Below I present a careful, formal discussion of the four families of "nave" candidate correspondences we tried as potential algebraic realizations of the T(Y)T(Y)T(Y)\otimes T(Y)T(Y)T(Y) summand in H2,2(YY)H^{2,2}(Y\times Y)H2,2(YY). For each candidate I: (i) give a precise geometric definition, (ii) describe its induced action on cohomology (how it projects to Knneth summands), (iii) explain why --- in the simple linear model or in generic geometric situations --- it may fail to generate the required transcendental components, and (iv) indicate what refinement or stronger construction would be needed to make it viable.
1. The diagonal YY\Delta\subset Y\times YYY
Definition.
={(y,y)YY}\displaystyle \Delta=\{(y,y)\in Y\times Y\}={(y,y)YY}. It is a smooth codimension--2 algebraic cycle; its cycle class []H4(YY,Q)[\Delta]\in H^4(Y\times Y,\mathbb Q)[]H4(YY,Q) is canonical.
Induced map on cohomology.
Let p1,p2p_1,p_2p1,p2 be the projections. For a correspondence ZYYZ\subset Y\times YZYY with class [Z][Z][Z], the standard induced map on H2(Y)H^2(Y)H2(Y) is
Z:H2(Y)p1H2(YY)[Z]H6(YY)p2H2(Y).\Phi_Z\;:\; H^2(Y)\xrightarrow{\;p_1^*\;} H^2(Y\times Y)\xrightarrow{\;\cup\,[Z]\;} H^6(Y\times Y)\xrightarrow{\;p_{2*}\;} H^2(Y).Z:H2(Y)p1H2(YY)[Z]H6(YY)p2H2(Y).
For the diagonal Z=Z=\DeltaZ=, =IdH2(Y)\Phi_\Delta=\mathrm{Id}_{H^2(Y)}=IdH2(Y) (the push--pull along the diagonal is the identity). Equivalently, under the canonical identification H4(YY)H2(Y)H2(Y)...H^4(Y\times Y)\cong H^2(Y)\otimes H^2(Y)\oplus\dotsH4(YY)H2(Y)H2(Y)... the diagonal class decomposes as
cl()=ieiei+(terms in H0H4 and H4H0),\operatorname{cl}(\Delta)\;=\;\sum_{i} e_i\otimes e_i \;+\; \text{(terms in } H^0\otimes H^4 \text{ and } H^4\otimes H^0),cl()=ieiei+(terms in H0H4 and H4H0),
where {ei}\{e_i\}{ei} is any orthonormal basis of H2(Y)H^2(Y)H2(Y) (over Q\mathbb QQ or R\mathbb RR).
Why it might plausibly hit TTT\otimes TTT.
Write the orthogonal decomposition H2(Y)=NS(Y)QT(Y)H^2(Y)=\operatorname{NS}(Y)\otimes\mathbb Q\oplus T(Y)H2(Y)=NS(Y)QT(Y). Expanding ieiei\sum_i e_i\otimes e_iieiei in the block decomposition shows it contains contributions in each of the blocks:
NSNS\operatorname{NS}\otimes\operatorname{NS}NSNS,
NST\operatorname{NS}\otimes TNST and TNST\otimes\operatorname{NS}TNS,
TTT\otimes TTT.
In particular, the TTT\otimes TTT component of cl()\operatorname{cl}(\Delta)cl() equals tBTtt\sum_{t\in\mathcal B_T} t\otimes ttBTtt for a basis BT\mathcal B_TBT of T(Y)T(Y)T(Y), which is nonzero provided T(Y)0T(Y)\neq 0T(Y)=0. Thus a priori [][\Delta][] has a nontrivial TTT\otimes TTT projection.
Why the diagonal may nevertheless "fail" in practice.
