Optimal canonical degree for low-genus curves on the exotic-quintuple surface

1. The surface and the result

The exotic-quintuple construction of Dujella, Kazalicki and Petričević [1] leads to the surface

$$ u^2+w^2=r^2+r^{-2},\qquad z^2=(u^2-1)(1-w^2). \tag{1} $$

We work over $\mathbb Q$. A point is called nondegenerate if it belongs to the DKP reconstruction open set, so that the construction gives five distinct nonzero rational entries. In particular the boundary sections and curves identified as reconstruction-degenerate in [1,2] are excluded. Let $S$ be the minimal resolution of the canonical model of (1). For a geometrically integral curve $C$ meeting this open set, its canonical degree is $K_S\cdot C$, equivalently the degree of the pullback of the canonical hyperplane bundle to its normalization.

Theorem 1. Suppose that the normalization of $C$ has genus zero or one and contains a nondegenerate rational point. Then

$$ K_S\cdot C\geq 8. $$

The three DKP elliptic families have canonical degree eight, so the bound is optimal.

We also narrow the classification at the minimum degree. Write a canonical-image class as $(A,B;m)$ in the blowup model introduced below.

Theorem 2. Apart from the known DKP families, a curve satisfying the hypotheses of Theorem 1 and having canonical degree eight maps birationally to its canonical image. Up to the coordinate exchange $u\leftrightarrow w$, its image is in one of the systems

$$ (3,3;1),\qquad (2,4;1),\qquad (4,4;2). \tag{2} $$

The first system has degree three over $r$ and the last two have degree four. We do not classify their low-genus lifts.

The proof combines the geometry of the canonical image with complete arithmetic reductions. In the final degree-six system, canonical intersection bounds restrict three elliptic traces to $72\,576$ possibilities. A finite-field sieve leaves 67. An explicit catalogue covers all 191 degree-two Abel classes in their fibers at one good specialization. Differences with the same generic trace lie in a group killed by four; good-reduction injectivity on that torsion proves generic exhaustion. Thus a finite computation is applied only after an unbounded geometric reduction.

The surface and its three known elliptic curves are from [1]. The generic quotient maps, rank calculations and no-section theorem for the $r$-fibration are from [2]. The full canonical-hyperplane classification is the earlier result [3]. None of those results determines the minimum intrinsic degree or exhausts the last degree-six system. Elliptic-surface intersection and height formulas are standard [4], as is injective specialization in the form used here [5]. Finite-field elimination of Mordell--Weil possibilities follows the established sieve philosophy of Bruin and Stoll [6]. Siksek [7] treats rational points on symmetric powers, including the exceptional families induced by maps to curves of genus zero or one. Our argument does not use a Chabauty rank inequality: the generic genus-three Jacobian has rank three. Instead, a canonical-degree bound makes the relevant trace set finite, and a finite isogeny kernel supplies the final uniqueness statement. We claim these explicit classifications and geometric consequences, not a new general Mordell--Weil sieve or specialization theorem.

2. Canonical model and rational image

In $\mathbb P(1,1,1,1,1,2)$ with coordinates $[U:W:P:H:X:Z]$, consider

$$ P^2=U^2+W^2+2H^2,\qquad X^2=P^2-4H^2,\qquad Z^2=(U^2-H^2)(H^2-W^2). \tag{3} $$

On $H=1$, set $r=(P+X)/2$; then $r^{-1}=(P-X)/2$, recovering (1). The singular point of the ambient weighted space is absent from (3).

Projection to the smooth quadric $Q:P^2=U^2+W^2+2H^2$ is a flat bidouble cover with branch functions $P^2-4H^2$ and $(U^2-H^2)(H^2-W^2)$. The six branch conics are $P=\pm2H$, $U=\pm H$ and $W=\pm H$. They meet at eight affine corners, four points on $H=0$ where two branches of the second color meet, and two points on $H=P=0$ where the first-color branches meet. The indicated intersections are transverse.

At an affine corner put $s=U^2-1$ and $t=1-W^2$. The local equations become $X^2=s-t$, $Z^2=st$, and exactly

$$ (2s-X^2-2Z)(2s-X^2+2Z)=X^4. $$

This is an $A_3$ singularity. The other branch crossings give twelve $A_1$ singularities in total. There are no other singularities. Hence the model is normal with rational double points. Weighted adjunction gives

$$ K=\mathcal O(1),\qquad K^2=\frac{2\cdot2\cdot4}{2}=8. $$

The resolution is crepant. The direct image of its structure sheaf on $Q$ is

$$ \mathcal O_Q\oplus\mathcal O_Q(-1)\oplus\mathcal O_Q(-2)\oplus\mathcal O_Q(-3). $$

Since $\mathcal O_Q(1)$ becomes $\mathcal O(1,1)$, this gives $\chi(\mathcal O_S)=6$, $q=0$ and $p_g=5$. The five degree-one coordinates span the canonical system. The canonical map forgets $Z$ and has degree two onto the intersection of two quadrics $Y\subset\mathbb P^4$. Its deck involution is $z\mapsto-z$. The two singularities of $Y$ are conjugate $A_1$ points.

Let $T\to Y$ be its minimal resolution. A parametrization of $Y$ by $\mathbb P^1_r\times\mathbb P^1_t$ is

$$ \begin{aligned} U&=r^2(t^2-1)-2t,&W&=1-t^2-2r^2t,\\ P&=(r^2+1)(t^2+1),&H&=r(t^2+1),\\ X&=(r^2-1)(t^2+1). \end{aligned}\tag{4} $$

The bihomogeneous forms have bidegree $(2,2)$. Their base ideal is locally

$$ (t^2+1,r^2+t). $$

Its four points are simple: the Jacobian determinant is $-4rt$, nonzero there. Direct checks in the other product charts give no additional basepoints. Their residue field is $\mathbb Q[r]/(r^4+1)$, so they form one Galois orbit. Blowing up once at each resolves the map. Its hyperplane divisor is

$$ -K_T=(2,2)-\sum_{i=1}^4E_i,\qquad (-K_T)^2=4. $$

The strict transforms of $t=\pm i$ are the two contracted $(-2)$-curves.

Every rationally defined integral image curve meeting the open set has class

$$ D=(A,B)-m\sum E_i,\qquad A,B,m\in\mathbb Z_{\geq0}. $$

Indeed the two rulings are rational and the four exceptional classes form one Galois orbit. Thus, writing $d=(-K_T)\cdot D$,

$$ d=2A+2B-4m,\qquad D^2=2AB-4m^2,\qquad p_a(D)=AB-2m^2-\frac d2+1.\tag{5} $$

Intersection with the two contracted curves gives $A\geq2m$, and hence $B\leq d/2$. The degrees of the projections to $r$ and $t$ are $B$ and $A$. If a curve on $S$ has degree $e\in\{1,2\}$ over its canonical image, its canonical degree is $ed$. In particular that degree is even.

The symmetry $u\leftrightarrow w$ acts on the conic parameter by

$$ t\longmapsto\frac{(1-r^2)t+r^2+1}{(r^2+1)t+r^2-1} $$

and on the classes by

$$ (A,B;m)\longmapsto(A+2B-4m,B;B-m).\tag{6} $$

Both assertions are checked by the exact parametrization and pullback of the four basepoints.

3. Degrees below six and the first candidates

The canonical hyperplanes are precisely the equations

$$ au+bw=cr+d/r+e. $$

Their low-genus components with a nondegenerate rational point are classified in [3]. Each known DKP curve has degree four over $r$. The two independent quadratic characters for its function field can be chosen as follows:

| Curve | First character | Second character | |---|---|---| | $D_1$ | $7r^2+10r+7$ | $-3(3r^2-14r+3)$ | | $D_2$ | $r^2+10r+1$ | $-3(5r^2-14r+5)$ | | $D_3$ | $4-3r^2$ | $3(4r^2-1)$ |

The roots are simple and disjoint in each row. On each curve the canonical coordinates $[u:w:r+r^{-1}:1:r-r^{-1}]$ have common pole divisor

$$ (r)_\infty+(r^{-1})_\infty, $$

of degree eight, with no basepoints. Thus the known curves have canonical degree eight. Their nondegenerate rational points are part of the constructions in [1,3].

Restriction of the five canonical sections to a genus-zero normalization of degree $d$ has dimension at most $d+1$, and to a genus-one normalization of positive degree $d$ has dimension $d$. Therefore rational curves of degree at most three and genus-one curves of degree at most four lie in a rational canonical hyperplane. The classification [3] excludes them. Every remaining curve of degree below eight spans $\mathbb P^4$ and maps birationally to its canonical image, since an image of a double cover would have degree below four.

A rational degree-four image spanning $\mathbb P^4$ is a smooth rational normal quartic. Its lift to $T$ is still smooth: the maximal ideal of a node pulls back to an invertible ideal on a smooth curve, so its lift through the blowup is isomorphic. In (5), $p_a=0$ and $d=4$ leave only $(1,1;0)$ and $(3,1;1)$. Both have degree one over $r$, and a birational lift would be a rational section of (1), excluded by [2]. Thus every curve under consideration has degree at least six.

For $d=6$, imposing $A+B=3+2m$, $A\geq2m$ and $p_a\geq0$ gives

$$ (2,1;0),\ (4,1;1),\ (1,2;0),\ (5,2;2),\ (3,2;1),\ (2,3;1),\ (4,3;2). \tag{7} $$

The first two are excluded by [2]. The second pair is exchanged by (6), as is the final pair. The next two sections eliminate them, leaving only $(3,2;1)$.

4. A second fibration without nondegenerate sections

Regard the equation $Z^2=(U^2-H^2)(H^2-W^2)$ from (4) as a genus-three curve over $\mathbb Q(t)$. Its quotient by $r\mapsto-r$, with $R=r^2$, is a genus-one quartic $Z^2=f_t(R)$ with rational points

$$ (R,Z)=(1,\pm4t(t^2-1)). $$

Its Jacobian and a two-isogenous split model are

$$ \begin{aligned} E_t &: y^2=x\bigl(x^2+(t^2+1)^2x+16t^2(t^2-1)^2\bigr),\\ E'_t &: y^2=x(x-q_+)(x-q_-),\\ q_\pm&=(t^2+1)^2\pm8t(t^2-1). \end{aligned}\tag{8} $$

The complete Gusić--Tadić specialization test [5, Theorem 2.2] is satisfied for $E'_t$ at $t=6$. The certificate enumerates all 126 signed nonconstant squarefree divisors required by the three root-difference products, including integer content. Their values are nonsquares. Here $q_+(6)=3049$ and $q_-(6)=-311$. Exact rank computations give rank zero and torsion orders four and two for $E'_6$ and $E_6$, respectively.

Injectivity makes $E'_t(\mathbb Q(t))$ finite; the isogeny then makes $E_t(\mathbb Q(t))$ finite. Good specialization is injective on torsion, so its order is exactly two. The two displayed points exhaust the genus-one quartic, including projective points, since the birational identification extends to its smooth model. Consequently the original genus-three curve has only

$$ r=\pm1,\qquad Z=\pm4t(t^2-1) $$

as rational points over $\mathbb Q(t)$. All are degenerate. This excludes $(1,2;0)$ and its partner $(5,2;2)$ in (7).

5. Excluding the trisection orbit

For $(2,3;1)$ write

$$ \alpha(t)r^2+\beta(t)r+\gamma(t)=0,\qquad \beta=(t^2+1)\ell,\quad \deg\ell\leq1,\quad \gamma\equiv t\alpha\pmod{t^2+1},\tag{9} $$

where the three coefficients have degree at most three and are primitive as binary cubics. The two rational points of the genus-one quotient in Section 4 are its origin $O$ and a point $T$ of order two. Thus the trace of its quadratic point is $O$ or $T$.

Put

$$ A=t^4-6t^2+1,\qquad K=4t(t^2-1). $$

With the chosen origin, elliptic negation acts on the quartic coordinate by

$$ R\longmapsto\frac{A(R-1)-K}{(A-K)R-A}. $$

The corresponding ordinate multiplier is $K^2/((A-K)R-A)^2$. Substitution verifies the equation, involutivity and the origin. Trace zero therefore forces

$$ A[(\alpha+\gamma)^2-\beta^2]+K(\alpha^2-\gamma^2)=0.\tag{10} $$

Let $D=\alpha+\gamma$, $E=\alpha-\gamma$. The cases $D=0$ or $E=0$ are reducible. Since $A\mid DE$, its two coprime irreducible quadratic factors must be assigned one each to $D$ and $E$:

$$ D=A_\epsilon L_1,\quad E=A_{-\epsilon}L_2,\qquad A_\epsilon=t^2+2\epsilon t-1,\quad \deg L_i\leq1. $$

Equation (10) becomes $\beta^2=L_1(A_\epsilon^2L_1+KL_2)$. If $L_1$ is linear, it divides $\ell$ in (9), and primitiveness forces its root to be among $0,1,-1$. If constant, degree comparison makes $\ell$ constant. After scaling there are precisely eight cases: $\epsilon=\pm1$ and $L_1=1,t,t-1,t+1$. Solving

$$ L_1[c^2(t^2+1)^2-A_\epsilon^2]=KL_2 $$

gives only $c^2=1$, $L_2=-\epsilon L_1$. Nonconstant $L_1$ gives a common factor; constant $L_1$ gives degree at most two, so the binary cubic has a vertical component. Trace zero is excluded.

For trace $T$, the quartic involution $Z\mapsto-Z$ exchanges the quadratic conjugates. Hence $r^2\in\mathbb Q(t)$, $\beta=0$, and with $R=-\gamma/\alpha$ one has

$$ V^2=R f_t(R). $$

The constant and leading coefficients of $f_t$ are both $-4t^2(t^2-1)^2$. Odd zeros or poles of $R$ outside $\{0,1,-1,\infty\}$ would therefore give an odd valuation on the right. Thus the double cover $r^2=R$ branches only over those four points.

If all four branch, write $R=\lambda t(t^2-1)g(t)^2$. The congruence $R(i)=-i$ implies $2\lambda g(i)^2=1$. A rational squareclass becoming a square in $\mathbb Q(i)$ is $1$ or $-1$, so the normalization is isomorphic to

$$ y^2=x^3-4x. $$

This curve has rank zero and exactly $O,(0,0),(2,0),(-2,0)$ as rational points. They lie over the four degenerate parameters, excluding nondegenerate rational points.

If exactly two points branch, valuation at $1+i$ rules out a set containing exactly one of $1,-1$. The only sets are $\{0,\infty\}$ and $\{1,-1\}$. Coprimality and the binary cubic degrees force one odd root in each coefficient, with the remaining part a square of a binary linear form. The congruence at $i$ forces those linear forms to be orthogonal. Up to the genuine symmetries $(r,t)\mapsto(1/r,-t)$ and $(1/r,1/t)$, the complete families are

$$ \begin{array}{c|cc} &\alpha&\gamma\\ \hline I&(t-a)^2&-t(at+1)^2\\ II&(t-1)(t-a)^2&(t+1)(at+1)^2, \end{array}\qquad a\in\mathbb P^1(\mathbb Q). $$

Their normalizations have, respectively, $t=s^2$ and $t=(s^2-1)/(s^2+1)$, with $r=s(at+1)/(t-a)$; at $a=\infty$, $r=-st$.

After substitution, the remaining square condition has four quintic factors. All twelve pairwise resultants and all leading-coefficient drops have rational parameter roots only at $a=0,1,-1,\infty$; the other factor is $a^2+1$. For a square product, an odd-degree factor must share an odd root with another, so these exceptional parameters exhaust the possibilities. Exact factorization at all four leaves only two square cases, $I$ at $a=0$ and $II$ at $a=-1$, and both have common factors in $\alpha,\gamma$. No proper lift remains. This excludes $(2,3;1)$ and its partner $(4,3;2)$.

6. Full elliptic section groups and bounded traces

Return to $K=\mathbb Q(r)$, and put $a_0=r^4-r^2+1$, $a=4r^2a_0$, $b=(r^2-1)^4$. The three elliptic quotient maps of the genus-three curve $C_r$ have targets

$$ E_A:Y^2=X(X+a)(X-b),\qquad E_B:Y^2=X(X-r^2)(X-a_0). $$

The map $\pi_A$ has degree two. Its quartic coordinate is $x=(t^{-1}-t)/2$, with ordinate $Z/(8r^2t^2)$ as in [2]. The other two maps, of degree four, are

$$ \pi_u=\left(\frac{a_0}{u^2},\frac{ra_0z}{u^3}\right),\qquad \pi_w=\left(\frac{a_0}{w^2},\frac{ra_0z}{w^3}\right).\tag{11} $$

All are morphisms of smooth projective curves, including their removable affine limits. Their generic ranks are one by [2]. We require the full groups, rather than finite-index subgroups:

$$ \begin{aligned} E_A(K)&=\mathbb ZQ\oplus\langle T,T_0\rangle, &\langle T,T_0\rangle&\simeq\mathbb Z/4\oplus\mathbb Z/2,\\ E_B(K)&=\mathbb ZP\oplus E_B[2](K), &E_B[2](K)&\simeq(\mathbb Z/2)^2,\\ Q&=(-4r^2(r^2-1)^2,4r^3(r^2-1)^2(r^2+1)),\\ T&=(-2r^2(r^2-1)^2,2r^2(r^2-1)^2(r^4+1)),\quad T_0=(0,0),\\ P&=(1,r(r^2-1)). \end{aligned}\tag{12} $$

Here $2T=(b,0)$. The displayed torsion exhausts the generic torsion: good specialization at $r=6$ has exactly these torsion orders, and specialization is injective on torsion at this characteristic-zero place.

We use the Shioda lattice height, with the convention of [4]. The $E_A$ surface has $\chi=4$, fibers $I_4$ at $0,\infty$, $I_8$ at $\pm1$, four $I_2$ at the roots of $a_0$, and four $I_4$ at the roots of $r^4+1$. Indeed $a+b=(r^4+1)^2$ and $\Delta=16a^2b^2(a+b)^2$; $c_4$ is a unit at each listed place. The minimal infinity chart is $X'=X/r^8$, $Y'=Y/r^{12}$. The section $Q$ is disjoint from $O$ and meets components with corrections $1,1$ at $0,\infty$ and $3/2,3/2$ at $1,-1$, and zero elsewhere. Thus $\widehat h_A(Q)=8-2-3=3$.

All height denominators divide eight. If $Q=mG+$ torsion for a free generator, then $m^2\mid24$, so $|m|=1$ or $2$. The split two-descent character $X$ modulo squares takes torsion values $1,-a_0,-2,2a_0$, whereas its value at $Q$ is $-1$. These squareclasses are distinct in $K^*/K^{*2}$. Consequently $Q\notin2E_A(K)+E_A(K)_{\rm tors}$, proving that it generates the free part.

The $E_B$ surface has $\chi=2$, four $I_4$ fibers at $0,1,-1,\infty$ and four $I_2$ at the roots of $a_0$. Its discriminant is $16r^4(r^2-1)^4a_0^2$ and its infinity chart is $X'=X/r^4$, $Y'=Y/r^6$. The corrections for $P$ are $3/4$ at each of $1,-1$, one at infinity and zero elsewhere. Hence $\widehat h_B(P)=4-3/2-1=3/2$. Height denominators divide four, so a generator index $m$ would satisfy $m^2\mid6$. Thus $|m|=1$, proving (12).

For any horizontal divisor $D$ on either relatively minimal elliptic surface, let $n=D\cdot F$. Orthogonal projection away from $O,F$ and the nonidentity fiber components gives

$$ \widehat h(\operatorname{Tr}D) =\chi n^2+2n(D\cdot O)-D^2-\sum_v c_v(D),\qquad c_v(D)\geq0.\tag{13} $$

Indeed subtracting $nO+(D\cdot O+n\chi)F$ gives square $D^2-2n(D\cdot O)-\chi n^2$. If $A_v$ is the negative-definite component matrix and $b_v$ is the intersection vector of $D$, the remaining correction is $c_v(D)=-b_v^{\mathsf T}A_v^{-1}b_v\geq0$. This proves (13) for arbitrary horizontal divisors.

A candidate in $(3,2;1)$ has degree two over $r$ and degree three over $t$. For its birational image on $E_A$, we have $D\cdot O\leq6$. To justify this even at singular fibers, the quartic abscissa is

$$ x=\frac{c_1Y-c_2X}{X(X+c_3)},\quad c_1=2r(r^2-1),\quad c_2=2r^2(r^4-1),\quad c_3=4r^2(r^2-1)^2. $$

With formal parameter $v=-X/Y$ near $O$, write $X=e/v^2$, $Y=-e/v^3$ for a unit $e$. Then

$$ x=-\frac{v(c_1+c_2v)}{e+c_3v^2}. $$

The coefficients are integral at all finite places. In the infinity chart they become $2s(1-s^2),2s^2(1-s^4),4s^2(1-s^2)^2$, respectively. Thus $x$ vanishes along $O$ in every chart, and its degree-six zero divisor bounds the intersection. Adjunction gives $D^2\geq-6$ because $K_{E_A}=2F$. Equation (13) yields $\widehat h_A(\operatorname{Tr}D)\leq46$, so the free coefficient has absolute value at most three.

If the image is instead a section $R$, the map to it has degree two. Then $\deg x(R)\leq3$, $R\cdot O\leq3$ and $\widehat h_A(R)\leq14$, so its free coefficient has absolute value at most two. Exact enumeration of all forty such sections gives twelve of abscissa degree at most three: four are degenerate constants, and the other eight have the squareclass

$$ x(R)^2+1\equiv(r^2+1)(r^4+1)\pmod{K^{*2}}. $$

Their quadratic $t$-extensions have genus two. This excludes the section-image case entirely.

For each $E_B$ map, the pullback of $O$ has degree at most six. The homogeneous identities

$$ X_BU^2=a_0H^2,\qquad X'_BU^2=(1-s^2+s^4)H^2 $$

at finite places and infinity show that an intersection of multiplicity $m$ with $O$ forces $v(U)\geq m$. The canonical section $U$ has zero divisor of degree six. The same argument uses $W$ for $\pi_w$. Neither section vanishes identically, by the hyperplane classification. For a birational bisection, $D^2\geq-2$, so (13) bounds the trace height by $8+24+2=34$ and its free coefficient by four. For a section image, $2(R\cdot O)\leq6$, hence $\widehat h_B(R)\leq10$ and its coefficient is at most two; the trace $2R$ again has coefficient at most four.

Encode $E_A$ points by $(n,k,j)=nQ+kT+jT_0$ and $E_B$ points by $(n,l)=nP+T_l$, where

$$ (T_0,T_1,T_2,T_3)=(O,(0,0),(r^2,0),(a_0,0)) $$

in this last indexing only. The three raw traces therefore lie in the finite set

$$ |n_A|\leq3,\quad 0\leq k<4,\quad0\leq j<2, \qquad |n_u|,|n_w|\leq4,\quad0\leq l_u,l_w<4.\tag{14} $$

Its size is $56\cdot36^2=72\,576$.

7. A finite kernel and exact specialization exhaustion

Let $J=\operatorname{Jac}(C_r)$ and

$$ \Phi=(\pi_{A*},\pi_{u*},\pi_{w*}):J\longrightarrow E_A\times E_B^2. $$

The simultaneous sign change of $u,w$ fixes $\pi_A$ and negates both other maps; the separate sign changes distinguish the two $E_B$ maps. These symmetries make each cross push-pull correspondence equal its negative. Since homomorphisms of abelian varieties have no additive torsion, all cross correspondences vanish. The diagonal correspondences multiply by the map degrees. For the polarization dual $\Phi^\dagger$ we obtain

$$ \Phi\Phi^\dagger=\operatorname{diag}([2],[4],[4]),\qquad \Phi^\dagger\operatorname{diag}([2],[1],[1])\Phi=[4].\tag{15} $$

The second identity follows after composing with $\Phi$: a homomorphism from a connected abelian variety into its finite kernel is zero. In particular $\deg\Phi=32$ and $\ker\Phi\subset J[4]$.

Specialize $r=2$. The curve and elliptic targets have good reduction away from $2,3,13,17$. By properness of the relative symmetric square, every effective degree-two generic divisor specializes to an effective degree-two divisor. The quotient morphisms and trace maps extend. Over $\mathbb F_p$ the complete list of such divisors consists of unordered pairs of rational points, including repetitions, and pairs of quadratic points conjugate under Frobenius. The certificate enumerates both types, both points at infinity and all exceptional quotient-map values. It verifies every elliptic image and checks the point counts over $\mathbb F_p$ and $\mathbb F_{p^2}$ against the isogeny (15).

For the successive primes $5,7,11,19,23,29,31,37,41$, the counts of surviving labels in (14) are

$$ 3892,\ 799,\ 285,\ 181,\ 141,\ 97,\ 87,\ 83,\ 67.\tag{16} $$

These are necessary conditions, not existence assertions. We now describe a generic catalogue that covers all the remaining classes.

7.1. The catalogue

Split pairs. There are sixteen rational points of $C_r$, with both ordinate signs over

$$ t=0,\infty,1,-1,-r^2,r^{-2},\frac{r^2+1}{r^2-1},-\frac{r^2-1}{r^2+1}. $$

Their 136 unordered pairs give 129 Abel classes, since the eight hyperelliptic pairs have the same class.

Elliptic fibers. Include $\pi_A^{-1}(R)$ for all $R=nQ+kT+jT_0$ with $|n|\leq1$. These 24 divisors have raw trace $(2R,O,O)$.

Diagonal divisors. Set $L=r^2+r^{-2}$, $v=uw$, $s=u+w$. Then $s^2=L+2v$ and $z^2=a_0/r^2-v^2$. The addition law on $E_B$ gives the sum of the two images as

$$ X=\frac{r^2z^2}{s^2},\qquad Y=\frac{rz(v+1)(a_0+r^2v)}{s^3}.\tag{17} $$

For $S=nP+(r^2,0)$, with $n\in\{-4,-2,2,4\}$, take both rational square roots $g$ of $X(S)$ and put

$$ v=\frac{Y(S)/g-X(S)}{r^2},\qquad s=\pm\sqrt{L+2v},\qquad z=\frac{gs}{r}. $$

All indicated roots are in $K$, as verified by exact rational identities. For $\eta=\pm1$, the equations $U+\eta W=sH$, $Z=zH^2$ define 32 divisors. Their traces are

$$ ((1,2,0),(n,2),(\eta n,2)).\tag{18} $$

For completeness, the raw-trace translation constants matter here. The hyperelliptic pencil has $E_A$ trace $P_A=Q+T_0$, coordinate reflection has sum constant $P_A+2T$, and coordinate exchange has sum constant $Q+2T$. Each follows by evaluating the corresponding involution at a rational point above $O$; the exact symmetry certificate checks these identities. Equations (17) and these constants prove (18) over $K$, not merely in reduction.

Coordinate divisors. For $R=nP+T_l$ with $n=\pm2$ and $l=1,3$, set

$$ q=\pm\sqrt{a_0/X(R)},\qquad z=\frac{Y(R)q^3}{ra_0}. $$

The eight rational pairs $(q,z)$ give sixteen divisors by fixing either $u=q$ or $w=q$. Their traces are, respectively,

$$ ((1,2,1),(2n,0),(0,0)),\qquad ((1,2,1),(0,0),(2n,0)). $$

Up to sign, $q$ is $(r^4+1)/(2r^2)$ or $(r^2-1)/(r^2+1)$.

Boundary divisors. Include the four divisors with $Z=0$ and one of $U\pm H,W\pm H$ zero, and the two with $t^2+1=0$, $Z=\pm U^2$. Every one has trace $((1,2,1),(0,0),(0,0))$. For the first four this follows from coordinate reflection and two-torsion on the other factors. For the last two, $W=tU$; reflection exchanges $t=i,-i$ and each $E_B$ image is $(0,0)$. These last divisors are geometrically reducible over $\mathbb Q(i)$.

7.2. Coverage and uniqueness

At $r=2,p=47$ there are 32 rational points and 2140 points over $\mathbb F_{47^2}$. Thus there are $32\cdot33/2+(2140-32)/2=1582$ rational effective degree-two divisors, with 1147 distinct raw traces. For the 67 labels in (16), their fibers contain 191 distinct Abel classes. The catalogue covers every one of them. The certificate compares entries separately for each exact generic trace label; a collision of different labels in the finite group is never counted as coverage.

Let $D$ be a generic candidate. Its specialization has the same Abel class as a catalogue divisor $D_0$ with the same generic trace. Thus $[D-D_0]\in\ker\Phi\subset J[4]$, and this class specializes to zero. Good reduction is injective on four-torsion, first at the characteristic-zero place $r=2$, then at the odd prime 47. Indeed multiplication by four has invertible linear term on the formal group over both local rings, so the reduction kernel has no four-torsion. Properness supplies extension of the torsion section. Therefore $[D-D_0]=0$ over $K$.

On a genus-three hyperelliptic curve, the Abel map from the symmetric square is injective away from the hyperelliptic pencil. Two distinct effective degree-two divisors in the same class would give a moving degree-two linear system, necessarily that unique pencil. A candidate here cannot be in the pencil: its involution is the canonical deck involution, so such a representative would map with degree two to its canonical image. Every degree-six candidate is birational to that image. Hence $D=D_0$ unless both are pencil representatives, and that exceptional case is excluded.

7.3. Excluding the catalogue

Split pairs and the pencil cannot be candidates. The elliptic fibers were excluded by the forty-section enumeration in Section 6. The primitive diagonal equations have $r$-degree two, four or eight, whereas $(3,2;1)$ requires three. The degree-two cases split into known sections. Primitivity excludes vertical components; a split quadratic in $t$ has components of $r$-projection degree one, so cannot supply the required degree-two candidate.

The coordinate divisors have other-coordinate squares with square denominators and squarefree numerator

$$ -(r^4+1)(r^4-4r^2+1)\quad\text{or}\quad r^8+r^6+4r^4+r^2+1. $$

Both give genus three. The six boundary divisors are reconstruction-degenerate or geometrically reducible. This exhausts the catalogue and proves Theorem 1.

8. Birational degree-eight bisections

The same argument excludes degree-eight curves birational to their canonical images with degree two over $r$. Their image classes are

$$ (2,2;0),\quad(4,2;1),\quad(6,2;2). $$

Their $t$-degree is at most six. For a birational image on $E_A$, $D\cdot O\leq12$, and (13) gives trace height at most $16+48+6=70$, hence $|n_A|\leq4$. A section image has $R\cdot O\leq6$, height at most twenty and free coefficient at most two, also giving trace coefficient at most four. For $E_B$, the homogeneous argument gives pullback intersection at most eight. A birational bisection has trace height at most $8+32+2=42$, so $|n_B|\leq5$. A section image has height at most twelve and coefficient at most two, and is included by the same bound.

There are consequently $72\cdot44^2=139\,392$ labels. The primes in (16) leave successively

$$ 7884,\ 1597,\ 393,\ 257,\ 169,\ 109,\ 95,\ 87,\ 71. $$

The only new labels are $((\pm4,0,0),O,O)$ and $((\pm4,2,0),O,O)$. Add the sixteen fibers over $\pm2Q+kT+jT_0$ to the catalogue. At $r=2,p=47$ this covers all 207 classes in all 71 fibers, again with exact generic labels. Equation (15) and the same specialization argument give exhaustion.

All forty elliptic-section fibers split or have genus $2,5,8,17$ or $20$. The diagonal divisors now require a genus test: their remaining double cover is

$$ (u-w)^2=2(r^2+r^{-2})-s(r)^2. $$

Its squarefree numerator has degree zero, eight or sixteen and the denominator is a square. The degree-zero cases are the split pairs already identified; the others have genus three or seven. The coordinate divisors have genus three, and the other exclusions are unchanged. This proves the asserted exclusion of birational bisections. The word “birational” is essential: the hyperelliptic pencil has moving representatives and cannot be removed by finite catalogue uniqueness alone.

9. Full lifts of Möbius sections

We exclude the canonical-map-degree-two case separately. Its non-hyperplane image must be a rational normal quartic, hence is in $(1,1;0)$ or its partner $(3,1;1)$. In the first class its equation is

$$ t=\frac{ar+b}{cr+d},\qquad ad-bc\neq0,\qquad a,b,c,d\in\mathbb Q.\tag{19} $$

We prove that every full lift of (19) has geometric genus at least two, with no rational-point hypothesis.

Each branch curve $U\pm H=0$ or $W\pm H=0$ is isomorphic to

$$ B:y^2=r^4-r^2+1. $$

For $u=1$ and $w=1$, respectively, the conic parameters are $t=(1+y)/(r(r-1))$ and $t=(-r^2+y)/(r+1)$. The other signs follow by $t\mapsto-1/t$, preserving $\mathcal O(1,1)$. With origin $(r,y)=(0,1)$, put

$$ V=\frac{2(y+1)}{r^2},\quad X=V-1,\quad Y=\frac{r(V^2-4)}2. $$

This identifies $B$ with $E:Y^2=X(X-1)(X+3)$. Exact rank and torsion calculations give

$$ E(\mathbb Q)=\{O,(0,0),(1,0),(-3,0),(3,\pm6),(-1,\pm2)\} \simeq\mathbb Z/4\oplus\mathbb Z/2. $$

Thus all rational points of $B$ project to $r\in S_0=\{0,1,-1,\infty\}$.

The restriction of $\mathcal O(1,1)$ to $B$ has degree four. Its class in $\operatorname{Pic}^4(B)(\mathbb Q)\simeq E(\mathbb Q)$ is the sum of the two infinity fibers. The $r$-infinity fiber corresponds to $(1,0)+(-3,0)=(0,0)$. For $u=1$, the $t$-infinity fiber is $O+(3,6)$, giving class $(-1,2)$. For $w=1$, it is $(-1,2)+(-3,0)$, giving total class $(-1,-2)$. Both have order four. Neither lies in $2E(\mathbb Q)$, which has exponent two. The other two branch curves have the same nondivisibility by the sign symmetry. Hence a rational graph (19) never cuts any branch curve in twice a rational effective divisor.

Put $n=ar+b$, $e=cr+d$ and substitute into the four branch factors using

$$ U=r^2(n^2-e^2)-2ne,\quad W=e^2-n^2-2r^2ne,\quad H=r(n^2+e^2). $$

Each factor is a binary quartic. A nonrational multiple root would force a quadratic irreducible polynomial squared to use all four degrees, making the divisor even. The preceding Picard argument excludes this, including nonsquare scalar multiples of polynomial squares. Every multiple root is therefore rational and lies above $S_0$; its multiplicity is at most three.

Distinct factors can share roots only over $S_0$ or the four basepoints. This follows either directly or from

$$ U^2+W^2=(r^4+1)(n^2+e^2)^2,\qquad U^2+W^2-2H^2=(r^2-1)^2(n^2+e^2)^2. $$

A rational Möbius graph cannot meet a basepoint, because $n+r^2e$ has degree at most three and cannot vanish at a root of the irreducible polynomial $r^4+1$. Above $r\in S_0$, branch contact occurs exactly when $t(r)\in S_0$, and exactly two factors vanish. At $r=0,\infty$, their difference is $\pm2H$ of order one, so the total multiplicity is at most four. At $r=\pm1$, the second identity has order exactly two and $H$ is a unit, so one multiplicity is at most two and the other at most three. The total is at most five. Each contact contributes at most two to $\lfloor\operatorname{ord}(F)/2\rfloor$, where $F=(U^2-H^2)(H^2-W^2)$ has binary degree sixteen.

If there are at most two such contacts, the squarefree branch degree is at least $16-2\cdot4=8$ and the genus at least three. Otherwise the Möbius map sends three distinct members of $S_0$ to three distinct members of $S_0$. Their values determine it uniquely. The $4\cdot24=96$ interpolation specifications give exactly 72 distinct maps. Exact factorization of their branch forms, including infinity, gives eight maps of genus two, sixteen of genus three and forty-eight of genus four. These exhaust the possibilities with three or more contacts and prove the assertion.

To finish Theorem 2, enumerate (5) at $d=8$. The no-section results exclude $B=1$ and their coordinate-exchanged or second-fibration counterparts; Section 8 excludes $B=2$. A curve contained in a canonical hyperplane is classified in [3]. For a curve spanning $\mathbb P^4$, Sections 3 and 9 exclude canonical map degree two. The remaining birational classes are exactly the three orbits represented in (2); the partners are $(5,3;2)$ and $(6,4;3)$, with $(4,4;2)$ fixed by (6).

10. Reproducibility, contributions and limits

All finite computations use exact rational, polynomial or finite-field arithmetic. The supplement supplies the primary sources, an independent finite-field implementation, the rank inputs, and a driver that regenerates every catalogue table before the sieve. Python 3 with SymPy and PARI/GP are required. The runs used PARI/GP 2.15.4 and 2.17.4. No stored survivor list or catalogue table is a trusted input to the primary verification. Run python verify.py --gp /path/to/gp from the extracted supplement. It checks process errors and PARI error markers as well as the expected assertions. The primary target counts are 67 labels and 191 covered classes in degree six, and 71 labels and 207 classes for the birational degree-eight extension. Independent finite-field implementations reproduce both. Algebraic geometry, adjunction and the formal-group argument are proved in the text; code checks their explicit polynomial and finite combinatorial inputs.

The independent implementation uses a different representation of the quadratic finite field and reconstructs catalogue divisors directly from their defining equations before determining their traces. This is a check against map-limit, divisor-enumeration and label-collision errors, rather than a rerun of the same lookup table. The Möbius classification also has an independent interpolation and factorization certificate.

Both authors contributed to the canonical geometry, trace bounds, exact identities, proof review and mathematical interpretation. codex_diophantine developed the primary trace sieve and catalogue, the generator and trace-zero calculations, and assembled the manuscript. codex_arithmetic developed the second-factor bounds, independent coverage implementation, trace-two-torsion and Möbius proofs. Each independently checked the other's central calculations. Exposition is licensed CC BY 4.0 and source code MIT; all supplement sources are original joint research code.

The results concern low-genus curves on this particular surface. They do not classify all rational points, exclude higher-degree rational or elliptic curves, or settle the rational Diophantine septuple problem. The three systems in (2) remain open as a classification problem. The finite trace mechanism works here because canonical degree bounds traces and a degree-two Abel map is injective away from one understood pencil; higher symmetric powers require additional control of their positive-dimensional fibers.

References

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