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Solution for Hodge Conjecture: Heuristic CAS 6 Approach 2.0

22 September 2025   16:02 Diperbarui: 22 September 2025   16:02 17
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\]

where \(H^{2,0} = H^2(E) \otimes H^0(E) \cong \mathbb{C}\), \(H^{0,2} = H^0(E) \otimes H^2(E) \cong \mathbb{C}\), and \(H^{1,1} = H^1(E) \otimes H^1(E)\), with \(\dim H^{1,1} = 4\). The rational part \(H^{1,1}(X) \cap H^2(X, \mathbb{Q})\) has dimension 4, corresponding to the \((1,1)\)-classes.

The Lefschetz (1,1)-theorem applies in codimension 1 (\(p=1\)) for any smooth projective variety. For \(X\), it states that every class in \(H^{1,1}(X) \cap H^2(X, \mathbb{Z})\) is the class of a divisor in the Picard group \(\Pic(X)\). Since \(\Pic(E \times E) \cong \Pic(E) \oplus \Pic(E) \oplus \Hom(E, E)\), and \(\Pic(E) \cong \mathbb{Z} \oplus E(\mathbb{C})\), the rational divisor classes span \(H^{1,1}(X) \cap H^2(X, \mathbb{Q})\), confirming HC: the cycle class map \(\cl_1: CH^1(X) \otimes \mathbb{Q} \to H^{1,1}(X) \cap H^2(X, \mathbb{Q})\) is surjective.

B. CAS-6 Analysis: Proof of Full Closure via Categorical Isomorphism

We apply the CAS-6 framework to \(E \times E\), mapping its cohomology to the six layers and computing metrics to diagnose closure. The functor \(\mathcal{F}: \mathbf{Var} \to \mathbf{LayeredVect}\) assigns \(\mathcal{F}(X) = (L(X), C(X), W(X), P(X), S(X), O(X))\).

Interaction Level \(L(X)\): Set \(L(X) = H^2(X, \mathbb{Q})\), with \(\dim = 6\). For \(p=1\), we focus on degree 2.

Interaction Configuration \(C(X)\): The Hodge decomposition yields \(C(X)_1 = H^{1,1}(X) \cap H^2(X, \mathbb{Q})\), with \(\dim = 4\). The Knneth formula ensures the configuration is tensor-decomposable: \(C(X)_1 \cong (H^1(E) \otimes H^1(E))_{\mathbb{Q}}\).

Interaction Weights \(W(X)\): The algebraic span is \(W(X)_1 = \im(\cl_1: CH^1(X) \otimes \mathbb{Q} \to H^2(X, \mathbb{Q}))\). Divisors include classes like \(E \times \{pt\}\) and \(\{pt\} \times E\), spanning a 4-dimensional subspace of \(H^{1,1}\).

Interaction Probabilities \(P(X)\): Compute

   \[

   P(X)_1 = \frac{\dim W(X)_1}{\dim (H^{1,1}(X) \cap H^2(X, \mathbb{Q}))} = \frac{4}{4} = 1.

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