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 to obtain a new 4-manifold:

$$(S^4 \smallsetminus D^2\times T^2) \cup (S^4 \smallsetminus D^2 \times T^2)=(S^3\times S^1)\#(S^2 \times S^2)\#(S^2\times S^2),$$

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 $S^3 \times S^1 \# S^2 \times S^2 \# S^2 \times S^2$ (by the gluing procedure in Step 1) are somehow "linked" in the 4-manifold $S^3 \times S^1 \# S^2 \times S^2 \# S^2 \times S^2$.

My question is that how are the four $T^2$ linked (or not) in the glued manifold 4-manifold $S^3 \times S^1 \# S^2 \times S^2 \# S^2 \times S^2$? 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 $S^3 \times S^1 \# S^2 \times S^2 \# S^2 \times S^2$.

  • 2
    $\begingroup$ I susoect part of the link for some pairs of $T^2$ 2-tori show the spun Hopf link. It is not arbitrary. I am looking forward to some insightful analysis or Ref exploring the topology of this configuration carefully -- with a great imagination. $\endgroup$ – wonderich Sep 22 '16 at 2:37

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