# RCHM × Hodge Conjecture — Research Round 24
## Küchle $c5$: Exceptional-Twist Correction, Exact Saturation, $A_1\oplus A_3$ Inherited Lattice, and Failure of the Direct Integral $K3^{[2]}$ Template

**Status:** Internal research note + exact lattice verifier  
**Version:** Round 24 / v2.4  
**Date:** 2026-09-17  
**Methodological base:** RCHM; X-Layer Mathematical Completeness; Ancestry Reduction; Bridge Search Compiler  
**Previous rounds:** RCHM × Hodge Conjecture — Rounds 1–23  
**Canonical source:** UTF-8 Markdown

---

# 0. Round 24 Objective

Rounds 22--23 identified a sextic-del-Pezzo root sector and attempted to compute its intersection form through the right exceptional divisor of Kuznetsov's double-blow-up model.

A global rank-consistency test revealed that one bundle class had been type-erased incorrectly.

The error was:

$$
v_{\bar E}=h
$$

was treated as if it implied

$$
c_1(N_{\bar E/\widetilde X})=-h.
$$

This is false.

The projective-bundle relative hyperplane class and the blow-up exceptional normal class differ by a base twist.

Round 24 corrects this, recomputes the root Gram matrix, determines the exact index of the explicit rank-$4$ sublattice, identifies the saturated inherited lattice, and runs the full integral $K3^{[2]}$ discriminant-form gate.

---

# 1. Correct Exceptional Normal Class

Kuznetsov's Picard relations are:

$$
\boxed{
\bar h=h-e,
}
$$

$$
\boxed{
\bar e=h-2e,
}
$$

equivalently:

$$
\boxed{
\bar e=2\bar h-h.
}
$$

On:

$$
\bar E=\mathbb P_Z(\mathcal V_Z),
$$

we have:

$$
\boxed{
v_{\bar E}=h,
}
$$

while on $Z$:

$$
\boxed{
\bar h=v_Z+\bar h_1+\bar h_2.
}
$$

Therefore:

$$
\boxed{
c_1(N_{\bar E/\widetilde X})
=
\bar e|_{\bar E}
=
2(v_Z+\bar h_1+\bar h_2)-h.
}
$$

This is the normal class that must enter exceptional self-intersection.

---

# 2. Why the Old Identification Was Invalid

The projectivization:

$$
\bar E\simeq\mathbb P_Z(\mathcal V_Z)
$$

is unchanged if:

$$
\mathcal V_Z
$$

is tensored by a line bundle.

But the relative:

$$
\mathcal O_{\mathbb P(\mathcal V_Z)}(1)
$$

changes by the pullback of that line bundle.

Therefore:

$$
\boxed{
\text{relative hyperplane}
\neq
-\text{exceptional normal}
}
$$

unless the chosen projective-bundle presentation has been normalized accordingly.

This is a bundle-twist type error.

---

# 3. Corrected Exceptional Pairing

Let:

$$
s=\bar h_1+\bar h_2,
$$

and write:

$$
\xi=h|_{\bar E}.
$$

Kuznetsov's Grothendieck relation is:

$$
\boxed{
\xi^2
=
(3v_Z+2s)\xi
-
4v_Zs.
}
$$

Hence:

$$
\bar p_*(\xi)=1,
$$

and:

$$
\boxed{
\bar p_*(\xi^2)
=
3v_Z+2s.
}
$$

For:

$$
\gamma_\eta
=
\bar j_*\bar p^*\eta,
$$

we obtain:

$$
\begin{aligned}
\langle\gamma_\eta,\gamma_\theta\rangle_h
&=
\int_{\bar E}
\eta\theta
\left(
2(v_Z+s)-\xi
\right)\xi\\
&=
\int_Z
\eta\theta
\left(
2(v_Z+s)-(3v_Z+2s)
\right).
\end{aligned}
$$

Therefore:

$$
\boxed{
\langle\gamma_\eta,\gamma_\theta\rangle_h
=
-\int_Z
\eta\theta\,v_Z.
}
$$

This replaces the incorrect Round-22 factor:

$$
-3.
$$

---

# 4. Corrected Root Gram Matrix

For root classes:

$$
r_1=E_1-E_2,
$$

$$
r_2=E_2-E_3
$$

on the sextic del Pezzo surface:

$$
(r_i\cdot r_j)
=
\begin{pmatrix}
-2&1\\
1&-2
\end{pmatrix}.
$$

Since:

$$
\sigma_*(v_Z)=1,
$$

the corrected exceptional pairing is:

$$
\boxed{
G_{\mathrm{root}}
=
-
(r_i\cdot r_j)
=
\begin{pmatrix}
2&-1\\
-1&2
\end{pmatrix}.
}
$$

Thus the descended root sector is:

$$
\boxed{
A_2
}
$$

with determinant:

$$
\boxed{
3.
}
$$

---

# 5. Global Regression Check with $v_Z$

The same formula gives:

$$
\langle\gamma_{v_Z},\gamma_{v_Z}\rangle_h
=
-\int_Zv_Z^3.
$$

Because:

$$
v_Z^2=-\bar h_1\bar h_2
$$

and:

$$
\bar h_1\bar h_2=2,
$$

we have:

$$
\int_Zv_Z^3=-2.
$$

Therefore:

$$
\boxed{
\langle\gamma_{v_Z},\gamma_{v_Z}\rangle_h=2.
}
$$

This is exactly the self-intersection of:

$$
[F]
=
\sigma_2-\sigma_{1,1}.
$$

The global contradiction that exposed the old normal-class error disappears.

---

# 6. Revalidated Descent Map

The descent obstruction is computed on the left exceptional divisor:

$$
E=\mathbb P_F(\mathcal V_F).
$$

There:

$$
\boxed{
v_E=-e
}
$$

does coincide with the negative exceptional normal class in the chosen presentation.

Thus the left blow-up projector remains valid.

The map:

$$
N:
\operatorname{Pic}(Z)
\to
\operatorname{Pic}(F)
$$

induced by the exceptional handoff can be recomputed independently.

On the sextic del Pezzo base:

$$
N(\bar h_1)=2h_1+h_2,
$$

$$
N(\bar h_2)=h_1+2h_2,
$$

and:

$$
\boxed{
N(r_1)=N(r_2)=0.
}
$$

Moreover:

$$
\boxed{
N(v_Z)=0.
}
$$

Thus:

$$
\boxed{
\ker N
=
\mathbb Zv_Z
\oplus
A_2(-1).
}
$$

The root descent conclusion survives the correction.

---

# 7. Surjectivity over the Full Del Pezzo Picard Group

The lattice:

$$
\langle
\bar h_1,\bar h_2,r_1,r_2
\rangle
$$

has index:

$$
3
$$

in:

$$
\operatorname{Pic}(S).
$$

For example:

$$
\boxed{
E_1
=
\frac{
2\bar h_1-\bar h_2+2r_1+r_2
}{3}.
}
$$

Applying $N$ gives:

$$
\boxed{
N(E_1)=h_1.
}
$$

Similarly, the full Picard lattice maps surjectively onto:

$$
\operatorname{Pic}(F).
$$

Hence:

$$
\boxed{
0
\to
A_2(-1)
\to
\operatorname{Pic}(S)
\to
\operatorname{Pic}(F)
\to
0
}
$$

is exact at the lattice level, with the $v_Z$ direction giving one additional kernel direction in $\operatorname{Pic}(Z)$.

---

# 8. Corrected Explicit Rank-$4$ Sublattice

The visible Grassmannian block remains:

$$
G_{\mathrm{vis}}
=
\begin{pmatrix}
20&16\\
16&14
\end{pmatrix},
$$

with determinant:

$$
24.
$$

The corrected descended root block is:

$$
G_{\mathrm{root}}
=
\begin{pmatrix}
2&-1\\
-1&2
\end{pmatrix},
$$

with determinant:

$$
3.
$$

The cross pairings vanish.

Therefore:

$$
\boxed{
G_0
=
\begin{pmatrix}
20&16&0&0\\
16&14&0&0\\
0&0&2&-1\\
0&0&-1&2
\end{pmatrix}.
}
$$

Its determinant is:

$$
\boxed{
\det G_0
=
72.
}
$$

---

# 9. Right-Side Base Generator

Let:

$$
\bar X
$$

be the smooth hyperplane section of:

$$
LG(3,6)
$$

appearing in the right blow-up diagram.

Kuznetsov proves:

$$
\boxed{
CH^2(\bar X)\simeq\mathbb Z
}
$$

and one may take:

$$
\boxed{
\alpha
=
c_2(\bar{\mathcal U}_3)
}
$$

as generator.

For the Lagrangian tautological bundle:

$$
0
\to
\bar{\mathcal U}_3
\to
V\otimes\mathcal O
\to
\bar{\mathcal U}_3^\vee
\to
0,
$$

the degree-two Chern relation gives:

$$
\boxed{
\bar h^2=2\alpha.
}
$$

Since:

$$
\deg LG(3,6)=16,
$$

we have on its hyperplane section:

$$
\boxed{
\int_{\bar X}
\alpha\bar h^3
=
8.
}
$$

---

# 10. Projection Coefficient of $\sigma_{1,1}$

To determine the integral index of the explicit sublattice, compute the coefficient of:

$$
\sigma_{1,1}
$$

along:

$$
CH^2(\bar X)=\mathbb Z\alpha.
$$

Since:

$$
\bar h=h-e,
$$

we have:

$$
\begin{aligned}
\int_{\widetilde X}
\pi^*\sigma_{1,1}\bar h^3
&=
\int_{\widetilde X}
\pi^*\sigma_{1,1}(h-e)^3.
\end{aligned}
$$

For the left codimension-two blow-up:

$$
\pi_*e=0,
$$

$$
\pi_*e^2=-[F].
$$

Also:

$$
c_1(N_{F/X})=0,
$$

so the cubic exceptional correction does not contribute.

Round 20--22 give:

$$
\int_X\sigma_{1,1}h^3=30,
$$

and:

$$
\int_F\sigma_{1,1}h=2.
$$

Therefore:

$$
\boxed{
\int_{\widetilde X}
\pi^*\sigma_{1,1}\bar h^3
=
30-3(2)
=
24.
}
$$

Thus its right-base coefficient is:

$$
\boxed{
24/8=3.
}
$$

---

# 11. The Same Coefficient for $\sigma_2$

Similarly:

$$
\int_X\sigma_2h^3=36,
$$

and:

$$
\int_F\sigma_2h=4.
$$

Therefore:

$$
\boxed{
\int_{\widetilde X}
\pi^*\sigma_2\bar h^3
=
36-3(4)
=
24.
}
$$

Hence:

$$
\boxed{
\sigma_2
\mapsto
3\alpha
}
$$

and:

$$
\boxed{
\sigma_{1,1}
\mapsto
3\alpha
}
$$

in the right-base quotient.

Consequently:

$$
\boxed{
\sigma_2-\sigma_{1,1}
}
$$

lies entirely in the right exceptional kernel, consistent with its identification with the $v_Z$ direction.

---

# 12. Exact Index of the Explicit Rank-$4$ Sublattice

The right blow-up decomposition gives a rank-$6$ group:

$$
CH^2(\bar X)
\oplus
\operatorname{Pic}(Z).
$$

The left exceptional projection is surjective onto:

$$
\operatorname{Pic}(F).
$$

Its kernel is:

$$
CH^2(X).
$$

Inside the kernel:

- the $v_Z$ direction is represented by
  $$
  \sigma_2-\sigma_{1,1};
  $$
- the two $A_2$ root directions descend exactly;
- the quotient by these three kernel directions is a free rank-$1$ group detected by the coefficient of $\alpha$.

Since:

$$
\sigma_{1,1}
$$

has coefficient:

$$
3
$$

in that quotient, the explicit four-class lattice has exact index:

$$
\boxed{
[
CH^2(X):
L_0
]
=
3.
}
$$

---

# 13. Saturated Determinant

Therefore:

$$
\boxed{
|\det L_{\mathrm{inh}}|
=
\frac{
|\det G_0|
}{
3^2
}
=
\frac{72}{9}
=
8.
}
$$

This is now an exact saturated determinant.

The earlier values:

$$
648
$$

and the provisional route to:

$$
72
$$

were artifacts of the incorrect right exceptional normal class.

---

# 14. Unique Integral Index-$3$ Overlattice

The discriminant group of:

$$
L_0
$$

has Smith invariants:

$$
\boxed{
(1,1,6,12).
}
$$

To construct an index-$3$ integral overlattice, look for:

$$
x\in
L_0^\vee/L_0
$$

of order $3$ such that:

$$
(x,x)\in\mathbb Z.
$$

Up to the symmetry exchanging the two simple roots of $A_2$, the unique possible subgroup is generated by:

$$
\boxed{
g
=
\frac{
e_1+e_2+e_3+2e_4
}{3},
}
$$

where:

$$
(e_1,e_2)
$$

are the visible basis vectors and:

$$
(e_3,e_4)
$$

are the root basis vectors.

Its norm is:

$$
\boxed{
(g,g)=8.
}
$$

Thus the saturated lattice is the unique integral index-$3$ overlattice of this type.

---

# 15. Explicit Saturated Gram Matrix

Using the basis:

$$
e_1,
\quad
e_2,
\quad
e_4,
\quad
g,
$$

the saturated Gram matrix is:

$$
\boxed{
G_{\mathrm{inh}}
=
\begin{pmatrix}
20&16&0&12\\
16&14&0&10\\
0&0&2&1\\
12&10&1&8
\end{pmatrix}.
}
$$

Its determinant is:

$$
\boxed{
8.
}
$$

Its Smith normal form is:

$$
\boxed{
\operatorname{diag}(1,1,2,4).
}
$$

Hence:

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

---

# 16. Identification with $A_1\oplus A_3$

There is a unimodular change of basis transforming:

$$
G_{\mathrm{inh}}
$$

to:

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

namely to:

$$
\boxed{
\begin{pmatrix}
2&0&0&0\\
0&2&-1&0\\
0&-1&2&-1\\
0&0&-1&2
\end{pmatrix}.
}
$$

Therefore:

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

This identifies the full primitive inherited rank-$4$ lattice.

---

# 17. Consequence for the Hard Complement

Let:

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

with its middle intersection form.

By integral weak Lefschetz:

$$
L_{\mathrm{inh}}
\hookrightarrow
L
$$

is primitive.

Since the fourfold middle cohomology is torsion-free, Poincaré duality makes:

$$
L
$$

unimodular.

Let:

$$
T_{c5}
=
L_{\mathrm{inh}}^\perp.
$$

Then discriminant-form theory gives:

$$
\boxed{
(A_{T_{c5}},q_{T_{c5}})
\simeq
(A_{L_{\mathrm{inh}}},-q_{L_{\mathrm{inh}}}).
}
$$

After the global sign change used to compare the Tate-twisted degree-$4$ structure with a Beauville--Bogomolov weight-$2$ lattice:

$$
\boxed{
q_{-T_{c5}}
\simeq
q_{L_{\mathrm{inh}}}.
}
$$

---

# 18. Discriminant Form of the Inherited Lattice

For:

$$
A_1,
$$

the nonzero discriminant element has value:

$$
\boxed{
\frac12
\pmod{2\mathbb Z}.
}
$$

For:

$$
A_3,
$$

a generator of the order-$4$ discriminant group has value:

$$
\boxed{
\frac34
\pmod{2\mathbb Z}.
}
$$

Thus:

$$
\boxed{
q_{L_{\mathrm{inh}}}
\simeq
\left\langle\frac12\right\rangle
\oplus
\left\langle\frac34\right\rangle.
}
$$

---

# 19. Determinant Filter for a Polarized $K3^{[2]}$ Bridge

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

$$
\Lambda
=
U^3
\oplus
E_8(-1)^2
\oplus
\langle-2\rangle,
$$

with:

$$
|\det\Lambda|=2.
$$

For a primitive polarization:

$$
h
$$

of square:

$$
h^2=2d
$$

and divisibility:

$$
\gamma,
$$

the orthogonal complement satisfies:

$$
\boxed{
|\det h^\perp|
=
\frac{
2(2d)
}{
\gamma^2
}
=
\frac{4d}{\gamma^2}.
}
$$

Set this equal to:

$$
8.
$$

Then the formal possibilities are:

$$
(d,\gamma)
=
(2,1),
$$

or:

$$
(8,2).
$$

---

# 20. Divisibility-$2$ Congruence Gate

For a primitive polarization on a $K3^{[2]}$-type fourfold:

$$
\gamma\in\{1,2\}.
$$

When:

$$
\gamma=2,
$$

the square must satisfy:

$$
\boxed{
2d\equiv6\pmod8,
}
$$

equivalently:

$$
d\equiv3\pmod4.
$$

The candidate:

$$
d=8
$$

fails this condition.

Therefore determinant alone leaves only:

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

---

# 21. Primitive Square-$4$ Complement

For a square-$4$, divisibility-$1$ polarization, one may choose a primitive vector in a hyperbolic plane.

Its orthogonal complement is isometric to:

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

Its discriminant group is:

$$
\boxed{
\mathbb Z/4
\oplus
\mathbb Z/2.
}
$$

The discriminant form is:

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

---

# 22. Finite Quadratic-Form Mismatch

The sign-reversed hard lattice has discriminant form:

$$
\boxed{
q_{-T_{c5}}
\simeq
\left\langle\frac12\right\rangle
\oplus
\left\langle\frac34\right\rangle.
}
$$

The square-$4$ primitive $K3^{[2]}$ complement has:

$$
\boxed{
q_{K3^{[2]},4}
\simeq
\left\langle-\frac12\right\rangle
\oplus
\left\langle-\frac14\right\rangle.
}
$$

These finite quadratic forms are not isomorphic.

For example, among elements of exact order $4$:

### $c5$ inherited-form side

the values are:

$$
\boxed{
\frac34,\frac34,\frac54,\frac54.
}
$$

### square-$4$ $K3^{[2]}$ primitive side

the values are:

$$
\boxed{
\frac74,\frac74,\frac54,\frac54.
}
$$

The multisets differ.

Therefore:

$$
\boxed{
q_{-T_{c5}}
\not\simeq
q_{K3^{[2]},4}.
}
$$

---

# 23. Integral $K3^{[2]}$ Gate Failure

The strongest direct template:

$$
\boxed{
T_{c5}(1)
\simeq
H^2_{\mathrm{prim}}(Y,\mathbb Z)
}
$$

for a primitively polarized:

$$
K3^{[2]}
$$

fourfold preserving the natural integral quadratic forms fails.

The gate history is:

```text
rank:
  PASS

Hodge numbers:
  PASS

real signature:
  PASS

absolute determinant:
  PASS

polarization degree/divisibility:
  forced to (4,1)

finite discriminant form:
  FAIL
```

This is a much stronger rejection than the earlier OG10 regression.

---

# 24. What Still Survives

The failure above does not rule out:

1. a rational Hodge similarity with nontrivial scaling;
2. a correspondence whose integral image has finite index;
3. a twisted/moduli realization with a modified Mukai lattice;
4. a larger hyperkähler lattice containing the hard block as a sublattice;
5. degeneration-induced or categorical bridges.

Thus the $K3^{[2]}$ geometric search is not discarded.

Only the naive **integral primitive-Hodge-isometry** template is rejected.

---

# 25. Methodological Lesson

The $c5$ experiment now exhibits a full hierarchy of increasingly strong bridge tests:

$$
\boxed{
\text{rank}
\to
\text{Hodge numbers}
\to
\text{real signature}
\to
\text{determinant}
\to
\text{finite discriminant form}.
}
$$

The candidate survives the first four structural appearances and dies only at the final finite quadratic-form gate.

This is exactly why mathematical relations must retain lattice-level type information.

---

# 26. Corrections to Rounds 22--23

The following Round-22/23 statements are superseded:

### Superseded

$$
G_{\mathrm{root}}=3A_2,
$$

and:

$$
\det G_{\mathrm{explicit}}=648.
$$

### Corrected

$$
\boxed{
G_{\mathrm{root}}=A_2,
}
$$

and:

$$
\boxed{
\det G_{\mathrm{explicit}}=72.
}
$$

The root descent conclusion survives after independent rederivation.

The old $3A_2$ factor arose solely from confusing the projective-bundle relative hyperplane on $\bar E$ with the blow-up exceptional normal class.

---

# 27. Round 24 Conclusion

Round 24 completes the integral inherited-lattice computation for Küchle $c5$ at the level needed for bridge discrimination.

The corrected exceptional normal class gives:

$$
\boxed{
G_{\mathrm{root}}=A_2.
}
$$

The explicit visible-plus-root sublattice has:

$$
\boxed{
\det=72.
}
$$

The double-blow-up kernel computation shows that this sublattice has exact index:

$$
\boxed{
3
}
$$

in the full inherited lattice.

Therefore:

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

The unique integral index-$3$ saturation is:

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

Consequently the ancestry-normalized hard complement has a completely determined finite discriminant form.

When compared with polarized $K3^{[2]}$ primitive lattices, determinant matching forces the only possible degree/divisibility candidate:

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

But its finite discriminant form does not match.

Hence:

$$
\boxed{
\text{direct integral primitive }K3^{[2]}\text{ Hodge isometry}
=
\text{FAIL}.
}
$$

The remaining frontier is therefore more subtle:

$$
\boxed{
\text{rational similarity / finite-index correspondence / moduli or categorical bridge}.
}
$$

This is the main closure of Round 24.

---

# References

1. Alexander Kuznetsov, **Küchle fivefolds of type c5**, Mathematische Zeitschrift 284 (2016), 1245--1278.  
   Relevant items: Proposition 4.6; Corollary 4.9; Lemmas 4.14--4.16; Theorem 4.2; Theorem 5.3; Corollary 5.4; Remark 6.6.

2. Degree of the Lagrangian Grassmannian:
   $$
   \deg LG(3,6)=16.
   $$

3. Standard symplectic tautological relation:
   $$
   c(\mathcal U)c(\mathcal U^\vee)=1,
   $$
   which gives:
   $$
   h^2=2c_2(\mathcal U).
   $$

4. Standard discriminant-form relation for primitive orthogonal complements in a unimodular lattice.

5. For primitively polarized $K3^{[2]}$-type fourfolds, divisibility is $1$ or $2$; in the divisibility-$2$ case:
   $$
   2d\equiv6\pmod8.
   $$

6. Round 20 exact localization:
   $$
   G_{\mathrm{vis}}
   =
   \begin{pmatrix}
   20&16\\
   16&14
   \end{pmatrix}.
   $$