# Perfect Cuboid Radical-FDRS R3

## Reciprocal Splitting-State, Rational Chart Completion, and the Symmetric Square-Lift Quotient Surface

**Series:** Perfect Cuboid Global Closure  
**Branch:** Radical-FDRS / Geometry-Coupling Branch R  
**Phase:** R3 — Reciprocal Root Algebra and Square-Lift Quotient  
**Date:** 2026-10-03  
**Status:** Exact structural reformulation — not a proof of Perfect Cuboid existence or nonexistence

---

# 0. Executive summary

R2 reduced the Perfect Cuboid problem to the Twin-Psi system:

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

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

where:

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

and:

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

R3 identifies three further structures.

First, the Twin-Psi outputs satisfy a new Pythagorean relation after one Cayley reciprocal transform.

Second, every face closure is equivalent to the splitting over:

$$
\mathbb Q
$$

of a one-dimensional reciprocal quadratic polynomial.

Third, the entire rational-edge shell admits a symmetric rational quotient surface whose only remaining Perfect-Cuboid condition is a simultaneous square lift of three rational coordinates.

For the current G2 seed, the quotient point is:

$$
\boxed{
\left(
\frac{625}{1521},
\frac{289}{784},
\frac{233}{1913}
\right).
}
$$

The first two coordinates are rational squares.

The third is rational but not a rational square.

Thus the old six-channel state:

$$
(1,1,1,0,1,1)
$$

compresses canonically to:

$$
\boxed{
(1,1,0)
}
$$

on the Cayley square-lift quotient.

The same defect:

$$
233\cdot1913
$$

appears as:

- the squarefree kernel of the missing face diagonal;
- the discriminant squareclass of the only nonsplit reciprocal quadratic;
- the squareclass of:
  $$
  \frac{233}{1913};
  $$
- the obstruction to completing the third rational FDRS chart.

---

# 1. Twin-Psi recap

Use the nested rational Pythagorean parameterization:

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

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

and:

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

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

Then:

$$
a^2+b^2=d^2,
$$

and:

$$
d^2+c^2=1.
$$

The remaining two face closures are:

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

and:

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

with:

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

Perfect Cuboid closure occurs exactly when:

$$
u,v\in\mathbb Q.
$$

---

# 2. Cayley coordinate functions

Define:

$$
\boxed{
X(t)
=
\frac{1-t^2}{1+t^2},
}
$$

and:

$$
\boxed{
Y(t)
=
\frac{2t}{1+t^2}.
}
$$

Then:

$$
X(t)^2+Y(t)^2=1.
$$

The Möbius involution:

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

exchanges the two unit-circle coordinates:

$$
\boxed{
X(\mu(t))
=
Y(t),
}
$$

and:

$$
\boxed{
Y(\mu(t))
=
X(t).
}
$$

This is the exact algebraic content of the half-angle complement operation.

---

# 3. Twin-Psi multiplicative collapse

Assume:

$$
u^2
=
\Psi(m,n)
=
\frac{m^2+n^2}{1+m^2n^2}.
$$

Then:

$$
1-u^2
=
\frac{
(1-m^2)(1-n^2)
}{
1+m^2n^2
},
$$

and:

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

Therefore:

$$
\boxed{
X(u)
=
X(m)X(n).
}
$$

Similarly:

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

gives:

$$
X(v)
=
X(\mu(m))X(n).
$$

Using:

$$
X(\mu(m))=Y(m),
$$

we obtain:

$$
\boxed{
X(v)
=
Y(m)X(n).
}
$$

Hence the complete Twin-Psi system becomes:

$$
\boxed{
X(u)
=
X(m)X(n),
}
$$

$$
\boxed{
X(v)
=
Y(m)X(n).
}
$$

---

# 4. Emergent Pythagorean triangle

Because:

$$
X(m)^2+Y(m)^2=1,
$$

the previous identities imply:

$$
X(u)^2+X(v)^2
=
X(n)^2.
$$

Therefore:

$$
\boxed{
X(u)^2+X(v)^2=X(n)^2.
}
$$

This is an exact rational Pythagorean relation whenever:

$$
m,n,u,v\in\mathbb Q.
$$

Thus two successful Twin-Psi gates automatically generate a new rational right triangle.

The Twin-Psi system is therefore self-similar:

$$
\boxed{
\text{closed rational-circle data}
\longrightarrow
\text{new rational Pythagorean closure}.
}
$$

This gives a precise recursive meaning to the earlier Radical-FDRS / Magic-Cube intuition.

---

# 5. Reconstruction of the inner parameter

From:

$$
X(u)=X(m)X(n),
$$

and:

$$
X(v)=Y(m)X(n),
$$

we have:

$$
\frac{X(u)}{X(n)}
=
X(m),
$$

and:

$$
\frac{X(v)}{X(n)}
=
Y(m).
$$

Therefore:

$$
m
$$

is the rational half-angle parameter of the new Pythagorean triangle:

$$
\bigl(
X(u),
X(v),
X(n)
\bigr).
$$

Explicitly:

$$
\boxed{
m
=
\frac{X(v)}{X(n)+X(u)}
}
$$

for the standard positive branch.

Thus the original inner chart parameter is reconstructible from the Twin-Psi outputs.

---

# 6. Reciprocal face-root parameter

Consider a normalized rational edge:

$$
x\in\mathbb Q,
\qquad
0<x<1.
$$

Let its complementary face diagonal be:

$$
h=\sqrt{1-x^2}.
$$

Define a half-angle parameter:

$$
\tau
=
\frac{x}{1+h}.
$$

Equivalently:

$$
\tau
=
\frac{1-h}{x}.
$$

Then:

$$
\boxed{
x
=
\frac{2\tau}{1+\tau^2},
}
$$

and:

$$
\boxed{
h
=
\frac{1-\tau^2}{1+\tau^2}.
}
$$

Therefore the rationality of the complementary face diagonal is equivalent to rationality of:

$$
\tau.
$$

---

# 7. Reciprocal quadratic polynomial

Eliminate:

$$
h
$$

from the previous equations.

The parameter:

$$
\tau
$$

satisfies:

$$
x(1+\tau^2)=2\tau.
$$

Therefore:

$$
\boxed{
x\tau^2-2\tau+x=0.
}
$$

For an unnormalized rational cuboid with body diagonal:

$$
g,
$$

the corresponding polynomial is:

$$
\boxed{
P_x(T)
=
xT^2-2gT+x.
}
$$

Its two roots are reciprocal:

$$
\boxed{
T_1T_2=1.
}
$$

Thus every complementary face channel naturally defines a reciprocal quadratic.

---

# 8. Splitting criterion

The discriminant of:

$$
P_x(T)
=
xT^2-2gT+x
$$

is:

$$
\Delta_x
=
4(g^2-x^2).
$$

If:

$$
h^2=g^2-x^2,
$$

then:

$$
\boxed{
\Delta_x=4h^2.
}
$$

Therefore:

$$
\boxed{
h\in\mathbb Q
\iff
P_x(T)
\text{ splits over }
\mathbb Q.
}
$$

So face closure is exactly polynomial splitting.

This provides a purely one-dimensional arithmetic formulation.

---

# 9. Squareclass of the discriminant

Since:

$$
4
$$

is a rational square:

$$
\boxed{
[\Delta_x]
=
[g^2-x^2]
}
$$

in:

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

Therefore the squareclass defect of a face diagonal is exactly the discriminant squareclass of the reciprocal quadratic.

This converts Radical-FDRS defect tracking into splitting-state tracking.

---

# 10. Three reciprocal quadratics for a cuboid

For rational edges:

$$
a,b,c
$$

and rational body diagonal:

$$
g,
$$

define:

$$
P_a(T)=aT^2-2gT+a,
$$

$$
P_b(T)=bT^2-2gT+b,
$$

$$
P_c(T)=cT^2-2gT+c.
$$

Then:

$$
P_a
$$

splits over:

$$
\mathbb Q
$$

if and only if:

$$
f^2=b^2+c^2
$$

has rational square root.

Likewise:

$$
P_b
$$

corresponds to:

$$
e^2=a^2+c^2,
$$

and:

$$
P_c
$$

corresponds to:

$$
d^2=a^2+b^2.
$$

Hence:

$$
\boxed{
\text{Perfect Cuboid}
}
$$

is equivalent to:

$$
\boxed{
P_a,
P_b,
P_c
\text{ all split over }\mathbb Q.
}
$$

---

# 11. G2 first polynomial

For:

$$
a=448,
\qquad
g=1073,
$$

we have:

$$
P_a(T)
=
448T^2-2146T+448.
$$

Divide by:

$$
2:
$$

$$
224T^2-1073T+224.
$$

Exact factorization gives:

$$
\boxed{
224T^2-1073T+224
=
(7T-32)(32T-7).
}
$$

Therefore the two reciprocal roots are:

$$
\boxed{
\frac7{32},
\qquad
\frac{32}{7}.
}
$$

The physical half-angle root is:

$$
\boxed{
\tau_a=\frac7{32}.
}
$$

---

# 12. G2 second polynomial

For:

$$
b=495,
\qquad
g=1073,
$$

we have:

$$
P_b(T)
=
495T^2-2146T+495.
$$

It factors exactly as:

$$
\boxed{
495T^2-2146T+495
=
(11T-45)(45T-11).
}
$$

Therefore:

$$
\boxed{
\tau_b=\frac{11}{45}
}
$$

on the physical branch.

Thus the two historical fixed-slice values:

$$
\frac7{32},
\qquad
\frac{11}{45}
$$

are simply the physical roots of the two split reciprocal quadratics.

---

# 13. G2 third polynomial

For:

$$
c=840,
\qquad
g=1073,
$$

we have:

$$
P_c(T)
=
840T^2-2146T+840.
$$

Divide by:

$$
2:
$$

$$
\boxed{
420T^2-1073T+420.
}
$$

Its discriminant is:

$$
1073^2
-
4\cdot420^2.
$$

Therefore:

$$
\boxed{
\Delta_c
=
445729
=
233\cdot1913.
}
$$

The roots are:

$$
\boxed{
\tau_c^\pm
=
\frac{
1073\pm\sqrt{445729}
}{
840
}.
}
$$

They are reciprocal:

$$
\tau_c^+\tau_c^-=1.
$$

Since:

$$
445729
$$

is not a rational square:

$$
P_c
$$

does not split over:

$$
\mathbb Q.
$$

Thus G2 is exactly a state with:

$$
\boxed{
2
\text{ split reciprocal quadratics}
+
1
\text{ nonsplit reciprocal quadratic}.
}
$$

---

# 14. Splitting field interpretation

The third polynomial splits over:

$$
\boxed{
K
=
\mathbb Q\left(
\sqrt{445729}
\right).
}
$$

Thus the entire remaining G2 arithmetic defect is concentrated in one quadratic extension.

The two rational face channels require no field extension.

The missing face channel requires:

$$
K/\mathbb Q.
$$

This gives the field-state representation:

$$
\boxed{
(\mathbb Q,\mathbb Q,K).
}
$$

Perfect Cuboid closure corresponds to collapse of all three splitting fields to:

$$
\mathbb Q.
$$

---

# 15. Norm-one property of the missing root

Take:

$$
\tau_c
=
\frac{
1073-\sqrt{445729}
}{
840
}.
$$

Its Galois conjugate is:

$$
\overline{\tau_c}
=
\frac{
1073+\sqrt{445729}
}{
840
}.
$$

Their product is:

$$
\boxed{
\tau_c\overline{\tau_c}=1.
}
$$

Therefore:

$$
\boxed{
N_{K/\mathbb Q}(\tau_c)=1.
}
$$

The missing chart parameter is a quadratic norm-one element.

This is the reciprocal-root version of the earlier conjugate-wall defect factorization.

---

# 16. Cayley square coordinate

For normalized edge:

$$
x
$$

define:

$$
\boxed{
R_x
=
\frac{1-x}{1+x}.
}
$$

Then:

$$
R_x\in\mathbb Q
$$

whenever:

$$
x\in\mathbb Q.
$$

The complementary face square is:

$$
1-x^2
=
\frac{4R_x}{(1+R_x)^2}.
$$

Therefore:

$$
\boxed{
\sqrt{1-x^2}\in\mathbb Q
\iff
R_x\in\mathbb Q^2.
}
$$

Hence each reciprocal polynomial splitting condition can be replaced by a square-lift condition on a rational Cayley coordinate.

---

# 17. Relation between $R_x$ and reciprocal roots

Let:

$$
r_x=\sqrt{R_x}.
$$

Then:

$$
x
=
\frac{1-r_x^2}{1+r_x^2}.
$$

Apply:

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

Using:

$$
Y(\mu(r_x))=X(r_x)=x,
$$

we obtain:

$$
\boxed{
\tau_x
=
\mu(r_x).
}
$$

Therefore:

$$
\boxed{
R_x
=
r_x^2
}
$$

is square exactly when the reciprocal quadratic has rational roots.

The three descriptions are identical:

$$
\boxed{
\text{face square closure}
}
$$

$$
\Longleftrightarrow
$$

$$
\boxed{
R_x\in\mathbb Q^2
}
$$

$$
\Longleftrightarrow
$$

$$
\boxed{
P_x(T)
\text{ splits over }\mathbb Q.
}
$$

---

# 18. Symmetric Cayley square-lift surface

Normalize:

$$
g=1.
$$

Let:

$$
a,b,c\in\mathbb Q_{>0}
$$

satisfy:

$$
a^2+b^2+c^2=1.
$$

Define:

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

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

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

Invert:

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

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

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

Therefore the rational-edge sphere is birationally rewritten as:

$$
\boxed{
\left(
\frac{1-R}{1+R}
\right)^2
+
\left(
\frac{1-S}{1+S}
\right)^2
+
\left(
\frac{1-T}{1+T}
\right)^2
=
1.
}
$$

---

# 19. The $\Phi$ form

Define:

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

Because:

$$
\left(
\frac{1-X}{1+X}
\right)^2
=
1-\Phi(X),
$$

the sphere equation becomes:

$$
3-
\Phi(R)
-
\Phi(S)
-
\Phi(T)
=
1.
$$

Hence:

$$
\boxed{
\Phi(R)
+
\Phi(S)
+
\Phi(T)
=
2.
}
$$

Define the rational surface:

$$
\boxed{
\mathcal B:
\quad
\Phi(R)+\Phi(S)+\Phi(T)=2.
}
$$

This is the **Cayley square-lift quotient surface**.

---

# 20. Perfect Cuboid as simultaneous square lift

For every rational-edge normalized cuboid:

$$
R,S,T\in\mathbb Q.
$$

The three complementary face diagonals are rational if and only if:

$$
R,S,T
$$

are rational squares.

Therefore:

## Square-Lift Theorem

A nondegenerate Perfect Cuboid exists if and only if there is a rational point:

$$
(R,S,T)\in\mathcal B(\mathbb Q)
$$

with:

$$
\boxed{
R,S,T\in\mathbb Q^2.
}
$$

Thus the classical problem becomes:

$$
\boxed{
\text{find an all-square rational point on }\mathcal B.
}
$$

No geometric square roots remain in the base equation.

All arithmetic difficulty has been transferred into the coordinate square-lift condition.

---

# 21. G2 point on the quotient surface

For:

$$
(a,b,c;g)
=
(448,495,840;1073),
$$

the normalized Cayley coordinates are:

$$
R
=
\frac{1073-448}{1073+448}
=
\boxed{
\frac{625}{1521}
}
=
\boxed{
\left(
\frac{25}{39}
\right)^2,
}
$$

$$
S
=
\frac{1073-495}{1073+495}
=
\boxed{
\frac{289}{784}
}
=
\boxed{
\left(
\frac{17}{28}
\right)^2,
}
$$

and:

$$
T
=
\frac{1073-840}{1073+840}
=
\boxed{
\frac{233}{1913}.
}
$$

Therefore:

$$
\boxed{
(R,S,T)
=
\left(
\frac{625}{1521},
\frac{289}{784},
\frac{233}{1913}
\right).
}
$$

This is an exact rational point on:

$$
\mathcal B.
$$

---

# 22. G2 square-lift state

The first two coordinates satisfy:

$$
R\in\mathbb Q^2,
$$

$$
S\in\mathbb Q^2.
$$

The third satisfies:

$$
T\notin\mathbb Q^2.
$$

Its squareclass is:

$$
[T]
=
[233\cdot1913].
$$

Thus the compressed G2 state is:

$$
\boxed{
\Sigma_{\mathcal B}
=
(1,1,0),
}
$$

or with full squareclass data:

$$
\boxed{
K_{\mathcal B}
=
(1,1,233\cdot1913).
}
$$

This is the three-channel canonical version of the earlier six-channel state.

---

# 23. Face-square interpretation of $\Phi$

For:

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

we have:

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

Hence:

$$
\Phi(R)
=
f^2,
$$

in normalized units.

Likewise:

$$
\Phi(S)
=
e^2,
$$

and:

$$
\Phi(T)
=
d^2.
$$

For G2:

$$
\Phi(R)
=
\frac{975^2}{1073^2},
$$

$$
\Phi(S)
=
\frac{952^2}{1073^2},
$$

and:

$$
\Phi(T)
=
\frac{445729}{1073^2}.
$$

Their sum is:

$$
\boxed{
2.
}
$$

This is simply:

$$
d^2+e^2+f^2=2g^2
$$

written in Cayley square coordinates.

---

# 24. Multidegree $(2,2,2)$ model

Clearing denominators in:

$$
\Phi(R)+\Phi(S)+\Phi(T)=2
$$

gives:

$$
4R(1+S)^2(1+T)^2
$$

$$
+
4S(1+R)^2(1+T)^2
$$

$$
+
4T(1+R)^2(1+S)^2
$$

$$
-
2(1+R)^2(1+S)^2(1+T)^2
=
0.
$$

This polynomial has degree:

$$
2
$$

in each of:

$$
R,S,T.
$$

Thus the natural denominator-cleared model is of multidegree:

$$
\boxed{
(2,2,2).
}
$$

---

# 25. Smoothness on the physical affine open

For:

$$
X\ne-1,
$$

we have:

$$
\Phi'(X)
=
\frac{
4(1-X)
}{
(1+X)^3
}.
$$

A singular point of:

$$
\Phi(R)+\Phi(S)+\Phi(T)=2
$$

inside the affine denominator-safe region would require:

$$
R=S=T=1.
$$

But:

$$
\Phi(1)=1,
$$

so:

$$
\Phi(1)+\Phi(1)+\Phi(1)=3\ne2.
$$

Therefore:

$$
\boxed{
\mathcal B
\text{ is smooth on the physical affine open }R,S,T\ne-1.
}
$$

In particular the positive chamber relevant to cuboids is smooth.

---

# 26. Boundary warning

The naive denominator-cleared compactification contains boundary loci with:

$$
R=-1,
$$

$$
S=-1,
$$

or:

$$
T=-1.
$$

For example, simultaneous conditions such as:

$$
R=S=-1
$$

produce boundary components where the cleared equation degenerates.

Therefore one must not automatically identify the naive projective closure with a smooth:

$$
(2,2,2)
$$

surface.

Any K3-type or other birational classification requires:

- saturation by denominator divisors;
- boundary analysis;
- singularity resolution;
- minimal-model verification.

R3 does not make an unverified surface-classification claim.

---

# 27. Reciprocal symmetry

The function:

$$
\Phi
$$

satisfies:

$$
\boxed{
\Phi(X)=\Phi(1/X).
}
$$

Therefore:

$$
\mathcal B
$$

has three independent involutions:

$$
R\mapsto\frac1R,
$$

$$
S\mapsto\frac1S,
$$

$$
T\mapsto\frac1T.
$$

The square-lift property is preserved because:

$$
X\in\mathbb Q^2
\iff
\frac1X\in\mathbb Q^2.
$$

Thus reciprocal orientation changes do not alter arithmetic closure.

---

# 28. Permutation symmetry

The equation:

$$
\Phi(R)+\Phi(S)+\Phi(T)=2
$$

is fully symmetric under:

$$
S_3.
$$

Therefore the exact symmetry group generated by reciprocal flips and coordinate permutations contains:

$$
\boxed{
(\mathbb Z/2\mathbb Z)^3
\rtimes
S_3.
}
$$

This group has:

$$
48
$$

elements.

It is the algebraic symmetry underlying:

- reciprocal root choice;
- edge permutation;
- face permutation;
- FDRS chart orientation.

This is the cleanest current algebraic realization of the earlier Magic-Cube symmetry intuition.

---

# 29. Physical fundamental chamber

For a positive normalized edge:

$$
0<a<1,
$$

the Cayley coordinate satisfies:

$$
0<R<1.
$$

Likewise:

$$
0<S,T<1.
$$

Thus physical cuboids lie in:

$$
\boxed{
0<R,S,T<1.
}
$$

Reciprocal involutions map this chamber to orientation-equivalent regions with coordinates greater than:

$$
1.
$$

Therefore:

$$
(0,1)^3
$$

can be used as a natural arithmetic fundamental chamber.

---

# 30. Symmetric outer-root surface

When a face is already closed, define the physical reciprocal root:

$$
\tau_x
=
\frac{x}{1+h},
$$

where:

$$
h=\sqrt{1-x^2}.
$$

Then:

$$
x
=
\frac{2\tau_x}{1+\tau_x^2}.
$$

Define:

$$
\boxed{
\eta(\tau)
=
\frac{2\tau}{1+\tau^2}.
}
$$

For a Perfect Cuboid all three:

$$
\tau_a,\tau_b,\tau_c
$$

are rational and satisfy:

$$
\boxed{
\eta(\tau_a)^2
+
\eta(\tau_b)^2
+
\eta(\tau_c)^2
=
1.
}
$$

Thus a second symmetric formulation is:

$$
\boxed{
\mathcal O:
\quad
\eta(p)^2+\eta(q)^2+\eta(r)^2=1.
}
$$

A nondegenerate Perfect Cuboid exists if and only if:

$$
\mathcal O
$$

contains a positive rational point.

Unlike:

$$
\mathcal B,
$$

the G2 seed does not give a fully rational point on:

$$
\mathcal O
$$

because its third reciprocal root is quadratic irrational.

---

# 31. Why $\mathcal B$ is the better defect quotient

The outer-root surface:

$$
\mathcal O
$$

absorbs face closure into coordinate rationality.

The quotient surface:

$$
\mathcal B
$$

keeps every rational-edge cuboid rational and records failure only as nonsquare coordinates.

Therefore:

$$
\mathcal B
$$

is better for defect dynamics.

The G2 defect becomes:

$$
\boxed{
(\text{square},\text{square},\text{nonsquare})
}
$$

instead of:

$$
(\text{rational},\text{rational},\text{quadratic irrational}).
$$

This is precisely the representation desired by Radical-FDRS.

---

# 32. Twin-Psi chart completion theorem

Return to a Twin-Psi chart:

$$
(m,n).
$$

It reconstructs:

$$
d=X(n),
$$

$$
c=Y(n),
$$

$$
a=X(n)X(m),
$$

$$
b=X(n)Y(m).
$$

Suppose:

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

and:

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

Then:

$$
a=X(u),
$$

and:

$$
b=X(v).
$$

The physical reciprocal roots associated with:

$$
a
$$

and:

$$
b
$$

are therefore:

$$
\boxed{
\tau_a=\mu(u),
}
$$

and:

$$
\boxed{
\tau_b=\mu(v).
}
$$

The remaining edge:

$$
c=Y(n)
$$

already has physical reciprocal root:

$$
\boxed{
\tau_c=n.
}
$$

Therefore the three outer chart parameters are:

$$
\boxed{
(\tau_a,\tau_b,\tau_c)
=
(\mu(u),\mu(v),n).
}
$$

---

# 33. Meaning of Twin-Psi closure

Because:

$$
n\in\mathbb Q
$$

is built into the chart:

$$
\tau_c
$$

is already rational.

Then:

$$
u\in\mathbb Q
$$

if and only if:

$$
\tau_a\in\mathbb Q,
$$

and:

$$
v\in\mathbb Q
$$

if and only if:

$$
\tau_b\in\mathbb Q.
$$

Therefore Twin-Psi is exactly a chart-completion problem:

$$
\boxed{
\text{one rational outer chart}
\longrightarrow
\text{construct the other two rational charts}.
}
$$

A Perfect Cuboid is achieved precisely when the rational chart triangle closes.

---

# 34. G2 chart transition

Use the G2 chart:

$$
(m,n)
=
\left(
\frac47,
\frac7{32}
\right).
$$

The successful Twin-Psi root is:

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

Therefore:

$$
\mu(u)
=
\frac{
1-\frac{17}{28}
}{
1+\frac{17}{28}
}
=
\boxed{
\frac{11}{45}.
}
$$

Thus the successful gate sends:

$$
\boxed{
\frac7{32}
\longrightarrow
\frac{11}{45}.
}
$$

This is exactly the second historical rational fixed slice.

---

# 35. Reverse G2 transition

Start from the other rational chart:

$$
(m,n)
=
\left(
\frac35,
\frac{11}{45}
\right).
$$

The successful gate root is:

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

Then:

$$
\mu(u)
=
\frac{
1-\frac{25}{39}
}{
1+\frac{25}{39}
}
=
\boxed{
\frac7{32}.
}
$$

Thus:

$$
\boxed{
\frac{11}{45}
\longrightarrow
\frac7{32}.
}
$$

The two rational charts form an exact two-way transition.

---

# 36. Missing third chart

The failed Twin-Psi root satisfies:

$$
v^2
=
\frac{233}{1913}.
$$

Therefore:

$$
v
=
\sqrt{
\frac{233}{1913}
}.
$$

The third outer chart parameter would be:

$$
\boxed{
\tau_{\rm miss}
=
\mu(v)
=
\frac{
1-\sqrt{233/1913}
}{
1+\sqrt{233/1913}
}.
}
$$

This is quadratic irrational.

It is equivalent to:

$$
\boxed{
\tau_{\rm miss}
=
\frac{
1073-\sqrt{445729}
}{
840
}.
}
$$

Therefore the G2 chart graph consists of:

- two rational vertices;
- one quadratic vertex.

Perfect closure would require rationalization of the third vertex.

---

# 37. Magic-triangle interpretation

There are three choices of outer edge / opposite face pair.

Associate one outer chart parameter to each:

$$
\tau_a,
\qquad
\tau_b,
\qquad
\tau_c.
$$

The chart graph is a triangle.

For a Perfect Cuboid:

$$
\boxed{
\tau_a,\tau_b,\tau_c\in\mathbb Q.
}
$$

For G2:

$$
\boxed{
\tau_a,\tau_b\in\mathbb Q,
\qquad
\tau_c\notin\mathbb Q.
}
$$

Thus the earlier six-channel Magic-Cube picture admits a smaller chart-level projection:

$$
\boxed{
\text{three-vertex rationality triangle}.
}
$$

The one-defect seed has exactly one unclosed vertex.

---

# 38. Relation to the old fixed slices

The values:

$$
\frac7{32}
$$

and:

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

were previously found as two useful one-axis fixed slices.

R3 now proves that they are not independent arithmetic accidents.

They are the two rational reciprocal roots of:

$$
P_a
$$

and:

$$
P_b
$$

for the G2 seed.

The third analogous parameter is precisely the nonsplit root of:

$$
P_c.
$$

Therefore the old statement:

> G2 has exactly two useful rational one-axis repair orbits

is equivalent to:

> G2 has exactly two split reciprocal face quadratics.

---

# 39. Fixed-coordinate quotient fibers

Fix:

$$
R\in\mathbb Q.
$$

Then:

$$
\Phi(S)+\Phi(T)
=
2-\Phi(R).
$$

After clearing denominators, this is a curve of bidegree:

$$
\boxed{
(2,2)
}
$$

in:

$$
(S,T)\in\mathbb P^1\times\mathbb P^1.
$$

For generic nonsingular specialization, a:

$$
(2,2)
$$

curve has genus:

$$
\boxed{
1.
}
$$

Therefore:

$$
\mathcal B
$$

has natural genus-$1$ quotient fibers.

This is lower genus than the full square-lift fixed fibers encountered previously.

---

# 40. Why this matters for future descent

The fixed-$n$ Twin-Psi model imposes square roots before quotienting and naturally produces genus:

$$
5.
$$

The:

$$
\mathcal B
$$

model first quotients reciprocal and sign data.

Its fixed-coordinate fibers are generically genus:

$$
1.
$$

The remaining difficulty is then the square-lift condition:

$$
S,T\in\mathbb Q^2.
$$

This suggests a revised proof architecture:

$$
\boxed{
\text{elliptic quotient fiber}
\longrightarrow
\text{square-lift covering analysis}
}
$$

instead of immediately constructing the genus-$5$ cover.

The two approaches are equivalent in information content but distribute complexity differently.

This is exactly the FDRS principle.

---

# 41. Reciprocal quotient variable

Because:

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

define:

$$
L_X
=
X+\frac1X.
$$

Then:

$$
\boxed{
\Phi(X)
=
\frac{4}{L_X+2}.
}
$$

Therefore:

$$
\mathcal B
$$

can be further quotiented by reciprocal involutions into:

$$
\boxed{
\frac{4}{L_R+2}
+
\frac{4}{L_S+2}
+
\frac{4}{L_T+2}
=
2.
}
$$

This quotient is rationally much simpler.

However, the lift back to:

$$
R,S,T
$$

carries the discriminant data.

---

# 42. Squarehood in reciprocal quotient coordinates

For rational:

$$
X\ne0,
$$

we have:

$$
L_X-2
=
X+\frac1X-2.
$$

Hence:

$$
\boxed{
L_X-2
=
\frac{(X-1)^2}{X}.
}
$$

Therefore:

$$
[L_X-2]
=
[X].
$$

Consequently:

$$
\boxed{
X\in\mathbb Q^2
\iff
L_X-2\in\mathbb Q^2.
}
$$

Thus the squareclass defect survives reciprocal quotienting in an especially simple shifted form.

---

# 43. Three levels of defect representation

The same face defect may now be represented in three exact ways.

## Level A — face squareclass

$$
[g^2-x^2].
$$

## Level B — Cayley coordinate squareclass

$$
\left[
\frac{g-x}{g+x}
\right].
$$

## Level C — reciprocal quotient shift

$$
[L_X-2].
$$

All three squareclasses are equal.

Therefore the defect is representation-invariant under the R3 quotient chain.

---

# 44. Canonical G2 defect chain

For the missing face opposite:

$$
c=840,
$$

we have:

$$
g^2-c^2
=
445729
=
233\cdot1913.
$$

The Cayley ratio is:

$$
T
=
\frac{g-c}{g+c}
=
\frac{233}{1913}.
$$

Its squareclass is:

$$
[T]
=
[233\cdot1913].
$$

The reciprocal quotient shift:

$$
L_T-2
=
\frac{(T-1)^2}{T}
$$

has the same squareclass.

Thus:

$$
\boxed{
[445729]
=
\left[
\frac{233}{1913}
\right]
=
[L_T-2].
}
$$

This is the canonical R3 defect invariant.

---

# 45. R3 main theorem package

R3 establishes the following equivalent conditions for each face channel.

For rational edge:

$$
x
$$

and rational body diagonal:

$$
g:
$$

the following are equivalent:

1. the complementary face diagonal is rational;

2. the reciprocal quadratic:
   $$
   xT^2-2gT+x
   $$
   splits over:
   $$
   \mathbb Q;
   $$

3. the Cayley ratio:
   $$
   \frac{g-x}{g+x}
   $$
   is a rational square;

4. the physical reciprocal root:
   $$
   \frac{x}{g+\sqrt{g^2-x^2}}
   $$
   is rational.

Thus Perfect Cuboid existence is equivalent to simultaneous closure of all three representations.

---

# 46. R3 global compression

The complete rational-edge shell is mapped birationally to:

$$
\boxed{
\mathcal B:
\Phi(R)+\Phi(S)+\Phi(T)=2.
}
$$

The Perfect Cuboid condition becomes:

$$
\boxed{
R,S,T\in\mathbb Q^2.
}
$$

The G2 seed is:

$$
\boxed{
\left(
\left(\frac{25}{39}\right)^2,
\left(\frac{17}{28}\right)^2,
\frac{233}{1913}
\right).
}
$$

Thus all previous one-defect formulations collapse to one all-square lifting problem on a symmetric rational surface.

---

# 47. Proposed R4 program

R4 should prioritize the square-lift quotient surface rather than another direct fixed genus-$5$ slice.

## R4-A — Elliptic quotient fibers

Fix one coordinate:

$$
R.
$$

Study the generic:

$$
(2,2)
$$

fiber in:

$$
S,T.
$$

Construct:

- explicit Weierstrass models;
- rational sections;
- discriminants;
- singular fibers;
- rank behavior.

## R4-B — Square-lift coverings

On the elliptic quotient fiber, impose:

$$
S=s^2,
$$

$$
T=t^2.
$$

Study these as explicit:

$$
2
$$

-cover conditions.

Compare the resulting coverings to the old genus-$5$ curves.

## R4-C — G2 local deformation

Start at:

$$
\left(
\frac{625}{1521},
\frac{289}{784},
\frac{233}{1913}
\right).
$$

Allow motion on:

$$
\mathcal B(\mathbb Q)
$$

while tracking coordinate squareclasses.

Search for exact rational trajectories:

$$
(1,1,\kappa)
\longrightarrow
(1,\kappa_2,\kappa_3)
\longrightarrow
(1,1,1).
$$

## R4-D — Splitting-polynomial dynamics

Track:

$$
P_a,
P_b,
P_c
$$

under rational-shell moves.

Study whether discriminant squareclasses admit:

- transfer;
- factor separation;
- cancellation;
- local obstruction invariants.

## R4-E — Chart graph dynamics

Treat the three reciprocal-root charts as vertices.

Study exact transition maps induced by axis permutations and Twin-Psi closure.

Determine whether an invariant prevents completion of the third rational vertex.

---

# 48. Final conclusion

R3 converts the Perfect Cuboid arithmetic problem into a splitting and square-lift problem.

The one-defect G2 seed is no longer described merely as:

$$
\text{one irrational face diagonal}.
$$

It is equivalently:

$$
\boxed{
\text{one nonsplit reciprocal quadratic},
}
$$

or:

$$
\boxed{
\text{one nonsquare Cayley coordinate},
}
$$

or:

$$
\boxed{
\text{one quadratic chart vertex}.
}
$$

The two historical rational slices:

$$
\frac7{32}
$$

and:

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

are the two rational roots belonging to the two already split face quadratics.

They transform into one another through successful Twin-Psi chart completion.

The missing third chart is:

$$
\frac{
1073-\sqrt{445729}
}{
840
},
$$

with norm:

$$
1
$$

over:

$$
\mathbb Q(\sqrt{445729}).
$$

At the global level, every rational-edge cuboid maps to a rational point on:

$$
\boxed{
\mathcal B:
\Phi(R)+\Phi(S)+\Phi(T)=2,
}
$$

and Perfect Cuboid closure is exactly the all-square lift:

$$
\boxed{
R,S,T\in\mathbb Q^2.
}
$$

This surface has:

- full:
  $$
  S_3
  $$
  permutation symmetry;
- three reciprocal involutions;
- generic genus-$1$ fixed-coordinate quotient fibers.

Therefore the next natural research object is not another isolated genus-$5$ curve.

It is the elliptic-fiber and square-lift structure of:

$$
\mathcal B.
$$

This is the canonical starting point for R4.