On tunnel numbers of a cable knot and its companion

Not only $A\cap V^{\star}$ but also $A\cap W^{\star}$ consist of finitely many essential annuli. For if not, one of them contains at least one boundary parallel annulus. Then we do an isotopy on $A$ to reduce $\mid S^{\star}\cap A\mid$, which is against the minimal assumption of $\mid S^{\star}\cap A\mid$. By Lemma 2.5, there is an essential disk $D\subset\overline{V^{\star}-A}$ (resp. $E\subset\overline{W^{\star}-A}$ ). Then $\partial D$ is contained in one subsurface of $S^{\star}-A$, says $S_{1}$. Since $V^{\star}\cup_{S^{\star}}W^{\star}$ is strongly irreducible, $\partial E\subset S_{1}$. Then any essential disk of $V^{\star}$ disjoint from $A\cap V^{\star}$ has its boundary curve in $S_{1}$. So any essential disk of $W^{\star}$ disjoint from $A\cap W^{\star}$ has its boundary curve in $S_{1}$. Hence $S_{1}$ is the only compressible subsurface in $\overline{S^{\star}-A}$. Since $S_{1}\subset S^{*}$ is essential, i.e., the inclusion map on its fundamental group is injective, $S_{1}$ is strongly irreducible.

∎

Since $S^{\star}\cap J$ consists of finitely many nested annuli in $J$ which are not parallel to $A$, by doing the Dehn surgery in $V^{\star}$, the solid torus $ST_{0}\subset J$ bounded by $A_{0}\cup A_{j}$ is changed into a new solid torus so that $A_{j}$ is parallel to $A_{0}$. It means that among of all these annuli $\{A_{1},..,A_{n}\}$, each one is parallel to $A_{0}$. So we do an isotopy on $S$ so that it is disjoint from $A$. Therefore, $A$ is in either $V$ or $W$. Since $A$ has the same core curve with $A_{0}$, $A$ is incompressible. Therefore, $A$ is boundary parallel in $V$ or $W$. So $A$ cuts out the I-bundle $A\times I$ in $M$. It means that $M$ is homeomorphic to $E(K)$. ∎

Since $\partial A^{{}^{\prime}}$ bounds a subannulus $A^{{}^{\prime\prime}}\subset A$, $A^{{}^{\prime}}$ is parallel to either $A^{{}^{\prime\prime}}$ or its complement annulus in $\partial J$.

(1) If $A^{{}^{\prime}}$ is parallel to a subannulus $A^{{}^{\prime\prime}}\subset A$, then $int(A^{{}^{\prime\prime}})\cap S^{\star}=\emptyset$ up to isotopy, see Figure 3. For if not, then there is an annulus in $S^{\star}\cap J$ parallel to $A$. So we do an isotopy along this annulus to reduce $\mid S^{\star}\cap A\mid$. Then $A^{{}^{\prime\prime}}$ is an essential annulus in one of $V^{\star}$ and $W^{\star}$. Without loss of generality, we assume that $A^{{}^{\prime\prime}}\subset V^{\star}$. The other case is similar. So we omit it. Then cutting $V^{\star}$ along $A^{{}^{\prime\prime}}$ produces two handlebodies or one handlebody and a compression body. Let $V^{{}^{\prime}}$ be the compression body or the handlebody containing no $S_{1}$.

By Claim 3.1, except $S_{1}$, all other components of $S^{\star}\cap J$ are nested annuli in $J$. If there is no annulus in $S^{\star}\cap J$, there are a pair of two spanning annuli, says $A^{2^{\prime\prime}}$ and $A^{3^{\prime\prime}}$ so that one component of $W^{\star}-A^{2^{\prime\prime}}\cup A^{3^{\prime\prime}}$, says $N$, lies in $E(K)$. It is not hard to see that $N$ is a handlebody and $\partial A^{{}^{\prime\prime}}\subset N$. Let $A^{{}^{\prime\prime}}\times I$ be I-bundle in $V^{{}^{\prime}}$. Then we remove this I-bundle from $V^{{}^{\prime}}$ and attach it to $N$ along these two annuli $\partial A^{{}^{\prime\prime}}\times I$. So $N$ is changed into a new 3-manifold, says $N_{0}$. In this case, $\partial N_{0}$ contains $\partial E(K)$. Let $a$ be a vertical arc connecting $A^{{}^{\prime\prime}}\times\{0,1\}$ in $A^{{}^{\prime\prime}}\times I$. So $a$ is a fiber arc in $N_{0}$ connecting $\partial E(K)$ and the other boundary surface of $N_{0}$. We remove the union of a regular neighborhood of $\partial E(K)$ and a regular neighborhood of $a$ in $N_{0}$ from $N_{0}$. Then $N_{0}$ is changed into a 3-manifold $W$, which is also the amalgamation of $N$ and a tubed annulus I-bundle along two essential annuli. It is not hard to see that $W$ is also a handlebody and furthermore $\overline{E(K)-W}$ is the disk sum of $V^{{}^{\prime}}$ and $\partial E(K)\times I$, denoted by $V$. So $V$ is a compression body and $V\cup_{\partial_{+}V=\partial W}W$ is a Heegaard splitting of $E(K)$. During the process of this surgery, $g(\partial W)=g(S^{\star})$. It means that $g(K^{\star})\geq g(K)$ and $t(K^{\star})\geq t(K)$.

So we assume that there is at least one annulus in $S^{\star}\cap J$. Let $A^{2}$ be the innermost annulus in $S^{\star}\cap J$, i.e., no other component of $S^{\star}\cap J$ lies between it and $S_{1}$. Then $A^{2}$ is not boundary parallel into $A$. For if not, then we can do an isotopy on $A$ to reduce $S^{\star}\cap J$. Let $V^{{}^{\prime}}$ be the handlebody or compression body as above. Then $\overline{E(K^{\star})-V^{{}^{\prime}}}$ contains a smaller copy of $J$. Then we do a Dehn surgery on this smaller copy of $J$ along its longitude so that $A^{2}$ is parallel to $A^{{}^{\prime}}$. Then $\overline{E(K^{\star})-V^{{}^{\prime}}}$ is changed into a new 3-manifold, denoted by $N$ again. By the same argument of the case that $S^{\star}\cap J$ is incompressible, $N\cup V^{{}^{\prime}}$ is homeomorphic to $E(K)$. Since $A^{2}\cup A^{{}^{\prime}}$ bounds an annulus I-bundle, let $a$ be a fiber arc connecting $A^{2}$ and $A^{{}^{\prime}}$ in this I-bundle. Then we remove a regular neighborhood of $a$ and then $N$ is changed into a 3-manifold $W$. By the same argument as above, $W$ is a compression body or handlebody. Without loss of generality, we assume that $W$ is a handlebody. Meanwhile, the union of $V^{{}^{\prime}}$ and the closed regular neighborhood of $a$ is a compression body. So $V\cup_{\partial_{+}V=\partial W}W$ is a Heegaard splitting of $E(K)$. By the same argument, $t(K^{\star})\geq t(K)$.

(2) If $A^{{}^{\prime}}$ is parallel to the complement annulus of $A^{{}^{\prime\prime}}$ in $\partial J$, then $int(A^{{}^{\prime\prime}})\cap S^{\star}=\emptyset$. For if not, by Claim 3.1, then there is an innermost annulus $A{{}^{\prime\prime\prime\prime}}$ of $S^{\star}\cap J$ so that its boundary curves lie in the interior of $A^{{}^{\prime\prime}}$. By minimality of $\mid S^{\star}\cap A\mid$, $A^{{}^{\prime\prime\prime\prime}}$ is not boundary parallel to $A$. Then $A^{{}^{\prime\prime\prime\prime}}\cup A^{{}^{\prime\prime}}$ bounds a smaller copy of $J$. By the similar argument as above, we do a Dehn surgery on this smaller copy of $J$ and get a Heegaard splitting of $E(K)$. Moreover, $t(K^{\star})\geq t(K)$.

It is not hard to see that the dual core curve of the tube in $S_{1}$ bounds an essential disk $E$, says in $W^{\star}$. Then we do a compression on $W^{\star}$ along the disk $E$ and get a new 3-manifold $W^{{}^{\prime}}$. Moreover, $S_{1}$ is changed into an annulus $A^{{}^{\prime}}$ in $J$. Then $A^{{}^{\prime}}\cup A^{{}^{\prime\prime}}$ bounds a smaller copy of $J$ in $\overline{E(K^{\star})-W^{{}^{\prime}}}$. Denote $N=\overline{E(K^{\star})-W^{{}^{\prime}}}$. If $A^{{}^{\prime\prime}}$ is an essential annulus in $V^{\star}$, then $N$ is an amalgamation of a handlebody or a compression body $V^{{}^{\prime}}$ and a smaller copy of $J$ along $A^{{}^{\prime\prime}}$. So we do a Dehn surgery as above so that the smaller copy of $J$ is changed into the I-bundle $A^{{}^{\prime\prime}}\times I$ and $N$ is changed into a 3-manifold $V$, where $V$ is the union of $V^{{}^{\prime}}$ and $A^{{}^{\prime\prime}}\times I$. Then $V$ is homeomorphic to $V^{{}^{\prime}}$. By the same argument as above, $V^{{}^{\prime}}\cup W$ is a Heegaard splitting of $E(K)$ and $t(K^{\star})\geq t(K)+1$.

Otherwise, $A^{{}^{\prime\prime}}$ is an essential annulus in $W^{\star}$. Then both of $A^{{}^{\prime}}$ and $A^{{}^{\prime\prime}}$ are in $W^{\star}$, where they bound a smaller copy of $J$. Then we do a Dehn surgery on this smaller copy of $J$ so that they are parallel. Hence $W^{\star}$ is changed into a compression body or handlebody $W$. Replace $V^{\star}$ by $V$. So $V\cup_{S^{\star}}W$ is a Heegaard splitting of $E(K)$ and $t(K^{\star})\geq t(K)$. ∎

Let $S_{1}$ be the tube sum of two annuli $A_{1}$ and $A_{2}$ in $J$. We say there is no other component of $S^{\star}\cap J$ in the region between $A_{1}$ and $A_{2}$ in $J$. For if not, then the tube would not connect them. It is not hard to see that $\partial A_{1}$ (resp. $\partial A_{2}$) bounds an annulus in $A$. Without loss of generality, we assume that the annulus bounded by $\partial A_{1}$ in $A$ doesn’t contain $\partial A_{2}$. Then we say that there is no other component of $S^{\star}\cap J$, of which boundary curves lies in the annulus bounded by $\partial A_{1}$ in $A$. Otherwise, we either do an isotopy on $A$ to reduce $\mid S^{\star}\cap A\mid$ or do a Dehn surgery as above and get $t(K^{\star})\geq t(K)$.

There are three types of $A_{1}$ and $A_{2}$ as follows:

• •

  (1) one of them is parallel to $A$, says $A_{1}$ while $A_{2}$ not;

• •

  (2) both of them are parallel to $A$;

• •

  (3) neither of them is parallel to $A$.

For the first case, either $\partial A_{1}$ separate $\partial A_{2}$ in $A$ or not. If $\partial A_{1}$ separate $\partial A_{2}$, then $A_{1}\cup A_{2}$ bounds a smaller copy of $J$. Then we do a compression on this tube along the disk $D$ bounded by the core curve. Without loss of generality, we assume that $D\subset V^{\star}$. So $V^{\star}$ is changed into a new 3-manifold, denoted by $V^{{}^{\prime}}$, which is also a compression body or handlebody. Denoted $\overline{E(K^{\star})-V^{{}^{\prime}}}$ by $N$. Again we do a Dehn surgery on $N$ as in the proof of Lemma 3.2 and so the smaller copy of $J$ is changed into the annulus I-bundle $A_{1}\times I$. So $N$ is changed into $N^{{}^{\prime}}$ and $N^{{}^{\prime}}\cup V^{{}^{\prime}}=E(K)$. Let $a$ be the vertical arc of $A_{1}\times I$, which connects $A_{1}\times\{0,1\}$. Then $W=N^{{}^{\prime}}-N(a)$ is a compression body or handlebody. Meanwhile, the union of $V^{{}^{\prime}}$ and the closed regular neighborhood of $a$ is a handlebody or compression body, denoted by $V$. Moreover, $g(\partial_{+}V)=g(S^{\star})$. So $t(K^{\star})\geq t(K)$. If $\partial A_{1}$ doesn’t separate $\partial A_{2}$, then there is a smaller copy of $J$ bounded by $A_{2}$. So we do a dehn surgery as above. Then $t(K^{\star})\geq t(K)$.

For the second case, $A_{1}$ and $A_{2}$ are not nested. Without loss of generality, we assume that the disk $D$ bounded by core curve of this tube lies in $V^{*}$. Then we do a compression on $S_{1}$ along an essential disk $E$ in $W^{*}$. So $W^{*}$ is changed into a new compression body or handlebody, says $W^{{}^{\prime}}$. Since $\partial A_{1}$ and $\partial A_{2}$ bound two essential annuli in $V^{*}$, cutting $V^{*}$ along them produce a handlebody or compression body, says $V^{{}^{\prime}}$. Then $E(K^{*})-W^{{}^{\prime}}$ is the amalgamation of $V^{{}^{\prime}}$ and an annulus I-bundle or a smaller copy of $J$. For the first case, removing a closed neighborhood of a fiber arc in this I-bundle changes $E(K^{*})-W^{{}^{\prime}}$ into a handlebody or compression body $V$. For the later case, we do a dehn surgery on this smaller copy of $J$ so that it is changed into an annulus I-bundle. In both of these two cases, $t(K^{\star})\geq t(K)$.

For the third case, $A_{1}$ and $A_{2}$ are parallel in $J$ and there is a smaller copy of $J$ bounded by a subannulus in $A$ and $A_{1}$. For if not, then both of $A_{1}$ and $A_{2}$ lie in an I-bundle of $A$. Then we do an isotopy to reduce $S^{\star}\cap A$. We do a compression on this tube along the disk $D$ bounded by the core curve. Without loss of generality, we assume that $D\subset V^{\star}$. So $V^{\star}$ is changed into a new 3-manifold, denoted by $V^{{}^{\prime}}$, which is also a compression body or handlebody. Denoted $\overline{E(K^{\star})-V^{{}^{\prime}}}$ by $N$. Since there is no other component of $S^{\star}\cap J$ lie between $A_{1}$ and $A$, this smaller copy of $J$ lies in $V^{{}^{\prime}}$. By the same argument in the proof of Lemma 3.2, we do a Dehn surgery on this smaller copy of $J$ so that it is changed into $A_{1}\times I$. So $V^{{}^{\prime}}$ is changed into a new compression body or handlebody $V^{{}^{\prime\prime}}$. Moreover, $N\cup V^{{}^{\prime\prime}}=E(K)$. On one side, we attach a closed regular neighborhood of a vertical fiber arc $a$ in the I-bundle bounded by $A_{1}$ and $A_{2}$ to $V^{{}^{\prime\prime}}$ so that $V^{{}^{\prime\prime}}$ is changed into a compression body or handlebody, says $V$. On the other side, $W=\overline{N-N(a)}$ is still a handlebody or compression body and $\partial_{+}W=\partial_{+}V$. During this process, $g(\partial_{+}V)=g(S^{\star})$ and $t(K^{\star})\geq t(K)$. ∎