# Perfect Cuboid G2E.2 — The Exact Square-Lift Tower

## Genus $0\rightarrow1\rightarrow3\rightarrow5$, rank–genus locking, and the intermediate quadratic-Chabauty target

**Series:** Perfect Cuboid Global Closure  
**Phase:** G2E.2  
**Fixed slice:** $k=7/32$  
**Status:** exact structural theorem  
**Date:** 2026-09-30

---

# 1. Main result

The fixed-$k$ simultaneous Perfect-Cuboid closure does not appear as one opaque genus-$5$ object.

It is naturally built by three successive quadratic lifts:

$$
\boxed{
C_0
\longleftarrow
C_1
\longleftarrow
C_2
\longleftarrow
C_3.
}
$$

Their genera are:

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

For the current slice:

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

the corresponding Jacobian ranks are:

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

Thus every nontrivial layer satisfies a rank–genus equality.

The first lift is exactly the already-certified elliptic factor $E_0$.

The second lift is the genus-$3$ quotient whose Jacobian is:

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

The final unramified lift restores the two-dimensional Prym:

$$
E_f^2.
$$

Therefore:

$$
\boxed{
\operatorname{Jac}(C_3)
\sim
E_f^2\times E_0^2\times E_-.
}
$$

The entire previous Jacobian decomposition is now explained by the geometry of the square-lift tower.

---

# 2. Base level $C_0$

G2E.1 introduced:

$$
Q=q^2,
\qquad
R=r^2,
$$

and the symmetric coordinates:

$$
S=Q+R,
$$

$$
P=QR.
$$

For:

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

the symmetric quotient is the rational conic:

$$
1352033(S^2+P^2+1)
-
1901250S(P+1)
-
15717198P
=
0.
$$

This is:

$$
\boxed{
C_0.
}
$$

Hence:

$$
g(C_0)=0.
$$

A rational parameter $t$ gives:

$$
P(t)
=
\frac{
49(t-2)^2+1024
}{
49+1024t^2
}.
$$

---

# 3. First square lift: $P=U^2$

The first arithmetic lift requires:

$$
P=QR=(qr)^2.
$$

Introduce:

$$
U=qr.
$$

Then:

$$
U^2=P(t).
$$

After clearing the square denominator, this becomes:

$$
\boxed{
C_1:
\qquad
Y^2
=
(1024t^2+49)
(49t^2-196t+1220).
}
$$

The quartic is separable.

Therefore:

$$
\boxed{
g(C_1)=1.
}
$$

---

# 4. Jacobian of the first lift

For the binary quartic

$$
(1024t^2+49)
(49t^2-196t+1220),
$$

the classical invariants are:

$$
I=1596916897473,
$$

and:

$$
J=-3695139075955421250.
$$

Its Jacobian can be written:

$$
y^2=x^3-27Ix-27J.
$$

The cubic factors exactly:

$$
\boxed{
y^2
=
(x-4658211)
(x-2851875)
(x+7510086).
}
$$

Translate:

$$
X=x-2851875.
$$

Then:

$$
Y^2
=
X
(X-1806336)
(X+10361961).
$$

But:

$$
1806336=1344^2=9\cdot448^2,
$$

and:

$$
10361961=3219^2=9\cdot1073^2.
$$

Now scale:

$$
X=9X_0,
$$

$$
Y=27Y_0.
$$

We obtain:

$$
\boxed{
Y_0^2
=
X_0
(X_0-448^2)
(X_0+1073^2).
}
$$

This is exactly the full-$2$-torsion model $E_0'$ used in G2D.

Therefore:

$$
\boxed{
\operatorname{Jac}(C_1)
\cong
E_0'.
}
$$

This identifies the first square lift with the previously discovered elliptic factor, rather than introducing a new curve.

---

# 5. Independent local rank certification for $C_1$

The integral Jacobian roots are:

$$
4658211,
$$

$$
2851875,
$$

$$
-7510086.
$$

The squareclass support is:

$$
\{
-1,2,3,7,29,37,73,18521
\}.
$$

Hence the raw full-$2$-descent space contains:

$$
256^2=65536
$$

coverings.

Exact local gates give:

$$
65536
\xrightarrow{2^6}
8192,
$$

$$
8192
\xrightarrow{3^3}
2048,
$$

$$
2048
\xrightarrow{7^2}
512,
$$

$$
512
\xrightarrow{29^2}
128,
$$

$$
128
\xrightarrow{37^2}
32,
$$

$$
32
\xrightarrow{73^2}
\boxed{8}.
$$

Thus:

$$
\#\operatorname{Sel}^{(2)}
\le8.
$$

The curve has full rational $2$-torsion.

Good reduction gives:

$$
\#E(\mathbb F_5)=8,
$$

$$
\#E(\mathbb F_{13})=8,
$$

$$
\#E(\mathbb F_{19})=20.
$$

Hence rational torsion order divides:

$$
\gcd(8,8,20)=4.
$$

Since full $2$-torsion already has order $4$:

$$
E(\mathbb Q)_{\rm tors}
\cong
(\mathbb Z/2\mathbb Z)^2.
$$

The exact integral point:

$$
\boxed{
(-5703750,12654532800)
}
$$

is not torsion.

Therefore:

$$
\operatorname{rank}E(\mathbb Q)\ge1.
$$

The $2$-Selmer bound gives:

$$
\operatorname{rank}E(\mathbb Q)\le1.
$$

Hence:

$$
\boxed{
\operatorname{rank}\operatorname{Jac}(C_1)=1.
}
$$

So the first nontrivial square lift has:

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

---

# 6. Second lift: split $Q$ and $R$

Once:

$$
P=U^2
$$

is imposed, the two squared coordinates $Q,R$ are the roots of:

$$
X^2-SX+P=0.
$$

They separate rationally exactly when:

$$
S^2-4P
$$

is a rational square.

Set:

$$
V^2=S^2-4P.
$$

Using the rational $t$-parameterization of $C_0$, exact simplification gives:

$$
S(t)
=
\frac{
589309841t^2
+
5165282168t
+
1226953309
}{
1352033(1024t^2+49)
}.
$$

Furthermore:

$$
S(t)^2-4P(t)
$$

equals:

$$
\left[
\frac{
975t+1171
}{
1352033(1024t^2+49)
}
\right]^2
$$

times:

$$
-20617451839t^2
+
7997395778678t
+
779078718161.
$$

Thus the second lift is birational to:

$$
\boxed{
C_2:
\begin{cases}
Y_1^2
=
(1024t^2+49)
(49t^2-196t+1220),\\
Y_2^2
=
-20617451839t^2
+
7997395778678t
+
779078718161.
\end{cases}
}
$$

---

# 7. Genus of $C_2$

The first quadratic cover of $\mathbb P^1_t$ has four branch locations.

The second has two branch locations.

Their branch sets are disjoint because:

$$
\operatorname{Res}(f_4,f_2)
=
29^4
37^4
73^8
18521^8
\neq0.
$$

Therefore the degree-$4$ biquadratic cover has six branch locations.

At every branch location the total ramification contribution is $2$.

Riemann–Hurwitz gives:

$$
2g(C_2)-2
=
4(-2)+6\cdot2.
$$

Hence:

$$
2g(C_2)-2=4,
$$

so:

$$
\boxed{
g(C_2)=3.
}
$$

---

# 8. Group-theoretic meaning of $C_2$

Return to the actual half-angle square roots:

$$
q,
\qquad
r.
$$

Define:

$$
U=qr,
$$

and:

$$
V=q^2-r^2.
$$

The simultaneous sign involution:

$$
(q,r)
\mapsto
(-q,-r)
$$

fixes both $U$ and $V$.

Thus $C_2$ is the quotient of the full genus-$5$ curve by simultaneous sign.

The additional exchange:

$$
q\leftrightarrow r
$$

fixes $U$ and sends:

$$
V\mapsto-V.
$$

Quotienting $C_2$ by this exchange gives $C_1$.

Therefore:

$$
\boxed{
C_2
\longrightarrow
C_1
\longrightarrow
C_0
}
$$

is the exact geometric realization of the first two square lifts.

---

# 9. Jacobian of the genus-$3$ layer

The quotient $C_2$ is the genus-$3$ curve denoted $H_k$ in G2C.

G2C established:

$$
\operatorname{Jac}(H_k)
\sim_{\mathbb Q}
E_0\times E_+\times E_-.
$$

For the fixed slice:

$$
E_+\cong E_0.
$$

Therefore:

$$
\boxed{
\operatorname{Jac}(C_2)
\sim_{\mathbb Q}
E_0^2\times E_-.
}
$$

G2D proved:

$$
\operatorname{rank}E_0=1,
$$

and:

$$
\operatorname{rank}E_-=1.
$$

Hence:

$$
\boxed{
\operatorname{rank}
\operatorname{Jac}(C_2)(\mathbb Q)
=
3.
}
$$

Since:

$$
g(C_2)=3,
$$

the second lift again satisfies:

$$
\boxed{
r=g=3.
}
$$

---

# 10. Third lift: recover $q$ and $r$

At the genus-$3$ level we know:

$$
S=Q+R,
$$

$$
P=QR=U^2,
$$

and:

$$
V^2=(Q-R)^2.
$$

Hence:

$$
Q
=
\frac{S+V}{2},
$$

$$
R
=
\frac{S-V}{2}.
$$

The final Perfect-Cuboid lift requires:

$$
Q=q^2.
$$

Once $q$ is chosen,

$$
r=\frac{U}{q}
$$

is automatically rational and satisfies:

$$
r^2=R.
$$

Thus the final lift is the quadratic extension:

$$
\boxed{
q^2=\frac{S+V}{2}.
}
$$

This recovers the full genus-$5$ curve $C_3=C_{7/32}$.

---

# 11. The final double cover is unramified

The deck transformation is:

$$
(q,r)
\mapsto
(-q,-r).
$$

A fixed point would require simultaneous branching of the two original square gates.

But the two original branch divisors are disjoint.

Equivalently, the simultaneous sign involution has no fixed point on the smooth projective normalization.

Therefore:

$$
\boxed{
C_3\to C_2
\text{ is an unramified double cover}.
}
$$

Riemann–Hurwitz gives:

$$
2g(C_3)-2
=
2\left(
2g(C_2)-2
\right).
$$

Since:

$$
g(C_2)=3,
$$

we obtain:

$$
2g(C_3)-2=8,
$$

and therefore:

$$
\boxed{
g(C_3)=5.
}
$$

The Prym has dimension:

$$
g(C_3)-g(C_2)=2.
$$

G2C identifies this Prym, up to isogeny, as:

$$
\boxed{
E_f^2.
}
$$

Hence:

$$
\boxed{
\operatorname{Jac}(C_3)
\sim
E_f^2
\times
\operatorname{Jac}(C_2).
}
$$

Substitution gives:

$$
\boxed{
\operatorname{Jac}(C_3)
\sim
E_f^2\times E_0^2\times E_-.
}
$$

This explains the G2C decomposition directly from the square-lift geometry.

---

# 12. Rank–genus locking

The tower now has the exact form:

| Layer | Arithmetic lift | Genus | Jacobian rank |
|---|---|---:|---:|
| $C_0$ | symmetric quotient | $0$ | $0$ |
| $C_1$ | $P=U^2$ | $1$ | $1$ |
| $C_2$ | $S^2-4P=V^2$ | $3$ | $3$ |
| $C_3$ | $Q=q^2$ | $5$ | $5$ |

Thus:

$$
\boxed{
r(C_i)=g(C_i)
}
$$

for every nontrivial level.

This explains why ordinary Chabauty never becomes available anywhere in the natural tower.

The failure is structural rather than accidental.

---

# 13. Quadratic Chabauty at the intermediate layer

Although:

$$
r(C_2)=g(C_2)=3,
$$

the Jacobian splitting gives:

$$
\operatorname{Jac}(C_2)
\sim
E_0^2\times E_-.
$$

The repeated block:

$$
E_0^2
$$

contributes at least three independent Néron–Severi classes.

The remaining elliptic factor contributes at least one.

Therefore:

$$
\boxed{
\rho(\operatorname{Jac}(C_2)/\mathbb Q)\ge4.
}
$$

Hence:

$$
g+\rho-1
\ge
3+4-1
=
6.
$$

Since:

$$
r=3,
$$

we have:

$$
\boxed{
3<6.
}
$$

Thus the genus-$3$ intermediate curve already satisfies the quadratic-Chabauty rank criterion.

This is strategically important.

Any rational point on the full genus-$5$ curve maps to a rational point on $C_2$.

Therefore, if the rational points on $C_2$ can be completely classified and shown to be only degenerate lift images, then the full fixed slice is eliminated without running quadratic Chabauty directly on genus $5$.

---

# 14. A favorable finite-field prime for $C_2$

At the good prime:

$$
p=31,
$$

exact enumeration gives:

$$
\boxed{
\#C_2(\mathbb F_{31})=16.
}
$$

Thus a genus-$3$ quadratic-Chabauty computation can begin with only sixteen residue disks.

This is currently the preferred local prime for the intermediate layer.

The full genus-$5$ curve has the even stronger boundary-only phenomenon at:

$$
p=11,19,31.
$$

Therefore both levels provide independent $p$-adic consistency checks.

---

# 15. New preferred proof route

The previous G2E.1 route proposed applying quadratic Chabauty directly to $C_3$.

The square-lift tower gives a lower-dimensional alternative:

## Step 1

Work first on:

$$
\boxed{
C_2,\quad g=3,\quad r=3,\quad \rho\ge4.
}
$$

## Step 2

Use:

- elliptic quotient maps to $E_0,E_0,E_-$;
- their certified rank-one Mordell–Weil groups;
- $p=31$ residue disks;
- quadratic-Chabauty height functions;
- Mordell–Weil compatibility.

## Step 3

Determine:

$$
C_2(\mathbb Q).
$$

## Step 4

For every surviving rational point on $C_2$, test the final unramified square lift:

$$
q^2=\frac{S+V}{2}.
$$

Only those that lift rationally can belong to $C_3(\mathbb Q)$.

Thus the full fixed-slice problem is replaced by:

$$
\boxed{
\text{genus-3 rational-point classification}
+
\text{one final exact square test}.
}
$$

This is strictly smaller than attacking genus $5$ directly.

---

# 16. Conceptual interpretation

The Perfect Cuboid arithmetic obstruction is now resolved into three successive geometric questions:

### Level 1

Can the product of the two missing half-angle squares be a square?

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

This creates the elliptic curve $E_0$.

### Level 2

Can the two squared coordinates be separated rationally?

$$
(Q-R)^2\in\mathbb Q^2.
$$

This creates the genus-$3$ intermediate curve.

### Level 3

Is one of the separated rational coordinates itself a square?

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

This restores the full genus-$5$ curve.

So the arithmetic complexity is not a single jump.

It grows by controlled square lifts:

$$
\boxed{
\text{product}
\rightarrow
\text{separation}
\rightarrow
\text{individual square}.
}
$$

This is the exact arithmetic analogue of the visual X-diagonal closure program.

---

# 17. Current theorem target

The preferred next theorem is now:

$$
\boxed{
C_2(\mathbb Q)
=
\text{known degenerate/intermediate points}.
}
$$

If that theorem is achieved and none of the nondegenerate survivors passes the final square lift, then:

$$
\boxed{
C_{7/32}(\mathbb Q)
=
\mathcal B.
}
$$

This would globally eliminate the entire current fixed first-channel slice.

---

**End of G2E.2**