# Examples of surfaces with canonical map of degree 4

Carlos Rito  Universidade de Trás-os-Montes e Alto Douro, Quinta de Prados,
5000-801 Vila Real, Portugal<EMAIL_ADDRESS>
  *     * scAbstract. We give two examples of surfaces with canonical map of degree 4 onto a canonical surface.

scKeywords. algebraic surface; surface of general type; abelian cover;
canonical map

sc2020 Mathematics Subject Classification. 14J29

  *
cFebruary 4, 2022Received by the Editors on June 22, 2021\.
Accepted on March 4, 2022.

Universidade de Trás-os-Montes e Alto Douro, Quinta de Prados, 5000-801 Vila
Real, Portugal

sce-mail<EMAIL_ADDRESS>
This research was partially supported by FCT — Fundação para a Ciência e a
Tecnologia, I.P. — under the fellowship SFRH/BPD/111131/2015 and CMUP (project
with reference UIDB/00144/2020).

© by the author(s) This work is licensed under
http://creativecommons.org/licenses/by-sa/4.0/

###### Contents

  1. 1 Introduction
  2. 2 $(\mathbb Z/2)^{r}$-coverings
  3. 3 Construction
  4. 4 Computations

## 1\. Introduction

Let $S$ be a smooth minimal surface of general type with geometric genus
$p_{g}\geq 3$. Denote by ${\phi:S\dashrightarrow\mathbb P^{p_{g}-1}}$ the
canonical map and let $d:=\deg(\phi).$ The following result of Beauville is
well-known.

######  Theorem 1.1 ([Bea79]).

If the canonical image $\Sigma:=\phi(S)$ is a surface, then either:

  1. $(A)$

$p_{g}(\Sigma)=0,$ or

  2. $(B)$

is a canonical surface (i.e. it is the canonical image of a surface with
birational canonical map), in particular $p_{g}(\Sigma)=p_{g}(S)$.

Moreover, in case $(A)$ $d\leq 36$ and in case $(B)$ $d\leq 9.$

The question of which pairs $(d,p_{g})$ can actually occur has been object of
study for some authors. Several examples were given for case $(A),$ but case
$(B)$ is still mysterious. It is known that if $d>3,$ then $p_{g}\leq 12,$ but
so far only the case $(d,p_{g})=(5,4)$ has been shown to exist (independently
by Tan [Tan92] and by Pardini [Par91b]). We refer the recent preprint by
Mendes Lopes and Pardini [MLP21] for a more detailed account on the subject.
They leave some open problems, this note is motivated by their last question.

######  Question.

For what pairs $(d,p_{g})$, with $d>3,$ are there examples of surfaces in case
$(B)$ of Theorem 1.1?

Here we give examples for the cases $(d,p_{g})=(4,5)$ and $(4,7)$, with
canonical images a 40-nodal complete intersection surface in $\mathbb P^{4}$
and a 48-nodal complete intersection surface in $\mathbb P^{6},$ respectively
(Beauville also paid some attention to such nodal surfaces, see [Bea17]).

The strategy for the construction is the following. If $X$ is a surface with
nodes admitting a Galois covering $Y\rightarrow X$ ramified over the nodes and
with Galois group $G$, a group with a “big” number of subgroups, then we have
a “big” number of intermediate coverings of $X.$ By computing the geometric
genus $p_{g}$ of all involved surfaces, we may hope to find some
$\rho:W\rightarrow Z$ with $p_{g}(W)=p_{g}(Z),$ hence such that the canonical
map of $W$ factors through $\rho.$

We work explicitely with the equations of a 40-nodal surface from [RRS19], all
computations are implemented with Magma [BCP97].

### Notation

As usual the holomorphic Euler characteristic of a surface $S$ is denoted by
$\chi(S),$ the geometric genus by $p_{g}(S),$ the irregularity by $q(S),$ and
a canonical divisor by $K_{S}.$ A $(-m)$-curve is a curve isomorphic to
$\mathbb P^{1}$ with self-intersection $-m$. A node of $S$ is an ordinary
double point of $S.$ We say that a set of nodes of $S$ is 2-divisible if the
sum $\sumop\displaylimits A_{i}$ of the corresponding $(-2)$-curves in the
smooth minimal model of $S$ is 2-divisible in the Picard group.

## 2\. $(\mathbb Z/2)^{r}$-coverings

The following result is taken from [Cat08, Proposition 7.6]. See also
[Par91a].

######  Proposition 2.1.

A normal finite $G\cong(\mathbb Z/2)^{r}$-covering $\pi:Y\rightarrow X$ of a
smooth variety $X$ is completely determined by the datum of

  1. (1)

reduced effective divisors $D_{\sigma},$ for all $\sigma\in G,$ with no common
components;

  2. (2)

divisor linear equivalence classes $L_{\chi_{1}},\ldots,L_{\chi_{r}},$ for
$\chi_{1},\ldots,\chi_{r}$ a basis of the group of characters $G^{\vee},$ such
that

$2L_{\chi_{i}}\equiv\sumop\displaylimits_{\chi_{i}(\sigma)=1}D_{\sigma}$

$($with additive notation for the characters$\,)$.

Conversely, given (1) and (2), one obtains a normal scheme $Y$ with a finite
$G\cong(\mathbb Z/2)^{r}$-covering $Y\rightarrow X,$ with branch curves the
divisors $D_{\sigma}.$

The scheme $Y$ is irreducible if $\\{\sigma|D_{\sigma}>0\\}$ generates $G$. We
have a splitting

$\pi_{*}\mathcal{O}_{Y}=\bigoplusop\displaylimits_{\chi\in
G^{\vee}}L_{\chi}^{-1}.$

From now on, we assume that $X$ and $Y$ are surfaces. If each $D_{\sigma}$ is
smooth and $\sumop\displaylimits D_{\sigma}$ has simple normal crossings, then
$Y$ is smooth and its invariants are

(2.1)
$\begin{split}\chi(\mathcal{O}_{Y})&=2^{r}\chi(\mathcal{O}_{X})+\frac{1}{2}\sumop\displaylimits_{\chi\in
G^{{\vee}*}}\left(L_{\chi}^{2}+K_{X}\cdot L_{\chi}\right),\\\
p_{g}(Y)&=p_{g}(X)+\sumop\displaylimits_{\chi\in
G^{{\vee}*}}h^{0}(X,\mathcal{O}_{X}(K_{X}+L_{\chi})).\end{split}$

Let $R_{\sigma}$ be the support of $\pi^{*}(D_{\sigma})$. The Hurwitz formula
gives

$K_{Y}\equiv\pi^{*}(K_{X})+\sumop\displaylimits_{\sigma\in G^{*}}R_{\sigma}.$

Now assume that the $D_{\sigma}$ are disjoint $(-2)$-curves. Then the
$R_{\sigma}$ are disjoint $(-1)$-curves, the canonical map of $Y$ factors
through the covering $Y\rightarrow X$ if and only if $p_{g}(Y)=p_{g}(X),$ and
one has a commutative diagram

${Y}$${Y^{\prime}}$${X}$${X^{\prime}}$

where $Y\rightarrow Y^{\prime}$ is the contraction of the $(-1)$-curves
$R_{\sigma},$ the surface $X^{\prime}$ has nodes corresponding to the
$(-2)$-curves of $X,$ and $Y^{\prime}\rightarrow X^{\prime}$ is a $(\mathbb
Z/2)^{r}$-covering ramified on those nodes. In this case Equation (2.1)
becomes

(2.2) $\chi(\mathcal{O}_{Y})=2^{r}\left(\chi(\mathcal{O}_{X})-m/8\right)$

where $m$ is the number of nodes of $X^{\prime}.$

## 3\. Construction

Let $X_{40}$ be the surface in $\mathbb P^{4}$ given by the equations

(3.1) $\displaystyle
5\left(x^{2}+y^{2}+z^{2}+w^{2}+t^{2}\right)-7\left(x+y+z+w+t\right)^{2}$
$\displaystyle=0$ $\displaystyle
4\left(x^{4}+y^{4}+z^{4}+w^{4}+t^{4}+h^{4}\right)-\left(x^{2}+y^{2}+z^{2}+w^{2}+t^{2}+h^{2}\right)^{2}$
$\displaystyle=0$

where

$h:=-(x+y+z+w+t).$

It is the canonical model of a surface with invariants $p_{g}=5,$ $q=0$ and
$K^{2}=8.$ The above quartic $I$ is classically known as the Igusa quartic;
its singular set is the union of 15 lines. The quadric meets these lines
transversally, and is tangent to $I$ at 10 smooth points, thus the singular
set of $X_{40}$ is the union of 40 nodes $N_{1},\ldots,N_{40}$ (for more
details see [RRS19]).

Let $\widetilde{X}_{40}$ be the smooth minimal model of $X_{40}$ and denote by
$A_{i}$ the $(-2)$-curves in $\widetilde{X}_{40}$ corresponding to the nodes
$N_{i},$ $i=1,\ldots,40.$ Let $a,b,c$ be the canonical generators of the group
$(\mathbb Z/2)^{3}$ and, for $i,j,k\in\mathbb Z/2,$ let $\chi_{ijk}$ denote
the character which takes the value $i,j,k$ on $a,b,c,$ respectively. We show
in Section 4.1 that one can write

$A_{1}+\cdots+A_{40}=D_{a}+D_{b}+D_{c}+D_{abc}+D_{bc}+D_{ac}+D_{ab}$

where each of $D_{a},D_{b},D_{c},D_{abc}$ is a sum of $4$ $(-2)$-curves, each
of $D_{bc},D_{ac},D_{ab}$ is a sum of $8$ $(-2)$-curves, and such that there
exist divisors $L_{100},L_{010},L_{001}$ satisfying:

(3.2) $\displaystyle D_{a}+D_{abc}+D_{ac}+D_{ab}$ $\displaystyle\equiv
2L_{100}$ $\displaystyle D_{b}+D_{abc}+D_{bc}+D_{ab}$ $\displaystyle\equiv
2L_{010}$ $\displaystyle D_{c}+D_{abc}+D_{bc}+D_{ac}$ $\displaystyle\equiv
2L_{001}.$

It follows from Proposition 2.1 that these data define a $(\mathbb
Z/2)^{3}$-covering $\pi:\widetilde{Y}\rightarrow\widetilde{X}_{40}$ branched
on the $(-2)$-curves $A_{i},$ equivalently a $(\mathbb Z/2)^{3}$-covering
$\psi:Y\rightarrow X_{40}$ branched on the nodes of $X_{40}$ (the surface $Y$
is minimal because $X_{40}$ is minimal and $\psi$ is étale in codimension 1).
In particular there exist divisors $L_{111},L_{110},L_{101},L_{011}$ such
that:

(3.3) $\displaystyle D_{a}+D_{b}+D_{c}+D_{abc}$ $\displaystyle\equiv 2L_{111}$
$\displaystyle D_{a}+D_{b}+D_{bc}+D_{ac}$ $\displaystyle\equiv 2L_{110}$
$\displaystyle D_{a}+D_{c}+D_{bc}+D_{ab}$ $\displaystyle\equiv 2L_{101}$
$\displaystyle D_{b}+D_{c}+D_{ac}+D_{ab}$ $\displaystyle\equiv 2L_{011}.$

One has

$2L_{ijk}\equiv\sumop\displaylimits_{\chi_{ijk}(\sigma)=1}D_{\sigma}.$

Since $\psi$ is ramified only on nodes, we have
$K_{Y}\equiv\psi^{*}(K_{X_{40}})$ and then $K_{Y}^{2}=8K_{X_{40}}^{2}=64.$ We
show in Section 4.1 that

$h^{0}\left({\widetilde{X}_{40}},\mathcal{O}_{\widetilde{X}_{40}}\left(K_{\widetilde{X}_{40}}+L_{111}\right)\right)=2$

and

$h^{0}\left({\widetilde{X}_{40}},\mathcal{O}_{\widetilde{X}_{40}}\left(K_{\widetilde{X}_{40}}+L_{ijk}\right)\right)=0\
\ {\rm for}\ \ {ijk}\neq 111,$

thus

$p_{g}(Y)=p_{g}(X_{40})+2+0+\cdots+0=7.$

We get from (2.2) that $\chi(Y)=8(6-5)=8,$ thus $q(Y)=0.$

The covering $\psi$ factors as

${Y}$${Y_{32}}$${Y_{48}}$${X_{16}}$${X_{32}}$${X_{40}}$

with $Y_{48}$ and $X_{16}$ given by the quotients by the groups $\langle
ab,ac\rangle$ and $\langle c\rangle,$ respectively (the subscript $n$ means a
surface with singular set the union of $n$ nodes). All these surfaces are
regular because $q(Y)=0.$

It follows from (2.2) that $\chi(X_{16})=4(6-36/8)=6,$ thus
$p_{g}(X_{16})=p_{g}(X_{40})=5,$ and we conclude that

the $(\mathbb Z/2)^{2}$-covering $X_{16}\rightarrow X_{40}$ is the canonical
map of $X_{16}.$

Analogously, $p_{g}(Y)=p_{g}(Y_{48})=7$ and we claim that

the $(\mathbb Z/2)^{2}$-covering $Y\rightarrow Y_{48}$ is the canonical map of
$Y.$

For this it suffices to show that $Y_{48}$ is a canonical surface.

Since the canonical system of $Y_{48}$ contains the pullback of the canonical
system of $X_{40}$ and since $p_{g}(Y_{48})>p_{g}(X_{40}),$ the canonical map
of $Y_{48}$ must be birational. But we can be more precise. We follow
Beauville [Bea17] and show that $Y_{48}$ can be embedded in $\mathbb P^{6}$ as
a complete intersection of 4 quadrics in the following way. The linear system
$L$ of quadrics through the branch locus of the covering $Y_{48}\rightarrow
X_{40}$ (16 nodes) is of dimension 2. Using computer algebra it is not
difficult to show that $L$ contains quadrics $B,C,D$ such that the surface
$X_{40}$ is given by $Q=0,$ $B^{2}-CD=0,$ where $Q$ is the quadric from (3.1)
(we write the quadrics as general elements of $L$, thus depending on some
parameters; then we obtain a variety on these parameters by imposing that the
hypersurfaces $Q=0$ and $B^{2}-CD=0$ are tangent at the 24 nodes of $X_{40}$
which are disjoint from the 16 nodes of $B^{2}-CD=0$; finally we compute
points in this variety).

Then $Y_{48}$ is given in $\mathbb P^{6}(x,y,z,w,t,u,v)$ by equations

$u^{2}-C=v^{2}-D=uv-B=Q=0.$

We give these equations in Section 4.2 and verify that $Y_{48}$ is as stated.

Let us explain how we find 2-divisible sets of nodes in $X_{40}.$ The surface
$X_{40}$ contains 40 tropes, which are hyperplane sections $H_{i}=2T_{i}$ with
$T_{i}\subset X_{40}$ a reduced curve through 12 nodes of $X_{40},$ and smooth
at these points. Thus in $\widetilde{X}_{40}$ the pullback of such a trope can
be written as

$\widetilde{H}_{i}=2\widehat{T}_{i}+\sumop\displaylimits_{j\in J}A_{j}\,,\
\text{with}\ \\#J=12.$

Thus for each pair of tropes the sum of nodes contained in their union and not
contained in their intersection is 2-divisible.

Using these 2-divisibilities, the strategy for finding configurations as in
(3.2) is simple: we have used a computer algorithm to list and check
possibilities.

## 4\. Computations

The computations below are implemented with Magma V2.26-5.

### 4.1. The covering $Y\rightarrow X_{40}$

We start by defining the surface $X_{40}$ and its singular set.

    K:=Rationals();
    R<r>:=PolynomialRing(K);
    K<r>:=ext<K|r^2 + 15>;
    P<x,y,z,w,t>:=ProjectiveSpace(K,4);
    h:=-x-y-z-w-t;
    Q:=5*(x^2+y^2+z^2+w^2+t^2)-7*(x+y+z+w+t)^2;
    I:=4*(x^4+y^4+z^4+w^4+t^4+h^4)-(x^2+y^2+z^2+w^2+t^2+h^2)^2;
    X40:=Surface(P,[Q,I]);
    SX40:=SingularSubscheme(X40);

The partition of the 40 nodes:

    Da:={P![3,3,-2,-2,3],P![4,-r+1,r-5,-r+1,4],
         P![-r+1,4,r-5,-r+1,4],P![r+1,r+1,-r-5,4,4]};
    Db:={P![2,-3,-3,-3,2],P![4,r+1,r+1,-r-5,4],
         P![-r-5,r-5,r-5,-r-5,10],P![r-5,-r+1,-r+1,4,4]};
    Dc:={P![-3,-3,2,-3,2],P![-r+1,-r+1,r-5,4,4],
         P![r-5,r-5,-r-5,-r-5,10],P![r+1,r+1,4,-r-5,4]};
    Dabc:={P![-2,3,3,-2,3],P![-r-5,r+1,r+1,4,4],
           P![r-5,4,-r+1,-r+1,4],P![r-5,-r+1,4,-r+1,4]};
    Dbc:={P![-2,-2,3,3,3],P![3,-2,-2,3,3],
          P![4,-r-5,r+1,r+1,4],P![4,-r+1,-r+1,r-5,4],
          P![4,r+1,-r-5,r+1,4],P![-r-5,r+1,4,r+1,4],
          P![-r+1,-r+1,4,r-5,4],P![r+1,-r-5,4,r+1,4]};
    Dac:={P![-3,2,-3,-3,2],P![3,-2,3,-2,3],
          P![4,r-5,-r+1,-r+1,4],P![-r+1,r-5,4,-r+1,4],
          P![-r+1,r-5,-r+1,4,4],P![r-5,-r-5,r-5,-r-5,10],
          P![r+1,4,r+1,-r-5,4],P![r+1,-r-5,r+1,4,4]};
    Dab:={P![-3,-3,-3,2,2],P![-2,3,-2,3,3],
          P![-r-5,4,r+1,r+1,4],P![-r+1,4,-r+1,r-5,4],
          P![-r-5,-r-5,r-5,r-5,10],P![-r-5,r-5,-r-5,r-5,10],
          P![r-5,-r-5,-r-5,r-5,10],P![r+1,4,-r-5,r+1,4]};

Verification that these are in fact the nodes:

    &join[Da,Db,Dc,Dabc,Dbc,Dac,Dab] eq SingularPoints(X40);
    HasSingularPointsOverExtension(X40) eq false;

Some of the tropes of $X_{40}:$

    tropes:=[
        6*x + (-r - 9)*y + (r - 9)*z + (r - 9)*w + (-r - 9)*t,
        16*x + (-r - 9)*y + 16*z + (3*r + 11)*w + (3*r + 11)*t,
        16*x + (r - 9)*y + 16*z + (-3*r + 11)*w + (-3*r + 11)*t,
        6*x + (r - 9)*y + (-r - 9)*z + (r - 9)*w + (-r - 9)*t,
        16*x + (3*r + 11)*y + 16*z + (3*r + 11)*w + (-r - 9)*t,
        16*x + (-3*r + 11)*y + (-3*r + 11)*z + (r - 9)*w + 16*t,
        x + y + w,
        16*x + (r - 9)*y + (-3*r + 11)*z + (-3*r + 11)*w + 16*t,
        x + z + w
    ];

The reduced subscheme of these tropes:

    red:=[ReducedSubscheme(Scheme(X40,q)):q in tropes];
    &and[Degree(q) eq 4:q in red];

They are smooth at the nodes of $X_{40}$:

    &and[Dimension(SingularSubscheme(q) meet SX40) eq -1:q in red];

Two 2-divisible disjoint sets of 20 nodes, which confirm that the 40 nodes are
2-divisible:

    s1:=Points(Scheme(SX40,tropes[1]*tropes[2])) diff
        Points(Scheme(SX40,[tropes[1],tropes[2]]));
    s2:=Points(Scheme(SX40,tropes[6]*tropes[7])) diff
        Points(Scheme(SX40,[tropes[6],tropes[7]]));
    &and[#s1 eq 20,#s2 eq 20,#(s1 join s2) eq 40];

We compute three 2-divisible sets of 24 nodes:

    Sets:=[];
    for q in [[2,5],[1,4],[3,8]] do
      pts:=Points(Scheme(SX40,tropes[q[1]]*tropes[q[2]])) diff
           Points(Scheme(SX40,[tropes[q[1]],tropes[q[2]]]));
      Append(~Sets,SingularPoints(X40) diff pts);
    end for;

and use these sets to check the divisibilities in (3.2):

    Da join Dabc join Dac join Dab eq Sets[1];
    Db join Dabc join Dbc join Dab eq Sets[2];
    Dc join Dabc join Dbc join Dac eq Sets[3];

Now we show that

$h^{0}\left(\widetilde{X}_{40},\mathcal{O}_{\widetilde{X}_{40}}\left(K_{\widetilde{X}_{40}}+L_{111}\right)\right)=2.$

Let $N_{1},\ldots,N_{16}$ be the nodes in $D_{a}+D_{b}+D_{c}+D_{abc}$ and
$A_{1},\ldots,A_{16}$ be the corresponding $(-2)$-curves. Let $H_{1},H_{2}$ be
the tropes whose pullback to $\widetilde{X}_{40}$ is

$\widetilde{H}_{1}+\widetilde{H}_{2}=2\widehat{T}_{1}+2\widehat{T}_{2}+\sumop\displaylimits_{i=1}^{16}A_{i}+2\sumop\displaylimits_{i=17}^{20}A_{i},$

with $A_{17},\ldots,A_{20}\in\widetilde{H}_{1}\cap\widetilde{H}_{2}.$ Then

$\sumop\displaylimits_{i=1}^{16}A_{i}\equiv 2L_{111},\ \ \ {\rm with}\ \ \
K_{\widetilde{X}_{40}}+L_{111}\equiv
2\widetilde{H}-\widehat{T}_{1}-\widehat{T}_{2}-\sumop\displaylimits_{i=17}^{20}A_{i}.$

We compute below that the system of quadrics through the curves
$T_{1},T_{2}\subset\mathbb P^{4}$ is generated by 2 elements, modulo the
quadric $Q$. For $i=17,\ldots 20,$ the fact
$\left(2\widetilde{H}-\widehat{T}_{1}-\widehat{T}_{2}\right)\cdot A_{i}<0$
implies that $A_{i}$ is contained in the base component of the linear system
$\left|2\widetilde{H}-\widehat{T}_{1}-\widehat{T}_{2}\right|.$ This gives
$h^{0}\left(\widetilde{X}_{40},\mathcal{O}_{\widetilde{X}_{40}}\left(K_{\widetilde{X}_{40}}+L_{111}\right)\right)=2.$

    T1:=ReducedSubscheme(Scheme(X40,tropes[2]));
    T2:=ReducedSubscheme(Scheme(X40,tropes[9]));
    pts:=Points(SX40 meet (T1 join T2)) diff
         Points(SX40 meet T1 meet T2);
    pts eq (Da join Db join Dc join Dabc);
    L:=LinearSystem(LinearSystem(P,2),T1 join T2);
    #Sections(LinearSystemTrace(L,X40)) eq 2;

Let us show that

$h^{0}\left(\widetilde{X}_{40},\mathcal{O}_{\widetilde{X}_{40}}\left(K_{\widetilde{X}_{40}}+L_{ijk}\right)\right)=0$

for $ijk\neq 111$. Suppose the opposite. Let $A_{1},\ldots,A_{24}$ be the
corresponding $(-2)$-curves. Then there is a curve
$E\in\left|K_{\widetilde{X}_{40}}+L_{ijk}\right|,$ and $E\cdot A_{i}=-1$
implies that the linear system
$\left|K_{\widetilde{X}_{40}}+L_{ijk}-\sumop\displaylimits_{j=1}^{24}A_{j}\right|=\left|K_{\widetilde{X}_{40}}-L_{ijk}\right|$
is nonempty. Therefore
$\left|2K_{\widetilde{X}_{40}}-\sumop\displaylimits_{j=1}^{24}A_{j}\right|$ is
nonempty, which implies that there is at least one quadric in $\mathbb P^{4}$
through the corresponding nodes $N_{1},\ldots,N_{24}$ (modulo the quadric
$Q$). We show below that this does not happen.

    Sets:=[
    Da join Dabc join Dac join Dab,
    Db join Dabc join Dbc join Dab,
    Dc join Dabc join Dbc join Dac,
    Da join Db join Dbc join Dac,
    Da join Dc join Dbc join Dab,
    Db join Dc join Dac join Dab
    ];
    for q in Sets do
      L:=LinearSystem(LinearSystem(P,2),[P!x:x in q]);
      #Sections(LinearSystemTrace(L,X40)) eq 0;
    end for;

### 4.2. The surface $Y_{48}$

Here we give the equations of $Y_{48}$ as a complete intersection of 4
quadrics in $\mathbb P^{6}.$ We start by defining $\mathbb P^{6}$ over a
certain number field.

    K:=Rationals(); R<x>:=PolynomialRing(K);
    K<r,m>:=ext<K|x^2 + 15,x^2 - 95/42*x + 2855/2646>;
    R<n>:=PolynomialRing(K);
    K<n>:=ext<K|
    n^2 + 443889677/206391214080000*r - 46942774543/619173642240000>;
    P6<x,y,z,w,t,u,v>:=ProjectiveSpace(K,6);

The three quadrics $B,C,D$:

    B:=(675/4802*r+334125/33614)*n*x*z+(-389475/67228*r+3266325/67228)*n*x*w+
     (34425/9604*r+451575/67228)*n*y*w+(-389475/67228*r+3266325/67228)*n*z*w+
     (-62100/16807*r+348300/16807)*n*w^2+(239625/33614*r+1541025/33614)*n*x*t
     +(-8100/2401*r+137700/16807)*n*y*t+(239625/33614*r+1541025/33614)*n*z*t
     +(6075/9604*r+3007125/67228)*n*w*t+(71550/16807*r+319950/16807)*n*t^2;
    C:=x*y+1/154*(126*m-181)*y^2+1/42*(-42*m+95)*x*z+y*z+(1/1540*(14*m-25)*r
     +1/924*(-798*m+1997))*x*w+(1/420*(42*m-65)*r+1/308*(-294*m+767))*y*w
     +(1/1540*(14*m-25)*r+1/924*(-798*m+1997))*z*w+(1/385*(-119*m+185)*r
     +1/462*(-168*m+311))*w^2+(1/1540*(-14*m+25)*r+1/924*(-798*m+
     1997))*x*t+(1/420*(-42*m+65)*r+1/308*(-294*m+767))*y*t+(1/1540*(-14*m
     +25)*r+1/924*(-798*m+1997))*z*t+1/154*(126*m-71)*w*t+(1/385*(119*m-
     185)*r+1/462*(-168*m+311))*t^2;
    D:=x*y+1/77*(-63*m+52)*y^2+m*x*z+y*z+(1/2310*(-21*m+10)*r+1/154*(133*m+
     32))*x*w+(1/70*(-7*m+5)*r+1/154*(147*m+51))*y*w+(1/2310*(-21*m+
     10)*r+1/154*(133*m+32))*z*w+(1/2310*(714*m-505)*r+1/154*(56*m-
     23))*w^2+(1/2310*(21*m-10)*r+1/154*(133*m+32))*x*t+(1/70*(7*m-5)*r
     +1/154*(147*m+51))*y*t+(1/2310*(21*m-10)*r+1/154*(133*m+32))*z*t+
     1/77*(-63*m+107)*w*t+(1/2310*(-714*m+505)*r+1/154*(56*m-23))*t^2;

We obtain alternative equations for $X_{40}$:

    F:=B^2-C*D;
    Q:=5*(x^2+y^2+z^2+w^2+t^2)-7*(x+y+z+w+t)^2;
    X:=Scheme(P6,[F,Q,u,v]);
    h:=-x-y-z-w-t;
    I:=4*(x^4+y^4+z^4+w^4+t^4+h^4)-(x^2+y^2+z^2+w^2+t^2+h^2)^2;
    X40:=Scheme(P6,[Q,I,u,v]);
    X eq X40;

And finally the equations of $Y_{48}$ in $\mathbb P^{6}$:

    Y48:=Surface(P6,[u^2-C,v^2-D,u*v-B,Q]);
    SY48:=SingularSubscheme(Y48);
    Dimension(SY48) eq 0;
    Degree(SY48) eq 48;
    Degree(ReducedSubscheme(SY48)) eq 48;

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