# Satellite Knots Obtained by Twisting Torus Knots: Hyperbolicity of Twisted Torus Knots

Satellite Knots Obtained by Twisting Torus Knots: Hyperbolicity of Twisted Torus Knots Abstract The twisted torus knot $$T(p,q,r,s)$$ is obtained from the torus knot $$T(p,q)$$ by twisting $$r$$ adjacent parallel strands fully $$s$$ times. In the case that $$r=p$$ or $$r$$ is a multiple of $$q$$, it is known that $$T(p,q,r,s)$$ is either a cable knot, a torus knot, or a trivial knot. Under the assumption that $$r\ne p$$ and $$r$$ is not a multiple of $$q$$, we show that if $$T(p,q,r,s)$$ is a satellite knot then $$|s|=1$$. As a consequence, we show that $$T(p,q,r,s)$$ is usually a hyperbolic knot for $$|s|\ge 2$$: exceptional cases arise only for restricted values of $$(p,q,r,s)$$. 1 Introduction A surface $$F$$, not a disk or sphere, properly embedded in a $$3$$-manifold $$X$$ is called an essential surface if it is incompressible, $$\partial$$-incompressible, and not parallel to a surface in $$\partial X$$. When $$F$$ is a disk, it is called an essential disk if $$\partial F$$ does not bound a disk in $$\partial X$$. When $$F$$ is a sphere, it is called an essential sphere if it does not bound a $$3$$-ball in $$X$$. Let $$k$$ be a knot in the $$3$$-sphere $$S^3$$ and let $$E(k)=S^3-\mathrm{int}N(k)$$ be its exterior, where $$N(k)$$ is a regular neighborhood of $$k$$ in $$S^3$$. Any slope$$\sigma$$ on $$\partial E(k)$$, the isotopy class of an essential simple closed curve on $$\partial E(k)$$, can be parameterized by $$\mathbb{Q}\cup\{1/0\}$$: if $$\sigma=a\mu+b\lambda$$, then we write $$\sigma=a/b$$, where $$(\mu,\lambda)$$ is a standard meridian-longitude system for $$H_1(\partial E(k))$$ (When we given a link in $$S^3$$, slopes on each boundary component of the exterior of the link can be similarly parameterized by $$\mathbb{Q}\cup\{1/0\}$$). See [15]. Given a slope $$\sigma$$, $$k(\sigma)$$ denotes the result of Dehn surgery on $$k$$ along $$\sigma$$, that is, $$k(\sigma)$$ is the closed $$3$$-manifold constructed by gluing a solid torus to $$E(k)$$ along $$\partial E(k)$$ so that the curve of slope $$\sigma$$ bounds a meridian disk of the solid torus. If $$F$$ is a properly embedded essential surface in $$E(k)$$ with $$\partial F$$ nonempty, then all components of $$\partial F$$ are parallel in $$\partial E(k)$$ and hence they determine a slope on $$\partial E(k)$$, called the boundary slope of $$F$$. Let $$S^3=V_1\cup_\Sigma V_2$$ be a decomposition of $$S^3$$ into two solid tori $$V_1$$ and $$V_2$$ such that $$\Sigma=V_1\cap V_2=\partial V_1=\partial V_2$$. Let $$K=T(p,q)$$ be the torus knot of type $$p,q$$ that lies on the torus $$\Sigma$$, where $$p$$ and $$q$$ are relatively prime integers with $$0< q< p$$. We assume that $$K$$ is homotopic to $$p$$ times the core of $$V_1$$ and $$q$$ times the core of $$V_2$$. Up to isotopy, the exterior $$E(K)$$ contains a unique essential annulus, which has boundary slope $$pq$$. See [20]. Take a trivial knot $$C$$ that is disjoint from $$K$$ and encircles $$r$$ adjacent parallel strands of $$K$$ ($$2\le r\le p+q$$). Suppose that $$C$$ intersects each solid torus $$V_i$$ in a trivial arc, a properly embedded arc in $$V_i$$ parallel to an arc in $$\Sigma$$. For a nonzero integer $$s$$, $$(-1/s)$$-Dehn surgery on $$C$$ results in a $$3$$-sphere, that is, $$C(-1/s)=S^3$$. In this new $$3$$-sphere, the image of $$K$$, denoted $$T(p,q,r,s)$$, is called a twisted torus knot. When $$r<p$$, a typical picture of $$T(p,q,r,s)$$ is shown in Figure 1, where a rectangle labeled by an ordered pair of integers, $$(a,b)$$, denotes an $$(a,b)$$-torus braid. See [8] for details. If $$r=p$$, then it is easy to see that $$T(p,q,r,s)$$ is a torus knot (or sometimes a trivial knot). Thus we always assume that $$r\ne p$$. Fig. 1. View largeDownload slide Twisted torus knot $$T(p,q,r,s)$$. Fig. 1. View largeDownload slide Twisted torus knot $$T(p,q,r,s)$$. In his doctoral thesis [3], Dean introduced twisted torus knots to study Dehn surgeries yielding Seifert fiber spaces. Since then, a substantial amount of research has been devoted to these knots (see references in [9]). In particular, the author obtained some results on the knot-types of twisted torus knots. In [7], he showed that if $$r=kq$$ for some integer $$k$$, then $$T(p,q,r,s)$$ is the $$(q, p+k^2qs)$$-cable on the torus knot $$T(k,ks+1)$$; thus hereafter we only need to investigate what happens when $$q$$ does not divide $$r$$. In [9], he determined the parameters $$p,q,r,s$$ for which $$T(p,q,r,s)$$ becomes a trivial knot. In [10], he showed that if $$T(p,q,r,s)$$ is a torus knot then $$|s|\le 2$$ and moreover, when $$|s|=2$$, he also showed that $$T(p,q,r,s)$$ is a torus knot if and only if $$(p,q,r,s)=(2n\pm 1,n,n\pm 1, -2)$$ for some positive integer $$n$$. Finally, in [8], he proved that if $$T(p,q,r,s)$$ is a satellite knot then $$|s|\le 2$$. In this paper, we continue this investigation, showing that if $$T(p,q,r,s)$$ is a satellite knot then $$|s|$$ must be equal to $$1$$. We remark that there are infinitely many examples of twisted torus knots which are satellite knots, constructed by Morimoto [12, 13]. Theorem 1.1. Let $$p,q,r,s$$ be positive integers satisfying the following conditions: $$p,q$$ are relatively prime, and $$1\le q<p$$; $$2\le r\le p+q$$, $$r\ne p$$, and $$r$$ is not a multiple of $$q$$; and $$s\ne 0$$. Suppose that $$T(p,q,r,s)$$ is a satellite knot. Then $$|s|=1$$. □ Combining Theorem 1.1 with earlier results in [8–10], Thurston’s classification of knots [19] enables us to conclude the following. Corollary 1.2. Let $$p,q,r,s$$ be as in Theorem 1.1. Then $$T(p,q,r,s)$$ is a hyperbolic knot unless $$s=\pm 1$$ or $$(p,q,r,s)=(2n\pm 1,n,n\pm 1, -2)$$ for some positive integer $$n$$. □ Throughout this paper, we assume that the twisted torus knot $$T(p,q,r,s)$$ is a satellite knot for the parameters $$p,q,r,s$$ satisfying the conditions in Theorem 1.1. In order to prove Theorem 1.1, we may also assume that $$|s|=2$$ by [8, Theorem 1]. This assumption will eventually yield a contradiction. Some figures in this paper are best viewed in color; readers confused by figures in a black-and-white version are recommended to view the electronic version. 2 Lemmas We start with the following lemma, which is useful in understanding the twisted torus knot $$T(p,q,r,s)$$ in the case that $$r>p$$. Lemma 2.1. If $$r>p$$, then there is an isotopy deforming $$T(p,q,r,s)$$ to the knot obtained by closing the braid in Figure 2(a). □ Fig. 2. View largeDownload slide $$T(p,q,r,s)$$ is the closure of one of these braids when $$r>p$$. Fig. 2. View largeDownload slide $$T(p,q,r,s)$$ is the closure of one of these braids when $$r>p$$. Proof See the first paragraph in the proof of [10, Lemma 2.4]. ■ Given a knot $$k$$ in $$S^3$$, let $$b(k)$$ and $$g(k)$$ denote the bridge index and the genus of $$k$$, respectively. Lemma 2.2. $$b(T(p,q,r,s))\le\mathrm{max}(p,r)$$. □ Proof If $$r<p$$, then the twisted torus knot $$T(p,q,r,s)$$ is the closure of a braid on $$p$$ strands (see Figure 1), which implies $$b(T(p,q,r,s))\le p$$. If $$r>p$$, then $$T(p,q,r,s)$$ is the closure of a braid on $$r$$ strands by Lemma 2.1 and hence $$b(T(p,q,r,s))\le r$$. ■ It is well-known that if a knot $$k$$ is presented as the closure of a braid all of whose crossings are of the same sign, then $$k$$ is a fibered knot and   g(k)=1−b+c2, where $$b$$ is the number of strands of the braid and $$c$$ is the number of crossings. See [18]. Lemma 2.3. If $$r=p+q$$, then there is an isotopy deforming $$T(p,q,r,s)$$ to the closure of the braid in Figure 2(b). Moreover, if $$s>0$$, then   g(T(p,q,r,s))=1+pq−r+(r−1)rs2. □ Proof The first statement immediately follows from Lemma 2.1. If $$s>0$$, then all crossings of the braid in Figure 2(b) are positive. The braid has $$r$$ strands and $$(r-1)rs+pq$$ crossings, so the genus of its closure equals $$(1-r+(r-1)rs+pq)/2$$. ■ Lemma 2.4. Suppose that $$r=p+q$$ and $$s=-2$$. Then $$T(p,q,r,s)$$ is isotopic to the mirror image of $$T(r,2r-q,q,1)$$ and its genus is given by   g(T(p,q,r,s))=(r−1)(2r−q−1)+q(q−1)2. □ Fig. 3. View largeDownload slide $$T(p,q,r,s)$$ is the closure of any of these braids when $$r=p+q$$ and $$s=-2$$. Fig. 3. View largeDownload slide $$T(p,q,r,s)$$ is the closure of any of these braids when $$r=p+q$$ and $$s=-2$$. Fig. 4. View largeDownload slide An isotopy. Fig. 4. View largeDownload slide An isotopy. Proof By Lemma 2.1, there is an isotopy deforming $$T(p,q,r,s)$$ to the closure of the first braid in Figure 3. It is clear that the first and second braids in the figure are the same. Applying the isotopy illustrated in Figure 4, one sees that the second braid is isotopic to the third. This shows that $$T(p,q,r,s)$$ is isotopic to the mirror image of $$T(r,2r-q,q,1)$$. All crossings of the third braid are negative and the braid has $$r$$ strands and $$(r-1)(2r-q)+(q-1)q$$ crossings, so the genus of $$T(p,q,r,s)$$ equals $$\frac{1}{2}(1-r+(r-1)(2r-q)+(q-1)q)=\frac{1}{2}((r-1)(2r-q-1)+q(q-1))$$. ■ Lemma 2.5. Suppose that $$r=p-1,r\ge 2q-1$$, and $$s=-2$$. Then $$T(p,q,r,s)$$ has braid index $$r$$. □ Proof The twisted torus knot $$T(p,q,r,s)$$ is the closure of the leftmost braid in Figure 5. One can see that the closures of the four braids in the figure are isotopic. For example, Figure 6 illustrates the case that $$(p,q,r,s)=(8,3,7,-2)$$. Consider the mirror image of the rightmost braid in Figure 5. The closure of the mirror image is the knot $$T((r-1, q-1),(r,2r-2q+1))$$ in the notation of [1]. Note that $$r-1\le 2r-2q+1$$. Thus the knot $$T((r-1,q-1),(r,2r-2q+1))$$ has braid index $$\mathrm{min}(2r-2q+1,r)=r$$ by [1, Corollary 8] and the result follows. ■ Fig. 5. View largeDownload slide The closures of these braids are isotopic. Fig. 5. View largeDownload slide The closures of these braids are isotopic. Fig. 6. View largeDownload slide An isotopy for $$T(8,3,7,-2)$$. Fig. 6. View largeDownload slide An isotopy for $$T(8,3,7,-2)$$. Let $$k$$ be a knot in $$S^3$$. An arc $$\tau$$ with $$k\cap \tau=\partial \tau$$ is called an unknotting tunnel for $$k$$ if $$S^3-k\cup \tau$$ is an open handlebody of genus two. Moreover, $$\tau$$ is called a $$(1,1)$$-tunnel if $$\partial \tau$$ divides $$k$$ into two subarcs $$\gamma_1$$ and $$\gamma_2$$ in such a way that for some $$i=1,2$$, $$S^3-\mathrm{int}N(\tau\cup \gamma_i)$$ is a solid torus intersecting $$k$$ in a trivial arc. See [14, Proposition 1.3]. Lemma 2.6. Let $$k_m$$ denote the knot obtained by closing the braid in Figure 7(a), where $$m(\ge 3)$$ is an odd integer. Then the arc $$\tau_m$$ indicated in Figure 7(b) is an unknotting tunnel for the knot $$k_m$$. Moreover, if $$m\ge 5$$, then $$\tau_m$$ is not a $$(1,1)$$-tunnel. □ Fig. 7. View largeDownload slide The knots $$k_m$$ and $$k'_n$$. Fig. 7. View largeDownload slide The knots $$k_m$$ and $$k'_n$$. Proof Let $$k'_n$$ denote the knot obtained by closing the braid in Figure 7(c) and let $$\tau'_n$$ be an arc intersecting $$k'_n$$ in its endpoints as shown in Figure 7(d). By using a deformation as shown in Figure 8, which illustrates the case $$m=7$$, one can see that the two spatial graphs $$k_m\cup \tau_m$$ and $$k'_\frac{m+1}{2}\cup\tau'_\frac{m+1}{2}$$ have homeomorphic complements. On the other hand, a deformation as shown in Figure 9, which illustrates the case $$n=4$$, shows that the two spatial graphs $$k'_n\cup\tau'_n$$ and $$k'_{n-1}\cup\tau'_{n-1}$$ have homeomorphic complements. It is clear that $$\tau'_2$$ is an unknotting tunnel for $$k'_2$$. This shows that $$\tau_m$$ is an unknotting tunnel for $$k_m$$. Suppose $$m\ge 5$$. Let $$\gamma_1$$ and $$\gamma_2$$ be two subarcs of $$k_m$$ cut off by $$\partial\tau_m$$. Consider the two knots $$\gamma_1\cup\tau_m$$ and $$\gamma_2\cup\tau_m$$. One can see that these knots are torus knots $$T(\frac{m-1}{2},m)$$ and $$T(\frac{m+1}{2},m+2)$$, both of which are nontrivial knots. See Figure 10, which illustrates the case $$m=7$$. It follows that $$\tau_m$$ is not a $$(1,1)$$-tunnel. ■ Fig. 8. View largeDownload slide The two spatial graphs $$k_m\cup \tau_m$$ and $$k'_\frac{m+1}{2}\cup\tau'_\frac{m+1}{2}$$ have homeomorphic complements. Fig. 8. View largeDownload slide The two spatial graphs $$k_m\cup \tau_m$$ and $$k'_\frac{m+1}{2}\cup\tau'_\frac{m+1}{2}$$ have homeomorphic complements. Fig. 9. View largeDownload slide The two spatial graphs $$k'_n\cup\tau'_n$$ and $$k'_{n-1}\cup\tau'_{n-1}$$ have homeomorphic complements. Fig. 9. View largeDownload slide The two spatial graphs $$k'_n\cup\tau'_n$$ and $$k'_{n-1}\cup\tau'_{n-1}$$ have homeomorphic complements. Fig. 10. View largeDownload slide The knots $$\gamma_1\cup\tau_m$$ and $$\gamma_2\cup\tau_m$$ are torus knots $$T(\frac{m-1}{2},m)$$ and $$T(\frac{m+1}{2},m+2)$$, respectively. Fig. 10. View largeDownload slide The knots $$\gamma_1\cup\tau_m$$ and $$\gamma_2\cup\tau_m$$ are torus knots $$T(\frac{m-1}{2},m)$$ and $$T(\frac{m+1}{2},m+2)$$, respectively. Lemma 2.7. $$q\ge 3$$. □ Proof Since $$q$$ does not divide $$r$$, $$q\ne 1$$ and hence $$q\ge 2$$. Assume that $$q=2$$. First, suppose $$r>p$$. Then $$r=p+1$$ or $$p+2$$ (since $$r\le p+q=p+2$$). In the former, $$q$$ divides $$r$$, a contradiction. Suppose $$r=p+2(=p+q)$$. Then $$r=p+q\ge (q+1)+q=5$$. By Lemma 2.3 there is an isotopy deforming $$T(p,q,r,s)$$ to the closure of the braid in Figure 2(b) ($$q$$ must be equal to $$2$$ in the figure). If $$s=2$$, then $$T(p,q,r,s)$$ is the knot $$k_r$$ defined in Lemma 2.6, and since $$r\ge 5$$, the lemma guarantees that $$k_r$$ admits an unknotting tunnel which is not a $$(1,1)$$-tunnel. However, since $$T(p,q,r,s)$$ is assumed to be a satellite knot, any of its unknotting tunnels is a $$(1,1)$$-tunnel by [14, Proposition 1.8 and Theorem 2.1], a contradiction. Suppose that $$s=-2$$. Then $$T(p,q,r,s)$$ is the knot obtained by closing the first braid in Figure 11, which is equivalent to the third braid. The arc $$\tau$$ indicated in the figure is an unknotting tunnel for the closed braid knot. As in the last paragraph of the proof of Lemma 2.6, the closed braid knot is split along the tunnel $$\tau$$ into two nontrivial torus knots $$T(\frac{r+1}{2},-r)$$ and $$T(\frac{r-1}{2},-(r-2))$$. Thus $$\tau$$ is not a $$(1,1)$$-tunnel by [14, Proposition 1.3] and we get a contradiction since any unknotting tunnel for a satellite knot is a $$(1,1)$$-tunnel by [14, Theorem 2.1]. Suppose $$r<p$$. Since $$q$$ does not divide $$r$$, $$r$$ is odd and hence $$p-r$$ is even. Let $$p-r=\ell q$$. Then an isotopy as in [9, Figure 10] deforms $$T(p,q,r,s)$$ to the closure of the braid in Figure 12. As above, one sees that the arc $$\tau'$$ indicated in the figure is an unknotting tunnel for the closed braid knot which is not a $$(1,1)$$-tunnel and this also gives a contradiction. ■ Fig. 11. View largeDownload slide An isotopy. Fig. 11. View largeDownload slide An isotopy. Fig. 12. View largeDownload slide $$\tau'$$ is not a $$(1,1)$$-tunnel. Fig. 12. View largeDownload slide $$\tau'$$ is not a $$(1,1)$$-tunnel. Given a knot $$k$$ in $$S^3$$, choose a standard longitude-meridian system $$\lambda,\mu$$ for $$H_1(\partial E(k))$$. We follow [15] to say that a knot $$k'$$ on the boundary of a regular neighborhood $$N(k)$$ is an $$(a,b)$$-cable of $$k$$ if $$[k']=a\lambda+b\mu\in H_1(\partial E(k))$$. (For nontriviality, we assume here that $$a\ge 2$$, but $$b$$ may have a negative value.) In particular, we use $$T_{a,b}(c,d)$$ to denote the $$(a,b)$$-cable of the $$(c,d)$$-torus knot. Lemma 2.8. Let $$(a,b)$$ and $$(c,d)$$ be pairs of relatively prime positive integers. Then   g(Ta,b(c,d))=1−ac+ab−ad+acd−b2. □ Proof This follows immediately from Satz 1 on page 247 of [16]. ■ Lemma 2.9. Let $$a,b,c,d$$ be positive integers for which the cable knot $$T_{a,b}(c,d)$$ is defined. Suppose that $$T(c,d)$$ is not a trivial knot. Then   b(Ta,b(c,d))=a⋅min(c,d). □ Proof Using a well-known fact that $$b(T(c,d))=\mathrm{min}(c,d)$$ (see [17]), one easily sees that the knot $$T_{a,b}(c,d)$$ has a bridge position with bridge number $$a\cdot\mathrm{min}(c,d)$$. Hence the lemma follows from another classical result of Schubert [17]: if $$k_1$$ is a satellite knot with companion $$k_2$$ and has wrapping number $$n$$ in a solid torus neighborhood of $$k_2$$, then $$b(k_1)\ge n\cdot b(k_2)$$. ■ 3 Primitive/Seifert-fibered knots Let $$\ell$$ be a simple closed curve on the boundary of a $$3$$-manifold $$Q$$. We use $$Q[\ell]$$ to denote the $$3$$-manifold constructed from $$Q$$ by attaching a $$2$$-handle along the curve $$\ell$$. For two disjoint nonisotopic simple closed curves $$\ell$$ and $$\ell'$$ on $$\partial Q$$, $$Q[\ell\cup\ell']$$ is similarly defined, that is, $$Q[\ell\cup\ell']=Q[\ell][\ell'](=Q[\ell'][\ell])$$. We say that a knot $$k$$ in $$S^3$$ is a primitive/Seifert-fibered knot if $$k$$ lies on a genus two surface, which splits $$S^3$$ into two handlebodies $$H$$ and $$H'$$, in such a way that $$H[k]$$ is a solid torus and $$H'[k]$$ is a Seifert fiber space over the disk with two exceptional fibers. The purpose of this section is to show that $$T(p,q,r,s)$$ is a primitive/Seifert-fibered knot. A result of Miyazaki and Motegi enables us to conclude that $$T(p,q,r,s)$$ is a cable of a torus knot. See [11, Theorem 1.2]. Throughout this section, we let $$K=T(p,q)$$ and $$K'=T(p,q,r,s)$$. Recall that there are two solid tori $$V_1,V_2$$ such that $$V_1\cup V_2=S^3$$ and $$\Sigma=V_1\cap V_2=\partial V_1=\partial V_2$$ is a torus containing the knot $$K$$. Here, $$V_1$$ and $$V_2$$ are indexed so that $$K$$ wraps around $$V_1$$$$p$$ times and $$V_2$$$$q$$ times. An unknotted circle $$C$$ is taken in $$S^3-K$$ so that it encircles $$r$$ adjacent parallel strands of $$K$$ and intersects each $$V_i$$ in a trivial arc. Then $$K'$$ is the image of $$K$$ after $$(-1/s)$$-surgery on $$C$$. The exterior of $$K$$ contains an essential annulus, say $$\widehat{P}(\subset \Sigma)$$, with boundary slope $$pq$$. We may assume that $$C$$ intersects $$\widehat{P}$$ transversely in two points. Let $$M=S^3-\mathrm{int}N(C\cup K)$$. Then $$M$$ is a hyperbolic $$3$$-manifold by [9, Proposition 5.7] and it is bounded by two tori, $$\partial_C M=\partial N(C)$$ and $$\partial_K M=\partial N(K)$$. Also, $$P=\widehat{P}\cap M$$ is a twice-punctured annulus with two boundary components in $$\partial_C M$$ and the other two boundary components in $$\partial_K M$$. Each component of $$\partial P\cap \partial_C M$$ has a trivial slope on $$\partial_C M$$. Let $$W_i=V_i\cap M$$ for each $$i=1,2$$ and let $$\partial_X W_i=W_i\cap \partial_X M$$ for each $$X=C,K$$. Then each $$W_i$$ is a genus two handlebody with $$\partial W_i=P\cup \partial_C W_i\cup \partial_K W_i$$, where $$\partial_X W_i$$ is an annulus for each $$X=C,K$$. Let $$X_i$$ denote the core of the annulus $$\partial_X W_i$$. Then $$C_i$$ and $$K_i$$ are disjoint nonisotopic curves on $$\partial W_i$$, and if we let $$U_i=W_i[C_i]$$, then $$U_i$$ is a solid torus such that $$U_1\cup_{\widehat{P}} U_2=E(K)$$. For a slope $$\alpha$$ on $$\partial_C M$$, let $$M(C;\alpha)$$ denote the result of $$\alpha$$-Dehn filling, that is, $$M(C;\alpha)=M\cup J_\alpha$$, where $$J_\alpha$$ is a solid torus glued along $$\partial_C M$$ so that $$\alpha$$ bounds a meridian disk of $$J_\alpha$$. Note that $$M(C;1/0)$$ and $$M(C;-1/s)$$ are the exteriors of $$K$$ and $$K'$$, respectively. If $$\beta$$ is a slope on $$\partial_K M$$, then $$M(C,K;\alpha,\beta)=M(K,C;\beta,\alpha)$$ is defined to be the result of $$\beta$$-Dehn filling on $$M(C;\alpha)$$ along the torus $$\partial_K M$$. The annulus $$\widehat{P}$$ intersects $$J_{1/0}$$ in two meridian disks, say $$u_1$$ and $$u_2$$. Also, $$P$$ is essential in $$M$$ by [9, Lemma 5.3]. Since $$K'$$ is a satellite knot, its exterior $$M(C;-1/s)$$ contains an essential torus. Among all essential tori in $$M(C;-1/s)$$, we choose an essential torus $$\widehat{T}$$ to intersect $$J_{-1/s}$$ in a minimal number of meridian disks, say $$v_1,\ldots,v_t$$, successively indexed along $$J_{-1/s}$$ (note that $$t$$ must be even, since $$\widehat{T}$$ is separating). We also assume that $$\widehat{T}$$ is chosen so that $$T=\widehat{T}\cap M$$ intersects $$P$$ transversely and minimally. Then $$\partial u_x\cap \partial v_y$$ consists of exactly two points for each pair of disks $$u_x,v_y$$ by the assumption $$|s|=2$$ at the beginning of the paper. We construct two labeled graphs $$G_P$$ on $$\widehat{P}$$ and $$G_T$$ on $$\widehat{T}$$ as follows. The edges of the graphs are the arc components of $$P\cap T$$, the (fat) vertices are the disks $$u_x$$ or $$v_y$$ on $$G_P$$ or $$G_T$$, and each point in $$\partial u_x\cap \partial v_y$$, which is an endpoint of an edge in both graphs, is labeled $$y$$ in $$G_P$$ and labeled $$x$$ in $$G_T$$. We assume familiarity with the definitions and terminology in [8]: positive/negative edges, the parity rule, $$S$$-cycles, etc. We remark that $$T$$ is incompressible and boundary-incompressible in $$M$$ by the minimality of $$t$$. It follows that $$G_P$$ contains no trivial loops. We also remark that any disk face of $$G_T$$ is contained in $$W_1$$ or $$W_2$$. Lemma 3.1. Suppose that a disk face of $$G_T$$ is contained in $$W_i(i=1,2)$$. Then $$W_i[K_i]$$ is a solid torus. Furthermore, if the disk face is nonseparating in $$W_i$$, then it is a meridian disk of the solid torus. □ Proof Let $$f$$ be a disk face of $$G_T$$ which is contained in $$W_i$$ for some $$i=1,2$$. Note that $$f$$ is a properly embedded disk in $$W_i$$ with its boundary disjoint from the annulus $$\partial_K W_i$$. Its boundary circle, $$\partial f$$, is an alternating sequence of properly embedded arcs in $$P$$ and spanning arcs in the annulus $$\partial_C W_i$$, the former being the edges of $$f$$ and the latter being the corners of $$f$$. We first claim that $$f$$ is an essential disk in $$W_i$$. Assume otherwise. Then $$\partial f$$ bounds a disk $$D$$ on $$\partial W_i$$. The core $$C_i$$ of the annulus $$\partial_C W_i$$ intersects $$D$$ in a set of arcs. An outermost arc together with a subarc of $$\partial D=\partial f$$ bounds a subdisk of $$D$$. This implies that the edge of $$f$$ corresponding to the subarc is a trivial loop in $$G_P$$, a contradiction. Since $$K_i$$ is disjoint from the essential disk $$f$$ in $$W_i$$, one can take a nonseparating essential disk in $$W_i$$ which is disjoint from $$K_i$$. (If $$f$$ is itself a nonseparating disk, then $$f$$ can be taken to be the nonseparating disk.) Thus $$W_i[K_i]$$ is $$\partial$$-reducible and it is a solid torus by [10, Lemma 2.1]. Furthermore, if $$f$$ is a nonseparating disk in $$W_i$$, then it is a meridian disk of the solid torus $$W_i[K_i]$$. ■ Lemma 3.2. $$G_T$$ contains no $$S$$-cycle. □ Proof Suppose that $$G_T$$ contains an $$S$$-cycle. Let $$f$$ be the disk face of $$G_T$$ bounded by the $$S$$-cycle and suppose that $$f$$ is contained in $$W_i$$. Then the oriented intersection of $$C_i$$ with $$\partial f$$ consists of two points of the same sign. This implies that $$f$$ is a nonseparating disk in $$W_i$$. By Lemma 3.1 $$W_i[K_i]$$ is a solid torus in which $$f$$ is a meridian disk. Since $$C_i$$ intersects $$\partial f$$ in two points, the fundamental group of $$W_i[K_i][C_i]$$ is a cyclic group of order $$2$$. Note that $$W_i[K_i][C_i]=W_i[K_i\cup C_i]=W_i[C_i][K_i]=U_i[K_i]$$ and that $$U_i$$ is a solid torus around which the curve $$K_i$$ wraps $$p$$ or $$q$$ times according to whether $$i=1$$ or $$2$$. Thus we have $$p=2$$ (and then $$q=1$$) or $$q=2$$, any of which contradicts Lemma 2.7. ■ Lemma 3.3. If all disk faces of $$G_T$$ lie on one side of $$P$$, then $$t=2$$ and the graphs $$G_P,G_T$$ and their edge correspondence are as shown in Figure 13(a and c). □ Fig. 13. View largeDownload slide Graphs of intersection. Fig. 13. View largeDownload slide Graphs of intersection. Proof Let $$V,E,F$$ be the number of vertices, edges, and disk faces of $$G_T$$, respectively. Every vertex of $$G_T$$ has valence $$4$$, so $$4V=2E$$ and hence $$0=\chi(\widehat{T})\le V-E+F= \frac{E}{2}-E+F=-\frac{E}{2}+F$$, giving $$2F\ge E(=2V=2t)$$. It follows that $$G_T$$ contains at least $$t$$ disk faces. Assume that all disk faces of $$G_T$$ are contained in $$W_i$$ for some fixed $$i\in\{1,2\}$$. Then since every disk face of $$G_T$$ has at least two sides, we have $$2F\le E$$. Combining the above inequalities, we get $$E=2F$$ and conclude that every disk face of $$G_T$$ is a bigon, which must be bounded by negative edges according to Lemma 3.2. Also, every edge of $$G_T$$ belongs to a bigon face. Therefore all edges of $$G_T$$ are negative. All edges of $$G_P$$ are positive by the parity rule and each vertex of $$G_P$$ is a base of $$t$$ parallel positive edges. If $$t\ge 4$$, then $$G_P$$ has an $$S$$-cycle immediately surrounded by two parallel edges, contradicting [8, Lemma 6(3)]. Suppose $$t=2$$. Then each of the graphs $$G_P$$ and $$G_T$$ has four edges, say, $$e_1,e_2,e_3,e_4$$. Up to homeomorphisms of $$\widehat{P}$$ and $$\widehat{T}$$, there are exactly two possibilities for each of $$G_P$$ and $$G_T$$ as shown in Figure 13. If $$G_P$$ is the graph in Figure 13(b), then one can see that for any possibility for $$G_T$$ (see Figure 13(c or d)), the corners of the two bigon faces of $$G_T$$ cannot be placed to be mutually disjoint and parallel in the annulus $$\partial_C W_i$$: take the four corners of the two bigon faces of $$G_T$$ and apply [5, Lemma 3.2] to these corners. Thus $$G_P$$ must be the graph in Figure 13(a). Suppose that $$G_T$$ is the graph in Figure 13(d). Let $$H_i$$ denote the $$1$$-handle $$J_{1/0}\cap U_i$$. Let $$f$$ be the bigon face of $$G_T$$ bounded by $$e_1$$ and $$e_3$$. The edges $$e_1,e_3$$ divide $$P$$ into three annuli $$A_1,A_2,A_3$$. One of these annuli, say, $$A_2$$ is disjoint from $$\partial \widehat{P}$$. By shrinking $$H_i$$ to its core, $$H_i\cup f\cup A_2$$ becomes a Klein bottle in the solid torus $$U_i$$. This is impossible. Thus $$G_T$$ must be the graph in Figure 13(c). ■ Lemma 3.4. $$W_i[K_i]$$ is a solid torus for each $$i=1,2$$. □ Proof The first paragraph in the proof of Lemma 3.3 shows that $$G_T$$ has disk faces. If, for each $$i=1,2$$, $$G_T$$ has a disk face contained in $$W_i$$, then the result follows from Lemma 3.1. Assume otherwise, that is, all disk faces of $$G_T$$ are contained in $$W_i$$ for some fixed $$i\in\{1,2\}$$ (then $$W_i[K_i]$$ is a solid torus) and let $$j=3-i$$ so that $$\{i,j\}=\{1,2\}$$. Then by Lemma 3.3, $$t=2$$ and the graphs $$G_P,G_T$$ and their edge correspondence are as shown in Figure 13(a and c). We will eventually show that $$W_j[K_j]$$ is also a solid torus. Note that $$G_T$$ has precisely one annulus face. Let it be $$A$$. We first claim that $$P\cap T$$ contains no circle components. Assume otherwise. If $$P\cap T$$ contains a circle component which is inessential in both $$P$$ and $$T$$, then applying a disk-swapping argument one can reduce $$|P\cap T|$$, contradicting the minimality of $$|P\cap T|$$. If $$P\cap T$$ contains a circle component which is inessential in one of $$P$$ and $$T$$ and essential in the other, then one of $$P$$ and $$T$$ is compressible, a contradiction. Thus all circle components of $$P\cap T$$ are essential on both $$P$$ and $$T$$. In particular, they are all parallel and essential in $$A$$. An outermost such circle in $$A$$, say, $$c$$ cuts off a subannulus $$A'$$ from $$A$$. We may assume that $$\partial A'-c$$ contains $$e_1\cup e_3$$ (otherwise, $$\partial A'-c$$ contains $$e_2\cup e_4$$ and the same argument applies). The annulus $$A'$$ is contained in $$W_j$$, and $$\partial A'-c$$ intersects the annulus $$\partial_C W_j$$ in two spanning arcs, cutting $$\partial_C W_j$$ into two rectangles. See Figure 14(a). Take any rectangle and let $$Q$$ be the union of $$A'$$ and the rectangle. Then $$Q$$ is a properly embedded thrice-punctured sphere in $$U_j$$, one of whose boundary components is $$c$$ and the other two contain edges $$e_1$$ and $$e_3$$, respectively. The boundary components of $$Q$$ are mutually parallel in $$\widehat{P}$$ and are isotopic to $$K_j$$ in the torus $$\partial U_j$$. Taking a subannulus of $$\widehat{P}$$ between any two adjacent boundary components of $$Q$$ in $$\widehat{P}$$ and attaching it to $$Q$$, we obtain a once-punctured torus. This implies that the knot $$K_j$$, which is isotopic to $$K$$ in $$S^3$$, has genus at most one. But, $$g(K)=(p-1)(q-1)/2>1$$ by Lemma 2.7, a contradiction. Hence $$P\cap T$$ consists only of arc components and $$A$$ is entirely contained in $$W_j$$. Since $$T$$ is incompressible and $$\partial$$-incompressible in $$M$$, $$A$$ is incompressible and not $$\partial$$-parallel in $$W_j$$: if $$A$$ were $$\partial$$-parallel in $$W_j$$, then a disk $$\Delta(\subset W_j)$$ would guide a $$\partial$$-compression of $$T$$ in $$M$$ as shown in Figure 15. Also, since $$\widehat{T}$$ is separating, $$A$$ is separating in $$W_j$$. Thus by [6, Lemma 3.2], $$A$$ splits $$W_j$$ into a genus two handlebody $$X$$ and a solid torus $$Y$$ so that $$A$$ wraps around $$Y$$ at least two times. See Figure 16. As shown in Figure 14(b), $$\partial A$$ intersects the annulus $$\partial_C W_j$$ in four spanning arcs, cutting $$\partial_C W_j$$ into four rectangles. Let $$a_1,a_2,a_3,a_4$$ be the spanning arcs, successively indexed along $$\partial_C W_j$$ as in Figure 14(b), and let $$R_k$$ be the rectangle between $$a_k$$ and $$a_{k+1}$$ (subscripts understood modulo $$4$$). Then the union $$S=A\cup R_1\cup R_3$$ is a properly embedded four-punctured sphere in the solid torus $$U_j$$, which must be separating and compressible in $$U_j$$. Note that $$S$$ divides $$U_j$$ into two genus two handlebodies $$X$$ and $$Z=Y\cup H_j$$, where $$H_j$$ is the $$1$$-handle given by $$H_j=J_{1/0}\cap U_j$$ and attached to $$Y$$ along the two rectangles $$R_2\cup R_4$$. We claim that $$S$$ is incompressible in $$Z$$ (and hence it must be compressible in $$X$$). Suppose that $$S$$ is compressible in $$Z$$. We may regard the $$1$$-handle $$H_j$$ as a product manifold $$(\text{a rectangle})\times [0,1]$$, that is, there is a rectangle $$R$$ in $$H_j$$, parallel to $$R_2$$ and $$R_4$$, such that $$H_j=R\times [0,1]$$ and $$R_{2k}=R\times\{k-1\}$$ for $$k=1,2$$. Choose a compressing disk $$D$$ for $$S$$ so that $$D$$ and $$R$$ intersect transversely and minimally. Then $$D\cap R$$ consists of a finite number of properly embedded arcs in $$D$$. If $$D$$ is disjoint from $$H_j$$ (and hence from $$R$$) then $$\partial D$$ lies in $$A$$ and $$D$$ compresses $$T$$, a contradiction. Otherwise, take an arc component $$a$$ of $$D\cap R$$ which is outermost in $$D$$. Then $$a\times [0,1]\subset D\cap H_j=(D\cap R)\times [0,1]$$ cuts off a subdisk from $$D$$, which serves as a $$\partial$$-compressing disk for $$T$$. This is a contradiction. Hence $$S$$ must be compressible in $$X$$. Each component of $$\partial S$$ is parallel to $$K_j$$ in $$\partial U_j$$, so it cannot bound a disk in $$U_j$$. It follows that $$S$$ compresses into two disjoint incompressible annuli, say, $$A_0$$ and $$A_1$$. Conversely, $$S$$ can be recovered from $$A_0$$ and $$A_1$$ by tubing along a (knotted) arc $$\alpha$$ connecting the two annuli. By [6, Lemma 3.1], each of these annuli is $$\partial$$-parallel in $$U_j$$. We divide our arguments into two cases depending on whether the annuli are nested or not. First, suppose that $$A_0$$ and $$A_1$$ are nested, that is, they cut off a product region $$A_0\times [0,1]$$ from $$U_j$$, where $$A_0=A_0\times\{0\}$$ and $$A_1=A_0\times\{1\}$$. The arc $$\alpha$$ is properly embedded in $$A_0\times [0,1]$$ and has one point in each end of the product. Removing an open regular neighborhood of $$\alpha$$ from the product, we obtain $$Z$$, that is, $$Z=A_0\times [0,1]-\eta(\alpha)$$, where $$\eta(\alpha)$$ is an open regular neighborhood of $$\alpha$$. See Figure 17 in which $$A_\alpha(\subset\partial\eta(\alpha))$$ denotes a tube along $$\alpha$$, connecting $$A_0$$ and $$A_1$$. The intersection $$Z\cap \partial U_j$$ consists of two annuli in $$(\partial A_0)\times [0,1]$$, one being the union of the disk $$u_1$$ and the bigon face of $$G_P$$ bounded by $$e_1\cup e_2$$ and the other being the union of the disk $$u_2$$ and the bigon face bounded by $$e_3\cup e_4$$ (see Figure 14(b)). Let $$\gamma_1,\gamma_2$$ be the cores of these two annuli. Then attaching two $$2$$-handles to $$A_0\times [0,1]$$ along $$\gamma_1\cup\gamma_2$$ gives rise to the product manifold $$S^2\times [0,1]$$ in which $$\alpha$$ is properly embedded, connecting $$S^2\times\{0\}$$ and $$S^2\times\{1\}$$. The classical light bulb trick (see [15, Exercise E4 in Chapter 9]) shows that $$Z[\gamma_1\cup\gamma_2]=S^2\times [0,1]-\eta(\alpha)$$ is homeomorphic to a $$3$$-ball, and each of the curves $$\gamma_1$$ and $$\gamma_2$$ intersects $$\partial R_2$$ (or $$\partial R_4$$) in a single point. It follows from [2, Lemma 2.3.2] that $$S$$ is compressible in $$Z$$, a contradiction. Next, suppose that $$A_0$$ and $$A_1$$ are not nested, that is, the annuli cut off two disjoint solid tori from $$U_j$$. These solid tori are connected by the arc $$\alpha$$ and the genus two handlebody $$X$$ is the union of these solid tori together with a regular neighborhood of $$\alpha$$. The boundary of $$X$$ is the union of $$S$$ and two annuli $$B_0\cup B_1$$ as shown in Figure 18(a), where $$B_0$$ and $$B_1$$ are subannuli of $$\partial U_j$$ such that $$\partial B_0\cup \partial B_1=\partial S$$ (see Figure 14(b)). We may assume that the annulus $$\partial_K W_j$$ is contained in $$B_1$$. Note that $$\partial X\cap H_j(=S\cap H_j)$$ consists of two rectangles $$R_1\cup R_3$$. The rectangle $$R_k(k=1,3)$$ is bounded by the two arcs $$a_k,a_{k+1}$$ and two other arcs in $$\partial u_1\cup\partial u_2$$, and we may regard $$R_k$$ as obtained by thickening $$a_k$$ properly in $$S=A\cup R_1\cup R_3$$. The genus two handlebody $$X$$ can be regarded as a $$3$$-ball with two unknotted holes (see Figure 18(a)) and the pair $$a_1\cup a_3$$ is a pair of arcs in $$S(\subset \partial X)$$ whose boundary has one endpoint in each component of $$\partial S$$. The core $$\gamma$$ of $$A(\subset S)$$ is an essential curve in $$S$$ and separates the arcs $$a_1$$ and $$a_3$$. There is a natural one-to-one correspondence between isotopy classes of essential curves in $$S$$ and the extended rationals in $$\mathbb{Q}\cup\{1/0\}$$: the boundary of a disk in $$X$$ separating the unknotted holes corresponds to $$1/0$$ and the equator $$\iota$$ shown in Figure 18(a) corresponds to $$0$$. Each arc in the pair $$a_1\cup a_3$$ has one endpoint on $$\partial B_0$$ and the other on $$\partial B_1$$ (see Figure 14(b)), so $$\gamma$$ corresponds to a rational number $$m/n$$ with $$n$$ odd. See Figure 18(b), which illustrates the case $$m/n=1/3$$. Recall that $$A=\overline{S-R_1\cup R_3}$$ is an incompressible, non-$$\partial$$-parallel, separating annulus in the handlebody $$W_j$$, splitting $$W_j$$ into $$X$$ and $$Y$$. It follows from [6, Lemma 3.2(i)] that $$\gamma$$ is a primitive curve on the genus two handlebody $$X$$, that is, $$X[\gamma]$$ is a solid torus. On the other hand, after an isotopy of $$X$$, we may consider $$X$$ as obtained from a $$(-m/n)$$-rational tangle by removing open regular neighborhoods of the stings of the rational tangle. See Figure 18(c), where $$\gamma$$ becomes a vertical circle separating the arcs $$a_1$$ and $$a_3$$. Using the figure, one easily sees that $$X[\gamma]$$ is the exterior of a $$2$$-bridge knot associated to the rational number $$-m/n$$. Since $$X[\gamma]$$ is a solid torus, $$-m/n$$ must be an integer and a further isotopy of $$X$$ allows us to assume $$-m/n=0$$. See Figure 18(d). Since $$K_j\subset\partial_K W_j\subset B_1$$, $$X[K_j]$$ is defined and it is a solid torus around which $$A$$ winds once. Thus $$W_j[K_j]=X[K_j]\cup_A Y$$ is a solid torus. This completes the proof. ■ Fig. 14. View largeDownload slide (a) $$\partial A'-c$$ and (b) $$\partial A$$. Fig. 14. View largeDownload slide (a) $$\partial A'-c$$ and (b) $$\partial A$$. Fig. 15. View largeDownload slide If $$A$$ were $$\partial$$-parallel in $$W_j$$, then $$T$$ would be $$\partial$$-compressible in $$M$$. Fig. 15. View largeDownload slide If $$A$$ were $$\partial$$-parallel in $$W_j$$, then $$T$$ would be $$\partial$$-compressible in $$M$$. Fig. 16. View largeDownload slide A separating essential annulus in a genus two handlebody. Fig. 16. View largeDownload slide A separating essential annulus in a genus two handlebody. Fig. 17. View largeDownload slide $$Z=A_0\times [0,1]-\eta (\alpha)$$. Fig. 17. View largeDownload slide $$Z=A_0\times [0,1]-\eta (\alpha)$$. Fig. 18. View largeDownload slide $$\partial X=S\cup B_0\cup B_1$$. Fig. 18. View largeDownload slide $$\partial X=S\cup B_0\cup B_1$$. Proposition 3.5. (1) $$r\equiv \pm 1$$ or $$\pm q\mod p$$ and $$r\equiv \pm 1$$ or $$\pm p\mod q$$. (2) There are only five possibilities for the value of $$r$$:   r=p−1,p+1,p−q,p+q, or 2p−q.$$\quad$$ Moreover, if $$r=2p-q$$, then $$p<2q$$. □ Proof (1) follows from Lemma 3.4 and [10, Lemma 2.1]. (2) Since $$2\le r\le p+q$$, the congruence equation $$r\equiv \pm 1$$ or $$\pm q\mod p$$ places restrictions on the value of $$r$$:   r=p−1,p+1,p−q,p+q, or 2p−q. In the case that $$r=2p-q$$, we obtain $$p<2q$$ by using the fact that $$r\le p+q$$ and $$p,q$$ are relatively prime. ■ Lemma 3.6. Let $$Z$$ be the $$3$$-manifold constructed by Dehn surgery on the cable knot $$T_{a,b}(c,d)$$ with an integral slope $$n$$, where $$c>|d|\ge 2$$. Suppose that $$Z$$ is a small Seifert fiber space. Then $$n=ab\pm 1$$ and $$Z$$ has base orbifold a $$2$$-sphere with three cone points of indices $$c,|d|$$, and $$|n-a^2cd|$$. □ Proof Let $$k=T(c,d)$$ and $$k'=T_{a,b}(c,d)$$. Note that $$T_{a,b}(c,d)$$ is $$C_{b,a}(T(d,c))$$ in the notation of [4, p. 692]. By [4, Corollary 7.3] $$n=ab\pm 1$$ and $$Z=k'(n)=k(n/a^2)$$. The integers $$n$$ and $$a^2$$ are relatively prime, so $$n/a^2$$ is a reduced fraction. By [4, Corollary 7.4] $$Z$$ has base orbifold a $$2$$-sphere with three cone points of indices $$c,|d|$$, and $$|n-a^2cd|$$. ■ Proposition 3.7. There are integers $$\varepsilon_1,\varepsilon_2,\varepsilon_3,\varepsilon_4=\pm 1$$ and $$m\ge 2$$ satisfying one of the following: (1) $$T(p,q,r,s)=T_{m, \varepsilon_1 mpq+\varepsilon_2}(p,\varepsilon_1 q)\text{ and } pq+r^2s-\varepsilon_1m^2pq=\varepsilon_2 m+\varepsilon_3=2\varepsilon_4;$$ (2) $$T(p,q,r,s)=T_{m,2\varepsilon_1 mp+\varepsilon_2}(p,2\varepsilon_1)\text{ and } pq+r^2s-2\varepsilon_1m^2p=\varepsilon_2 m+\varepsilon_3=\varepsilon_4 q;\text{ or}$$ (3) $$T(p,q,r,s)=T_{m,2\varepsilon_1 mq+\varepsilon_2}(q,2\varepsilon_1)\text{ and } pq+r^2s-2\varepsilon_1m^2q=\varepsilon_2 m+\varepsilon_3=\varepsilon_4 p.$$ □ Proof Note that $$M(K,C;1/0,-1/s)=S^3$$ and that the exterior of $$K'=T(p,q,r,s)$$ is given as   M(C;−1/s)=M∪∂CMJ−1/s=(W1∪PW2)∪∂CMJ−1/s=W1∪P1(W2∪∂CW2J−1/s), where $$P_1=P\cup \partial_C W_1$$ is a twice punctured torus. We may consider that $$P_1$$ is properly embedded in the exterior of $$K$$ as well as in the exterior of $$K'$$. Recall that $$W_1$$ is a genus two handlebody. Since the core $$C_2$$ of the splitting annulus $$\partial_C W_2$$ of the union $$W_2\cup J_{-1/s}$$ is a primitive curve in $$W_2$$ (i.e., $$W_2[C_2]=U_2$$ is a solid torus), $$W_2\cup J_{-1/s}$$ is also a genus two handlebody. Since the slope of the boundary of $$P_1$$ in the exterior of $$K$$ is $$pq$$, it follows from [15, Proposition H2 in Chapter 9] that the slope is changed to $$pq+r^2s$$, an integral slope, in the exterior of $$K'$$. Thus if we let $$N(K')=S^3-\mathrm{int}(M(C;-1/s))$$ be a regular neighborhood of $$K'$$, then each component of $$\partial P_1$$ (and hence the curve $$K_1$$) wraps around the solid torus $$N(K')$$ once in the longitudinal direction. Thus $$N(K')\cup_{\partial_K W_1} W_1(\cong W_1)$$ is a genus two handlebody. Also,   S3=M(K,C;1/0,−1/s)=N(K′)∪M(C;−1/s)=(N(K′)∪∂KW1W1)∪(W2∪∂CW2J−1/s). Here, $$S^3=(N(K')\cup_{\partial_K W_1} W_1)\cup (W_2\cup_{\partial_C W_2} J_{-1/s})$$ is a genus two Heegaard decomposition of $$S^3$$ and $$P_1\cup \partial_K W_2$$ is the splitting surface of the decomposition. The splitting surface can be isotoped fixing $$P_1$$ so that it contains $$K'$$ and intersects $$\partial N(K')$$ in two curves in $$\partial P_1$$. Consider $$K'(pq+r^2s)$$. It is given as   K′(pq+r2s)=M(K,C;pq,−1/s)=W1[K1]∪(W2[K2]∪J−1/s). By Lemma 3.4, $$W_1[K_1]$$ and $$W_2[K_2]$$ are solid tori. Recall that the splitting surface of the union $$W_2[K_2]\cup J_{-1/s}$$ is the annulus $$\partial_C W_2$$. Its core, $$C_2$$, wraps around the solid torus $$J_{-1/s}$$$$|s|$$ times in the longitudinal direction. Since $$W_2[K_2][C_2]=W_2[K_2\cup C_2]=W_2[C_2][K_2]=U_2[K_2]$$ is a once punctured lens space $$L(q,p)$$, $$C_2$$ wraps around the solid torus $$W_2[K_2]$$$$q$$ times in the longitudinal direction. This implies that the union $$W_2[K_2]\cup J_{-1/s}$$ is a Seifert fiber space over the disk with two exceptional fibers of indices $$|s|$$ and $$q$$. The common boundary torus of $$W_1[K_1]$$ and $$W_2[K_2]\cup J_{-1/s}$$ contains the annulus $$\partial_C W_1$$. The core of this annulus, $$C_1$$, is a Seifert fiber of the Seifert fiber space $$W_2[K_2]\cup J_{-1/s}$$, since it is isotopic to $$C_2$$ in $$\partial J_{-1/s}=\partial_C M$$. Also, $$C_1$$ wraps around the solid torus $$W_1[K_1]$$$$p$$ times in the longitudinal direction, since $$W_1[K_1][C_1]=W_1[K_1\cup C_1]=W_1[C_1][K_1]=U_1[K_1]$$ is a once punctured lens space $$L(p,q)$$. The Seifert fibration of $$W_2[K_2]\cup J_{-1/s}$$ extends to the solid torus $$W_1[K_1]$$ so that the curve $$C_1$$ becomes a regular fiber, wrapping around $$W_1[K_1]$$$$p$$ times. It follows that $$K'(pq+r^2s)=W_1[K_1]\cup (W_2[K_2]\cup J_{-1/s})$$ is a Seifert fiber space over the $$2$$-sphere with three exceptional fibers of indices $$p,q$$, and $$|s|=2$$. Hence $$K'=T(p,q,r,s)$$ is an $$(m,mcd+\varepsilon)$$-cable knot of a $$(c,d)$$-torus knot by [11, Theorem 1.2], where $$m\ge 2,c>|d|\ge 2$$, and $$\varepsilon=\pm 1$$. By Lemma 3.6   pq+r2s=m2cd+εm+ε′(ε′=±1) and one of the following holds:   (c,d,pq+r2s−m2cd)=(p,ε″q,2ε‴);=(p,2ε″,ε‴q); or=(q,2ε″,ε‴p) for some integers $$\varepsilon'',\varepsilon'''=\pm 1$$. If we let $$\varepsilon_1=\varepsilon'',\varepsilon_2=\varepsilon,\varepsilon_3=\varepsilon'$$, and $$\varepsilon_4=\varepsilon'''$$, then the result follows. ■ According to Proposition 3.7, we divide our arguments into three cases as follows: Case A: $$T(p,q,r,s)=T_{m, \varepsilon_1 mpq+\varepsilon_2}(p,\varepsilon_1 q)\text{ and } pq+r^2s-\varepsilon_1m^2pq=\varepsilon_2 m+\varepsilon_3=2\varepsilon_4;$$ Case B: $$T(p,q,r,s)=T_{m,2\varepsilon_1 mp+\varepsilon_2}(p,2\varepsilon_1)\text{ and } pq+r^2s-2\varepsilon_1m^2p=\varepsilon_2 m+\varepsilon_3=\varepsilon_4 q;\text{ or}$$ Case C: $$T(p,q,r,s)=T_{m,2\varepsilon_1 mq+\varepsilon_2}(q,2\varepsilon_1)\text{ and } pq+r^2s-2\varepsilon_1m^2q=\varepsilon_2 m+\varepsilon_3=\varepsilon_4 p.$$ In the following three sections, we will show any of these cases gives a contradiction, which completes the proof of Theorem 1.1. 4 Case A Throughout this section, we assume that there are integers $$\varepsilon_i=\pm 1(i=1,2,3,4)$$ and $$m\ge 2$$ such that $$T(p,q,r,s)=T_{m, \varepsilon_1 mpq+\varepsilon_2}(p,\varepsilon_1 q)$$ and $$pq+r^2s-\varepsilon_1m^2pq=\varepsilon_2 m+\varepsilon_3=2\varepsilon_4$$. Multiplying both sides of the equation $$\varepsilon_2 m+\varepsilon_3=2\varepsilon_4$$ by $$\varepsilon_2$$, we get $$m+\varepsilon_2\varepsilon_3=2\varepsilon_2\varepsilon_4$$. Since $$m\ge 2$$, $$m=2\varepsilon_2\varepsilon_4-\varepsilon_2\varepsilon_3$$ implies that $$m=3$$ and $$\varepsilon_2\varepsilon_4=1,\varepsilon_2\varepsilon_3=-1$$. Let $$s=2\varepsilon_5(\varepsilon_5=\pm 1)$$. Then $$2\varepsilon_4=pq+r^2s-\varepsilon_1m^2pq=pq+2r^2\varepsilon_5-9\varepsilon_1pq$$ and hence we get   2r2+(ε5−9ε1ε5)pq=2ε4ε5. (4.1) If $$\varepsilon_1\varepsilon_5=-1$$, then $$2r^2+(\varepsilon_5-9\varepsilon_1\varepsilon_5)pq=2r^2+(9+\varepsilon_5)pq\ge 2r^2+8pq>8>2\varepsilon_4\varepsilon_5$$, which implies that equation (4.1) has no solution. Thus $$\varepsilon_1\varepsilon_5=1$$; $$(\varepsilon_1,\varepsilon_5)=(1,1)$$ or $$(\varepsilon_1,\varepsilon_5)=(-1,-1)$$. 4.1 Subcase $$\boldsymbol{r=p\pm 1}$$ In this subsection, we assume that $$r=p\pm 1$$. Let $$r=p+\varepsilon(\varepsilon=\pm 1)$$. From equation (4.1), we get   2ε4ε5=2(p+ε)2+(ε5−9ε1ε5)pq=2p2+4εp+2+(ε5−9)pq=2p(p+2ε+ε5−92q)+2, or   ε4ε5−1=p(p+2ε+ε5−92q). Since $$p(>q\ge 3)$$ divides $$\varepsilon_4\varepsilon_5-1$$, we must have $$\varepsilon_4\varepsilon_5-1=0$$ and $$p+2\varepsilon+\frac{\varepsilon_5-9}{2}q=0$$. Hence $$(\varepsilon_4,\varepsilon_5)=(1,1)$$ or $$(\varepsilon_4,\varepsilon_5)=(-1,-1)$$. First, suppose that $$(\varepsilon_4,\varepsilon_5)=(1,1)$$. Then $$p=\frac{9-\varepsilon_5}{2}q-2\varepsilon=4q-2\varepsilon,r=p+\varepsilon=4q-\varepsilon,s=2\varepsilon_5=2$$, and we get $$\varepsilon_1=\varepsilon_2=1$$ from the equations $$\varepsilon_1\varepsilon_5=1$$ and $$\varepsilon_2\varepsilon_4=1$$. The relation $$T(p,q,r,s)=T_{m, \varepsilon_1 mpq+\varepsilon_2}(p,\varepsilon_1 q)$$ yields   T(4q−2ε,q,4q−ε,2)=T3,3(4q−2ε)q+1(4q−2ε,q). If $$\varepsilon=-1$$, then $$T(p,q,r,s)=T(4q+2,q,4q+1,2)$$ is the knot $$T((4q+1,8q+2),(4q+2,q))$$ in the notation of [1] and has braid index $$\mathrm{min}(9q+2,4q+1)=4q+1$$ by [1, Corollary 8], but $$T_{m, \varepsilon_1 mpq+\varepsilon_2}(p,\varepsilon_1 q)=T_{3,3(4q+2)q+1}(4q+2,q)$$ has braid index $$3q$$ by [21, Theorem 1], and hence we get $$4q+1=3q$$, which is impossible for $$q\ge 3$$. Suppose that $$\varepsilon=1$$. Then $$T(p,q,r,s)=T(4q-2,q,4q-1,2)$$ is isotopic to the closure of the braid in Figure 19 by Lemma 2.1. The braid has $$4q-1$$ strands and $$(4q-2)(8q-1)+(q-2)(q-1)+(3q-1)(q-1)$$ crossings of the same sign. Hence the genus of $$T(p,q,r,s)$$ is given by $$g(T(p,q,r,s))=\frac{1}{2}(36q^2-31q+7)$$. On the other hand, $$T_{m, \varepsilon_1 mpq+\varepsilon_2}(p,\varepsilon_1 q)=T_{3,3(4q-2)q+1}(4q-2,q)$$ has genus $$\frac{1}{2}(36q^2-33q+9)$$ by Lemma 2.8. Hence we get $$36q^2-31q+7=36q^2-33q+9$$, which gives $$q=1$$ and contradicts Lemma 2.7. Fig. 19. View largeDownload slide $$T(4q-2,q,4q-1,2)$$ is isotopic to the closure of the braid. Fig. 19. View largeDownload slide $$T(4q-2,q,4q-1,2)$$ is isotopic to the closure of the braid. Suppose that $$(\varepsilon_4,\varepsilon_5)=(-1,-1)$$. Then $$p=\frac{9-\varepsilon_5}{2}q-2\varepsilon=5q-2\varepsilon,r=p+\varepsilon=5q-\varepsilon,s=2\varepsilon_5=-2$$, and we get $$\varepsilon_1=\varepsilon_2=-1$$ from the equations $$\varepsilon_1\varepsilon_5=1$$ and $$\varepsilon_2\varepsilon_4=1$$. The relation $$T(p,q,r,s)=T_{m, \varepsilon_1 mpq+\varepsilon_2}(p,\varepsilon_1 q)$$ yields   T(5q−2ε,q,5q−ε,−2)=T3,−3(5q−2ε)q−1(5q−2ε,−q). If $$\varepsilon=-1$$, then $$T(p,q,r,s)=T(5q+2,q,5q+1,-2)$$ has braid index $$5q+1$$ by Lemma 2.5, but $$T_{m, \varepsilon_1 mpq+\varepsilon_2}(p,\varepsilon_1 q)=T_{3,-3(5q+2)q-1}(5q+2,-q)$$ has braid index $$3q$$ by [21, Theorem 1], and hence we get $$5q+1=3q$$, which is impossible for $$q\ge 3$$. Suppose that $$\varepsilon=1$$. Then $$T(p,q,r,s)=T(5q-2,q,5q-1,-2)$$ is isotopic to the closure of the first braid in Figure 20 by Lemma 2.1. It is easy to see that the five braids in Figure 20 are isotopic. The last braid has $$5q-1$$ stands and $$(5q-2)(9q-3)+(q-1)$$ crossings of the same sign. Hence the genus of $$T(p,q,r,s)$$ is given by $$\frac{1}{2}(45q^2-37q+7)$$. On the other hand, $$T_{m, \varepsilon_1 mpq+\varepsilon_2}(p,\varepsilon_1 q)=T_{3,-3(5q-2)q-1}(5q-2,-q)$$ is the mirror image of $$T_{3,3(5q-2)q+1}(5q-2,q)$$, which has genus $$\frac{1}{2}(45q^2-36q+9)$$ by Lemma 2.8. Thus we get $$45q^2-37q+7=45q^2-36q+9$$, contradicting Lemma 2.7. Fig. 20. View largeDownload slide $$T(5q-2,q,5q-1,-2)$$ is isotopic to the closure of any of these braids. Fig. 20. View largeDownload slide $$T(5q-2,q,5q-1,-2)$$ is isotopic to the closure of any of these braids. 4.2 Subcase $$\boldsymbol{r=p+q}$$ In this subsection, we assume that $$r=p+q$$. Recall that either $$(\varepsilon_1,\varepsilon_5)=(1,1)$$ or $$(\varepsilon_1,\varepsilon_5)=(-1,-1)$$. First, suppose that $$(\varepsilon_1,\varepsilon_5)=(1,1)$$. Then equation (4.1) gives us $$2(p+q)^2-8pq=2\varepsilon_4$$ or equivalently $$(p-q)^2=\varepsilon_4$$, so $$\varepsilon_4=1$$ and $$p=q+1$$. Therefore $$T(p,q,r,s)=T(q+1,q,2q+1,2)$$ and hence $$b(T(p,q,r,s))\le 2q+1$$ by Lemma 2.2. On the other hand, $$T(p,q,r,s)=T_{m, \varepsilon_1 mpq+\varepsilon_2}(p,\varepsilon_1 q)=T_{3,3q(q+1)+\varepsilon_2}(q+1, q)$$ and $$b(T_{3,3q(q+1)+\varepsilon_2}(q+1, q))=3q$$ by Lemma 2.9. This yields $$3q\le 2q+1$$ or $$q\le 1$$, contradicting Lemma 2.7. Hence $$(\varepsilon_1,\varepsilon_5)=(-1,-1)$$ and equation (4.1) gives us $$p^2+q^2-3pq=-\varepsilon_4$$. It follows from [10, Lemma 4.1] that $$p=f_{n+1}$$ and $$q=f_{n-1}$$ for some integer $$n$$, where $$f_n$$ is the $$n$$th Fibonacci number. By Lemma 2.4 $$T(p,q,r,s)=T(f_{n+1},f_{n-1},f_{n-1}+f_{n+1},-2)$$ is isotopic to the mirror image of $$T(f_{n-1}+f_{n+1},f_{n-1}+2f_{n+1},f_{n-1},1)$$ because $$2r-q=2f_{n-1}+2f_{n+1}-f_{n-1}=f_{n-1}+2f_{n+1}$$. In the notation of [1], we have $$T(f_{n-1}+f_{n+1},f_{n-1}+2f_{n+1},f_{n-1},1) =T((f_{n-1},f_{n-1}),(f_{n-1}+f_{n+1},f_{n-1}+2f_{n+1}))$$. The braid index of this knot is equal to $$\mathrm{min}(f_{n-1}+f_{n+1},f_{n-1}+2f_{n+1})=f_{n-1}+f_{n+1}$$ by [1, Corollary 8]. On the other hand, it is easy to see that $$T(p,q,r,s)=T_{m, \varepsilon_1 mpq+\varepsilon_2}(p,\varepsilon_1 q)=T_{3,-3pq+\varepsilon_2}(p,-q)$$ has braid index $$3q=3f_{n-1}$$ by [21, Theorem 1]. This observation yields $$f_{n-1}+f_{n+1}=3f_{n-1}$$, or equivalently $$f_n=f_{n-1}(=1=q)$$, a contradiction. 4.3 Subcase $$\boldsymbol{r=2p-q}$$ In this subsection, we assume that $$r=2p-q$$. Then $$p<2q$$. Recall that $$r\equiv \pm 1$$ or $$\pm p\mod q$$ by Proposition 3.5(1). If $$r=2p-q\equiv p\mod q$$, then $$q$$ must divide $$p$$, a contradiction. If $$r=2p-q\equiv -p\mod q$$, then $$3p\equiv 0\mod q$$ and hence $$q=3$$ and $$p=4,5$$ (since $$q<p<2q$$), but any of $$(p,q,r)=(4,3,5),(5,3,7)$$ does not satisfy equation (4.1). Suppose $$r=2p-q\equiv \pm 1\mod q$$. Since $$q<p<2q$$, it is easy to see that $$2p=3q+\varepsilon(\varepsilon=\pm 1)$$. Multiplying both sides of equation (4.1) by $$2$$ and then substituting $$r=2p-q=2q+\varepsilon$$, we get   4(2q+ε)2+(ε5−9ε1ε5)(3q2+εq)=4ε4ε5. Recall that $$\varepsilon_1\varepsilon_5=1$$, so   4=|4ε4ε5|=|4(2q+ε)2+(ε5−9)(3q2+εq)|≥|(ε5−9)q(3q+ε)|−|4(2q+ε)2|≥8q(3q+ε)−4(2q+ε)2=8q(q−ε)−4≥8q(q−1)−4≥8⋅3⋅2−4=44, a contradiction. 4.4 Subcase $$\boldsymbol{r=p-q}$$ In this subsection, we assume that $$r=p-q$$. Recall that $$(\varepsilon_1,\varepsilon_5)=(1,1)$$ or $$(-1,-1)$$. First, suppose $$(\varepsilon_1,\varepsilon_5)=(1,1)$$. From equation (4.1), we get $$2(p-q)^2-8pq=2\varepsilon_4$$ or   p2−6pq+q2=ε4. Dividing both sides by $$pq$$, we get   p/q+1p/q=6+ε4pq≥6−1pq≥6−14⋅3(>5+15). Noting that $$t+1/t$$ is an increasing function for $$t\ge 1$$, one sees that $$p/q>5$$ or   p>5q. In the notation of [1], $$T(p,q,r,s)=T(p,q,p-q,2)$$ is the knot $$T((p-q,2(p-q)),(p,q))$$ and that $$p-q\ge q$$ (since $$p>5q$$). By [1, Corollary 8] the braid index of $$T(p,q,r,s)$$ is equal to $$\mathrm{min}(2(p-q)+q,p-q)=p-q$$. On the other hand, the cable knot $$T(p,q,r,s)=T_{m, \varepsilon_1 mpq+\varepsilon_2}(p,\varepsilon_1 q)=T_{3, 3pq+\varepsilon_2}(p, q)$$ has braid index $$3q$$ by [21, Theorem 1]. Thus $$p-q=3q$$, giving $$p=4q$$. This contradicts the above observation $$p>5q$$. Suppose $$(\varepsilon_1,\varepsilon_5)=(-1,-1)$$. From equation (4.1), we get $$2(p-q)^2-10pq\,{=} -2\varepsilon_4$$ or   p2−7pq+q2=−ε4. Dividing both sides by $$pq$$, we get   p/q+1p/q=7−ε4pq≥7−1pq≥7−14⋅3(>6+16) and see that   p>6q. Thus, by [10, Lemma 2.6], $$T(p,q,r,s)$$ is a fibered knot and its fiber surface has Euler characteristic $$\chi$$ given by   χ=(p−q)−((p−q−1)(2p−4q)+(p−2q)q)=−(2p2−5pq+2q2−3p+5q), so the genus of the knot is given by   g(T(p,q,r,s))=12(1−χ)=12(2p2−5pq+2q2−3p+5q+1). On the other hand, $$T(p,q,r,s)=T_{m, \varepsilon_1 mpq+\varepsilon_2}(p,\varepsilon_1 q)=T_{3, -3pq+\varepsilon_2}(p,-q)$$ is the mirror image of $$T_{3, 3pq-\varepsilon_2}(p,q)$$. The genus of $$T_{3, 3pq-\varepsilon_2}(p,q)$$ is equal to   12(9pq−3p−3q−2ε2+1) by Lemma 2.8. Thus $$\frac{1}{2}(2p^2-5pq+2q^2-3p+5q+1)=\frac{1}{2}(9pq-3p-3q-2\varepsilon_2+1)$$, giving $$-\varepsilon_2=p^2-7pq+q^2+4q=-\varepsilon_4+4q\ge -1+4q\ge -1+4\cdot 3=11$$. This gives a contradiction. 5 Case B Throughout this section, we assume that there are integers $$\varepsilon_i=\pm 1(i=1,2,3,4)$$ and $$m\ge 2$$ such that $$T(p,q,r,s)=T_{m,2\varepsilon_1 mp+\varepsilon_2}(p,2\varepsilon_1)$$ and $$pq+r^2s-2\varepsilon_1m^2p=\varepsilon_2 m+\varepsilon_3=\varepsilon_4 q.$$ Multiplying both sides of the equation $$\varepsilon_2 m+\varepsilon_3=\varepsilon_4 q$$ by $$\varepsilon_2$$, we get $$m+\varepsilon_2\varepsilon_3=\varepsilon_2\varepsilon_4q$$. Since $$m\ge 2$$, $$m=\varepsilon_2\varepsilon_4 q-\varepsilon_2\varepsilon_3$$ implies that   m=q+ε(ε=±1) and ε2ε4=1,ε2ε3=−ε. Let $$s=2\varepsilon_5(\varepsilon_5=\pm 1)$$. Then $$\varepsilon_4 q=pq+r^2s-2\varepsilon_1m^2p=pq+2\varepsilon_5r^2-2\varepsilon_1(q+\varepsilon)^2p$$ and hence we get   2r2+ε5pq−2ε1ε5p(q+ε)2=ε4ε5q. Note that $$2q^2-5q+2> 1$$ for $$q\ge 3$$. Hence if $$\varepsilon_1\varepsilon_5=-1$$, then   q≥ε4ε5q=2r2+ε5pq−2ε1ε5p(q+ε)2=2r2+ε5pq+2p(q+ε)2≥2r2−pq+2p(q−1)2=2r2+p(2q2−5q+2)>2r2+p>p>q, a contradiction. Hence   ε1ε5=1 and we get   2r2+ε5pq−2p(q+ε)2=ε4ε5q. (5.1) 5.1 Subcase $$\boldsymbol{r=p\pm 1}$$ In this subsection, we assume that $$r=p\pm 1$$ and let $$r=p+\varepsilon_6(\varepsilon_6=\pm 1)$$. From equation (5.1), we get $$\varepsilon_4\varepsilon_5q=2(p+\varepsilon_6)^2+\varepsilon_5pq-2p(q+\varepsilon)^2=2p^2+4\varepsilon_6p+2+\varepsilon_5pq-2p(q+\varepsilon)^2$$, giving   ε4ε5q−2=p(2p+4ε6+ε5q−2(q+ε)2). This implies that $$p$$ divides $$\varepsilon_4\varepsilon_5q-2(=q-2\text{ or } -q-2)$$. Since $$p>q$$, it is easy to see that $$q-2=0$$ or $$p=q+2$$ according to whether $$\varepsilon_4\varepsilon_5=1$$ or $$\varepsilon_4\varepsilon_5=-1$$. But $$q\ne 2$$ by Lemma 2.7. Hence $$\varepsilon_4\varepsilon_5=-1,p=q+2$$, and $$-1=2p+4\varepsilon_6+\varepsilon_5q-2(q+\varepsilon)^2=2(q+2)+4\varepsilon_6+\varepsilon_5q-2(q+\varepsilon)^2$$, giving   q(2+ε5−2q−4ε)=−3−4ε6. Since $$q$$ divides $$-3-4\varepsilon_6$$, we have $$\varepsilon_6=1,q=3+4\varepsilon_6=7$$, and   −1=2+ε5−2q−4ε=−12+ε5−4ε, which is impossible. 5.2 Subcase $$\boldsymbol{r=p-q,2p-q}$$ or $$\boldsymbol{p+q}$$ In this subsection, we assume that $$r=p-q,2p-q$$ or $$p+q$$. Let $$r=\delta p+\varepsilon_6q$$, where $$(\delta,\varepsilon_6)=(1,-1),(2,-1)$$ or $$(1,1)$$. From equation (5.1), we get $$\varepsilon_4\varepsilon_5q=2(\delta p+\varepsilon_6q)^2+\varepsilon_5pq-2p(q+\varepsilon)^2=2\delta^2p^2 +4\delta\varepsilon_6pq+2q^2+\varepsilon_5pq-2p(q+\varepsilon)^2$$, giving   −q(2q−ε4ε5)=p(2δ2p+4δε6q+ε5q−2(q+ε)2). Since $$p$$ is relatively prime to $$q$$, $$p$$ must divide $$2q-\varepsilon_4\varepsilon_5$$. Since $$p>q$$, it is easy to see that   p=2q−ε4ε5. Hence $$-q=2\delta^2p+4\delta\varepsilon_6q+\varepsilon_5q-2(q+\varepsilon)^2 =2\delta^2(2q-\varepsilon_4\varepsilon_5)+4\delta\varepsilon_6q+\varepsilon_5q-2(q+\varepsilon)^2$$, giving   q(q+2ε−2δ2−2δε6−1+ε52)=−1−δ2ε4ε5. (5.2) Note that $$q+2\varepsilon-2\delta^2-2\delta\varepsilon_6-\frac{1+\varepsilon_5}{2}$$ is an integer, so $$q$$ is a divisor of the integer $$-1-\delta^2\varepsilon_4\varepsilon_5$$. First, suppose $$(\delta,\varepsilon_6)=(1,-1)$$. Then $$r=p-q$$ and equation (5.2) becomes   q(q+2ε−1+ε52)=−1−ε4ε5. Since $$|-1-\varepsilon_4\varepsilon_5|\le 2$$, we must have $$q+2\varepsilon-\frac{1+\varepsilon_5}{2}=-1-\varepsilon_4\varepsilon_5=0$$. This implies that $$q=3,\varepsilon=-1,\varepsilon_5=1,\varepsilon_4=-1$$ and hence $$p=2q-\varepsilon_4\varepsilon_5=7, r=p-q=4, s=2\varepsilon_5=2,m=q+\varepsilon=2$$, and $$\varepsilon_1=1,\varepsilon_2=-1$$ from $$\varepsilon_1\varepsilon_5=1$$ and $$\varepsilon_2\varepsilon_4=1$$. Therefore   T(7,3,4,2)=T(p,q,r,s)=Tm,2ε1mp+ε2(p,2ε1)=T2,27(7,2). Since $$T(7,3,4,2)$$ is the closure of a positive braid with $$7$$ strands and $$42$$ crossings, $$g(T(7,3,4,2))=\frac{1-7+42}{2}=18$$. However, $$g(T_{2,27}(7,2))=19$$ by Lemma 2.8. This gives a contradiction. Next, suppose $$(\delta,\varepsilon_6)=(2,-1)$$. Then $$r=2p-q$$ and equation (5.2) becomes   q(q+2ε−9+ε52)=−1−4ε4ε5. Note that $$-1-4\varepsilon_4\varepsilon_5$$ equals $$3$$ or $$-5$$. If $$-1-4\varepsilon_4\varepsilon_5=3$$, then $$q=3$$ and $$q+2\varepsilon-\frac{9+\varepsilon_5}{2}=1$$, so $$\varepsilon=1,\varepsilon_5=-1,\varepsilon_4=1,$$ and $$p=2q-\varepsilon_4\varepsilon_5=7$$, contradicting $$p<2q$$. If $$-1-4\varepsilon_4\varepsilon_5=-5$$, then $$q=5$$ and $$q+2\varepsilon-\frac{9+\varepsilon_5}{2}=-1$$, so $$\varepsilon=-1,\varepsilon_5=-1,\varepsilon_4=-1,$$ and $$p=2q-\varepsilon_4\varepsilon_5=9,r=2p-q=13$$, contradicting Proposition 3.5(1). Finally, suppose $$(\delta,\varepsilon_6)=(1,1)$$. Then $$r=p+q$$ and equation (5.2) becomes   q(q+2ε−9+ε52)=−1−ε4ε5. Since $$|-1-\varepsilon_4\varepsilon_5|\le 2$$, we must have $$q+2\varepsilon-\frac{9+\varepsilon_5}{2}=-1-\varepsilon_4\varepsilon_5=0$$. This implies that $$(q,\varepsilon,\varepsilon_4,\varepsilon_5)=(3,1,-1,1),(2,1,1,-1)(7,-1,-1,1)$$, or $$(6,-1,1,-1)$$. Lemma 2.7 rules out the second possibility. Using the equalities $$p=2q-\varepsilon_4\varepsilon_5,r=p+q,s=2\varepsilon_5, m=q+\varepsilon, \varepsilon_1\varepsilon_5=1, \varepsilon_2\varepsilon_4=1$$, we get   (p,q,r,s;m,ε1,ε2)={(7,3,10,2;4,1,−1)if (ε,ε4,ε5)=(1,−1,1),(15,7,22,2;6,1,−1)if (ε,ε4,ε5)=(−1,−1,1),(13,6,19,−2;5,−1,1)if (ε,ε4,ε5)=(−1,1,−1). Since $$T(p,q,r,s)=T_{m,2\varepsilon_1 mp+\varepsilon_2}(p,2\varepsilon_1)$$, we have   T(7,3,10,2)=T4,55(7,2)if (ε,ε4,ε5)=(1,−1,1),T(15,7,22,2)=T6,179(15,2)if (ε,ε4,ε5)=(−1,−1,1),T(13,6,19,−2)=T5,−129(13,−2)if (ε,ε4,ε5)=(−1,1,−1). The calculation of the genera of these knots gives a contradiction as follows. We have $$g(T(7,3,10,2))=96$$ and $$g(T_{4,55}(7,2))=93$$ by Lemmas 2.3 and 2.8, so $$T(7,3,10,2)\ne T_{4,55}(7,2)$$. We also have $$g(T(15,7,22,2))=504$$ and $$g(T_{6,179}(15,2))=487$$, so $$T(15,7,22,2)\ne T_{6,179}(15,2)$$. Note that $$T_{5,-129}(13,-2)$$ is the mirror image of $$T_{5,129}(13,2)$$, so $$g(T_{5,-129}(13,-2))=g(T_{5,129}(13,2))=286$$. But $$g(T(13,6,19,-2))=294$$ by Lemma 2.4. Thus $$T(13,6,19,-2)\ne T_{5,-129}(13,-2)$$. 6 Case C Throughout this section, we assume that there are integers $$\varepsilon_i=\pm 1(i=1,2,3,4)$$ and $$m\ge 2$$ such that $$T(p,q,r,s)=T_{m,2\varepsilon_1 mq+\varepsilon_2}(q,2\varepsilon_1)$$ and $$pq+r^2s-2\varepsilon_1m^2q=\varepsilon_2 m+\varepsilon_3=\varepsilon_4 p.$$ Multiplying both sides of the equation $$\varepsilon_2 m+\varepsilon_3=\varepsilon_4p$$ by $$\varepsilon_2$$, we get $$m+\varepsilon_2\varepsilon_3=\varepsilon_2\varepsilon_4p$$. Since $$m\ge 2$$, $$m=\varepsilon_2\varepsilon_4p-\varepsilon_2\varepsilon_3$$ implies that $$m=p+\varepsilon(\varepsilon=\pm 1)$$ and $$\varepsilon_2\varepsilon_4=1,\varepsilon_2\varepsilon_3=-\varepsilon$$. Note that $$T_{m,2\varepsilon_1 mq+\varepsilon_2}(q,2\varepsilon_1)$$ is either $$T_{m,2 mq+\varepsilon_2}(q,2)$$ or the mirror image of $$T_{m,2 mq-\varepsilon_2}(q,2)$$, depending on whether $$\varepsilon_1=1$$ or $$\varepsilon_1=-1$$. Hence $$b(T_{m,2\varepsilon_1 mq+\varepsilon_2}(q,2\varepsilon_1))=2m$$ by Lemma 2.9. If $$r<p$$, then $$2(p-1)\le 2(p+\varepsilon)=2m=b(T_{m,2\varepsilon_1 mq+\varepsilon_2}(q,2\varepsilon_1))=b(T(p,q,r,s))\le p$$ by Lemma 2.2 and hence $$(q<)p\le 2$$, contradicting Lemma 2.7. Hence $$r>p$$ and $$2(p-1)\le 2(p+\varepsilon)=b(T(p,q,r,s))\le r$$, which implies $$r\ne p+1$$ and $$r\ne 2p-q$$ by Lemma 2.7. Hence $$r=p+q$$ and $$\varepsilon=-1$$. Let $$s=2\varepsilon_5(\varepsilon_5=\pm 1)$$. Then $$\varepsilon_4 p=pq+r^2s-2\varepsilon_1m^2q=pq+2\varepsilon_5(p+q)^2-2\varepsilon_1(p-1)^2q$$ and hence we get   2(p+q)2+ε5pq−2ε1ε5(p−1)2q=ε4ε5p. If $$\varepsilon_1\varepsilon_5=-1$$, then   p≥ε4ε5p=2(p+q)2+ε5pq+2(p−1)2q≥2(p+q)2−pq+2(p−1)2q=2p2+3pq+2q2+2(p−1)2q>p, a contradiction. Hence $$\varepsilon_1\varepsilon_5=1$$ and we get   2(p+q)2+ε5pq−2(p−1)2q=ε4ε5p, which yields   p(2p+8q+ε5q−ε4ε5−2pq)=−2(q−1)q. Since $$p$$ is relatively prime to $$q$$, $$p$$ must divide $$2(q-1)$$. Since $$p>q$$, it is easy to see that $$p=2(q-1)$$. Hence $$-q=2p+8q+\varepsilon_5q-\varepsilon_4\varepsilon_5-2pq =4(q-1)+8q+\varepsilon_5q-\varepsilon_4\varepsilon_5-4(q-1)q$$, giving   2q(2q−17+ε52)=−4−ε4ε5. The left-hand side of this equation is even, while the right-hand side is odd. This is impossible. Funding This work was supported by the National Research Foundation of Korea Grant funded by the Korean Government [NRF-2013R1A1A2A10064864]. Acknowledgment The author would like to thank the referee for a careful reading and helpful comments. References [1] Birman J. and Kofman. I. “A new twist on Lorenz links.” Journal of Topology  2, no. 2 ( 2009): 227– 48. Google Scholar CrossRef Search ADS   [2] Culler M. Gordon C. Luecke J. and Shalen. P. “Dehn surgery on knots.” Annals of Mathematics  125, no. 2 ( 1987): 237– 300. Google Scholar CrossRef Search ADS   [3] Dean J. C. “Hyperbolic knots with small Seifert-fibered Dehn surgeries.” PhD thesis, University of Texas at Austin, 1996. Google Scholar CrossRef Search ADS   [4] Gordon C. McA. “Dehn surgery and satellite knots.” Transactions of the American Mathematical Society  275, no. 2 ( 1983): 687– 708. Google Scholar CrossRef Search ADS   [5] Gordon C. McA., and Luecke J. “Non-integral toroidal Dehn surgeries.” Communications in Analysis and Geometry  12, no. 1–2 ( 2004): 417– 85. 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# Satellite Knots Obtained by Twisting Torus Knots: Hyperbolicity of Twisted Torus Knots

, Volume 2018 (3) – Feb 1, 2018
31 pages

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Oxford University Press
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1073-7928
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1687-0247
D.O.I.
10.1093/imrn/rnw255
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### Abstract

Abstract The twisted torus knot $$T(p,q,r,s)$$ is obtained from the torus knot $$T(p,q)$$ by twisting $$r$$ adjacent parallel strands fully $$s$$ times. In the case that $$r=p$$ or $$r$$ is a multiple of $$q$$, it is known that $$T(p,q,r,s)$$ is either a cable knot, a torus knot, or a trivial knot. Under the assumption that $$r\ne p$$ and $$r$$ is not a multiple of $$q$$, we show that if $$T(p,q,r,s)$$ is a satellite knot then $$|s|=1$$. As a consequence, we show that $$T(p,q,r,s)$$ is usually a hyperbolic knot for $$|s|\ge 2$$: exceptional cases arise only for restricted values of $$(p,q,r,s)$$. 1 Introduction A surface $$F$$, not a disk or sphere, properly embedded in a $$3$$-manifold $$X$$ is called an essential surface if it is incompressible, $$\partial$$-incompressible, and not parallel to a surface in $$\partial X$$. When $$F$$ is a disk, it is called an essential disk if $$\partial F$$ does not bound a disk in $$\partial X$$. When $$F$$ is a sphere, it is called an essential sphere if it does not bound a $$3$$-ball in $$X$$. Let $$k$$ be a knot in the $$3$$-sphere $$S^3$$ and let $$E(k)=S^3-\mathrm{int}N(k)$$ be its exterior, where $$N(k)$$ is a regular neighborhood of $$k$$ in $$S^3$$. Any slope$$\sigma$$ on $$\partial E(k)$$, the isotopy class of an essential simple closed curve on $$\partial E(k)$$, can be parameterized by $$\mathbb{Q}\cup\{1/0\}$$: if $$\sigma=a\mu+b\lambda$$, then we write $$\sigma=a/b$$, where $$(\mu,\lambda)$$ is a standard meridian-longitude system for $$H_1(\partial E(k))$$ (When we given a link in $$S^3$$, slopes on each boundary component of the exterior of the link can be similarly parameterized by $$\mathbb{Q}\cup\{1/0\}$$). See [15]. Given a slope $$\sigma$$, $$k(\sigma)$$ denotes the result of Dehn surgery on $$k$$ along $$\sigma$$, that is, $$k(\sigma)$$ is the closed $$3$$-manifold constructed by gluing a solid torus to $$E(k)$$ along $$\partial E(k)$$ so that the curve of slope $$\sigma$$ bounds a meridian disk of the solid torus. If $$F$$ is a properly embedded essential surface in $$E(k)$$ with $$\partial F$$ nonempty, then all components of $$\partial F$$ are parallel in $$\partial E(k)$$ and hence they determine a slope on $$\partial E(k)$$, called the boundary slope of $$F$$. Let $$S^3=V_1\cup_\Sigma V_2$$ be a decomposition of $$S^3$$ into two solid tori $$V_1$$ and $$V_2$$ such that $$\Sigma=V_1\cap V_2=\partial V_1=\partial V_2$$. Let $$K=T(p,q)$$ be the torus knot of type $$p,q$$ that lies on the torus $$\Sigma$$, where $$p$$ and $$q$$ are relatively prime integers with $$0< q< p$$. We assume that $$K$$ is homotopic to $$p$$ times the core of $$V_1$$ and $$q$$ times the core of $$V_2$$. Up to isotopy, the exterior $$E(K)$$ contains a unique essential annulus, which has boundary slope $$pq$$. See [20]. Take a trivial knot $$C$$ that is disjoint from $$K$$ and encircles $$r$$ adjacent parallel strands of $$K$$ ($$2\le r\le p+q$$). Suppose that $$C$$ intersects each solid torus $$V_i$$ in a trivial arc, a properly embedded arc in $$V_i$$ parallel to an arc in $$\Sigma$$. For a nonzero integer $$s$$, $$(-1/s)$$-Dehn surgery on $$C$$ results in a $$3$$-sphere, that is, $$C(-1/s)=S^3$$. In this new $$3$$-sphere, the image of $$K$$, denoted $$T(p,q,r,s)$$, is called a twisted torus knot. When $$r<p$$, a typical picture of $$T(p,q,r,s)$$ is shown in Figure 1, where a rectangle labeled by an ordered pair of integers, $$(a,b)$$, denotes an $$(a,b)$$-torus braid. See [8] for details. If $$r=p$$, then it is easy to see that $$T(p,q,r,s)$$ is a torus knot (or sometimes a trivial knot). Thus we always assume that $$r\ne p$$. Fig. 1. View largeDownload slide Twisted torus knot $$T(p,q,r,s)$$. Fig. 1. View largeDownload slide Twisted torus knot $$T(p,q,r,s)$$. In his doctoral thesis [3], Dean introduced twisted torus knots to study Dehn surgeries yielding Seifert fiber spaces. Since then, a substantial amount of research has been devoted to these knots (see references in [9]). In particular, the author obtained some results on the knot-types of twisted torus knots. In [7], he showed that if $$r=kq$$ for some integer $$k$$, then $$T(p,q,r,s)$$ is the $$(q, p+k^2qs)$$-cable on the torus knot $$T(k,ks+1)$$; thus hereafter we only need to investigate what happens when $$q$$ does not divide $$r$$. In [9], he determined the parameters $$p,q,r,s$$ for which $$T(p,q,r,s)$$ becomes a trivial knot. In [10], he showed that if $$T(p,q,r,s)$$ is a torus knot then $$|s|\le 2$$ and moreover, when $$|s|=2$$, he also showed that $$T(p,q,r,s)$$ is a torus knot if and only if $$(p,q,r,s)=(2n\pm 1,n,n\pm 1, -2)$$ for some positive integer $$n$$. Finally, in [8], he proved that if $$T(p,q,r,s)$$ is a satellite knot then $$|s|\le 2$$. In this paper, we continue this investigation, showing that if $$T(p,q,r,s)$$ is a satellite knot then $$|s|$$ must be equal to $$1$$. We remark that there are infinitely many examples of twisted torus knots which are satellite knots, constructed by Morimoto [12, 13]. Theorem 1.1. Let $$p,q,r,s$$ be positive integers satisfying the following conditions: $$p,q$$ are relatively prime, and $$1\le q<p$$; $$2\le r\le p+q$$, $$r\ne p$$, and $$r$$ is not a multiple of $$q$$; and $$s\ne 0$$. Suppose that $$T(p,q,r,s)$$ is a satellite knot. Then $$|s|=1$$. □ Combining Theorem 1.1 with earlier results in [8–10], Thurston’s classification of knots [19] enables us to conclude the following. Corollary 1.2. Let $$p,q,r,s$$ be as in Theorem 1.1. Then $$T(p,q,r,s)$$ is a hyperbolic knot unless $$s=\pm 1$$ or $$(p,q,r,s)=(2n\pm 1,n,n\pm 1, -2)$$ for some positive integer $$n$$. □ Throughout this paper, we assume that the twisted torus knot $$T(p,q,r,s)$$ is a satellite knot for the parameters $$p,q,r,s$$ satisfying the conditions in Theorem 1.1. In order to prove Theorem 1.1, we may also assume that $$|s|=2$$ by [8, Theorem 1]. This assumption will eventually yield a contradiction. Some figures in this paper are best viewed in color; readers confused by figures in a black-and-white version are recommended to view the electronic version. 2 Lemmas We start with the following lemma, which is useful in understanding the twisted torus knot $$T(p,q,r,s)$$ in the case that $$r>p$$. Lemma 2.1. If $$r>p$$, then there is an isotopy deforming $$T(p,q,r,s)$$ to the knot obtained by closing the braid in Figure 2(a). □ Fig. 2. View largeDownload slide $$T(p,q,r,s)$$ is the closure of one of these braids when $$r>p$$. Fig. 2. View largeDownload slide $$T(p,q,r,s)$$ is the closure of one of these braids when $$r>p$$. Proof See the first paragraph in the proof of [10, Lemma 2.4]. ■ Given a knot $$k$$ in $$S^3$$, let $$b(k)$$ and $$g(k)$$ denote the bridge index and the genus of $$k$$, respectively. Lemma 2.2. $$b(T(p,q,r,s))\le\mathrm{max}(p,r)$$. □ Proof If $$r<p$$, then the twisted torus knot $$T(p,q,r,s)$$ is the closure of a braid on $$p$$ strands (see Figure 1), which implies $$b(T(p,q,r,s))\le p$$. If $$r>p$$, then $$T(p,q,r,s)$$ is the closure of a braid on $$r$$ strands by Lemma 2.1 and hence $$b(T(p,q,r,s))\le r$$. ■ It is well-known that if a knot $$k$$ is presented as the closure of a braid all of whose crossings are of the same sign, then $$k$$ is a fibered knot and   g(k)=1−b+c2, where $$b$$ is the number of strands of the braid and $$c$$ is the number of crossings. See [18]. Lemma 2.3. If $$r=p+q$$, then there is an isotopy deforming $$T(p,q,r,s)$$ to the closure of the braid in Figure 2(b). Moreover, if $$s>0$$, then   g(T(p,q,r,s))=1+pq−r+(r−1)rs2. □ Proof The first statement immediately follows from Lemma 2.1. If $$s>0$$, then all crossings of the braid in Figure 2(b) are positive. The braid has $$r$$ strands and $$(r-1)rs+pq$$ crossings, so the genus of its closure equals $$(1-r+(r-1)rs+pq)/2$$. ■ Lemma 2.4. Suppose that $$r=p+q$$ and $$s=-2$$. Then $$T(p,q,r,s)$$ is isotopic to the mirror image of $$T(r,2r-q,q,1)$$ and its genus is given by   g(T(p,q,r,s))=(r−1)(2r−q−1)+q(q−1)2. □ Fig. 3. View largeDownload slide $$T(p,q,r,s)$$ is the closure of any of these braids when $$r=p+q$$ and $$s=-2$$. Fig. 3. View largeDownload slide $$T(p,q,r,s)$$ is the closure of any of these braids when $$r=p+q$$ and $$s=-2$$. Fig. 4. View largeDownload slide An isotopy. Fig. 4. View largeDownload slide An isotopy. Proof By Lemma 2.1, there is an isotopy deforming $$T(p,q,r,s)$$ to the closure of the first braid in Figure 3. It is clear that the first and second braids in the figure are the same. Applying the isotopy illustrated in Figure 4, one sees that the second braid is isotopic to the third. This shows that $$T(p,q,r,s)$$ is isotopic to the mirror image of $$T(r,2r-q,q,1)$$. All crossings of the third braid are negative and the braid has $$r$$ strands and $$(r-1)(2r-q)+(q-1)q$$ crossings, so the genus of $$T(p,q,r,s)$$ equals $$\frac{1}{2}(1-r+(r-1)(2r-q)+(q-1)q)=\frac{1}{2}((r-1)(2r-q-1)+q(q-1))$$. ■ Lemma 2.5. Suppose that $$r=p-1,r\ge 2q-1$$, and $$s=-2$$. Then $$T(p,q,r,s)$$ has braid index $$r$$. □ Proof The twisted torus knot $$T(p,q,r,s)$$ is the closure of the leftmost braid in Figure 5. One can see that the closures of the four braids in the figure are isotopic. For example, Figure 6 illustrates the case that $$(p,q,r,s)=(8,3,7,-2)$$. Consider the mirror image of the rightmost braid in Figure 5. The closure of the mirror image is the knot $$T((r-1, q-1),(r,2r-2q+1))$$ in the notation of [1]. Note that $$r-1\le 2r-2q+1$$. Thus the knot $$T((r-1,q-1),(r,2r-2q+1))$$ has braid index $$\mathrm{min}(2r-2q+1,r)=r$$ by [1, Corollary 8] and the result follows. ■ Fig. 5. View largeDownload slide The closures of these braids are isotopic. Fig. 5. View largeDownload slide The closures of these braids are isotopic. Fig. 6. View largeDownload slide An isotopy for $$T(8,3,7,-2)$$. Fig. 6. View largeDownload slide An isotopy for $$T(8,3,7,-2)$$. Let $$k$$ be a knot in $$S^3$$. An arc $$\tau$$ with $$k\cap \tau=\partial \tau$$ is called an unknotting tunnel for $$k$$ if $$S^3-k\cup \tau$$ is an open handlebody of genus two. Moreover, $$\tau$$ is called a $$(1,1)$$-tunnel if $$\partial \tau$$ divides $$k$$ into two subarcs $$\gamma_1$$ and $$\gamma_2$$ in such a way that for some $$i=1,2$$, $$S^3-\mathrm{int}N(\tau\cup \gamma_i)$$ is a solid torus intersecting $$k$$ in a trivial arc. See [14, Proposition 1.3]. Lemma 2.6. Let $$k_m$$ denote the knot obtained by closing the braid in Figure 7(a), where $$m(\ge 3)$$ is an odd integer. Then the arc $$\tau_m$$ indicated in Figure 7(b) is an unknotting tunnel for the knot $$k_m$$. Moreover, if $$m\ge 5$$, then $$\tau_m$$ is not a $$(1,1)$$-tunnel. □ Fig. 7. View largeDownload slide The knots $$k_m$$ and $$k'_n$$. Fig. 7. View largeDownload slide The knots $$k_m$$ and $$k'_n$$. Proof Let $$k'_n$$ denote the knot obtained by closing the braid in Figure 7(c) and let $$\tau'_n$$ be an arc intersecting $$k'_n$$ in its endpoints as shown in Figure 7(d). By using a deformation as shown in Figure 8, which illustrates the case $$m=7$$, one can see that the two spatial graphs $$k_m\cup \tau_m$$ and $$k'_\frac{m+1}{2}\cup\tau'_\frac{m+1}{2}$$ have homeomorphic complements. On the other hand, a deformation as shown in Figure 9, which illustrates the case $$n=4$$, shows that the two spatial graphs $$k'_n\cup\tau'_n$$ and $$k'_{n-1}\cup\tau'_{n-1}$$ have homeomorphic complements. It is clear that $$\tau'_2$$ is an unknotting tunnel for $$k'_2$$. This shows that $$\tau_m$$ is an unknotting tunnel for $$k_m$$. Suppose $$m\ge 5$$. Let $$\gamma_1$$ and $$\gamma_2$$ be two subarcs of $$k_m$$ cut off by $$\partial\tau_m$$. Consider the two knots $$\gamma_1\cup\tau_m$$ and $$\gamma_2\cup\tau_m$$. One can see that these knots are torus knots $$T(\frac{m-1}{2},m)$$ and $$T(\frac{m+1}{2},m+2)$$, both of which are nontrivial knots. See Figure 10, which illustrates the case $$m=7$$. It follows that $$\tau_m$$ is not a $$(1,1)$$-tunnel. ■ Fig. 8. View largeDownload slide The two spatial graphs $$k_m\cup \tau_m$$ and $$k'_\frac{m+1}{2}\cup\tau'_\frac{m+1}{2}$$ have homeomorphic complements. Fig. 8. View largeDownload slide The two spatial graphs $$k_m\cup \tau_m$$ and $$k'_\frac{m+1}{2}\cup\tau'_\frac{m+1}{2}$$ have homeomorphic complements. Fig. 9. View largeDownload slide The two spatial graphs $$k'_n\cup\tau'_n$$ and $$k'_{n-1}\cup\tau'_{n-1}$$ have homeomorphic complements. Fig. 9. View largeDownload slide The two spatial graphs $$k'_n\cup\tau'_n$$ and $$k'_{n-1}\cup\tau'_{n-1}$$ have homeomorphic complements. Fig. 10. View largeDownload slide The knots $$\gamma_1\cup\tau_m$$ and $$\gamma_2\cup\tau_m$$ are torus knots $$T(\frac{m-1}{2},m)$$ and $$T(\frac{m+1}{2},m+2)$$, respectively. Fig. 10. View largeDownload slide The knots $$\gamma_1\cup\tau_m$$ and $$\gamma_2\cup\tau_m$$ are torus knots $$T(\frac{m-1}{2},m)$$ and $$T(\frac{m+1}{2},m+2)$$, respectively. Lemma 2.7. $$q\ge 3$$. □ Proof Since $$q$$ does not divide $$r$$, $$q\ne 1$$ and hence $$q\ge 2$$. Assume that $$q=2$$. First, suppose $$r>p$$. Then $$r=p+1$$ or $$p+2$$ (since $$r\le p+q=p+2$$). In the former, $$q$$ divides $$r$$, a contradiction. Suppose $$r=p+2(=p+q)$$. Then $$r=p+q\ge (q+1)+q=5$$. By Lemma 2.3 there is an isotopy deforming $$T(p,q,r,s)$$ to the closure of the braid in Figure 2(b) ($$q$$ must be equal to $$2$$ in the figure). If $$s=2$$, then $$T(p,q,r,s)$$ is the knot $$k_r$$ defined in Lemma 2.6, and since $$r\ge 5$$, the lemma guarantees that $$k_r$$ admits an unknotting tunnel which is not a $$(1,1)$$-tunnel. However, since $$T(p,q,r,s)$$ is assumed to be a satellite knot, any of its unknotting tunnels is a $$(1,1)$$-tunnel by [14, Proposition 1.8 and Theorem 2.1], a contradiction. Suppose that $$s=-2$$. Then $$T(p,q,r,s)$$ is the knot obtained by closing the first braid in Figure 11, which is equivalent to the third braid. The arc $$\tau$$ indicated in the figure is an unknotting tunnel for the closed braid knot. As in the last paragraph of the proof of Lemma 2.6, the closed braid knot is split along the tunnel $$\tau$$ into two nontrivial torus knots $$T(\frac{r+1}{2},-r)$$ and $$T(\frac{r-1}{2},-(r-2))$$. Thus $$\tau$$ is not a $$(1,1)$$-tunnel by [14, Proposition 1.3] and we get a contradiction since any unknotting tunnel for a satellite knot is a $$(1,1)$$-tunnel by [14, Theorem 2.1]. Suppose $$r<p$$. Since $$q$$ does not divide $$r$$, $$r$$ is odd and hence $$p-r$$ is even. Let $$p-r=\ell q$$. Then an isotopy as in [9, Figure 10] deforms $$T(p,q,r,s)$$ to the closure of the braid in Figure 12. As above, one sees that the arc $$\tau'$$ indicated in the figure is an unknotting tunnel for the closed braid knot which is not a $$(1,1)$$-tunnel and this also gives a contradiction. ■ Fig. 11. View largeDownload slide An isotopy. Fig. 11. View largeDownload slide An isotopy. Fig. 12. View largeDownload slide $$\tau'$$ is not a $$(1,1)$$-tunnel. Fig. 12. View largeDownload slide $$\tau'$$ is not a $$(1,1)$$-tunnel. Given a knot $$k$$ in $$S^3$$, choose a standard longitude-meridian system $$\lambda,\mu$$ for $$H_1(\partial E(k))$$. We follow [15] to say that a knot $$k'$$ on the boundary of a regular neighborhood $$N(k)$$ is an $$(a,b)$$-cable of $$k$$ if $$[k']=a\lambda+b\mu\in H_1(\partial E(k))$$. (For nontriviality, we assume here that $$a\ge 2$$, but $$b$$ may have a negative value.) In particular, we use $$T_{a,b}(c,d)$$ to denote the $$(a,b)$$-cable of the $$(c,d)$$-torus knot. Lemma 2.8. Let $$(a,b)$$ and $$(c,d)$$ be pairs of relatively prime positive integers. Then   g(Ta,b(c,d))=1−ac+ab−ad+acd−b2. □ Proof This follows immediately from Satz 1 on page 247 of [16]. ■ Lemma 2.9. Let $$a,b,c,d$$ be positive integers for which the cable knot $$T_{a,b}(c,d)$$ is defined. Suppose that $$T(c,d)$$ is not a trivial knot. Then   b(Ta,b(c,d))=a⋅min(c,d). □ Proof Using a well-known fact that $$b(T(c,d))=\mathrm{min}(c,d)$$ (see [17]), one easily sees that the knot $$T_{a,b}(c,d)$$ has a bridge position with bridge number $$a\cdot\mathrm{min}(c,d)$$. Hence the lemma follows from another classical result of Schubert [17]: if $$k_1$$ is a satellite knot with companion $$k_2$$ and has wrapping number $$n$$ in a solid torus neighborhood of $$k_2$$, then $$b(k_1)\ge n\cdot b(k_2)$$. ■ 3 Primitive/Seifert-fibered knots Let $$\ell$$ be a simple closed curve on the boundary of a $$3$$-manifold $$Q$$. We use $$Q[\ell]$$ to denote the $$3$$-manifold constructed from $$Q$$ by attaching a $$2$$-handle along the curve $$\ell$$. For two disjoint nonisotopic simple closed curves $$\ell$$ and $$\ell'$$ on $$\partial Q$$, $$Q[\ell\cup\ell']$$ is similarly defined, that is, $$Q[\ell\cup\ell']=Q[\ell][\ell'](=Q[\ell'][\ell])$$. We say that a knot $$k$$ in $$S^3$$ is a primitive/Seifert-fibered knot if $$k$$ lies on a genus two surface, which splits $$S^3$$ into two handlebodies $$H$$ and $$H'$$, in such a way that $$H[k]$$ is a solid torus and $$H'[k]$$ is a Seifert fiber space over the disk with two exceptional fibers. The purpose of this section is to show that $$T(p,q,r,s)$$ is a primitive/Seifert-fibered knot. A result of Miyazaki and Motegi enables us to conclude that $$T(p,q,r,s)$$ is a cable of a torus knot. See [11, Theorem 1.2]. Throughout this section, we let $$K=T(p,q)$$ and $$K'=T(p,q,r,s)$$. Recall that there are two solid tori $$V_1,V_2$$ such that $$V_1\cup V_2=S^3$$ and $$\Sigma=V_1\cap V_2=\partial V_1=\partial V_2$$ is a torus containing the knot $$K$$. Here, $$V_1$$ and $$V_2$$ are indexed so that $$K$$ wraps around $$V_1$$$$p$$ times and $$V_2$$$$q$$ times. An unknotted circle $$C$$ is taken in $$S^3-K$$ so that it encircles $$r$$ adjacent parallel strands of $$K$$ and intersects each $$V_i$$ in a trivial arc. Then $$K'$$ is the image of $$K$$ after $$(-1/s)$$-surgery on $$C$$. The exterior of $$K$$ contains an essential annulus, say $$\widehat{P}(\subset \Sigma)$$, with boundary slope $$pq$$. We may assume that $$C$$ intersects $$\widehat{P}$$ transversely in two points. Let $$M=S^3-\mathrm{int}N(C\cup K)$$. Then $$M$$ is a hyperbolic $$3$$-manifold by [9, Proposition 5.7] and it is bounded by two tori, $$\partial_C M=\partial N(C)$$ and $$\partial_K M=\partial N(K)$$. Also, $$P=\widehat{P}\cap M$$ is a twice-punctured annulus with two boundary components in $$\partial_C M$$ and the other two boundary components in $$\partial_K M$$. Each component of $$\partial P\cap \partial_C M$$ has a trivial slope on $$\partial_C M$$. Let $$W_i=V_i\cap M$$ for each $$i=1,2$$ and let $$\partial_X W_i=W_i\cap \partial_X M$$ for each $$X=C,K$$. Then each $$W_i$$ is a genus two handlebody with $$\partial W_i=P\cup \partial_C W_i\cup \partial_K W_i$$, where $$\partial_X W_i$$ is an annulus for each $$X=C,K$$. Let $$X_i$$ denote the core of the annulus $$\partial_X W_i$$. Then $$C_i$$ and $$K_i$$ are disjoint nonisotopic curves on $$\partial W_i$$, and if we let $$U_i=W_i[C_i]$$, then $$U_i$$ is a solid torus such that $$U_1\cup_{\widehat{P}} U_2=E(K)$$. For a slope $$\alpha$$ on $$\partial_C M$$, let $$M(C;\alpha)$$ denote the result of $$\alpha$$-Dehn filling, that is, $$M(C;\alpha)=M\cup J_\alpha$$, where $$J_\alpha$$ is a solid torus glued along $$\partial_C M$$ so that $$\alpha$$ bounds a meridian disk of $$J_\alpha$$. Note that $$M(C;1/0)$$ and $$M(C;-1/s)$$ are the exteriors of $$K$$ and $$K'$$, respectively. If $$\beta$$ is a slope on $$\partial_K M$$, then $$M(C,K;\alpha,\beta)=M(K,C;\beta,\alpha)$$ is defined to be the result of $$\beta$$-Dehn filling on $$M(C;\alpha)$$ along the torus $$\partial_K M$$. The annulus $$\widehat{P}$$ intersects $$J_{1/0}$$ in two meridian disks, say $$u_1$$ and $$u_2$$. Also, $$P$$ is essential in $$M$$ by [9, Lemma 5.3]. Since $$K'$$ is a satellite knot, its exterior $$M(C;-1/s)$$ contains an essential torus. Among all essential tori in $$M(C;-1/s)$$, we choose an essential torus $$\widehat{T}$$ to intersect $$J_{-1/s}$$ in a minimal number of meridian disks, say $$v_1,\ldots,v_t$$, successively indexed along $$J_{-1/s}$$ (note that $$t$$ must be even, since $$\widehat{T}$$ is separating). We also assume that $$\widehat{T}$$ is chosen so that $$T=\widehat{T}\cap M$$ intersects $$P$$ transversely and minimally. Then $$\partial u_x\cap \partial v_y$$ consists of exactly two points for each pair of disks $$u_x,v_y$$ by the assumption $$|s|=2$$ at the beginning of the paper. We construct two labeled graphs $$G_P$$ on $$\widehat{P}$$ and $$G_T$$ on $$\widehat{T}$$ as follows. The edges of the graphs are the arc components of $$P\cap T$$, the (fat) vertices are the disks $$u_x$$ or $$v_y$$ on $$G_P$$ or $$G_T$$, and each point in $$\partial u_x\cap \partial v_y$$, which is an endpoint of an edge in both graphs, is labeled $$y$$ in $$G_P$$ and labeled $$x$$ in $$G_T$$. We assume familiarity with the definitions and terminology in [8]: positive/negative edges, the parity rule, $$S$$-cycles, etc. We remark that $$T$$ is incompressible and boundary-incompressible in $$M$$ by the minimality of $$t$$. It follows that $$G_P$$ contains no trivial loops. We also remark that any disk face of $$G_T$$ is contained in $$W_1$$ or $$W_2$$. Lemma 3.1. Suppose that a disk face of $$G_T$$ is contained in $$W_i(i=1,2)$$. Then $$W_i[K_i]$$ is a solid torus. Furthermore, if the disk face is nonseparating in $$W_i$$, then it is a meridian disk of the solid torus. □ Proof Let $$f$$ be a disk face of $$G_T$$ which is contained in $$W_i$$ for some $$i=1,2$$. Note that $$f$$ is a properly embedded disk in $$W_i$$ with its boundary disjoint from the annulus $$\partial_K W_i$$. Its boundary circle, $$\partial f$$, is an alternating sequence of properly embedded arcs in $$P$$ and spanning arcs in the annulus $$\partial_C W_i$$, the former being the edges of $$f$$ and the latter being the corners of $$f$$. We first claim that $$f$$ is an essential disk in $$W_i$$. Assume otherwise. Then $$\partial f$$ bounds a disk $$D$$ on $$\partial W_i$$. The core $$C_i$$ of the annulus $$\partial_C W_i$$ intersects $$D$$ in a set of arcs. An outermost arc together with a subarc of $$\partial D=\partial f$$ bounds a subdisk of $$D$$. This implies that the edge of $$f$$ corresponding to the subarc is a trivial loop in $$G_P$$, a contradiction. Since $$K_i$$ is disjoint from the essential disk $$f$$ in $$W_i$$, one can take a nonseparating essential disk in $$W_i$$ which is disjoint from $$K_i$$. (If $$f$$ is itself a nonseparating disk, then $$f$$ can be taken to be the nonseparating disk.) Thus $$W_i[K_i]$$ is $$\partial$$-reducible and it is a solid torus by [10, Lemma 2.1]. Furthermore, if $$f$$ is a nonseparating disk in $$W_i$$, then it is a meridian disk of the solid torus $$W_i[K_i]$$. ■ Lemma 3.2. $$G_T$$ contains no $$S$$-cycle. □ Proof Suppose that $$G_T$$ contains an $$S$$-cycle. Let $$f$$ be the disk face of $$G_T$$ bounded by the $$S$$-cycle and suppose that $$f$$ is contained in $$W_i$$. Then the oriented intersection of $$C_i$$ with $$\partial f$$ consists of two points of the same sign. This implies that $$f$$ is a nonseparating disk in $$W_i$$. By Lemma 3.1 $$W_i[K_i]$$ is a solid torus in which $$f$$ is a meridian disk. Since $$C_i$$ intersects $$\partial f$$ in two points, the fundamental group of $$W_i[K_i][C_i]$$ is a cyclic group of order $$2$$. Note that $$W_i[K_i][C_i]=W_i[K_i\cup C_i]=W_i[C_i][K_i]=U_i[K_i]$$ and that $$U_i$$ is a solid torus around which the curve $$K_i$$ wraps $$p$$ or $$q$$ times according to whether $$i=1$$ or $$2$$. Thus we have $$p=2$$ (and then $$q=1$$) or $$q=2$$, any of which contradicts Lemma 2.7. ■ Lemma 3.3. If all disk faces of $$G_T$$ lie on one side of $$P$$, then $$t=2$$ and the graphs $$G_P,G_T$$ and their edge correspondence are as shown in Figure 13(a and c). □ Fig. 13. View largeDownload slide Graphs of intersection. Fig. 13. View largeDownload slide Graphs of intersection. Proof Let $$V,E,F$$ be the number of vertices, edges, and disk faces of $$G_T$$, respectively. Every vertex of $$G_T$$ has valence $$4$$, so $$4V=2E$$ and hence $$0=\chi(\widehat{T})\le V-E+F= \frac{E}{2}-E+F=-\frac{E}{2}+F$$, giving $$2F\ge E(=2V=2t)$$. It follows that $$G_T$$ contains at least $$t$$ disk faces. Assume that all disk faces of $$G_T$$ are contained in $$W_i$$ for some fixed $$i\in\{1,2\}$$. Then since every disk face of $$G_T$$ has at least two sides, we have $$2F\le E$$. Combining the above inequalities, we get $$E=2F$$ and conclude that every disk face of $$G_T$$ is a bigon, which must be bounded by negative edges according to Lemma 3.2. Also, every edge of $$G_T$$ belongs to a bigon face. Therefore all edges of $$G_T$$ are negative. All edges of $$G_P$$ are positive by the parity rule and each vertex of $$G_P$$ is a base of $$t$$ parallel positive edges. If $$t\ge 4$$, then $$G_P$$ has an $$S$$-cycle immediately surrounded by two parallel edges, contradicting [8, Lemma 6(3)]. Suppose $$t=2$$. Then each of the graphs $$G_P$$ and $$G_T$$ has four edges, say, $$e_1,e_2,e_3,e_4$$. Up to homeomorphisms of $$\widehat{P}$$ and $$\widehat{T}$$, there are exactly two possibilities for each of $$G_P$$ and $$G_T$$ as shown in Figure 13. If $$G_P$$ is the graph in Figure 13(b), then one can see that for any possibility for $$G_T$$ (see Figure 13(c or d)), the corners of the two bigon faces of $$G_T$$ cannot be placed to be mutually disjoint and parallel in the annulus $$\partial_C W_i$$: take the four corners of the two bigon faces of $$G_T$$ and apply [5, Lemma 3.2] to these corners. Thus $$G_P$$ must be the graph in Figure 13(a). Suppose that $$G_T$$ is the graph in Figure 13(d). Let $$H_i$$ denote the $$1$$-handle $$J_{1/0}\cap U_i$$. Let $$f$$ be the bigon face of $$G_T$$ bounded by $$e_1$$ and $$e_3$$. The edges $$e_1,e_3$$ divide $$P$$ into three annuli $$A_1,A_2,A_3$$. One of these annuli, say, $$A_2$$ is disjoint from $$\partial \widehat{P}$$. By shrinking $$H_i$$ to its core, $$H_i\cup f\cup A_2$$ becomes a Klein bottle in the solid torus $$U_i$$. This is impossible. Thus $$G_T$$ must be the graph in Figure 13(c). ■ Lemma 3.4. $$W_i[K_i]$$ is a solid torus for each $$i=1,2$$. □ Proof The first paragraph in the proof of Lemma 3.3 shows that $$G_T$$ has disk faces. If, for each $$i=1,2$$, $$G_T$$ has a disk face contained in $$W_i$$, then the result follows from Lemma 3.1. Assume otherwise, that is, all disk faces of $$G_T$$ are contained in $$W_i$$ for some fixed $$i\in\{1,2\}$$ (then $$W_i[K_i]$$ is a solid torus) and let $$j=3-i$$ so that $$\{i,j\}=\{1,2\}$$. Then by Lemma 3.3, $$t=2$$ and the graphs $$G_P,G_T$$ and their edge correspondence are as shown in Figure 13(a and c). We will eventually show that $$W_j[K_j]$$ is also a solid torus. Note that $$G_T$$ has precisely one annulus face. Let it be $$A$$. We first claim that $$P\cap T$$ contains no circle components. Assume otherwise. If $$P\cap T$$ contains a circle component which is inessential in both $$P$$ and $$T$$, then applying a disk-swapping argument one can reduce $$|P\cap T|$$, contradicting the minimality of $$|P\cap T|$$. If $$P\cap T$$ contains a circle component which is inessential in one of $$P$$ and $$T$$ and essential in the other, then one of $$P$$ and $$T$$ is compressible, a contradiction. Thus all circle components of $$P\cap T$$ are essential on both $$P$$ and $$T$$. In particular, they are all parallel and essential in $$A$$. An outermost such circle in $$A$$, say, $$c$$ cuts off a subannulus $$A'$$ from $$A$$. We may assume that $$\partial A'-c$$ contains $$e_1\cup e_3$$ (otherwise, $$\partial A'-c$$ contains $$e_2\cup e_4$$ and the same argument applies). The annulus $$A'$$ is contained in $$W_j$$, and $$\partial A'-c$$ intersects the annulus $$\partial_C W_j$$ in two spanning arcs, cutting $$\partial_C W_j$$ into two rectangles. See Figure 14(a). Take any rectangle and let $$Q$$ be the union of $$A'$$ and the rectangle. Then $$Q$$ is a properly embedded thrice-punctured sphere in $$U_j$$, one of whose boundary components is $$c$$ and the other two contain edges $$e_1$$ and $$e_3$$, respectively. The boundary components of $$Q$$ are mutually parallel in $$\widehat{P}$$ and are isotopic to $$K_j$$ in the torus $$\partial U_j$$. Taking a subannulus of $$\widehat{P}$$ between any two adjacent boundary components of $$Q$$ in $$\widehat{P}$$ and attaching it to $$Q$$, we obtain a once-punctured torus. This implies that the knot $$K_j$$, which is isotopic to $$K$$ in $$S^3$$, has genus at most one. But, $$g(K)=(p-1)(q-1)/2>1$$ by Lemma 2.7, a contradiction. Hence $$P\cap T$$ consists only of arc components and $$A$$ is entirely contained in $$W_j$$. Since $$T$$ is incompressible and $$\partial$$-incompressible in $$M$$, $$A$$ is incompressible and not $$\partial$$-parallel in $$W_j$$: if $$A$$ were $$\partial$$-parallel in $$W_j$$, then a disk $$\Delta(\subset W_j)$$ would guide a $$\partial$$-compression of $$T$$ in $$M$$ as shown in Figure 15. Also, since $$\widehat{T}$$ is separating, $$A$$ is separating in $$W_j$$. Thus by [6, Lemma 3.2], $$A$$ splits $$W_j$$ into a genus two handlebody $$X$$ and a solid torus $$Y$$ so that $$A$$ wraps around $$Y$$ at least two times. See Figure 16. As shown in Figure 14(b), $$\partial A$$ intersects the annulus $$\partial_C W_j$$ in four spanning arcs, cutting $$\partial_C W_j$$ into four rectangles. Let $$a_1,a_2,a_3,a_4$$ be the spanning arcs, successively indexed along $$\partial_C W_j$$ as in Figure 14(b), and let $$R_k$$ be the rectangle between $$a_k$$ and $$a_{k+1}$$ (subscripts understood modulo $$4$$). Then the union $$S=A\cup R_1\cup R_3$$ is a properly embedded four-punctured sphere in the solid torus $$U_j$$, which must be separating and compressible in $$U_j$$. Note that $$S$$ divides $$U_j$$ into two genus two handlebodies $$X$$ and $$Z=Y\cup H_j$$, where $$H_j$$ is the $$1$$-handle given by $$H_j=J_{1/0}\cap U_j$$ and attached to $$Y$$ along the two rectangles $$R_2\cup R_4$$. We claim that $$S$$ is incompressible in $$Z$$ (and hence it must be compressible in $$X$$). Suppose that $$S$$ is compressible in $$Z$$. We may regard the $$1$$-handle $$H_j$$ as a product manifold $$(\text{a rectangle})\times [0,1]$$, that is, there is a rectangle $$R$$ in $$H_j$$, parallel to $$R_2$$ and $$R_4$$, such that $$H_j=R\times [0,1]$$ and $$R_{2k}=R\times\{k-1\}$$ for $$k=1,2$$. Choose a compressing disk $$D$$ for $$S$$ so that $$D$$ and $$R$$ intersect transversely and minimally. Then $$D\cap R$$ consists of a finite number of properly embedded arcs in $$D$$. If $$D$$ is disjoint from $$H_j$$ (and hence from $$R$$) then $$\partial D$$ lies in $$A$$ and $$D$$ compresses $$T$$, a contradiction. Otherwise, take an arc component $$a$$ of $$D\cap R$$ which is outermost in $$D$$. Then $$a\times [0,1]\subset D\cap H_j=(D\cap R)\times [0,1]$$ cuts off a subdisk from $$D$$, which serves as a $$\partial$$-compressing disk for $$T$$. This is a contradiction. Hence $$S$$ must be compressible in $$X$$. Each component of $$\partial S$$ is parallel to $$K_j$$ in $$\partial U_j$$, so it cannot bound a disk in $$U_j$$. It follows that $$S$$ compresses into two disjoint incompressible annuli, say, $$A_0$$ and $$A_1$$. Conversely, $$S$$ can be recovered from $$A_0$$ and $$A_1$$ by tubing along a (knotted) arc $$\alpha$$ connecting the two annuli. By [6, Lemma 3.1], each of these annuli is $$\partial$$-parallel in $$U_j$$. We divide our arguments into two cases depending on whether the annuli are nested or not. First, suppose that $$A_0$$ and $$A_1$$ are nested, that is, they cut off a product region $$A_0\times [0,1]$$ from $$U_j$$, where $$A_0=A_0\times\{0\}$$ and $$A_1=A_0\times\{1\}$$. The arc $$\alpha$$ is properly embedded in $$A_0\times [0,1]$$ and has one point in each end of the product. Removing an open regular neighborhood of $$\alpha$$ from the product, we obtain $$Z$$, that is, $$Z=A_0\times [0,1]-\eta(\alpha)$$, where $$\eta(\alpha)$$ is an open regular neighborhood of $$\alpha$$. See Figure 17 in which $$A_\alpha(\subset\partial\eta(\alpha))$$ denotes a tube along $$\alpha$$, connecting $$A_0$$ and $$A_1$$. The intersection $$Z\cap \partial U_j$$ consists of two annuli in $$(\partial A_0)\times [0,1]$$, one being the union of the disk $$u_1$$ and the bigon face of $$G_P$$ bounded by $$e_1\cup e_2$$ and the other being the union of the disk $$u_2$$ and the bigon face bounded by $$e_3\cup e_4$$ (see Figure 14(b)). Let $$\gamma_1,\gamma_2$$ be the cores of these two annuli. Then attaching two $$2$$-handles to $$A_0\times [0,1]$$ along $$\gamma_1\cup\gamma_2$$ gives rise to the product manifold $$S^2\times [0,1]$$ in which $$\alpha$$ is properly embedded, connecting $$S^2\times\{0\}$$ and $$S^2\times\{1\}$$. The classical light bulb trick (see [15, Exercise E4 in Chapter 9]) shows that $$Z[\gamma_1\cup\gamma_2]=S^2\times [0,1]-\eta(\alpha)$$ is homeomorphic to a $$3$$-ball, and each of the curves $$\gamma_1$$ and $$\gamma_2$$ intersects $$\partial R_2$$ (or $$\partial R_4$$) in a single point. It follows from [2, Lemma 2.3.2] that $$S$$ is compressible in $$Z$$, a contradiction. Next, suppose that $$A_0$$ and $$A_1$$ are not nested, that is, the annuli cut off two disjoint solid tori from $$U_j$$. These solid tori are connected by the arc $$\alpha$$ and the genus two handlebody $$X$$ is the union of these solid tori together with a regular neighborhood of $$\alpha$$. The boundary of $$X$$ is the union of $$S$$ and two annuli $$B_0\cup B_1$$ as shown in Figure 18(a), where $$B_0$$ and $$B_1$$ are subannuli of $$\partial U_j$$ such that $$\partial B_0\cup \partial B_1=\partial S$$ (see Figure 14(b)). We may assume that the annulus $$\partial_K W_j$$ is contained in $$B_1$$. Note that $$\partial X\cap H_j(=S\cap H_j)$$ consists of two rectangles $$R_1\cup R_3$$. The rectangle $$R_k(k=1,3)$$ is bounded by the two arcs $$a_k,a_{k+1}$$ and two other arcs in $$\partial u_1\cup\partial u_2$$, and we may regard $$R_k$$ as obtained by thickening $$a_k$$ properly in $$S=A\cup R_1\cup R_3$$. The genus two handlebody $$X$$ can be regarded as a $$3$$-ball with two unknotted holes (see Figure 18(a)) and the pair $$a_1\cup a_3$$ is a pair of arcs in $$S(\subset \partial X)$$ whose boundary has one endpoint in each component of $$\partial S$$. The core $$\gamma$$ of $$A(\subset S)$$ is an essential curve in $$S$$ and separates the arcs $$a_1$$ and $$a_3$$. There is a natural one-to-one correspondence between isotopy classes of essential curves in $$S$$ and the extended rationals in $$\mathbb{Q}\cup\{1/0\}$$: the boundary of a disk in $$X$$ separating the unknotted holes corresponds to $$1/0$$ and the equator $$\iota$$ shown in Figure 18(a) corresponds to $$0$$. Each arc in the pair $$a_1\cup a_3$$ has one endpoint on $$\partial B_0$$ and the other on $$\partial B_1$$ (see Figure 14(b)), so $$\gamma$$ corresponds to a rational number $$m/n$$ with $$n$$ odd. See Figure 18(b), which illustrates the case $$m/n=1/3$$. Recall that $$A=\overline{S-R_1\cup R_3}$$ is an incompressible, non-$$\partial$$-parallel, separating annulus in the handlebody $$W_j$$, splitting $$W_j$$ into $$X$$ and $$Y$$. It follows from [6, Lemma 3.2(i)] that $$\gamma$$ is a primitive curve on the genus two handlebody $$X$$, that is, $$X[\gamma]$$ is a solid torus. On the other hand, after an isotopy of $$X$$, we may consider $$X$$ as obtained from a $$(-m/n)$$-rational tangle by removing open regular neighborhoods of the stings of the rational tangle. See Figure 18(c), where $$\gamma$$ becomes a vertical circle separating the arcs $$a_1$$ and $$a_3$$. Using the figure, one easily sees that $$X[\gamma]$$ is the exterior of a $$2$$-bridge knot associated to the rational number $$-m/n$$. Since $$X[\gamma]$$ is a solid torus, $$-m/n$$ must be an integer and a further isotopy of $$X$$ allows us to assume $$-m/n=0$$. See Figure 18(d). Since $$K_j\subset\partial_K W_j\subset B_1$$, $$X[K_j]$$ is defined and it is a solid torus around which $$A$$ winds once. Thus $$W_j[K_j]=X[K_j]\cup_A Y$$ is a solid torus. This completes the proof. ■ Fig. 14. View largeDownload slide (a) $$\partial A'-c$$ and (b) $$\partial A$$. Fig. 14. View largeDownload slide (a) $$\partial A'-c$$ and (b) $$\partial A$$. Fig. 15. View largeDownload slide If $$A$$ were $$\partial$$-parallel in $$W_j$$, then $$T$$ would be $$\partial$$-compressible in $$M$$. Fig. 15. View largeDownload slide If $$A$$ were $$\partial$$-parallel in $$W_j$$, then $$T$$ would be $$\partial$$-compressible in $$M$$. Fig. 16. View largeDownload slide A separating essential annulus in a genus two handlebody. Fig. 16. View largeDownload slide A separating essential annulus in a genus two handlebody. Fig. 17. View largeDownload slide $$Z=A_0\times [0,1]-\eta (\alpha)$$. Fig. 17. View largeDownload slide $$Z=A_0\times [0,1]-\eta (\alpha)$$. Fig. 18. View largeDownload slide $$\partial X=S\cup B_0\cup B_1$$. Fig. 18. View largeDownload slide $$\partial X=S\cup B_0\cup B_1$$. Proposition 3.5. (1) $$r\equiv \pm 1$$ or $$\pm q\mod p$$ and $$r\equiv \pm 1$$ or $$\pm p\mod q$$. (2) There are only five possibilities for the value of $$r$$:   r=p−1,p+1,p−q,p+q, or 2p−q.$$\quad$$ Moreover, if $$r=2p-q$$, then $$p<2q$$. □ Proof (1) follows from Lemma 3.4 and [10, Lemma 2.1]. (2) Since $$2\le r\le p+q$$, the congruence equation $$r\equiv \pm 1$$ or $$\pm q\mod p$$ places restrictions on the value of $$r$$:   r=p−1,p+1,p−q,p+q, or 2p−q. In the case that $$r=2p-q$$, we obtain $$p<2q$$ by using the fact that $$r\le p+q$$ and $$p,q$$ are relatively prime. ■ Lemma 3.6. Let $$Z$$ be the $$3$$-manifold constructed by Dehn surgery on the cable knot $$T_{a,b}(c,d)$$ with an integral slope $$n$$, where $$c>|d|\ge 2$$. Suppose that $$Z$$ is a small Seifert fiber space. Then $$n=ab\pm 1$$ and $$Z$$ has base orbifold a $$2$$-sphere with three cone points of indices $$c,|d|$$, and $$|n-a^2cd|$$. □ Proof Let $$k=T(c,d)$$ and $$k'=T_{a,b}(c,d)$$. Note that $$T_{a,b}(c,d)$$ is $$C_{b,a}(T(d,c))$$ in the notation of [4, p. 692]. By [4, Corollary 7.3] $$n=ab\pm 1$$ and $$Z=k'(n)=k(n/a^2)$$. The integers $$n$$ and $$a^2$$ are relatively prime, so $$n/a^2$$ is a reduced fraction. By [4, Corollary 7.4] $$Z$$ has base orbifold a $$2$$-sphere with three cone points of indices $$c,|d|$$, and $$|n-a^2cd|$$. ■ Proposition 3.7. There are integers $$\varepsilon_1,\varepsilon_2,\varepsilon_3,\varepsilon_4=\pm 1$$ and $$m\ge 2$$ satisfying one of the following: (1) $$T(p,q,r,s)=T_{m, \varepsilon_1 mpq+\varepsilon_2}(p,\varepsilon_1 q)\text{ and } pq+r^2s-\varepsilon_1m^2pq=\varepsilon_2 m+\varepsilon_3=2\varepsilon_4;$$ (2) $$T(p,q,r,s)=T_{m,2\varepsilon_1 mp+\varepsilon_2}(p,2\varepsilon_1)\text{ and } pq+r^2s-2\varepsilon_1m^2p=\varepsilon_2 m+\varepsilon_3=\varepsilon_4 q;\text{ or}$$ (3) $$T(p,q,r,s)=T_{m,2\varepsilon_1 mq+\varepsilon_2}(q,2\varepsilon_1)\text{ and } pq+r^2s-2\varepsilon_1m^2q=\varepsilon_2 m+\varepsilon_3=\varepsilon_4 p.$$ □ Proof Note that $$M(K,C;1/0,-1/s)=S^3$$ and that the exterior of $$K'=T(p,q,r,s)$$ is given as   M(C;−1/s)=M∪∂CMJ−1/s=(W1∪PW2)∪∂CMJ−1/s=W1∪P1(W2∪∂CW2J−1/s), where $$P_1=P\cup \partial_C W_1$$ is a twice punctured torus. We may consider that $$P_1$$ is properly embedded in the exterior of $$K$$ as well as in the exterior of $$K'$$. Recall that $$W_1$$ is a genus two handlebody. Since the core $$C_2$$ of the splitting annulus $$\partial_C W_2$$ of the union $$W_2\cup J_{-1/s}$$ is a primitive curve in $$W_2$$ (i.e., $$W_2[C_2]=U_2$$ is a solid torus), $$W_2\cup J_{-1/s}$$ is also a genus two handlebody. Since the slope of the boundary of $$P_1$$ in the exterior of $$K$$ is $$pq$$, it follows from [15, Proposition H2 in Chapter 9] that the slope is changed to $$pq+r^2s$$, an integral slope, in the exterior of $$K'$$. Thus if we let $$N(K')=S^3-\mathrm{int}(M(C;-1/s))$$ be a regular neighborhood of $$K'$$, then each component of $$\partial P_1$$ (and hence the curve $$K_1$$) wraps around the solid torus $$N(K')$$ once in the longitudinal direction. Thus $$N(K')\cup_{\partial_K W_1} W_1(\cong W_1)$$ is a genus two handlebody. Also,   S3=M(K,C;1/0,−1/s)=N(K′)∪M(C;−1/s)=(N(K′)∪∂KW1W1)∪(W2∪∂CW2J−1/s). Here, $$S^3=(N(K')\cup_{\partial_K W_1} W_1)\cup (W_2\cup_{\partial_C W_2} J_{-1/s})$$ is a genus two Heegaard decomposition of $$S^3$$ and $$P_1\cup \partial_K W_2$$ is the splitting surface of the decomposition. The splitting surface can be isotoped fixing $$P_1$$ so that it contains $$K'$$ and intersects $$\partial N(K')$$ in two curves in $$\partial P_1$$. Consider $$K'(pq+r^2s)$$. It is given as   K′(pq+r2s)=M(K,C;pq,−1/s)=W1[K1]∪(W2[K2]∪J−1/s). By Lemma 3.4, $$W_1[K_1]$$ and $$W_2[K_2]$$ are solid tori. Recall that the splitting surface of the union $$W_2[K_2]\cup J_{-1/s}$$ is the annulus $$\partial_C W_2$$. Its core, $$C_2$$, wraps around the solid torus $$J_{-1/s}$$$$|s|$$ times in the longitudinal direction. Since $$W_2[K_2][C_2]=W_2[K_2\cup C_2]=W_2[C_2][K_2]=U_2[K_2]$$ is a once punctured lens space $$L(q,p)$$, $$C_2$$ wraps around the solid torus $$W_2[K_2]$$$$q$$ times in the longitudinal direction. This implies that the union $$W_2[K_2]\cup J_{-1/s}$$ is a Seifert fiber space over the disk with two exceptional fibers of indices $$|s|$$ and $$q$$. The common boundary torus of $$W_1[K_1]$$ and $$W_2[K_2]\cup J_{-1/s}$$ contains the annulus $$\partial_C W_1$$. The core of this annulus, $$C_1$$, is a Seifert fiber of the Seifert fiber space $$W_2[K_2]\cup J_{-1/s}$$, since it is isotopic to $$C_2$$ in $$\partial J_{-1/s}=\partial_C M$$. Also, $$C_1$$ wraps around the solid torus $$W_1[K_1]$$$$p$$ times in the longitudinal direction, since $$W_1[K_1][C_1]=W_1[K_1\cup C_1]=W_1[C_1][K_1]=U_1[K_1]$$ is a once punctured lens space $$L(p,q)$$. The Seifert fibration of $$W_2[K_2]\cup J_{-1/s}$$ extends to the solid torus $$W_1[K_1]$$ so that the curve $$C_1$$ becomes a regular fiber, wrapping around $$W_1[K_1]$$$$p$$ times. It follows that $$K'(pq+r^2s)=W_1[K_1]\cup (W_2[K_2]\cup J_{-1/s})$$ is a Seifert fiber space over the $$2$$-sphere with three exceptional fibers of indices $$p,q$$, and $$|s|=2$$. Hence $$K'=T(p,q,r,s)$$ is an $$(m,mcd+\varepsilon)$$-cable knot of a $$(c,d)$$-torus knot by [11, Theorem 1.2], where $$m\ge 2,c>|d|\ge 2$$, and $$\varepsilon=\pm 1$$. By Lemma 3.6   pq+r2s=m2cd+εm+ε′(ε′=±1) and one of the following holds:   (c,d,pq+r2s−m2cd)=(p,ε″q,2ε‴);=(p,2ε″,ε‴q); or=(q,2ε″,ε‴p) for some integers $$\varepsilon'',\varepsilon'''=\pm 1$$. If we let $$\varepsilon_1=\varepsilon'',\varepsilon_2=\varepsilon,\varepsilon_3=\varepsilon'$$, and $$\varepsilon_4=\varepsilon'''$$, then the result follows. ■ According to Proposition 3.7, we divide our arguments into three cases as follows: Case A: $$T(p,q,r,s)=T_{m, \varepsilon_1 mpq+\varepsilon_2}(p,\varepsilon_1 q)\text{ and } pq+r^2s-\varepsilon_1m^2pq=\varepsilon_2 m+\varepsilon_3=2\varepsilon_4;$$ Case B: $$T(p,q,r,s)=T_{m,2\varepsilon_1 mp+\varepsilon_2}(p,2\varepsilon_1)\text{ and } pq+r^2s-2\varepsilon_1m^2p=\varepsilon_2 m+\varepsilon_3=\varepsilon_4 q;\text{ or}$$ Case C: $$T(p,q,r,s)=T_{m,2\varepsilon_1 mq+\varepsilon_2}(q,2\varepsilon_1)\text{ and } pq+r^2s-2\varepsilon_1m^2q=\varepsilon_2 m+\varepsilon_3=\varepsilon_4 p.$$ In the following three sections, we will show any of these cases gives a contradiction, which completes the proof of Theorem 1.1. 4 Case A Throughout this section, we assume that there are integers $$\varepsilon_i=\pm 1(i=1,2,3,4)$$ and $$m\ge 2$$ such that $$T(p,q,r,s)=T_{m, \varepsilon_1 mpq+\varepsilon_2}(p,\varepsilon_1 q)$$ and $$pq+r^2s-\varepsilon_1m^2pq=\varepsilon_2 m+\varepsilon_3=2\varepsilon_4$$. Multiplying both sides of the equation $$\varepsilon_2 m+\varepsilon_3=2\varepsilon_4$$ by $$\varepsilon_2$$, we get $$m+\varepsilon_2\varepsilon_3=2\varepsilon_2\varepsilon_4$$. Since $$m\ge 2$$, $$m=2\varepsilon_2\varepsilon_4-\varepsilon_2\varepsilon_3$$ implies that $$m=3$$ and $$\varepsilon_2\varepsilon_4=1,\varepsilon_2\varepsilon_3=-1$$. Let $$s=2\varepsilon_5(\varepsilon_5=\pm 1)$$. Then $$2\varepsilon_4=pq+r^2s-\varepsilon_1m^2pq=pq+2r^2\varepsilon_5-9\varepsilon_1pq$$ and hence we get   2r2+(ε5−9ε1ε5)pq=2ε4ε5. (4.1) If $$\varepsilon_1\varepsilon_5=-1$$, then $$2r^2+(\varepsilon_5-9\varepsilon_1\varepsilon_5)pq=2r^2+(9+\varepsilon_5)pq\ge 2r^2+8pq>8>2\varepsilon_4\varepsilon_5$$, which implies that equation (4.1) has no solution. Thus $$\varepsilon_1\varepsilon_5=1$$; $$(\varepsilon_1,\varepsilon_5)=(1,1)$$ or $$(\varepsilon_1,\varepsilon_5)=(-1,-1)$$. 4.1 Subcase $$\boldsymbol{r=p\pm 1}$$ In this subsection, we assume that $$r=p\pm 1$$. Let $$r=p+\varepsilon(\varepsilon=\pm 1)$$. From equation (4.1), we get   2ε4ε5=2(p+ε)2+(ε5−9ε1ε5)pq=2p2+4εp+2+(ε5−9)pq=2p(p+2ε+ε5−92q)+2, or   ε4ε5−1=p(p+2ε+ε5−92q). Since $$p(>q\ge 3)$$ divides $$\varepsilon_4\varepsilon_5-1$$, we must have $$\varepsilon_4\varepsilon_5-1=0$$ and $$p+2\varepsilon+\frac{\varepsilon_5-9}{2}q=0$$. Hence $$(\varepsilon_4,\varepsilon_5)=(1,1)$$ or $$(\varepsilon_4,\varepsilon_5)=(-1,-1)$$. First, suppose that $$(\varepsilon_4,\varepsilon_5)=(1,1)$$. Then $$p=\frac{9-\varepsilon_5}{2}q-2\varepsilon=4q-2\varepsilon,r=p+\varepsilon=4q-\varepsilon,s=2\varepsilon_5=2$$, and we get $$\varepsilon_1=\varepsilon_2=1$$ from the equations $$\varepsilon_1\varepsilon_5=1$$ and $$\varepsilon_2\varepsilon_4=1$$. The relation $$T(p,q,r,s)=T_{m, \varepsilon_1 mpq+\varepsilon_2}(p,\varepsilon_1 q)$$ yields   T(4q−2ε,q,4q−ε,2)=T3,3(4q−2ε)q+1(4q−2ε,q). If $$\varepsilon=-1$$, then $$T(p,q,r,s)=T(4q+2,q,4q+1,2)$$ is the knot $$T((4q+1,8q+2),(4q+2,q))$$ in the notation of [1] and has braid index $$\mathrm{min}(9q+2,4q+1)=4q+1$$ by [1, Corollary 8], but $$T_{m, \varepsilon_1 mpq+\varepsilon_2}(p,\varepsilon_1 q)=T_{3,3(4q+2)q+1}(4q+2,q)$$ has braid index $$3q$$ by [21, Theorem 1], and hence we get $$4q+1=3q$$, which is impossible for $$q\ge 3$$. Suppose that $$\varepsilon=1$$. Then $$T(p,q,r,s)=T(4q-2,q,4q-1,2)$$ is isotopic to the closure of the braid in Figure 19 by Lemma 2.1. The braid has $$4q-1$$ strands and $$(4q-2)(8q-1)+(q-2)(q-1)+(3q-1)(q-1)$$ crossings of the same sign. Hence the genus of $$T(p,q,r,s)$$ is given by $$g(T(p,q,r,s))=\frac{1}{2}(36q^2-31q+7)$$. On the other hand, $$T_{m, \varepsilon_1 mpq+\varepsilon_2}(p,\varepsilon_1 q)=T_{3,3(4q-2)q+1}(4q-2,q)$$ has genus $$\frac{1}{2}(36q^2-33q+9)$$ by Lemma 2.8. Hence we get $$36q^2-31q+7=36q^2-33q+9$$, which gives $$q=1$$ and contradicts Lemma 2.7. Fig. 19. View largeDownload slide $$T(4q-2,q,4q-1,2)$$ is isotopic to the closure of the braid. Fig. 19. View largeDownload slide $$T(4q-2,q,4q-1,2)$$ is isotopic to the closure of the braid. Suppose that $$(\varepsilon_4,\varepsilon_5)=(-1,-1)$$. Then $$p=\frac{9-\varepsilon_5}{2}q-2\varepsilon=5q-2\varepsilon,r=p+\varepsilon=5q-\varepsilon,s=2\varepsilon_5=-2$$, and we get $$\varepsilon_1=\varepsilon_2=-1$$ from the equations $$\varepsilon_1\varepsilon_5=1$$ and $$\varepsilon_2\varepsilon_4=1$$. The relation $$T(p,q,r,s)=T_{m, \varepsilon_1 mpq+\varepsilon_2}(p,\varepsilon_1 q)$$ yields   T(5q−2ε,q,5q−ε,−2)=T3,−3(5q−2ε)q−1(5q−2ε,−q). If $$\varepsilon=-1$$, then $$T(p,q,r,s)=T(5q+2,q,5q+1,-2)$$ has braid index $$5q+1$$ by Lemma 2.5, but $$T_{m, \varepsilon_1 mpq+\varepsilon_2}(p,\varepsilon_1 q)=T_{3,-3(5q+2)q-1}(5q+2,-q)$$ has braid index $$3q$$ by [21, Theorem 1], and hence we get $$5q+1=3q$$, which is impossible for $$q\ge 3$$. Suppose that $$\varepsilon=1$$. Then $$T(p,q,r,s)=T(5q-2,q,5q-1,-2)$$ is isotopic to the closure of the first braid in Figure 20 by Lemma 2.1. It is easy to see that the five braids in Figure 20 are isotopic. The last braid has $$5q-1$$ stands and $$(5q-2)(9q-3)+(q-1)$$ crossings of the same sign. Hence the genus of $$T(p,q,r,s)$$ is given by $$\frac{1}{2}(45q^2-37q+7)$$. On the other hand, $$T_{m, \varepsilon_1 mpq+\varepsilon_2}(p,\varepsilon_1 q)=T_{3,-3(5q-2)q-1}(5q-2,-q)$$ is the mirror image of $$T_{3,3(5q-2)q+1}(5q-2,q)$$, which has genus $$\frac{1}{2}(45q^2-36q+9)$$ by Lemma 2.8. Thus we get $$45q^2-37q+7=45q^2-36q+9$$, contradicting Lemma 2.7. Fig. 20. View largeDownload slide $$T(5q-2,q,5q-1,-2)$$ is isotopic to the closure of any of these braids. Fig. 20. View largeDownload slide $$T(5q-2,q,5q-1,-2)$$ is isotopic to the closure of any of these braids. 4.2 Subcase $$\boldsymbol{r=p+q}$$ In this subsection, we assume that $$r=p+q$$. Recall that either $$(\varepsilon_1,\varepsilon_5)=(1,1)$$ or $$(\varepsilon_1,\varepsilon_5)=(-1,-1)$$. First, suppose that $$(\varepsilon_1,\varepsilon_5)=(1,1)$$. Then equation (4.1) gives us $$2(p+q)^2-8pq=2\varepsilon_4$$ or equivalently $$(p-q)^2=\varepsilon_4$$, so $$\varepsilon_4=1$$ and $$p=q+1$$. Therefore $$T(p,q,r,s)=T(q+1,q,2q+1,2)$$ and hence $$b(T(p,q,r,s))\le 2q+1$$ by Lemma 2.2. On the other hand, $$T(p,q,r,s)=T_{m, \varepsilon_1 mpq+\varepsilon_2}(p,\varepsilon_1 q)=T_{3,3q(q+1)+\varepsilon_2}(q+1, q)$$ and $$b(T_{3,3q(q+1)+\varepsilon_2}(q+1, q))=3q$$ by Lemma 2.9. This yields $$3q\le 2q+1$$ or $$q\le 1$$, contradicting Lemma 2.7. Hence $$(\varepsilon_1,\varepsilon_5)=(-1,-1)$$ and equation (4.1) gives us $$p^2+q^2-3pq=-\varepsilon_4$$. It follows from [10, Lemma 4.1] that $$p=f_{n+1}$$ and $$q=f_{n-1}$$ for some integer $$n$$, where $$f_n$$ is the $$n$$th Fibonacci number. By Lemma 2.4 $$T(p,q,r,s)=T(f_{n+1},f_{n-1},f_{n-1}+f_{n+1},-2)$$ is isotopic to the mirror image of $$T(f_{n-1}+f_{n+1},f_{n-1}+2f_{n+1},f_{n-1},1)$$ because $$2r-q=2f_{n-1}+2f_{n+1}-f_{n-1}=f_{n-1}+2f_{n+1}$$. In the notation of [1], we have $$T(f_{n-1}+f_{n+1},f_{n-1}+2f_{n+1},f_{n-1},1) =T((f_{n-1},f_{n-1}),(f_{n-1}+f_{n+1},f_{n-1}+2f_{n+1}))$$. The braid index of this knot is equal to $$\mathrm{min}(f_{n-1}+f_{n+1},f_{n-1}+2f_{n+1})=f_{n-1}+f_{n+1}$$ by [1, Corollary 8]. On the other hand, it is easy to see that $$T(p,q,r,s)=T_{m, \varepsilon_1 mpq+\varepsilon_2}(p,\varepsilon_1 q)=T_{3,-3pq+\varepsilon_2}(p,-q)$$ has braid index $$3q=3f_{n-1}$$ by [21, Theorem 1]. This observation yields $$f_{n-1}+f_{n+1}=3f_{n-1}$$, or equivalently $$f_n=f_{n-1}(=1=q)$$, a contradiction. 4.3 Subcase $$\boldsymbol{r=2p-q}$$ In this subsection, we assume that $$r=2p-q$$. Then $$p<2q$$. Recall that $$r\equiv \pm 1$$ or $$\pm p\mod q$$ by Proposition 3.5(1). If $$r=2p-q\equiv p\mod q$$, then $$q$$ must divide $$p$$, a contradiction. If $$r=2p-q\equiv -p\mod q$$, then $$3p\equiv 0\mod q$$ and hence $$q=3$$ and $$p=4,5$$ (since $$q<p<2q$$), but any of $$(p,q,r)=(4,3,5),(5,3,7)$$ does not satisfy equation (4.1). Suppose $$r=2p-q\equiv \pm 1\mod q$$. Since $$q<p<2q$$, it is easy to see that $$2p=3q+\varepsilon(\varepsilon=\pm 1)$$. Multiplying both sides of equation (4.1) by $$2$$ and then substituting $$r=2p-q=2q+\varepsilon$$, we get   4(2q+ε)2+(ε5−9ε1ε5)(3q2+εq)=4ε4ε5. Recall that $$\varepsilon_1\varepsilon_5=1$$, so   4=|4ε4ε5|=|4(2q+ε)2+(ε5−9)(3q2+εq)|≥|(ε5−9)q(3q+ε)|−|4(2q+ε)2|≥8q(3q+ε)−4(2q+ε)2=8q(q−ε)−4≥8q(q−1)−4≥8⋅3⋅2−4=44, a contradiction. 4.4 Subcase $$\boldsymbol{r=p-q}$$ In this subsection, we assume that $$r=p-q$$. Recall that $$(\varepsilon_1,\varepsilon_5)=(1,1)$$ or $$(-1,-1)$$. First, suppose $$(\varepsilon_1,\varepsilon_5)=(1,1)$$. From equation (4.1), we get $$2(p-q)^2-8pq=2\varepsilon_4$$ or   p2−6pq+q2=ε4. Dividing both sides by $$pq$$, we get   p/q+1p/q=6+ε4pq≥6−1pq≥6−14⋅3(>5+15). Noting that $$t+1/t$$ is an increasing function for $$t\ge 1$$, one sees that $$p/q>5$$ or   p>5q. In the notation of [1], $$T(p,q,r,s)=T(p,q,p-q,2)$$ is the knot $$T((p-q,2(p-q)),(p,q))$$ and that $$p-q\ge q$$ (since $$p>5q$$). By [1, Corollary 8] the braid index of $$T(p,q,r,s)$$ is equal to $$\mathrm{min}(2(p-q)+q,p-q)=p-q$$. On the other hand, the cable knot $$T(p,q,r,s)=T_{m, \varepsilon_1 mpq+\varepsilon_2}(p,\varepsilon_1 q)=T_{3, 3pq+\varepsilon_2}(p, q)$$ has braid index $$3q$$ by [21, Theorem 1]. Thus $$p-q=3q$$, giving $$p=4q$$. This contradicts the above observation $$p>5q$$. Suppose $$(\varepsilon_1,\varepsilon_5)=(-1,-1)$$. From equation (4.1), we get $$2(p-q)^2-10pq\,{=} -2\varepsilon_4$$ or   p2−7pq+q2=−ε4. Dividing both sides by $$pq$$, we get   p/q+1p/q=7−ε4pq≥7−1pq≥7−14⋅3(>6+16) and see that   p>6q. Thus, by [10, Lemma 2.6], $$T(p,q,r,s)$$ is a fibered knot and its fiber surface has Euler characteristic $$\chi$$ given by   χ=(p−q)−((p−q−1)(2p−4q)+(p−2q)q)=−(2p2−5pq+2q2−3p+5q), so the genus of the knot is given by   g(T(p,q,r,s))=12(1−χ)=12(2p2−5pq+2q2−3p+5q+1). On the other hand, $$T(p,q,r,s)=T_{m, \varepsilon_1 mpq+\varepsilon_2}(p,\varepsilon_1 q)=T_{3, -3pq+\varepsilon_2}(p,-q)$$ is the mirror image of $$T_{3, 3pq-\varepsilon_2}(p,q)$$. The genus of $$T_{3, 3pq-\varepsilon_2}(p,q)$$ is equal to   12(9pq−3p−3q−2ε2+1) by Lemma 2.8. Thus $$\frac{1}{2}(2p^2-5pq+2q^2-3p+5q+1)=\frac{1}{2}(9pq-3p-3q-2\varepsilon_2+1)$$, giving $$-\varepsilon_2=p^2-7pq+q^2+4q=-\varepsilon_4+4q\ge -1+4q\ge -1+4\cdot 3=11$$. This gives a contradiction. 5 Case B Throughout this section, we assume that there are integers $$\varepsilon_i=\pm 1(i=1,2,3,4)$$ and $$m\ge 2$$ such that $$T(p,q,r,s)=T_{m,2\varepsilon_1 mp+\varepsilon_2}(p,2\varepsilon_1)$$ and $$pq+r^2s-2\varepsilon_1m^2p=\varepsilon_2 m+\varepsilon_3=\varepsilon_4 q.$$ Multiplying both sides of the equation $$\varepsilon_2 m+\varepsilon_3=\varepsilon_4 q$$ by $$\varepsilon_2$$, we get $$m+\varepsilon_2\varepsilon_3=\varepsilon_2\varepsilon_4q$$. Since $$m\ge 2$$, $$m=\varepsilon_2\varepsilon_4 q-\varepsilon_2\varepsilon_3$$ implies that   m=q+ε(ε=±1) and ε2ε4=1,ε2ε3=−ε. Let $$s=2\varepsilon_5(\varepsilon_5=\pm 1)$$. Then $$\varepsilon_4 q=pq+r^2s-2\varepsilon_1m^2p=pq+2\varepsilon_5r^2-2\varepsilon_1(q+\varepsilon)^2p$$ and hence we get   2r2+ε5pq−2ε1ε5p(q+ε)2=ε4ε5q. Note that $$2q^2-5q+2> 1$$ for $$q\ge 3$$. Hence if $$\varepsilon_1\varepsilon_5=-1$$, then   q≥ε4ε5q=2r2+ε5pq−2ε1ε5p(q+ε)2=2r2+ε5pq+2p(q+ε)2≥2r2−pq+2p(q−1)2=2r2+p(2q2−5q+2)>2r2+p>p>q, a contradiction. Hence   ε1ε5=1 and we get   2r2+ε5pq−2p(q+ε)2=ε4ε5q. (5.1) 5.1 Subcase $$\boldsymbol{r=p\pm 1}$$ In this subsection, we assume that $$r=p\pm 1$$ and let $$r=p+\varepsilon_6(\varepsilon_6=\pm 1)$$. From equation (5.1), we get $$\varepsilon_4\varepsilon_5q=2(p+\varepsilon_6)^2+\varepsilon_5pq-2p(q+\varepsilon)^2=2p^2+4\varepsilon_6p+2+\varepsilon_5pq-2p(q+\varepsilon)^2$$, giving   ε4ε5q−2=p(2p+4ε6+ε5q−2(q+ε)2). This implies that $$p$$ divides $$\varepsilon_4\varepsilon_5q-2(=q-2\text{ or } -q-2)$$. Since $$p>q$$, it is easy to see that $$q-2=0$$ or $$p=q+2$$ according to whether $$\varepsilon_4\varepsilon_5=1$$ or $$\varepsilon_4\varepsilon_5=-1$$. But $$q\ne 2$$ by Lemma 2.7. Hence $$\varepsilon_4\varepsilon_5=-1,p=q+2$$, and $$-1=2p+4\varepsilon_6+\varepsilon_5q-2(q+\varepsilon)^2=2(q+2)+4\varepsilon_6+\varepsilon_5q-2(q+\varepsilon)^2$$, giving   q(2+ε5−2q−4ε)=−3−4ε6. Since $$q$$ divides $$-3-4\varepsilon_6$$, we have $$\varepsilon_6=1,q=3+4\varepsilon_6=7$$, and   −1=2+ε5−2q−4ε=−12+ε5−4ε, which is impossible. 5.2 Subcase $$\boldsymbol{r=p-q,2p-q}$$ or $$\boldsymbol{p+q}$$ In this subsection, we assume that $$r=p-q,2p-q$$ or $$p+q$$. Let $$r=\delta p+\varepsilon_6q$$, where $$(\delta,\varepsilon_6)=(1,-1),(2,-1)$$ or $$(1,1)$$. From equation (5.1), we get $$\varepsilon_4\varepsilon_5q=2(\delta p+\varepsilon_6q)^2+\varepsilon_5pq-2p(q+\varepsilon)^2=2\delta^2p^2 +4\delta\varepsilon_6pq+2q^2+\varepsilon_5pq-2p(q+\varepsilon)^2$$, giving   −q(2q−ε4ε5)=p(2δ2p+4δε6q+ε5q−2(q+ε)2). Since $$p$$ is relatively prime to $$q$$, $$p$$ must divide $$2q-\varepsilon_4\varepsilon_5$$. Since $$p>q$$, it is easy to see that   p=2q−ε4ε5. Hence $$-q=2\delta^2p+4\delta\varepsilon_6q+\varepsilon_5q-2(q+\varepsilon)^2 =2\delta^2(2q-\varepsilon_4\varepsilon_5)+4\delta\varepsilon_6q+\varepsilon_5q-2(q+\varepsilon)^2$$, giving   q(q+2ε−2δ2−2δε6−1+ε52)=−1−δ2ε4ε5. (5.2) Note that $$q+2\varepsilon-2\delta^2-2\delta\varepsilon_6-\frac{1+\varepsilon_5}{2}$$ is an integer, so $$q$$ is a divisor of the integer $$-1-\delta^2\varepsilon_4\varepsilon_5$$. First, suppose $$(\delta,\varepsilon_6)=(1,-1)$$. Then $$r=p-q$$ and equation (5.2) becomes   q(q+2ε−1+ε52)=−1−ε4ε5. Since $$|-1-\varepsilon_4\varepsilon_5|\le 2$$, we must have $$q+2\varepsilon-\frac{1+\varepsilon_5}{2}=-1-\varepsilon_4\varepsilon_5=0$$. This implies that $$q=3,\varepsilon=-1,\varepsilon_5=1,\varepsilon_4=-1$$ and hence $$p=2q-\varepsilon_4\varepsilon_5=7, r=p-q=4, s=2\varepsilon_5=2,m=q+\varepsilon=2$$, and $$\varepsilon_1=1,\varepsilon_2=-1$$ from $$\varepsilon_1\varepsilon_5=1$$ and $$\varepsilon_2\varepsilon_4=1$$. Therefore   T(7,3,4,2)=T(p,q,r,s)=Tm,2ε1mp+ε2(p,2ε1)=T2,27(7,2). Since $$T(7,3,4,2)$$ is the closure of a positive braid with $$7$$ strands and $$42$$ crossings, $$g(T(7,3,4,2))=\frac{1-7+42}{2}=18$$. However, $$g(T_{2,27}(7,2))=19$$ by Lemma 2.8. This gives a contradiction. Next, suppose $$(\delta,\varepsilon_6)=(2,-1)$$. Then $$r=2p-q$$ and equation (5.2) becomes   q(q+2ε−9+ε52)=−1−4ε4ε5. Note that $$-1-4\varepsilon_4\varepsilon_5$$ equals $$3$$ or $$-5$$. If $$-1-4\varepsilon_4\varepsilon_5=3$$, then $$q=3$$ and $$q+2\varepsilon-\frac{9+\varepsilon_5}{2}=1$$, so $$\varepsilon=1,\varepsilon_5=-1,\varepsilon_4=1,$$ and $$p=2q-\varepsilon_4\varepsilon_5=7$$, contradicting $$p<2q$$. If $$-1-4\varepsilon_4\varepsilon_5=-5$$, then $$q=5$$ and $$q+2\varepsilon-\frac{9+\varepsilon_5}{2}=-1$$, so $$\varepsilon=-1,\varepsilon_5=-1,\varepsilon_4=-1,$$ and $$p=2q-\varepsilon_4\varepsilon_5=9,r=2p-q=13$$, contradicting Proposition 3.5(1). Finally, suppose $$(\delta,\varepsilon_6)=(1,1)$$. Then $$r=p+q$$ and equation (5.2) becomes   q(q+2ε−9+ε52)=−1−ε4ε5. Since $$|-1-\varepsilon_4\varepsilon_5|\le 2$$, we must have $$q+2\varepsilon-\frac{9+\varepsilon_5}{2}=-1-\varepsilon_4\varepsilon_5=0$$. This implies that $$(q,\varepsilon,\varepsilon_4,\varepsilon_5)=(3,1,-1,1),(2,1,1,-1)(7,-1,-1,1)$$, or $$(6,-1,1,-1)$$. Lemma 2.7 rules out the second possibility. Using the equalities $$p=2q-\varepsilon_4\varepsilon_5,r=p+q,s=2\varepsilon_5, m=q+\varepsilon, \varepsilon_1\varepsilon_5=1, \varepsilon_2\varepsilon_4=1$$, we get   (p,q,r,s;m,ε1,ε2)={(7,3,10,2;4,1,−1)if (ε,ε4,ε5)=(1,−1,1),(15,7,22,2;6,1,−1)if (ε,ε4,ε5)=(−1,−1,1),(13,6,19,−2;5,−1,1)if (ε,ε4,ε5)=(−1,1,−1). Since $$T(p,q,r,s)=T_{m,2\varepsilon_1 mp+\varepsilon_2}(p,2\varepsilon_1)$$, we have   T(7,3,10,2)=T4,55(7,2)if (ε,ε4,ε5)=(1,−1,1),T(15,7,22,2)=T6,179(15,2)if (ε,ε4,ε5)=(−1,−1,1),T(13,6,19,−2)=T5,−129(13,−2)if (ε,ε4,ε5)=(−1,1,−1). The calculation of the genera of these knots gives a contradiction as follows. We have $$g(T(7,3,10,2))=96$$ and $$g(T_{4,55}(7,2))=93$$ by Lemmas 2.3 and 2.8, so $$T(7,3,10,2)\ne T_{4,55}(7,2)$$. We also have $$g(T(15,7,22,2))=504$$ and $$g(T_{6,179}(15,2))=487$$, so $$T(15,7,22,2)\ne T_{6,179}(15,2)$$. Note that $$T_{5,-129}(13,-2)$$ is the mirror image of $$T_{5,129}(13,2)$$, so $$g(T_{5,-129}(13,-2))=g(T_{5,129}(13,2))=286$$. But $$g(T(13,6,19,-2))=294$$ by Lemma 2.4. Thus $$T(13,6,19,-2)\ne T_{5,-129}(13,-2)$$. 6 Case C Throughout this section, we assume that there are integers $$\varepsilon_i=\pm 1(i=1,2,3,4)$$ and $$m\ge 2$$ such that $$T(p,q,r,s)=T_{m,2\varepsilon_1 mq+\varepsilon_2}(q,2\varepsilon_1)$$ and $$pq+r^2s-2\varepsilon_1m^2q=\varepsilon_2 m+\varepsilon_3=\varepsilon_4 p.$$ Multiplying both sides of the equation $$\varepsilon_2 m+\varepsilon_3=\varepsilon_4p$$ by $$\varepsilon_2$$, we get $$m+\varepsilon_2\varepsilon_3=\varepsilon_2\varepsilon_4p$$. Since $$m\ge 2$$, $$m=\varepsilon_2\varepsilon_4p-\varepsilon_2\varepsilon_3$$ implies that $$m=p+\varepsilon(\varepsilon=\pm 1)$$ and $$\varepsilon_2\varepsilon_4=1,\varepsilon_2\varepsilon_3=-\varepsilon$$. Note that $$T_{m,2\varepsilon_1 mq+\varepsilon_2}(q,2\varepsilon_1)$$ is either $$T_{m,2 mq+\varepsilon_2}(q,2)$$ or the mirror image of $$T_{m,2 mq-\varepsilon_2}(q,2)$$, depending on whether $$\varepsilon_1=1$$ or $$\varepsilon_1=-1$$. Hence $$b(T_{m,2\varepsilon_1 mq+\varepsilon_2}(q,2\varepsilon_1))=2m$$ by Lemma 2.9. If $$r<p$$, then $$2(p-1)\le 2(p+\varepsilon)=2m=b(T_{m,2\varepsilon_1 mq+\varepsilon_2}(q,2\varepsilon_1))=b(T(p,q,r,s))\le p$$ by Lemma 2.2 and hence $$(q<)p\le 2$$, contradicting Lemma 2.7. Hence $$r>p$$ and $$2(p-1)\le 2(p+\varepsilon)=b(T(p,q,r,s))\le r$$, which implies $$r\ne p+1$$ and $$r\ne 2p-q$$ by Lemma 2.7. Hence $$r=p+q$$ and $$\varepsilon=-1$$. Let $$s=2\varepsilon_5(\varepsilon_5=\pm 1)$$. Then $$\varepsilon_4 p=pq+r^2s-2\varepsilon_1m^2q=pq+2\varepsilon_5(p+q)^2-2\varepsilon_1(p-1)^2q$$ and hence we get   2(p+q)2+ε5pq−2ε1ε5(p−1)2q=ε4ε5p. If $$\varepsilon_1\varepsilon_5=-1$$, then   p≥ε4ε5p=2(p+q)2+ε5pq+2(p−1)2q≥2(p+q)2−pq+2(p−1)2q=2p2+3pq+2q2+2(p−1)2q>p, a contradiction. Hence $$\varepsilon_1\varepsilon_5=1$$ and we get   2(p+q)2+ε5pq−2(p−1)2q=ε4ε5p, which yields   p(2p+8q+ε5q−ε4ε5−2pq)=−2(q−1)q. Since $$p$$ is relatively prime to $$q$$, $$p$$ must divide $$2(q-1)$$. Since $$p>q$$, it is easy to see that $$p=2(q-1)$$. Hence $$-q=2p+8q+\varepsilon_5q-\varepsilon_4\varepsilon_5-2pq =4(q-1)+8q+\varepsilon_5q-\varepsilon_4\varepsilon_5-4(q-1)q$$, giving   2q(2q−17+ε52)=−4−ε4ε5. The left-hand side of this equation is even, while the right-hand side is odd. This is impossible. Funding This work was supported by the National Research Foundation of Korea Grant funded by the Korean Government [NRF-2013R1A1A2A10064864]. 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International Mathematics Research NoticesOxford University Press

Published: Feb 1, 2018

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