On the Quantum SU(2) Invariant at $q\,{\hbox {=}}\,\exp (4\pi \sqrt{-1}/N)$ and the Twisted Reidemeister Torsion for Some Closed 3-Manifolds

Abstract

The perturbative expansion of the Chern–Simons path integral predicts a formula of the asymptotic expansion of the quantum invariant of a 3-manifold. When $q=\exp (2 \pi \sqrt{-1}/N)$, there have been some researches where the asymptotic expansion of the quantum $\mathrm{SU}(2)$ invariant is presented by a sum of contributions from $\mathrm{SU}(2)$ flat connections whose coefficients are square roots of the Reidemeister torsions. When $q=\exp (4 \pi \sqrt{-1}/N)$, it is conjectured recently that the quantum $\mathrm{SU}(2)$ invariant of a closed hyperbolic 3-manifold M is of exponential order of N whose growth is given by the complex volume of M. The first author showed in the previous work that this conjecture holds for the hyperbolic 3-manifold $M_p$ obtained from $S^3$ by p surgery along the figure-eight knot. From the physical viewpoint, we use the (formal) saddle point method when $q=\exp (4 \pi \sqrt{-1}/N)$, while we have used the stationary phase method when $q=\exp ({2 \pi \sqrt{-1}}/N)$, and these two methods give quite different resulting formulas from the mathematical viewpoint. In this paper, we show that a square root of the Reidemeister torsion appears as a coefficient in the semi-classical approximation of the asymptotic expansion of the quantum $\mathrm{SU}(2)$ invariant of $M_p$ at $q = \exp (4 \pi \sqrt{-1}/N)$. Further, when $q = \exp (4 \pi \sqrt{-1}/N)$, we show that the semi-classical approximation of the asymptotic expansion of the quantum $\mathrm{SU}(2)$ invariant of some Seifert 3-manifolds M is presented by a sum of contributions from some of $\mathrm{SL}_2 {\mathbb C}$ flat connections on M, and square roots of the Reidemeister torsions appear as coefficients of such contributions.

Appendices

The Quantum SU(2) and SO(3) Invariants

In this section, we review the definition and basic properties of the quantum $\mathrm{SU}(2)$ and $\mathrm{SO}(3)$ invariants. We recall that N is an odd integer $\ge 3$, and we put $A = e^{\frac{\pi \sqrt{-1}}{N}}$, $a = e^{\frac{{2 \pi \sqrt{-1}}}{N}}$ and $[n] = \frac{a^n-a^{-n}}{a-a^{-1}}$.

In Sect. A.1, we review the definition of the quantum $\mathrm{SU}(2)$ and $\mathrm{SO}(3)$ invariants, and gives a formula of the quantum $\mathrm{SO}(3)$ invariant of the Seifert 3-manifold $M_{p_1,p_2}$. In Sect. A.2, we review the equivalence between the quantum $\mathrm{SU}(2)$ and $\mathrm{SO}(3)$ invariants at $q = e^{\frac{4\pi \sqrt{-1}}{N}}$.

1.1 Review of the quantum SU(2) and SO(3) invariants

In this section, we briefly review the definition of the quantum $\mathrm{SU}(2)$ and $\mathrm{SO}(3)$ invariant of [19, 39] at $q=e^{\frac{{4\pi \sqrt{-1}}}{N}}$, following the construction of [25]; we note that these two invariants are equivalent as we see in Sect. A.2 later. Further, we give a formula of the quantum $\mathrm{SO}(3)$ invariant of the Seifert 3-manifold $M_{p_1,p_2}$, which we use in Sect. 4.2.

We briefly review the definition of the quantum $\mathrm{SU}(2)$ and $\mathrm{SO}(3)$ invariant at $q=e^{\frac{{4\pi \sqrt{-1}}}{N}}$ following the construction of Lickorish [25], noting that we put $A = e^{\frac{\pi \sqrt{-1}}{N}}$ in this paper unlike the usual case where A is a primitive 4N-th root of unity. In order to state the symmetry principle of our case later in Sect. A.2, we use the notation of [19, 39], that is, the notation in the right-hand side of the following formula,

where the left-hand side denotes the box of the Jones-Wenzl idempotent in the Lickorish notation, and the right-hand side is a strand associated with the n-dimensional irreducible representation of $U_q({\mathfrak {s}}{\mathfrak {l}}_2)$. It is known, see [17, 25], that

(38)

For an $\ell$-component framed link L whose components are associated with $V_{n_1}, \cdots , V_{n_\ell }$, we denote by $Q(L; \, V_{n_1}, \cdots , V_{n_\ell })$ the invariant of L defined by the linear skein of Lickorish [25]. Let M be the 3-manifold obtained from $S^3$ by surgery along L. Then, the quantum$\mathrm{SU}(2)$invariant$\hat{\tau }_{{}_N}^{\mathrm{SU(2)}}(M)$ and the quantum$\mathrm{SO}(3)$invariant$\hat{\tau }_{{}_N}(M)$ are defined by

$$\begin{aligned} \begin{aligned} \hat{\tau }_{{}_N}^{\mathrm {SU(2)}}(M)&= {c'_+}^{-\sigma _+} {c'_-}^{-\sigma _-} \sum _{1 \le n_1,\cdots ,n_\ell< N} \big ( (-1)^{n_1-1}[n_1] \big ) \cdots \big ( (-1)^{n_\ell -1}[n_\ell ] \big ) \, \\ {}&\qquad \qquad \qquad \qquad \qquad \times Q(L; \, V_{n_1}, \cdots , V_{n_\ell }) ,\\ \hat{\tau }_{{}_N}(M)&= c_+^{\,\,-\sigma _+} c_-^{\,\,-\sigma _-} \sum _ {\begin{array}{c} 1 \le n_1,\cdots ,n_\ell < N \\ n_1,\cdots ,n_\ell \mathrm {\ are \ odd} \end{array}} [n_1] \cdots [n_\ell ] \,\, Q(L; \, V_{n_1}, \cdots , V_{n_\ell }) , \end{aligned} \end{aligned}$$

where $\sigma _+$ and $\sigma _-$ are the numbers of the positive and negative eigenvalues of the linking matrix of L, and $c'_{\pm }$ and $c_{\pm }$ are constants given by

$$\begin{aligned} c'_{\pm } = \sum _{1 \,\le \, n \,<\, N} (-1)^{n-1} A^{\pm (n^2-1)} [n]^2 , \qquad c_{\pm } = \sum _{\begin{array}{c} 1 \,\le \, n \,<\, N \\ n \mathrm{\ is \ odd} \end{array}} A^{\pm (n^2-1)} [n]^2 . \end{aligned}$$

We calculate the value of the constant $c_+$, by using Lemma 4.1, as follows,

$$\begin{aligned} c_+&= \sum _{\begin{array}{c} 1 \,\le \, n \,<\, N \\ n \mathrm{\ is \ odd} \end{array}} A^{n^2-1} [n]^2 = \frac{-A^{-1}}{(a-a^{-1})^2} \sum _{j \,\in \, {\mathbb Z}/N{\mathbb Z}} a^{2 j^2} \big ( a^{4 j} -1 \big ) \nonumber \\&= \frac{-A^{-1}(a^{-2}-1)}{(a-a^{-1})^2} \sum _{j \,\in \, {\mathbb Z}/N{\mathbb Z}} a^{2 j^2} = \frac{A^{-3}}{a-a^{-1}} \sum _{j \,\in \, {\mathbb Z}/N{\mathbb Z}} a^{2 j^2} \nonumber \\&= \frac{A^{-3}}{a-a^{-1}} \big ( -\sqrt{-1}\big )^{\frac{N-1}{2}} \sqrt{N} , \end{aligned}$$

(39)

where we obtain the last equality by [34, Lemma A.1]. The constant $c_-$ is the complex conjugate of $c_+$,

$$\begin{aligned} c_- = \frac{A^{3}}{a^{-1}-a} \sqrt{-1}^{\frac{N-1}{2}} \sqrt{N} . \end{aligned}$$

Hence,

$$\begin{aligned} \frac{c_+}{c_-} = A^{-6} (-1)^{\frac{N+1}{2}} . \end{aligned}$$

(40)

We give a formula of the quantum $\mathrm{SO}(3)$ invariant of the Seifert 3-manifold given in Sect. 4.2. Let L be the framed link in (26). When i, j, k, m are odd, it follows from (38) that the invariant of L is given by

$$\begin{aligned} Q(L; \, V_i,V_j,V_k,V_m) = A^{2(i^2-1)+p_1(j^2-1)+p_2(k^2-1)+(m^2-1)} \, \frac{[m i][m j][m k]}{[m]^2} \, . \end{aligned}$$

Hence, the quantum $\mathrm{SO}(3)$ invariant of $M_{p_1,p_2}$ is given by

$$\begin{aligned} \hat{\tau }_{{}_N}(M_{p_1,p_2})=&{} c_+^{\,\, -\sigma _+} c_-^{\,\, -\sigma _-} \sum _{\begin{array}{c} 1 \,\le \, i,j,k,m \,<\, N \\ i,j,k,m \mathrm {\ are \ odd} \end{array}} A^{2(i^2-1)+p_1(j^2-1)+p_2(k^2-1)+(m^2-1)} \, \nonumber \\ {}&\qquad \qquad \qquad \qquad \times \frac{[m i][m j][m k]}{[m]} \, [i][j][k] , \end{aligned}$$

(41)

where $\sigma _+$ and $\sigma _-$ are the numbers of the positive and negative eigenvalues of the linking matrix of L.

1.2 Equivalence of the quantum SU(2) and SO(3) invariants

It is known that the quantum $\mathrm{SU}(2)$ and $\mathrm{SO}(3)$ invariants are equivalent at $q = e^{\frac{{4\pi \sqrt{-1}}}{N}}$ in the sense of Proposition A.2, unlike the usual case where $q = e^{\frac{2\pi \sqrt{-1}}{N}}$. This equivalence is mentioned in the first paragraph of Section 1 of [26]. In this section, we briefly review this equivalence.

Before we review this equivalence, we review the symmetry principle at $q = e^{\frac{{4\pi \sqrt{-1}}}{N}}$ in the following proposition.

Proposition A.1

When $A = e^{\frac{\pi \sqrt{-1}}{N}}$, for an $\ell$-component framed link L,

$$\begin{aligned} Q(L; \, V_{n_1}, V_{n_2}, \ldots , V_{n_\ell }) = Q(L; \, V_{N-n_1}, V_{n_2}, \ldots , V_{n_\ell }) . \end{aligned}$$

Proof

The symmetry principle is shown in [24, Proposition 9] when A is a primitive 4N-th root of unity. When $A = e^{\frac{\pi \sqrt{-1}}{N}}$, we can show the symmetry principle in the same way as in the proof of [24, Proposition 9]. As shown in the proof of [24, Proposition 9], we can reduce the proof to the case where L is a framed Hopf link of the following form.

In this case, it follows from (38) that the invariants of $H_{f,0}$ are given by

$$\begin{aligned}&Q(H_{f,0}; \, V_n, V_m) = \big ( (-1)^{n-1} A^{n^2-1} \big )^f \, (-1)^{n+m} [n m] , \\&Q(H_{f,0}; \, V_{N-n}, V_m) = \big ( (-1)^{N-n-1} A^{(N-n)^2-1} \big )^f \, (-1)^{N-n+m} [(N - n) m] , \end{aligned}$$

and it is sufficient to show that these values are equal.

In fact, these values are equal, since

$$\begin{aligned} (-1)^{n-1} A^{n^2-1}= & {} (-1)^{N-n-1} A^{(N-n)^2-1} \ \ \\&\hbox { and } \ \ (-1)^{n+m} [n m] \, = \, (-1)^{N-n+m} [(N - n) m] , \end{aligned}$$

when $A = e^{\frac{\pi \sqrt{-1}}{N}}$. Therefore, we obtain the proposition. $\quad \square$

By using the symmetry principle (the above proposition), we can obtain the following proposition.

Proposition A.2

(see [26]). When $A = e^{\frac{\pi \sqrt{-1}}{N}}$, the quantum $\mathrm{SU}(2)$ and $\mathrm{SO}(3)$ invariants of a closed 3-manifold M are related by

$$\begin{aligned} \hat{\tau }_{{}_N}^{\,\mathrm SU(2)}(M) = 2^{b_1(M)} \, \hat{\tau }_{{}_N}(M) , \end{aligned}$$

where $b_1(M)$ denotes the first Betti number of M.

Proof

We have that

$$\begin{aligned} c'_{\pm }&= \sum _{\begin{array}{c} 1 \,\le \, n \,<\, N \\ n \mathrm{\ is \ odd} \end{array}} A^{\pm (n^2-1)} [n]^2 + \sum _{\begin{array}{c} 1 \,\le \, m \,<\, N \\ m \mathrm{\ is \ even} \end{array}} (-1) A^{\pm (m^2-1)} [m]^2 \\&= \sum _{\begin{array}{c} 1 \,\le \, n \,<\, N \\ n \mathrm{\ is \ odd} \end{array}} A^{\pm (n^2-1)} [n]^2 + \sum _{\begin{array}{c} 1 \,\le \, m \,<\, N \\ m \mathrm{\ is \ odd} \end{array}} A^{\pm (m^2-1)} [m]^2 = 2 \, c_{\pm } \, , \end{aligned}$$

where we obtain the second equality by replacing m with $N - m$. In a similar way, by Proposition A.1, we can obtain that

$$\begin{aligned}&\sum _{1 \le n_1,\ldots ,n_\ell< N} \big ( (-1)^{n_1-1}[n_1] \big ) \cdots \big ( (-1)^{n_\ell -1}[n_\ell ] \big ) \, Q(L; \, V_{n_1}, \ldots , V_{n_\ell }) \\&= 2^{\ell } \sum _{\begin{array}{c} 1 \le n_1,\ldots ,n_\ell < N \\ n_1,\cdots ,n_\ell \mathrm{\ are \ odd} \end{array}} [n_1] \cdots [n_\ell ] \,\, Q(L; \, V_{n_1}, \ldots , V_{n_\ell }) . \end{aligned}$$

Therefore, since $\ell = \sigma _+ + \sigma _- + b_1(M)$, we obtain the proposition from the definitions of the quantum $\mathrm{SU}(2)$ and $\mathrm{SO}(3)$ invariants. $\quad \square$

The Case of Lens Spaces L(p, 1) for Odd p

In this section, we calculate the semi-classical limit of the quantum invariant of the lens space L(p, 1) for odd $p \ge 3$. We recall that $A = e^{\frac{\pi \sqrt{-1}}{N}}$, $a = e^{\frac{{2 \pi \sqrt{-1}}}{N}}$ and $[n] = \frac{a^n-a^{-n}}{a-a^{-1}}$.

Let p be an odd integer $\ge 3$. We denote by L(p, 1) the lens space obtained from $S^3$ by p surgery along the trivial knot; we note that this L(p, 1) has the opposite orientation of L(p, 1) of [15] because of the difference of the convention of an orientation of a lens space.

By definition, the quantum invariant of L(p, 1) is presented by

$$\begin{aligned} \begin{aligned} \hat{\tau }_{{}_N}\big ( L(p,1) \big )&= \frac{1}{c_+} \sum _{\begin{array}{c} 1 \le m < N \\ m \mathrm {\ is \ odd} \end{array}} A^{p(m^2-1)} [m]^2 \\ {}&= \frac{-A^{-p}}{c_+ \, (a-a^{-1})^2} \sum _{j \,\in \, {\mathbb Z}/N {\mathbb Z}} a^{2 p j^2} (a^{4j} -1) . \end{aligned} \end{aligned}$$

By using (24) with $M=4p$ and $\ell = 0,4$, the last sum is calculated as

$$\begin{aligned} \sum _{j \,\in \, {\mathbb Z}/N {\mathbb Z}} a^{2 p j^2} (a^{4j} -1)&= \frac{\sqrt{N}}{2 \sqrt{p}} \, e^{\frac{\pi \sqrt{-1}}{4}} \sum _{m \,\in \, {\mathbb Z}/4 p {\mathbb Z}} \big ( e^{-\frac{\pi \sqrt{-1}}{4 p N} (N m +4)^2} - e^{-\frac{\pi \sqrt{-1}}{4 p N} N^2 m^2} \big )\\&\ \sim \ \frac{\sqrt{N}}{2 \sqrt{p}} \, e^{\frac{\pi \sqrt{-1}}{4}} \sum _{m \,\in \, {\mathbb Z}/4 p {\mathbb Z}} e^{-\frac{\pi \sqrt{-1}}{4p} N m^2} \big ( e^{-\frac{2\pi \sqrt{-1}}{p} m} -1 \big ) . \end{aligned}$$

By replacing m with $m+2p$, we can show that the last sum satisfies that

$$\begin{aligned} \begin{aligned}&\sum _{m \,\in \, {\mathbb Z}/4 p {\mathbb Z}} e^{-\frac{\pi \sqrt{-1}}{4p} N m^2} \big ( e^{-\frac{2\pi \sqrt{-1}}{p} m} -1 \big ) \\ {}&= \sum _{m \,\in \, {\mathbb Z}/4 p {\mathbb Z}} (-1)^{m+1} e^{-\frac{\pi \sqrt{-1}}{4p} N m^2} \big ( e^{-\frac{2\pi \sqrt{-1}}{p} m} -1 \big ) . \end{aligned} \end{aligned}$$

Hence, we can restrict this sum for odd m. Therefore, by putting $m=2n+p$, we have that

$$\begin{aligned}&\sum _{j \,\in \, {\mathbb Z}/N {\mathbb Z}} a^{2 p j^2} (a^{4j} -1) \\&\quad \sim \frac{\sqrt{N}}{2 \sqrt{p}} \, e^{\frac{\pi \sqrt{-1}}{4}} \sum _{\begin{array}{c} m \,\in \, {\mathbb Z}/4 p {\mathbb Z}\\ m \mathrm{\ is \ odd} \end{array}} e^{-\frac{\pi \sqrt{-1}}{4p} N m^2} \big ( e^{-\frac{2\pi \sqrt{-1}}{p} m} -1 \big ) \\&\quad = \frac{\sqrt{N}}{2 \sqrt{p}} \, e^{\frac{\pi \sqrt{-1}}{4}} \, e^{-\frac{\pi \sqrt{-1}}{4} p N} \sum _{n \,\in \, {\mathbb Z}/2 p {\mathbb Z}} (-1)^n \, e^{-\frac{\pi \sqrt{-1}}{p} N n^2} \big ( e^{-\frac{4\pi \sqrt{-1}}{p} n} - 1 \big ) \\&\quad = \frac{\sqrt{N}}{\sqrt{p}} \, e^{\frac{\pi \sqrt{-1}}{4}} \, e^{-\frac{\pi \sqrt{-1}}{4} p N} \sum _{n \,\in \, {\mathbb Z}/p {\mathbb Z}} (-1)^n \, e^{-\frac{\pi \sqrt{-1}}{p} N n^2} \big ( e^{-\frac{4\pi \sqrt{-1}}{p} n} - 1 \big ) \\&\quad = \frac{\sqrt{N}}{2 \sqrt{p}} \, e^{\frac{\pi \sqrt{-1}}{4}} \, e^{-\frac{\pi \sqrt{-1}}{4} p N} \sum _{k \,\in \, {\mathbb Z}/p {\mathbb Z}} (-1)^k \, e^{-\frac{\pi \sqrt{-1}}{p} N k^2} \big ( e^{\frac{4\pi \sqrt{-1}}{p} k} - 2 + e^{-\frac{4\pi \sqrt{-1}}{p} k} \big ) \\&\quad = \frac{\sqrt{N}}{2 \sqrt{p}} \, e^{\frac{\pi \sqrt{-1}}{4}} \, e^{-\frac{\pi \sqrt{-1}}{4} p N} \sum _{k \,\in \, {\mathbb Z}/p {\mathbb Z}} (-1)^k \, e^{-\frac{\pi \sqrt{-1}}{p} N k^2} \big ( e^{\frac{2\pi \sqrt{-1}}{p} k} - e^{-\frac{2\pi \sqrt{-1}}{p} k} \big )^2 \\&\quad = \frac{-2\sqrt{N}}{\sqrt{p}} \, e^{\frac{\pi \sqrt{-1}}{4}} \, e^{-\frac{\pi \sqrt{-1}}{4} p N} \sum _{k \,\in \, {\mathbb Z}/p {\mathbb Z}} (-1)^k \, e^{-\frac{\pi \sqrt{-1}}{p} N k^2} \sin ^2 \frac{2 \pi k}{p} \\&\quad = \frac{4\sqrt{N}}{\sqrt{p}} \, e^{\frac{\pi \sqrt{-1}}{4}} \, e^{-\frac{\pi \sqrt{-1}}{4} p N} \sum _{\begin{array}{c} 1 \le k < p \\ k \mathrm{\ is \ odd} \end{array}} e^{-\frac{\pi \sqrt{-1}}{p} N k^2} \sin ^2 \frac{2 \pi k}{p} \, , \end{aligned}$$

where we obtain the fourth line by making the average of the third line for $n=\pm k$, and obtain the last equality by replacing even k with $p-k$. Therefore, by (39), we have that

$$\begin{aligned}&\hat{\tau }_{{}_N}\big ( L(p,1) \big )\ \nonumber \\ {}&\sim \ \frac{-1}{a-a^{-1}} \, e^{\frac{\pi \sqrt{-1}}{2} \frac{N-1}{2}} \, \frac{4}{\sqrt{p}} \, e^{\frac{\pi \sqrt{-1}}{4}} \, e^{-\frac{\pi \sqrt{-1}}{4} p N} \sum _{\begin{array}{c} 1 \le k< p \\ k \mathrm {\ is \ odd} \end{array}} e^{-\frac{\pi \sqrt{-1}}{p} N k^2} \sin ^2 \frac{2 \pi k}{p} \nonumber \\ {}&\ \sim \ \frac{N}{\pi \sqrt{p}} \, e^{\frac{\pi \sqrt{-1}}{2}} \, e^{-\frac{\pi \sqrt{-1}}{2} \frac{p-1}{2} N} \sum _{\begin{array}{c} 1 \le k< p \\ k \mathrm {\ is \ odd} \end{array}} e^{-\frac{\pi \sqrt{-1}}{p} N k^2} \sin ^2 \frac{2 \pi k}{p} \nonumber \\ {}&\ \sim \ \frac{-N}{\pi \sqrt{p}} \, e^{-\frac{\pi \sqrt{-1}}{2} \, \frac{p+1}{2}} \, (-1)^{\frac{p-1}{2} \, \frac{N-1}{2}} \sum _{\begin{array}{c} 1 \le k < p \\ k \mathrm {\ is \ odd} \end{array}} e^{-\pi \sqrt{-1}\, \frac{k^2}{p} \, N} \sin ^2 \frac{2 \pi k}{p} \, , \end{aligned}$$

(42)

where we obtain the second approximation since $a-a^{-1} = 2 \sqrt{-1}\sin \frac{2\pi }{N} \sim \frac{4\pi \sqrt{-1}}{N}$.

Let $\rho _k$ be a representation of $\pi _1 \big ( L(p,1) \big ) = {\mathbb Z}/p{\mathbb Z}$ to $\mathrm{SL}_2 {\mathbb C}$ such that the eigenvalues of $\rho _k(\hbox {generator})$ are $\exp (\pm \, 2 \pi \sqrt{-1}\, \frac{k}{p} )$. Then, we obtain the following proposition from the above formula.

Proposition B.1

Let p be an odd integer $\ge 3$. Then, the quantum invariant $\hat{\tau }_{{}_N}\big ( L(p,1) \big )$ of L(p, 1) for odd N is expanded as $N\rightarrow \infty$ in the following form,

$$\begin{aligned} \hat{\tau }_{{}_N}\big ( L(p,1) \big )\sim&{} - \, e^{-\frac{\pi \sqrt{-1}}{2} \, \frac{p+1}{2}} \, (-1)^{\frac{p-1}{2} \, \frac{N-1}{2}} N\\ {}&\times \sum _{\begin{array}{c} 1 \le k < p \\ k \mathrm {\ is \ odd} \end{array}} e^{\pi \sqrt{-1}\, \mathrm {CS}( L(p,1); \, \mathrm {ad}\circ \rho _k ) \, N} \omega \big ( L(p,1); \, \rho _k \big ), \end{aligned}$$

where we put

$$\begin{aligned} \omega \big ( L(p,1); \, \rho _k \big ) = \frac{1}{\pi \sqrt{p}} \, \sin ^2 \frac{2 \pi k}{p} \, . \end{aligned}$$

Further, we have that

$$\begin{aligned} \omega \big ( L(p,1); \, \rho _k \big )^2 = \pm \, \frac{1}{32 \pi ^2} \, \mathrm{Tor}\big ( L(p,1); \, \mathrm{ad}\circ \rho _k \big ) . \end{aligned}$$

Proof

It is known, see [15], that the Chern–Simons invariant and the twisted Reidemeister torsion of L(p, 1) are given by

$$\begin{aligned}&\mathrm{CS}\big ( L(p,1); \, \mathrm{ad} \!\circ \! \rho _k \big ) = -\frac{k^2}{p} \, ,\\&\mathrm{Tor}\big ( L(p,1); \, \mathrm{ad}\circ \rho _k \big ) = \pm \, \frac{32}{p} \, \sin ^4 \frac{2\pi k}{p} \, . \end{aligned}$$

Hence, by (42), we obtain the proposition. $\quad \square$

Equivalences Between Some Seifert 3-Manifolds

In this appendix, we review some homeomorphisms between different presentations of some Seifert 3-manifolds in Remarks C.1 and C.2 below.

We recall that, for an integer p, we denote by $M_p$ the 3-manifold obtained from $S^3$ by p surgery along the figure-eight knot. We also recall that, for coprime odd integers $p_1,p_2$, we denote by $M_{p_1,p_2}$ the Seifert 3-manifold obtained from $S^3$ by surgery along the framed link (26).

Remark C.1

$M_{p_1,p_2}$ is homeomorphic to $X(-2,p_1,p_2)$ of [41].

Proof

By definition, our $M_{p_1,p_2}$ is given by a surgery presentation the left picture below. Further, as in [41], $X(-2,p_1,p_2)$ is given by a surgery presentation the right picture below. They are related by Kirby moves, as follows.

Hence, $M_{p_1,p_2}$ is homeomorphic to $X(-2,p_1,p_2)$, as required. $\quad \square$

Remark C.2

(see e.g. [18, Remark (5) of Problem 1.77]). $M_{3,7}$ is homeomorphic to $M_{-1}$, which is homeomorphic to $M_1$ with opposite orientation.

Proof

We show a surgery presentation of $M_{3,7}$ is related to a surgery presentation of $M_{-1}$ by Kirby moves, as follows. A surgery presentation of $M_{3,7}$ is given by the left picture below, which is calculated as follows, where “an integer n in a box” means n full twists.

Further, the last picture is calculated by Kirby moves, as follows.

The last picture is a surgery presentation of $M_{-1}$. Hence, $M_{3,7}$ is homeomorphic to $M_{-1}$.

Further, since the figure-eight knot is isotopic to its mirror image, $M_{-1}$ is homeomorphic to $M_1$ with the opposite orientation. Therefore, we obtain the statement of the remark. $\quad \square$