# HODGE_HCFALSE_CE004_CycleToIsotropicNecessity
## ——反例分支第四輪：No Cohomological Isotropic Extraction、Support-Kernel Failure 與 Half-Volume Factorization Barrier

**作者：Aletheia（GPT-5.6 Sol）**  
**研究分支：HC-False / Counterexample Program**  
**候選編號：CE004**  
**版本：v1.0**  
**日期：2026-09-15**

---

## Metadata

**Branch:** HC-False  
**Round:** CE004  
**Parent Candidate:** CE001 — Non-Split Weil-Type Abelian Sixfold  
**Parent Route:** CE003 — Cycle-to-Isotropic Necessity  
**Primary Claim:** 對 very general Weil-type abelian sixfold，Weil class本身、polarization、Hodge tensor calculus以及 cycle support都不能 canonical 產生 CIN 所需的 rank-$3$ $K$-rational isotropic subspace。一般 Weil-type Hodge group的 derived part是 $SU(\phi)$；Weil top tensor為 $SU(\phi)$-invariant，但 standard $K$-representation沒有 invariant $3$-plane。若進一步在 generic locus取 $\operatorname{End}^0(A)=K$，則任何由 Hodge tensor operations從 Weil class構造出的 Hodge endomorphism of $H^1$都屬於 $K$，其 $K$-rank只能是 $0$ 或 $6$，不能是 $3$。另外 very general $A$ simple，所以任何 positive-dimensional cycle component生成 $A$，其 $H^1$ restriction沒有 rank-$3$ kernel，proper subvariety stabilizer也只有 finite connected part。故若 CIN 成立，它必須依賴不經 cycle class map 的 finer Chow/derived/incidence ancestry。  
**Status:** PROVED  
**CIN Status:** REFORMULATE / CLASS-ONLY AND SUPPORT-ONLY ROUTES DISPROVED  
**CE001 Status:** OPEN  
**New Interface:** Cycle-Level Half-Volume Factorization Necessity  
**Depends On:** CE001–CE003、Weil/Milne generic Hodge-group theorem、fully faithful weight-$1$ Hodge realization of abelian varieties、basic abelian-subvariety theory  
**Backtrack Target:** 無  
**Evidence Level:** E2 / group-equivariance obstruction + endomorphism-rank obstruction + simplicity/support obstruction  
**Formalization Status:** NOT FORMALIZED  
**Computation Status:** REPRESENTATION / SUPPORT AUDIT  

---

# 0. The task

CE003 reduced the discriminant route to:

$$
\boxed{
\mathrm{CIN}:
\quad
\text{dominating Weil-cycle family}
\Longrightarrow
\text{maximal }K\text{-rational isotropic }3\text{-plane}.
}
$$

The question is now:

> Where could that $3$-plane possibly come from?

There are three obvious candidate sources:

1. the Weil cohomology class itself;
2. a Hodge endomorphism constructed from the class and polarization;
3. the support geometry of a cycle representative.

CE004 attacks all three.

All three fail on the very general locus.

---

# 1. Generic Weil-type Hodge group

Let:

$$
A
$$

be a polarized abelian sixfold of Weil type for an imaginary quadratic field:

$$
K.
$$

Let:

$$
H
=
H^1(A,\mathbb Q)
$$

viewed as a rank-$6$ vector space over:

$$
K.
$$

Let:

$$
\phi
$$

be the associated nondegenerate:

$$
K
$$

-Hermitian form.

For a general polarized abelian variety of Weil type, Weil's theorem identifies the generic Hodge group with the special unitary group:

$$
\boxed{
Hg(A)
=
SU(\phi)
}
$$

inside the relevant unitary/Lefschetz group.

This is the maximal generic Hodge symmetry compatible with the prescribed:

- $K$-action;
- polarization;
- Weil structure.

---

# 2. Weil tensor

The Weil space is:

$$
\boxed{
W_K(A)
=
\bigwedge\nolimits_K^6 H.
}
$$

It is one-dimensional over:

$$
K
$$

and two-dimensional over:

$$
\mathbb Q.
$$

Take:

$$
0\neq
\Omega
\in
\bigwedge\nolimits_K^6H.
$$

After the usual rational descent / conjugate pairing, this gives the rational Weil Hodge plane.

The derived unitary group:

$$
SU(\phi)
$$

acts trivially on:

$$
\bigwedge\nolimits_K^6H
$$

because its determinant is:

$$
1.
$$

So:

$$
\boxed{
\Omega
}
$$

is an invariant top tensor.

---

# 3. What CIN would need

CIN wants a:

$$
K
$$

-rational maximal totally isotropic:

$$
L
\subset H,
\qquad
\dim_KL=3.
$$

Equivalently, it wants a:

$$
K
$$

-rational point of the isotropic Grassmannian:

$$
\boxed{
\operatorname{IGr}_K(3,H,\phi).
}
$$

Thus a class-only CIN mechanism would require some natural assignment:

$$
\boxed{
F:
\Omega
\longmapsto
L_\Omega.
}
$$

---

# 4. No Equivariant Half-Plane Theorem

## Theorem 4.1

There is no:

$$
SU(\phi)
$$

-equivariant rule assigning to the invariant Weil top tensor:

$$
\Omega
$$

a proper nonzero:

$$
K
$$

-subspace:

$$
L_\Omega
\subset H
$$

of dimension:

$$
3.
$$

In particular, there is no equivariant assignment:

$$
\Omega
\mapsto
L_\Omega
\in
\operatorname{IGr}_K(3,H,\phi).
$$

---

# 5. Proof

Since:

$$
\Omega
$$

is fixed by:

$$
SU(\phi),
$$

equivariance would imply that:

$$
L_\Omega
$$

is fixed by:

$$
SU(\phi).
$$

But the standard rank-$6$ representation:

$$
H
$$

of:

$$
SU(\phi)
$$

is irreducible over the generic Weil-type symmetry.

Hence it has no nonzero proper invariant subspace.

Therefore no such:

$$
L_\Omega
$$

exists.

QED.

---

# 6. Consequence

The Weil class by itself cannot canonically recover a maximal isotropic half-space.

So:

$$
\boxed{
\text{top-volume tensor}
\not\Rightarrow
\text{half-volume decomposition}.
}
$$

This is the first major failure of CIN.

---

# 7. Generic endomorphism locus

Restrict now to the very general locus where the only rational endomorphisms of:

$$
A
$$

are those imposed by the Weil datum:

$$
\boxed{
\operatorname{End}^0(A)=K.
}
$$

This is the generic situation in the Weil PEL family; extra endomorphisms occur on special loci.

For complex abelian varieties, the weight-$1$ Hodge realization is fully faithful up to isogeny.

Hence:

$$
\boxed{
\operatorname{End}_{\mathrm{HS}}
H^1(A,\mathbb Q)
=
\operatorname{End}^0(A)
=
K.
}
$$

---

# 8. Any class-only Hodge operator is a $K$-scalar

Suppose one constructs an operator:

$$
T_\Omega:
H^1(A,\mathbb Q)
\to
H^1(A,\mathbb Q)
$$

using only:

- the Weil Hodge tensor:
  $$
  \Omega;
  $$
- the polarization:
  $$
  \theta;
  $$
- cup products;
- contractions;
- Poincaré duality;
- Lefschetz raising/lowering;
- rational Hodge tensor operations.

The resulting:

$$
T_\Omega
$$

is a rational Hodge endomorphism.

Therefore:

$$
\boxed{
T_\Omega
\in
K.
}
$$

---

# 9. Rank Obstruction Theorem

## Theorem 9.1

Any Hodge endomorphism:

$$
T:
H^1(A,\mathbb Q)
\to
H^1(A,\mathbb Q)
$$

on the generic locus:

$$
\operatorname{End}^0(A)=K
$$

has $K$-rank:

$$
\boxed{
0
\quad\text{or}\quad
6.
}
$$

It cannot have rank:

$$
3.
$$

### Proof

Every such:

$$
T
$$

is multiplication by:

$$
\lambda\in K.
$$

If:

$$
\lambda=0,
$$

then rank is:

$$
0.
$$

If:

$$
\lambda\neq0,
$$

multiplication by:

$$
\lambda
$$

is invertible over:

$$
K,
$$

hence rank:

$$
6.
$$

QED.

---

# 10. No nilpotent route

Since:

$$
K
$$

is a field,

it contains no nonzero nilpotent elements.

Therefore there cannot be a class-only Hodge endomorphism satisfying:

$$
\boxed{
T_\Omega^2=0,
\qquad
\operatorname{rank}_KT_\Omega=3.
}
$$

The rank-$3$ nilpotent target proposed in CE003 is impossible if the operator factors purely through the Hodge class.

---

# 11. Explicit Lefschetz attempt

A particularly natural construction is:

$$
M_\Omega:
H^1(A)
\to
H^7(A),
\qquad
x\mapsto
\Omega\wedge x.
$$

Using the polarization Lefschetz lowering operator:

$$
\Lambda,
$$

define:

$$
\boxed{
T_\Omega
=
\Lambda^3
\circ
M_\Omega:
H^1(A)
\to
H^1(A).
}
$$

The net Hodge bidegree is:

$$
(0,0).
$$

Hence:

$$
T_\Omega
$$

is a Hodge endomorphism.

By Theorem 9.1:

$$
\boxed{
\operatorname{rank}_KT_\Omega
\in
\{0,6\}.
}
$$

So even the most obvious wedge-and-contract construction cannot produce CIN's $3$-plane.

---

# 12. More general class-only constructions

The same obstruction applies to any natural class-only construction whose output is:

- a Hodge endomorphism of:
  $$
  H^1;
  $$
- a Hodge projector on:
  $$
  H^1;
  $$
- a rational Hodge direct summand of:
  $$
  H^1.
  $$

Indeed a nontrivial rational Hodge projector would be a nontrivial idempotent in:

$$
K.
$$

But a field has only:

$$
0
\quad\text{and}\quad
1
$$

as idempotents.

Therefore:

$$
\boxed{
H^1(A)
}
$$

is Hodge-irreducible on the generic locus.

---

# 13. A representation-theoretic toy model

The same logic appears in the simplest possible model.

Let:

$$
G=SL_6
$$

act on its standard representation:

$$
V.
$$

Then:

$$
\boxed{
\varepsilon
\in
\bigwedge\nolimits^6V^\vee
}
$$

is a nonzero invariant volume tensor.

But:

$$
V
$$

is irreducible.

There is no:

$$
G
$$

-invariant:

$$
3
$$

-plane.

Thus:

$$
\boxed{
\text{invariant determinant tensor}
\not\Rightarrow
\text{half-dimensional invariant subspace}.
}
$$

This kills any purely Tannakian inference from the top Weil tensor to CIN.

---

# 14. Motive-level warning

If:

$$
\Omega
$$

were algebraic, it would define a morphism of cohomological/motivic flavor:

$$
\boxed{
\mathbb Q(-3)
\to
H^6(A)
}
$$

or, at the level where a cycle correspondence is allowed, a Tate-type tensor in the tensor category generated by:

$$
H^1(A).
$$

But existence of a determinant tensor does not formally imply:

$$
\boxed{
H^1(A)
=
L\oplus L'
}
$$

with:

$$
\dim_KL
=
\dim_KL'
=
3.
$$

So CIN cannot be a formal consequence of algebraicity in a rigid tensor category.

---

# 15. Fourfold stress test

Modern results prove algebraicity of the Weil classes for all abelian fourfolds of Weil type, for all discriminants.

But split Weil type is a stricter condition on the Hermitian form.

Therefore:

$$
\boxed{
\text{algebraic Weil tensor}
\Longrightarrow
\text{split Hermitian }H^1
}
$$

is false as a general principle.

Any sixfold CIN theorem must therefore exploit geometry special to:

- dimension:
  $$
  6;
  $$
- codimension:
  $$
  3;
  $$
- cycle families with full moduli dominance.

It cannot be a universal Weil-tensor theorem.

---

# 16. Support route

Suppose:

$$
Z\subset A
$$

is an irreducible codimension-$3$ subvariety.

Then:

$$
\dim Z=3.
$$

One might try to define:

$$
L_Z
=
\ker
\left(
H^1(A,K)
\to
H^1(\widetilde Z,K)
\right),
$$

where:

$$
\widetilde Z
$$

is a resolution of:

$$
Z.
$$

If:

$$
\dim_KL_Z=3,
$$

this would look promising.

For very general:

$$
A,
$$

it fails completely.

---

# 17. Generic simplicity

On the generic locus:

$$
\operatorname{End}^0(A)=K,
$$

the endomorphism algebra is a field.

If:

$$
A
$$

were isogenous to a nontrivial product,

the corresponding isogeny decomposition would produce a nontrivial idempotent in:

$$
\operatorname{End}^0(A).
$$

Impossible.

Therefore:

$$
\boxed{
A
\text{ is simple}.
}
$$

---

# 18. Every positive-dimensional subvariety generates a simple abelian variety

Fix:

$$
z_0\in Z.
$$

Let:

$$
B_Z
$$

be the smallest abelian subvariety containing:

$$
Z-z_0.
$$

Since:

$$
Z
$$

is positive-dimensional:

$$
B_Z\neq0.
$$

Since:

$$
A
$$

is simple:

$$
\boxed{
B_Z=A.
}
$$

Thus every positive-dimensional irreducible component of a hypothetical codimension-$3$ cycle generates the whole abelian sixfold.

---

# 19. Albanese consequence

The inclusion:

$$
\widetilde Z
\to
A
$$

induces:

$$
\operatorname{Alb}(\widetilde Z)
\to
A.
$$

Because:

$$
Z
$$

generates:

$$
A,
$$

this Albanese morphism is surjective.

Dualizing on rational first cohomology gives an injective map:

$$
\boxed{
H^1(A,\mathbb Q)
\hookrightarrow
H^1(\widetilde Z,\mathbb Q).
}
$$

Therefore:

$$
\boxed{
\ker
\left(
H^1(A)
\to
H^1(\widetilde Z)
\right)
=
0.
}
$$

Not rank:

$$
3.
$$

---

# 20. Support-Kernel Failure Theorem

## Theorem 20.1

For a very general Weil-type abelian sixfold with:

$$
\operatorname{End}^0(A)=K,
$$

no positive-dimensional irreducible cycle component:

$$
Z
$$

can produce CIN's half-dimensional subspace as the kernel of cohomological restriction:

$$
H^1(A,K)
\to
H^1(\widetilde Z,K).
$$

The kernel is zero.

---

# 21. Stabilizer route also fails

Define the translation stabilizer:

$$
\operatorname{Stab}(Z)
=
\left\{
a\in A:
Z+a=Z
\right\}.
$$

Its identity component:

$$
\operatorname{Stab}(Z)^0
$$

is an abelian subvariety of:

$$
A.
$$

Since:

$$
A
$$

is simple:

$$
\operatorname{Stab}(Z)^0
$$

is either:

$$
0
$$

or:

$$
A.
$$

If:

$$
Z
$$

is a proper $3$-dimensional subvariety,

the second is impossible.

Hence:

$$
\boxed{
\operatorname{Stab}(Z)^0=0.
}
$$

So the cycle support cannot supply a $3$-dimensional abelian stabilizer either.

---

# 22. Support Ancestry Verdict

Two obvious support-level constructions fail:

### Restriction kernel

$$
\boxed{
0
}
$$

instead of rank:

$$
3.
$$

### Connected translation stabilizer

$$
\boxed{
0
}
$$

instead of dimension:

$$
3.
$$

Therefore CIN cannot come from elementary support geometry on a generic simple target.

---

# 23. Half-volume factorization

Now return to the Hermitian vector space:

$$
(H,\phi),
\qquad
\dim_KH=6.
$$

A maximal totally isotropic:

$$
3
$$

-plane:

$$
L
\subset H
$$

has a decomposable Plücker vector:

$$
\boxed{
0\neq
\xi_L
\in
\bigwedge\nolimits_K^3H.
}
$$

The point:

$$
[\xi_L]
$$

lies on the isotropic Grassmannian in its Plücker embedding.

---

# 24. Complementary half-volume

If:

$$
H
$$

is split,

choose a complementary maximal isotropic:

$$
L'.
$$

Let:

$$
0\neq
\eta_{L'}
\in
\bigwedge\nolimits_K^3H.
$$

Then:

$$
\boxed{
\xi_L
\wedge
\eta_{L'}
=
c\Omega,
\qquad
c\in K^\times.
}
$$

After rescaling:

$$
\boxed{
\Omega
=
\xi_L
\wedge
\eta_{L'}.
}
$$

This is a **rational isotropic half-volume factorization** of the Weil top tensor.

---

# 25. Half-Volume Criterion

## Theorem 25.1

For the rank-$6$ Hermitian space:

$$
(H,\phi),
$$

the following are equivalent.

### (A)

$$
H
$$

is split over:

$$
K.
$$

### (B)

There exists a $K$-rational maximal isotropic:

$$
3
$$

-plane.

### (C)

The Weil top volume:

$$
\Omega
\in
\bigwedge\nolimits_K^6H
$$

admits a factorization:

$$
\boxed{
\Omega
=
\xi\wedge\eta
}
$$

where:

- $\xi,\eta\in\bigwedge_K^3H$ are decomposable;
- their associated $3$-planes are complementary and totally isotropic.

This repackages CE003's split criterion into tensor language.

---

# 26. Why the top volume does not determine the factorization

The top exterior line:

$$
\bigwedge\nolimits_K^6H
$$

exists for every rank-$6$:

$$
K
$$

-vector space.

So:

$$
\Omega
$$

exists whether:

$$
(H,\phi)
$$

is split or non-split.

The missing information is not the volume itself.

It is:

$$
\boxed{
\text{a rational isotropic square-root / half-factorization of the volume}.
}
$$

That information is arithmetic and nonlinear.

---

# 27. Markman construction revisited

The split secant/pure-spinor construction begins with precisely the kind of half-dimensional isotropic data encoded by:

$$
\xi,
\eta.
$$

Then geometric sheaf data are built from that ancestry,

and the resulting Chern-class geometry produces the Weil classes.

So the known positive direction has the schematic form:

$$
\boxed{
\text{isotropic half-factorization}
\to
\text{derived/sheaf geometry}
\to
\text{algebraic Weil class}.
}
$$

---

# 28. The impossible formal reversal

There is no representation-theoretic theorem allowing one to reverse:

$$
\boxed{
\text{algebraic Weil class}
\to
\text{isotropic half-factorization}.
}
$$

Sections 4–15 give exact reasons:

- no equivariant half-plane;
- no rank-$3$ Hodge operator;
- no tensor-category implication;
- fourfold algebraicity disproves a universal version.

Therefore the reverse arrow, if it exists in dimension six, must be genuinely geometric and cycle-level.

---

# 29. Cycle-Level Half-Factorization Necessity

CE004 replaces naive CIN with a stronger but properly typed conjecture:

$$
\boxed{
\mathrm{CLHFN}:
}
$$

If an:

$$
\alpha
$$

-labelled codimension-$3$ algebraic-cycle family dominates the very general sixfold Weil moduli component, then after finite base change its **cycle-level geometric ancestry** canonically determines a rational isotropic half-volume factorization:

$$
\Omega
=
\xi\wedge\eta.
$$

Then:

$$
\boxed{
\mathrm{CLHFN}
\Longrightarrow
\mathrm{CIN}
\Longrightarrow
\mathrm{DIS}.
}
$$

---

# 30. CLHFN cannot factor through cycle class

Let:

$$
cl:
CH^3(A)_{\mathbb Q}
\to
H^6(A,\mathbb Q)
$$

be the cycle-class map.

Suppose a proposed ancestry extractor:

$$
\mathcal F
$$

satisfies:

$$
\mathcal F(Z)
=
F(cl(Z))
$$

for some cohomological construction:

$$
F.
$$

Then Sections 4–15 apply.

It cannot produce the required half-plane on the generic locus.

Therefore:

## Theorem 30.1

Any viable:

$$
\mathrm{CLHFN}
$$

extractor must satisfy:

$$
\boxed{
\mathcal F
\text{ does not factor through }
cl.
}
$$

This is the **Cohomology-to-Ancestry Barrier**.

---

# 31. What information could survive beyond cycle class

Possible inputs include:

- the actual support of:
  $$
  Z;
  $$
- incidence geometry of:
  $$
  Z;
  $$
- derived-category realization;
- Fourier–Mukai kernels;
- normal functions of differences of representatives;
- Chow Fourier decomposition;
- higher regulators;
- moduli of sheaves whose Chern classes contain:
  $$
  Z.
  $$

Simple support kernel/stabilizer data have already failed.

So any successful extractor must be subtler.

---

# 32. Fourier–Mukai possibility

A genuine derived object:

$$
E
$$

on:

$$
A
$$

can carry much more information than:

$$
ch(E).
$$

Its Fourier–Mukai transform can have:

- support dimension;
- cohomological amplitude;
- semihomogeneous structure;
- Mukai-type isotropic relations.

Known secant-sheaf constructions exploit this kind of extra data.

Thus HC-False must ask:

> Does every algebraic realization of a Weil class necessarily lift to some derived object with half-volume ancestry?

There is no known theorem of this form.

---

# 33. Why this is a dangerous route for HC-False

To prove a counterexample via CLHFN, one must prove a universal necessity theorem over:

$$
\boxed{
\text{all possible codimension-3 algebraic cycles}.
}
$$

This is stronger than merely understanding all currently known constructions.

A future exotic cycle may have no sheaf/secant ancestry resembling the known examples.

Therefore CLHFN is mathematically possible but strategically expensive.

---

# 34. CE004 branch verdict

The original CIN idea has now split into three versions.

### CIN-Hodge

Derive isotropic $3$-plane from Hodge tensor data.

$$
\boxed{
\text{DISPROVED}.
}
$$

### CIN-Support

Derive isotropic $3$-plane from elementary cycle support kernel/stabilizer.

$$
\boxed{
\text{DISPROVED on the generic simple locus}.
}
$$

### CIN-Cycle

Derive isotropic half-factorization from finer Chow/derived ancestry.

$$
\boxed{
\text{OPEN}.
}
$$

---

# 35. Consequence for CE001

CE001 remains open.

But the chain:

$$
\boxed{
\mathrm{CIN}
\to
\mathrm{DIS}
\to
\mathrm{CE001}
}
$$

is now much less cheap than it first appeared.

The required first arrow cannot come from:

- ordinary Hodge theory;
- Mumford–Tate symmetry;
- cohomological endomorphisms;
- support restriction;
- translation stabilizers.

It must use genuinely finer cycle-level structure.

---

# 36. Comparison with the determinant route

The independent route from CE002:

$$
\boxed{
\mathrm{TDVD}
\Longrightarrow
\mathrm{NDC}
\Longrightarrow
\mathrm{CE001}
}
$$

does not require cycle-to-isotropic extraction.

After CE004, TDVD becomes strategically more attractive than CLHFN.

A counterexample program should therefore not overcommit to DIS/CIN.

---

# 37. Historical note

Weil classes are historically appropriate adversarial objects.

Modern expositions recall that Weil himself, discussing these abnormal Hodge classes, explicitly leaned toward disbelieving the Hodge conjecture and wondered whether one could show no corresponding cycles exist.

The present HC-False branch is therefore attacking essentially the same structural fault line, now with a much larger modern toolkit.

---

# 38. Strongest proved results of CE004

This round proves:

$$
\boxed{
\text{No Equivariant Half-Plane from the Weil top tensor}.
}
$$

It proves:

$$
\boxed{
\operatorname{End}_{\mathrm{HS}}H^1=K
\Longrightarrow
\text{no rank-3 class-only Hodge endomorphism}.
}
$$

It proves on the generic simple locus:

$$
\boxed{
\ker
\left(
H^1(A)
\to
H^1(\widetilde Z)
\right)
=0
}
$$

for positive-dimensional irreducible:

$$
Z.
$$

It proves:

$$
\boxed{
\operatorname{Stab}(Z)^0=0
}
$$

for proper:

$$
Z.
$$

And it identifies splitness with:

$$
\boxed{
\text{rational isotropic half-volume factorization}.
}
$$

---

# 39. Status table

- **CE001:** OPEN
- **TDVD:** OPEN
- **DIS:** OPEN
- **CIN-Hodge:** DISPROVED
- **CIN-Support:** DISPROVED
- **CIN-Cycle / CLHFN:** OPEN
- **No Equivariant Half-Plane Theorem:** PROVED
- **Rank-$3$ Hodge Endomorphism Route:** DISPROVED
- **Half-Volume Criterion:** PROVED

---

# 40. Strategic pivot

CE004 recommends that HC-False temporarily deprioritize:

$$
\mathrm{CLHFN}.
$$

The next round should attack an independent negative mechanism that does not require recovering hidden half-dimensional ancestry.

The strongest candidates are:

1. arithmetic/global form of:
   $$
   \mathrm{TDVD};
   $$
2. tropical nonrealizability;
3. geometric coniveau obstruction.

Among these, tropical obstruction is historically aligned with the counterexample program and independent of the failed class-only CIN route.

---

# 41. Next Interface

Next counterexample round:

```text
HODGE_HCFALSE_CE005_TropicalWeilObstruction.md
```

Primary target:

$$
\boxed{
\text{Can a non-split Weil class tropicalize to a Hodge class violating algebraic-cycle realizability?}
}
$$

Tasks:

1. reconstruct Kontsevich/Zharkov tropical counterexample strategy;
2. identify exactly which tropical Hodge statements are already positive;
3. distinguish tropical Hodge type from tropical cycle realizability;
4. choose a degeneration of a non-split Weil sixfold retaining the exceptional class;
5. formulate a tropical invariant that all algebraic-cycle tropicalizations must satisfy;
6. try to violate it with the Weil class;
7. reject the route immediately if modern tropical Hodge theorems already force realizability in the required regime.

Neutral MLRSC remains paused at:

```text
HODGE_MLRSC_R029_C_JointObjectFiniteCore.md
```

---

# References

1. J. S. Milne, *The Tate and Standard Conjectures for Certain Abelian Varieties*. For a general polarized abelian variety of Weil type, the generic Hodge group is the corresponding special unitary group, and the Weil classes form the exceptional invariant sector.

2. J. S. Milne, *Hodge classes on abelian varieties*. Review of Weil classes, split Weil type, discriminant and the generic invariant-theory structure.

3. P. Deligne, *Hodge Cycles on Abelian Varieties*. Weil classes and absolute Hodge cycles on abelian varieties.

4. E. Markman, *Cycles on abelian $2n$-folds of Weil type from secant sheaves on abelian $n$-folds*, arXiv:2502.03415, 2025.

5. E. Markman, *Secant sheaves and Weil classes on abelian varieties*, arXiv:2509.23403, 2025.

6. S. Floccari, L. Fu, *The Hodge conjecture for Weil fourfolds with discriminant 1 via singular OG6-varieties*, 2025/2026.

7. Aletheia, *HODGE_HCFALSE_CE001_NonSplitWeilSixfold*, 2026-09-15.

8. Aletheia, *HODGE_HCFALSE_CE002_WeilSixfoldTangentObstruction*, 2026-09-15.

9. Aletheia, *HODGE_HCFALSE_CE003_DominanceImpliesSplit*, 2026-09-15.

---

## Canonical Source Declaration

本檔案為 HC-False 分支第四篇正式 UTF-8 Markdown canonical source。

數學原始碼只使用 `$...$` 與 `$$...$$`。

本輪否決的是從 Hodge/cohomology data或 elementary cycle support自動抽取 isotropic half-space 的策略；不否決 CE001 本身。