# Perfect Cuboid Radical-FDRS R4

## Conic Base Fibers, Elliptic Square-Lift Gates, and the Five-Elliptic Decomposition of the Fixed Fiber

**Series:** Perfect Cuboid Global Closure  
**Branch:** Radical-FDRS / Geometry-Coupling Branch R  
**Phase:** R4 — Quotient-Fiber Correction and Elliptic Decomposition  
**Date:** 2026-10-03  
**Status:** Exact structural analysis and corrigendum — not a proof of Perfect Cuboid existence or nonexistence

---

# 0. Abstract

R3 introduced the Cayley square-lift quotient surface

$$
\mathcal B:
\qquad
\Phi(R)+\Phi(S)+\Phi(T)=2,
$$

where

$$
\Phi(X)
=
\frac{4X}{(1+X)^2}.
$$

R3 also stated that fixing one Cayley coordinate produces a generic bidegree-$(2,2)$ genus-$1$ fiber.

R4 corrects that statement.

The fixed-coordinate curve is indeed represented by a bidegree-$(2,2)$ equation after denominators are cleared, but that model contains a boundary singularity. Its normalization is birational to a conic and therefore has genus

$$
\boxed{
0.
}
$$

The first genuine genus-$1$ object appears only after imposing one square-lift condition.

The second simultaneous square lift raises the complete fixed fiber to genus

$$
\boxed{
5.
}
$$

This gives the exact FDRS complexity ladder:

$$
\boxed{
\text{quotient base}
\;g=0
\longrightarrow
\text{one square lift}
\;g=1
\longrightarrow
\text{two simultaneous square lifts}
\;g=5.
}
$$

The two individual square-lift quartics admit explicit elliptic cubic models with full rational $2$-torsion.

The genus-$3$ product subcover has a second Klein-four involution structure generated by

$$
m\mapsto-m
$$

and

$$
m\mapsto\frac1m.
$$

Its Jacobian therefore splits, up to isogeny, into three elliptic quotient factors.

Combining the two gate elliptics with these three genus-$3$ quotient elliptics yields a generic five-elliptic decomposition of the genus-$5$ fixed fiber.

This gives a structural explanation for why the previously studied $n=7/32$ fixed slice admitted a five-dimensional elliptic decomposition of the form

$$
E_f^2\times E_0^2\times E_-.
$$

R4 does not identify the five generic factors with those historical named curves without an additional explicit isogeny check, but it explains why exactly five elliptic factors should appear.

---

# 1. Corrigendum to the R3 fixed-coordinate genus statement

R3 considered the surface

$$
\mathcal B:
\qquad
\Phi(R)+\Phi(S)+\Phi(T)=2.
$$

Fix

$$
R=R_0\in\mathbb Q.
$$

After clearing denominators, the remaining equation in

$$
S,T
$$

has bidegree

$$
(2,2).
$$

A smooth bidegree-$(2,2)$ curve in

$$
\mathbb P^1\times\mathbb P^1
$$

has genus

$$
1.
$$

However, the fixed-$R$ model arising here is not a generic smooth bidegree-$(2,2)$ curve.

Its denominator-cleared model contains a boundary singularity at

$$
S=T=-1.
$$

The correct geometric genus is obtained after normalization.

---

# 2. Direct birational reduction to a conic

Define

$$
a
=
\frac{1-R_0}{1+R_0},
$$

$$
b
=
\frac{1-S}{1+S},
$$

$$
c
=
\frac{1-T}{1+T}.
$$

Because

$$
\Phi(R)=1-a^2,
$$

$$
\Phi(S)=1-b^2,
$$

and

$$
\Phi(T)=1-c^2,
$$

the fixed-fiber equation becomes

$$
(1-a^2)+(1-b^2)+(1-c^2)=2.
$$

Hence

$$
\boxed{
b^2+c^2=1-a^2.
}
$$

Therefore the normalization of the fixed-$R$ fiber is a conic.

Whenever the fiber contains one rational point, it is rationally parameterizable.

Thus:

$$
\boxed{
g_{\rm base}=0.
}
$$

---

# 3. Why the bidegree count overestimated the genus

The denominator-cleared fixed-fiber model has arithmetic genus

$$
1,
$$

but the boundary singularity lowers its geometric genus by

$$
1.
$$

Therefore:

$$
\boxed{
g_{\rm geom}
=
1-1
=
0.
}
$$

This is a concrete example of the FDRS warning that representation degree alone does not determine true structural complexity.

A singular compactification can make a rational object appear elliptic until the boundary is normalized.

---

# 4. Rational parameterization of the conic

Assume

$$
1-a^2=d^2
$$

for rational

$$
d>0.
$$

This is exactly the case when the fixed Cayley coordinate

$$
R_0
$$

is already a rational square.

Parameterize

$$
b^2+c^2=d^2
$$

by

$$
m\in\mathbb Q:
$$

$$
\boxed{
b
=
d
\frac{1-m^2}{1+m^2},
}
$$

$$
\boxed{
c
=
d
\frac{2m}{1+m^2}.
}
$$

The fixed closed complementary pair can itself be written as

$$
a
=
\frac{2n}{1+n^2},
$$

$$
d
=
\frac{1-n^2}{1+n^2},
$$

or with the complementary convention exchanged.

This recovers exactly the nested-Pythagorean coordinates used in R2.

---

# 5. Return of the Twin-Psi gates

After the conic is parameterized by

$$
m,
$$

the remaining two face closures become the R2 Twin-Psi conditions.

Define

$$
\Psi(x,n)
=
\frac{x^2+n^2}{1+x^2n^2},
$$

and

$$
\mu(m)
=
\frac{1-m}{1+m}.
$$

Then the two remaining square gates are

$$
\boxed{
\Psi(m,n)\in\mathbb Q^2,
}
$$

and

$$
\boxed{
\Psi(\mu(m),n)\in\mathbb Q^2.
}
$$

Thus the R3 quotient surface and the R2 Twin-Psi model are related by

$$
\boxed{
\text{quotient conic}
\longrightarrow
\text{one-dimensional rational parameter }m
\longrightarrow
\text{two square-lift covers}.
}
$$

---

# 6. First square-lift quartic

For fixed rational

$$
n,
$$

write

$$
u^2
=
\Psi(m,n).
$$

Define

$$
Y_1
=
u(1+m^2n^2).
$$

Then

$$
\boxed{
Y_1^2
=
F_1(m;n),
}
$$

where

$$
\boxed{
F_1(m;n)
=
(m^2+n^2)(1+m^2n^2).
}
$$

Expanded:

$$
\boxed{
F_1(m;n)
=
n^2m^4
+
(1+n^4)m^2
+
n^2.
}
$$

This is an even reciprocal quartic.

---

# 7. Second square-lift quartic

The second gate is

$$
v^2
=
\Psi(\mu(m),n).
$$

After clearing the square denominator, define

$$
Y_2
$$

so that

$$
\boxed{
Y_2^2
=
F_2(m;n),
}
$$

with

$$
\boxed{
F_2(m;n)
=
(1+n^2)^2(m^4+1)
-
2(1-6n^2+n^4)m^2.
}
$$

Equivalently,

$$
F_2
=
A(m;n)B(m;n),
$$

where

$$
A(m;n)
=
(1-m)^2+n^2(1+m)^2,
$$

and

$$
B(m;n)
=
(1+m)^2+n^2(1-m)^2.
$$

This is also an even reciprocal quartic.

---

# 8. Exact discriminants

The discriminant of the first quartic is

$$
\boxed{
\operatorname{Disc}_m(F_1)
=
16n^4
(n-1)^4
(n+1)^4
(n^2+1)^4.
}
$$

The discriminant of the second quartic is

$$
\boxed{
\operatorname{Disc}_m(F_2)
=
65536n^4
(n-1)^4
(n+1)^4
(n^2+1)^4.
}
$$

Thus for rational physical fibers

$$
0<n<1,
$$

both quartics are separable.

---

# 9. Branch-locus intersection

The resultant of the two quartics is

$$
\boxed{
\operatorname{Res}_m(F_1,F_2)
=
(n-1)^8
(n+1)^8
(n^4+6n^2+1)^4.
}
$$

For real

$$
n,
$$

the factor

$$
n^4+6n^2+1
$$

is strictly positive.

Therefore for

$$
0<n<1,
$$

the two branch loci are disjoint.

Hence the generic physical fixed fiber has the expected smooth simultaneous-cover structure.

---

# 10. Complexity ladder

The fixed-$n$ arithmetic problem has three levels.

## Level 0 — quotient base

The normalized fixed-coordinate geometry is a conic:

$$
\boxed{
g=0.
}
$$

## Level 1 — one square gate

Either

$$
Y_1^2=F_1
$$

or

$$
Y_2^2=F_2
$$

is a separable quartic double cover:

$$
\boxed{
g=1.
}
$$

## Level 2 — both square gates

The simultaneous fiber product

$$
C_n:
\qquad
Y_1^2=F_1,
\qquad
Y_2^2=F_2
$$

has

$$
\boxed{
g=5.
}
$$

Therefore the correct FDRS hierarchy is

$$
\boxed{
0\rightarrow1\rightarrow5.
}
$$

---

# 11. Quartic-to-elliptic lemma

Consider the symmetric quartic

$$
\boxed{
y^2
=
\alpha^2m^4
+
\beta m^2
+
\alpha^2.
}
$$

Assume

$$
\alpha\ne0.
$$

Define

$$
X
=
\frac{y+\alpha}{m^2}.
$$

Then

$$
y
=
Xm^2-\alpha.
$$

Substitution gives

$$
(X^2-\alpha^2)m^2
=
2\alpha X+\beta.
$$

Now define

$$
Y
=
m(X^2-\alpha^2).
$$

Then

$$
\boxed{
Y^2
=
(X^2-\alpha^2)
(2\alpha X+\beta).
}
$$

Thus every reciprocal even quartic of the current form has an explicit elliptic cubic model with three rational roots whenever

$$
\alpha,\beta\in\mathbb Q.
$$

---

# 12. Normalized first gate elliptic family

For

$$
F_1,
$$

take

$$
\alpha=n,
$$

and

$$
\beta=1+n^4.
$$

After the rational scaling

$$
X=n\xi,
$$

the elliptic model becomes

$$
\boxed{
\mathcal E_1(n):
\qquad
\eta^2
=
(\xi^2-1)
\left(
2n^2\xi+1+n^4
\right).
}
$$

Its three rational $2$-torsion $x$-coordinates are

$$
\boxed{
\xi=1,
}
$$

$$
\boxed{
\xi=-1,
}
$$

and

$$
\boxed{
\xi
=
-\frac{1+n^4}{2n^2}.
}
$$

Therefore

$$
\boxed{
\mathcal E_1(n)
}
$$

has full rational $2$-torsion for every nondegenerate rational

$$
n.
$$

---

# 13. Normalized second gate elliptic family

For

$$
F_2,
$$

take

$$
\alpha=1+n^2,
$$

and

$$
\beta=-2(1-6n^2+n^4).
$$

After scaling

$$
X=(1+n^2)\xi,
$$

we obtain

$$
\boxed{
\mathcal E_2(n):
\qquad
\eta^2
=
(\xi^2-1)
\left[
2(1+n^2)^2\xi
-
2(1-6n^2+n^4)
\right].
}
$$

The three rational $2$-torsion $x$-coordinates are

$$
\boxed{
\xi=1,
}
$$

$$
\boxed{
\xi=-1,
}
$$

and

$$
\boxed{
\xi
=
\frac{
1-6n^2+n^4
}{
(1+n^2)^2
}.
}
$$

Thus

$$
\boxed{
\mathcal E_2(n)
}
$$

also has full rational $2$-torsion.

---

# 14. Why $2$-descent is structurally natural

The repeated appearance of full-$2$ descent in the historical fixed-slice program is not accidental.

At the first square-lift level, both natural elliptic gate families already possess full rational

$$
2
$$

-torsion.

Therefore:

- Kummer maps;
- $2$-isogenies;
- full-$2$ descent;
- local squareclass sieves;

are intrinsic to the geometry of the Twin-Psi lift.

They are not merely computational tricks chosen after the fact.

---

# 15. Example: the $n=7/32$ closed first gate

Take

$$
n=\frac7{32},
$$

and the G2 point

$$
m=\frac47.
$$

The first Twin-Psi gate is

$$
u=\frac{17}{28}.
$$

The quartic point satisfies

$$
Y_1
=
u(1+m^2n^2)
=
\boxed{
\frac{1105}{1792}.
}
$$

Under the quartic-to-cubic map, one obtains the rational point

$$
\boxed{
X
=
\frac{10479}{4096},
}
$$

$$
\boxed{
Y
=
\frac{15572375}{4194304}
}
$$

on the unscaled cubic model.

This serves as an exact sanity check for the elliptic conversion.

---

# 16. Example: the $n=11/45$ closed first gate

Take

$$
n=\frac{11}{45},
$$

and

$$
m=\frac35.
$$

The closed first gate is

$$
u=\frac{25}{39}.
$$

Then

$$
Y_1
=
\boxed{
\frac{442}{675}.
}
$$

The associated cubic point is

$$
\boxed{
X
=
\frac{607}{243},
}
$$

$$
\boxed{
Y
=
\frac{9123016}{2460375}.
}
$$

Thus both historical rational fibers map cleanly into the same generic elliptic gate family.

---

# 17. The genus-$3$ product subcover

Define

$$
\boxed{
H_n:
\qquad
Y^2
=
F_1(m;n)F_2(m;n).
}
$$

Since

$$
\deg_m(F_1F_2)=8,
$$

and the branch locus is separable for physical nondegenerate

$$
n,
$$

we have

$$
\boxed{
g(H_n)=3.
}
$$

Its discriminant is

$$
\boxed{
\operatorname{Disc}_m(F_1F_2)
=
1048576
n^8
(n-1)^{24}
(n+1)^{24}
(n^2+1)^8
(n^4+6n^2+1)^8.
}
$$

Thus the physical interval

$$
0<n<1
$$

contains no generic discriminant degeneration.

---

# 18. Palindromic structure of the genus-$3$ curve

Write

$$
F_1
=
a_1m^4+b_1m^2+a_1,
$$

with

$$
a_1=n^2,
$$

$$
b_1=1+n^4,
$$

and

$$
F_2
=
a_2m^4+b_2m^2+a_2,
$$

with

$$
a_2=(1+n^2)^2,
$$

$$
b_2=-2(1-6n^2+n^4).
$$

Therefore

$$
F_i(1/m)
=
\frac{
F_i(m)
}{
m^4
}.
$$

The product curve has two visible non-hyperelliptic involutions:

$$
\boxed{
\sigma:
m\mapsto-m,
}
$$

and

$$
\boxed{
\rho:
m\mapsto\frac1m,
\qquad
Y\mapsto\frac{Y}{m^4}.
}
$$

They commute.

Their product is

$$
\boxed{
\sigma\rho:
m\mapsto-\frac1m.
}
$$

Thus the genus-$3$ product cover carries a Klein-four symmetry.

---

# 19. First genus-$3$ elliptic quotient

Quotient by

$$
\sigma:
m\mapsto-m.
$$

Set

$$
x=m^2.
$$

Then

$$
\boxed{
\mathcal Q_0(n):
}
$$

is

$$
Y^2
=
\left[
n^2x^2
+
(1+n^4)x
+
n^2
\right]
$$

$$
\qquad\qquad\cdot
\left[
(1+n^2)^2x^2
-
2(1-6n^2+n^4)x
+
(1+n^2)^2
\right].
$$

This is a quartic genus-$1$ curve.

Its discriminant is

$$
\boxed{
-64n^2
(n-1)^{12}
(n+1)^{12}
(n^2+1)^2
(n^4+6n^2+1)^4.
}
$$

---

# 20. Reciprocal genus-$3$ elliptic quotient

Quotient by

$$
\rho:
m\mapsto\frac1m.
$$

Set

$$
z=m+\frac1m,
$$

and

$$
W=\frac{Y}{m^2}.
$$

Since

$$
m^2+\frac1{m^2}=z^2-2,
$$

the quotient becomes

$$
\boxed{
\mathcal Q_+(n):
}
$$

$$
W^2
=
\left[
n^2z^2
+
(1-n^2)^2
\right]
$$

$$
\qquad\qquad\cdot
\left[
(1+n^2)^2z^2
-
4(1-n^2)^2
\right].
$$

This is again genus

$$
1.
$$

Its discriminant is

$$
\boxed{
-64n^2
(n-1)^{12}
(n+1)^{12}
(n^2+1)^2
(n^4+6n^2+1)^4.
}
$$

---

# 21. Anti-reciprocal genus-$3$ elliptic quotient

Quotient by

$$
\sigma\rho:
m\mapsto-\frac1m.
$$

Set

$$
w=m-\frac1m,
$$

and again

$$
W=\frac{Y}{m^2}.
$$

Since

$$
m^2+\frac1{m^2}=w^2+2,
$$

we obtain

$$
\boxed{
\mathcal Q_-(n):
}
$$

$$
W^2
=
\left[
n^2w^2
+
(1+n^2)^2
\right]
$$

$$
\qquad\qquad\cdot
\left[
(1+n^2)^2w^2
+
16n^2
\right].
$$

Its discriminant is

$$
\boxed{
256n^4
(n-1)^8
(n+1)^8
(n^2+1)^4
(n^4+6n^2+1)^4.
}
$$

For

$$
0<n<1,
$$

it is nonsingular and genus

$$
1.
$$

---

# 22. Genus-$3$ Jacobian splitting

The genus-$3$ curve

$$
H_n
$$

has the commuting involutions

$$
\sigma,
\qquad
\rho,
\qquad
\sigma\rho.
$$

The three nontrivial quotient curves are

$$
\mathcal Q_0(n),
$$

$$
\mathcal Q_+(n),
$$

and

$$
\mathcal Q_-(n),
$$

all of genus

$$
1.
$$

Therefore, under the standard smooth nondegenerate hypotheses, the Jacobian decomposes up to isogeny as

$$
\boxed{
\operatorname{Jac}(H_n)
\sim_{\mathbb Q}
\mathcal Q_0(n)
\times
\mathcal Q_+(n)
\times
\mathcal Q_-(n).
}
$$

The dimensions match:

$$
3=1+1+1.
$$

---

# 23. First Klein-four decomposition of the genus-$5$ fiber

The full fixed fiber is

$$
C_n:
\qquad
Y_1^2=F_1,
\qquad
Y_2^2=F_2.
$$

Its deck transformations independently change the signs of

$$
Y_1
$$

and

$$
Y_2.
$$

The three nontrivial quadratic quotients are:

1. the first gate elliptic curve:
   $$
   Y_1^2=F_1;
   $$

2. the second gate elliptic curve:
   $$
   Y_2^2=F_2;
   $$

3. the product genus-$3$ curve:
   $$
   Y^2=F_1F_2.
   $$

Therefore

$$
\boxed{
\operatorname{Jac}(C_n)
\sim_{\mathbb Q}
\mathcal E_1(n)
\times
\mathcal E_2(n)
\times
\operatorname{Jac}(H_n).
}
$$

The dimensions are

$$
5=1+1+3.
$$

---

# 24. Five-elliptic decomposition

Insert the genus-$3$ splitting:

$$
\operatorname{Jac}(H_n)
\sim
\mathcal Q_0(n)
\times
\mathcal Q_+(n)
\times
\mathcal Q_-(n).
$$

Then

$$
\boxed{
\operatorname{Jac}(C_n)
\sim_{\mathbb Q}
\mathcal E_1(n)
\times
\mathcal E_2(n)
\times
\mathcal Q_0(n)
\times
\mathcal Q_+(n)
\times
\mathcal Q_-(n).
}
$$

Thus the generic genus-$5$ fixed fiber is controlled by five explicit elliptic factors.

This is the central structural result of R4.

---

# 25. Relation to the historical $7/32$ decomposition

The previous exact fixed-slice program at

$$
n=\frac7{32}
$$

found a Jacobian decomposition of the form

$$
\boxed{
E_f^2
\times
E_0^2
\times
E_-.
}
$$

R4 independently predicts five elliptic factors at every generic fixed

$$
n.
$$

Therefore the historical multiplicity pattern can be interpreted as special isogeny identifications among the five generic factors at

$$
n=\frac7{32}.
$$

However, R4 does not yet identify which of

$$
\mathcal E_1,
\mathcal E_2,
\mathcal Q_0,
\mathcal Q_+,
\mathcal Q_-
$$

corresponds to

$$
E_f,
E_0,
E_-.
$$

That exact isogeny matching is reserved for the next phase.

No unverified factor identification is promoted here.

---

# 26. Why the five-factor view is useful

Direct genus-$5$ analysis asks for rational points on

$$
C_n.
$$

The five-factor view instead allows the problem to be decomposed into:

- elliptic Mordell-Weil groups;
- $2$-Selmer data;
- local images;
- compatibility among quotient maps;
- simultaneous lifting conditions.

Thus the fixed-fiber problem can be reorganized as

$$
\boxed{
\text{five elliptic arithmetic systems}
+
\text{compatibility constraints}.
}
$$

This does not eliminate the arithmetic difficulty.

It redistributes it into lower-genus components.

That is exactly the FDRS objective.

---

# 27. Bad fibers

The individual quartics degenerate at

$$
n=0,
$$

or

$$
n=\pm1.
$$

The branch loci of

$$
F_1
$$

and

$$
F_2
$$

intersect when

$$
n=\pm1,
$$

or when

$$
n^4+6n^2+1=0.
$$

The latter has no real solution.

Therefore the physical rational region

$$
\boxed{
0<n<1
}
$$

avoids all generic bad fibers.

This means the entire positive Perfect-Cuboid search region lies inside one smooth algebraic regime of the fixed-fiber family.

---

# 28. Global family interpretation

The previous high-genus studies fixed

$$
n
$$

first.

R4 suggests reversing the order.

Treat

$$
n
$$

as a rational base parameter.

Then

$$
\mathcal E_1(n),
\quad
\mathcal E_2(n),
\quad
\mathcal Q_0(n),
\quad
\mathcal Q_+(n),
\quad
\mathcal Q_-(n)
$$

form five elliptic families over the rational

$$
n
$$

-line.

The Perfect Cuboid problem becomes a problem of simultaneous compatible specialization across these five families.

This is potentially more informative than solving isolated genus-$5$ fibers one by one.

---

# 29. A revised fixed-fiber proof architecture

For a selected rational

$$
n,
$$

the preferred proof architecture is now:

$$
\boxed{
\text{conic base}
}
$$

$$
\Downarrow
$$

$$
\boxed{
\mathcal E_1(n),
\mathcal E_2(n)
}
$$

$$
\Downarrow
$$

$$
\boxed{
\mathcal Q_0(n),
\mathcal Q_+(n),\mathcal Q_-(n)
}
$$

$$
\Downarrow
$$

$$
\boxed{
\text{Mordell-Weil / Selmer / local compatibility}
}
$$

$$
\Downarrow
$$

$$
\boxed{
\text{simultaneous lift back to }C_n.
}
$$

The genus-$5$ curve remains the final exact object, but it need not be the first computational object.

---

# 30. Revised global search architecture

Instead of

$$
n
\longrightarrow
\text{construct genus-5 curve}
\longrightarrow
\text{solve from scratch},
$$

use

$$
\boxed{
n
\longrightarrow
\text{five elliptic fingerprints}.
}
$$

For each rational

$$
n,
$$

record:

- discriminants;
- torsion;
- rank bounds;
- $2$-Selmer dimensions;
- local solubility;
- rational sections;
- specialization behavior;
- compatibility with the Twin-Psi reconstruction maps.

This may identify structurally exceptional fibers before heavy rational-point machinery is used.

---

# 31. Five-family arithmetic fingerprint

Define the R4 elliptic fingerprint

$$
\boxed{
\mathfrak E(n)
=
\left(
\mathcal E_1(n),
\mathcal E_2(n),
\mathcal Q_0(n),
\mathcal Q_+(n),
\mathcal Q_-(n)
\right).
}
$$

A fixed slice is no longer characterized only by its genus-$5$ equation.

It is characterized by the five-component elliptic packet

$$
\mathfrak E(n).
$$

The historical fibers

$$
n=\frac7{32}
$$

and

$$
n=\frac{11}{45}
$$

can now be compared at this lower-genus packet level.

---

# 32. G2 as a two-fiber comparison problem

The G2 seed supplies two rational outer parameters:

$$
\boxed{
n_1=\frac7{32},
}
$$

and

$$
\boxed{
n_2=\frac{11}{45}.
}
$$

Both correspond to different nested-face charts of the same geometric seed.

Therefore their elliptic packets

$$
\mathfrak E\left(\frac7{32}\right)
$$

and

$$
\mathfrak E\left(\frac{11}{45}\right)
$$

should encode the same canonical missing squareclass

$$
233\cdot1913
$$

through different quotient maps.

This gives a precise R5 comparison target.

---

# 33. Expected permutation-induced relation

The two G2 fibers arise by permuting which already closed face is used as the nested outer face.

Therefore one expects an algebraic correspondence between

$$
C_{7/32}
$$

and

$$
C_{11/45}
$$

induced by cuboid-axis permutation.

At the five-elliptic level, this should induce a permutation or isogeny correspondence among the factors in

$$
\mathfrak E\left(\frac7{32}\right)
$$

and

$$
\mathfrak E\left(\frac{11}{45}\right).
$$

R4 does not assume this correspondence is trivial.

It becomes an explicit object to derive.

---

# 34. Relation to reciprocal splitting polynomials

R3 described the three face channels using reciprocal quadratics

$$
P_a,
\quad
P_b,
\quad
P_c.
$$

R4 shows that fixing one split face polynomial corresponds to choosing the rational outer parameter

$$
n.
$$

The remaining two splitting problems become the two elliptic square-lift gates

$$
\mathcal E_1(n),
$$

and

$$
\mathcal E_2(n).
$$

Thus:

$$
\boxed{
\text{reciprocal splitting-state}
}
$$

and

$$
\boxed{
\text{Twin-Psi elliptic gate-state}
}
$$

are the same arithmetic problem at different FDRS levels.

---

# 35. Why the genus-$3$ subcover matters

The product condition

$$
F_1F_2\in\mathbb Q^2
$$

is weaker than requiring

$$
F_1\in\mathbb Q^2
$$

and

$$
F_2\in\mathbb Q^2
$$

separately.

Therefore the genus-$3$ curve

$$
H_n
$$

is a necessary intermediate closure layer.

Its three elliptic quotients encode combined squareclass information before the individual lifts are separated.

This explains why the historical genus-$3$ sieve could eliminate large portions of the search space before full genus-$5$ analysis.

---

# 36. Squareclass interpretation of the product cover

If

$$
[F_1]=[F_2]
$$

in

$$
\mathbb Q^\times/\mathbb Q^{\times2},
$$

then

$$
F_1F_2
$$

is a square.

Thus

$$
H_n
$$

detects equality of the two gate squareclasses.

The full fiber

$$
C_n
$$

requires the stronger condition

$$
[F_1]=[F_2]=1.
$$

Therefore the hierarchy is:

$$
\boxed{
[F_1]=[F_2]
}
$$

at genus

$$
3,
$$

followed by

$$
\boxed{
[F_1]=[F_2]=1
}
$$

at genus

$$
5.
$$

This is a direct squareclass interpretation of the intermediate curve.

---

# 37. A new proof-safe workflow

For each rational physical

$$
n:
$$

1. certify that:
   $$
   F_1,F_2
   $$
   are separable;

2. compute:
   $$
   \mathcal E_1(n),
   \mathcal E_2(n);
   $$

3. compute:
   $$
   \mathcal Q_0(n),
   \mathcal Q_+(n),
   \mathcal Q_-(n);
   $$

4. determine Mordell-Weil and Selmer information;

5. apply local compatibility across all quotient maps;

6. lift surviving points to:
   $$
   H_n;
   $$

7. finally lift to:
   $$
   C_n.
   $$

No rational point should be promoted to a Perfect Cuboid candidate until all lifts are exact.

---

# 38. R4 main theorem package

For every rational

$$
n
$$

with

$$
0<n<1,
$$

the Twin-Psi fixed fiber satisfies:

1. the quotient base is genus:
   $$
   0;
   $$

2. each single square-lift gate is genus:
   $$
   1;
   $$

3. the product squareclass cover is genus:
   $$
   3;
   $$

4. the simultaneous double-square fiber is genus:
   $$
   5;
   $$

5. both single gates have explicit full-rational-$2$-torsion elliptic models;

6. the genus-$3$ product cover admits three explicit genus-$1$ quotient curves;

7. the genus-$5$ Jacobian is generically isogenous to the product of five explicit elliptic factors.

Symbolically:

$$
\boxed{
\operatorname{Jac}(C_n)
\sim
\mathcal E_1
\times
\mathcal E_2
\times
\mathcal Q_0
\times
\mathcal Q_+
\times
\mathcal Q_-.
}
$$

---

# 39. Conceptual result

The apparent jump from a simple cuboid to genus-$5$ arithmetic is not one indivisible obstruction.

It is a layered lift process:

$$
\boxed{
\text{rational conic geometry}
}
$$

$$
\longrightarrow
$$

$$
\boxed{
\text{elliptic square lift}
}
$$

$$
\longrightarrow
$$

$$
\boxed{
\text{genus-3 squareclass compatibility}
}
$$

$$
\longrightarrow
$$

$$
\boxed{
\text{genus-5 simultaneous exact lift}.
}
$$

This is the precise FDRS decomposition of the high-genus complexity.

---

# 40. Proposed R5 program

R5 should pursue four exact targets.

## R5-A — Identify the five factors at $n=7/32$

Compute explicit Weierstrass models and isogenies connecting

$$
\mathcal E_1,
\mathcal E_2,
\mathcal Q_0,
\mathcal Q_+,
\mathcal Q_-
$$

to the historical curves

$$
E_f,
\quad
E_0,
\quad
E_-.
$$

Determine exactly which factors occur twice.

## R5-B — Compare the two G2 fibers

Construct an explicit correspondence between

$$
n=\frac7{32}
$$

and

$$
n=\frac{11}{45}.
$$

Track the squareclass

$$
233\cdot1913
$$

through the five elliptic quotient maps.

## R5-C — Elliptic-family invariants over $\mathbb Q(n)$

Compute:

- discriminants;
- $j$-invariants;
- torsion sections;
- isogeny relations;
- generic ranks if accessible;
- exceptional specializations.

Search for structural reasons a simultaneous lift might be impossible or possible.

## R5-D — Global compatibility before Chabauty

Use the five elliptic families to build a global compatibility sieve in the parameter

$$
n
$$

before specializing to expensive high-genus arithmetic.

The goal is to determine whether large classes of rational fibers can be eliminated or prioritized algebraically.

---

# 41. Final conclusion

R4 corrects the quotient-fiber geometry and decomposes the fixed-slice complexity completely.

The correct genus ladder is

$$
\boxed{
0
\rightarrow
1
\rightarrow
3
\rightarrow
5.
}
$$

The quotient base is rational.

A single square lift is elliptic.

The product squareclass condition is genus-$3$.

The full simultaneous square lift is genus-$5$.

Moreover, the genus-$5$ Jacobian is generically controlled by five explicit elliptic factors.

Thus the historical appearance of five elliptic dimensions on the

$$
n=\frac7{32}
$$

slice is now structurally explained.

The next phase should no longer treat the fixed genus-$5$ curve as a monolithic object.

It should identify and compare the five elliptic factors, especially across the two G2 fibers

$$
\frac7{32}
$$

and

$$
\frac{11}{45}.
$$

This is the canonical starting point for R5.