Brunnian links of 3-balls in the 4-sphere

Abstract

For each integer $n\ge 2$, we construct infinitely many $n$-component Brunnian links of 3-balls in $S^4$. Our main tool is the third author's result on the existence of splitting spheres for the trivial two-component link of $2$-spheres in $S^{4}$; we also give a new proof of this.

Figures (8)

• Figure 2.1: (a): The arcs $A_{1}$ and $A_{2}$, the 2-spheres $S_{1}$ and $S_{2}$, and the 2-disks $D_{1}$ and $D_{2}$ in the $B^{3}\times0$ slice of $B^{4}=B^{3}\times[-1,1]$. Note that $D_{1}$ appears fully in this 3-dimensional slice, while $D_{2}$ intersects it in an arc; similarly, $S_{2}$ appears fully in this slice, while $S_{1}$ intersects it in an equatorial $S^{1}$. (b): The barbell $\mathcal{B}$.
• Figure 3.1: Barbells $\delta_{5}$ and $\delta_{5}'$; only their intersections with $S^{1}\times S^{2}\times0$ are drawn. The arrows on the cuffs specify the orientation of the oriented intersection of the cuffs with $S^{1}\times S^{2}\times0$. The arrows on the bars specify the orientation of the bars. Note (Remark \ref{['rem:twist-barbell']}) that the barbells drawn in (a) and (b) are isotopic.
• Figure 3.2: An embedding of an $S^{1}$ (red) in the complement of a trivial two-component $2$-link $K_1 \sqcup K_2 \subset S^{4}$. Only the intersection with the equatorial $S^3 \times 0 \subset B^{4}\cup S^{3}\times[-1,1]\cup \overline{B^{4}} = S^4$ is drawn.
• Figure 3.3: View $S^{1}\times S^{3}$ as $(S^{1}\times B^3)\cup(S^{1}\times S^{2}\times[-1,1])\cup(\overline{S^{1}\times B^3})$. (a) and (b): The two lifts of the barbell $\beta_{5}$ to $S^{1}\times S^{3}$; their intersections with the time slices $S^{1}\times S^{2}\times\pm\varepsilon$ are drawn. (c): An isotopic copy of the barbell from (b)
• Figure 4.1: (a): A trivial two-component $2$-link $K_{1}\sqcup K_{2}$ in $S^{4}$, a red circle $r$ and two blue circles $b_1 ,b_2$ in $S^{4}\setminus(K_{1}\sqcup K_{2})$. Pushing forward $\boldsymbol{\delta_{k}}$ along $r$ and $\boldsymbol{\delta_{k}}^{-1}$ along $b_1$ and $b_2$ gives rise to two-component Brunnian links of $3$-balls in $S^{4}$. (b): If we ignore $K_{2}$, then only $r$ and $b_1$ intersect $B_{1}$, and the corresponding diffeomorphisms cancel out. Note that $B_{1}$ intersects the $S^{3}\times0$ time slice in the purple 2-disk. (c): (a) together with the splitting sphere $\Sigma=\partial N(B_{1})$, which intersects the $S^3\times 0$ slice in the green 2-sphere. Note that the blue circles do not intersect $\Sigma$.

Theorems & Definitions (22)

• Definition 1.1: Brunnian links of $3$-balls
• Theorem 1.2
• Definition 1.3
• Theorem 1.4: tatsuoka2025splittingspheresunlinkeds2s; restated in Theorem \ref{['thm:another-proof']}
• Remark 1.5: Unknotting and unlinking $3$-balls in the $5$-ball
• Remark 1.6
• Definition 2.1: Barbells
• Lemma 2.2
• proof
• Definition 3.1