Explicit Nikulin configurations on Kummer surfaces

TL;DR

This paper explores explicit constructions of non-isomorphic Kummer structures on Kummer surfaces, extending previous methods to broader cases and addressing limitations in earlier approaches.

Contribution

It generalizes explicit constructions of non-isomorphic Kummer structures on Kummer surfaces with weaker polarization restrictions.

Findings

Constructed explicit non-isomorphic Kummer structures on certain Kummer surfaces.

Extended previous constructions to cases with lower polarization degrees.

Identified scenarios where earlier methods are not applicable.

Abstract

A Nikulin configuration is the data of $16$ disjoint smooth rational curves on a K3 surface. According to results of Nikulin, the existence of a Nikulin configuration means that the K3 surface is a Kummer surface, moreover the abelian surface from the Kummer structure is determined by the $16$ curves. A classical question of Shioda is about the existence of non isomorphic Kummer structures on the same Kummer K3 surface. The question was studied by several authors, and it was shown that the number of non-isomorphic Kummer structures is finite, but no explicit geometric construction of such structures was given. In a previous paper, we constructed explicitly non isomorphic Kummer structures on some Kummer surfaces. In this paper we generalise the construction to Kummer surfaces with a weaker restriction on the degree of the polarization and we describe some cases where the previous construction does not work.

Tables

Table 1. Table 1. Fundamental solutions of the Pell-Fermat equations

$2t$	2*	4	6*	$8$	10’	12*	14	16	18	20*	22	$24$	26’	28	30*
${\alpha }_0$	3	-	5	3	19	7	15	-	17	9	197	5	51	127	11
${\beta }_0$	2	-	2	1	6	2	4	-	4	2	42	1	10	24	2
$2t$	$32$	34	36	38	$40$	42*	44	46	$48$	50’	52	54	56*	58’	60
${\alpha }_0$	17	35	-	37	19	13	199	24335	7	99	649	485	15	19603	31
${\beta }_0$	3	6	-	6	3	2	30	3588	1	14	90	66	2	2574	4

Equations

$$\{\text{mbox}Kummerstructures\}⟷\{\text{mbox}Nikulinconfigurations\}/\text{mbox}Aut(X)$$

$${\alpha }^2-2t{\beta }^2=1$$

$$x_k+y_k\sqrt{2t}=({\alpha }_0+\sqrt{2t}{\beta }_0{)}^k.$$

$$\beta =\pm 1,\pm 3\text{ mod }8,$$

$$x_n+y_n\sqrt{2t}=(x_1+y_1\sqrt{2t}{)}^n$$

$$x^2-2t{y}_n^2{y}^2=1$$

$$x^2-2t{z}_n^2{y}^2=1$$

$$\{7,9,14,23,30,31,33,34,46,47,56,57,62,63,69,71,73,75,77,79,81,82,89,90,94\}.$$

$${\alpha }^2-2t{\beta }^2=-1$$

$$1,5,13,25,29,37,41,53,61,65,85,101,\mathrm{109...}$$

$$L^2=4t$$

$$\mathbb{Z}L\oplus K\subset \mathrm{NS}(X),$$

$$\begin{array}[]{c}A^{\prime}_{1}=\beta L-\alpha A_{1}\\ L^{\prime}=bL-aA_{1}\end{array},$$

$$\begin{array}[]{c}\alpha^{2}-2t\beta^{2}=1\\ 2tb\beta=a\alpha\\ a^{2}=2t(b^{2}-1)\end{array},$$

$$(a,b)=(2t\beta ,\alpha ).$$

$$4t^2{b}^2{\beta }^2=2t{\alpha }^2(b^2-1),$$

$$b^2({\alpha }^2-2t{\beta }^2)={\alpha }^2,$$

$$A_1^{\prime }={\beta }_{0}L-{\alpha }_0{A}_1,$$

$$\mathcal{C}=\sum _{i=1}^{16}{A}_i,{\mathcal{C}}^{\prime }=A_1^{\prime }+\sum _{i=2}^{16}{A}_i.$$

$$\frac{1}{2}(A_1+A_1^{\prime })=\frac{{\beta }_0}{2}L-\frac{{\alpha }_0-1}{2}{A}_1$$

$$A_1^{\prime }={\beta }_{0}L-{\alpha }_0{A}_1,L^{\prime }={\alpha }_{0}L-2t{\beta }_0{A}_1.$$

$$\Gamma =uL-\sum _{i=1}^{16}{v}_i{A}_i$$

$$\Gamma L^{\prime }\le 0,$$

$$u{\alpha }_0\le v_1{\beta }_0,$$

$$\sum _{i\ge 1}{v}_i^2=2t{u}^2+1\le 2t\left(\frac{{\beta }_0^2}{{\alpha }_0^2}{v}_1^2\right)+1=2t{\beta }_0^2\left(\frac{v_1^2}{{\alpha }_0^2}\right)+1$$

$$\sum _{i\ge 1}{v}_i^2\le ({\alpha }_0^2-1)\left(\frac{v_1^2}{{\alpha }_0^2}\right)+1$$

$$\sum _{i\ge 2}{v}_i^2\le 1-\frac{v_1^2}{{\alpha }_0^2}.$$

$$0<v_1\le \frac{1}{2}{\alpha }_0$$

$$(2v_1{)}^2-2t(2u{)}^2=1.$$

$$\Gamma L^{\prime }\le 0$$

Full text

1. Introduction

A (projective, as always in this paper) Kummer surface is obtained as the desingularization of the quotient of an abelian surface by an involution with $16$ isolated fixed points. It is well known that Kummer surfaces are K3 surfaces and that their Picard number is at least $17$, the rank $17$ sub-group being generated by the $16$ rational curves in the resolution of the $16$ nodes and by the polarization. In [12], Nikulin showed the converse, i.e. that a K3 surface containing $16$ disjoint smooth rational curves, or $(-2)$-curves, is the Kummer surface associated to an abelian surface. Let $X$ be a K3 surface; we call a Kummer structure on $X$ an abelian surface $A$ (up to isomorphism) such that $X\simeq Km(A)$, and we call a Nikulin configuration a set of $16$ disjoint smooth rational curves on $X$. By the result of Nikulin we have a bijection:

[TABLE]

In 1977, see [20, Question 5], T. Shioda raised the following question :

Is it possible to have non-isomorphic abelian surfaces $A$ and $B$, such that $Km(A)$ and $Km(B)$ are isomorphic?

Shioda and Mitani in [10, Theorem 5.1] answer negatively the question if $\rho(Km(A))=20$, where $\rho(Km(A))$ is the Picard number of $Km(A)$, i.e. the rank of the Néron-Severi group of $Km(A)$. The answer is also negative if $A$ is a generic principally polarized abelian surface, i.e. $A$ is the jacobian of a curve of genus $2$ and $\rho(A)=1$. Then in [7, Theorem 1.5], Gritsenko and Hulek answered positively the question. They showed that if $A$ is a generic $(1,t)$-polarized abelian surface with $t>1$ then the abelian surface $A$ and its dual $\hat{A}$, though not isomorphic, satisfy $Km(A)\cong Km(\hat{A})$. In [9, Theorem 0.1], Hosono, Lian, Oguiso and Yau, by using lattice theory, showed that the number of Kummer structures is finite and for each integer $N\in\mathbb{N}^{*}$, they construct a Kummer surface of Picard number $18$ with at least $N$ Kummer structures.

In [13, Example 4.16], Orlov showed that if $A$ is a generic abelian surface (i.e. $\rho(Km(A))=17$) then the number of abelian surfaces (up to isomorphism) with equivalent bounded derived categories is $2^{\nu}$, where $\nu$ is the number of prime divisors of $\frac{1}{2}M^{2}$, for $M$ an ample generator of the Néron-Severi group of $A$. By [9, Theorem 0.1], there is a one-to-one correspondence between these equivalent bounded derived categories of $A$ and the Kummer structures on the Kummer surface ${\rm Km(}A)$ associated to $A$. Thus, for example if $A$ is principally polarized we have that $M^{2}=2$ so that $\nu=0$ and we find again the fact that in this case there is only one Kummer structure on ${\rm Km(}A)$. Observe that $\nu$ can be also defined as the number of prime divisors of $\frac{1}{4}L^{2}$, where $L$ is the polarization induced by $M$ on $Km(A)$, (in particular $L$ is orthogonal to the $16$ rational curves ; it is easy to see that by changing the $16$ rational disjoint curves, the number $\nu$ does not change).

In [15, Theorem 1], we constructed explicit examples of two Nikulin configurations $\mathcal{C}$, $\mathcal{C}^{\prime}$ on some K3 surface $X$ such that the abelian surfaces $A$ and $A^{\prime}$ associated to these two configurations are not isomorphic. This was the first geometric construction of two distinct Kummer structures. These examples are for generic Kummer surfaces, such that the orthogonal complement of the $16$ rational curves in $\mathcal{C}$ is generated by a class $L$ such that $L^{2}=2k(k+1)$ for some integer $k$ (we give a motivation for this restriction in the Appendix of this paper).

The main goal of this paper is to provide a generalization of that result to other Kummer surfaces. For that aim, let $t\in\mathbb{N}^{*}$ be an integer and let $X$ be a general Kummer surface with a Nikulin configuration $\mathcal{C}$ such that the orthogonal complement of the $16$ $(-2)$-curves $A_{1},\dots,A_{16}$ in $\mathcal{C}$ is generated by $L$ with $L^{2}=4t$. A class $C$ of the form $C=\beta L-\alpha A_{1}$ with $\beta\in\mathbb{N}^{*}$ has self-intersection $C^{2}$ equals to $-2$ if and only if the coefficients $(\alpha,\beta)$ satisfy the Pell-Fermat equation $\alpha^{2}-2t\beta^{2}=1.$ There is a non-trivial solution if and only if $2t$ is not a square. Let us suppose that this is the case. Then there exists a so-called fundamental solution which we denote by $(\alpha_{0},\beta_{0})$. Our main result is as follows:

Theorem 1**.**

*Suppose that $\beta_{0}$ is even. Then $\beta_{0}L-\alpha_{0}A_{1}$ is the class of an irreducible $(-2)$-curve $A_{1}^{\prime}$, which curve is disjoint from $A_{2},\dots,A_{16}$.

The Nikulin configurations $\mathcal{C}=\sum_{i=1}^{16}A_{i}$ and $\mathcal{C}^{\prime}=A_{1}^{\prime}+\sum_{i=2}^{16}A_{i}$ define the same Kummer structure on the Kummer surface $X$ if and only if the negative Pell-Fermat equation $\alpha^{2}-2t\beta^{2}=-1$ has a solution.

Suppose that this is the case. Then there exists a double cover map $X\to\mathbb{P}^{2}$ branched over $6$ lines $L_{1},\dots,L_{6}$, contracting the $15$ $(-2)$-curves $A_{j},j\geq 2$ to the singularities of $\sum_{i=1}^{6}L_{i}$, and such that the induced involution exchanges the curves $A_{1},A_{1}^{\prime}$ and therefore the configurations $\mathcal{C},\mathcal{C}^{\prime}$.*

The integers $t\in\mathbb{N}$ such that $\beta_{0}$ is even have density at least $\frac{3}{4}$ ; among these integers, at least $\frac{2}{3}$ are such that the negative Pell-Fermat equation have no solution, and therefore give examples of two distinct Kummer structures (see Remark 6 for a precise meaning of that affirmation, and also the table in the Appendix). As a by-product of our study, let us mention the following result (see Proposition 17), which we believe can be of independent interest: *Suppose that the equation $2\mu^{2}-t\nu^{2}=-1$ has a solution. Then there exists a model of the K3 surface $X$ as a quartic surface in $\mathbb{P}^{3}$ with $15$ nodes. *

One could also raise a weaker question than Shioda’s question by asking if $Km(A)\cong Km(B)$ implies $A$ and $B$ must be isogenous ? The answer is positive and the result was surely known, but we could not find an explicit proof in the literature, hence we recall it in Section 2 and we show how it can be obtained as a direct consequence of a result of Stellari [21, Theorem 1.2]. In the rest of the paper, we point out why the construction in Theorem 1 can not work for $\beta_{0}$ odd, moreover we study examples of Nikulin configurations in the case that $\beta_{0}$ is odd or $2t$ is a square.

Acknowledgements: We thank P. Stellari for pointing out his paper [21]. We also thank K. Hulek, H. Lange, K. Oguiso, M. Ramponi, J. Rivat and T. Shioda for useful discussions. We are very grateful to the referee for the many questions, remarks and comments that improved substantially this article.

2. Construction of Nikulin configurations

2.1. Preliminaries: the Pell-Fermat

equation and its negative

The aim of this first sub-section is to recall results on Pell-Fermat equations. We give various criteria when the fundamental solution $(\alpha_{0},\beta_{0})$ of it is such that $\beta_{0}$ is even, and when the negative Pell-Fermat solution has no solution.

2.1.1. The Pell-Fermat equation

For $t\in\mathbb{N}^{*}$, the Pell-Fermat equation

[TABLE]

has a non-trivial solution $(\alpha,\beta)\in\mathbb{Z}^{2}$ if and only if $2t$ is not a square. Then there exists a fundamental solution $(\alpha_{0},\beta_{0})\in\mathbb{N}$, such that for every other solution $(\alpha,\beta)$, there exists $k\in\mathbb{Z}$ with $\alpha+\beta\sqrt{2t}=\pm(\alpha_{0}+\sqrt{2t}\beta_{0})^{k}$.

Remark 2*.*

For $k\in\mathbb{Z}$, let $(x_{k},y_{k})\in\mathbb{Z}^{2}$ be such that

[TABLE]

Using that $\alpha_{0}\geq 1,\beta_{0}\geq 1$ and an induction, one can check that the sequences $(x_{k})_{k\in\mathbb{N}},\,(y_{k})_{k\in\mathbb{N}}$ are strictly increasing with $k\in\mathbb{N}$. Therefore, if $(\alpha,\beta)\in\mathbb{N}^{2}$ is a solution different from $(1,0)$ and with $\alpha\leq\alpha_{0}$ or $\beta\leq\beta_{0}$, then $(\alpha,\beta)$ is the fundamental solution.

We observe moreover that for a solution $(\alpha,\beta)$ of equation (2.1), the integer $\alpha$ is necessarily odd.

For $t$ a positive integer such that $2t$ is not a square, we denote by $(\alpha_{0},\beta_{0})$ the fundamental solution of $\alpha^{2}-2t\beta^{2}=1$. Part a) of the following Lemma shows that the density of integers $t$ such that $\beta_{0}$ is even is at least $\frac{3}{4}$ (we thank Joël Rivat for useful discussions on that question, and also Lemma 5 below):

Lemma 3**.**

*a) Suppose that $t\neq 0\text{ mod }4$. Then $\beta_{0}$ is even.

b) There is an infinite number of integers $s$ such that the fundamental solution $(\alpha_{0},\beta_{0})$ of $\alpha^{2}-8s^{2}\beta^{2}=1$ has odd $\beta_{0}$.

c) There is an infinite number of integers $s$ such that the fundamental solution $(\alpha_{0},\beta_{0})$ of $\alpha^{2}-8s^{2}\beta^{2}=1$ has even $\beta_{0}$.*

Proof.

Let $(\alpha,\beta)$ be a solution of equation $\alpha^{2}-2t\beta^{2}=1$. Suppose that $\beta$ is odd. Then

[TABLE]

and one has $\beta^{2}=1\,\text{mod}\,8$. Since $\alpha^{2}-2t\beta^{2}=1$, one has $\alpha^{2}=1+2t$ mod $8$. Since $\alpha$ is also odd, $\alpha^{2}=1\,\text{mod }8$, thus $2t=0\text{ mod }8$ and therefore $t=0\text{ mod 4}$. That proves part a).

Let $(x_{1},y_{1})$ be the fundamental solution of $x^{2}-2ty^{2}=1$. For $n\in\mathbb{Z}$, the integers $\pm x_{n},\pm y_{n}$ defined by

[TABLE]

are the solutions of equation $x^{2}-2ty^{2}=1$. The sequence $(y_{n})_{n\geq 1}$ is strictly increasing and we see that the fundamental solution of

[TABLE]

is $(x_{n},1)$. Using part a), we remark that always $ty_{n}^{2}=0\text{ mod 4}$. Take now $t=4$, we therefore obtain result b). For $n$ even, $y_{n}$ is even; let $z_{n}$ be such that $y_{n}=2z_{n}$. The fundamental solution of

[TABLE]

is $(x_{n},2)$ ; taking $t=4$ as in the previous case, one obtains result c). ∎

Example 4**.**

For $1\leq s\leq 100$ such that $8s$ is not a square, i.e. for $s\not\in\{2,8,18,32,50,72,98\}$, the fundamental solution $(\alpha_{0},\beta_{0})$ of equation $\alpha^{2}-8s\beta^{2}=1$ is such that $\beta_{0}$ is even if and only if $s$ is in

[TABLE]

2.1.2. The negative Pell-Fermat equation

The equation

[TABLE]

is called the negative Pell-Fermat equation. If $(x,y)$ is a solution, then $(\alpha,\beta)=(x^{2}-2ty^{2},2xy)$ is a solution of the Pell-Fermat equation (2.1), with $\beta$ even. The negative Pell-Fermat equation can be solved by the method of continued fractions and it has solutions if and only if the period of the continued fraction has odd length. A necessary (but not sufficient) condition for solvability is that $t$ is not divisible by a prime of form $4k+3$. The following Lemma implies that the density of integers $t$ such that the negative Pell-Fermat equation (2.2) has no solution is at least $\frac{5}{6}$:

Lemma 5**.**

Suppose that the negative Pell-Fermat equation (2.2) has a solution. Then $t=1\,mod\,4$ and $t\neq 0\,mod\,3$, in other words: $t=1$ or $5\,mod\,12$.

Proof.

Suppose that $(\alpha,\beta)$ is a solution of equation (2.2). Since $\alpha^{2}-2t\beta^{2}=-1$, the integer $\alpha$ is odd, thus $\alpha^{2}=1\,\text{mod}\,8$, and $2t\beta^{2}=\alpha^{2}+1=2\,\text{mod}\,8$, which implies that $\beta$ is odd (otherwise $2t\beta^{2}=0\,\text{mod}\,8$), thus $\beta^{2}=1\,\text{mod}\,8$. In that way $2t=2\,\text{mod}\,8$ hence $t=1\,\text{mod}\,4$.

Since $\alpha^{2}=0$ or $1\,\text{mod}\,3$, one has $2tb^{2}=1$ or $2\,\text{mod}\,3$, thus $t\neq 0\,\text{mod}\,3$. ∎

The first few numbers $t$ for which equation (2.2) is solvable are

[TABLE]

Remark 6*.*

From Lemmas 3 and 5, we conclude that the density of integers $t$ such that the negative Pell-Fermat equation (2.2) has no solution and the Pell-Fermat equation (2.1) has a solution $(\alpha_{0},\beta_{0})$ with $\beta_{0}$ even is at least $\frac{7}{12}$.

2.2. The general problem

2.2.1. Isogenies

Before to state our results about the question of Shioda [20, Question 5] recalled in the Introduction, we can generalize the problem to the following question:

Given two abelian surfaces $A$ and $B$ such that $Km(A)\cong Km(B)$ are then $A$ and $B$ isogenous ?

The answer is positive and certainly well known, in particular to people working on derived categories on abelian surfaces. For convenience we give here a short proof:

Proposition 7**.**

Let $A$ and $B$ be abelian surfaces such that the associated Kummer surfaces are isomorphic, then $A$ and $B$ are isogenous abelian surfaces.

Proof.

Since $Km(A)\cong Km(B)$ then the derived categories $D^{b}(Km(A))$ and $D^{b}(Km(B))$ are equivalent. Thus by [21, Theorem 1.2], the abelian surfaces are isogenous. ∎

2.2.2. Notations and known results on the Néron-Severi group of a Kummer

surface.

Let $t\in\mathbb{N}$ be an integer and let $B$ be a generic Abelian surface with (primitive) polarization $M$ such that $M^{2}=2t$. Let $X={\rm Km(}B)$ be the associated Kummer surface. Let $A_{1},\dots,A_{16}$ be the $16$ disjoint $(-2)$-curves on $X$ that are resolution of the singularities of the quotient $B/[-1]$. By [11, Proposition 3.2], [6, Proposition 2.6], corresponding to the polarization $M$ on $B$, there is a primitive big and nef divisor $L$ on ${\rm Km(}B)$ such that

[TABLE]

and $LA_{i}=0,\,i\in\{1,\dots,16\}$. The Néron-Severi group of $X={\rm Km(}B)$ satisfies:

[TABLE]

where $K$ denotes the Kummer lattice (the saturated lattice containing the $16$ disjoint $(-2)$-curves $A_{i},\,i=1,\dots,16$) which is a negative definite lattice of rank $16$ and discriminant $2^{6}$. For $B$ generic among polarized Abelian surfaces ${\rm rk(}{\rm NS(}X))=17$ and ${\rm NS(}X)$ is an over-lattice of index two of $\mathbb{Z}L\oplus K$ which is described precisely in [6, Theorem 2.7], in particular we will repeatedly use the following result:

Lemma 8**.**

([6, Remarks 2.3 & 2.10]) An element $\Gamma\in{\rm NS(}X)$ has the form $\Gamma=\alpha L-\sum\beta_{i}A_{i}$ with $\alpha,\beta_{i}\in\frac{1}{2}\mathbb{Z}$. If $\alpha$ or $\beta_{i}$ for some $i$ is in $\frac{1}{2}\mathbb{Z}\setminus\mathbb{Z},$ then at least $4$ of the $\beta_{j}$’s are in $\frac{1}{2}\mathbb{Z}\setminus\mathbb{Z}$. If $\alpha\in\mathbb{Z}$, then at least $8$ of the $\beta_{j}$’s are in $\frac{1}{2}\mathbb{Z}\setminus\mathbb{Z}$ or $\forall j,\,\beta_{j}\in\mathbb{Z}$.

2.2.3. The Pell-Fermat equation and construction of $(-2)$-classes

We are looking for a polarization $L^{\prime}$ and a class $A_{1}^{\prime}$ of the form

[TABLE]

with $\alpha,\beta,a,b\in\mathbb{N}\setminus\{0\}$ such that one has $A_{1}^{\prime 2}=-2$, $L^{\prime}A_{1}^{\prime}=0$ and $L^{\prime 2}=L^{2}=4t$. These three conditions are respectively

[TABLE]

the first expresses that $A_{1}^{\prime}$ is a $(-2)$-class, the second that this $(-2)$-class is disjoint from the polarisation $L^{\prime}$, the third that $L^{2}=L^{\prime 2}$. We will use the divisor $L^{\prime}$ and the property that $L^{\prime}A_{1}^{\prime}=0$ in order to show that $A_{1}^{\prime}$ can be represented by an irreducible curve.

Lemma 9**.**

There are non-trivial solutions to the three equations (2.3) if and only if $2t$ is not a square. In that case, if $(\alpha,\beta)$ is a solution of the first equation in (2.3), one has

[TABLE]

Proof.

In order that the Pell-Fermat equation (2.1) admits a solution, we need that $2t$ is not a square. Let us suppose that this is the case and let $(\alpha,\beta)$ be such a solution, which we can suppose with $\alpha>0,\,\beta>0$. By replacing $a=2t\frac{\beta}{\alpha}b$ in the third equation, one gets

[TABLE]

which is equivalent to

[TABLE]

since $\alpha^{2}-2t\beta^{2}=1$ and we search solutions with $b>0$, we obtain $b=\alpha$. Then by the third equality, we get $a^{2}=2t(\alpha^{2}-1)$ and equality $\alpha^{2}-1=2t\beta^{2}$ implies $a=2t\beta$. ∎

2.3. The $\beta_{0}$ odd case

Suppose $2t$ is not a square and let $(\alpha_{0},\beta_{0})$ be a solution of equation (2.1). Let us suppose that $\beta_{0}$ is odd and let us define

[TABLE]

which is a $(-2)$-class. Then

Proposition 10**.**

The $(-2)$-class $A_{1}^{\prime}=\beta_{0}L-\alpha_{0}A_{1}$ cannot be the class of an irreducible rational curve.

Proof.

Suppose that $A_{1}^{\prime}$ is irreducible. Then we have two Nikulin configurations

[TABLE]

Since Nikulin configurations are $2$-divisible (see [12]), the divisor $A_{1}+A_{1}^{\prime}$ is $2$-divisible and

[TABLE]

is an integral class. Since $\beta_{0}$ is odd by assumption and $\alpha_{0}$ must be odd by the equality $\alpha_{0}^{2}-2t\beta_{0}^{2}=1$, it follows that $\frac{L}{2}\in{\rm NS(}X)$, which contradicts $L$ being a primitive class. ∎

We will come back to this case (when $\beta_{0}$ is odd) in Subsection 3.1.

2.4. The $\beta_{0}$ even case: $\beta_{0}L-\alpha_{0}A_{1}$

is the class of a $(-2)$-curve

Assume $2t$ is not a square. Let $(\alpha_{0},\beta_{0})$ be the fundamental solution of the Pell-Fermat equation (2.1). We assume in this section that $\beta_{0}$ is even and we define as in Subsection 2.2.3 the classes:

[TABLE]

One has $A_{1}^{\prime 2}=-2$, $L^{\prime}A_{1}^{\prime}=0,$ $L^{\prime 2}=L^{2}=4t$.

Proposition 11**.**

Suppose that $\beta_{0}$ is even. The class $L^{\prime}$ is big and nef and the classes $A_{1}^{\prime},A_{2}\dots,A_{16}$ are the only $(-2)$-classes contracted by $L^{\prime}$.

Proof.

The class $L^{\prime}$ is nef if and only if for any $(-2)$-curves $\Gamma$, one has $\Gamma L^{\prime}\geq 0$. Let

[TABLE]

be a $(-2)$-curve (thus $\sum v_{i}^{2}-2tu^{2}=1$) ; we recall that by Lemma 8, if one coefficient $u$ or $v_{i}$ is in $\frac{1}{2}\mathbb{Z}\setminus\mathbb{Z}$, then at least four of the $v_{i}$’s are in $\frac{1}{2}\mathbb{Z}\setminus\mathbb{Z}$. Suppose that

[TABLE]

this is equivalent to

[TABLE]

in other words $u\leq\frac{\beta_{0}}{\alpha_{0}}v_{1}$, thus

[TABLE]

and therefore using the relation $\alpha_{0}^{2}-2t\beta_{0}^{2}=1$, one obtains

[TABLE]

and therefore

[TABLE]

Apart from the trivial cases of curves $\Gamma=A_{i}$ for $i>1$, one can suppose $u>0$. If $v_{1}>\frac{1}{2}\alpha_{0}$ then $\sum_{i\geq 2}v_{i}^{2}<\frac{3}{4}$, then by Lemma 8, one gets $v_{2}=\dots=v_{16}=0$, $v_{1}=\alpha_{0}$, $u=\beta_{0}$, so that $\Gamma=A_{1}^{\prime}$, for which $\Gamma L^{\prime}=0$. Thus one can suppose that

[TABLE]

(if $v_{1}=0$, then $u=0$, which we already excluded) and, up to permutation of the indices: $v_{2}=v_{3}=v_{4}=\frac{1}{2}$ (since $\sum_{i\geq 2}v_{i}^{2}<1$ and by the structure of the Néron-Severi group as described in Lemma 8). The relation $\sum v_{i}^{2}-2tu^{2}=1$ is now $v_{1}^{2}-2tu^{2}=\frac{1}{4}$, which is

[TABLE]

Defining $V=2v_{1}\in\mathbb{N}$ and $U=2u\in\mathbb{N}$, we see that $(U,V)$ is a solution of the Pell-Fermat equation $\beta^{2}-2t\alpha^{2}=1$. Moreover, by Equation (2.4) we know that $0<V\leq\alpha_{0}$. Since by hypothesis $(\alpha_{0},\beta_{0})$ is the primitive solution, and $V>0$, we have $\alpha_{0}\leq V$. Therefore, by remark 2: $V=\alpha_{0}$, which implies that $U=\beta_{0}$, thus $v_{1}=\frac{1}{2}\alpha_{0}$, $u=\frac{1}{2}\beta_{0}$, and thus (for $\Gamma\neq A_{2},\dots,A_{16}$) we have

[TABLE]

if and only if $\Gamma L^{\prime}=0$ and $\Gamma$ has the form $\Gamma=\frac{1}{2}(\beta_{0}L-\alpha_{0}A_{1}-A_{2}-A_{3}-A_{4})$. But by Lemma 8, in order for $\Gamma$ to be in ${\rm NS(}X)$, the integer $\beta_{0}$ must be odd, which is impossible by our assumption on $\beta_{0}$.

In conclusion, we obtain that $L^{\prime}$ is big and nef, and if $\beta_{0}$ is even, then the only $(-2)$-classes $\Gamma$ such that $\Gamma L^{\prime}=0$ are $A_{1}^{\prime},A_{2},\dots,A_{16}$. ∎

Let us prove the following result:

Proposition 12**.**

Suppose that $\beta_{0}$ is even. The line bundle $3L^{\prime}$ (where $L^{\prime}=\alpha_{0}L-2t\beta_{0}A_{1}$) defines a morphism $\phi_{3L^{\prime}}:X\to\mathbb{P}^{N}$ which is birational onto its image and contracts exactly the divisor $A_{1}^{\prime}=\beta_{0}L-\alpha_{0}A_{1}$ and the $15$ $(-2)$-curves $A_{i},\,i\geq 2$.

Proof.

By [14, Section 3.8] either $|3L^{\prime}|$ has no fixed part or $3L^{\prime}=aE+\Gamma$, where $|E|$ is a free pencil, and $\Gamma$ is a $(-2)$-curve with $E\Gamma=1$. However if $E\Gamma=1$, then $3L^{\prime}E=aE^{2}+1$, but since $E^{2}=0$ this is impossible. Thus $|3L^{\prime}|$ has no fixed part; moreover by [18, Corollary 3.2], it has then no base points.

Let us prove that the morphism $\phi_{3L^{\prime}}$ has degree one, i.e. that $|3L^{\prime}|$ is not hyperelliptic (see [18, Section 4]). By loc. cit., $|3L^{\prime}|$ is hyperelliptic only if there exists a genus $2$ curve $C$ such that $3L^{\prime}=2C$ or there exists an elliptic curve $E$ such that $(3L^{\prime})E=2$. Suppose we are in the first case. Since $C^{2}=2$, one has $9\cdot 4t=8$, which is impossible. The second alternative is also readily impossible. Thus the morphism $\phi_{3L^{\prime}}$ has degree one. Moreover since $3L^{\prime}A_{1}^{\prime}=3L^{\prime}A_{2}=\dots=3L^{\prime}A_{16}=0$, the $16$ divisors are contracted by $\phi_{3L^{\prime}}$. ∎

We obtain:

Corollary 13**.**

Suppose that $\beta_{0}$ is even. The divisor $A_{1}^{\prime}$ is an irreducible $(-2)$-curve.

Proof.

Since $A_{1}^{\prime 2}=-2$ and $LA_{1}^{\prime}\geq 0$, by Riemann-Roch Theorem we can assume it is effective. Let $B$ be one of the divisors $A_{1}^{\prime},A_{2},\dots,A_{16}$. One has $3L^{\prime}B=0$, thus the linear system $|3L^{\prime}|$ contracts $B$ to a singular point. Since the Picard number of the K3 surface $X={\rm Km(}B)$ is $17$, that singularity must be a node and therefore $A_{1}^{\prime}$ is irreducible. ∎

2.5. Two Kummer structures in case

$\beta_{0}$ even and the negative Pell-Fermat equation is not solvable

Suppose that $2t$ is not a square and let $(\alpha_{0},\beta_{0})$ be the fundamental solution of the Pell-Fermat equation

[TABLE]

We suppose that $\beta_{0}$ is even, and we recall that $A_{1}^{\prime}=\beta_{0}L-\alpha_{0}A_{1}$ in ${\rm NS(}X)$. The aim of this Section is to prove the following:

Theorem 14**.**

Suppose that $t\geq 2$ and the negative Pell-Fermat equation

[TABLE]

has no solution. There is no automorphism $f$ of $X$ sending the configuration $\mathcal{C}=\sum_{i=1}^{16}A_{i}$ to the configuration $\mathcal{C}^{\prime}=A_{1}^{\prime}+\sum_{i=2}^{16}A_{i}$.

Remark*.*

We recall that the Nikulin configurations $\mathcal{C}$ and $\mathcal{C}^{\prime}$ define two distinct Kummer structures if and only if there is no automorphism sending $\mathcal{C}$ to $\mathcal{C}^{\prime}$ (see [9]; a proof is given in [15, Proposition 21]).

In order to prove Theorem 14, let us suppose that such an automorphism $f$ exists. The group of translations by the $2$-torsion points on $B$ acts on $X={\rm Km(}B)$ and that action is transitive on the set of curves $A_{1},\dots,A_{16}$. Thus, up to changing $f$ by $f\circ t$ (where $t$ is such a translation), one can suppose that the image of $A_{1}$ is $A_{1}^{\prime}$. Then the automorphism $f$ induces a permutation of the curves $A_{2},\dots,A_{16}$. The $(-2)$-curve $A_{1}^{\prime\prime}=f^{2}(A_{1})=f(A_{1}^{\prime})$ is orthogonal to the $15$ curves $A_{i},\,i>1$ and therefore its class is in the group generated by $L$ and $A_{1}$. Let $\lambda,\mu\in\mathbb{Z}$ such that $A_{1}^{\prime\prime}=\lambda A_{1}+\mu L$, so that $(\lambda,\mu)$ is a solution of the Pell-Fermat equation (2.5). Let us prove:

Lemma 15**.**

Let $C=\lambda A_{1}+\mu L$ be an effective $(-2)$-class. Then there exists $u,v\in\mathbb{N}$ such that $C=uA_{1}+vA_{1}^{\prime}$, in particular the only $(-2)$-curves in the lattice generated by $L$ and $A_{1}$ are $A_{1}$ and $A_{1}^{\prime}$.

Proof.

If $(\lambda,\mu)$ is a solution of equation (2.5), then so are $(\pm\lambda,\pm\mu)$. We say that a solution is positive if $\lambda\geq 0$ and $\mu\geq 0$. Let us identify $\mathbb{Z}^{2}$ with $\mathbb{Z}[\sqrt{2t}]$ by sending $(\lambda,\mu)$ to $\lambda+\mu\sqrt{2t}$. The solutions of equation 2.5 are units of the ring $\mathbb{Z}[\sqrt{2t}]$. Let $\alpha_{0}+\beta_{0}\sqrt{2t}$ ($\alpha_{0},\beta_{0}\in\mathbb{N}^{*}$) be the fundamental solution to equation (2.5). The solutions with positive coefficients are the elements of the form

[TABLE]

An effective $(-2)$-class $C=\lambda A_{1}+\mu L$ either equals $A_{1}$ or satisfies $CL>0$ and $CA_{1}>0$, therefore $\mu>0$ and $\lambda<0$. Thus if $C\neq A_{1}$, there exists $m$ such that $C=-\lambda_{m}A_{1}+\mu_{m}L$. Since $A_{1}^{\prime}=\beta_{0}L-\alpha_{0}A_{1}$ corresponds to the fundamental solution of equation (2.5), we have $L=\frac{1}{\beta_{0}}(A_{1}^{\prime}+\alpha_{0}A_{1})$ and we obtain

[TABLE]

and the Lemma is proved if the coefficients $u_{m}=\frac{\mu_{m}}{\beta_{0}}$ and $v_{m}=\frac{\alpha_{0}}{\beta_{0}}\mu_{m}-\lambda_{m}$ are both positive and in $\mathbb{Z}$. Using the fact that

[TABLE]

we obtain

[TABLE]

Then we compute that

[TABLE]

and by induction we conclude that $u_{m},v_{m}$ are in $\mathbb{N}$ for any $m\geq 1$. ∎

Lemma 15 implies that $A_{1}^{\prime\prime}=A_{1}$ i.e. $f$ permutes $A_{1}$ and $A_{1}^{\prime}$. Let us continue the proof of Theorem 14:

Since the automorphism $f$ preserves the set

[TABLE]

it acts with finite order $n_{0}$ on $B$. Since $B$ is a $\mathbb{Q}$-basis of ${\rm NS(}X)\otimes\mathbb{Q}$, the automorphism $f^{n_{0}}$ acts trivially on ${\rm NS(}X),$ thus it preserves an ample class, and by [8, Proposition 5.3.3], the automorphism $f^{n_{0}}$ has finite order, which proves that $f$ has finite order. Up to taking an odd power of $f$, one can suppose that $f$ has order $2^{m}$ for some $m\in\mathbb{N}^{*}$. Suppose $m=1$, i.e. $f$ is an involution. Then the integral class

[TABLE]

(recall that $\beta_{0}$ is even and $\alpha_{0}$ is odd) is fixed by $f$. Let us define

[TABLE]

then the class $D^{\prime}=\frac{1}{d_{0}}D$ is primitive in ${\rm NS(}X)$. We have

[TABLE]

and in fact, since ${\rm NS(}X)$ is an even lattice, $D^{\prime 2}$ is even, so that $\frac{\alpha_{0}-1}{2d_{0}^{2}}\in\mathbb{Z}$. Let us define

[TABLE]

Let ${\rm NS(}X)^{f}$ be the sub-lattice of ${\rm NS(}X)$ fixed by $f$. By Lemma 8, for any class $E$ in ${\rm NS(}X)^{f}$, there exists $a,b_{2},\dots,b_{16}\in\mathbb{Z}$ such that

[TABLE]

Since $WD^{\prime}=2$, we get $WE=a\in\mathbb{Z}$, therefore $W$ is an element of the dual of ${\rm NS(}X)^{f}$, and the discriminant group of ${\rm NS(}X)^{f}$ contains the sub-group isomorphic to $\mathbb{Z}/\frac{\alpha_{0}-1}{2d_{0}^{2}}\mathbb{Z}$ generated by the class of $W$.

Case when $f$ is a non-symplectic involution

Suppose that $f$ is non-symplectic. Then (see e.g. [1]) ${\rm NS(}X)^{f}$ is a $2$-elementary lattice, which means that the discriminant group of ${\rm NS(}X)^{f}$ is isomorphic to $(\mathbb{Z}/2\mathbb{Z})^{h}$ for some positive integer $h$. Since $\mathbb{Z}/\frac{\alpha_{0}-1}{2d_{0}^{2}}\mathbb{Z}$ is a sub-group of the discriminant group, and $f$ is supposed to be non-symplectic, we get two cases:

[TABLE]

Sub-case $\frac{\alpha_{0}-1}{2d_{0}^{2}}=1$. If $\frac{\alpha_{0}-1}{2d_{0}^{2}}=1$, then $\alpha_{0}=1+2d_{0}^{2}$. Since $d_{0}|\beta_{0}/2$, there exists $e_{0}\in\mathbb{Z}$ such that $\beta_{0}=2e_{0}d_{0}$. From the relation $\alpha_{0}^{2}-2t\beta_{0}^{2}=1$, it follows that $d_{0}$ and $e_{0}$ satisfy the negative Pell-Fermat equation

[TABLE]

Conversely, suppose that $(d_{0},e_{0})$ is the primitive solution to the equation $d^{2}-2te^{2}=-1$. The fundamental solution to the Pell-Fermat equation is

[TABLE]

and $\frac{\alpha_{0}-1}{2d_{0}^{2}}=1$. Since, in the hypothesis of Theorem 14, we supposed that the negative Pell-Fermat equation has no solution, the case $\frac{\alpha_{0}-1}{2d_{0}^{2}}=1$ is excluded.

Remark 16*.*

In [15], we studied the cases with $2t=k(k+1)$. Then $D=2D^{\prime}=2L-2kA_{1}$, which gives $d_{0}=1$ and $\alpha_{0}=2k+1$. The proof of [15, Theorem 19] implies that the negative Pell-Fermat equation $x^{2}-k(k+1)y^{2}=-1$ has no solution for $k\geq 2$.

Sub-case $\frac{\alpha_{0}-1}{2d_{0}^{2}}=2$. If $\frac{\alpha_{0}-1}{2d_{0}^{2}}=2$, then $\alpha_{0}=1+4d_{0}^{2}$. Since $d_{0}|\beta_{0}/2$, there exists $e_{0}\in\mathbb{Z}$ such that $\beta_{0}=2e_{0}d_{0}$. From the relation $\alpha_{0}^{2}-2t\beta_{0}^{2}=1$, it follows that $d_{0}$ and $e_{0}$ satisfy the relation

[TABLE]

(conversely, if $(d_{0},e_{0})$ is a solution of (2.7), then $\alpha_{0}=1+4d_{0}^{2}$ and $\beta_{0}=2e_{0}d_{0}$ is a solution of the Pell-Fermat equation (2.5)). We have

[TABLE]

with $D^{\prime 2}=4,D^{\prime}A_{1}=D^{\prime}A_{1}^{\prime}=4d_{0},D^{\prime}A_{j}=0$ for $j\in\{2,\dots,16\}$. Let us prove that

Proposition 17**.**

The divisor $D^{\prime}$ is nef, the linear system $|D^{\prime}|$ is base point free, non hyperelliptic and defines a morphism $\varphi:X\to\mathbb{P}^{3}$ such that $\varphi(X)=Y$ is a quartic surface with $15$ nodes, which are images of the disjoint curves $A_{j},j\geq 2$.

Proof.

Let us prove that $D^{\prime}$ is nef. Let $\Gamma=\alpha L-\sum_{i=1}^{16}\beta_{i}A_{i}$ be a $(-2)$-curve:

[TABLE]

where $\alpha,\beta_{i}\in\frac{1}{2}\mathbb{Z}$ are subject to the restrictions in Lemma 8. Suppose that $D^{\prime}\Gamma\leq 0$, which is equivalent to

[TABLE]

Using this relation in equation (2.8), we get

[TABLE]

By using the relation (2.7), this is equivalent to

[TABLE]

Suppose that $\alpha$ is an integer. Then from Lemma 8 and equation (2.9), we have either $\forall j\geq 2,\,\beta_{j}=0$ or $\exists k\geq 2,\,\beta_{k}=1$ and $\forall j\geq 2,\,j\neq k,\,\beta_{j}=0$. In the first case $\Gamma=\alpha L-\beta_{1}A_{1}$ with $\beta_{1}\in\mathbb{Z}$, and from Lemma 15, either $\Gamma=A_{1}$ or $\Gamma=A_{1}^{\prime}$. Since $D^{\prime}A_{1}=D^{\prime}A_{1}^{\prime}=4d_{0}>0$, this is impossible. In the second case, $\sum_{i=2}^{16}\beta_{i}^{2}=1$ implies $\alpha=0$ and $\Gamma=A_{k}$ for $k\geq 2$, and indeed $D^{\prime}A_{k}=0$. It remains the case when $\alpha$ is an half-integer. Then three of the $\beta_{i}$ with $i>1$ are equal to $\frac{1}{2}$, the others are [math], and $\beta_{1}$ is in $\frac{1}{2}\mathbb{Z}\setminus\mathbb{Z}$. Let $a,b\in\mathbb{Z}$ be the odd integers such that $\alpha=\frac{a}{2},\,\beta_{1}=\frac{b}{2}.$ Equation (2.8) becomes $b^{2}-2ta^{2}=1.$ By hypothesis, the fundamental solution $(\alpha_{0},\beta_{0})$ of that Pell-Fermat equation $\alpha^{2}-2t\beta^{2}=1$ is such that $\beta_{0}$ is even. Then an easy induction shows that every solution $(\alpha,\beta)$ is also such that $\beta$ is even. Hence that case is also impossible, and we obtain that $D^{\prime}$ is nef with $D^{\prime}\Gamma=0$ for a $(-2)$-curve $\Gamma$ if and only if $\Gamma=A_{k}$ for $k\geq 2$.

Let us prove that the linear system $|D^{\prime}|=|e_{0}L-2d_{0}A_{1}|$ is base point free. Suppose that this is not the case. Then (see [18]) there exist an elliptic curve $E$ and a $(-2)$-curve $\Gamma=\alpha L-\sum_{i=1}^{16}\beta_{i}A_{i},\,\,\alpha,\beta_{i}\in\frac{1}{2}\mathbb{Z},$ such that $E\Gamma=1$ and $D^{\prime}=3E+\Gamma$. One has $D^{\prime}\Gamma=(3E+\Gamma)\Gamma=1$ but

[TABLE]

This is a contradiction, and we conclude that $|D^{\prime}|$ is base-point free.

Let us study the degree of the morphism defined by the linear system $|D^{\prime}|$. By [18], the morphism has degree $2$ if and only if there exists en elliptic curve

[TABLE]

such that $D^{\prime}E=2$. Equality $D^{\prime}E=2$ is equivalent to

[TABLE]

Since $E$ is an elliptic curve, one has

[TABLE]

Let us define $S=\sum_{i=2}^{16}b_{i}^{2}\in\mathbb{N}$. The equations (2.10) and (2.11) give

[TABLE]

therefore

[TABLE]

which is equivalent to

[TABLE]

Since $e_{0}^{2}t-2d_{0}^{2}=1$, the reduced discriminant of equation (2.12) in $b_{1}$ is

[TABLE]

Since $b_{1}$ is an integral solution, that implies $S\in\{0,1,2\}$. Lemma 8 implies that cases $S=1$ and $S=2$ are not possible. The case $S=0$ is not possible either since $2(1+2d_{0}^{2})$ is not a square. We thus proved that the morphism defined by $|D^{\prime}|$ has degree $1$ and that concludes the proof of Proposition 17. ∎

Using that model $Y\hookrightarrow\mathbb{P}^{3}$, we can use the same reasoning as in [15, Proof of Theorem 19] (which was made for the case $t=3$, with $d_{0}=e_{0}=1$) in order to prove that sub-case $\frac{\alpha_{0}-1}{2d_{0}^{2}}=2$ is also impossible if we suppose that the non-symplectic automorphism $f$ exists.

Case when $f$ is a symplectic involution.

Suppose $f$ is a symplectic involution. It fixes $A_{1}+A_{1}^{\prime}$, it permutes the classes $A_{j}$ ($j>1$) by pairs, thus the number $s$ of fixed $A_{j}$ is odd. A symplectic automorphism acts trivially on the transcendental lattice $T_{X}$, which in our situation has rank $5$. Therefore the trace of $f$ on $H^{2}(X,\mathbb{Z})$ equals $5+{\rm rk(}{\rm NS(}X)^{f})\geq 6+s>6$. But the trace of a symplectic involution equals $6$ (see e.g. [19, Section 1.2]). This is a contradiction, thus $f$ cannot have order $2$ and the integer $m$ (such that the order of $f$ is $2^{m}$) is larger than $1$.

Remaining cases

We know that $f$ has order $2^{m}>1$. The automorphism $g=f^{2^{m-1}}$ has order $2$ and $g(A_{1})=A_{1},\,g(A_{1}^{\prime})=A_{1}^{\prime}$, thus $g(L)=L$. There are curves $A_{i},\,i>1$ such that $f(A_{i})=A_{i}$ (say $s$ of such curves, $s$ is odd since $A_{1}$ is fixed) and the remaining curves $A_{j}$ are permuted $2$ by $2$ (there are $s^{\prime}=\frac{1}{2}(15-s)$ such pairs). Let $\mathcal{L}^{\prime}$ be the sub-lattice generated by $L,A_{1}$ and the fix classes $A_{i}$, $A_{j}+g(A_{j})$. It is a finite index sub-lattice of the fixed lattice ${\rm NS(}X)^{g}$ and its discriminant group is

[TABLE]

By the same reasoning as before, the involution $g$ must be symplectic as soon as $t>1$. However the trace of $g$ is $8+s>6$, thus $g$ cannot be symplectic either. Therefore we conclude that such an automorphism $f$ does not exist, which conclude the proof of Theorem 14.

Remark 18*.*

In the course of the proof of Theorem 14, using the divisor $D^{\prime}$, we obtained a model of our K3 surface as a quartic in $\mathbb{P}^{3}$ with $15$ nodes, as soon as Equation (2.7) has a solution. This is the case for example for $t=3,9,11,19,27$…

2.6. When the negative Pell-Fermat equation has a solution

Suppose that the negative Pell-Fermat equation $\lambda^{2}-2t\mu^{2}=-1$ has a solution and let $(d_{0},e_{0})$ be the fundamental solution. Then $(\alpha_{0},\beta_{0})=(1+2d_{0}^{2},2e_{0}d_{0})$ is the fundamental solution of the Pell-Fermat equation $\lambda^{2}-2t\mu^{2}=1$, in particular $\beta_{0}$ is even. Moreover we have $A_{1}+A_{1}^{\prime}=2d_{0}D^{\prime}$ (we keep the notation of the previous Section) with

[TABLE]

such that $D^{\prime 2}=2$. Let us prove that

Proposition 19**.**

*The divisor $D^{\prime}$ is nef. We have $D^{\prime}\Gamma=0$ for a $(-2)$-curve $\Gamma$ if and only if $\Gamma=A_{j},$ for $j\in\{2,\dots,16\}$. The linear system $|D^{\prime}|$ is base point free. It defines a double cover $\varphi:X\to\mathbb{P}^{2}$ which contracts exactly the $15$ curves $A_{j},j\geq 2$ to $15$ singular points of a sextic curve which is the union of $6$ lines.

The involution $\sigma$ defined by the double cover $\varphi$ exchanges $A_{1}$ and $A_{1}^{\prime}$ and the two Nikulin configurations $\mathcal{C},\mathcal{C}^{\prime}$, which therefore give the same Kummer structure.*

Proof.

Let us prove that $D^{\prime}$ is nef. Let $\Gamma=\alpha L-\sum_{i=1}^{16}\beta_{i}A_{i}$ be a $(-2)$-curve: $2t\alpha^{2}-\sum_{i=1}^{16}\beta_{i}^{2}=-1$, where $\alpha,\beta_{i}\in\frac{1}{2}\mathbb{Z}$ are subject to the restrictions in Lemma 8. We suppose that $D^{\prime}\Gamma\leq 0$. This is equivalent to:

[TABLE]

Then $\beta_{1}\geq\frac{2t\alpha e_{0}}{d_{0}}$, thus $-\left(\frac{2t\alpha e_{0}}{d_{0}}\right)^{2}\geq-\beta_{1}^{2}$. From the relation $2t\alpha^{2}+1-\beta_{1}^{2}=S$, where $S=\sum_{i=2}^{16}\beta_{i}^{2}$, we get

[TABLE]

By using the relation $d_{0}^{2}-2te_{0}^{2}=-1$, we obtain

[TABLE]

We can then follow the same proof as for the divisor in Proposition 17, and we conclude that $D^{\prime}$ is nef, with $D^{\prime}\Gamma=0$ for a $(-2)$-curve $\Gamma$ if and only if $\Gamma=A_{k}$ for $k\in\{2,\dots,16\}$.

Let us prove that the linear system $|D^{\prime}|=|e_{0}L-d_{0}A_{1}|$ is base point free. Suppose that this is not the case. Then (see [18]) there exist an elliptic curve $E$ and a $(-2)$-curve $\Gamma$ such that $D^{\prime}=2E+\Gamma$ and $E\Gamma=1$. Since we have $D^{\prime}\Gamma=(2E+\Gamma)\Gamma=0$, there exists $k\geq 2$ such that $\Gamma=A_{k}$. Thus $D^{\prime}=2E+A_{k}$, and then $e_{0}L-d_{0}A_{1}-A_{k}=2E$, which is impossible by Lemma 8. Therefore $|D^{\prime}|$ is base point free and defines a double cover of the plane that contracts the $15$ disjoint $(-2)$-curves $A_{j},j\geq 2$ to points. Since $A_{1}+A_{1}^{\prime}=2d_{0}D^{\prime}$, the divisor $A_{1}+A_{1}^{\prime}$ is the pull-back of a plane rational cuspidal curve of degree $2d_{0}$, and the involution exchanges the two curves $A_{1},A_{1}^{\prime}$. It is well-known that a sextic curve with $15$ singular points is the union of $6$ lines in general position. ∎

3. Further examples

3.1. An example of a Nikulin configuration

when $\beta_{0}$ is odd

Let us study the $t=4$ case. Then the fundamental solution $(\alpha_{0},\beta_{0})$ equals $(3,1)$. This is the first case with $\beta_{0}$ odd (see Table in the Appendix). We have

[TABLE]

and we already know that $A_{1}^{\prime}$ is not irreducible. In order to understand better what is happening, let us define

[TABLE]

where the classes $A_{2},A_{3},A_{4}$ are chosen so that the classes $A_{j}^{\prime\prime}$ exists in ${\rm NS(}X)$ (which is possible by [6, Theorem 2.7], since $t=0\text{ mod }2$). We compute that these are $(-2)$-classes i.e. $A_{j}^{\prime\prime 2}=-2$. Moreover we have

[TABLE]

(but $A_{i}^{\prime\prime}L^{\prime}\not=0$, for $i=2,3,4$). Let us also define

[TABLE]

We remark that $L_{1}^{2}=L^{2}=16$, $L_{1}A_{i}^{\prime\prime}=0$ and $A_{i}^{\prime\prime}A_{j}^{\prime\prime}=0$ for $i\neq j$ in $\{1,2,3,4\}$.

Lemma 20**.**

The class $L_{1}$ is big and nef. Moreover if $\Gamma$ is an effective $(-2)$-class, we have $L_{1}\Gamma\geq 0$ and $L_{1}\Gamma=0$ if and only if $\Gamma$ is one of the classes $A_{1}^{\prime\prime},...,A_{4}^{\prime\prime},A_{5},\dots,A_{16}$.

Proof.

Let

[TABLE]

be an effective $(-2)$-class (thus $\sum b_{i}^{2}-8a^{2}=1$). One has

[TABLE]

if and only if

[TABLE]

Suppose $\Gamma\not\in\{A_{5},\dots,A_{16}\}$. Then $a>0,\,b_{i}\geq 0$ and equation $L_{1}\Gamma\leq 0$ is equivalent to

[TABLE]

Using

[TABLE]

we get

[TABLE]

and since $\sum b_{i}^{2}=8a^{2}+1$, we have

[TABLE]

thus $a^{2}\leq 1$ and $a\in\{\frac{1}{2},1\}$. Suppose that $a=\frac{1}{2}$. Then $\sum_{i=1}^{i=16}b_{i}^{2}=3$ and either there are $12$ $b_{i}$’s equal to $\frac{1}{2}$ or (up to permutation of the indices) $b_{1}=\frac{3}{2}$, $b_{2}=b_{3}=b_{4}=\frac{1}{2}$. The first case is impossible since $3=6a>b_{1}+b_{2}+b_{3}+b_{4}$. The second case corresponds to $A_{1}^{\prime\prime},\dots,A_{4}^{\prime\prime}$, and then $L_{1}A_{j}^{\prime\prime}=0$.

It remains to study the case $a=1$, then $\sum b_{i}^{2}=9$ and

[TABLE]

That implies $b_{i}\leq\frac{5}{2}$. Up to permutation we can suppose that the largest $b_{i}$ with $i\in\{1,2,3,4\}$ is $b_{1}$. Suppose $b_{1}=\frac{5}{2}$, then $\sum_{i\geq 2}b_{i}^{2}=\frac{11}{4}$ and $b_{2}+b_{3}+b_{4}\geq\frac{7}{2}$. One can suppose that $b_{2}$ is the largest among $b_{2},b_{3},b_{4}$, then there are two cases : $b_{2}=2$ or $b_{2}=\frac{3}{2}$. The first case is impossible since one would obtain $\sum_{i\geq 2}b_{i}^{2}>\frac{11}{4}$. Suppose $b_{2}=\frac{3}{2}$, then $b_{3}^{2}+b_{4}^{2}=\frac{1}{2}$ and $b_{3}+b_{4}\geq 2$, but this is also impossible.

Suppose that $b_{1}=2$. Then $\sum_{i\geq 2}b_{i}^{2}=5$ and $b_{2}+b_{3}+b_{4}\geq 4$. The largest $b_{i}$ among $b_{2},b_{3},b_{4}$ (say it is $b_{2}$) is $2$ or $\frac{3}{2}$. If $b_{2}=2$, then $b_{3}^{2}+b_{4}^{2}=1$ and $b_{3}+b_{4}\geq 2$, which is impossible. If $b_{2}=\frac{3}{2}$, then $b_{3}+b_{4}\geq\frac{5}{2}$ and $b_{3}^{2}+b_{4}^{2}=\frac{11}{4},$thus $b_{3}=\frac{3}{2}$ and $b_{4}\geq 1$ gets a contradiction.

It remains $b_{1}=\frac{3}{2}$, but then $b_{2}=b_{3}=b_{4}=\frac{3}{2}$. That implies $b_{j}=0$ for $j\geq 5$. But $L-\frac{3}{2}(A_{1}+A_{2}+A_{3}+A_{4})$ is not in the Néron-Severi group of the surface (see Lemma 8).

We thus proved that the only effective $(-2)$-classes $\Gamma$ such that $L_{1}\Gamma\leq 0$ are $A_{1}^{\prime\prime},\dots,A_{4}^{\prime\prime},A_{5},\dots,A_{16}$ and moreover $L_{1}\Gamma=0$ for these classes. Thus $L_{1}$ is nef and big. ∎

As before, one can prove that the linear system $3L_{1}$ define a degree $1$ morphism which contracts $A_{1}^{\prime\prime},\dots,A_{4}^{\prime\prime},A_{5},\dots,A_{16}$ onto singularities. Since we assume that the Kummer surface is generic, it has Picard number $17$ and we conclude that the divisors $A_{1}^{\prime\prime},\dots,A_{4}^{\prime\prime}$ are irreducible. Therefore:

Corollary 21**.**

The $16$ $(-2)$-curves $A_{1}^{\prime\prime},\dots,A_{4}^{\prime\prime},A_{5},\dots,A_{16}$ form a Nikulin configuration $\mathcal{C}^{\prime}$ on the K3 surface $X$. The $(-2)$-class $A_{1}^{\prime}$ is not irreducible and $A_{1}^{\prime}=2A_{1}^{\prime\prime}+A_{2}+A_{3}+A_{4}$.

Remark 22*.*

i) One can check that the class $L^{\prime}$ is big and nef; the image of $X$ by the linear system $|3L^{\prime}|$ is a surface with $12$ nodal singularities and one $D_{4}$ singularity obtained by contracting $A_{1}^{\prime\prime},A_{2},A_{3},A_{4}$.

ii) We do not know yet if $\mathcal{C}^{\prime}$ is another Kummer structure on the Kummer surface $X$, we intend to study that problem in a forthcoming paper.

3.2. An example of a Nikulin configuration

when $2t$ is a square

Let us consider the case $t=2$ i.e. $A$ is a $(1,2)$-polarized abelian surface. Then $2t$ is a square and the method in Section 2.4 does not apply. We start by recalling the following

Remark 23*.*

Since $t$ is even, by [6, Theorem 2.7 and Remark 2.10], up to re-labelling the $16$ curves $(-2)$-curves $A_{j}$, we can suppose that the classes

[TABLE]

are contained in ${\rm NS(}X)$. For $j\in\{1,...,4\}$, we define

[TABLE]

and for $j\in\{5,...,8\}$, we define

[TABLE]

These are $(-2)$-classes; they are effective since $LB_{j}>0$. We check moreover that

[TABLE]

where $\delta_{jk}$ is the Kronecker symbol. Let us prove the following result:

Proposition 24**.**

The classes $B_{1},\dots,B_{8},A_{9},\dots,A_{16}$ are $16$ disjoint $(-2)$-curves.

Proof.

We have $B_{k}A_{j}=0$ for $k\in\{1,\dots,8\}$ and $j\in\{9,\dots,16\}$. It remains to prove that $B_{1},\dots,B_{8}$ are irreducible. We compute that $F_{1}^{2}=0=F_{2}^{2}$, $F_{1}F_{2}=2$. Since $LF_{1}=4,$ the divisor $F_{1}$ is effective. The fact that $F_{1}$ is nef can be found in [6, Proposition 4.6], but for completeness, let us prove it here. Suppose that $F_{1}$ is not nef. Then there exist a $(-2)$-curve $\Gamma$ such that $F_{1}\Gamma=-a<0$. By [8, Remark 8.2.13], the divisor $E=F_{1}-a\Gamma$ is effective, moreover

[TABLE]

Let us write $E=\alpha L-\sum\beta_{i}A_{i}$ with $\alpha\in\frac{1}{2}\mathbb{Z},\,\text{and }\alpha>0$ since $E$ is effective and the lattice $L^{\perp}$ is negative definite. Since $L$ is nef and

[TABLE]

with $a>0$, that forces $\alpha=\frac{1}{2}$ and $L\Gamma=0$. Thus $\Gamma$ is one of the curves $A_{j}$. But for these curves $F_{1}A_{j}\in\{0,1\}$ is not negative, a contradiction. Therefore $F_{1}$ is nef and the linear system $|F_{1}|$ has no base points. We prove similarly that all the linear system $|F_{k}|,\,k\in\{1,2,3,4\}$ have no base points, and then defines a fibration $\psi_{k}:X\to\mathbb{P}^{1}$. Since $F_{1}A_{j}=1$ for $j\in\{1,...,4\}$, the fibration $\psi_{1}$ has connected fibers. By the same kind of argument so is $\psi_{2}$. For $k\in\{5,...,16\}$, let us define

[TABLE]

(so that in fact $B_{k}=C_{k}$ for $k\in\{5,...,8\}$). The divisor $C_{k}$ is an effective $(-2)$-class and the $12$ divisors

[TABLE]

are distinct singular fibers of $\psi_{1}$, with $A_{k}C_{k}=2$. We now use [3, Proposition 11.4, Chapter III]: the Euler characteristic of $X$ (equal to $24$) is the sum $\sum_{s}e(f_{s})$ of the Euler numbers of all the singular fibers. By the Kodaira classification of singular fibers of elliptic fibrations (see e.g. [3, Table 3, Chapter V, Section 7]), the reducible fibers $f_{s}=C_{k}+A_{k}$ for $k\geq 5$ satisfy $e(f_{s})\geq 2$. Moreover, by the above cited Table, a singular fiber $f_{s}$ containing a smooth rational curve satisfies $e(f_{s})=2$ if and only if it is the union of two $(-2)$-curves $D_{1},D_{2}$ with $D_{1}D_{2}=2$ and meeting transversally. Computing the Euler characteristic of $X$, we see that necessarily $e(C_{k}+A_{k})=2$, for $k\in\{5,\dots,16\}$ and therefore the curves $B_{k}=C_{k}$, $k\in\{5,...,8\}$ are irreducible $(-2)$-curves. We proceed in a similar way with $\psi_{2}$ for the curves $B_{k}$ with $k\in\{1,...,4\}$, and we thus obtain the claimed result. ∎

Remark 25*.*

i) By the Proposition 24, we see that the elliptic fibration defined by $F_{1}$ contains $12$ fibers of type $I_{2}$. By general results on elliptic K3 surfaces, the rank $\rho$ of the Néron-Severi group is $14=12+2$ plus the rank of the Mordell-Weil group, which is the group generated by the zero section (we can take $A_{1}$ as zero section) and the sections of infinite order. Since we know that $\rho=17$ we get that the rank of the Mordell-Weil group is three. That group contains the disjoint sections $A_{2}$, $A_{3}$ and $A_{4}$. The remark is similar for the fibration defined by $F_{2}$.

ii) On the K3 surface $X$ we have two Nikulin configurations

[TABLE]

We do not know if these configurations define two Kummer structures on $X$. We intend to come back on the subject later.

iii) It is also possible to check that the divisor $L^{\prime}=3L-2(A_{1}+\dots+A_{8})$ is big and nef, $L^{\prime 2}=8$ and $L^{\prime}\Gamma=0$ for an effective $(-2)$-class $\Gamma$ if and only if $\Gamma$ is in $\{B_{1},\dots,B_{8},A_{9},\dots,A_{16}\}$.

Appendix

Why it was natural to study the case $t=\frac{1}{2}k(k+1)$

in the paper [15].

Since $\alpha^{2}=1+2t\beta^{2}$, the integer $\alpha$ is odd. Let $k\in\mathbb{N}$ be such that $\alpha=2k+1$ (then one has $A_{1}A_{1}^{\prime}=4k+2$). The integer $\beta$ is then solution of the equation

[TABLE]

which is equivalent to

[TABLE]

Then

[TABLE]

are solutions of the three conditions in (2.3). Since $a^{2}=2t(b^{2}-1)$, one gets

[TABLE]

Thus $2t\cdot 4k(k+1)$ must be the square of an integer and it is therefore natural to define

[TABLE]

Then one computes easily that $a=2k(k+1)$ and $\beta=2$. Then one has

[TABLE]

thus as soon as $\alpha_{0}>5,$ i.e. $k>2$, one can apply Theorem 14. That were the cases we studied in [15].

A table

We resume in the following table the fundamental solutions $(\alpha{}_{0},\beta{}_{0})$ of the Pell-Fermat equation $\alpha^{2}-2t\beta^{2}=1$ for $2t\leq 60$. Recall that there are non-trivial solutions if and only if $2t$ is not a square. Observe that when $2t=k(k+1)$ the minimal solution is $(2k+1,2)$, these correspond to Nikulin configurations studied in the paper [15], we put a $*$ close to these cases. We put a box around the cases with $\beta_{0}$ odd, and a prime ′ when $\beta_{0}$ is even but such that the negative Pell-Fermat equation has a solution: these cases are left out in this paper.