Step 1: We glue two sets of complement space of $D^2\times T^2$ out of the 4-sphere $S^4$, through their $T^3$ boundary with their three $S^1$ boundaries of $T^3$ cyclic permuted to obtain a new 4-manifold:
$$(S^4 \smallsetminus D^2\times T^2) \cup_{\text{cyclic permutation}} (S^4 \smallsetminus D^2 \times T^2)=(S^2\times S^2),$$
Step 2: Meanwhile we insert two $T^2$ 2-tori along the generator of homology group of the first 4-manifold $(S^4 \smallsetminus D^2\times T^2)$, $$H_2[(S^4 \smallsetminus D^2 \times T^2),\mathbb{Z}]=\mathbb{Z}^2$$ and we insert two additional $T^2$ 2-tori along the two generators of the homology group $H_2[(S^4 \smallsetminus D^2 \times T^2),\mathbb{Z}]=\mathbb{Z}^2$ for the other $(S^4 \smallsetminus D^2\times T^2)$.
Step 3: Now you can imagine that the four 2-tori $T^2$ (inserted in Step 2) sit in $S^2 \times S^2$ (by the gluing procedure in Step 1) are somehow "linked" in the 4-manifold $S^2 \times S^2$.
My question is that how are the four $T^2$ linked (or not) in the glued manifold 4-manifold $S^2 \times S^2$? And are the four $T^2$ linked together?
Or are there any three of $T^2$ linked together? In the sense of the triple linking (defined for example in arXiv:math/0007141)?
Please describe what is the mathematical link of four $T^2$ tori in $S^2 \times S^2$.