While the CAS-6 reading yields a satisfying explanation of why E4E^4E4 exhibits closure, it also highlights the fragility of such closure in general:
Closure here rests on the product structure and the simplicity of factor cohomology; it does not generalize to arbitrary varieties where transcendental subspaces can contribute.
Stability claims rely on deformations that preserve the product-type combinatorics; more general deformations can reduce algebraic cycles or alter Picard ranks, affecting closure.
Finally, equality of dimensions and existence of explicit algebraic generators suffice to establish surjectivity of the cycle class map in this context, but in other settings one must address subtle rationality and integrality issues (rational vs integral Hodge conjectures) and the possibility of nontrivial relations among cycles.
Conclusion. For the case X=E4X=E^4X=E4, the CAS-6 perspective accurately captures the algebraic-topological geometry triad: the system exhibits exact dimension closure, no transcendental residue remains, and the algebraic interaction cycles are stable under natural deformations. This positive benchmark strengthens the heuristic claim that CAS-6 can meaningfully categorize where the Hodge Conjecture is likely to hold and where substantive obstructions are to be expected.
V. Heuristic Experiment C --- The Case of K3K3K3\times K3K3K3
A. Dimensional analysis: 404404404 versus 400400400
Let YYY be a complex K3 surface and consider the product variety
X=YY,X \;=\; Y \times Y,X=YY,
which is a smooth projective variety of complex dimension 444. In this subsection we perform a precise dimensional analysis of the Hodge subspace H2,2(X)H^{2,2}(X)H2,2(X) and of the obvious algebraic subspace generated by products of divisor classes; the resulting discrepancy identifies the precise locus of difficulty for the Hodge Conjecture in this example.
1. Hodge numbers of a K3 surface
Recall the Hodge numbers of a complex K3 surface YYY:
h0,0(Y)=1,h2,0(Y)=h0,2(Y)=1,h1,1(Y)=20,h^{0,0}(Y)=1,\qquad h^{2,0}(Y)=h^{0,2}(Y)=1,\qquad h^{1,1}(Y)=20,h0,0(Y)=1,h2,0(Y)=h0,2(Y)=1,h1,1(Y)=20,
and all other hp,q(Y)h^{p,q}(Y)hp,q(Y) vanish except those determined by conjugation and symmetry. Consequently
