In this post I would like to propose a quartic 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 quartic link 
$$\text{Q}(\Sigma^3_W{^{(i)}}, \Sigma^3_W{^{(ii)}}, \Sigma^3_W{^{(iii)}}, \Sigma^2_U{})$$ 
in a 5-sphere $S^5$, in my proposal that

$\bullet$ $\Sigma^3_W{^{(i)}}$, $\Sigma^3_W{^{(ii)}}$, $\Sigma^3_W{^{(iii)}}$ are three insertions along the $H_3({S^5 \smallsetminus D^2 \times T^3},\mathbb Z)=\mathbb Z^3$
We can choose  $T^3_{xyw}$, $T^3_{yzw}$, $T^3_{xzw}$ for $H_3({S^5 \smallsetminus D^2 \times T^3},\mathbb Z)=\mathbb Z^3$. I view these three 3-torus as the Alexander dual to 1-circle $S^1_z,S^1_x,S^1_y$.

$\bullet$ $\Sigma^2_U$  is an insertion along a $\mathbb Z$ subgroup of ${H_2}(D^2 \times T^3,\mathbb Z)=\mathbb Z^3$.
For ${H_2}(D^2 \times T^3,\mathbb Z)=\mathbb Z^3$, with $xy$, $yz$, $zx$ 2-torus. I think it is fair to choose any of $T^2$, such as $T^2_{xy}$.

>So my proposal is that
$$\text{Q}(\Sigma^3_W{^{(i)}}=T^3_{xyw},\; \Sigma^3_W{^{(ii)}}=T^3_{yzw},\; 
 \Sigma^3_W{^{(iii)}}=T^3_{xzw},\;  \Sigma^2_U=T^2_{xy}) = \pm 1.$$



More precisely, to detect the link invariant, I can use the intersection formula for their Seifert surfaces:

- Here $V^4_W{^{(i)}},V^4_W{^{(ii)}},V^4_W{^{(iii)}}, V^3_U$ are the Seifert volumes (in 4,4,4,3 dims) bounded by the surfaces $\Sigma^3_W,\Sigma^2_U,\Sigma^2_U{'}$ (in 3,3,3,2 dims).

>$${\#(V^4_W\cap V^4_W\cap V^4_W\cap V^3_U)}:=\text{Q}(\Sigma^3_W{^{(i)}}, \Sigma^3_W{^{(ii)}}, \Sigma^3_W{^{(iii)}}, \Sigma^2_U{})$$


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>**My question** here is that is this link invariant or similar kind derived in the past literature? And based on what techniques, how are this invariant derived?