Write Betti (singular) cohomology with rational coefficients. By the Knneth formula and the known cohomology of EEE,
Hk(E,Q){Q,k=0,2,Q2,k=1,0,otherwise.H^k(E,\mathbb{Q}) \simeq \begin{cases} \mathbb{Q}, & k=0,2,\\ \mathbb{Q}^2, & k=1,\\ 0, & \text{otherwise.} \end{cases}Hk(E,Q)Q,Q2,0,k=0,2,k=1,otherwise.
Hence for the surface X=EEX=E\times EX=EE we have
H2(X,Q)H2(E,Q)H0(E,Q)H1(E,Q)H1(E,Q)H0(E,Q)H2(E,Q).H^2(X,\mathbb{Q}) \;\simeq\; H^2(E,\mathbb{Q})\otimes H^0(E,\mathbb{Q}) \;\oplus\; H^1(E,\mathbb{Q})\otimes H^1(E,\mathbb{Q}) \;\oplus\; H^0(E,\mathbb{Q})\otimes H^2(E,\mathbb{Q}).H2(X,Q)H2(E,Q)H0(E,Q)H1(E,Q)H1(E,Q)H0(E,Q)H2(E,Q).
Concretely, writing uuu for a generator of H2(E,Q)H^2(E,\mathbb{Q})H2(E,Q) and {a,b}\{a,b\}{a,b} for a basis of H1(E,Q)H^1(E,\mathbb{Q})H1(E,Q) (dual to a choice of homology cycles), a convenient Q \mathbb{Q}Q-basis for H2(X,Q)H^2(X,\mathbb{Q})H2(X,Q) may be taken as
{u1,1u,aa,ab,ba,bb},\{\,u\otimes 1,\; 1\otimes u,\; a\otimes a,\; a\otimes b,\; b\otimes a,\; b\otimes b\,\},{u1,1u,aa,ab,ba,bb},
so that dimQH2(X,Q)=6\dim_{\mathbb Q} H^2(X,\mathbb{Q}) = 6dimQH2(X,Q)=6.
Passing to complex coefficients and Hodge decomposition, recall that for the elliptic curve EEE,
H1(E,C)=H1,0(E)H0,1(E),H^1(E,\mathbb{C}) = H^{1,0}(E)\oplus H^{0,1}(E),H1(E,C)=H1,0(E)H0,1(E),
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. Using the Knneth decomposition of Hodge types on X=EEX=E\times EX=EE, the Hodge summands of H2(X,C)H^2(X,\mathbb{C})H2(X,C) are
H2,0(X)=H1,0(E)H1,0(E),H1,1(X)H1,0(E)H0,1(E)H0,1(E)H1,0(E)H2,0H0,2,H^{2,0}(X)=H^{1,0}(E)\otimes H^{1,0}(E),\qquad H^{1,1}(X) \cong H^{1,0}(E)\otimes H^{0,1}(E)\;\oplus\;H^{0,1}(E)\otimes H^{1,0}(E)\;\oplus\; H^{2,0}\oplus H^{0,2},H2,0(X)=H1,0(E)H1,0(E),H1,1(X)H1,0(E)H0,1(E)H0,1(E)H1,0(E)H2,0H0,2,
