Kashaev invariants of twice-iterated torus knots

TL;DR

This paper analyzes the asymptotic behavior of Kashaev invariants for twice-iterated torus knots, linking the results to topological invariants like Chern--Simons and Reidemeister torsion.

Contribution

It provides a new asymptotic formula for Kashaev invariants of a specific class of knots with a topological interpretation.

Findings

Asymptotic formula for Kashaev invariants derived

Topological interpretation involving Chern--Simons invariant

Connection to twisted Reidemeister torsion established

Abstract

We calculate the asymptotic behavior of the Kashaev invariant of a twice-itarated torus knot and obtain topological interpretation of the formula in terms of the Chern--Simons invariant and the twisted Reidemeister torsion.

Equations

$$\frac{\log \left|J_N(K;\exp (2\pi i/N))\right|}{N}=\frac{\mathrm{Vol}(S^3\setminus K)}{2\pi },$$

$$J_N(4_1;\exp (2\pi i/N))\underset{N\to \infty }{\sim }2{\pi }^{3/2}{\left(\frac{N}{2\pi i}\right)}^{3/2}\tau (4_1)\exp \left(\frac{N}{2\pi i}S(4_1)\right),$$

$$J_N(K;\exp (2\pi i/N))\underset{N\to \infty }{\sim }2{\pi }^{3/2}{\left(\frac{N}{2\pi i}\right)}^{3/2}\tau (K)\exp \left(\frac{N}{2\pi i}S(K)\right),$$

$$\begin{split}&J_{N}\bigl{(}T(c,d);e^{2\pi i/N}\bigr{)}\\ \underset{N\to\infty}{\sim}&\frac{\pi^{3/2}}{2cd}\left(\frac{N}{2\pi i}\right)^{3/2}\sum_{k=1}^{cd-1}(-1)^{k+1}k^{2}\tau(k)\exp\left(S(k)\frac{N}{2\pi i}\right)+O(1),\end{split}$$

$$S(k)$$

$$\tau (k)$$

$$\tilde{S}(k):=\frac{k^2{\pi }^2}{cd},$$

$$\frac{1}{2\sqrt{\pi }}{\left(\frac{N}{2\pi i}\right)}^{3/2}\sum _{k=1}^{cd-1}(-1{)}^{k(N+1)}{i}^{-cdN}\tau (k)\tilde{S}(k)\exp \left(\tilde{S}(k)\frac{N}{2\pi i}\right)+O(1).$$

$$\begin{split}&J_{N}\bigl{(}T(2,2a+1)^{(2,2b+1)},e^{2\pi i/N}\bigr{)}\\ \underset{N\to\infty}{\sim}&\frac{1}{2\sqrt{\pi}}\left(\frac{N}{2\pi i}\right)^{3/2}\sum_{l=0}^{2b}\tau_{1}(l)S_{1}(l)\,e^{\frac{N}{2\pi i}S_{1}(l)}\\ &+(-1)^{N}\frac{1}{2\sqrt{\pi}}\left(\frac{N}{2\pi i}\right)^{3/2}\sum_{m=0}^{4a+1}\tau_{2}(m)S_{2}(m)\,e^{\frac{N}{2\pi i}S_{2}(m)}\\ &-\frac{1}{2}\left(\frac{N}{2\pi i}\right)^{2}\sum_{(j,k)\in\mathcal{B}}\tau_{3}(j,k)S_{3}(j,k)\,e^{\frac{N}{2\pi i}S_{3}(j,k)}+O(N^{1/2}),\end{split}$$

$${\tau }_1(l)$$

$$S_1(l)$$

$${\tau }_2(m)$$

$$S_2(m)$$

$${\tau }_3(j,k)$$

$$S_3(j,k)$$

$$\begin{split}&(-1)^{n}\exp\left(\frac{(2b+1-3(2a+1))\pi i}{4(n+1)}\right)(-n-1)\left(\frac{\langle e_{n}\rangle_{\mathcal{T}^{2(2b+1)}}}{[n+1]}\right)\Biggm{|}_{A=\exp(\pi i/(2(n+1)))}\\ =&\frac{2(n+1)^{3}i}{\pi^{4}\sqrt{(2a+1)(2b+1-4(2a+1))}}\exp\left(\frac{-\pi i}{(2a+1)(n+1)}\right)\\ &\times\int_{C_{\pi/4}}\int_{C_{\pi/4}}dz_{1}dz_{2}\psi_{1}(z_{1})\psi_{2}(z_{2})\delta(z_{1},z_{2})\,e^{(n+1)\theta(z_{1},z_{2})},\end{split}$$

$$[k]$$

$$\theta (z_1,z_2)$$

$$\delta (z_1,z_2)$$

$${\psi }_1(z_1)$$

$${\psi }_2(z_2)$$

$$J_{N}(K;q)=\frac{\langle e_{N-1}\rangle_{\mathcal{K}^{0}}}{\langle e_{N-1}\rangle_{\mathcal{U}^{0}}}=\bigl{(}(-1)^{N-1}A^{N^{2}-1}\bigr{)}^{-\sigma}\frac{\langle e_{N-1}\rangle_{\mathcal{K}^{\sigma}}}{(-1)^{N+1}[N]},$$

$$\begin{split}&J_{N}\bigl{(}T(2,2a+1)^{(2,2b+1)};e^{2\pi i/N}\bigr{)}\\ =&(-1)^{N-1}\frac{1}{-N}\exp\left(-\frac{(2b+1-3(2a+1))\pi i}{4N}\right)\exp\left(\frac{-\pi i}{(2a+1)N}\right)\\ &\left((-1)^{N-1}\exp\left(\frac{(N^{2}-1)\pi i}{2N}\right)\right)^{-2(2b+1)}(-1)^{N+1}\\ &\times\frac{2N^{3}i}{\pi^{4}\sqrt{(2a+1)(2b+1-4(2a+1))}}\\ &\times\int_{C_{\pi/4}}\int_{C_{\pi/4}}dz_{1}dz_{2}\psi_{1}(z_{1})\psi_{2}(z_{2})\delta(z_{1},z_{2})\,e^{N\theta(z_{1},z_{2})}\\ =&(-1)^{N+1}e^{-\frac{\pi i}{4N}(2b+1-3(2a+1)+\frac{4}{2a+1})}e^{\frac{(2b+1)\pi i}{N}}\\ &\times\left(\frac{2N^{2}i}{\sqrt{(2a+1)(2b+1-4(2a+1))}\,\pi^{4}}\right)\\ &\times\int_{C_{\pi/4}}\int_{C_{\pi/4}}dz_{1}dz_{2}\psi_{1}(z_{1})\psi_{2}(z_{2})\delta(z_{1},z_{2})\,e^{N\theta(z_{1},z_{2})},\end{split}$$

$$\theta (z_1,z_2)$$

$$\delta (z_1,z_2)$$

$${\psi }_1(z_1)$$

$${\psi }_2(z_2)$$

$$(w_1,w_2)=\left(\frac{(2b+1)\pi i}{2},2(2a+1)\pi i\right).$$

$$\mathrm{Res}({\psi }_1,{\xi }_l)=(-1{)}^{l-1}\frac{i}{4}.$$

$$\mathrm{Res}({\psi }_2,{\eta }_m)=\frac{(-1{)}^m}{2}\sin (\frac{(2m+1)\pi }{2a+1}).$$

$$F_N(z_1,z_2)$$

$$I_N$$

$${\mathcal{J}}_N$$

$$I_N=I_{0,N}+I_{1,N}+I_{2,N}+I_{3,N}$$

$$I_{0,N}$$

$$I_{1,N}$$

$$I_{2,N}$$

$$I_{3,N}$$

$${\mathcal{J}}_N={\mathcal{J}}_{0,N}+{\mathcal{J}}_{1,N}+{\mathcal{J}}_{2,N}+{\mathcal{J}}_{3,N}.$$

Full text

H.M. was supported by JSPS KAKENHI Grant Numbers 26400079, 17K05239. A.T. has been partially supported by a grant from the Simons Foundation (#354595 to AT).

1. Introduction

For a knot $K$ in the three-sphere $S^{3}$ and an integer $N\geq 2$, let $\langle K\rangle_{N}\in\mathbb{C}$ be the Kashaev invariant [6]. In [7], he conjectured that $|\langle K\rangle_{N}|$ grows exponentially with growth rate $\operatorname{Vol}(S^{3}\setminus{K})/(2\pi)$ for $N\to\infty$ when $K$ is hyperbolic, where $\operatorname{Vol}(S^{3}\setminus{K})$ is the hyperbolic volume of $S^{3}\setminus{K}$.

J. Murakami and the first author proved that the Kashaev invariant coincides with $J_{N}(K;\exp(2\pi i/N))$, where $J_{N}(K;q)$ is the colored Jones polynomial of a knot $K$ in the three-sphere $S^{3}$ associated with the $N$-dimensional irreducible representation of the Lie algebra $\mathfrak{sl}_{2}(\mathbb{C})$, normalized so that $J_{N}(U;q)=1$ for the unknot $U$ [13]. They also proposed the following conjecture:

Conjecture 1.1** (Volume Conjecture).**

For any knot $K$, we have

[TABLE]

where $\operatorname{Vol}(S^{3}\setminus{K})$ is the simplicial volume of $S^{3}\setminus{K}$ normalized so that the simplicial volume of a hyperbolic knot complement equals its hyperbolic volume. In particular, when $K$ is hyperbolic, Kashaev’s conjecture holds.

This conjecture was first proved for torus knots by Kashaev and O. Tirkkonen [8]. Note that since the complement of a torus knot is a Seifert fibered space, its simplicial volume is zero. A proof of the conjecture for the figure-eight knot was given by T. Ekholm (see for example [14] for the proof).

J. Andersen and S. Hansen proved for the figure-eight knot $4_{1}$ the following asymptotic equivalence holds [1].

[TABLE]

where $\tau(4_{1}):=\left(\frac{2}{\sqrt{-3}}\right)^{1/2}$ and $S(4_{1}):=\sqrt{-1}\operatorname{Vol}(S^{3}\setminus{4_{1}})$. It is known that $\tau(4_{1})^{-2}$ is the homological Reidemeister tosion twisted by the holonomy representation of $\pi_{1}(S^{3}\setminus{4_{1}})$ associated with the meridian, and $S(4_{1})$ is the $\mathrm{SL}(2;\mathbb{C})$ Reidemeister torsion of the holonomy representation. It is also conjectured that for any hyperbolic knot $K$, we have

[TABLE]

where $\tau(K)^{-2}$ and $S(K)$ are defined as above. See [4] and [2]. See also [15, 16].

Let $T(c,d)$ be the torus knot of type $(c,d)$ for coprime integers $c$ and $d$. J. Dubois and Kashaev [3] obtained the following formula:

[TABLE]

where

[TABLE]

See also [5] for the formulation above. They also show that $S(k)$ is the Chern–Simons invariant and $\tau(k)^{-2}$ is the homological twisted Reidemeister torsion both associated with suitable irreducible representation from $\pi_{1}(S^{3}\setminus{T(c,d)})$ to $\mathrm{SL}(2,\mathbb{C})$. See also [8].

Remark 1.2*.*

If we define

[TABLE]

the right hand side of (1.1) becomes

[TABLE]

Note that the Chern–Simons invariant is defined modulo $\pi^{2}\mathbb{Z}$ and that $S(k)\equiv\tilde{S}(k)\pmod{\pi^{2}\mathbb{Z}}$.

In [11] and [12], the first author obtained a similar asymptotic formula for J_{N}\bigl{(}T(2,2a+1)^{(2,2b+1)};\exp(\xi/N)\bigr{)} with $\xi\neq 2\pi i$, where $T(2,2a+1)^{(2,2b+1)}$ is the $(2,2b+1)$-cable of $T(2,2a+1)$. The purpose of this paper is to give an asymptotic formula for J_{N}\bigl{(}T(2,2a+1)^{(2,2b+1)};\exp(2\pi i/N)\bigr{)}.

Theorem 1.3**.**

If $2b+1>4(2a+1)>0$, then we have

[TABLE]

where

[TABLE]

and $\mathcal{B}$ is the set of all pairs of integers $(j,k)$ such that $0\leq k\leq 4a+1$, $0\leq j\leq 2b-4(2a+1)$, and $(2b+1-4(2a+1))(2k+1)<2(2a+1)(2j+1)$.

We can also prove that $\tau_{1}(l)^{-2}$, $\tau_{2}(m)^{-2}$, and $\tau_{3}(l,m)^{-2}$ are the homological twisted Reidemeister torsions of certain representations of the fundamental group to $\rm{SL}(2;\mathbb{C})$, and that $S_{1}(l)$, $S_{2}(m)$, and $S_{3}(l,m)$ are the Chern–Simons invariants of these representations.

2. Proof of the asymptotic formula

We will follow [10, Section 5.2]. Let $e_{n}$ be the $n$-th Jones–Wenzl idempotent in the Kauffman bracket skein algebra of an annulus defined by $z_{1}e_{n}=e_{n-1}+e_{n+1}$ with $e_{0}=1,e_{1}=z_{1}$, where $z_{1}$ is the circle around the annulus. Let $\langle e_{n}\rangle_{\mathcal{K}^{\sigma}}$ be the Kauffman bracket of the element obtained from a framed knot $\mathcal{K}^{\sigma}$ by replacing a diagram of $\mathcal{K}^{\sigma}$ with $e_{n}$, where $\sigma$ is the framing. If $\mathcal{T}^{2(2b+1)}$ is the $(2,2b+1)$-cable of the torus knot $T(2,2a+1)$ with framing $2(2b+1)$, then in [10, Proposition 4] Q. Liu proved

[TABLE]

where

[TABLE]

Since the colored Jones polynomial $J_{N}(K;q)$ is normalized so that $J_{N}(U;q)=1$ with $U$ the unknot, if $K$ is the knot obtained from a framed knot $\mathcal{K}^{\sigma}$ by forgetting the framing, we have

[TABLE]

where $\mathcal{K}^{0}$ is the [math]-framed knot obtained from $\mathcal{K}^{\sigma}$ by changing the framing and $\mathcal{U}^{0}$ is the framed unknot with framing [math]. Therefore we have

[TABLE]

where

[TABLE]

Remark 2.1*.*

We need to multiply by $(-1)^{N-1}\left((-1)^{N-1}\exp(\frac{(N^{2}-1)\pi i}{2N})\right)^{-2(2b+1)}$. The first $(-1)^{N-1}$ is because Liu normalized the colored Jones polynomial by dividing the Kauffman bracket by $\frac{q^{N/2}-q^{-N/2}}{q^{1/2}-q^{-1/2}}$ but we need to divide it by $(-1)^{N-1}\frac{q^{N/2}-q^{-N/2}}{q^{1/2}-q^{-1/2}}$, which is the Kauffman bracket for the unknot. The second one is because Liu’s formula is for a framed knot with framing $2(2b+1)$.

Remark 2.2*.*

1. (1)

$\theta(z_{1},z_{2})$ has a unique critical point

[TABLE] 2. (2)

The poles of $\psi_{1}(z_{1})$ between $C_{\pi/4}$ and $C_{\pi/4}+w_{1}$ are $\xi_{l}=\frac{2l+1}{4}\pi i$ ($0\leq l\leq 2b$), where $C_{\pi/4}+w_{1}$ is the line passing through $w_{1}$ that is parallel to $C_{\pi/4}$. Moreover, we have

[TABLE] 3. (3)

The poles of $\psi_{2}(z_{2})$ between $C_{\pi/4}$ and $C_{\pi/4}+w_{2}$ are $\eta_{m}=\frac{2m+1}{2}\pi i$ ($0\leq m\leq 4a+1$), where $C_{\pi/4}+w_{2}$ is the line passing through $w_{2}$ that is parallel to $C_{\pi/4}$. Moreover, we have

[TABLE]

Put

[TABLE]

Then $e^{-\frac{\pi i}{4N}\left(2b+1-3(2a+1)+\frac{4}{2a+1}\right)}\,J_{N}(K,e^{2\pi i/N})=\mathcal{J}_{N}$.

By shifting the paths of integrations from $C_{\pi/4}$ to $C_{\pi/4}+w_{1}$ for the integration with respect to $z_{1}$, and from $C_{\pi/4}$ to $C_{\pi/4}+w_{2}$ for the integration with respect to $z_{2}$, we have

[TABLE]

with

[TABLE]

Put $\mathcal{J}_{k,N}:=(-1)^{N-1}\left(\frac{2N^{2}i}{\sqrt{(2a+1)(2b+1-4(2a+1))}\,\pi^{4}}\right)I_{k,N}$ so that

[TABLE]

2.1. $I_{3,N}$

In this subsection we calculate $I_{3,N}$.

Since $\xi_{l}=\frac{2l+1}{4}\pi i$ and $\eta_{m}=\frac{2m+1}{2}\pi i$, we have

[TABLE]

where we put

[TABLE]

Hence $\theta(\xi_{l},\eta_{m})=-(\pi i)^{-1}\delta(\xi_{l},\eta_{m})+4\xi_{l}=\frac{1}{2\pi i}\tilde{S}_{3}(l,m)+(2l+1)\pi i$ and

[TABLE]

Since $\operatorname{Res}(\psi_{1},\xi_{l})\operatorname{Res}(\psi_{2},\eta_{m})=(-1)^{l+m-1}\,\frac{i}{8}\sin(\frac{(2m+1)\pi}{2a+1})$, we obtain

[TABLE]

Hence

[TABLE]

2.2. $I_{2,N}$

Next we calculate $I_{2,N}$.

Consider the integral

[TABLE]

The polynomial $\theta(z_{1},\eta_{m})=-\frac{\eta_{m}^{2}}{(2a+1)\pi i}-\frac{4(z_{1}-\eta_{m})^{2}}{(2b+1-4(2a+1))\pi i}+4z_{1}$ has a unique critical point

[TABLE]

and

[TABLE]

Note that $\zeta_{m}$ is not a pole of $\psi_{1}(z_{1})$ for any integer $m$.

We have

[TABLE]

Note that the poles of $\psi(z_{1})$ between $C_{\pi/4}+w_{1}$ and $C_{\pi/4}+\zeta_{m}$ are $\xi_{l^{\prime}}=\frac{2l^{\prime}+1}{r}\pi i$ ($(2b+1)-4(2a+1)+(2m+1)\leq l^{\prime}\leq 2b$).

Since $F_{N}(z_{1},\eta_{m})=e^{N\theta(\zeta_{m},\eta_{m})}\,\delta(z_{1},\eta_{m})\,e^{-\frac{4N}{(2b+1-4(2a+1))\pi i}(z_{1}-\zeta_{m})^{2}}$ we have

[TABLE]

By the saddle point method (see, for example, [10, Lemma 1]) we have

[TABLE]

Hence

[TABLE]

Since $\zeta_{m}=\frac{(2b+1)-4(2a+1)+(2m+1)}{2}\pi i$ and $\eta_{m}=\frac{2m+1}{2}\pi i$ we have

[TABLE]

Hence

[TABLE]

and

[TABLE]

Since $\psi_{1}(\zeta_{m})=\frac{1}{2\cosh(2\zeta_{m})}=\frac{1}{2}$, we obtain

[TABLE]

From (2.1) we have

[TABLE]

Since $S_{2}(m+2a+1)=S_{2}(m)+4(a+m+1)\pi^{2}$, we have

[TABLE]

Therefore we have

[TABLE]

and so

[TABLE]

2.3. $I_{1,N}$

Now we calculate $I_{1,N}$.

Consider the integral

[TABLE]

The polynomial $\theta(\xi_{l},z_{2})=-\frac{z_{2}^{2}}{(2a+1)\pi i}-\frac{4(\xi_{l}-z_{2})^{2}}{(2b+1-4(2a+1))\pi i}+4\xi_{l}$ has a unique critical point

[TABLE]

and

[TABLE]

Note that ${\zeta^{\prime}}_{l}$ is not a pole of $\psi_{2}(z_{2})$ for any $l$.

We have

[TABLE]

Note that the pole of $\psi_{2}(z_{2})$ between $C_{\pi/4}+w_{2}$ and $C_{\pi/4}+{\zeta^{\prime}}_{l}$ are $\eta_{m^{\prime}}$ ($\frac{(2a+1)(2l+1)}{2b+1}-\frac{1}{2}<m^{\prime}<4a+\frac{3}{2}$, that is, $\lfloor\frac{(2a+1)(2l+1)}{2b+1}+\frac{1}{2}\rfloor\leq m^{\prime}\leq 4a+1$).

By the same argument as in the previous case, we have

[TABLE]

Since $\xi_{l}=\frac{2l+1}{4}\pi i$ and ${\zeta^{\prime}}_{l}=\frac{(2a+1)(2l+1)}{2b+1}\pi i$ we have

[TABLE]

Hence $\theta(\xi_{l},{\zeta^{\prime}}_{l})=-(\pi i)^{-1}\delta(\xi_{l},{\zeta^{\prime}}_{l})+4\xi_{l}=\frac{1}{2\pi i}S_{1}(l)+(2l+1)\pi i$ and

[TABLE]

Since

[TABLE]

we obtain

[TABLE]

From (2.1) we have

[TABLE]

Hence

[TABLE]

2.4. $I_{0,N}$

In this subsection we calculate $I_{0,N}$.

We have

[TABLE]

Note that $\psi_{1}(z_{1}+w_{1})=-\psi_{1}(z_{1})$ and $\psi_{2}(z_{2}+w_{2})=\psi_{2}(z_{2})$. Moreover, we have

[TABLE]

and

[TABLE]

Hence

[TABLE]

Since $\psi_{1}(z_{1})\psi_{2}(z_{2})$ is an odd function in $(z_{1},z_{2})$ and $\delta(z_{1},z_{2})$ is even, we obtain

[TABLE]

From the saddle point method (see, for example, [10, Lemma 3]), we have

[TABLE]

Hence

[TABLE]

Finally we have

[TABLE]

The sum of the three double summations becomes

[TABLE]

where

[TABLE]

Putting $j:=l-(2m+1)$ and replacing $m$ with $k$, the summation above becomes

[TABLE]

where

[TABLE]

and

[TABLE]

The theorem follows.

3. Topological interpresentation

In this section we give a toplogical interpretation of the right hand side of (1.2).

3.1. Fundamental group

We calculate the fundamental group of the complement of the twice-iterated torus knot $T(2,2a+1)^{(2,2b+1)}$.

Put $E:=S^{3}\setminus\operatorname{Int}{N\left(T(2,2a+1)^{(2,2b+1)}\right)}$. Then $E$ can be decomposed into C:=S^{3}\setminus\operatorname{Int}N\bigl{(}T(2,2a+1)\bigr{)} (Figure 1) and $D$, where $N$ meas the regular neighborhood, $\operatorname{Int}$ means the interior, and $D$ is the complement of $(2,2b+1)$ torus knot in the solid torus (Figure 2).

Note that $D$ is homeomorphic to S^{3}\setminus\operatorname{Int}N\bigl{(}T(2,4)\bigr{)} (see (3.1)).

[TABLE]

If we choose $x$ and $y$ as generators of $\pi_{1}(C)$ as indicated in Figure 1, we have

[TABLE]

where we choose the basepoint on the boundary of N\bigl{(}T(2,2a+1)\bigr{)}. Let $p$ and $t$ be generators of \pi_{1}\left(S^{3}\setminus\operatorname{Int}{N\bigl{(}T(2,4)\bigr{)}}\right) as in Figure 3. Then we have

[TABLE]

where the basepoint is at the bottom-right of the torus.

Let $q\in\pi_{1}(D)$ be the element as indicated in Figure 4. Then we see that $q=tpt^{-1}$

From van Kampen’s theorem we have

[TABLE]

Moreover we can see that the meridian $\mu$ and the preferred longitude $\lambda$ of $T(2,2a+1)^{(2,2b+1)}$ are given as follows:

[TABLE]

3.2. Representation

In this subsection, we construct non-Abelian representations from $\pi_{1}(E)$ to $\mathrm{SL}(2;\mathbb{C})$. Note that we do not know whether we exhaust all such representations.

Put $\omega_{1}:=\exp\left(\frac{\pi i}{2b+1}\right)$, $\omega_{2}:=\exp\left(\frac{\pi i}{2a+1}\right)$, and $\omega_{3}:=\exp\left(\frac{\pi i}{2b+1-4(2a+1)}\right)$. Let $\rho^{\rm{AN}}_{u;j}$ be the representation $\pi_{1}(E)\to\mathrm{SL}(2;\mathbb{C})$ defined by

[TABLE]

where

[TABLE]

and $l=0,1,\dots,b-1$. Note that

[TABLE]

and that $\rho^{\rm{AN}}_{u,2b-l}$ is conjugate to $\rho^{\rm{AN}}_{u,l}$ by

[TABLE]

The longitude $\lambda$ is sent to

[TABLE]

Let $\rho^{\rm{NA}}_{u;m}$ be the representation with

[TABLE]

with $m=0,1,\dots,a-1$. The longitude $\lambda$ is sent to

[TABLE]

Note the following symmetries:

[TABLE]

Let $\rho^{\rm{NN}}_{u;j,k}$ be the representation with

[TABLE]

where

[TABLE]

$=0,1,\dots,2b-4(2a+1)$, and $k=0,1,\dots,a-1$. The longitude $\lambda$ is sent to

[TABLE]

Note the following symmetries:

[TABLE]

Remark 3.1*.*

Note that $\rho^{\rm{NN}}_{0;j,k}$ and $\rho^{\rm{NN}}_{0;2b-4(2a+1)-j,k}$ are not conjugate because $\operatorname{Tr}\rho^{\rm{NN}}_{0;j,k}(py)=-(\omega_{3}^{2j+1}-1)^{2}+\omega_{2}^{2k+1}+\omega_{2}^{-(2k+1)}$.

3.3. Chern–Simons invariant

For a knot $K$ in $S^{3}$, let $M$ be the complement of the interior of the regular neighborhood of $K$. Denote by $\mu$ and $\lambda$ the meridian and the preferred longitude of $M$, respectively. By a conjugation we may assume that a representation $\rho\colon M\to\mathrm{SL}(2;\mathbb{Z})$ sends $\mu$ and $\lambda$ to

[TABLE]

respectively. The $\mathrm{SL}(2;\mathbb{C})$ Chern–Simons invariant is a map from $X(M)$ to $\mathbb{C}$ modulo $\pi^{2}\mathbb{Z}$, where $X(M)$ is the $\mathrm{SL}(2;\mathbb{C})$ character variety of $M$. Note that we need to fix log branches of the eigenvalues $e^{u/2}$ and $e^{v/2}$. See [9] for details.

In [11, § 2.5] the first author proved that the Chern–Simons invariants of $\rho^{\rm{AN}}_{u;l}$, $\rho^{\rm{NA}}_{u;m}$, $\rho^{\rm{NN}}_{u;j,k}$ are given as follows.

Theorem 3.2** ([11]).**

Let $\rho^{\rm{AN}}_{u;l}$, $\rho^{\rm{NA}}_{u;m}$, and $\rho^{\rm{NN}}_{u;j,k}$ be representations given in Subsection 3.2. Then we have

[TABLE]

for an odd integer $d_{1}$, and integers $d_{2}$ and $d_{3}$. Here we choose $-2(2b+1)u+2d_{1}\pi i$, $-8(2a+1)u+4d_{2}\pi i$, and $-2(2b+1)u+2d_{3}\pi i$ as lifts of the eigenvalues of $\rho^{\rm{AN}}_{u;l}(\lambda)$, $\rho^{\rm{NA}}_{u;m}(\lambda)$, and $\rho^{\rm{NN}}_{u;j,k}(\lambda)$, respectively.

Therefore we conclude

[TABLE]

3.4. Reidemeister torsion

Let $M$ be the complement of the interior of the regular neighborhood of a knot, and $\tilde{M}$ be its universal covering space. Then the chain group $C_{j}(\tilde{M};\mathbb{Z})$ can be regarded as a $\mathbb{Z}[\pi_{1}(M)]$-module. Given a representation $\rho\colon\pi_{1}(M)\to\mathrm{SL}(2;\mathbb{C})$, we can also regard the Lie algebra $\mathfrak{sl}(2;\mathbb{C})$ as a $\mathbb{Z}[\pi_{1}(M)]$-module by the adjoint action. So we can define the tensor product $C_{j}(\tilde{M};\mathbb{Z})\otimes_{\mathbb{Z}[\pi_{1}(M)]}\mathfrak{sl}(2;\mathbb{C})$ and denote it by $C_{j}(M;\rho)$. The Reidemeister torsion of the corresponding chain complex $\{C_{j}(M;\rho),\partial_{j}\}$ is denoted by $\mathbb{T}(M;\rho)$ and called the homological twisted Reidemeister torsion of $M$ associated with $\rho$. Note that we need to specify bases of $H_{j}(M;\rho)$ unless $\{C_{j}(M;\rho),\partial_{j}\}$ is acyclic, where $H_{j}(M;\rho)$ is the $j$-th homology group of the chain complex $\{C_{j}(M;\rho),\partial_{j}\}$.

In [12], the first author calculated the homplogical twisted Reidemeister torsions of $S^{3}\setminus\operatorname{Int}{N\left(T(2,2a+1)^{(2,2b+1)}\right)}$ associated with $\rho^{\rm{AN}}_{u,l}$, $\rho^{\rm{NA}}_{u;m}$ and $\rho^{\rm{NN}}_{u;j,k}$.

Theorem 3.3** ([12]).**

Put $M:=S^{3}\setminus\operatorname{Int}{N\left(T(2,2a+1)^{(2,2b+1)}\right)}$. The homological twisted Reidemeister torsions of $M$ associated with the representations defined in Subsection 3.2 are given as follows.

[TABLE]

Note that since we have

[TABLE]

we need to specify bases of non-trivial homology groups. See [12] for details.