The most favorable algebraic scenario occurs when YYY is a singular K3 surface in the sense of Shioda--Inose, i.e. =20\rho=20=20 (the Nron--Severi rank is maximal). In that case the algebraically generated subspace from divisor-products attains its maximal possible dimension
2=202=400.\rho^2 \;=\; 20^2 \;=\; 400.2=202=400.
5. The four-dimensional transcendental residue
Comparing the two counts in the maximal algebraic situation, we obtain
h2,2(X)=404,dimQ(NSNS)=400,h^{2,2}(X) \;=\; 404, \qquad \dim_{\mathbb Q} \big(\operatorname{NS}\otimes\operatorname{NS}\big) \;=\; 400,h2,2(X)=404,dimQ(NSNS)=400,
and therefore a four-dimensional complementary subspace
(H2,2(X)H4(X,Q)) / spanQ{NSNS}\big(H^{2,2}(X)\cap H^4(X,\mathbb Q)\big)\ /\ \operatorname{span}_{\mathbb Q}\{\operatorname{NS}\otimes\operatorname{NS}\}(H2,2(X)H4(X,Q)) / spanQ{NSNS}
of dimension exactly 444. This complementary subspace is accounted for precisely by the tensor-square of the transcendental part:
T(Y)QT(Y),T(Y)\otimes_{\mathbb Q} T(Y),T(Y)QT(Y),
because dimQT(Y)=22=2\dim_{\mathbb Q} T(Y) = 22-\rho = 2dimQT(Y)=22=2 when =20\rho=20=20, and hence
dimQ(T(Y)T(Y))=22=4.\dim_{\mathbb Q} \big( T(Y)\otimes T(Y) \big) \;=\; 2^2 \;=\; 4.dimQ(T(Y)T(Y))=22=4.
