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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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\((2,2,0,0)\): \(H^2 \otimes H^2 \otimes H^0 \otimes H^0\), dimension \(1 \times 1 \times 1 \times 1 = 1\) (and permutations, 6 terms).

Total dimension: \(16 + 4 \times 4 + 6 \times 1 = 16 + 16 + 6 = 38\). The Hodge decomposition is:

\[

H^4(X, \mathbb{C}) = \bigoplus_{r+s=4} H^{r,s}(X),

\]

with \(H^{2,2}(X) = (H^{1,0} \otimes H^{1,0} \otimes H^{1,0} \otimes H^{1,0}) \oplus (H^{0,1} \otimes H^{0,1} \otimes H^{1,0} \otimes H^{1,0}) \oplus \cdots\), but we focus on rational classes. The dimension of \(H^{2,2}(X) \cap H^4(X, \mathbb{Q})\) is computed via the Knneth component \((1,1,1,1)\), with \(\dim = \binom{4}{2} = 6\), as it counts choices of two \((1,0)\) and two \((0,1)\) forms.

B. Exhaustion by Divisor Products

For \(E^4\), codimension-2 cycles (\(p=2\)) include products of divisors. Since \(E\) is a curve, \(\Pic(E) \cong \mathbb{Z} \oplus E(\mathbb{C})\), and divisors on \(E^4\) arise from:

Pullbacks of divisors from each factor, e.g., \(E \times E \times \{pt\} \times \{pt\}\),

Diagonal cycles, e.g., \(\Delta_{12} = \{(x, x, y, z)\}\).

The Chow group \(CH^2(E^4) \otimes \mathbb{Q}\) is generated by such cycles. By the Knneth formula and properties of abelian varieties, the cycle class map \(\cl_2: CH^2(E^4) \otimes \mathbb{Q} \to H^{2,2}(X) \cap H^4(X, \mathbb{Q})\) is surjective, as divisor products exhaust the \((2,2)\)-classes. This is a known result for abelian varieties, where HC holds up to codimension \(n-1\) for dimension \(n\) (here, \(n=4\)).

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