# RCHM × Hodge Conjecture — Research Round 25
## Odd-Unimodular Gluing Correction for Küchle $c5$: the Integral $K3^{[2]}$ Primitive Lattice Reopens

**Status:** Internal research note + exact finite-quadratic-form verifier  
**Version:** Round 25 / v2.5  
**Date:** 2026-09-18  
**Methodological base:** RCHM; X-Layer Mathematical Completeness; Ancestry Reduction; Bridge Search Compiler  
**Previous rounds:** RCHM × Hodge Conjecture — Rounds 1–24  
**Canonical source:** UTF-8 Markdown

---

# 0. Round 25 Objective

Round 24 completed the inherited rank-$4$ lattice of the Küchle $c5$ fivefold:

$$
\boxed{
L_{\mathrm{inh}}
\simeq
A_1\oplus A_3,
}
$$

with:

$$
\boxed{
|\det L_{\mathrm{inh}}|=8,
}
$$

and:

$$
\boxed{
A_{L_{\mathrm{inh}}}
\simeq
\mathbb Z/2
\oplus
\mathbb Z/4.
}
$$

Round 24 then rejected a direct integral primitive-$K3^{[2]}$ bridge by assigning the negative discriminant **quadratic** form of $L_{\mathrm{inh}}$ to the hard orthogonal complement.

That last step was incorrect.

The full middle lattice of the $c5$ fourfold is **odd unimodular**, not even unimodular.

For a primitive orthogonal decomposition inside an odd unimodular lattice:

- the discriminant **bilinear** forms are anti-isometric;
- the discriminant **quadratic** refinements need not be negatives of one another.

The missing refinement is determined by parity and Milgram's formula.

The corrected refinement is exactly the one carried by the primitive lattice of a degree-$4$, divisibility-$1$ polarized $K3^{[2]}$ fourfold.

Hence the direct integral lattice template reopens.

---

# 1. The Full Middle Lattice Is Unimodular

Let:

$$
X=X^4_{c5}.
$$

Round 21 established that:

$$
H^4(X,\mathbb Z)
$$

is torsion-free.

By Poincaré duality, the middle intersection pairing is unimodular.

The middle Hodge numbers are:

$$
(1,24,1),
$$

so the real signature is:

$$
\boxed{
(24,2).
}
$$

Let:

$$
L
=
H^4(X,\mathbb Z)
$$

with the cup-product pairing.

Then:

$$
\boxed{
L
\text{ is unimodular of signature }(24,2).
}
$$

---

# 2. The Full Middle Lattice Is Odd

An even unimodular lattice must satisfy:

$$
\sigma_+-\sigma_-
\equiv0
\pmod8.
$$

But:

$$
24-2=22.
$$

Since:

$$
22\not\equiv0\pmod8,
$$

the lattice cannot be even.

Therefore:

$$
\boxed{
L
\text{ is odd unimodular}.
}
$$

By the classification of indefinite odd unimodular lattices:

$$
\boxed{
L
\simeq
I_{24,2}.
}
$$

This parity fact is the key correction to Round 24.

---

# 3. The Inherited Primitive Lattice

Round 24 gives:

$$
\boxed{
A
=
L_{\mathrm{inh}}
\simeq
A_1\oplus A_3.
}
$$

It is:

- even;
- positive definite;
- rank $4$;
- primitive in $L$.

Its discriminant quadratic form is:

$$
\boxed{
q_A
=
\left\langle\frac12\right\rangle
\oplus
\left\langle\frac34\right\rangle
}
$$

on:

$$
A_A
\simeq
\mathbb Z/2
\oplus
\mathbb Z/4.
$$

---

# 4. The Hard Orthogonal Complement

Define:

$$
\boxed{
T=A^\perp\subset L.
}
$$

Then:

$$
\boxed{
\operatorname{rank}T=22,
}
$$

and:

$$
\boxed{
\operatorname{sig}(T)=(20,2).
}
$$

The Tate-twisted/sign-reversed K3-type comparison lattice is:

$$
\boxed{
-T,
}
$$

with:

$$
\boxed{
\operatorname{sig}(-T)=(2,20).
}
$$

---

# 5. The Hard Complement Is Even

For any oriented smooth $8$-manifold, the middle intersection pairing satisfies the Wu congruence:

$$
x^2
\equiv
x\cdot v_4
\pmod2.
$$

For a complex fourfold:

$$
\boxed{
v_4
=
w_4+w_2^2
\equiv
c_2+c_1^2
\pmod2.
}
$$

The fourfold is a smooth hyperplane section:

$$
i:X^4\hookrightarrow X^5.
$$

From:

$$
c(TX^4)
=
\frac{i^*c(TX^5)}{1+h},
$$

both:

$$
c_1(X^4)
$$

and:

$$
c_2(X^4)
$$

are restrictions of expressions in Chern and hyperplane classes on the fivefold.

Hence:

$$
\boxed{
v_4(X^4)
\in
i^*H^4(X^5,\mathbb Z/2).
}
$$

But:

$$
i^*H^4(X^5,\mathbb Z)
=
A.
$$

Therefore:

$$
v_4
\in
A\otimes\mathbb Z/2.
$$

For:

$$
t\in T=A^\perp,
$$

we get:

$$
t^2
\equiv
t\cdot v_4
=
0
\pmod2.
$$

Thus:

$$
\boxed{
T
\text{ is even}.
}
$$

This legitimizes a discriminant quadratic form:

$$
q_T.
$$

---

# 6. What Unimodularity Actually Gives

Because:

$$
A
$$

is primitive in the unimodular lattice:

$$
L,
$$

we have an isomorphism of discriminant groups:

$$
\boxed{
A_A
\simeq
A_T.
}
$$

At the level of discriminant **bilinear forms**:

$$
\boxed{
b_T
\simeq
-b_A.
}
$$

After changing sign:

$$
\boxed{
b_{-T}
\simeq
b_A.
}
$$

This is true in an odd or even unimodular ambient lattice.

What fails in the odd case is the stronger shortcut:

$$
q_{-T}\stackrel{?}{=}q_A.
$$

The quadratic refinement must be determined separately.

---

# 7. Quadratic Refinements of the Fixed Bilinear Form

Let:

$$
a
$$

generate the $\mathbb Z/2$ factor and:

$$
b
$$

generate the $\mathbb Z/4$ factor.

Take the reference refinement:

$$
q_{00}
=
\left\langle\frac12\right\rangle
\oplus
\left\langle\frac34\right\rangle.
$$

All quadratic refinements with the same bilinear form differ by a:

$$
\mathbb Z/2
$$

-valued character.

Thus there are exactly four:

$$
\boxed{
q_{\epsilon,\eta}(x,y)
=
\frac{x^2}{2}
+
\frac{3y^2}{4}
+
\epsilon x
+
\eta y
\pmod{2\mathbb Z},
}
$$

where:

$$
\epsilon,\eta\in\{0,1\}.
$$

---

# 8. Exact Gauss Sums

For a nondegenerate finite quadratic form:

$$
q,
$$

define:

$$
\boxed{
G(q)
=
\frac1{\sqrt{|A|}}
\sum_{z\in A}
e^{\pi iq(z)}.
}
$$

For the four refinements the exact sums are:

$$
\boxed{
G(q_{00})=-1,
}
$$

$$
\boxed{
G(q_{01})=1,
}
$$

$$
\boxed{
G(q_{10})=i,
}
$$

$$
\boxed{
G(q_{11})=-i.
}
$$

Therefore their Brown invariants are:

$$
\boxed{
4,\quad0,\quad2,\quad6
\pmod8.
}
$$

The accompanying verifier computes these sums symbolically.

---

# 9. Milgram Selects the Correct Refinement

For an even lattice $M$:

$$
\boxed{
\operatorname{Br}(q_M)
\equiv
\operatorname{sig}(M)
\pmod8.
}
$$

For:

$$
-T,
$$

the signature difference is:

$$
2-20=-18.
$$

Hence:

$$
\boxed{
\operatorname{Br}(q_{-T})
\equiv6
\pmod8.
}
$$

Among the four possible refinements, exactly one has Brown invariant $6$:

$$
\boxed{
q_{-T}=q_{11}.
}
$$

On the generators:

$$
q_{11}(a)
=
\frac32
\equiv
-\frac12
\pmod{2\mathbb Z},
$$

and:

$$
q_{11}(b)
=
\frac74
\equiv
-\frac14
\pmod{2\mathbb Z}.
$$

Therefore:

$$
\boxed{
q_{-T}
\simeq
\left\langle-\frac12\right\rangle
\oplus
\left\langle-\frac14\right\rangle.
}
$$

---

# 10. Primitive Degree-$4$ $K3^{[2]}$ Lattice

The Beauville--Bogomolov lattice of $K3^{[2]}$ type is:

$$
\boxed{
\Lambda
=
U^3
\oplus
E_8(-1)^2
\oplus
\langle-2\rangle.
}
$$

Choose a primitive polarization:

$$
h
$$

with:

$$
\boxed{
h^2=4,
\qquad
\operatorname{div}(h)=1.
}
$$

Inside one copy of:

$$
U,
$$

take:

$$
h=e+2f.
$$

Then:

$$
h^\perp
$$

contains:

$$
e-2f
$$

of square:

$$
-4.
$$

Therefore:

$$
\boxed{
K_4
=
h^\perp
\simeq
U^2
\oplus
E_8(-1)^2
\oplus
\langle-4\rangle
\oplus
\langle-2\rangle.
}
$$

Its signature is:

$$
\boxed{
(2,20).
}
$$

Its discriminant form is:

$$
\boxed{
q_{K_4}
=
\left\langle-\frac14\right\rangle
\oplus
\left\langle-\frac12\right\rangle.
}
$$

Hence:

$$
\boxed{
q_{K_4}
\simeq
q_{-T}.
}
$$

---

# 11. Nikulin Uniqueness

Both:

$$
-T
$$

and:

$$
K_4
$$

are:

- even;
- indefinite;
- rank $22$;
- signature $(2,20)$;
- discriminant group length $2$;
- equipped with the same discriminant quadratic form.

Nikulin's uniqueness criterion applies because:

$$
\boxed{
22
\ge
2+2.
}
$$

Therefore their genus contains one class.

Hence:

$$
\boxed{
-T
\simeq
K_4
}
$$

as integral quadratic lattices.

Equivalently:

$$
\boxed{
-T_{c5,\mathbb Z}
\simeq
H^2_{\mathrm{prim}}
(Y,\mathbb Z)
}
$$

at the lattice level for the degree-$4$, divisibility-$1$ $K3^{[2]}$ polarization type.

---

# 12. Round-24 Reversal

Round 24 concluded:

```text
rank: PASS
Hodge: PASS
real signature: PASS
determinant: PASS
finite discriminant form: FAIL
```

The corrected result is:

```text
rank: PASS
Hodge: PASS
real signature: PASS
determinant: PASS
parity of ambient lattice: ODD
Wu-evenness of hard block: PASS
Brown invariant refinement: UNIQUE
finite discriminant form: PASS
Nikulin integral lattice gate: PASS
```

Thus the Round-24 final rejection is superseded.

The inherited lattice computation:

$$
L_{\mathrm{inh}}\simeq A_1\oplus A_3
$$

remains valid.

Only the discriminant-form transfer to the orthogonal complement was wrong.

---

# 13. Why the Odd Ambient Lattice Changes the Answer

If the ambient unimodular lattice were even, the gluing condition would require:

$$
\boxed{
q_T=-q_A.
}
$$

But the ambient lattice is odd.

The gluing vectors are allowed to have odd norm.

Therefore the graph of the discriminant-group identification is isotropic only for the discriminant **bilinear** pairing, while:

$$
q_A+q_T
$$

may take odd integral values.

The difference between the two quadratic refinements is exactly the parity character carried by the odd gluing.

This is the missing X-layer datum.

---

# 14. Hodge-Theoretic Period Realization

The lattice isometry alone does not yet produce a geometric correspondence.

However it sharpens the period problem dramatically.

The Tate-twisted $c5$ hard block:

$$
T_{c5}(1)
$$

has a K3-type Hodge line:

$$
H^{2,0}.
$$

Choose an integral lattice isometry:

$$
\boxed{
\phi:
-T_{c5}
\overset{\sim}{\longrightarrow}
K_4.
}
$$

Transport the Hodge line via:

$$
\phi.
$$

This produces a period point in the type-IV period domain for degree-$4$, divisibility-$1$ polarized $K3^{[2]}$ type.

The polarized period map is an open embedding whose image is the complement of finitely many Heegner divisors.

Therefore the remaining geometric realization gate is:

$$
\boxed{
\text{does the transported }c5\text{ period avoid the excluded Heegner divisors?}
}
$$

For a period outside those divisors, a polarized $K3^{[2]}$ fourfold realizes the Hodge structure.

Thus lattice existence is no longer an obstruction.

---

# 15. Very-General Interpretation

If the very general Küchle $c5$ hard block has no additional rational $(1,1)$ classes beyond the inherited sector already removed, the transported period has primitive Picard rank zero.

In that generic situation it avoids additional Noether--Lefschetz conditions, and the expected polarized realization is:

$$
\boxed{
(Y,H),
\qquad
q(H)=4,
\quad
\operatorname{div}(H)=1.
}
$$

This statement is a period-domain consequence once the relevant Heegner exclusions are checked.

It is not yet an algebraic-correspondence theorem.

---

# 16. The Actual Remaining Bridge Gate

The strongest unresolved question is now:

$$
\boxed{
\text{Is the integral Hodge isometry }
T_{c5}(1)
\to
H^2_{\mathrm{prim}}(Y)
\text{ induced by an algebraic correspondence?}
}
$$

This is categorically different from the previous lattice-search problem.

The target is no longer:

> find the deformation type.

The target is:

> construct the correspondence.

This aligns with Kuznetsov's historical expectation that moduli spaces of sheaves on $c5$ should produce hyperkähler varieties.

---

# 17. Kuga--Satake Secondary Consequence

Varesco proves that Hodge similarities of K3-type structures induce isogenies of their Kuga--Satake varieties.

Since the present lattice match has multiplier:

$$
\boxed{
1,
}
$$

the induced relation is an isometry-level instance of that functoriality.

Therefore, once a $K3^{[2]}$ Hodge realization $Y$ is fixed:

$$
\boxed{
KS(T_{c5}(1))
\sim
KS(H^2_{\mathrm{prim}}(Y)).
}
$$

This is a secondary Hodge-theoretic bridge.

It does not by itself prove algebraicity of the desired $c5\times Y$ correspondence.

---

# 18. New Methodological Principle — Parity Before Discriminant Quadratics

Before transporting discriminant quadratic forms across a unimodular orthogonal decomposition, record:

$$
\boxed{
\text{ambient parity}.
}
$$

The correct order is:

$$
\boxed{
\text{primitive embedding}
\to
\text{ambient parity}
\to
\text{bilinear gluing}
\to
\text{complement evenness}
\to
\text{quadratic refinement}.
}
$$

Skipping ambient parity can reverse a final bridge decision.

---

# 19. New Methodological Principle — Brown-Invariant Refinement Gate

When:

- the discriminant group is known;
- the bilinear form is known;
- several even quadratic refinements remain;

use:

$$
\boxed{
\operatorname{Br}(q)
\equiv
\operatorname{sig}(L)
\pmod8
}
$$

to select the legal refinement.

This converts parity ambiguity into a finite exact gate.

---

# 20. Round 25 Conclusion

Round 25 reverses the final conclusion of Round 24.

The Küchle $c5$ middle lattice is:

$$
\boxed{
\text{odd unimodular of signature }(24,2).
}
$$

The inherited lattice is:

$$
\boxed{
A_1\oplus A_3.
}
$$

Its orthogonal hard complement is nevertheless even, because the Wu class lies in the inherited sector.

The odd ambient gluing fixes only the discriminant bilinear form.

Milgram's formula selects the unique legal quadratic refinement:

$$
\boxed{
q_{-T_{c5}}
=
\left\langle-\frac12\right\rangle
\oplus
\left\langle-\frac14\right\rangle.
}
$$

This is exactly the discriminant form of the primitive degree-$4$, divisibility-$1$ $K3^{[2]}$ lattice:

$$
\boxed{
K_4
=
U^2
\oplus
E_8(-1)^2
\oplus
\langle-4\rangle
\oplus
\langle-2\rangle.
}
$$

Nikulin uniqueness therefore gives:

$$
\boxed{
-T_{c5,\mathbb Z}
\simeq
K_4.
}
$$

The direct **integral lattice** bridge is not obstructed.

The remaining problem is genuinely geometric:

$$
\boxed{
\text{period realization}
+
\text{algebraic correspondence}.
}
$$

This is the main closure of Round 25.

---

# References

1. Alexander Kuznetsov, **Küchle fivefolds of type c5**, Mathematische Zeitschrift 284 (2016), 1245--1278.

2. V. V. Nikulin, results on discriminant forms and uniqueness of indefinite even lattices; in the present rank-$22$, length-$2$ case the uniqueness criterion applies immediately.

3. Milgram formula for the Brown invariant of the discriminant quadratic form of an even lattice.

4. Standard Wu formula in real dimension $8$:
   $$
   x^2\equiv x\cdot v_4\pmod2,
   \qquad
   v_4=w_4+w_2^2.
   $$

5. For a complex manifold:
   $$
   w_{2i}\equiv c_i\pmod2.
   $$

6. The $K3^{[2]}$ Beauville--Bogomolov lattice:
   $$
   U^3\oplus E_8(-1)^2\oplus\langle-2\rangle.
   $$

7. Polarized $K3^{[2]}$ period-map results: the period map for fixed degree/divisibility is an open embedding with image the complement of finitely many Heegner divisors.

8. Mauro Varesco, **Hodge similarities, algebraic classes, and Kuga--Satake varieties**, Mathematische Zeitschrift 305 (2023), article 69.