with dimH1,0(E)=dimH0,1(E)=1\dim H^{1,0}(E)=\dim H^{0,1}(E)=1dimH1,0(E)=dimH0,1(E)=1. Write \alpha for a basis element of H1,0(E)H^{1,0}(E)H1,0(E) and \overline\alpha for its complex conjugate in H0,1(E)H^{0,1}(E)H0,1(E).
For the product X=E4X=E^4X=E4, the Knneth decomposition gives a canonical identification
H(X,C)i=14H(Ei,C),H^{\ast}(X,\mathbb C) \;\cong\; \bigotimes_{i=1}^4 H^\ast(E_i,\mathbb C),H(X,C)i=14H(Ei,C),
and the Hodge decomposition on XXX is obtained by taking tensor products of the Hodge types on each factor.
A general element of H2,2(X)H^{2,2}(X)H2,2(X) arises as a linear combination of pure tensors whose bi-degrees on the four factors sum to (2,2)(2,2)(2,2). Concretely, a pure tensor has type
(p1,q1)(p2,q2)(p3,q3)(p4,q4)(\,p_1,q_1\,)\otimes(\,p_2,q_2\,)\otimes(\,p_3,q_3\,)\otimes(\,p_4,q_4\,)(p1,q1)(p2,q2)(p3,q3)(p4,q4)
with ipi=2\sum_i p_i = 2ipi=2 and iqi=2\sum_i q_i = 2iqi=2. Because each Hr,s(E)H^{r,s}(E)Hr,s(E) is nonzero only for (r,s){(0,0),(1,0),(0,1),(1,1)}(r,s)\in\{(0,0),(1,0),(0,1),(1,1)\}(r,s){(0,0),(1,0),(0,1),(1,1)} (and H1,1(E)H^{1,1}(E)H1,1(E) is trivial except in degree 2), the nontrivial contributions to H2,2(X)H^{2,2}(X)H2,2(X) come from selecting exactly two factors to contribute a (1,0)(1,0)(1,0)-piece and two factors to contribute a (0,1)(0,1)(0,1)-piece (up to permutation), or from combinations involving H2(E)H^2(E)H2(E) on a factor together with H0(E)H^0(E)H0(E) on others --- but the latter locates in different bi-degree totals and, after accounting for degrees, reduces to the same combinatorial count described next.
2. Combinatorial count: h2,2(E4)=(42)=6\;h^{2,2}(E^4)=\binom{4}{2}=6h2,2(E4)=(24)=6
To produce type (2,2)(2,2)(2,2) one must choose exactly two of the four factors to contribute a (1,0)(1,0)(1,0)-form and the complementary two factors to contribute (0,1)(0,1)(0,1)-forms. The number of unordered choices of two factors from four is
(42)=6.\binom{4}{2} \;=\; 6.(24)=6.
Thus
