by Huybrechts [Huy19].

###  3.4. More neighbors

In this section we show one way to generalize Kneser correspondence to the
case when $d$ and $\mathrm{discr}(L)$ are not coprime. Let $L$ be an even
lattice, for any prime $p$, we denote by $L_{p}=L\otimes\mathbf{Z}_{p}$. We
observe that the notion of neighbors can be considered also for $p$-adic
lattices (even, if $p=2$) and in general over every principal ideal domain
(see [DF21, Definition 3.1.1]). To ‘glue’ the local notion to the global one,
one proceeds as follows. For any prime $p$ such that $L_{p}$ is unimodular
(that is, for any prime $p$ coprime to $\mathrm{discr}(L)$) the
$p^{n}$-neighbors of $L_{p}$ correspond to isotropic lines in
$L_{p}\otimes\mathbf{Z}_{p}/p^{n}\mathbf{Z}_{p}\cong L/p^{n}L.$ Decompose
$d=\prod_{p}p^{n_{p}}$ and let $\ell_{p}\subset L_{p}/p^{n_{p}}L_{p}$ be
isotropic lines. Let $L^{\prime}_{p}\subset L_{p}\otimes\mathbf{Q}_{p}$ be the
neighbor of $L_{p}$ corresponding to $\ell_{p}$, where we put
$L^{\prime}_{p}=L_{p}$ if $\ell_{p}=0$. By strong approximation, there is a
unique lattice $L^{\prime}\subset L_{\mathbf{Q}}$ such that
$L^{\prime}\otimes\mathbf{Z}_{p}=L^{\prime}_{p}$ for every $p$. This is the
$d$-neighbor associated to $\ell=\prod_{p}\ell_{p}\subset L/dL$. For an even
lattice $L$ and a prime number $p$ one has a modular decomposition

$L_{p}\cong U_{0}\perp U_{1}(p)\perp\cdots\perp U_{n}(p^{n})$

where $U_{i}$ is a $p$-adic unimodular lattice (even, if $p=2$) and
$U_{i}(p^{n})$ denotes the lattice $U_{i}$ with form scaled by a factor of
$p^{n}$. The lattice $U_{0}$ is a maximal unimodular sublattice of $L_{p}$,
and it follows that one has an isomorphism

$A_{L_{p}}=A_{U_{1}(p)}\perp\cdots\perp A_{U_{n}(p^{n})}.$

Since a $p$-adic lattice (even, if $p=2$) is uniquely determined by its rank,
its discriminant form, and its discriminant modulo squares, one deduces that
all the maximal unimodular sublattices of a $p$-adic lattice are isometric.

For any integer $k\geq 0$, we say that a subgroup of $L_{p}/p^{k}L_{p}$ is
non-degenerate if it is the form $U_{0}/p^{n}U_{0}$ for some maximal
unimodular sublattice $U_{0}\subset L_{p}$. Similarly, a subgroup of
$\operatorname{Hom}(L_{p},\mathbf{Z}/p^{n}\mathbf{Z})$ is non-degenerate if it
is the image of $U_{0}$ under the natural map $L_{p}\rightarrow
L_{p}/p^{n}L_{p}\rightarrow\operatorname{Hom}(L_{p},\mathbf{Z}/p^{n}\mathbf{Z})$.

######  Definition 3.20.

  * •

A line $\ell\cong\mathbf{Z}/p^{n}\mathbf{Z}\subset L_{p}/p^{n}L_{p}$ is
generalized isotropic if it is isotropic with respect to the natural pairing
on $L_{p}/p^{n}L_{p}$ and if it is contained in a non-degenerate subgroup. The
analogue definition applies to a line
$\ell^{\prime}\subset\operatorname{Hom}(L_{p},\mathbf{Z}/p^{n})$. If $L$ is an
even lattice and $d$ is an integer, we say that a line $\ell\subset L/dL$ is
generalized isotropic if $\ell_{p}$ is generalized isotropic for every prime
$p$, where, as before, $\ell_{p}$ denotes the $p$-part of
$\ell=\prod_{p|d}\ell_{p}$. The analogous definition applies for a line in
$\operatorname{Hom}(L,\mathbf{Z}/d\mathbf{Z})$.

  * •

Let $L_{p}$ be a $p$-adic lattice (even, if $p=2$). A generalized
$p^{n}$-neighbor is a lattice $L^{\prime}_{p}\subset
L_{p}\otimes\mathbf{Q}_{p}$ that can be written as
$L^{\prime}_{p}=U_{0}^{\prime}\oplus(U_{0})^{\perp}$ for some maximal
unimodular sublattice $U_{0}\subset L_{p}$, where $U_{0}^{\prime}$ is a
$p^{n}$-neighbor of $U_{0}$. From the global point of view, let $L$ be an even
lattice and let $L^{\prime}\subset L\otimes\mathbf{Q}$. Then $L^{\prime}$ is a
generalized $\prod_{p}p^{n_{p}}=d$-neighbor of $L$ if
$L^{\prime}\otimes\mathbf{Z}_{p}$ is a generalized $p^{n_{p}}$-neighbor of
$L\otimes\mathbf{Z}_{p}$ for every $p$.

  * •

A sublattice $M_{p}\subset L_{p}$ is said to be generalized $p^{n}$-split if
it is of the form $M^{\prime}_{p}\perp U_{0}^{\perp}$ for some maximal
unimodular sublattice $U_{0}\subset L_{p}$ and some $p^{n}$-split sublattice
$M^{\prime}_{p}$ of $U_{0}$. Similarly, one has this notion in the global
case.

Clearly, one has a bijection between generalized isotropic lines in $L/dL$ and
generalized isotropic lines in $\operatorname{Hom}(L,\mathbf{Z}/d\mathbf{Z})$
for every $d>1$, and, if $d$ is coprime to $\mathrm{discr}(L)$, all these
notions recover the original ones of Kneser. We need the following
generalization of Lemma 3.11:

######  Lemma 3.21.

Let $M_{p}$ be a generalized $p^{n}$-split sublattice of a $p$-adic lattice
$L_{p}$. Let $H_{p}\cong\mathbf{Z}/p^{n}\mathbf{Z}\subset A_{M_{p}}$ be the
isotropic subgroup associated to the overlattice $M_{p}\subset L_{p}$. There
exists a unique $H^{\prime}_{p}\cong\mathbf{Z}/p^{n}\mathbf{Z}\subset
A_{M_{p}}$ such that $H_{p}\cap H^{\prime}_{p}=0$, $H^{\prime}_{p}$ is
isotropic, and one has an isomorphism of quadratic forms
$(H^{\prime}_{p})^{\perp}\cong H^{\prime}_{p}\perp A_{L_{p}}$.

###### Proof.

By definition, we can write $M_{p}=M^{\prime}_{p}\perp U_{0}^{\perp}$ for some
maximal unimodular sublattice $U_{0}\subset L_{p}$ and some $p^{n}$-split
sublattice $M_{p}^{\prime}$ of $U_{0}$. We write
$A_{M_{p}}=A_{M^{\prime}_{p}}\perp A_{L_{p}}$. From Lemma 3.11 we know that
there exists a unique $H^{\prime}_{p}\subset A_{M^{\prime}_{p}}$ such that
$H^{\prime}_{p}$ is isotropic, $H^{\prime}_{p}\cap H_{p}=0$ and
$A_{M^{\prime}_{p}}=H_{p}\oplus H^{\prime}_{p}$. It follows that
$H^{\prime}_{p}$ satisfies the properties required in the statement. To show
the uniqueness of $H^{\prime}_{p}$, let $H^{\prime\prime}_{p}$ be another
cyclic subgroup generated by $a+b+c\in A_{M_{p}}=H_{p}\oplus
H^{\prime}_{p}\perp A_{L_{p}}$. Since $(H^{\prime\prime}_{p})^{\perp}\cong
H^{\prime\prime}_{p}\perp A_{L_{p}}$ by assumption, we see that for every
$d\in A_{L_{p}}$ one has $(c,d)=0$. But then $c=0$, and from the fact that
$(a,b,0)$ is isotropic one concludes that
$H^{\prime}_{p}=H^{\prime\prime}_{p}$. ∎

Our generalization of Kneser neighbors is then rephrased in the following.

######  Proposition 3.22.

Let $L$ be an even lattice and let $d$ be any integer. Then, there is a
natural one-to-one correspondence between:

  1. (1)

Generalized $d$-neighbors of $L$;

  2. (2)

Generalized isotropic lines in $\operatorname{Hom}(L,\mathbf{Z}/d\mathbf{Z})$;

  3. (3)

Generalized $d$-split sublattices of $L$.

Moreover, if $L^{\prime}$ is a generalized $d$-neighbor of $L$, there is a
natural isomorphism $A_{L^{\prime}}\cong A_{L}$.

###### Proof.

It is enough to show this correspondence for $p$-adic lattices. Let
$L^{\prime}_{p}$ be a generalized $p^{n}$-neighbor of a $p$-adic lattice
$L_{p}$ (even, if $p=2$). By definition, we can write
$L^{\prime}_{p}=U_{0}^{\prime}\perp U_{0}^{\perp}$ where $U_{0}$ is a maximal
unimodular sublattice and $U_{0}^{\prime}$ is a classical $p^{n}$-neighbor of
$U_{0}$. Let $\ell_{p}=\operatorname{Im}(p^{n}L^{\prime}_{p}\rightarrow
L_{p}\rightarrow L_{p}/p^{n}L_{p})$. Since we know that
$\operatorname{Im}(p^{n}U^{\prime}_{0}\rightarrow U_{0}\rightarrow
U_{0}/p^{n}U_{0})$ is an isotropic line due to Kneser theory, we deduce that
$\ell_{p}\subset U_{0}/p^{n}U_{0}\subset L_{p}/p^{n}L_{p}$ is a generalized
isotropic line. Similarly, let
$\ell_{p}\subset\operatorname{Hom}(L_{p},\mathbf{Z}/p^{n}\mathbf{Z})$ be a
generalized isotropic line belonging to some non-degenerate subgroup
$U_{0}/p^{n}U_{0}$. Denote by
$\ell^{\prime}_{p}\subset\operatorname{Hom}(U_{0},\mathbf{Z}/p^{n}\mathbf{Z})$
the same isotropic line. Then $\ker(\ell_{p})=\ker(\ell^{\prime}_{p})\perp
U_{0}^{\perp}$ is a generalized $p^{n}$-split sublattice of $L_{p}$. Finally,
let $M_{p}\subset L_{p}$ be a generalized $p^{n}$-split sublattice and let
$H^{\prime}_{p}\subset A_{M_{p}}$ be the unique isotropic subgroup given by
the lemma above. Then, the associated overlattice $L^{\prime}_{p}$ is easily
seen to be a generalized $p^{n}$-neighbor.

The natural isomorphism between $A_{L^{\prime}}$ and $A_{L}$ also follows from
Lemma 3.21. ∎

######  Remark 3.23.

Let $L$ be a lattice that satisfies $\ell(A_{L})\leq\mathrm{rank}(L)-3$ and
let $N_{d}(L)$ be the number of its generalized $d$-neighbors. Under these
assumption, for every prime $p$, $L/p^{n}L$ contains a non-degenerate subgroup
of the form $(\mathbf{Z}/p^{n}\mathbf{Z})^{3}$. It follows that $N_{d}(L)\geq
d$ for every $d$ and every such $L$.

We can extend this to K3 surfaces in an analogue way as we did before. Then
only difference is about the existence of $Y_{\ell}$ for a generalized
isotropic line: the condition $\rho(X)\geq 3$ must be replaced by
$\ell(A_{L})+3\leq\rho(X)$ in order to ensure that every generalized $d$-split
sublattice of $\mathrm{T}(X)$ embedds primitively into the K3 lattice.

###  3.5. An example

In this section we give an explicit example of the theory just exposed by
proving the following proposition:

######  Proposition 3.24.

Let $X=\mathrm{Km}(E_{1}\times E_{2})$ be the Kummer surface associated to the
product of two non-isogenous, non-CM elliptic curves $E_{1}$ and $E_{2}$.
Then, for any odd integer $d$, the $d$-neighbors of $X$ are all of the form
$\mathrm{Km}(E_{1}/C_{1}\times E_{2}/C_{2})$ where $C_{1}\cong
C_{2}\cong\mathbf{Z}/d\mathbf{Z}$.

We recall that the Kummer surface $X$ associated to an abelian surface $A$ is
the blow up of $A/\\{\pm 1\\}$ at its sixteen notes. We also recall that
$\mathrm{T}(X)$ uniquely determines $X$ in this case, and that one always has
a natural Hodge isometry between $\mathrm{T}(A)(2)$ and $\mathrm{T}(X)$, where
the former denotes the transcendental lattice of $A$ with form scaled by a
factor of $2$.

Let $E_{1},E_{2}$ be two non-isogenous, non-CM elliptic curves and let
$X=\mathrm{Km}(E_{1}\times E_{2})$ be the associated Kummer surface. One has
$\mathrm{T}(A)\cong\mathrm{H}^{1}(E_{1},\mathbf{Z})\otimes\mathrm{H}^{1}(E_{2},\mathbf{Z})^{\vee}.$
Here, the dual $\mathrm{H}^{1}(E_{2},\mathbf{Z})^{\vee}$ plays a role mostly
due to a reason of weights in Hodge theory. Let $\\{e_{1},f_{2}\\}$ and
$\\{e_{2},f_{2}\\}$ be basis of $\mathrm{H}^{1}(E_{1},\mathbf{Z})$ and
$\mathrm{H}^{1}(E_{2},\mathbf{Z})$ respectively, such that $(e_{i},f_{i})=1$.
Then, $v_{1}=e_{1}\otimes e_{2}$, $v_{2}=e_{1}\otimes f_{2}$,
$v_{3}=f_{1}\otimes e_{1}$ and $v_{4}=e_{2}\otimes f_{2}$ form a basis of
$\mathrm{T}(A)$. With respect to this basis, the Gram matrix of
$\mathrm{T}(A)$ takes the following form:

$\begin{bmatrix}0&0&0&1\\\ 0&0&-1&0\\\ 0&-1&0&0\\\ 1&0&0&0\end{bmatrix}$

Under the natural isomorphism

$\mathrm{T}(A)\cong\operatorname{Hom}(\mathrm{H}^{1}(E_{1},\mathbf{Z}),\mathrm{H}^{1}(E_{2},\mathbf{Z})),$

an element
$f\in\operatorname{Hom}(\mathrm{H}^{1}(E_{1},\mathbf{Z}),\mathrm{H}^{1}(E_{2},\mathbf{Z}))$
represented by the matrix $\begin{bmatrix}a&b\\\ c&d\end{bmatrix}$ with
respect to the basis $\\{e_{i},f_{i}\\}$ is sent to
$bv_{1}+dv_{2}-av_{3}-cv_{4}$. Thus, one sees that $(f,f)=\det(f)$. Since
$\mathrm{T}(A)$ is unimodular, we can work with any integer $d>1$. From the
computation above, an isotropic line
$\ell\subset\mathrm{T}(A)\otimes\mathbf{Z}/d\mathbf{Z}$ is generated by an
endormorphism
$f\colon\mathrm{H}^{1}(E_{1},\mathbf{Z}/d\mathbf{Z})\rightarrow\mathrm{H}^{1}(E_{2},\mathbf{Z}/d\mathbf{Z})$
such that $\mathrm{ker}(f)=\mathrm{im}(f)\cong\mathbf{Z}/d\mathbf{Z}$. In
particular, we have a one-to-one correspondence

$\\{d-\text{isotropic lines of
}\mathrm{T}(A)\otimes\mathbf{Z}/d\mathbf{Z}\\}\leftrightarrow\\{(C_{1},C_{2})\colon
C_{i}\subset E_{i}(\mathbf{C})_{\mathrm{tors}},C_{1}\cong
C_{2}\cong\mathbf{Z}/d\mathbf{Z}\\}.$

We show now how to associate the corresponding neighbor of $\mathrm{T}(A)$.
Let $f\in\mathrm{T}(A)\otimes\mathbf{Z}/d\mathbf{Z}$ be a generator of an
isotropic line, and let $C_{1}=\ker(f)$ and $C_{2}=\mathrm{im}(f)$, when we
see $f$ as a map
$\mathrm{H}^{1}(E_{1},\mathbf{Z}/d\mathbf{Z})\rightarrow\mathrm{H}^{1}(E_{2},\mathbf{Z}/d\mathbf{Z}))$.
Let $E_{i}^{\prime}\coloneqq E_{i}/C_{i}$ and put
$H_{i}=\mathrm{im}(\mathrm{H}^{1}(E_{i}^{\prime},\mathbf{Z})\rightarrow\mathrm{H}^{1}(E_{i},\mathbf{Z}))$,
where the images are taken under the pullback in cohomology of the natural
quotient map. Up to changing the bases, we can assume that
$H_{i}=\mathbf{Z}de_{i}\oplus\mathbf{Z}f_{i}$ for $i=1,2$. Consider
$\mathrm{T}^{\prime}\coloneqq\frac{1}{d}(H_{1}\otimes
H_{2})\subset\mathrm{T}(A)\otimes\mathbf{Q}$, which is generated by
$dv_{1},v_{2},v_{3},d^{-1}v_{4}$. The lattice $\mathrm{T}^{\prime}$ is
isometric to $\mathrm{T}(A)$ and it is clearly a $d$-neighbor of
$\mathrm{T}(A)$. Also, one has that $\mathrm{T}^{\prime}\cap\mathrm{T}(A)$ is
generated by $dv_{1},v_{2},v_{3},v_{4}$, and a quick computation shows that
this is the kernel of the isotropic Brauer line.

The only thing left to show is that the K3 surface associated to
$\mathrm{T}^{\prime}$ is isomorphic to
$X^{\prime}=\mathrm{Km}(E_{1}^{\prime}\times E_{2}^{\prime})$, i.e., that
$\mathrm{T}^{\prime}$ and $\mathrm{T}(X^{\prime})$ are Hodge isometric. If we
endow $H_{i}$ with the polarization coming from the inclusion into
$\mathrm{H}^{1}(E_{i},\mathbf{Z})$ we see that
$H_{i}\cong\mathrm{H}^{1}(E_{i}^{\prime},\mathbf{Z})(d)$. It follows that
$H_{1}\otimes H_{2}$ is Hodge isometric to
$\mathrm{H}^{1}(E_{1}^{\prime},\mathbf{Z})(d)\otimes\mathrm{H}^{1}(E_{2}^{\prime},\mathbf{Z})(d)$
and therefore that $\frac{1}{d}(H_{1}\otimes H_{2})$ is Hodge isometric to
$\mathrm{H}^{1}(E_{1}^{\prime},\mathbf{Z})\otimes\mathrm{H}^{1}(E_{2}^{\prime},\mathbf{Z})$.

##  4\. Fields of definitions

In this section we determine a field of definition for every K3 surface in
$\mathcal{H}_{X}$. We start with a K3 surface $X/K$ over a number field
$K\subset\mathbf{C}$, we pick a K3 surface $(Y,f)\in\mathcal{H}^{m}_{\bar{X}}$
and we determine whether $(Y,f)$ can be defined over $K$ as well. This means
that there should exist a K3 surface $Y^{\prime}/K$ with a Hodge isometry
$f^{\prime}\colon\mathrm{T}(Y^{\prime}_{\mathbf{C}},\mathbf{Q})\rightarrow\mathrm{T}(X_{\mathbf{C}},\mathbf{Q})$
such that $f^{\prime}$ is defined over $K$ and such that there is an
isomorphism $h\colon\overline{Y^{\prime}}\xrightarrow{\sim}Y$ such that
$f\circ h^{*}=f^{\prime}$, where $h^{*}$ denotes the natural pullback map on
the transcendental lattices.

######  Theorem 4.1.

Let $(Y,f)\in\tilde{\mathcal{H}}_{X}$ and assume that $X$ is defined over a
number field $K$. Denote by $\rho_{X}$ the Galois representation associated to
$T(\overline{X},\mathbf{A}_{f})$. Then, the data $(Y,f,\operatorname{NS}(Y))$
can defined over $K$ if and only

  1. (1)

The $\widehat{\mathbf{Z}}$-lattice
$f(\mathrm{T}(Y))\otimes\widehat{\mathbf{Z}}$ is stable under $\rho_{X}$;

  2. (2)

The induced $G_{K}$-representation on $A_{Y}$ is trivial.

Note that the analogue result for abelian varieties is almost tautological,
and the only fundamental difference between the two situations is point (2).

###### Proof.

Let $X/\overline{\mathbf{Q}}$ be a smooth projective variety and let
$\sigma\in\operatorname{Aut}(\overline{\mathbf{Q}})$. Denote by $X^{\sigma}$
the fibre product of $X$ along $\sigma$, and denote by $\sigma_{X}\colon
X\rightarrow X^{\sigma}$ the induced isomorphism of schemes. Note that
$X^{\sigma}$ is the scheme obtained by applying $\sigma$ to the defying
equation of $X$. When $X$ is a K3 surface, we denote by
$\sigma_{X}^{*}\colon\mathrm{H}^{2}_{\textrm{\'{e}t}}(X^{\sigma},\widehat{\mathbf{Z}}(1))\rightarrow\mathrm{H}^{2}_{\textrm{\'{e}t}}(X,\widehat{\mathbf{Z}}(1))$
the induced isomorphism in étale cohomology. Moreover, we write
$\sigma_{X,\mathrm{alg}}^{*}$ and $\sigma_{X,\mathrm{tr}}^{*}$ to mean
respectively the restriction of $\sigma_{X}^{*}$ to the algebraic and
transcendental part of
$\mathrm{H}^{2}_{\textrm{\'{e}t}}(X^{\sigma},\widehat{\mathbf{Z}}(1))$. To
produce a model of $X$ over $K$ we use Galois descent and the global Torelli
theorem for K3 surfaces. We refer the reader to [Poo17] for a general
introduction to Galois descent, and also to [Val22], where a similar argument
is used. Essentially, we will produce a Galois descent data for $X$ over $K$,
which consists, for every $\sigma\in G_{K}$, of an isomorphism
$f_{\sigma}\colon X^{\sigma}\rightarrow X$ such that the assignment
$\sigma\mapsto f_{\sigma}$ satisfies the cocycle condition
$f_{\sigma\tau}=f_{\sigma}\circ(f_{\tau})^{\sigma}$. Corollary 4.4.6 together
with Remark 4.4.8 of [Poo17] says that if such data exists, then $X$ admits a
model $X_{0}/K$. This model is characterized by the fact that the Galois
action on its $\overline{\mathbf{Q}}$-points is given by the composition

$X(\overline{\mathbf{Q}})\xrightarrow{\sigma}X^{\sigma}(\overline{\mathbf{Q}})\xrightarrow{f_{\sigma}}X(\overline{\mathbf{Q}}),$

for each $\sigma\in G_{K}$. Similarly, one has a natural identification
$\overline{X}_{0}\cong X$ and then the Galois action on
$\mathrm{H}^{2}_{\textrm{\'{e}t}}(\overline{X}_{0},\widehat{\mathbf{Z}})$ is
given by the composition

(4.1)
$\mathrm{H}^{2}_{\textrm{\'{e}t}}(X,\widehat{\mathbf{Z}}(1))\xrightarrow{f_{\sigma}^{*}}\mathrm{H}^{2}_{\textrm{\'{e}t}}(X^{\sigma},\widehat{\mathbf{Z}}(1))\xrightarrow{\sigma_{X}^{*}}\mathrm{H}^{2}_{\textrm{\'{e}t}}(X,\widehat{\mathbf{Z}}(1)).$

Now, if $(Y,f,\operatorname{NS}(Y))$ is defined over $K$, then (1) and (2) are
satisfied. To prove the other direction we argue as follows. For $\sigma\in
G_{K}$, we write the diagram 2.3, taking into account that
$\mathrm{T}(X,\widehat{\mathbf{Z}})=\mathrm{T}(X^{\sigma},\widehat{\mathbf{Z}})$
because $X$ is defined over $K$:

${\mathrm{T}(Y^{\sigma},\mathbf{A}_{f})}$${\mathrm{T}(X,\mathbf{A}_{f})}$${\mathrm{T}(Y,\mathbf{A}_{f})}$${\mathrm{T}(X,\mathbf{A}_{f}).}$$\scriptstyle{f^{\sigma}}$$\scriptstyle{\sigma_{Y,\mathrm{tr}}^{*}}$$\scriptstyle{\sigma_{X,\mathrm{tr}}^{*}}$$\scriptstyle{f}$

In particular, we have that

(4.2) $\sigma_{X,\mathrm{tr}}^{*}\circ
f^{\sigma}=f\circ\sigma_{Y,\mathrm{tr}}^{*}.$

Due to the fact that $f(\mathrm{T}(Y,\widehat{\mathbf{Z}}))$ is
$G_{K}$-stable, and that $\sigma_{Y,\mathrm{tr}}^{*}$ induces an
identification between $\mathrm{T}(Y^{\sigma},\widehat{\mathbf{Z}})$ and
$\mathrm{T}(Y,\widehat{\mathbf{Z}})$, we deduce from the diagram above that

(4.3)
$f^{\sigma}(\mathrm{T}(Y^{\sigma},\widehat{\mathbf{Z}}))=f(\mathrm{T}(Y,\widehat{\mathbf{Z}})).$

Hence

$f^{\sigma}(\mathrm{T}(Y^{\sigma}))=f^{\sigma}(\mathrm{T}(Y^{\sigma},\widehat{\mathbf{Z}}))\cap\mathrm{T}(X,\mathbf{Q})=f(\mathrm{T}(Y,\widehat{\mathbf{Z}}))\cap\mathrm{T}(X,\mathbf{Q})=f(\mathrm{T}(Y)).$

In particular, the map $(f^{\sigma})^{-1}\circ
f\colon\mathrm{T}(Y,\mathbf{Q})\rightarrow\mathrm{T}(Y^{\sigma},\mathbf{Q})$
sends $\mathrm{T}(Y)$ to $\mathrm{T}(Y^{\sigma})$, i.e., it is a integral
Hodge isometry. Our aim is to show that the isometry

(4.4) $((f^{\sigma})^{-1}\circ
f)\oplus(\sigma_{Y,\mathrm{alg}}^{*})^{-1}\colon\mathrm{T}(Y)\oplus\operatorname{NS}(Y)\rightarrow\mathrm{T}(Y^{\sigma})\oplus\operatorname{NS}(Y^{\sigma})$

extends to a global Hodge isometry
$\mathrm{H}^{2}_{\text{sing}}(Y,\mathbf{Z})\rightarrow\mathrm{H}^{2}_{\text{sing}}(Y^{\sigma},\mathbf{Z}).$
To this purpose, denote by
$\tilde{\sigma}_{Y^{\sigma},\mathrm{tr}}^{*}\colon\mathrm{T}(Y,\widehat{\mathbf{Z}})\rightarrow\mathrm{T}(Y,\widehat{\mathbf{Z}})$
the unique map that satisfies

(4.5)
$f^{\sigma}\circ\tilde{\sigma}_{Y^{\sigma},\mathrm{tr}}^{*}=\sigma_{X,\mathrm{tr}}^{*}\circ
f^{\sigma};$

this is nothing but the restriction of $\sigma_{X,\mathrm{tr}}^{*}$ to
$f^{\sigma}(\mathrm{T}(Y^{\sigma},\widehat{\mathbf{Z}}))$ seen as an isometry
of $\mathrm{T}(Y^{\sigma},\widehat{\mathbf{Z}})$, and it exists because of
(4.3) and the fact that $f(\mathrm{T}(Y,\widehat{\mathbf{Z}}))$ is
$G_{K}$-invariant. By point (2) $\tilde{\sigma}_{Y^{\sigma},\mathrm{tr}}^{*}$
acts as the identity on $A_{Y^{\sigma}}\cong A_{Y}$. From equation (4.2) and
(4.5) we have that
$f^{\sigma}\circ\tilde{\sigma}_{Y^{\sigma},\mathrm{tr}}^{*}=f\circ\sigma_{Y,\mathrm{tr}}^{*}$,
hence that $(f^{\sigma})^{-1}\circ
f=\tilde{\sigma}_{Y^{\sigma},\mathrm{tr}}^{*}\circ(\sigma_{Y,\mathrm{tr}}^{*})^{-1}.$
From this we conclude that the maps $A_{Y}\rightarrow A_{Y^{\sigma}}$ induced
respectively by $(f^{\sigma})^{-1}\circ f$ and
$(\sigma_{Y,\mathrm{tr}}^{*})^{-1}$ coincide. But also the maps
$A_{Y}\rightarrow A_{Y^{\sigma}}$ induced by
$(\sigma_{Y,\mathrm{tr}}^{*})^{-1}$ and $(\sigma_{Y,\mathrm{alg}}^{*})^{-1}$
coincide, because both are restrictions of global
$\widehat{\mathbf{Z}}$-isometry $(\sigma_{Y}^{*})^{-1}$. Using Proposition
2.2, we conclude that there exists a global Hodge isometry
$\mathrm{H}^{2}_{\text{sing}}(Y^{\sigma},\mathbf{Z})\rightarrow\mathrm{H}^{2}_{\text{sing}}(Y,\mathbf{Z})$
which extends (4.4), and since $(\sigma_{Y,\mathrm{alg}}^{*})^{-1}$ sends
ample line bundles to ample line bundles, this is in turn induced by a unique
isomorphism $g_{\sigma}\colon Y\xrightarrow{\sim}Y^{\sigma}$ by the global
Torelli theorem. Then, we are left to show that the association
$\sigma\mapsto(g_{\sigma}\colon Y\rightarrow Y^{\sigma})$ satisfies the
cocycle condition, i.e., that
$g_{\sigma}\circ(g_{\tau})^{\sigma}=g_{\sigma\tau}$ for any $\sigma,\tau\in
G_{K}$. It is sufficient to check this identity cohomologically. At the level
of Néron-Severi lattices, we have that
$g_{\tau}^{*}=(\tau_{Y,\mathrm{alg}}^{*})^{-1}$, hence that
$((g_{\tau})^{\sigma})^{*}=(\sigma_{Y^{\tau},\mathrm{alg}}^{*})^{-1}\circ
g_{\tau}^{*}\circ\sigma_{Y}^{*}=(\sigma_{Y^{\tau},\mathrm{alg}}^{*})^{-1}\circ(\tau_{Y,\mathrm{alg}}^{*})^{-1}\circ\sigma_{Y,\mathrm{alg}}^{*}$.
Since $(\sigma\tau)_{Y}=\sigma_{Y^{\tau}}\circ\tau_{Y}$, we readily conclude
in this case. At the level of transcendental lattices, we have that
$g_{\tau}^{*}=(f^{\tau})^{-1}\circ f$ and hence that
$((g_{\tau})^{\sigma})^{*}=(\sigma_{Y^{\tau},\mathrm{tr}}^{*})^{-1}\circ[(f^{\tau})^{-1}\circ
f]\circ\sigma_{Y,\mathrm{tr}}^{*}$. Note that
$f^{\sigma}=(\sigma_{X,\mathrm{tr}}^{*})^{-1}\circ
f\circ\sigma_{Y,\mathrm{tr}}^{*}$ from which we see that
$((g_{\tau})^{\sigma})^{*}\circ
g_{\sigma}^{*}=(\sigma_{Y^{\tau},\mathrm{tr}}^{*})^{-1}\circ(f^{\tau})^{-1}\circ\sigma_{X,\mathrm{tr}}^{*}\circ
f=(f^{\sigma\tau})^{-1}\circ f$. Thus, the cocycle condition is proved, and we
check finally that both $f$ and $\operatorname{NS}(\overline{Y})$ are defined
over $K$. Consider the model $Y_{0}/K$ constructed with such Galois descent
data. We need to show that $G_{K}$ acts trivially on
$\operatorname{NS}(\overline{Y}_{0})$. Using (4.1), we know that $\sigma\in
G_{K}$ acts on
$\mathrm{H}^{2}_{\textrm{\'{e}t}}(\overline{Y}_{0},\widehat{\mathbf{Z}}(1))\cong\mathrm{H}^{2}_{\textrm{\'{e}t}}(Y,\widehat{\mathbf{Z}}(1))$
as the composition $\sigma_{Y}^{*}\circ g_{\sigma}^{*}$. It is clear that this
is the identity on the Néron-Severi group, and we are only left to check that
$f\circ(\sigma_{Y,\mathrm{tr}}^{*}\circ(g_{\sigma}^{*})_{\mathrm{tr}})=\sigma_{X,\mathrm{tr}}^{*}\circ
f$ and since $(g_{\sigma}^{*})_{\mathrm{tr}}=(f^{\sigma})^{-1}\circ f$ this
becomes equation (4.2). ∎

######  Remark 4.2.

Note that the model of $Y$ thus constructed depends on the chosen model of $X$
over $K$.

Let $K^{\prime}/K$ be the smallest extension that satisfies (1) and (2) in the
theorem above. Then, this is also the smallest extension over which
$(Y,f,\operatorname{NS}(Y))$ can be defined. Hence, the data
$(Y,f,\operatorname{NS}(Y))$ admits a smallest field of definition over $K$.

A direct consequence of Proposition 2.7 and Lemma 2.3 is that a minimal field
of definition for $(Y,f,\operatorname{NS}(Y))$ is not too far away from a
field of definition for $Y$ of smallest degree:

######  Corollary 4.3.

Let $L$ be a field of definition for $Y$, and let $K^{\prime}/K$ be the
minimal field extension of $K$ that satisfies (1) and (2) of the theorem
above. Then, there exists an extension $L^{\prime}/L$ of degree
$[L^{\prime}\colon L]\leq C_{n}\cdot C_{h}$ such that $K^{\prime}\subset
L^{\prime}$.

Another corollary comes from Lemma 3.15:

######  Corollary 4.4.

Let $(Y,f)\in\tilde{\mathcal{H}}_{X}$, and assume that
$(X,\operatorname{NS}(X))$ can be defined over a number field $K$. Assume that
$f(\mathrm{T}(Y))$ and $\mathrm{T}(X)$ belong to the same genus, then:

  1. (1)

If $X$ (and hence also $Y$) is maximal, $(Y,f,\operatorname{NS}(Y))$ is
defined over $K$ if and only if $f(\mathrm{T}(Y))\otimes\widehat{\mathbf{Z}}$
is $G_{K}$-stable.

  2. (2)

If $f$ is such that $[\mathrm{T}(X)\colon f(\mathrm{T}(Y))\cap\mathrm{T}(X)]$
is coprime to $\mathrm{discr}(\mathrm{T}(X))$, then the same conclusion of (a)
is true.

###### Proof.

All the cases follow in the same way, as an application of Lemma 3.15 and
Theorem 4.1. To prove case (1), for example, one notes that if $G_{K}$
stabilizes $f(\mathrm{T}(Y))\otimes\widehat{\mathbf{Z}}=:\mathrm{T}$, the the
naturality in Lemma 3.15 implies that the isomorphism between
$A_{X}=\mathrm{T}(X)^{\vee}/\mathrm{T}(X)$ and
$A_{\mathrm{T}}=\mathrm{T}^{\vee}/\mathrm{T}$ is $G_{K}$-equivariant, and
since $G_{K}$ acts trivially on $A_{X}$ by assumption, the statement follows
from Theorem 4.1. ∎

Note in particular that maximal K3 surfaces behave, in this respect, like
abelian varieties. Concerning generalized neighboring K3 surfaces, we have the
following

######  Corollary 4.5.

Let $X/K$ be a K3 surface over a number field $K$ and let
$\ell\subset\operatorname{Br}(\overline{X})[d]$ be a generalized isotropic
line. Assume that $G_{K}$ acts trivially on $\operatorname{NS}(\overline{X})$.

  1. (1)

If $\ell$ is $G_{K}$-stable, then there is a K3 surface $X_{\ell}/K$ with
$\operatorname{NS}(\overline{X}_{\ell})=\operatorname{NS}(\overline{X}_{\ell})^{G_{K}}$
and a Hodge isometry
$f\colon\mathrm{T}(X_{\ell,\mathbf{C}},\mathbf{Q})\rightarrow\mathrm{T}(X,\mathbf{Q})$
defined over $K$ such that $f(\mathrm{T}(X_{\ell,\mathbf{C}}))$ is the
$d$-neighbor of $\mathrm{T}(X_{\mathbf{C}})$ associated to $\ell$.

  2. (2)

Assume also that $G_{K}$ acts trivially on $\ell$, and that
$\rho_{\bar{X}}\geq 3$. Then $(Y_{\ell},g_{\ell},\operatorname{NS}(Y_{\ell}))$
can be defined over $K$.

###### Proof.

Point (1) follows directly from the last part of Proposition 3.22. To prove
point (2), we need to prove that if $G_{K}$ acts trivially on $\ell$ then it
also acts trivially on the discriminant form of $\ker(\ell)$. This follows
from Lemma 3.21: let $A$ be the discriminant form of $\ker(\ell)$ and let
$H\subset A$ be the isotropic subgroup associated to
$\ker(\ell)\subset\mathrm{T}(X)$ and let $H^{\prime}\subset A$ its unique
transverse as in 3.21. It follows that
$H^{\prime}\cong\operatorname{Hom}(H,\mathbf{Q}/\mathbf{Z})\cong\ell$, and
therefore, if $G_{K}$ acts trivially on $\ell$ and on
$\operatorname{NS}(\overline{X})$ then it also acts trivially on $A$. ∎

Our last corollary of the theorem concerns moduli spaces of sheaves on K3
surfaces. Let $Y$ be a complex K3 surface, let
$v\in\mathrm{H}^{*}(Y,\mathbf{Z})$ be a primitive and isotropic Mukai vector,
and let $X=M(v)$ be the associated moduli space. Assume that $X$ is a K3
surface. Let
$f\colon\mathrm{T}(Y,\mathbf{Q})\rightarrow\mathrm{T}(X,\mathbf{Q})$ be the
isogeny induced by the quasi-universal family, so that
$f(\mathrm{T}(Y))\subset\mathrm{T}(X)$.

######  Corollary 4.6.

Assume that $X$ has a model $X^{\prime}$ over a number field
$K\subset\mathbf{C}$ and assume that $G_{K}$ acts trivially on
$\operatorname{NS}(\overline{X^{\prime}})$ and on $A_{Y}$ as in point (2) of
1.1. Then, there exists a model $Y^{\prime}/K$ of $Y$ such that $X^{\prime}$
is the moduli space of stable sheaves on $Y$.

###### Proof.

We know by 1.1 that under these conditions there exists a K3 surface
$Y^{\prime}/K$ together with a $K$-Hodge isometry
$f^{\prime}\colon\mathrm{T}(Y^{\prime}_{\mathbf{C}},\mathbf{Q})\rightarrow\mathrm{T}(X^{\prime}_{\mathbf{C}},\mathbf{Q})$
and an isomorphism $h\colon Y^{\prime}_{\mathbf{C}}\rightarrow Y$ such that
$f^{\prime}\circ h^{*}_{\mathrm{tr}}=f$.

Let
$h^{*}\colon\mathrm{H}^{*}(Y,\mathbf{Z})\rightarrow\mathrm{H}^{*}(Y^{\prime}_{\mathbf{C}},\mathbf{Z})$
be the induced map on cohomology, and let $v^{\prime}=h^{*}(v)$. Then,
$v^{\prime}$ is defined over $K$ because $G_{K}$ acts trivially on
$\operatorname{NS}(\overline{Y^{\prime}})$. It follows that we can consider
the moduli space $M(v^{\prime})$ and we have a canonical isomorphism $g\colon
M(v^{\prime})_{\mathbf{C}}\xrightarrow{\sim}X$. We are only left to show that
$M(v^{\prime})\cong X^{\prime}$ already over $K$, which follows from the fact
that
$g^{*}\colon\mathrm{H}_{\textrm{\'{e}t}}^{2}(\overline{X^{\prime}},\widehat{\mathbf{Z}})\rightarrow\mathrm{H}_{\textrm{\'{e}t}}^{2}(\overline{Y^{\prime}},\widehat{\mathbf{Z}})$
commutes with the Galois action.

∎

##  5\. Finiteness and uniformity

We begin by proving Theorem 1.5, which is straightforward at this point.

###### Proof of Theorem 1.5.

For $i\in\mathbb{N}$, let $X_{i}$ be K3 surfaces defined over number fields
$K_{i}$ with $[K_{i}\colon\mathbf{Q}]\leq N$ and let $d_{i}>0$ be a sequence
of integers such that
$\operatorname{Br}(\overline{X}_{i})[d_{i}]\subset\operatorname{Br}(\overline{X}_{i})^{G_{K}}$.
We can assume without loss of generality that
$\operatorname{NS}(\overline{X}_{i})^{G_{K_{i}}}=\operatorname{NS}(\overline{X}_{i})$.
Assume that every $X_{i}$ satisfies $\ell(A_{X})\leq 19-\rho(X)$.

  1. (1)

Assume that $3+\ell(A)\leq\rho_{\bar{X}_{i}}$ for every $i$. Since Shafarevich
conjecture holds by assumption, let $\mathcal{L}_{N}$ be the finite set of
lattices that can appear as $\operatorname{NS}(\overline{X})$ for $X/K$ a K3
surface with $[K\colon\mathbf{Q}]\leq N$. Let

$M=\max\\{d\in\mathbb{N}\colon\,\,\text{there
is}\,\,L_{1},L_{2}\in\mathcal{L}_{N}\,\,\text{such
that}\,\,d^{2}\mathrm{discr}(L_{1})=\mathrm{discr}(L_{2})\\},$

then $M<\infty$. Due to Remark 3.23 we conclude that there are generalized
$d_{i}$-isotropic lines in $\operatorname{Br}(\overline{X}_{i})[d_{i}]$ for
every $i>0$. Choose one of them,
$\ell_{i}\subset\operatorname{Br}(\overline{X})[d_{i}]$. Since $G_{K_{i}}$
acts trivially on $\ell_{i}\subset\operatorname{Br}(\overline{X})[d_{i}]$ we
have a K3 surface $Y_{i}/K_{i}$ together with a Hodge isometry that identifies
$\mathrm{T}(Y_{i,\mathbf{C}})$ with $\ker(\ell)$. But since
$\mathrm{discr}(\operatorname{NS}(\overline{Y}_{i}))=d_{i}^{2}\cdot\mathrm{discr}(\operatorname{NS}(\overline{X}_{i}))$
one deduces that $d_{i}\leq M<\infty$.

  2. (2)

The proof uses the same ideas, but only part (1) of Proposition 4.5. Let
$N_{i}$ be the number of generalized $d_{i}$-isotropic lines in
$\mathrm{T}(X_{i,\mathbf{C}})$. Due again to Remark 3.23 we know that
$N_{i}\geq d_{i}$ for every $i$. Since $E_{X_{i\mathbf{C}}}$ is totally real,
we conclude that there are at least $N_{i}$ geometric isomorphism classes of
K3 surfaces that are isogenous to $X_{i}$ and that can be defined over
$K_{i}$. Thus, if Conjecture 1.4 is true, we conclude that the $d_{i}$ are
again bounded.

If $\rho(X_{i})<3+\ell(A_{X_{i}})$ then one can use the classical form of
Kneser correspondence to conclude analogously for $d_{i}$ coprime to
$\mathrm{discr}(\operatorname{NS}(X_{i}))$. In this case, we only need
$3\leq\rho_{\bar{X}_{i}}$ to conclude in the first point. ∎

In this last part of the paper, we prove the finiteness stated at the end of
the introduction. We use some ideas from the previous sections, but the
finiteness itself is independent from 4.1. Our main ingredient is Theorem 5.2
of Cadoret and Moonen; we remark that the results in [CM20] builds on the
Mumford-Tate conjecture for K3 surfaces, whose proof uses the Tate-conjecture
for divisors of abelian varieties in a crucial way (see [Tan95] and [And96]).
Thus, our finiteness is not independent from Faltings’ celebrated work
[Fal83].

We shall use the following notation: let $V$ be a rational vector space, and
let $G\subset\operatorname{GL}(V)$ be a linear algebraic group. For any
maximal rank free $\mathbf{Z}$ module $L\subset V$ we denote by
$\mathbf{G}_{L}$ the integral form of $G$ induced by $L$. This is constructed
as the Zariski closure of $G$ seen inside the generic fibre of
$\operatorname{GL}(L)$. It follows that $\mathbf{G}_{L}$ is a flat affine
group scheme over $\operatorname{Spec}(\mathbf{Z})$ whose generic fibre
recovers $G$. In case $L=\mathrm{T}(X)$ we simply write $\mathbf{G}_{X}$ for
$\mathbf{G}_{L}$.

We begin with a lemma:

######  Lemma 5.1.

Let $Y\in\mathcal{H}^{\mathrm{low}}_{X}$ and assume that $X$ can be defined
over a field $K$ and that $Y$ can defined over a field $K^{\prime}$. Choose
any Hodge isometry
$f\colon\mathrm{T}(Y,\mathbf{Q})\xrightarrow{\sim}\mathrm{T}(X,\mathbf{Q})$
such that $f(\mathrm{T}(Y))\subset\mathrm{T}(X)$. Then, there exists an
extension $K^{\prime\prime}/KK^{\prime}$ of degree $[K^{\prime\prime}\colon
K^{\prime}K]\leq C_{n}\cdot C_{h}$ such that
$f(\mathrm{T}(Y))^{\vee}/\mathrm{T}(X)\subset\operatorname{Br}(\overline{X})^{G_{K^{\prime\prime}}}.$

###### Proof.

In fact, let $K^{\prime\prime}/K^{\prime}K$ be an extension of degree
$[K^{\prime\prime}\colon K^{\prime}K]\leq C_{h}\cdot C_{n}$ over which both
$\operatorname{NS}(\overline{Y})$ and $f$ are defined. It follows that
$f(\mathrm{T}(Y))^{\vee}/\mathrm{T}(X)\subset\operatorname{Br}(\overline{X})^{G_{K}}$.
∎

The next is a direct consequence of Theorem 2.4, see also Corollary 5.3. of
[OS18].

######  Proposition 5.2.

Let $X$ be a K3 surface defined over a number field $K$ and let $N>0$ be an
integer. There exists a constant $C=C(N,X)$ such that for any field extension
$K^{\prime}/K$ of degree bounded by $N$, one has

$|\operatorname{Br}(\bar{X})^{G_{K^{\prime}}}|\leq C.$

###### Proof.

Let $\mathbf{MT}_{X}$ be the integral form of $\operatorname{MT}_{X}$ induced
by the lattice $\mathrm{T}(X)\subset\mathrm{T}(X,\mathbf{Q})$. Hence,
$\mathbf{MT}_{X}(\widehat{\mathbf{Z}})$ naturally acts on
$\mathrm{T}(X)^{\vee}\otimes\mathbf{Q}/\mathbf{Z}.$ If
$\Gamma\subset\mathbf{MT}_{X}(\widehat{\mathbf{Z}})$ is a finite-index group,
then the fixed points
$(\mathrm{T}(X)^{\vee}\otimes\mathbf{Q}/\mathbf{Z})^{\Gamma}$ are finitely
many. Moreover, it is well-known that the group
$\mathbf{MT}_{X}(\widehat{\mathbf{Z}})$ has only finitely many subgroups of a
certain index, and for each $M>0$ we define

$C(X,M)\coloneqq\max\\{|(\mathrm{T}(X)^{\vee}\otimes\mathbf{Q}/\mathbf{Z})^{\Gamma}|\colon\Gamma\,\,\text{has
index}\,\,\leq M\\}.$

Let now $\rho_{X}\colon
G_{K}\rightarrow\operatorname{GL}(\mathrm{T}(X,\mathbf{Q}))(\mathbf{A}_{f})\cong\operatorname{GL}_{X}(\mathbf{A}_{f})$
be the Galois representation associated to $T(\overline{X},\mathbf{A}_{f})$.
Then, Theorem 2.4 says that the index

$M\coloneqq[\mathbf{MT}_{X}(\widehat{\mathbf{Z}})\colon\rho_{X}^{\vee}(G_{K^{\prime}})\cap\mathbf{MT}_{X}(\widehat{\mathbf{Z}})]$

is finite. For any extension $K^{\prime}/K$ of degree $[K^{\prime}\colon
K]\leq N$ we then have

$|\operatorname{Br}(\bar{X})^{G_{K^{\prime}}}|=|(\mathrm{T}(\overline{X})^{\vee}\otimes\mathbf{Q}/\mathbf{Z})^{\rho_{X}^{\vee}(G_{K^{\prime}})}|\leq|(\mathrm{T}(\overline{X})^{\vee}\otimes\mathbf{Q}/\mathbf{Z})^{\rho_{X}^{\vee}(G_{K^{\prime}})\cap\mathbf{MT}_{X}(\widehat{\mathbf{Z}})}|,$

and this last quantity is always smaller than $C(X,NM)$. ∎

We recall that $\mathcal{H}^{\mathrm{low}}_{X}$ was defined in 3.2.

######  Proposition 5.3.

Let $N>0$ be an integer and let $X/\overline{\mathbf{Q}}$ be a K3 surface.
Then, only finitely many elements of $\mathcal{H}^{\mathrm{low}}_{X}$ can be
defined over some number field $K$ of degree $[K\colon\mathbf{Q}]\leq N.$

###### Proof.

Assume that $X$ is defined over a number field $K$ and suppose that there were
infinitely many, pairwise distinct K3 surfaces
$X_{i}\in\mathcal{H}^{\mathrm{low}}_{X}$, for $i\in\mathbb{N}$, that can be
defined over field extensions $K_{i}/K$ of degree $[K_{i}\colon K]\leq N$. The
index of $\mathrm{T}(X_{i})$ in $\mathrm{T}(X)$ does not depend on a chosen
rational Hodge isometry. In fact, for any inclusion of lattices $N\subset L$
of finite index, one has $\mathrm{discr}(N)=\mathrm{discr}(N)\cdot[L\colon
N]^{2}$. From the fact that $\mathrm{T}(X)$ has only finitely many subgroups
of a given index, and that a K3 surface has only finitely many Fourier-Mukai
partners, we conclude that the index of $\mathrm{T}(X_{i})$ in $\mathrm{T}(X)$
must be unbounded for $i\in\mathbb{N}$. For any $i$, denote by $\tilde{X}_{i}$
a chosen model over $K_{i}$ and also choose a Hodge isometry
$f_{i}\colon\mathrm{T}(X_{i})\rightarrow\mathrm{T}(X)$ such that
$f_{i}(\mathrm{T}(X_{i}))\subset\mathrm{T}(X)$, and let $K_{i}^{\prime}/K_{i}$
be an extension of universally bounded degree over which both
$\operatorname{NS}(\overline{X}_{i})$ and $f_{i}$ are defined. Write
$A_{i}\subset\operatorname{Br}(X)$ for the group
$f_{i}(\mathrm{T}(X_{i}))^{\vee}/\mathrm{T}(X)^{\vee}$. From Lemma 5.1 we see
that

$|\operatorname{Br}(\overline{X})^{G_{K_{i}^{\prime}}}|\geq|A_{i}|=[\mathrm{T}(X)\colon
f_{i}(\mathrm{T}(X_{i}))].$

But since the indices $[\mathrm{T}(X)\colon f_{i}(\mathrm{T}(X_{i}))]$ are
unbounded for $i\in\mathbb{N}$, we find a contradiction to Proposition 5.2. ∎

######  Proposition 5.4.

The two following assertions are equivalent.

  1. (1)

Let $X/\mkern 1.5mu\overline{\mkern-1.5mu\mathbf{Q}\mkern-1.5mu}\mkern 1.5mu$.
Then, only finitely many elements of $\mathcal{H}_{X}$ can be defined over a
field extension $K/\mathbf{Q}$ of a given degree.

  2. (2)

Let $X/\mkern 1.5mu\overline{\mkern-1.5mu\mathbf{Q}\mkern-1.5mu}\mkern 1.5mu$.
Then, only finitely many elements of $\mathcal{H}^{\mathrm{max}}_{X}$ can be
defined over a field extension $K/\mathbf{Q}$ of a given degree.

###### Proof.

Clearly (1)$\implies$(2). To show the reverse implication one argues as
follows. Suppose that (2) is true but (1) is not. Then, there are K3 surfaces
$X_{i}/K_{i}$ defined over fields $K_{i}$ such that $[K_{i}\colon\mathbf{Q}]$
is bounded and $\overline{X}_{i}\in\mathcal{H}_{X}$ are pairwise disjoint. We
can assume without loss of generality that $G_{K_{i}}$ acts trivially on
$\operatorname{NS}(\overline{X}_{i})$. Let
$\mathrm{T}(X_{i,\mathbf{C}})\subset\mathrm{T}_{i}$ be a maximal lattice
containing $\mathrm{T}(X_{i,\mathbf{C}})$. Since $G_{K_{i}}$ acts trivially on
$A_{X_{i}}$ one shows that $G_{K_{i}}$ stabilizes
$\mathrm{T}_{i}\otimes\widehat{\mathbf{Z}}$ and that moreover it acts
trivially on $A_{\mathrm{T}_{i}}$. It follows that there are K3 surfaces
$(Y_{i},f_{i},\operatorname{NS}(Y_{i}))$ defined over $K_{i}$ with a Hodge
isometry
$f_{i}\colon\mathrm{T}(Y_{i,\mathbf{C}},\mathbf{Q})\rightarrow\mathrm{T}(X_{i,\mathbf{C}},\mathbf{Q})$
such that $f_{i}(\mathrm{T}(Y_{i,\mathbf{C}}))=\mathrm{T}_{i}$. In particular,
these are maximal K3 surfaces, and by $(2)$, they can be only finitely many
geometrically. The results now follow from 5.3. ∎

Up until now, our results work for every K3 surface. In the next theorem we
specialize to the generic case when
$\operatorname{End}_{\operatorname{Hdg}}(\mathrm{T}(X))=\mathbf{Z}$.

######  Theorem 5.5.

Let $N>0$ be an integer and assume that
$\operatorname{End}_{\operatorname{Hdg}}(\mathrm{T}(X))=\mathbf{Z}$. Then,
only finitely many elements of $\mathcal{H}_{X}^{\mathrm{max}}$ can be defined
over a field extension $K/\mathbf{Q}$ of degree $[K\colon\mathbf{Q}]\leq N$.

###### Proof.

We can assume without loss of generality that $X$ is maximal. As a consequence
of Shimura characterization of maximal lattices there is an isomorphism

$\mathrm{SO}_{X}(\mathbf{A}_{f})/\mathbf{SO}_{X}(\widehat{\mathbf{Z}})\xrightarrow{\sim}\mathcal{L}^{\text{max}}.$

Assume that $X$ can be defined over a number field $M/\mathbf{Q}$, and assume
without loss of generality that
$\rho_{X}(G_{M})\subset\mathrm{SO}_{X}(\mathbf{A}_{f})$ (this can always be
achieved after a finite extension). Then by 5.2 $\rho_{X}(G_{M})$ is a
compact-open subgroup. Let $C_{h}>0$ be the constant from Proposition 2.7, and
let $\mathcal{V}\subset\rho_{X}(G_{M})$ be the intersection of all the
subgroups of $\rho_{X}(G_{M})$ of index smaller than $C_{h}\cdot N$. Since
there are only finitely many such subgroups, also $\mathcal{V}$ is a compact-
open subgroup of $\mathrm{SO}_{X}(\mathbf{A}_{f})$. Let now
$(Y,f)\in\tilde{\mathcal{H}}^{\mathrm{max}}$, so that
$f(\mathrm{T}(Y))\in\mathcal{L}^{\mathrm{max}}$, and assume that $Y$ can be
defined over an extension $K/\mathbf{Q}$ of degree $[K\colon\mathbf{Q}]\leq
N$. Then, there exists an extension $K^{\prime}/KM$ of degree
$[K^{\prime}\colon KM]\leq C_{h}$ over which $f$ is defined too. Since $f$ is
defined over $K^{\prime}$, it follows that
$f(T(\overline{Y},\widehat{\mathbf{Z}}))$ is $G_{K^{\prime}}$-stable, hence
that $\mathcal{V}\subset\mathbf{SO}_{Y}(\widehat{\mathbf{Z}})$. Thus, to prove
the statement, it suffices to show that only finitely many elements
$g\in\mathrm{SO}_{X}(\mathbf{A}_{f})/\mathbf{SO}_{X}(\widehat{\mathbf{Z}})$
satisfy

(5.1)
$\mathcal{V}\subset\mathrm{SO}_{g\mathrm{T}(X)}(\widehat{\mathbf{Z}})=g\mathrm{SO}_{X}(\widehat{\mathbf{Z}})g^{-1}.$

A compact-open subgroup $\mathcal{U}\subset\mathrm{SO}_{X}(\mathbf{A}_{f})$ is
said to be maximal if it is not properly contained in another compact-open
subgroup. A similar definition applies to compact open subgroups
$\mathcal{U}_{p}\subset\mathrm{SO}_{X}(\mathbf{Q}_{p})$.

######  Lemma 5.6.

If $\mathcal{U}\subset\mathrm{SO}_{X}(\mathbf{A}_{f})$ is any compact open
subgroup, and $\mathcal{U}_{p}\subset\mathrm{SO}_{X}(\mathbf{Q}_{p})$ denotes
its $p$-component, then $\mathcal{U}_{p}$ is maximal for almost every prime
$p$.

###### Proof.

Consider the natural embedding $\mathbf{SO}_{X}\subset\mathbf{GL}_{X}$. By the
definition of the adelic topology, for almost every $p$ we have that
$\mathcal{U}_{p}=\mathbf{GL}_{X}(\mathbf{Z}_{p})\cap\mathbf{SO}_{X}(\mathbf{Q}_{p})$.
If moreover $p$ is a prime of good reduction for $\mathbf{SO}_{X}$, which
happens for example whenever $p$ is odd and does not divide
$\mathrm{discr}(\mathrm{T}(X))$, then it follows from Bruhat-Tits theory that
$\mathbf{GL}_{X}(\mathbf{Z}_{p})\cap\mathbf{SO}_{X}(\mathbf{Q}_{p})$ is
maximal (for an elementary proof of this fact one can consult [Mac17]). ∎

Since $\mathrm{SO}_{X}$ is reductive, [PR94, Proposition 3.16] and the lemma
above say that every compact-open subgroup of
$\mathrm{SO}_{X}(\mathbf{A}_{f})$ is contained in at least one maximal
compact-open subgroup, and in at most finitely many such subgroups. Assume by
contradiction that there are infinitely many
$g_{i}\in\mathrm{SO}_{X}(\mathbf{A}_{f})/\mathbf{SO}_{X}(\widehat{\mathbf{Z}})$,
for $i\in\mathbb{N}$, such that $\mathcal{V}\subset
g_{i}\mathrm{SO}_{X}(\widehat{\mathbf{Z}})g_{i}^{-1}$. Since $\mathcal{V}$ is
contained in only finitely many maximal compact-open subgroups, and since
every compact-open subgroup of $\mathrm{SO}_{X}(\widehat{\mathbf{Z}})$ has
only finitely many subgroups of a given index, we can assume without loss of
generality that
$g_{i}\mathrm{SO}_{X}(\widehat{\mathbf{Z}})g_{i}^{-1}=g_{j}\mathrm{SO}_{X}(\widehat{\mathbf{Z}})g_{j}^{-1}$
for $i,j\in\mathbf{N}$. Let $\mathcal{N}$ be the normalizer of
$\mathrm{SO}_{X}(\widehat{\mathbf{Z}})$ in $\mathrm{SO}_{X}(\mathbf{A}_{f})$.
Then we must have that $g_{1}g_{i}^{-1}$ belongs to the image of $\mathcal{N}$
in $\mathrm{SO}_{X}(\mathbf{A}_{f})/\mathrm{SO}_{X}(\widehat{\mathbf{Z}})$ for
any $i\in\mathbb{N}$. But since $\mathrm{SO}_{X}$ is semisimple, from the last
part of [PR94, Proposition 3.16] one has that the natural inclusion

(5.2) $\mathrm{SO}_{X}(\widehat{\mathbf{Z}})\subset\mathcal{N}$

is of finite-index. Thus, we find a contradiction to the infinitude of the
$g_{i}$’s.

∎

######  Remark 5.7.

When $E_{X}\neq\mathbf{Q}$ we do not have anymore the equality (5.1) for any
$g\in\mathrm{SO}_{X}(\mathbf{A}_{f})$, but only for
$g\in\operatorname{MT}_{X}(\mathbf{A}_{f})\subset\mathrm{SO}_{X}(\mathbf{A}_{f})$.
Using the same ideas of above and the description of Mumford-Tate groups of K3
surfaces of Zarhin [Zar83], we can prove the finiteness of the smaller
$\mathrm{H}^{\operatorname{MT}_{X}}_{X}\subset\mathrm{H}_{X}$ given by all the
K3 surfaces $Y$ with a Hodge isometry
$f\colon\mathrm{T}(Y,\mathbf{Q})\rightarrow\mathrm{T}(X,\mathbf{Q})$ such that
$f(\mathrm{T}(Y))$ and $\mathrm{T}(X)$ belong to the same
$\operatorname{MT}_{X}$-genus.

## References

  * [And96] Yves André.  On the Shafarevich and Tate conjectures for hyperkähler varieties.  Math. Ann., 305(2):205–248, 1996.
  * [Bus19] Nikolay Buskin.  Every rational Hodge isometry between two K3 surfaces is algebraic.  Journal für die reine und angewandte Mathematik, 2019(755):127 – 150, 2019.
  * [Cha16] François Charles.  Birational boundedness for holomorphic symplectic varieties, Zarhin’s trick for $K3$ surfaces, and the Tate conjecture.  Ann. of Math. (2), 184(2):487–526, 2016.
  * [CM20] Anna Cadoret and Ben Moonen.  Integral and adelic aspects of the Mumford-Tate conjecture.  J. Inst. Math. Jussieu, 19(3):869–890, 2020.
  * [C0̆2] Andrei Căldăraru.  Nonfine moduli spaces of sheaves on $K3$ surfaces.  Int. Math. Res. Not., (20):1027–1056, 2002.
  * [DF21] Neil Dummigan and Dan Fretwell.  Automorphic forms for some even unimodular lattices.  Abh. Math. Semin. Univ. Hamb., 91(1):29–67, 2021.
  * [Fal83] G. Faltings.  Endlichkeitssätze für abelsche Varietäten über Zahlkörpern.  Invent. Math., 73:349–366, 1983.
  * [Huy19] Daniel Huybrechts.  Motives of isogenous K3 surfaces.  Comment. Math. Helv., 94(3):445–458, 2019.
  * [Kne57] Martin Kneser.  Klassenzahlen definiter quadratischer Formen.  Arch. Math., 8:241–250, 1957.
  * [LP92] M. Larsen and R. Pink.  On $\ell$-independence of algebraic monodromy groups in compatible systems of representations.  Invent. Math., 107(3):603–636, 1992.
  * [Mac17] Marco Maculan.  Maximality of hyperspecial compact subgroups avoiding Bruhat-Tits theory.  Ann. Inst. Fourier (Grenoble), 67(1):1–21, 2017.
  * [MSTVA17] Kelly McKinnie, Justin Sawon, Sho Tanimoto, and Anthony Várilly-Alvarado.  Brauer groups on K3 surfaces and arithmetic applications.  In Brauer groups and obstruction problems, volume 320 of Progr. Math., pages 177–218. Birkhäuser/Springer, Cham, 2017.
  * [Muk87] S. Mukai.  On the moduli space of bundles on $K3$ surfaces. I.  In Vector bundles on algebraic varieties (Bombay, 1984), volume 11 of Tata Inst. Fund. Res. Stud. Math., pages 341–413. Tata Inst. Fund. Res., Bombay, 1987.
  * [MW93] David Masser and Gisbert Wüstholz.  Isogeny estimates for abelian varieties, and finiteness theorems.  Ann. of Math. (2), 137(3):459–472, 1993.
  * [Nik79] Viacheslav Nikulin.  Integer symmetric bilinear forms and some of their geometric applications.  Izvestiya Akademii Nauk SSSR. Seriya Matematicheskaya, 43(1):111–177, 238, 1979.
  * [Nik07] Viacheslav Nikulin.  On correspondences between K3 surfaces.  Mathematics of the USSR-Izvestiya, 30:375, 2007.
  * [OS18] Martin Orr and Alexei N. Skorobogatov.  Finiteness theorems for K3 surfaces and Abelian varieties of CM type.  Compositio Mathematica, 154(8):1571–1592, 2018.
  * [OSZ19] Martin Orr, Alexei N. Skorobogatov, and Yuri G. Zarhin.  On uniformity conjectures for Abelian varieties and K3 surfaces.  American Journal of Mathematics, October 2019.
  * [Poo17] Bjorn Poonen.  Rational Points on Varieties.  Graduate Studies in Mathematics. American Mathematical Society, 2017.
  * [PR94] Vladimir Platonov and Andrei Rapinchuk.  Algebraic groups and number theory. Transl. from the Russian by Rachel Rowen.  Boston, MA: Academic Press, 1994.
  * [Pvv71] Ilija I. Pjateckiĭ-Šapiro and Igor’ R. Šafarevič.  Torelli’s theorem for algebraic surfaces of type ${\textrm{K}}3$.  Izvestiya Akademii Nauk SSSR. Seriya Matematicheskaya, 35:530–572, 1971.
  * [R1́8] Gaël Rémond.  Conjectures uniformes sur les variétés abéliennes.  Q. J. Math., 69(2):459–486, 2018.
  * [Shi10] Goro Shimura.  Arithmetic of quadratic forms.  Springer Monographs in Mathematics. Springer, New York, 2010.
  * [Tan95] S. G. Tankeev.  Surfaces of type K3 over number fields and the Mumford-Tate conjecture. II.  Izv. Math., 59(3):619–646, 1995.
  * [Val21] Domenico Valloni.  Complex multiplication and Brauer groups of K3 surfaces.  Advances in Mathematics, 385:107772, 2021.
  * [Val22] Domenico Valloni.  Fields of definition of k3 surfaces with complex multiplication.  Journal of Number Theory, 2022.
  * [van05] Bert van Geemen.  Some remarks on Brauer groups of $K3$ surfaces.  Adv. Math., 197(1):222–247, 2005.
  * [Zar83] Yuri G. Zarhin.  Hodge groups of $K3$ surfaces.  Journal für die Reine und Angewandte Mathematik, 341:193–220, 1983.