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Elliptic surfaces for 6/9 patterns · 5.13

The ABCDGJ Pattern: One Progression and Two Gaussian Norms

The red BDJ progression fixes the difference of squares in both yellow quadrics. After parametrizing it, the system reduces to an even genus-one curve. Its Jacobian is a split 12I₂ elliptic K3 surface with a non-torsion section.

ABCDEFGHJ
ABCDGJ: red BDJ; yellow ACGJ and BCDG

1. The exact red-yellow-yellow system

Let a,b,c,d,g,j be rational square roots of the selected entries. The ABCDGJ pattern is defined by three independent conditions

b2+d2=2j2,a2+j2=c2+g2,b2+c2=d2+g2. \begin{aligned} b^2+d^2&=2j^2,\\ a^2+j^2&=c^2+g^2,\\ b^2+c^2&=d^2+g^2. \end{aligned}

The first equation is the red BDJ progression. The other two are equalities of Gaussian norms on ACGJ and BCDG. Together they are necessary and sufficient for the six selected entries.

To recover the whole square, put

E0=a2+j22,x=a2−j22,y=c2−g22. E_0=\frac{a^2+j^2}{2},\qquad x=\frac{a^2-j^2}{2},\qquad y=\frac{c^2-g^2}{2}.(E0+xE0−x−yE0+yE0−x+yE0E0+x−yE0−yE0+x+yE0−x). \begin{pmatrix} E_0+x&E_0-x-y&E_0+y\\ E_0-x+y&E_0&E_0+x-y\\ E_0-y&E_0+x+y&E_0-x \end{pmatrix}.

2. The red BDJ progression

Use the standard triple

L(t)=t2−2t−1,C(t)=t2+1,R(t)=t2+2t−1. L(t)=t^2-2t-1,\qquad C(t)=t^2+1,\qquad R(t)=t^2+2t-1.

The identity L²+R²=2C² parametrizes the red condition. On the chosen affine chart take

b=L(t),j=C(t),d=R(t). b=L(t),\qquad j=C(t),\qquad d=R(t).

The difference of the endpoint squares is

δ(t):=R(t)2−L(t)2=8t(t2−1). \delta(t):=R(t)^2-L(t)^2=8t(t^2-1).

The third quadric now simply says c²−g²=δ. Thus the same quantity δ links the red progression to both yellow norm equalities.

3. The residual even quartic

Put v=c+g. For v≠0 the equality c²−g²=δ is inverted completely:

c=12(v+δv),g=12(v−δv). c=\frac12\left(v+\frac{\delta}{v}\right),\qquad g=\frac12\left(v-\frac{\delta}{v}\right).

Substitution into the other yellow quadric and the change Y=2av give

Ct:Y2=2v4−4C(t)2v2+2δ(t)2. \mathcal C_t:\qquad Y^2=2v^4-4C(t)^2v^2+2\delta(t)^2.

Exact lift back

Every rational point (v,Y) with v≠0 lifts to an ABCDGJ solution:

a=Y2v,b=L(t),c=12(v+δv),d=R(t),g=12(v−δv),j=C(t). \begin{aligned} a&=\frac{Y}{2v},& b&=L(t),& c&=\frac12\left(v+\frac{\delta}{v}\right),\\ d&=R(t),& g&=\frac12\left(v-\frac{\delta}{v}\right),& j&=C(t). \end{aligned}

The quartic has the natural rational points (R+L,2C(R+L)) and (R−L,±2C(R−L)). They correspond to the diagonal solutions a²=j², c²=d², g²=b². Hence the generic smooth fiber is a pointed genus-one curve.

4. The Jacobian and full 2-torsion

For the binary quartic in the preceding section, the classical invariants are

I=16(t8+196t6−378t4+196t2+1),J=128C(t)2(t4−24t3+2t2+24t+1)⋅(t4+24t3+2t2−24t+1). \begin{aligned} I={}&16(t^8+196t^6-378t^4+196t^2+1),\\ J={}&128C(t)^2 (t^4-24t^3+2t^2+24t+1)\\ &\qquad\cdot(t^4+24t^3+2t^2-24t+1). \end{aligned}

Take the short Jacobian model

Et:Y2=X3−27I(t)X−27J(t). \mathcal E_t:\qquad \mathsf Y^2=\mathsf X^3-27I(t)\mathsf X-27J(t).

Its cubic splits completely over ℚ(t):

Y2=(X−24C2)⋅(X+12(t4−24t3+2t2+24t+1))⋅(X+12(t4+24t3+2t2−24t+1)). \begin{aligned} \mathsf Y^2={}&(\mathsf X-24C^2)\\ &\cdot\bigl(\mathsf X+12(t^4-24t^3+2t^2+24t+1)\bigr)\\ &\cdot\bigl(\mathsf X+12(t^4+24t^3+2t^2-24t+1)\bigr). \end{aligned}

Hence the Jacobian has full rational 2-torsion. The two yellow quadrics introduce no new quadratic extension: their symmetry is already visible in the three linear factors.

5. The K3 surface passport

Up to a nonzero constant, the discriminant is

Δ(t)∼t2(t−1)2(t+1)2⋅(t4−8t3+2t2+8t+1)2⋅(t4+8t3+2t2−8t+1)2. \begin{aligned} \Delta(t)\sim{}& t^2(t-1)^2(t+1)^2\\ &\cdot(t^4-8t^3+2t^2+8t+1)^2\\ &\cdot(t^4+8t^3+2t^2-8t+1)^2. \end{aligned}

The eleven finite roots of the reduced discriminant support are simple, while c₄ does not vanish there: these are eleven I₂ fibers. After the minimal transformation in the chart s=1/t, one further I₂ appears at infinity.

fiber configuration=12I2.\text{fiber configuration}=12I_2.

The total Euler number is 24, so the minimal elliptic surface is K3.

6. A non-torsion section and the rank

The covariant map of the complete intersection of two quadrics sends the diagonal point (a,c,g)=(C,R,L) to the section

P(t)=(24U16(t)(LCR)2,3456t2(t2−1)2V1V2V3V4(LCR)3), P(t)=\left( \frac{24U_{16}(t)}{(LCR)^2}, \frac{3456t^2(t^2-1)^2V_1V_2V_3V_4}{(LCR)^3} \right),
U16=t16−8t14+380t12−1464t10+2438t8−1464t6+380t4−8t2+1,V1=t4−6t3+2t−1,V2=t4−2t3+6t−1,V3=t4+2t3−6t−1,V4=t4+6t3−2t−1. \begin{aligned} U_{16}={}&t^{16}-8t^{14}+380t^{12}-1464t^{10}+2438t^8\\ &-1464t^6+380t^4-8t^2+1,\\ V_1={}&t^4-6t^3+2t-1,& V_2={}&t^4-2t^3+6t-1,\\ V_3={}&t^4+2t^3-6t-1,& V_4={}&t^4+6t^3-2t-1. \end{aligned}

At t=2 this gives the curve and point

E2:y2=x3−3255984x+1737590400,P2=(126789361225,4449103718442875). \begin{aligned} E_2:\quad y^2&=x^3-3255984x+1737590400,\\ P_2&=\left(\frac{12678936}{1225}, \frac{44491037184}{42875}\right). \end{aligned}

At the good reductions #E₂(𝔽₁₁)=16 and #E₂(𝔽₁₃)=20, so rational torsion has order dividing 4. Yet the reduction of P₂ modulo 13 has order 5. Therefore P₂, and hence the generic section P(t), is non-torsion.

Mordell–Weil bound

The non-torsion section gives the lower bound 1. For the configuration 12I₂ the trivial lattice has rank 14; using ρ≤20 for a K3 surface, the Shioda–Tate formula gives

1≤rank⁡E(Q‾(t))≤6.1\le \operatorname{rank}\mathcal E(\overline{\mathbb Q}(t))\le6.

7. An explicit infinite family

Pass through (R+L,2C(R+L)) a parabola tangent to the quartic and require it to pass through (R−L,−2C(R−L)). The fourth intersection point has coordinates

v=−4tN8Q8,Y=8tC M16Q82, v=-\frac{4tN_8}{Q_8},\qquad Y=\frac{8tC\,M_{16}}{Q_8^2},
Q8=t8−20t6+22t4−20t2+1,N8=t8+12t6−42t4+12t2+1,M16=t16−72t14+636t12−2936t10+4998t8−2936t6+636t4−72t2+1. \begin{aligned} Q_8={}&t^8-20t^6+22t^4-20t^2+1,\\ N_8={}&t^8+12t^6-42t^4+12t^2+1,\\ M_{16}={}&t^{16}-72t^{14}+636t^{12}-2936t^{10}+4998t^8\\ &-2936t^6+636t^4-72t^2+1. \end{aligned}

After lifting back and clearing the common denominator, one obtains degree-18 roots:

a=−C M16,b=L Q8N8,d=R Q8N8,j=C Q8N8,c=−2tN82−(t2−1)Q82,g=−2tN82+(t2−1)Q82. \begin{aligned} a&=-C\,M_{16},& b&=L\,Q_8N_8,& d&=R\,Q_8N_8,\\ j&=C\,Q_8N_8,& c&=-2tN_8^2-(t^2-1)Q_8^2,\\ g&=-2tN_8^2+(t^2-1)Q_8^2. \end{aligned}

Identity-level correctness

Substitution makes all three ABCDGJ quadrics vanish in ℤ[t]. The ratio a/j is nonconstant, so outside a finite set of degenerate parameters the family contains infinitely many projectively distinct rational solutions.

8. An exact positive example

At t=−3, after removing the common factor of the roots, one obtains the following magic square. Root signs do not affect the entries.

A52 · 432 · 96292
B72 · 4012 · 7512
C10487992
D4012 · 7512
E3276585537625
F6462479150449
G72 · 2712 · 12312
H2109266760001
J52 · 4012 · 7512
Square entries are written as prime factorizations.

The magic sum is 9,829,756,612,875. All nine entries are positive and pairwise distinct; exactly A,B,C,D,G,J are perfect squares.

9. Exact scope of the result

The quartic model describes the generic nondegenerate open part of ABCDGJ. The cases v=0, a constant red progression, zeros of Q₈N₈, and the chart t=∞ require adjacent projective charts, but do not change the generic Jacobian.

The K3 passport, existence of a non-torsion section, geometric rank bound, and an explicit infinite family are proved. The exact rank, completeness of the displayed family, and positivity of every rational specialization are not asserted.

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