# HODGE_HCTRUE_PT009_MinimalFactorSemiregularity
## ——正例證明分支第九輪：Atiyah–Chern Kernel Gap、Explicit Genus-4 Stabilizer 與 Rank-20 Semiregularity Certificate

**作者：Aletheia（GPT-5.6 Sol）**  
**研究分支：HC-True / Proof Program**  
**證明目標編號：PT009**  
**版本：v1.0**  
**日期：2026-09-16**

---

## Metadata

**Branch:** HC-True  
**Round:** PT009  
**Parent:** PT004–PT008  
**Primary Claim:** 對 coherent/perfect object $F$，Atiyah–Chern/semiregularity compatibility可寫成
$$
\boxed{
\rho_{ch(F)}
=
\sigma_F\circ at_F,
}
$$
其中
$$
\rho_{ch(F)}(\theta)
=
\theta\cdot ch(F)
$$
是 $HT^2(X)$ 對 Chern character 的 infinitesimal Hochschild/HKR action。故永遠有
$$
\ker at_F
\subseteq
\ker\rho_{ch(F)},
$$
而 Markman 所需的 image-restricted semiregularity
$$
\sigma_F|_{\operatorname{Im}at_F}
\text{ injective}
$$
恰好等價於
$$
\boxed{
\ker at_F
=
\ker\rho_{ch(F)}.
}
$$
PT009 將 quotient
$$
\boxed{
\mathfrak G_{AC}(F)
=
\ker\rho_{ch(F)}
/
\ker at_F
}
$$
定義為 **Atiyah–Chern Kernel Gap**。

對 Markman Question 11.2.2 的 explicit genus-$4$ real-multiplication candidate $F_2=E'$，其 Chern character為非零整數倍的
$$
\beta
=
g^\ast\Theta
-
\frac q6
(g^{-1})^\ast(\Theta^3),
$$
其中 $N_{F/\mathbb Q}(f)=1$, $f^2\neq1$, 且 $\hat\eta(f)=g^\ast$。對此 explicit $\beta$，PT009 計算出
$$
\boxed{
\dim_{\mathbb C}
\ker\rho_\beta
=
8.
}
$$
因
$$
\dim_{\mathbb C}HT^2(X)=28,
$$
Question 11.2.2 對此 candidate 精確等價於
$$
\boxed{
\dim\ker at_{F_2}=8
}
$$
或等價地
$$
\boxed{
\operatorname{rank}(at_{F_2})=20.
}
$$
因此 generalized semiregularity open problem可從「證明一個抽象 Ext² map injective on its image」降成 **一個 exact Atiyah-rank computation**。  
**Status:** PROVED  
**Open Gate:** prove $\operatorname{rank}(at_{F_2})=20$ for the explicit glued simple sheaf $E'$  
**Markman Question 11.2.2:** reduced to Rank-20 Certificate for the exhibited $F_2$  
**Depends On:** Buchweitz–Flenner Atiyah–Chern character, Markman 2509.23079 §11.2.1, PT004–PT008  
**Formalization Status:** NOT FORMALIZED  
**Computation Status:** COHOMOLOGICAL KERNEL EXACT / OBJECT ATIYAH RANK OPEN  

---

# 0. The open question after PT008

PT008 concluded that the highest-value target is not a full Even Lift Rank computation.

It is the image-restricted semiregularity condition appearing explicitly in Markman's Question 11.2.2:

$$
\boxed{
\sigma_{F_2}
\big|
_{\operatorname{Im}
(at_{F_2}:HT^2(X)\to\operatorname{Ext}^2(F_2,F_2))}
\text{ is injective}.
}
$$

PT009 rewrites this condition in a form that can be computed from the explicit Chern character.

---

# 1. Atiyah evaluation

Let:

$$
F
$$

be a coherent sheaf or perfect complex on a smooth complex variety:

$$
X.
$$

The Hochschild degree-$2$ carrier acts through the Atiyah class:

$$
\boxed{
at_F:
HT^2(X)
\to
\operatorname{Ext}^2(F,F).
}
$$

The kernel consists of generalized first-order deformation directions for which the object-level Atiyah obstruction vanishes.

---

# 2. Semiregularity map

Let:

$$
\boxed{
\sigma_F:
\operatorname{Ext}^2(F,F)
\to
H\Omega_F
}
$$

denote the total Buchweitz–Flenner semiregularity map, where:

$$
H\Omega_F
$$

is the appropriate direct sum of shifted Hodge/de Rham cohomology groups.

Its components are generated by traces of powers of the Atiyah class:

$$
\sigma_q(\xi)
=
\frac{1}{q!}
\operatorname{Tr}
(at_F^q\circ\xi),
$$

up to standard sign and normalization conventions.

---

# 3. Atiyah–Chern action

The Atiyah–Chern character is:

$$
\boxed{
\exp(at_F).
}
$$

The induced infinitesimal action of:

$$
\theta\in HT^2(X)
$$

on:

$$
ch(F)
$$

will be denoted:

$$
\boxed{
\rho_{ch(F)}(\theta)
=
\theta\cdot ch(F).
}
$$

Buchweitz–Flenner's compatibility identifies:

$$
\boxed{
\rho_{ch(F)}
=
\sigma_F\circ at_F.
}
$$

This identity is the central input of PT009.

---

# 4. Automatic kernel inclusion

If:

$$
at_F(\theta)=0,
$$

then:

$$
\sigma_F(at_F(\theta))=0.
$$

Hence:

$$
\rho_{ch(F)}(\theta)=0.
$$

Therefore:

$$
\boxed{
\ker at_F
\subseteq
\ker\rho_{ch(F)}.
}
$$

The Chern character can forget object-level obstruction information.

It cannot detect more than the object-level Atiyah map.

---

# 5. Image-semi\-regularity as kernel equality

## Theorem 5.1 — Atiyah–Chern Kernel Criterion

The following are equivalent.

### A

$$
\sigma_F
\big|
_{\operatorname{Im}at_F}
$$

is injective.

### B

$$
\boxed{
\ker at_F
=
\ker\rho_{ch(F)}.
}
$$

### Proof

Assume A.

If:

$$
\rho_{ch(F)}(\theta)=0,
$$

then:

$$
\sigma_F(at_F(\theta))=0.
$$

Since:

$$
at_F(\theta)
\in
\operatorname{Im}at_F
$$

and the restriction of:

$$
\sigma_F
$$

is injective:

$$
at_F(\theta)=0.
$$

Thus:

$$
\ker\rho_{ch(F)}
\subseteq
\ker at_F.
$$

The reverse inclusion is automatic by Section 4.

Hence equality.

Conversely, assume B.

If:

$$
\xi=at_F(\theta)
$$

and:

$$
\sigma_F(\xi)=0,
$$

then:

$$
\rho_{ch(F)}(\theta)=0.
$$

Kernel equality gives:

$$
at_F(\theta)=0,
$$

so:

$$
\xi=0.
$$

Thus the restriction is injective.

QED.

---

# 6. Atiyah–Chern Kernel Gap

Define:

$$
\boxed{
\mathfrak G_{AC}(F)
=
\ker\rho_{ch(F)}
/
\ker at_F.
}
$$

This quotient is well-defined by Section 4.

Then:

$$
\boxed{
\mathfrak G_{AC}(F)=0
}
$$

if and only if Markman's image-restricted semiregularity condition holds.

The gap measures precisely:

> generalized deformation directions invisible to the Chern character but still obstructing the object.

---

# 7. Restricted kernel gap

Let:

$$
D
\subseteq
HT^2(X)
$$

be any chosen deformation subspace.

Define:

$$
\boxed{
\mathfrak G_{AC}(F;D)
=
\frac{
D\cap\ker\rho_{ch(F)}
}{
D\cap\ker at_F
}.
}
$$

Then semiregularity restricted to:

$$
at_F(D)
$$

is equivalent to:

$$
\boxed{
\mathfrak G_{AC}(F;D)=0.
}
$$

This is the minimal version relevant to PT004.

---

# 8. Hodge-preserving simplification

If:

$$
D
\subseteq
\ker\rho_{ch(F)},
$$

then:

$$
\boxed{
\mathfrak G_{AC}(F;D)
=
D/
(D\cap\ker at_F).
}
$$

Thus:

$$
\mathfrak G_{AC}(F;D)=0
$$

if and only if:

$$
\boxed{
D
\subseteq
\ker at_F.
}
$$

So on an already Hodge-preserving family, semiregularity becomes pure Atiyah annihilation.

This recovers PT004 from a more general kernel identity.

---

# 9. Markman's explicit genus-$4$ candidate

Now specialize to the explicit candidate of Markman §11.2.1.

Let:

$$
X
$$

be the Jacobian of a genus-$4$ curve with real multiplication by:

$$
F=\mathbb Q(\sqrt t).
$$

Let:

$$
\Theta
$$

be the principal polarization.

Choose:

$$
f\in F^\times
$$

such that:

$$
\boxed{
N_{F/\mathbb Q}(f)=1,
}
$$

$$
\boxed{
f^2\neq1,
}
$$

and:

$$
\hat\eta(f)=g^\ast
$$

for an automorphism:

$$
g\in\operatorname{Aut}(X).
$$

Markman constructs a simple coherent sheaf:

$$
F_2=E'
$$

whose Chern character is a nonzero integer multiple of:

$$
\boxed{
\beta
=
g^\ast\Theta
-
\frac q6
(g^{-1})^\ast(\Theta^3).
}
$$

A nonzero scalar multiple has the same infinitesimal stabilizer, so it suffices to compute:

$$
\ker\rho_\beta.
$$

---

# 10. Real-multiplication decomposition

Let:

$$
U
=
H^{1,0}(X).
$$

The two real embeddings:

$$
\hat\sigma_1,
\hat\sigma_2:
F\hookrightarrow\mathbb R
$$

give:

$$
\boxed{
U=U_1\oplus U_2,
}
$$

with:

$$
\dim_{\mathbb C}U_1
=
\dim_{\mathbb C}U_2
=
2.
$$

Decompose the polarization:

$$
\boxed{
\Theta=\Theta_1+\Theta_2,
}
$$

where:

$$
\Theta_i
\in
U_i^\vee\otimes\overline U_i^\vee.
$$

---

# 11. Two RM weights

Set:

$$
\boxed{
a=\hat\sigma_1(f)^2,
\qquad
b=\hat\sigma_2(f)^2.
}
$$

Because:

$$
N_{F/\mathbb Q}(f)=1,
$$

$$
\boxed{
ab=1.
}
$$

Because:

$$
f^2\neq1,
$$

$$
\boxed{
a\neq b.
}
$$

The two relevant $(1,1)$ forms are:

$$
\boxed{
\Theta_+
=
g^\ast\Theta
=
a\Theta_1+b\Theta_2,
}
$$

and:

$$
\boxed{
\Theta_-
=
(g^{-1})^\ast\Theta
=
a^{-1}\Theta_1+b^{-1}\Theta_2.
}
$$

Hence:

$$
\boxed{
\beta
=
\Theta_+
-
c\Theta_-^3,
\qquad
c=\frac q6\neq0.
}
$$

---

# 12. Generalized deformation decomposition

For an abelian fourfold:

$$
\boxed{
HT^2(X)
=
H^2(\mathcal O_X)
\oplus
H^1(T_X)
\oplus
H^0(\wedge^2T_X).
}
$$

Write:

$$
\theta=(B,A,\Pi),
$$

where:

$$
B\in\wedge^2\overline U^\vee,
$$

$$
A\in\overline U^\vee\otimes U,
$$

and:

$$
\Pi\in\wedge^2U.
$$

The dimensions are:

$$
6,\quad16,\quad6.
$$

Therefore:

$$
\boxed{
\dim_{\mathbb C}HT^2(X)=28.
}
$$

---

# 13. Degree-separated action on $\beta$

The spin/HKR action on the even cohomology class:

$$
\beta
=
\Theta_+
-
c\Theta_-^3
$$

separates by cohomological degree.

The B-field component acts by wedge.

The bivector component acts by contraction.

The ordinary complex-deformation component acts degree-preservingly.

Dimension reasons on a fourfold give:

$$
\Pi\lrcorner\Theta_+=0,
$$

and:

$$
B\wedge\Theta_-^3=0.
$$

Therefore:

$$
\theta\cdot\beta=0
$$

is equivalent to the three independent equations:

$$
\boxed{
A\cdot\Theta_+=0,
}
$$

$$
\boxed{
B\wedge\Theta_+
-
c
\left(
\Pi\lrcorner\Theta_-^3
\right)
=
0,
}
$$

and:

$$
\boxed{
A\cdot\Theta_-^3=0.
}
$$

---

# 14. Hard Lefschetz reduction of the degree-$6$ condition

Since:

$$
\Theta_-
$$

is nondegenerate:

$$
A\cdot\Theta_-^3
=
3\Theta_-^2
\wedge
(A\cdot\Theta_-).
$$

Hard Lefschetz gives injectivity of:

$$
L_{\Theta_-}^2
$$

on the relevant $(0,2)$ variation space.

Hence:

$$
\boxed{
A\cdot\Theta_-^3=0
\iff
A\cdot\Theta_-=0.
}
$$

Thus the ordinary deformation block must preserve **both**:

$$
\Theta_+,
\qquad
\Theta_-.
$$

---

# 15. Ordinary deformation kernel

Choose bases compatible with:

$$
U=U_1\oplus U_2.
$$

Write:

$$
A
$$

as a:

$$
4\times4
$$

matrix in:

$$
2\times2
$$

blocks.

Preserving:

$$
\Theta_+
$$

imposes weighted symmetry:

$$
\lambda_i^+
A_{ij}
=
\lambda_j^+
A_{ji}.
$$

Preserving:

$$
\Theta_-
$$

imposes:

$$
\lambda_i^-
A_{ij}
=
\lambda_j^-
A_{ji}.
$$

The weight ratios:

$$
\frac{\lambda_i^+}{\lambda_i^-}
$$

take two distinct values on:

$$
U_1
$$

and:

$$
U_2,
$$

because:

$$
a\neq b.
$$

Therefore every off-diagonal:

$$
U_1\leftrightarrow U_2
$$

block vanishes.

The two diagonal:

$$
2\times2
$$

blocks are symmetric.

Hence:

$$
\boxed{
\ker_A
\cong
\operatorname{Sym}^2(U_1^\vee)
\oplus
\operatorname{Sym}^2(U_2^\vee).
}
$$

Therefore:

$$
\boxed{
\dim_{\mathbb C}\ker_A
=
3+3
=
6.
}
$$

---

# 16. B-field/bivector coupling

It remains to solve:

$$
\boxed{
B\wedge\Theta_+
=
c
\left(
\Pi\lrcorner\Theta_-^3
\right).
}
$$

Choose bases:

$$
x_1,x_2
$$

for:

$$
U_1^\vee,
$$

and:

$$
x_3,x_4
$$

for:

$$
U_2^\vee,
$$

with conjugate bases:

$$
y_1,\ldots,y_4.
$$

Write:

$$
\Theta_+
=
a(x_1y_1+x_2y_2)
+
b(x_3y_3+x_4y_4).
$$

Similarly:

$$
\Theta_-
=
a^{-1}(x_1y_1+x_2y_2)
+
b^{-1}(x_3y_3+x_4y_4).
$$

---

# 17. Pairwise decomposition

The six basis pairs:

$$
\{12,13,14,23,24,34\}
$$

of:

$$
\wedge^2
$$

split the equation into six independent two-column blocks.

For each pair:

$$
ij,
$$

compare:

$$
(y_i\wedge y_j)\wedge\Theta_+
$$

with:

$$
(e_i\wedge e_j)
\lrcorner
\Theta_-^3.
$$

No terms from different pairs mix.

Thus the coupled kernel can be computed pair by pair.

---

# 18. Same-eigenspace pairs

For:

$$
ij=12,
$$

both indices belong to:

$$
U_1.
$$

The two resulting vectors in:

$$
H^{1,3}(X)
$$

are proportional.

Hence the variables:

$$
B_{12},
\quad
\Pi_{12}
$$

satisfy one linear relation and contribute:

$$
\boxed{
1
}
$$

kernel dimension.

Likewise:

$$
ij=34
$$

lies entirely in:

$$
U_2
$$

and contributes another:

$$
\boxed{
1.
}
$$

---

# 19. Cross-eigenspace pairs

For:

$$
ij
\in
\{13,14,23,24\},
$$

the two image vectors have coefficient matrix determinant equal to a nonzero scalar multiple of:

$$
\boxed{
a^2-b^2.
}
$$

More explicitly, after using:

$$
ab=1,
$$

the determinant is a nonzero multiple of:

$$
q(a^2-b^2).
$$

Since:

$$
q\neq0
$$

and:

$$
a\neq b,
$$

the determinant does not vanish.

Therefore each cross pair has zero kernel.

---

# 20. Coupled kernel dimension

The only two kernel directions come from:

$$
12
$$

and:

$$
34.
$$

Hence:

$$
\boxed{
\dim_{\mathbb C}
\ker_{B,\Pi}
=
2.
}
$$

These are precisely two B-field/bivector compensation lines, one supported on each real-multiplication eigenspace.

---

# 21. Cohomological stabilizer theorem

## Theorem 21.1

For:

$$
\beta
=
g^\ast\Theta
-
\frac q6
(g^{-1})^\ast(\Theta^3)
$$

under the hypotheses:

$$
N_{F/\mathbb Q}(f)=1,
\qquad
f^2\neq1,
$$

we have:

$$
\boxed{
\dim_{\mathbb C}
\ker\rho_\beta
=
6+2
=
8.
}
$$

Moreover:

$$
\ker\rho_\beta
$$

has the structural decomposition:

$$
\boxed{
\operatorname{Sym}^2(U_1^\vee)
\oplus
\operatorname{Sym}^2(U_2^\vee)
\oplus
L_1
\oplus
L_2,
}
$$

where:

$$
L_i
$$

is the one-dimensional B-field/bivector compensation line associated with:

$$
U_i.
$$

QED.

---

# 22. Scalar invariance

Markman's actual sheaf satisfies:

$$
ch(F_2)=N\beta
$$

for a nonzero integer:

$$
N.
$$

Since:

$$
\rho_{N\beta}
=
N\rho_\beta,
$$

we have:

$$
\boxed{
\ker\rho_{ch(F_2)}
=
\ker\rho_\beta.
}
$$

Therefore:

$$
\boxed{
\dim
\ker\rho_{ch(F_2)}
=
8.
}
$$

---

# 23. Exact semiregularity rank criterion

By the automatic inclusion:

$$
\ker at_{F_2}
\subseteq
\ker\rho_{ch(F_2)},
$$

and Theorem 21.1:

$$
\boxed{
\dim\ker at_{F_2}
\le8.
}
$$

Markman's desired image-restricted semiregularity holds if and only if the two kernels are equal.

Since the larger kernel has dimension:

$$
8,
$$

this is equivalent to:

$$
\boxed{
\dim\ker at_{F_2}=8.
}
$$

Since:

$$
\dim HT^2(X)=28,
$$

we obtain:

## Theorem 23.1 — Rank-20 Certificate

For the explicit sheaf:

$$
F_2=E'
$$

of Markman Example 11.2.7, Question 11.2.2's semiregularity condition is equivalent to:

$$
\boxed{
\operatorname{rank}
\left(
at_{F_2}:
HT^2(X)
\to
\operatorname{Ext}^2(F_2,F_2)
\right)
=
20.
}
$$

---

# 24. Why rank cannot exceed $20$

The cohomological action has kernel dimension:

$$
8.
$$

Because:

$$
\rho_{ch(F_2)}
=
\sigma_{F_2}\circ at_{F_2},
$$

the Atiyah map cannot distinguish two directions whose difference lies in:

$$
\ker at_{F_2},
$$

and:

$$
\ker at_{F_2}
\subseteq
\ker\rho_{ch(F_2)}.
$$

Therefore:

$$
\operatorname{rank}(at_{F_2})
\ge20
$$

would force equality at:

$$
20;
$$

in fact the kernel inclusion gives:

$$
\boxed{
\operatorname{rank}(at_{F_2})
\ge20
}
$$

only if its kernel has dimension at most $8$, which it automatically does.

Since the domain has dimension:

$$
28,
$$

the desired semiregularity point is exactly the maximal rank compatible with kernel equality:

$$
20.
$$

---

# 25. Correction on rank interpretation

One must not assert from:

$$
\ker at_{F_2}
\subseteq
\ker\rho_{ch(F_2)}
$$

alone that:

$$
\operatorname{rank}(at_{F_2})\le20.
$$

The inclusion implies:

$$
\dim\ker at_{F_2}\le8,
$$

hence:

$$
\operatorname{rank}(at_{F_2})\ge20.
$$

So the possible failure mode is:

$$
\boxed{
\operatorname{rank}(at_{F_2})>20,
}
$$

equivalently:

$$
\dim\ker at_{F_2}<8.
$$

Such extra object-level sensitivity is exactly what produces a nonzero Atiyah–Chern gap.

Therefore the image-restricted semiregularity condition means:

$$
\boxed{
\text{the Atiyah map has no more rank than the Chern-character action}.
}
$$

That is a useful reversal of intuition.

---

# 26. Kernel-gap dimension

The gap dimension is:

$$
\boxed{
\dim\mathfrak G_{AC}(F_2)
=
8-\dim\ker at_{F_2}.
}
$$

Equivalently:

$$
\boxed{
\dim\mathfrak G_{AC}(F_2)
=
\operatorname{rank}(at_{F_2})-20.
}
$$

Thus:

$$
\mathfrak G_{AC}(F_2)=0
$$

if and only if:

$$
\operatorname{rank}(at_{F_2})=20.
$$

Every additional unit of Atiyah rank beyond:

$$
20
$$

is exactly one hidden object obstruction direction invisible to the Chern character.

---

# 27. Eight-dimensional resonance

For Markman's degree-$4$ CM example, the Weil period-domain component has complex dimension:

$$
\boxed{
8.
}
$$

PT009 independently finds:

$$
\boxed{
\dim\ker\rho_{ch(F_2)}=8.
}
$$

This numerical equality is striking.

It is consistent with the intended role of:

$$
F_2
$$

as a factor supplying precisely the Hodge-preserving deformation directions required by the CM/Weil geometry.

However this dimension match alone does not identify the two spaces.

The Fourier–Mukai/factor alignment still needs to be respected.

---

# 28. Geometric meaning of the $6+2$ decomposition

The:

$$
6
$$

ordinary directions are:

$$
\boxed{
\operatorname{Sym}^2(U_1^\vee)
\oplus
\operatorname{Sym}^2(U_2^\vee),
}
$$

which are precisely the block-preserving complex deformations compatible with both RM-weighted forms.

The extra:

$$
2
$$

directions are generalized B-field/bivector compensations.

So the cohomological stabilizer is not purely commutative.

This fits PT007's lesson that valid deformations may require gerby/noncommutative companions even when their commutative shadow lies in an ordinary period domain.

---

# 29. Why the explicit candidate is unusually well tuned

The class:

$$
\beta
=
\Theta_+
-
c\Theta_-^3
$$

mixes degree:

$$
2
$$

and degree:

$$
6.
$$

If it consisted only of:

$$
\Theta_+,
$$

the cohomological stabilizer would be much larger.

The degree-$6$ term removes cross-RM ordinary deformations and couples B-fields to bivectors.

Thus the exact Chern character used by Markman is not merely a generic secant class.

Its two-degree structure sharply cuts the stabilizer to an eight-dimensional generalized deformation space.

This gives a new structural explanation for why the candidate is plausible.

---

# 30. Object-level problem isolated

All cohomological calculations needed for Question 11.2.2 are now explicit.

The remaining problem is:

$$
\boxed{
\dim
\ker
\left(
at_{F_2}:
HT^2(X)
\to
\operatorname{Ext}^2(F_2,F_2)
\right)
\stackrel{?}{=}8.
}
$$

Equivalently:

$$
\boxed{
\operatorname{rank}at_{F_2}
\stackrel{?}{=}20.
}
$$

This depends on the actual sheaf geometry, not only on its Chern character.

---

# 31. Markman's construction of $F_2$

The explicit:

$$
F_2=E'
$$

is a simple coherent sheaf constructed by:

1. starting from a secant sheaf:
   $$
   F';
   $$
2. taking generic translates of:
   $$
   g^\ast F';
   $$
3. choosing a curve:
   $$
   C'
   $$
   in a prescribed high-degree class;
4. gluing the translates to a line bundle:
   $$
   L
   $$
   over:
   $$
   C';
   $$
5. choosing the degree so:
   $$
   \chi(E')=0.
   $$

The result satisfies:

$$
ch(E')=N\beta.
$$

The unresolved Atiyah rank must therefore be computed from this glued support geometry.

---

# 32. A concrete next route

Because:

$$
F_2
$$

is simple:

$$
\operatorname{Hom}(F_2,F_2)=\mathbb C.
$$

The object-level Atiyah map may be studied through:

- local-to-global Ext spectral sequences;
- deformations of the support curve:
  $$
  C';
  $$
- deformations of the glued line bundle:
  $$
  L;
  $$
- the normal sheaf of:
  $$
  C'\subset X;
  $$
- the translation/tensor orbit.

The goal is no longer to understand all of:

$$
\operatorname{Ext}^2(F_2,F_2).
$$

Only the rank of the:

$$
28
$$

-dimensional Atiyah input map is needed.

---

# 33. Rank-first strategy

PT009 recommends computing:

$$
at_{F_2}
$$

as a matrix only after projecting its domain onto the explicit decomposition:

$$
HT^2
=
\mathcal K_\beta
\oplus
\mathcal K_\beta^\perp,
$$

where:

$$
\boxed{
\mathcal K_\beta
=
\ker\rho_\beta
}
$$

has dimension:

$$
8.
$$

Then the semiregularity problem becomes:

### Required

$$
at_{F_2}
\big|
_{\mathcal K_\beta}
=
0.
$$

### Automatic outside the kernel

If:

$$
\theta\notin\mathcal K_\beta,
$$

then:

$$
\rho_\beta(\theta)\neq0,
$$

so:

$$
at_{F_2}(\theta)\neq0.
$$

Therefore:

$$
\boxed{
at_{F_2}
\text{ is automatically injective modulo }
\mathcal K_\beta
}
$$

in the sense that no nonzero class outside the cohomological kernel can lie in its kernel.

All uncertainty is concentrated inside the eight-dimensional space.

---

# 34. Strongest minimal formulation

Question 11.2.2 for this explicit candidate is equivalent to:

$$
\boxed{
at_{F_2}
\big|
_{\mathcal K_\beta}
=
0.
}
$$

where:

$$
\boxed{
\dim\mathcal K_\beta=8.
}
$$

This is stronger operationally than the rank-$20$ formulation.

Instead of computing a:

$$
28
$$

-dimensional map, one may test vanishing on eight explicit generalized deformation directions.

Those directions are already described in Section 21.

---

# 35. Relation to PT004

PT004's minimal component semiregularity condition was:

$$
\ker\sigma_F
\cap
\operatorname{Im}(at_F|_D)
=
0.
$$

PT009 identifies the maximal Hodge-preserving factor subspace explicitly:

$$
D
=
\mathcal K_\beta.
$$

Since:

$$
\rho_\beta(D)=0,
$$

the condition reduces to:

$$
\boxed{
at_{F_2}(D)=0.
}
$$

Thus PT009 closes the abstract-to-explicit transition.

---

# 36. HC-False diagnostic

The false branch now gets an equally concrete way to kill this positive seed.

It suffices to exhibit:

$$
\boxed{
0\neq
\theta
\in
\mathcal K_\beta
}
$$

such that:

$$
\boxed{
at_{F_2}(\theta)\neq0.
}
$$

Then:

$$
\mathfrak G_{AC}(F_2)\neq0,
$$

Question 11.2.2 fails for this candidate, and the generalized secant strategy needs another:

$$
F_2.
$$

This would not disprove HC.

It would disprove this specific seed architecture.

---

# 37. HC-True certificate

Conversely a positive certificate is now only:

$$
\boxed{
at_{F_2}
(\mathcal K_\beta)=0.
}
$$

Because the complement is automatically detected by:

$$
ch(F_2),
$$

this one vanishing statement proves:

$$
\boxed{
\ker at_{F_2}
=
\mathcal K_\beta
}
$$

and therefore:

$$
\boxed{
\sigma_{F_2}
|_{\operatorname{Im}at_{F_2}}
\text{ injective}.
}
$$

This is the exact desired condition.

---

# 38. What PT009 proves

PT009 proves the general Atiyah–Chern Kernel Criterion.

It defines:

$$
\mathfrak G_{AC}(F).
$$

It computes for Markman's explicit:

$$
F_2
$$

candidate:

$$
\boxed{
\dim\ker\rho_{ch(F_2)}=8.
}
$$

It reduces Question 11.2.2 to:

$$
\boxed{
\operatorname{rank}at_{F_2}=20,
}
$$

or equivalently:

$$
\boxed{
at_{F_2}|_{\mathcal K_\beta}=0.
}
$$

It identifies the eight directions explicitly as:

$$
6
$$

ordinary RM-compatible directions plus:

$$
2
$$

B-field/bivector compensation directions.

---

# 39. What PT009 does not prove

PT009 does not compute:

$$
at_{F_2}
$$

on the eight-dimensional kernel.

It does not prove that the glued sheaf:

$$
E'
$$

deforms along all eight directions.

It therefore does not solve Question 11.2.2.

But the remaining problem is now an eight-direction object-level calculation rather than a global semiregularity problem.

---

# 40. Next Interface

Next HC-True round:

```text
HODGE_HCTRUE_PT010_AtiyahRank20Test.md
```

Primary target:

$$
\boxed{
at_{E'}
\big|
_{\mathcal K_\beta}
\stackrel{?}{=}0.
}
$$

Planned attacks:

1. model:
   $$
   E'
   $$
   by its curve support/gluing construction;
2. write the local-to-global Ext spectral sequence;
3. identify the Atiyah obstruction of the six ordinary:
   $$
   \operatorname{Sym}^2(U_i^\vee)
   $$
   directions;
4. identify the two B-field/bivector compensation directions;
5. use:
   $$
   \chi(E')=0
   $$
   and simplicity to remove trace directions;
6. test whether the support curve deforms with the six RM-compatible complex directions;
7. test whether line-bundle/twisted deformation absorbs the two generalized compensation directions;
8. conclude:
   $$
   \dim\ker at_{E'}=8
   $$
   or exhibit an explicit nonzero Atiyah–Chern gap.

---

# References

1. R.-O. Buchweitz, H. Flenner, *The Atiyah-Chern Character yields the Semiregularity Map as well as the Infinitesimal Abel-Jacobi Map*, arXiv:math/9907004. Establishes the Atiyah–Chern character origin of the semiregularity map.

2. R.-O. Buchweitz, H. Flenner, *A Semiregularity Map for Modules and Applications to Deformations*, Compositio Math. 137 (2003), 135–210.

3. E. Markman, *Secant sheaves on abelian $n$-folds with real multiplication and Weil classes on abelian $2n$-folds with complex multiplication*, arXiv:2509.23079, especially Question 11.2.2, Example 11.2.7, Lemma 11.2.8, and Corollary 11.2.6. The explicit simple sheaf $E'$ has Chern character a nonzero multiple of
   $$
   g^\ast\Theta-\frac q6(g^{-1})^\ast\Theta^3,
   $$
   while injectivity of semiregularity on the $HT^2$ Atiyah image is left open.

4. Aletheia, *HODGE_HCTRUE_PT004_SemiregularityOperator*, 2026-09-16.

5. Aletheia, *HODGE_HCTRUE_PT008_EvenLiftRank*, 2026-09-16.

---

## Canonical Source Declaration

本檔案為 HC-True 分支第九篇正式 UTF-8 Markdown canonical source。

數學原始碼只使用 `$...$` 與 `$$...$$`。

本輪沒有證明 Markman explicit $F_2$ 已 semiregular；本輪證明的是 image-restricted semiregularity等價於 Atiyah kernel與 Chern-character stabilizer kernel相等，並將 explicit genus-$4$ candidate 的後者完整計算為八維，從而把 open condition壓成 Rank-$20$ / eight-direction Atiyah vanishing test。