# About isogenies between some $K3$ surfaces

Marie José BERTIN Sorbonne Université
Institut de Mathématiques de Jussieu-Paris Rive Gauche
Case 247
4 Place Jussieu, 75252 PARIS, Cedex 85, France<EMAIL_ADDRESS>and  Odile LECACHEUX Sorbonne Université
Institut de Mathématiques de Jussieu-Paris Rive Gauche
Case 247
4 Place Jussieu, 75252 PARIS, Cedex 85, France<EMAIL_ADDRESS>
###### Abstract.

We study some $2$ and $3$-isogenies on the singular $K3$ surface $Y_{10}$ of
discriminant $72$ and belonging to the Apéry-Fermi pencil ($Y_{k}$), and find
on it many interesting properties. For example some of its elliptic fibrations
with $3$-torsion section induce by $3$-isogeny either an elliptic fibration of
$Y_{2}$, the unique $K3$ surface of discriminant $8$, or an elliptic fibration
of other $K3$ surfaces of discriminant $72$.

###### Key words and phrases:

Elliptic fibrations of $K3$ surfaces, Isogenies between Elliptic Fibrations,
Transcendental lattices

###### 2010 Mathematics Subject Classification:

11F23, 11G05, 14J28 (Primary); 14J27

##  1\. Introduction

In the Apéry-Fermi pencil $Y_{k}$ defined by the equation

$(Y_{k})\qquad X+\frac{1}{X}+Y+\frac{1}{Y}+Z+\frac{1}{Z}=k,$

two $K3$ surfaces, namely $Y_{2}$ and $Y_{10}$ retain our attention. We
observe first the relation between their transcendental lattices $T(Y_{2})$
and $T(Y_{10})$ (see [1], [2])

$T(Y_{2})=\begin{pmatrix}2&0\\\ 0&4\\\ \end{pmatrix}\qquad
T(Y_{10})=\begin{pmatrix}6&0\\\ 0&12\\\ \end{pmatrix}=T(Y_{2})[3].$

Even more, from results of Kuwata [10] and Shioda [18] (theorem 2.1), the
equality $T\left(Y_{10}\right)=T\left(Y_{2}\right)\left[3\right]$ reveals a
relation between $Y_{2}$ and $Y_{10}.$ Indeed, starting with an elliptic
fibration of $Y_{2}$ with two singular fibers $II^{*}$ and Weierstrass
equation

$(Y_{2})_{h}\qquad y^{2}=x^{3}+\alpha x+h+\frac{1}{h}+\beta,$

a base change $h=u^{3}$ gives a Weierstrass equation denoted
$(Y_{2})_{h}^{(3)}$ of an elliptic fibration of a $K3$ surface with
transcendental lattice $T(Y_{2})[3]$, which is precisely $Y_{10}.$ If, instead
of the previous base change, we use the base change $h=u^{2}$, we obtain a
Weierstrass equation $(Y_{2})_{h}^{(2)}$ of an elliptic fibration of a $K3$
surface with transcendental lattice $T(Y_{2})[2]$ which is the Kummer surface
$K_{2}$. The idea, previously developped in [3] when searching $2$-isogenies
between some elliptic fibrations of $Y_{2}$ and its Kummer $K_{2}$, suggests
possible $3$-isogenies between some elliptic fibrations of $Y_{2}$ and
$Y_{10}$. Indeed, in a recent paper, Bertin and Lecacheux [4] obtained
Weierstrass equations of two rank $0$ elliptic fibrations of $Y_{10}$ by
$3$-isogenies from Weierstrass equations of rank $0$ elliptic fibrations of
$Y_{2}$.

In [3], Bertin and Lecacheux obtained all elliptic fibrations, called generic,
of the Apéry-Fermi pencil together with a Weierstrass equation. Some of these
fibrations are endowed with a $2$ or a $3$-torsion section. It was also proved
that the quotient $K3$ surface by a $2$-torsion section is either the Kummer
surface $K_{k}$ of its Shioda-Inose structure or a non Kummer $K3$ surface
$S_{k}$ with transcendental lattice

$\begin{pmatrix}-2&0&0\\\ 0&2&0\\\ 0&0&6\\\ \end{pmatrix}.$

More precisely the $2$-isogenies of $Y_{k}$ are divided in two classes, the
Morrison-Nikulin ones, i.e. $2$-isogenies from $Y_{k}$ to its Kummer $K_{k}$,
and the others, called van Geemen-Sarti involutions. It was also proved that
by specialization $S_{2}=Y_{2}$. But there exist also other non specialized
$2$-isogenies on $Y_{2}$, some of them being Morrison-Nikulin, the others
being called ”self-isogenies”, meaning either they preserve the same elliptic
fibration (”PF self-isogenies”) or they exchange two elliptic fibrations of
$Y_{2}$ (”EF self-isogenies”).

We shall prove in section 3 a similar result for the specializations $K_{10}$
and $S_{10}$, that is $K_{10}$ is the Kummer surface with transcendental
lattice $\begin{pmatrix}12&0\\\ 0&24\\\ \end{pmatrix}$ and $S_{10}=Y_{10}$. We
exhibit also a Morrison-Nikulin involution of $Y_{10}$ not coming by
specialization and ”self isogenies” of rank $0$ or positive rank not coming
from specialization.

We end this section with the following application that is the determination
of the Mordell-Weil group of a certain specialized fibration of $Y_{10}$.
After expliciting the Kummer surface $K_{10}=\text{Kum}(E_{1},E_{2})$ where
$E_{1}$ has complex multiplication we exhibit an infinite section on a
fibration of $K_{10}$ giving, by a 2-isogeny, a section on the fibration of
$Y_{10}$.

The situation is quite different concerning the $3$-isogenies.

We prove in section 4 that the quotient $K3$ surface of a generic member
$(Y_{k})$ by any $3$-torsion section is a $K3$ surface $N_{k}$ with
transcendental lattice $\begin{pmatrix}0&0&3\\\ 0&4&0\\\ 3&0&0\\\
\end{pmatrix}$ whose specializations satisfy $N_{2}=Y_{10}$ while $N_{10}$ is
a $K3$ surface of discriminant $72$ and transcendental lattice $[4\quad 0\quad
18]:=\begin{pmatrix}4&0\\\ 0&18\\\ \end{pmatrix}$.

We prove also the main result, our first motivation, that is, all the
$3$-isogenies from elliptic fibrations of $Y_{2}$ are $3$-isogenies from
$Y_{2}$ to $Y_{10}$. It remains a natural question: what about the other
$3$-isogenies from elliptic fibrations of $Y_{10}$? Indeed we found
$3$-isogenies from $Y_{10}$ to two other $K3$ surfaces with respective
transcendental lattices $[4\quad 0\quad 18]$ and $[2\quad 0\quad 36]$.

In the same section we use the elliptic fibration $(Y_{2})_{h}^{(3)}$ to
construct elliptic fibrations of $Y_{10}$ of high rank (namely $7$ the highest
we found) and by the $2$-neighbour method, a rank $4$ elliptic fibration with
a $2$-torsion section defining the Morrison-Nikulin involution exhibited in
section 3.

In the last section 5, we prove that the $L$-series of the transcendental
lattice of a certain singular $K3$ surface is inchanged by a $2$ or a
$3$-isogeny. This result explains why the isogenous surfaces found in the
previous sections have equal discriminants up to square.

Finally we put our results on $2$ and $3$-isogenies on $Y_{2}$ and $Y_{10}$ in
the perspective of a result of Bessière, Sarti and Veniani [5].

Computations were performed using partly the computer algebra system PARI
[13], partly Sage [14] and mostly the computer algebra system MAPLE and the
Maple Library “Elliptic Surface Calculator” written by Kuwata [9].

##  2\. Background

###  2.1. Discriminant forms

Let $L$ be a non-degenerate lattice. The dual lattice $L^{*}$ of $L$ is
defined by

$L^{*}:=\text{Hom}(L,\mathbb{Z})=\\{x\in
L\otimes\mathbb{Q}/\,\,\,b(x,y)\in\mathbb{Z}\hbox{ for all }y\in L\\}.$

and the discriminant group $G_{L}$ by

$G_{L}:=L^{*}/L.$

This group is finite if and only if $L$ is non-degenerate. In the latter case,
its order is equal to the absolute value of the lattice determinant
$\mid\det(G(e))\mid$ for any basis $e$ of $L$. A lattice $L$ is unimodular if
$G_{L}$ is trivial.

Let $G_{L}$ be the discriminant group of a non-degenerate lattice $L$. The
bilinear form on $L$ extends naturally to a $\mathbb{Q}$-valued symmetric
bilinear form on $L^{*}$ and induces a symmetric bilinear form

$b_{L}:G_{L}\times G_{L}\rightarrow\mathbb{Q}/\mathbb{Z}.$

If $L$ is even, then $b_{L}$ is the symmetric bilinear form associated to the
quadratic form defined by

$\begin{matrix}q_{L}:G_{L}&\rightarrow&\mathbb{Q}/2\mathbb{Z}\\\
q_{L}(x+L)&\mapsto&x^{2}+2\mathbb{Z}.\end{matrix}$

The latter means that $q_{L}(na)=n^{2}q_{L}(a)$ for all $n\in\mathbb{Z}$,
$a\in G_{L}$ and
$b_{L}(a,a^{\prime})=\frac{1}{2}(q_{L}(a+a^{\prime})-q_{L}(a)-q_{L}(a^{\prime}))$,
for all $a,a^{\prime}\in G_{L}$, where
$\frac{1}{2}:\mathbb{Q}/2\mathbb{Z}\rightarrow\mathbb{Q}/\mathbb{Z}$ is the
natural isomorphism. The pair $\boldsymbol{(G_{L},b_{L})}$ (resp.
$\boldsymbol{(G_{L},q_{L})}$) is called the discriminant bilinear (resp.
quadratic) form of $L$.

When the even lattice $L$ is given by its Gram matrix, we can compute its
discriminant form using the following lemma as explained in Shimada [16].

######  Lemma 2.1.

Let $A$ be the Gram matrix of $L$ and $U$, $V\in Gl_{n}(\mathbb{Z})$ such that

$UAV=D=\begin{pmatrix}d_{1}&&0\\\ &\ddots&\\\ 0&&d_{n}\\\ \end{pmatrix}$

with $1=d_{1}=\ldots=d_{k}<d_{k+1}\leq\ldots\leq d_{n}$. Then

$G_{L}\simeq\oplus_{i>k}\mathbb{Z}/(d_{i}).$

Moreover the $i$th row vector of $V^{-1}$, regarded as an element of $L^{*}$
with respect to the dual basis $e_{1}^{*}$, …, $e_{n}^{*}$ generate the cyclic
group $\mathbb{Z}/(d_{i})$.

###  2.2. Nikulin’s results

######  Lemma 2.2 (Nikulin [12], Proposition 1.6.1).

Let $L$ be an even unimodular lattice and $T$ a primitive sublattice. Then we
have

$G_{T}\simeq G_{T^{\perp}}\simeq L/(T\oplus
T^{\perp}),\,\,\,\,\,\,\,q_{T^{\perp}}=-q_{T}.$

In particular, $|\det T|=|\det T^{\perp}|=[L:T\oplus T^{\perp}]$.

###  2.3. Shioda’s results

Let $(S,\Phi,\mathbb{P}^{1})$ be an elliptic surface with a section $\Phi$,
without exceptional curves of first kind.

Denote by $NS(S)$ the group of algebraic equivalence classes of divisors of
$S$ .

Let $u$ be the generic point of $\mathbb{P}^{1}$ and $\Phi^{-1}(u)=E$ the
elliptic curve defined over $K=\mathbb{C}(u)$ with a $K$-rational point
$o=o(u)$. Then, $E(K)$ is an abelian group of finite type provided that $j(E)$
is transcendental over $\mathbb{C}$. Let $r$ be the rank of $E(K)$ and
$s_{1},...,s_{r}$ be generators of $E(K)$ modulo torsion. Besides, the torsion
group $E(K)_{tors}$ is generated by at most two elements $t_{1}$ of order
$e_{1}$ and $t_{2}$ of order $e_{2}$ such that $1\leq e_{2}$, $e_{2}|e_{1}$
and $\mid E(K)_{tors}\mid=e_{1}e_{2}$.

The group $E(K)$ of $K$-rational points of $E$ is canonically identified with
the group of sections of $S$ over $\mathbb{P}^{1}(\mathbb{C})$.

For $s\in E(K)$, we denote by $(s)$ the curve image in $S$ of the section
corresponding to $s$.

Let us define

$D_{\alpha}:=(s_{\alpha})-(o)\,\,\,\,\,\,1\leq\alpha\leq r$
$D^{\prime}_{\beta}:=(t_{\beta})-(o)\,\,\,\,\,\,\beta=1,2.$

Consider now the singular fibers of $S$ over $\mathbb{P}^{1}$. We set

$\Sigma:=\\{v\in\mathbb{P}^{1}/C_{v}=\Phi^{-1}(v)\,\,\,\,{\hbox{be a singular
\ fiber}}\\}$

and for each $v\in\Sigma$, $\Theta_{v,i}$, $0\leq i\leq m_{v}-1$, the $m_{v}$
irreducible components of $C_{v}$.

Let $\Theta_{v,0}$ be the unique component of $C_{v}$ passing through $o(v)$.

One gets

$C_{v}=\Theta_{v,0}+\sum_{i\geq
1}\mu_{v,i}\Theta_{v,i},\,\,\,\,\,\,\,\mu_{v,i}\ \geq 1.$

Let $A_{v}$ be the matrix of order $m_{v}-1$ whose entry of index $(i,j)$ is
$(\Theta_{v,i}\Theta_{v,j})$, $i,j\geq 1$, where $(DD^{\prime})$ is the
intersection number of the divisors $D$ et $D^{\prime}$ along $S$. Finally $f$
will denote a non singular fiber, i.e. $f=C_{u_{0}}$ for $u_{0}\notin\Sigma$.

######  Theorem 2.1.

The Néron-Severi group $NS(S)$ of the elliptic surface $S$ is generated by the
following divisors

$f,\Theta_{v,i}\,\,\,\,\,\,(1\leq i\leq m_{v}-1,\,\,\,\,v\in\Sigma)$
$(o),D_{\alpha}\,\,\,\,\,\,1\leq\alpha\leq
r,\,\,\,\,D^{\prime}_{\beta}\,\,\,\,\beta=1,2.$

The only relations between these divisors are at most two relations

$e_{\beta}D^{\prime}_{\beta}\approx
e_{\beta}(D^{\prime}_{\beta}(o))f+\sum_{v\in\Sigma}(\Theta_{v,1},...,\Theta_{v,m_{v}-1})e_{\beta}A_{v}^{-1}\left(\begin{array}[]{l}(D^{\prime}_{\beta}\Theta_{v,1})\\\
.\\\ .\\\ .\\\ (D^{\prime}_{\beta}\Theta_{v,m_{v}-1})\end{array}\right)$

where $\approx$ stands for the algebraic equivalence.

###  2.4. Transcendental lattice

Let $X$ be an algebraic $K3$ surface; the group $H^{2}(X,\mathbb{Z})$, with
the intersection pairing, has a structure of a lattice and by Poincaré duality
is unimodular. The Néron-Severi lattice $NS(X):=H^{2}(X,\mathbb{Z})\cap
H^{1,1}(X)$ and the transcendental lattice $T(X)$, its orthogonal complement
in $H^{2}(X,\mathbb{Z})$, are primitive sublattices of $H^{2}(X,\mathbb{Z})$
with respective signatures $(1,\rho-1)$ and $(2,20-\rho)$ where $\rho$ is the
rank of the Néron-Severi lattice.

By Nikulin’s lemma, their discriminant forms differ just by the sign, that is

$(G_{T(X)},q_{T(X)})\equiv(G_{NS(X)},-q_{NS(X)}).$

###  2.5. $3$-isogenous curves

####  2.5.1. Method

Let $E$ be an elliptic curve with a $3$-torsion point
$\omega=\left(0,0\right)$

$E:Y^{2}+AYX+BY=X^{3}$

and $\phi$ the isogeny of kernel $<\omega>.$

To determine a Weierstrass equation for the elliptic curve $E/<\omega>$ we
need two functions $x_{1}$ of degree $2$ and $y_{1}$ of degree $3$ invariant
by $M\rightarrow M+\omega$ where $M=\left(X_{M},Y_{M}\right)$ is a general
point on $E.$ We compute $M+\omega$ and $M+2\omega\left(=M-\omega\right)$ and
can choose

$\displaystyle x_{1}$
$\displaystyle=X_{M}+X_{M+\omega}+X_{M+2\omega}=\frac{X^{3}+ABX+B^{2}}{X^{2}}$
$\displaystyle y_{1}$ $\displaystyle=Y_{M}+Y_{M+\omega}+Y_{M+2\omega}$
$\displaystyle=\frac{Y\left(X^{3}-AXB-2B^{2}\right)-B\left(X^{3}+A^{2}X^{2}+2AXB+B^{2}\right)}{X^{3}}.$

The relation between $x_{1}$ and $y_{1}$ gives a Weierstrass equation for
$E/<\omega>$

$y_{1}^{2}+\left(Ax_{1}+3B\right)y_{1}=x_{1}^{3}-6ABx_{1}-B\left(A^{3}+9B\right).$

Notice that the points with $x_{1}=-\frac{A^{2}}{3}$ are $3$-torsion points.
Taking one of these points to origin and after some transformation we can
obtain a Weierstrass equation $y^{2}+ayx+by=x^{3}$ with the following
transformations.

####  2.5.2. Formulae

If $j^{3}=1$ then we define

$\displaystyle S_{1}$
$\displaystyle=2\left(j^{2}-1\right)y+6Ax-2\left(j-1\right)\left(A^{3}-27B\right)$
$\displaystyle S_{2}$
$\displaystyle=2\left(j-1\right)y+6Ax-2\left(j^{2}-1\right)\left(A^{3}-27B\right)$

and

$X=\frac{-1}{324}\frac{S_{1}S_{2}}{x^{2}},\qquad
Y=\frac{1}{5832}\frac{S_{1}^{3}}{x^{3}}$

then we have

$E/<\omega>:y^{2}+\left(-3A\right)yx+\left(27B-A^{3}\right)y=x^{3}.$

If $A_{1}=-3A$, $B_{1}=27B-A^{3},$ then we define

$\displaystyle\sigma_{1}$
$\displaystyle=2\left(j^{2}-1\right)3^{6}Y+6A_{1}3^{4}X-2\left(j-1\right)\left(A_{1}^{3}-27B_{1}\right)$
$\displaystyle\sigma_{2}$
$\displaystyle=2\left(j-1\right)3^{6}Y+6A_{1}3^{4}X-2\left(j^{2}-1\right)\left(A_{1}^{3}-27B_{1}\right)$

and then

$x=\frac{-1}{324}\frac{\sigma_{1}\sigma_{2}}{3^{8}X^{2}}=-{\frac{3\,{X}^{3}+{A}^{2}{X}^{2}+3\,BAX+3\,{B}^{2}}{{X}^{2}}},\qquad
y=\frac{1}{5832}\frac{\sigma_{1}^{3}}{3^{12}X^{3}}.$

####  2.5.3. Other properties of isogenies

The divisor of the function $Y$ is equal to $-3\left(0\right)+3\omega$ so
$Y=W^{3}$ where $W$ is a function on the curve $E/<\omega>$. If $X=WZ$, the
function field of $E/<\omega>$ is generated by $W$ and $Z.$ So replacing in
the equation of $E$ we obtain the relation between $Z$ and $W$

$W^{3}+AZW+B-Z^{3}=0.$

This cubic equation, with a rational point at infinity with $W=Z$ can be
transformed to obtain a Weierstrass equation in the coordinates $X_{2}$ and
$Y_{2}$:

$\displaystyle W$
$\displaystyle=\frac{1}{9}\frac{(-243B-3X_{2}A+9A^{3}-Y_{2})}{X_{2}},\quad
Z=-\frac{1}{9}\frac{Y_{2}}{X_{2}}$ $\displaystyle\text{of inverse}\quad X_{2}$
$\displaystyle=3\frac{A^{3}-27B}{3\left(W-Z\right)+A},\quad
Y_{2}=-27Z\frac{A^{3}-27B}{3\left(W-Z\right)+A}$ $\displaystyle
Y_{2}^{2}+3AY_{2}X_{2}+\left(-9A^{3}+243B\right)Y_{2}=$ $\displaystyle
X_{2}^{3}-9X_{2}^{2}A^{2}+27A\left(A^{3}-27B\right)X_{2}-27\left(A^{3}-27B\right)^{2}.$

The points of $X_{2}$-coordinate equal to $0$ are $3-$torsion points and
easily we recover the previous formulae.

###  2.6. Notation

The singular fibers of type $\,I_{n},D_{m},IV^{\ast},...$ at $t=t_{1},.,t_{m}$
or at roots of a polynomial $p(t)$ of degree $m$ are denoted
$mI_{n}(t_{1},..,t_{m})$ or $mI_{n}(p(t)).$ The zero component of a reducible
fiber is the component intersecting the zero section and is denoted
$\theta_{0}$ or $\theta_{t_{0},0}.$ The other components denoted
$\theta_{t_{0},i}$ satisfy the property
$\theta_{t_{0},i}\cdot\theta_{t_{0},i+1}=1.$

##  3\. $2$-isogenies of $Y_{10}$

In [3], Bertin and Lecacheux classified all the $2$-isogenies of $Y_{2}$ in
two sets, the first defining Morrison-Nikulin involutions, that is from
$Y_{2}$ to its Kummer surface $K_{2}$ and the second giving van Geemen-Sarti
involutions that is exchanging two elliptic fibrations (different or the same)
of $Y_{2}$ named ”self-isogenies”.

Since we have no exhaustive list of elliptic fibrations of $Y_{10}$ with
$2$-torsion sections, we cannot give such a classification. However we found
on $Y_{10}$ Morrison-Nikulin involutions from $Y_{10}$ to its Kummer surface
$K_{10}$.

###  3.1. Morrison-Nikulin involutions of $Y_{10}$

####  3.1.1. Specialized Morrison-Nikulin involutions

We first recall a result concerning the involutions on the generic member
$(Y_{k})$ of the Apéry-Fermi pencil.

######  Theorem 3.1.

(Bertin and Lecacheux [3])

Suppose $Y_{k}$ is a generic $K3$ surface of the family with Picard number
$19$.

Let $\pi:Y_{k}\rightarrow\mathbb{P}^{1}$ be an elliptic fibration with a
torsion section of order $2$ which defines an involution $i$ of $Y_{k}$ (van
Geemen-Sarti involution) then the minimal resolution of the quotient $Y_{k}/i$
is either the Kummer surface $K_{k}$ associated to $Y_{k}$ given by its
Shioda-Inose structure or a surface $S_{k}$ with transcendental lattice
$T(S_{k})=\langle-2\rangle\oplus\langle 2\rangle\oplus\langle 6\rangle$ and
Néron-Severi lattice $NS(S_{k})=U\oplus E_{8}[-1]\oplus
E_{7}[-1]\oplus\langle(-2)\rangle\oplus\langle(-6)\rangle$, which is not a
Kummer surface. Thus, $\pi$ leads to an elliptic fibration either of $K_{k}$
or of $S_{k}$. Moreover there exist some genus $1$ fibrations $\theta:K_{k}$
$\rightarrow\mathbb{P}^{1}$ without section such that their Jacobian variety
satisfies $J_{\theta}\left(K_{k}\right)=S_{k}$.

More precisely, among the elliptic fibrations of $Y_{k}$ (up to automorphisms)
$12$ of them have a two-torsion section. And only $7$ of them possess a
Morrison-Nikulin involution $i$ such that $Y_{k}/i=K_{k}$.

With the same argument as for specialization to $Y_{2}$, Morrison-Nikulin
involutions specialized to $Y_{10}$ remain Morrison-Nikulin involutions of
$Y_{10}$. Hence we obtain in the following Table 1 the corresponding
Weierstrass equations of such elliptic fibrations of the Kummer surface
$K_{10}$ with transcendental lattice $\begin{pmatrix}12&0\\\ 0&24\\\
\end{pmatrix}.$

$\begin{array}[c]{|c|c|}\hline\cr\text{No}&\text{Weierstrass Equation}\\\
\hline\cr\\#4&\begin{array}[c]{c}y^{2}=x(x-\frac{1}{8}(t-a_{1})(t-a_{2})(t-a_{3}))(x-\frac{1}{8}(t-b_{1})(t-b_{2})(t-b_{3}))\\\
a_{1}=-\frac{58}{3}-4\sqrt{6}+8\sqrt{3}-4\sqrt{2},a_{2}=-\frac{58}{3}-4\sqrt{6}-8\sqrt{3}+4\sqrt{2},\\\
b_{1}=\frac{2}{3}+4\sqrt{6}+24\sqrt{2}+8\sqrt{3},b_{2}=\frac{2}{3}+4\sqrt{6}-24\sqrt{2}-8\sqrt{3}\\\
a_{3}=\frac{116}{3}+8\sqrt{6},b_{3}=-\frac{4}{3}-8\sqrt{6}\\\
I_{6}^{*}(\infty),6I_{2}(a_{i},b_{i}),(r=2)\end{array}\\\
\hline\cr\hline\cr\\#8&\begin{array}[c]{c}y^{2}=x^{3}+2(t^{3}-25t^{2}+50t-24)x^{2}+(t^{2}-24t+36)(t^{2}-24t+16)(t-1)^{2}x\\\
I_{3}(0),I_{4}(1),4I_{2}(t^{2}-24t+16,t^{2}-24t+36),I_{3}^{*}(\infty),(r=2)\end{array}\\\
\hline\cr\hline\cr\\#16&\begin{array}[c]{c}y^{2}=x(x+t(t^{2}+(238-96\sqrt{6})t+1)(x+t(t^{2}+(42-16\sqrt{6})t+1))\\\
2I_{2}^{*}(0,\infty),4I_{2}(t^{2}+(-16\sqrt{6}+42)t+1,t^{2}+(-96\sqrt{6}+238)t+1),(r=2)\end{array}\\\
\hline\cr\hline\cr\\#17&\begin{array}[c]{c}y^{2}=x(x-(t^{2}-48)(t^{2}-8))(x-(t^{2}+4t-20)(t^{2}-4t-20))\\\
I_{8}(\infty),8I_{2}(t^{2}-4t-20,t^{2}+4t-20,t^{2}-8,t^{2}-48),(r=3)\end{array}\\\
\hline\cr\hline\cr\\#23&\begin{array}[c]{c}y^{2}=x(x+\frac{1}{4}(4t^{2}-12t+1)(8t-1))(x+(12t-1)(4t^{2}-8t+1))\\\
I_{0}^{*}(\infty),I_{6}(0),6I_{2}(\frac{1}{8},\frac{1}{12},4t^{2}-8t+1,4t^{2}-12t+1),(r=3)\\\
\end{array}\\\
\hline\cr\hline\cr\\#24&\begin{array}[c]{c}y^{2}=x^{3}+2t(t^{2}-22t+1)x^{2}+t^{2}(t^{2}-14t+1)(t^{2}-34t+1)x\\\
2I_{1}^{*}(0,\infty),5I_{2}(-1,t^{2}-14t+1,t^{2}-34t+1),(r=3)\end{array}\\\
\hline\cr\hline\cr\\#26&\begin{array}[c]{c}y^{2}=x(x-\frac{1}{4}(t^{2}-14t+1)(t^{2}-6t+1))(x-\frac{1}{4}(t^{2}-10t+1)^{2})\\\
4I_{4}(0,\infty,t^{2}-10t+1),4I_{2}(t^{2}-14t+1,t^{2}-6t+1),(r=2)\end{array}\\\
\hline\cr\end{array}$ Table 1. Specialized fibrations of the Kummer $K_{10}$

####  3.1.2. A non specialized Morrison-Nikulin involution of $Y_{10}$

######  Theorem 3.2.

The rank $4$ elliptic fibration of $Y_{10}$ (4.3), with Weierstrass equation

$E_{t}\qquad y^{2}=x^{3}-(t^{3}+5t^{2}-2)x^{2}+(t^{3}+1)^{2}x$

and singular fibers $I_{0}^{*}(\infty)$, $3I_{4}(t^{3}+1)$, $I_{2}(0)$,
$4I_{1}(1,-5/3,t^{2}-4t-4)$, has a $2$-torsion section defining a Morrison-
Nikulin involution from $Y_{10}$ to $K_{10}$, that is
$F_{t}=E_{t}/\langle(0,0)\rangle$ is a rank $4$ elliptic fibration of $K_{10}$
with Weierstrass equation

$F_{t}\qquad Y^{2}=X^{3}+2(t^{3}+5t^{2}-2)X^{2}-(t^{2}-4t-4)(t-1)(3t+5)t^{2}X$

and singular fibers $I_{0}^{*}(\infty)$, $I_{4}(0)$, $7I_{2}(\pm
1,-5/3,t^{2}-t+1,t^{2}-4t-4)$.

###### Proof.

Starting from $F_{t}$ and taking the new parameter
$p=\frac{X}{(t^{2}-4t-4)(t-1)(3t+5)}$, we get a rank $1$ elliptic fibration
with Weierstrass equation

$\displaystyle F_{p}$
$\displaystyle:Y^{2}=X^{3}+\frac{3}{4}p\left(5p-1\right)^{2}X^{2}+\frac{1}{6}p^{2}\left(2p-1\right)\left(5p-1\right)\left(49p^{2}-13p+1\right)X$
$\displaystyle+\frac{1}{108}p^{3}\left(2p-1\right)^{2}\left(49p^{2}-13p+1\right)^{2}$

and singular fibers $I_{0}^{*}(\infty)$, $I_{3}^{*}(0)$,
$3I_{3}(\frac{1}{2},49p^{2}-13p+1)$. The infinite section
$P=(-\frac{1}{12}p(49p^{2}-13p+1),\frac{1}{8}p^{2}(49p^{2}-13p+1))$ is of
height $h(P)=\frac{2}{3}$, is not equal to $2Q$ or $3Q$, hence the
discriminant of the Néron-Severi lattice satisfies $\Delta=4\times 4\times
3^{3}\times\frac{2}{3}=72\times 4$.

Now we are going to compute its Gram matrix and deduce its transcendental
lattice.

To compute the Néron-Severi lattice we order the generators as $(0)$ the zero
section, $(f)$ the generic fiber, $\theta_{i},1\leq i\leq 4$, $\eta_{i},1\leq
i\leq 7$, $\gamma_{i}$, $\delta_{i}$, $\epsilon_{i}$, $1\leq i\leq 2$ the
rational components of respectively $D_{4}$, $D_{7}$ and the three $A_{2}$ and
finally the infinite section $(P)$.

$NS=\left(\setcounter{MaxMatrixCols}{20}\begin{smallmatrix}-2&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1\\\
0&0&-2&0&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&0&-2&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&1&1&-2&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&0&0&1&-2&0&0&0&0&0&0&0&0&0&0&0&0&0&1\\\
0&0&0&0&0&0&-2&0&1&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&-2&1&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&1&1&-2&1&0&0&0&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&1&-2&1&0&0&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&0&1&-2&1&0&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&0&0&1&-2&1&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&0&0&0&1&-2&0&0&0&0&0&0&1\\\
0&0&0&0&0&0&0&0&0&0&0&0&0&-2&1&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&0&0&0&0&0&1&-2&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&-2&1&0&0&1\\\
0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&-2&0&0&0\\\
0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&-2&1&1\\\
0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&-2&0\\\
0&1&0&0&0&1&0&0&0&0&0&0&1&0&0&1&0&1&0&-2\\\ \end{smallmatrix}\right).$

Fom lemma 2.1 we deduce that
$G_{NS}=\mathbb{Z}/12\mathbb{Z}\oplus\mathbb{Z}/24\mathbb{Z}$ and we get
generators $f_{1}$ and $f_{2}$ with respective norms
$q(f_{1})=-\frac{41}{12}$, $q(f_{2})=-\frac{59}{24}$ modulo $2$ and scalar
product $f_{1}.f_{2}=\frac{7}{4}$ modulo $1$.

In order to prove that the transcendental lattice corresponds to the Gram
matrix $\begin{pmatrix}12&0\\\ 0&24\\\ \end{pmatrix}$ we must find for the
corresponding quadratic form generators $g_{1}$ and $g_{2}$ satisfying
$q(g_{1})=\frac{41}{12}$, $q(g_{2})=\frac{59}{24}$ modulo $2$ and scalar
product $g_{1}.g_{2}=-\frac{7}{4}$ modulo $1$. This is obtained with
$g_{1}=\begin{pmatrix}\frac{1}{4}\\\ \frac{1}{3}\\\ \end{pmatrix}$ and
$g_{2}=\begin{pmatrix}\frac{5}{12}\\\ \frac{1}{8}\\\ \end{pmatrix}$. Thus
$F_{p}$, hence $F_{t}$, are elliptic fibrations of the Kummer surface
$K_{10}$.

∎

In a previous paper, Bertin and Lecacheux [3] explained that, in the Apéry-
Fermi family, only $Y_{2}$ and $Y_{10}$ may have ”self-isogenies”. A ”self-
isogeny” is a van Geemen-Sarti involution which either preserve an elliptic
fibration (called ”PF self-isogeny” for more precision) or exchanges two
elliptic fibrations ”EF self-isogeny”.

Moreover, in the same paper, all the ”self-isogenies” of $Y_{2}$ were listed.
Since it is quite difficult to get all the elliptic fibrations of $Y_{10}$
with $2$-torsion sections, we shall give ”self-isogenies” of $Y_{10}$ obtained
either as specializations or from rank $0$ fibrations or from non specialized
positive rank elliptic fibrations.

###  3.2. Specialized ”self-isogenies”

In [3] we characterized the surface $S_{k}$ obtained by $2$-isogeny deduced
from van Geemen-Sarti involutions of $Y_{k}$ which are not Morrison-Nikulin.
Let us recall the specialized Weierstrass equations of $S_{10}$. We denote
$E{\\#n}$ (resp. $EE{\\#n}$) a Weierstrass equation of a fibration of $Y_{10}$
(resp. $S_{10}$).

The specialization of $S_{k}$ for $k=10$ has the following five elliptic
fibrations given on Table 2. The first Weierstrass equation concerns $Y_{10}$
and the second $S_{10}$ obtained as its $2$-isogenous curve.

$\begin{array}[c]{|c|c|}\hline\cr\text{No}&\text{Weierstrass Equation}\\\
\hline\cr\\#7&\begin{array}[c]{c}E7:{y}^{2}={x}^{3}+2{x}^{2}t\left(11t+1\right)-{t}^{2}\left(t-1\right)^{3}x\\\
III^{*}({\infty}),I_{1}^{*}(0),I_{6}(1),2I_{1}(t^{2}+118t+25)\\\
EE7:={Y}^{2}={X}^{3}-4{X}^{2}t\left(11t+1\right)+4{t}^{3}\left(118t+25+{t}^{2}\right)X\\\
III^{*}({\infty}),I_{2}^{*}(0),I_{3}(1),2I_{2}(t^{2}+118t+25)\end{array}\\\
\hline\cr\hline\cr\\#9&\begin{array}[c]{c}E9:{y}^{2}={x}^{3}+28{t}^{2}{x}^{2}+{t}^{3}\left({t}^{2}+98t+1\right)x\\\
2III^{*}(0,{\infty}),2I_{2}(t^{2}+98t+1),2I_{1}(t^{2}-98t+1)\\\
EE9:{Y}^{2}={X}^{3}-56{t}^{2}{X}^{2}-4{t}^{3}\left({t}^{2}-98t+1\right)X\\\
2III^{*}(0,{\infty}),2I_{2}(t^{2}-98t+1),2I_{1}(t^{2}+98t+1)\\\
\end{array}\par\\\
\hline\cr\hline\cr\\#14&\begin{array}[c]{c}E14:{y}^{2}={x}^{3}+t\left(98{t}^{2}+28t+1\right){x}^{2}+{t}^{6}x\par\\\
I_{8}^{*}(0),I_{0}^{*}(\infty),I_{1}(4t+1),I_{1}(24t+1),2I_{1}(100t^{2}+28t+1)\\\
EE14:{Y}^{2}=X\left(X-96{t}^{3}-28{t}^{2}-t\right)\left(X-100{t}^{3}-28{t}^{2}-t\right)\\\
I_{4}^{*}(0),I_{0}^{*}(\infty),I_{2}(4t+1),I_{2}(24t+1),2I_{2}(100t^{2}+28t+1)\\\
\end{array}\\\
\hline\cr\hline\cr\\#15&\begin{array}[c]{c}E15:{y}^{2}={x}^{3}-t\left(2+{t}^{2}-22t\right){x}^{2}+{t}^{2}\left(t+1\right)^{2}x\\\
I_{1}^{*}(0),I_{4}^{*}(\infty),I_{4}(-1),I_{1}(24),2I_{1}(t^{2}-20t+4)\\\
EE15:{Y}^{2}=X\left(X+t^{3}-24t^{2}\right)\left(X+t^{3}-20t^{2}+4t\right)\\\
I_{2}^{*}(0),I_{2}^{*}(\infty),I_{2}(-1),I_{2}(24),2I_{2}(t^{2}-20t+4)\\\
\end{array}\\\
\hline\cr\hline\cr\\#20&\begin{array}[c]{c}E20:{y}^{2}={x}^{3}+\left(\frac{1}{4}{t}^{4}-5{t}^{3}+{\frac{53}{2}}{t}^{2}-15t-\frac{3}{4}\right){x}^{2}-t\left(t-10\right)x\\\
I_{2}(0),I_{12}(\infty),2I_{3}(t^{2}-10t+1),I_{2}(10),I_{1}(1),I_{1}(9)\\\
EE20:{Y}^{2}={X}^{3}+\left(-\frac{1}{2}{t}^{4}+10{t}^{3}-53{t}^{2}+30t+\frac{3}{2}\right){X}^{2}\\\
+\frac{1}{16}\left(t-1\right)\left(t-9\right)\left({t}^{2}-10t+1\right)^{3}X\\\
I_{1}(0),I_{6}(\infty),2I_{6}(t^{2}-10t+1),I_{2}(1),I_{2}(9),I_{1}(10)\\\
\end{array}\par\\\ \hline\cr\end{array}$ Table 2. Fibrations $E{\\#i}$ of
$Y_{10}$ and $EE{\\#i}$ of $S_{10}$

######  Theorem 3.3.

The previous $2$-isogenies are in fact ”self-isogenies”, the surface $S_{10}$
being equal to $Y_{10}$.

###### Proof.

We observe that $E9$ and $EE9$ have the same singular fibers. In fact these
two fibrations are isomorphic, the isomorphism being defined by $t=-T$,
$x=-\frac{X}{2}$, $y=\frac{Y}{2\sqrt{-2}}$. This property is sufficient to
identify $S_{10}$ to $Y_{10}$.

Among these ”self-isogenies” only the number $\\#9$ is ”PF”.

∎

###  3.3. Rank $0$ ”self-isogenies”

######  Theorem 3.4.

There are four $2$-isogenies from $Y_{10}$ to $Y_{10}$ defined by extremal
elliptic fibrations with $2$-torsion sections, denoted number $8$, $87$,
$153$, $262$ as in Shimada and Zhang’s paper [17]. They are all ”PF self-
isogenies”.

###### Proof.

We write below Weierstrass equation $E_{n}$, its $2$-isogenous $EE_{n}$ and
the corresponding isomorphism.

$E_{262}\qquad y^{2}=x^{3}+x^{2}(9(t+5)(t+3)+(t+9)^{2})-t^{3}(t+5)^{2}x$
---
$III^{*}(\infty)$, $I_{6}(0)$, $I_{4}(-5)$, $I_{3}(-9)$, $I_{2}(-4)$
$EE_{262}\qquad
Y^{2}=X^{3}-2(9(T+5)(T+3)+(T+9)^{2})X^{2}+4(T+4)^{2}(T+9)^{3}X$
$III^{*}(\infty)$, $I_{6}(-9)$, $I_{4}(-4)$, $I_{3}(0)$, $I_{2}(-5)$
Isomorphism: $t=-T-9$, $x=-\frac{X}{2}$, $y=\frac{Y}{2\sqrt{-2}}$.

$E_{153}\qquad y^{2}=x^{3}+t(t^{2}+10t-2)x^{2}+(2t+1)^{3}t^{2}x$
---
$I_{3}(4)$, $I_{6}(-1/2)$, $I_{1}^{*}(0)$, $I_{2}^{*}(\infty)$
$EE_{153}\qquad Y^{2}=X^{3}-2T(T^{2}+10T-2)X^{2}+T^{3}(T-4)^{3}X$
$I_{6}(4)$, $I_{3}(-1/2)$, $I_{2}^{*}(0)$, $I_{1}^{*}(\infty)$
Isomorphism: $t=-\frac{2}{T}$, $x=-\frac{2X}{T^{4}}$,
$y=-\frac{2\sqrt{-2}Y}{T^{6}}$.

$E_{87}\qquad y^{2}=x^{3}-(9t^{4}+9t^{3}+6t^{2}-6t+4)x^{2}+(21t^{2}-12t+4)x$
---
$I_{12}(\infty)$, $I_{6}(0)$, $2I_{2}(21t^{2}-12t+4)$, $2I_{1}(3t^{2}+6t+7)$
$EE_{87}\qquad Y^{2}=X^{3}-(9T^{4}+9T^{3}+6T^{2}-6T+4)X^{2}+(21T^{2}-12T+4)X$
$I_{12}(0)$, $I_{6}(\infty)$, $2I_{2}(3t^{2}+6t+7)$, $2I_{1}(21t^{2}-12t+4)$
Isomorphism: $t=-\frac{2}{T}$, $x=-\frac{2X}{9T^{4}}$,
$y=\frac{2\sqrt{-2}}{27T^{6}}$.

$E_{8}\qquad y^{2}=x^{3}-(3t^{4}-60t^{2}-24)x^{2}-144(t^{2}-1)^{3}x$
---
$I_{2}(0)$, $2I_{3}(t^{2}+8)$, $I_{4}(\infty)$, $2I_{6}(t^{2}-1)$
$EE_{8}\qquad y^{2}=x^{3}+2(3t^{4}-60t^{2}-24)x^{2}+9t^{2}(t^{2}+8)^{3}x$
$I_{2}(\infty)$, $2I_{3}(t^{2}-1)$, $I_{4}(0)$, $2I_{6}(t^{2}+8)$
Isomorphism: $t=\frac{2\sqrt{-2}}{T}$, $x=\frac{4X}{T^{4}}$,
$y=\frac{8Y}{T^{6}}$.

∎

###  3.4. Positive rank non specialized ”EF” and ”PF self-isogenies”

Using the $2$-neighbor method we found many examples of $2$-torsion elliptic
fibrations of $Y_{10}$.

Denote $E$, $E_{1}$, $E_{2}$, $E_{3}$, $E_{4}$ the following elliptic
fibrations of $Y_{10}$ obtained in the following way. Starting from $E_{153}$
and new parameter $\frac{x}{t(2t+1)^{2}}$ we get $E$. Starting from $EE15$ we
get successively $E_{2}$, $E_{3}$, $E_{4}$ with the successive parameters
$\frac{x}{t^{2}(t-24)}$, $\frac{x}{t^{2}(t+1)}$, $\frac{x}{t(t-4)(t-24)}$. And
from $EE_{2}=E_{2}/\langle(0,0)\rangle$ we get $E_{1}$ with the new parameter
$\frac{x}{t(t-1)(t-4)}$.

######  Theorem 3.5.

  1. (1)

The $2$-isogenies, from $E_{3}$ to $EE_{3}$, from $E_{4}$ to $EE_{4}$, from
$E_{1}$ to $EE_{1}$ are ”PF self-isogenies”.

  2. (2)

The $2$-isogenies from $E$ to $EE$, from $EE14$ to
$EE14/\langle(100t^{2}+28t+1,0)\rangle$ and from $E_{2}$ to $EE_{2}$ are ”EF
self-isogenies”.

###### Proof.

  1. (1)

We only need to give the respective Weierstrass equations, singular fibers and
isomorphisms concerning the $2$-isogenies from $E_{i}$ to $EE_{i}$.

$E_{3}\qquad y^{2}=x^{3}-2t(t^{2}-14t-2)x^{2}+t^{4}(t-4)(t-24)x$
---
$I_{4}^{*}(0)$, $2I_{2}(4,24)$, $2I_{1}(-1/2,-1/12)$, $I_{2}^{*}(\infty)$
$EE_{3}=E_{3}/\langle(0,0)\rangle\hfill$
$Y^{2}=X^{3}+T(T^{2}-14T-2)X^{2}+T^{2}(2T+1)(12T+1)X$
$I_{2}^{*}(0)$, $2I_{1}(4,24)$, $2I_{2}(-1/2,-1/12)$, $I_{4}^{*}(\infty)$
Isomorphism: $t=-\frac{2}{T},\qquad x=-\frac{8X}{T^{4}}\qquad
y=\frac{16\sqrt{-2}Y}{T^{6}}$

$E_{4}\qquad y^{2}=x^{3}-28t^{2}(t-1)x^{2}+4t^{3}(t-1)^{2}(24t+1)x$
---
$III^{*}(0)$, $I_{0}^{*}(1)$, $I_{2}(-1/24)$, $I_{1}(1/25)$,
$I_{0}^{*}(\infty)$
$EE_{4}=E_{4}/\langle(0,0)\rangle\hfill$
$Y^{2}=X^{3}+56T^{2}(T-1)X^{2}+16T^{3}(T-1)^{2}(25T-1)X$
$III^{*}(0)$, $I_{0}^{*}(1)$, $I_{2}(1/25)$, $I_{1}(-1/24)$,
$I_{0}^{*}(\infty)$
Isomorphism: $t=\frac{T}{T-1},\qquad x=-\frac{X}{2(T-1)^{4}}\qquad
y=-\frac{\sqrt{-2}Y}{4(T-1)^{6}}$

$E_{1}\qquad y^{2}=x^{3}-t(5t^{2}+56t+160)x^{2}+4t^{2}(t+6)^{2}(t+4)^{2}x$
---
$I_{0}^{*}(0)$, $2I_{4}(-4,-6)$, $2I_{2}(-8,-16/3)$, $I_{0}^{*}(\infty)$
$EE_{1}=E_{1}/\langle(0,0)\rangle\hfill$
$Y^{2}=X^{3}+2T(5T^{2}+56T+160)X^{2}+T^{2}(T+8)^{2}(3T+16)^{2}X$
$I_{0}^{*}(0)$, $2I_{4}(-8,-16/3)$, $2I_{2}(-6,-4)$, $I_{0}^{*}(\infty)$
Isomorphism: $t=\frac{32}{T},\qquad x=-\frac{2^{9}X}{T^{4}}\qquad
y=\frac{2^{13}\sqrt{-2}Y}{T^{6}}$

  2. (2)

Let us give Weierstrass equations and singular fibers of $E$ and $EE$.

$E\qquad y^{2}=x^{3}+2t(2t^{2}+5t+1)x^{2}+t^{3}(4t+1)(t-1)^{2}x$
---
$I_{2}^{*}(0)$, $I_{4}(1)$, $I_{3}(-1/3)$, $I_{2}(-1/4)$, $I_{1}^{*}(\infty)$
$EE=E/\langle(0,0)\rangle\hfill$
$Y^{2}=X^{3}-4T(2T^{2}+5T+1)X^{2}+4T^{2}(3T+1)^{3}X$
$I_{1}^{*}(0)$, $I_{6}(-1/3)$, $I_{2}(1)$, $I_{1}(-1/4)$, $I_{2}^{*}(\infty)$

The fibration $EE$ is a fibration of $Y_{10}$, since with the new parameter
$\frac{X}{(3T+1)^{3}}$, we get the rank $0$ elliptic fibration $E_{252}$, that
is the extremal elliptic fibration numbered $252$ in Shimada and Zhang’s paper
[17].

We also obtain

$E_{2}\qquad y^{2}=x^{3}-4t(t+1)(6t+5)x^{2}+4t^{2}(t+1)^{3}x$
---
$I_{0}^{*}(0)$, $I_{2}^{*}(-1)$, $2I_{1}(-8/9,-3/4)$
$EE_{2}=E_{2}/\langle(0,0)\rangle\hfill$
$Y^{2}=X^{3}+8T(T+1)(6T+5)X^{2}$
$+16T^{2}(T+1)^{2}(9T+8)(4T+3)X$
$I_{1}^{*}(-1)$, $I_{0}^{*}(0)$, $2I_{2}(-3/4,-8/9)$, $I_{1}^{*}(\infty)$

To prove that $EE_{2}$ is an elliptic fibration of $Y_{10}$ first we change
the parameter $T=1/u-1$ to get the new equation $EE_{2}(1)$

$EE_{2}(1)\qquad y^{2}=x^{3}+8u(u-1)(u-6)x^{2}+16u^{2}(u-4)(u-9)(u-1)^{2}x.$

Now with the new parameter $\frac{x}{u(u-1)(u-4)(u-9)}$, we obtain

$EE_{2}(2)\qquad y^{2}=x^{3}-t(59t^{2}-88t+32)x^{2}+32t^{2}(t-1))(3t-2)^{3}x.$

Again, from $EE_{2}(2)$, the parameter $\frac{x}{t^{2}(t-1)}$ leads to the
rank $0$ fibration $E_{252}$ of $Y_{10}$.

Finally, the fibration $EE14/\langle(100t^{2}+28t+1,0)\rangle$ with
Weierstrass equation

$y^{2}=x^{3}-2t(104t^{2}+28t+1)x^{2}+t^{2}(4t+1)^{2}(24t+1)^{2}x$

is a fibration of $Y_{10}$, since with the new parameter
$\frac{x}{t(4t+1)^{2}}$ we obtain $EE_{2}$.

∎

In the previous theorem we gave ”self-isogenies” of elliptic fibrations with
rank less than $2$. However we found in section 4 an interesting $2$-torsion
rank $4$ fibration. We present it in the following theorem.

######  Theorem 3.6.

The rank $4$ elliptic fibration of $Y_{10}$ (4.2) with singular fibers
$3I_{4}$, $3I_{2}$, $2III$, Weierstrass equation

$F\qquad y^{2}=x^{3}+4t^{2}x^{2}+t(t^{3}+1)^{2}x$

and its $2$-isogenous $F/\langle(0,0)\rangle$ are ”PF self-isogenous”.

###### Proof.

We get

$F/\langle(0,0)\rangle\qquad Y^{2}=X^{3}-8T^{2}X^{2}-4T(T^{3}-1)^{2}X$

with the same type of singular fibers. The isomorphism is given by

$T=-t,\qquad Y=-2\sqrt{-2}y,\qquad X=-2x.$

∎

###  3.5. Generators for specialization of #16 fibration on $Y_{10}$

The rank of the specializations for $k=10$ of generic elliptic fibrations
increases by one ([4], Theorem 4.1), so we have to determine one more
generator for the Mordell-Weil group. We give an example where the computation
is easy using a 2-isogeny between an elliptic fibration of $Y_{10}$ and an
elliptic fibration of the Kummer surface $K_{10}=\text{Kum}(E_{1},E_{2})$
associated to $Y_{10}$, where $E_{1},E_{2}$ are elliptic curves with complex
multiplication. Then using the method developped in [20] and [11], we
determine a section on an elliptic fibration of $K_{10}$.

From [3] Corollary 4.1, the two elliptic curves $E_{1}$ and $E_{2}$ have
respective invariants $j_{1}=8000$ and
$j_{2}=188837384000-77092288000\sqrt{6}.$ Take

$E_{1}:Y^{2}=X\left(X^{2}+4X+2\right)$

as a model of the first curve. The $2$-torsion sections have $X$-coordinates
$0,$ $-2\pm\sqrt{2},$ the $3-$ torsion sections have $X$-coordinates
$\frac{1}{3}\left(1\pm i\sqrt{2}\right)$ and $-1\pm\sqrt{6}$ that are roots of
$(3X^{2}-2X+1)(X^{2}+2X-5).$ The elliptic curve $E_{1}$ has complex
multiplication by $m_{2}=\sqrt{-2}$ defined by

$\left(X,Y\right)\overset{m_{2}}{\mapsto}\left(-\frac{1}{2}\frac{X^{2}+4X+2}{X},\frac{i\sqrt{2}}{4}\frac{Y\left(X^{2}-2\right)}{X^{2}}\right).$

Let $C_{3}$ and $\widetilde{C_{3}}$ the two groups of order $3$ generated by
the points of respective $X$-coordinates
$\frac{1}{3}\left(-2+i\sqrt{2}\right)$ and
$\frac{1}{3}\left(-2-i\sqrt{2}\right).$ These groups are fixed by $m_{2}$
while the two order $3$ groups $\Gamma_{3}$ and $\widetilde{\Gamma_{3}}$
generated by the points of respective $X$-coordinates $-2+\sqrt{6}$ and
$-2-\sqrt{6}$ are exchanged by $m_{2}.$

If $M=\left(X,Y\right)$ is a general point on $E_{1}$, the $3$-isogenous curve
by the isogeny $w_{3}$ of kernel $\Gamma_{3}$ is thus obtained with
$X_{2}=\sum_{S\in\Gamma_{3}}X_{M+S}+c$ and
$Y_{2}=\sum_{S\in\Gamma_{3}}Y_{M+S}$ where $c$ can be chosen so that the image
of $\left(0,0\right)$ is $X_{2}=0.$ It follows the $3$-isogeny $w_{3}$

$w_{3}:X_{2}=\frac{X\left(X-2-\sqrt{6}\right)^{2}}{\left(X+2-\sqrt{6}\right)^{2}},\quad
Y_{2}=-\frac{Y\left(X^{2}+\left(8-2\sqrt{6}\right)X+2\right)\left(X-2-\sqrt{6}\right)}{\left(X+2-\sqrt{6}\right)^{3}}$

and its $3$-isogenous curve $E_{2}$

$\displaystyle E_{2}$
$\displaystyle:Y_{2}^{2}=X_{2}^{3}+28X_{2}^{2}+\left(98+40\sqrt{6}\right)X_{2}$
$\displaystyle j\left(E_{2}\right)$
$\displaystyle=188837384000-77092288000\sqrt{6}.$

An equation for the Kummer surface $K_{10}$ is therefore

$K_{10}:X\left(X^{2}+4X+2\right)=y^{2}X_{2}\left(X_{2}^{2}+28X_{2}+98+40\sqrt{6}\right).$

####  3.5.1. Elliptic fibrations of $K_{10}$ and $Y_{10}$

We use the following units of $\mathbb{Q}\left(\sqrt{2},\sqrt{3}\right)$

$\displaystyle r_{1}$ $\displaystyle=1+\sqrt{2}+\sqrt{6},\quad
r_{1}^{\prime}=1-\sqrt{2}+\sqrt{6},\quad
r_{1}r_{1}^{\prime}=s=\left(\sqrt{2}+\sqrt{3}\right)^{2}$ $\displaystyle
r_{2}$ $\displaystyle=1-2\sqrt{3}-\sqrt{6},\quad
r_{2}^{\prime}=1+2\sqrt{3}-\sqrt{6}.$

In this paragraph we construct an elliptic fibration of $K_{10}$ giving after
a two-isogeny the specialization of the elliptic fibration #16 on $Y_{10}$.

We consider the fibration

$\displaystyle K_{10}$ $\displaystyle\rightarrow\mathbb{P}^{1}$
$\displaystyle\left(X,X_{2},y\right)$ $\displaystyle\mapsto
t=\frac{X_{2}}{X}.$

Notice that $X_{2}=tX$ and a Weierstrass equation $(K_{10})_{t}$ for this
fibration is obtained with the following transformation

$X=-\sqrt{2}\left(1+\sqrt{2}\right)\frac{X_{1}-2t\left(t-r_{1}^{2}\right)\left(t-r_{2}^{2}\right)}{X_{1}\left(3+2\sqrt{2}\right)-2t\left(t-r_{1}^{2}\right)\left(t-r_{2}^{2}\right)},\qquad
y=2\sqrt{2}\frac{X_{1}}{Y_{1}}$

(3.1)
$(K_{10})_{t}:Y_{1}^{2}=X_{1}\left(X_{1}-2t\left(t-r_{1}^{2}\right)\left(t-r_{1}^{\prime
2}\right)\right)\left(X_{1}-2t\left(t-r_{2}^{2}\right)\left(t-r_{2}^{\prime
2}\right)\right).$

The singular fibers are in $t=0$ and $\infty$ of type $I_{2}^{\ast}$ and at
$t=r_{1}^{2},r_{2}^{2},r_{1}^{\prime 2},r_{2}^{\prime 2}$ of type $I_{2}.$ The
rank of the Mordell-Weil group is $2.$

######  Remark 3.1.

We can show that $(K_{10})_{t}$ is the fibration of line $\\#16$ of Table 1 .
More precisely if $t=-st_{0}$ and $X=b^{2}x$, $Y=b^{3}y$ with
$b=\sqrt{2}(\sqrt{2}+\sqrt{3})^{3}$ we get exactly the fibration $\\#16$ with
parameter $t_{0}$.

####  3.5.2. Sections on the elliptic fibration $(K_{10})_{t}$

In many papers ([18] [19] [20] Th 1.2. and [8], [11]) results on the Mordell-
Weil lattice of the Inose fibration are given. We follow the same idea here,
with the previous fibration $(K_{10})_{t}$ of parameter $t$.

We find a section on this fibration using
$w_{3}\in\text{Hom}\left(E_{1},E_{2}\right).$ The graph of $w_{3}$ on
$E_{1}\times E_{2}$ and the image on $K_{10}=E_{1}\times E_{2}/\pm 1$
correspond to
$X_{2}=\frac{X\left(X-2-\sqrt{6}\right)^{2}}{\left(X+2-\sqrt{6}\right)^{2}}$
or $t=\frac{\left(X-2-\sqrt{6}\right)^{2}}{\left(X+2-\sqrt{6}\right)^{2}}.$ If
we consider the base-change of the fibration $u^{2}=t$ we obtain a section
defined by $u=\frac{\left(X-2-\sqrt{6}\right)}{\left(X+2-\sqrt{6}\right)}$ or
$X=\frac{\left(-2+\sqrt{6}\right)\left(u-s\right)}{u-1},$ that is
$P_{u}=\left(X_{1}\left(u\right),Y_{1}\left(u\right)\right)$ on the
Weierstrass equation $(K_{10})_{u^{2}}$

$\displaystyle X_{1}\left(u\right)$
$\displaystyle=\frac{2}{s_{2}^{2}s_{3}^{2}}u^{2}\left(u^{2}-r_{2}^{2}\right)\left(u+r_{1}\right)\left(u-r_{1}^{\prime}\right)$
$\displaystyle Y_{1}\left(u\right)$
$\displaystyle=2uX_{1}\left(u\right)\left(\left(\sqrt{3}-\sqrt{2}\right)u^{2}+\left(-2\sqrt{3}+\sqrt{2}\right)u+\sqrt{3}+\sqrt{2}\right),$

where $s_{2}=\frac{\sqrt{2}}{2}\left(-\sqrt{3}+1\right),s_{3}=\sqrt{2}+1$ . If
$\widetilde{P_{u}}=\left(X_{1}\left(-u\right),Y_{1}\left(-u\right)\right),$
then $\widetilde{P_{u}}\in(K_{10})_{u^{2}}$ and
$P=\widetilde{P_{u}}+P_{u}\in(K_{10})_{t}$, thus

$\displaystyle P$ $\displaystyle=\left(x_{P},y_{P}\right)$ $\displaystyle
x_{P}$
$\displaystyle=\frac{1}{s}\left(t+s\right)^{2}\left(t-r_{1}^{2}\right)\left(t-r_{1}^{\prime
2}\right),y_{P}=x_{P}\frac{2-\sqrt{6}}{2}\left(\frac{t-s}{t+s}\right)\left(t^{2}-14t-4\sqrt{6}t+s^{2}\right)$

so we recover an infinite section $P$ on the fibration $(K_{10})_{t}$ of the
Kummer surface $K_{10}$.

####  3.5.3. Sections on the fibration #16 of $Y_{10}$

The $2$-isogenous elliptic curve to (3.1) in the isogeny of kernel
$\left(0,0\right)$ has a Weierstrass equation

(3.2) $\displaystyle Y_{3}^{2}$
$\displaystyle=X_{3}\left(X_{3}^{2}+8\,t\left({t}^{2}-28\,t+{s}^{2}\right)X_{3}+64\,{\frac{{t}^{4}}{{s}^{2}}}\right)$
$\displaystyle X_{3}$
$\displaystyle=\left(\frac{Y_{1}}{X_{1}}\right)^{2},\qquad
Y_{3}=\frac{Y_{1}(B-X_{1}^{2})}{{X_{1}}}$

where $B$ is the coefficient of $X_{1}$ in (3.1). Singular fibers are in $t=0$
and $\infty$ of type $I_{4}^{\ast}$ and of type $I_{1}$ at
$t=r_{1}^{2},r_{2}^{2},r_{1}^{\prime 2},r_{2}^{\prime 2}$ .

Using the remark 3.1 this is fibration $\\#16$ of $Y_{10}$. The image of $P$
by this isogeny, in the Weierstrass equation (3.2), is
$Q=\left(\xi_{Q},\eta_{Q}\right)$ with

$\displaystyle\xi_{Q}$
$\displaystyle=\frac{1}{2s}\frac{\left(t^{2}-14t-4\sqrt{6}t+s^{2}\right)^{2}\left(t-s\right)^{2}}{(t+s)^{2}}$
$\displaystyle\eta_{Q}$
$\displaystyle=-\frac{\left(-2+\sqrt{6}\right)}{4s}\frac{\left(t^{2}-14t-4\sqrt{6}t+s^{2}\right)\left(t-s\right)L_{t}}{(t+s)^{3}}$
$\displaystyle\text{where}\quad L_{t}$
$\displaystyle=t^{6}+2\left(1+2\sqrt{6}\right)t^{5}-\left(993+404\sqrt{6}\right)t^{4}+\left(17820+7272\sqrt{6}\right)t^{3}$
$\displaystyle-\left(97137+39656\sqrt{6}\right)t^{2}+\left(56642+23124\sqrt{6}\right)t+s^{6}.$

Recall that by specialization of the generic case [3] we have also a point
$P^{\prime}=\left(\xi^{\prime},\eta^{\prime}\right)$ of $X_{3\text{
}}$-coordinate

$\displaystyle\xi^{\prime}$
$\displaystyle=-8\frac{t^{3}\left(t-1\right)^{2}}{\left(t-s^{2}\right)^{2}}$
$\displaystyle\eta^{\prime}$
$\displaystyle=\frac{i32}{19}\frac{\left(5\sqrt{2}+2\sqrt{3}\right)t^{4}\left(19t^{2}-(326+140\sqrt{6})t+931+380\sqrt{6}\right)}{\left(t-s^{2}\right)^{3}}.$

We verify using definitions that $<P^{\prime},Q>=0$ and
$h\left(P^{\prime}\right).h\left(Q\right)=18$. So by Shioda-Tate formula ([21]
[22]) $P^{\prime}$ and $Q$ and $(0,0)$ generate the Mordell-Weil group.

##  4\. $3$-isogenies from $Y_{2}$ and from $Y_{10}$

###  4.1. Generic 3-isogenies

In [3], Bertin and Lecacheux exhibited all the elliptic fibrations of a
generic member of the Apéry-Fermi pencil, called generic elliptic fibrations
and found two $3$-torsion elliptic fibrations defined by a Weierstrass
equation, namely $E_{\\#19}$ with rank $1$ and $E_{\\#20}$ with rank $0$. We
are giving their $3$-isogenous $K3$ surface.

######  Theorem 4.1.

The $3$-isogenous elliptic fibrations of fibration $\\#19$ (resp. $\\#20$)
defined by Weierstrass equations $H_{\\#19}(k)$ (resp. $H_{\\#20}(k)$) are
elliptic fibrations of the same $K3$ surface $N_{k}$ with transcendental
lattice $\big{(}\begin{smallmatrix}0&0&3\\\ 0&4&0\\\ 3&0&0\\\
\end{smallmatrix}\big{)}$ and discriminant form of its Néron-Severi lattice
$G_{NS}=\mathbb{Z}/3\mathbb{Z}(-\frac{2}{3})\oplus\mathbb{Z}/12\mathbb{Z}(\frac{5}{12})$.

###### Proof.

The $6$-torsion elliptic fibration $\\#20$ has a Weierstrass equation

$E_{\\#20}(k)\qquad y^{2}-(t^{2}-tk+3)xy-(t^{2}-tk+1)y=x^{3}$

with singular fibers $I_{12}(\infty)$, $2I_{3}(t^{2}-kt+1)$, $2I_{2}(0,k)$,
$2I_{1}(t^{2}-kt+9)$ and $3$-torsion point $(0,0)$. Using 2.5.2, it follows
the Weierstrass equation of its $3$-isogenous fibration

$H_{\\#20}(k)=E_{\\#20}(k)/\langle(0,0)\rangle\qquad
Y^{2}+3(t^{2}-tk+3)XY+t^{2}(t^{2}-tk+9)(t-k)^{2}Y=X^{3}$

with singular fibers $2I_{6}(0,k)$, $I_{4}(\infty)$, $2I_{3}(t^{2}-kt+9)$,
$2I_{1}(t^{2}-kt+1)$. Thus it is a rank $0$ and $6$-torsion elliptic fibration
of a $K3$-surface with Picard number $19$ and discriminant $\frac{6\times
6\times 3\times 3\times 4}{6\times 6}=12\times 3$.

Now we shall compute the Gram matrix $NS(20)$ of the Néron-Severi lattice of
the $K3$ surface with elliptic fibration $H_{\\#20}(k)$ in order to deduce its
discriminant form.

Applying Shioda’s result 2.3, we order the following elements as, $s_{0}$,
$F$, $\theta_{0,i}$, $1\leq i\leq 4$, $s_{3}$, $\theta_{k,i}$, $1\leq i\leq
5$, $\theta_{\infty,i}$, $1\leq i\leq 3$, $\theta_{t_{0},i}$, $1\leq i\leq 2$,
$\theta_{t_{1},i}$, $1\leq i\leq 2$, where $s_{0}$ and $s_{3}$ denotes
respectively the zero and $3$-torsion section, $F$ the generic section,
$\theta_{k,i}$ the components of reducible singular fiber s, $t_{0}$ and
$t_{1}$ being roots of $t^{2}-kt+9$. We obtain

$NS(20)=\left(\setcounter{MaxMatrixCols}{19}\begin{smallmatrix}-2&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
1&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&-2&1&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&1&-2&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&0&1&-2&1&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&0&0&1&-2&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&1&1&0&0&0&-2&1&0&0&0&0&0&1&0&0&1&0&1\\\
0&0&0&0&0&0&1&-2&1&0&0&0&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&1&-2&1&0&0&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&1&-2&1&0&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&0&1&-2&1&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&0&0&1&-2&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&0&0&0&0&-2&1&0&0&0&0&0\\\
0&0&0&0&0&0&1&0&0&0&0&0&1&-2&1&0&0&0&0\\\
0&0&0&0&0&0&0&0&0&0&0&0&0&1&-2&0&0&0&0\\\
0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&-2&1&0&0\\\
0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&1&-2&0&0\\\
0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&-2&1\\\
0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&1&-2\\\ \end{smallmatrix}\right).$

We get $\det(NS(20))=12\times 3$ and applying Shimada’s lemma 2.1, the
discriminant form $G_{NS(20)}\simeq\mathbb{Z}/3\oplus\mathbb{Z}/12$ is
generated by vectors $L_{1}$ and $L_{2}$ satisfying $q_{L_{1}}=0$,
$q_{L_{2}}=-\frac{11}{12}$ and $b(L_{1},L_{2})=\frac{1}{3}$. Denoting $M(20)$
the following Gram matrix of the lattice $U(3)\oplus\langle 4\rangle$,

$M(20)=\begin{pmatrix}0&0&3\\\ 0&4&0\\\ 3&0&0\\\ \end{pmatrix},$

we find for generators of its discriminant form the vectors

$g_{1}=\begin{pmatrix}0\\\ 0\\\ \frac{1}{3}\\\ \end{pmatrix}\qquad
g_{2}=\begin{pmatrix}\frac{1}{3}\\\ \frac{1}{4}\\\ \frac{1}{3}\\\
\end{pmatrix}$

satisfying $q_{g_{1}}=0$, $q_{g_{2}}=\frac{11}{12}$ and
$b(g_{1},g_{2})=\frac{1}{3}$. We deduce that $M(20)$ is the transcendental
lattice of the $K3$ surface with elliptic fibration $H_{\\#20}(k)$.

A Weierstrass equation of the $3$-torsion, rank $1$, elliptic fibration
$\\#19$ can be written as

$E_{\\#19}(k)\qquad y^{2}+ktxy+t^{2}(t^{2}+kt+1)y=x^{3}$

with singular fibers $2IV^{*}(0,\infty)$, $2I_{3}(t^{2}+kt+1)$,
$2I_{1}(k^{3}t-27kt-27t^{2}-27)$ and infinite order point $P=(-t^{2},-t^{2})$
of height $h(P)=\frac{4}{3}$. Using 2.5.2, its $3$-isogenous elliptic
fibration $E_{\\#19}(k)/\langle(0,0)\rangle$ has a Weierstrass equation

$H_{\\#19}(k)\qquad Y^{2}-3ktXY-Yt^{2}(27t^{2}-k(k^{2}-27)t+27)=X^{3}.$

It is a $3$-torsion, rank $1$, elliptic fibration of a $K3$-surface with
Picard number $19$ and singular fibers $2IV^{*}(0,\infty)$,
$2I_{3}(27t^{2}-k(k^{2}-27)t+27)$, $2I_{1}(t^{2}+kt+1)$ and infinite order
point $Q$ with $X$-coordinate $X_{Q}=-3-3kt-(k^{2}+3)t^{2}-3kt^{3}-3t^{4}$ and
height $h(Q)=4$. This point $Q$ is the image of the point $P$ in the
$3$-isogeny and non $3$-divisible, hence generator of the non torsion part of
the Mordell-Weil lattice. We deduce the discriminant of this $K3$-surface
$\frac{3\times 3\times 3\times 3\times 4}{3\times 3}=12\times 3$.

Applying Shioda’s result 2.3, we order the components of the singular fibers
as, $s_{0}$, $F$, $\theta_{0,i}$, $1\leq i\leq 6$, $\theta_{\infty,i}$, $1\leq
i\leq 6$, $\theta_{t_{0},i}$, $1\leq i\leq 2$, $s_{3}$, $\theta_{t_{1},2}$,
$s_{\infty}$, where $s_{0}$, $s_{3}$, $s_{\infty}$ denotes respectively the
zero, $3$-torsion and infinite section, $F$ the generic section, $t_{0}$ and
$t_{1}$ being roots of $27t^{2}-k(k^{2}-27)t+27$. The numbering of components
of $IV^{*}$ is done using Bourbaki’s notations [6]. It follows the Gram matrix
$NS(19)$ of the corresponding $K3$ surface

$NS(19)=\left(\setcounter{MaxMatrixCols}{19}\begin{smallmatrix}-2&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&1\\\
0&0&-2&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&0&-2&1&0&0&0&0&0&0&0&0&0&0&0&1&0&0\\\
0&0&0&1&-2&1&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&1&0&1&-2&1&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&0&0&0&1&-2&1&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&1&-2&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&-2&0&0&1&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&0&-2&1&0&0&0&0&0&1&0&0\\\
0&0&0&0&0&0&0&0&0&1&-2&1&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&1&0&1&-2&1&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&0&0&0&1&-2&1&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&0&0&0&0&1&-2&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&0&0&0&0&0&0&-2&1&0&0&0\\\
0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&-2&1&0&0\\\
0&1&0&1&0&0&0&0&0&1&0&0&0&0&0&1&-2&1&2\\\
0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&-2&0\\\
0&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&2&0&-2\ \end{smallmatrix}\right).$

Its determinant satisfies $\det(NS(19))=12\times 3$ and according to Shimada’s
lemma 2.1, $G_{NS(19)}\simeq\mathbb{Z}/3\oplus\mathbb{Z}/12$ is generated by
vectors $M_{1}$ and $M_{2}$ satisfying $q_{M_{1}}=-\frac{2}{3}$,
$q_{M_{2}}=\frac{5}{12}$ and $b(M_{1},M_{2})=0$. We find also generators for
the transcendental discriminant form $M(20)$

$h_{1}=\begin{pmatrix}\frac{1}{3}\\\ 0\\\ \frac{1}{3}\ \end{pmatrix}\qquad
h_{2}=\begin{pmatrix}\frac{1}{3}\\\ \frac{1}{4}\\\ -\frac{1}{3}\\\
\end{pmatrix}$

satisfying $q_{h_{1}}=\frac{2}{3}$, $q_{h_{2}}=-\frac{5}{12}$ and
$b(h_{1},h_{2})=0$. We deduce that $M(20)$ is also the transcendental lattice
of the $K3$ surface with elliptic fibration $H_{\\#19}(k)$.

It follows the discriminant form of its Néron-Severi lattice,

$G_{NS(19)}=\mathbb{Z}/3\mathbb{Z}(-\frac{2}{3})\oplus\mathbb{Z}/12\mathbb{Z}(\frac{5}{12})$,
which is also $G_{NS(20)}$ since the generators
$L^{\prime}_{1}=15L_{1}+4L_{2}$, $L^{\prime}_{2}=L_{1}-L_{2}$ satisfy
$q_{L^{\prime}_{1}}=-\frac{2}{3}$, $q_{L^{\prime}_{2}}=\frac{5}{12}$,
$b(L^{\prime}_{1},L^{\prime}_{2})=0$.

∎

We can prove the following specializations of $N_{k}$ for $k=2$ and $k=10$.

######  Theorem 4.2.

For $k=2$, the $K3$ surface $N_{2}$ is $Y_{10}$ with transcendental lattice
$[6\quad 0\quad 12]=T(Y_{2})[3]$.

For $k=10$, the $K3$ surface $N_{10}$ is the $K3$-surface with discriminant
$72$ and transcendental lattice $[4\quad 0\quad 18\ ]$.

###### Proof.

To prove that $N_{2}=Y_{10}$ it is sufficient to prove that $H_{\\#20}(2)$ is
an elliptic fibration of $Y_{10}$ since, by the previous theorem,
$H_{\\#19}(2)$ is another fibration of the same $K3$-surface.

But we see easily that $H_{\\#20}(2)$ is the $6$-torsion extremal fibration of
$Y_{10}$ numbered $8$ in Shimada and Zhang [17].

Similarly, $N_{10}$ is the $K3$ surface with transcendental lattice $[4\quad
0\quad 18]$ since $H_{\\#20}(10)$ is a fibration of that surface according to
the following proof.

A Weierstrass equation for $H_{\\#20}\left(10\right)$ has a Weierstrass
equation

$\displaystyle
Y^{2}+3\left(t^{2}-22\right)YX+\left(t^{2}-25\right)^{2}\left(t^{2}-16\right)Y=X^{3}$

with singular fibers $I_{4}\left(\infty\right),2I_{6}\left(\pm
5\right),2I_{3}\left(\pm 4\right),2I_{1}\left(t^{2}-24\right).$ We have a
$2$-torsion section
$s_{2}=\left(-\left(t^{2}-25\right)^{2},\left(t^{2}-25\right)^{3}\right)$ and
a $6$-torsion section

$s_{6}=\left(-\left(t^{2}-25\right)\left(t^{2}-16\right),-\left(t^{2}-25\right)\left(t^{2}-16\right)^{2}\right)$.

The section
$P_{w}=\left(4\left(t+5\right)^{2}\left(t-4\right),-\left(t+5\right)^{4}\left(t-4\right)^{2}\right)$
is of infinite order and is a generator of the Mordell Weil lattice. Moreover
we have $\theta_{\pm 5,1}.s_{6}=1,\theta_{\infty,2}.s_{6}=1,\theta_{\pm
3,1}.s_{6}=1$. So the following divisor is $6$ divisible

$\sum_{i=1}^{5}\left(6-i\right)\theta_{\pm
5,i}+3\theta_{\infty,1}+3\theta_{\infty,3}+4\theta_{\pm,4,1}+2\theta_{\pm
4,2}\approx 6s_{6}$

So we can replace $\theta_{5,5}$ by $s_{6}.$

Moreover we can compute $s_{6}.P_{w}=1$ (for $t=-3).$ We have also
$\theta_{5,0}.P_{w}=1,\theta_{-5,4}.P_{w}=1$ and
$\theta_{\infty,0}.P_{w}=1,\theta_{4,1}.P_{w}=1,\theta_{-4,0}.P_{w}=1.$ All
these computations give the Gram matrix of the Néron-Severi lattice of
discriminant $-72$

$\left(\setcounter{MaxMatrixCols}{20}\begin{smallmatrix}-2&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
1&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&1\\\
0&0&-2&1&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0\\\
0&0&1&-2&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&0&1&-2&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&0&0&1&-2&1&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&0&0&0&1&-2&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&-2&1&0&0&1&0&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&1&-2&1&0&0&0&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&1&-2&1&0&0&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&0&1&-2&0&0&0&0&0&0&0&0&1\\\
0&1&1&0&0&0&0&1&0&0&0&-2&0&1&0&1&0&1&0&1\\\
0&0&0&0&0&0&0&0&0&0&0&0&-2&1&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&0&0&0&1&1&-2&1&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&0&0&0&0&0&1&-2&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&-2&1&0&0&1\\\
0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&-2&0&0&0\\\
0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&-2&1&0\\\
0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&-2&0\\\
0&1&0&0&0&0&0&0&0&0&1&1&0&0&0&1&0&0&0&-2\end{smallmatrix}\right).$

According to Shimada’s lemma 2.1,
$G_{NS}=\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/36$ is generated by the vectors
$L_{1}$ and $L_{2}$ satisfying
$q_{L_{1}}=\frac{-1}{2},q_{L_{2}}=\frac{-35}{36}$ and
$b(L_{1},L_{2})=\frac{-1}{2}.$

Moreover the following generators of the discriminant group of the lattice
with Gram matrix $M_{18}=\left(\begin{smallmatrix}4&0\\\
0&18\end{smallmatrix}\right)$, namely
$f_{1}=(0,\frac{1}{2})$,$f_{2}=(\frac{1}{4},\frac{7}{18})$ verify
$q_{f_{1}}=\frac{1}{2}$, $q_{f_{2}}=\frac{35}{36}$ and
$b(f_{1},f_{2})=\frac{1}{2}.$

So the Gram matrix of the transcendental lattice is
$\left(\begin{array}[c]{cc}4&0\\\ 0&18\end{array}\right).$

######  Remark 4.1.

Instead of proving that $H_{\\#20}(10)$ is a fibration of the $K3$ surface
with transcendental lattice $[4\quad 0\quad 18]$, we may prove that
$H_{\\#19}(10)$ is a fibration of the same $K3$ surface.

We have the Weierstrass equation

$H_{\\#19}\left(10\right):Y^{2}-30tYX+t^{2}\left(t-27\right)\left(27t-1\right)Y=X^{3}$

with the two sections

$P^{\prime}=(-100t^{2}-3-3t(t+10)(1+t^{2}),\frac{1}{72}i\sqrt{3}(6t^{2}+10i\sqrt{3}t+30t+3i\sqrt{3}+3)^{3})$

image of the point $\left(-t^{2},-t^{2}\right)$ on $E_{\\#19}\left(10\right),$
and

$P^{\prime\prime}=\left(-\left(t+1\right)\left(27t-1\right),-\left(27t-1\right)\left(t+1\right)^{3}\right).$

The Néron-Severi lattice, with the following basis
$(s_{0},F,\theta_{\infty,i}$ $1\leq i\leq 6$ , $\theta_{0},i,1\leq i\leq 6$,
$\theta_{t_{0},1},\theta_{t_{0},2},$
$s_{3},\theta_{t_{1},2},P^{\prime},P^{\prime\prime})$ has for Gram matrix

$\left(\setcounter{MaxMatrixCols}{20}\begin{smallmatrix}-2&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&1&1\\\
0&0&-2&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&0&-2&1&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0\\\
0&0&0&1&-2&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&1&0&1&-2&1&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&0&0&0&1&-2&1&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&1&-2&0&0&0&0&0&0&0&0&0&0&0&1\\\
0&0&0&0&0&0&0&0&-2&0&0&1&0&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&0&-2&1&0&0&0&0&0&1&0&0&0\\\
0&0&0&0&0&0&0&0&0&1&-2&1&0&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&1&0&1&-2&1&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&0&0&0&1&-2&1&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&0&0&0&0&1&-2&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&0&0&0&0&0&0&-2&1&0&0&0&1\\\
0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&-2&1&0&0&0\\\
0&1&0&1&0&0&0&0&0&1&0&0&0&0&0&1&-2&1&2&1\\\
0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&-2&0&0\\\
0&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&2&0&-2&2\\\
0&1&0&0&0&0&0&1&0&0&0&0&0&0&1&0&1&0&2&-2\end{smallmatrix}\right).$

According to Shimada’s lemma 2.1,
$G_{NS}=\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/36$ is generated by the vectors
$L_{1}$ and $L_{2}$ satisfying
$q_{L_{1}}=\frac{-1}{2},q_{L_{2}}=\frac{13}{36}$ and
$b(L_{1},L_{2})=\frac{-1}{2}.$

Moreover the following generators of the discriminant group of the lattice
with Gram matrix $M_{18}=\left(\begin{smallmatrix}4&0\\\
0&18\end{smallmatrix}\right)$ namely
$f_{1}=(0,\frac{1}{2})$,$f_{2}=(\frac{1}{4},\frac{5}{18})$ verify
$q_{f_{1}}=\frac{1}{2}$, $q_{f_{2}}=-\frac{13}{36}$ and
$b(f_{1},f_{2})=\frac{1}{2}.$

So the transcendental lattice is $M_{18}=\left(\begin{array}[c]{cc}4&0\\\
0&18\end{array}\right).$

An alternative proof: From the equation $H_{\\#19}\left(10\right)$ and with
the parameter $m=\frac{Y}{\left(t-27\right)^{2}}$ we obtain another elliptic
fibration defined by the following cubic equation in $W$ and $t$ with
$X=W\left(t-27\right)$

$W^{3}+30Wtm-m\left(t^{2}\left(27t-1\right)+m\left(t-27\right)\right)=0$

and the rational point

$W=676\frac{m\left(m-1\right)}{m^{2}-648m+27},t=\frac{27m^{2}-648+1}{m^{2}-648m+27}.$

This fibration has a Weierstrass equation of the form

$y^{2}=x^{3}-3ax+\left(m+\frac{1}{m}-2b\right)$

with $a=38425/9,b=-7521598/27$. So the Kummer surface associated is the
product of two elliptic curves with $J,J^{\prime}=2950584125/27\pm
1204567000/27\sqrt{6}$ that is $j,j^{\prime}=188837384000\pm
77092288000\sqrt{6}.$ So the fibration $H_{\\#19}\left(10\right)$ corresponds
to a surface with transcendental lattice of Gram matrix
$\left(\begin{array}[c]{cc}4&0\\\ 0&18\end{array}\right).$

∎

######  Theorem 4.3.

Define $Y_{k}^{\left(3\right)}$ the elliptic surface obtained by the base
change $\tau$ of the elliptic fibration of $Y_{k}$ with two singular fibers of
type $II^{\ast},$ where $\tau$ is the morphism given by $u\mapsto h=u^{3}.$
Then the $K3$ surface $Y_{k}^{\left(3\right)}$ has a genus one fibration
without section such that its Jacobian variety satisfies
$J\left(Y_{k}^{\left(3\right)}\right)=N_{k}.$

###### Proof.

Recall a Weierstrass equation for fibration $\\#19$ (see [3], Table 4)

$(*)\qquad Y^{2}+tkYX+t^{2}\left(t+s\right)\left(t+1/s\right)Y=X^{3}$

where $k=s+\frac{1}{s}.$ The fibration of $Y_{k}$ with two singular fibers
$II^{\ast}$ can be obtained from (*) with the parameter
$h_{k}=\frac{Y}{\left(t+s\right)^{2}}$ ([3], Table 3). The surface
$Y_{k}^{\left(3\right)}$ is defined by $h_{k}=u^{3}$ and has then the
following equation

$u^{3}s\left(t+s\right)+tkuWs+t^{2}\left(ts-1\right)-W^{3}s=0,$

where $X=\left(t+s\right)uW$.

We consider the fibration

$\displaystyle Y_{k}^{\left(3\right)}$
$\displaystyle\rightarrow\mathbb{P}^{1}$ $\displaystyle\left(u,t,W\right)$
$\displaystyle\mapsto t;$

this is a genus one fibration since we have a cubic equation in $u,W.$

However, this fibration seems to have no section. Nevertheless, taking its
Jacobian fibration produces an elliptic fibration with section and the same
fiber type.

If we make a base change of this fibration: $\left(t+s\right)=m^{3}$ then we
obtain the following elliptic fibration with $U=um.$

$U^{3}sm-\left(s-m^{3}\right)ksWU-W^{3}sm-\left(s-m^{3}\right)^{2}\left(s^{2}-sm^{3}-1\right)m=0.$

The transformation

$\displaystyle W$
$\displaystyle=\frac{-24y+12(s^{2}+1)(s-m^{3})x+\left(s-m^{3}\right)^{2}Q}{18sm(4x-m^{2}(s^{2}+1)^{2})}$
$\displaystyle U$
$\displaystyle=\frac{-24y-12(s^{2}+1)(s-m^{3})x+\left(s-m^{3}\right)^{2}Q}{18sm(4x-m^{2}(s^{2}+1)^{2})}$

where
$Q=\left(108m^{6}s^{3}+(s^{6}+105s^{2}-111s^{4}-1)m^{3}+s(s^{2}+1)^{3}\right)$
gives a Weierstrass equation, the point $\pi_{3}$ of $x$ coordinate
$\frac{1}{4}m^{2}(s^{2}+1)^{2}$ is a $3$-torsion point. Taking again
$m^{3}=\left(t+s\right),$ and $\pi_{3}=\left(X=0,Y=0\right),$ we recover a
Weierstrass equation for the $3$-isogenous fibration $\\#19$

$Y^{2}-3tkYX-t^{2}\left(27t^{2}-k\left(k^{2}-27\right)t+27\right)Y=X^{3}$

hence a fibration of $N_{k}.$ Recall that the transcendental lattice of
$Y_{k}^{(3)}$ is $T(Y_{k})[3]$ [18]. ∎

###  4.2. 3-isogenies of $Y_{2}$

Recall the results, given in [2], about the $4$ elliptic fibrations of $Y_{2}$
with $3$-torsion.

$\begin{array}[c]{ccc}&\begin{array}[c]{c}\text{Weierstrass Equation}\\\
\text{Singular Fibers}\end{array}&\text{Rank}\\\
\hline\cr\\#20\left(7-w\right)&\begin{array}[c]{c}Y^{2}-\left(w^{2}+2\right)YX-w^{2}Y=X^{3}\\\
I_{12}\left(\infty\right),\quad I_{6}\left(0\right),\quad 2I_{2}\left(\pm
1\right),\quad 2I_{1}\end{array}&0\\\
\hline\cr\\#19\left(8-b\right)&\begin{array}[c]{c}Y^{2}+2bYX+b^{2}\left(b+1\right)^{2}Y=X^{3}\\\
2IV^{\ast}\left(\infty,0\right),\quad I_{6}\left(-1\right),\quad
2I_{1}\end{array}&1\\\ \hline\cr
20-j&\begin{array}[c]{c}Y^{2}-4\left(j^{2}-1\right)YX+4\left(j+1\right)^{2}Y=X^{3}\\\
I_{12}\left(\infty\right),\quad IV^{\ast}\left(-1\right),\quad
I_{2}\left(-\frac{1}{2}\right),\quad 2I_{1}\end{array}&0\\\ \hline\cr
21-c&\begin{array}[c]{c}Y^{2}+\left(c^{2}+5\right)YX+Y=X^{3}\\\
I_{18}\left(\infty\right),\quad 6I_{1}\end{array}&1\\\ \hline\cr\end{array}$

######  Theorem 4.4.

The $K3$ surface $Y_{2}$ has $4$ elliptic fibrations with $3$-torsion, two of
them being specializations. The $3$-isogenies induce elliptic fibrations of
$Y_{10}.$

###### Proof.

Using 2.5.2 we compute the $3$-isogenous elliptic fibrations named
$H_{w},H_{b}$ $,H_{j}$ and $H_{c}$ and given in the next table. To simplify we
denote $H_{w}$ (resp. $H_{b}$) the specialized elliptic fibration
$H_{\\#20}(2)$ (resp. $H_{\\#19}(2)).$

We know from Theorem 4.2 that $H_{w}$ and $H_{b}$ are elliptic fibrations of
$Y_{10}$; we present here a proof for $H_{j}$ and $H_{c}$ together with
remarks using ideas from [10], [8], [18],[19].

$\begin{array}[c]{cc}&\begin{array}[c]{c}\text{Weierstrass Equation}\\\
\text{Singular Fibers}\end{array}\\\ \hline\cr
H_{w}&\begin{array}[c]{c}Y^{2}+3\left(w^{2}+2\right)YX+\left(w^{2}+8\right)\left(w^{2}-1\right)^{2}Y=X^{3}\\\
I_{4}\left(\infty\right),\quad 2I_{6}\left(\pm 1\right),\quad
I_{2}\left(0\right),\quad 2I_{3}\end{array}\\\ \hline\cr
H_{b}&\begin{array}[c]{c}Y^{2}-6bYX+b^{2}\left(27b^{2}+46b+27\right)Y=X^{3}\\\
2IV^{\ast}\left(\infty,0\right),\quad 2I_{3},\quad
I_{2}\left(-1\right)\end{array}\\\ \hline\cr
H_{j}&\begin{array}[c]{c}Y^{2}+12\left(j^{2}-1\right)YX+4\left(4j^{2}-12j+11\right)\left(j+1\right)^{2}\left(2j+1\right)^{2}Y=X^{3}\\\
I_{4}\left(\infty\right),\quad IV^{\ast}\left(-1\right),\quad
I_{6}\left(\frac{-1}{2}\right),\quad 2I_{3}\end{array}\\\ \hline\cr
H_{c}&\begin{array}[c]{c}Y^{2}-3\left(c^{2}+5\right)YX-\left(c^{2}+2\right)\left(c^{2}+c+7\right)\left(c^{2}-c+7\right)Y=X^{3}\\\
I_{6}\left(\infty\right),\quad 6I_{3}\end{array}\\\ \hline\cr\end{array}$

Recall that the transcendental lattice of $Y_{10}$ is
$T\left(Y_{2}\right)[3]=T\left(Y_{10}\right).$ Notice that the surface $Y_{2}$
has an elliptic fibration with singular fibers
$2II^{\ast}\left(\infty,0\right),I_{2},2I_{1}$ and Weierstrass equation

$E_{h}\qquad
y^{2}=x^{3}-\frac{25}{3}x+h+\frac{1}{h}-\frac{196}{27}\quad\text{or}\quad
y^{2}=z^{3}-5z^{2}+\frac{(h+1)^{2}}{h}\quad\text{with}\quad x=z-\frac{5}{3}.$

The base change of degree $3,$ $h=u^{3}$ ramified at the two fibers
$II^{\ast}$ induces an elliptic fibration of the resulting $K3$ surface named
$Y_{2}^{\left(3\right)}$ in [10]. As the transcendental lattice of the surface
$Y_{2}^{\left(3\right)}$ is $T\left(Y_{2}\right)[3]$ [18], this surface
$Y_{2}^{\left(3\right)}$ is $Y_{10.}$ Moreover we can precise the fibration
obtained: a Weierstrass equation is

(4.1) $\displaystyle E_{u}$
$\displaystyle:Y^{2}=X^{3}-5u^{2}X^{2}+u^{3}(u^{3}+1)^{2}$ $\displaystyle
2I_{0}^{\ast}\left(\infty,0\right),3I_{2}\left(u^{3}+1\right),6I_{1}\text{
rank }7.$

Now we are going to show that every elliptic fibration of $Y_{2}$ with
$3$-torsion is linked to the elliptic fibration of $Y_{2}$ with
$2II^{\ast}\left(\infty,0\right),I_{2},2I_{1}.$

For the fibration $20-j$ we can obtain the elliptic fibration
$2II^{\ast}\left(\infty,0\right),I_{2},2I_{1}$ from the Weierstrass equation
given in the table and the elliptic parameter $h=Y.$ So the fibration $20-j$
induces a fibration on $Y_{10}$ with parameter $j$ and an equation obtained
after substitution of $Y$ by $u^{3}$. So, with the previous computations
2.5.3, this is the $3$-isogenous to $20-j.$

The same proof can be done for fibration $21-c.$

######  Remark 4.2.

Moreover we can remark using 2.5.3 that the $3$-isogenous to $21-c$ fibration
has an equation

$W^{3}+\left(c^{2}+5\right)ZW+1-Z^{3}=0.$

Since the general elliptic surface with
$\left(\mathbb{Z}/3\mathbb{Z}\right)^{2}$ torsion is
$x^{3}+y^{3}+t^{3}+3kxyt=0$, we deduce that the torsion on the fibration
$H_{c}$ induced on $Y_{10}$ is $\left(\mathbb{Z}/3\mathbb{Z}\right)^{2}.$

For the fibration $\\#19\left(8-b\right)$ we obtain the elliptic fibration
$2II^{\ast}\left(\infty,0\right),I_{2},2I_{1}$ from the Weierstrass equation
given in the table and the elliptic parameter
$h=\frac{Y}{\left(b+1\right)^{2}}.$ Substituing $h$ by $u^{3}$ and defining
$W$ as $X=\left(b+1\right)^{2}uW$ we obtain a cubic equation in $u$ and $W$
with a rational point $u=1,W=1,$ so an elliptic fibration of $Y_{10}$

$\displaystyle Y_{10}$ $\displaystyle\rightarrow\mathbb{P}_{1}$
$\displaystyle\left(u,b,W\right)$ $\displaystyle\mapsto b.$

Computation gives the $3$-isogenous elliptic curve to $\\#19\left(8-b\right).$

For the last fibration $\\#20\left(7-w\right)$ the relation with the fibration
$2II^{*},I_{2},2I_{1}$ is less direct. ∎

######  Remark 4.3.

With the previous method we can construct two elliptic fibrations of $Y_{10}$
of rank $4$.

First from Weierstrass equation $\\#20\left(7-w\right)$ and with the parameter
$m=Y$ we obtain the fibration $\\#1\left(11-f\right)$ of $Y_{2}$ ( [3] last
table) with singular fibers
$II^{\ast}\left(\infty\right),III^{\ast}\left(0\right),$
$I_{4}\left(1\right),I_{1}\left(\frac{32}{27}\right).$ A Weierstrass equation

$E_{m}:Y_{1}^{2}=X_{1}^{3}-m(2m-3)X_{1}^{2}+3m^{2}\left(m-1\right)^{2}X_{1}+m^{3}\left(m-1\right)^{4}$

is obtained with the following transformations

$\displaystyle m$ $\displaystyle=Y,\quad
X_{1}=-\frac{Y\left(Y-1\right)^{2}}{X+1},\quad
Y_{1}=w\frac{Y^{2}\left(Y-1\right)^{2}}{X+1}$ $\displaystyle w$
$\displaystyle=\frac{-Y_{1}}{X_{1}m},\quad
X=-\frac{X_{1}+m\left(m-1\right)^{2}}{X_{1}},\quad Y=m.$

The base change $m=u^{\prime 3}$ gives an elliptic fibration of $Y_{10}$ with
singular fibers $I_{0}^{\ast}\left(\infty\right)$, $III\left(0\right)$,
$3I_{4}\left(1,j,j^{2}\right)$, $3I_{1}$ , rank $4$, a Weierstrass equation
and sections

$\displaystyle{y^{\prime}}^{2}$ $\displaystyle=x^{\prime 3}+u^{\prime
2}x^{\prime 2}+2u^{\prime}\left(u^{\prime 3}-1\right)x^{\prime}+u^{\prime
3}\left(u^{\prime 3}-1\right)^{2}$ $\displaystyle P$
$\displaystyle=\left(x_{P}\left(u^{\prime}\right),y_{P}\left(u^{\prime}\right)\right)=\left(-\left(u^{\prime
3}-1\right),\left(u^{\prime}-1\right)^{2}\left(u^{\prime
2}+u^{\prime}+1\right)\right)$ $\displaystyle Q$
$\displaystyle=\left(x_{Q}\left(u^{\prime}\right),y_{Q}\left(u^{\prime}\right)\right)=\left(-\left(u^{\prime}+2\right)\left(u^{\prime
2}+u^{\prime}+1\right),2i\sqrt{2}\left(u^{\prime
2}+u^{\prime}+1\right)^{2}\right).$

Also we have the points $P^{\prime}$ with
$x_{P^{\prime}}=jx_{P}\left(ju^{\prime}\right)$ and $Q^{\prime}$ with
$x_{Q^{\prime}}=jx_{Q}\left(ju^{\prime}\right)$ (with $j^{3}=1$). As explained
in the next paragraph we can compute the height matrix and show that the
Mordell-Weil lattice is generated by $P,P^{\prime},Q,Q^{\prime}$ and is equal
to $A_{2}\left(\frac{1}{4}\right)\oplus A_{2}\left(\frac{1}{2}\right).$

The second example is obtained from $\\#20\left(7-w\right)$ with the parameter
$n=\frac{Y}{t^{2}}$ we obtain the fibration $\\#9\left(12-g\right)$ of $Y_{2}$
([3] last table)

$\displaystyle E_{n}$ $\displaystyle:y^{2}=x^{3}+4x^{2}n^{2}+n^{3}(n+1)^{2}x$

with the following transformation

$\displaystyle X$
$\displaystyle=\frac{x^{2}\left(x+2n^{2}\right)\left(n-1\right)}{y^{2}},Y=\frac{1}{n}\frac{x^{2}\left(x+2n^{2}\right)^{2}}{y^{2}},t=\frac{x\left(x+2n^{2}\right)}{ny}$
$\displaystyle x$
$\displaystyle=\frac{Y^{2}\left(Y-2X-t^{2}\right)}{Xt^{4}},y=\frac{Y^{3}\left(Y-t^{2}\right)\left(Y-2X-t^{2}\right)}{X^{2}t^{7}},n=\frac{Y}{t^{2}}.$

Notice that if $n=\frac{Y}{t^{2}}=v^{3}$ in $E_{w}$ then we have the equation
of $H_{w}.$ More precisely if $X=tQv$, the equation becomes

$-tv^{3}+t^{2}Qv+2Qv+Q^{3}+t=0,$

a cubic equation in $Q$ and $v$ with a rational point $v=1,Q=0.$ Easily we
obtain $H_{w}.$ So in the Weierstrass equation $E_{n}$, if we replace the
parameter $n$ by $g^{3}$ we obtain the following fibration of $Y_{10}$

(4.2) $\displaystyle y^{2}$
$\displaystyle=x^{3}+4g^{2}x^{2}+g\left(g+1\right)^{2}\left(g^{2}-g+1\right)^{2}x$

with singular fibers
$2III\left(0,\infty\right),3I_{4}\left(-1,g^{2}-g+1\right),3I_{1}\left(1,g^{2}+g+1\right)$
and rank $4.$ Notice the two infinite sections with $x$ coordinates
$(t+1)^{2}(t^{2}-t+1)$ and $-\frac{1}{3}(t-1)^{2}(t^{2}-t+1).$

###  4.3.  Mordell-Weil group of $E_{u}$

The aim of this paragraph is to construct generators of the Mordell-Weil
lattice of the previous fibration of rank $7$ with Weierstrass equation

$\displaystyle E_{u}$
$\displaystyle:Y^{2}=X^{3}-5u^{2}X^{2}+u^{3}(u^{3}+1)^{2}$ $\displaystyle
2I_{0}^{\ast}\left(\infty,0\right),3I_{2}\left(u^{3}+1\right),6I_{1}.$

Notice that the $j$-invariant of $E_{u}$ is invariant by the two
transformations $u\mapsto\frac{1}{u}$ and $u\mapsto ju.$ These automorphisms
of the base $\mathbb{P}^{1}$ of the fibration $E_{u}$ can be extended to the
sections as explained below.

Let $S_{3}=<\gamma,\tau;\gamma^{3}=1,\tau^{2}=1>$ be the non abelian group of
order $6$ and define an action of $S_{3}$ on the sections of $E_{u}$ by

$\displaystyle\left(X(u),Y(u)\right)\overset{\tau}{\mapsto}\left(u^{4}X\left(\frac{1}{u}\right),u^{6}Y\left(\frac{1}{u}\right)\right)$
$\displaystyle\left(X\left(u\right),Y\left(u\right)\right)\overset{\gamma}{\mapsto}\left(jX\left(ju\right),Y\left(ju\right)\right).$

To obtain generators of $E_{u}$ following Shioda [18] we use the rational
elliptic surface $X^{\left(3\right)+}$ with $\sigma=u+\frac{1}{u}$ and a
Weierstrass equation

$\displaystyle E_{\sigma}$
$\displaystyle:y^{2}=x^{3}-5x^{2}+\left(\sigma-1\right)^{2}\left(\sigma+2\right)$
$\displaystyle I_{0}^{\ast}\left(\infty\right),I_{2}\left(-1\right),4I_{1}$

of rank $3.$

The Mordell-Weil lattice of a rational elliptic surface is generated by
sections of the form
$(a+b\sigma+c\sigma^{2},d+e\sigma+f\sigma^{2}+g\sigma^{3}).$ Moreover since we
have a singular fiber of type $I_{0}^{\ast}$ at $\infty$ the coefficients $c$
and $f,g$ are $0$ [8]. So after an easy computation we find the $3$ sections
(with $j^{3}=1,i^{2}=-1).$

$\displaystyle q_{1}$
$\displaystyle=\left(-\left(\sigma-1\right),i\sqrt{2}\left(\sigma-1\right)\right)\quad$
$\displaystyle q_{2}$
$\displaystyle=\left(-j\left(\sigma-1\right),\left(3+j\right)\left(\sigma-1\right)\right)\quad
q_{3}=\left(-j^{2}\left(\sigma-1\right),\left(3+j^{2}\right)\left(\sigma-1\right)\right).$

These sections give the sections $\pi_{1},\pi_{2},\pi_{3}$ on $E_{u}$ which
are fixed by $\tau$.

$\begin{array}[c]{c}\pi_{1}=\left(-u\left(u^{2}-u+1\right),i\sqrt{2}u^{2}\left(u^{2}-u+1\right)\right)\\\
\pi_{2}=\left(-ju\left(u^{2}-u+1\right),\left(3+j\right)u^{2}\left(u^{2}-u+1\right)\right)\\\
\pi_{3}=\left(-j^{2}u\left(u^{2}-u+1\right),\left(3+j^{2}\right)u^{2}\left(u^{2}-u+1\right)\right).\end{array}$

We notice $\rho_{i}=\gamma\left(\pi_{i}\right)$ and
$\mu_{i}=\gamma^{2}\left(\pi_{i}\right)$ for $1\leq i\leq 3$ which give $9$
rational sections with some relations.

Moreover we have another section from the fibration $E_{h}$ of rank $1.$

The point of $x$ coordinate
$\frac{1}{16}\left(h^{2}+\frac{1}{h^{2}}\right)-h-\frac{1}{h}+\frac{29}{24}$
is defined on $\mathbb{Q}\left(h\right).$ Passing to $E_{u}$ we obtain
$\omega=$

$\left(\frac{1}{16}\frac{(1-16u^{3}+46u^{6}-16u^{9}+u^{12})}{u^{4}},-\frac{1}{64}\frac{(u^{6}-1)(1-24u^{3}+126u^{6}-24u^{9}+1)}{u^{6}}\right).$

We hope to get a generator system with $\pi_{i},\rho_{i}$ and $\omega$ so we
have to compute the height matrix. The absolute value of its determinant is
$\frac{81}{16}.$ Since the discriminant of the surface is $72,$ we obtain a
subgroup of index $a$ with $\frac{81}{16}\times\frac{1}{a^{2}}\times
2^{3}4^{2}=72$ so $a=3$.

After some specializations of $u$ $\in\mathbb{Z}$ (for example if $u=11$,
$E_{u}$ has rank $3$ on $\mathbb{Q})$ we find other sections with $x$
coordinate of the shape $\ \left(au+b\right)\left(u^{2}-u+1\right)$

$\displaystyle\mu$
$\displaystyle=\left(-\left(u-1\right)\left(u^{2}-u+1\right),-\left(u^{2}-u+1\right)\left(u^{2}+2u-1\right)\right)$
$\displaystyle\mu_{1}$
$\displaystyle=\left(-\left(u-9\right)\left(u^{2}-u+1\right),\left(u^{2}-u+1\right)\left(5u^{2}-18u+27\right)\right)$
$\displaystyle\mu_{2}$
$\displaystyle=\left(-\left(u+\frac{1}{3}\right)\left(u^{2}-u+1\right),\frac{i\sqrt{3}}{9}\left(u^{2}-u+1\right)\left(9u^{2}+4u+1\right)\right).$

We deduce the relations

$\displaystyle 3\mu$
$\displaystyle=\omega+\pi_{2}-\gamma\left(\pi_{3}\right)+\pi_{3}-\gamma^{2}\left(\pi_{2}\right)$
$\displaystyle=\omega+2\pi_{2}+\gamma\left(\pi_{2}\right)+\pi_{3}-\gamma\left(\pi_{3}\right)$

so, the Mordell-Weil lattice is generated by
$\pi_{j},\rho_{j}=\gamma\left(\pi_{j}\right)$ for $1\leq j\leq 3$ and $\mu$
with Gram matrix

$\left(\begin{array}[c]{ccccccc}1&-\frac{1}{2}&0&0&0&0&0\\\
-\frac{1}{2}&1&0&0&0&0&0\\\ 0&0&1&-\frac{1}{2}&0&0&\frac{1}{2}\\\
0&0&-\frac{1}{2}&1&0&0&0\\\ 0&0&0&0&1&-\frac{1}{2}&\frac{1}{2}\\\
0&0&0&0&-\frac{1}{2}&1&\frac{-1}{2}\\\
0&0&\frac{1}{2}&0&\frac{1}{2}&\frac{-1}{2}&2\end{array}\right).$

###  4.4. A fibration for Theorem 3.2

From the previous fibration $E_{u}$ we construct by a $2$-neigbour method a
fibration with a $2$-torsion section used in Theorem 3.2.

We start from the Weierstrass equation (4.1)

$Y^{2}=X^{3}-5u^{2}X^{2}+u^{3}\left(u^{3}+1\right)^{2}$

and obtain another elliptic fibration with the parameter
$m=\frac{X}{u\left(u^{2}-u+1\right)}$, which gives the Weierstrass equation

(4.3) $\displaystyle E_{m}$
$\displaystyle:y^{2}=x^{3}-\left(m^{3}+5m^{2}-2\right)x^{2}+\left(m^{3}+1\right)^{2}x$

with singular fibers $I_{0}^{\ast}\left(\infty\right)$,
$3I_{4}\left(m^{3}+1\right)$, $I_{2}\left(0\right)$,
$4I_{1}\left(1,-\frac{5}{3},m^{2}-4m-4\right)$ and rank $4.$

######  Remark 4.4.

From this fibration with the parameter $q=\frac{y}{xm}$ we recover the
fibration $H_{c.}$

###  4.5. $3$-isogenies from $Y_{10}$

######  Theorem 4.5.

Consider the two $K3$ surfaces of discriminant $72$ and of transcendental
lattice $[4\quad 0\quad 18]$ or $[2\quad 0\quad 36]$. There exist elliptic
fibrations of $Y_{10}$ with a $3$-torsion section inducing by $3$-isogeny
elliptic fibrations of one or the other previous surface.

###### Proof.

In Bertin and Lecacheux [4] we observe that the fibration numbered $89$ of
rank $0$ has a $3$-torsion section. A Weierstrass equation for the
$3$-isogenous fibration is

$\displaystyle
Y^{2}+\left(-27t^{2}-18t+27\right)YX+27\left(4t+3\right)\left(5t-3\right)^{2}Y=X^{3}$

with singular fibers
$I_{9}\left(\infty\right),I_{6}\left(\frac{3}{5}\right),I_{4},\left(0\right),I_{3}\left(-\frac{3}{4}\right),2I_{1}.$

From singular fibers, torsion and rank we see in Shimada and Zhang table [17]
that it is the $n^{\circ}48$ case. So the transcendental lattice of the
surface is $\left(\begin{smallmatrix}4&0\\\ 0&18\end{smallmatrix}\right).$

In Bertin and Lecacheux [4] is given also a rank $2$ elliptic fibration of
$Y_{10}$ numbered $(11)$ with a $3$-torsion section. We shall prove that this
$3$-torsion section induces by $3$-isogeny an elliptic fibration of the $K3$
surface with transcendental lattice $[4\quad 0\quad 18]$.

Starting with the Weierstrass equation given in [4], after a translation to
put the $3$-torsion section in $\left(0,0\right)$ we obtain the following
Weierstrass equation and $p_{1}$ and $p_{2}$ generators of the Mordell-Weil
lattice

(4.4)
$\begin{array}[c]{c}E_{11}:Y^{2}+\left(t^{2}-4\right)YX+t^{2}\left(2t^{2}-3\right)Y=X^{3}\\\
p_{1}=(6t^{2},27t^{2}),\qquad p_{2}=(6i\sqrt{3}t-3t^{2},27t^{2})\\\
2I_{6}\left(\infty,0\right),\quad 2I_{3}\left(2t^{2}-3\right),\quad
2I_{2}\left(\pm 1\right),\quad 2I_{1}\left(\pm 8\right).\end{array}$

We see that the $3$-isogenous elliptic fibration has a Weierstrass equation,
generators of Mordell-Weil lattice and singular fibers

$\displaystyle H_{11}:$ $\displaystyle
Y^{2}-3\left(t^{2}-4\right)YX-\left(t^{2}-1\right)^{2}\left(t^{2}-64\right)Y=X^{3}$
$\displaystyle 2I_{6}\left(\pm 1\right),\quad 2I_{3}\left(\pm 8\right),\quad
2I_{2}\left(\infty,0\right),\quad 2I_{1}\left(2t^{2}-3\right)\text{ of rank
}2.$

Notice the two sections

$\displaystyle\pi_{1}$
$\displaystyle=\left(-\frac{1}{4}\left(t^{2}-1\right)\left(t^{2}-64\right),\frac{1}{8}\left(t-8\right)\left(t-1\right)\left(t+1\right)^{2}\left(t+8\right)^{2}\right)\quad$
$\displaystyle\omega$
$\displaystyle=\left(-7\left(t^{2}-1\right)^{2},49\alpha\left(t^{2}-1\right)^{3}\right)\quad\text{where
}49\alpha^{2}+20\alpha+7=0.$

So, computing the height matrix of $\pi_{1}$ and $\omega$, we see the
discriminant is $72.$

For each reducible fiber at $t=i$ we denote $\left(X_{i},Y_{i}\right)$ the
singular point of $H_{11}$

$\begin{array}[c]{ccc}t=\pm 1\quad t=\pm 8&t=0&t=\infty\\\ (X_{\pm 1}=0,Y_{\pm
1}=0)&(X_{0}=-16,Y_{0}=64)&(x_{\infty}=-1,y_{\infty}=-1)\\\ (X_{\pm
8}=0,Y\pm_{8}=0)&&\end{array}$

where if $t=\infty$ we substitute $t=\frac{1}{T}$, $x=T^{4}X,y=T^{6}Y.$ We
notice also $\theta_{i,j}$ the $j$-th component of the reducible fiber at
$t=i.$ A section $M=\left(X_{M},Y_{M}\right)$ intersects the component
$\theta_{i,0}$ if and only if
$\left(X_{M},Y_{M}\right)\not\equiv\left(X_{i},Y_{i}\right)\operatorname{mod}\left(t-i\right).$
Using the additivity on the component, we deduce that $\omega$ does not
intersect $\theta_{i,0},2\omega$ intersects $\theta_{i,0}$ and so $\omega$
intersects $\theta_{i,3}$ for $i=\pm 1.$ Also $\omega$ intersects
$\theta_{i,0}$ for $i=\pm 8$ $,i=0$ and $\infty.$

For $\pi_{1}$ we compute $k\pi_{1}$ with $2\leq k\leq 6.$ For $i=\pm 1$, only
$6\pi_{1}$ intersects $\theta_{i,0}$ so $\pi_{1}$ intersects $\theta_{i,1}.$
(this choice 1, not 5, fixes the numbering of components). For $i=\pm 8$, only
$3\pi_{1}$ intersects $\theta_{i,0},$ so $\pi_{i}$ intersects $\theta_{i,1}.$
Modulo $t$, we get $\pi_{1}=\left(-16,64\right),$ so $\pi_{1}$ intersects
$\theta_{0,1},$ and $\pi_{1}$ intersects $\theta_{\infty,0}$.

As for the $3$-torsion section $s_{3}=\left(0,0\right)$, $s_{3}$ intersects
$\theta_{i,2}$ or $\theta_{i,4}$ if $i=\pm 1$. Computing $2\pi_{1}-s_{3}$, we
see that $s_{3}$ intersects $\theta_{1,2}$ and $\theta_{-1,4}$.

For $i=\pm 8$, we compute $\pi_{1}-s_{3},$ for $t=8$ and show that $s_{3}$
intersects $\theta_{8,2}$ and $\theta_{-8,1}.$ For $t=0$ and $t=\infty\ s_{3}$
intersects the $0$ component.

So we can compute the relation between the section $s_{3}$ and the
$\theta_{i,j}$ and find that
$3s_{3}\approx-2\theta_{1,1}-4\theta_{1,2}-3\theta_{1,3}-2\theta_{1,4}-\theta_{1,5}$.
Thus, we can choose the following base of the Néron-Severi lattice ordered as
$s_{0},F,\theta_{1,j},$ with $1\leq j\leq 4,$ $s_{3},\theta_{-1,k}$ with
$1\leq k\leq 5,\theta_{8,k},k=1,2,$ $\theta_{-8,k},k=1,2$ and $\theta_{0,1},$
$\theta_{\infty,1},\omega,\pi_{1}.$

The last remark is that only the two sections $\omega$ and $\pi_{1}$
intersect. So we can write the Gram matrix $NS$ of the Néron-Severi lattice,

$\left(\setcounter{MaxMatrixCols}{20}\begin{smallmatrix}-2&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
1&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&1&1\\\
0&0&-2&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1\\\
0&0&1&-2&1&0&1&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&0&1&-2&1&0&0&0&0&0&0&0&0&0&0&0&0&1&0\\\
0&0&0&0&1&-2&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&1&0&1&0&0&-2&0&0&0&1&0&0&1&1&0&0&0&0&0\\\
0&0&0&0&0&0&0&-2&1&0&0&0&0&0&0&0&0&0&0&1\\\
0&0&0&0&0&0&0&1&-2&1&0&0&0&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&1&-2&1&0&0&0&0&0&0&0&1&0\\\
0&0&0&0&0&0&1&0&0&1&-2&1&0&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&0&0&1&-2&0&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&0&0&0&0&-2&1&0&0&0&0&0&1\\\
0&0&0&0&0&0&1&0&0&0&0&0&1&-2&0&0&0&0&0&0\\\
0&0&0&0&0&0&1&0&0&0&0&0&0&0&-2&1&0&0&0&1\\\
0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&-2&0&0&0&0\\\
0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&-2&0&0&1\\\
0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&-2&0&0\\\
0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&-2&1\\\
0&1&1&0&0&0&0&1&0&0&0&0&1&0&1&0&1&0&1&-2\end{smallmatrix}\right).$

According to Shimada’s lemma 2.1,
$G_{NS}\equiv\mathbb{Z/}2\mathbb{Z}\oplus\mathbb{Z/}36\mathbb{Z}$ is generated
by the vectors $L_{1}$ and $L_{2}$ satisfying $q_{L_{1}}=-\frac{1}{2},$
$q_{L_{2}}=\frac{37}{36},$ and $b(L_{1},L_{2})=\frac{1}{2}.$

Moreover the following generators of the discriminant group of the lattice
with Gram matrix $M_{18}=\left(\begin{smallmatrix}4&0\\\
0&18\end{smallmatrix}\right)$ namely $f_{1}=(0,\frac{1}{2})$,
$f_{2}=(\frac{1}{4},\frac{7}{18})$ verify $q_{f_{1}}=\frac{1}{2}$,
$q_{f_{2}}=-\frac{37}{36}$ and $b(f_{1},f_{2})=-\frac{1}{2}$.

So the Gram matrix of the transcendental lattice of the surface is $M_{18}$.

A $3$-isogeny between two elliptic fibrations of $Y_{10}$ and the $K3$ surface
with transcendental lattice $[2\quad 0\quad 36]$ can also be obtained. We have
shown in remark 4.2 that $H_{c}$ has a $(Z/3Z)^{2}$\- torsion group and
exhibited a 3-isogeny between some elliptic fibrations of $Y_{10}$ and
$Y_{2}$. Notice that with the Weierstrass equation $H_{c}$ the point
$\sigma_{3}$ of $X$ coordinate $-\left(c^{2}+c+7\right)\left(c^{2}-c+7\right)$
defines a $3$-torsion section. After a translation to put this point in
$(0,0)$ and scaling, we obtain a Weierstrass equation

$Y^{\prime
2}-(t^{2}+11)Y^{\prime}X^{\prime}-(t^{2}+t+7)(t^{2}-t+7)Y^{\prime}=X^{\prime
3}.$

The $3$-isogenous curve of kernel $<\sigma_{3}>$ has a Weierstrass equation

$\displaystyle y^{2}+3\left(t^{2}+11\right)xy+\left(t^{2}+2\right)^{3}y$
$\displaystyle=x^{3},$

with singular fibers $I_{2}\left(\infty\right)$, $2I_{9}\left(t^{2}+2\right)$,
$4I_{1}\left(t^{4}+13t^{2}+49\right).$

The section
$P_{c}=\left(-\frac{1}{4}\left(t^{4}+t^{2}+1\right),-\frac{1}{8}\left(t-j\right)^{3}\left(t+j^{2}\right)^{3}\right)$
where $j=\frac{-1+i\sqrt{3}}{2},$ of infinite order, generates the Mordell-
Weil lattice.

We consider the components of the reducible fibers in the following order
$\theta_{i\sqrt{2},j}$, $j\leq 1\leq 8,$ $\theta_{-i\sqrt{2},k}$, $1\leq k\leq
8$, $\theta_{\infty,1}$.

The $3$-torsion section $s_{3}=\left(0,0\right)$ and the previous components
are linked by the relation

$3s_{3}\approx-\theta_{i\sqrt{2},8}+\sum a_{i,j}\theta_{i,j}.$

So we can replace, in the previous ordered sequence of components, the element
$\theta_{i\sqrt{2},8}$ by $s_{3}.$ We notice that
$\left(s_{3}.P_{c}\right)=2,\left(P_{c}.\theta_{\pm i\sqrt{2},0}\right)=1$ and
$\left(P_{c}.\theta_{\infty,0}\right)=1,$ so the Gram matrix of the Néron-
Severi lattice $NS$ is

$\left(\setcounter{MaxMatrixCols}{20}\begin{smallmatrix}-2&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
1&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&1\\\
0&0&-2&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&1&-2&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&0&1&-2&1&0&0&0&1&0&0&0&0&0&0&0&0&0&0\\\
0&0&0&0&1&-2&1&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&0&0&0&1&-2&1&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&1&-2&1&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&1&-2&0&0&0&0&0&0&0&0&0&0&0\\\
0&1&0&0&1&0&0&0&0&-2&0&0&1&0&0&0&0&0&0&2\\\
0&0&0&0&0&0&0&0&0&0&-2&1&0&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&0&0&1&-2&1&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&0&1&0&1&-2&1&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&0&0&0&0&1&-2&1&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&0&0&0&0&0&1&-2&1&0&0&0&0\\\
0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&-2&1&0&0&0\\\
0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&-2&1&0&0\\\
0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&-2&0&0\\\
0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&-2&0\\\
0&1&0&0&0&0&0&0&0&2&0&0&0&0&0&0&0&0&0&-2\end{smallmatrix}\right).$

Its determinant is $-72.$ According to Shimada’s lemma 2.1,
$G_{NS}=\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/36$ is generated by the vectors
$L_{1}$ and $L_{2}$ satisfying $q_{L_{1}}=\frac{-1}{2},q_{L_{2}}=\frac{5}{36}$
and $b(L_{1},L_{2})=\frac{1}{2}.$

Moreover the following generators of the discriminant group of the lattice
with Gram matrix $M_{36}=\left(\begin{smallmatrix}2&0\\\
0&36\end{smallmatrix}\right)$, namely $f_{1}=(\frac{1}{2},0)$,
$f_{2}=(\frac{-1}{2},\frac{-7}{36})$ verify $q_{f_{1}}=\frac{1}{2}$,
$q_{f_{2}}=-\frac{5}{36}$ and $b(f_{1},f_{2})=\frac{-1}{2}.$

So the transcendental lattice is $M_{36}=\left(\begin{array}[c]{cc}2&0\\\
0&36\end{array}\right).$

∎

##  5\. Isogenies and $L$-series of singular $K3$ surfaces

We notice that, along all the previous computations, the discriminants of the
$K3$ surface $Y_{2}$ (resp. $Y_{10}$) and their $2$ or $3$ -isogenous $K3$
surfaces are the same up to square. It is indeed a corollary of the following
theorem about the $L$-series of a singular $K3$ surface and some of its $2$-or
$3$-isogenous.

Let us recall first the following results.

######  Theorem 5.1 (Tate’s isogeny theorem).

[23] The fact that two elliptic curves $E_{1}$ and $E_{2}$ defined over
$\mathbb{F}_{q}$ are isogenous is equivalent to the fact they have the same
number of $\mathbb{F}_{q}$ points.

######  Lemma 5.1.

[1] Let $Y$ an elliptic $K3$-surface defined over $\mathbb{Q}$ by a
Weierstrass equation $Y(t)$.

If rank $(Y(t))=r$ and the $r$ infinite sections generating the Mordell-Weil
lattice are defined respectively over $\mathbb{Q}(\sqrt{d_{i}})$, $i=1,...,r$,
then

$A_{p}=-\sum_{t\in\mathbb{P}^{1}(\mathbb{F}_{p}),\quad
Y(t)\quad{\text{smooth}}}a_{p}(t)-\sum_{t\in\mathbb{P}^{1}(\mathbb{F}_{p}),\quad
Y(t)\quad{\text{singular}}}\epsilon_{p}(t)-\sum_{i=1}^{r}\left(\frac{d_{i}}{p}\right)p$

where

$a_{p}(t)=p+1-\\#Y(t)(\mathbb{F}_{p})$

and $\epsilon_{p}(x)$ defined by

$\epsilon_{p}(t)=\begin{cases}0,&\text{if the reduction of $Y(t)$ is
additive}\\\ 1,&\text{if the reduction of $Y(t)$ is split multiplicative}\\\
-1,&\text{if the reduction of $Y(t)$ is non split multiplicative}\end{cases}.$

######  Theorem 5.2.

The $L$-series of the transcendental lattice of a singular $K3$ surface $Y$
defined over $\mathbb{Q}$ is inchanged by a $2$ or a $3$-isogeny whose kernel
is defined over $\mathbb{Q}(t)$ and obtained from an elliptic fibration whose
infinite sections (if any) are defined on $\mathbb{Q}$ or on a quadratic
number field.

###### Proof.

Denote $Y(t)$ (resp. $\widetilde{Y(t)}$) a Weierstrass equation of a singular
$K3$ surface (resp. of its $2$ or $3$ isogenous curve).

The coefficients of the newform associated to the L-series of the K3 surface
are given in the previous lemma.

1) Suppose $\widetilde{Y(t)}$ is the Weierstrass equation of its
$2$-isogenous.

We get

$(Y(t))\qquad y^{2}=x^{3}+a(t)x^{2}+b(t)x\qquad\qquad(\widetilde{Y(t)})\qquad
Y^{2}=X^{3}-2a(t)X^{2}+(a(t)^{2}-4b(t))X$

and

$(X,Y)=(\frac{x^{2}+a(t)x+b(t)}{x},y\frac{b(t)-x^{2}}{x^{2}}).$

Hence, since the ranks of $Y(t)$ and its $2$-isogenous are the same, if $Y(t)$
has $r$ infinite sections defined on $\mathbb{Q}(\sqrt{d_{i}})$ it is similar
for $\widetilde{Y(t)}$.

If $t\in\mathbb{P}^{1}(\mathbb{F}_{p})$ satisfies $Y(t)$ smooth, $t$ is not a
root of $\Delta=16b^{2}(a^{2}-4b)$ and also not a root of
$\widetilde{\Delta}=256b(a^{2}-4b)^{2}$; hence $\widetilde{Y(t)}$ is also
smooth. For these $t$, using Tate’s isogeny theorem, we find
$a_{p}(t)=\widetilde{a_{p}(t)}$.

Suppose $Y(t_{0})$ singular i.e. either $b(t_{0})=0$ or
$a(t_{0})^{2}=4b(t_{0})$ (in these cases the reduction of $Y(t_{0})$ is
multiplicative).

Suppose $b(t_{0})=0$, we get

$(Y(t_{0}))\qquad
y^{2}=x^{2}(x+a(t_{0}))\qquad\qquad(\widetilde{Y(t_{0})})\qquad
Y^{2}=(X-a(t_{0}))^{2}X=U^{2}(U+a(t_{0})).$

Hence $Y(t_{0})$ and $\widetilde{Y(t_{0})}$ have the same multiplicatice
reduction, either split if $a(t_{0})$ is a square modulo $p$ or non split if
$a(t_{0})$ is not a square modulo $p$.

Suppose now $a(t_{0})^{2}=4b(t_{0})$, we get

$(Y(t_{0}))\qquad
y^{2}=x(x+a(t_{0})/2)^{2}=U^{2}(U-a(t_{0})/2)\qquad\qquad(\widetilde{Y(t_{0})})\qquad
Y^{2}=X^{2}(X-2a(t_{0})).$

Similarly, if $-a(t_{0})/2$ or equivalently $-2a(t_{0})$ is a square (resp.
not a square) modulo $p$, the reduction is split (resp. non split)
multiplicative. Thus $Y(t_{0})$ and $\widetilde{Y(t_{0})}$ have the same type
of multiplicative reduction.

Finally when both $a(t_{0})=0$ and $b(t_{0})=0$, the reduction of $Y(t_{0})$
and $\widetilde{Y(t_{0})}$ is additive.

Thus we have proved that $A_{p}=\widetilde{A_{p}}$, that is the $2$-isogenous
$K3$ surface has the same transcendental $L$-series as $Y$.

2) Suppose $\widetilde{Y(t)}$ is the Weierstrass equation of its
$3$-isogenous.

Since we want to apply Tate’s isogeny theorem we need a $3$-isogeny defined
over $\mathbb{Q}$ whose kernel is defined over $\mathbb{Q}(t)$. Using the
formulae of $2.5.1$, we get

$(Y(t))\qquad y^{2}+a(t)xy+b(t)y=x^{3}$ $(\widetilde{Y(t)})\qquad
Y^{2}+a(t)XY+3b(t)Y=X^{3}-6a(t)b(t)X-b(t)(a(t)^{3}+9b(t))$

and

$X=\frac{x^{3}+a(t)b(t)x+b(t)^{2}}{x^{2}}$
$Y=\frac{y(x^{3}-a(t)b(t)x-2b(t)^{2})-b(t)(x^{3}+a(t)^{2}x^{2}+2a(t)b(t)x+b(t)^{2})}{x^{3}}.$

Hence, since the ranks of $Y(t)$ and its $3$-isogenous curve are the same, if
$Y(t)$ has $r$ infinite sections defined on $\mathbb{Q}(\sqrt{d_{i}})$ it is
similar for $\widetilde{Y(t)}$.

If $t\in\mathbb{P}^{1}(\mathbb{F}_{p})$ satisfies $Y(t)$ smooth, $t$ is not a
root of $\Delta=b^{3}(a^{3}-27b)$ and also not a root of
$\widetilde{\Delta}=b(a^{3}-27b)^{3}/16$; hence $\widetilde{Y(t)}$ is also
smooth. For these $t$, using Tate’s isogeny theorem, we find
$a_{p}(t)=\widetilde{a_{p}(t)}$.

Suppose $Y(t_{0})$ singular i.e. either $b(t_{0})=0$ or
$a(t_{0})^{3}=27b(t_{0})$ (in these cases the reduction of $Y(t_{0})$ is
multiplicative).

Suppose $b(t_{0})=0$. We get

$(Y(t_{0}))\qquad
y^{2}+a(t_{0})xy=x^{3}\qquad\qquad\widetilde{(Y(t_{0}))}\qquad
Y^{2}+a(t_{0})XY=X^{3}.$

Hence the two curves have the same multiplicative reduction.

Suppose $b(t_{0})=a(t_{0})^{3}/27$. Putting at the origin the singular point
$(-\frac{a(t_{0})^{2}}{9},\frac{a(t_{0})^{3}}{27})$ of $Y(t_{0})$ (resp.
$(-\frac{a(t_{0})^{2}}{3},\frac{a(t_{0})^{3}}{9})$ of $\widetilde{Y(t_{0})}$,
it follows

$(Y(t_{0}))\qquad
y_{1}^{2}+a(t_{0})x_{1}y_{1}=x_{1}^{3}-a(t_{0})^{2}x_{1}^{2}/3\qquad(\widetilde{(Y(t_{0}))}\qquad
y_{2}^{2}+a(t_{0})x_{2}y_{2}=x_{2}^{3}-a(t_{0})^{2}x_{2}^{2}.$

Since their respective discriminants are $x_{1}^{2}(4x_{1}-a(t_{0})^{2}/3)$
and $x_{2}^{2}(4x_{2}-3a(t_{0})^{2})$, the two curves have the same
multiplicative reduction.

Thus we have proved that $A_{p}=\widetilde{A_{p}}$, that is the $3$-isogenous
$K3$ surface has the same transcendental $L$-series as $Y$.

∎

######  Corollary 5.1.

A singular $K3$ surface $Y$ as in Theorem 5.2 and its $2$ or $3$-isogenous
surface have their discriminants equal up to square.

###### Proof.

This is a consequence of a Schütt’s theorem.

######  Theorem 5.3.

(Schütt’s classification) [15] Consider the following classification of
singular $K3$-surfaces over $\mathbb{Q}$

  1. (1)

by the discriminant $d$ of the transcendental lattice of the surface up to
squares,

  2. (2)

by the associated newform up to twisting,

  3. (3)

by the level of the associated newform up to squares,

  4. (4)

by the $CM$-field $\mathbb{Q}(\sqrt{-d})$ of the associated newform.

Then, all these classifications are equivalent. In particuliar,
$\mathbb{Q}(\sqrt{-d})$ has exponent $1$ or $2$.

∎

###  5.1. Isogenies as isometries of the rational transcendental lattice

Denoting the rational transcendental lattice
$T(X)_{\mathbb{Q}}:=T(X)\otimes\mathbb{Q}$, we recall that $T(X)_{\mathbb{Q}}$
and $T(Y)_{\mathbb{Q}}$ are isometric if they define congruent lattices, that
is if there exists $M\in Gl_{n}(\mathbb{Q})$ satisfying
$T(X)_{\mathbb{Q}}=^{t}MT(Y)_{\mathbb{Q}}M$.

Bessière, Sarti and Veniani proved the following theorem [5].

######  Theorem 5.4.

[5] Let $\gamma:X\rightarrow Y$ be a $p$-isogeny between complex projective
$K3$ surfaces $X$ and $Y$. Then
$\text{rk}(T(Y)_{\mathbb{Q}})=\text{rk}(T(X)_{\mathbb{Q}})=:r$ and

  1. (1)

If $r$ is odd, there is no isometry between $T(Y)_{\mathbb{Q}}$ and
$T(X)_{\mathbb{Q}}$.

  2. (2)

If $r$ is even, there exists an isometry between $T(Y)_{\mathbb{Q}}$ and
$T(X)_{\mathbb{Q}}$ if and only if $T(Y)_{\mathbb{Q}}$ is isometric to
$T(Y)_{\mathbb{Q}}(p)$. This property is equivalent to the following:

a) If $p=2$, for every prime number $q$ congruent to $3$ or $5$ modulo $8$,
the $q$-adic valuation $\nu_{q}(\det T(Y))$ is even.

b) If $p>2$, for every prime number $q>2$, $q\neq p$, such that $p$ is not a
square in $\mathbb{F}_{q}$, the number $\nu_{q}(det(T_{Y})$ is even and the
following equation holds in $\mathbb{F}_{p}^{*}/(\mathbb{F}_{p}^{*})^{2}$

$res_{p}(det(T_{y})=(-1)^{\frac{p(p-1)}{2}+\nu_{p}(det(T_{Y}))}$

where $res_{p}(det(T_{Y}))=\frac{det(T_{Y})}{p^{\nu_{p}(det(T_{Y}))}}$.

This theorem allows us to find $2$-isogenies as self isogenies on $Y_{2}$ and
$Y_{10}$. In a previous paper we gave all the $2$-isogenies of $Y_{2}$ and
exhibited self isogenies on $Y_{2}$. In section $3$ we also exhibited
$2$-isogenies as self isogenies on $Y_{10}$.

In section $4$ we proved that all the $3$\- isogenies on $Y_{2}$ are between
$Y_{2}$ and $Y_{10}$ and some $3$-isogenies on $Y_{10}$ are between $Y_{10}$
and other $K3$ surfaces with dicriminant $72$, namely $[4\quad 0\quad 18]$ or
$[2\quad 0\quad 36]$. These results illustrate Bessière, Sarti, Veniani’s
theorem. Indeed $det(T(Y_{2}))=8$, hence $res_{3}(8)=8$ which is congruent
modulo $3$ to $(-1)^{3}$ and $det(T(Y_{10}))=8\times 9$, hence
$res_{3}(8\times 9)=8$ which is congruent modulo $3$ to $(-1)^{3+2}$. And,
since

$T_{\mathbb{Q}}(Y_{2})=\begin{pmatrix}2&0\\\ 0&1\\\ \end{pmatrix}\qquad
T_{\mathbb{Q}}(Y_{10})=\begin{pmatrix}6&0\\\ 0&3\\\ \end{pmatrix}\qquad
T_{\mathbb{Q}}([4\quad 0\quad 18])=\begin{pmatrix}1&0\\\ 0&2\\\
\end{pmatrix}\qquad T_{\mathbb{Q}}([2\quad 0\quad 36])=\begin{pmatrix}2&0\\\
0&1\\\ \end{pmatrix}$

we find, as expected, these matrices are isometric since

$\begin{pmatrix}6&0\\\ 0&3\\\ \end{pmatrix}=\begin{pmatrix}2&1\\\ -1&1\\\
\end{pmatrix}\begin{pmatrix}1&0\\\ 0&2\\\ \end{pmatrix}\begin{pmatrix}2&-1\\\
1&1\\\ \end{pmatrix}\qquad\begin{pmatrix}2&0\\\ 0&1\\\
\end{pmatrix}=\begin{pmatrix}0&1\\\ 1&0\\\ \end{pmatrix}\begin{pmatrix}1&0\\\
0&2\\\ \end{pmatrix}\begin{pmatrix}0&1\\\ 1&0\\\ \end{pmatrix}.$

Some remarks

As a consequence of Bessière, Sarti and Veniani’s theorem, there could be $2$
or $3$-self (either PF or EF) elliptic fibrations on $Y_{2}$ and on $Y_{10}$.
Indeed we found $2$-self isogenies on both $Y_{2}$ and $Y_{10}$. As for
$3$-isogenies, there is no self-isogeny on $Y_{2}$ and also probably none on
$Y_{10}$. Concerning rank $0$ elliptic fibrations, using Shimada and Zhang’s
table [17], we recover easily all our results without using Weierstrass
equations. We have only to know the transform by a $2$\- or a $3$-isogeny of a
type of singular fiber. This can be obtained using Tate’s algorithm [24] and
an analog of Dockchitzer’s remark [7].

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