2.1 Algebraic torsion of a cochain complex

Let $C^*$ be a bounded cochain complex consisting of finite-dimensional vector spaces over a commutative field ${\Bbb F}$ , that is,

$$ \begin{align*} C^*=(0\rightarrow C^0\xrightarrow{\delta^0}C^{1} \xrightarrow{\delta^{1}}\cdots \xrightarrow{\delta^{m-1}} C^m\rightarrow 0). \end{align*} $$

Let $H^i=H^i(C^*)$ be the ith cohomology group. Choose a basis ${\mathbf {c}}^i$ of $C^i$ and a basis ${\mathbf {h}}^i$ of $H^i$ . The Reidemeister torsion $\mathop {\mathrm {Tor}}\nolimits (C^*,{\mathbf {c}}^*,{\mathbf {h}}^*)$ is defined as follows.

Let $\widetilde {{\mathbf {h}}}^i\subset C^i$ be a representative cocycle of ${\mathbf {h}}^i$ in $C^i$ . Let ${\mathbf {b}}^i$ be a tuple of vectors in $C^i$ such that $\delta ^i({\mathbf {b}}^i)$ is a basis of $B^{i+1}=\mathop {\mathrm {Im}}\nolimits {\delta ^i}$ . Then the union of the sequences of the vectors $\delta ^{i-1}({\mathbf {b}}^{i-1})\widetilde {{\mathbf {h}}}^i {\mathbf {b}}^i$ gives a basis of $C^i$ . We write $[\delta ^{i-1}({\mathbf {b}}^{i-1})\widetilde {{\mathbf {h}}}^i {\mathbf {b}}^i/{\mathbf {c}}^i] \in {\Bbb F}^\times ={\Bbb F}\backslash \{0\}$ for the determinant of the transition matrix that takes ${\mathbf {c}}^i$ to $\delta ^{i-1}({\mathbf {b}}^{i-1})\widetilde {{\mathbf {h}}}^i {\mathbf {b}}^i$ . Let $|C^*|$ be $\sum _{i=0}^{m}\alpha _i(C^*)\beta _i(C^*),$ where and . Let ${\mathbf {c}}^*$ be $({\mathbf {c}}^0,\dots ,{\mathbf {c}}^{m})$ for $C^*$ and ${\mathbf {h}}^*$ be $({\mathbf {h}}^0,\dots ,{\mathbf {h}}^{m})$ for $H^*$ . Then, the torsion is defined to be the alternating product of the form

It is known that the torsion $\mathop {\mathrm {Tor}}\nolimits (C^*,{\mathbf {c}}^*,{\mathbf {h}}^*)$ does not depend on the choices of $\widetilde {{\mathbf {h}}}^i$ and ${\mathbf {b}}^i$ , but depends only on ${\mathbf {c}}^*$ and ${\mathbf {h}}^*$ . We refer to [Reference Milnor7, Reference Turaev15] for the details. Note that, if $C^*$ is acyclic (i.e., $H^*(C^*)=0$ ), then the torsion $\mathop {\mathrm {Tor}}\nolimits (C^*,{\mathbf {c}}^*,{\mathbf {h}}^*)$ is usually denoted by $\mathop {\mathrm {Tor}}\nolimits (C^*,{\mathbf {c}}^*)$ .

2.2 Adjoint Reidemeister torsion of a $3$ -manifold

Let M be a connected oriented closed $3$ -manifold, and let G be a semisimple Lie group with Lie algebra ${\mathfrak {g}}$ . Let $\varphi :\pi _1(M)\rightarrow G$ be a representation, that is, a group homomorphism. Suppose that G injects $\mathop {\mathrm {SL}}\nolimits _n({\Bbb C})$ for some $n\in {\Bbb N}$ .

First, we introduce the cochain complex. Choose a finite cellular decomposition of M and consider the universal covering space $\widetilde {M}$ . We can canonically obtain a cellular structure of $\widetilde {M}$ as a lift of the decomposition of M, and define the cellular complex $(C_*(\widetilde {M};{\Bbb Z}),\partial _*)$ . We regard the covering transformation of M as a left action of $\pi _1(M)$ on $\widetilde {M}$ , and naturally regard $C_*(\widetilde {M};{\Bbb Z})$ as a left ${\Bbb Z}[\pi _1(M)]$ -module. Since ${\mathfrak {g}}$ is a left ${\Bbb Z}[\pi _1(M)]$ -module via the composite of $\varphi $ and the adjoint action $G\rightarrow \mathop {\mathrm {Aut}}\nolimits ({\mathfrak {g}})$ , we have the cochain complex of the form

where $\delta ^i$ is defined by $\delta ^i(f)=f\circ \partial _{i{+1}}$ .

Next, we define an ordered basis of $C^i_\varphi ({M};{\mathfrak {g}})$ . Let ${\mathbf {c}}_i\kern1.3pt{=}\kern1.3pt(c_{i,1},c_{i,2},\ldots ,c_{i,\mathop {\mathrm {rank}}\nolimits _{\Bbb Z}{C_i({M};{\Bbb Z})}}\kern-1.2pt)$ be a basis of $C_i({M};{\Bbb Z})$ derived from the i-cells. Then, $\widetilde {{\mathbf {c}}}_i\kern1.3pt{=}\kern1.3pt(\widetilde {c}_{i,1},\widetilde {c}_{i,2},\ldots ,\widetilde {c}_{i,\mathop {\mathrm {rank}}\nolimits _{\Bbb Z}{C_i({M};{\Bbb Z})}}\kern-1.2pt)$ is a basis of the free ${\Bbb Z}[\pi _1(M)]$ -module $C_i(\widetilde {M};{\Bbb Z})$ . Here, $\widetilde {c}_{i,j}$ is a lift of $c_{i,j}$ to $\widetilde {M}$ . Since ${\mathfrak {g}}$ is semisimple, the Killing form B is nondegenerate, and we can fix an ordered basis ${\mathcal {B}}=(e_1,e_2,\ldots ,e_{\mathop {\mathrm {dim}}\nolimits {{\mathfrak {g}}}})$ of ${\mathfrak {g}}$ that is orthogonal with respect to B. Let $c_{i,j}^k\in C^i_\varphi ({M};{\mathfrak {g}})$ be a ${\Bbb Z}[\pi _1(M)]$ -homomorphism defined by $c_{i,j}^k(\widetilde {c}_{i,\ell })=\delta _{j,\ell }e_k\in {\mathfrak {g}}$ for any $i\in \{0,1,2,3\}$ , $j,\ell \in \{1,2,\ldots ,\mathop {\mathrm {rank}}\nolimits _{\Bbb Z}{C_i({M};{\Bbb Z})}\}$ , and $k\in \{1,2,\ldots ,\mathop {\mathrm {dim}}\nolimits {{\mathfrak {g}}}\}$ . Here, $\delta _{j,\ell }$ is the Kronecker delta. Then the tuple

$$ \begin{align*}{\mathbf{c}}^i&=(c_{i,1}^1,c_{i,1}^2,\ldots,c_{i,1}^{\mathop{\mathrm{dim}}\nolimits{{\mathfrak{g}}}},c_{i,2}^1,c_{i,2}^2,\ldots,c_{i,2}^{\mathop{\mathrm{dim}}\nolimits{{\mathfrak{g}}}},\ldots,c_{i,\mathop{\mathrm{rank}}\nolimits_{\Bbb Z}{C_i(M;{\Bbb Z})}}^1,\\&\quad c_{i,\mathop{\mathrm{rank}}\nolimits_{\Bbb Z}{C_i(M;{\Bbb Z})}}^2,\ldots,c_{i,\mathop{\mathrm{rank}}\nolimits_{\Bbb Z}{C_i(M;{\Bbb Z})}}^{\mathop{\mathrm{dim}}\nolimits{{\mathfrak{g}}}})\end{align*} $$

provides an ordered basis of $C^i_\varphi (M;{\mathfrak {g}})$ as desired.

We next consider the cellular cochain complex $C^*(M;{\Bbb R})$ with the real coefficient. Let $c^i_j:C_i(M;{\Bbb Z})\rightarrow {\Bbb R}$ be a homomorphism defined by $c^i_j(c_{i,k})=\delta _{j,k}$ for any ${i\in \{0,1,2,3\}}$ and $j,k\in \{1,2,\ldots ,\mathop {\mathrm {rank}}\nolimits _{\Bbb Z}{C_i({M};{\Bbb Z})}\}$ . Then, ${\mathbf {c}}^i_{\Bbb R}=(c^i_1,\ldots ,c^i_{\mathop {\mathrm {rank}}\nolimits _{\Bbb Z}{C_i(M;{\Bbb Z})}})$ is a basis of $C^i(M;{\Bbb R})$ . By Poincaré duality, we can naturally fix a homology orientation $\sigma _M$ of $H^*(M;{\Bbb R})= \bigoplus _{i = 0}^3 H^i(M;\mathbb {R})$ . Let ${\mathbf {h}}^*_{\Bbb R} $ be a basis of $H^*(M;{\Bbb R})$ such that the exterior product of ${\mathbf {h}}^*_{\Bbb R}$ coincides with $\sigma _M$ . The Reidemeister torsion of $C^*(M;{\Bbb R})$ associated with ${\mathbf {c}}^*_{\Bbb R}$ and ${\mathbf {h}}^*_{\Bbb R}$ lies in ${\Bbb R}^\times $ . Therefore, we can define the sign

Then, the adjoint Reidemeister torsion of M associated with $\varphi $ is defined to be

if $C^*_\varphi ({M};{\mathfrak {g}})$ is acyclic. If $C^*_\varphi ({M};{\mathfrak {g}})$ is not acyclic, then we define $\tau _{\varphi }(M)=1$ . As is known [Reference Dubois, Huynh and Yamaguchi3, Reference Porti11], the definition of $\tau _\varphi (M)$ does not depend on the choices of the orthogonal basis ${\mathcal {B}}$ , finite cellular decompositions of M, $\widetilde {{\mathbf {c}}_i}$ , and ${\mathbf {h}}^i_{\Bbb R}$ , but depends only on M and the conjugacy class of $\varphi $ .

Finally, we give a sufficient condition for the acyclicity, which might be known.

2.3 Presentations of the cellular complexes of M

From now on, we assume that $G=\mathop {\mathrm {SL}}\nolimits _2({\Bbb C})$ and M is one of $S^3_{p/1}(4_1)$ and $S^3_{1/q}(4_1)$ for some integers p and $q\neq 0$ as in Theorem 1.2. According to [Reference Nosaka8], group presentations of $\pi _1(M)$ are given as follows:

(2) $$ \begin{align} \pi_1(S^3_{p/1}(4_1))&\cong\langle x_1,x_2,{\mathfrak{m}}\, |\, {\mathfrak{m}}x_1x_2{\mathfrak{m}}^{-1}x_1^{-1}, {\mathfrak{m}}x_2x_1x_2{\mathfrak{m}}^{-1}x_2^{-1}, [x_1,x_2]{\mathfrak{m}}^p \rangle, \\ \pi_1(S^3_{1/q}(4_1))&\cong\langle x_1,x_2,{\mathfrak{m}},{\mathfrak{m}}'\, |\, {\mathfrak{m}}x_1x_2{\mathfrak{m}}^{-1}x_1^{-1}, {\mathfrak{m}}x_2x_1x_2{\mathfrak{m}}^{-1}x_2^{-1}, {\mathfrak{m}}[x_1,x_2]^q, {\mathfrak{m}}'[x_1,x_2]^{-1} \rangle.\nonumber \end{align} $$

Here, $[x,y]$ is $xyx^{-1}y^{-1}$ . Let g be the number of generators of the group presentation above. Replace ${\mathfrak {m}}$ by $x_3$ , ${\mathfrak {m}}'$ by $x_4$ , and let $r_i$ denote the ith relator in (2). Under the identifications $C^i_\varphi ({M};{\mathfrak {g}})=\mathop {\mathrm {Hom}}\nolimits _{{\Bbb Z}[\pi _1(M)]}(C_i(\widetilde {M};{\Bbb Z}), {\mathfrak {g}}) = \mathop {\mathrm {Hom}}\nolimits _{{\mathfrak {g}}}({\mathfrak {g}}^{\mathop {\mathrm {rank}}\nolimits _{\Bbb Z} C_i(M;{\Bbb Z})},{\mathfrak {g}})$ , the cochain complex $(C^*_\varphi ({M};{\mathfrak {g}}),\delta ^*)$ is isomorphic to the dual of the following chain complex:

(3) $$ \begin{align} 0\rightarrow {\mathfrak{g}}\xrightarrow{\delta^3}{\mathfrak{g}}^{g} \xrightarrow{\delta^2} {\mathfrak{g}}^{g}\xrightarrow{\delta^1} {\mathfrak{g}}\rightarrow 0. \end{align} $$

We now describe the differentials $\delta ^*$ in detail. Let F and P be the free groups $\langle x_1, \dots , x_g\ | \ \rangle $ and $ \langle \rho _1,\ldots ,\rho _g \ | \ \rangle $ , respectively. We define the homomorphism $\psi : P *F \rightarrow F$ by setting $\psi (\rho _j)=r_j$ and $\psi (x_i)=x_i.$ Let $\mu $ denote the natural surjection from F to $\pi _1(M)$ . According to [Reference Nosaka8, Section 3.1], we can describe $\delta ^*$ by the words of the presentations (2) as follows: let $W\in P*F$ be

$$ \begin{align*}\rho_1\cdot x_1\rho_2 x_1^{-1}\cdot (x_1x_2x_1^{-1})\rho_1^{-1}(x_1x_2x_1^{-1})^{-1}\cdot ([x_1,x_2])\rho_2^{-1}([x_1,x_2])^{-1}\cdot \rho_3 \cdot {\mathfrak{m}}\rho_3^{-1}{\mathfrak{m}}^{-1},\end{align*} $$

if $M=S^3_{p/1}(4_1)$ . Let $W \in P*F$ be

$$ \begin{align*}\rho_1\cdot x_1\rho_2 x_1^{-1}\cdot (x_1x_2x_1^{-1})\rho_1^{-1}(x_1x_2x_1^{-1})^{-1}\cdot ([x_1,x_2])\rho_2^{-1}([x_1,x_2])^{-1}\cdot \rho_4^{-1}\cdot {\mathfrak{m}}'\rho_3{\mathfrak{m}}'^{-1}\cdot \rho_4 \cdot \rho_3^{-1}, \end{align*} $$

if $M=S^3_{1/q}(4_1)$ . Then, each $\delta ^*$ can be written as the matrices

(4) $$ \begin{align} \delta^1=\left(1-x_j\right)_{j=1,\ldots,g},\,\,\, \delta^2=\left({\small \frac{\partial r_j}{\partial x_i}}\right)_{i,j=1,\ldots,g},\,\,\, \delta^3=\mu \circ \psi\left({\small \frac{\partial W}{\partial \rho_i}}\right)_{i=1,\ldots,g}, \end{align} $$

where $\frac {\partial *}{\partial *}$ is Fox derivative (see [Reference Turaev15, Section 16] for the definition). Although each entry of the matrices is described in ${\Bbb Z}[\pi _1(M)]$ , we regard the entry as an automorphism of ${\mathfrak {g}}$ via the adjoint action.

3.1 Preliminary

To state Propositions 3.1 and 3.2 and Theorems 3.3 and 3.4, let us consider a domain D in ${\Bbb C}$ of the form

as in Figure 1, and define the Laurent polynomial $Q_M(x)\in {\Bbb Z}[x{,x^{-1}}]$ by setting

Let $Q_M^{-1}(0){\in {\Bbb C}}$ denote the zero set of the Laurent polynomial $Q_M$ .

Figure 1

$D\subset {\Bbb C}$ .

Proposition 3.1 Let $M=S^3_{p/1}(4_1)$ for some integer p. If $p\neq 0$ , then there is a bijection $\Phi _{M}:R^{\mathrm {irr}}_G(M)\rightarrow Q_M^{-1}(0)\cap D$ . Here, for $[\varphi ]\in R^{\mathrm {irr}}_G(M)$ , we define

(5)

when the eigenvalues of $\varphi ({\mathfrak {m}})$ are not $\pm \sqrt {-1}$ . If $\varepsilon \sqrt {-1}\in Q_M^{-1}(0)$ for some $\varepsilon \in \{\pm 1\}$ , $\Phi _M^{-1}(\varepsilon \sqrt {-1})$ is a conjugacy class with a representation $\varphi $ defined by

(6) $$ \begin{align}\varphi({\mathfrak{m}})=\left(\begin{array}{cc} \varepsilon \sqrt{-1} & 0 \\ 0 & -\varepsilon \sqrt{-1}\\ \end{array} \right),\,\,\,\, \varphi(x_1)=\left(\begin{array}{cc} \frac{1}{4} (-1+\varepsilon\sqrt{5}) & 1 \\ \frac{1}{8} (-5-\varepsilon\sqrt{5}) & \frac{1}{4} (-1+\varepsilon\sqrt{5}) \\ \end{array} \right). \end{align} $$

If $p=0$ , then there is a bijection $\Phi _{M}:R^{\mathrm {irr}}_{G}(M)\rightarrow \{\pm \sqrt {-1} ,\pm (1-\sqrt {5})/2\}$ .

Proof of Proposition 3.1

Let $p\neq 0$ . For an irreducible representation $\varphi :\pi _1(M)\rightarrow \mathop {\mathrm {SL}}\nolimits _2({\Bbb C})$ , take $x,y,z,w\in {\Bbb C}$ so that $\varphi (x_1)=\left ( {\small \begin {array}{cc}x&y\\z&w\\ \end {array}}\right )$ and $xw-yz=1$ . We first claim that $\varphi ({\mathfrak {m}})$ is diagonalizable. In fact, if not so, we may suppose $\varphi ({\mathfrak {m}})=\left ({\small \begin {array}{cc}\eta &b\\0&\eta \\ \end {array}} \right )$ for some $b\in {\Bbb C}^\times $ and $\eta \in \{\pm 1\}$ . Since $\varphi (r_1)=E_2$ , we have

(7) $$ \begin{align} \varphi(x_2)=\varphi(x_1){^{-1}}\varphi({\mathfrak{m}})^{-1}\varphi(x_1)\varphi({\mathfrak{m}})= \left({\small \begin{array}{cc} 1-\eta b w z & - b \left(b w z+\eta w^2-\eta\right) \\ \eta b z^2 & b^2 z^2+\eta b w z+1 \\ \end{array}} \right). \end{align} $$

It follows from (7) that

(8) $$ \begin{align} \varphi(r_3)&=\varphi(x_1)\varphi(x_2)\varphi(x_1)^{-1}\varphi(x_2)^{-1}\varphi({\mathfrak{m}})^p\nonumber\\ &=\eta^p{\small \left( \begin{array}{cc} b^4 z^4+\eta b^3 w z^3-b^2 z^2 {\left( x^2+xw-3\right)} +\eta b z (w-x)+1 & * \\ -\eta b^3 z^4-b^2 z^3 (w+x)-\eta 2 b z^2 & {*} \\ \end{array} \right).} \end{align} $$

Then, the condition $\varphi (r_3)=E_2$ and $b\neq 0$ leads to $z=0$ . In fact, if $z\neq 0$ , the (2,1)-entries of $\varphi (r_3)=E_2$ yields $x=-\eta 2 b^{-1}z^{-1}-w-\eta b z$ by (8). Thus, the (1,1)-entry of (8) equals $-1$ , resulting in a contradiction.

By substituting $z=0$ into $\varphi (r_2)$ , we obtain

$$ \begin{align*} E_2=\varphi(r_2)&=\varphi({\mathfrak{m}})\varphi(x_2)\varphi(x_1)\varphi(x_2)\varphi({\mathfrak{m}})^{-1}\varphi(x_2)^{-1} =\left({\small \begin{array}{cc} x & y-\eta b \left(w^3-2 w+x\right) \\ 0 & w \\ \end{array}} \right). \end{align*} $$

Thus, $x=w=1$ and $y=0$ ; therefore, $\varphi (x_1)$ and $\varphi (x_2)$ are upper triangular matrices, which leads to a contradiction to the irreducibility.

By the above claim, we may suppose $\varphi ({\mathfrak {m}})=\left ({\small \begin {array}{cc}a & 0 \\ 0&a^{-1}\\ \end {array}}\right )$ for some $a\in D{\backslash \{0\}}$ . Since we consider $\varphi $ up to conjugacy, we may suppose $y=1$ . Thus, $z=xw-1$ . Since ${\varphi (r_1)=\varphi (r_2)=\varphi (r_3)=E_2}$ , with the help of a computer program of Mathematica, we have

(9) $$ \begin{align} x&={\small \frac{1+a^2-a^4+\eta(1-2 a^2-a^4-2 a^6+a^8)^{1/2}}{2(1- a^2)}},\nonumber\\ z&={\small -\frac{1-3 a^2+a^4+\eta(1-2 a^2-a^4-2 a^6+a^8)^{1/2}}{2 (a^2-1)^2 }},\nonumber\\ w&={\small \frac{-1+a^2+a^4+\eta(1-2 a^2-a^4-2 a^6+a^8)^{1/2}}{2 a^2 (a^2-1)}}, \end{align} $$

and $Q_M(a)=0$ when $a\neq \pm \sqrt {-1}$ . Here, we fix a branch of the $1/2$ th power on ${\Bbb C}^\times \backslash {\Bbb R}$ , and define the signs $\eta \in \{\pm 1\}$ by setting

$$ \begin{align*} \eta= \begin{cases} +1, & {\mathrm{ if } }\,\, -1+a^2+2a^4+a^6-a^8+2 a^{p+4}=(a^4-1)(1-2 a^2-a^4-2 a^6+a^8)^{1/2},\\ -1, & {\mathrm{ if } } \,\,-1+a^2+2a^4+a^6-a^8+2 a^{p+4}=-(a^4-1)(1-2 a^2-a^4-2 a^6+a^8)^{1/2}.\\ \end{cases} \end{align*} $$

When $a=\varepsilon \sqrt {-1}$ for some $\varepsilon \in \{\pm 1\}$ , we have

(10) $$ \begin{align} x= \frac{-1+\varepsilon \sqrt{5}}{4},\,\,\,\,\, z=\frac{-5-\varepsilon\sqrt{5}}{8},\,\,\,\,\, w= \frac{-1+\varepsilon\sqrt{5}}{4}\end{align} $$

by the condition $\varphi (r_1)=\varphi (r_2)=\varphi (r_3)=E_2$ . In summary, the map $\Phi _M$ is well-defined and injective. Finally, we can easily show the surjectivity of $\Phi _M$ by following the reverse process of the above calculation.

In the remaining case of $p=0$ , define $\Phi _M$ as follows: for each $\varepsilon \in \{\pm 1\}$ , let $\Phi _M^{-1}(\varepsilon \sqrt {-1})$ be a representation $\varphi $ defined by (6). For $\varepsilon ' \in \{\pm 1\}$ , let $\Phi _M^{-1}(\varepsilon ' (1-\sqrt {5})/2)$ has a representation $\varphi $ defined by

$$ \begin{align*}\varphi({\mathfrak{m}})= \left(\begin{array}{cc} \varepsilon' \frac{1-\sqrt{5}}{2} & 0 \\ 0 & \varepsilon' \frac{-1-\sqrt{5}}{2}\\ \end{array} \right),\,\,\,\, \varphi(x_1)=\left(\begin{array}{cc} 1 & 1 \\ 0 & 1\\ \end{array} \right), \end{align*} $$

respectively. Then, by $\varphi (r_1)=\varphi (r_2)=\varphi (r_3)=E_2$ , we can show the well-definedness and injectivity of $\Phi _M$ as in the case of $p\neq 0$ . By following the reverse process, we can check the surjectivity of $\Phi _M$ as well.

Proof of Proposition 3.2

It can be proved in the same manner as Proposition 3.1. In this case, instead of (9), we have

(11) $$ \begin{align} &x=\frac{a^{-2 q} \left(2 a^{6 q}+2 a^{4 q+1}\right)}{2 \left(a^{2 q}-1\right)^2 \left(a^{2 q}+1\right)}, \ z=-\frac{4 a^{2 q}+2 a^{4 q}-2 a^{6 q}+2 a^{4 q+1}-2}{2 y \left(a^{2 q}-1\right)^3 \left(a^{2 q}+1\right)},\nonumber\\ &w=-\frac{4 a^{4 q}+2 a^{6 q}-2 a^{8 q}+2 a^{4 q+1}-2}{2 \left(a^{2 q}-1\right)^2 \left(a^{2 q}+1\right)}.\\[-42pt]\nonumber \end{align} $$

Theorem 3.3 Let $M=S^3_{p/1}(4_1)$ for some integer $p\neq 0$ . For $a\in Q_M^{-1}(0)\cap D$ as in Proposition 3.1, we denote the representative $\mathop {\mathrm {SL}}\nolimits _2({\Bbb C})$ -representation of $\Phi _M^{-1}(a)$ by $\varphi _a$ . Then, the adjoint Reidemeister torsion of M with respect to $\varphi _a$ is computed as

(12)(13) $$ \begin{align}\tau_{\varphi_a}(M)=\left\{\begin{array}{@{}l@{}ll}& -\dfrac{ 4-p+(-2+p)a^2+2pa^4+(2+p)a^6-(4+p)a^8+2pa^{4+p}}{2(a^2-1)^3(1+a^2)}, &\text{if } \,a\notin\{\pm \sqrt{-1}\},\\[4pt]& \dfrac{1}{8} (10+ a p \sqrt{-5}), &\text{if }\,a\in\{\pm \sqrt{-1}\}. \end{array}\right.\end{align} $$

Theorem 3.4 Let $M=S^3_{1/q}(4_1)$ for some integer $q\neq 0$ . For $a\in Q_M^{-1}(0)\cap D$ as in Proposition 3.2, we denote the representative $\mathop {\mathrm {SL}}\nolimits _2({\Bbb C})$ -representation of $\Phi _M^{-1}(a)$ by $\varphi _a$ . Then, the adjoint Reidemeister torsion of M with respect to $\varphi _a$ is computed as

(14) $$ \begin{align} \tau_{\varphi_a}(M)= -\frac{a^{6q}(-1+4q+(1-2q)a^{2q}+2(1+a)a^{4q}+(1+2q)a^{6q}-(1+4q)a^{8q})}{2(a^{4q}-1)^3(1-2a^{2q}-a^{4q}-2a^{6q}+a^{8q})}. \end{align} $$

3.2 Two key lemmas

As preliminaries of the proof of Theorem 1.2, we prepare two lemmas.

Lemma 3.6 Define a polynomial $\kappa _p(x) \in {\Bbb Z}[x]$ by setting

$$ \begin{align*} \kappa_p(x) =\left\{ \begin{array}{ll} (1+x)^2 ,& \mathrm{if} \ \ p= 2m+1, \\ (1+x^2)^2 ,& \mathrm{if} \ \ p= 4m, \\ 1 ,& \mathrm{if} \ \ p= 4m+2, \\ \end{array} \right.\end{align*} $$

for some $m \in {\Bbb Z}$ . Then, $Q_{M}(x)$ with $M=S^3_{p/1}(4_1) $ is divisible by $\kappa _p(x) $ , and the quotient $ Q_{M}(x)/\kappa _p(x)$ has no repeated roots. On the other hand, $Q_{M}(x)$ with $M=S^3_{1/q}(4_1) $ is divisible by $(1+x)^2 $ , and the quotient $ Q_{M}(x)/(1+x)^2 $ also has no repeated roots.

Proof The required statement with $|p| \leq 4$ and $|q| \leq 4$ can be directly shown, we may assume $|p| \geq 5 $ and $|q| \geq 5.$ We first focus on the case $M=S^3_{p/1}(4_1) $ . By a computation of $ \frac {\mathrm {d}^n \ }{\mathrm {d}x^n} (Q_M(x))\mid _{x=b}$ with $b= \pm 1 , \pm \sqrt {-1}$ , we can easily verify the multiplicity of $Q_M(x)$ . To elaborate, if $p=2m+1$ , then

$$ \begin{align*}Q_M(1)=4,\ \ Q_M(\sqrt{-1})=-2(-1)^m\sqrt{-1},\ \ Q_M(-\sqrt{-1})=2(-1)^m\sqrt{-1}\end{align*} $$

are all nonzero, which implies that $1,\pm \sqrt {-1}$ are not roots of $Q_M(x)$ . Furthermore,

$$ \begin{align*}Q_M(-1)=Q_M'(-1)=0,\ \ \ Q_M^{(2)}(-1)=-2(-12-p^2)\neq0,\end{align*} $$

indicates that $-1$ is a root of $Q_M(x)$ with multiplicity $2$ . When $p=4m$ or $4m+2$ , we can analogously determine the multiplicity of $Q_M(x)$ with $b=\pm 1, \pm \sqrt {-1}$ . Thus, $Q_M(x)$ is divisible by $\kappa _p(x)$ , and $Q_M(x)/\kappa _p(x)$ is not divisible by $x \pm 1$ and $x^2+1$ .

Next, suppose $Q_{M}(x) $ has a repeated root $a\in {\Bbb C}$ with $a \neq \pm 1 , \pm \sqrt {-1}$ . Then, $Q_{M}(a) =0$ and $ Q_{M}'(a) =0,$ which are equivalent to

(15) $$ \begin{align} 1-a^p(a^{-4}+a^{-2}+2+a^{2}-a^{4})+(a^p)^{2} =0,\end{align} $$

(16) $$ \begin{align} (p-4)a^{-4}+(p-2)a^{-2}+2p+(p+2)a^{2}-(p+4)a^{4}= - 2pa^{p}. \end{align} $$

Applying (16) to (15) to kill the term $a^p$ , we equivalently have

$$ \begin{align*} (1+a)^2(1+a^2)^2 \bigl(p^2 - 16 +(16-2p^2) a^2-(36 +p^2)a^4 +(16-2p^2)a^6 +(p^2 -16)a^8 \bigr) =0.\end{align*} $$

Since $a^2 \neq \pm 1$ , the last quartic term equation can be solved as

$$ \begin{align*}a^2 = \frac{ p^2-8 + 2 \eta p\sqrt{p^2 -15} + \varepsilon \sqrt{(40-3p^2 +2\eta p \sqrt{p^2 -15}) (p^2-24 +2\eta p \sqrt{p^2 -15})}}{2p^2 -32},\end{align*} $$

for some $\varepsilon , \eta \in \{ \pm 1\}.$ Let F be the field extension ${\Bbb Q}(a)$ of degree 8. Let us regard (15) as a quadratic equation in F of $a^p$ . Since the discriminant is not zero and $|p|>4$ , $a^p $ does not lie in F. This is a contradiction. In summary, $Q_M(x)/\kappa _p(x)$ has no repeated roots as required.

On the other hand, if $M=S^3_{1/q}(4_1) $ , we can easily show that $Q_M(x)$ is divisible not by $(1+x)^3$ but by $(1+x)^2$ . Similarly, suppose $Q_{M}(x) $ has a repeated root $a\in {\Bbb C}$ with $a \neq \pm 1 $ . Then, $Q_{M}(a) = Q_{M}'(a) =0$ . We can easily see $Q_{M}(1/a) = Q_{M}'(1/ a) =0$ by reciprocity of $Q_M$ . Thus, we obtain $ ( x^{-4q} Q_{M})' (a) = ( x^{-4q} Q_{M})' (1/a) =0,$ which are equivalent to

(17) $$ \begin{align}2(1+a)= (2q+1)a^{2q}+ (-2q+1)a^{-2q}-(4q+1)a^{4q} - (-4q+1)a^{-4q}, \end{align} $$

(18) $$ \begin{align} 2(1+a^{-1})= (2q+1)a^{-2q}+ (-2q+1)a^{2q}-(4q+1)a^{-4q} - (-4q+1)a^{4q}.\end{align} $$

Since $a^{-4q}Q_{M}(a) = 0$ is equivalent to

(19) $$ \begin{align} 2(1+a) 2(1+a^{-1})= 4( a^{4q}-a^{2q}-a^{-2q} + a^{-4q}),\end{align} $$

the substitution of (17) and (18) into (19) gives the equation

(20) $$ \begin{align} (1-b)^2 (1+b)^2 (-1 +2b+b^2+2b^3-b^4+16q^2 -16bq^2+36b^2q^2 -16b^3q^2+16b^4 q^2)=0 ,\end{align} $$

where we replace $a^{2q}$ by b. If $\omega ^{2q}= \pm 1$ and $\omega \in {\Bbb C}$ , we can easily check $ Q_M( \omega ) \neq 0$ by definition. Thus, $a^q$ is a solution of the quartic equation in (20) and does not lie in ${\Bbb Q}$ , for any $q \in {\Bbb Z}$ . Let $F /{\Bbb Q}$ be the field extension by the quartic equation. By definition, F does not contain $ a $ and $2+a+a^{-1}$ , which contradicts (19) since $|q|>4$ . In summary, $Q_M(x)/(1+x)^2$ has no repeated roots as required.

Next, we should mention a slight modification of Jacobi’s residue theorem.

Lemma 3.7 Fix $ {\zeta } \in {\{ 0, 2\}}$ and ${\theta } \in \{ 1,2\}$ . Suppose a polynomial $k(x) \in {\Bbb Q}[x]$ has no repeated roots and $k(0)\neq 0$ . Take another polynomial $g(x)\in {\Bbb Q}[x] $ such that $\mathrm {deg}(g) \leq \mathrm {deg}(k) - {\theta } {\zeta } -2$ . Then, the following sum is zero:

(21) $$ \begin{align} \sum_{a \in k^{-1}(0)} \frac{ (1 +a^{{\theta}})^{ {\zeta} } g (a)}{ \frac{\mathrm{d}}{ \mathrm{d}x }((1+x^{{\theta}})^{\zeta} k(x))|_{x=a}} =0.\end{align} $$

3.3 Proof of Theorem 1.2 with $n=-1$

We suppose $n=-1$ and give the proof of Theorem 1.2. Recall the fact that $M=S^3_{p/1}(4_1)$ and $M=S^3_{1/q}(4_1)$ are hyperbolic if and only if $|p|\geq 5$ and $|q|\geq 2$ , respectively (see, e.g., Theorem 4.7 of [Reference Thurston13]).

First, we focus on the case where $p\geq 5$ , $M=S^3_{p/1}(4_1)$ , and p is not divisible by $4$ . From the definition of $Q_M(x)$ and Theorem 3.3, we can easily verify

(22) $$ \begin{align} \frac{1}{\tau_{\varphi_a}(M)}= \frac{2(1-a^2)^3( 1+a^2)a^{p-5} }{Q_M ' (a)} \qquad \textrm{for any } a \in \left(Q_M^{-1}(0)\cap D\right)\backslash \{\pm \sqrt{-1}\}. \end{align} $$

If $p -2$ is divisible by $4$ , we replace $g(x)$ and $k(x)$ by $2(1-x^2)^3( 1+x^2)x^{p-5} $ and $Q_M(x)$ , respectively. Then, Lemma 3.7 with $\zeta =0$ deduces to the required conclusion as

(23) $$ \begin{align} 0 = \sum_{ a \in Q_M^{-1} (0) } \frac{g(a)}{Q_M'(a)} =\sum_{ a \in Q_M^{-1} (0) } \frac{1}{\tau_{\varphi_a}(M)}= 2 \sum_{ a \in Q_M^{-1} (0) \cap D } \frac{1}{\tau_{\varphi_a}(M)} = \sum_{ \varphi \in R_{G}^{\mathrm{irr}} } \frac{2}{\tau_{\varphi}(M)}.\end{align} $$

Here, the second, third, and fourth equalities immediately follow from (22), Remark 3.5(iii), and Proposition 3.1, respectively. Meanwhile, when $p -1$ is divisible by $2$ , we replace $g(x)$ and $k(x)$ by $2(x-1)( x^4 -1 )x^{p-5} $ and $Q_M(x)/(1+x)^2$ , respectively. Then, we can readily show similar equalities to (23).

We further discuss the case of $p/4 \in {\Bbb Z}$ . By Lemma 3.6, $ Q_{M}(x)/(1+x^2) $ lies in ${\Bbb Z}[x]$ , and has no double roots. We let $g(x)$ and $k(x)$ be $2(1-x^2)^3 x^{p-5} $ and $Q_M(x)/(1+x^2)$ , respectively. By Lemma 3.7 with $\zeta =1$ and $\theta =2$ , we have

$$ \begin{align*}0 &= \sum_{ a \in k^{-1} (0) } \frac{g(a)}{k'(a)} = \frac{g( \sqrt{-1})}{k'( \sqrt{-1})}+ \frac{g(- \sqrt{-1})}{k'(- \sqrt{-1})}+\sum_{ a \in Q_M^{-1} (0) \cap D \backslash \{ \pm \sqrt{-1}\} } \frac{2}{\tau_{\varphi_a}(M)}\\ &= \frac{32 \sqrt{-1}}{20-p^2} + \sum_{ a \in Q_M^{-1} (0) \cap D \backslash \{ \pm \sqrt{-1}\} } \frac{2}{\tau_{\varphi_a}(M)} \\ &= \frac{2}{\tau_{\varphi_{\sqrt{-1}}}(M)} +\frac{2}{\tau_{\varphi_{-\sqrt{-1}}}(M)} + \sum_{ a \in Q_M^{-1} (0) \cap D \backslash \{ \pm \sqrt{-1}\} } \frac{2}{\tau_{\varphi_a}(M)} \\ &= \sum_{ a \in Q_M^{-1} (0) \cap D } \frac{2}{\tau_{\varphi_a}(M)} = \sum_{ \varphi \in R_{G}^{\mathrm{irr}} } \frac{2}{\tau_{\varphi}(M)}, \end{align*} $$

which is the required vanishing identity. Here, the second, fourth, and sixth equalities follow from (22), Theorem 3.3, and Proposition 3.1, respectively.

Next, we focus on the case of $q\geq 2$ and $M=S^3_{1/q}(4_1)$ . Similarly to (22), we can show

(24) $$ \begin{align} \frac{1}{\tau_{\varphi_a}(M)}= \frac{2 (a^{4 q}-1)^3 (a^{4 q}-(a^2+a+1) a^{2 q-1}+1) }{ \frac{\mathrm{d}}{\mathrm{d}x }(x^{4q+1} Q_M (x))|_{x=a}} \qquad \textrm{for any } a \in Q_M^{-1}(0)\cap D. \end{align} $$

By a Euclidean Algorithm, we can choose a polynomial $h(x) \in {\Bbb Q}[x]$ such that

$$ \begin{align*} 2 (x^{4 q}-1)^3 (x^{4 q}-(x^2+x+1) x^{2 q-1}+1) \equiv x^{4q+1} h (x) \qquad (\mathrm{modulo} \ Q_M(x)), \end{align*} $$

and $\mathrm {deg}h(x ) < 8q -2$ . Recall from Lemma 3.6 that $Q_M(x) $ is divisible by $(1+x)^2 $ ; thus so is $h(x)$ . In summary, we can define polynomials $g(x)$ and $k(x)$ to be $h(x)/ (1+x)^2$ and $Q_M(x)/(1+x)^2$ , respectively. Then, Lemma 3.7 with $\zeta =2$ and $\theta =1$ readily leads to the same equalities as (23).

The proof of the cases of $p\leq -5$ and $q \leq -2$ can be shown in the same manner; so we here do not carry out the detailed proof.

Finally, in the remaining cases of $|p|\leq 4$ for $M=S^3_{p/1}(4_1)$ , we can obtain the following by a direct calculation:

$$ \begin{align*} \sum_{\varphi\in R^{\mathrm{irr}}_G(M)}\frac{1}{\tau_{\varphi}(M)}&= \begin{cases} 2, & \text{ if }p\in \{0,\pm1,\pm2,\pm3\},\\ 8, & \text{ if }p\in \{\pm4\}. \end{cases} \end{align*} $$

For example, we now discuss the detail in the case $p=4$ for $M=S^3_{p/1}(4_1)$ . The roots of $Q_M(x)=x^2 + 2 x^4 + x^6=0$ are $x=\pm \sqrt {-1}$ . By Theorem 3.3, we have $\tau _{\varphi _{\sqrt {-1}}}(M)= (5-2 \sqrt {5})/4$ and $\tau _{\varphi _{-\sqrt {-1}}}(M)=(5+2 \sqrt {5})/4$ , leading to $\sum _{\varphi \in R^{\mathrm {irr}}_G(M)}\tau _{\varphi }(M)^{-1}=8$ . Similarly, the computations in the other cases run well.