Irreducible $SL(2,\mathbb{C})$-metabelian representations of branched twist spins

TL;DR

This paper extends Lin's result on irreducible $SL(2,b{C})$-metabelian representations from knots to branched twist spins, showing their count depends on the original knot's determinant.

Contribution

It proves that the number of such representations for branched twist spins is determined by the determinant of the original knot, generalizing Lin's result.

Findings

Number of irreducible $SL(2,b{C})$-metabelian representations depends on the knot determinant.

Comparison of group presentations links the representations of branched twist spins to the original knot.

The result applies to $(m,n)$-branched twist spins, broadening understanding of their algebraic properties.

Abstract

An $(m,n)$-branched twist spin is a fibered $2$-knot in $S^4$ which is determined by a $1$-knot $K$ and coprime integers $m$ and $n$. For a $1$-knot, Lin proved that the number of irreducible $SL(2,\mathbb{C})$-metabelian representations of the knot group of a $1$-knot up to conjugation is determined by the knot determinant of the $1$-knot. In this paper, we prove that the number of irreducible $SL(2,\mathbb{C})$-metabelian representations of the knot group of an $(m,n)$-branched twist spin up to conjugation is determined by the determinant of a $1$-knot in the orbit space by comparing a presentation of the knot group of the branched twist spin with the Lin's presentation of the knot group of the $1$-knot.

Equations

$$\left\{\begin{split}&\frac{|\Delta_{K}(-1)|-1}{2}&\hskip 20.0pt(m:{\rm even})\\ &0&(m:{\rm odd}),\end{split}\right.$$

$$S^3=(S^3\setminus \mathrm{int}N(K))\cup (E_m^{c\ast }\times D^2)\cup (E_n^{c\ast }\times D^2)\cup (D_1^{3\ast }\bigsqcup D_2^{3\ast }),$$

$$S^4=(\left(S^3\setminus \mathrm{int}N(K)\right)\times S^1)\cup (E_m\times D^2)\cup (E_n\times D^2)\times (D_1^4\bigsqcup D_2^4).$$

$${\pi }_1(S^4\setminus \text{int}N(K^{m,n}))\cong ⟨y_1,\dots ,y_s,h|r_1,\dots ,r_t,y_{i}h{y}_i^{-1}{h}^{-1},y_1^{|m|}{h}^{\beta }⟩,$$

$${\pi }_1(S^4\setminus \mathrm{int}N(K^{m,n}))\cong {\pi }_1(\mathrm{punc}(M_K))\ast ⟨\eta ⟩/⟨\eta ({\tau }^{n}z){\eta }^{-1}=z\mathrm{for}\mathrm{all}z\in {\pi }_1(M_K)⟩,$$

$${\pi }_1(S^3\setminus \mathrm{int}N(K))\cong ⟨x_1,\dots ,x_{2g},\mu |\mu {\alpha }_i{\mu }^{-1}={\beta }_i⟩,$$

$${\rho }_0(x_i)=\left(\begin{array}{cc}{\lambda }_i& 0\\ 0& {{\lambda }_i}^{-1}\end{array}\right)\mathrm{or}\pm \left(\begin{array}{cc}1& {\omega }_i\\ 0& 1\end{array}\right)(i=1,\dots ,2g),$$

$$\rho (x_i)=\left(\begin{array}{cc}{\lambda }_i& 0\\ 0& {{\lambda }_i}^{-1}\end{array}\right),\rho (\mu )=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)(i=1,\dots ,2g),$$

$$\frac{|{\Delta }_K(-1)|-1}{2}.$$

$${\pi }_1(M_K)\cong ⟨{\tau }^0{\tilde{x}}_1,\dots ,{\tau }^0{\tilde{x}}_{2g},\dots ,{\tau }^{m-1}{\tilde{x}}_1,\dots ,{\tau }^{m-1}{\tilde{x}}_{2g}|{\tilde{\alpha }}_i^{(j)}={\tilde{\beta }}_i^{(j-1)}⟩,$$

$$\begin{array}{rl}⟨{\tau }^0{\tilde{x}}_1,\dots ,{\tau }^0{\tilde{x}}_{2g},\dots ,{\tau }^{m-1}{\tilde{x}}_1,\dots ,& {\tau }^{m-1}{\tilde{x}}_{2g},\eta |\\ & {\tilde{\alpha }}_i^{(j)}={\tilde{\beta }}_i^{(j-1)},\eta {\tau }^{j+n}{\tilde{x}}_i{\eta }^{-1}={\tau }^j{\tilde{x}}_i⟩.\end{array}$$

$$\rho ({\tau }^j{\tilde{x}}_i)=\left(\begin{array}{cc}{\lambda }_i^{(j)}& 0\\ 0& {{\lambda }_i^{(j)}}^{-1}\end{array}\right)\rho (\eta )=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right),$$

$$\rho ({\tau }^j{\tilde{x}}_i)=\left(\begin{array}{cc}{\lambda }_i^{(j)}& 0\\ 0& {{\lambda }_i^{(j)}}^{-1}\end{array}\right)({\lambda }_i^{(j)}\in \mathbb{C}\setminus 0),$$

$$\begin{array}{rl}& \rho (\eta {\tau }^{j+n}{\tilde{x}}_i)=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\left(\begin{array}{cc}{\lambda }_i^{(j+n)}& 0\\ 0& {{\lambda }_i^{(j+n)}}^{-1}\end{array}\right)=\left(\begin{array}{cc}a{\lambda }_i^{(j+n)}& b{{\lambda }_i^{(j+n)}}^{-1}\\ c{\lambda }_i^{(j+n)}& d{{\lambda }_i^{(j+n)}}^{-1}\end{array}\right),\\ & \rho ({\tau }^j{\tilde{x}}_i\eta )=\left(\begin{array}{cc}{\lambda }_i^{(j)}& 0\\ 0& {{\lambda }_i^{(j)}}^{-1}\end{array}\right)\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)=\left(\begin{array}{cc}a{\lambda }_i^{(j)}& b{\lambda }_i^{(j)}\\ c{{\lambda }_i^{(j)}}^{-1}& d{{\lambda }_i^{(j)}}^{-1}\end{array}\right).\end{array}$$

$${\tau }^j{\tilde{x}}_i=\eta {\tau }^{j+n}{\tilde{x}}_i{\eta }^{-1}={\eta }^q{\tau }^{j+nq}{\tilde{x}}_i{\eta }^{-q}={\eta }^q{\tau }^{j+1}{\tilde{x}}_i{\eta }^{-q},$$

$$\left(\begin{array}{cc}{\lambda }_i^{(j+1)}& 0\\ 0& {{\lambda }_i^{(j+1)}}^{-1}\end{array}\right)=\{\begin{array}{r}\left(\begin{array}{cc}{\lambda }_i^{(j)}& 0\\ 0& {{\lambda }_i^{(j)}}^{-1}\end{array}\right)(q:\mathrm{even})\\ \left(\begin{array}{cc}{{\lambda }_i^{(j)}}^{-1}& 0\\ 0& {\lambda }_i^{(j)}\end{array}\right)(q:\mathrm{odd}).\end{array}$$

$$\overline{\rho }(x)=\rho (\eta x{\eta }^{-1})=\rho (\eta )\rho (x)\rho ({\eta }^{-1})$$

$$\rho ({\tilde{\alpha }}_i^{(j+1)})=\overline{\rho }({\tilde{\alpha }}_i^{(j)}),\rho ({\tilde{\beta }}_i^{(j+1)})=\overline{\rho }({\tilde{\beta }}_i^{(j)}).$$

$${\tilde{\beta }}_i^{(j)}={\tilde{\alpha }}_i^{(j+1)}={\eta }^{-q}{\tilde{\alpha }}_i^{(j)}{\eta }^q$$

$$\rho ({\tilde{\beta }}_i^{(j)})=\overline{\rho }({\tilde{\alpha }}_i^{(j)}).$$

$$⟨{\tau }^0{\tilde{x}}_1,\dots ,{\tau }^0{\tilde{x}}_{2g},\dots ,{\tau }^{m-1}{\tilde{x}}_1,\dots ,{\tau }^{m-1}{\tilde{x}}_{2g},\eta |\eta {\tilde{\alpha }}_i^{(0)}{\eta }^{-1}={\tilde{\beta }}_i^{(0)},\eta {\tau }^{j+1}{\tilde{x}}_i{\eta }^{-1}={\tau }^j{\tilde{x}}_i⟩.$$

$$⟨{\tau }^0{\tilde{x}}_1,\dots ,{\tau }^0{\tilde{x}}_{2g},\eta |\eta {\tilde{\alpha }}_i^{(0)}{\eta }^{-1}={\tilde{\beta }}_i^{(0)}⟩.$$

$$\rho ({\tau }^0{\tilde{x}}_i)=\left(\begin{array}{cc}{\lambda }_i^{(0)}& 0\\ 0& {{\lambda }_i^{(0)}}^{-1}\end{array}\right),\rho (\eta )=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right),$$

$$\overline{\rho }({\tau }^0{\tilde{x}}_i)=\left(\begin{array}{cc}{{\lambda }_i^{(0)}}^{-1}& 0\\ 0& {\lambda }_i^{(0)}\end{array}\right),\overline{\rho }(\eta )=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$$

$${\mathrm{e}}^{\sqrt{-1}}{r_1}^{{\omega }_{i1}{\theta }_i}\cdots {r_{2g}}^{{\omega }_{i2g}{\theta }_{2g}}=1,$$

$${r_1}^{{\omega }_{i1}}\cdots {r_{2g}}^{{\omega }_{i2g}}=1,$$

$${\tau }^j{\tilde{x}}_i={\eta }^q{\tau }^{j+1}{\tilde{x}}_i{\eta }^{-q}={\eta }^{mq}{\tau }^{j+m}{\tilde{x}}_i{\eta }^{-mq}={\eta }^{mq}{\tau }^j{\tilde{x}}_i{\eta }^{-mq}.$$

Full text

1. Introduction

A $2$-knot is a smoothly embedded $2$-sphere in $S^{4}$. A $2$-knot is said to be fibered if its complement admits a fibration structure over the circle with some natural structure in a tubular neighborhood of the $2$-knot. Although it is very difficult to see how the $2$-knot is embedded in $S^{4}$, the idea of admitting a fibration helps us to construct many examples of $2$-knots, such as spun knots, twist spun knots, rolling, deformed spun knots, branched twist spins, fibered homotopy-ribbon knots, etc [1, 2, 5, 7, 9, 16, 17]. A branched twist spin is a 2-knot which admits an $S^{1}$-action in its exterior. The terminology “branched twist spin” appears in the book of Hillman [5]. It is known by Pao and Plotnick that a fibered $2$-knot is a branched twist spin if and only if its monodromy is periodic [14]. Therefore, this class has special importance among other known classes of fibered $2$-knots. Note that spun knots and twist spun knots are included in the class of branched twist spins.

We give here a short introduction of branched twist spins based on the classification of locally smooth $S^{1}$-actions on the $4$-sphere. Montgomery and Yang showed that effective locally smooth $S^{1}$-actions are classified into four types [10] and Fintushel and Pao showed that there is a bijection between orbit data and weak equivalence classes of $S^{1}$-actions on $S^{4}$ [3, 13]. Suppose that $S^{1}$ acts locally smoothly and effectively on $S^{4}$ and the orbit space is $S^{3}$. Then there are at most two types of exceptional orbits called $\mathbb{Z}_{m}$-type and $\mathbb{Z}_{n}$-type, where $m,n$ are coprime positive integers. Let $E_{m}$ (resp. $E_{n}$) be the set of exceptional orbits of $\mathbb{Z}_{m}$-type (resp. $\mathbb{Z}_{n}$-type) and $F$ be the fixed point set. The image of the orbit map of $E_{n}$, denoted by $E^{\ast}_{n}$, is an open arc in the orbit space $S^{3}$, and that of $F$, denoted by $F^{\ast}$, is the two points in $S^{3}$ which are the end points of $E^{\ast}_{n}$. It is known that $E^{\ast}_{m}\cup E^{\ast}_{n}\cup F^{\ast}$ constitutes a $1$-knot $K$ in $S^{3}$ and $E_{n}\cup F$ is diffeomorphic to the 2-sphere. The $(m,n)$-branched twist spin of $K$ is defined as $E_{n}\cup F$. Note that the $(m,1)$-branched twist spin is the $m$-twist spun knot and the $(0,1)$-branched twist spin is the spun knot. If $K$ is a torus knot or a hyperbolic knot then its $(m,n)$-branched twist spins with $m>n$ and $m\geq 3$ are non-trivial. This follows from the fact that $K^{m,n}$ is not reflexive known by Hillman and Plotnick [6].

An oriented $k$-knot $K$ is said to be equivalent to another oriented $k$-knot $K^{\prime}$, denoted by $K\sim K^{\prime}$, if there exists a smooth isotopy $H_{t}:S^{k+2}\to S^{k+2}$ such that $H_{0}={\rm id}$ and $H_{1}(K)=K^{\prime}$ as oriented $k$-knots. In [4], the author studied the elementary ideal of the fundamental group of the complement of a branched twist spin and gave a criterion to detect if two branched twist spins $K_{1}^{m_{1},n_{1}}$ and $K_{2}^{m_{2},n_{2}}$ are inequivalent by the knot determinants $\Delta_{K_{1}}(-1)$ and $\Delta_{K_{2}}(-1)$, where $\Delta_{K}(t)$ is the Alexander polynomial of a $1$-knot $K$ in $S^{3}$. Note that the definition of an $(m,n)$-branched twist spin is generalized to $(m,n)\in\mathbb{Z}\times\mathbb{N}$ by taking orientations into account, see Section $2.1$.

The knot determinant is related to the number of irreducible $SL(2,\mathbb{C})$-metabelian representations of the fundamental group of the knot complement [8, 12]. The aim of this paper is to count the number of such representations for a branched twist spin. Similar to the results in [8, 12], the number of such representations is given by the knot determinant as follows:

Theorem 1.1**.**

The number of irreducible $SL(2,\mathbb{C})$-metabelian representations of $\pi_{1}(S^{4}\setminus{\rm int}N(K^{m,n}))$ is

[TABLE]

where $N(K^{m,n})$ is a compact tubular neighborhood of $K^{m,n}$ in $S^{4}$.

As an immediate corollary, we obtain the same criterion as in [4].

Corollary 1.2** (F. [4]).**

Branched twist spins $K^{m_{1},n_{1}}$ and $K^{m_{2},n_{2}}$ are inequivalent if one of the following holds:

• (1)

$m_{1}$ and $m_{2}$ are even and $|\Delta_{K_{1}}(-1)|\neq|\Delta_{K_{2}}(-1)|$,

• (2)

$m_{1}$ is even, $m_{2}$ is odd and $|\Delta_{K_{1}}(-1)|\neq 1$.

This paper is organized as follows: In Section 1, we define an $(m,n)$-branched twist spin $K^{m,n}$ as an oriented $2$-knot and introduce Plotnick’s presentation of $\pi_{1}(S^{4}\setminus{\rm int}N(K^{m,n}))$. In Section 2, we state the Lin’s presentation of a $1$-knot and the Nagasato-Yamaguchi’s presentation of the $m$-fold cyclic branched cover of $S^{3}$ along $K$. In Section 3, we observe irreducible $SL(2,\mathbb{C})$-metabelian representations of $\pi_{1}(S^{4}\setminus{\rm int}N(K^{m,n}))$ and prove Theorem 1.1.

Acknowledgments**.**

The author is grateful to his supervisor, Masaharu Ishikawa, for many helpful suggestions.

2. Two presentations of branched twist spins

2.1. The $(m,n)$-branched twist spin

Suppose that $S^{4}$ has an effective locally smooth $S^{1}$-action. Let $E_{m}$ be the set of exceptional orbits of $\mathbb{Z}_{m}$-type, where $m$ is a positive integer, and $F$ be the fixed point set. Set $E^{\ast}_{m}$ and $F^{\ast}$ to be the image of $E_{m}$ and $F$ by the orbit map, respectively. Montgomery and Yang showed that effective locally smooth $S^{1}$-actions are classified into the following four types: (1) $\{D^{3}\}$,(2) $\{S^{3}\}$, (3) $\{S^{3},m\}$, (4) $\{(S^{3},K),m,n\}$, which are called orbit data [10]. The $3$-ball and the $3$-sphere in these notations represent the orbit spaces. In case (4), the union $E^{\ast}_{m}\cup E^{\ast}_{n}\cup F^{\ast}$ constitutes a $1$-knot $K$ in the orbit space $S^{3}$ and the union $E_{n}\cup F$ is diffeomorphic to the $2$-sphere. This $2$-sphere is embedded in $S^{4}$, and is called the $(m,n)$-branched twist spin of $K$, denoted by $K^{m,n}$. In case (3), for an arc $A^{\ast}$ in $S^{3}$ whose end points are $F^{\ast}$, the preimage of $A^{\ast}$ is denoted by $A$. Then the union $A\cup F$ is diffeomorphic to the $2$-sphere, and is called a twist spun knot. We may regard an $m$-twist spun knot as $K^{m,1}$, where $K$ is $A^{\ast}\cup E^{\ast}_{m}\cup F^{\ast}$.

We recall the definition of $(m,n)$-branched twist spins for $(m,n)\in\mathbb{Z}\times\mathbb{N}$ in [4]. First, we remark that the definition in  [4] depends on the choice of the orientation of $K$. Actually, in the definition we fixed a preferred meridian-longitude $(\theta,\phi)$ of $S^{3}\setminus{\rm int}N(K)$, where $N(K)$ is a compact tubular neighborhood of $K$, and replacing $(\theta,\phi)$ by $(-\theta,-\phi)$ may change the equivalence class of $K^{m,n}$.

We give the definition of $K^{m,n}$. Let $K$ be a $1$-knot in $S^{3}$ and $(m,n)$ be a pair of integers in $(\mathbb{Z}\setminus\{0\})\times\mathbb{N}$ such that $|m|$ and $n$ are coprime. We decompose the orbit space $S^{3}$ into five pieces as follows:

[TABLE]

where $D^{3\ast}_{1}\sqcup D^{3\ast}_{2}$ is a compact neighborhood of $F^{\ast}$ and $E^{c\ast}_{m}$ and $E^{c\ast}_{n}$ are the connected components of $K\setminus\text{int}(D^{3\ast}_{1}\cup D^{3\ast}_{2})$ such that $E_{m}^{c\ast}\subset E_{m}^{\ast}$ and $E_{n}^{c\ast}\subset E_{n}^{\ast}$, see Figure 1. Considering the preimage of the orbit map, we decompose $S^{4}$ as follows:

[TABLE]

Let $p$ denote the orbit map. Choosing a point $z^{\ast}_{m}$ in $E^{c\ast}_{m}$, let $D^{2\ast}_{z^{\ast}_{m}}$ be a 2-disk in $S^{3}$ centered at $z^{\ast}_{m}\in E^{c\ast}_{m}$ and transversal to $E^{c\ast}_{m}$. The preimage $p^{-1}(D^{2\ast}_{z^{\ast}_{m}})$ is a solid torus $V_{m}$ whose core is the exceptional orbit of $\mathbb{Z}_{m}$-type.

Now we discuss the orientations of $V_{m}$ and $E^{c\ast}_{m}$. Let $K$ be an oriented $1$-knot in $S^{3}$. First, fix the orientation of $S^{4}$ and those of orbits such that they coincide with the direction of the $S^{1}$-action. These orientations determine the orientation of $V_{m}\times E_{m}^{c*}$. Let $(\theta,\phi)$ be the preferred meridian-longitude pair of $K$ such that the orientation of the longitude $\phi$ coincides the orientation of $K$. From the decomposition (2.1), we can see that $\phi$ is regarded as a coordinate of the second factor of $V_{m}\times E^{c\ast}_{m}$. We assign the orientation of $V_{m}$ so that the orientation of $V_{m}\times E^{c\ast}_{m}$ coincides with the given orientation of $S^{4}$. Finally, we choose the meridian and longitude pair $(\Theta,H)$ of $V_{m}\cong D^{2}\times S^{1}$ such that $H$ becomes the meridian of $V_{n}$ in the decomposition $V_{m}\cup V_{n}=p^{-1}(\partial D_{i}^{3*})$ and the orbits of the $S^{1}$-action are in the direction $\varepsilon n\Theta+|m|H$ with $n>0$, where $\varepsilon=1$ if $m\geq 0$ and $\varepsilon=-1$ if $m<0$.

Definition 2.1** (Branched twist spin).**

Let $K$ be an oriented knot in $S^{3}$. For each pair $(m,n)\in\mathbb{Z}\times\mathbb{N}$ with $m\neq 0$ such that $|m|$ and $n$ are coprime, let $K^{m,n}$ denote the $2$-knot $E_{n}\cup F$. If $(m,n)=(0,1)$ then define $K^{0,1}$ to be the spun knot of $K$. The $2$-knot $K^{m,n}$ is called an $(m,n)$-$branched\ twist\ spin$ of $K$.

Note that the branched twist spin $K^{m,1}$ constructed from $\{(S^{3},K),m,1\}$ is an $m$-twist spun knot of $K$.

Remark 2.2**.**

Let $-K$ be an oriented knot obtained from $K$ by reversing the orientation of $K$. From the construction of $K^{m,n}$, we see that $K^{m,n}$ is equivalent to $-(-K)^{-m,n}$.

Let $K$ be a $k$-knot in $S^{k+2}$. The fundamental group of the knot complement $S^{k+2}\setminus\text{int}N(K)$ is called the knot group of $K$, where $N(K)$ is a compact tubular neighborhood of $K$.

Lemma 2.3** ([4]).**

Let $K$ be an oriented $1$-knot and $K^{m,n}$ be the $(m,n)$-branched twist spin of $K$ with $(m,n)\in\mathbb{Z}\times\mathbb{N}$, where $|m|$ and $n$ are coprime. Let $\langle y_{1},\ldots,y_{s}\mid r_{1},\ldots,r_{t}\rangle$ be a presentation of the knot group of $K$ such that $y_{1}$ is a meridian. Then the knot group of $K^{m,n}$ has the presentation

[TABLE]

where $\beta$ is an integer such that $n\beta\equiv\varepsilon$ (mod $m$) if $m$ is non-zero and $\beta=1$ if $m=0$. Recall that $\varepsilon=1$ if $m\geq 0$ and $\varepsilon=-1$ if $m<0$.

Note that $\pi_{1}(S^{4}\setminus{\rm int}N(K^{m,n}))$ is isomorphic to $\pi_{1}(S^{4}\setminus\text{int}N((-K)^{-m,n}))$ by Remark 2.2.

2.2. Plotnick’s presentation

Assume that $m\neq 0$. We ignore the orientation of $K^{m,n}$ since we are interested in the fundamental group of its complement. By Remark 2.2, changing the orientation of $K$ and the sign of $m$ if necessary, we can assume that $m$ is positive. Pao constructed the knot complement of $K^{m,n}$ as follows [13]: Let $M_{K}$ be the $m$-fold cyclic branched cover of $S^{3}$ along $K$ and $\tau:M_{K}\to M_{K}$ be the diffeomorphism associated with the canonical deck transformation of $M_{K}$. Let $M_{K}\times_{\tau^{n}}S^{1}$ be the manifold obtained from $M_{K}\times I$ by identifying $M_{K}\times\{0\}$ with $M_{K}\times\{1\}$ by $(z,1)\mapsto(\tau^{n}z,0)$, where $\tau^{n}$ means the $n$-th power of composite of $\tau$. Note that $M_{K}\times_{\tau^{n}}S^{1}$ has the natural $S^{1}$-action $\varphi_{s}\langle y,t\rangle=\langle y,t+s\rangle$, where $\langle y,t\rangle$ denotes the image of $(y,t)\in M_{K}\times I$ by the identification. Let $x$ be a branch point of $M_{K}$. Then the orbit of $\langle x,0\rangle$ is a circle in $M_{K}\times_{\tau^{n}}S^{1}$. There is a neighborhood of the orbit which is invariant by the $S^{1}$-action, denoted by $T$. It is known in [13] that the knot complement of $K^{m,n}$ is diffeomorphic to ($M_{K}\times_{\tau^{n}}S^{1})\setminus{\rm int}T$, which is also diffeomorphic to ${\rm punc}(M_{K})\times_{\tau^{n}}S^{1}$, where ${\rm punc}(M_{K})=M_{K}\setminus{\rm int}D^{3}$ with $D^{3}$ being a $3$-ball in $M_{K}$. Note that $K^{m,n}$ is regarded as the branch set of the $n$-fold cyclic branched cover of $S^{4}$ along the $m$-twist spun knot of $K$.

The following lemma is shown by Plotnick in [15].

Lemma 2.4** (Plotnick [15]).**

Let $K^{m,n}$ be a branched twist spin of $K$. Then the following holds:

[TABLE]

where $\eta$ is a meridian of $K^{m,n}$.

2.3. Lin’s presentation

Let $K$ be a $1$-knot in $S^{3}$. A Seifert surface $S$ of $K$ is called free if $S^{3}=N(S)\cup(S^{3}\setminus{\rm int}N(S))$ gives a Heegaard splitting of $S^{3}$. It is known that any $1$-knot has a free Seifert surface. A presentation of $\pi_{1}(S^{3}\setminus{\rm int}N(K))$ is obtained from the Heegaard splitting associated to a free Seifert surface as follows: Let $S$ be a free Seifert surface of $K$ of genus $g$ and $W$ be a spine of $S$. Then $H_{1}=S\times[-1,1]$ and $H_{2}=S^{3}\setminus{\rm int}H_{1}$ is a Heegaard splitting of $S^{3}$. Let $K^{\prime}$ be a simple closed curve obtained from $K$ by pushing it into $H_{1}$ slightly. Choose a base point $\ast$ in $W\subset S\times\{0\}$ such that $\ast$ does not on $K$ and $K^{\prime}$. Since $H_{1}$ and $H_{2}$ are handlebodies with genus $2g$, we may choose generators $a_{1},\ldots,a_{2g}$ of $\pi_{1}(H_{1})$ and generators $x_{1},\ldots,x_{2g}$ of $\pi_{1}(H_{2})$. Let $a^{+}_{1},\ldots,a^{+}_{2g}$ and $a^{-}_{1},\ldots,a^{-}_{2g}$ denote the loops $a_{1}\times\{1\},\ldots,a_{2g}\times\{1\}$ and $a_{1}\times\{-1\},\ldots,a_{2g}\times\{-1\}$. Each $a^{+}_{i}$ (resp. $a^{-}_{i}$) is written in a word of $x_{1},\ldots,x_{2g}$ by the homeomorphism from $\partial H_{2}$ to $\partial H_{1}$. The words of $a^{+}_{i}$ (resp. $a^{-}_{i}$) are denoted by $\alpha_{i}$ (resp. $\beta_{i}$) for $i=1,\ldots,2g$. There is a unique arc $c$, up to isotopy, such that $(\ast\times[-1,1])\cup c$ is a meridian of $K^{\prime}$. The homotopy class of this loop is denoted by $\mu$. From van Kampen theorem, the following theorem holds:

Lemma 2.5** (Lin [8]).**

Let $K$ be a $1$-knot in $S^{3}$ and $S$ be a free Seifert surface of $K$. Let $S^{3}=H_{1}\cup H_{2}$ be the Heegaard splitting associated to $S$. For generators $x_{1},\ldots,x_{2g}$ of $\pi_{1}(H_{2})$, $\pi_{1}(E_{K})$ has the following presentation:

[TABLE]

where $g$ is the genus of $S$, and $\alpha_{i}$, $\beta_{i}$ are the words in $x_{1},\ldots,x_{2g}$ determined above.

Let $\langle x_{1},\ldots,x_{2g},\mu\ |\ \mu\alpha_{i}\mu^{-1}=\beta_{i}\rangle$ be a Lin’s presentation of $\pi_{1}(S^{3}\setminus{\rm int}N(K))$. Denote the sum of indices of $x_{j}$ in $\alpha_{i}$ by $v_{ij}$ and that in $\beta_{i}$ by $u_{ij}$. Then the $2g\times 2g$ matrix $V=(v_{ij})$ is defined. The matrix V is called a Seifert matrix and $\det(V+{{}^{t}}V)$ is called the knot determinant of $K$, which equals to $\Delta_{K}(-1)$. Note that all generators $x_{1},\ldots,x_{2g}$ are commutators of $\pi_{1}(S^{3}\setminus{\rm int}N(K))$. Let $\rho_{0}:\pi_{1}(S^{3}\setminus{\rm int}N(K))\to SL(2,\mathbb{C})$ be an $SL(2,\mathbb{C})$-metabelian representation. Since all $x_{1},\ldots,x_{2g}$ are commutators of $\pi_{1}(S^{3}\setminus{\rm int}N(K))$, we can assume that

[TABLE]

up to conjugation. If there exists $i\in\{1,\ldots,2g\}$ such that $\rho_{0}(x_{i})$ is $\pm\left(\begin{matrix}1&\omega_{i}\\ 0&1\end{matrix}\right)$, then all $\rho(x_{j})$ are of the forms $\pm\left(\begin{matrix}1&\omega_{j}\\ 0&1\end{matrix}\right).$ This is implied by all $x_{j}$ are conjugate to each other and such a representation is abelian, especially reducible. Therefore we can assume that an irreducible $SL(2,\mathbb{C})$-metabelian representation satisfies

[TABLE]

up to conjugation. Since $\alpha_{i}$ and $\beta_{i}$ are written in words $x_{1},\ldots,x_{2g}$, each $\rho_{0}(\alpha_{i})$ and $\rho_{0}(\beta_{i})$ is a diagonal matrix. From (2.5), Lin checked directly the number of irreducible $SU(2,\mathbb{C})$-metabelian representations of $\pi_{1}(S^{3}\setminus{\rm int}N(K))$.

Theorem 2.6** (Lin [8]).**

The number of conjugacy classes of irreducible $SU(2,\mathbb{C})$-metabelian representations of $\pi_{1}(S^{3}\setminus{\rm int}N(K))$ is

[TABLE]

Remark 2.7**.**

In  [11], Nagasato showed that the same statement holds for irreducible $SL(2,\mathbb{C})$-metabelian representations of $\pi_{1}(S^{3}\setminus{\rm int}N(K))$.

2.4. Nagasato-Yamaguchi’s presentation

Let $M_{K}$ be the $m$-fold cyclic branched cover of $S^{3}$ along $K$ and $\tau$ be the canonical deck transformation on $M_{K}$. The fundamental region of $M_{K}$ contains a free Seifert surface of $K$. Nagasato and Yamaguchi gave a presentation of $\pi_{1}(M_{K})$ from the Lin’s presentation of $\pi_{1}(S^{3}\setminus{\rm int}N(K))$.

Theorem 2.8** (Nagasato,Yamaguchi [12]).**

Let $\langle x_{1},\ldots,x_{2g},\mu\ |\ \mu\alpha_{i}\mu^{-1}=\beta_{i}\rangle$ be a Lin’s presentation of a $1$-knot $K$. Then $\pi_{1}(M_{K})$ has the following presentation:

[TABLE]

where $\tilde{x}_{i}$ is the lift of $x_{i}$ to $M_{K}$, and $\tilde{\alpha}^{(j)}_{i},\tilde{\beta}^{(j)}_{i}$ are the words obtained from $\alpha_{i},\beta_{i}$ by replacing $x_{1},\ldots x_{2g}$ with $\tau^{j}\tilde{x}_{1},\ldots,\tau^{j}\tilde{x}_{2g}$ for $i=1,\ldots,2g$ and $j\equiv 0,\ldots,m-1\ ({\rm mod}\ m)$.

We can rewrite the presentation in Lemma 2.4 by applying a Nagasato-Yamaguchi’s presentation to ${\rm punc}(M_{K})$ as follows:

[TABLE]

3. Proof of Theorem 1.1

We first introduce a property of irreducible $SL(2,\mathbb{C})$-metabelian representations of $\pi_{1}(S^{4}\setminus{\rm int}N(K^{m,n}))$ from (2.6).

Lemma 3.1**.**

Let $\rho$ be an irreducible $SL(2,\mathbb{C})$-metabelian representation of $\pi_{1}(S^{4}\setminus{\rm int}N(K^{m,n}))$. Then, up to conjugation, $\rho$ is of the form

[TABLE]

where $i=1,\ldots 2g,j\equiv 0,\ldots,m-1\ ({\rm mod}\ m)$, and $\lambda^{(j)}_{i}\neq{\lambda^{(j)}_{i}}^{-1}$ for some $i,j$.

Proof.

Since $\rho$ is a metabelian representation, $\rho([\pi_{1}(S^{4}\setminus{\rm int}N(K^{m,n})),\pi_{1}(S^{4}\setminus{\rm int}N(K^{m,n}))])$ is an abelian group. Up to conjugation of $\rho$, we can assume that $\rho(x)$ is a diagonal matrix for any $x\in[\pi_{1}(S^{4}\setminus{\rm int}N(K^{m,n})),\pi_{1}(S^{4}\setminus{\rm int}N(K^{m,n}))]$. Since the generators $\tau^{j}\tilde{x}_{i}$ are on the Seifert surface of $K^{m,n}$, all $\tau^{j}\tilde{x}_{i}$ are commutators in $\pi_{1}(S^{4}\setminus{\rm int}N(K^{m,n}))$. Then $\rho(\tau^{j}\tilde{x}_{i})$ are of the forms

[TABLE]

see the observation of (2.5). The matrix $\rho(\eta)$ is determined by the relations $\eta\tau^{j+n}\tilde{x}_{i}\eta^{-1}=\tau^{j}\tilde{x}_{i}$ as follows. Set $\rho(\eta)=\left(\begin{matrix}a&b\\ c&d\end{matrix}\right)\in SL(2,\mathbb{C}).$ Then $\rho(\eta\tau^{j+n}\tilde{x}_{i})$ and $\rho(\tau^{j}\tilde{x}_{i}\eta)$ are given as

[TABLE]

These two matrices must be the same. Assume that $\lambda^{(j+n)}_{i}=\lambda^{(j)}_{i}$ for all $i,j$. Since $m$ and $n$ are coprime, $\lambda^{(j)}_{i}=\lambda^{(0)}_{i}$ for any $i,j$. If $\lambda^{(0)}_{i}={{\lambda}^{(0)}_{i}}^{-1}$ for any $i$, then $\rho(\tau^{(j)}\tilde{x}_{i})=\left(\begin{matrix}\pm 1&0\\ 0&\pm 1\end{matrix}\right)$ for any $i,j$. Then $\rho$ is not irreducible. If $\lambda^{(0)}_{i}\neq{{\lambda}^{(0)}_{i}}^{-1}$ for some $i$, then $\rho(\eta)$ is a diagonal matrix and $\rho(\pi_{1}(S^{4}\setminus{\rm int}N(K^{m,n})))$ becomes an abelian group. It also contradicts the irreducibility of $\rho$. Therefore $\lambda^{(j+n)}_{i}\neq\lambda^{(j)}_{i}$ for some $i,j$. In this case, $a=d=0$ and $\rho(\eta)=\left(\begin{matrix}0&b\\ -b^{-1}&0\end{matrix}\right).$ Set $B=\left(\begin{matrix}b^{\frac{1}{2}}&0\\ 0&b^{-\frac{1}{2}}\end{matrix}\right).$ Since $B\rho(\eta)B^{-1}=\left(\begin{matrix}0&-1\\ 1&0\end{matrix}\right)$ and $B\rho(\tau^{j}\tilde{x}_{i})B^{-1}=\rho(\tau^{j}\tilde{x}_{i})$, we have $\rho(\eta)=\left(\begin{matrix}0&-1\\ 1&0\end{matrix}\right)$ up to conjugation. ∎

Let $S_{0}$ be a free Seifert surface of $K$ contained in a fundamental reagion of $M_{K}$, where $M_{K}$ is a fiber of $K^{m,n}$. Let $S_{1},\ldots,S_{m-1}$ be copies of $S_{0}$ by the deck transformations. We want to know relation between $\lambda^{(j)}_{i}$ and $\lambda^{(j+1)}_{i}$ for $j\equiv 1,\ldots,m-1\ ({\rm mod}\ m)$. The relation $\eta\tau^{j+n}\tilde{x}_{i}\eta^{-1}=\tau^{j}\tilde{x}_{i}$ means that the conjugation by $\eta$ brings $\tau^{j+n}\tilde{x}_{i}$ on $S_{j+n}$ to $\tau^{j}\tilde{x}_{i}$ on $S_{j}$. Let $q$ be an integer such that $nq\equiv 1({\rm mod}\ m)$ and take conjugation of $\tau^{j}\tilde{x}_{i}$ by $\eta^{q}$. Then we obtain the relation

[TABLE]

which brings $\tau^{j+1}\tilde{x}_{i}$ on $S_{j+1}$ to $\tau^{j}\tilde{x}_{i}$ on $S_{j}$, where we used $nq\equiv 1\ ({\rm mod}\ m)$.

Let $\rho$ be an irreducible $SL(2,\mathbb{C})$-metabelian representation of $\pi_{1}(S^{4}\setminus{\rm int}N(K^{m,n}))$ in Lemma 3.1. From the relation (3.1) and Lemma 3.1,

[TABLE]

Suppose that $m$ is even. Then $q$ is odd since $m$ and $q$ are coprime. We define the representation $\overline{\rho}$ by

[TABLE]

for all $x\in\pi_{1}(S^{4}\setminus{\rm int}N(K^{m,n}))$. Note that $\rho(\eta)=\left(\begin{matrix}0&-1\\ 1&0\end{matrix}\right)$ by Lemma 3.1. In particular, $\overline{\overline{\rho}}(x)=\rho(x)$. By (3.2), $\rho(\tau^{j+1}\tilde{x}_{i})=\overline{\rho}(\tau^{j}\tilde{x}_{i})$ for all $i,j$. Since $\tilde{\alpha}^{(j)}_{i}$ and $\tilde{\beta}^{(j)}_{i}$ are words written in $\tau^{j}\tilde{x}_{1},\ldots\tau^{j}\tilde{x}_{2g}$, we have

[TABLE]

On the other hand,

[TABLE]

holds, where the relation (3.1) is applied to the second equality. Since $q$ is odd, $\rho(\eta^{q}x\eta^{-q})=\rho(\eta^{q})\rho(x)\rho(\eta^{-q})=\rho(\eta)\rho(x)\rho(\eta^{-1})=\overline{\rho}(x)$. Hence, by (3.4), we have

[TABLE]

From (3.3) and (3.5), one can see that the relations of representations of the first relations $\tilde{\alpha}^{(j)}_{i}=\tilde{\beta}^{(j-1)}_{i}$ in (2.6) are equivalent to $\rho(\tilde{\beta}^{(0)}_{i})=\rho(\eta\tilde{\alpha}^{(0)}_{i}\eta^{-1})$.

The second relations $\eta\tau^{j+n}\tilde{x}_{i}\eta^{-1}=\tau^{j}\tilde{x}_{i}$ in (2.6) are equivalent to $\eta^{q}\tau^{j+1}\tilde{x}_{i}\eta^{-q}=\tau^{j}\tilde{x}_{i}$ for all $j$ as checked in  (3.1). Therefore $\rho(\eta\tau^{j+n}\tilde{x}_{i}\eta^{-1})=\rho(\tau^{j}\tilde{x}_{i})$ are equivalent to $\rho(\eta\tau^{j+1}\tilde{x}_{i}\eta^{-1})=\rho(\tau^{j}\tilde{x}_{i})$. Hence the number of irreducible $SL(2,\mathbb{C})$-representations of the presentation (2.6) is equal to that of representations of the group presented by

[TABLE]

Now, we reduce the generators $\tau^{1}\tilde{x}_{1},\ldots,\tau^{1}\tilde{x}_{2g},\ldots,\tau^{m-1}\tilde{x}_{1},\ldots,\tau^{m-1}\tilde{x}_{2g}$ and the relations $\eta\tau^{j+1}\tilde{x}_{i}\eta^{-1}=\tau^{j}\tilde{x}_{i}$ from the above presentation to simplify counting the number of irreducible $SL(2,\mathbb{C})$-metabelian representations of $\pi_{1}(S^{4}\setminus{\rm int}N(K^{m,n}))$.

Lemma 3.2**.**

Let $m$ be an even integer. Then the number of irreducible $SL(2,\mathbb{C})$-metabelian representations of $\pi_{1}(S^{4}\setminus{\rm int}N(K^{m,n}))$ coincides that of the group $G$ presented by

[TABLE]

Proof.

A representation of (3.6) is a representation of  (3.7). So, we prove the converse. The representation of $\tau^{j}\tilde{x}_{i}$ for $j\equiv 1,\ldots,m-1\ ({\rm mod}\ m)$ is determined by the equality $\rho(\tau^{j+1}\tilde{x}_{i})=\overline{\rho}(\tau^{j}\tilde{x}_{i})$ obtained from (3.2). Hence, it is enough to prove that any irreducible $SL(2,\mathbb{C})$-metamerian representation $\rho$ of (3.7) has the property $\rho(\eta\tau^{0}\tilde{x}_{i}\eta^{-1})=\overline{\rho}(\tau^{0}\tilde{x}_{i})$. Since the presentation in (3.7) is exactly of the same form as the Lin’s presentation (2.3), all $\rho(\tau^{0}\tilde{x}_{i})$ and $\rho(\eta)$ are of the forms

[TABLE]

for $i=1,\ldots,2g$ up to conjugation, see [11]. Then, by the definition of $\overline{\rho}$,

[TABLE]

hold. Therefore $\rho(\eta\tau^{0}\tilde{x}_{i}\eta^{-1})=\left(\begin{matrix}{{\lambda}^{(0)}_{i}}^{-1}&0\\ 0&\lambda^{(0)}_{i}\end{matrix}\right),$ and this is $\overline{\rho}(\tau^{0}\tilde{x}_{i})$. ∎

Proof of  Theorem 1.1.

We decompose the proof into two cases: (1) $m$ is even or (2) $m$ is odd. In case (1), by Lemma 3.2, we only need to count the number of irreducible $SL(2,\mathbb{C})$-metabelian representations of $G$ in the Lemma. Since the presentation in (3.7) is exactly of the same form as the Lin’s presentation (2.3), each $\lambda^{(0)}_{1},\ldots,\lambda^{(0)}_{2g}$ must satisfy the following equations as explained in [8]:

[TABLE]

[TABLE]

where $\lambda^{(0)}_{i}=r_{i}\mathrm{e}^{\sqrt{-1}\theta_{i}}$ for $i=1,\ldots,2g$ and the matrix $(\omega_{ij})=V+{{}^{t}}V$ is defined in Section 2.3. With some linear algebra, one can see that the solution of (3.9) is $(r_{1},\ldots,r_{2g})=(1,\ldots,1)$ and the number of non-trivial solutions of (3.8) is $|{\det}(V+{{}^{t}}V)|-1=|\Delta_{K}(-1)|-1$. If $\{\gamma_{i}\}_{0\leq i\leq 2g}$ is a solution of (3.8) and (3.9), then $\{{\gamma_{i}}^{-1}\}_{0\leq i\leq 2g}$ is also, which is given by the conjugation of $\rho$. Therefore the number of irreducible $SL(2,\mathbb{C})$-metabelian representations of $\pi_{1}(S^{4}\setminus{\rm int}N(K^{m,n}))$ is $\frac{|\Delta_{K}(-1)|-1}{2}$.

In case (2), if $q$ is even, then $\lambda^{(j)}_{i}=\lambda^{(0)}_{i}$ for all $j$ by (3.2). Then the relation $\eta\tau^{j+n}\tilde{x}_{i}\eta^{-1}=\tau^{j}\tilde{x}_{i}$ in (2.6) gives $\lambda^{(j)}_{i}={{\lambda}^{(j)}_{i}}^{-1}$ for all $i,j$. In this case, there is no irreducible $SL(2,\mathbb{C})$-metabelian representation of $\pi_{1}(S^{4}\setminus{\rm int}N(K^{m,n}))$ by Lemma 3.1. Suppose that $q$ is odd. From the relations (3.1), we have

[TABLE]

Then we have $\lambda^{(j)}_{i}={{\lambda}^{(j)}_{i}}^{-1}$ for all $i,j$ since $mq$ is odd, and hence there is no irreducible $SL(2,\mathbb{C})$-metabelian representation of $\pi_{1}(S^{4}\setminus{\rm int}N(K^{m,n}))$ by Lemma 3.1. Thus the assertion holds. ∎