When is a band-connected sum equal to the connected sum?

TL;DR

This paper characterizes when a band-connected sum of two knots equals their connected sum, showing it occurs precisely when the band is trivial, thus clarifying the relationship between these knot operations.

Contribution

The paper provides a complete characterization of trivial bands in the context of band-connected sums equaling connected sums of knots.

Findings

A band-connected sum equals the connected sum if and only if the band is trivial.

Provides a necessary and sufficient condition for the equivalence of band-connected sum and connected sum.

Clarifies the relationship between band-connected sums and connected sums in knot theory.

Abstract

We show that a band-connected sum of knots $K_0$ and $K_1$ along a band $b$ is equal to the connected sum $K_0\# K_1$ if and only if $b$ is a trivial band.

Full text

1. Introduction

Let $K_{0}\cup K_{1}$ be a $2$-component split link in $S^{3}$, and $b:I\times I\to S^{3}$ an embedding satisfying $b(I\times I)\cap K_{i}=b(\{i\}\times I)$ for $i=0,1$. The image $b(I\times I)$ is called a band connecting $K_{0}$ and $K_{1}$; by abuse of notation, $b$ also denotes $b(I\times I)$. Then the knot $K_{0}\cup K_{1}\cup\partial b-\mathrm{int}((K_{0}\cup K_{1})\cap b)$ is called the band-connected sum of $K_{0}$ and $K_{1}$ along $b$, and denoted by $K_{0}\#_{b}K_{1}$, or $K_{b}$ for short.

A core of a band $b$ is $b(I\times\{*\})$, and a cocore of $b$ is $b(\{*\}\times I)$, where $*\in\mathrm{int}{I}$. A band $b$ is trivial if there is a splitting sphere for $K_{0}\cup K_{1}$ meeting $b$ in a cocore of $b$. Obviously, if $b$ is a trivial band, then $K_{0}\#_{b}K_{1}\cong K_{0}\#K_{1}$. We prove the converse.

Theorem 1**.**

If a band-connected sum $K_{0}\#_{b}K_{1}$ is equal to the connected sum $K_{0}\#K_{1}$, then $b$ is a trivial band.

Regarding the problem when a band-connected sum is composite, Eudave-Muñoz proved the following.

Theorem 2** ([EM92]).**

If $K_{b}$ is a composite knot, then there is a decomposing sphere for $K_{b}$ which is either disjoint from $b$ or intersects $b$ in a core of $b$.

To prove Theorem 1 we use Theorem 2 and the fact that if $K_{b}\cong K_{0}\#K_{1}$, $K_{b}$ bounds a Seifert surface of type 1a (Lemma 5), which is defined below. Any band-connected sum $K_{b}$ bounds a compact connected surface $S$ containing $b$. Such a surface $S$ is called type 1 if $S-b$ is disconnected, and type 2 otherwise. In other words, a type 1 surface for $K_{0}\#_{b}K_{1}$ is a compact connected, possibly nonorientable, surface that is the union of $b$ and two connected surfaces bounded by $K_{i}$ $(i=0,1)$. A surface $S$ of type 1 is called type 1a if there is a splitting sphere for $K_{0}\cup K_{1}$ that is disjoint from $S-b$, and* type 1b* otherwise.

For a composite band-connected sum bounding a surface of type 1a, we study the configuration of a splitting sphere, a decomposing sphere, and a type 1a surface, and obtain Theorem 3 below. Theorem 1 follows from Theorem 3.

Theorem 3**.**

If $K_{b}$ is a composite knot and bounds a surface of type $1$a, then there is a decomposing sphere disjoint from $b$.

As another corollary to Theorem 3, we give sufficient conditions for band-connected sums to be prime.

Proposition 4**.**

Assume that $K_{b}=K_{0}\#_{b}K_{1}$ bounds a surface $S_{0}\cup b\cup S_{1}$ of type $1$a, where $S_{i}$ is a surface bounded by $K_{i}$ $(i=0,1)$. Then $K_{b}$ is prime if $b$ is a nontrivial band and either $(1)$ or $(2)$ below holds.

1. $(1)$

$K_{i}$* is a trivial knot for $i=0,1$.* 2. $(2)$

$K_{i}$* is prime and $S_{i}$ is an incompressible Seifert surface for $i=0,1$.*

2. Proofs

Lemma 5**.**

If $K_{b}\cong K_{0}\#K_{1}$, then $K_{b}$ bounds a Seifert surface of type $1$a.

Proof of Lemma 5. The assumption implies $g(K_{b})=g(K_{0})+g(K_{1})$, where $g(\cdot)$ denotes the genus of a knot. Then, by [Gab87, Sch87] $K_{b}$ bounds a Seifert surface $S$ of type 1 that is the union of $b$ and minimal genus Seifert surfaces $S_{i}$ for $K_{i}$ $(i=0,1)$. Let $P$ be a splitting sphere for $K_{0}\cup K_{1}$ such that $|P\cap(S-b)|$ is minimal among all splitting spheres. Suppose $P\cap(S-b)\neq\emptyset$ for a contradiction; then $P\cap(S-b)$ consists of simple closed curves. Let $\Delta_{1}$ be an innermost one among disks in $P$ bounded by components of $P\cap(S-b)$. Without loss of generality $\partial\Delta_{1}\subset S_{0}$. If $\partial\Delta_{1}$ is essential in $S_{0}$, then surgery of $S_{0}$ along $\Delta_{1}$ yields a Seifert surface for $K_{0}$ with fewer genus than $S_{0}$ and possibly a closed surface. Since $S_{0}$ has minimal genus, we see $\partial\Delta_{1}$ bounds a disk $\Delta_{2}$ in $S_{0}$.

Each component of $\Delta_{2}\cap P$ bounds a disk in $\Delta_{2}$, and let $\Delta_{3}(\subset S_{0})$ be an innermost one among all such disks in $\Delta_{2}$; note $\Delta_{3}\cap P=\partial\Delta_{3}$. Let $B$ be the $3$-ball satisfying $\partial B=P$ and $\Delta_{3}\subset B$, and $B^{\prime}$ the closure of a component of $B-\Delta_{3}$ containing $K_{0}$ or $K_{1}$. Shrinking $B^{\prime}$ slightly, we obtain a splitting sphere $\partial B^{\prime}$ with $|\partial B^{\prime}\cap(S-b)|<|P\cap(S-b)|$. This contradicts the minimality of $|P\cap(S-b)|$. It follows $P\cap(S-b)=\emptyset$, so that $S$ is type 1a $\square$

Proof of Theorem 3. By Theorem 2 we may assume that there is a decomposing sphere $Q$ for $K_{b}=K_{0}\#_{b}K_{1}$ intersecting $b$ in a core of $b$. Take a splitting sphere $P$ for $K_{0}\cup K_{1}$, a decomposing sphere $Q$ for $K_{b}$, and a type 1a surface $S$ bounded by $K_{b}$ that satisfy conditions (1)—(4) below.

1. (1)

$P,Q,S$ are in general position, i.e. any two of these are in general position, and the intersection of any two of these and the rest are in general position. 2. (2)

$P\cap(S-b)=\emptyset$. 3. (3)

$Q\cap b$ is a core of $b$, and $P\cap b$ consists of cocores of $b$. 4. (4)

$|Q\cap S|+|P\cap Q|+|P\cap Q\cap S|$ is minimal among all $P,Q,S$ satisfying (1)—(3) above.

We first study the configuration of $P\cap Q$ and $Q\cap S$ on $Q$. Note that $Q\cap S$ consists of an arc connecting the two points $K_{b}\cap Q$ and possibly simple closed curves. Let $\alpha$ be the arc component of $Q\cap S$, and set $c=Q\cap b$, a core of $b$. Since $c$ connects $K_{0}$ and $K_{1}$, and $S-b$ is disconnected, $c$ is contained in $\alpha$. Note also that $P\cap Q$, consisting of simple closed curves, is disjoint from any circle component of $Q\cap S$ because $P\cap(S-b)=\emptyset$. For the same reason $P\cap Q$ intersects $\alpha$ only in $c$. See Figure 1.

Assume for a contradiction that either $Q\cap S$ or $P\cap Q$ has a circle component disjoint from $\alpha$. Each such component bounds a disk in $Q$ disjoint from $\alpha$. Let $\Delta$ be an innermost one among all these disks. If $\partial\Delta$ is a component of $Q\cap S$, surger $S$ along $\Delta$ to yield a bounded surface $S^{\prime}$ and possibly a closed surface. Since $\Delta\cap P=\emptyset$, we see that $P\cap(S^{\prime}-b)=\emptyset$ and $|Q\cap S^{\prime}|<|Q\cap S|$. This contradicts the minimality assumption (4).

If $\partial\Delta$ is a component of $P\cap Q$, surger $P$ along $\Delta$ to yield two spheres, one of which is a splitting sphere $P^{\prime}$. Since $\Delta\cap S=\emptyset$, we see that $P^{\prime}\cap(S-b)=\emptyset$ and $|P^{\prime}\cap Q|<|P\cap Q|$. This contradicts the minimality assumption. We thus obtain the following.

Claim 6**.**

$Q\cap S=\alpha$, and each component of $P\cap Q$ intersects $\alpha$. $\square$

Assume for a contradiction that there is a component of $P\cap Q$ intersecting $\alpha$ more than once. Each such component bounds a disk in $Q$ containing at most one endpoint of $\alpha$. Let $\Delta_{1}$ be an innermost one among all such disks. Then, $\Delta_{1}\cap\alpha$ consists of at most one non-properly embedded arc and at least one properly embedded arc (Figure 2(1)). Each properly embedded arc of $\Delta_{1}\cap\alpha$ and a subarc of $\partial\Delta_{1}$ cobound a disk in $\Delta_{1}$ containing no endpoint of $\alpha$. Let $\Delta_{2}$ be an innermost one among all such disks; note $\Delta_{2}\cap(S-b)=\emptyset$. We can isotop $b$ along $\Delta_{2}$ to reduce $|P\cap Q\cap S|$ (Figure 2(2)), a contradiction to the minimality. Thus Claim 7 below is proved.

Claim 7**.**

Each component of $P\cap Q$ intersects $\alpha$ in one point of the core $c$. $\square$

Let us take a look at $P\cap Q$ on the splitting sphere $P$. Let $\Delta_{3}$ be an innermost one among all disks in $P$ bounded by components of $P\cap Q$; see Figure 3. Surger the decomposing sphere $Q$ along $\Delta_{3}$ to obtain two spheres, at least one of which is a decomposing sphere for $K_{b}$. Let $Q^{\prime}$ be a decomposing sphere obtained from $Q$ after the surgery. Without loss of generality $Q^{\prime}$ is disjoint from $K_{0}$ and intersects $K_{1}$ in two points. Let $\Delta_{4}$ be the closure of the component of $b-(b\cap Q^{\prime})$ disjoint from $K_{0}$ (Figure 4). Sliding $Q^{\prime}$ along the disk $\Delta_{4}$, we obtain a decomposing sphere for $K_{b}$ disjoint from $b$ as claimed. This completes the proof of Theorem 3. $\square$

Proof of Theorem 1. By Lemma 5 $K_{b}$ bounds a Seifert surface that is the union of minimal genus Seifert surfaces $S_{i}$ for $K_{i}$ $(i=0,1)$ and $b$. We proceed by induction on the product of genera $g(K_{0})g(K_{1})$.

Case $1$. $g(K_{0})g(K_{1})=0$.

Since $K_{0}$ or $K_{1}$ is a trivial knot, $S_{0}$ or $S_{1}$ is a disk. The boundary of a regular neighborhood of the disk is a splitting sphere for $K_{0}\cup K_{1}$ intersecting $b$ in a single cocore of $b$. It follows that $b$ is a trivial band.

Case $2$. $g(K_{0})g(K_{1})\geq 1$.

By Theorem 3 there is a decomposing sphere $Q$ for $K_{b}$ disjoint from $b$. Without loss of generality $Q$ intersects $K_{1}$ in two points. Let $B$ be the $3$-ball which is bounded by $Q$ and disjoint from $K_{0}$. Then, $B\cap K_{1}$ is an arc knotted in $B$. Let $K^{\prime}_{1}$ be the knot obtained from $K_{1}$ by replacing $(B,B\cap K_{1})$ with $(D^{2}\times I,\{0\}\times I)$. We can regard $b$ as a band connecting $K_{0}$ and $K^{\prime}_{1}$. If $b$ is a trivial band for $K_{0}\#_{b}K^{\prime}_{1}$, then it is trivial for $K_{0}\#_{b}K_{1}$. Since $g(K_{0})g(K^{\prime}_{1})<g(K_{0})g(K_{1})$, Theorem 1 is proved inductively. $\square$

Proof of Proposition 4. Assume for a contradiction that $K_{b}$ is not prime, but satisfies either condition (1) or (2) in Proposition 4. If $K_{b}$ is a trivial knot, then $K_{0}$ and $K_{1}$ are trivial knots and especially $b$ is a trivial band ([Sch85]), a contradiction to the assumption of Proposition 4. It follows that $K_{b}$ is a composite knot.

Then, following the arguments in Case 2 in the proof of Theorem 1, without loss of generality we can take a $3$-ball $B$ such that $B\cap(K_{0}\cup b)=\emptyset$ but $B\cap K_{1}$ is an arc knotted in $B$. This implies that $K_{1}$ is not a trivial knot, which violates condition (1). It follows that $K_{b}$ satisfies condition (2).

Let $K^{\prime}_{1}$ be $K_{1}$ with $(B,B\cap K_{1})$ replaced by $(D^{2}\times I,\{0\}\times I)$. Since $K_{1}$ is prime and $B\cap K_{1}$ is a knotted arc in $B$, we see $K^{\prime}_{1}$ is a trivial knot. The Seifert surface $S_{1}$ for $K_{1}$ intersects $\partial B$ in an arc connecting the two points $\partial B\cap K_{1}$ and possibly simple closed curves. However, using the incompressibility of $S_{1}$, we can isotop $S_{1}$ without meeting $S_{0}\cup b$ so that $S_{1}\cap\partial B$ is just one arc. The arc $S_{1}\cap\partial B$ in $\partial B=\partial(D^{2}\times I)$ and $\{0\}\times I$ cobound a disk in $D^{2}\times I$. Then, the union $S^{\prime}$ of the disk and $S_{1}-\mathrm{int}B$ is an incompressible Seifert surface for the trivial knot $K^{\prime}_{1}$. It follows that $S^{\prime}$ and thus $S_{1}-\mathrm{int}B$ are disks. Isotopy of $\partial B$ along the disk $S_{1}-\mathrm{int}B$ yields a splitting sphere for $K_{0}\cup K_{1}$ intersecting $b$ in a cocore of $b$. This contradicts the assumption that $b$ is a nontrivial band. $\square$