# Perfect Cuboid C1 — Conservation Move Algebra

## Exact Move Generators, Rational Pair Rotations, and One-Axis Repair Orbits

**Series:** Perfect Cuboid Global Closure  
**Branch:** Conservation–Duality Branch C  
**Phase:** C1 — Move Algebra  
**Date:** 2026-10-02

---

# Abstract

The C0 Conservation–Duality Methodology rewrites Perfect Cuboid geometry as a six-cell conserved state

$$
X=(A,B,C,D,E,F)
$$

satisfying

$$
A+B+C=1,
$$

$$
A+F=1,
$$

$$
B+E=1,
$$

$$
C+D=1.
$$

The geometric state is therefore an affine two-dimensional conservation manifold.

This paper develops the exact move algebra on that manifold.

The main conclusions are:

1. every infinitesimal or finite conservation move lies in a two-dimensional move space;
2. the three visually natural pair-transfer moves satisfy one linear relation;
3. repairing the present defect in $D$ requires moving at least four cells;
4. from the current seed there are exactly two useful one-axis repair orbits;
5. the previously studied $k=7/32$ nested-X / genus-$5$ family is precisely one of those one-axis conservation orbits;
6. a second equally natural one-axis family exists at
   $$
   k=\frac{11}{45};
   $$
7. both one-axis descriptions expose the same arithmetic defect
   $$
   \frac{233}{1913};
   $$
8. rational-square-preserving dual-pair moves form a rational rotation group.

Thus the previous arithmetic machinery can be reinterpreted as the analysis of particular move orbits inside a simpler conservation algebra.

---

# 1. Six-cell conservation state

Use the normalized squared variables

$$
A=\frac{a^2}{g^2},
\qquad
B=\frac{b^2}{g^2},
\qquad
C=\frac{c^2}{g^2},
$$

and

$$
D=\frac{d^2}{g^2},
\qquad
E=\frac{e^2}{g^2},
\qquad
F=\frac{f^2}{g^2}.
$$

The exact geometric conservation laws are

$$
\boxed{
A+B+C=1,
}
$$

$$
\boxed{
A+F=1,
}
$$

$$
\boxed{
B+E=1,
}
$$

$$
\boxed{
C+D=1.
}
$$

Hence

$$
D=1-C,
$$

$$
E=1-B,
$$

$$
F=1-A.
$$

The geometric state is therefore determined by only two independent real variables.

---

# 2. Affine conservation manifold

Define

$$
\mathcal M_C
=
\left\{
(A,B,C,D,E,F):
A+B+C=1,\;
A+F=1,\;
B+E=1,\;
C+D=1
\right\}.
$$

For nondegenerate cuboids impose

$$
0<A,B,C,D,E,F<1.
$$

Then

$$
\mathcal M_C
$$

is an affine two-dimensional manifold.

A tangent displacement

$$
\Delta X
=
(\delta_A,\delta_B,\delta_C,\delta_D,\delta_E,\delta_F)
$$

must satisfy

$$
\delta_A+\delta_B+\delta_C=0,
$$

$$
\delta_F=-\delta_A,
$$

$$
\delta_E=-\delta_B,
$$

$$
\delta_D=-\delta_C.
$$

Therefore

$$
\boxed{
\Delta X
=
(
\delta_A,
\delta_B,
\delta_C,
-\delta_C,
-\delta_B,
-\delta_A
),
}
$$

with

$$
\delta_A+\delta_B+\delta_C=0.
$$

---

# 3. Three visible move generators

Define

$$
\boxed{
M_{AB}
=
(1,-1,0,0,1,-1),
}
$$

$$
\boxed{
M_{BC}
=
(0,1,-1,1,-1,0),
}
$$

and

$$
\boxed{
M_{CA}
=
(-1,0,1,-1,0,1).
}
$$

They obey

$$
\boxed{
M_{AB}+M_{BC}+M_{CA}=0.
}
$$

Thus there are three visually natural turns but only two independent move directions.

A general conservation move may be written

$$
\boxed{
\Delta X
=
\lambda M_{AB}
+
\mu M_{BC}.
}
$$

---

# 4. Explicit finite move

Starting from

$$
X=(A,B,C,D,E,F),
$$

define

$$
X'
=
X+\lambda M_{AB}+\mu M_{BC}.
$$

Then

$$
A'=A+\lambda,
$$

$$
B'=B-\lambda+\mu,
$$

$$
C'=C-\mu,
$$

$$
D'=D+\mu,
$$

$$
E'=E+\lambda-\mu,
$$

$$
F'=F-\lambda.
$$

Direct substitution gives

$$
A'+B'+C'=1,
$$

$$
A'+F'=1,
$$

$$
B'+E'=1,
$$

$$
C'+D'=1.
$$

Therefore every such move preserves all geometric closure laws exactly.

---

# 5. Move group

Ignoring positivity boundaries, conservation moves form the additive group

$$
\boxed{
(\mathbb R^2,+).
}
$$

For rational moves,

$$
\lambda,\mu\in\mathbb Q,
$$

the rational conservation states form an affine action of

$$
\boxed{
(\mathbb Q^2,+).
}
$$

Composition is simply

$$
T_{\lambda_1,\mu_1}
\circ
T_{\lambda_2,\mu_2}
=
T_{\lambda_1+\lambda_2,\mu_1+\mu_2}.
$$

Thus the geometric move algebra itself is linear and abelian.

The nonlinearity of Perfect Cuboid enters only through arithmetic squarehood.

---

# 6. Dual-pair square locus

Define the rational dual-pair square locus

$$
\boxed{
\mathcal P
=
\left\{
X\in\mathbb Q\cap(0,1):
X\in\mathbb Q^2,\;
1-X\in\mathbb Q^2
\right\}.
}
$$

A value

$$
X\in\mathcal P
$$

means the dual pair

$$
(X,1-X)
$$

is arithmetically closed.

Every rational point on the unit circle gives such a value.

Write

$$
r\in\mathbb Q.
$$

Then

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

and

$$
\boxed{
1-X(r)
=
\left(
\frac{2r}{1+r^2}
\right)^2.
}
$$

Hence

$$
\mathcal P
$$

is rationally parameterized.

---

# 7. Minimal conservation form of Perfect Cuboid

Perfect Cuboid existence is equivalent to finding

$$
A,B,C\in\mathcal P
$$

such that

$$
\boxed{
A+B+C=1.
}
$$

Indeed, once

$$
A,B,C\in\mathcal P,
$$

their complements

$$
F=1-A,
$$

$$
E=1-B,
$$

$$
D=1-C
$$

are rational squares automatically.

Thus the complete six-square problem becomes

$$
\boxed{
(A,B,C)\in\mathcal P^3,
\qquad
A+B+C=1.
}
$$

This is the arithmetic core of the Conservation–Duality Branch.

---

# 8. Current seed

For

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

we have

$$
A
=
\frac{200704}{1151329}
=
\left(
\frac{448}{1073}
\right)^2,
$$

$$
B
=
\frac{245025}{1151329}
=
\left(
\frac{495}{1073}
\right)^2,
$$

$$
C
=
\frac{705600}{1151329}
=
\left(
\frac{840}{1073}
\right)^2.
$$

The dual cells are

$$
F
=
\frac{950625}{1151329}
=
\left(
\frac{975}{1073}
\right)^2,
$$

$$
E
=
\frac{906304}{1151329}
=
\left(
\frac{952}{1073}
\right)^2,
$$

and

$$
D
=
\frac{445729}{1151329}.
$$

Since

$$
445729
=
233\cdot1913,
$$

$D$ is rational but not a rational square.

Therefore:

$$
\boxed{
A,B\in\mathcal P,
}
$$

while

$$
\boxed{
C\notin\mathcal P.
}
$$

More precisely, $C$ itself is a square but its dual complement $D$ is not.

---

# 9. Arithmetic state of the seed

Using cell order

$$
(A,B,C,D,E,F),
$$

the seed has square-status vector

$$
\boxed{
\boldsymbol\Sigma_{\rm seed}
=
(1,1,1,0,1,1).
}
$$

The defect set is

$$
\boxed{
\mathcal D_{\rm seed}
=
\{D\}.
}
$$

The geometry already satisfies

$$
\boxed{
\mathbf T=\mathbf1_6.
}
$$

Hence only arithmetic repair remains.

---

# 10. Minimum-support repair theorem

## Theorem

Any conservation move that changes the defective cell $D$ must move at least four of the six cells.

### Proof

Since

$$
D=1-C,
$$

changing $D$ requires

$$
\delta_D\neq0,
$$

hence

$$
\delta_C=-\delta_D\neq0.
$$

But global conservation requires

$$
\delta_A+\delta_B+\delta_C=0.
$$

Therefore at least one of

$$
\delta_A,
\delta_B
$$

must also be nonzero.

Its dual complement then changes with opposite sign.

Thus at least the four cells

$$
(C,D)
$$

plus one additional dual pair must move.

Therefore the minimum support is

$$
\boxed{4}.
$$

---

# 11. Consequence for the three visible generators

The move

$$
M_{AB}
$$

leaves

$$
C
$$

and

$$
D
$$

unchanged.

Therefore:

$$
\boxed{
M_{AB}
\text{ can never repair the present defect.}
}
$$

The only useful one-generator repair directions are:

$$
\boxed{
M_{BC}
}
$$

and

$$
\boxed{
M_{CA}.
}
$$

Thus the current one-defect seed has exactly two natural one-axis repair orbits.

---

# 12. First repair orbit: fix $(A,F)$

Under

$$
M_{BC},
$$

the pair

$$
(A,F)
$$

is fixed.

The moving cells are

$$
B,\ C,\ D,\ E.
$$

Since

$$
B+C=F
$$

is conserved, a fully square repair requires:

$$
\boxed{
B',C'\in\mathcal P
}
$$

with

$$
\boxed{
B'+C'=F.
}
$$

This is the exact one-axis fixed-$A$ repair problem.

---

# 13. Nested rational parameter on the fixed-$(A,F)$ orbit

Because

$$
F
=
\left(
\frac{975}{1073}
\right)^2,
$$

write

$$
\sqrt{B'}
=
\frac{975}{1073}
\frac{1-h^2}{1+h^2},
$$

and

$$
\sqrt{C'}
=
\frac{975}{1073}
\frac{2h}{1+h^2},
$$

with

$$
h\in\mathbb Q.
$$

Then automatically:

$$
B'+C'=F.
$$

The remaining arithmetic conditions are only:

$$
E'=1-B'\in\mathbb Q^2,
$$

and

$$
D'=1-C'\in\mathbb Q^2.
$$

Thus a four-cell repair has become a two-gate square problem.

---

# 14. Recovery of the previous $k=7/32$ theory

The fixed dual pair

$$
(A,F)
$$

has half-angle parameter

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

Apply the Möbius coordinate

$$
k_A
=
\frac{1-p_A}{1+p_A}.
$$

Then

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

The seed split

$$
(B,C)
=
\left(
\left(\frac{495}{1073}\right)^2,
\left(\frac{840}{1073}\right)^2
\right)
$$

corresponds to

$$
\boxed{
h_A=\frac47.
}
$$

Define

$$
F_k(h)
=
\frac{h^2+k^2}{1+k^2h^2},
$$

and

$$
M(h)
=
\frac{1-h}{1+h}.
$$

Then the two repair gates are exactly:

$$
\boxed{
F_{7/32}\left(\frac47\right)
=
\left(
\frac{17}{28}
\right)^2,
}
$$

and

$$
\boxed{
F_{7/32}
\left(
M\left(\frac47\right)
\right)
=
\frac{233}{1913}.
}
$$

Thus the entire previous G2A–G2E fixed-slice program is exactly the analysis of the single conservation orbit generated by

$$
M_{BC}
$$

through the current seed.

This is a major conceptual reinterpretation.

---

# 15. Second repair orbit: fix $(B,E)$

The move

$$
M_{CA}
$$

fixes

$$
(B,E).
$$

The moving cells are

$$
A,\ C,\ D,\ F.
$$

Now

$$
A+C=E
$$

is conserved.

A fully square repair requires:

$$
\boxed{
A',C'\in\mathcal P
}
$$

with

$$
\boxed{
A'+C'=E.
}
$$

This is the second one-axis repair orbit.

It is algebraically parallel to the first but has not been explored to the same depth.

---

# 16. Exact parameters of the second repair orbit

The fixed pair

$$
(B,E)
$$

has half-angle parameter

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

Its Möbius coordinate is

$$
k_B
=
\frac{1-p_B}{1+p_B}.
$$

Hence

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

The current split

$$
(A,C)
$$

inside

$$
E=\left(\frac{952}{1073}\right)^2
$$

is the normalized Pythagorean pair

$$
\frac{448}{952}
=
\frac8{17},
$$

$$
\frac{840}{952}
=
\frac{15}{17}.
$$

Therefore the nested parameter is

$$
\boxed{
h_B=\frac35.
}
$$

---

# 17. The same defect reappears

For the second repair orbit:

$$
\boxed{
F_{11/45}\left(\frac35\right)
=
\left(
\frac{25}{39}
\right)^2.
}
$$

The complementary gate is:

$$
\boxed{
F_{11/45}
\left(
M\left(\frac35\right)
\right)
=
\frac{233}{1913}.
}
$$

Thus both useful one-axis repair orbits expose exactly the same unresolved arithmetic defect:

$$
\boxed{
\frac{233}{1913}.
}
$$

This is not an accident.

It is the half-angle form of the single defective dual pair

$$
(C,D).
$$

---

# 18. Central defect invariant at the seed

The defective cell is

$$
D
=
1-C
=
\frac{445729}{1073^2}.
$$

Its numerator factors as

$$
445729
=
233\cdot1913.
$$

The half-angle defect is

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

Hence the two factors

$$
233
$$

and

$$
1913
$$

are the additive dual decomposition of the same arithmetic square defect.

The previous shared-core method, CARCE half-angle method, and the present conservation method are all observing the same defect in different coordinates.

---

# 19. Rational rotation operator on a closed dual pair

The linear conservation move is the simplest geometric representation.

For arithmetic square preservation, there is a more natural nonlinear move.

Suppose:

$$
X=u^2,
$$

$$
1-X=v^2,
$$

with

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

and

$$
u^2+v^2=1.
$$

Choose:

$$
r\in\mathbb Q.
$$

Define the rational rotation coefficients

$$
c_r
=
\frac{1-r^2}{1+r^2},
$$

and

$$
s_r
=
\frac{2r}{1+r^2}.
$$

Then:

$$
c_r^2+s_r^2=1.
$$

Define:

$$
\boxed{
u'
=
u c_r-v s_r,
}
$$

$$
\boxed{
v'
=
u s_r+v c_r.
}
$$

Since all quantities are rational,

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

And:

$$
u'^2+v'^2=1.
$$

Therefore:

$$
\boxed{
X'=u'^2,
\qquad
1-X'=v'^2
}
$$

remains a fully rational-square dual pair.

This is the exact arithmetic version of a “turn”.

---

# 20. Rational turn group law

Let:

$$
R(r)
=
\begin{pmatrix}
c_r & -s_r\\
s_r & c_r
\end{pmatrix}.
$$

Then:

$$
R(r_1)R(r_2)
=
R(r_1\oplus r_2),
$$

where:

$$
\boxed{
r_1\oplus r_2
=
\frac{r_1+r_2}{1-r_1r_2}
}
$$

whenever the denominator is nonzero.

Thus rational square-preserving dual-pair turns form the rational unit-circle rotation group.

This is the precise algebraic analogue of a reversible cube turn.

---

# 21. Pair rotations versus linear conservation moves

There are now two complementary descriptions.

## Linear move coordinates

$$
(\lambda,\mu)
$$

act directly on squared cell masses.

They make global conservation transparent.

## Rational rotation coordinates

$$
(r_A,r_B,r_C)
$$

act on square roots inside each dual pair.

They make arithmetic square preservation transparent.

The full problem asks for transformations that satisfy both descriptions simultaneously.

This dual coordinate system is one of the main benefits of the Conservation–Duality Branch.

---

# 22. Two-pair repair strategy

At the current seed:

$$
(A,F)
$$

and

$$
(B,E)
$$

are already fully square-closed.

Therefore one can apply rational rotations independently:

$$
(A,F)\mapsto(A',F'),
$$

$$
(B,E)\mapsto(B',E').
$$

Then define the third pair by conservation:

$$
C'
=
1-A'-B',
$$

$$
D'
=
A'+B'.
$$

The first four cells remain rational squares automatically.

The entire Perfect Cuboid problem is reduced to the final requirement:

$$
\boxed{
C'\in\mathbb Q^2,
\qquad
D'\in\mathbb Q^2.
}
$$

Since:

$$
C'+D'=1,
$$

this means simply:

$$
\boxed{
D'\in\mathcal P.
}
$$

Thus a general two-turn repair is:

> rotate the two currently good dual pairs rationally until their top-cell sum itself becomes a dual-pair square value.

---

# 23. Two-turn repair equation

Let:

$$
A'=X(r_A),
$$

$$
B'=X(r_B),
$$

where:

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

after absorbing the current pair phases into the rational rotation parameters.

Then the repair condition is:

$$
\boxed{
A'+B'\in\mathcal P.
}
$$

Equivalently, there exists rational:

$$
r_C
$$

such that:

$$
\boxed{
X(r_A)+X(r_B)+X(r_C)=1.
}
$$

This is exactly CARCE in conservation coordinates.

Hence the CARCE surface is the global two-turn square-preserving repair space.

---

# 24. Meaning of the previous fixed slice

The previous deep $k=7/32$ analysis held the first pair

$$
(A,F)
$$

fixed.

In the new language, it set:

$$
r_A=0
$$

and explored only a one-dimensional section of the full two-turn repair space.

Therefore failure or extreme arithmetic sparsity on that fixed slice does not imply failure of the complete conservation method.

The new branch makes explicit the extra freedom:

$$
\boxed{
r_A\neq0,
\qquad
r_B\neq0.
}
$$

This is precisely the freedom that a fixed-slice approach suppresses.

---

# 25. Seed-local move hierarchy

Starting from the current seed, there are three levels of repair.

## Level 0 — no move

Arithmetic state:

$$
111011.
$$

## Level 1 — one-axis move

Possible useful axes:

$$
M_{BC},
$$

or:

$$
M_{CA}.
$$

Each gives a one-dimensional fixed-pair family.

## Level 2 — two-axis move

Use both independent move directions.

No dual pair needs to remain fixed.

This accesses the full two-dimensional conservation simplex.

The Level-2 search is the natural next target if both one-axis families remain arithmetically obstructed.

---

# 26. C1 theorem summary

The current seed has one arithmetic defect.

The defect cannot be repaired by a move supported on fewer than four cells.

There are exactly two useful one-generator repair lines:

$$
\boxed{
\text{Fix }(A,F):
\quad
k=\frac7{32},
\quad
h_{\rm seed}=\frac47,
}
$$

and:

$$
\boxed{
\text{Fix }(B,E):
\quad
k=\frac{11}{45},
\quad
h_{\rm seed}=\frac35.
}
$$

Both expose the same residual half-angle defect:

$$
\boxed{
\frac{233}{1913}.
}
$$

The first line is exactly the previously studied G2A–G2E family.

The second line is its conservation-dual sibling.

---

# 27. Next phase

The natural next phase is:

$$
\boxed{
\text{C2 — Two-Turn Square-Preserving Repair}.
}
$$

Instead of fixing one good dual pair, rotate both:

$$
(A,F)
$$

and:

$$
(B,E)
$$

through rational unit-circle turns.

Then enforce only:

$$
\boxed{
A'+B'\in\mathcal P.
}
$$

The research question becomes:

> Can the extra second move direction close the $233/1913$ defect without entering the high-genus fixed-slice obstruction?

This should be attacked algebraically before any large search.

---

## Canonical C1 Result

$$
\boxed{
\text{Previous fixed-slice arithmetic}
=
\text{one-axis orbit of the conservation move algebra}.
}
$$

and:

$$
\boxed{
\text{Full new search}
=
\text{two-axis rational rotation on the conservation simplex}.
}
$$

---

**End of C1**