# RCHM × Hodge Conjecture — Research Round 22
## Küchle $c5$: Porteous Collapse of the Flag Center and the Sextic-del-Pezzo $A_2$ Root-Sector Candidate

**Status:** Internal research note + exact symbolic verifier  
**Version:** Round 22 / v2.2  
**Date:** 2026-09-17  
**Methodological base:** RCHM; X-Layer Mathematical Completeness; Bridge Search Compiler; Ancestry Reduction  
**Previous rounds:** RCHM × Hodge Conjecture — Rounds 1–21  
**Canonical source:** UTF-8 Markdown

---

# 0. Round 22 Objective

Round 21 established that the Küchle $c5$ fivefold contributes a primitive rank-$4$ Tate lattice to the middle cohomology of its hyperplane-section fourfold.

Round 20 had already computed a visible rank-$2$ Grassmannian sublattice generated by:

$$
\sigma_2=c_2(\mathcal Q),
\qquad
\sigma_{1,1}=c_2(\mathcal U^\vee),
$$

with hyperplane-section Gram matrix:

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

Two inherited integral directions are therefore still missing from the explicit basis.

The natural first candidate is the blow-up center:

$$
F\simeq\operatorname{Fl}(1,2;3).
$$

Round 22 shows that this candidate collapses:

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

Thus $F$ contributes no third direction.

The search then moves to the opposite side of Kuznetsov's double-blow-up diagram, where a canonical rank-$2$ root sector:

$$
A_2(-1)
$$

appears inside the Picard lattice of the sextic del Pezzo surface.

On the common resolution its exceptional lifts have exact Lefschetz pairing:

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

Whether this root sector descends without correction to the two hidden generators of $CH^2(X^5)$ is left as an explicit descent gate.

---

# 1. Kuznetsov's Double Blow-Up

For a general Küchle $c5$ fivefold:

$$
X=X^5_{\lambda,\mu},
$$

Kuznetsov constructs a smooth common resolution:

$$
\widetilde X
$$

and two blow-ups:

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

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

where:

- $\pi$ has center
  $$
  F\simeq\operatorname{Fl}(1,2;3);
  $$
- $\bar X$ is a smooth hyperplane section of the Lagrangian Grassmannian;
- $\bar\pi$ has center
  $$
  Z;
  $$
- $Z$ is a $\mathbb P^1$-bundle over a sextic del Pezzo surface
  $$
  \sigma:Z\to S.
  $$

The integral Chow-motive calculation gives:

$$
\operatorname{rank}CH^2(X)=4.
$$

The purpose of this round is to locate those four directions geometrically.

---

# 2. The Visible Grassmannian Plane

Let:

$$
\mathcal U=\mathcal U_3,
$$

and:

$$
\mathcal Q=W/\mathcal U.
$$

Define:

$$
\sigma_2=c_2(\mathcal Q)|_X,
$$

$$
\sigma_{1,1}=c_2(\mathcal U^\vee)|_X.
$$

Round 20 computed:

$$
\int_X\sigma_2^2h=20,
$$

$$
\int_X\sigma_2\sigma_{1,1}h=16,
$$

$$
\int_X\sigma_{1,1}^2h=14.
$$

Thus:

$$
\boxed{
B_{\mathrm{vis}}
=
\langle\sigma_2,\sigma_{1,1}\rangle
}
$$

is rank $2$.

---

# 3. The Flag Center as a Degeneracy Locus

On the odd symplectic Grassmannian, the skew form induces:

$$
\boxed{
\mu:
\mathcal Q
\longrightarrow
\mathcal U^\vee,
}
$$

a map of ranks:

$$
4\to3.
$$

Its first degeneracy locus has codimension $2$, and the higher degeneracy locus is empty.

Intersecting with the Küchle fivefold gives the center:

$$
F.
$$

Therefore its class can be calculated by Thom--Porteous.

---

# 4. Thom--Porteous Calculation

For a rank-$4$ to rank-$3$ map, the rank-$\le2$ locus has partition:

$$
(1,1).
$$

Hence:

$$
[D_2]
=
\det
\begin{pmatrix}
c_1&c_2\\
1&c_1
\end{pmatrix}
=
c_1^2-c_2
$$

for:

$$
c_i=c_i(\mathcal U^\vee-\mathcal Q).
$$

Let:

$$
h=c_1(\mathcal U^\vee)=c_1(\mathcal Q).
$$

Then:

$$
c_1(\mathcal U^\vee-\mathcal Q)=0.
$$

Moreover:

$$
\begin{aligned}
c_2(\mathcal U^\vee-\mathcal Q)
&=
c_2(\mathcal U^\vee)
-c_1(\mathcal U^\vee)c_1(\mathcal Q)
+c_1(\mathcal Q)^2
-c_2(\mathcal Q)\\
&=
\sigma_{1,1}-\sigma_2.
\end{aligned}
$$

Therefore:

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

This identity holds in the codimension-$2$ Chow/cohomology class relevant to the fivefold.

---

# 5. First Consequence

The most geometrically obvious blow-up center is not one of the two missing directions.

Indeed:

$$
\boxed{
[F]\in
\operatorname{span}
\{\sigma_2,\sigma_{1,1}\}.
}
$$

Thus:

$$
\boxed{
\operatorname{rank}
\langle
\sigma_2,\sigma_{1,1},[F]
\rangle
=
2.
}
$$

This is a concrete example of a visually distinct geometric carrier producing no new cohomological direction.

---

# 6. Independent Pairing Check on $F$

Write:

$$
F=\operatorname{Fl}(1,2;3)
\subset
\mathbb P^2\times\mathbb P^2
$$

as the incidence divisor of class:

$$
h_1+h_2.
$$

Kuznetsov gives:

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

On $F$:

$$
\mathcal U^\vee|_F
\simeq
\mathcal O
\oplus
\mathcal O(h_1)
\oplus
\mathcal O(h_2).
$$

Therefore:

$$
\boxed{
\sigma_{1,1}|_F=h_1h_2.
}
$$

Using:

$$
c(\mathcal U)c(\mathcal Q)=1,
$$

we obtain:

$$
\boxed{
\sigma_2|_F
=
h_1^2+h_1h_2+h_2^2.
}
$$

Exact integration gives:

$$
\boxed{
\int_F\sigma_2h=4,
}
$$

and:

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

---

# 7. Self-Intersection of the Flag Center

The normal bundle of $F$ in $X$ is the rank-$2$ bundle:

$$
\mathcal V_F
$$

appearing in the blow-up construction.

Kuznetsov's Grothendieck relation:

$$
v_E^2=-h_1h_2
$$

for:

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

gives:

$$
c_2(\mathcal V_F)=h_1h_2
$$

in the convention used here.

Hence:

$$
\begin{aligned}
\int_X[F]^2h
&=
\int_Fc_2(N_{F/X})h\\
&=
\int_Fh_1h_2(h_1+h_2)\\
&=
\boxed{2}.
\end{aligned}
$$

---

# 8. Consistency with the Round-20 Matrix

If:

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

then the Round-20 matrix predicts:

$$
\langle\sigma_2,[F]\rangle
=
20-16
=
4,
$$

$$
\langle\sigma_{1,1},[F]\rangle
=
16-14
=
2,
$$

and:

$$
\langle[F],[F]\rangle
=
20-2(16)+14
=
2.
$$

These exactly match the independent flag-variety calculations.

Therefore the Porteous identity passes a nontrivial pairing-level cross-check.

---

# 9. Where Can the Missing Two Directions Live?

The motive calculation gives:

$$
\operatorname{rank}CH^2(X)=4.
$$

But:

$$
\operatorname{rank}B_{\mathrm{vis}}=2,
$$

and:

$$
[F]\in B_{\mathrm{vis}}.
$$

Thus the two additional directions must arise elsewhere in the double-blow-up geometry.

The natural candidate is the:

$$
Z/S
$$

side.

---

# 10. Sextic del Pezzo Geometry

Kuznetsov proves that:

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

is the intersection of two divisors of bidegree:

$$
(1,1).
$$

The two projections exhibit:

$$
S
$$

as the blow-up of:

$$
\mathbb P^2
$$

at three points.

Choose a basis:

$$
H,E_1,E_2,E_3
$$

of:

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

with:

$$
H^2=1,
$$

$$
E_i^2=-1,
$$

$$
H\cdot E_i=0,
$$

and:

$$
E_i\cdot E_j=0
\quad(i\ne j).
$$

One may choose the two natural hyperplane classes as:

$$
\boxed{
\bar h_1=H,
}
$$

and:

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

Then:

$$
\bar h_1^2=\bar h_2^2=1,
$$

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

and:

$$
\bar h_1+\bar h_2=-K_S.
$$

---

# 11. The Hidden Root Sector

Consider:

$$
R
=
\langle
\bar h_1,\bar h_2
\rangle^\perp
\subset
\operatorname{Pic}(S).
$$

It has rank:

$$
2.
$$

A convenient basis is:

$$
\boxed{
r_1=E_1-E_2,
}
$$

$$
\boxed{
r_2=E_2-E_3.
}
$$

Their Gram matrix is:

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

Therefore:

$$
\boxed{
R\simeq A_2(-1).
}
$$

Moreover:

$$
r_i\cdot\bar h_1
=
r_i\cdot\bar h_2
=
0.
$$

This is the first canonical rank-$2$ object in the double-blow-up geometry with exactly the missing rank.

---

# 12. Exceptional Lift of the Root Sector

Let:

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

be the exceptional divisor of:

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

Let:

$$
\bar p:\bar E\to Z
$$

be the projection, and:

$$
\sigma:Z\to S.
$$

For:

$$
r\in R,
$$

define the exceptional codimension-$2$ class:

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

These classes are exact algebraic lifts of the del Pezzo root directions to the common resolution.

---

# 13. Exceptional Lefschetz Pairing

Kuznetsov gives:

$$
h|_{\bar E}=v_{\bar E},
$$

and:

$$
v_{\bar E}^2
=
(3v_Z+2\bar h_1+2\bar h_2)v_{\bar E}
-
4v_Z(\bar h_1+\bar h_2).
$$

The exceptional normal bundle has:

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

Therefore:

$$
\begin{aligned}
\int_{\widetilde X}
\gamma_r\gamma_s h
&=
-\int_{\bar E}
\bar p^*\sigma^*(rs)
v_{\bar E}^2\\
&=
-\int_Z
\sigma^*(rs)
(3v_Z+2\bar h_1+2\bar h_2).
\end{aligned}
$$

Since:

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

is a $\mathbb P^1$-bundle:

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

while:

$$
\sigma_*(1)=0
$$

in the required codimension.

Hence:

$$
\boxed{
\int_{\widetilde X}
\gamma_r\gamma_s h
=
-3
\int_Srs.
}
$$

---

# 14. Exact Root-Sector Gram Matrix on the Resolution

Applying the previous formula to:

$$
r_1,r_2,
$$

gives:

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

Thus:

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

with:

$$
\boxed{
\det G_{\mathrm{root,res}}=27.
}
$$

This calculation is exact on the common resolution.

---

# 15. What Has Been Proved

The following statements are now exact:

1. the flag center satisfies
   $$
   [F]=\sigma_2-\sigma_{1,1};
   $$
2. the flag center gives no new inherited cohomological direction;
3. the sextic del Pezzo surface contains a canonical orthogonal root sector
   $$
   A_2(-1);
   $$
4. its exceptional lifts to the common resolution have Lefschetz Gram matrix
   $$
   3A_2.
   $$

These results identify a geometrically natural rank-$2$ candidate for the missing inherited sector.

---

# 16. What Has Not Yet Been Proved

It has **not** yet been proved that:

$$
\boxed{
\gamma_{r_1},
\gamma_{r_2}
}
$$

are the pullbacks of the two missing classes in:

$$
CH^2(X).
$$

The common resolution also contains the exceptional sector for:

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

A class on the right-side exceptional divisor may acquire a correction in:

$$
\bar j_*\bar p^*\operatorname{Pic}(Z)
\longrightarrow
j_*p^*\operatorname{Pic}(F)
$$

when rewritten in the left blow-up decomposition.

Therefore:

$$
\boxed{
G_{\mathrm{root,res}}
}
$$

must not yet be identified with the missing $2\times2$ block of the fivefold Gram matrix.

---

# 17. Blow-Up Decomposition as a Handoff Problem

The two codimension-$2$ Chow decompositions are:

$$
\boxed{
CH^2(\widetilde X)
\simeq
\pi^*CH^2(X)
\oplus
j_*p^*\operatorname{Pic}(F),
}
$$

and:

$$
\boxed{
CH^2(\widetilde X)
\simeq
\bar\pi^*CH^2(\bar X)
\oplus
\bar j_*\bar p^*\operatorname{Pic}(Z).
}
$$

The ranks are:

$$
4+2=6,
$$

and:

$$
1+5=6.
$$

Thus the explicit generator problem is exactly a change-of-decomposition problem inside a rank-$6$ lattice.

This is a concrete RCHM handoff.

---

# 18. Descent Obstruction Map

For:

$$
\eta\in\operatorname{Pic}(Z),
$$

write:

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

Its left-side decomposition has the form:

$$
\boxed{
\gamma_\eta
=
\pi^*\delta_\eta
+
j_*p^*\beta_\eta,
}
$$

where:

$$
\delta_\eta\in CH^2(X),
$$

and:

$$
\beta_\eta\in\operatorname{Pic}(F).
$$

Define:

$$
\boxed{
\mathfrak D(\eta)
=
\beta_\eta.
}
$$

Call:

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

the **root-sector descent obstruction map**.

The exceptional root class descends without left-side correction exactly when:

$$
\boxed{
\mathfrak D(\eta)=0.
}
$$

---

# 19. Next Exact Gate

The decisive next calculation is:

$$
\boxed{
\mathfrak D(r_1),
\qquad
\mathfrak D(r_2).
}
$$

If:

$$
\mathfrak D(r_1)
=
\mathfrak D(r_2)
=
0,
$$

then the root sector lies directly in:

$$
\pi^*CH^2(X),
$$

and the missing inherited rank-$2$ lattice is explicitly recovered.

If not, the correction terms must be computed and the corrected Gram matrix descended to $X$.

Either outcome is finite and exact.

---

# 20. Why the Root Sector Is Structurally Plausible

The motive bookkeeping gives:

$$
\operatorname{rank}CH^2(\widetilde X)=6.
$$

On the left:

$$
4
+
\operatorname{rank}\operatorname{Pic}(F)
=
4+2.
$$

On the right:

$$
1
+
\operatorname{rank}\operatorname{Pic}(Z)
=
1+5.
$$

The sextic del Pezzo base contributes four Picard directions.

Two are the natural classes:

$$
\bar h_1,\bar h_2.
$$

The remaining two form precisely:

$$
A_2(-1).
$$

Thus the root sector has exactly the correct rank to account for the two inherited classes invisible from the naive Grassmannian plane.

This is strong structural evidence, not yet a descent theorem.

---

# 21. A New False-Generator Regression Test

The flag center gives a useful benchmark:

```text
candidate_generator:
  class: [F]

geometric_distinctness:
  PASS

Porteous_test:
  [F] = sigma2 - sigma11

new_lattice_direction:
  FAIL
```

Thus:

$$
\boxed{
\text{new geometric subvariety}
\not\Rightarrow
\text{new cohomological generator}.
}
$$

This is another relation/type distinction worth preserving in AMRAL.

---

# 22. Updated Integral-Lattice Program

The rank-$4$ inherited lattice should now be approached as:

$$
\boxed{
B_{\mathrm{vis}}
+
\text{descended }A_2\text{ root sector}.
}
$$

But the plus sign is provisional.

The remaining tasks are:

1. compute the descent map
   $$
   \mathfrak D;
   $$
2. determine correction terms;
3. obtain four explicit integral classes in $CH^2(X)$;
4. compute their full $4\times4$ Gram matrix;
5. verify saturation;
6. compute the discriminant group and quadratic form;
7. compare the orthogonal complement with primitive polarized
   $K3^{[2]}$ lattices.

---

# 23. Round 22 Conclusion

Round 22 makes two concrete advances toward the full inherited integral lattice of Küchle $c5$.

First, the obvious flag-center candidate collapses:

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

This is independently verified by three hyperplane-section pairings:

$$
\boxed{
(4,2,2).
}
$$

Second, the opposite blow-up center reveals a canonical hidden rank-$2$ sector.

The sextic del Pezzo base has:

$$
\operatorname{Pic}(S)
\simeq
\langle H,E_1,E_2,E_3\rangle,
$$

and the orthogonal complement of the two natural projective classes is:

$$
\boxed{
A_2(-1).
}
$$

Its exceptional lifts have exact common-resolution pairing:

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

The outstanding issue is no longer “where are the missing two dimensions?”

It is the sharper finite question:

$$
\boxed{
\text{how does the }A_2\text{ root sector transform across the double blow-up?}
}
$$

Equivalently:

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

This is the main closure of Round 22.

---

# References

1. Alexander Kuznetsov, **Küchle fivefolds of type c5**, Mathematische Zeitschrift 284 (2016), 1245--1278.  
   DOI: https://doi.org/10.1007/s00209-016-1707-9  
   Relevant material: Theorem 4.2; Proposition 4.6; Remark 4.7; Lemmas 4.14--4.16; Theorem 5.3; Corollary 5.4.

2. Standard Thom--Porteous formula for degeneracy loci of vector-bundle morphisms.

3. Standard blow-up formula:
   $$
   CH^2(\operatorname{Bl}_Z Y)
   \simeq
   CH^2(Y)
   \oplus
   CH^1(Z)
   $$
   for a smooth codimension-$2$ center.

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