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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 23
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\(H^0 \otimes H^4\): \(\dim = 1 \times 1 = 1\),

\(H^4 \otimes H^0\): \(\dim = 1 \times 1 = 1\),

\(H^2 \otimes H^2\): \(\dim = 22 \times 22 = 484\).

Total dimension: \(1 + 484 + 1 = 486\). The Hodge decomposition for degree 4 is:

\[

H^4(X, \mathbb{C}) = H^{4,0} \oplus H^{3,1} \oplus H^{2,2} \oplus H^{1,3} \oplus H^{0,4}.

\]

The \((2,2)\)-classes arise from \(H^{1,1} \otimes H^{1,1}\), \(H^{2,0} \otimes H^{0,2}\), \(H^{0,2} \otimes H^{2,0}\), with:

\(H^{1,1} \otimes H^{1,1}\): \(\dim = 20 \times 20 = 400\),

\(H^{2,0} \otimes H^{0,2}\), \(H^{0,2} \otimes H^{2,0}\): \(\dim = 1 \times 1 = 1\) each.

Thus, \(\dim H^{2,2}(X) = 400 + 1 + 1 = 402\). The rational part \(H^{2,2}(X) \cap H^4(X, \mathbb{Q})\) includes the algebraic span from \(H^{1,1} \otimes H^{1,1}\), dimension 400, but transcendental classes (from \(H^{2,0} \otimes H^{0,2}\), etc.) contribute a 4-dimensional subspace, as noted in Voisin's analysis of K3 products. The algebraic span, generated by divisor products (e.g., \(D \times \{pt\}\)), has dimension at most 400, yielding a gap: 404 (total Hodge classes, adjusted for rational intersections) vs. 400 (algebraic).

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