# HODGE_HCTRUE_PT006_WeilTangentChannelDecomposition
## ——正例證明分支第六輪：Split-Sixfold Mixed-Channel Vanishing、Finite-Stabilizer Criterion 與 Weil-Tangent Alignment Gate

**作者：Aletheia（GPT-5.6 Sol）**  
**研究分支：HC-True / Proof Program**  
**證明目標編號：PT006**  
**版本：v1.0**  
**日期：2026-09-16**

---

## Metadata

**Branch:** HC-True  
**Round:** PT006  
**Parent:** PT004、PT005  
**Primary Claim:** 在 Markman 已成功的 split Weil sixfold secant construction 中，actual $9$-dimensional unobstructed source deformation space在
$$
HT^2(X\times X)
=
C_{20}\oplus C_{11}\oplus C_{02}
$$
中完全位於 even channels：
$$
\boxed{
D_{11}=0.
}
$$
更一般地，對 outer-product seed $G=F_1\boxtimes F_2^\vee$，若兩個 factor 的 identity-component translation/tensor stabilizer皆為有限群，則
$$
HT^1(X)\to\operatorname{Ext}^1(F_i,F_i)
$$
皆 injective，從而 mixed obstruction channel
$$
C_{11}
=
HT^1(X)\otimes HT^1(X)
\to
B_{11}
=
\operatorname{Ext}^1(F_1,F_1)\otimes\operatorname{Ext}^1(F_2,F_2)
$$
injective。由 PT005 的 channel separation，任何 unobstructed source deformation必須有零 $C_{11}$ component。因此，若某 outer-product seed要 deform over the full Weil tangent $T_{\mathrm{Weil}}$, 必須滿足 **Weil-Tangent Alignment Gate**
$$
\boxed{
\operatorname{pr}_{11}
\left(
\Phi_{HT}^{-1}(T_{\mathrm{Weil}})
\right)
=
0.
}
$$
在原始 split sixfold構造中此 gate 已被 diagonal deformation theorem滿足；在 2025 generalized higher-CM construction中，現有 paper建立 Hodge permanence與 period domain，但沒有計算 full Weil tangent 的 Hochschild three-channel preimage，因此 generalized $D_{11}=0$ 目前不能視為已證。  
**Status:** PROVED  
**Split Sixfold:** $D_{11}=0$ PROVED  
**General Outer-Product Criterion:** finite stabilizers + full deformability imply $D_{11}=0$  
**Generalized Higher-CM Alignment:** OPEN  
**Open Gate:** compute $\Phi_{HT}^{-1}(T_{\mathrm{Weil}})$ and test the mixed projection before attacking even-channel semiregularity  
**Collision Target:** HC-False may kill an outer-product seed by proving unavoidable mixed projection  
**Depends On:** PT005、Markman 2502.03415 Sections 8.4–8.5、Orlov/Toda Hochschild transform  
**Formalization Status:** NOT FORMALIZED  
**Computation Status:** EXACT CHANNEL ALIGNMENT REDUCTION  

---

# 0. The question left by PT005

PT005 decomposed:

$$
HT^2(X\times X)
=
C_{20}\oplus C_{11}\oplus C_{02},
$$

with:

$$
C_{20}
=
HT^2(X)\otimes HT^0(X),
$$

$$
C_{11}
=
HT^1(X)\otimes HT^1(X),
$$

and:

$$
C_{02}
=
HT^0(X)\otimes HT^2(X).
$$

The matching self-Ext target is:

$$
B_{20}\oplus B_{11}\oplus B_{02}.
$$

The next question is:

> Where does the actual Weil period-domain tangent live under the inverse Hochschild transform?

PT006 answers this completely in the successful split-sixfold model and extracts a general necessary criterion.

---

# 1. Split sixfold control model

Let:

$$
X
$$

be the genus-$3$ Jacobian used in the split secant construction.

Then:

$$
\dim_{\mathbb C}X=3.
$$

Let:

$$
F_1,
F_2
$$

be the two secant sheaves.

Consider:

$$
G
=
\pi_1^\ast F_1
\otimes
\pi_2^\ast F_2
$$

on:

$$
X\times X.
$$

After the duality involution on the second Hochschild factor and Orlov transform, this yields the object:

$$
E
$$

on:

$$
X\times\hat X.
$$

---

# 2. Split self-Ext decomposition

As in PT005:

$$
\operatorname{Ext}^2(G,G)
=
B_{20}\oplus B_{11}\oplus B_{02},
$$

where:

$$
B_{20}
=
\operatorname{Ext}^2(F_1,F_1)
\otimes
\operatorname{Hom}(F_2,F_2),
$$

$$
B_{11}
=
\operatorname{Ext}^1(F_1,F_1)
\otimes
\operatorname{Ext}^1(F_2,F_2),
$$

$$
B_{02}
=
\operatorname{Hom}(F_1,F_1)
\otimes
\operatorname{Ext}^2(F_2,F_2).
$$

---

# 3. Mixed obstruction is injective in the split model

Markman's split calculation proves that:

$$
\exp(at_{F_i})
:
HT^1(X)
\to
\operatorname{Ext}^1(F_i,F_i)
$$

is injective for:

$$
i=1,2.
$$

The product obstruction map on:

$$
C_{11}
=
HT^1(X)\otimes HT^1(X)
$$

is the tensor product of these degree-$1$ actions.

Therefore:

$$
\boxed{
ob_G\big|_{C_{11}}
\text{ is injective}.
}
$$

Hence:

$$
\boxed{
\ker ob_G
\cap
C_{11}
=
0.
}
$$

---

# 4. Exact kernel formula in the split model

Markman obtains:

$$
\boxed{
\ker ob_G
=
\left[
\pi_1^\ast\ker(ob_{F_1})
\otimes
\pi_2^\ast HT^0(X)
\right]
\oplus
\left[
\pi_1^\ast HT^0(X)
\otimes
\pi_2^\ast\ker(ob_{F_2})
\right].
}
$$

Thus:

$$
\boxed{
\ker ob_G
\subset
C_{20}\oplus C_{02}.
}
$$

No mixed Hochschild deformation is unobstructed.

This is stronger than merely saying a chosen tangent subspace has no mixed component.

---

# 5. Dimension of the factor kernel

For the genus-$3$ secant sheaves:

$$
ob_{F_i}
:
HT^2(X)
\to
\operatorname{Ext}^2(F_i,F_i)
$$

has rank:

$$
6.
$$

Now:

$$
\dim HT^2(X)
=
h^2(\mathcal O_X)
+
h^1(T_X)
+
h^0(\wedge^2T_X).
$$

For an abelian threefold:

$$
h^2(\mathcal O_X)=3,
$$

$$
h^1(T_X)=9,
$$

$$
h^0(\wedge^2T_X)=3.
$$

Hence:

$$
\dim HT^2(X)=15.
$$

Therefore:

$$
\boxed{
\dim\ker ob_{F_i}=9.
}
$$

---

# 6. Duality involution

On:

$$
HT^2(X)
=
H^2(\mathcal O_X)
\oplus
H^1(T_X)
\oplus
H^0(\wedge^2T_X),
$$

the duality involution used by Markman is:

$$
\boxed{
(\alpha,\beta,\gamma)^\ast
=
(-\alpha,\beta,-\gamma).
}
$$

The split secant sheaves satisfy:

$$
\boxed{
\ker ob_{F_1}
=
\ker ob_{F_2^\vee}
}
$$

after this involution.

This makes the following diagonal embedding natural.

---

# 7. Diagonal even-channel space

Define:

$$
\boxed{
\Delta_{\mathrm{sec}}
:
\ker ob_{F_1}
\to
HT^2(X\times X)
}
$$

by:

$$
\boxed{
\Delta_{\mathrm{sec}}(\alpha)
=
\pi_1^\ast\alpha
+
\pi_2^\ast\alpha^\ast.
}
$$

Every element of:

$$
\Delta_{\mathrm{sec}}
\left(
\ker ob_{F_1}
\right)
$$

lies in:

$$
C_{20}\oplus C_{02}.
$$

Therefore:

$$
\boxed{
\operatorname{pr}_{11}
\Delta_{\mathrm{sec}}(\alpha)
=
0.
}
$$

The subspace has dimension:

$$
9.
$$

---

# 8. Orlov Hochschild transform

Let:

$$
\Phi_{HT}
:
HT^2(X\times X)
\overset{\sim}{\longrightarrow}
HT^2(X\times\hat X)
$$

be Orlov's Hochschild transform.

Markman proves that:

$$
\boxed{
\Phi_{HT}
\circ
(id\otimes(\bullet)^\ast)
}
$$

maps the diagonal embedding of:

$$
HT^2(X)
$$

into:

$$
H^1(T_{X\times\hat X})
\oplus
H^2(\mathcal O_{X\times\hat X}).
$$

Thus diagonal generalized deformations become commutative-gerby deformations.

---

# 9. The nine-dimensional split space

Restrict the above map to:

$$
\ker ob_{F_1}.
$$

Then:

$$
\boxed{
D_{\mathrm{split}}
=
\Phi_{HT}
\left(
\Delta_{\mathrm{sec}}
(\ker ob_{F_1})
\right)
}
$$

is a:

$$
9
$$

-dimensional subspace of:

$$
H^1(T_{X\times\hat X})
\oplus
H^2(\mathcal O_{X\times\hat X}).
$$

Moreover:

$$
D_{\mathrm{split}}
\subset
\ker ob_E.
$$

So these are actual unobstructed generalized deformations of the seed object.

---

# 10. Projection to the Weil tangent

Markman further proves that:

$$
D_{\mathrm{split}}
$$

projects onto the tangent space in:

$$
H^1(T_{X\times\hat X})
$$

of the moduli space of abelian sixfolds of Weil type.

The relevant tangent space has complex dimension:

$$
9.
$$

Since:

$$
D_{\mathrm{split}}
$$

is itself:

$$
9
$$

-dimensional and the commutative projection is injective in the diagonal transform calculation, it supplies the full Weil tangent.

Thus the split construction realizes every Weil first-order deformation via the diagonal even-channel source space.

---

# 11. Split Mixed-Channel Vanishing Theorem

## Theorem 11.1

For the successful split-sixfold secant construction:

$$
\boxed{
D_{\mathrm{Weil}}^{\mathrm{source}}
=
\Delta_{\mathrm{sec}}
(\ker ob_{F_1})
\subset
C_{20}\oplus C_{02}.
}
$$

Hence:

$$
\boxed{
D_{11}=0.
}
$$

The actual nine-dimensional Weil tangent is therefore obtained with no mixed:

$$
HT^1\otimes HT^1
$$

source contribution.

QED.

---

# 12. Why this mattered to the original proof

The vanishing:

$$
D_{11}=0
$$

is not cosmetic.

The mixed channel obstruction map is injective.

Therefore any source deformation with:

$$
D_{11}\neq0
$$

would be obstructed already at first order.

The diagonal-even alignment is precisely what allows the nine required Weil directions to survive.

---

# 13. General meaning of $HT^1$

For an abelian variety:

$$
X,
$$

we have:

$$
\boxed{
HT^1(X)
=
H^1(\mathcal O_X)
\oplus
H^0(T_X).
}
$$

This is the Lie algebra of the identity component:

$$
\boxed{
X\times\operatorname{Pic}^0(X)
}
$$

of standard translation/tensor autoequivalences of:

$$
D^b(X).
$$

Thus the degree-$1$ Atiyah/Hochschild action has a direct geometric meaning.

---

# 14. Orbit differential

For:

$$
F\in D^b(X),
$$

consider the orbit map:

$$
\boxed{
\mathcal O_F:
X\times\operatorname{Pic}^0(X)
\to
\mathfrak M,
}
$$

$$
(x,L)
\mapsto
\tau_x^\ast F\otimes L.
$$

Its differential at the identity is:

$$
\boxed{
d\mathcal O_F:
HT^1(X)
\to
\operatorname{Ext}^1(F,F).
}
$$

Under the standard Hochschild–Atiyah identification:

$$
d\mathcal O_F
$$

is the degree-$1$ evaluation:

$$
\boxed{
at_F:
HT^1(X)
\to
\operatorname{Ext}^1(F,F).
}
$$

---

# 15. Stabilizer criterion

Let:

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

be the identity component of the translation/tensor stabilizer of:

$$
F.
$$

Then:

$$
\boxed{
\ker
\left(
at_F:
HT^1(X)
\to
\operatorname{Ext}^1(F,F)
\right)
=
\operatorname{Lie}
\operatorname{Stab}^0(F).
}
$$

Therefore:

## Lemma 15.1

If:

$$
\operatorname{Stab}^0(F)=0,
$$

then:

$$
\boxed{
at_F:
HT^1(X)
\to
\operatorname{Ext}^1(F,F)
}
$$

is injective.

This is the **Finite-Stabilizer Criterion**.

---

# 16. General mixed-channel injectivity

Let:

$$
G
=
F_1\boxtimes F_2^\vee.
$$

Assume:

$$
\operatorname{Stab}^0(F_1)
=
\operatorname{Stab}^0(F_2)
=
0.
$$

Then both maps:

$$
at_{F_i}:
HT^1(X)
\to
\operatorname{Ext}^1(F_i,F_i)
$$

are injective.

The product obstruction map on:

$$
C_{11}
$$

is:

$$
\boxed{
at_{F_1}
\otimes
at_{F_2^\vee}.
}
$$

A tensor product of injective linear maps over:

$$
\mathbb C
$$

is injective.

Hence:

## Theorem 16.1

$$
\boxed{
ob_G\big|_{C_{11}}
\text{ is injective}.
}
$$

Therefore:

$$
\boxed{
\ker ob_G
\cap
C_{11}
=
0.
}
$$

---

# 17. Channel separation is essential

PT005 proved that:

$$
C_{20},
C_{11},C_{02}
$$

map into distinct:

$$
B_{20},B_{11},B_{02}
$$

Ext channels.

Therefore a nonzero mixed obstruction cannot cancel against an even-channel obstruction.

Thus if:

$$
d
=
d_{20}+d_{11}+d_{02}
$$

is unobstructed, then:

$$
\boxed{
d_{11}=0
}
$$

whenever the mixed map is injective.

This gives a general channel-alignment theorem.

---

# 18. Unobstructed Mixed-Channel Exclusion Theorem

## Theorem 18.1

Let:

$$
G=F_1\boxtimes F_2^\vee
$$

with:

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

Then every first-order source deformation:

$$
d\in HT^2(X\times X)
$$

satisfying:

$$
ob_G(d)=0
$$

has:

$$
\boxed{
\operatorname{pr}_{11}(d)=0.
}
$$

Equivalently:

$$
\boxed{
\ker ob_G
\subseteq
C_{20}\oplus C_{02}.
}
$$

This generalizes the structural reason behind Markman's split calculation.

---

# 19. Weil-Tangent Alignment Gate

Let:

$$
T_{\mathrm{Weil}}
\subset
HT^2(X\times\hat X)
$$

denote the required commutative Weil period-domain tangent space.

Define its source preimage:

$$
\boxed{
D_{\mathrm{Weil}}
=
\Phi_{HT}^{-1}(T_{\mathrm{Weil}})
\subset
HT^2(X\times X).
}
$$

For the seed:

$$
E=\Phi(G)
$$

to deform along **every** Weil tangent direction, every element of:

$$
D_{\mathrm{Weil}}
$$

must be unobstructed for:

$$
G.
$$

Under the finite-stabilizer hypothesis, Theorem 18.1 therefore implies the necessary condition:

$$
\boxed{
\operatorname{pr}_{11}
(D_{\mathrm{Weil}})
=
0.
}
$$

Call this the:

$$
\boxed{
\text{Weil-Tangent Alignment Gate}.
}
$$

---

# 20. Necessity theorem

## Theorem 20.1

Assume:

1. $E=\Phi(F_1\boxtimes F_2^\vee)$;
2. both factor identity-component stabilizers are trivial;
3. $E$ deforms first-order along the full Weil period-domain tangent.

Then:

$$
\boxed{
D_{\mathrm{Weil}}
\subseteq
C_{20}\oplus C_{02}.
}
$$

In particular:

$$
\boxed{
D_{11}=0.
}
$$

Thus mixed-channel vanishing is a **necessary condition for dominance**, not merely a desirable simplification.

---

# 21. A positive gate and a falsifier

Theorem 20.1 can be used in two directions.

### Positive use

If direct calculation gives:

$$
D_{11}=0,
$$

the seed passes the mixed-channel gate and one may continue to even-channel semiregularity.

### Negative use

If:

$$
D_{11}\neq0
$$

while the factor stabilizers are finite, then:

$$
\boxed{
\text{this outer-product seed cannot deform over the full Weil component}.
}
$$

No improvement of even-channel semiregularity can repair this seed.

This makes channel alignment a strong early falsifier.

---

# 22. Original split sixfold passes the gate

The original construction satisfies:

$$
D_{\mathrm{Weil}}
=
\Delta_{\mathrm{sec}}
(\ker ob_{F_1}),
$$

so:

$$
\boxed{
D_{11}=0.
}
$$

Thus the known successful proof passes the alignment gate exactly.

This provides a nontrivial control test of the criterion.

---

# 23. Relation to PT005

PT005 left three possible obstruction blocks:

$$
J_{20},
\quad
J_{11},
\quad
J_{02}.
$$

PT006 shows that under the finite-stabilizer alignment hypothesis:

$$
\boxed{
J_{11}=0
}
$$

for any seed capable of full Weil-tangent deformation.

Hence a successful outer-product positive proof reduces from three channels to at most:

$$
\boxed{
J_{20}\oplus J_{02}.
}
$$

The mixed odd channel is eliminated structurally.

---

# 24. Old and new even channels

For the generalized secant program:

$$
F_1
$$

is chosen from an existing secant construction.

The new object to be found is:

$$
F_2.
$$

Conceptually:

$$
J_{20}
$$

is the old factor even channel.

$$
J_{02}
$$

is the new factor even channel.

If:

$$
F_1
$$

already satisfies the required image-semiregularity/obstruction-annihilation property, the genuinely new gate is:

$$
\boxed{
J_{02}.
}
$$

This is the architecture suggested by Markman's Question 11.2.2.

---

# 25. Why Question 11.2.2 focuses on $F_2$

Markman's generalized paper asks for:

$$
F_2
$$

such that:

$$
\sigma_{F_2}
$$

is injective on:

$$
\boxed{
\operatorname{Im}
\left(
at_{F_2}:
HT^2(X)
\to
\operatorname{Ext}^2(F_2,F_2)
\right).
}
$$

PT006 explains how such a one-factor condition can be sufficient **provided**:

1. the actual Weil tangent satisfies the alignment gate;
2. the old:
   $$
   F_1
   $$
   even channel is already controlled.

The paper's formulation should not be read as a formal proof that all other channels always disappear.

Those surrounding hypotheses must be checked.

---

# 26. Generalized higher-CM scope

In the 2025 generalized construction, Markman proves:

- existence of a period domain:
  $$
  \Omega_B
  $$
  of Weil-type deformations;
- Hodge permanence of:
  $$
  \kappa(E);
  $$
- criteria for nonzero projection to the Hodge-Weil sector.

However, the generalized paper does not carry out the split-paper calculation:

$$
\boxed{
\Phi_{HT}^{-1}(T_{\mathrm{Weil}})
=
\text{explicit diagonal even source space}.
}
$$

Nor does it establish generalized semiregularity.

Therefore:

$$
\boxed{
D_{11}=0
}
$$

for the full generalized higher-CM Weil tangent is currently an **open alignment calculation**.

---

# 27. No illegitimate extrapolation

The following inference is invalid:

> Split sixfold Weil tangents are diagonal even-channel, therefore all CM Weil tangents are diagonal even-channel.

The first statement uses a specific Orlov/Hochschild calculation tied to the split secant model.

The generalized period domain is built via:

$$
{\rm Spin}(V_{\mathbb R})_B
$$

and may have a different source-channel realization.

So PT006 keeps:

$$
\boxed{
\text{split alignment theorem}
}
$$

separate from:

$$
\boxed{
\text{generalized alignment conjecture}.
}
$$

---

# 28. Dimension sanity check

The generalized higher-CM example does not face an immediate dimension obstruction to even-channel alignment.

For an abelian fourfold:

$$
X,
$$

$$
\dim HT^2(X)
=
\binom42
+
4^2
+
\binom42
=
6+16+6
=
28.
$$

Therefore:

$$
\dim(C_{20}\oplus C_{02})
=
56.
$$

In the quartic-CM example of:

$$
[K:\mathbb Q]=4
$$

with:

$$
d=4,
$$

the Weil period-domain component has dimension:

$$
\boxed{
d^2[K:\mathbb Q]/2
=
32.
}
$$

Thus a:

$$
32
$$

-dimensional Weil tangent could in principle lie entirely inside the:

$$
56
$$

-dimensional even source sector.

So:

$$
D_{11}=0
$$

is dimensionally possible but not automatic.

---

# 29. Finite-stabilizer test for candidate factors

Before computing the full Hochschild transform, HC-True can test each factor:

$$
F_i
$$

for its translation/tensor stabilizer.

If:

$$
\operatorname{Stab}^0(F_i)\neq0,
$$

then:

$$
HT^1(X)\to\operatorname{Ext}^1(F_i,F_i)
$$

has kernel and the mixed-channel injectivity theorem may fail.

If:

$$
\operatorname{Stab}^0(F_i)=0,
$$

mixed injectivity follows immediately.

This gives a cheap preliminary certificate.

---

# 30. Stabilizer certificate

Define:

$$
\boxed{
\mathsf{FinStab}(F)
:
\quad
\operatorname{Stab}^0(F)=0.
}
$$

Then:

$$
\boxed{
\mathsf{FinStab}(F_1)
+
\mathsf{FinStab}(F_2)
\Longrightarrow
ob_G|_{C_{11}}
\text{ injective}.
}
$$

For any candidate outer-product seed intended to dominate a full Weil component, this turns:

$$
D_{11}=0
$$

into a compulsory alignment check.

---

# 31. Actual reduced positive architecture

A general outer-product seed now faces the gates:

### Gate A — Factor stabilizers

$$
\mathsf{FinStab}(F_1),
\quad
\mathsf{FinStab}(F_2).
$$

### Gate B — Weil tangent alignment

$$
\boxed{
D_{11}=0.
}
$$

### Gate C — Old even channel

Control:

$$
J_{20}.
$$

### Gate D — New even channel

Control:

$$
J_{02}.
$$

### Gate E — Higher-order effectivity

Upgrade first-order deformability to formal/algebraic dominance.

This is a much sharper checklist than generic semiregularity.

---

# 32. Collision with HC-False

HC-False now has a new way to kill a positive seed.

It may prove:

$$
\boxed{
\operatorname{pr}_{11}
\Phi_{HT}^{-1}
(T_{\mathrm{Weil}})
\neq0
}
$$

for the proposed seed architecture.

If factor stabilizers are finite, this gives unavoidable first-order obstruction.

Then that entire outer-product construction cannot dominate the Weil moduli component.

Conversely HC-True can prove alignment and proceed to even channels.

This is a clean adversarial interface.

---

# 33. What PT006 proves

PT006 proves for the split sixfold model:

$$
\boxed{
D_{11}=0.
}
$$

It proves generally:

$$
\boxed{
\mathsf{FinStab}(F_1)
+
\mathsf{FinStab}(F_2)
\Longrightarrow
ob_G|_{C_{11}}\text{ injective}.
}
$$

It proves that full first-order Weil deformability then requires:

$$
\boxed{
D_{11}=0.
}
$$

It therefore turns mixed-channel vanishing into a necessary dominance gate.

---

# 34. What PT006 does not prove

PT006 does not compute:

$$
D_{\mathrm{Weil}}
$$

for the generalized higher-CM construction.

It does not prove finite stabilizers for every generalized secant sheaf.

It does not prove:

$$
J_{20}=0
$$

or:

$$
J_{02}=0
$$

there.

It therefore does not establish generalized semiregularity.

The next unknown is now an explicit Hochschild-linear alignment calculation.

---

# 35. Triple-program interpretation

### HC-True

Wants:

$$
\boxed{
D_{11}=0
}
$$

and both even obstruction channels controlled.

### HC-False

Can attack:

$$
\boxed{
D_{11}\neq0
}
$$

or a surviving even obstruction.

### Neutral MLRSC

Interprets this as a legality/category coupling defect:

$$
\boxed{
\text{period tangent}
\not\Rightarrow
\text{deformable seed tangent}.
}
$$

All three branches now share a concrete three-channel deformation object.

---

# 36. Next Interface

Next HC-True round:

```text
HODGE_HCTRUE_PT007_MixedChannelAlignment.md
```

Primary target:

$$
\boxed{
\operatorname{pr}_{11}
\left(
\Phi_{HT}^{-1}(T_{\mathrm{Weil}})
\right)
\stackrel{?}{=}0
}
$$

for the generalized CM/real-multiplication secant architecture.

Planned attacks:

1. write Orlov's action on:
   $$
   HT^1
   $$
   and:
   $$
   HT^2
   $$
   as an explicit block matrix;
2. express the unitary Weil tangent via the Lie algebra of:
   $$
   {\rm Spin}(V_{\mathbb R})_B;
   $$
3. pull its commutative tangent directions through:
   $$
   \Phi_{HT}^{-1};
   $$
4. project onto:
   $$
   C_{11};
   $$
5. determine whether the mixed projection vanishes identically;
6. if not, test whether candidate factors have positive-dimensional stabilizers capable of absorbing it;
7. if the mixed projection is nonzero and factor stabilizers are finite, reject that seed architecture for full dominance;
8. if it vanishes, advance immediately to:
   $$
   J_{20},J_{02}.
   $$

---

# References

1. E. Markman, *Cycles on abelian $2n$-folds of Weil type from secant sheaves on abelian $n$-folds*, arXiv:2502.03415, especially Sections 8.4–8.5. The product obstruction kernel has no mixed $HT^1\otimes HT^1$ component, and Corollary 8.5.2 maps a diagonal $9$-dimensional source kernel to unobstructed commutative-gerby deformations projecting to the Weil moduli tangent.

2. E. Markman, *Secant sheaves on abelian $n$-folds with real multiplication and Weil classes on abelian $2n$-folds with complex multiplication*, arXiv:2509.23079. Constructs the generalized CM period domain and proves Hodge permanence, but leaves semiregularity open.

3. E. Markman, *Secant sheaves and Weil classes on abelian varieties*, arXiv:2509.23403. Surveys the split sixfold proof and explains how the semiregular secant object deforms over the split Weil component.

4. Y. Toda, *Deformations and Fourier-Mukai transforms*, J. Differential Geom. 81 (2009), 197–224. Provides the Hochschild deformation interpretation transported by Fourier–Mukai equivalences.

5. Aletheia, *HODGE_HCTRUE_PT005_SecantExtRepresentation*, 2026-09-16.

---

## Canonical Source Declaration

本檔案為 HC-True 分支第六篇正式 UTF-8 Markdown canonical source。

數學原始碼只使用 `$...$` 與 `$$...$$`。

本輪沒有證明 generalized higher-CM Weil tangent已通過 mixed-channel alignment；本輪證明的是 split sixfold 的 $D_{11}=0$、一般 finite-stabilizer mixed-injectivity theorem，以及 full Weil dominance 必須滿足的 alignment gate。