# Perfect Cuboid Radical-FDRS R16 ## Explicit Canonical Second Elliptic Quotient, Exact Identification with Historical $E_-$, and Dual-Sieve Saturation **Series:** Perfect Cuboid Global Closure **Branch:** Radical-FDRS / Geometry-Coupling Branch R **Phase:** R16 — Canonical Second Quotient and Dual Compatibility **Date:** 2026-10-05 **Status:** Exact quotient-map theorem + finite-prime dual-sieve checkpoint — not a proof of Perfect Cuboid existence or nonexistence --- # 0. Executive summary R15 fixed one canonical historical genus-$2$ curve: $$ \boxed{ \mathcal H_{\rm can}: \qquad \mathcal Y^2 = (2-r^2) \left( M^2r^4 - 2M^2r^2 + L^2 \right), } $$ where $$ \boxed{ M=950625, } $$ and $$ \boxed{ L=1352033. } $$ Its first elliptic quotient is the historical rank-one curve $$ \boxed{ E_0: \qquad y^2 = x^3 - 2x^2 + \left( \frac LM \right)^2x, } $$ through $$ \boxed{ x=2-r^2, \qquad y=\frac{\mathcal Y}{M}. } $$ R16 makes the complementary elliptic quotient completely explicit. The second nonhyperelliptic involution is $$ \boxed{ (r,\mathcal Y) \longmapsto (-r,-\mathcal Y). } $$ Its invariant coordinates are $$ \boxed{ z=r^2, \qquad W=r\mathcal Y. } $$ The quotient is the genus-$1$ quartic $$ \boxed{ W^2 = z(2-z) \left( M^2z^2 - 2M^2z + L^2 \right). } $$ This quartic admits the direct birational cubic model $$ \boxed{ \eta^2 = (\xi-L^2) \left( \xi^2 - 4M^2\xi + 4L^2M^2 \right), } $$ using $$ \boxed{ \xi = \frac{2L^2}{z}, } $$ and $$ \boxed{ \eta = \frac{2L^2W}{z^2}. } $$ Shift $$ u=\xi-L^2. $$ Then $$ \boxed{ \eta^2 = u \left[ u^2 + (2L^2-4M^2)u + L^4 \right]. } $$ The standard $2$-isogeny with kernel $$ (0,0) $$ gives the full-$2$-torsion curve $$ \boxed{ v^2 = U (U-16AB) (U+4M^2), } $$ where $$ A=448^2, \qquad B=1073^2. $$ The isogeny map is $$ \boxed{ U = \frac{\eta^2}{u^2}, } $$ $$ \boxed{ v = \frac{ \eta(L^4-u^2) }{ u^2 }. } $$ This curve is not merely in the historical $E_-$ isogeny class. It is directly $\mathbb Q$-isomorphic to the exact historical integral full-$2$-torsion model $$ \boxed{ E_-: \qquad y_-^2 = (x_--e_1) (x_--e_2) (x_--e_3), } $$ with $$ e_1=-8195043371142, $$ $$ e_2=8256895726659, $$ and $$ e_3=-61852355517. $$ The exact isomorphism is $$ \boxed{ x_- = e_3+\frac94U, } $$ $$ \boxed{ y_- = \frac{27}{8}v. } $$ The root correspondence is: $$ \boxed{ U=-4M^2 \longmapsto e_1, } $$ $$ \boxed{ U=16AB \longmapsto e_2, } $$ and $$ \boxed{ U=0 \longmapsto e_3. } $$ Equivalently, $$ \boxed{ e_1=e_3-9M^2, } $$ and $$ \boxed{ e_2=e_3+36AB. } $$ Because $$ L^2-M^2=4AB, $$ we also have $$ \boxed{ e_2=e_1+9L^2, } $$ and $$ \boxed{ e_3=e_1+9M^2. } $$ Composing every step yields a direct canonical map $$ \boxed{ \pi_-: \mathcal H_{\rm can} \longrightarrow E_-. } $$ In the original canonical coordinates it is: $$ \boxed{ x_- = e_1 + \frac{ 9L^2 }{ r^2(2-r^2) }, } $$ and $$ \boxed{ y_- = \frac{ 27L^2\mathcal Y(r^2-1) }{ r^3(2-r^2)^2 }. } $$ Since $$ x=2-r^2, $$ the $x_-$ formula may also be written: $$ \boxed{ x_- = e_1 + \frac{ 9L^2 }{ x(2-x) }. } $$ Therefore one canonical genus-$2$ point now produces its two rank-one elliptic images immediately: $$ \boxed{ P_0\in E_0(\mathbb Q), } $$ and $$ \boxed{ P_-\in E_-(\mathbb Q), } $$ without returning to the older $h,t,s$ coordinate chains. The boundary anchor $$ x=1 $$ or $$ r=\pm1 $$ maps to $$ \boxed{ (e_2,0)\in E_-(\mathbb Q)[2]. } $$ Thus the boundary is simultaneously: - the point: $$ x=1 $$ on $E_0$; - and the rational $2$-torsion point: $$ (e_2,0) $$ on $E_-$. R16 then tests the newly explicit dual quotient at the two strongest high-order primes found in R15: $$ p_1=11{,}532{,}251, $$ and $$ p_2=15{,}721{,}883. $$ For the historical $E_-$ curve, exact Hasse-certified point counts give: $$ \boxed{ \#E_-(\mathbb F_{p_1}) = 11{,}533{,}200 = 2^4 \cdot3 \cdot5^2 \cdot7 \cdot1373, } $$ and $$ \boxed{ \#E_-(\mathbb F_{p_2}) = 15{,}728{,}976 = 2^4 \cdot3^2 \cdot109229. } $$ At both primes the Sylow-$2$ subgroup therefore has order $$ 16. $$ The historical odd-index non-torsion surrogate $$ \boxed{ P_- = \left( \frac{5132247118878268203}{606841}, \frac{2520702383000286338446320000}{472729139} \right) } $$ together with the rational torsion subgroup $$ \boxed{ E_-(\mathbb Q)_{\rm tors} \cong \mathbb Z/2\mathbb Z \times \mathbb Z/4\mathbb Z } $$ projects to a subgroup of exactly $$ \boxed{ 8 } $$ points inside each $16$-point Sylow-$2$ group. Thus the second quotient is locally nontrivial: half of the $2$-primary elliptic group is excluded by global Mordell-Weil compatibility. Nevertheless, after correctly enumerating the odd-primary preimages of the R11/R15 coefficient states on $E_0$, imposing the square gate $$ 2-x\in\mathbb F_p^2, $$ mapping every lift through the exact R16 formula to $E_-$, and then projecting to the $E_-$ Sylow-$2$ subgroup, all sixteen modulo-$2^{19}$ coefficient lifts remain compatible at both primes. More strongly, at these two primes every odd-primary branch that passes the first square gate also lands in the allowed projected $E_-$ Mordell-Weil image. Hence: $$ \boxed{ \text{the R16 dual finite-prime sieve gives no new elimination at }p_1,p_2. } $$ This is an important saturation checkpoint. The obstruction is no longer missing geometry, an implicit quotient, or oversized local Mordell-Weil subgroups. The relevant local branches themselves are simultaneously compatible with both rank-one elliptic images at these primes. Therefore the next genuine escalation is: $$ \boxed{ \text{elliptic Chabauty / quadratic Chabauty / global height compatibility}. } $$ --- # 1. Canonical genus-$2$ curve Use $$ \boxed{ \mathcal H_{\rm can}: \qquad \mathcal Y^2 = (2-r^2) \left( M^2r^4 - 2M^2r^2 + L^2 \right). } $$ The constants are $$ M=950625, $$ and $$ L=1352033. $$ The curve is even in $$ r. $$ Therefore it has the involutions: $$ \iota_0: (r,\mathcal Y) \mapsto (-r,\mathcal Y), $$ $$ \iota_h: (r,\mathcal Y) \mapsto (r,-\mathcal Y), $$ and $$ \iota_-: (r,\mathcal Y) \mapsto (-r,-\mathcal Y). $$ The first and third are the two nonhyperelliptic involutions. --- # 2. First elliptic quotient The quotient by $$ \iota_0 $$ uses $$ r^2=2-x. $$ The map is $$ \boxed{ x=2-r^2, } $$ $$ \boxed{ y=\frac{\mathcal Y}{M}. } $$ It lands on $$ \boxed{ E_0: \qquad y^2 = x^3 - 2x^2 + \left( \frac LM \right)^2x. } $$ The exact rank theorem gives $$ \boxed{ \operatorname{rank}E_0(\mathbb Q)=1. } $$ --- # 3. Second elliptic quotient coordinates For the quotient by $$ \iota_-, $$ define $$ \boxed{ z=r^2, } $$ and $$ \boxed{ W=r\mathcal Y. } $$ Both are invariant under: $$ (r,\mathcal Y) \mapsto (-r,-\mathcal Y). $$ The quotient equation is $$ \boxed{ W^2 = z(2-z) \left( M^2z^2 - 2M^2z + L^2 \right). } $$ This is a genus-$1$ quartic. --- # 4. Quartic-to-cubic map The quartic has the rational point $$ (z,W)=(0,0). $$ Define $$ \boxed{ \xi = \frac{2L^2}{z}, } $$ and $$ \boxed{ \eta = \frac{2L^2W}{z^2}. } $$ Then direct substitution gives $$ \boxed{ \eta^2 = (\xi-L^2) \left( \xi^2 - 4M^2\xi + 4L^2M^2 \right). } $$ This is the first compact cubic model of the second quotient. --- # 5. Shift the rational $2$-torsion point Set $$ \boxed{ u=\xi-L^2. } $$ Then: $$ \boxed{ \eta^2 = u \left[ u^2 + (2L^2-4M^2)u + L^4 \right]. } $$ The point: $$ \boxed{ (u,\eta)=(0,0) } $$ is rational $2$-torsion. --- # 6. Standard $2$-isogeny For a curve: $$ \eta^2 = u(u^2+au+b), $$ the standard $2$-isogeny with kernel: $$ \{O,(0,0)\} $$ is: $$ \boxed{ U = \frac{\eta^2}{u^2}, } $$ $$ \boxed{ v = \frac{ \eta(b-u^2) }{ u^2 }. } $$ Here: $$ a=2L^2-4M^2, $$ and: $$ b=L^4. $$ The target is: $$ v^2 = U^3 - 2aU^2 + (a^2-4b)U. $$ --- # 7. Factorization of the isogenous target Using: $$ L^2-M^2=4AB, $$ one obtains: $$ \boxed{ a^2-4b = -64ABM^2. } $$ The target factors exactly as: $$ \boxed{ v^2 = U (U-16AB) (U+4M^2). } $$ Thus the $2$-isogenous second quotient has full rational $2$-torsion. Its three roots are: $$ \boxed{ -4M^2, \qquad 0, \qquad 16AB. } $$ --- # 8. Historical integral $E_-$ The exact historical model is: $$ \boxed{ E_-: \qquad y_-^2 = (x_--e_1) (x_--e_2) (x_--e_3), } $$ where: $$ \boxed{ e_1=-8195043371142, } $$ $$ \boxed{ e_2=8256895726659, } $$ and: $$ \boxed{ e_3=-61852355517. } $$ The historical root differences are: $$ \boxed{ e_2-e_1 = 4056099^2, } $$ $$ \boxed{ e_3-e_1 = 2851875^2, } $$ and: $$ \boxed{ e_2-e_3 = 2884224^2. } $$ --- # 9. Root-difference matching The new full-$2$-torsion model has differences: $$ 16AB, $$ $$ 4M^2, $$ and: $$ 4L^2. $$ Their square roots are: $$ 4\sqrt{AB} = 1922816, $$ $$ 2M = 1901250, $$ and: $$ 2L = 2704066. $$ Multiplying each by: $$ \frac32 $$ gives: $$ 2884224, $$ $$ 2851875, $$ and: $$ 4056099. $$ These are exactly the historical root-difference square roots. Thus the $x$-scaling factor is: $$ \boxed{ \left( \frac32 \right)^2 = \frac94. } $$ --- # 10. Exact historical root identities The historical roots satisfy: $$ \boxed{ e_1=e_3-9M^2, } $$ and: $$ \boxed{ e_2=e_3+36AB. } $$ Equivalently: $$ \boxed{ e_2=e_1+9L^2, } $$ and: $$ \boxed{ e_3=e_1+9M^2. } $$ These identities are exact integer equalities. --- # 11. Direct isomorphism to historical $E_-$ Define: $$ \boxed{ x_- = e_3+\frac94U, } $$ and: $$ \boxed{ y_- = \frac{27}{8}v. } $$ Then: $$ U=-4M^2 $$ maps to: $$ e_1, $$ $$ U=0 $$ maps to: $$ e_3, $$ and: $$ U=16AB $$ maps to: $$ e_2. $$ The cubic equation becomes exactly: $$ \boxed{ y_-^2 = (x_--e_1) (x_--e_2) (x_--e_3). } $$ Therefore the full-$2$-torsion isogenous quotient is $\mathbb Q$-isomorphic to the historical integral: $$ E_-. $$ This upgrades the older G2E.5 statement from: $$ \text{same isogeny class} $$ to an explicit composition of: 1. quotient; 2. birational cubic conversion; 3. degree-$2$ isogeny; 4. rational isomorphism. --- # 12. Direct composite map: $x_-$ Now eliminate all intermediate coordinates. Recall: $$ z=r^2. $$ The isogenous coordinate satisfies: $$ \boxed{ U = \frac{ 4 \left( M^2z^2 - 2M^2z + L^2 \right) }{ z(2-z) }. } $$ Using: $$ L^2-M^2=4AB, $$ the historical coordinate simplifies to: $$ \boxed{ x_- = e_1 + \frac{ 9L^2 }{ z(2-z) }. } $$ Since: $$ z=r^2, $$ we obtain: $$ \boxed{ x_- = e_1 + \frac{ 9L^2 }{ r^2(2-r^2) }. } $$ --- # 13. Direct composite map: $y_-$ The complete $y$-coordinate simplifies to: $$ \boxed{ y_- = \frac{ 27L^2\mathcal Y(r^2-1) }{ r^3(2-r^2)^2 }. } $$ Therefore the canonical second quotient map is: $$ \boxed{ \pi_-(r,\mathcal Y) = \left( e_1+ \frac{ 9L^2 }{ r^2(2-r^2) }, \; \frac{ 27L^2\mathcal Y(r^2-1) }{ r^3(2-r^2)^2 } \right). } $$ This is defined on the natural affine open where the denominators are nonzero. The finite exceptional points are treated on the projective normalization. --- # 14. Direct $E_0$-to-$E_-$ $x$ relation Because: $$ x=2-r^2, $$ we have: $$ r^2=2-x. $$ Thus: $$ \boxed{ x_- = e_1 + \frac{ 9L^2 }{ x(2-x) }. } $$ This is one of the simplest formulas in the entire fixed-slice program. A point on the canonical square-value cover determines the second elliptic $x$-coordinate by one rational function of the first elliptic $x$-coordinate. --- # 15. Factor proof of the direct map Let: $$ z=r^2, $$ and: $$ x=2-z. $$ Then: $$ x_--e_1 = \frac{ 9L^2 }{ z(2-z) }. $$ Also: $$ \boxed{ x_--e_2 = \frac{ 9L^2(z-1)^2 }{ z(2-z) }. } $$ Finally: $$ \boxed{ x_--e_3 = \frac{ 9\mathcal Y^2 }{ z(2-z)^2 }. } $$ Therefore: $$ (x_--e_1) (x_--e_2) (x_--e_3) $$ is exactly: $$ \boxed{ \left[ \frac{ 27L^2\mathcal Y(z-1) }{ z^{3/2}(2-z)^2 } \right]^2, } $$ which is: $$ y_-^2. $$ This directly verifies the composite map. --- # 16. Boundary synchronization At the historical boundary: $$ r=\pm1. $$ Then: $$ x=1. $$ The second quotient gives: $$ \boxed{ x_- = e_1+9L^2 = e_2. } $$ Also: $$ \boxed{ y_-=0. } $$ Therefore: $$ \boxed{ r=\pm1 \longmapsto (e_2,0). } $$ The canonical boundary is: - a non-torsion rank-one anchor on $E_0$; - but a rational $2$-torsion anchor on $E_-$. This asymmetry is useful for local compatibility calculations. --- # 17. Rank and torsion data on $E_-$ The historical exact results are: $$ \boxed{ \operatorname{rank}E_-(\mathbb Q)=1, } $$ and: $$ \boxed{ E_-(\mathbb Q)_{\rm tors} \cong \mathbb Z/2\mathbb Z \times \mathbb Z/4\mathbb Z. } $$ An explicit order-$4$ point is: $$ \boxed{ Q_- = ( 19955593808835, 81192743369781982848 ), } $$ with: $$ \boxed{ 2Q_-=(e_2,0). } $$ Thus the boundary second-quotient image is precisely the double of the historical rational order-$4$ torsion point. --- # 18. Odd-index free surrogate on $E_-$ Use the historical non-torsion point: $$ \boxed{ P_- = \left( \frac{ 5132247118878268203 }{ 606841 }, \frac{ 2520702383000286338446320000 }{ 472729139 } \right). } $$ Its Kummer class is: $$ \boxed{ (1,406,406). } $$ The exact descent and Cassels-pairing computation proves that this class spans the free Mordell-Weil direction modulo $2$. Therefore $P_-$ differs from a primitive free generator only by odd index and torsion. It is a proof-safe surrogate for all Sylow-$2$ Mordell-Weil compatibility tests. --- # 19. Canonical dual-quotient condition A rational point: $$ R\in\mathcal H_{\rm can}(\mathbb Q) $$ produces: $$ \boxed{ \pi_0(R)\in E_0(\mathbb Q), } $$ and: $$ \boxed{ \pi_-(R)\in E_-(\mathbb Q). } $$ At a good prime: $$ p, $$ write: $$ \#E_0(\mathbb F_p) = 2^{s_0}m_0, $$ and: $$ \#E_-(\mathbb F_p) = 2^{s_-}m_-, $$ with: $$ m_0, m_- $$ odd. A necessary condition is simultaneous membership: $$ \boxed{ [m_0]\pi_0(R) \in \langle [m_0]Q, E_0(\mathbb Q)_{\rm tors} \rangle, } $$ and: $$ \boxed{ [m_-]\pi_-(R) \in \langle [m_-]P_-, E_-(\mathbb Q)_{\rm tors} \rangle. } $$ The candidate points themselves must be projected by the same maps. --- # 20. R15 high-order prime $p_1$ Use: $$ \boxed{ p_1=11{,}532{,}251. } $$ R15 established: $$ \boxed{ \#E_0(\mathbb F_{p_1}) = 11\cdot2^{20}, } $$ and: $$ \boxed{ \operatorname{ord}([11]Q)=2^{19}. } $$ For the second quotient, exact Hasse-certified counting gives: $$ \boxed{ \#E_-(\mathbb F_{p_1}) = 11{,}533{,}200. } $$ Its factorization is: $$ \boxed{ 11{,}533{,}200 = 2^4 \cdot3 \cdot5^2 \cdot7 \cdot1373. } $$ Hence: $$ \boxed{ \#E_-(\mathbb F_{p_1})[2^\infty] = 16. } $$ --- # 21. Exact $E_-$ count certificate at $p_1$ The point: $$ \boxed{ (2,9036800) \in E_-(\mathbb F_{p_1}) } $$ has exact order: $$ \boxed{ 5{,}766{,}600. } $$ The Hasse interval length is less than: $$ 4\sqrt{p_1} < 13{,}584. $$ Because: $$ 5{,}766{,}600 > 4\sqrt{p_1}, $$ there is at most one multiple of the certified point order inside the Hasse interval. The displayed group order is that unique multiple. Thus: $$ \#E_-(\mathbb F_{p_1}) = 11{,}533{,}200 $$ is exact. --- # 22. Projected $E_-$ Mordell-Weil image at $p_1$ The odd cofactor is: $$ \boxed{ m_-=720825. } $$ Project: $$ P_- $$ by: $$ [m_-]. $$ Its projected order is: $$ \boxed{ 2. } $$ Together with the eight rational torsion reductions, the projected global Mordell-Weil image contains exactly: $$ \boxed{ 8 } $$ points. Thus it is a proper index-$2$ subgroup of the $16$-point Sylow-$2$ group. The second quotient therefore supplies a genuine local compatibility restriction at this prime. --- # 23. Dual test at $p_1$ The R11 packet has eight classes modulo: $$ 2^{18}. $$ Lift them to all sixteen classes modulo: $$ 2^{19}. $$ For each coefficient class: 1. reconstruct all eleven odd-primary preimages on: $$ E_0(\mathbb F_{p_1}); $$ 2. retain preimages satisfying: $$ 2-x\in\mathbb F_{p_1}^2; $$ 3. choose both square roots: $$ r=\pm\sqrt{2-x}; $$ 4. map each lift through: $$ \pi_-; $$ 5. project the result to: $$ E_-(\mathbb F_{p_1})[2^\infty]; $$ 6. require membership in the eight-point projected global Mordell-Weil subgroup. The result is: $$ \boxed{ \text{all sixteen coefficient lifts survive}. } $$ --- # 24. Stronger statement at $p_1$ At this prime, the number of odd-primary $E_0$ branches passing the first square gate ranges from: $$ 2 $$ to: $$ 7 $$ per coefficient state. After imposing the second elliptic quotient compatibility, the counts are unchanged. Therefore: $$ \boxed{ \text{every first-gate branch already satisfies the projected }E_-\text{ condition}. } $$ This is stronger than merely saying that each coefficient class has one surviving branch. --- # 25. R15 high-order prime $p_2$ Use: $$ \boxed{ p_2=15{,}721{,}883. } $$ R15 established: $$ \boxed{ \#E_0(\mathbb F_{p_2}) = 15\cdot2^{20}, } $$ and: $$ \boxed{ \operatorname{ord}([15]Q)=2^{19}. } $$ For: $$ E_-, $$ exact Hasse-certified counting gives: $$ \boxed{ \#E_-(\mathbb F_{p_2}) = 15{,}728{,}976. } $$ Factorization: $$ \boxed{ 15{,}728{,}976 = 2^4 \cdot3^2 \cdot109229. } $$ Again: $$ \boxed{ \#E_-(\mathbb F_{p_2})[2^\infty] = 16. } $$ --- # 26. Exact $E_-$ count certificate at $p_2$ The point: $$ \boxed{ (5,14406551) \in E_-(\mathbb F_{p_2}) } $$ has exact order: $$ \boxed{ 7{,}864{,}488. } $$ The Hasse interval length is less than: $$ 4\sqrt{p_2} < 15{,}861. $$ Because: $$ 7{,}864{,}488 > 4\sqrt{p_2}, $$ the group order is uniquely determined in the Hasse interval. Hence: $$ \boxed{ \#E_-(\mathbb F_{p_2}) = 15{,}728{,}976 } $$ is exact. --- # 27. Projected $E_-$ Mordell-Weil image at $p_2$ The odd cofactor is: $$ \boxed{ m_-=983061. } $$ The projected non-torsion surrogate has order: $$ \boxed{ 2. } $$ Together with rational torsion, the projected global Mordell-Weil image again has exactly: $$ \boxed{ 8 } $$ points inside a $16$-point Sylow-$2$ group. Thus the local second-quotient condition is again genuinely nontrivial. --- # 28. Dual test at $p_2$ Repeat the exact R11-to-$2^{19}$ lift enumeration. The odd-primary kernel on the first elliptic quotient has order: $$ 15. $$ For every one of the sixteen coefficient lifts: - at least one odd-primary preimage satisfies: $$ 2-x\in\mathbb F_{p_2}^2; $$ - every such first-gate branch maps into the allowed projected: $$ E_- $$ Mordell-Weil image. Therefore: $$ \boxed{ \text{all sixteen coefficient lifts survive}. } $$ --- # 29. Stronger statement at $p_2$ The number of first-gate-admissible odd-primary branches ranges from: $$ 5 $$ to: $$ 12. $$ The dual $E_-$ compatibility test leaves the same counts. Thus the second quotient again produces: $$ \boxed{ \text{zero branch-level elimination}. } $$ at this prime. --- # 30. Meaning of the dual-sieve saturation The second quotient is not vacuous globally. Its projected Mordell-Weil image occupies only half of the local Sylow-$2$ group at both new primes. Nevertheless the specific local branches arising from the canonical square-value cover all land inside that allowed half. Therefore the current failure to eliminate candidates is not caused by: - a missing explicit quotient; - an oversized rational torsion group; - a full local Mordell-Weil image; - or insufficient coefficient resolution. It is caused by genuine local compatibility of the surviving branches. This is a substantially stronger saturation statement than R15. --- # 31. Historical clean primes revisited At the historical clean primes: $$ 11, 59, 181, 197, 1669, $$ the projected $E_0$ order of: $$ Q $$ is at most: $$ 2^3. $$ All R11 coefficient classes therefore collapse to the boundary sign classes on the first quotient. The boundary genus-$2$ lifts map to: $$ (e_2,0) $$ on the second quotient. Since this is rational torsion, it automatically lies in the projected global Mordell-Weil image. Therefore the canonical genus-$2$ dual quotient alone cannot use those low-order primes to resolve the high bits of the R11 packet. The earlier clean-prime strength came from the larger genus-$3$ compatibility architecture, not from this single genus-$2$ quotient pair alone. --- # 32. Canonical bielliptic diagram The complete canonical genus-$2$ picture is now: $$ \boxed{ \mathcal H_{\rm can} } $$ with: $$ \boxed{ \pi_0: \mathcal H_{\rm can} \to E_0, } $$ and: $$ \boxed{ \pi_-: \mathcal H_{\rm can} \to E_-. } $$ The Jacobian split is: $$ \boxed{ \operatorname{Jac} (\mathcal H_{\rm can}) \sim_{\mathbb Q} E_0 \times E_-. } $$ Both factors have exact rank: $$ \boxed{ 1. } $$ Thus: $$ \boxed{ \operatorname{rank} \operatorname{Jac} (\mathcal H_{\rm can})(\mathbb Q) = 2 = g(\mathcal H_{\rm can}). } $$ Ordinary Chabauty does not apply directly. --- # 33. Quadratic-Chabauty criterion The two elliptic factors are nonisogenous as the named distinct factors of the historical decomposition. Therefore the Néron-Severi rank satisfies at least: $$ \boxed{ \rho \left( \operatorname{Jac} (\mathcal H_{\rm can}) \right) \ge2. } $$ Hence: $$ g+\rho-1 \ge 2+2-1 = 3. $$ Since: $$ r=2, $$ we have: $$ \boxed{ r