Heuristics for the Hodge Conjecture

The best answer I can imagine for a question like this is to quote the man himself: "The second result of Lefschetz tells us that a necessary and sufficient condition that a 2-cycle $\Gamma_2$ in $V_2$ be algebraic... This result has many geometrical applications... It is clearly a matter of great importance to extend Lefschetz's condition for a 2-cycle to be algebraic. The general problem is as follows...."

See page 184 of the Proceedings of the ICM 1950 for the full statement:

Hodge, W. V. D., The topological invariants of algebraic varieties, Proc. Intern. Congr. Math. (Cambridge, Mass., Aug. 30-Sept. 6, 1950) 1, 182-192 (1952). ZBL0048.41701.

Edited: One point is that Hodge's original version of the conjecture was wrong, and in a couple of ways. You do need rational coefficients (integral is too much to ask for, see ref below). Also a more general conjecture of Hodge fails: see

http://people.math.jussieu.fr/~leila/grothendieckcircle/HodgeConj.pdf

for one of all the all-time great disrespectful titles. (Hodge was a major innovator in algebraic geometry and complex manifold theory: but his theory needed plenty of work, particularly I think from Kodaira, before it became clear foundationally. The same is true, perhaps in spades, for Lefschetz.)

Edit: At the level of linear algebra, Hodge theory deals with a vector space with a double grading. Using the heuristic grading = concept of homogeneity, the subspaces of (p,p) type in Hodge theory can be picked out by the use of the circle group, as Freed does. Really (pun intended) there is a bigger group with two dimensions involved, the group of non-zero complex numbers under multiplication. A shallow remark is that these two groups control the linear algebra in Hodge theory. A deeper remark is that the vector spaces involved also have a rational structure, coming from the concept of integral cycle in topology. The interaction of the groups with the rational structure isn't shallow at all.