Complexified tetrahedrons, fundamental groups, and volume conjecture for double twist knots

Jun Murakami [email protected] Department of Mathematics, Faculty of Fundamental Science and Engineering, Waseda University, 1-4-1 Ohkubo, Shinjuku-ku, Tokyo, 169-9555, JAPAN

Abstract.

In this paper, the volume conjecture for double twist knots are proved. The main tool is the complexified tetrahedron and the associated $\mathrm{SL}(2,\mathbb{C})$ representation of the fundamental group. A complexified tetrahedron is a version of a truncated or a doubly truncated tetrahedron whose edge lengths and the dihedral angles are complexified. The colored Jones polynomial is expressed in terms of the quantum $6j$ symbol, which corresponds to the complexified tetrahedron.

Key words and phrases:

Double twist knot, hyperbolic structure, volume conjecture, knot group

2020 Mathematics Subject Classification:

57K20,57K32,57M05

This work wassupported by JSPS KAKENHI Grant Numbers JP20H01803, JP20K20881.

Introduction

Let $K$ be a framed knot or link in $S^{3}$ . In the following, knots include links unless otherwise described. Let $V_{N}(K)$ be the colored Jones polynomial of $K$ which corresponds to the $N+1$ dimensional irreducible representation of $\mathcal{U}_{q}(sl_{2})$ . Here $V_{N}(K)$ is normalized to satisfy $V_{N}(\phi)=1$ and $V_{N}(\bigcirc)=-(q^{N+1}-q^{-N-1})/(q-q^{-1})$ for the trivial knot. The parameter $q$ corresponds to $A^{2}$ where $A$ is the parameter used for defining the Kauffman bracket polynomial. Let $J_{N-1}(K)=V_{N-1}(K)/V_{N-1}(\bigcirc)$ where $q=\exp(\pi i/N)$ for $i=\sqrt{-1}$ . The volume conjecture predicts that certain limit of the colored Jones polynomial gives Gromov’s simplicial volume $||S^{3}\setminus K||$ of the complement of $K$ as follows.

Conjecture 1 (Volume conjecture [9]).

For a knot or link $K$ ,

2\,\pi\lim_{N\to\infty}\frac{\left|J_{N-1}(K)\right|}{N}=v_{3}\,||S^{3}\setminus K||

where $v_{3}$ is the hyperbolic volume of the regular ideal tetrahedron.

If $S^{3}\setminus K$ admits the hyperbolic structure, in other words, $K$ is a hyperbolic knot or link, then $v_{3}\,||S^{3}\setminus K||=\mathrm{Vol}(K)$ where $\mathrm{Vol}(K)$ is the hyperbolic volume of $S^{3}\setminus K$ . For hyperbolic knots and links, the following is also conjectured.

Conjecture 2 (Complexified volume conjecture [10]).

For a hyperbolic knot or link $K$ ,

2\,\pi\,\lim_{N\to\infty}\frac{J_{N-1}(K)}{N}=\mathrm{Vol}(K)+\mathrm{CS}(K)\,\sqrt{-1}\quad(\mathrm{mod}\ \pi^{2}\sqrt{-1}\,\mathbb{Z})

where $\mathrm{CS}(K)$ is $2\pi^{2}$ times the Chern-Simons invariant $cs(S^{3}\setminus K)$ , which is a real number between $0$ and $1/2$ .

For prime hyperbolic knots, this conjecture is proved for knots with less than or equal to seven crossings. Here, we prove Conjecture 1 for all hyperbolic double twist knots.

Refer to caption — Figure 1. Knots and links handled in this paper.

Theorem 1.

Let $K$ be a hyperbolic double twist knot. Then the following holds.

2\,\pi\,\lim_{N\to\infty}\frac{J_{N-1}(K)}{N}=\mathrm{Vol}(K)+\mathrm{CS}(K)\,\sqrt{-1}\quad(\mathrm{mod}\ \pi^{2}\sqrt{-1}\,\mathbb{Z}).

Remark 1.

The volume conjecture for hyperbolic knot with crossing number less than or equal to 7 are proved in [15], [17] and [16]. That for the twist knot $T_{p}$ for $p\geq 6$ is proved in [2].

The main tool is the complexified tetrahedron. Volume formulas of hyperbolic tetrahedrons are given in [3], [13] in terms of dihedral angles at edges and in [12] in terms of edge lengths. The formulas in [13] and [12] are based on the volume conjecture for the quantum $6j$ symbol, and they are analytic functions on the parameters. These formulas are also work for truncated tetrahedra as shown in [18] and for doubly truncated tetrahedra as in [8].

\begin{matrix}\begin{matrix}\includegraphics[scale={0.8}]{tetrahedron1}\end{matrix}&&\begin{matrix}\includegraphics[scale={0.8}]{tetrahedron2}\end{matrix}&&\begin{matrix}\includegraphics[scale={0.8}]{tetrahedron3}\end{matrix}\\ \text{usual tetrahedron}&&\text{truncated tetrahedron}&&\text{doubly truncated tetrahedron}\end{matrix}

Figure 2. A usual tetrahedron, a truncated tetrahedron and a doubly truncated tetrahedron. Any face which truncate a vertex is perpendicular to the original three faces of the tetrahedron which are adjacent to the vertex.

Here the length considered to be a real number and the angle considered to be a pure imaginary number. Now let us complexify these numbers of a truncated tetrahedron and a doubly truncated tetrahedron as in Figure 3.

The adjacent edges at an endpoint of the edge are rotated by $\theta_{\ell}$ and then faces (no more planner) glued at the edge are shifted by $\ell_{\theta}$ . Then the angle parameter $i\theta$ is generalized to $\ell_{\theta}+i\theta$ and the length parameter $\ell$ is generalized to $\ell+i\theta_{\ell}$ . After such deformation, the faces of the tetrahedron is no more planner. But, by assigning elements of $\mathrm{PSL}(2,\mathbb{C})$ to the edges of the tetrahedron, we can define the volume of such generalized tetrahedron by considering the fundamental domain of the action by such group elements. For the complexified tetrahedron, the Schläfli differential formula is generalized to the differential equation satisfied by the Neumann-Zagier function.

The difficulty for proving the volume conjecture is to check the condition to apply the saddle point method. The colored Jones polynomial is given by a sum of terms which consists of a product of quantum factorials and some power of $q$ . For the large $N$ case, this sum is converted to an integral, and the range of the sum is changed to the range of the integral. To apply the saddle point method, this range must be wide enough to surround the saddle point. This range corresponds to the range of the sum in the colored Jones polynomial, and to check that the range is wide enouph is very hard for knots with many crossings. Here we reformulate the colored Jones polynomial by using the quantum $6j$ symbol, and is expressed by parameters which also called colors assigned to the edges of one tetrahedron. Then the range for sum is rather simple and it is easy to see that we can apply the saddle point method. The edge parameters correspond to the saddle point are complex numbers, and the corresponding geometric object is the complexified tetrahedron, while the expression of the colored Jones polynomial from the quantum $R$ matrix corresponds to an ideal tetrahedral decomposition of the knot complement.

The new idea of this article is to introduce the complexified tetrahedron which is constructed from the geometric $\mathrm{SL}(2,\mathbb{C})$ representation of the fundamental group of the complement. For the techniques to apply the saddle point method and the Poisson sum formula, we just follow the arguments developed in papers [15, 16, 17] to prove the volume conjecture for hyperbolic knots with small crossing numbers.

The paper organized as follows. In Section 1, we explain the volume conjecture for Borromean rings. In this case, volume conjecture is already solved, but we reconsider it by using the expression of the colored Jones invariant in terms of the quantum $6j$ symbol. In Section 2, we tread twisted Whitehead links. The volume conjecture is also solved for this case, but here reprove it by using the complexified tetrahedron and the quantum $6j$ symbol. For the Whitehead link case, we first use a complexified tetrahedron which appears as a deformation of the regular ideal octahedron. In Section 3, the double twist knots are investigated. The method to prove the volume conjecture is same as for the twisted Whitehead links explained in Section 2.

Some notions and detailed computations are given in appendices. Especially, in Appendix B, colored Jones invariants are reformulated by using the ADO invariants. This part is the most complicated part of this paper, but the reformulation of the colored Jones polynomial explained here simplifies the rest of the proof of the volume conjecture.

Acknowledgment. The author was strongly encouraged to pursuit this research when I attended “Winter School on Low-dimensional Topology and Related Topics” at IBS-CGP in Pohang, Korea in December 2023, and he would like to thank all the participants of the school, especially Jessica Purcell, Seonhwa Kim, Thiago de Paiva Souza, and the organizer Anderson Vera. He also would like to thank Anh Tran for giving me a lot of information about $SL(2,\mathbb{C})$ representations of the double twist knots and two-bridge knots.

1. Borromean rings

The volume conjecture for the Borromean rings is easily proved elementary, but here we recall the proof to see its corresponds to the $\mathrm{PSL}(2,\mathbb{C})$ representation of the fundamental group of the complement. Throughout this paper, $N$ is assumed to be an odd positive integer greater than or equal to $3$ .

1.1. Representation matrix

Let $B$ be the Borromean rings in Figure 1. We first construct the parabolic $\mathrm{SL}(2,\mathbb{C})$ representation $\rho$ of $\pi_{1}(S^{3}\setminus B)$ which corresponds to the hyperbolic structure of $S^{3}\setminus B$ . In other words, let $\Gamma$ be the image of $\rho$ , then $S^{3}\setminus B$ is isomorphic to $\mathbb{H}^{3}/\Gamma$ , where $\mathbb{H}^{3}$ is the hyperbolic three space. Here we use the upper half model, so $\mathbb{H}^{3}$ is identified with $\mathbb{C}\times\mathbb{R}_{>0}$ and $\partial\mathbb{H}^{3}$ is identified with $\mathbb{C}$ . To assign elements of $\pi_{1}(S^{3}\setminus B)$ , we draw $B$ as in Figure 4 and assign the elements $g_{1}$ , $\cdots$ , $g_{4}$ , $h_{1}$ , $h_{2}$ as in the figure.

Then the relations of $\pi_{1}(S^{3}\setminus B)$ are given as follows.

\pi_{1}(S^{3}\setminus B)=\left<g_{1},g_{2},g_{3},g_{4},h_{1},h_{2}\mid g_{2}=h_{1}\,g_{1}^{-1}\,h_{1}^{-1},\ g_{3}=h_{1}\,g_{4}^{-1}\,h_{1}^{-1},\ g_{3}^{-1}=h_{2}\,g_{2}\,h_{2}^{-1}\right>.

(1)

Now let us consider parabolic representation $\rho$ . Here we assume that the eigenvalues of $\rho(g_{i})$ are all $-1$ . Recall that any parabolic matrix of $\mathrm{SL}(2,\mathbb{C})$ with eigenvalue $-1$ is represented as $\begin{pmatrix}-1+\alpha\beta&\beta^{2}\\ -\alpha^{2}&-1-\alpha\beta\end{pmatrix}$ for some complex numbers $\alpha$ and $\beta$ . So, up to the conjugation, we can assign

\rho(g_{1})=\begin{pmatrix}-1&x\\ 0&-1\end{pmatrix},\quad\rho(g_{2})=\begin{pmatrix}-1+y&y\\ -y&-1-y\end{pmatrix},\quad\rho(g_{3})=\begin{pmatrix}-1&0\\ -z&-1\end{pmatrix}.

Let $g_{12}=g_{1}\,g_{2}$ and $g_{23}=g_{2}\,g_{3}$ . Since $h_{1}$ and $g_{23}$ are commutative and $\rho(h_{1})$ is parabolic, $\rho(g_{23})$ must be parabolic with eigenvalue $1$ or $-1$ . Hence $\operatorname{trace}\rho(g_{23})$ must be $2$ or $-2$ . On the other hand, $\operatorname{trace}\rho(g_{23})=2-yz$ , so if $\operatorname{trace}\rho(g_{23})=2$ , then $y$ or $z$ is zero, which contradict the assumption that the representation $\rho$ is non-abelian. Therefore, $\operatorname{trace}\rho(g_{23})=-2$ and $yz=4$ . Similar argument for $g_{12}$ and $h_{2}$ implies that $xy=4$ . We also have $g_{4}=(g_{1}\,g_{2}\,g_{3})^{-1}$ and $\operatorname{trace}(g_{4})=-2$ . This means that $xy+xz+yz+xyz=0$ and we get the following two solutions.

	$\displaystyle x$	$\displaystyle=-2+2i,\quad y=-1-i,\quad z=-2+2i,$		(2)
	$\displaystyle x$	$\displaystyle=-2-2i,\quad y=-1+i,\quad z=-2-2i.$		(3)

Choose the solution (3) for $\rho$ and let $p_{i}$ , $p_{ij}$ be the fixed points of $\rho(g_{i})$ , $\rho(g_{ij})$ in $\mathbb{C}$ . Then

p_{1}=\infty,\ \ p_{2}=-1,\ \ p_{3}=0,\ \ p_{4}=1,\ \ p_{12}=i,\ \ p_{23}=-i,

and these points are the vertices of a regular ideal octahedron $O_{1}$ in $\mathbb{H}^{3}$ . The action of $\rho(g_{1})$ to $\mathbb{C}$ is the translation by $2+2i$ . Let $O_{2}$ be another regular ideal octahedron with vertices

q_{1}=\infty,\ \ q_{2}=i,\ \ q_{3}=1+i,\ \ q_{4}=2+i,\ \ q_{12}=1+2i,\ \ q_{23}=1,

then $O_{1}\cup O_{2}$ is the fundamental domain of the action of $\mathrm{Im}\,\rho$ .

1.2. Volume conjecture

The colored Jones polynomial $J_{N-1}(B)$ is computed in Appendix A, and given by (45), that is the following.

J_{N-1}(B)=N^{2}\sum_{0\leq k,l\leq N-1}\sum_{\max(k,l)\leq s\leq\min(k+l,N-1)}\!\!\dfrac{\{s\}!^{2}}{\{s-k\}!^{2}\,\{s-l\}!^{2}\,\{k+l-s\}!^{2}}.

(4)

Now we prove the volume conjecture for $B$ by using (4). The idea of proof is the same as that in [6]. The terms in the sum are all positive and the limit $2\pi\lim_{N\to\infty}\frac{\log\left|J_{N-1}(B)\right|}{N}$ is given by the largest term in the sum. The maximal is attained at $k=l=\lfloor\frac{N-1}{2}\rfloor$ and $s=\lfloor\frac{3(N-1)}{4}\rfloor$ and the maximal value is $2\left(-\Lambda(\frac{3\pi}{4})+7\Lambda(\frac{\pi}{4})\right)=16\Lambda(\frac{\pi}{4})=7.3277...$ , which is equal to the twice of the volume of the regular ideal octahedron and is equal to the volume of $S^{3}\setminus B$ . Here $\Lambda(x)$ is the Lobachevsky function given by $\Lambda(x)=-\int_{0}^{x}\log|2\sin t|\,dt$ .

1.3. Regular ideal octahedron

The regular ideal octahedron

\begin{matrix}\begin{matrix}\includegraphics[scale={0.8}]{tetrahedron2}\end{matrix}&&\begin{matrix}\includegraphics[scale={0.8}]{octahedron}\end{matrix}\\ \text{truncated tetrahedron}&&\text{regular ideal octahedron}\end{matrix}

Figure 6. Recular ideal octahedron is an extremal truncated tetrahedron. The faces have checkerboard coloring, and the white faces corresponds to the faces of the original tetrahedron, and the vertices corresponds to the edges of the original tetrahedron.

can be thought as an extremal case of the truncated tetrahedron whose dihedral angles at edges are all zero. In this case, the length of edges are also zero.

1.4. Variations of the Borromean rings

Here we investigate variation $B_{1}$ and $B_{1,1}$ of the Borromean rings $B$ . Let $g_{1}$ , $\cdots$ , $g_{4}$ , $h_{1}$ , $h_{2}$ be the elements of $\pi_{1}(S^{3}\setminus B_{1})$ given in Figure 7.

The fundamental groups $\pi_{1}(S^{3}\setminus B_{1})$ and $\pi_{1}(S^{3}\setminus B_{1,1})$ are presented by

	$\displaystyle\pi_{1}(S^{3}\setminus B_{1})=$
	$\displaystyle\qquad\left<g_{1},g_{2},g_{3},g_{4},h_{1},h_{2}\mid g_{2}=h_{1}g_{4}^{-1}h_{1}^{-1},\ g_{3}=h_{1}g_{4}g_{1}g_{4}^{-1}h_{1}^{-1},\ g_{3}^{-1}=h_{2}g_{2}h_{2}^{-1}\right>,$		(5)
	$\displaystyle\pi_{1}(S^{3}\setminus B_{1,1})=$
	$\displaystyle\quad\left<g_{1},g_{2},g_{3},g_{4},h_{1},h_{2}\mid g_{2}=h_{1}g_{4}^{-1}h_{1}^{-1},\ g_{3}=h_{1}g_{4}g_{1}g_{4}^{-1}h_{1}^{-1},\ g_{3}^{-1}=h_{2}g_{2}^{-1}g_{1}g_{2}h_{2}^{-1}\right>.$		(6)

Let $\rho^{\prime}$ , $\rho^{\prime}_{1}$ , $\rho^{\prime}_{1,1}$ be the geometric $\mathrm{SL}(2,\mathbb{C})$ representations of $\pi_{1}(S^{3}\setminus B)$ , $\pi_{1}(S^{3}\setminus B_{1})$ , $\pi_{1}(S^{3}\setminus B_{1,1})$ respectively so that

	$\displaystyle\rho^{\prime}(g_{23})$	$\displaystyle=\rho^{\prime}_{1}(g_{23})=\rho^{\prime}_{1,1}(g_{23})=\begin{pmatrix}-1&x\\ 0&-1\end{pmatrix},$
	$\displaystyle\rho^{\prime}(g_{1})$	$\displaystyle=\rho^{\prime}_{1}(g_{1})=\rho^{\prime}_{1,1}(g_{1})=\begin{pmatrix}-1+y&y\\ -y&-1-y\end{pmatrix},$
	$\displaystyle\rho^{\prime}(g_{2})$	$\displaystyle=\rho^{\prime}_{1}(g_{2})=\rho^{\prime}_{1,1}(g_{2})=\begin{pmatrix}-1&0\\ -z&-1\end{pmatrix}.$

Let $\tau$ be one of $\rho^{\prime}$ , $\rho^{\prime}_{1}$ , $\rho^{\prime}_{1,1}$ , then $\tau$ must satisufy $\operatorname{trace}\tau(g_{2})=\operatorname{trace}\tau(g_{3})=\operatorname{trace}\tau(g_{4})=-2$ , and we get

x=2i,\ \ y=-2i\ \ z=2i,\quad\text{or}\quad x=-2i,\ \ y=2i,\ \ z=-2i

for all $\rho^{\prime}$ , $\rho^{\prime}_{1}$ , $\rho^{\prime}_{1,1}$ . By choosing the first solution for $x$ , $y$ , $z$ , the representation matrices for $h_{1}$ are given as follows from the relations (1), (5), (6).

\displaystyle\rho^{\prime}(h_{1})=\begin{pmatrix}-1&-1\\ 0&-1\end{pmatrix},\quad\rho^{\prime}_{1}(h_{1})=\rho^{\prime}_{1,1}(h_{1})=\begin{pmatrix}-1&-1+i\\ 0&-1\end{pmatrix}.

The fixed points $r_{1}$ , $r_{2}$ , $r_{3}$ , $r_{4}$ , $r_{23}$ , $r_{12}$ of $g_{1}$ , $g_{2}$ , $g_{3}$ , $g_{4}$ , $g_{23}$ , $g_{12}$ are given as follows.

r_{1}=-1,\ r_{2}=0,\ r_{3}=i,\ r_{4}=-1+i,\ r_{23}=\infty,\ r_{12}=\frac{-1+i}{2}.

Let $O_{1}$ be the regular ideal octahedron with vertices $r_{1}$ , $\cdots$ , $r_{4}$ , $r_{23}$ , $r_{12}$ , and let $O_{2}$ be that with vertices $s_{1}=-1+i$ , $s_{2}=i$ , $s_{3}=2i$ , $s_{4}=-1+2i$ , $s_{23}=\infty$ , then $O_{1}\cup O_{2}$ is the fundamental domain for the actions of $\pi_{1}(S^{3}\setminus B)$ , $\pi_{1}(S^{3}\setminus B_{1})$ , $\pi_{1}(S^{3}\setminus B_{1,1})$ . By doing such computation for $h_{2}$ instead of $h_{1}$ , we get the similar result. Here we get the same fundamental domain for the actons of the fundamental groups $\pi_{1}(S^{3}\setminus B)$ , $\pi_{1}(S^{3}\setminus B_{1})$ , $\pi_{1}(S^{3}\setminus B_{1,1})$ . However, the actions of $h_{1}$ and $h_{2}$ are different as in Figure 8 while the actions of $g_{1}$ , $\cdots$ , $g_{4}$ coincide respectively for $B$ , $B_{1}$ and $B_{1,1}$ .

The colored Jones polynomials of $B_{1}$ and $B_{1,1}$ are given by (46), (47) as follows.

J_{N-1}(B_{1})=N^{2}\,q^{\frac{(N-1)^{2}}{4}}\sum_{k,l=0}^{N-1}\sum_{m=\max(k,l)}^{\min(k+l,N-1)}\dfrac{q^{\left(k-\frac{N-1}{2}\right)^{2}}\{m\}!^{2}}{\{m-k\}!^{2}\,\{m-l\}!^{2}\,\{k+l-m\}!^{2}},

J_{N-1}(B_{1,1})=N^{2}\,q^{\frac{(N-1)^{2}}{2}}\sum_{k,l=0}^{N-1}\sum_{m=\max(k,l)}^{\min(k+l,N-1)}\dfrac{q^{\left(k-\frac{N-1}{2}\right)^{2}}q^{\left(l-\frac{N-1}{2}\right)^{2}}\{m\}!^{2}}{\{m-k\}!^{2}\,\{m-l\}!^{2}\,\{k+l-m\}!^{2}}.

These formulas have phase factors $q^{\left(k-\frac{N-1}{2}\right)^{2}}$ and $q^{\left(k-\frac{N-1}{2}\right)^{2}}q^{\left(l-\frac{N-1}{2}\right)^{2}}$ added to $J_{N-1}(B)$ , and no more real numbers. For $J_{N-1}(B_{1})$ , the term with $k=(N-1)/2$ , $l=(N-1)/2$ , $s=\lfloor 3(N-1)/4\rfloor$ have the maximal modulus among the terms in the sums and the oscillation at $k=(N-1)/2$ is stopped, so we have

\lim_{N\to\infty}\frac{2\pi}{N}\log|J_{N-1}(B_{1})|=\lim_{N\to\infty}\frac{2\pi}{N}\left(\log|J_{N-1}(B)|+\frac{\sqrt{-1}\pi}{4}\right).

Similarly, for $J_{N-1}(B_{1,1})$ , the term with $k=(N-1)/2$ , $l=(N-1)/2$ , $s=\lfloor 3(N-1)/4\rfloor$ have the maximal modulus among the terms in the sums and the oscillation around $k=(N-1)/2$ and $l=(N-1)/2$ is very small, so we have

\lim_{N\to\infty}\frac{2\pi}{N}\log|J_{N-1}(B_{1,1})|=\lim_{N\to\infty}\frac{2\pi}{N}\log|J_{N-1}(B)|.

The above rough argument can be replaced by a rigorous argument by using the Poisson sum formula and the saddle point method as in [15]. The hyperbolic volumes of the complements of $B_{1}$ and $B_{1,1}$ are equal to that of the complement of $B$ since these complements are both split into two regular ideal tetrahedrons. We also see the Chern-Simons invariant by SnapPy and $\operatorname{CS}(B)=\operatorname{CS}(B_{1,1})=0$ , $\operatorname{CS}(B_{1})=\pi/4$ . Therefore, we have

Theorem 2.

Conjecture 2 holds for $B_{1}$ and $B_{1,1}$ .

2. Twisted Whitehead links

In this section, we introduce the complexified tetrahedron by using $\mathrm{SL}(2,\mathbb{C})$ representation of $\pi_{1}(S^{3}\setminus W_{p})$ for the twisted Whitehead link $W_{p}$ with $|p|\geq 2$ . Then we prove Conjecture 1 for $W_{p}$ with the help of the complexified tetrahedron, which is a deformation of the regular ideal octahedron used in the previous section. Conjecture 1 is already proved by [21], and here we explain how the hyperbolic volume relates to the complexified tetrahedron, especially to its complexified length and angle, which corresponds to the eigenvalues of representation matrices of certain elements of $\pi_{1}(S^{3}\setminus W_{p})$ . Note that the Whitehead link $W$ is equal to $W_{2}$ , and $W_{-2}$ is the mirror image of $W_{2}$ , We exclude $W_{0}$ and $W_{\pm 1}$ since they are not hyperbolic.

2.1. Representation matrices

Assign the generators of $\pi_{1}(S^{3}\setminus W_{p})$ as in Figure 9. These generators satisfy the following relations.

	$\displaystyle g_{4}=hg_{1}h^{-1},\ \ g_{3}^{-1}=hg_{2}h^{-1},\ \ g_{1}g_{2}g_{3}g_{4}=1,$		(7)
	$\displaystyle g_{4}^{-1}=(g_{2}g_{3})^{\frac{p}{2}}g_{3}(g_{2}g_{3})^{-\frac{p}{2}},\ \ g_{1}^{-1}=(g_{2}g_{3})^{\frac{p}{2}}g_{2}(g_{2}g_{3})^{-\frac{p}{2}},\quad\text{($p$ : even)}$
	$\displaystyle g_{4}=hg_{1}h^{-1},\ \ g_{3}^{-1}=hg_{2}h^{-1},\ \ g_{1}g_{2}g_{3}g_{4}=1,$
	$\displaystyle g_{4}^{-1}=(g_{2}g_{3})^{\frac{p-1}{2}}g_{2}(g_{2}g_{3})^{-\frac{p-1}{2}},\ \ g_{1}^{-1}=(g_{2}g_{3})^{\frac{p+1}{2}}g_{3}(g_{2}g_{3})^{-\frac{p+1}{2}}.\quad\text{($p$ : odd)}$

Now we construct the geometric representation $\rho:\pi_{1}(S^{3}\setminus W)\to\mathrm{SL}(2,\mathbb{C})$ . The matrices corresponding to the meridians are all parabolic. Note that any parabolic matrix is of the form $\begin{pmatrix}\pm 1+\alpha\beta&\beta^{2}\\ -\alpha^{2}&\pm 1-\alpha\beta\end{pmatrix}$ for some $\alpha$ , $\beta\in\mathbb{C}$ . Let $g_{12}=g_{1}g_{2}$ , $g_{23}=g_{2}g_{3}$ . For geometric representation, it is known that the matrix $\rho(g_{23})$ is diagonalizable. By applying conjugation, we may assume that $\rho(g_{23})$ is a diagonal matrix and an off-diagonal element of $\rho(g_{1})$ is the minus of the other off-diagonal element of $\rho(g_{1})$ . Now we put

	$\displaystyle\rho(g_{1})$	$\displaystyle=\begin{pmatrix}-1+x&-x\\ x&-1-x\end{pmatrix},\qquad\rho(g_{2})=\begin{pmatrix}-1+a_{2}b_{2}&b_{2}^{2}\\ -a_{2}^{2}&-1-a_{2}b_{2}\end{pmatrix},$
	$\displaystyle\rho(g_{3})$	$\displaystyle=\begin{pmatrix}-1+a_{3}b_{3}&b_{3}^{2}\\ -a_{3}^{2}&-1-a_{3}b_{3}\end{pmatrix},\quad\rho(g_{23})=\begin{pmatrix}u^{-1}&0\\ 0&u\end{pmatrix}.$

Since $g_{12}$ commutes with the parabolic matrix $\rho(h)$ and $\rho$ is a non-abelian representation, we get $\operatorname{trace}\rho(g_{12})=-2$ . From the relations

	$\displaystyle\operatorname{trace}\rho(g_{23})=u+u^{-1},\quad\text{$\rho(g_{23})$ is a diagonal matrix,}$
	$\displaystyle\operatorname{trace}\rho(g_{1}g_{2}g_{3})=\operatorname{trace}\rho(g_{12})=-2,$

we get the following matrices.

	$\displaystyle\rho(g_{1})$	$\displaystyle=\left(\begin{array}[]{cc}-\frac{2u}{u+1}&\frac{u-1}{u+1}\\ -\frac{u-1}{u+1}&-\frac{2}{u+1}\\ \end{array}\right),\ \ \rho(g_{2})=\left(\begin{array}[]{cc}-\frac{2}{u+1}&\frac{\left(\sqrt{u}+1\right)^{3}}{\left(\sqrt{u}-1\right)(u+1)}\\ -\frac{\left(\sqrt{u}-1\right)^{3}}{\left(\sqrt{u}+1\right)(u+1)}&-\frac{2u}{u+1}\\ \end{array}\right),$		(8)
	$\displaystyle\rho(g_{3})$	$\displaystyle=\left(\begin{array}[]{cc}-\frac{2}{u+1}&-\frac{\left(\sqrt{u}+1\right)^{3}u}{\left(\sqrt{u}-1\right)(u+1)}\\ \frac{\left(\sqrt{u}-1\right)^{3}}{\left(\sqrt{u}+1\right)u(u+1)}&-\frac{2u}{u+1}\\ \end{array}\right),\ \ \rho(g_{4})=\left(\begin{array}[]{cc}-\frac{2u}{u+1}&-\frac{(u-1)u}{u+1}\\ \frac{u-1}{u(u+1)}&-\frac{2}{u+1}\\ \end{array}\right).$		(8)

Let $p_{i}$ be the fixed point of $\rho(g_{i})$ for $i=1$ , $2$ , $3$ , $4$ and $p_{12}$ be the fixed point of $g_{23}$ , and let $p_{23}^{0}$ and $p_{23}^{1}$ be the ratios of the elements of the eigenvectors of $\rho(g_{23})$ . Since $p_{i}$ is the ratio of the elements of the eigenvector of $\rho(g_{i})$ , $p_{23}^{0}$ and $p_{23}^{1}$ are fixed points of $\rho(g_{23})$ , and $\rho(g_{23})$ maps the geodesic line connecting $p_{23}^{0}$ and $p_{23}^{1}$ to itself, so we get

	$\displaystyle p_{1}$	$\displaystyle=1,\quad p_{2}=-\frac{\left(\sqrt{u}+1\right)^{2}}{\left(\sqrt{u}-1\right)^{2}},\quad p_{3}=\frac{\left(\sqrt{u}+1\right)^{2}u}{\left(\sqrt{u}-1\right)^{2}},\quad p_{4}=-u,\quad$
	$\displaystyle p_{12}$	$\displaystyle=\frac{\left(\sqrt{u}+1\right)\sqrt{u}}{\sqrt{u}-1},\qquad p_{23}^{0}=0,\qquad p_{23}^{1}=\infty.$

By the relation (7), we have

{g_{23}}^{\frac{p}{2}}\cdot p_{3}=p_{4},\qquad{g_{23}}^{\frac{p}{2}}\cdot p_{2}=p_{1},\qquad\text{($p$ : even)}

{g_{23}}^{\frac{p-1}{2}}\cdot p_{2}=p_{4},\qquad{g_{23}}^{\frac{p+1}{2}}\cdot p_{3}=p_{1}.\qquad\text{($p$ : odd)}

Since $\rho(g_{23})=\begin{pmatrix}u^{-1}&0\\ 0&u\end{pmatrix}$ , $p_{4}=-up_{1}$ , $p_{3}=-up_{2}$ , we have

(-u)^{-p}p_{3}=p_{4},\qquad(-u)^{-p}p_{2}=p_{1}.

These two equations are equal to the following equation.

-(-u)^{-p}\frac{\left(\sqrt{u}+1\right)^{2}}{\left(\sqrt{u}-1\right)^{2}}=1.

(9)

For the Whitehead link $W=W_{2}$ , the above equation is

(u+1)\left(u^{2}-2u^{3/2}+2\sqrt{u}+1\right)=0.

The solutions are $u=1.78615\pm 2.27202i$ and $u=-1$ , and the first two solutions give the geometric representations. For generic $p$ , there are many solutions for $u$ satisfying (9), and to find the geometric solution, we consider the complexified tetrahedron and the developing map associated with this tetrahedron as in the following subsection.

2.2. Complexified terahedron

Here we construct the complexified tetrahedron for a twisted Whitehead link with respect to $\rho(g_{1})$ , $\cdots$ , $\rho(g_{23})$ . At first, we assign points on the complex plane associated with $\rho(g_{1})$ , $\cdots$ , $\rho(g_{23})$ . Let $p_{i}$ be the fixed point of $\rho(g_{i})$ for $i=1$ , $2$ , $3$ , $4$ and $p_{12}$ be the fixed point of $g_{23}$ . Let $p_{23}^{0}$ and $p_{23}^{1}$ be the ratios of the elements of the eigenvectors of $\rho(g_{23})$ . Note that $p_{i}$ is the ratio of the elements of the eigenvector of $\rho(g_{i})$ , $p_{23}^{0}$ and $p_{23}^{1}$ are fixed points of $\rho(g_{23})$ , and $\rho(g_{23})$ maps the geodesic line connecting $p_{23}^{0}$ and $p_{23}^{1}$ to itself.

For the Whitehead link case with $u=1.78615-2.27202i$ ,

	$\displaystyle p_{1}$	$\displaystyle=1,\quad p_{2}=-1.97174-8.11634i,\quad p_{3}=21.9623+10.0172i,\quad$
	$\displaystyle p_{4}$	$\displaystyle=-1.78615+2.27202i,\quad p_{12}=4.80111+1.04322i,\quad p_{23}^{0}=0,\quad p_{23}^{1}=\infty.$

Here we see that $p_{3}=-u\,p_{2}$ , $p_{4}=-u\,p_{1}$ , $p_{2}=u^{2}\,p_{1}$ and $p_{3}=u^{2}\,p_{4}$ as in Figure 10.

Let $p_{i}^{\prime}$ be the point on the line $p_{23}^{0}p_{23}^{1}$ such that the geodesic line $p_{i}p_{i}^{\prime}$ is perpendicular to $p_{23}^{0}p_{23}^{1}$ . Then construct four geodesic triangles $F_{j}$ whose vertices are $p_{12}$ , $p_{j}$ , $p_{j+1}$ for $j=1$ , $2$ , $3$ , $4$ . Here $j+1$ means $j+1\mod 4$ . Now we choose two surfaces $F_{5}$ , $F_{8}$ where the boundary of $F_{1}$ is $p_{1}p_{1}^{\prime}\cup p_{1}^{\prime}p_{2}^{\prime}\cup p_{2}^{\prime}p_{2}\cup p_{1}p_{1}$ and the boundary of $F_{2}$ is $p_{4}p_{4}^{\prime}\cup p_{4}^{\prime}p_{1}^{\prime}\cup p_{1}^{\prime}p_{1}\cup p_{1}p_{4}$ . Let $\rho(g_{23})^{1/2}=\begin{pmatrix}iu^{-1/2}&0\\ 0&-iu^{1/2}\end{pmatrix}$ . Let $F_{6}=\rho(g_{23})\,F_{8}$ and $F_{7}=\rho(g_{23})^{1/2}\,F_{5}$ . Now we introduce the complexified tetrahedra $T$ , which is the hyperbolic solid surrounded by $F_{1}$ , $\cdots$ , $F_{8}$ . The surfaces $F_{2}$ , $F_{4}$ , $F_{5}$ , $F_{7}$ correspond to the faces and the surfaces $F_{1}$ , $F_{3}$ , $F_{6}$ , $F_{8}$ shaded in Figure 11 correspond to the vertices of the tetrahedral graph in Figure 9. The solid $T$ is considered to be a deformation of the regular ideal octahedron. There are many ways to take $F_{5}$ and $F_{8}$ , and here we choose them so that $T\cup\rho(g_{23})^{1/2}T$ is a fundamental domain of the action of $\rho(\pi_{1}(S^{3}\setminus W))$ .

For general $p$ , the solution of (9) corresponding to the geometric representation is the one satisfying

p\arg(-u)+\arg(p_{2}^{-1})=\pm 2\pi i.

For this solution, $\pi_{1}(S^{3}\setminus W_{p})(T\cup\rho(g_{23})^{1/2}T)$ covers the hyperbolic space $\mathbb{H}^{3}$ evenly.

2.3. Poisson sum formula

From now on, we prove the volume conjecture for $W_{p}$ . The colored Jones polynomial $J_{N}(W_{p})$ is given in (53) as follows.

J_{N-1}(W_{p})=N\frac{q^{p\frac{(N-1)^{2}}{4}}}{4\pi i}\sum_{k=0}^{N-1}\frac{d}{dx}\!\!\left(q^{p(x-\frac{N-1}{2})^{2}}\{2x+1\}\hfill\left.\sum_{l=0}^{N-1}\sum_{s-\frac{x-k}{2}=\max(k,l)}^{\min(k+l,N-1)}\xi_{N}(x,l,s)\!\right)\right|_{x=k}.

The function $\xi_{N}(x,l,s)$ is real valued and it takes the maximal at $s_{0}$ given in (55) and $l=\frac{N-1}{2}$ . Hence

J_{N-1}(W_{p})=N\frac{q^{p\frac{(N-1)^{2}}{4}}}{4\pi i}\sum_{k=0}^{N-1}\left.\frac{d}{dx}\Big{(}q^{p(x-\frac{N-1}{2})^{2}}\{2x+1\}D_{N}\xi_{N}(x,\tfrac{N-1}{2},s_{0})\Big{)}\right|_{x=k}

where $D_{N}$ is a constant with polynomial growth and

s_{0}=\frac{N}{2\pi i}\log w_{0},\quad w_{0}=\frac{(u+1)(v+1)-\sqrt{(u+1)^{2}(v+1)^{2}-16uv}}{4}=-\sqrt{u},

u=q^{2x+1},\qquad v=q^{2l+1}=-1

as shown in Appendix C. Let $N\alpha=x+\frac{1}{2}$ , $N\gamma_{0}=s_{0}+\frac{1}{2}$ and

\Psi_{W}(\alpha)=-4\pi^{2}\big{(}\gamma_{0}^{2}-2(\alpha+\tfrac{1}{2})\gamma_{0}+\alpha^{2}+\tfrac{1}{2}\alpha+\tfrac{1}{4}\big{)}\\ -2\mathrm{Li}_{2}(e^{2\pi i\gamma_{0}})+2\mathrm{Li}_{2}(e^{2\pi i(\gamma_{0}-\alpha)})+2\mathrm{Li}_{2}(-e^{2\pi i\gamma_{0}})+2\mathrm{Li}_{2}(-e^{2\pi i(\alpha-\gamma_{0})})-\frac{2\pi^{2}}{3}.

Then

J_{N-1}(W_{p})=E_{N}q^{p\frac{(N-1)^{2}}{4}}\sum_{k=0}^{N-1}\left.\!\!\frac{d}{d\alpha}\{2N\alpha\}\exp\Big{(}\!\tfrac{N}{2\pi i}\big{(}-2\pi^{2}p(\alpha-\tfrac{1}{2})^{2}+\Psi_{W}(\alpha)\big{)}\!\Big{)}\right|_{\alpha=\frac{2k+1}{2N}=\alpha},

where $E_{N}$ is a constant which grows at most polynomially with respect to $N$ .

To see the asymptotics of $J_{N-1}(W_{p})$ , we use the Poisson sum formula. Let $f$ be a rapidly decreasing function, then

\sum_{k\in\mathbb{Z}}f(k)=\sum_{k\in\mathbb{Z}}\hat{f}(k)

where $\hat{f}$ is the Fourier transform of $f$ given by

\hat{f}(x)=\int_{\mathbb{R}}e^{-2\pi ikt}f(t)\,dt.

To apply this to the parameter $l$ , we extend the function $\Psi_{W}$ by $0$ for $\alpha\leq 0$ and $\alpha\geq 1$ . Then

J_{N-1}(W_{p})=E_{N}\,q^{p\frac{(N-1)^{2}}{4}}\,\times\\ \left.\sum_{k\in\mathbb{Z}}\int_{0}^{N}e^{-2\pi ikt}\frac{d}{d\alpha}\{2N\alpha\}\exp\Big{(}\tfrac{N}{2\pi i}\big{(}-2\pi^{2}p(\alpha-\tfrac{1}{2})^{2}+\Psi_{W}(\alpha))\Big{)}\right|_{\alpha=\frac{2t+1}{2N}}\!\!dt\\ =NE_{N}\,q^{p\frac{(N-1)^{2}}{4}}\sum_{k\in\mathbb{Z}}\int_{0}^{1}e^{-2\pi ikN\alpha+\pi ik}\frac{d}{d\alpha}\{2N\alpha\}\exp\Big{(}\tfrac{N}{2\pi i}\big{(}-2\pi^{2}p(\alpha-\tfrac{1}{2})^{2}+\Psi_{W}(\alpha))\Big{)}d\alpha.

Now we apply integral by part and we get

J_{N-1}(W_{p})=NE_{N}\,q^{p\frac{(N-1)^{2}}{4}}\,\times\\ \left.\sum_{k\in\mathbb{Z}}(-1)^{k}e^{-2\pi iNk\alpha}\{2N\alpha\}\exp\Big{(}\tfrac{N}{2\pi i}\big{(}-2\pi^{2}p(\alpha-\tfrac{1}{2})^{2}+\Psi_{W}(\alpha))\Big{)}\right|_{-\infty}^{\infty}-\\ E_{N}q^{p\frac{(N-1)^{2}}{4}}2\pi i\sum_{k\neq 0}(-1)^{k+1}k\!\int_{0}^{1}\!e^{-2\pi iNk\alpha}\{2N\alpha\}\exp\Big{(}\tfrac{N}{2\pi i}\big{(}-2\pi^{2}p(\alpha-\tfrac{1}{2})^{2}+\Psi_{W}(\alpha))\!\Big{)}d\alpha\\ =2\pi iE_{N}\,q^{p\frac{(N-1)^{2}}{4}}\sum_{k\in\mathbb{Z}}(-1)^{k}k\int_{0}^{1}e^{-2\pi iNk\alpha}\{2N\alpha\}\exp\Big{(}\tfrac{N}{2\pi i}\big{(}-2\pi^{2}p(\alpha-\tfrac{1}{2})^{2}+\Psi_{W}(\alpha))\Big{)}d\alpha.

Let

\Phi_{W_{p}}(\alpha)=-2\pi^{2}p(\alpha-\tfrac{1}{2})^{2}+\Phi_{W}(\alpha).

Then

J_{N-1}(W_{p})=E_{N}q^{p\frac{(N-1)^{2}}{4}}\sum_{k\neq 0}(-1)^{k}k\int_{0}^{1}e^{-2\pi iNk\alpha}\{2N\alpha\}e^{\frac{N}{2\pi i}\Phi_{W_{p}}(\alpha)}d\alpha,

In the rest, we follow the method in [15]. Let

\Phi_{W_{p}}^{+}(\alpha)=\Phi_{W_{p}}(\alpha)-\frac{4\pi^{2}}{N}\alpha,\qquad\Phi_{W_{p}}^{-}(\alpha)=\Phi_{W_{p}}(\alpha)+\frac{4\pi^{2}}{N}\alpha.

2.4. Saddle point method

Here we investigate

\lim_{N\to\infty}\frac{2\pi}{N}\log\int_{0}^{1}e^{-2\pi iNk\alpha}E^{\prime}_{N}e^{N\Phi_{W_{p}}^{\pm}(\alpha)}\,d\alpha

with the help of the saddle point method. We first compute for $k=1$ . Choose a small positive $\delta$ so that $|\mathrm{Im}\Phi_{W_{p}}^{\pm}(\alpha)|<{v_{W}}$ for $\alpha\in[0.\delta]$ and $[1-\delta,1]$ and we devide the integral in the above formula into three parts.

\int_{0}^{1}e^{-2\pi iN\alpha}E^{\prime}_{N}e^{\frac{N}{2\pi i}\Phi_{W_{p}}^{\pm}(\alpha)}\,d\alpha=\int_{0}^{\delta}e^{-2\pi iN\alpha}E^{\prime}_{N}e^{\frac{N}{2\pi i}\Phi_{W_{p}}^{\pm}(\alpha)}\,d\alpha+\\ \int_{\delta}^{1-\delta}e^{-2\pi iN\alpha}E^{\prime}_{N}e^{\frac{N}{2\pi i}\Phi_{W_{p}}^{\pm}(\alpha)}\,d\alpha+\int_{1-\delta}^{1}e^{-2\pi iN\alpha}E^{\prime}_{N}e^{\frac{N}{2\pi i}\Phi_{W_{p}}^{\pm}(\alpha)}\,d\alpha.

Then,

\left|\int_{0}^{\delta}e^{-2\pi iN\alpha}E^{\prime}_{N}e^{\frac{N}{2\pi i}\Phi_{W_{p}}^{\pm}(\alpha)}\,d\alpha\right|<E^{\prime\prime}_{N}e^{N\frac{v_{W}}{2\pi}}

and

\left|\int_{1-\delta}^{1}e^{-2\pi iN\alpha}E^{\prime}_{N}e^{\frac{N}{2\pi i}\Phi_{W_{p}}^{\pm}(\alpha)}\,d\alpha\right|<E^{\prime\prime}_{N}e^{N\frac{v_{W}}{2\pi}}

for some factors $E_{N}^{\prime\prime}$ with polynomial growth. The remaining integral is estimated by the value at the saddle point, where the saddle point $\alpha_{0}$ is the point that the differential of $\Phi_{W_{p}}^{\pm}(\alpha)$ vanishes.

Now let us consider the Whitehead link case, i.e. $p=2$ . Let $\alpha_{0}$ be the solution of

\frac{1}{2\pi i}\frac{d}{d\alpha}\left(4\pi^{2}\alpha+\Phi_{W_{2}}(\alpha)\right)=0.

By taking the exponential of this equation, we get

-\frac{(1+e^{2\pi i\frac{\alpha+1}{2}})^{2}}{(1-e^{2\pi i\frac{\alpha+1}{2}})^{2}}e^{4\pi i\frac{\alpha+1}{2}}=-\frac{(1-e^{\pi i\alpha})^{2}}{(1+e^{\pi i\alpha})^{2}}e^{4\pi i\alpha}=1.

(10)

Note that this equation is equal to (9) by putting $u=e^{2\pi i\alpha}$ , and is an algebraic equation. So it has several solutions and they satisfy

\frac{1}{2\pi i}\frac{d}{d\alpha}\left(\Phi_{W_{2}}(\alpha)\right)=2\pi ik.\qquad(k\in\mathbb{Z})

Then $\alpha_{0}$ is one of the solutions of (10) satisfying

\frac{1}{2\pi i}\frac{d}{d\alpha}\left(\Phi_{W_{2}}(\alpha_{0})\right)=2\pi i.

We actually have such solution $\alpha_{0}=0.856035...-0.168907...i=\frac{1}{2\pi i}\log(1-i+\sqrt{-1-2i})$ . We can see this solution as the saddle point in the contour graph of $\operatorname{Re}\Phi_{W_{2}}(\alpha)$ given in Figure 12.

In this case, the end points $\alpha=0$ and $\alpha=1$ of the integral path are located on the different regions of $\operatorname{Re}\frac{1}{2\pi i}\Phi_{W_{2}}(\alpha)\leq 0.57$ and we can apply the saddle point method by deforming the integral path to the dashed line in Figure 12. Therefore,

\int_{\delta}^{1-\delta}e^{-2\pi iN\alpha}E^{\prime}_{N}e^{N\Phi_{W_{2}}(\alpha)}\,d\alpha\underset{N\to\infty}{\sim}\frac{1}{2\pi i}e^{\frac{N}{2\pi i}\left(4\pi^{2}\alpha_{0}+\Phi_{W_{2}}(\alpha_{0})\right)}.

Let $\alpha_{0}^{\pm}$ be the solution of

\frac{d}{d\alpha}\left(4\pi^{2}\alpha+\Phi_{W_{2}}^{\pm}(\alpha)\right)=0.

Then, since $\mathrm{\alpha_{0}}>0$ , there is positive constants $c$ such that $\alpha_{0}^{\pm}=\alpha_{0}\pm\frac{c}{N}+O(\frac{1}{N^{2}})$ , and

\int_{\delta}^{1-\delta}e^{-2\pi iN\alpha}D_{N}e^{\frac{N}{2\pi i}\Phi_{W_{2}}^{\pm}(\alpha)}\,d\alpha\underset{N\to\infty}{\sim}D_{N}e^{\frac{N}{2\pi i}\left(4\pi^{2}\alpha_{0}^{\pm}+\Phi_{W_{2}}^{\pm}(\alpha_{0}^{\pm})\right)}\\ =D_{N}e^{\frac{N}{2\pi i}\left(4\pi^{2}(\alpha_{0}\pm\frac{c}{N})+\Phi_{W_{2}}^{\pm}(\alpha_{0}\pm\frac{c}{N}+O(\frac{1}{N^{2}}))\right)}=D_{N}e^{\frac{N}{2\pi i}\left(4\pi^{2}\alpha_{0}+\Phi_{W_{2}}(\alpha_{0})\pm\frac{d}{N}+O(\frac{1}{N^{2}})\right)}

for some constant $d$ . Therefore,

\lim_{N\to\infty}\frac{2\pi}{N}\log|J_{N-1}(W_{2})|=\\ \lim_{N\to\infty}\frac{2\pi}{N}\log\left|D_{N}\left(e^{\frac{N}{2\pi i}\left(4\pi^{2}\alpha_{0}+\Phi_{W_{2}}(\alpha_{0})+\frac{d}{N}+O(\frac{1}{N^{2}})\right)}-e^{\frac{N}{2\pi i}\left(4\pi^{2}\alpha_{0}+\Phi_{W_{2}}(\alpha_{0})-\frac{d}{N}+O(\frac{1}{N^{2}})\right)}\right)\right|=\\ \lim_{N\to\infty}\frac{2\pi}{N}\log\left|D_{N}\left((e^{d}-e^{-d})e^{\frac{N}{2\pi i}\left(4\pi^{2}\alpha_{0}+\Phi_{W_{2}}(\alpha_{0})+O(\frac{1}{N^{2}})\right)}\right)\right|=2\pi\operatorname{Re}\Phi_{W_{2}}(\alpha_{0}).

For $p>2$ , the contour graph is similar to the case $p=2$ and we can apply the similar argument to get

\lim_{N\to\infty}\frac{2\pi}{N}\log|J_{N-1}(W_{p})|=2\pi\operatorname{Im}\Phi_{W_{p}}(\alpha_{0}^{(p)}),

where $\alpha_{0}^{(p)}$ is the solution of

\frac{d}{d\alpha}\left(4\pi^{2}\alpha+\Phi_{W_{p}}(\alpha)\right)=0.

(11)

For positive $p$ , $1/2<\operatorname{Re}\alpha_{0}^{(p)}<1$ and so

e^{\gamma_{0}(\alpha_{0}^{(p)},1/2)}=e^{2\pi i(\alpha_{0}^{(p)}+1)/2}=-e^{\pi i\alpha_{0}^{(p)}}=\sqrt{e^{2\pi i\alpha_{0}^{(p)}}}.

By taking the exponential of this equation, we see that $\alpha_{0}^{(p)}$ is a solution of

-\frac{(1-e^{\pi i\alpha})^{2}}{(1+e^{\pi i\alpha})^{2}}(-e^{2\pi i\alpha})^{p}=1

(12)

satisfying (11). For such $\alpha_{0}^{(p)}$ , the value $\mathrm{Im}\Phi_{W_{p}}(\alpha_{0}^{(p)})$ satisfies

\mathrm{Im}(4\pi^{2}\alpha_{0}+\Phi_{W_{2}}(\alpha_{0}))<\mathrm{Im}(4\pi^{2}\alpha_{0}^{(p)}+\Phi_{W_{p}}(\alpha_{0}^{(p)}))<v_{B},

and the condition to apply the saddle point method is also fulfilled. Actually, the contour graph for $p=5$ , $20$ is given in Figure 13. If $p$ becomes large, then the term $\operatorname{Re}(2\pi ip(\alpha-\frac{1}{2})$ becomes dominant.

The saddle points $\alpha_{0}^{(p)}$ for $2\leq|p|\leq 100$ are given in Figure 14.

The contribution of the term $k=-1$ is the same as $k=1$ term.

We have to check the contribution of the term $k$ with $|k|\geq 2$ is negrigible. In such case, the saddle point moves and the imaginary part of the value at the saddle point is smaller than $v_{W_{p}}$ . If $|k|$ is large, then there is no saddle points and the integral path can be moved to the path on which the imaginary part of the value is 0. See Figure 15.

2.5. Volume of the complement

Here we show the following.

Theorem 3.

The value $\frac{1}{i}\big{(}4\pi^{2}\alpha_{0}^{(p)}+\Phi_{W_{p}}(\alpha_{0}^{(p)})\big{)}$ is equal to the complex volume of the complement of $W_{p}$ ,

Proof.

The key is the coincidence of $e^{2\pi i\alpha_{0}^{(p)}}$ with the eigenvalue $u$ of $\rho(g_{23})$ in §2.1. To prove the theorem, we compare $\zeta(\alpha,\frac{1}{2},\frac{\alpha+1}{2})$ with the Neumann-Zagier potential function, which relates to the hyperbolic volume of the deformation of the complement of the Borromean rings $B$ and its variation $B_{1}$ . For even $p$ , $W_{p}$ is obtained from the Borromean rings $B$ by the $2/p$ surgery along the component $C$ which corresponds to $h_{1}$ in Figure 4, and for odd $p$ , $W_{p}$ is obtained from $B_{1}$ by $2/(p-1)$ surgery. We deform the complement of $B$ by changing the cusp shape of $C$ .

First we prove for positive even $p$ case. Let $\rho_{\mu,\lambda}:\pi_{1}(B\setminus B)\to\mathrm{SL}(2,\mathbb{C})$ be the non-parabolic representation of $\pi_{1}(B\setminus B)$ where $\mu$ and $\lambda$ are eigenvalues of $h_{1}$ and $g_{23}$ respectively. Let $m$ and $l$ be the the dilatations with respect to the meridian and the longitude of the cusp along $C$ respectively, then it is known that $e^{m}=\mu^{2}$ and $e^{l}=\lambda^{2}$ , and $\rho_{\mu,\lambda}$ gives a deformed hyperbolic structure to the complement of $B$ such that the cusp shape along $C$ matches $\mu$ and $\lambda$ . For such deformation, the volume of the complement with respect to this deformed hyperbolic structure is studied by Neumann and Zagier [14]. Let $f(m)$ be the Neumann-Zagier function for the complement of $B$ given in [14]. The function $f(m)$ is determined by the following differential equation.

\frac{d}{dm}f(m)=-\frac{1}{2}l,\qquad f(0)=0.

Such deformation is actually realized as a deformation of a union of two ideal regular octahedrons which form the complement of $B$ . Let $p_{i}$ be the fixed point of $\rho(g_{i})$ given by (8) for $i=1$ , $2$ , $3$ , $4$ . Since $h_{1}$ commute with $g_{23}$ and $\rho(g_{23})$ is a diagonal matrix, $\rho(h_{1})$ is also a diagonal matrix, and the action of $\rho(h_{1})$ sends $p_{1}$ to $p_{2}$ and $p_{4}$ to $p_{3}$ . These points satisfy

p_{2}=-\frac{\left(\sqrt{u}+1\right)^{2}}{\left(\sqrt{u}-1\right)^{2}}\,p_{1},\qquad p_{4}=-\frac{\left(\sqrt{u}+1\right)^{2}}{\left(\sqrt{u}-1\right)^{2}}\,p_{3},

so $e^{\pm m}=\mu^{\pm 2}=-\frac{\left(\sqrt{u}+1\right)^{2}}{\left(\sqrt{u}-1\right)^{2}}$ , and we choose

m=\log\left(-\frac{\left(\sqrt{u}-1\right)^{2}}{\left(\sqrt{u}+1\right)^{2}}\right),\qquad\mu=-e^{\frac{m}{2}}.\qquad(u=e^{2\pi i\alpha})

On the other hand, the eigenvalues of $\rho(g_{23})$ are $\lambda=u^{\pm 1}=e^{\pm 2\pi i\alpha}$ , $e^{l}=\lambda^{2}=u^{\pm 2}=e^{\pm 4\pi i\alpha}$ and we put

l=4\pi i\alpha-2\pi i.

For positive $p$ , $\operatorname{Re}\alpha_{0}^{(p)}>1/2$ , so we adjust $l$ so that $0\leq\arg l<2\pi$ by subracting $2\pi i$ . Note that $l=2\xi=4\pi i\alpha$ and $m=2\eta$ for $\xi$ , $\eta$ in [19]. The function $\Phi_{W}(\alpha)$ satisfies

\frac{d}{dl}\Phi_{W}(\alpha)=\frac{1}{4\pi i}\frac{d}{d\alpha}\Phi_{W}(\alpha)=\frac{1}{2}\log\big{(}-\frac{\left(\sqrt{u}-1\right)^{2}}{\left(\sqrt{u}+1\right)^{2}}\big{)}=\frac{1}{2}m.

(13)

Let $H(m)=\Phi_{W}(\alpha)-\frac{1}{2}ml$ where $\lambda$ and $\mu$ satisfies (13), then we have

\frac{\partial}{\partial m}H(m)=\frac{\partial}{\partial\alpha}\Phi_{W}(\alpha)\frac{\partial\alpha}{\partial m}-\frac{1}{2}l-\frac{1}{2}m\frac{\partial\alpha}{\partial m}=-\frac{1}{2}l.

(14)

The differential equation (14) for $H(m)$ is the same differential equation for the Neumann-Zagier function $\Phi(m)$ in [11], which is explained in Appendix D. Note that $u$ , $v$ in [11] are equal to $m/2$ , $l/2$ . Let

h(m)=H(m)+\frac{1}{4}ml.

If $m=0$ , then $\mu=1$ , $u=-1$ , $\alpha=\pi$ and $h(0)$ coincides with $\mathrm{Vol}(S^{3}\setminus B)$ . Therefore, $h(\mu)-\mathrm{Vol}(S^{3}\setminus B)$ equals to the function $f(m)$ in [14]. Moreover, the length and the torsion of the core geodesic of the surgery component is given by the real part and the imaginary part of $l$ . Hence we have

\mathrm{Vol}(S^{3}\setminus W_{p})+i\,\mathrm{CS}(S^{3}\setminus W_{p})=\frac{1}{i}\big{(}h(m)-\frac{\pi i}{2}\log l\big{)}.

Since $m+\frac{p}{2}l=2\pi i$ , we have

i\left(\mathrm{Vol}(S^{3}\setminus W_{p})+i\,\mathrm{CS}(S^{3}\setminus W_{p})\right)=h(m)-\frac{\pi i}{2}l\\ =\Phi_{W}(\alpha_{0}^{(p)})-\frac{1}{4}ml-\frac{\pi i}{2}l\\ =\Phi_{W}(\alpha_{0}^{(p)})-\frac{1}{4}\big{(}2\pi i-\frac{p}{2}l\big{)}l-\frac{\pi i}{2}l\\ =\Phi_{W}(\alpha_{0}^{(p)})-4\pi^{2}\frac{p}{2}\big{(}\alpha_{0}^{(p)}-\frac{1}{2}\big{)}^{2}+4\pi^{2}\big{(}\alpha_{0}^{(p)}-\frac{1}{2}\big{)}.

The last formula coincides with $\Phi_{W_{p}}(\alpha_{0}^{(p)})-2\pi i\left(2\pi i(\alpha_{0}^{(p)}-\frac{1}{2})\right)$ at the saddle point $\alpha_{0}^{(p)}$ and so we get

\frac{1}{i}\left(\Phi_{W_{p}}(\alpha_{0}^{(p)})-2\pi i\left(2\pi i(\alpha_{0}^{(p)}-\frac{1}{2})\right)\right)=\mathrm{Vol}(S^{3}\setminus W_{p})+i\,\mathrm{CS}(S^{3}\setminus W_{p}).

For positive odd $p$ , $W_{p}$ is obtained by applying $2/(p-1)$ surgery to the middle complent of $B_{1}$ in Figure 1. We assign $m$ and $l$ along the component getting the surgery, then we get similar function $h(m)$ which corresponds to the Neumann-Zagier function. The only difference is that $h(0)=\mathrm{Vol}(S^{3}\setminus B_{1})+i\,\mathrm{CS}(S^{3}\setminus B_{1})$ , which implies that $\frac{1}{i}\Phi_{W_{p}}(\alpha_{0}^{(p)})=\mathrm{Vol}(S^{3}\setminus W_{p})+i\,\mathrm{CS}(S^{3}\setminus W_{p})$ .

The proof for negative $p$ case is similar. ∎

3. Double twist knots

We explain the complexified tetrahedron coming from $\mathrm{SL}(2,\mathbb{C})$ representation of $\pi_{1}(S^{3}\setminus D_{p,r})$ for the hyperbolic double twist knot $D_{p,r}$ ., and we prove Conjecture 1 for $D_{p,r}$ with the help of the complexified tetrahedron as in the previous section for the twisted Whitehead link. Note that the twist knot $T_{p}$ is equal to $D_{p,2}$ , and $D_{-p,-r}$ is the mirror image of $D_{p,r}$ ,

3.1. Representation matrices

We first construct $\mathrm{SL}(2,\mathbb{C})$ representation. Let $g_{1}$ , $g_{2}$ , $g_{3}$ , $g_{4}$ , $g_{12}$ , $g_{23}$ be elements of $\pi_{1}(S^{3}\setminus D_{p,r})$ as in Figure 16.

Then $g_{1}$ , $\cdots$ , $g_{4}$ , $g_{12}$ , $g_{23}$ satisfy the following relation.

	$\displaystyle g_{12}=g_{1}g_{2},\ \ g_{23}=g_{2}g_{3},\ \ g_{1}g_{2}g_{3}g_{4}=1,$		(15)
	$\displaystyle\begin{cases}g_{1}^{-1}={g_{23}}^{\frac{p}{2}}g_{2}{g_{23}}^{-\frac{p}{2}}\\ g_{4}^{-1}={g_{23}}^{\frac{p}{2}}g_{3}{g_{23}}^{-\frac{p}{2}}\end{cases}\text{if $p$ is even},\ \ \begin{cases}g_{1}^{-1}={g_{23}}^{\frac{p+1}{2}}g_{3}{g_{23}}^{-\frac{p+1}{2}}\\ g_{4}^{-1}={g_{23}}^{\frac{p-1}{2}}g_{2}{g_{23}}^{-\frac{p-1}{2}}\end{cases}\text{if $p$ is odd},$		(16)
	$\displaystyle\begin{cases}g_{4}^{-1}={g_{12}}^{\frac{r}{2}}g_{1}{g_{12}}^{-\frac{r}{2}}\\ g_{3}^{-1}={g_{12}}^{\frac{r}{2}}g_{2}{g_{12}}^{-\frac{r}{2}}\end{cases}\text{if $r$ is even},\ \ \begin{cases}g_{4}^{-1}={g_{12}}^{\frac{r+1}{2}}g_{2}{g_{12}}^{-\frac{r+1}{2}}\\ g_{3}^{-1}={g_{12}}^{\frac{r-1}{2}}g_{1}{g_{12}}^{-\frac{r-1}{2}}\end{cases}\text{if $r$ is odd}.$		(17)

Let $\rho$ be the geometric $\mathrm{SL}(2,\mathbb{C})$ of $\pi_{1}(S^{3}\setminus D_{p,r})$ , then $\rho(g_{1})$ , $\cdots$ , $\rho(g_{4})$ are parabolic matrices. Here we assume that the eigenvalue of $\rho(g_{i})$ is $-1$ . We also assume that $\rho(g_{23})=\begin{pmatrix}u^{-1}&0\\ 0&u\end{pmatrix}$ and the eigenvalues of $\rho(g_{12})$ are $v$ and $v^{-1}$ . Then, up to the conjugation, $\rho$ is given as follows.

	$\displaystyle\rho(g_{1})$	$\displaystyle=\left(\begin{matrix}-\frac{2u}{{u}+1}&\frac{{u}-1}{{u}+1}\\[3.0pt] -\frac{{u}-1}{{u}+1}&-\frac{2}{{u}+1}\end{matrix}\right),$
	$\displaystyle\rho(g_{2})$	$\displaystyle=\begin{pmatrix}-\frac{2}{{u}+1}&-\frac{(u+1)^{2}(v^{2}+1)-8uv+(u+1)(v-1)\sqrt{D}}{2v({u}-1)({u}+1)}\\ \frac{(u+1)^{2}(v^{2}+1)-8uv-(u+1)(v-1)\sqrt{D}}{2v({u}-1)({u}+1)}&-\frac{2u}{{u}+1}\end{pmatrix},$
	$\displaystyle\rho(g_{3})$	$\displaystyle=\left(\begin{matrix}-\frac{2}{{u}+1}&u\frac{(u+1)^{2}(v^{2}+1)-8uv+(u+1)(v-1)\sqrt{D}}{2v({u}-1)({u}+1)}\\ -\frac{(u+1)^{2}(v^{2}+1)-8uv-(u+1)(v-1)\sqrt{D}}{2uv({u}-1)({u}+1)}&-\frac{2u}{{u}+1}\end{matrix}\right),$
	$\displaystyle\rho(g_{4})$	$\displaystyle=\begin{pmatrix}-\frac{2u}{u+1}&-\frac{u(u-1)}{u+1}\\[3.0pt] \frac{u-1}{u(u+1)}&-\frac{2}{u+1}\end{pmatrix},\qquad\text{where $D=(u+1)^{2}(v+1)^{2}-16uv$.}$

Let $p_{1}$ , $p_{2}$ , $p_{3}$ , $p_{4}$ be the fixed points of $\rho(g_{1})$ , $\rho(g_{2})$ , $\rho(g_{3})$ , $\rho(g_{4})$ on $\partial\mathbb{H}^{2}$ . Then they are given as follows.

	$\displaystyle p_{1}$	$\displaystyle=1,\qquad p_{2}=\frac{(u+1)^{2}(v^{2}+1)-8uv+(u+1)(v-1)\sqrt{D}}{2v({u}-1)^{2}},$
	$\displaystyle p_{3}$	$\displaystyle=-u\frac{(u+1)^{2}(v^{2}+1)-8uv-(u+1)(v-1)\sqrt{D}}{2v({u}-1)^{2}},\qquad p_{4}=-u.$

Let $p_{23}^{0}$ , $p_{23}^{1}$ be the fixed points of $\rho(g_{23})$ , then $p_{23}^{0}=0$ and $p_{23}^{1}=\infty$ , and let $p_{12}^{0}$ , $p_{12}^{1}$ be the fixed points of $\rho(g_{23})$ , then they are

	$\displaystyle p_{12}^{0}$	$\displaystyle=-\frac{(u+1)^{2}(v+1)-8u+(u+1)\sqrt{D}}{4(u-1)},$
	$\displaystyle p_{12}^{1}$	$\displaystyle=-\frac{(u+1)^{2}(v+1)-8uv-(u+1)\sqrt{D}}{4(u-1)v}.$

Let $\rho^{\prime}$ be the representation similar to $\rho$ where $g_{12}$ is mapped to the diagonal matrix $\rho^{\prime}(g_{12})=\begin{pmatrix}v^{-1}&0\\ 0&v\end{pmatrix}$ instead of $g_{23}$ . Such $\rho^{\prime}$ is obtained by the transformation matrix

Q=\begin{pmatrix}-\frac{(u+1)^{2}(v+1)-8u+(u+1)\sqrt{D}}{4(u-1)}&-\frac{(u+1)(v+1)(u+v)-8uv+(uv-1)\sqrt{D}}{2(u-1)(v-1)}\\[3.0pt] 1&\frac{(u+1)(v+1)^{2}-8v+(v+1)\sqrt{D}}{4(v-1)}\end{pmatrix}.

For $g\in\pi_{1}(S^{3}\setminus K)$ , let $\rho^{\prime}(g)=Q^{-1}\rho(g)\,Q$ , then we have

	$\displaystyle\rho^{\prime}(g_{1})$	$\displaystyle=\left(\begin{matrix}-\frac{2}{v+1}&-\frac{v-1}{v+1}\\[3.0pt] \frac{v-1}{v+1}&-\frac{2v}{v+1}\\ \end{matrix}\right),\qquad\rho^{\prime}(g_{2})=\begin{pmatrix}-\frac{2}{v+1}&\frac{v-1}{v(v+1)}\\[3.0pt] -\frac{v(v-1)}{v+1}&-\frac{2v}{v+1}\end{pmatrix},$
	$\displaystyle\rho^{\prime}(g_{3})$	$\displaystyle=\left(\begin{matrix}-\frac{2v}{{v}+1}&-v\frac{\left((u^{2}+1)(v+1)^{2}-8uv+(u-1)(v+1)\sqrt{D}\right)}{2u({v}-1)({v}+1)}\\ \frac{(u^{2}+1)(v+1)^{2}-8uv-(u-1)(v+1)\sqrt{D}}{2u({v}-1)({v}+1)}&-\frac{2}{{v}+1}\end{matrix}\right),$
	$\displaystyle\rho^{\prime}(g_{4})$	$\displaystyle=\begin{pmatrix}-\frac{2v}{{v}+1}&\frac{\left((u^{2}+1)(v+1)^{2}-8uv+(u-1)(v+1)\sqrt{D}\right)}{2u({v}-1)({v}+1)}\\ -\frac{(u^{2}+1)(v+1)^{2}-8uv-(u-1)(v+1)\sqrt{D}}{2uv({v}-1)({v}+1)}&-\frac{2}{{v}+1}\end{pmatrix}.$

The fixed points $p^{\prime}_{1}$ , $p^{\prime}_{2}$ , $p^{\prime}_{3}$ , $p^{\prime}_{4}$ of $\rho^{\prime}(g_{1})$ , $\rho^{\prime}(g_{2})$ , $\rho^{\prime}(g_{3})$ , $\rho^{\prime}(g_{4})$ on $\partial\mathbb{H}^{3}$ are

	$\displaystyle p^{\prime}_{1}$	$\displaystyle=1,\qquad p^{\prime}_{2}=-{v},\qquad p^{\prime}_{3}=-v\frac{(u^{2}+1)(v+1)^{2}-8uv+(u-1)(v+1)\sqrt{D}}{2u({v}-1)^{2}},$
	$\displaystyle p^{\prime}_{4}$	$\displaystyle=\frac{(u^{2}+1)(v+1)^{2}-8uv+(u-1)(v+1)\sqrt{D}}{2u({v}-1)^{2}}.$

The fixed points ${p_{12}^{0}}^{\prime}$ and ${p_{12}^{1}}^{\prime}$ of $\rho^{\prime}(g_{12})$ are ${p_{12}^{0}}^{\prime}=0$ and ${p_{12}^{1}}^{\prime}=\infty$ , and the fixed points ${p_{23}^{0}}^{\prime}$ and ${p_{23}^{1}}^{\prime}$ of $\rho^{\prime}(g_{23})$ are

	$\displaystyle{p_{23}^{0}}^{\prime}$	$\displaystyle=-\frac{(u+1)(v+1)^{2}-8v+(v+1)\sqrt{D}}{4(v-1)},$
	$\displaystyle{p_{23}^{1}}^{\prime}$	$\displaystyle=-\frac{(u+1)(v+1)^{2}-8uv-(v+1)\sqrt{D}}{4u(v-1)}.$

The eigenvalues $u$ and $v$ are determined by the relations (16) and (17). They satisfy

(-u)^{p}=p_{2},\qquad(-v)^{-r}=q_{4}.

(18)

Moreover, the geometric representation is given by a solution among the solutions of (18) satisfying

p\log(-u)+\log p_{2}^{-1}=\pm 2\pi\sqrt{-1},\qquad-r\log(-v)+\log q_{4}^{-1}=\pm 2\pi\sqrt{-1}.

(19)

3.2. Complexified tetrahedron

Here we explain the complexified tetrahedron $T$ determined by the fixed points $p_{1}$ , $\cdots$ , $p_{12}^{1}$ , which is congruent to the complexified tetrahedron $T^{\prime}$ determined by $p^{\prime}_{1}$ , $\cdots$ , ${p_{12}^{1}}^{\prime}$ .

For $D_{6,2}$ , the solution of equations (18) and (19) where the sums are both $+2\pi i$ is given by

u=-0.619307-0.884567i,\qquad v=1.72565+2.06055i,

The fixed points are given as follows.

	$\displaystyle p_{1}$	$\displaystyle=1,\ \ p_{2}=\ \ 1.3731-0.7921i,\ \ p_{3}=\ \ 1.5511+0.7240i,\ \ p_{4}=\ \ 0.6193+0.8845i,$
	$\displaystyle p^{\prime}_{1}$	$\displaystyle=1,\ \ p^{\prime}_{2}=-1.7256-2.0605i,\ \ p^{\prime}_{3}=-0.2388+0.2852i,\ \ p^{\prime}_{4}=-0.0242-0.1362i.$

Then, $T$ and $T^{\prime}$ in $\mathbb{H}^{3}$ corresponding to $D_{6,2}$ are given as in Figure 17.

The elements $\rho(g_{23})$ , $\rho(g_{12})$ have axes $l_{23}$ , $l_{12}$ , so we assign complex parameters to these axes $u$ , $v$ , which is the eigenvalues of $g_{23}$ , $g_{12}$ . Let $r_{1}$ , $r_{2}$ , $r_{3}$ , $r_{4}$ be the foots of perpendicular on $l_{23}$ from $p_{1}$ , $p_{2}$ , $p_{3}$ , $p_{4}$ . Similarly, Let $q_{1}$ , $q_{2}$ , $q_{3}$ , $q_{4}$ be the foots of perpendicular on $l_{12}$ from $p_{1}$ , $p_{2}$ , $p_{3}$ , $p_{4}$ . Let us define eight faces $p_{1}p_{2}r_{2}r_{1}$ , $p_{2}p_{3}r_{3}r_{2}$ , $p_{3}p_{4}r_{4}r_{3}$ , $p_{4}p_{1}r_{1}r_{4}$ , $p_{1}p_{2}q_{2}q_{1}$ , $p_{2}p_{3}q_{3}q_{2}$ , $p_{3}p_{4}q_{4}q_{3}$ , $p_{4}p_{1}q_{1}q_{4}$ . These faces are not flat and are not defined uniquely, but the edges of the faces are straight lines and we define these faces topologically. Let $T$ be the subset of $\mathbb{H}^{3}$ surrounded by these eight faces, and this is the complexified tetrahedron corresponding to the representation $\rho$ . Let $T_{1}$ be similar complexified tetrahedron constructed from $(-u)p_{1}$ , $(-u)p_{2}$ , $(-u)p_{3}$ , $(-u)p_{4}$ , $(-u)l_{12}$ and $(-u)l_{23}=l_{23}$ . Then $T$ and $T_{1}$ are adjacent at the face $p_{3}p_{4}r_{4}r_{3}$ and $T\cup T_{1}$ is a fundamental domain of the action of $\pi_{1}(S^{3}\setminus D_{6,2})$ to $\mathbb{H}^{3}$ given by $\rho$ .

The action of $\rho(g_{23})$ on $\partial\mathbb{H}^{3}$ corresponds to the multiplication of $u^{2}$ , so we get the picture in Figure 18. Similarly, the action of $\rho^{\prime}(g_{12})$ corresponds to the multiplication of $v^{-2}$ and is also explained in the figure.

3.3. Poisson sum formula

We reformulate the colored Jones polynomial $J_{N-1}(D_{p,r})$ into integral form by using the Poisson sum formula. The colored Jones polynomial $J_{N-1}(D_{p,r})$ is given by (54) in Appendix C as follows.

J_{N-1}(D_{p,r})=-\frac{N^{2}q^{(p-r)\frac{(N-1)^{4}}{4}}}{16\pi^{2}}\,\times\\ \sum_{k,l=0}^{N-1}\frac{\partial^{2}}{\partial x\partial y}q^{p(x-\frac{N-1}{2})^{2}-r(y-\frac{N-1}{2})^{2}}\{2x+1\}\{2y+1\}\left.\sum_{s}\xi_{N}(x,y,s)\right|_{\text{$\scriptstyle\begin{matrix}x=k\\[0.0pt] y=l\end{matrix}$}}.

Since $\zeta_{N}(x,y,k,l,s)$ is a real positive number, it takes the maximal at $s_{0}$ given in (55). Hence

J_{N-1}(D_{p,r})=-\frac{N^{2}q^{(p-r)\frac{(N-1)^{4}}{4}}}{16\pi^{2}}\,\times\\ \sum_{k,l=0}^{N-1}\frac{\partial^{2}}{\partial x\partial y}q^{p(x-\frac{N-1}{2})^{2}-r(y-\frac{N-1}{2})^{2}}\{2x+1\}\{2y+1\}\left.F_{N}\xi_{N}(x,y,s_{0})\right|_{\text{$\scriptstyle\begin{matrix}x=k\\[-5.0pt] y=l\end{matrix}$}}.

where $F_{N}$ is a constant with polynomial growth and

s_{0}=\frac{N}{2\pi i}\log w_{0},\quad w_{0}=\frac{(u+1)(v+1)-\sqrt{(u+1)^{2}(v+1)^{2}-16uv}}{4},

u=q^{2x+1},\qquad v=q^{2y+1}

as shown in Appendix C. Let $N\alpha=x+\frac{1}{2}$ , $N\beta=y+\frac{1}{2}$ , $N\gamma_{0}=s_{0}+\frac{1}{2}$ and

\Psi_{D}(\alpha,\beta)=-4\pi^{2}\big{(}\gamma_{0}^{2}-2(\alpha+\beta)\gamma_{0}+\alpha^{2}+\alpha\beta+\beta^{2}\big{)}\\ -2\mathrm{Li}_{2}(e^{2\pi i\gamma_{0}})+2\mathrm{Li}_{2}(e^{2\pi i(\gamma_{0}-\alpha)})+2\mathrm{Li}_{2}(e^{2\pi i(\gamma_{0}-\beta)})+2\mathrm{Li}_{2}(-e^{2\pi i(\alpha+\beta-\gamma_{0})})-\frac{2\pi^{2}}{3}.

Then

J_{N-1}(D_{p,r})=G_{N}\frac{N^{2}q^{(p-r)\frac{(N-1)^{4}}{4}}}{16\pi^{2}}\,\times\sum_{k,l=0}^{N-1}\frac{\partial^{2}}{\partial x\partial y}q^{p(k-\frac{N-1}{2})^{2}-r(l-\frac{N-1}{2})^{2}}\{2N\alpha\}\{2N\beta\}\\ \left.\exp\Big{(}\tfrac{N}{2\pi i}\big{(}-2\pi^{2}p(\alpha-\tfrac{1}{2})^{2}+2\pi^{2}r(\eta-\tfrac{1}{2})^{2}+\Psi_{D}(\alpha,\eta,\alpha,\beta)\big{)}\Big{)}\right|_{\text{$\scriptstyle\begin{matrix}\alpha=\frac{2k+1}{2N}=\alpha\\[0.0pt] \eta=\frac{2l+1}{2N}=\beta\end{matrix}$}},

where $G_{N}$ is a constant with polynomial growth.

Now we apply the Poisson sum formula for $k$ and $l$ . Let

	$\displaystyle\Phi_{D_{p,r}}(\alpha,\beta)$	$\displaystyle=\frac{1}{2\pi i}\left(-2\pi^{2}p(\alpha-\frac{1}{2})^{2}+4\pi^{2}r(\beta-\frac{1}{2})^{2}+\Psi_{D}(\alpha,\beta)\right),$
	$\displaystyle\Phi_{D_{p,r}}^{\varepsilon_{1},\varepsilon_{2}}(\alpha,\beta)$	$\displaystyle=$
	$\displaystyle\frac{1}{2\pi i}$	$\displaystyle\left(-2\pi^{2}p((\alpha-\frac{1}{2})^{2}+\varepsilon_{1}\frac{\alpha}{N})+4\pi^{2}(r(\beta-\frac{1}{2})^{2}-\varepsilon_{2}\frac{\beta}{N})+\Psi_{D}(\alpha,\beta)\right),$

where $\varepsilon_{1},\varepsilon_{2}=\pm 1$ . Then

J_{N-1}(D_{p,r})=-\frac{N^{2}q^{(p-r)\frac{(N-1)^{4}}{4}}}{16\pi^{2}}\sum_{\varepsilon_{1},\varepsilon_{2}\in\{-,+\}}\\ \sum_{k,l=0}^{N-1}\frac{\partial}{\partial\alpha}\left.C_{N}e^{\frac{N}{2\pi i}\left(\Phi_{D_{p,r}}^{\varepsilon_{1},\varepsilon_{2}}(\alpha,\beta)+O(\frac{1}{N})\right)}\right|_{\text{$\scriptstyle\begin{matrix}\alpha=\frac{2k+1}{2N}\\[0.0pt] \beta=\frac{2l+1}{2N}\end{matrix}$}}.

As in the case of twisted Whitehead links, the Poisson sum formula yields

J_{N-1}(D_{p,r})=q^{(p-r)\frac{(N-1)^{4}}{4}}\,\times\\ \sum_{\varepsilon_{1},\varepsilon_{2}\in\{-,+\}}\sum_{m,n\in\mathbb{Z}}(-1)^{m+n}\iint_{D}C^{\prime}_{N}e^{-2\pi i(k\alpha+l\beta)}\frac{\partial}{\partial\alpha}e^{\frac{N}{2\pi i}\left(\Phi_{D_{p,r}}^{\varepsilon_{1},\varepsilon_{2}}(\alpha,\beta)+O(\frac{1}{N})\right)}d\alpha d\beta.

Hence, by reformulate as before, we get

J_{N-1}(D_{p,r})\underset{N\to\infty}{\sim}\iint_{D}C_{N}^{\prime\prime}e^{\frac{N}{2\pi i}\big{(}\pm 4\pi^{2}\alpha\pm 4\pi^{2}\beta+\Phi_{D_{p,r}}(\alpha,\beta)\big{)}}d\alpha d\beta.

Every choice of the signature gives the same asymptotics.

3.4. Saddle point method

Here we investigate the integral

\iint_{D}e^{\frac{N}{2\pi i}\big{(}-4\pi^{2}\alpha-4\pi^{2}\beta+\Phi_{D_{p,r}}(\alpha,\beta)\big{)}}d\alpha d\beta

where $D=[0,1]^{2}$ .

Proposition 3.1.

Let $p$ , $r$ be integers atisfying $p,r\geq 2$ and $p+r\geq 8$ , or $p,-q\geq 3$ and $p-q\geq 9$ . The asymptotics of the following integral is given by its value at the saddle point as follows.

\iint_{D}e^{\frac{N}{2\pi i}\big{(}-4\pi^{2}\alpha-4\pi^{2}\beta+\Phi_{D_{p,r}}(\alpha,\beta)\big{)}}d\alpha d\beta\underset{N\to\infty}{\longrightarrow}e^{\frac{N}{2\pi i}\big{(}-4\pi^{2}\alpha_{0}-4\pi^{2}\beta_{0}+\Phi_{D_{p,r}}(\alpha_{0},\beta_{0})\big{)}},

where $(\alpha_{0}$ , $\beta_{0})$ is the solution of

	$\displaystyle\frac{\partial}{\partial\alpha}\Big{(}-4\pi^{2}\alpha-4\pi^{2}\beta+\Phi_{D_{p,r}}(\alpha,\beta)\Big{)}$	$\displaystyle=0,$		(20)
	$\displaystyle\frac{\partial}{\partial\beta}\Big{(}-4\pi^{2}\alpha-4\pi^{2}\beta+\Phi_{D_{p,r}}(\alpha,\beta)\Big{)}$	$\displaystyle=0.$		(20)

This system of equations is called the saddle point equation.

Proof.

We can push the integral region inside the contour of $\mathrm{Im}\big{(}-4\pi^{2}\alpha_{0}-4\pi^{2}\beta_{0}+\Phi_{D_{p,r}}(\alpha_{0},\beta_{0})\big{)}=v_{D_{p,r}}$ to the saddle point as in Figure 19. The contours os the boundary of the gray regions show the level indicating the hyperbolic volume of $S^{3}\setminus D_{p,r}$ . Therefore, we can apply the saddle point method. In the figures, we see the contours of the function at planes parallel to the real plane including the original integral region. In the function $-4\pi^{2}\alpha-4\pi^{2}\beta+\Phi_{D_{p,r}}(\alpha,\beta)$ , we can deform $p$ and $r$ continuously. Therefore, we can also deform the integral region continuously from small $p$ , $|r|$ to large $p$ , $|r|$ , where the saddle point converges to $\alpha=\beta=1/2$ as in Figure 22. In this deformation, the saddle point is getting closer to $(1/2,1/2)$ and the value of $-4\pi^{2}\alpha-4\pi^{2}\beta+\Phi_{D_{p,r}}(\alpha,\beta)$ at the saddle point increases. ∎

\begin{matrix}\includegraphics[scale={0.7}]{contour0}\\ \alpha=x-i0\\ \beta=y-i0\end{matrix}\qquad\begin{matrix}\includegraphics[scale={0.7}]{contour621}\\ \alpha=x-0.08i\\ \beta=y-0.006i\end{matrix}\qquad\begin{matrix}\includegraphics[scale={0.7}]{contour622}\\ \alpha=x-0.15i\\ \beta=y-0.01i\end{matrix}\qquad\begin{matrix}\includegraphics[scale={0.7}]{contour623}\\ \alpha=x-0.012i\\ \beta=y-0.157i\end{matrix}

$D_{6,2}\qquad\mathbb{C}^{2}\ \ \raisebox{0.0pt}{\ \ $\begin{matrix}\uparrow\\ \mathbb{R}^{2}&\\ \ \ \ \searrow\end{matrix}$}\begin{matrix}\includegraphics[scale={0.8}]{contours}\end{matrix}$
$\longleftarrow(i\,\mathbb{R})^{2}$

Figure 19. Push the integral region for

D_{6,2}

to the imaginary direction.

\begin{matrix}D_{5,3}&\begin{matrix}\includegraphics[scale={0.6}]{contour0}\\ \alpha=x-0i\\ \beta=y-0i\end{matrix}&\begin{matrix}\includegraphics[scale={0.6}]{contour531}\\ \alpha=x-0.0125i\\ \beta=y-0.04i\end{matrix}&\begin{matrix}\includegraphics[scale={0.6}]{contour532}\\ \alpha=x-0.02i\\ \beta=y-0.08i\end{matrix}&\begin{matrix}\includegraphics[scale={0.6}]{contour533}\\ \alpha=x-0.025i\\ \beta=y-0.088i\end{matrix}\\ {}\\ D_{4,4}&\begin{matrix}\includegraphics[scale={0.6}]{contour0}\\ \alpha=x-0i\\ \beta=y-0i\end{matrix}&\begin{matrix}\includegraphics[scale={0.6}]{contour441}\\ \alpha=x-0.02i\\ \beta=y-0.02i\end{matrix}&\begin{matrix}\includegraphics[scale={0.6}]{contour442}\\ \alpha=x-0.04i\\ \beta=y-0.04i\end{matrix}&\begin{matrix}\includegraphics[scale={0.6}]{contour444}\\ \alpha=x-0.0477i\\ \beta=y-0.0477i\end{matrix}\\ {}\\ D_{6,-3}&\begin{matrix}\includegraphics[scale={0.6}]{contour0}\\ \alpha=x-0i\\ \beta=y-0i\end{matrix}&\begin{matrix}\includegraphics[scale={0.6}]{contour6m31}\\ \alpha=x-0.01i\\ \beta=y-0.05i\end{matrix}&\begin{matrix}\includegraphics[scale={0.6}]{contour6m32}\\ \alpha=x-0.02i\\ \beta=y-0.09i\end{matrix}&\begin{matrix}\includegraphics[scale={0.6}]{contour6m33}\\ \alpha=x-0.0206i\\ \beta=y-0.0935i\end{matrix}\\ {}\\ D_{5,-4}&\begin{matrix}\includegraphics[scale={0.6}]{contour0}\\ \alpha=x-0i\\ \beta=y-0i\end{matrix}&\begin{matrix}\includegraphics[scale={0.6}]{contour5m41}\\ \alpha=x-0.02i\\ \beta=y-0.03i\end{matrix}&\begin{matrix}\includegraphics[scale={0.6}]{contour5m42}\\ \alpha=x-0.03i\\ \beta=y-0.05i\end{matrix}&\begin{matrix}\includegraphics[scale={0.6}]{contour5m43}\\ \alpha=x-0.033i\\ \beta=y-0.055i\end{matrix}\end{matrix}

Figure 20. Push the integral region for

D_{p,r}

to the imaginary direction.

Remark 2.

The above method also work for $D_{p,r}$ with small $p$ , $|r|$ . For example, the integral region for the figure-eight knot $D_{2,2}$ can be deformed as in Figure 21. In this case, the cutting planes are no more parallel to the real plane.

\begin{matrix}\includegraphics[scale={0.35}]{contour22}\\ \alpha=x+0i\\ y=0y+0i\end{matrix}\begin{matrix}\includegraphics[scale={0.35}]{contour220}\\ \alpha=x-(0.2+0.1(\alpha-0.861))i\\ \beta=y-(0.2-0.1(\beta-5.421))i\end{matrix}\quad\begin{matrix}\includegraphics[scale={0.35}]{contour221}\\ \alpha=x-(0.4+0.3(\alpha-0.861))i\\ \beta=y-(0.4-0.3(\beta-5.421))i\end{matrix}\quad\begin{matrix}\includegraphics[scale={0.35}]{contour222}\\ \alpha=x-(0.5435+0.3(\alpha-0.861))i\\ \beta=y-(0.5435-0.3(\beta-5.421))i\end{matrix}

$\qquad D_{2,2}\qquad\mathbb{C}^{2}\ \ \raisebox{0.0pt}{\ \ $\begin{matrix}\uparrow\\ \mathbb{R}^{2}&\\ \ \ \ \searrow\end{matrix}$}\!\!\!\!\!\!\!\begin{matrix}\includegraphics[scale={0.8}]{contours22}\end{matrix}$
$\longleftarrow(i\,\mathbb{R})^{2}$

Figure 21. Push the integral region for

D_{2,2}

to the imaginary direction.

3.5. Volume of the complement

By comparing with the Neumann-Zagier function as in the case of the twisted Whitehead link, we get the following.

\frac{1}{i}\left(\Phi_{D_{p,r}}(\alpha_{0},\beta_{0})-2\pi i\Big{(}2\pi i(\alpha_{0}-\frac{1}{2})+2\pi i(\beta_{0}-\frac{1}{2})\Big{)}\right)=\mathrm{Vol}(S^{3}\setminus D_{p,r})+i\,\mathrm{CS}(S^{3}\setminus D_{p,r}).

Therefore, the volume conjecture holds for $D_{p,r}$ . The volume conjecture for the double twist knots $D_{p,r}$ with the integers $p$ , $r$ excluded in the above proposition is already proved in [15], [17], [16].

Appendices

Appendix A. ADO invariants for colored knotted graphs

Here we recall two quantum invariants defined for colored knotted graphs, which is also known as the quantum spin network. The first one is the Kirillov-Reshetikhin invariant introduced in [7], which is a generalization of the colored Jones polynomial, and the second one is the ADO invariant, which is also related to quantum $sl_{2}$ as the colored Jones polynomial, but this invariant is defined for the case that the quantum parameter $q$ is a root of unity. The ADO invariant was introduced in [1] for knots and links, and generalized to colored knotted graphs in [4]. The colored Jones invariant $J_{N-1}(K)$ is equal to $(-1)^{N-1}\operatorname{ADO}_{N}(K)$ , and is equal to $\operatorname{ADO}_{N}(K)$ for odd $N$ , where all the components of $K$ are colored by $(N-1)/2$ . Here we compute $\operatorname{ADO}_{N}(K)$ instead of $J_{N-1}(K)$ to get the desired form of the invariant which fits to the investigation of the asymptitics of the invariant.

A.1. ADO invariant for colored knots and links

We use the following notations.

	$\displaystyle q^{a}$	$\displaystyle=\exp\big{(}\frac{\pi ia}{N}\big{)}\ \ (a\in\mathbb{C}),\qquad\{a\}=q^{a}-q^{-a},\ \ \{a,k\}=\prod_{j=0}^{k-1}\{a-j\},$
	$\displaystyle\left[\,\begin{matrix}a\\ b\end{matrix}\,\right]$	$\displaystyle=\prod_{j=0}^{a-b-1}\frac{\{a-j\}}{\{a-b-j\}}\ \ (a-b\in\{0,1,\ldots,N-1\}),$
	$\displaystyle t_{a}$	$\displaystyle=a(a+1-N)=(a-\frac{N-1}{2})^{2}-\frac{(N-1)^{2}}{4}.$

Let $\mathcal{U}_{q}(sl_{2})$ is the quantum $sl_{2}$ at the $2N$ -th root of unity $q$ and let $V_{a}$ be the highest weight irreducible module with the highest weight $q^{a}$ . For $a\in(\mathbb{C}\setminus\mathbb{Z}/2)\cup(N\mathbb{Z}-1)/2$ , $\dim V_{a}=N$ .

Let $K=K_{1}\cup K_{2}\cup\cdots\cup K_{\ell}$ be a $\ell$ component oriented link diagram whose components are labeled by $c_{1}$ , $\cdots$ , $c_{\ell}$ where $c_{i}\in(\mathbb{C}\setminus\mathbb{Z}/2)\cup(N\mathbb{Z}-1)/2$ . The label $c_{i}$ is called the color of the $i$ -th component $K_{i}$ . Let $T_{K}$ be a $(1,1)$ tangle obtained by cutting the $j$ -th component of $K$ . Then, by assigning the quantum $R$ matrix to the crossings, evaluation map to the maximal points and coevaluation map to the minimal points given in [4], we get a scalar matrix of size $N$ . This scalar depends on the color $c_{j}$ for the $j$ -th component, and by multiplying $\left[\begin{matrix}2c_{j}+N\\ 2c_{j}+1\end{matrix}\right]^{-1}$ , we get the ADO invariant $\operatorname{ADO}_{N}(K^{c_{1},\cdots,c_{\ell}})$ corresponding to the blackboard framing of $K$ . Especially, the framings of a link diagram $K$ are all zero, them $\operatorname{ADO}_{N}(K^{c_{1},\cdots,c_{\ell}})$ is a link invariant of $K$ .

A.2. ADO invariant for colored knotted graphs

By introducing operators corresponding to trivalent vertices, the ADO invariant is generalized to colored knotted graphs as in [4]. The ADO invariant is defined for a root of unity $q=e^{2\pi i/N}$ and the colors assigned to edges must contained in $\left(\mathbb{C}\setminus\mathbb{Z}/2\right)\cup N\mathbb{Z}/2$ . In the following, we sometimes consider colors in $\mathbb{Z}/2$ , and in such case, the corresponding invariant is considered to be a limit of the invariants with non-half-integer colors.

Definition A.1.

A coloring of a knotted graph is admissible if the three colors $a$ , $b$ , $c$ of three edges around a vertex must satisfy the following condition.

	$\displaystyle\begin{matrix}\includegraphics[scale={0.8}]{admissible0}\end{matrix}$	$\displaystyle\qquad a+b+c=-2N+2,-2N+3,\cdots,-N+1,$
	$\displaystyle\begin{matrix}\includegraphics[scale={0.8}]{admissible1}\end{matrix}$	$\displaystyle\qquad a+b-c=-N+1,-N+2,\cdots,0,$
	$\displaystyle\begin{matrix}\includegraphics[scale={0.8}]{admissible21}\end{matrix}$	$\displaystyle\qquad a+b-c=0,1,\cdots,N-1,$
	$\displaystyle\begin{matrix}\includegraphics[scale={0.8}]{admissible3}\end{matrix}$	$\displaystyle\qquad a+b+c=N-1,N,\cdots,2N-2.$

In the rest, we only consider admissible colorings.

The ADO invariant for knotted graphs satisfies the following relations.

	$\displaystyle\mathrm{ADO}_{N}(\bigcirc^{a})=\left[\begin{matrix}2a+N\\ 2a+1\end{matrix}\right]^{-1},$		(21)
	$\displaystyle\mathrm{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.8}]{bubble}\end{matrix}\right)=\delta_{ad}\left[\begin{matrix}2a+N\\ 2a+1\end{matrix}\right]\ \mathrm{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.8}]{bubble0}\end{matrix}\right),$		(22)
	$\displaystyle\mathrm{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.8}]{parallel2}\end{matrix}\right)=\sum_{a+b-c=0,1,\cdots,N-1}\left[\begin{matrix}2c+N\\ 2c+1\end{matrix}\right]^{-1}\,\mathrm{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.8}]{parallel3}\end{matrix}\right),$		(23)
	$\displaystyle\mathrm{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.8}]{fp}\end{matrix}\right)=q^{2t_{a}}\,\mathrm{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.8}]{bubble0}\end{matrix}\right),$		(24)
	$\displaystyle\mathrm{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.8}]{fm}\end{matrix}\right)=q^{-2t_{a}}\mathrm{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.8}]{bubble0}\end{matrix}\right),$		(25)
	$\displaystyle\mathrm{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.8}]{twistp}\end{matrix}\right)=q^{t_{a}-t_{b}-t_{c}}\mathrm{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.8}]{twist0}\end{matrix}\right),$		(26)
	$\displaystyle\mathrm{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.8}]{twistm}\end{matrix}\right)=q^{-(t_{a}-t_{b}-t_{c})}\mathrm{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.8}]{twist0}\end{matrix}\right),$		(27)
	$\displaystyle\mathrm{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.8}]{bubble0}\end{matrix}\right)=\mathrm{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.8}]{bubble1}\end{matrix}\right)\quad\text{(dual representation)}.$		(28)

By using the above relations, we get the following relation.

Lemma A.1.

We can remove a circle around an edge as follows.

\mathrm{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.8}]{hopfab}\end{matrix}\right)=i^{N-1}q^{(2a+1-N)(2b+1-N)}\{2a+N,N-1\}\,\mathrm{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.8}]{bubble0}\end{matrix}\right).

(29)

Proof.

The lefthand side of the formula is computed as follows.

	$\displaystyle\mathrm{ADO}_{N}$	$\displaystyle\left(\begin{matrix}\includegraphics[scale={0.8}]{hopfab}\end{matrix}\right)\underset{\eqref{eq:ADOparallel}}{=}\sum_{a+b-c=0,1,\cdots,N-1}\left[\begin{matrix}2c+N\\ 2c+1\end{matrix}\right]^{-1}\,\mathrm{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.8}]{hopfab1}\end{matrix}\right)$
	$\displaystyle\underset{\eqref{eq:ADOtwist3p}}{=}$	$\displaystyle\sum_{k=0}^{N-1}q^{2(t_{a+b-k}-t_{a}-t_{b})}\left[\begin{matrix}2(a+b-k)+N\\ 2(a+b-k)+1\end{matrix}\right]^{-1}\,\mathrm{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.8}]{hopfab2}\end{matrix}\right)\hfill$
	$\displaystyle=$	$\displaystyle\frac{\{N-1\}!\,q^{-t_{a}-t_{b}}}{\{2(a+b)+N\}\cdots\{2(a+b)+1\}}\times\hfill$
		$\displaystyle\qquad\qquad\qquad\qquad\sum_{k=0}^{N-1}\{2(a+b-k)+1\}\,q^{t_{a+b-k}}\mathrm{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.8}]{hopfab2}\end{matrix}\right)$
	$\displaystyle\underset{\eqref{eq:ADOtheta}}{=}$	$\displaystyle\frac{\{N-1\}!\,q^{-t_{a}-t_{b}}}{\{2(a+b)+N\}\cdots\{2(a+b)+1\}}\left[\begin{matrix}2a+N\\ 2a+1\end{matrix}\right]\times\hfill$
		$\displaystyle\qquad\qquad\qquad\qquad\sum_{k=0}^{N-1}\{2(a+b-k)+1\}\,q^{t_{a+b-k}}\mathrm{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.8}]{bubble0}\end{matrix}\right)$
	$\displaystyle{=}$	$\displaystyle\frac{q^{-t_{a}-t_{b}}\,\{2a+N,N-1\}}{\{2(a+b)+N,N\}}\sum_{k=0}^{N-1}\{2(a+b-k)+1\}\,q^{t_{a+b-k}}\mathrm{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.8}]{bubble0}\end{matrix}\right).$

Now we compute

	$\displaystyle\sum_{k=0}^{N-1}\{2(a+b-k)+1\}\,q^{t_{a+b-k}}=\sum_{k=0}^{N-1}(q^{2(a+b-k)+1}-q^{-2(a+b-k)-1})\,q^{2\left((a+b-k-\frac{N-1}{2})^{2}-\frac{(N-1)^{2}}{4}\right)}$
	$\displaystyle=-q^{-\frac{(N-1)^{2}}{2}}\sum_{k=0}^{N-1}(q^{2(a+b-k)+1-N}-q^{-2(a+b-k)-1+N})\,q^{\frac{1}{2}(2(a+b-k)+1-N)^{2}}\hfill$
	$\displaystyle=-q^{-\frac{(N-1)^{2}}{2}}\sum_{k=0}^{N-1}\left(q^{\frac{1}{2}(2(a+b-k)+2-N)^{2}-\frac{1}{2}}-q^{\frac{1}{2}(2(a+b-k)-N)^{2}-\frac{1}{2}}\right)\hfill$
	$\displaystyle=-q^{-\frac{(N-1)^{2}+1}{2}}\left(q^{2(a+b+1-N+\frac{N}{2})^{2}}-q^{2(a+b+1-N-\frac{N}{2})^{2}}\right)\hfill$
	$\displaystyle=-q^{-\frac{(N-1)^{2}+1}{2}}\left(q^{2\left((a+b+1-N)^{2}+N(a+b+1-N)+\frac{N^{2}}{4}\right)}-q^{2\left((a+b+1-N)^{2}-N(a+b+1-N)+\frac{N^{2}}{4}\right)}\right)$
	$\displaystyle=q^{2a^{2}+2b^{2}+4ab+4a+4b+1-4Na-4Nb}\{2N(a+b)\}.$

Since

2a^{2}+2b^{2}+4ab+4a+4b+1-4Na-4Nb-2t_{a}-2t_{b}=\\ (2a+1-N)(2b+1-N)-N^{2}

and

\{2(a+b)+N,N\}=-i^{N-1}\{2N(a+b)\},

we have

\frac{q^{-t_{a}-t_{b}}\,\{2a+N,N-1\}}{\{2(a+b)+N,N\}}\sum_{k=0}^{N-1}\{2(a+b-k)+1\}\,q^{t_{a+b-k}}\mathrm{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.8}]{bubble0}\end{matrix}\right)\\ =q^{(2a+1-N)(2b+1-N)-N^{2}}\{2a+N,N-1\}\frac{\{2N(a+b)\}}{-i^{N-1}\{2N(a+b)\}}\\ =\frac{(-1)^{N-1}}{i^{N-1}}q^{(2a+1-N)(2b+1-N)}\{2a+N,N-1\}\\ =i^{N-1}q^{(2a+1-N)(2b+1-N)}\{2a+N,N-1\},

and we get (29). ∎

A.3. Quantum $6j$ symbol

The quantum $6j$ symbol of the ADO invariant is the ADO invariant for the tetrahedral graph labeled as in Figure 23.

Figure 23. The oriented tetrahedral graph labeled by

a

b

c

d

e

f

The quantum $6j$ symbol $\left\{\begin{matrix}a&b&e\\ d&c&f\end{matrix}\right\}_{q}$ is given in [4] as follows. Let

A_{xyz}=x+y+z,\qquad B_{xyz}=x+y-z.

Then

\left\{\begin{matrix}a&b&e\\ d&c&f\end{matrix}\right\}_{q}=(-1)^{N-1}\frac{\{B_{dec}\}!\{B_{abe}\}!}{\{B_{bdf}\}!\{B_{afc}\}!}\left[\,\begin{matrix}2e\\ A_{abe}+1-N\end{matrix}\right]\left[\,\begin{matrix}2e\\ B_{ced}\end{matrix}\,\right]^{-1}\,\times\\ \sum_{s=\max(0,-B_{bdf}+B_{dec})}^{\min(B_{dec},B_{afc})}\left[\,\begin{matrix}A_{acf}+1-N\\ 2c+s+1-N\end{matrix}\right]\left[\,\begin{matrix}B_{acf}+s\\ B_{acf}\end{matrix}\,\right]\left[\,\begin{matrix}B_{bfd}+B_{dec}-s\\ B_{bfd}\end{matrix}\,\right]\left[\,\begin{matrix}B_{cde}+s\\ B_{dfb}\end{matrix}\,\right].

(30)

Lemma A.2.

By using the quantum $6j$ symbol, we can remove a triangle in the colored knotted graph as follows.

\operatorname{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.8}]{triangle0}\end{matrix}\right)=\left\{\begin{matrix}a&b&e\\ d&c&f\end{matrix}\right\}_{q}\operatorname{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.8}]{vertex0}\end{matrix}\right).

(31)

\operatorname{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.8}]{triangle}\end{matrix}\right)=\left\{\begin{matrix}a&b&e\\ d&c&f\end{matrix}\right\}_{q}\operatorname{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.8}]{vertex}\end{matrix}\right).

(32)

Proof.

The above two relations comes from the following formulas.

\operatorname{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.8}]{alexandertet}\end{matrix}\right)=\left\{\begin{matrix}a&b&e\\ d&c&f\end{matrix}\right\}_{q},\quad\operatorname{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.8}]{theta1}\end{matrix}\right)=\operatorname{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.8}]{theta}\end{matrix}\right)=1.

The second formula comes from (21) and (22). ∎

Lemma A.3.

The ADO invariant of the colored tetrahedral graph given in Figure 23 with colors

a=\frac{N-1}{2},\ b=\frac{N-1}{2},\ c=\frac{N-1}{2}+\varepsilon,\ d=\frac{N-1}{2}+\varepsilon,\ e=l+\varepsilon,\ f=k+\varepsilon

is the following.

\left\{\begin{matrix}\frac{N-1}{2}&\frac{N-1}{2}&l\\ \frac{N-1}{2}+\varepsilon&\frac{N-1}{2}+\varepsilon&k+\varepsilon\end{matrix}\right\}_{q}=\frac{\{N-1+2\varepsilon,N-1\}}{\{N-1\}!}\,\times\\ \hfill\sum_{s=\max(k,l)}^{\min(k+l,N-1)}\frac{\{s\}!^{2}}{\{s-k\}!\{s-l\}!^{2}\{k+l-s\}!\{s-k-2\varepsilon,s-k\}\{k+l-s+2\varepsilon,k+l-s\}}.

(33)

Especially, if $\varepsilon=0$ , then we have

\left\{\begin{matrix}\frac{N-1}{2}&\frac{N-1}{2}&l\\ \frac{N-1}{2}&\frac{N-1}{2}&k\end{matrix}\right\}_{q}=\sum_{s=\max(0,l+k-N+1)}^{\min(k,l)}\frac{\{s\}!^{2}}{\{s-k\}!^{2}\{s-l\}!^{2}\{k+l-s\}!^{2}}.

(34)

Moreover, we have the following.

\left\{\begin{matrix}\frac{N-1}{2}+\delta&\frac{N-1}{2}+\varepsilon&l+\varepsilon+\delta\\ \frac{N-1}{2}-\delta&\frac{N-1}{2}+\varepsilon&k+\varepsilon-\delta\end{matrix}\right\}_{q}=\\ \hfill\frac{\{N-1-2\delta,N-1\}}{\{N-1\}!}\sum_{s=\max(k,l)}^{\min(k+l,N-1)}\frac{\{s\}!}{\{s-k\}!\{s-l\}!\{k+l-s\}!}\,\times\hfill\\ \frac{\{s+2\varepsilon,s\}}{\{s-k+2\delta,s-k\}\{s-l-2\delta,s-l\}\{k+l-s+2\varepsilon,k+l-s\}},

(35)

$\displaystyle\left\{\begin{matrix}l&\frac{N-1}{2}&\frac{N-1}{2}\\[4.0pt] \varepsilon&\frac{N-1}{2}+\varepsilon&\frac{N-1}{2}+\varepsilon\end{matrix}\right\}_{q}$	$\displaystyle=\frac{\{l+2\varepsilon,l\}}{\{l\}!},$	(36)
$\displaystyle\left\{\begin{matrix}l&\frac{N-1}{2}-\frac{\varepsilon}{2}&\frac{N-1}{2}-\frac{\varepsilon}{2}\\[4.0pt] \varepsilon&\frac{N-1}{2}+\frac{\varepsilon}{2}&\frac{N-1}{2}+\frac{\varepsilon}{2}\end{matrix}\right\}_{q}$	$\displaystyle=\frac{\{N-1-\varepsilon,N-1\}}{\{N-1\}!},$	(37)
$\displaystyle\left\{\begin{matrix}-\varepsilon&\frac{N-1}{2}&\frac{N-1}{2}-\varepsilon\\[4.0pt] N-1-l+\varepsilon&\frac{N-1}{2}&\frac{N-1}{2}+\varepsilon\end{matrix}\right\}_{q}$	$\displaystyle=\frac{\{l-2\varepsilon,l\}}{\{l\}!},\qquad\quad$	(38)
$\displaystyle\left\{\begin{matrix}\frac{N-1}{2}-\frac{\varepsilon+\delta}{2}&\frac{N-1}{2}+\frac{\varepsilon-\delta}{2}&-\delta\\[4.0pt] \frac{N-1}{2}+\frac{\varepsilon+\delta}{2}&\frac{N-1}{2}+\frac{\varepsilon-\delta}{2}&k+\varepsilon\end{matrix}\right\}_{q}$	$\displaystyle=\frac{\{k+\varepsilon-\delta,k\}\{N-1+\varepsilon+\delta\}}{\{k+\varepsilon+\delta,k\}!\{N-1\}!},\qquad\quad$	(39)
$\displaystyle\left\{\begin{matrix}l+\delta&\frac{N-1}{2}-\frac{\varepsilon+\delta}{2}&\frac{N-1}{2}-\frac{\varepsilon-\delta}{2}\\[4.0pt] \varepsilon&\frac{N-1}{2}+\frac{\varepsilon+\delta}{2}&\frac{N-1}{2}+\frac{\varepsilon-\delta}{2}\end{matrix}\right\}_{q}$	$\displaystyle=\frac{\{l+\varepsilon+\delta,l\}}{\{l-\varepsilon+\delta,l\}}.\qquad\quad$	(40)

Proof.

First we prove (33). We have $B_{dec}=l$ , $B_{abe}=N-1-l$ , $B_{bdf}=N-1-k$ , $B_{afc}=k$ and

\left\{\begin{matrix}\frac{N-1}{2}&\frac{N-1}{2}&l\\ \frac{N-1}{2}+\varepsilon&\frac{N-1}{2}+\varepsilon&k+\varepsilon\end{matrix}\right\}_{q}=\\ (-1)^{N-1}\frac{\{{l}\}!\{{N-1-l}\}!}{\{{N-1-k}\}!\{{k}\}!}\left[\,\begin{matrix}2l\\ l\end{matrix}\right]\left[\,\begin{matrix}2l\\ l\end{matrix}\,\right]^{-1}\,\times\\ \sum_{s=\max(0,l+k-N+1)}^{\min(k,l)}\left[\,\begin{matrix}k+2\varepsilon\\ s+2\varepsilon\end{matrix}\,\right]\left[\,\begin{matrix}N-1-k+s\\ N-1-k\end{matrix}\,\right]\left[\,\begin{matrix}k+l-s\\ k\end{matrix}\,\right]\left[\,\begin{matrix}N-1-l+s+2\varepsilon\\ k+2\varepsilon\end{matrix}\,\right]\\ =\sum_{s=\max(0,l+k-N+1)}^{\min(k,l)}\left[\,\begin{matrix}k+2\varepsilon\\ s+2\varepsilon\end{matrix}\,\right]\left[\,\begin{matrix}N-1-k+s\\ N-1-k\end{matrix}\,\right]\left[\,\begin{matrix}k+l-s\\ k\end{matrix}\,\right]\left[\,\begin{matrix}N-1-l+s+2\varepsilon\\ k+2\varepsilon\end{matrix}\,\right].

By replacing $s$ to $k+l-s$ , we get

\left\{\begin{matrix}\frac{N-1}{2}&\frac{N-1}{2}&l\\ \frac{N-1}{2}+\varepsilon&\frac{N-1}{2}+\varepsilon&k+\varepsilon\end{matrix}\right\}_{q}=\\ \sum_{s=\max(k,l)}^{\min(k+l,N-1)}\left[\,\begin{matrix}k+2\varepsilon\\ k+l-s+2\varepsilon\end{matrix}\,\right]\left[\,\begin{matrix}N-1+l-s\\ N-1-k\end{matrix}\,\right]\left[\,\begin{matrix}s\\ k\end{matrix}\,\right]\left[\,\begin{matrix}N-1+k-s+2\varepsilon\\ k+2\varepsilon\end{matrix}\,\right]\\[6.0pt] =\frac{\{k+2\varepsilon,k\}}{\{N-1-k\}!\{k\}!\{k+2\varepsilon,k\}}\,\times\hfill\\ \sum_{s=\max(k,l)}^{\min(k+l,N-1)}\frac{\{N-1+l-s\}!\{s\}!\{N-1+k-s+2\varepsilon,N-1+k-s\}}{\{s-l\}!\{k+l-s+2\varepsilon,k+l-s\}\{k+l-s\}!\{s-k\}!\{N-1-s\}!}\\[6.0pt] =\frac{\{N-1+2\varepsilon,N-1\}}{\{N-1\}!}\sum_{s=\max(k,l)}^{\min(k+l,N-1)}\frac{\{s\}!^{2}}{\{s-k\}!\{s-l\}!^{2}\{k+l-s\}!}\,\times\hfill\\ \frac{1}{\{s-k-2\varepsilon,s-k\}\{k+l-s+2\varepsilon,k+l-s\}}.

Next, we prove (35). We have $B_{dec}=l$ , $B_{abe}=N-1-l$ , $B_{bdf}=N-1-k$ , $B_{afc}=k$ and

\left\{\begin{matrix}\frac{N-1}{2}+\delta&\frac{N-1}{2}+\varepsilon&l+\varepsilon+\delta\\ \frac{N-1}{2}-\delta&\frac{N-1}{2}+\varepsilon&k+\varepsilon-\delta\end{matrix}\right\}_{q}=\\ (-1)^{N-1}\frac{\{{l}\}!\{{N-1-l}\}!}{\{{N-1-k}\}!\{{k}\}!}\left[\,\begin{matrix}2l+2\varepsilon+2\delta\\ l+2\varepsilon+2\delta\end{matrix}\right]\left[\,\begin{matrix}2l+2\varepsilon+2\delta\\ l+2\varepsilon+2\delta\end{matrix}\,\right]^{-1}\,\times\\ \sum_{s=\max(0,l+k-N+1)}^{\min(k,l)}\left[\,\begin{matrix}k+2\varepsilon\\ s+2\varepsilon\end{matrix}\,\right]\left[\,\begin{matrix}N-1-k+s+2\delta\\ N-1-k+2\delta\end{matrix}\,\right]\left[\,\begin{matrix}k+l-s+2\varepsilon\\ k+2\varepsilon\end{matrix}\,\right]\left[\,\begin{matrix}N-1-l+s-2\delta\\ k-2\delta\end{matrix}\,\right]\\ =\sum_{s=\max(0,l+k-N+1)}^{\min(k,l)}\text{\small$\displaystyle\left[\,\begin{matrix}k+2\varepsilon\\ s+2\varepsilon\end{matrix}\,\right]\left[\,\begin{matrix}N-1-k+s+2\delta\\ N-1-k+2\delta\end{matrix}\,\right]\left[\,\begin{matrix}k+l-s+2\varepsilon\\ k+2\varepsilon\end{matrix}\,\right]\left[\,\begin{matrix}N-1-l+s-2\delta\\ k-2\delta\end{matrix}\,\right]$}.

By replacing $s$ to $k+l-s$ , we get

\left\{\begin{matrix}\frac{N-1}{2}+\delta&\frac{N-1}{2}+\varepsilon&l+\varepsilon+\delta\\ \frac{N-1}{2}-\delta&\frac{N-1}{2}+\varepsilon&k+\varepsilon-\delta\end{matrix}\right\}_{q}=\\ \sum_{s=\max(k,l)}^{\min(k+l,N-1)}\left[\,\begin{matrix}k+2\varepsilon\\ k+l-s+2\varepsilon\end{matrix}\,\right]\left[\,\begin{matrix}N-1+l-s+2\delta\\ N-1-k+2\delta\end{matrix}\,\right]\left[\,\begin{matrix}s+2\varepsilon\\ k+2\varepsilon\end{matrix}\,\right]\left[\,\begin{matrix}N-1+k-s-2\delta\\ k-2\delta\end{matrix}\,\right]\\[6.0pt] =\frac{\{k+2\varepsilon,k\}}{\{N-1-k+2\delta,N-1-k\}\{k+2\varepsilon,k\}\{k-2\delta,k\}}\,\times\hfill\\ \sum_{s=\max(k,l)}^{\min(k+l,N-1)}{\text{\footnotesize$\frac{\{N-1+l-s+2\delta,N-1+l-s\}\{s+2\varepsilon,s\}\{N-1+k-s-2\delta,N-1+k-s\}}{\{s-l\}!\{k+l-s+2\varepsilon,k+l-s\}\{k+l-s\}!\{s-k\}!\{N-1-s\}!}$ }}\\[6.0pt] =\frac{\{N-1-2\delta,N-1\}}{\{N-1\}!}\sum_{s=\max(k,l)}^{\min(k+l,N-1)}\frac{\{s\}!}{\{N-1\}!\{s-k\}!\{s-l\}!\{k+l-s\}!}\,\times\hfill\\ \frac{\{s+2\varepsilon,s\}}{\{s-k+2\delta,s-k\}\{s-l-2\delta,s-l\}\{k+l-s+2\varepsilon,k+l-s\}}.

The relations (36), (37), (38), (39) and (40) are proved as follows.

\left\{\begin{matrix}l&\frac{N-1}{2}&\frac{N-1}{2}\\[4.0pt] \varepsilon&\frac{N-1}{2}+\varepsilon&\frac{N-1}{2}+\varepsilon\end{matrix}\right\}_{q}=\frac{\{0\}!\{l\}!}{\{0\}!\{l\}!}\left[\begin{matrix}N-1\\ l\end{matrix}\right]\left[\begin{matrix}N-1\\ N-1\end{matrix}\right]^{-1}\left[\begin{matrix}l+2\varepsilon\\ 2\varepsilon\end{matrix}\right]\left[\begin{matrix}N-1\\ N-1\end{matrix}\right]\left[\begin{matrix}2\varepsilon\\ 2\varepsilon\end{matrix}\right]\\ =\frac{\{l+2\varepsilon,l\}}{\{l\}!},

\left\{\begin{matrix}l&\frac{N-1}{2}-\frac{\varepsilon}{2}&\frac{N-1}{2}-\frac{\varepsilon}{2}\\[4.0pt] \varepsilon&\frac{N-1}{2}+\frac{\varepsilon}{2}&\frac{N-1}{2}+\frac{\varepsilon}{2}\end{matrix}\right\}_{q}\\ =\frac{\{0\}!\{l\}!}{\{0\}!\{l\}!}\left[\begin{matrix}N-1-\varepsilon\\ l-\varepsilon\end{matrix}\right]\left[\begin{matrix}N-1-\varepsilon\\ N-1-\varepsilon\end{matrix}\right]^{-1}\left[\begin{matrix}l+\varepsilon\\ \varepsilon\end{matrix}\right]\left[\begin{matrix}N-1-\varepsilon\\ N-1-\varepsilon\end{matrix}\right]\left[\begin{matrix}2\varepsilon\\ 2\varepsilon\end{matrix}\right]\\ =\frac{\{N-1-\varepsilon,N-1\}}{\{N-1\}!},

\left\{\begin{matrix}-\varepsilon&\frac{N-1}{2}&\frac{N-1}{2}-\varepsilon\\[4.0pt] N-1-l+\varepsilon&\frac{N-1}{2}&\frac{N-1}{2}+\varepsilon\end{matrix}\right\}_{q}=\\ \frac{\{N-1-l\}!\{0\}!}{\{N-1-l\}!\{0\}!}\left[\begin{matrix}N-1-2\varepsilon\\ -2\varepsilon\end{matrix}\right]\left[\begin{matrix}N-1-2\varepsilon\\ l-2\varepsilon\end{matrix}\right]^{-1}\left[\begin{matrix}0\\ 0\end{matrix}\right]\left[\begin{matrix}N-1\\ l\end{matrix}\right]\left[\begin{matrix}N-1-l+2\varepsilon\\ N-1-l+2\varepsilon\end{matrix}\right]\\ =\frac{\{l-2\varepsilon,l\}}{\{l\}!}.

\left\{\begin{matrix}\frac{N-1}{2}-\frac{\varepsilon+\delta}{2}&\frac{N-1}{2}+\frac{\varepsilon-\delta}{2}&-\delta\\[4.0pt] \frac{N-1}{2}+\frac{\varepsilon+\delta}{2}&\frac{N-1}{2}+\frac{\varepsilon-\delta}{2}&k+\varepsilon\end{matrix}\right\}_{q}=\\ \frac{\{0\}!\{N-1\}!}{\{N-1-k\}!\{k\}!}\left[\begin{matrix}-2\delta\\ -2\delta\end{matrix}\right]\left[\begin{matrix}-2\delta\\ -2\delta\end{matrix}\right]^{-1}\left[\begin{matrix}k+\varepsilon-\delta\\ \varepsilon-\delta\end{matrix}\right]\left[\begin{matrix}N-1+\varepsilon+\delta\\ k+\varepsilon+\delta\end{matrix}\right]\\ =\frac{\{k+\varepsilon-\delta,k\}\{N-1+\varepsilon+\delta,N-1\}}{\{k+\varepsilon+\delta,k\}\{N-1\}!}.

\left\{\begin{matrix}l+\delta&\frac{N-1}{2}-\frac{\varepsilon+\delta}{2}&\frac{N-1}{2}-\frac{\varepsilon-\delta}{2}\\[4.0pt] \varepsilon&\frac{N-1}{2}+\frac{\varepsilon+\delta}{2}&\frac{N-1}{2}+\frac{\varepsilon-\delta}{2}\end{matrix}\right\}_{q}=\\ \frac{\{0\}!\{l\}!}{\{0\}!\{l\}!}\left[\begin{matrix}N-1-\varepsilon+\delta\\ l-\varepsilon+\delta\end{matrix}\right]\left[\begin{matrix}N-1-\varepsilon+\delta\\ -\varepsilon+\delta\end{matrix}\right]^{-1}\left[\begin{matrix}l+\varepsilon+\delta\\ \varepsilon+\delta\end{matrix}\right]\left[\begin{matrix}2\varepsilon\\ 2\varepsilon\end{matrix}\right]=\frac{\{l+\varepsilon+\delta,l\}}{\{l-\varepsilon+\delta,l\}}.

∎

A.4. Symmetry

Here we introduce the notion of symmetry for a function defined on the set $\{0,1,2,\cdots,N-1\}$ .

Definition A.2.

A function $f$ defined on $\{0,1,2,\cdots,N-1\}$ is called symmetric if $f(k)=f(N-1-k)$ , and is called anti-symmetric if $f(k)=-f(N-1-k)$ .

Lemma A.4.

Let

\xi_{N}(k,l,s)=\dfrac{\{s\}!^{2}}{\{s-k\}!^{2}\,\{s-l\}!^{2}\,\{k+l-s\}!^{2}}.

(41)

Then it satisfies

\xi_{N}(k,l,s)=\xi_{N}(N-1-k,l,N-1-s+l)=\xi_{N}(k,N-1-l,N-1-s+k)\\ =\xi_{N}(N-1-k,N-1-l,N-1-k-l+s).

(42)

Proof.

We have

\xi_{N}(N-k,l,N-1-s+l)=\dfrac{\{N-1-s+l\}!^{2}}{\{k+l-s\}!^{2}\{N-1-s\}!^{2}\{s-k\}!^{2}}\\ =\dfrac{\{s\}!^{2}}{\{s-l\}!^{2}\,\{k+l-s\}!^{2}\,\{s-k\}!^{2}}=\xi_{N}(k,l,s).

Similarly, we have

\xi_{N}(k,N-l,N-1-s+k)=\dfrac{\{N-1-s+k\}!^{2}}{\{N-1-s\}!^{2}\{k+l-s\}!^{2}\{s-l\}!^{2}}\\ =\dfrac{\{s\}!^{2}}{\{s-k\}!^{2}\,\{k+l-s\}!^{2}\,\{s-l\}!^{2}}=\xi_{N}(k,l,s).

Combining these two, we get the last equality. ∎

These relations imply the following symmetry of the quantum $6j$ symbols.

Proposition A.1.

The quantum $6j$ symbol defined by the ADO invariant satisfies the following symmetry.

\left\{\begin{matrix}\frac{N-1}{2}&\frac{N-1}{2}&l\\ \frac{N-1}{2}&\frac{N-1}{2}&k\end{matrix}\right\}_{q}=\left\{\begin{matrix}\frac{N-1}{2}&\frac{N-1}{2}&l\\ \frac{N-1}{2}&\frac{N-1}{2}&N-1-k\end{matrix}\right\}_{q}=\\ \left\{\begin{matrix}\frac{N-1}{2}&\frac{N-1}{2}&N-1-l\\ \frac{N-1}{2}&\frac{N-1}{2}&k\end{matrix}\right\}_{q}=\left\{\begin{matrix}\frac{N-1}{2}&\frac{N-1}{2}&N-1-l\\ \frac{N-1}{2}&\frac{N-1}{2}&N-1-k\end{matrix}\right\}_{q}.

(43)

In other words, $\left\{\begin{matrix}\frac{N-1}{2}&\frac{N-1}{2}&l\\ \frac{N-1}{2}&\frac{N-1}{2}&k\end{matrix}\right\}_{q}$ is symmetric with respect to $k$ and $l$ .

Proof.

We prove the first equality.

\left\{\begin{matrix}\frac{N-1}{2}&\frac{N-1}{2}&l\\ \frac{N-1}{2}&\frac{N-1}{2}&k\end{matrix}\right\}_{q}=\sum_{s=\max(k,l)}^{\min(N-1,k+l)}\xi_{N}(k,l,s)\underset{\eqref{eq:klsymmetry}}{=}\\ \sum_{s=\max(k,l)}^{\min(N-1,k+l)}\xi_{N}(N-1-k,l,N-1+l-s)=\sum_{s=\max(N-1-k,l)}^{\min(N-1,N-1-k+l)}\xi_{N}(N-1-k,l,s)\\ =\left\{\begin{matrix}\frac{N-1}{2}&\frac{N-1}{2}&l\\ \frac{N-1}{2}&\frac{N-1}{2}&N-1-k\end{matrix}\right\}_{q}.

The other equalities are proved similarly. ∎

Appendix B. Colored Jones invariants of some links

Here we compute the colored Jones invariant $J_{N-1}(K)$ for $K=B$ , $B_{1}$ , $B_{1,1}$ , $W$ , $W_{P}$ , $T_{p}$ and $D_{p,r}$ given in Figure 1.

B.1. Colored Jones invariants and ADO invariants

We compute $J_{N-1}(K)$ by using the ADO invariant.

Proposition B.1.

For a framed link $K$ , the following holds.

J_{N-1}(K)=(-1)^{N-1}\operatorname{ADO}_{N}(K^{\frac{N-1}{2},\cdots,\frac{N-1}{2}}).

Proof.

The invariants $J_{N-1}(K)$ and $\operatorname{ADO}_{N}(K^{\frac{N-1}{2},\cdots,\frac{N-1}{2}})$ are constructed from the same $R$ matrix since $J_{N-1}(K)$ is the colored Jones invariant corresponding to the $N$ dimensional representation $V^{(N)}$ of $\mathcal{U}_{q}(sl_{2})$ at $q=e^{\pi i/N}$ . Let $T$ be a $(1,1)$ tangle whose closure is isotopic to $K$ , then $T$ determines a scalar operator $\alpha\,\mathrm{id}:V^{(N)}\to V^{(N)}$ by assigning the $R$ matrix to each crossing of $T$ and the factor for the minimal and maximal points. Then $J_{N-1}(K)=\alpha$ . On the other hand,

\operatorname{ADO}_{(N)}(K)=\left[\begin{matrix}2N-1\\ N\end{matrix}\right]^{-1}\alpha=\frac{\{2N-1\}\{2N-2\}\cdots\{N+1\}}{\{N-1\}\{N-2\}\cdots\{1\}}=(-1)^{N-1}\alpha.

Hence we have $J_{N-1}(K)=(-1)^{N-1}\operatorname{ADO}_{N}(K)$ . ∎

In this paper, $N$ is assumed to be odd and we have

J_{N-1}(K)=ADO_{N}(K).

(44)

Remark 3.

The knots treated in this paper is all colored by $N-1$ and their colored Jones polynomial $J_{N-1}$ and their ADO invariant $\operatorname{ADO}_{N}$ are not depend on the framings of them.

B.2. Borromean rings and their variants

Here we compute the ADO invariants of the Borromean rings $B$ and its variants $B_{1}$ , $B_{1,1}$ .

Proposition B.2.

The ADO invariants of the Borromean rings $B$ and its variants $B_{1}$ , $B_{1,1}$ are given as follows.

J_{N-1}(B)=N^{2}\sum_{k,l=0}^{N-1}\sum_{s=\max(k,l)}^{\min(k+l,N-1)}\dfrac{\{s\}!^{2}}{\{s-k\}!^{2}\,\{s-l\}!^{2}\,\{k+l-s\}!^{2}},

(45)

J_{N-1}(B_{1})=N^{2}q^{-\frac{(N-1)^{2}}{2}}\sum_{k,l=0}^{N-1}\sum_{s=\max(k,l)}^{\min(k+l,N-1)}\dfrac{q^{\left(k-\frac{N-1}{2}\right)^{2}}\{s\}!^{2}}{\{s-k\}!^{2}\,\{s-l\}!^{2}\,\{k+l-s\}!^{2}},

(46)

J_{N-1}(B_{1,1})=N^{2}q^{-\frac{(N-1)^{2}}{2}}\sum_{k,l=0}^{N-1}\sum_{s=\max(k,l)}^{\min(k+l,N-1)}\dfrac{q^{\left(k-\frac{N-1}{2}\right)^{2}}q^{\left(l-\frac{N-1}{2}\right)^{2}}\{s\}!^{2}}{\{s-k\}!^{2}\,\{s-l\}!^{2}\,\{k+l-s\}!^{2}}.

(47)

Proof.

We compute the ADO invariants instead of the colored Jones invariant. For $B$ , $\operatorname{ADO}^{(N)}(B)$ is computed as follows.

	$\displaystyle\operatorname{ADO}^{(N)}(B)=\operatorname{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.8}]{borromeanADO}\end{matrix}\right)$
	$\displaystyle=\sum_{k,l=0}^{N-1}\left[\begin{matrix}2k+N\\ 2k+1\end{matrix}\right]^{-1}\left[\begin{matrix}2l+N\\ 2l+1\end{matrix}\right]^{-1}\operatorname{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.8}]{borromeanADO1}\end{matrix}\right)$
	$\displaystyle=\ \sum_{k,l=0}^{N-1}\left[\begin{matrix}2k+N\\ 2k+1\end{matrix}\right]^{-1}\left[\begin{matrix}2l+N\\ 2l+1\end{matrix}\right]^{-1}\left<\begin{matrix}\includegraphics[scale={0.8}]{hopfk}\end{matrix}\right>\left<\begin{matrix}\includegraphics[scale={0.8}]{hopfl}\end{matrix}\right>\left<\begin{matrix}\includegraphics[scale={0.8}]{borromeanADO2}\end{matrix}\right>$
	$\displaystyle=\sum_{k=0}^{N-1}\left[\begin{matrix}2k+N\\ 2k+1\end{matrix}\right]^{-1}i^{N-1}\{2k+N,N-1\}\,\times$
	$\displaystyle\qquad\qquad\qquad\qquad\sum_{l=0}^{N-1}\left[\begin{matrix}2l+N\\ 2l+1\end{matrix}\right]^{-1}i^{N-1}\{2l+N,N-1\}\,\left\{\begin{matrix}\frac{N-1}{2}&\frac{N-1}{2}&l\\[4.0pt] \frac{N-1}{2}&\frac{N-1}{2}&k\end{matrix}\right\}_{q}$
	$\displaystyle=\sum_{k,l=0}^{N-1}\{N-1\}!^{2}\sum_{s=m}^{M}\dfrac{\{s\}!^{2}}{\{s-k\}!^{2}\,\{s-l\}!^{2}\,\{k+l-s\}!^{2}}$
	$\displaystyle\qquad\qquad\qquad\qquad\qquad\big{(}m=\max(k,l),\ \ M=\min(k+l,N-1)\big{)}$
	$\displaystyle=(-1)^{N-1}N^{2}\sum_{k,l=0}^{N-1}\sum_{s=m}^{M}\dfrac{\{s\}!^{2}}{\{s-k\}!^{2}\,\{s-l\}!^{2}\,\{k+l-s\}!^{2}}.$

For $B_{1}$ , $\operatorname{ADO}(B_{1})$ is computed as follows.

\operatorname{ADO}^{(N)}(B_{1})=\sum_{k,l=0}^{N-1}q^{(k-\frac{N-1}{2})^{2}-\frac{(N-1)^{2}}{4}}\,\times\\[-12.0pt] \left[\begin{matrix}2k+N\\ 2k+1\end{matrix}\right]^{-1}\left[\begin{matrix}2l+N\\ 2l+1\end{matrix}\right]^{-1}\operatorname{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.8}]{borromeanADO11}\end{matrix}\right)\\ =(-1)^{N-1}N^{2}q^{-\frac{(N-1)^{2}}{4}}\sum_{k,l=0}^{N-1}\sum_{s=\max(k,l)}^{\min(k+l,N-1)}\dfrac{q^{\left(k-\frac{N-1}{2}\right)^{2}}\{s\}!^{2}}{\{s-k\}!^{2}\,\{s-l\}!^{2}\,\{k+l-s\}!^{2}}.

For $B_{1,1}$ , similar computation leads to (47). ∎

B.3. Twisted Whitehead link

For the twisted Whitehead link $W_{p}$ , the ADO invariant $\mathrm{ADO}^{(N)}(W_{p})$ is computed as follows.

\mathrm{ADO}^{(N)}(W_{p})=\mathrm{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.6}]{whiteheadADO}\end{matrix}\right)\\ =\sum_{k,l=0}^{N-1}q^{p(k-\frac{N-1}{2})^{2}-p\frac{(N-1)^{2}}{4}}\left[\begin{matrix}2k+N\\ 2k+1\end{matrix}\right]^{-1}\left[\begin{matrix}2l+N\\ 2l+1\end{matrix}\right]^{-1}\mathrm{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.6}]{whiteheadADO1}\end{matrix}\right)\\ =q^{-p\frac{(N-1)^{2}}{4}}\sum_{k,l=0}^{N-1}q^{p(k-\frac{N-1}{2})^{2}}\left[\begin{matrix}2k+N\\ 2k+1\end{matrix}\right]^{-1}\left[\begin{matrix}2l+N\\ 2l+1\end{matrix}\right]^{-1}i^{N-1}\{2l+N,N-1\}\,\left\{\begin{matrix}\frac{N-1}{2}&\frac{N-1}{2}&l\\[4.0pt] \frac{N-1}{2}&\frac{N-1}{2}&k\end{matrix}\right\}_{q}\\ =q^{-p\frac{(N-1)^{2}}{4}}\sum_{k,l=0}^{N-1}q^{p(k-\frac{N-1}{2})^{2}}\frac{\{N-1\}!^{2}\{2k+1\}}{\{2k+N,N\}}i^{N-1}\sum_{s=m}^{M}\dfrac{\{s\}!^{2}}{\{s-k\}!^{2}\,\{s-l\}!^{2}\,\{k+l-s\}!^{2}}\hfill\\ \hfill\big{(}m=\max(k,l),\ \ M=\min(k+l,N-1)\big{)}\\ =-q^{-p\frac{(N-1)^{2}}{4}}\sum_{k,l=0}^{N-1}q^{p(k-\frac{N-1}{2})^{2}}\,N^{2}\,\frac{\{2k+1\}}{\{2Nk\}}\sum_{s=m}^{M}\dfrac{\{s\}!^{2}}{\{s-k\}!^{2}\,\{s-l\}!^{2}\,\{k+l-s\}!^{2}}.\hfill

(48)

The denominator $\{2Nk\}$ of this formula is zero for integer $k$ , but the numerator is also equal to zero and it must be well-defined since $J_{N-1}(W_{p})$ is well-defined. Here we reformulate (48) as a limit of certain colored knotted graph. We prepare a lemma to treat such perturbation of colors of a knotted graph.

Lemma B.1.

For $a\in(\mathbb{C}\setminus\mathbb{Z}/2)\cup(N\mathbb{Z}-1)/2$ and $\varepsilon\in\mathbb{C}$ near $0$ , the following holds.

\lim_{\varepsilon\to 0}\operatorname{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.8}]{small}\end{matrix}\right)=\operatorname{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.8}]{small1}\end{matrix}\right).

(49)

Proof.

Recall that the ADO invariant is defined by using the quantum $R$ matrix associated with the non-integral highest weight representation of $\mathcal{U}_{q}(sl_{2})$ where $q$ is a root of unity. Let $V_{a}$ is the highest weight representation with the highest weight $a$ . Then $\dim V_{a}=N$ if the weight $\lambda$ is in $(\mathbb{C}\setminus\mathbb{Z}/2)\cup(N\mathbb{Z}-1)/2$ . The trivalent vertex in the lefthand side of (49) represents the projection operator $V_{\varepsilon}\otimes V_{a}\to V_{a+\varepsilon}$ . The limit

\lim_{\varepsilon\to 0}V_{\varepsilon}=V_{0}=V^{(0)}\oplus V^{\prime}

where $V^{(0)}$ is the trivial $1$ -dimensional representation and $V^{\prime}$ is the $N-1$ dimensional representation with highest weight $-1$ . Then

\lim_{\varepsilon\to 0}V_{\varepsilon}\otimes V_{a}=(V^{(0)}\oplus V^{\prime})\otimes V_{a}=V_{a}\oplus(V^{\prime}\otimes V_{a}).

The projection operator corresponding to the vertex picks up $V_{a}$ part and discards $V^{\prime}\otimes V_{a}$ part. So, to see the limiting case, we can replace the representation $V_{\varepsilon}$ on the thin line by the trivial representation $V^{(0)}$ , which does not contribute to the invariant. Hence we can remove the thin line at the limit. ∎

Now we compute $J_{N-1}(w_{p})$ .

Proposition B.3.

For the twisted whitehead link $W_{p}$ , $J_{N-1}(W_{p})$ is given as follows.

J_{N-1}(W_{p})=N\frac{q^{p\frac{(N-1)^{2}}{4}}}{4\pi i}\sum_{l=0}^{N-1}\frac{d}{dx}\left(\sum_{k=0}^{N-1}q^{p(x-\frac{N-1}{2})^{2}}\{2x+1\}\right.\,\times\hfill\\ \hfill\left.\left.\sum_{s-\frac{x-k}{2}=\max(k,l)}^{\min(k+l,N-1)}\frac{\{s,s-\frac{x-k}{2}\}^{2}}{\{s-x,s-\frac{x+k}{2}\}^{2}\{s-l,s-l-\frac{x-k}{2}\}^{2}\{x+l-s,\frac{x+k}{2}+l-s_{1}\}^{2}}\right)\right|_{x=k}.

(50)

Proof.

We first compute $\operatorname{ADO}_{N}(W_{p})$ for even $p$ , which is the limit of the knotted graph in Figure 24 at $\varepsilon\to 0$ .

J_{N-1}(W_{p})=\operatorname{ADO}^{(N)}(W_{p})\underset{\eqref{eq:small}}{=}\lim_{\varepsilon\to 0}\operatorname{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.6}]{whiteheadADOe}\end{matrix}\right)\\ \underset{\eqref{eq:ADOparallel},\eqref{eq:ADOinverse}}{=}\lim_{\varepsilon\to 0}\sum_{k,l=0}^{N-1}q^{p(k+\varepsilon-\frac{N-1}{2})^{2}-p\varepsilon^{2}+p\frac{(N-1)^{2}}{4}}\left[\begin{matrix}2k+2\varepsilon+N\\ 2k+2\varepsilon+1\end{matrix}\right]^{-1}\left[\begin{matrix}2l+N\\ 2l+1\end{matrix}\right]^{-1}\,\times\hfill\\ \hfill\mathrm{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.6}]{whiteheadADOo0}\end{matrix}\right)\\ \underset{\eqref{eq:removetriangle2}}{=}\lim_{\varepsilon\to 0}\sum_{k,l=0}^{N-1}q^{p(k+\varepsilon-\frac{N-1}{2})^{2}-p\varepsilon^{2}+p\frac{(N-1)^{2}}{4}}\left[\begin{matrix}2k+2\varepsilon+N\\ 2k+2\varepsilon+1\end{matrix}\right]^{-1}\left[\begin{matrix}2l+N\\ 2l+1\end{matrix}\right]^{-1}\,\times\hfill\\ \hfill\left\{\begin{matrix}l&\frac{N-1}{2}-\frac{\varepsilon}{2}&\frac{N-1}{2}-\frac{\varepsilon}{2}\\[4.0pt] \varepsilon&\frac{N-1}{2}+\frac{\varepsilon}{2}&\frac{N-1}{2}+\frac{\varepsilon}{2}\end{matrix}\right\}_{q}\mathrm{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.6}]{whiteheadADOo1}\end{matrix}\right)\\ \underset{\eqref{eq:hopf},\eqref{eq:ADO6j6}}{=}q^{p\frac{(N-1)^{2}}{4}}\lim_{\varepsilon\to 0}\sum_{k,l=0}^{N-1}q^{p(k+\varepsilon-\frac{N-1}{2})^{2}-p\varepsilon^{2}}\frac{\{N-1\}!^{2}\{2k+2\varepsilon+1\}}{\{2k+2\varepsilon+N,N\}}i^{N-1}\,\times\hfill\\ \hfill\frac{\{N-1-\varepsilon,N-1\}}{\{N-1\}!}\left\{\begin{matrix}\frac{N-1}{2}-\frac{\varepsilon}{2}&\frac{N-1}{2}+\frac{\varepsilon}{2}&l\\[4.0pt] \frac{N-1}{2}+\frac{\varepsilon}{2}&\frac{N-1}{2}+\frac{\varepsilon}{2}&k+\varepsilon\end{matrix}\right\}_{q}\\ \underset{\eqref{eq:ADO6j3}}{=}N\,q^{p\frac{(N-1)^{2}}{4}}\lim_{\varepsilon\to 0}\sum_{l=0}^{N-1}\frac{\{N-1-\varepsilon,N-1\}}{\{N-1\}!}\,\times\hfill\\ \hfill\sum_{k=0}^{N-1}q^{p(k+\varepsilon-\frac{N-1}{2})^{2}-p\varepsilon^{2}}\frac{\{2k+2\varepsilon+1\}}{\{2N(k+\varepsilon)\}}\left\{\begin{matrix}\frac{N-1}{2}-\frac{\varepsilon}{2}&\frac{N-1}{2}+\frac{\varepsilon}{2}&l\\[4.0pt] \frac{N-1}{2}+\frac{\varepsilon}{2}&\frac{N-1}{2}+\frac{\varepsilon}{2}&k+\varepsilon\end{matrix}\right\}_{q}\\ =N\,q^{p\frac{(N-1)^{2}}{4}}\lim_{\varepsilon\to 0}\frac{1}{\{2N\varepsilon\}}\sum_{l=0}^{N-1}\frac{\{N-1-\varepsilon,N-1\}}{\{N-1\}!}\,\times\hfill\\ \hfill\sum_{k=0}^{N-1}q^{p(k+\varepsilon-\frac{N-1}{2})^{2}-p\varepsilon^{2}}\{2k+2\varepsilon+1\}\left\{\begin{matrix}\frac{N-1}{2}-\frac{\varepsilon}{2}&\frac{N-1}{2}+\frac{\varepsilon}{2}&l\\[4.0pt] \frac{N-1}{2}+\frac{\varepsilon}{2}&\frac{N-1}{2}+\frac{\varepsilon}{2}&k+\varepsilon\end{matrix}\right\}_{q}\\ =N\,\frac{q^{p\frac{(N-1)^{2}}{4}}}{4\pi i}\frac{d}{d\varepsilon}\left(\frac{q^{-p\varepsilon^{2}}\{N-1-\varepsilon,N-1\}}{\{N-1\}!}\right.\,\times\hfill\\ \hfill\left.\left.\sum_{k,l=0}^{N-1}q^{p(k+\varepsilon-\frac{N-1}{2})^{2}}\{2k+2\varepsilon+1\}\left\{\begin{matrix}\frac{N-1}{2}-\frac{\varepsilon}{2}&\frac{N-1}{2}+\frac{\varepsilon}{2}&l\\[4.0pt] \frac{N-1}{2}+\frac{\varepsilon}{2}&\frac{N-1}{2}+\frac{\varepsilon}{2}&k+\varepsilon\end{matrix}\right\}_{q}\right)\right|_{\varepsilon=0}\\ =N\,\frac{q^{p\frac{(N-1)^{2}}{4}}}{4\pi i}\left(\frac{d}{d\varepsilon}\frac{q^{-p\varepsilon^{2}}\{N-1-\varepsilon,N-1\}}{\{N-1\}!}\right)\sum_{k,l=0}^{N-1}q^{p(k-\frac{N-1}{2})^{2}}\{2k+1\}\left\{\begin{matrix}\frac{N-1}{2}&\frac{N-1}{2}&l\\[4.0pt] \frac{N-1}{2}&\frac{N-1}{2}&k\end{matrix}\right\}_{q}\hfill\\ +N\frac{q^{p\frac{(N-1)^{2}}{4}}}{4\pi i}\left.\frac{d}{d\varepsilon}\left(\sum_{k,l=0}^{N-1}q^{p(k+\varepsilon-\frac{N-1}{2})^{2}}\{2k+2\varepsilon+1\}\left\{\begin{matrix}\frac{N-1}{2}-\frac{\varepsilon}{2}&\frac{N-1}{2}+\frac{\varepsilon}{2}&l\\[4.0pt] \frac{N-1}{2}+\frac{\varepsilon}{2}&\frac{N-1}{2}+\frac{\varepsilon}{2}&k+\varepsilon\end{matrix}\right\}_{q}\right)\right|_{\varepsilon=0}\\ \underset{\eqref{eq:6jsymmetry}}{=}N\ \frac{q^{p\frac{(N-1)^{2}}{4}}}{4\pi i}\sum_{l=0}^{N-1}\left.\!\!\frac{d}{d\varepsilon}\!\!\left(\!\sum_{k=0}^{N-1}q^{p(k+\varepsilon-\frac{N-1}{2})^{2}}\{2k+2\varepsilon+1\}\left\{\begin{matrix}\frac{N-1}{2}-\frac{\varepsilon}{2}&\frac{N-1}{2}+\frac{\varepsilon}{2}&l\\[4.0pt] \frac{N-1}{2}+\frac{\varepsilon}{2}&\frac{N-1}{2}+\frac{\varepsilon}{2}&k+\varepsilon\end{matrix}\right\}_{q}\right)\right|_{\varepsilon=0}\\ \underset{\eqref{eq:ADO6j3}}{=}N\frac{q^{p\frac{(N-1)^{2}}{4}}}{4\pi i}\sum_{l=0}^{N-1}\frac{d}{d\varepsilon}\left(\sum_{k=0}^{N-1}\{2k+2\varepsilon+1\}\sum_{s=\max(k,l)}^{\min(k+l,N-1)}\frac{q^{p(k+\varepsilon-\frac{N-1}{2})^{2}}\{s\}!}{\{s-k\}!\{s-l\}!\{k+l-s\}!}\right.\,\times\hfill\\ \hfill\left.\left.\frac{\{s+\varepsilon,s\}}{\{s-k-\varepsilon,s-k\}\{s-l+\varepsilon,s-l\}\{k+\varepsilon+l-s,k+l-s\}}\right)\right|_{\varepsilon=0}.

Now we replace $s$ by $s_{1}+\frac{k+l}{2}$ and then use $\frac{d}{d\varepsilon}f(x)f(x+2\varepsilon)=\frac{d}{d\varepsilon}f(x+\varepsilon)^{2}$ , we get

N\frac{q^{p\frac{(N-1)^{2}}{4}}}{4\pi i}\!\sum_{l=0}^{N-1}\!\frac{d}{d\varepsilon}\!\!\left(\sum_{k=0}^{N-1}\{2k+2\varepsilon+1\}\!\!\!\!\sum_{s_{1}+\frac{k+l}{2}=\max(k,l)}^{\min(k+l,N-1)}\!\!\frac{q^{p(k+\varepsilon-\frac{N-1}{2})^{2}}\{s_{1}+\frac{k+l}{2}\}!}{\{s_{1}+\frac{-k+l}{2}\}!\{s_{1}+\frac{k-l}{2}\}!\{\frac{k+l}{2}-s_{1}\}!}\right.\times\\ \hfill\left.\left.\frac{\{s_{1}+\frac{k+2\varepsilon+l}{2},s_{1}+\frac{k+l}{2}\}}{\{s_{1}+\frac{-k-2\varepsilon+l}{2},s_{1}+\frac{-k+l}{2}\}\{s_{1}+\frac{k+2\varepsilon-l}{2},s_{1}+\frac{k-l}{2}\}\{\frac{k+2\varepsilon+l}{2}-s_{1},\frac{k+l}{2}-s_{1}\}}\right)\right|_{\varepsilon=0}\\ =N\frac{q^{p\frac{(N-1)^{2}}{4}}}{4\pi i}\sum_{l=0}^{N-1}\frac{d}{d\varepsilon}\left(\sum_{k=0}^{N-1}q^{p(k+\varepsilon-\frac{N-1}{2})^{2}}\{2k+2\varepsilon+1\}\right.\,\times\hfill\\ \left.\left.\sum_{s_{1}+\frac{k+l}{2}=\max(k,l)}^{\min(k+l,N-1)}\!\!\!\!\frac{\{s_{1}+\frac{k+\varepsilon+l}{2},s_{1}+\frac{k+l}{2}\}^{2}}{\{s_{1}+\frac{-k-\varepsilon+l}{2},s_{1}+\frac{-k+l}{2}\}^{2}\{s_{1}+\frac{k+\varepsilon-l}{2},s_{1}+\frac{k-l}{2}\}^{2}\{\frac{k+\varepsilon+l}{2}-s_{1},\!\frac{k+l}{2}-s_{1}\}^{2}}\!\!\right)\!\!\right|_{\varepsilon=0}\!\!\!.

Then, by replacing $k+\varepsilon$ by $x$ and $s_{1}$ by $s-\frac{x+l}{2}$ , we get

J_{N-1}(W_{p})=N\frac{q^{p\frac{(N-1)^{2}}{4}}}{4\pi i}\sum_{l=0}^{N-1}\frac{d}{dx}\left(\sum_{k=0}^{N-1}q^{p(k+\varepsilon-\frac{N-1}{2})^{2}}\{2x+1\}\right.\,\times\hfill\\ \hfill\left.\left.\sum_{s-\frac{x-k}{2}=\max(k,l)}^{\min(k+l,N-1)}\frac{\{s,s-\frac{x-k}{2}\}^{2}}{\{s-x,s-\frac{x+k}{2}\}^{2}\{s-l,s-l-\frac{x-k}{2}\}^{2}\{x+l-s,\frac{x+k}{2}+l-s_{1}\}^{2}}\right)\right|_{x=k}.

Hence we obtained (50).

Next, we prove for odd case. For odd $p$ , $W_{p}$ is considered as the limiting case of the knotted graph in Figure 25 at $\varepsilon=0$ by (49), and $J_{N-1}(W_{p})$ is computed as follows.

J_{N-1}(W_{p})=\operatorname{ADO}_{N}(W_{p})\underset{\eqref{eq:small}}{=}\lim_{\varepsilon\to 0}\operatorname{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.6}]{whiteheadADOo}\end{matrix}\right)\underset{\eqref{eq:ADOparallel}}{=}\\ \text{\small$\lim_{\varepsilon\to 0}\sum_{k,l=0}^{N-1}\!q^{p(k+\varepsilon-\frac{N-1}{2})^{2}-p\varepsilon^{2}+p\frac{(N-1)^{2}}{4}}\!\!\left[\begin{matrix}2k+2\varepsilon+N\\ 2k+2\varepsilon+1\end{matrix}\right]^{-1}\left[\begin{matrix}2l+N\\ 2l+1\end{matrix}\right]^{-1}$}\!\!\!\mathrm{ADO}_{N}\!\left(\begin{matrix}\includegraphics[scale={0.5}]{whiteheadADOo0}\end{matrix}\right).

Then the rest of the computation is the same as the even $p$ case and we get (50). ∎

B.4. Twist knots and double twist knots

Here we compute the colored Jones polynomial $J_{N}(D_{p,r})$ for the double twist knot $D_{p,r}$ . Note that, if $p$ and $r$ are both odd, then $D_{p,r}$ is a two-component link. The following formula also holds for double twist links.

Proposition B.4.

For the double twist knot $D_{p,r}$ , $J_{N-1}(D_{p,r})$ is given as follows.

J_{N-1}(D_{p,r})=\operatorname{ADO}_{N}(D_{p,r})=\\ -\frac{N^{2}\,q^{(p-r)\frac{(N-1)^{2}}{4}}}{16\pi^{2}}\frac{\partial^{2}}{\partial x\partial y}\sum_{k,l=0}^{N-1}q^{p(x-\frac{N-1}{2})^{2}-r(y-\frac{N-1}{2})^{2}}\{2x+1\}\{2y+1\}\,\times\\ \left.\sum_{s-\frac{x-k+y-l}{2}=\max(k,l)}^{\min(k+l,N-1)}\!\!\!\frac{\{s,s-\frac{x-k+y-l}{2}\}^{2}}{\{s-x,s-\frac{x+k+y-l}{2}\}^{2}\{s-y,s-\frac{x-k+y+l}{2}\}^{2}\{x\!+\!y-\!s,\!\frac{x+k+y+l}{2}-\!s\}^{2}}\right|_{\text{$\begin{matrix}x\!=\!k\\[-4.0pt] y\!=\!l\end{matrix}$}}\!\!\!\!.

(51)

Proof.

First we prove for the case that $p$ and $r$ are both even. For this case, we compute the ADO invariant of $D_{p,r}$ as a limit $\varepsilon,\delta\to 0$ of the knotted graph in Figure 26.

\operatorname{ADO}_{N}(D_{p,r})=\lim_{\varepsilon,\delta\to 0}\operatorname{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.6}]{doubletwistADOee}\end{matrix}\right)\\ =q^{(p-r)\frac{(N-1)^{2}}{4}}\lim_{\varepsilon,\delta\to 0}\sum_{k,l=0}^{N-1}q^{p(k+\varepsilon-\frac{N-1}{2})^{2}-p(\frac{\varepsilon+\delta}{2})^{2}-p(\frac{\varepsilon-\delta}{2})^{2}}q^{-r(l+\delta-\frac{N-1}{2})^{2}+r(\frac{\varepsilon+\delta}{2})^{2}+r(\frac{\varepsilon-\delta}{2})^{2}}\,\times\\ \left[\begin{matrix}2k+2\varepsilon+N\\ 2k+2\varepsilon+1\end{matrix}\right]^{-1}\left[\begin{matrix}2l+2\delta+N\\ 2l+2\delta+1\end{matrix}\right]^{-1}\left\{\begin{matrix}\frac{N-1}{2}-\frac{\varepsilon+\delta}{2}&\frac{N-1}{2}+\frac{\varepsilon-\delta}{2}&-\delta\\[4.0pt] \frac{N-1}{2}+\frac{\varepsilon+\delta}{2}&\frac{N-1}{2}+\frac{\varepsilon-\delta}{2}&k+\varepsilon\end{matrix}\right\}_{q}\,\times\\ \hfill\left\{\begin{matrix}l+\delta&\frac{N-1}{2}-\frac{\varepsilon+\delta}{2}&\frac{N-1}{2}-\frac{\varepsilon-\delta}{2}\\[4.0pt] \varepsilon&\frac{N-1}{2}+\frac{\varepsilon+\delta}{2}&\frac{N-1}{2}+\frac{\varepsilon-\delta}{2}\end{matrix}\right\}_{q}\operatorname{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.6}]{doubletwistADOee1}\end{matrix}\right)\\ \underset{\eqref{eq:ADO6j7},\eqref{eq:ADO6j8}}{=}q^{(p-r)\frac{(N-1)^{2}}{4}}\lim_{\varepsilon,\delta\to 0}\sum_{k,l=0}^{N-1}q^{p(k+\varepsilon-\frac{N-1}{2})^{2}-r(l+\delta-\frac{N-1}{2})^{2}-(p-r)(\frac{\varepsilon^{2}+\delta^{2}}{2})}\,\times\hfill\\ \frac{\{N-1\}!^{2}\{2k+2\varepsilon+1\}\{2l+2\delta+1\}}{\{2k+2\varepsilon+N,N\}\{2l+2\delta+N,N\}}\frac{\{k+\varepsilon-\delta,k\}\{N-1+\varepsilon+\delta\}}{\{k+\varepsilon+\delta,k\}\{N-1\}!}\,\times\\ \hfill\frac{\{l+\varepsilon+\delta,l\}}{\{l-\varepsilon+\delta,l\}}\operatorname{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.6}]{doubletwistADOee1}\end{matrix}\right)\\ =N^{2}\,q^{(p-r)\frac{(N-1)^{2}}{4}}\lim_{\varepsilon,\delta\to 0}\sum_{k,l=0}^{N-1}q^{p(k+\varepsilon-\frac{N-1}{2})^{2}-r(l+\delta-\frac{N-1}{2})^{2}-(p-r)(\frac{\varepsilon^{2}+\delta^{2}}{2})}\,\times\hfill\\ \frac{\{2k+2\varepsilon+1\}\{2l+2\delta+1\}}{\{2N(k+\varepsilon)\}\{2N(l+\delta)\}}\frac{\{k+\varepsilon-\delta,k\}\{N-1+\varepsilon+\delta\}}{\{k+\varepsilon+\delta,k\}\{N-1\}!}\,\times\\ \hfill\frac{\{l+\varepsilon+\delta,l\}}{\{l-\varepsilon+\delta,l\}}\operatorname{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.6}]{doubletwistADOee1}\end{matrix}\right)\\ \underset{\eqref{eq:ADO6j3}}{=}N^{2}\,q^{(p-r)\frac{(N-1)^{2}}{4}}\frac{\partial^{2}}{\partial\varepsilon\partial\delta}\frac{q^{-(p-r)(\frac{\varepsilon^{2}+\delta^{2}}{2})}}{\{2N\varepsilon\}\{2N\delta\}}\frac{\{N-1-2\delta,N-1\}}{\{N-1\}!}\frac{\{N-1+\varepsilon+\delta\}}{\{N-1\}!}\,\times\hfill\\ \hfill\sum_{k,l=0}^{N-1}q^{p(k+\varepsilon-\frac{N-1}{2})^{2}-r(l+\delta-\frac{N-1}{2})^{2}}\frac{\{2k+2\varepsilon+1\}\{2l+2\delta+1\}}{\{2N(k+\varepsilon)\}\{2N(l+\delta)\}}\frac{\{k+\varepsilon-\delta,k\}}{\{k+\varepsilon+\delta,k\}}\frac{\{l+\varepsilon+\delta,l\}}{\{l-\varepsilon+\delta,l\}}\,\times\\ \qquad\sum_{s=\max(k,l)}^{\min(k+l,N-1)}\frac{\{s\}!}{\{s-k\}!\{s-l\}!\{k+l-s\}!}\,\times\hfill\\ \hfill\left.\frac{\{s+\varepsilon+\delta,s\}}{\{s-k-\varepsilon+\delta,s-k\}\{s-l+\varepsilon-\delta,s-l\}\{k+l-s+\varepsilon+\delta,k+l-s\}}\right|_{\varepsilon=\delta=0}\\ =\frac{N^{2}\,q^{(p-r)\frac{(N-1)^{2}}{4}}}{-16\pi^{2}}\frac{\partial^{2}}{\partial\varepsilon\partial\delta}q^{-(p-r)(\frac{\varepsilon^{2}+\delta^{2}}{2})}\frac{\{N-1+\varepsilon+\delta\}}{\{N-1\}!}\frac{\{N-1-2\delta,N-1\}}{\{N-1\}!}\,\times\hfill\\ \sum_{k,l=0}^{N-1}q^{p(k+\varepsilon-\frac{N-1}{2})^{2}-r(l+\delta-\frac{N-1}{2})^{2}}\frac{\{2k+2\varepsilon+1\}\{2l+2\delta+1\}\{k+\varepsilon-\delta,k\}\{l+\varepsilon+\delta,l\}}{\{k+\varepsilon+\delta,k\}\{l-\varepsilon+\delta,l\}}\,\times\\ \sum_{s=\max(k,l)}^{\min(k+l,N-1)}\frac{\{s\}!}{\{s-k\}!\{s-l\}!\{k+l-s\}!}\,\times\hfill\\ \hfill\left.\frac{\{s+\varepsilon+\delta,s\}}{\{s-k-\varepsilon+\delta,s-k\}\{s-l+\varepsilon-\delta,s-l\}\{k+l-s+\varepsilon+\delta,k+l-s\}}\right|_{\varepsilon=\delta=0}.

By substituting $s_{1}+\frac{k+l}{2}$ intto $s$ , we have

\operatorname{ADO}_{N}(D_{p,r})=\\ \frac{N^{2}\,q^{(p-r)\frac{(N-1)^{2}}{4}}}{-16\pi^{2}}\frac{\partial^{2}}{\partial\varepsilon\partial\delta}q^{(r-p)(\frac{\varepsilon^{2}+\delta^{2}}{2})}\frac{\{N-1+\varepsilon+\delta\}}{\{N-1\}!}\frac{\{N-1-2\delta,N-1\}}{\{N-1\}!}\,\times\\ \sum_{k,l=0}^{N-1}q^{p(k+\varepsilon-\frac{N-1}{2})^{2}-r(l+\delta-\frac{N-1}{2})^{2}}\frac{\{2k+2\varepsilon+1\}\{2l+2\delta+1\}\{k+\varepsilon-\delta,k\}\{l+\varepsilon+\delta,l\}}{\{k+\varepsilon+\delta,k\}\{l-\varepsilon+\delta,l\}}\,\times\hfill\\ \sum_{s_{1}+\frac{k+l}{2}=\max(k,l)}^{\min(k+l,N-1)}\frac{\{s_{1}+\frac{k+l}{2}\}!}{\{s_{1}+\frac{-k+l}{2}\}!\{s_{1}+\frac{k-l}{2}\}!\{\frac{k+l}{2}-s_{1}\}!}\,\times\hfill\\ \hfill\left.\frac{\{s_{1}+\frac{k+2\varepsilon+l+2\delta}{2},s_{1}+\frac{k+l}{2}\}}{\{s_{1}+\frac{-k-2\varepsilon+l+2\delta}{2},s_{1}+\frac{-k+l}{2}\}\{s_{1}+\frac{k+2\varepsilon-l-2\delta}{2},s_{1}+\frac{k-l}{2}\}\{\frac{k+2\varepsilon+l+2\delta}{2}-s_{1},\frac{k+l}{2}-s_{1}\}}\right|_{\varepsilon=\delta=0}.

Let

f(k,l,\varepsilon,\delta)=q^{p(k+\varepsilon-\frac{N-1}{2})^{2}-r(l+\delta-\frac{N-1}{2})^{2}}\{2k+2\varepsilon+1\}\{2l+2\delta+1\}\,\times\\ \sum_{s_{1}+\frac{k+l}{2}=\max(k,l)}^{\min(k+l,N-1)}\frac{\{s_{1}+\frac{k+l}{2}\}!}{\{s_{1}+\frac{-k+l}{2}\}!\{s_{1}+\frac{k-l}{2}\}!\{\frac{k+l}{2}-s_{1}\}!}\,\times\hfill\\ \hfill\frac{\{s_{1}+\frac{k+2\varepsilon+l+2\delta}{2},s_{1}+\frac{k+l}{2}\}}{\{s_{1}+\frac{-k-2\varepsilon+l+2\delta}{2},s_{1}+\frac{-k+l}{2}\}\{s_{1}+\frac{k+2\varepsilon-l-2\delta}{2},s_{1}+\frac{k-l}{2}\}\{\frac{k+2\varepsilon+l+2\delta}{2}-s_{1},\frac{k+l}{2}-s_{1}\}}.

Note that $f(k,l,0,0)$ is anti-symmetric with respect to $k$ and $l$ . Moreover, $f(k,l,0,\delta)$ and $f(k,l,\varepsilon,0)$ are anti-symmetric with respect to $k$ and $l$ respectively. Therefore, we have

\operatorname{ADO}_{N}(D_{p,r})=\frac{N^{2}\,q^{(p-r)\frac{(N-1)^{2}}{4}}}{-16\pi^{2}}\frac{\partial^{2}}{\partial\varepsilon\partial\delta}\left.\sum_{k,l=0}^{N-1}f(k,l,\varepsilon,\delta)\right|_{\varepsilon=\delta=0}.

By using the definition of the derivation, we have

\frac{\partial^{2}}{\partial\varepsilon\partial\delta}\left.\sum_{k,l=0}^{N-1}f(k,l,\varepsilon,\delta)\right|_{\varepsilon=\delta=0}=\\ \lim_{\varepsilon,\delta\to 0}\frac{1}{\varepsilon\delta}\sum_{k,l=0}^{N-1}f(k,l,\varepsilon,\delta)-f(k,l,0,\delta)-f(k,l,\varepsilon,0)+f(k,l,0,0)=\\ \lim_{\varepsilon,\delta\to 0}\frac{1}{\varepsilon\delta}\sum_{k,l=0}^{N-1}f(k-\varepsilon,l-\delta,\varepsilon,\delta)-f(k-\varepsilon,l-\delta,0,\delta)-f(k-\varepsilon,l-\delta,\varepsilon,0)+f(k-\varepsilon,l-\delta,0,0)\\[-5.0pt] =\lim_{\varepsilon,\delta\to 0}\frac{1}{\varepsilon\delta}\sum_{k,l=0}^{N-1}f(k-\varepsilon,l-\delta,0,0).

Here we use that $\frac{\partial^{2}}{\partial\varepsilon\partial\delta}f(k,l,\varepsilon,\delta)$ is continuous with respect to $\varepsilon$ and $\delta$ for the second equality, and use that $f(k-\varepsilon,l-\delta,\varepsilon,\delta)$ , $f(k-\varepsilon,l-\delta,0,\delta)$ and $f(k-\varepsilon,l-\delta,\varepsilon,0)$ are anti-symmetric with respect to $k$ or $l$ for the last equality.

On the other hand, let

g(k,l,\varepsilon,\delta)=q^{p(k+\varepsilon-\frac{N-1}{2})^{2}-r(l+\delta-\frac{N-1}{2})^{2}}\{2k+2\varepsilon+1\}\{2l+2\delta+1\}\,\times\\ \sum_{s_{1}+\frac{k+l}{2}=\max(k,l)}^{\min(k+l,N-1)}\frac{\{s_{1}\!+\!\frac{k+\varepsilon+l+\delta}{2},s_{1}\!+\!\frac{k+l}{2}\}^{2}}{\{s_{1}\!+\!\frac{-k-\varepsilon+l+\delta}{2},s_{1}\!+\!\frac{-k+l}{2}\}^{2}\{s_{1}\!+\!\frac{k+\varepsilon-l-\delta}{2},s_{1}\!+\!\frac{k-l}{2}\}^{2}\{\frac{k+\varepsilon+l+\delta}{2}\!-\!s_{1},\frac{k+l}{2}\!-\!s_{1}\}^{2}}.

Then $g(k,l,0,0)$ , $g(k,l,\varepsilon,0)$ and $g(k,l,0,\delta)$ are anti-symmetric with respect to $k$ or $l$ , we have

\frac{\partial^{2}}{\partial\varepsilon\partial\delta}\left.\sum_{k,l=0}^{N-1}g(k,l,\varepsilon,\delta)\right|_{\varepsilon=\delta=0}\\ =\lim_{\varepsilon,\delta\to 0}\frac{1}{\varepsilon\delta}\sum_{k,l=0}^{N-1}g(k,l,0,0)-g(k,l,-\varepsilon,0)-g(k,l,0,-\delta)+g(k,l,-\varepsilon,-\delta)\\[-5.0pt] =\lim_{\varepsilon,\delta\to 0}\frac{1}{\varepsilon\delta}\sum_{k,l=0}^{N-1}g(k,l,-\varepsilon,-\delta).

Now look at $f(k-\varepsilon,l-\delta,0,0)$ and $g(k,l,-\varepsilon,-\delta)$ . We have

f(k-\varepsilon,l-\delta,0,0)=g(k,l,-\varepsilon,-\delta)=\\ q^{p(k-\varepsilon-\frac{N-1}{2})^{2}-r(l-\delta-\frac{N-1}{2})^{2}-(p-r)(\frac{\varepsilon^{2}+\delta^{2}}{2})}\{2k-2\varepsilon+1\}\{2l-2\delta+1\}\,\times\\ \sum_{s_{1}+\frac{k+l}{2}=\max(k,l)}^{\min(k+l,N-1)}\!\!\frac{\{s_{1}+\frac{k-\varepsilon+l-\delta}{2},s_{1}+\frac{k+l}{2}\}^{2}}{\{s_{1}\!+\!\frac{-k+\varepsilon+l-\delta}{2},s_{1}\!+\!\frac{-k+l}{2}\}^{2}\{s_{1}\!+\!\frac{k-\varepsilon-l+\delta}{2},s_{1}\!+\!\frac{k-l}{2}\}^{2}\{\frac{k-\varepsilon+l-\delta}{2}\!-\!s_{1},\!\frac{k+l}{2}\!-\!s_{1}\}^{2}}.

Therefore,

\left.\frac{\partial^{2}}{\partial\varepsilon\partial\delta}\sum_{k,l=0}^{N-1}f(k,l,\varepsilon,\delta)\right|_{\varepsilon=\delta=0}=\left.\frac{\partial^{2}}{\partial\varepsilon\partial\delta}\sum_{k,l=0}^{N-1}g(k,l,\varepsilon,\delta)\right|_{\varepsilon=\delta=0}

and we have

\operatorname{ADO}_{N}(D_{p,r})=\left.\frac{N^{2}\,q^{(p-r)\frac{(N-1)^{2}}{4}}}{-16\pi^{2}}\frac{\partial^{2}}{\partial\varepsilon\partial\delta}\sum_{k,l=0}^{N-1}g(k,l,\varepsilon,\delta)\right|_{\varepsilon=\delta=0}.

In $g(k,l,\varepsilon,\delta)$ , $k$ and $\varepsilon$ appear as $k+\varepsilon$ , and $l$ and $\delta$ appear as $l+\delta$ , by putting $x=k+\varepsilon$ , $y=l+\delta$ , we get

J_{N-1}(D_{p,r})=-\frac{N^{2}\,q^{(p-r)\frac{(N-1)^{2}}{4}}}{16\pi^{2}}\frac{\partial^{2}}{\partial x\partial y}\sum_{k,l=0}^{N-1}q^{p(x-\frac{N-1}{2})^{2}-r(y-\frac{N-1}{2})^{2}}\{2x+1\}\{2y+1\}\,\times\\ \left.\sum_{s_{1}+\frac{k+l}{2}=\max(k,l)}^{\min(k+l,N-1)}\frac{\{s_{1}+\frac{x+y}{2},s_{1}+\frac{k+l}{2}\}^{2}}{\{s_{1}+\frac{-x+y}{2},s_{1}+\frac{-k+l}{2}\}^{2}\{s_{1}+\frac{x-y}{2},s_{1}+\frac{k-l}{2}\}^{2}\{\frac{x+y}{2}-s_{1},\frac{k+l}{2}-s_{1}\}^{2}}\right|_{\text{\scriptsize$\begin{matrix}x=k\\[-3.0pt] y=l\end{matrix}$}}.

By replacing $s_{1}$ by $s-\frac{x+y}{2}$ , we get (51).

The case for even $p$ and odd $r$ is computed as the limit $\varepsilon,\delta\to 0$ of the colored knotted graph in Figure 27.

\operatorname{ADO}_{N}(D_{p,r})=\lim_{\varepsilon,\delta\to 0}\operatorname{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.6}]{doubletwistADOeo}\end{matrix}\right)\\ =q^{(p-r)\frac{(N-1)^{2}}{4}}\lim_{\varepsilon,\delta\to 0}\sum_{k,l=0}^{N-1}q^{p(k+\varepsilon-\frac{N-1}{2})^{2}-p(\frac{\varepsilon+\delta}{2})^{2}-p(\frac{\varepsilon-\delta}{2})^{2}}q^{-r(l+\delta-\frac{N-1}{2})^{2}+r(\frac{\varepsilon+\delta}{2})^{2}+r(\frac{\varepsilon-\delta}{2})^{2}}\,\times\\ \left[\begin{matrix}2k+2\varepsilon+N\\ 2k+2\varepsilon+1\end{matrix}\right]^{-1}\left[\begin{matrix}2l+2\delta+N\\ 2l+2\delta+1\end{matrix}\right]^{-1}\left\{\begin{matrix}\frac{N-1}{2}-\frac{\varepsilon+\delta}{2}&\frac{N-1}{2}+\frac{\varepsilon-\delta}{2}&-\delta\\[4.0pt] \frac{N-1}{2}+\frac{\varepsilon+\delta}{2}&\frac{N-1}{2}+\frac{\varepsilon-\delta}{2}&k+\varepsilon\end{matrix}\right\}_{q}\,\times\\ \hfill\operatorname{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.6}]{doubletwistADOee1}\end{matrix}\right)\\ \underset{\eqref{eq:ADO6j7},\eqref{eq:ADO6j8}}{=}q^{(p-r)\frac{(N-1)^{2}}{4}}\lim_{\varepsilon,\delta\to 0}\sum_{k,l=0}^{N-1}q^{p(k+\varepsilon-\frac{N-1}{2})^{2}-r(l+\delta-\frac{N-1}{2})^{2}-(p-r)(\frac{\varepsilon^{2}+\delta^{2}}{2})}\,\times\hfill\\ \frac{\{N-1\}!^{2}\{2k+2\varepsilon+1\}\{2l+2\delta+1\}}{\{2k+2\varepsilon+N,N\}\{2l+2\delta+N,N\}}\frac{\{k+\varepsilon-\delta,k\}\{N-1+\varepsilon+\delta\}}{\{k+\varepsilon+\delta,k\}\{N-1\}!}\,\times\\ \hfill\operatorname{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.6}]{doubletwistADOee1}\end{matrix}\right).

Then the rest of the computation to get (51) is almost the same as even $p$ , $r$ case.

The case for odd $p$ and $r$ is computed by the limit $\varepsilon,\delta\to 0$ of the ADO invariant of the colored knotted graph in Figure 28.

\operatorname{ADO}_{N}(D_{p,r})=\lim_{\varepsilon,\delta\to 0}\operatorname{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.6}]{doubletwistADOoo}\end{matrix}\right)\\ =q^{(p-r)\frac{(N-1)^{2}}{4}}\lim_{\varepsilon,\delta\to 0}\sum_{k,l=0}^{N-1}q^{p(k+\varepsilon-\frac{N-1}{2})^{2}-p(\frac{\varepsilon+\delta}{2})^{2}-p(\frac{\varepsilon-\delta}{2})^{2}}q^{-r(l+\delta-\frac{N-1}{2})^{2}+r(\frac{\varepsilon+\delta}{2})^{2}+r(\frac{\varepsilon-\delta}{2})^{2}}\,\times\\ \left[\begin{matrix}2k+2\varepsilon+N\\ 2k+2\varepsilon+1\end{matrix}\right]^{-1}\left[\begin{matrix}2l+2\delta+N\\ 2l+2\delta+1\end{matrix}\right]^{-1}\operatorname{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.6}]{doubletwistADOee1}\end{matrix}\right)\\ \underset{\eqref{eq:ADO6j7},\eqref{eq:ADO6j8}}{=}q^{(p-r)\frac{(N-1)^{2}}{4}}\lim_{\varepsilon,\delta\to 0}\sum_{k,l=0}^{N-1}q^{p(k+\varepsilon-\frac{N-1}{2})^{2}-r(l+\delta-\frac{N-1}{2})^{2}-(p-r)(\frac{\varepsilon^{2}+\delta^{2}}{2})}\,\times\hfill\\ \frac{\{N-1\}!^{2}\{2k+2\varepsilon+1\}\{2l+2\delta+1\}}{\{2k+2\varepsilon+N,N\}\{2l+2\delta+N,N\}}\operatorname{ADO}_{N}\left(\begin{matrix}\includegraphics[scale={0.6}]{doubletwistADOee1}\end{matrix}\right).

Then the rest of the computation to get (51) is almost same as even $p$ , $r$ case. ∎

The twist knot $T_{p}$ is equal to $D_{p,2}$ , so (51) also gives a formula for $J_{N-1}(T_{p})$ .

C. Asymptotics

Here we investigate the asymptotic behavior of the colored Jones invariant for large $N$ . We also reformulate the sum over the parameter $s$ inside the quantum $6j$ symbol.

C.1. Quantum dilogarithm function

As a continuous version of the quantum factorial $\{n\}!$ , we use Fateev’s quantum dilogarithm function $\varphi_{N}(x)$ , which is the analytic continuation of the following function defined for $0<x<1$ .

\varphi_{N}(x)=\int_{-\infty}^{\infty}\frac{e^{(2x-1)t}\,dt}{4t\sinh t\sinh(t/N)}.

It is shown in [5] that

\varphi_{N}({\alpha+\frac{1}{2N}})-\varphi_{N}({\alpha-\frac{1}{2N}})=-\log(1-e^{2\pi i\alpha}).

This implies that

\{x,n\}=\{x\}\{x-1\}\cdots\{x-n+1\}=(-1)^{n}q^{-\frac{(2x-n+1)n}{2}}(1-q^{2x})(1-q^{2x-2})\cdots(1-q^{2x-2n+2})\\ =(-1)^{n}q^{-\frac{(2x-n+1)n}{2}}e^{\varphi(\frac{2x-2n+1}{2N})-\varphi(\frac{2x+1}{2N})}.

(52)

For fixed any sufficient small $\delta>$ and any $M>0$ ,

\varphi_{N}(t)=\frac{N}{2\pi i}\mathrm{Li}_{2}(e^{2\pi it})+O(\frac{1}{N})

in the domain

\{t\in\mathbb{C}\mid\delta<\mathrm{Re}\,t<1-\delta,\ \left|\mathrm{Im}\,t\right|<M\}

by Proposition A.1 of [15]. It is also shown by Lemma A of [15] that

\varphi_{N}(\frac{1}{2N})=\frac{N}{2\pi i}\frac{\pi^{2}}{6}+O(\log N),\quad\varphi_{N}(1-\frac{1}{2N})=\frac{N}{2\pi i}\frac{\pi^{2}}{6}+O(\log N).

C.2. Reformulation of the colored Jones polynomials

Here we reformulate the colored Jones polynomials (50) and (51) by using the dilogarithm function. We first reformulate $J_{N-1}(W_{p})$ . Let

\xi_{N}(x,k,l,s)=\frac{\{s,s-\frac{x-k}{2}\}^{2}}{\{s-x,s-\frac{x+k}{2}\}^{2}\{s-l,s-l-\frac{x-k}{2}\}^{2}\{x+l-s,\frac{x+k}{2}+l-s_{1}\}^{2}}.

Then

\left.\frac{d}{dx}\xi_{N}(x,k,l,s)\right|_{x=k}=\frac{d}{dx}q^{(4s^{2}-8(l+x)s+3x^{2}+2kx-k^{2}+4lx+4l^{2})/2}\,\times\hfill\\ \exp\Big{(}-2\varphi_{N}(\tfrac{2s+1}{2N})+2\varphi_{N}(\tfrac{2s-2x+1}{2N})+2\varphi_{N}(\tfrac{2s-2l+1}{2N})\hfill\\ \left.+2\varphi_{N}(\tfrac{2x+2l-2s+1}{2N})-2\varphi_{N}(\tfrac{x-k+1}{2N})-2\varphi_{N}(\tfrac{k-x+1}{2N})\Big{)}\right|_{x=k}\\ =\frac{d}{dx}q^{2(s^{2}-2(l+x)s+x^{2}+lx+l^{2})}\exp\Big{(}-2\varphi_{N}(\tfrac{2s+1}{2N})+2\varphi_{N}(\tfrac{2s-2x+1}{2N})+\hfill\\ \left.2\varphi_{N}(\tfrac{2s-2l+1}{2N})+2\varphi_{N}(\tfrac{2x+2l-2s+1}{2N})-4\varphi_{N}(\tfrac{1}{2N})\Big{)}\right|_{x=k}

since

\left.\frac{d}{dx}\Big{(}\varphi_{N}(\tfrac{x-k+1}{2N})+\varphi_{N}(\tfrac{k-x+1}{2N})\Big{)}\right|_{x=k}=0,

\left.2kx-k^{2}\right|_{x=k}=\left.x^{2}\right|_{x=k}=x^{2},\qquad\left.\frac{d}{dx}(2kx-k^{2})\right|_{x=k}=\left.\frac{d}{dx}x^{2}\right|_{x=k}=2k.

By using the relation between $\frac{2\pi i}{N}\varphi_{N}(t)$ and $\mathrm{Li}_{2}(e^{2\pi it})$ , we have

\xi_{N}(x,l,s)=E_{N}(x,l,s)q^{2(s^{2}-2(l+x)s+x^{2}+lx+l^{2})}\exp\Big{(}\frac{N}{2\pi i}\big{(}-2\mathrm{Li}_{2}(q^{2s+1})\\ \hfill+2\mathrm{Li}_{2}(q^{2s-2x+1})+2\mathrm{Li}_{2}(q^{2s-2l+1})+2\mathrm{Li}_{2}(q^{2x+2l-2s+1})-\frac{\pi^{2}}{3}\big{)}\Big{)}

where $E_{N}(x,l,s)$ is a function which grows at most a polynomially with respect to $N$ . Therefore,

J_{N-1}(W_{p})=N\frac{q^{p\frac{(N-1)^{2}}{4}}}{4\pi i}\sum_{k,l=0}^{N-1}\frac{d}{dx}\!\left(\!q^{p(x-\frac{N-1}{2})^{2}}\{2x+1\}\left.\sum_{s-\frac{x-k}{2}=\max(k,l)}^{\min(k+l,N-1)}\!\!\!\!\xi_{N}(x,l,s)\!\right)\!\right|_{x=k}.

(53)

Similarly, we have

J_{N-1}(D_{p,r})=N\frac{q^{p\frac{(N-1)^{2}}{4}-r\frac{(N-1)^{2}}{4}}}{4\pi i}\,\times\\ \sum_{k,l=0}^{N-1}\frac{\partial^{2}}{\partial x\partial y}\left(q^{p(x-\frac{N-1}{2})^{2}-r(y-\frac{N-1}{2})^{2}}\{2x+1\}\{2y+1\}\!\!\!\!\!\left.\sum_{s-\frac{x-k+y-l}{2}=\max(k,l)}^{\min(k+l,N-1)}\!\!\!\!\!\!\xi_{N}(x,y,s)\right)\right|_{\text{\scriptsize$\begin{matrix}x=k\\[-3.0pt] y=l\end{matrix}$}}.

(54)

C.3. Saddle points

We investigate the sum $\sum_{s}\xi_{N}(x,y,s)$ . Since $\xi_{N}(x,y,s)$ is non-negative for each $s$ and there exists $s_{0}$ such that $\xi_{N}(x,y,s_{0})$ is the maximal among $\xi_{N}(x,y,s)$ . In this case, there is some number $C_{N}$ satisfying $1\leq C_{N}\leq N$ satisfying

\sum_{s}\xi_{N}(x,y,s)=C_{N}\,\xi_{N}(x,y,s_{0}),

(55)

where $s_{0}$ satisfies

\frac{\partial}{\partial s}\xi_{N}(x,y,s)=0.

Now we compute the maximal point $s_{0}$ of $\xi_{N}(x,y,s)$ . Let $N\alpha=x+\frac{1}{2}$ , $N\eta=y+\frac{1}{2}$ , $N\gamma=s+\frac{1}{2}$ and $u=e^{2\pi i\alpha}$ , $v=e^{2\pi i\eta}$ , $w=e^{2\pi i\gamma}$ . Then the equation for obtaining the maximal point is

\log\frac{(w-1)^{2}(w-uv)^{2}}{(w-u)^{2}(w-v)^{2}}=0.

To solve it, we first solve

(w-1)^{2}(w-uv)^{2}-(w-u)^{2}(w-v)^{2}=\\ \big{(}(w-1)(w-uv)-(w-u)(w-v)\big{)}\big{(}(w-1)(w-uv)+(w-u)(w-v)\big{)}=\\ (-1-uv)w\big{(}2w^{2}-(u+1)(v+1)z+2uv\big{)}=0.

The solution is $w=\frac{(u+1)(v+1)\pm\sqrt{(u+1)^{2}(v+1)^{2}-16uv}}{4}$ . The solution corresponding to $q^{2s_{0}+1}$ is

w_{0}=\frac{(u+1)(v+1)-\sqrt{(u+1)^{2}(v+1)^{2}-16uv}}{4},

(56)

and $s_{0}=\frac{N}{2\pi i}\log w_{0}-\frac{1}{2}$ .

D. Neumann-Zagier function

Here we recall some properties of the Neumann-Zagier function developed in [14] and [20]. The relation between this function and the potential function coming from the quantum invariant is observed in [19].

D.1. Neumann-Zagier potential function

To prove the volume conjecture for double twist knots, we extend the argument in [19] to links. Let $L=L_{1}\cup L_{2}\cup\cdots\cup L_{k}$ be a link with connected components. Let $\rho$ be an $\mathrm{SL}(2,\mathbb{C})$ representation of $\pi_{1}(S^{3}\setminus L)$ , $\mu_{i}$ , $\lambda_{i}\in\pi_{1}(S^{3}\setminus L)$ are elements corresponding to the meridian and longitude of $L_{i}$ , and $\xi_{i}$ , $\eta_{i}$ are the eigenvalues of $\rho(\mu_{i})$ and $\rho(\lambda_{i})$ respectively. Then there is an analytic function $f(\xi_{1},\cdots,\xi_{k})$ satisfying the following differential equation.

\frac{\partial}{\partial\xi_{i}}f(\xi_{1},\cdots,\xi_{k})=-2\log\eta_{i}.\quad(i=1,2,\cdots,l)

Now we assume that $f(1,\cdots,1)=0$ . For an integer $l$ satisfying $0\leq l\leq k$ and rational numbers $p_{i}/q_{i}$ for $i=1,2,\cdots,l$ , let $M$ be a three manifold obtained by rational $p_{i}/q_{i}$ surgeries along the components $L_{1}$ , $L_{2}$ , $\cdots$ , $L_{l}$ , and $\rho$ be the representation of $\pi_{1}(S^{3}\setminus L)$ corresponding to this surgery. Then

2p_{i}\log\xi_{i}+2q_{i}\log\eta_{i}=2\pi\sqrt{-1}.\quad(i=1,2,\cdots,l)

This function corresponds to the deformation of the hyperbolic structure of the complement of $L$ .

D.2. Complex volume

Let $M$ be the manifold obtained by this surgery. Assume that $M$ is a hyperbolic manifold. Then the complex volume of $M$ is given by $f(\xi_{1},\cdots,\xi_{k})$ with a small modification. The complex volume of $M$ is

\mathrm{Vol}(M)+\sqrt{-1}\,\mathrm{CS}(M)

where $\mathrm{Vol}(M)$ be the hyperbolic volume and $\mathrm{CS}$ is the Chern-Simons invariant of $M$ . Let $\gamma_{i}$ be the core geodesic of $L_{i}$ for this surgery, them

\gamma_{i}=2(r_{i}\log\xi_{i}+s_{i}\log\eta_{i})

where $r_{i}$ , $s_{i}$ are integers satisfying $p_{i}\,s_{i}-r_{i}\,q_{i}=1$ .

Theorem 4.

([20, Theorem 2]) The complex volume of $M$ is given by

\mathrm{Vol}(M)+\sqrt{-1}\,\mathrm{CS}(M)=\frac{1}{i}\left(f(\xi_{1},\cdots,\xi_{k})+\sum_{i=1}^{k}\log\xi_{i}\,\log\eta_{i}-\frac{\pi i}{2}\sum_{i=1}^{k}\gamma_{i}\right).

(57)

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	truncated edge:	$\displaystyle\quad\begin{matrix}\includegraphics[scale={0.9}]{edge1}\end{matrix}\longrightarrow\ \ \begin{matrix}\includegraphics[scale={0.9}]{edge2}\end{matrix}$
	doubly truncated edge:	$\displaystyle\quad\begin{matrix}\includegraphics[scale={0.9}]{edge3}\end{matrix}\longrightarrow\ \ \begin{matrix}\includegraphics[scale={0.9}]{edge4}\end{matrix}$

Complexified tetrahedrons, fundamental groups, and volume conjecture for double twist knots

Abstract.

Key words and phrases:

2020 Mathematics Subject Classification:

Introduction

Conjecture 1 (Volume conjecture [9]).

Conjecture 2 (Complexified volume conjecture [10]).

Theorem 1.

Remark 1.

1. Borromean rings

1.1. Representation matrix

1.2. Volume conjecture

1.3. Regular ideal octahedron

1.4. Variations of the Borromean rings

Theorem 2.

2. Twisted Whitehead links

2.1. Representation matrices

2.2. Complexified terahedron

2.3. Poisson sum formula

2.4. Saddle point method

2.5. Volume of the complement

Theorem 3.

Proof.

3. Double twist knots

3.1. Representation matrices

3.2. Complexified tetrahedron

3.3. Poisson sum formula

3.4. Saddle point method

Proposition 3.1.

Proof.

Remark 2.

3.5. Volume of the complement

Appendix A. ADO invariants for colored knotted graphs

A.1. ADO invariant for colored knots and links

A.2. ADO invariant for colored knotted graphs

Definition A.1.

Lemma A.1.

Proof.

A.3. Quantum 6​j6j symbol

Lemma A.2.

Proof.

Lemma A.3.

Proof.

A.4. Symmetry

Definition A.2.

Lemma A.4.

Proof.

Proposition A.1.

Proof.

Appendix B. Colored Jones invariants of some links

B.1. Colored Jones invariants and ADO invariants

Proposition B.1.

Proof.

Remark 3.

B.2. Borromean rings and their variants

Proposition B.2.

Proof.

B.3. Twisted Whitehead link

Lemma B.1.

Proof.

Proposition B.3.

Proof.

B.4. Twist knots and double twist knots

Proposition B.4.

Proof.

C. Asymptotics

C.1. Quantum dilogarithm function

C.2. Reformulation of the colored Jones polynomials

C.3. Saddle points

D. Neumann-Zagier function

D.1. Neumann-Zagier potential function

D.2. Complex volume

Theorem 4.

References

A.3. Quantum $6j$ symbol