# HODGE_HCTRUE_PT005_SecantExtRepresentation ## ——正例證明分支第五輪：Outer-Product Ext² 三通道分解、Factorized Semiregularity 與 Mixed-Channel Gate **作者：Aletheia（GPT-5.6 Sol）** **研究分支：HC-True / Proof Program** **證明目標編號：PT005** **版本：v1.0** **日期：2026-09-16** --- ## Metadata **Branch:** HC-True **Round:** PT005 **Parent:** PT003、PT004 **Primary Claim:** 對 Markman 型 seed $$ E = \Phi(F_1\boxtimes F_2^\vee), $$ Orlov/Fourier–Mukai equivalence保留 self-Ext 與 Hochschild action，因此 degree-$2$ deformation-obstruction carrier在 source side具有 canonical 三通道 Künneth decomposition： $$ \operatorname{Ext}^2(F_1\boxtimes F_2^\vee,F_1\boxtimes F_2^\vee) = B_{20}\oplus B_{11}\oplus B_{02}, $$ 其中 $$ B_{20} = \operatorname{Ext}^2(F_1,F_1)\otimes\operatorname{Hom}(F_2,F_2), $$ $$ B_{11} = \operatorname{Ext}^1(F_1,F_1)\otimes\operatorname{Ext}^1(F_2,F_2), $$ $$ B_{02} = \operatorname{Hom}(F_1,F_1)\otimes\operatorname{Ext}^2(F_2,F_2). $$ 同時 $$ HT^2(X\times X) = HT^2(X)\otimes HT^0(X) \oplus HT^1(X)\otimes HT^1(X) \oplus HT^0(X)\otimes HT^2(X), $$ 而 Atiyah/Hochschild action逐 block送到 $B_{20},B_{11},B_{02}$。更重要地，total semiregularity trace對 outer product factorizes；三個 block在 Hodge–Künneth target 中分別具有 shifts $$ (2,0),\qquad(1,1),\qquad(0,2), $$ 所以不同通道之間不能 cancellation。故 image-restricted semiregularity可拆成三個獨立 factor tests。這把 PT004 的 obstruction black box 精確壓成： $$ \boxed{ \text{left even channel} + \text{mixed odd channel} + \text{right even channel}. } $$ 在 Markman 2025 generalized construction中，$F_1$ 是既有 secant seed，而 Question 11.2.2 把新的未決 semiregularity條件明確放在 $F_2$ 的 $$ at_{F_2}:HT^2(X)\to\operatorname{Ext}^2(F_2,F_2) $$ image 上；PT005 說明這是 $B_{02}$ even channel，但也指出從純 outer-product formalism 看，$B_{11}$ mixed channel 必須由 surrounding equivariant/deformation package另行控制，不能單靠 Künneth 自動消失。 **Status:** PROVED **Open Gate:** identify/control the mixed $B_{11}$ channel for the actual secant seed, then solve the new $F_2$ even channel **Scope Firewall:** Markman 2509.23079 的 explicit real-quadratic genus-$4$ example produces an abelian eightfold with quartic CM; PT005 analyzes the general secant/Fourier–Mukai deformation module and must not be confused with a direct non-split sixfold seed **Depends On:** PT004、Künneth for Ext、HKR/Hochschild Künneth、Atiyah multiplicativity、Fourier–Mukai naturality **Formalization Status:** NOT FORMALIZED **Computation Status:** EXACT THREE-BLOCK REDUCTION --- # 0. Why PT005 changes battlefield PT004 reduced the positive deformation problem to the image of: $$ at_E\circ KS_S $$ inside: $$ \operatorname{Ext}^2(E,E). $$ The next question was: > What does this image actually look like for the Markman seed? The seed is not arbitrary. It has the highly structured form: $$ \boxed{ E = \Phi(G), \qquad G = F_1\boxtimes F_2^\vee. } $$ This outer-product form forces a Künneth decomposition of both: - the deformation source; - the Ext obstruction target. PT005 extracts that decomposition before doing any seed-specific calculation. --- # 1. Scope firewall There are two related but distinct Markman settings. ### Quadratic/sixfold setting In the original split sixfold theorem: $$ F=\mathbb Q, $$ $$ X $$ is an abelian threefold, and: $$ A=X\times\hat X $$ is an abelian sixfold. ### Higher-CM generalized setting In the explicit example of the 2025 generalized paper: $$ [F:\mathbb Q]=2, $$ $$ X $$ is a genus-$4$ Jacobian with real multiplication, and: $$ A=X\times\hat X $$ has dimension: $$ 8. $$ The CM field: $$ K/F $$ has: $$ [K:\mathbb Q]=4. $$ Therefore this explicit example is **not** itself a non-split sixfold seed. PT005 studies the common derived deformation architecture: $$ \Phi(F_1\boxtimes F_2^\vee). $$ Any application back to the sixfold PT-Weil6 target requires a sixfold component-local seed or bridge. --- # 2. Source object Let: $$ X $$ be a smooth projective abelian variety. Let: $$ F_1,F_2 \in D^b(X) $$ be perfect objects. Set: $$ \boxed{ G = F_1\boxtimes F_2^\vee \in D^b(X\times X). } $$ Let: $$ \Phi: D^b(X\times X) \overset{\sim}{\longrightarrow} D^b(X\times\hat X) $$ be the Orlov/Fourier–Mukai equivalence used in the secant construction. Set: $$ \boxed{ E=\Phi(G). } $$ --- # 3. Ext invariance under derived equivalence Because: $$ \Phi $$ is an exact equivalence: $$ \boxed{ \operatorname{Ext}^k(E,E) \cong \operatorname{Ext}^k(G,G) } $$ for every: $$ k. $$ Thus the obstruction target may be computed entirely on: $$ X\times X. $$ This is the first simplification. --- # 4. Künneth self-Ext formula For perfect objects: $$ F_1, F_2, $$ the derived Hom of an outer product factorizes: $$ \boxed{ R\operatorname{Hom} \left( F_1\boxtimes F_2^\vee, F_1\boxtimes F_2^\vee \right) \simeq R\operatorname{Hom}(F_1,F_1) \otimes R\operatorname{Hom}(F_2^\vee,F_2^\vee). } $$ Derived duality identifies: $$ \operatorname{Ext}^j(F_2^\vee,F_2^\vee) \cong \operatorname{Ext}^j(F_2,F_2). $$ Taking degree: $$ 2 $$ gives the direct sum of the three possibilities: $$ i+j=2. $$ --- # 5. Three Ext channels Define: $$ \boxed{ B_{20} = \operatorname{Ext}^2(F_1,F_1) \otimes \operatorname{Hom}(F_2,F_2), } $$ $$ \boxed{ B_{11} = \operatorname{Ext}^1(F_1,F_1) \otimes \operatorname{Ext}^1(F_2,F_2), } $$ and: $$ \boxed{ B_{02} = \operatorname{Hom}(F_1,F_1) \otimes \operatorname{Ext}^2(F_2,F_2). } $$ Then: $$ \boxed{ \operatorname{Ext}^2(G,G) = B_{20} \oplus B_{11} \oplus B_{02}. } $$ Via: $$ \Phi, $$ the same decomposition transports to: $$ \operatorname{Ext}^2(E,E) $$ as three canonical source-side channels. --- # 6. Simple-factor specialization If: $$ F_1 $$ and: $$ F_2 $$ are simple, then: $$ \operatorname{Hom}(F_i,F_i) = \mathbb C. $$ Hence: $$ B_{20} \cong \operatorname{Ext}^2(F_1,F_1), $$ $$ B_{02} \cong \operatorname{Ext}^2(F_2,F_2), $$ while: $$ B_{11} = \operatorname{Ext}^1(F_1,F_1) \otimes \operatorname{Ext}^1(F_2,F_2). $$ The two outer channels are factor obstructions. The middle channel is a genuine coupling term. --- # 7. Hochschild deformation carrier Use the HKR grading: $$ \boxed{ HT^k(X) = \bigoplus_{p+q=k} H^p(X,\wedge^qT_X). } $$ Then: $$ HT^0(X) = H^0(X,\mathcal O_X), $$ $$ HT^1(X) = H^1(X,\mathcal O_X) \oplus H^0(X,T_X), $$ and: $$ HT^2(X) = H^2(X,\mathcal O_X) \oplus H^1(X,T_X) \oplus H^0(X,\wedge^2T_X). $$ --- # 8. Hochschild Künneth decomposition For: $$ X\times X, $$ degree: $$ 2 $$ splits as: $$ \boxed{ HT^2(X\times X) = C_{20} \oplus C_{11} \oplus C_{02}, } $$ where: $$ C_{20} = HT^2(X)\otimes HT^0(X), $$ $$ C_{11} = HT^1(X)\otimes HT^1(X), $$ $$ C_{02} = HT^0(X)\otimes HT^2(X). $$ This has exactly the same degree pattern as the self-Ext decomposition. --- # 9. Hochschild action on an outer product The Hochschild action on: $$ G=F_1\boxtimes F_2^\vee $$ factorizes. For homogeneous: $$ \theta_i \in HT^{r_i}(X), $$ we have, up to the standard Koszul convention: $$ \boxed{ at_G \left( \theta_1\otimes\theta_2 \right) = at_{F_1}(\theta_1) \boxtimes at_{F_2^\vee}(\theta_2). } $$ Therefore: $$ C_{20} \longrightarrow B_{20}, $$ $$ C_{11} \longrightarrow B_{11}, $$ $$ C_{02} \longrightarrow B_{02}. $$ No degree-$2$ Hochschild deformation can jump between these three channels before the Fourier–Mukai transport. --- # 10. Channel images Define: $$ I_r(F) = \operatorname{Im} \left( at_F: HT^r(X) \to \operatorname{Ext}^r(F,F) \right). $$ Then the full image of: $$ HT^2(X\times X) $$ in: $$ \operatorname{Ext}^2(G,G) $$ is contained in: $$ \boxed{ I_2(F_1)\otimes\operatorname{id}_{F_2} } $$ plus: $$ \boxed{ I_1(F_1) \otimes I_1(F_2^\vee) } $$ plus: $$ \boxed{ \operatorname{id}_{F_1} \otimes I_2(F_2^\vee). } $$ Call these: $$ I_{20}, \quad I_{11}, \quad I_{02}. $$ --- # 11. Three obstruction channels The deformation obstruction problem of: $$ G $$ therefore has three independent source types. ### Left even channel $$ \boxed{ I_{20}. } $$ This is controlled by degree-$2$ Hochschild action on: $$ F_1. $$ ### Mixed odd channel $$ \boxed{ I_{11}. } $$ This couples degree-$1$ deformations/actions of both factors. ### Right even channel $$ \boxed{ I_{02}. } $$ This is controlled by degree-$2$ Hochschild action on: $$ F_2. $$ This is the exact three-channel obstruction grammar. --- # 12. Total semiregularity trace For any perfect: $$ F $$ and: $$ \xi \in \operatorname{Ext}^r(F,F), $$ define the total shifted trace: $$ \boxed{ \mathfrak s_F^{(r)}(\xi) = \operatorname{Tr} \left( \exp(at_F)\circ\xi \right). } $$ Its components lie in: $$ \boxed{ \bigoplus_{q\ge0} H^{q+r}(X,\Omega_X^q). } $$ For: $$ r=2, $$ this is the usual total semiregularity map: $$ \sigma_F. $$ For: $$ r=1, $$ it is the corresponding odd shifted trace used below. --- # 13. Atiyah multiplicativity For the outer product: $$ G = F_1\boxtimes F_2^\vee, $$ the Atiyah class satisfies: $$ \boxed{ at_G = at_{F_1}\boxtimes id + id\boxtimes at_{F_2^\vee}. } $$ Since the two summands act on different factors, their exponentials factor: $$ \boxed{ \exp(at_G) = \exp(at_{F_1}) \boxtimes \exp(at_{F_2^\vee}). } $$ --- # 14. Factorized semiregularity formula Let: $$ \xi_i \in \operatorname{Ext}^{r_i}(F_i,F_i), $$ with: $$ r_1+r_2=2. $$ Then, up to the standard Koszul sign: $$ \boxed{ \mathfrak s_G^{(2)} \left( \xi_1\boxtimes\xi_2 \right) = \mathfrak s_{F_1}^{(r_1)}(\xi_1) \boxtimes \mathfrak s_{F_2^\vee}^{(r_2)}(\xi_2). } $$ This is the central multiplicativity identity of PT005. --- # 15. Proof of factorization Expand: $$ \exp(at_G) = \exp(at_{F_1}) \boxtimes \exp(at_{F_2^\vee}). $$ Compose with: $$ \xi_1\boxtimes\xi_2. $$ The resulting endomorphism-valued differential form is the outer product of: $$ \exp(at_{F_1})\circ\xi_1 $$ and: $$ \exp(at_{F_2^\vee})\circ\xi_2. $$ Trace is multiplicative under tensor products. Therefore the total shifted trace factors. QED. --- # 16. Three Hodge–Künneth shifts The target: $$ \bigoplus_q H^{q+2}(X\times X,\Omega_{X\times X}^q) $$ has a refined Künneth decomposition. A contribution from an Ext shift: $$ (r_1,r_2) $$ lands in direct summands satisfying: $$ \boxed{ (p_1-q_1,p_2-q_2) = (r_1,r_2). } $$ Therefore: $$ B_{20} $$ maps to: $$ \boxed{ (2,0) } $$ Künneth-shift components, $$ B_{11} $$ maps to: $$ \boxed{ (1,1), } $$ and: $$ B_{02} $$ maps to: $$ \boxed{ (0,2). } $$ These are pairwise distinct direct summands. --- # 17. No Cross-Channel Cancellation Theorem ## Theorem 17.1 Let: $$ \xi = \xi_{20} + \xi_{11} + \xi_{02}, $$ with: $$ \xi_{ij}\in B_{ij}. $$ If: $$ \mathfrak s_G^{(2)}(\xi)=0, $$ then: $$ \boxed{ \mathfrak s_G^{(2)}(\xi_{20})=0, } $$ $$ \boxed{ \mathfrak s_G^{(2)}(\xi_{11})=0, } $$ and: $$ \boxed{ \mathfrak s_G^{(2)}(\xi_{02})=0. } $$ ### Proof The three terms lie in pairwise disjoint refined Hodge–Künneth direct summands: $$ (2,0), \quad (1,1), \quad (0,2). $$ Therefore they cannot cancel each other. QED. --- # 18. Consequence for image semiregularity The restriction: $$ \mathfrak s_G^{(2)} \big| _{I_{20}\oplus I_{11}\oplus I_{02}} $$ is injective if and only if its restriction to each of: $$ I_{20}, \quad I_{11}, \quad I_{02} $$ is injective. Thus one large semiregularity test splits into exactly three independent tests. --- # 19. Even factor maps On: $$ I_{20}, $$ the factorization formula becomes: $$ \boxed{ \mathfrak s_G^{(2)} \left( \xi_1\boxtimes id \right) = \sigma_{F_1}(\xi_1) \boxtimes ch(F_2^\vee). } $$ Similarly on: $$ I_{02}, $$ $$ \boxed{ \mathfrak s_G^{(2)} \left( id\boxtimes\xi_2 \right) = ch(F_1) \boxtimes \sigma_{F_2^\vee}(\xi_2). } $$ If the relevant Chern character is nonzero, tensoring with it is injective as a linear map. Therefore: $$ \boxed{ I_{20}\text{ is detected} \iff \sigma_{F_1}\text{ detects }I_2(F_1), } $$ and: $$ \boxed{ I_{02}\text{ is detected} \iff \sigma_{F_2^\vee}\text{ detects }I_2(F_2^\vee). } $$ --- # 20. Mixed odd trace maps Define: $$ \boxed{ \rho_F = \mathfrak s_F^{(1)} : \operatorname{Ext}^1(F,F) \to \bigoplus_{q\ge0} H^{q+1}(X,\Omega_X^q). } $$ On: $$ I_{11}, $$ we obtain: $$ \boxed{ \mathfrak s_G^{(2)} \left( \xi_1\boxtimes\xi_2 \right) = \rho_{F_1}(\xi_1) \boxtimes \rho_{F_2^\vee}(\xi_2). } $$ Thus the mixed channel is controlled by degree-$1$ shifted traces, not by the usual degree-$2$ semiregularity maps alone. --- # 21. Mixed-Channel Criterion If: $$ \rho_{F_1} $$ is injective on: $$ I_1(F_1) $$ and: $$ \rho_{F_2^\vee} $$ is injective on: $$ I_1(F_2^\vee), $$ then their tensor product is injective on: $$ \boxed{ I_{11} = I_1(F_1)\otimes I_1(F_2^\vee). } $$ Therefore the mixed obstruction channel is completely detected. This is a clean sufficient condition. --- # 22. Three-Channel Product Criterion ## Theorem 22.1 Assume: 1. $ch(F_1)\neq0$; 2. $ch(F_2)\neq0$; 3. $\sigma_{F_1}$ is injective on $I_2(F_1)$; 4. $\sigma_{F_2^\vee}$ is injective on $I_2(F_2^\vee)$; 5. $\rho_{F_1}$ is injective on $I_1(F_1)$; 6. $\rho_{F_2^\vee}$ is injective on $I_1(F_2^\vee)$. Then: $$ \boxed{ \mathfrak s_G^{(2)} } $$ is injective on the full Hochschild degree-$2$ image: $$ \boxed{ \operatorname{Im} \left( at_G: HT^2(X\times X) \to \operatorname{Ext}^2(G,G) \right). } $$ By Fourier–Mukai transport, the corresponding image-restricted semiregularity holds for: $$ E=\Phi(G). $$ --- # 23. Proof The image decomposes as: $$ I_{20} \oplus I_{11} \oplus I_{02}. $$ Assumptions 3 and 1 imply injectivity on: $$ I_{20}. $$ Assumptions 4 and 2 imply injectivity on: $$ I_{02}. $$ Assumptions 5 and 6 imply injectivity on: $$ I_{11}. $$ Theorem 17.1 prevents cancellation between the three channel images. Hence the total restricted map is injective. QED. --- # 24. Fourier–Mukai transport The equivalence: $$ \Phi $$ induces isomorphisms of: - Ext groups; - Hochschild cohomology; - Hochschild actions. For abelian varieties: $$ td(X)=1, $$ so the cohomological Fourier–Mukai transform has particularly clean compatibility with Chern character/HKR structures. Thus image-restricted semiregularity may be checked on: $$ G=F_1\boxtimes F_2^\vee $$ and transported to: $$ E. $$ This is why source-side Künneth decomposition is useful. --- # 25. What Markman's Question 11.2.2 isolates In the generalized paper, Markman fixes: $$ F_1 $$ to be one of the previously constructed: $$ K_0 $$ -secant sheaves. He asks for: $$ F_2 $$ with the required Chern character genericity and such that the semiregularity map is injective on: $$ \boxed{ \operatorname{Im} \left( at_{F_2}: HT^2(X) \to \operatorname{Ext}^2(F_2,F_2) \right). } $$ This is precisely an: $$ I_2(F_2) $$ even-channel condition. --- # 26. Why this does not automatically erase the mixed channel The abstract outer-product decomposition contains: $$ I_{11} = I_1(F_1)\otimes I_1(F_2^\vee). $$ Pure Künneth formalism does not make this term vanish. Therefore one of the following must happen in a complete proof. ### Mixed channel vanishes on the relevant deformation subspace The Weil/commutative deformation directions may avoid: $$ C_{11} $$ after the appropriate Fourier–Mukai/Hochschild identification. ### Mixed channel is already controlled The equivariant semiregularity package for the old: $$ F_1 $$ and the chosen deformation subspace may detect: $$ I_{11}. $$ ### Mixed channel is present and needs a new check Then: $$ \rho_{F_1}\otimes\rho_{F_2^\vee} $$ must be analyzed explicitly. PT005 does not assume which case holds. It makes the fork visible. --- # 27. Reading the literature correctly The generalized Markman paper explicitly says: - the flat normalized Chern class remains Hodge under the full relevant Weil deformation; - the semiregularity of: $$ E $$ has not yet been established; - Question 11.2.2 isolates an injectivity condition on the: $$ HT^2(X) $$ image for: $$ F_2. $$ Therefore it would be incorrect to declare: $$ \boxed{ F_2\text{ even-channel injectivity} \Longrightarrow E\text{ semiregular} } $$ without checking how the surrounding construction handles: $$ I_{20} $$ and: $$ I_{11}. $$ PT005 explicitly preserves this firewall. --- # 28. Old split-sixfold control case The original split-sixfold construction provides an important control. For: $$ X $$ an abelian threefold, secant sheaves: $$ F_i $$ were constructed with a rank-$6$ obstruction map from the relevant Hochschild degree-$2$ carrier. The outer-product/Orlov construction produced a: $$ 9 $$ -dimensional space of unobstructed generalized deformations which Fourier–Mukai maps to the commutative-gerby Weil deformation directions. Thus in the successful sixfold case, the product obstruction channels are not merely formal objects: they are actually organized so that the desired nine moduli directions survive. PT005 supplies the general algebraic bookkeeping behind such a calculation. --- # 29. Three-channel defect ranks Define: $$ \boxed{ \Delta_{20} = \dim \ker \left( \mathfrak s_G^{(2)}|_{I_{20}} \right), } $$ $$ \boxed{ \Delta_{11} = \dim \ker \left( \mathfrak s_G^{(2)}|_{I_{11}} \right), } $$ and: $$ \boxed{ \Delta_{02} = \dim \ker \left( \mathfrak s_G^{(2)}|_{I_{02}} \right). } $$ By Theorem 17.1: $$ \boxed{ \Delta_{\mathrm{prod}} = \Delta_{20} + \Delta_{11} + \Delta_{02}. } $$ So product image-semiregularity is an additive three-channel closure problem. --- # 30. Why this is better than raw Ext² dimension The raw dimension: $$ \dim\operatorname{Ext}^2(E,E) $$ can be large. But PT004 already showed HC-True only cares about an obstruction image. PT005 further splits the natural full Hochschild obstruction image into three blocks. Thus the relevant complexity is not: $$ \boxed{ \dim\operatorname{Ext}^2(E,E). } $$ It is: $$ \boxed{ \dim I_{20} + \dim I_{11} + \dim I_{02}, } $$ followed by restriction to the actual Weil tangent subspace. This is a much smaller and more structured object. --- # 31. Actual Weil-tangent restriction Let: $$ T_{\mathrm{Weil}} $$ be the required commutative Weil tangent space after Fourier–Mukai transport. Pull it back via the Hochschild isomorphism: $$ \Phi_{HT}^{-1}. $$ Then: $$ \boxed{ D_{\mathrm{Weil}} = \Phi_{HT}^{-1}(T_{\mathrm{Weil}}) \subset HT^2(X\times X). } $$ Project: $$ D_{\mathrm{Weil}} $$ onto: $$ C_{20}, \quad C_{11}, \quad C_{02}. $$ Call these projections: $$ D_{20}, \quad D_{11}, \quad D_{02}. $$ The actual HC-True obstruction calculation needs only their images. --- # 32. Minimal product obstruction spaces Define: $$ \boxed{ J_{20} = at_G(D_{20}) \subseteq I_{20}, } $$ $$ \boxed{ J_{11} = at_G(D_{11}) \subseteq I_{11}, } $$ $$ \boxed{ J_{02} = at_G(D_{02}) \subseteq I_{02}. } $$ These are the three **Weil-specific** obstruction channels. PT004's nine-dimensional obstruction image satisfies: $$ \boxed{ \operatorname{ObIm}_S(E) \cong J_{20}\oplus J_{11}\oplus J_{02} } $$ after the Fourier–Mukai identification, modulo any linear relations imposed by the embedding of the nine-dimensional tangent space. --- # 33. True minimal criterion The actual positive goal is therefore not even Theorem 22.1. It is only: $$ \boxed{ \mathfrak s_G^{(2)} \text{ injective on } J_{20}, J_{11}, J_{02}. } $$ Because Hodge permanence places the full actual obstruction image in the semiregularity kernel, such injectivity forces: $$ \boxed{ J_{20}=J_{11}=J_{02}=0. } $$ This recovers PT004's: $$ \mathcal O_{E,S}=0. $$ --- # 34. A new priority order The next calculation should proceed in this order. ### First Compute: $$ D_{\mathrm{Weil}} \subset HT^2(X\times X). $$ ### Second Determine whether: $$ D_{11}=0. $$ If yes, the mixed channel disappears entirely for the actual Weil tangent problem. ### Third Determine whether the old: $$ F_1 $$ construction already controls: $$ J_{20}. $$ ### Fourth Only then attack the genuinely new: $$ J_{02} $$ condition for: $$ F_2. $$ This is more efficient than computing all Ext groups first. --- # 35. Best-case collapse The ideal outcome is: $$ \boxed{ D_{11}=0 } $$ and: $$ \boxed{ J_{20}=0 } $$ by the established old secant geometry. Then the whole generalized positive gate becomes: $$ \boxed{ J_{02}=0. } $$ By PT004, it would be enough to prove semiregularity injectivity on: $$ J_{02} \subseteq I_2(F_2). $$ That would justify precisely the intuition behind Question 11.2.2. But PT005 does not assume this best-case collapse. It identifies it as the next theorem to test. --- # 36. Worst-case finite reduction Even if: $$ D_{11}\neq0, $$ the problem remains finite and block-structured. One needs at most: 1. an even test for: $$ F_1; $$ 2. an odd mixed test: $$ \rho_{F_1}\otimes\rho_{F_2^\vee}; $$ 3. an even test for: $$ F_2. $$ This is still a major reduction from the raw semiregularity problem. --- # 37. Collision with HC-False HC-False could attack the same decomposition by showing: $$ \boxed{ J_{20}\neq0 } $$ or: $$ \boxed{ J_{11}\neq0 } $$ or: $$ \boxed{ J_{02}\neq0 } $$ for every candidate seed and that the resulting obstruction survives the Hodge-kernel constraints. HC-True needs only one seed for which all three vanish. Thus the True/False collision now has three named obstruction channels rather than one opaque deformation rank. --- # 38. What PT005 proves PT005 proves the exact self-Ext decomposition: $$ \boxed{ B_{20}\oplus B_{11}\oplus B_{02}. } $$ It proves the matching Hochschild decomposition: $$ \boxed{ C_{20}\oplus C_{11}\oplus C_{02}. } $$ It proves factorization of total shifted semiregularity on outer products. It proves no semiregularity cancellation can occur between the three channels. It gives a factorwise sufficient criterion for full image-restricted semiregularity. It reduces the actual Weil problem further to: $$ \boxed{ J_{20}, J_{11}, J_{02}. } $$ --- # 39. What PT005 does not prove PT005 does not prove: $$ D_{11}=0. $$ It does not prove the old: $$ F_1 $$ sector automatically kills: $$ J_{20}. $$ It does not prove: $$ F_2 $$ satisfies Markman's image-restricted semiregularity question. It therefore does not establish algebraicity in the generalized CM setting. The unresolved problem is now block-localized. --- # 40. Next Interface Next HC-True round: ```text HODGE_HCTRUE_PT006_WeilTangentChannelDecomposition.md ``` Primary target: $$ \boxed{ D_{\mathrm{Weil}} = D_{20}\oplus D_{11}\oplus D_{02} } $$ for the actual Fourier–Mukai pullback of the Weil period-domain tangent space. Planned attacks: 1. write Orlov's Hochschild transform: $$ \Phi_{HT}; $$ 2. identify the commutative tangent: $$ H^1(T_{X\times\hat X}) $$ inside: $$ HT^2(X\times\hat X); $$ 3. pull the unitary/Weil tangent representation back to: $$ HT^2(X\times X); $$ 4. project it onto: $$ C_{20},C_{11},C_{02}; $$ 5. determine whether: $$ D_{11}=0 $$ or is essential; 6. compare with the original split-sixfold statement that diagonal generalized deformations map to commutative-gerby Weil deformations; 7. identify which channel is genuinely new in the quartic-CM construction; 8. reduce the next semiregularity calculation to the smallest surviving block. --- # References 1. E. Markman, *Secant sheaves on abelian $n$-folds with real multiplication and Weil classes on abelian $2n$-folds with complex multiplication*, arXiv:2509.23079. The paper constructs $$ E=\Phi(F_1\boxtimes F_2^\vee), $$ proves Hodge permanence of its normalized Chern class, and in Question 11.2.2 isolates injectivity of semiregularity on $$ \operatorname{Im} \left( at_{F_2}:HT^2(X)\to\operatorname{Ext}^2(F_2,F_2) \right). $$ 2. E. Markman, *Cycles on abelian $2n$-folds of Weil type from secant sheaves on abelian $n$-folds*, arXiv:2502.03415. In the split sixfold case the secant-sheaf construction produces a nine-dimensional space of unobstructed generalized deformations which Orlov's Hochschild transform maps to the Weil deformation directions. 3. R.-O. Buchweitz, H. Flenner, *A Semiregularity Map for Modules and Applications to Deformations*, Compositio Math. 137 (2003), 135–210. 4. D. Orlov, *Derived Categories of Coherent Sheaves on Abelian Varieties and Equivalences Between Them*, arXiv:alg-geom/9712017. 5. Aletheia, *HODGE_HCTRUE_PT004_SemiregularityOperator*, 2026-09-16. --- ## Canonical Source Declaration 本檔案為 HC-True 分支第五篇正式 UTF-8 Markdown canonical source。 數學原始碼只使用 `$...$` 與 `$$...$$`。 本輪沒有證明 generalized secant seed 已 semiregular；本輪證明的是 outer-product/Fourier–Mukai obstruction problem可精確拆成三個互不 cancellation 的 Künneth channels，並將下一步縮成 actual Weil tangent 在三通道中的分量計算。