# HODGE_HCTRUE_PT002_OneCycleBinaryClosure
## ——正例證明分支第二輪：一個非零 Weil Cycle 即閉合整個 Weil Plane

**作者：Aletheia（GPT-5.6 Sol）**  
**研究分支：HC-True / Proof Program**  
**證明目標編號：PT002**  
**版本：v1.0**  
**日期：2026-09-16**

---

## Metadata

**Branch:** HC-True  
**Round:** PT002  
**Parent:** PT001 — Weil Sixfold Primitive Core  
**Primary Claim:** 對任何 polarized abelian sixfold of Weil type，令
$$
W_K(A)
=
\bigwedge\nolimits_K^6H^1(A,\mathbb Q).
$$
存在實際 algebraic endomorphism
$$
u\in K\cap\operatorname{End}(A)
$$
使
$$
u^6\notin\mathbb Q.
$$
其 pullback在 $W_K(A)$ 上為 multiplication by $u^6$，而 $u^6$ 在 $\mathbb Q$ 上具有 irreducible quadratic minimal polynomial。由於 algebraic cycle classes在 $u^\ast$ 下保持 algebraic，故
$$
\boxed{
W_K(A)\cap\operatorname{Alg}^3(A)
=
0
\quad\text{or}\quad
W_K(A).
}
$$
因此只要存在一個 nonzero algebraic Weil class，就自動 algebraize 整個二維 rational Weil plane。對 very general Weil sixfold，再結合 PT001：
$$
\boxed{
\text{one nonzero Weil cycle}
\Longrightarrow
W_K(A)\text{ algebraic}
\Longrightarrow
\operatorname{HC}(A^r)\text{ for every }r.
}
$$  
**Status:** PROVED  
**Genericity Needed:** binary theorem不需要；all-powers propagation使用 PT001 的 generic $Hg(A)=SU(\phi)$ hypothesis  
**Collision Target:** HC-False CE001  
**New Positive Bottleneck:** existence of one nonzero algebraic Weil class  
**Depends On:** PT001、algebraic endomorphism functoriality、$K$-determinant action  
**Formalization Status:** NOT FORMALIZED  
**Computation Status:** EXACT LINEAR ALGEBRA  

---

# 0. The compression target

PT001 reduced the very-general positive problem to:

$$
\boxed{
W_K(A)
\subseteq
\operatorname{Alg}^3(A).
}
$$

A priori this is a two-dimensional rational algebraicity problem because:

$$
\dim_{\mathbb Q}W_K(A)=2.
$$

PT002 asks:

> Do we really need to construct two independent algebraic Weil classes?

The answer is:

$$
\boxed{
\text{No}.
}
$$

One nonzero algebraic Weil class is enough.

---

# 1. Weil-type setup

Let:

$$
A
$$

be a polarized complex abelian sixfold of Weil type.

Let:

$$
K/\mathbb Q
$$

be its imaginary quadratic field acting through:

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

Set:

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

Then:

$$
\dim_KV=6.
$$

Define:

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

This is one-dimensional over:

$$
K
$$

and therefore:

$$
\boxed{
\dim_{\mathbb Q}W_K(A)=2.
}
$$

---

# 2. Algebraic Weil part

Define:

$$
\boxed{
\mathcal A_W(A)
=
W_K(A)
\cap
\operatorname{Alg}^3(A).
}
$$

Thus:

$$
\mathcal A_W(A)
$$

is the rational vector space of algebraic cohomology classes lying in the Weil plane.

A priori:

$$
\dim_{\mathbb Q}\mathcal A_W
\in
\{0,1,2\}.
$$

PT002 proves that the middle value:

$$
1
$$

cannot occur.

---

# 3. Actual endomorphism order

The embedding:

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

contains an order:

$$
\mathcal O
\subseteq
K
$$

with:

$$
\boxed{
\mathcal O
\subseteq
\operatorname{End}(A).
}
$$

Therefore every:

$$
u\in\mathcal O
$$

defines an actual algebraic endomorphism:

$$
[u]:
A\to A.
$$

We will choose:

$$
u
$$

inside this order.

---

# 4. Need for a non-rational sixth power

The key property is:

$$
\boxed{
u^6\notin\mathbb Q.
}
$$

Why exponent:

$$
6?
$$

Because:

$$
W_K(A)
=
\bigwedge\nolimits_K^6V.
$$

Multiplication by:

$$
u
$$

on:

$$
V
$$

induces multiplication by:

$$
u^6
$$

on the top:

$$
K
$$

-exterior power.

---

# 5. Existence Lemma

## Lemma 5.1

There exists:

$$
u\in\mathcal O
$$

such that:

$$
u^6\notin\mathbb Q.
$$

### Proof

Suppose:

$$
u\neq0.
$$

If:

$$
u^6\in\mathbb Q,
$$

then:

$$
u^6
=
\bar u^6,
$$

so:

$$
\boxed{
\left(
\frac{u}{\bar u}
\right)^6
=
1.
}
$$

Hence:

$$
u/\bar u
$$

must be a sixth root of unity.

An imaginary quadratic field contains only finitely many roots of unity:

- generally:
  $$
  \{\pm1\};
  $$
- for:
  $$
  K=\mathbb Q(i),
  $$
  also:
  $$
  \{\pm i\};
  $$
- for:
  $$
  K=\mathbb Q(\sqrt{-3}),
  $$
  the sixth roots of unity.

But the order:

$$
\mathcal O
$$

contains infinitely many nonzero elements and the condition:

$$
u/\bar u
\in\mu_6
$$

holds only on finitely many rational rays in:

$$
K\otimes_{\mathbb Q}\mathbb R\simeq\mathbb C.
$$

Choose:

$$
u
$$

outside these rays.

Then:

$$
u^6\neq\bar u^6,
$$

so:

$$
u^6\notin\mathbb Q.
$$

QED.

---

# 6. Explicit control example

Suppose:

$$
K=\mathbb Q(i)
$$

and the endomorphism order contains:

$$
\mathbb Z[i].
$$

Choose:

$$
u=2+i.
$$

Then:

$$
u^3
=
2+11i,
$$

so:

$$
\boxed{
u^6
=
-117+44i.
}
$$

Hence:

$$
u^6\notin\mathbb Q.
$$

Its conjugate is:

$$
\bar u^6
=
-117-44i.
$$

The norm is:

$$
N(u)^6
=
5^6
=
15625.
$$

Therefore the minimal polynomial of:

$$
u^6
$$

over:

$$
\mathbb Q
$$

is:

$$
\boxed{
m_u(x)
=
x^2+234x+15625.
}
$$

This is irreducible over:

$$
\mathbb Q.
$$

---

# 7. Action on $H^1$

The algebraic endomorphism:

$$
[u]:
A\to A
$$

induces:

$$
u^\ast:
H^1(A,\mathbb Q)
\to
H^1(A,\mathbb Q).
$$

Under the fixed:

$$
K
$$

-module structure on:

$$
V,
$$

this is the action of:

$$
u
$$

up to the fixed convention identifying:

$$
K
$$

with its cohomological action.

The convention does not affect the argument: if the cohomological action uses the conjugate embedding, replace:

$$
u
$$

by:

$$
\bar u.
$$

Either way the determinant scalar is a non-rational conjugate pair.

---

# 8. Determinant action

On:

$$
W_K(A)
=
\bigwedge\nolimits_K^6V,
$$

we have:

$$
\boxed{
u^\ast|_{W_K(A)}
=
u^6\cdot\operatorname{id}
}
$$

as a:

$$
K
$$

-linear statement.

Over:

$$
\mathbb Q,
$$

this is a:

$$
2\times2
$$

rational linear operator with minimal polynomial:

$$
\boxed{
m_{u^6}(x)
=
(x-u^6)(x-\bar u^6).
}
$$

Since:

$$
u^6\notin\mathbb Q,
$$

this quadratic is irreducible over:

$$
\mathbb Q.
$$

---

# 9. Rational irreducibility of the Weil plane

## Theorem 9.1

The rational representation:

$$
W_K(A)
$$

of the operator:

$$
u^\ast
$$

has no nonzero proper rational invariant subspace.

### Proof

A nonzero proper rational subspace of a two-dimensional rational vector space is a line:

$$
L\simeq\mathbb Q.
$$

If:

$$
L
$$

were invariant under:

$$
u^\ast,
$$

then:

$$
u^\ast|_L
$$

would be multiplication by some rational scalar:

$$
q\in\mathbb Q.
$$

Thus:

$$
q
$$

would be a rational eigenvalue of:

$$
u^\ast|_{W_K(A)}.
$$

But the only complex eigenvalues are:

$$
u^6,
\bar u^6,
$$

neither rational.

Contradiction.

QED.

---

# 10. Algebraic classes are stable under pullback

If:

$$
Z
\in
CH^3(A)_{\mathbb Q},
$$

then:

$$
[u]^\ast Z
\in
CH^3(A)_{\mathbb Q}.
$$

The cycle class map is functorial:

$$
\boxed{
cl
\left(
[u]^\ast Z
\right)
=
u^\ast cl(Z).
}
$$

Therefore:

$$
\operatorname{Alg}^3(A)
$$

is stable under:

$$
u^\ast.
$$

Since:

$$
W_K(A)
$$

is also stable under:

$$
u^\ast,
$$

their intersection:

$$
\boxed{
\mathcal A_W(A)
}
$$

is an invariant rational subspace of:

$$
W_K(A).
$$

---

# 11. Binary Algebraicity Theorem

## Theorem 11.1

For any polarized abelian sixfold of Weil type:

$$
\boxed{
\mathcal A_W(A)
=
0
\quad\text{or}\quad
W_K(A).
}
$$

Equivalently:

$$
\boxed{
W_K(A)
\cap
\operatorname{Alg}^3(A)
=
0
\quad\text{or}\quad
W_K(A).
}
$$

### Proof

By Section 10:

$$
\mathcal A_W(A)
$$

is invariant under:

$$
u^\ast.
$$

By Theorem 9.1:

$$
W_K(A)
$$

has no nonzero proper rational:

$$
u^\ast
$$

-invariant subspace.

Hence the only possibilities are:

$$
0
$$

and:

$$
W_K(A).
$$

QED.

---

# 12. No rank-one algebraic Weil sector

A priori one might imagine:

$$
\dim_{\mathbb Q}\mathcal A_W(A)=1.
$$

Theorem 11.1 rules this out.

Thus:

$$
\boxed{
\dim_{\mathbb Q}
\mathcal A_W(A)
\in
\{0,2\}.
}
$$

Correspondingly the residual:

$$
\mathfrak R_{\mathrm{Weil}}
=
W_K(A)/\mathcal A_W(A)
$$

satisfies:

$$
\boxed{
\dim_{\mathbb Q}
\mathfrak R_{\mathrm{Weil}}
\in
\{0,2\}.
}
$$

The positive/negative question is genuinely Boolean.

---

# 13. One-Cycle Closure Theorem

## Theorem 13.1

Suppose there exists:

$$
Z
\in
CH^3(A)_{\mathbb Q}
$$

such that:

$$
\boxed{
0\neq
cl(Z)
\in
W_K(A).
}
$$

Then:

$$
\boxed{
W_K(A)
\subseteq
\operatorname{Alg}^3(A).
}
$$

### Proof

The nonzero class:

$$
cl(Z)
$$

shows:

$$
\mathcal A_W(A)\neq0.
$$

By Theorem 11.1:

$$
\mathcal A_W(A)=W_K(A).
$$

QED.

---

# 14. Explicit generation of the second direction

The proof can be made constructive.

Let:

$$
\alpha
=
cl(Z)
\neq0.
$$

Then:

$$
\boxed{
u^\ast\alpha
=
cl([u]^\ast Z)
}
$$

is also algebraic and lies in:

$$
W_K(A).
$$

Suppose:

$$
u^\ast\alpha
=
q\alpha
$$

for some:

$$
q\in\mathbb Q.
$$

Then:

$$
\alpha
$$

would span a rational invariant line.

Theorem 9.1 forbids this.

Therefore:

$$
\boxed{
\alpha,
u^\ast\alpha
}
$$

are rationally linearly independent.

Hence:

$$
\boxed{
W_K(A)
=
\operatorname{span}_{\mathbb Q}
\left\{
cl(Z),
cl([u]^\ast Z)
\right\}.
}
$$

So the second algebraic Weil direction is obtained by an actual algebraic pullback.

---

# 15. No abstract projector is used

This closure uses only:

1. an actual algebraic endomorphism:
   $$
   [u]:A\to A;
   $$
2. pullback of algebraic cycles;
3. rational linear combinations.

There is no:

- Hodge projector;
- Standard Conjecture assumption;
- motivic idempotent;
- auxiliary variety.

Thus the proof is fully compatible with the R028 projector-hygiene rules.

---

# 16. Validity beyond the very-general locus

The binary theorem does **not** require:

$$
Hg(A)=SU(\phi).
$$

It uses only:

- the Weil-type:
  $$
  K
  $$
  -action;
- the definition:
  $$
  W_K(A)=\bigwedge_K^6H^1(A);
  $$
- algebraic functoriality under:
  $$
  K
  $$
  -endomorphisms.

Therefore Theorems 11.1 and 13.1 hold for:

$$
\boxed{
\text{every polarized abelian sixfold of Weil type}.
}
$$

Even CM or special-Hodge-group fibers satisfy the binary Weil-plane result.

---

# 17. Where genericity re-enters

What may fail on special loci is the statement:

> the Weil plane is the only exceptional primitive Hodge sector.

A CM sixfold can have additional Hodge classes.

Therefore on a special fiber:

$$
W_K(A)
\text{ algebraic}
$$

need not by itself settle **all** Hodge classes on all powers.

But for a very-general Weil sixfold, PT001 applies.

Thus:

$$
\boxed{
\text{one-cycle closure}
+
\text{generic invariant generation}
}
$$

gives the all-powers theorem.

---

# 18. One Cycle Implies All-Powers HC

## Theorem 18.1

Let:

$$
A
$$

be a very general polarized abelian sixfold of Weil type.

Suppose there exists:

$$
Z
\in
CH^3(A)_{\mathbb Q}
$$

with:

$$
0\neq
cl(Z)
\in
W_K(A).
$$

Then for every:

$$
r\ge1
$$

and every codimension:

$$
p,
$$

$$
\boxed{
H^{2p}(A^r,\mathbb Q)
\cap
H^{p,p}(A^r)
=
\operatorname{Alg}^p(A^r).
}
$$

### Proof

By Theorem 13.1:

$$
W_K(A)
\subseteq
\operatorname{Alg}^3(A).
$$

By PT001, algebraicity of the Weil primitive core on the very-general:

$$
SU(\phi)
$$

locus implies all-power Hodge.

QED.

---

# 19. Collision with HC-False CE001

HC-False CE001 seeks:

$$
0\neq
\alpha
\in
W_K(A)
$$

with:

$$
\alpha
\notin
\operatorname{Alg}^3(A).
$$

PT002 strengthens the dichotomy.

For every Weil-type sixfold:

$$
\boxed{
\text{either no nonzero Weil class is algebraic}
}
$$

or:

$$
\boxed{
\text{every Weil class is algebraic}.
}
$$

There is no mixed state.

Therefore CE001 cannot be rescued by saying:

> perhaps one Weil direction is algebraic but the other is not.

That possibility is now eliminated.

---

# 20. Collision becomes sharper

The two branches now make exact opposite Boolean predictions.

### HC-False

For the target non-split component:

$$
\boxed{
\mathcal A_W=0.
}
$$

### HC-True

For every Weil-type sixfold:

$$
\boxed{
\mathcal A_W=W_K(A).
}
$$

The unknown is no longer the dimension of:

$$
\mathcal A_W.
$$

It is one bit:

$$
\boxed{
0
\quad\text{vs}\quad
2.
}
$$

---

# 21. Mirror of the tropical false branch

HC-False CE006 found an analogous binary phenomenon for the tropical Weil plane:

$$
\boxed{
\mathcal A_{\mathrm{trop}}^4
\cap
W_{\mathrm{trop}}
=
0
\quad\text{or}\quad
W_{\mathrm{trop}}.
}
$$

PT002 proves the classical positive analog:

$$
\boxed{
\operatorname{Alg}^3(A)
\cap
W_K(A)
=
0
\quad\text{or}\quad
W_K(A).
}
$$

The mechanisms are parallel.

### Tropical false branch

A genuine:

$$
K
$$

-endomorphism produces a spectral irreducibility argument.

### Classical true branch

A genuine algebraic:

$$
K
$$

-endomorphism produces a determinant irreducibility argument.

This symmetry is structurally significant.

---

# 22. Positive certificate compression

PT001 required a positive certificate spanning:

$$
W_K(A).
$$

PT002 reduces the required certificate size.

A valid:

$$
\mathsf{PosCert}_1
$$

now only needs:

1. target:
   $$
   (A,K,\lambda);
   $$
2. one cycle:
   $$
   Z\in CH^3(A)_{\mathbb Q};
   $$
3. proof:
   $$
   0\neq cl(Z)\in W_K(A);
   $$
4. one endomorphism:
   $$
   u\in\mathcal O\subseteq\operatorname{End}(A);
   $$
5. proof:
   $$
   u^6\notin\mathbb Q.
   $$

Then:

$$
Z,
\quad
[u]^\ast Z
$$

supply a full basis of the Weil plane.

---

# 23. Projective cycle-level nature

This is not merely a statement about cohomology.

If:

$$
Z
$$

is an actual cycle,

then:

$$
[u]^\ast Z
$$

is an actual cycle.

So the binary closure is realized entirely inside:

$$
CH^3(A)_{\mathbb Q}.
$$

No passage through a larger motivated category is required.

This is exactly the type of legality-safe closure desired by MLRSC.

---

# 24. Pushforward variant

Since nonzero:

$$
u\in K
$$

defines an isogeny after choosing an integral representative, one may equally use pushforward:

$$
[u]_\ast.
$$

Its action on:

$$
W_K(A)
$$

is a nonzero rational multiple of the inverse pullback action determined by Poincaré duality and the degree of the isogeny.

For the binary theorem, pullback is conceptually simpler.

But both are algebraic operations.

---

# 25. General $2n$-dimensional version

The same proof is not special to dimension six.

Let:

$$
A
$$

be a Weil-type abelian:

$$
2n
$$

-fold with:

$$
\dim_KH^1(A,\mathbb Q)=2n.
$$

Then:

$$
W_K(A)
=
\bigwedge\nolimits_K^{2n}H^1(A,\mathbb Q)
$$

is one-dimensional over:

$$
K.
$$

Choose:

$$
u
$$

with:

$$
u^{2n}\notin\mathbb Q.
$$

Then:

$$
\boxed{
W_K(A)\cap\operatorname{Alg}^n(A)
=
0
\quad\text{or}\quad
W_K(A).
}
$$

Thus the one-cycle binary closure is a general Weil-type phenomenon.

---

# 26. Existence of suitable $u$ in general dimension

For fixed:

$$
2n,
$$

if:

$$
u^{2n}\in\mathbb Q,
$$

then:

$$
\left(
u/\bar u
\right)^{2n}=1.
$$

Thus:

$$
u/\bar u
$$

lies in the finite root-of-unity group of:

$$
K.
$$

An order in:

$$
K
$$

contains infinitely many elements outside those finitely many rays.

Therefore suitable:

$$
u
$$

always exists.

So Theorem 25 generalizes uniformly.

---

# 27. Positive route after PT002

The sixfold positive problem is now compressed to:

$$
\boxed{
\text{Construct one nonzero codimension-3 algebraic cycle whose class lies in }W_K(A).
}
$$

That is all.

The second Weil direction is free.

The all-power invariant closure is free on the generic locus.

Thus the real positive engineering problem is now one-cycle existence.

---

# 28. Existing split construction reinterpreted

Markman's split/discriminant-$-1$ theorem constructs enough geometry to algebraize the Weil classes.

PT002 says that from a logical minimum perspective, only one nonzero Weil class was needed.

Once one appears, the:

$$
K
$$

-action closes the full rational plane.

So future generalizations of Markman's geometry should optimize for:

$$
\boxed{
\text{one nonzero Weil projection},
}
$$

not for an a priori full basis.

This may materially simplify the search.

---

# 29. Relative-family consequence

Let:

$$
\pi:\mathcal A\to S
$$

be a family of Weil-type sixfolds with a flat Weil local subsystem:

$$
\mathbb W_K.
$$

Suppose a relative algebraic cycle:

$$
\mathcal Z
$$

has fiberwise class:

$$
\alpha_s
\in
W_K(A_s)
$$

and:

$$
\alpha_s
$$

is nonzero on a dense open subset.

Then PT002 gives, fiberwise on that open:

$$
\boxed{
W_K(A_s)
\subseteq
\operatorname{Alg}^3(A_s).
}
$$

Thus one relative Weil cycle family is enough to close the whole Weil plane on every fiber where its class does not vanish.

This is the natural bridge to the next positive round.

---

# 30. Relative nonvanishing is easier than full basis transport

A two-cycle construction would require tracking two independent classes under deformation.

PT002 reduces this to tracking one section:

$$
\alpha
\in
\Gamma(S,\mathbb W_K).
$$

The only conditions are:

1. algebraic relative origin;
2. nonzero Weil projection.

Then:

$$
K
$$

-endomorphism closure generates the second direction.

This cuts the relative-cycle complexity roughly in half.

---

# 31. Candidate positive deformation strategy

The next positive architecture is:

$$
\boxed{
\text{one seed cycle}
\to
\text{relative deformation}
\to
\text{one nonzero Weil section}
\to
\text{binary closure}
\to
\text{all-power HC}.
}
$$

The new hard step is:

$$
\boxed{
\text{relative deformation of one nonzero Weil cycle across non-split sixfold moduli}.
}
$$

This is now much sharper than the original question.

---

# 32. Relation to HC-False UCDRD

This creates a direct collision with CE001's original negative strategy.

HC-False attempted to prove:

$$
\boxed{
\text{no Weil-cycle parameter component dominates the non-split moduli}.
}
$$

HC-True now needs only prove:

$$
\boxed{
\text{one nonzero Weil-cycle parameter component dominates}.
}
$$

If such a component exists, binary closure makes its generic fibers carry the whole Weil plane algebraically.

Thus the branches meet exactly at:

$$
\boxed{
\text{dominance of one Weil-cycle family}.
}
$$

This is an excellent future collision interface.

---

# 33. Status

PT002 proves:

$$
\boxed{
\operatorname{Alg}^3(A)\cap W_K(A)
=
0
\quad\text{or}\quad
W_K(A)
}
$$

for every polarized Weil-type abelian sixfold.

It also proves:

$$
\boxed{
\text{one nonzero algebraic Weil class}
\Longrightarrow
\text{the full Weil plane is algebraic}.
}
$$

And on the very-general locus:

$$
\boxed{
\text{one nonzero Weil cycle}
\Longrightarrow
\text{all-power rational Hodge conjecture}.
}
$$

Therefore:

$$
\boxed{
\mathrm{Status}
=
\mathrm{PROVED}.
}
$$

The unresolved statement remains:

$$
\boxed{
\text{does every Weil-type sixfold possess one nonzero algebraic Weil class?}
}
$$

---

# 34. Next Interface

Next HC-True round:

```text
HODGE_HCTRUE_PT003_OneRelativeWeilCycle.md
```

Primary target:

$$
\boxed{
\text{Can one nonzero Weil cycle be spread over a non-split Weil sixfold moduli component?}
}
$$

Planned attacks:

1. fix an irreducible:
   $$
   \mathcal M_{K,\delta};
   $$
2. identify the rank-$2$ Weil local system:
   $$
   \mathbb W_K;
   $$
3. choose one algebraic seed fiber with a known/nonzero Weil cycle;
4. study the deformation space of the pair:
   $$
   (Z\subset A);
   $$
5. derive a semiregularity criterion sufficient for dominance;
6. use PT002 to show that only one relative nonzero section is needed;
7. compare directly with HC-False CE001's no-dominance/UCDRD conjecture;
8. if dominance is obtained, conclude:
   $$
   \text{generic full Weil algebraicity};
   $$
9. combine with PT001:
   $$
   \text{all-power HC on the very-general component}.
   $$

---

# References

1. B. J. J. Moonen, Yu. G. Zarhin, *Weil classes on abelian varieties*, arXiv:alg-geom/9612017. Gives the Weil class construction
   $$
   W_K(A)=\bigwedge_K^{2n}H^1(A,\mathbb Q)
   $$
   and its natural interaction with the imaginary-quadratic endomorphism field.

2. J. S. Milne, *Hodge classes on abelian varieties*, arXiv:2010.08857. Reviews Weil-type abelian varieties, endomorphism actions and the role of Weil classes in the Hodge problem.

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

4. Aletheia, *HODGE_HCTRUE_PT001_WeilSixfoldPrimitiveCore*, 2026-09-16.

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

6. Aletheia, *HODGE_HCFALSE_CE006_WeilEightfoldRationalDefect*, 2026-09-16.

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## Canonical Source Declaration

本檔案為 HC-True 分支第二篇正式 UTF-8 Markdown canonical source。

數學原始碼只使用 `$...$` 與 `$$...$$`。

本輪沒有證明所有 Weil sixfold 都存在 nonzero algebraic Weil cycle；本輪證明的是只要存在一個，整個 Weil plane立即由 algebraic $K$-endomorphism closure完成。