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1 Introduction↩︎

An \(n\)-dimensional cone-manifold is a simplicial complex \(M\) which can be triangulated so that the link of each simplex is piecewise-linear homeomorphic to a standard \((n-1)\)-sphere and \(M\) is equipped with a complete path metric such that the restriction of the metric to each simplex is isometric to a geodesic simplex of constant curvature \(\kappa\). The cone-manifold is hyperbolic, Euclidean, or spherical if \(\kappa\) is \(-1\), \(0\), or \(+1\) respectively.

The singular locus \(\Sigma\) of a cone-manifold \(M\) consists of the points in \(M\) with no neighborhood isometric to a ball in a Riemannian manifold. Then \(\Sigma\) is a union of totally geodesic closed simplices of dimension \(n-2\). At each point of \(\Sigma\) in an open \((n-2)\)-simplex, there is a cone angle which is the sum of dihedral angles of \(n\)-simplices containing the point. In general, the cone angle may vary from point to point within a simplex. The regular set \(M \setminus \Sigma\) is a dense open subset of \(M\) and has a smooth Riemannian metric of constant curvature \(\kappa\), but this metric is incomplete if \(\Sigma \not= \emptyset\).

In this paper, we will only consider \(3\)-dimensional cone-manifolds whose underlying space \(M\) is the \(3\)-sphere \(\mathbb{S}^3\) and whose singular set is a knot \(K\) with constant cone angle \(\alpha \in (0,2\pi]\). We will denote these cone manifolds by \(K(\alpha)\).

A two-bridge knot, also known as a rational knot, is a knot that admits a projection with two maxima and two minima. In the Conway notation, a two-bridge knot corresponds to a continued fraction \[[a_1, a_2, \dots, a_k] = a_1 + \dfrac{1}{a_2 + \dfrac{1}{\ddots + \dfrac{1}{a_k}}}\] and denoted by \(C(a_1, a_2, \dots, a_k)\). Its diagram is shown in Figure 1. In the \(a_i\) box, \(|a_i|\) denotes the number of signed half-twists and the sign of each half-twist is the same as the sign of \(a_i \in \mathbb{Z}\). Here, we use the convention that the sign of the right-handed half-twist \(\raisebox{-5pt}{ \epsfig{file=leftcross.pdf}}\) in the \(a_i\) box is positive for odd \(i\) and negative for even \(i\). In the Schubert notation, \(C(a_1, a_2, \dots, a_k)\) is the two-bridge knot \(\mathfrak b(p,q)\) where \(\frac{p}{q} = [a_1, a_2, \dots, a_k]\).

Geometric transitions of cone-manifolds↩︎

For a two-bridge knot \(K\), Kojima [1] and Porti [2] established that there exists a critical angle \(\alpha_K \in [\frac{2\pi}{3}, \pi)\) such that the cone manifold \(K(\alpha)\) undergoes geometric transitions:

• hyperbolic structure for \(\alpha \in (0, \alpha_K)\),

• Euclidean structure for \(\alpha = \alpha_K\),

• spherical structure for \(\alpha \in (\alpha_K, 2\pi - \alpha_K)\).

Previous work on volume formulas↩︎

Hilden, Lozano, and Montesinos-Amilibia [3] introduced a method for calculating volumes of two-bridge knot cone manifolds, but without providing explicit formulas. Integral formulas for hyperbolic volumes have been obtained for specific families:

• \(C(2n, 2)\) (twist knots) by Ham, Mednykh, and Petrov [4],

• \(C(2n,3)\) by Ham and Lee [5],

• \(C(2n,k)\) (double twist knots \(J(-2n, k)\)) by Tran [6].

However, these formulas involve implicitly defined integrands and are primarily useful for numerical approximation. The only exact integral formulas previously known were those given by Mednykh [7] for two-bridge knots with up to seven crossings.

Our contribution↩︎

In this paper, we extend Mednykh’s approach to obtain exact integral formulas for infinite families of two-bridge knots. We will give exact integral formulas for hyperbolic and spherical volumes of cone-manifolds along two-bridge knots \(C(2n, 2)\), \(C(2n, 3)\) and \(C(2n, -2n)\), where \(n\) is a non-zero integer.

To state our main result, we introduce the Chebychev polynomials of the second kind \(S_k(z)\). They are recursively defined by \(S_0(z)=1\), \(S_1(z)=z\) and \(S_{k}(z) = z S_{k-1}(z) - S_{k-2}(z)\) for all integers \(k\).

For \(K=C(2n,3)\), \(C(2n, 2)\) or \(C(2n, -2n)\) we let \[\begin{align} f_n(y) &=& \frac{2S_n(y)-yS_{n-1}(y)}{(y-2)S_{n-1}(y)}, \\ g_n(y) &=& \begin{cases} - \displaystyle{\frac{(S_n(y) - S_{n-1}(y))^2}{(y-2)^3S^4_{n-1}(y)}} & \quad \text{if } K=C(2n,3), \\ - \displaystyle{\frac{S_n(y) - S_{n-1}(y)}{(y-2)^2S^3_{n-1}(y)}} & \quad \text{if } K=C(2n,2), \\ \displaystyle{\frac{1}{(y-2)^2S^4_{n-1}(y)}} & \quad \text{if } K=C(2n, -2n). \end{cases} \end{align}\] Then the hyperbolic and spherical volumes of the cone-manifold \(K(\alpha)\) are given as follows.

As in [7], the proofs of Theorems 1 and 2 are based on

• trigonometric identity between the cone angle \(\alpha\) and the complex length \(\gamma_\alpha\) of the singular geodesic \(K\) in the cone-manifold \(K(\alpha)\), and

• the Schläfli formula \[\kappa \, d \mathrm{Vol}(K(\alpha)) = \frac{1}{2} l_\alpha d\alpha,\] where \(l_\alpha = \mathrm{Re} \, \gamma_\alpha > 0\) is the real length of \(K\).

The paper is organized as follows. In Section 2 we briefly review holonomy representations of hyperbolic and spherical knot cone-manifolds. In Section 3 we first study \(\mathrm{SL}_2(\mathbb{C})\)-representations of \(C(2n, 2p+1)\), then prove trigonometric identity between the cone angle and the complex length of the singular geodesic, and finally give a proof of Theorems 1 and 2 for \(C(2n,3)\). In Section 4 we carry out the same things for \(C(2n,2)\) and \(C(2n, -2n)\).

2 Knot cone-manifolds↩︎

Recall that \(K(\alpha)\) denotes the \(3\)-dimensional cone manifolds whose underlying space \(M\) is the \(3\)-sphere \(S^3\) and whose singular set is a knot \(K\) with constant cone angle \(\alpha \in (0,2\pi]\). Let \(G(K) := \pi_1(S^3 \setminus K)\) be the knot group, which is the fundamental group of the knot exterior. Choose the canonical meridian-longitude pair \((\mu, \lambda)\) in \(G(K)\) such that \(\mu\) is an oriented boundary of meridian disk of \(K\) and \(\lambda\) is null-homologous outside \(K\).

If \(K(\alpha)\) is hyperbolic, then let \(\rho_\alpha: G(K) \to \mathrm{Isom}^+(\mathbb{H}^3) \cong \mathrm{PSL}_2(\mathbb{C})\) be the holonomy representation. Then \(\rho_\alpha\) admits two liftings to \(\mathrm{SL}_2(\mathbb{C})\). Up to conjugation in \(\mathrm{SL}_2(\mathbb{C})\), we can assume that \[\rho_\alpha (\mu) = \pm \left[ \begin{array}{cc} e^{i\alpha/2} & 0 \\ 0 & e^{-i\alpha/2} \end{array} \right], \quad \rho_\alpha(\lambda) = \left[ \begin{array}{cc} e^{\gamma_\alpha/2} & 0\\ 0 & e^{-\gamma_\alpha/2} \end{array} \right]\] where \(\gamma_\alpha = l_\alpha + i \varphi_\alpha\), \(l_\alpha\) is the length of \(K\), and \(\varphi_\alpha \in [-2\pi, 2\pi)\) is the angle of the lifted holonomy of \(K\). We call \(\gamma_\alpha = l_\alpha + i \varphi_\alpha\) the complex length of the singular geodesic \(K\).

If \(K(\alpha)\) is spherical, then let \(\rho_\alpha: G(K) \to \mathrm{Isom}^+(\mathbb{S}^3) \cong\mathrm{SO}(4)\) be the holonomy representation. Then \(\rho_\alpha\) admits two liftings to \(\mathrm{SU}(2) \times \mathrm{SU}(2)\). Up to conjugation in \(\mathrm{SU}(2) \times \mathrm{SU}(2)\), we can assume that \[\begin{align} \rho_\alpha(\mu) &=& \left( \pm \left[ \begin{array}{cc} e^{i\alpha/2} & 0 \\ 0 & e^{-i\alpha/2} \end{array} \right], \pm \left[ \begin{array}{cc} e^{i\alpha/2} & 0 \\ 0 & e^{-i\alpha/2} \end{array} \right] \right), \\ \rho_\alpha(\lambda) &=& \left( \left[ \begin{array}{cc} e^{i\gamma} & 0 \\ 0 & e^{-i\gamma} \end{array} \right], \left[ \begin{array}{cc} e^{i\phi} & 0 \\ 0 & e^{-i\phi} \end{array} \right]\right). \end{align}\] In this case \(l_\alpha = \gamma - \phi\) is the length of the knot \(K\), and \(\varphi_\alpha = \gamma + \phi \in [-2\pi, 2\pi)\) is the angle of the lifted holonomy of \(K\). Note that \(\gamma = \frac{1}{2}(\varphi_\alpha + l_\alpha)\) and \(\phi = \frac{1}{2}(\varphi_\alpha - l_\alpha)\).

3 \(C(2n, 2p+1)\)↩︎

3.1 Knot group↩︎

Note that the knot group presentation in Proposition 1 is different from the one in [8]–[10] (where \(C(2n, 2p+1)\) is denoted by \(J(-2n, 2p+1)\)), but it can be applied to find exact integral formulas for volumes of cone-manifolds along \(C(2n, 2p+1)\).

3.2 \(\mathrm{SL}_2(\mathbb{C})\)-representations↩︎

Suppose \(\rho\colon G(C(2n, 2p+1))\to \mathrm{SL}_2(\mathbb{C})\) is a nonabelian representation. Up to conjugation, we may assume that \[\label{repn} A:=\rho(a) = \left[ \begin{array}{cc} m & 1\\ 0 & m^{-1} \end{array} \right] \quad \text{and} \quad B:=\rho(b) = \left[ \begin{array}{cc} m & 0\\ y - m^2 - m^{-2} & m^{-1} \end{array} \right]\tag{1}\] where \((m,y) \in \mathbb{C}^2\) satisfies \(\rho(\omega a) = \rho(b \omega)\). Note that \(y=\mathop{\mathrm{\mathrm tr}}\rho(ab)\).

We now solve the matrix equation \(\rho(\omega a) = \rho(b \omega)\). Recall that \(S_k(z)\)’s are the Chebychev polynomials defined by \(S_0(z)=1\), \(S_1(z)=z\) and \(S_{k}(z) = z S_{k-1}(z) - S_{k-2}(z)\) for \(k \in \mathbb{Z}\). Note that \(S_k(z) = (s^{k+1} - s^{-k-1})/(s - s^{-1})\) if \(z = s + s^{-1}\).

The following lemmas are elementary, see e.g. [10] and references therein.

Let \(x= \mathop{\mathrm{\mathrm tr}}\rho(a)=\mathop{\mathrm{\mathrm tr}}\rho(b) = m+m^{-1}\). Let \(U=\rho((a^{-1} b^{-1})^{n} (ab)^{n})\) and \(u = \mathop{\mathrm{\mathrm tr}}U\).

Proposition 3 implies that \(\rho(\omega a) = \rho(b \omega)\) if and only if \(\Phi_{C(2n, 2p+1)}(x,y) =0\).

3.3 Longitude and trignometric identity↩︎

If we choose the meridian to be \(\mu=a\) then the canonical longitude is \(\lambda = \omega \omega^* a^{-4n}\), where \(\omega^*\) is the word obtained from \(\omega\) by writing the letters in \(\omega\) in reversed order.

Since \(\rho(\mu) = \left[ \begin{array}{cc} m & 1\\ 0 & m^{-1} \end{array} \right]\) we have \(\rho(\lambda) = \left[ \begin{array}{cc} l & *\\ 0 & l^{-1} \end{array} \right]\). By [8] we have \(lm^{4n} = - \frac{\widetilde{W_{12}}}{W_{12}}\), where \(W_{12}\) is the \((1,2)\)-entry of \(W=\rho(\omega)\) and \(\widetilde{W_{12}}\) is obtained from \(W_{12}\) by replacing \(m\) by \(m^{-1}\). Note that \(W_{12}\) is a function in \(m\) and \(y\).

Hyperbolic case: Let \(K(\alpha)\) be a hyperbolic 3-dimensional cone-manifold whose singular set is \(K=C(2n, 2p+1)\) with cone angle \(\alpha \in (0, 2\pi]\). Up to conjugation in \(\mathrm{SL}_2(\mathbb{C})\), \[\rho_\alpha (\mu) = \pm \left[ \begin{array}{cc} e^{i\alpha/2} & 0 \\ 0 & e^{-i\alpha/2} \end{array} \right], \quad \rho_\alpha(\lambda) = \left[ \begin{array}{cc} e^{\gamma_\alpha/2} & 0\\ 0 & e^{-\gamma_\alpha/2} \end{array} \right]\] where \(\gamma_\alpha = l_\alpha + i \varphi_\alpha\) is the complex length of the singular geodesic \(K\) in \(K(\alpha)\), \(l_\alpha > 0\) is the real length of \(K\), and \(\varphi_\alpha \in [-2\pi, 2\pi)\) is the angle of the lifted holonomy of \(K\).

Spherical case: Let \(K(\alpha)\) be a spherical 3-dimensional cone-manifold whose singular set is \(K=C(2n, 2p+1)\) with cone angle \(\alpha \in (0, 2\pi]\). Up to conjugation in \(\mathrm{SU}(2) \times \mathrm{SU}(2)\), we can assume that \[\begin{align} \rho_\alpha(\mu) &=& \left( \pm \left[ \begin{array}{cc} e^{i\alpha/2} & 0 \\ 0 & e^{-i\alpha/2} \end{array} \right], \pm \left[ \begin{array}{cc} e^{i\alpha/2} & 0 \\ 0 & e^{-i\alpha/2} \end{array} \right] \right), \\ \rho_\alpha(\lambda) &=& \left( \left[ \begin{array}{cc} e^{i\gamma} & 0 \\ 0 & e^{-i\gamma} \end{array} \right], \left[ \begin{array}{cc} e^{i\phi} & 0 \\ 0 & e^{-i\phi} \end{array} \right]\right). \end{align}\] In this case \(l_\alpha = \gamma - \phi\) is the length of the knot \(K\), and \(\varphi_\alpha = \gamma + \phi \in [-2\pi, 2\pi)\) is the angle of the lifted holonomy of \(K\). Note that \(\gamma = \frac{1}{2}(\varphi_\alpha + l_\alpha)\) and \(\phi = \frac{1}{2}(\varphi_\alpha - l_\alpha)\). Hence \(m = e^{i\alpha/2}\) and \(\ell=e^{i(\varphi_\alpha \pm l_\alpha)/2}\).

3.4 Proof of Theorems 1 and 2 for \(C(2n,3)\)↩︎

Suppose \(p=1\). By Propositions 2 and 3 we have \(u = 2 + (y-2) (y+2-x^2) S^2_{n-1}(y)\) and \[\begin{align} \Phi_{C(2n,3)}(x,y) &=& (S_n(y) - S_{n-1}(y)) u - (S_n(y) - (y-1)S_{n-1}(y)) \\ &=& (y-2) (y+2-x^2) S^2_{n-1}(y) (S_n(y) - S_{n-1}(y)) \\ && + \, 2 (S_n(y) - S_{n-1}(y)) - (S_n(y) - (y-1)S_{n-1}(y)). \end{align}\]

We write \(\Phi_{C(2n,3)}(x,y) = b - a x^2\), where \[\begin{align} a &=& (y-2) S^2_{n-1}(y) (S_n(y) - S_{n-1}(y)), \\ b &=& S_n(y) + (y-3) S_{n-1}(y) + (y-2) (y+2) S^2_{n-1}(y) (S_n(y) - S_{n-1}(y)). \end{align}\] Note that \(b - 4a = S_n(y) + (y-3) S_{n-1}(y) + (y-2)^2 S^2_{n-1}(y) (S_n(y) - S_{n-1}(y))\).

Choose \(c = f_n(y)\). By Lemma 3, the equation \(\Phi_{C(2n,3)}(2\cos \frac{\alpha}{2}, y)=0\) is equivalent to \(c^2+ A^2= (1+A^2)d\) where \(d = 1 + (c^2-1)(1 - \frac{b}{4a})\). Since \[c^2 -1 = \frac{4(S_n(y)-S_{n-1}(y)) (S_n(y) - (y-1) S_{n-1}(y))}{(y-2)^2 S^2_{n-1}(y)},\] we have \[\begin{align} (c^2-1) \left( 1 - \frac{b}{4a} \right) &=& - \frac{S_n(y) - (y-1) S_{n-1}(y)}{(y-2)^3 S^4_{n-1}(y)} [ S_n(y) + (y-3) S_{n-1}(y) \\ && \qquad \qquad \qquad \qquad + (y-2)^2 S^2_{n-1}(y) (S_n(y) - S_{n-1}(y)) ] \\ &=& - \frac{1}{(y-2)^3 S^4_{n-1}(y)} [ (S_n(y) - S_{n-1}(y))^2 - (y-2)^2 S^2_{n-1}(y) \\ && + \, (y-2)^2 S^2_{n-1}(y) \left( S^2_n(y) - y S_n(y) S_{n-1}(y) + (y-1) S^2_{n-1}(y) \right) ]. \end{align}\]

By Lemma 1 we have \(S^2_n(y) - y S_n(y) S_{n-1}(y) + S^2_{n-1}(y) = 1\). This implies that \(S^2_n(y) - y S_n(y) S_{n-1}(y) + (y-1) S^2_{n-1}(y) = 1 + (y-2) S^2_{n-1}(y)\). Hence \[\begin{align} (c^2-1) \left( 1 - \frac{b}{4a} \right) &=& - \frac{(S_n(y) - S_{n-1}(y))^2 + (y-2)^3 S^4_{n-1}(y)}{(y-2)^3 S^4_{n-1}(y)} \end{align}\] and \(d = 1 + (c^2-1) \left( 1 - \frac{b}{4a} \right) = - \frac{(S_n(y) - S_{n-1}(y))^2}{(y-2)^3 S^4_{n-1}(y)}\).

In summary, we have proved the following.

For a two-bridge knot \(K\), there exists an angle \(\alpha_K \in [\frac{2\pi}{3},\pi)\) such that \(K(\alpha)\) is hyperbolic for \(\alpha \in (0, \alpha_K)\), Euclidean for \(\alpha =\alpha_K\), and spherical for \(\alpha \in (\alpha_K, 2\pi - \alpha_K)\).

3.4.1 Hyperbolic case↩︎

For \(\alpha \in (0, \alpha_K)\), by the Schläfli formula we have \[\frac{d\mathrm{Vol}(K(\alpha))}{d\alpha} = - \frac{1}{2} l_\alpha\] where \(l_\alpha = \mathrm{Re}(\gamma_\alpha) > 0\) is the real length of \(K \subset K(\alpha)\). Note that \(K(\alpha)\) is Euclidean at \(\alpha = \alpha_K\), so \(\mathrm{Vol}(K(\alpha)) \to 0\) as \(\alpha \to \alpha_K\). Let \[\label{F} F(\alpha)= i \int_{\overline{y_0}}^{y_0} \log \left( \frac{f^2_n(y)+ A^2}{(1+A^2)g_n(y)} \right) \frac{f'_n(y)dy}{f^2_n(y)-1}.\tag{2}\] Then Theorem 1 is equivalent to \(\mathrm{Vol}(K(\alpha)) = F(\alpha)\).

We first claim that \(F(\alpha) \to 0\) as \(\alpha \to \alpha_K\). Indeed, as \(\alpha \to \alpha_K\) we have \(l_\alpha \to 0\) and so \(\gamma_\alpha = \ell_\alpha + i \varphi_\alpha \to i \varphi_{\alpha_K}\). Then, by the trigonometric identity (Proposition 7) we obtain \[\begin{align} f_n(y_0) &=& i \coth \left( \frac{\gamma_\alpha + 4n i \alpha}{4} \right) \cot \left( \frac{\alpha}{2} \right) \\ &\to& i \coth\left( \frac{i \varphi_{\alpha_K} + 4n i \alpha_K}{4} \right) \cot \left( \frac{\alpha_K}{2} \right) \\ &=& \cot \left( \frac{\varphi_{\alpha_K} + 4n \alpha_K}{4} \right) \cot \left( \frac{\alpha_K}{2} \right), \end{align}\] where we used \(\coth (iz) = -i \cot(z)\). Then \(\mathrm{Im} \, f_n(y_0) \to 0\). Hence \(f_n(\overline{y_0}) - f_n(y_0) = \overline{ f_n(y_0)} - f_n(y_0) = - 2 i \, \mathrm{Im} \, f_n(y_0) \to 0\).

For \(\alpha \in (\alpha_K - \varepsilon, \alpha_K)\), with \(\varepsilon\) a sufficiently small positive real number, we let \(s := f_n(y)\). Since \(f_n(y)\) is a rational function \(y\), we can write \(y = h(s)\) for some continuous function \(h(s)\) in a small open neighborhood of \(f_n(y_0)\). Then, by changing variable we have \[F(\alpha)= i \int_{f_n(\overline{y_0})}^{f_n(y_0)} \log \left( \frac{s^2+ A^2}{(1+A^2) (g_n \circ h)(s)} \right) \frac{ds}{s^2-1}.\] As \(\alpha \to \alpha_K\), since \(f_n(\overline{y_0}) - f_n(y_0) \to 0\) we obtain \(F(\alpha) \to 0\).

Note that we also have \(\mathrm{Vol}(K(\alpha)) \to 0\) as \(\alpha \to \alpha_K\). Hence \(\mathrm{Vol}(K(\alpha)) = F(\alpha)\) if we can show that \[\frac{dF(\alpha)}{d\alpha} = \frac{d\mathrm{Vol}(K(\alpha))}{d\alpha} = - \frac{1}{2} l_\alpha.\].

By taking derivative of 2 and noting that \(dA/d\alpha=-(1+A^2)/2\), we have \[\begin{align} \frac{dF(\alpha)}{d\alpha} &=& \log \left( \frac{f^2_n(y_0)+ A^2}{(1+A^2)g_n(y_0)} \right) \frac{i f'_n(y_0)}{f^2_n(y_0)-1} \frac{dy_0}{d\alpha} - \log \left( \frac{f^2_n(\overline{y_0})+ A^2}{(1+A^2)g_n(\overline{y_0})} \right) \frac{i f'_n(\overline{y_0})}{f^2_n(\overline{y_0})-1} \frac{d\overline{y_0}}{d\alpha} \\ && + \, i \int_{\overline{y_0}}^{y_0} \frac{\partial}{\partial A} \left( \frac{f'_n(y)}{f^2_n(y)-1} \log \left( \frac{f^2_n(y)+ A^2}{(1+A^2)g_n(y)} \right) \right) \frac{dA}{d\alpha} \, dy \\ &=& i \int_{\overline{y_0}}^{y_0}\frac{f'_n(y)}{f^2_n(y)-1} \left( \frac{2A}{f^2_n(y)+ A^2} - \frac{2A}{1+ A^2} \right) \frac{-(1+A^2)}{2} dy \\ &=& i \int_{\overline{y_0}}^{y_0} \frac{f'_n(y) A}{f^2_n(y)+ A^2} dy \\ &=&i \left( \mathrm{arccot} \frac{f_n(\overline{y_0})}{A} - \mathrm{arccotn} \frac{f_n(y_0)}{A}\right). \end{align}\]

Since \(f_n(y_0) = i \coth \left( \frac{\gamma_\alpha + 4n i \alpha}{4} \right) \cot \left( \frac{\alpha}{2} \right)\) we have \(\frac{f_n(y_0)}{A} = i \coth \left( \frac{\gamma_\alpha + 4n i \alpha}{4} \right) = \cot \left( \frac{\gamma_\alpha + 4n i \alpha}{4i} \right)\) and \(\frac{f_n(\overline{y_0})}{A} = \overline{\frac{f_n(y_0)}{A} } = \cot \left( \frac{\overline{\gamma_\alpha + 4n i \alpha}}{-4i} \right)\). Hence \[\begin{align} \frac{dF(\alpha)}{d\alpha} &=& i \left( \mathrm{arccot} \frac{f_n(\overline{y_0})}{A} - \mathrm{arccotn} \frac{f_n(y_0)}{A}\right) \\ &=& i \left( \frac{\overline{\gamma_\alpha + 4n i \alpha} }{-4i} - \frac{\gamma_\alpha + 4n i \alpha}{4i} \right) \\ &=& -\frac{\overline{\gamma_\alpha} + \gamma_\alpha }{4} = - \frac{l_\alpha}{2}. \end{align}\] This proves Theorem 1 for \(C(2n, 3)\) in the hyperbolic case.

3.4.2 Spherical case↩︎

For \(\alpha \in (\alpha_K, 2\pi - \alpha_K)\), by the Schläfli formula we have \[\frac{d\mathrm{Vol}(K(\alpha))}{d\alpha} = \frac{1}{2} l_\alpha.\]

Let \[\label{G} G(\alpha)= \int_{y_+}^{y_-} \log \left( \frac{f^2_n(y)+ A^2}{(1+A^2)g_n(y)} \right) \frac{f'_n(y)dy}{f^2_n(y)-1}.\tag{3}\] Then Theorem 2 is equivalent to \(\mathrm{Vol}(K(\alpha)) = G(\alpha)\).

We first claim that \(G(\alpha) \to 0\) as \(\alpha \to \alpha_K\). Indeed, as \(\alpha \to \alpha_K\), we have \(l_\alpha \to 0\) and \[f_n(y_{\pm}) = \cot \left( \frac{\varphi_\alpha \pm l_\alpha+ 4n \alpha}{4} \right) \cot \left( \frac{\alpha}{2} \right) \to \cot \left( \frac{\varphi_{\alpha_K} + 4n \alpha_K}{4} \right) \cot \left( \frac{\alpha_K}{2} \right).\]

For \(\alpha \in (\alpha_K - \varepsilon, \alpha_K)\), with \(\varepsilon\) a sufficiently small positive real number, we let \(s := f_n(y)\). Since \(f_n(y)\) is a rational function \(y\), we can write \(y = h(s)\) for some continuous function \(h(s)\) in a small open neighborhood of \(f_n(y_{\pm})\). Then, by changing variable we have \[G(\alpha)= \int_{f_n(y_+)}^{f_n(y_-)} \log \left( \frac{s^2+ A^2}{(1+A^2) (g_n \circ h)(s)} \right) \frac{ds}{s^2-1}.\] As \(\alpha \to \alpha_K\), since \(f_n(y_{\pm}) \to \cot \left( \frac{\varphi_{\alpha_K} + 4n \alpha_K}{4} \right) \cot \left( \frac{\alpha_K}{2} \right)\), we obtain \(G(\alpha) \to 0\). Note that \(\mathrm{Vol}(K(\alpha)) \to 0\) as \(\alpha \to \alpha_K\). Hence \(\mathrm{Vol}(K(\alpha)) = G(\alpha)\) if we can show that \[\frac{dG(\alpha)}{d\alpha} = \frac{d\mathrm{Vol}(K(\alpha))}{d\alpha} = \frac{1}{2} l_\alpha.\]

By taking derivative of 3 and noting that \(dA/d\alpha=-(1+A^2)/2\), we have \[\begin{align} \frac{dG(\alpha)}{d\alpha} &=& \log \left( \frac{f^2_n(y_-)+ A^2}{(1+A^2)g_n(y_-)} \right) \frac{f'_n(y_-)}{f^2_n(y_-)-1} \frac{dy_-}{d\alpha} - \log \left( \frac{f^2_n(y_+)+ A^2}{(1+A^2)g_n(y_+)} \right) \frac{f'_n(y_+)}{f^2_n(y_+)-1} \frac{dy_+}{d\alpha} \\ && + \, \int_{y_+}^{y_-} \frac{\partial}{\partial A} \left( \frac{f'_n(y)}{f^2_n(y)-1} \log \left( \frac{f^2_n(y)+ A^2}{(1+A^2)g_n(y)} \right) \right) \frac{dA}{d\alpha} \, dy \\ &=& \int_{y_+}^{y_-}\frac{f'_n(y)}{f^2_n(y)-1} \left( \frac{2A}{f^2_n(y)+ A^2} - \frac{2A}{1+ A^2} \right) \frac{-(1+A^2)}{2} dy \\ &=& \int_{y_+}^{y_-} \frac{f'_n(y) A}{f^2_n(y)+ A^2} dy \\ &=& \mathrm{arccot} \frac{f_n(y_+)}{A} - \mathrm{arccot} \frac{f_n(y_-)}{A}. \end{align}\]

By the trigonometric identity (Proposition 7) we have \(f_n(y_{\pm}) = \cot \left( \frac{\varphi_\alpha \pm l_\alpha+ 4n \alpha}{4} \right) \cot \left( \frac{\alpha}{2} \right)\). This implies that \(\frac{f_n(y_{\pm})}{A} = \cot \left( \frac{\varphi_\alpha \pm l_\alpha+ 4n \alpha}{4} \right)\). Hence \[\begin{align} \frac{dG(\alpha)}{d\alpha} &=& - \mathrm{arccot} \frac{f_n(y_-)}{A} + \mathrm{arccot} \frac{f_n(y_+)}{A} \\ &=& - \frac{\varphi_\alpha - l_\alpha+ 4n \alpha}{4} +\frac{\varphi_\alpha + l_\alpha+ 4n \alpha}{4} \\ &=& \frac{l_\alpha}{2}. \end{align}\] This proves Theorem 2 for \(C(2n, 3)\) in the spherical case.

4 \(C(2n, 2p)\)↩︎

4.1 Knot group↩︎

Note that \(C(2n, 2p)\) is the double twist knot \(J(-2n, 2p)\), so by [8] its knot group has the following presentation.

We can also prove the above proposition by the same structure as Proposition 1 for \(C(2n, 2p + 1)\), with appropriate modifications for the even case. Starting from the knot diagram and using the Wirtinger presentation, we trace through the crossings to obtain the stated relation. See also [9].

4.2 \(\mathrm{SL}_2(\mathbb{C})\)-representations↩︎

Suppose \(\rho\colon G(C(2n, 2p))\to \mathrm{SL}_2(\mathbb{C})\) is a nonabelian representation. Up to conjugation, we may assume that \[A:=\rho(a) = \left[ \begin{array}{cc} m & 1\\ 0 & m^{-1} \end{array} \right] \quad \text{and} \quad B:=\rho(b) = \left[ \begin{array}{cc} m & 0\\ 2-z & m^{-1} \end{array} \right]\] where \((m,z) \in \mathbb{C}^2\) satisfies \(\rho(\omega' a) = \rho(b \omega')\). Note that \(z=\mathop{\mathrm{\mathrm tr}}\rho(ab^{-1})\).

The following propositions are proved in [9], [10].

4.3 Longitude and trignometric identity↩︎

If we choose the meridian to be \(\mu=a\) then the canonical longitude is \(\lambda = \omega' (\omega')^*\), where \((\omega')^*\) is the word obtained from \(\omega'\) by writing the letters in \(\omega'\) in reversed order.

Since \(\rho(\mu) = \left[ \begin{array}{cc} m & 1\\ 0 & m^{-1} \end{array} \right]\) we have \(\rho(\lambda) = \left[ \begin{array}{cc} l & *\\ 0 & l^{-1} \end{array} \right]\), where \(m = e^{i \alpha/2}\) and \(l = e^{\gamma_\alpha/2}\). By [8] we have \(l = - \frac{\widetilde{W'_{12}}}{W'_{12}}\), where \(W'_{12}\) is the \((1,2)\)-entry of \(W'=\rho(\omega')\) and \(\widetilde{W'_{12}}\) is obtained from \(W'_{12}\) by replacing \(m\) by \(m^{-1}\). Note that \(W'_{12}\) is a function in \(m\) and \(z\).

Similar to Propositions 5 and 6 we have the following propositions.

4.4 Proof of Theorems 1 and 2 for \(C(2n,2)\)↩︎

Suppose \(p=1\). By Propositions 11 and 12 we have \(v = 2 + (z-2) (z+2-x^2) S^2_{n-1}(z)\) and \[\begin{align} \Phi_{C(2n, 2)}(x,z) &=& 1 + (z+2-x^2) S_{n-1}(z) (S_{n}(z) - S_{n-1}(z)). \end{align}\]

We write \(\Phi_{C(2n,2)}(x,z) = b - a x^2\), where \[\begin{align} a &=& S_{n-1}(z) (S_{n}(z) - S_{n-1}(z)), \\ b &=& 1 + (z+2) S_{n-1}(z) (S_{n}(z) - S_{n-1}(z)). \end{align}\] Since \(S^2_n(z) - z S_n(z) S_{n-1}(z) + S^2_{n-1}(z) = 1\) we have \[\begin{align} b - 4a &=& 1 + (z-2) S_{n-1}(z) (S_{n}(z) - S_{n-1}(z)) \\ &=& S^2_{n}(z) - 2 S_{n}(z) S_{n-1}(z) + (3-z)S^2_{n-1}(z). \end{align}\]

Choose \(c = f_n(z)\). By Lemma 3, the equation \(\Phi_{C(2n, 2)}(2\cos \frac{\alpha}{2}, z)=0\) is equivalent to \(c^2+ A^2= (1+A^2)d\) where \(d = 1 + (c^2-1)(1 - \frac{b}{4a})\). Since \[c^2 -1 = \frac{4(S_n(z)-S_{n-1}(z)) (S_n(z) - (z-1) S_{n-1}(z))}{(z-2)^2 S^2_{n-1}(z)},\] by a direct calculation we have \[\begin{align} d &=& 1 + (c^2-1) \left( 1 - \frac{b}{4a} \right) \\ &=& 1 - \frac{\left( S_n(z) - (z-1) S_{n-1}(z) \right) \left( S^2_{n}(z) - 2 S_{n}(z) S_{n-1}(z) + (3-z)S^2_{n-1}(z) \right)}{(z-2)^2 S^3_{n-1}(z)} \\ &=& - \frac{\left( S_n(z) -S_{n-1}(z) \right)\left( S^2_n(z) + S^2_{n-1}(z) - z S_n(z) S_{n-1}(z) \right)}{(z-2)^2 S^3_{n-1}(z)} \\ &=& - \frac{ S_n(z) -S_{n-1}(z)}{(z-2)^2 S^3_{n-1}(z)}. \end{align}\]

Hence we have proved the following.

By using the trigonometric identity (Proposition 14) and Proposition 15, the proof of Theorems 1 and 2 for \(C(2n,2)\) is similar to that for \(C(2n,3)\).

4.5 Proof of Theorems 1 and 2 for \(C(2n, -2n)\)↩︎

Suppose \(p=-2n\). Note that \(C(2n, -2n)\) is the mirror image of the double twist knot \(J(2n, 2n)\) in [8]. By [9] the component of \(\Phi_{C(2n, -2n)}(x,z)\) containing the holonomy representation is a factor of \(v - z = (z-2) \left( -1 + (z+2-x^2) S^2_{n-1}(z) \right)\). The factor \(z-2\) corresponds to reducible representations, hence the factor \[\Phi^{\mathrm{hol}}_{C(2n, -2n)}(x,z) := -1 + (z+2-x^2) S^2_{n-1}(z)\] determines the component containing the holonomy representation.

We write \(\Phi^{\mathrm{hol}}_{C(2n, -2n)}(x,z) = b - a x^2\), where \(a = S^2_{n-1}(z)\) and \(b = -1 + (z+2) S^2_{n-1}(z)\). Since \(S^2_n(z) - z S_n(z) S_{n-1}(z) + S^2_{n-1}(z) = 1\) we have \[b - 4a = - 1 +(z-2) S^2_{n-1}(z) = - S^2_{n}(z) + z S_{n}(z) S_{n-1}(z) + (z-3)S_{n-1}(z).\]

Choose \(c = f_n(z)\). By Lemma 3, the equation \(\Phi^{\mathrm{hol}}_{C(2n, -2n)}(2\cos \frac{\alpha}{2}, z)=0\) is equivalent to \(c^2+ A^2= (1+A^2)d\) where \(d = 1 + (c^2-1)(1 - \frac{b}{4a})\). Since \[c^2 -1 = \frac{4(S_n(z)-S_{n-1}(z)) (S_n(z) - (z-1) S_{n-1}(z))}{(z-2)^2 S^2_{n-1}(z)},\] by a direct calculation we have \[\begin{align} d &=& 1 + (c^2-1) \left( 1 - \frac{b}{4a} \right) \\ &=& 1 - \frac{(S_n(z)-S_{n-1}(z)) (S_n(z) - (z-1) S_{n-1}(z)) }{(z-2)^2 S^4_{n-1}(z)} \\ && \qquad \times \left( - S^2_{n}(z) + z S_{n}(z) S_{n-1}(z) + (z-3)S_{n-1}(z) \right) \\ &=& \frac{\left( S^2_n(z) + S^2_{n-1}(z) - z S_n(z) S_{n-1}(z) \right)^2}{(z-2)^2 S^4_{n-1}(z)} \\ &=& \frac{1}{(z-2)^2 S^4_{n-1}(z)}. \end{align}\]

Hence we have proved the following.

By using the trigonometric identity (Proposition 14) and Proposition 16, the proof of Theorems 1 and 2 for \(C(2n,-2n)\) is similar to that for \(C(2n,3)\).

Acknowledgements↩︎

The first author has been supported by a grant from the Simons Foundation (#708778).

References↩︎