The representativity of pretzel knots

TL;DR

This paper characterizes when certain pretzel knots have a specific property called representativity 3, and shows that large algebraic knots have representativity at most 3, contributing to knot theory classification.

Contribution

It provides a complete characterization of pretzel knots with representativity 3 and bounds the representativity of large algebraic knots, advancing understanding in knot invariants.

Findings

Pretzel knots with parameters ±(-2,3,3) or ±(-2,3,5) have representativity 3.

Large algebraic knots have representativity ≤ 3.

The results classify knots based on their representativity values.

Abstract

In the present paper, we will show that a $(p,q,r)$-pretzel knot has the representativity 3 if and only if $(p,q,r)$ is either $\pm(-2,3,3)$ or $\pm(-2,3,5)$. We also show that a large algebraic knot has the representativity less than or equal to 3.

Figures

Equations

$$r(K)=\underset{F\in \mathcal{F}}{\max }\underset{D\in {\mathcal{D}}_F}{\min }|\partial D\cap K|,$$

$$\frac{1}{p^{\prime }}+\frac{1}{q^{\prime }}+\frac{1}{r^{\prime }}=0,(\ast )$$

$$-\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0(2\le a<b\le c).$$

$$a=k(l-k)d,b=l(l-k)d,c=kld.$$

$$-\frac{1}{km}+\frac{1}{lm}+\frac{1}{c}=0$$

$$-\frac{l-k}{klm}+\frac{1}{c}=0$$

$$-\frac{1}{kld}+\frac{1}{c}=0$$

$$-\frac{1}{k(l-k)d}+\frac{1}{l(l-k)d}+\frac{1}{kld}=0$$

$$-\frac{l-k}{kl(l-k)d}+\frac{1}{kld}=0$$

$$\chi (F)=(2+1+1)-(4+4)+4=0$$

$$\chi (F)=(3+2+1)-(6+6)+6=0$$

Full text

[TABLE]

where $\mathcal{F}$ is the set of all closed surfaces of positive genus containing $K$ and $\mathcal{D}_{F}$ is the set of all compressing disks for $F$ in $S^{3}$. Then an inequality $r(K)\leq b(K)$ has shown, where $b(K)$ denotes the bridge number of $K$. This shows that a $(p,q,r)$-pretzel knot has the representativity less than or equal to $3$.

In the present paper, we will show that a $(p,q,r)$-pretzel knot has the representativity $3$ if and only if $(p,q,r)$ is either $\pm(-2,3,3)$ or $\pm(-2,3,5)$. We also show that a large algebraic knot has the representativity less than or equal to $3$.

1. Introduction

Let $K$ be a knot in the 3-sphere $S^{3}$ and $F$ a closed surface of positive genus containing $K$. In [2], the representativity $r(F,K)$ of a pair $(F,K)$ is defined as the minimal number of intersecting points of $K$ and $\partial D$, where $D$ ranges over all compressing disks for $F$ in $S^{3}$. Furthermore, we defined in [3] the representativity $r(K)$ of a knot $K$ as the maximal number of $r(F,K)$ over all closed surfaces $F$ of positive genus containing $K$. The representativity $r(K)$ of a knot $K$ is a new knot invariant.

The following inequality is the first fundamental result for the representativity. Let $b(K)$ denote the bridge number of $K$.

Theorem 1.1** ([3, Theorem 1.2]).**

For a knot $K$, $r(K)\leq b(K)$.

This shows that a $(p,q,r)$-pretzel knot has the representativity less than or equal to $3$. In the present paper, we characterize $(p,q,r)$-pretzel knots with the representativity $3$.

Theorem 1.2**.**

Let $K$ be a $(p,q,r)$-pretzel knot in $S^{3}$. Then $r(K)=3$ if and only if $(p,q,r)=$ $\pm(-2,3,3)$ or $\pm(-2,3,5)$.

Remark 1.3*.*

It was shown in [1, Theorem III] that a $(p,q,r)$-pretzel knot is a torus knot if and only if $(p,q,r)=$ $\pm(1,1,1)$, $\pm(-2,3,3)$ or $\pm(-2,3,5)$, and observed that a $\pm(-2,3,3)$-pretzel knot is a $(3,\pm 4)$-torus knot and a $\pm(-2,3,5)$-pretzel knot is a $(3,\pm 5)$-torus knot. In the proof of Theorem 1.2, we will give unknotted tori containing these pretzel knots directly.

Let $K$ be a tangle composite knot, that is, which admits at least one essential tangle decomposition. We define the tangle string number $ts(K)$ of $K$ as the minimal number of $n$ over all $n$-string essential tangle decompositions of $K$.

Proposition 1.4** ([3, Example 3.6]).**

For a tangle composite knot $K$, $r(K)\leq 2ts(K)$.

By Proposition 1.4, $r(K)\leq 2$ if $K$ is a composite knot, and $r(K)\leq 4$ if $K$ has an essential Conway sphere. Nonetheless, the representativity of a large algebraic knot (i.e. algebraic knot with an essential Conway sphere) has more sharp upper bound.

Theorem 1.5**.**

Let $K$ be a large algebraic knot. Then $r(K)\leq 3$.

2. Proofs

Proof.

(of Theorem 1.2) Let $K$ be a $(p,q,r)$-pretzel knot with a form of a Montesinos knot of type $(1/p,1/q,1/r)$. Thus $K$ is obtained from three rational tangles $(B_{1},T_{1})$, $(B_{2},T_{2})$ and $(B_{3},T_{3})$ of slopes $1/p$, $1/q$ and $1/r$ respectively by connecting them in series. Let $A$ be the “axis” of the Montesinos knot $K$, that is, an unknotted loop in the complement of $K$ such that there exist three meridian disks $D_{1}$, $D_{2}$ and $D_{3}$ in the solid torus $V=S^{3}-\text{int}N(A)$ which cut $(V,K)$ into three rational tangles $(B_{i},T_{i})$, $(B_{2},T_{2})$ and $(B_{3},T_{3})$, where $B_{1}\cap B_{2}=D_{1}$, $B_{2}\cap B_{3}=D_{2}$ and $B_{3}\cap B_{1}=D_{3}$. See Figure 1.

Suppose that $r(K)=3$ and let $F$ be a closed surface containing $K$ which satisfies $r(F,K)=3$. We assume that $|F\cap A|$ is minimal up to isotopy of $F$. Moreover, we assume that $|F\cap(D_{1}\cup D_{2}\cup D_{3})|$ is minimal up to isotopy of $F$. Then, each component of $F\cap\partial V$ is a longitude of $V$ which intersects $\partial D_{i}$ in one point for each $i$, and $F\cap D_{i}$ consists of $|F\cap\partial V|/2$-arcs and loops.

Claim 2.1**.**

$|p|,\ |q|,\ |r|\geq 2$.

Proof.

If $|p|\leq 1$, then $K$ is a connected sum of two torus knots, a torus knot, or a 2-bridge knot. In either cases, we have $r(K)\leq 2$, a contradiction. ∎

In the following, we show that $F\cap D_{i}$ has no loop and consists of mutually parallel $|F\cap\partial V|/2$-arcs.

Claim 2.2**.**

There exists no outermost arc $\alpha$ of $F\cap D_{i}$ with $|\alpha\cap K|=0$.

Proof.

Let $\alpha$ be an outermost arc of $F\cap D_{i}$ with $|\alpha\cap K|=0$, and $\delta$ be the corresponding outermost disk. Since each component of $F\cap\partial V$ intersects $\partial D_{i}$ in one point for each $i$, $\delta\cap\partial V$ connects two longitudinal components of $F\cap\partial V$. Hence by an isotopy of $F$ along $\delta$, $|F\cap A|$ can be reduced, a contradiction. ∎

Claim 2.3**.**

There exists no innermost loop $\beta$ of $F\cap D_{i}$ with $|\beta\cap K|\leq 1$.

Proof.

Let $\beta$ be an innermost loop of $F\cap D_{i}$ with $|\beta\cap K|=0$, and $\delta$ be the corresponding innermost disk. If $\partial\delta$ is essential in $F$, then $r(F,K)=0$, a contradiction. Otherwise, $\partial\delta$ bounds a disk $\delta^{\prime}$ in $F$ and we can remove $\alpha$ from $D_{i}$ by isotoping $\delta^{\prime}$ into $\delta$.

Next let $\beta$ be an innermost loop of $F\cap D_{i}$ with $|\beta\cap K|=1$, and $\delta$ be the corresponding innermost disk. Since $|\beta\cap K|=1$, $\beta$ is essential in $F$. Then $r(F,K)\leq 1$, a contradiction. ∎

Claim 2.4**.**

There exists no innermost loop $\beta$ of $F\cap D_{i}$ with $|\beta\cap K|=2$.

Proof.

Suppose there exists an innermost loop $\beta$ of $F\cap D_{i}$ with $|\beta\cap K|=2$. Then $\beta$ bounds a disk $\delta$ in $F$ since $r(K)=3$. Let $\beta^{\prime}$ be a loop of $\delta\cap(D_{1}\cup D_{2}\cup D_{3})$ which is innermost in $\delta$, and $\delta^{\prime}$ be the corresponding innermost disk in $\delta$. By Claim 2.3, $|\beta^{\prime}\cap K|=2$ and $\delta^{\prime}$ contains one subarc of $K$. Since $\delta^{\prime}$ is parallel into some $D_{i}$, at least one of $p,\ q,\ r$ is equal to [math]. This contradicts Claim 2.1. ∎

By Claims 2.2, 2.3 and 2.4, $F\cap D_{i}$ has one of the following configurations. See Figure 2.

• (1)

$F\cap D_{i}$ consists of mutually parallel arcs whose two outermost arcs $\alpha_{i1}$, $\alpha_{i2}$ satisfying $|\alpha_{ij}\cap K|=1$ for $j=1,2$.

• (2)

$F\cap D_{i}$ consists of one arc $\alpha_{i}$ satisfying $|\alpha_{i}\cap K|=2$.

In the following, we show that Configuration (2) does not exist.

Lemma 2.5**.**

Let $(B,T)$ be a trivial tangle and $F$ is a properly embedded surface in $B$ which contains $T$. Let $h:B\to\mathbb{R}_{\geq 0}$ be the standard Morse function with one critical point and suppose that each component of $T$ has exactly one maximal point with respect to $h$. Then a pair $(F,T)$ can be isotoped so that $F$ has no inessential saddle point and each component of $T$ has one maximal point with respect to $h$.

Proof.

This lemma essentially follows [3, Lemma 2.1, 2.2]. ∎

Claim 2.6**.**

In Configuration (2), $F\cap B_{1}$ consists of a disk.

Proof.

In Configuration (2), $F\cap\partial B_{1}$ consists of one loop $\beta$ satisfying $|\beta\cap K|=4$. If $F\cap B_{1}$ was not a disk, then by Lemma 2.5, there exists an essential saddle point of $F\cap B_{1}$ and this shows that $r(F\cap B_{1},T_{1})\leq 2$ as the proof of [3, Theorem 1.2]. ∎

By Claim 2.6, we have $|p|\leq 1$. This contradicts Claim 2.1. Hence Configuration (2) does not exist. Hereafter, we assume Configuration (1). Since each component of $F\cap\partial V$ is a longitude of $V$ which intersects $\partial D_{i}$ in one point for each $i$, $F\cap\partial B_{i}$ has either following two cases. See Figure 3.

• (A)

$F\cap\partial B_{i}$ consists of mutually parallel loops whose two innermost loops $\beta_{i1},\ \beta_{i2}$ satisfying $|\beta_{ij}\cap K|=2$ for $j=1,2$.

• (B)

$F\cap\partial B_{i}$ consists of one loop $\beta_{i}$ satisfying $|\beta_{i}\cap K|=4$.

Claim 2.7**.**

In either cases, $F\cap B_{i}$ consists of disks.

Proof.

In Case (A), without loss of generality, let $\delta$ be the innermost disk in $\partial B_{1}$ bounded by $\beta_{11}$. Since $r(F,K)=3$, $\beta_{11}$ bounds a disk $\delta^{\prime}$ in $F$. Let $\gamma$ be a loop of $\delta^{\prime}\cap\partial B_{1}$ which is innermost in $\delta^{\prime}$, and $\epsilon$ be the corresponding innermost disk in $\delta^{\prime}$. Note that $\gamma$ is a loop of $F\cap\partial B_{1}$, and $\epsilon$ contains exactly one subarc of $K$ if and only if $\gamma$ is either $\beta_{11}$ or $\beta_{12}$.

If $\epsilon$ is contained in $B_{2}\cup B_{3}$, then an algebraic tangle $(B_{2}\cup B_{3},T_{2}\cup T_{3})$ is inessential. It follows that $|q|\leq 1$ or $|r|\leq 1$ and it contradicts Claim 2.1. Hence $\epsilon$ is contained in $B_{1}$.

If $\gamma=\beta_{11}$ and it bounds a disk $\epsilon$ in $B_{1}$, then a loop of $F\cap\partial B_{i}$ which is next to $\beta_{11}$ on $\partial B_{i}$, say $\beta^{\prime}$, also bounds a disk $\epsilon^{\prime}$ in $B_{1}$ such that $\epsilon^{\prime}\cap F=\beta^{\prime}$. Since $r(F,K)=3$, $\beta^{\prime}$ bounds a disk $\epsilon^{\prime\prime}$ in $F$ and there is an isotopy sending $\epsilon^{\prime\prime}$ to $\epsilon^{\prime}$. By the minimality of $|F\cap A|$ and $|F\cap D_{i}|$, $\beta^{\prime}$ bounds a disk $\epsilon^{\prime}$ of $F\cap B_{1}$. Continuing this argument, each loop of $F\cap\partial B_{1}$ except for $\beta_{12}$ bounds a disk of $F\cap B_{1}$ in $B_{1}$. Now we apply Lemma 2.5 to $(B_{1},T_{1})$ and $F\cap B_{1}$. If $\beta_{12}$ does not bound a disk of $F\cap B_{1}$, then by Lemma 2.5, there exists an essential saddle point of $F\cap B_{1}$ and this shows that $r(K)\leq 1$, a contradiction. Hence each loop of $F\cap\partial B_{1}$ bounds a disk of $F\cap B_{1}$ if $\gamma=\beta_{11}$.

If $\gamma\neq\beta_{1j}$, then by the same argument as above, we can show that each loop of $F\cap\partial B_{1}$ bounds a disk of $F\cap B_{1}$.

In Case (B), we apply Lemma 2.5 to $(B_{i},T_{i})$ and $F\cap B_{i}$. If $F\cap B_{i}$ was not a disk, then by Lemma 2.5, there exists an essential saddle point of $F\cap B_{i}$ and this shows that $r(F\cap B_{1},T_{1})\leq 2$, a contradiction. ∎

We call $F\cap B_{i}$ *Type *(A) or *Type *(B) if $F\cap\partial B_{i}$ has Case (A) or Case (B) respectively. See Figure 4.

• (A)

$F\cap B_{i}$ consists of mutually parallel $k$-disks whose two outermost disks $E_{i1}$ and $E_{i2}$ contain one string of $T_{i}$.

• (B)

$F\cap B_{i}$ consists of one disk which contains two strings of $T_{i}$.

Claim 2.8**.**

For Type (A), the boundary slope of $F\cap B_{i}$ is equal to the rational tangle slope of $(B_{i},T_{i})$.

Proof.

This claim follows the definition of the slope of rational tangle. ∎

Claim 2.9**.**

For Type (B), let $1/m$ be the slope of a rational tangle and $a/b$ be the boundary slope of $F\cap B_{i}$. Then one of the following conditions holds.

• (I)

$am=b-1$.

• (II)

$a>1$ and $am=b+1$.

Proof.

There are two possibility of the arrangement of $T_{i}$ in $F\cap B_{i}$. Depending on them, we have two conditions (I) and (II). See Figure 5. We remark that it is required $a>1$ in Condition (II). ∎

Claim 2.10**.**

In Claim 2.9, $a=1$.

Proof.

Since $F\cap B_{i}$ is a disk in $B_{i}$, for each region $R$ between two consecutive arcs of $(F\cap B_{i})\cap D_{i}$, there exists a disk $\delta$ in $B_{i}$ such that $\delta\cap R$ is an arc connecting the two consecutive arcs and $\delta\cap(F\cap B_{i})=\partial\delta-\text{int}(\delta\cap R)$. If $a\geq 2$, then $|\partial\delta\cap T_{i}|\leq 1$.

For the next tangle $(B_{j},T_{j})$ which has a common disk $D_{i}$, we have a similar disk $\delta^{\prime}$ in $B_{j}$ as $\delta$, for some region between two consecutive arcs of $(F\cap B_{i})\cap D_{i}$, such that $|\delta\cap T_{j}|=1$.

Then we have a disk $\Delta=\delta\cup\delta^{\prime}$ such that $|\Delta\cap K|\leq 2$, a contradiction. ∎

Hence, only (I) in Claim 2.9 occurs, and we have the following claim.

Claim 2.11**.**

For Type (B), the boundary slope of $F\cap B_{i}$ is equal to $1/(m+1)$ if the slope of a rational tangle $(B_{i},T_{i})$ is $1/m$.

Since $F\cap\partial V$ consists of longitudes, the next equation holds.

[TABLE]

where $p^{\prime}$, $q^{\prime}$ and $r^{\prime}$ denote the boundary slopes of $F\cap B_{1}$, $F\cap B_{2}$ and $F\cap B_{3}$ respectively. Without loss of generality, we assume that $p^{\prime}<0$ and $q^{\prime},r^{\prime}>0$. We note that the common denominator in the equation (*) is equal to $|F\cap D_{i}|=p^{\prime}|F\cap B_{1}|=q^{\prime}|F\cap B_{2}|=r^{\prime}|F\cap B_{3}|$. In the following, we will give a condition for $p^{\prime},q^{\prime},r^{\prime}$.

Lemma 2.12**.**

Generally, suppose that

[TABLE]

Then there exists a pair of coprime integers such that $k<l\leq 2k$ and an integer $d>0$ such that

[TABLE]

Proof.

Put $a=km$, $b=lm$, where $k<l$ are coprime integers. Then

[TABLE]

[TABLE]

Since $l-k$ divides $m$, putting $d=m/(l-k)$,

[TABLE]

Hence, $a=k(l-k)d$, $b=l(l-k)d$ and $c=kld$. Moreover, since $b\leq c$, we have $l\leq 2k$. ∎

Claim 2.13**.**

If $l=2k$, then there exists a positive integer $d$ such that $(p,q,r)=(p^{\prime},q^{\prime}-1,r^{\prime}-1)=(-d,2d-1,2d-1)$. Here, $F\cap B_{1}$ is of Type (A) with $|F\cap B_{1}|=2$, and $F\cap B_{2}$ and $F\cap B_{3}$ are of Type (B).

Proof.

If $l=2k$, then $l=2$, $k=1$ since $l$ and $k$ are coprime. Then we have $a=d$, $b=2d$ and $c=2d$ in Lemma 2.12. Since $K$ is a knot, it follows that $q=2d-1$ and $r=2d-1$, thus $F\cap B_{2}$ and $F\cap B_{3}$ are of Type (B) by Claim 2.11. Then the common denominator in the equation (*) must be $2d$, $F\cap B_{1}$ is of Type (A) with $|F\cap B_{1}|=2$. ∎

Example 2.14**.**

A $(-2,3,3)$-pretzel knot is of this type in Claim 2.13.

Claim 2.15**.**

If $l<2k$, then there exists positive integers $k,d$ such that $(p,q,r)=(p^{\prime},q^{\prime},r^{\prime}-1)=(-kd,(k+1)d,k(k+1)d-1)$. Here, $F\cap B_{1}$ is of Type (A) with $|F\cap B_{1}|=k+1$, $F\cap B_{2}$ is of Type (A) with $|F\cap B_{2}|=k$, and $F\cap B_{3}$ is of Type (B).

Proof.

If $l-k$ is even, then both of $a$ and $b$ are also even. Since $b<c$, both of $F\cap B_{1}$ and $F\cap B_{2}$ are of Type (A) to have a common denominator in the equation (*). Thus by Claim 2.8, the slopes of rational tangles $(B_{1},T_{1})$ and $(B_{2},T_{2})$ coincide with those boundary slopes of $F\cap B_{1}$ and $F\cap B_{2}$. Therefore, $p$ and $q$ are even, this shows that $K$ has more than two components.

Hence $l-k$ is odd, and one of $l$ and $k$ is even. It follows that one of $a$ and $b$ is even and $c$ is even. Since $a<b<c$ and $K$ has exactly one component, $F\cap B_{3}$ is of Type (B) by Claim 2.11. Then,

[TABLE]

[TABLE]

Since $|F\cap D_{2}|=kl(l-k)d=kld$, $l-k=1$ and $l=k+1$. Hence $a=kd$, $b=(k+1)d$ and $c=k(k+1)d$, and $F\cap B_{1}$ is of Type (A) with $|F\cap B_{1}|=k+1$, $F\cap B_{2}$ is of Type (A) with $|F\cap B_{2}|=k$ and $F\cap B_{3}$ is of Type (B). ∎

Example 2.16**.**

A $(-2,3,5)$-pretzel knot is of this type in Claim 2.15, where $k=2$ and $d=1$.

Claim 2.17**.**

In Claim 2.15, $d=1$.

Proof.

Suppose $d>1$. There are $k(k+1)d$-arcs of $F\cap D_{1}$. Let $\alpha$ be the $k(k+1)$-th arc and $\alpha^{\prime}$ be the $k(k+1)+1$-th arc, and $R$ be the region of $D_{1}$ between $\alpha$ and $\alpha^{\prime}$. Then there exists a disk $\delta$ in $B_{1}$ such that $\delta\cap(F\cap B_{1})$ is an subarc $\beta$ of $\partial\delta$ intersecting $T_{1}$ in one point and $\delta\cap R$ is an arc $\partial\delta-\text{int}\beta$ connecting $\alpha$ and $\alpha^{\prime}$. There also exists a disk $\delta^{\prime}$ in $B_{2}$ such that $\delta^{\prime}\cap(F\cap B_{2})$ is an subarc $\beta^{\prime}$ of $\partial\delta^{\prime}$ intersecting $T_{2}$ in one point and $\delta^{\prime}\cap R$ is an arc $\partial\delta^{\prime}-\text{int}\beta^{\prime}$ connecting $\alpha$ and $\alpha^{\prime}$. Then a disk obtained from $\delta$ and $\delta^{\prime}$ gives a compressing disk for $F$ whose boundary intersects $K$ in two points. This shows that $r(F,K)\leq 2$, a contradiction. ∎

Therefore, we have two Types for the triple $(p,q,r)$.

• (1)

$(p,q,r)=(-d,2d-1,2d-1)$ for some integer $d\geq 2$.

• (2)

$(p,q,r)=(-k,(k+1),k(k+1)-1)$ for some integer $k\geq 2$.

Claim 2.18**.**

In Type (1), $d=2$.

Proof.

If $d\geq 3$, then we can find a compressing disk $\delta$ for $F$ such that $|\partial D\cap K|=2$, a contradiction. See Figure 6. ∎

Claim 2.19**.**

In Type (2), $k=2$.

Proof.

If $k\geq 3$, then we can find a compressing disk $\delta$ for $F$ such that $|\partial D\cap K|=2$, a contradiction. See Figure 7 for a $(-3,4,11)$-pretzel knot. See Figure 8 for a $(-4,5,19)$-pretzel knot. For $k\geq 5$, there exists a disk similar to Figure 8. ∎

Conversely, let $K$ be a $(-2,3,3)$-pretzel knot. Then $K$ has a closed surface $F$ of Type (1), where $F\cap B_{1}$ is of Type (A) with $|F\cap B_{1}|=2$ and $F\cap B_{2}$ and $F\cap B_{3}$ are of Type (B). See Figure 9. By the calculation of the Euler characteristic of $F$,

[TABLE]

we know that $F$ is a torus, and there exist a meridian disks $D$ and $D^{\prime}$ for each solid torus component obtained by splitting $S^{3}$ along $F$ such that $|\partial D\cap K|=3$ and $|\partial D^{\prime}\cap K|=4$. By checking that the algebraic intersection number of $\partial D\cap K$ and $\partial D^{\prime}\cap K$ are equal to $3$ and $4$ respectively, it follows that $K$ is a $(3,4)$-torus knot and $F$ is an unknotted torus for $K$.

Next, let $K$ be a $(-2,3,5)$-pretzel knot. Then $K$ has a closed surface $F$ of Type (2), where $F\cap B_{1}$ is of Type (A) with $|F\cap B_{1}|=3$, $F\cap B_{2}$ is of Type (A) with $|F\cap B_{2}|=2$ and $F\cap B_{3}$ are of Type (B). See Figure 10. By the calculation of the Euler characteristic of $F$,

[TABLE]

we know that $F$ is a torus, and there exist a meridian disks $D$ and $D^{\prime}$ for each solid torus component obtained by splitting $S^{3}$ along $F$ such that $|\partial D\cap K|=3$ and $|\partial D^{\prime}\cap K|=5$. By checking that the algebraic intersection number of $\partial D\cap K$ and $\partial D^{\prime}\cap K$ are equal to $3$ and $5$ respectively, it follows that $K$ is a $(3,5)$-torus knot and $F$ is an unknotted torus for $K$.

∎

Proof.

(of Theorem 1.5) Let $K$ be a large algebraic knot. Then $K$ admits an essential $2$-string tangle decomposition $(B,T)\cup(B^{\prime},T^{\prime})$ by a Conway sphere $S$ such that $(B,T)$ is obtained from two rational tangles $(B_{1},T_{1})$ and $(B_{2},T_{2})$ by gluing along a disk $D$. Suppose that $r(K)=4$ and let $F$ be a closed surface containing $K$ so that $r(F,K)=4$.

We assume that $|F\cap S|$ is minimal up to isotopy of $F$. Then $F\cap S$ consists of one loop intersecting $K$ in four points. Moreover, we assume that $|(F\cap B)\cap D|$ is minimal up to isotopy of $F\cap B$. Then $(F\cap B)\cap D$ consists of mutually parallel arcs whose two outermost arcs intersect $T$ in one point (c.f. Configuration (1)) or one arc intersecting $T$ in two points (c.f. Configuration (2)). These are similar configurations as in Figure 2.

In Configuration (1), let $\delta$ be the outermost disk in $D$ bounded by an outermost arc of $(F\cap B)\cap D$. By cutting a disk in $S$ bounded by $F\cap S$ along an arc $\delta\cap S$, and pasting one of the resultant subdisks and $\delta$, we obtain a compressing disk $\delta^{\prime}$ for $F$ such that $|\partial\delta^{\prime}\cap K|\leq 3$, a contradiction.

In Configuration (2), $F\cap\partial B_{i}$ consists of one loop. By applying Lemma 2.5 to $F\cap B_{i}$, there exists a compressing disk $\epsilon$ for $F\cap B_{i}$ such that $|\partial\epsilon\cap K|\leq 2$ or $F\cap B_{i}$ is a disk. In the formar case, we have a contradiction. Hence $F\cap B_{i}$ is a disk and $F\cap B$ is also a disk. However, this shows that a tangle $(B,T)$ is inessential, a contradiction. ∎