Proof. Figure 1 shows a strongly invertible position of the link $K_0\cup c$ . The dotted line is the axis of the involution. Performing $(-1/n)$ -surgery on c changes $K_0$ into $K_n$ . This observation confirms that $K_n$ is strongly invertible for any $n\ge 0$ .

Proof. This is confirmed by Kirby–Rolfsen calculus. See Figure 2.

Figure 2

Kirby–Rolfsen calculus for $(8,0)$ -surgery on $K_0\cup c$ . The result is a lens space $L(9,7)$ .

Figure 2 Long description

The diagram consists of two large panels at the top and a sequence of five smaller diagrams at the bottom.

* Top-Left Panel: A complex knot labeled K sub 0 with an integer coefficient 8. Inside the knot is a smaller loop labeled c with a coefficient 0 and directional arrows.

* Top-Right Panel: An arrow points from the first panel to this modified version where the coefficient 8 has changed to 2. Perform a handle slide of the knot K over the inner loop c.

* Bottom Row Sequence: An arrow points from the top-right panel down to the start of the bottom row.

1. The first diagram is a vertical twisted knot with coefficients 0 and 2.

2. An arrow points to a link of two circles with coefficients minus 3 and minus 1, connected by a horizontal segment labeled minus 1 over 3.

3. An arrow points to a two-component link with coefficients minus 3 and 2 over 3.

4. An arrow points to a two-component link with coefficients minus 1 and 2 over 7.

5. A final arrow points to a single oval with the fraction 9 over 7, representing the final lens space L(9,7).

Proof. First, $K_0$ is a torus knot $T(3,4)$ of genus three. By Lemma 2.2, $(8,0)$ -surgery on $K_0\cup c$ yields an L-space, because a lens space is an L-space. Thus, our link $K_0\cup c$ satisfies the assumption of Lemma 2.3.

Proof. For the link $K_0\cup c$ given in Figure 1, consider $(9,-1/n)$ -surgery. Performing n-twists on c shows that it represents $(9n+9)$ -surgery on $K_n$ .

Take the quotient under the involution along the axis depicted there. Then, we have a diagram as shown in Figure 3 (top left) after the tangle replacements. (Since $K_0$ has writhe $9$ in the diagram of Figure 1, the tangle surgery for the quotient of $K_0$ is done by the $0$ -tangle.)

Figure 3

The diagram after the tangle replacements (top left) and its deformation. The box with integer n contains right handed horizontal n half twists.

Figure 3 Long description

The diagram consists of four panels arranged in a Z-shaped sequence connected by arrows.

1. Top-left panel: A complex knot diagram features a large vertical rectangular block on the left. From this block, multiple strands emerge. At the top, strands form a series of crossings. At the bottom, a small horizontal box labeled with the variable n contains right-handed horizontal n half twists. A small arrow points right along one of the lower strands.

2. Top-right panel: Following a horizontal arrow, the diagram shows a slight deformation of the top crossings. The strands are pulled tighter, and the relationship between the upper loops and the lower n-twist box remains connected via the large rectangular boundary.

3. Bottom-left panel: Following a diagonal arrow pointing down and left, the diagram simplifies. The number of parallel strands is reduced, and the top crossing is shown as a single intertwined knot. The n-twist box remains at the bottom-left.

4. Bottom-right panel: Following a final horizontal arrow, the diagram shows the most simplified state. The strands have been deformed such that the upper knot is more isolated, and the connection to the n-twist box is streamlined into a single continuous loop structure that wraps back toward the left boundary.

After a series of deformations in Figures 3 and 4, the resulting link (right in Figure 4) consists of two Montesinos tangles $[-1/3,1/3]$ and $[-1/2,-1/(n-1)]$ . By taking the double branched cover, which gives M, we have the first conclusion.

Figure 4

Further deformation. The resulting link (right) consists of two Montesinos tangles $[-1/3,1/3]$ and $[-1/(n-1), -1/2]$ .

Figure 4 Long description

The diagram consists of two panels separated by a right-pointing arrow.

Left panel: A single continuous knot diagram. At the top, there is a horizontal loop with several crossings. A vertical strand descends from the left side of this loop, leading to a smaller horizontal loop at the bottom left. This bottom loop contains a rectangular box labeled n minus 1.

Right panel: The knot has been deformed into two distinct regions highlighted by dashed elliptical boundaries.

- The upper dashed ellipse contains a complex tangle of overlapping loops and crossings, representing the first Montesinos tangle.

- The lower dashed ellipse contains a simpler loop structure that includes the rectangular box labeled n minus 1, representing the second Montesinos tangle.

- Strands connect the upper and lower tangles to form a closed link system.

To confirm that this graph manifold M has Heegaard genus at least three, we follow the argument in [Reference Anderson1, Section 2.3], which is based on the classification by Kobayashi [Reference Kobayashi12] of closed orientable $3$ -manifolds with Heegaard genus two that contain an essential torus. It claims that if a graph manifold obtained from gluing two Seifert fibred manifolds along an incompressible boundary torus has Heegaard genus two and contains a unique essential torus, then the regular fibres of the two pieces on the common torus intersect just once.

First, the cases where $n=3$ and $4$ are treated in [Reference Anderson1, Section 2.3.2]. Thus, we assume $n\ge 5$ . We remark that each piece of M admits a unique Seifert fibration. (In fact, this holds when $n\ge 4$ .) From Figure 4, we see that the regular fibres of the two pieces on the common torus intersect twice (see the proof of [Reference Burde, Zieschang and Heusener7, Proposition 12.28]). This implies that M has Heegaard genus at least three.

Proof. First, the two arcs $t_1$ and $t_2$ , as shown in Figure 5, give an unknotting tunnel system for $K_n$ . It is straightforward to verify that the exterior of the neighbourhood of $K_n\cup t_1\cup t_2$ is a handlebody of genus three. Hence, $K_n$ has tunnel number at most two.

Figure 5

An unknotting tunnel system $\{t_1,t_2\}$ for $K_n$ .

Figure 5 Long description

A line drawing depicts a horizontal braid consisting of four parallel strands that undergo a series of repeating crossings.

* The entire structure is labeled K sub n at the top center.

* Moving from left to right, the strands weave over and under each other in three distinct repeating blocks.

* Towards the right side, a vertical segment connects two strands. This segment contains two points labeled t sub 1 and t sub 2.

* Arrows point to these specific points: t sub 1 is the upper point on the vertical segment, and t sub 2 is the lower point.

* Below this specific section, a bracket spans one braid cycle and is labeled 1 full twist.

* To the far right, the strands transition into horizontal dashed lines, indicating the remaining full twists.

If $K_n$ has tunnel number one, then any resulting manifold by Dehn surgery on $K_n$ has Heegaard genus at most two. This is impossible by Lemma 3.1. Hence, $K_n$ has tunnel number two.

Proof. We start from the link $K_0\cup c$ used in Section 2. The orientation of each component is shown in Figure 1. The multivariable Alexander polynomial $\Delta _{K_0\cup c}(x,y)$ is

$$ \begin{align*} (x^8-x^5+x^4)y^2+(x^5-x^4+x^3)y+x^4-x^3+1. \end{align*} $$

(We used Kodama’s knot [Reference Kodama13] for the calculation.)

By performing $(-1/n)$ -surgery on c, the link $K_0\cup c$ is changed into the new link $K_n\cup c_n$ , where $c_n$ is the unknot. We set $c=c_0$ . Clearly, two links $K_0\cup c$ and $K_n\cup c_n$ have homeomorphic exteriors. Then, the induced isomorphism of the homeomorphism on their homology groups relates the Alexander polynomials of these two links [Reference Fox10, Reference Morton20].

Let $\mu _{K_n}$ and $\lambda _{K_n}$ (respectively $\mu _{c_n}$ and $\lambda _{c_n}$ ) be the oriented meridian and longitude of $K_n$ (respectively $c_n$ ). Then, $\mu _{K_n}=\mu _{K_0}$ , $\mu _{c_n}=\mu _{c}-n\lambda _{c}$ and $\lambda _{c}=3\mu _{K_0}$ . These give $\mu _{c}=\mu _{c_n}+3n\mu _{K_n}$ .

Hence, $ \Delta _{K_n\cup c_n}(x,y)=\Delta _{K_0\cup c}(x,yx^{3n}). $ Since $\mathrm {lk}(K_n,c_n)=\mathrm {lk}(K_0,c)=3$ , the Torres condition [Reference Torres25] implies

$$ \begin{align*} \Delta_{K_n\cup c_n}(x,1)=\frac{x^3-1}{x-1} \Delta_{K_n}(x). \end{align*} $$

Thus,

$$ \begin{align*} \Delta_{K_n}(t)&=\frac{t-1}{t^3-1} \Delta_{K_n\cup c_n}(t,1)=\frac{t-1}{t^3-1} \Delta_{K_0\cup c}(t,t^{3n})\\ &=\frac{t-1}{t^3-1}(t^{6n+8}-t^{6n+5}+t^{6n+4}+t^{3n+5}-t^{3n+4}+t^{3n+3}+t^4-t^3+1). \end{align*} $$

Here,

$$ \begin{align*} t^{6n+8}-t^{6n+5}=t^{6n+5}(t^3-1), \quad t^{6n+4}-t^{3n+4}=t^{3n+4}(t^{3n}-1) \end{align*} $$

and

$$ \begin{align*} t^{3n+5}+t^{3n+3}+t^4=(t^{3n+5}+t^{3n+4}+t^{3n+3})-t^{3n+4}+t^4 =t^{3n+3}(t^2+t+1)-t^4(t^{3n}-1). \end{align*} $$

Therefore,

$$ \begin{align*} \Delta_{K_n}(t)&=t^{6n+5}(t-1)+t^{3n+4}(t-1)\sum_{i=0}^{n-1}t^{3i}+t^{3n+3}-t^4(t-1)\sum_{i=0}^{n-1}t^{3i}-t+1\\ &=(t^{6n+6}-t^{6n+5})+\sum_{i=0}^{n-1}(t^{3n+3i+5}-t^{3n+3i+4})+t^{3n+3}-\sum_{i=0}^{n-1}(t^{3i+5}-t^{3i+4}) -t+1, \end{align*} $$

as desired.

Proof. The first conclusion immediately follows from Lemma 4.1. If $n\ge 2$ , then ${4\in \mathcal {S}}$ , but $8\not \in \mathcal {S}$ . Hence, $\mathcal {S}$ is not closed under addition.

Proof. By Lemma 4.2, the formal semigroup $\mathcal {S}$ of $K_n$ is not a semigroup. Hence, $K_n$ is not a torus knot, since the formal semigroup of a torus knot is a genuine semigroup [Reference Borodzik and Livingston6, Example 1.10].

Suppose that $K_n$ is a satellite knot for a contradiction. First, by Lemma 2.4, $K_n$ is prime because an L-space knot is prime [Reference Krcatovich14, Theorem 1.2]. Since $K_n$ is the closure of a $4$ -braid, it has bridge index at most four. Hence, the companion is a $2$ -bridge knot and the pattern has wrapping number two by [Reference Schubert23].

In addition, the companion is an L-space knot and the pattern is braided [Reference Baker and Motegi3, Theorems 1.17 and 7.2]. This implies that the companion is a $2$ -bridge torus knot [Reference Ozsváth and Szabó21, Theorem 1.5] and the pattern is $2$ -cabled. Then, $K_n$ is an iterated torus L-space knot, so its formal semigroup is closed under addition [Reference Wang26, Theorem 1.4]. This contradicts Lemma 4.2. Hence, $K_n$ is hyperbolic.

Proof. We know that $K_1$ is the $(2,9)$ -cable of the trefoil, so it has bridge index four. Thus, $K_1$ has braid index four.

Next, consider $K_2$ , which is s682. The Morton–Franks–Williams inequality [Reference Franks and Williams11, Reference Morton19] gives a lower bound on the braid index in terms of its HOMFLY-PT polynomial. For the $P(v,z)$ version of the HOMFLY-PT polynomial, let $d_+$ and $d_-$ be the max and min degrees of v. Then, $(d_+-d_-)/2+1$ gives the lower bound for the braid index. For $K_2$ , $d_+=24$ and $d_-=18$ by SageMath [22]. This shows that $K_2$ has braid index four.

Assume $n\ge 3$ . If $K_n$ has braid index at most three, then it has tunnel number one as mentioned in Section 1. By Lemma 3.2, $K_n$ has tunnel number two. Hence, it has braid index four.

Proof of Theorem 1.3.

This follows from Lemmas 2.1, 2.4, 3.2, 4.3 and 4.4.