Twist spun knots of twist spun knots of classical knots

We prove that if either $K$ is a trivial knot or $m=1$ then $\tau_{m_{2}}(\tau_{m_{1}}(K))$ is a trivial $3$-knot in $S^{5}$. It is known that if $K^{n}$ in $S^{n+2}$ is trivial then $\tau_{k}(K^{n})$ in $S^{n+3}$ is also trivial for any $k\geq 0$ [19, Corollary 3]. Therefore, if $K$ is trivial then $\tau_{m_{2}}(\tau_{m_{1}}(K))$ is also.

Assume that $m=1$. The fiber of $\tau_{m_{2}}(\tau_{m_{1}}(K))$ is the $m_{2}$-fold cyclic branched cover $M_{m_{2}}(\tau_{m_{1}}(K))$ of $S^{4}$ along $\tau_{m_{1}}(K)$ with removing one open $4$-ball. We will prove that this is diffeomorphic to the unit $4$-ball.

As mentioned in Theorem 2.2, there exists a locally trivial fibration

whose fiber $F$ is the $m_{1}$-fold cyclic branched cover $M_{m_{1}}(K)$ of $S^{3}$ along $K$ with removing one open $3$-ball and its monodromy is the $1$-shift of the sheets of the cyclic branched cover. The manifold $M_{m_{2}}(\tau_{m_{1}}(K))$ is obtained from the exterior $S^{4}\setminus\text{\rm Int\,Nbd}(\tau_{m_{1}}(K);S^{4})$ by cutting it by a fiber $F$, making $m_{2}$ copies, gluing them in order, and filling the boundary by $S^{2}\times D^{2}$ canonically. The manifold before the final filling is the manifold whose fiber is $F$ and monodromy is the $m_{2}$-shift of the sheets of the cyclic branched cover. This is the exterior of a branched twist spin. Remark that the $(m_{1},m_{2})$-branched twist spin is defined for positive integers $m_{1}$ and $m_{2}$ satisfying $\gcd(m_{1},m_{2})=1$ or $(m_{1},m_{2})=(0,1)$. Hence the assumption $m=\gcd(m_{1},m_{2})=1$ is needed. By filling the boundary of the manifold by $S^{2}\times D^{2}$, we obtain the manifold $M_{m_{2}}(\tau_{m_{1}}(K))$. There are two way of filling $S^{2}\times D^{2}$. The difference is the so-called Gluck twist ([9], cf. [6]). In either case, this is nothing but the construction of a branched twist spin. In particular, by [15, Theorem 4], $M_{m_{2}}(\tau_{m_{1}}(K))$ is diffeomorphic to $S^{4}$. Since the fiber of $\tau_{m_{2}}(\tau_{m_{1}}(K))$ is obtained from $M_{m_{2}}(\tau_{m_{1}}(K))$ by removing one open ball, it is diffeomorphic to the unit $4$-ball. Thus the $3$-knot $\tau_{m_{2}}(\tau_{m_{1}}(K))$, which is the boundary of the $4$-ball, is a trivial $3$-knot in $S^{5}$. ∎

By Lemma 2.1,

Then, the presentation in the assertion is obtained by applying Lemma 2.1 to this presentation again.

The decomposition of $S^{4}$ used to define the $k$-twist-spinning is as in Figure 3. We now describe the decomposition of $S^{5}$ to read off the meridian $\mu_{2}$ of $\tau_{m_{2}}(\tau_{m_{1}}(K))$. The first factor $(Y^{4},Y^{2})$ of the second piece of the decomposition of $S^{5}$ is obtained from $S^{4}$ by removing the interior of a $4$-dimensional ball $B^{4}$ intersecting $\tau_{m_{1}}(K)$ trivially. We set this $B^{4}$ to be the product of the upper hemisphere $H$ of $\partial X^{3}$ and the second factor $D^{2}$ of $(\partial X^{3},\partial X^{1})\times D^{2}$. Set $B^{2}$ to be the intersection of $B^{4}$ with $\tau_{m_{1}}(K)$. Note that $B^{2}=(H\cap\partial X^{1})\times D^{2}$, where $H\cap\partial X^{1}$ is the north pole of the sphere on the left in Figure 3. The first piece $(\partial X^{4},\partial X^{2})\times(D^{2})^{\prime}$ of the decomposition of $S^{5}$ is $(\partial B^{4},\partial B^{2})\times(D^{2})^{\prime}$, where we used the notation $(D^{2})^{\prime}$ instead of $D^{2}$ since it is different from $D^{2}$ in Figure 3. The meridian $\mu_{2}$ of $\tau_{m_{2}}(\tau_{m_{1}}(K))$ is a loop on $(\partial B^{4}\setminus\partial B^{2})\times\partial(D^{2})^{\prime}$ that goes around $\partial B^{2}\times\partial(D^{2})^{\prime}$ once and does not go around $\partial(D^{2})^{\prime}$. Hence it is a loop on $H\setminus\partial X^{1}$ that goes around $H\cap\partial X^{1}$ once. This is nothing but the loop $\mu_{1}$ in Figure 3. As explained in the proof of Lemma 2.1, $\mu_{1}$ is a meridian of $K$. Hence we have $\mu_{2}=\mu_{1}=x_{1}$. ∎

Take a Wirtinger presentation $\langle x_{1},\ldots,x_{u}\mid\;r_{1},\ldots,r_{v}\rangle$ of $\pi_{1}(S^{3}\setminus K)$. Set $\mu_{1}=\mu_{2}=x_{1}$ and eliminate $h_{1}$ and $h_{2}$ from the presentation (3.1) by using the relations $h_{1}=\mu_{1}^{-m_{1}}=x_{1}^{-m_{1}}$ and $h_{2}=\mu_{2}^{-m_{2}}=x_{1}^{-m_{2}}$ as

This group is the quotient group of $\pi_{1}(S^{3}\setminus K)$ by its normal subgroup $N$ generated by $x_{k}x_{1}^{-m_{1}}x_{k}^{-1}x_{1}^{m_{1}}$ and $x_{k}x_{1}^{-m_{2}}x_{k}^{-1}x_{1}^{m_{2}}$ for $k=2,\ldots,u$. Let $G$ denote $\pi_{1}(S^{5}\setminus\tau_{m_{2}}(\tau_{m_{1}}(K)))$ for simplicity. Then the above isomorphism is denoted as $G\cong\pi_{1}(S^{3}\setminus K)/N$. Since $N$ is contained in the normal closure $\langle\langle x_{1}^{m}\rangle\rangle$ of the group $\langle x_{1}^{m}\rangle$ generated by $x_{1}^{m}$, we have

Here the last isomorphism follows from the fact that $\langle x_{1}^{m}\rangle$ is central and hence normal in $G$.

Let $p:G\to G/\langle x_{1}^{m}\rangle$ be the projection and let $Z(G)$ denote the center of $G$. Since $p$ is a group epimorphism, $p(Z(G))$ is contained in the center of $G/\langle x_{1}^{m}\rangle$. Then, the assumption of the triviality of the center of $\pi_{1}^{orb}(\mathcal{O}(K,m))$ implies that $p(Z(G))$ is trivial. Thus we have $Z(G)=\langle x_{1}^{m}\rangle$. ∎

Suppose that $K$ is non-trivial, $m=\gcd(m_{1},m_{2})\geq 2$, and the center of $\pi_{1}^{orb}(\mathcal{O}(K,m))$ is trivial. By Lemma 3.2, the quotient group of $\pi_{1}(S^{5}\setminus\tau_{m_{2}}(\tau_{m_{1}}(K)))$ by its center is isomorphic to $\pi_{1}^{orb}(\mathcal{O}(K,m))$. Since the abelianization of $\pi_{1}^{orb}(\mathcal{O}(K,m))$ is isomorphic to $\mathbb{Z}/m\mathbb{Z}$, if $m\geq 2$ then $\pi_{1}^{orb}(\mathcal{O}(K,m))$ is non-trivial. On the other hand, for a trivial $3$-knot $O$, $\pi_{1}(S^{5}\setminus O)$ is isomorphic to $\mathbb{Z}$ and hence the quotient group of $\pi_{1}(S^{5}\setminus O)$ by its center is trivial. Therefore, $\tau_{m_{2}}(\tau_{m_{1}}(K))$ is non-trivial. ∎

Let $A:\pi_{1}^{orb}(\mathcal{O}(K,m))\to\mathbb{Z}/m\mathbb{Z}$ be the abelianization map. Since $x^{p}$, which is equal to $y^{q}$, corresponds to an element in the center of $\pi_{1}(S^{3}\setminus K)$, $x^{p}$ is an element in the center of $\pi_{1}^{orb}(\mathcal{O}(K,m))$. It is suffice to show that $x^{p}$ is non-trivial in $\pi_{1}^{orb}(\mathcal{O}(K,m))$. For $n\in\mathbb{Z}$, let $\bar{n}\in\mathbb{Z}/m\mathbb{Z}$ denote the image of $n$ by $A$. Set $A(x)=\bar{a}$ and $A(y)=\bar{b}$, where $a,b\in\mathbb{Z}$. Since $x^{s}y^{r}$ is a meridian of $K$, $A(x^{s}y^{r})\equiv\bar{a}s+\bar{b}r\equiv 1\mod m$. Since $pr+qs=1$, the solution of $as+br=1$ is given in the form $a=q-tr$ and $b=p+ts$ for $t\in\mathbb{Z}$. Since $x^{p}=y^{q}$, we have $\overline{p(q-tr)}=A(x^{p})=A(y^{q})=\overline{q(p+ts)}$. This and $pr+sq=1$ imply that $t=0\mod m$. Hence we have $A(x)=\bar{q}$ and $A(y)=\bar{p}$. Since $\bar{p}\bar{q}\neq\bar{0}$ by the assumption, we have $A(x^{p})\neq\bar{0}$. Hence $x^{p}$ is non-trivial in $\pi_{1}^{orb}(\mathcal{O}(K,m))$. ∎