In this post I would like to propose a triple link in a 5-sphere.
Let us start with the following gluing into a 5-sphere: $$S^5=(D^2_{} \times T^3_{}) \cup_{T^4} ({S^5 \smallsetminus D^2 \times T^3})$$
I write $(D^2_{} \times T^3_{})=D^2_{rw} \times T^3_{xyz}$.
We denote the meridian as the $S^1_w$ (the boundary of the disk factor $D^2_{rw}$)
We denote the longitude $S^1_x,S^1_y,S^1_z$ from the $T^3_{xyz}$.
I propose a triple link $$\text{Tlk}(\Sigma^3_W,\Sigma^2_U,\Sigma^2_U{'})$$ in a 5-sphere $S^5$, in my proposal that
$\bullet$ $\Sigma^3_W$ is insertion along the ${H_3}(D^2 \times T^3,\mathbb Z) =\mathbb Z.$
We can choose $T^3_{xyz}$ for ${H_3}(D^2 \times T^3,\mathbb Z) =\mathbb Z$.
$\bullet$ $\Sigma^2_U$ and ${\Sigma^2_U}'$ are two insertions along the $\mathbb Z^2$ subgroups of $H_2({S^5 \smallsetminus D^2 \times T^3},\mathbb Z)=\mathbb Z^3$.
For ${H_2}(D^2 \times T^3,\mathbb Z)=\mathbb Z^3$, with $xy$, $yz$, $zx$ 2-torus written as $T^2_{xy}, T^2_{yz}, T^2_{zx}$,
I write the Alexander-dual 2-torus in the complement space in ${S^5 \smallsetminus D^2 \times T^3}$ generating $H_2({S^5 \smallsetminus D^2 \times T^3},\mathbb Z)=\mathbb Z^3$ as: $zw$, $xw$, $yw$ of three 2-torus: $T^2_{zw}, T^2_{xw}, T^2_{yw}$.
My question/proposal: Is this true that the following defining a nontrivial link invariant Tlk by the intersecting number $\#$: $$ {\#(V^4_W\cap V^3_U\cap V^3_U{'})} =\text{Tlk}(\Sigma^3_W=T^3_{xyz}, \Sigma^2_U=T^2_{xw}, \Sigma^2_U{'}= T^2_{yw})= \pm 1 ? $$ if I design the submanifold insertions $T^2$ and $T^3$ along the $H_2$ and $H_3$ described above?
- Here $V^4_W, V^3_U, V^3_U{'}$ are the Seifert volumes (in 4,3,3 dims) bounded by the surfaces $\Sigma^3_W,\Sigma^2_U,\Sigma^2_U{'}$ (in 3,2,2 dims).
My question 2: I have not yet found related literature about this link invariant. Are there existing pieces of literature that I was not aware of but should I pay attention? Please give Refs. Can the linking form http://www.map.mpim-bonn.mpg.de/Linking_form provide any useful information for my proposal above?
Thank you for your feedback and encouragement!
