# Perfect Cuboid G2C — Five-Elliptic Decomposition of the Fixed-$k$ Genus-5 Closure Curve

## Exact symmetry decomposition

For a fixed nondegenerate rational parameter

$$
0<k<1,
$$

define

$$
f(h)
=
(h^2+k^2)(1+k^2h^2).
$$

The first square gate is represented by

$$
Y_1^2=f(h).
$$

After applying

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

to the second gate and clearing the square denominator, define

$$
g(h)
=
(1+h)^4 f(M(h)).
$$

Exact simplification gives

$$
g(h)
=
\left(
(1+k^2)(h^2+1)
+
2(1-k^2)h
\right)
$$

$$
\qquad\cdot
\left(
(1+k^2)(h^2+1)
-
2(1-k^2)h
\right).
$$

The simultaneous Perfect Cuboid closure curve is the biquadratic curve

$$
C_k:
\qquad
Y_1^2=f(h),
$$

$$
Y_2^2=g(h).
$$

Its smooth projective normalization has genus

$$
5.
$$

---

# 1. First $V_4$ decomposition

The sign changes

$$
(Y_1,Y_2)
\mapsto
(-Y_1,Y_2),
$$

and

$$
(Y_1,Y_2)
\mapsto
(Y_1,-Y_2)
$$

generate a Klein four group.

The three nontrivial quotients are:

## First elliptic quotient

$$
E_f:
\qquad
Y_1^2=f(h).
$$

## Second elliptic quotient

$$
E_g:
\qquad
Y_2^2=g(h).
$$

Because $g$ is obtained from $f$ by the Möbius involution $M$ and a square clearing factor,

$$
\boxed{
E_g\cong E_f
}
$$

over $\mathbb Q$.

## Genus-$3$ quotient

Using

$$
Y=Y_1Y_2,
$$

the third quotient is

$$
H_k:
\qquad
Y^2=f(h)g(h).
$$

Hence the quotient dimensions are

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

The pullbacks of holomorphic differentials from these three quotients span the full five-dimensional differential space of $C_k$.

Therefore the induced map on Jacobians has finite kernel and

$$
\boxed{
\operatorname{Jac}(C_k)
\sim_{\mathbb Q}
E_f
\times
E_g
\times
\operatorname{Jac}(H_k).
}
$$

---

# 2. The genus-$3$ polynomial is even and palindromic

Write

$$
P(h)=f(h)g(h).
$$

Exact expansion gives

$$
P(h)
=
a h^8
+
b h^6
+
c h^4
+
b h^2
+
a,
$$

where

$$
a
=
k^2(1+k^2)^2,
$$

$$
b
=
k^8+14k^4+1,
$$

and

$$
c
=
-2k^8+14k^6+14k^2-2.
$$

Thus

$$
P(-h)=P(h)
$$

and

$$
h^8P(1/h)=P(h).
$$

The genus-$3$ curve therefore carries the commuting involutions

$$
\sigma:
h\mapsto-h,
$$

$$
\tau:
h\mapsto\frac1h,
$$

and

$$
\sigma\tau:
h\mapsto-\frac1h.
$$

These form another Klein four symmetry.

---

# 3. Elliptic quotient $E_0$

Quotient by

$$
\sigma:h\mapsto-h.
$$

Set

$$
X=h^2.
$$

Then:

$$
\boxed{
E_0:
\quad
Y^2
=
aX^4+bX^3+cX^2+bX+a.
}
$$

At

$$
X=1,
$$

the right side equals

$$
16k^2(1+k^2)^2.
$$

Hence $E_0$ has the rational point

$$
\left(
1,
4k(1+k^2)
\right).
$$

Its smooth projective model is genus $1$.

---

# 4. Elliptic quotient $E_+$

Quotient by

$$
\tau:
h\mapsto\frac1h.
$$

Use the invariant coordinates

$$
U=h+\frac1h,
$$

and

$$
V=\frac{Y}{h^2}.
$$

Then:

$$
\boxed{
E_+:
\quad
V^2
=
aU^4
+
(1-k^2)^4U^2
-
4(1-k^2)^4.
}
$$

The point induced by $h=1$ is

$$
\left(
2,
4k(1+k^2)
\right).
$$

Thus $E_+$ is again an elliptic curve over $\mathbb Q$.

---

# 5. Elliptic quotient $E_-$

Quotient by

$$
\sigma\tau:
h\mapsto-\frac1h.
$$

Use

$$
W=h-\frac1h,
$$

and

$$
V=\frac{Y}{h^2}.
$$

Then:

$$
\boxed{
E_-:
\quad
V^2
=
aW^4
+
\left(
(1+k^2)^4+16k^4
\right)W^2
+
16a.
}
$$

At

$$
W=0,
$$

we obtain the rational point

$$
\left(
0,
4k(1+k^2)
\right).
$$

Hence $E_-$ is also elliptic.

---

# 6. Exact Jacobian splitting of $H_k$

A basis for the holomorphic differentials of

$$
H_k:
Y^2=P(h)
$$

is

$$
\omega_0=\frac{dh}{Y},
$$

$$
\omega_1=\frac{h\,dh}{Y},
$$

$$
\omega_2=\frac{h^2\,dh}{Y}.
$$

The quotient map to $E_0$ pulls back its elliptic differential to a nonzero multiple of

$$
\omega_1.
$$

The quotient map to $E_+$ pulls back to

$$
\omega_2-\omega_0.
$$

The quotient map to $E_-$ pulls back to

$$
\omega_2+\omega_0.
$$

The three forms

$$
\omega_1,
\qquad
\omega_2-\omega_0,
\qquad
\omega_2+\omega_0
$$

are linearly independent and span

$$
H^0(H_k,\Omega^1).
$$

Therefore the product of the three quotient maps induces an isogeny

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

---

# 7. Five-elliptic decomposition of the Perfect Cuboid closure curve

Combining the two decompositions:

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

Since:

$$
E_g\cong E_f,
$$

we obtain:

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

This is an exact structural reduction of the genus-$5$ simultaneous square-closure problem into five elliptic factors.

---

# 8. Why this matters

The fixed-$k$ problem no longer needs to be treated as an opaque genus-$5$ curve.

Its Jacobian arithmetic is controlled by five elliptic curves.

Therefore the next proof program can target:

1. exact Mordell-Weil rank of $E_f$;
2. exact rank of $E_0$;
3. exact rank of $E_+$;
4. exact rank of $E_-$;
5. compatibility of their quotient images with points on $C_k$.

If the total Jacobian rank can be certified below the genus threshold, Chabauty-type methods become available.

Even when the total rank is larger, the explicit split permits:

- elliptic Chabauty;
- Mordell-Weil sieve;
- covering collections;
- quotient-pullback compatibility tests;
- independent rank bounds on low-dimensional factors.

This is much more tractable than treating the original seven-integer problem directly.

---

# 9. Current fixed slice

For the current seed:

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

the first factor is exactly the elliptic curve used in G2B.

G2B proves:

$$
E_f(\mathbb Q)_{\mathrm{tors}}
\cong
\mathbb Z/4\mathbb Z
\times
\mathbb Z/2\mathbb Z,
$$

and:

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

The exact ranks of the five factors have not yet been certified.

That is now the highest-priority algebraic task for the fixed-$k$ slice.

---

# 10. Canonical next theorem target

The immediate next theorem should be one of the following:

### Target A

Certify all five elliptic ranks for

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

### Target B

Use the quotient maps plus local information to determine all rational points on

$$
C_{7/32}.
$$

### Target C

Prove the positive rational locus of

$$
C_{7/32}
$$

contains only degenerate boundary points.

Any of these would globally eliminate the entire current fixed-$k$ slice, rather than merely increasing a denominator or elliptic-index bound.

---

**End of G2C**