# RCHM × Hodge Conjecture — Research Round 23
## Closing the $A_2$ Descent Gate for Küchle $c5$: Blow-Up Intersection Geometry, Exact Root Descent, and a Rank-4 Gram Sublattice

**Status:** Internal research note + exact symbolic verifier  
**Version:** Round 23 / v2.3  
**Date:** 2026-09-17  
**Methodological base:** RCHM; Typed Relation & Constraint Algebra; X-Layer Mathematical Completeness; Ancestry Reduction; Bridge Search Compiler  
**Previous rounds:** RCHM × Hodge Conjecture — Rounds 1–22  
**Canonical source:** UTF-8 Markdown

---

# 0. Round 23 Objective

Round 22 found a canonical root sector

$$
R\simeq A_2(-1)
$$

inside the sextic del Pezzo geometry on the right side of Kuznetsov's double-blow-up model.

Its exceptional lifts on the common resolution have Gram matrix

$$
\boxed{
3A_2
=
\begin{pmatrix}
6&-3\\
-3&6
\end{pmatrix}.
}
$$

The unresolved question was whether these classes descend to the Küchle fivefold without a correction in the left exceptional sector.

Round 23 proves:

$$
\boxed{
\mathfrak D|_{A_2(-1)}=0.
}
$$

Thus the root sector descends exactly.

It also proves that the descended root sector is orthogonal, for the hyperplane-section intersection form, to the visible Grassmannian rank-$2$ sector.

Therefore we obtain an explicit rank-$4$ integral sublattice with Gram matrix

$$
\boxed{
\begin{pmatrix}
20&16&0&0\\
16&14&0&0\\
0&0&6&-3\\
0&0&-3&6
\end{pmatrix}.
}
$$

Its determinant is

$$
\boxed{
648.
}
$$

A final integral gluing/saturation gate remains.

---

# 1. Two Exceptional Divisors

The common resolution

$$
\widetilde X
$$

contains two distinct exceptional divisors:

$$
E
$$

for

$$
\pi:\widetilde X\to X,
$$

and

$$
\bar E
$$

for

$$
\bar\pi:\widetilde X\to\bar X.
$$

They satisfy:

$$
E\simeq\mathbb P_F(\mathcal V_F),
$$

and

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

Let:

$$
D=E\cap\bar E.
$$

---

# 2. Divisor Relations

Kuznetsov proves:

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

and:

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

On:

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

the relative hyperplane class is:

$$
v_E=-e.
$$

Hence:

$$
\bar e|_E
=
h+2v_E.
$$

Therefore:

$$
D\to F
$$

has relative degree $2$.

---

On:

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

the relative hyperplane class is:

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

Since:

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

the intersection divisor has class:

$$
\boxed{
[D]_{\bar E}
=
h-\bar h.
}
$$

---

# 3. Sequence (44) and the Intersection Divisor

On $Z$, Kuznetsov gives:

$$
0
\to
\mathcal O(-2v_Z-\bar h_1-\bar h_2)
\to
\mathcal V_Z
\to
\mathcal O(-\bar h)
\to
\mathcal O_{C_2}(-2)\oplus\mathcal O_{C_1}(-2)
\to0.
$$

Thus the morphism:

$$
\mathcal V_Z
\to
\mathcal O(-\bar h)
$$

has degeneracy locus:

$$
\boxed{
C_1\sqcup C_2.
}
$$

Its projectivized zero locus inside:

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

has divisor class:

$$
h-\bar h.
$$

This is exactly the class of:

$$
E\cap\bar E.
$$

Geometrically the zero-locus condition is the condition that the corresponding lift also lies in the left exceptional divisor.

Therefore:

$$
\boxed{
D
=
E\cap\bar E
}
$$

is the zero locus associated with sequence (44).

---

# 4. Blow-Up Description of $D$

Apply Kuznetsov's blow-up Lemma 2.1 to:

$$
\mathcal V_Z
\to
\mathcal O(-\bar h).
$$

Since the degeneracy locus is the disjoint union:

$$
C_1\sqcup C_2,
$$

we obtain:

$$
\boxed{
q:D
\simeq
\operatorname{Bl}_{C_1\sqcup C_2}Z
\to Z.
}
$$

Let:

$$
R_1,
\qquad
R_2
$$

be the two exceptional divisors.

---

# 5. The Map to the Flag Center

The inclusion:

$$
D\hookrightarrow E
$$

followed by:

$$
p:E\to F
$$

defines:

$$
\boxed{
f:D\to F.
}
$$

Since:

$$
[D]_E=h+2v_E,
$$

the map is generically degree $2$ over $F$.

---

# 6. Resolving the Two Kernel-Line Maps

On $Z$, Corollary 4.9 gives maps:

$$
\mathcal U_{2,A_1}
\to
\mathcal O(v_Z),
$$

and:

$$
\mathcal U_{2,A_2}
\to
\mathcal O(v_Z),
$$

with zero loci:

$$
C_1
$$

and:

$$
C_2
$$

respectively.

After blowing up, their images become:

$$
\mathcal O(v_Z-R_1),
$$

and:

$$
\mathcal O(v_Z-R_2).
$$

Let their kernel line bundles be:

$$
L_1,
\qquad
L_2.
$$

Since:

$$
c_1(\mathcal U_{2,A_i})=-\bar h_i,
$$

we obtain:

$$
c_1(L_i)
=
-\bar h_i-v_Z+R_i.
$$

The maps to:

$$
\mathbb P(A_i)
$$

are defined by these kernel lines.

Hence:

$$
\boxed{
f^*h_i
=
q^*(v_Z+\bar h_i)-R_i.
}
$$

---

# 7. Global Consistency Check

Summing gives:

$$
f^*(h_1+h_2)
=
2v_Z+\bar h_1+\bar h_2-R_1-R_2.
$$

But:

$$
h|_F=h_1+h_2.
$$

The blow-up lemma applied to sequence (44) gives total exceptional class:

$$
R_1+R_2
=
2v_Z+\bar h_1+\bar h_2-h.
$$

Therefore:

$$
\boxed{
f^*(h_1+h_2)=h.
}
$$

This is exactly the restriction of the global hyperplane class.

Thus the two individual pullback formulas are compatible with the global Picard relations.

---

# 8. The Del Pezzo Root Sector

Let:

$$
S=\operatorname{Bl}_{p_1,p_2,p_3}\mathbb P^2.
$$

Use:

$$
H,E_1,E_2,E_3.
$$

Set:

$$
\bar h_1=H,
$$

and:

$$
\bar h_2=2H-E_1-E_2-E_3.
$$

Define:

$$
r_1=E_1-E_2,
$$

and:

$$
r_2=E_2-E_3.
$$

Then:

$$
\boxed{
r_i\cdot\bar h_1
=
r_i\cdot\bar h_2
=
0.
}
$$

The root Gram matrix is:

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

---

# 9. Descent Obstruction

For:

$$
r\in R,
$$

Round 22 defined the right exceptional class:

$$
\gamma_r
=
\bar j_*
\bar p^*
\sigma^*r
\in
CH^2(\widetilde X).
$$

In the left blow-up decomposition write:

$$
\gamma_r
=
\pi^*\delta_r
+
j_*p^*\beta_r,
$$

where:

$$
\delta_r\in CH^2(X),
$$

and:

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

For a codimension-two blow-up:

$$
\boxed{
\beta_r
=
-
p_*j^*\gamma_r.
}
$$

Intersecting the two exceptional divisors gives:

$$
\boxed{
\beta_r
=
-
f_*q^*\sigma^*r.
}
$$

Thus the descent problem is now a pushforward problem on:

$$
f:D\to F.
$$

---

# 10. Pairing Test for the Pushforward

Write:

$$
\beta
=
f_*q^*r
\in
\operatorname{Pic}(F).
$$

Since:

$$
\operatorname{Pic}(F)
=
\mathbb Zh_1
\oplus
\mathbb Zh_2,
$$

write:

$$
\beta=ah_1+bh_2.
$$

On the flag variety:

$$
F\subset\mathbb P^2\times\mathbb P^2
$$

is the incidence divisor of class:

$$
h_1+h_2.
$$

Therefore:

$$
\boxed{
\int_F\beta h_1^2=b,
}
$$

and:

$$
\boxed{
\int_F\beta h_2^2=a.
}
$$

It is enough to prove that both integrals vanish.

---

# 11. Blow-Up Pushforward

Let:

$$
A_i=v_Z+\bar h_i.
$$

Then:

$$
f^*h_i=q^*A_i-R_i.
$$

For the blow-up of a smooth curve in a threefold:

$$
q_*R_i=0,
$$

and:

$$
\boxed{
q_*R_i^2=-[C_i].
}
$$

Therefore:

$$
\begin{aligned}
\int_Dq^*r(f^*h_i)^2
&=
\int_ZrA_i^2
-
\int_{C_i}r.
\end{aligned}
$$

---

# 12. Vanishing of the Base Terms

Kuznetsov gives on:

$$
Z=\mathbb P_S(\mathcal V_S):
$$

$$
\boxed{
v_Z^2=-\bar h_1\bar h_2.
}
$$

Thus:

$$
A_i^2
=
v_Z^2
+
2v_Z\bar h_i
+
\bar h_i^2.
$$

For:

$$
r\in R,
$$

the only term capable of integrating nontrivially over the $\mathbb P^1$-bundle is:

$$
2v_Zr\bar h_i.
$$

But:

$$
r\cdot\bar h_i=0.
$$

Hence:

$$
\boxed{
\int_ZrA_i^2=0.
}
$$

Moreover:

$$
C_i\simeq\bar C_i\in|\bar h_i|,
$$

so:

$$
\boxed{
\int_{C_i}r
=
r\cdot\bar h_i
=
0.
}
$$

Therefore:

$$
\boxed{
\int_Dq^*r(f^*h_i)^2=0.
}
$$

---

# 13. Descent Gate Closure

By projection formula:

$$
\int_F\beta h_i^2
=
\int_Dq^*r(f^*h_i)^2
=
0.
$$

Thus:

$$
a=b=0.
$$

Therefore:

$$
\boxed{
f_*q^*r=0
}
$$

for every:

$$
r\in R.
$$

Hence:

$$
\boxed{
\mathfrak D|_R=0.
}
$$

The descent obstruction vanishes on the entire $A_2(-1)$ root sector.

---

# 14. Exact Descent

We conclude:

$$
\boxed{
\gamma_r
=
\pi^*\delta_r.
}
$$

Thus each root direction gives an actual integral codimension-two class:

$$
\boxed{
\delta_r\in CH^2(X).
}
$$

No left-exceptional correction is required.

This closes the Round-22 descent gate.

---

# 15. Descended Root Gram Matrix

Round 22 proved on the common resolution:

$$
\int_{\widetilde X}
\gamma_r\gamma_s h
=
-3(r\cdot s)_S.
$$

Since:

$$
\gamma_r=\pi^*\delta_r,
$$

we now obtain on the Küchle fivefold:

$$
\boxed{
\int_X
\delta_r\delta_s h
=
-3(r\cdot s)_S.
}
$$

For:

$$
r_1,r_2,
$$

this gives:

$$
\boxed{
G_{\mathrm{root}}
=
\begin{pmatrix}
6&-3\\
-3&6
\end{pmatrix}.
}
$$

---

# 16. Cross Pairing with the Hyperplane Square

We next compare the root sector with the visible Grassmannian sector.

On:

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

write:

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

Lemma 4.16 gives:

$$
h^2
=
(3v_Z+2s)h
-
4v_Zs.
$$

Let:

$$
A=3v_Z+2s,
$$

and:

$$
B=4v_Zs.
$$

Then:

$$
\bar p_*h^3
=
A^2-B.
$$

Therefore:

$$
\begin{aligned}
\int_X\delta_rh^3
&=
\int_{\bar E}r h^3\\
&=
\int_Zr(A^2-B).
\end{aligned}
$$

Every surviving fiber term is proportional to:

$$
r\cdot s.
$$

Since:

$$
r\cdot s=0,
$$

we get:

$$
\boxed{
\int_X\delta_rh^3=0.
}
$$

Thus the root sector is orthogonal to:

$$
h^2.
$$

---

# 17. Cross Pairing with $\sigma_{1,1}$

On the right side, equation (34) gives:

$$
0
\to
\mathcal O(-\bar h_1)\oplus\mathcal O(-\bar h_2)
\to
\bar{\mathcal U}_3
\to
\mathcal O(-v_Z)
\to0.
$$

The original rank-$3$ tautological bundle satisfies:

$$
0
\to
\mathcal U_3
\to
\mathcal U_4
\to
\mathcal O(h-\bar h)
\to0.
$$

A direct Chern-class computation gives:

$$
\boxed{
\sigma_{1,1}
=
\bar h_1\bar h_2
+
v_Zs
+
h(h-\bar h).
}
$$

Thus:

$$
\bar p_*(\sigma_{1,1}h)
$$

is a polynomial in:

$$
v_Z,\bar h_1,\bar h_2
$$

whose only terms capable of pairing with a pulled-back divisor $r$ contain:

$$
v_Zs.
$$

Since:

$$
r\cdot s=0,
$$

we obtain:

$$
\boxed{
\int_X\delta_r\sigma_{1,1}h=0.
}
$$

---

# 18. Cross Pairing with $\sigma_2$

On the Grassmannian:

$$
\boxed{
h^2=\sigma_2+\sigma_{1,1}.
}
$$

Since the root sector is orthogonal to both:

$$
h^2
$$

and:

$$
\sigma_{1,1},
$$

it is also orthogonal to:

$$
\boxed{
\sigma_2.
}
$$

Therefore the visible and root sectors are orthogonal.

---

# 19. Rank-4 Explicit Gram Sublattice

Take the ordered basis:

$$
\boxed{
(\sigma_2,\sigma_{1,1},\delta_{r_1},\delta_{r_2}).
}
$$

The hyperplane-section Gram matrix is:

$$
\boxed{
G_0
=
\begin{pmatrix}
20&16&0&0\\
16&14&0&0\\
0&0&6&-3\\
0&0&-3&6
\end{pmatrix}.
}
$$

Therefore:

$$
\boxed{
\det G_0
=
24\cdot27
=
648.
}
$$

This is now an exact integral rank-$4$ sublattice of the inherited middle sector.

---

# 20. The Integral Gluing Alert

The computation above does **not** yet prove that the four displayed classes form a saturated basis of:

$$
CH^2(X)\simeq\mathbb Z^4.
$$

There is a specific reason for caution.

In:

$$
\operatorname{Pic}(S)
=
\mathbb ZH
\oplus
\mathbb ZE_1
\oplus
\mathbb ZE_2
\oplus
\mathbb ZE_3,
$$

the four classes:

$$
\bar h_1,
\quad
\bar h_2,
\quad
r_1,
\quad
r_2
$$

span a sublattice of index:

$$
\boxed{
3.
}
$$

Indeed the change-of-basis determinant is:

$$
\boxed{
\pm3.
}
$$

Thus the orthogonal decomposition:

$$
\langle\bar h_1,\bar h_2\rangle
\oplus
A_2(-1)
$$

is not the full integral Picard lattice.

There is nontrivial $3$-gluing.

---

# 21. Candidate Saturations

If the same $3$-gluing survives under the double-blow-up handoff, then:

$$
G_0
$$

would be an index-$3$ sublattice of the full inherited lattice.

In that case:

$$
\boxed{
|\det G_{\mathrm{inh}}|
=
\frac{648}{3^2}
=
72.
}
$$

If no such gluing survives, the determinant remains:

$$
648.
$$

At present the correct statement is therefore:

$$
\boxed{
\text{explicit sublattice determinant}=648,
}
$$

while:

$$
\boxed{
\text{full inherited determinant}
\text{ requires the gluing gate}.
}
$$

The value $72$ is a structurally motivated candidate, not yet a theorem.

---

# 22. Why the Gluing Gate Is Now the Only Integral Obstruction

The root sector:

- exists integrally;
- descends exactly;
- has exact Gram matrix;
- is orthogonal to the visible sector.

The visible sector:

- exists integrally;
- has exact Gram matrix.

Therefore the remaining problem is no longer geometric existence or pairing computation.

It is purely:

$$
\boxed{
\text{saturation / overlattice gluing}.
}
$$

This is a finite lattice problem.

---

# 23. New Methodological Principle

Round 23 separates three distinct questions that are often conflated:

### Existence

Does the candidate class exist?

### Descent

Does it survive the birational handoff without an exceptional correction?

### Saturation

Do the resulting integral classes generate the full integral lattice?

The logical chain is:

$$
\boxed{
\text{existence}
\to
\text{descent}
\to
\text{saturation}.
}
$$

Passing the first two does not imply the third.

---

# 24. Round 23 Conclusion

Round 23 closes the $A_2$ descent problem.

The exceptional intersection is:

$$
\boxed{
D
\simeq
\operatorname{Bl}_{C_1\sqcup C_2}Z.
}
$$

The map:

$$
f:D\to F
$$

has pullback formulas:

$$
\boxed{
f^*h_i
=
q^*(v_Z+\bar h_i)-R_i.
}
$$

For every root:

$$
r\in A_2(-1),
$$

projection formula and root orthogonality imply:

$$
\boxed{
f_*q^*r=0.
}
$$

Therefore:

$$
\boxed{
\mathfrak D|_{A_2(-1)}=0.
}
$$

The root sector descends exactly to $CH^2(X)$ and has Gram matrix:

$$
\boxed{
3A_2
=
\begin{pmatrix}
6&-3\\
-3&6
\end{pmatrix}.
}
$$

It is orthogonal to the visible Grassmannian sector, producing the exact rank-$4$ sublattice:

$$
\boxed{
G_0
=
\begin{pmatrix}
20&16&0&0\\
16&14&0&0\\
0&0&6&-3\\
0&0&-3&6
\end{pmatrix},
\qquad
\det G_0=648.
}
$$

The only remaining integral question is the $3$-gluing/saturation gate.

This is the main closure of Round 23.

---

# References

1. Alexander Kuznetsov, **Küchle fivefolds of type c5**, Mathematische Zeitschrift 284 (2016), 1245--1278.  
   Relevant items: Lemma 2.1; Proposition 4.6; Corollary 4.9; Lemma 4.14; Lemma 4.15; Lemma 4.16; Theorem 4.2.

2. Standard blow-up formulas:
   $$
   q_*R=0,
   \qquad
   q_*R^2=-[C]
   $$
   for the exceptional divisor of the blow-up of a smooth curve in a smooth threefold.

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

4. Round 22 exact del-Pezzo root calculation:
   $$
   A_2(-1)
   \mapsto
   3A_2
   $$
   on the common resolution.