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...

Figure 1. The k-twisted Wk.

Theorem 1.1. If k =2n − 1 then A1(M, L)=(L − 1)F (M, L) where n n +1+ i n − 1+ i L − 1 2i F (M, L) = − (M − M −1)2i 2i +1 2i +1 L +1 i=0 X       n−1 n + i L − 1 2i+1 + (M + M −1)(M − M −1)2i+1 . 2i +1 L +1 i=0 X     2 If k =2n then A1(M, L)=(LM − 1)G(M, L) where n n +1+ i n + i LM 2 − 1 2i+1 G(M, L) = + (M − M −1)2i+1 2i +1 2i +1 LM 2 +1 i=0 X       n n + i LM 2 − 1 2i + (M + M −1)(M − M −1)2i . 2i LM 2 +1 i=0 X     In both cases we have A1(M, L)= A2(M, L). It should be noted that in the definition of the A-polynomial m-tuple of an m-component link, we discard the irreducible component of the SL2(C)-character variety containing only characters of reducible representations and therefore consider only the nonabelian SL2(C)-character variety of the . 3 3 For a link L in S , let EL = S \L denote the link exterior. The link group of L is defined to be the π1(EL) of EL. It is known that the link group of the twisted Whitehead link Wk has a standard two-generator presentation of a two-bridge link group π1(EWk )= ha, b | aw = wai, where a, b are meridians depicted in Figure 1 and w is a word in the letters a, b. More precisely, we have w = (bab−1a−1)na(a−1b−1ab)n if k =2n − 1 and w =(bab−1a−1)nbab(a−1b−1ab)n if k =2n. C For a presentation ρ : π1(EWk ) → SL2( ) we let x, y, z denote the traces of the images of a, b, ab respectively. We also let v denote the trace of the image of bab−1a−1. Explicitly, 2 2 2 we have v = x + y + z − xyz − 2. It was shown in [Tr1] that the nonabelian SL2(C)- k character variety of Wk has exactly ⌊ 2 ⌋ irreducible components. For a , there are distinguished components of the SL2(C)-character variety called the canonical components. They contain important information about the hyperbolic structure of the link. By using the formula of the A-polynomial in Theorem 1.1, we can determine the canonical component of the hyperbolic twisted Whitehead link Wk (with k ≥ 1) as follows. THEA-POLYNOMIALOFTWISTEDWHITEHEADLINKS 3

Theorem 1.2. If k = 2n − 1 then the canonical component of Wk is the zero set of the polynomial (xy − vz)Sn−1(v) − (xy − 2z)Sn−2(v). If k =2n then the canonical component of Wk is the zero set of the polynomial zSn(v)− (xy − z)Sn−1(v).

Here the Sk(v) are the Chebychev polynomials of the second kind defined by S0(v) = 1, S1(v)= v and Sk(v)= vSk−1(v) − Sk−2(v) for all integers k. Explicitly, we have k − i S (v)= (−1)i vk−2i k k − 2i 0≤Xi≤k/2   for k ≥ 0, Sk(v)= −S−k−2(v) for k ≤ −2, and S−1(v) = 0. 3 For a hyperbolic link L ⊂ S , let ρhol be a holonomy representation of π1(EL) into P SL2(C). Thurston [Th] showed that ρhol can be deformed into an m-parameter family

{ρα1,··· ,αm } of representations to give a corresponding family {EL(α1, ··· ,αm)} of singular complete hyperbolic manifolds, where m is the number of components of L. In this paper we will consider only the case where all of αi’s are equal to a single parameter α. In which case we also denote EL(α1, ··· ,αm) by EL(α). These α’s and EL(α)’s are called the cone-angles and hyperbolic cone-manifolds of L, respectively. We consider the complete hyperbolic structure on a link complement as the cone- manifold structure of cone-angle zero. It is known that for a two-bridge knot or link L 2π there exists an angle αL ∈ [ 3 , π) such that EL(α) is hyperbolic for α ∈ (0,αL), Euclidean for α = αL, and spherical for α ∈ (αL, π) [HLM, Ko1, Po, PW]. Explicit volume formulas for hyperbolic cone-manifolds of knots were known for 41 [HLM, Ko1, Ko2, MR], 52 [Me1], twist knots [HMP] and two-bridge knots C(2n, 3) [HL]. Recently, the volume of double cone-manifolds has been computed in [Tr3]. We should remark that a formula for the volume of the cone-manifold of C(2n, 4) has just been given in [HLMR1]. However, with an appropriate change of variables, this formula has already obtained in [Tr3], since C(2n, 4) is the double twist knot J(2n, −4). In this paper we are interested in hyperbolic cone-manifolds of links. Explicit volume 2 2 formulas for hyperbolic cone-manifolds of links have been only known for 51 [MV], 62 2 2 [Me2], 63 [DMM] and 73 [HLMR2]. From the proof of Theorem 1.1, we can compute the volume of the hyperbolic cone-manifold of the twisted Whitehead link Wk as follows. Let 2 2 x Sn−1(v) − zSn(v) − (x − z)Sn−2(v) if k =2n − 1 RWk (s, z)= 2 (zSn(v) − (x − z)Sn−1(v) if k =2n, where x = s + s−1 and v =2x2 + z2 − x2z − 2.

Theorem 1.3. For α ∈ (0,αWk ) we have π z − (s−2 + 1) Vol E (α)= 2 log dω Wk z − (s2 + 1) Zα where s = eiω/2 and z (with Im z ≥ 0) satisfy R (s, z)=0. Wk

Note that the above volume formula for the hyperbolic cone-manifold EWk (α) depends iω/2 on the choice of a root z of RWk (e , z) = 0. In practice, we choose the root z which gives the maximal volume. As a direct consequence of Theorem 1.3, we obtain the following. 4 ANH T. TRAN

Corollary 1.4. The of the r-fold cyclic covering over the twisted White- head link Wk, with r ≥ 3, is given by the following formula 2π π z − (s−2 + 1) rVol EWk ( )= r 2 log 2 dω r 2π z − (s + 1) Z r where s = eiω/2 and z (with Im z ≥ 0) satisfy R (s, z)=0. W k 2 Remark 1.5. The link 73 is the twisted Whitehead link W3. With an appropriate change 2 of variables, we obtain the volume formula for the hyperbolic cone-manifold of 73 in [HLMR2, Theorem 1.1] by taking k = 3 in Theorem 1.3. The paper is organized as follows. In Section 2 we review the definition of the A- polynomial m-tuple of an m-component link in S3. In Section 3 we review the nonabelian SL2(C)-representations of a two-bridge link and compute them explicitly for twisted Whitehead links. We also give proofs of Theorems 1.1–1.3 in Section 3.

2. The A-polynomial 2.0.1. Character varieties. The set of characters of representations of a finitely generated group G into SL2(C) is known to be a complex algebraic set, called the character variety of G and denoted by χ(G) (see [CS, LM]). For a manifold Y we also use χ(Y ) to denote 2 χ(π1(Y )). Suppose G = Z , the free abelian group with 2 generators. Every pair of generators µ,λ will define an isomorphism between χ(G) and (C∗)2/τ, where (C∗)2 is the set of non-zero complex pairs (M, L) and τ : (C∗)2 → (C∗)2 is the involution defined by τ(M, L) := (M −1, L−1), as follows. Every representation is conjugate to an upper diagonal one, with M and L being the upper left entry of µ and λ respectively. The isomorphism does not change if we replace (µ,λ) with (µ−1,λ−1).

3 2.0.2. The A-polynomial. Suppose L = K1 ⊔···⊔ Km be an m-component link in S . Let 3 EL = S \L be the link exterior and T1,...,Tm the boundary tori of EL corresponding to K1,...,Km respectively. Each Ti is a torus whose fundamental group is free abelian of rank two. An orientation of Ki will define a unique pair of an oriented meridian µi and an oriented longitude λi such that the between the longitude λi and the knot C∗ 2 C∗ 2 Ki is 0. The pair provides an identification of χ(π1(Ti)) and ( )i /τi, where ( )i is the −1 −1 set of non-zero complex pairs (Mi, Li) and τi is the involution τ(Mi, Li)=(Mi , Li ), which actually does not depend on the orientation of Ki. The inclusion Ti ֒→ EL induces the restriction map C∗ 2 ρi : χ(EL) −→ χ(Ti) ≡ ( )i /τi.

For each γ ∈ π1(EL) let fγ be the regular function on the character variety χ(EL) defined 2 by fγ(χρ)=(χρ(γ)) −4, where χρ denotes the character of a representation ρ : π1(EL) → C SL2( ). Let χi(X) be the subvariety of χ(X) defined by fµj = 0, fλj = 0 for all j =6 i. ˆ C∗ 2 Let Zi be the image of χi(X) under ρi and Zi ⊂ ( )i the lift of Zi under the projection C∗ 2 C∗ 2 ˆ C∗ 2 C2 C2 ( )i → ( )i /τi. It is known that the Zariski closure of Zi ⊂ ( )i ⊂ i in i is an algebraic set consisting of components of dimension 0 or 1 [Zh]. The union of all the 1- ′ Z dimension components is defined by a single polynomial Ai ∈ [Mi, Li] whose coefficients ′ ′ are co-prime. Note that Ai is defined up to ±1. It is also known that Ai always contains the ′ factor Li−1 coming from the characters of reducible representations, hence Ai =(Li−1)Ai THEA-POLYNOMIALOFTWISTEDWHITEHEADLINKS 5 for some Ai ∈ Z[Mi, Li]. As in [Zh], we will call [A1(M1, L1), ··· , Am(Mm, Lm)] the A- polynomial m-tuple of L. For brevity, we also write Ai(M, L) for Ai(Mi, Li). We refer the reader to [Zh] for properties of the A-polynomial m-tuple. i j Recall that the Newton polygon of a two-variable polynomial aijM L is the convex 2 hull in R of the set {(i, j): aij =06 }. The slope of a side of the Newton polygon is called a boundary slope of the polygon. The following proposition isP useful for determining canonical components of hyperbolic links. Proposition 2.1. [Zh, Theorem 3(2)] Suppose L ⊂ S3 is a hyperbolic m-component link. can For each j =1, ··· , m, the factor Aj (M, L) of the A-polynomial Aj(M, L) corresponding to a canonical component has an irreducible factor whose Newton polygon has at least two distinct boundary slopes. In particular, this irreducible factor contains at least 3 monomials in M, L.

3. Proofs of Theorems 1.1–1.3

In this section we review the nonabelian SL2(C)-representations of a two-bridge link from [Ri] and compute them explicitly for twisted Whitehead links. Finally, we give proofs of Theorems 1.1–1.3. 3.1. Two-bridge links. Two-bridge links are those links admitting a projection with only two maxima and two minima. The double branched cover of S3 along a two-bridge link is a lens space L(2p, q), which is obtained by doing a 2p/q surgery on the unknot. Such a two-bridge link is denoted by b(2p, q). Here q is an odd integer co-prime with 2p and 2p > |q| ≥ 1. It is known that b(2p′, q′) is ambient isotopic to b(2p, q) if and only if p′ = p and q′ ≡ q±1 (mod 2p), see e.g. [BuZ]. The link group of the two-bridge link b(2p, q) has a standard two-generator presentation π1(Eb(2p,q)) = ha, b | wa = awi where ε1 ε2 εp−2 ε2p−1 ⌊iq/(2p)⌋ a, b are meridians, w = b a ··· a b and εi =(−1) for 1 ≤ i ≤ 2p − 1. Note that εi = ε2p−i. We now study representations of link groups into SL2(C). A representation is called nonabelian if its image is a nonabelian subgroup of SL2(C). Let L = b(2p, q). Suppose ρ : EL → SL2(C) is a nonabelian representation. Up to conjugation, we may assume that s 1 s 0 ρ(a)= 1 and ρ(b)= 2 0 s−1 u s−1  1   2  3 where (u,s1,s2) ∈ C satisfies the matrix equation ρ(aw)= ρ(wa). r r For any word r, we write ρ(r)= 11 12 . By induction on the word length, we can r21 r22 C ±1 ±1 ′ show that r21 is a multiple of u in [u,s1 ,s2 ]. Hence we can write r21 = ur21 for some ′ C ±1 ±1 r21 ∈ [u,s1 ,s2 ]. A word is said to be palindromic if it reads the same backward or forward. By Lemma 1 in [Ri] we have −1 −1 ′ r22 − r11 +(s1 − s1 )r12 =(s2 − s2 )r21 for any palindromic word r of odd length. The matrix equation ρ(aw)= ρ(wa) is easily seen to be equivalent to the two equations ′ −1 −1 ′ uw21 = 0 and w22 − w11 +(s1 − s1 )w12 =(s2 − s2 )w21. Since w is a palindromic word of odd length, we conclude that the matrix equation ρ(aw)= ρ(wa) is equivalent to a single ′ ′ equation w21 = 0. The polynomial w21 is called the Riley polynomial of a two-bridge link, and it determines the nonabelian representations of the link. 6 ANH T. TRAN

Remark 3.1. Other approaches, using character varieties and skein modules, to the Riley polynomial of a two-bridge link can be found in [Qa, Tr2, LT].

3.2. Chebyshev polynomials. Recall that the Sk(v) are the Chebychev polynomials of the second kind defined by S0(v) = 1, S1(v) = v and Sk(v) = vSk−1(v) − Sk−2(v) for all integers k. The following results are elementary, see e.g. [Tr3]. Lemma 3.2. For any integer k we have 2 2 Sk(v)+ Sk−1(v) − vSk(v)Sk−1(v)=1.

Lemma 3.3. Suppose V ∈ SL2(C) and v = tr V . For any integer k we have k V = Sk−1(v)V − Sk−2(v)1 where 1 denotes the 2 × 2 identity matrix. We will need the following lemma in the proof of Theorem 1.1. Lemma 3.4. Suppose v =2+ q. For k ≥ 0 we have k k +1+ i (3.1) S (v)= qi. k 2i +1 i=0 X   Proof. We use induction on k ≥ 0. The cases k =0, 1 are clear. Suppose k ≥ 2 and (3.1) holds true for k − 2 and k − 1. Since Sk(v)= vSk−1(v) − Sk−2(v), we have k−1 k + i k−2 k − 1+ i S (v) = (2+ q) qi − qi k 2i +1 2i +1 i=0 i=0 X   X   k k + i k + i − 1 k − 1+ i = 2 + − qi. 2i +1 2i − 1 2i +1 i=0 X        It remains to show the following identity k +1+ i k + i k + i − 1 k − 1+ i =2 + − . 2i +1 2i +1 2i − 1 2i +1         c c c+1  This follows by applying the equality d + d+1 = d+1 three times. 3.3. Twisted Whitehead links. In this subsection  we compute nonabelian representa- tions of twisted Whitehead links explicitly. We first consider the case of W2n−1. Recall that the link group of W2n−1 has a presen- tation π1(EW2n−1 ) = ha, b | aw = wai, where a, b are meridians depicted in Figure 1 and w =(bab−1a−1)na(a−1b−1ab)n. C Suppose ρ : π1(EW2n−1 ) → SL2( ) is a nonabelian representation. Up to conjugation, we may assume that s 1 s 0 ρ(a)= 1 and ρ(b)= 2 0 s−1 u s−1  1   2  3 where (u,s1,s2) ∈ C satisfies the matrix equation ρ(aw)= ρ(wa). −1 −1 Recall from the Introduction that x = tr ρ(a) = s1 + s1 , y = tr ρ(b) = s2 + s2 , −1 −1 z = tr ρ(ab)= u + s1s2 + s1 s2 and −1 −1 −1 −1 −1 −1 v = tr ρ(bab a )= u(u + s1s2 + s1 s2 − s1s2 − s1 s2)+2. THEA-POLYNOMIALOFTWISTEDWHITEHEADLINKS 7

Let c = bab−1a−1 and d = a−1b−1ab. We have c c d d ρ(c)= 11 12 and ρ(d)= 11 12 c c d d  21 22   21 22  where

−1 c11 = 1 − s1 s2u, 2 c12 = −s1 + s1s2 + s2u, −2 −1 −1 −1 c21 = u(−s1 s2 + s2 − s1 u), −1 −1 −1 2 c22 = 1+(s1 s2 − s1s2 + s1s2)u + u , −1 −1 −1 2 d11 = 1+(s1 s2 − s1 s2 + s1s2)u + u , −1 −2 −1 −1 d12 = s1 s2 − s1 + s2 u, 2 d21 = u(s2 − s1s2 − s1u), −1 d22 = 1 − s1s2 u. Since v = tr ρ(c)=tr ρ(d), by Lemma 3.3 we have c S (v) − S (v) c S (v) ρ(cn) = 11 n−1 n−2 12 n−1 , c S (v) c S (v) − S (v)  21 n−1 22 n−1 n−2  d S (v) − S (v) d S (v) ρ(dn) = 11 n−1 n−2 12 n−1 . d S (v) d S (v) − S (v)  21 n−1 22 n−1 n−2  w ∗ This implies that ρ(w)= 11 = ρ(cnadn) with w ∗  21  −1 2 2 w11 = (c12d21s1 + c11d21 + c11d11s1)Sn−1(v)+ s1Sn−2(v)

− (d21 + c11s1 + d11s1)Sn−1(v)Sn−2(v), −1 −1 w21 = (c22d21s1 + c21d21 + c21d11s1)Sn−1(v) − (d21s1 + c21s1)Sn−2(v) Sn−1(v). By direct calculations we have 

−1 c22d21s1 + c21d21 + c21d11s1 = u(xy − vz), −1 d21s1 + c21s1 = u(xy − 2z). ′ Hence w21 = uw21 where ′ w21(W2n−1)= (xy − vz)Sn−1(v) − (xy − 2z)Sn−2(v) Sn−1(v), which is the Riley polynomial of W2n−1.  Similarly, the Riley polynomial of W2n is given by the following formula

′ w21(W2n)= zSn(v) − (xy − z)Sn−1(v) Sn(v) − Sn−1(v) . Remark 3.5. The above formulas for the Riley polynomials of twisted Whitehead links were already obtained in [Tr1] using character varieties. It was also shown in [Tr1] that ′ C the Riley polynomial w21(W2n−1) ∈ [x, y, z] is factored into exactly n irreducible factors jπ ′ (xy−vz)Sn−1(v)−(xy−2z)Sn−2(v) and v−2 cos n (1 ≤ j ≤ n−1). Similarly, w21(W2n) ∈ C[x, y, z] is factored into exactly n + 1 irreducible factors zSn(v) − (xy − z)Sn−1(v) and 8 ANH T. TRAN

(2j−1)π v − 2 cos 2n+1 (1 ≤ j ≤ n). Note that n−1 jπ n (2j − 1)π S (v)= v − 2 cos and S (v) − S (v)= v − 2 cos . n−1 n n n−1 2n +1 j=1 j=1 Y  Y  3.4. Proof of Theorem 1.1. We first consider the case of W2n−1. The canonical longi- −1 −1 tudes corresponding to the meridians a and b are respectively λa = wa and λb = wb , where w is the word obtained from w by exchanging a and b. Precisely, we have w = (aba−1b−1)nb(b−1a−1ba)n. C Suppose ρ : π1(EW2n−1 ) → SL2( ) is a nonabelian representation. With the same notations as in the previous subsection, we have w ∗ w 0 ρ(w)= 11 and ρ(w)= 11 . 0 (w )−1 ∗ (w )−1  11   11  jπ Proposition 3.6. On v − 2 cos n =0 we have

w11 = s1 and w11 = s2. Proof. Recall from the previous subsection that −1 2 2 w11 = (c12d21s1 + c11d21 + c11d11s1)Sn−1(v)+ s1Sn−2(v)

− (d21 + c11s1 + d11s1)Sn−1(v)Sn−2(v). jπ If v = 2 cos n then Sn−1(v) = 0 then Sn−2(v)= ±1. Hence w11 = s1. The proof for w11 = s2 is similar. 

Proposition 3.7. On (xy − vz)Sn−1(v) − (xy − 2z)Sn−2(v)=0 we have 2 2 −1 − s2 + s1s2z −1 − s1 + s1s2z w11 = 2 and w11 = 2 . s1 + s1s2 − s2z s2 + s1s2 − s1z 2 2 Proof. By Lemma 3.2 we have Sk(v)+ Sk−1(v) − vSk(v)Sk−1(v) = 1 for all integers k. Hence, on (xy − vz)Sn−1(v) − (xy − 2z)Sn−2(v) = 0 we have 2 2 −1 Sn−1(v)= 1 − rv + r where r =(xy − vz)/(xy − 2z). By a direct calculation we have −1 2 2 −1 w11 = (c12d21s1 + c11d21 + c11d11s1)+ s1r − (d21 + c11s1 + d11s1)r 1 − rv + r 2 2 s1s2(s1s2 + s1u − s2) −1 − s2 + s1s2z   = 2 = 2 . s1 − s1s2u − 1 s1 + s1s2 − s2z

The proof for w11 is similar. 

La ∗ −1 We now prove Theorem 1.1 for W2n−1. Let ρ(λa)= −1 . Since λa = wa , we 0 La −1   have La = w11s1 . On jπ C := {v − 2 cos =0}∩{s2 =(w )2 =1} j n 2 11   by Proposition 3.6 we have w11 = s1. Hence La − 1=0on Cj for all 1 ≤ j ≤ n − 1. By Proposition 3.7, on

2 2 C := {(xy − vz)Sn−1(v) − (xy − 2z)Sn−2(v)=0}∩{s2 =(w11) =1}   THEA-POLYNOMIALOFTWISTEDWHITEHEADLINKS 9

2 −1−s2+s1s2z 2 2 we have w11 = 2 . Note that on C, we have s = 1 implies (w11) = 1. s1+s1s2−s2z 2 −2+s1z 2(1+s1w11) 2 With s2 = 1 we then have w11 = 2s1−z . Hence z = s1+w11 . Note that v = 2+(z−x) . Let z − x = t. Then

2(1 + s1w11) −1 −1 w11 − s1 −1 La − 1 t = − (s1 + s1 )=(s1 − s1 ) =(s1 − s1 ) s1 + w11 w11 + s1 La +1 2 −1    and v =2+ t . Since y = s2 + s2 = 2 and vSn−1(v)= Sn(v)+ Sn−2(v) we have

0 = 2xSn−1(v) − zSn(v) − (2x − z)Sn−2(v)

= 2xSn−1(v) − (x + t)Sn(v) − (x − t)Sn−2(v) 2 = −xt Sn−1(v) − t Sn(v) − Sn−2(v) .

−2+s1x If t = 0 then w11 = 2s1−x = s1. In this case we have La = 1. We now consider the case xtSn−1(v)+ Sn(v) − Sn−2(v) = 0. Then, by Lemma 3.4 we have n n +1+ i n − 1+ i n−1 n + i 0 = − t2i + x t2i+1 2i +1 2i +1 2i +1 i=0 i=0 X     X   n n +1+ i n − 1+ i L − 1 2i = − (s − s−1)2i a 2i +1 2i +1 1 1 L +1 i=0 a X       n−1 n + i L − 1 2i+1 + (s + s−1)(s − s−1)2i+1 a . 2i +1 1 1 1 1 L +1 i=0 a X     The last expression in the above equalities is exactly the polynomial F (Ma, La), with Ma = s1, defined in Theorem 1.1. Hence, with s2 = 1, we have (La − 1)F (Ma, La)=0on C. Similarly, with s2 = −1, we obtain the same equation (La − 1)F (Ma, La)=0on C. This proves the formula of the A-polynomial for the component of W2n−1 corresponding to the meridian a. The one for the component corresponding b is exactly the same. 2 For W2n and its component corresponding to a, we have LaMa − 1=0on (2j − 1)π D := {v − 2 cos =0}∩{s2 =(w )2 =1} j 2n +1 2 11   for 1 ≤ j ≤ n. Moreover, on

2 2 D := {zSn(v) − (xy − z)Sn−1(v)=0}∩{s2 =(w11) =1}   we have G(Ma, La) = 0 where G is the polynomial defined in Theorem 1.1. The same formulas can be obtained for the component of W2n corresponding to b. This completes the proof of Theorem 1.1.

3.5. Proof of Theorem 1.2. We make use of Proposition 2.1 which implies that for can each j = 1, 2 the factor Aj (M, L) of the A-polynomial Aj(L, M) corresponding to a canonical component of Wk (with k ≥ 1) has an irreducible factor containing at least 3 monomials in M, L. This irreducible factor cannot be L − 1 or LM 2 − 1. Hence, from the proof of Theorem 1.1 we conclude that the canonical components of W2n−1 and W2n are respectively the zero sets of the polynomials (xy − vz)Sn−1(v) − (xy − 2z)Sn−2(v) and zSn(v) − (xy − z)Sn−1(v). This completes the proof of Theorem 1.2. 10 ANH T. TRAN

3.6. Proof of Theorem 1.3. Recall that EWk (α) is the cone-manifold of Wk with cone 2π angles α1 = α2 = α. There exists an angle αWk ∈ [ 3 , π) such that EWk (α) is hyperbolic for α ∈ (0,αWk ), Euclidean for α = αWk , and spherical for α ∈ (αWk , π). We first consider the case of W2n−1. By Theorem 1.2 the canonical component of W2n−1 is determined by (xy − vz)Sn−1(v) − (xy − 2z)Sn−2(v) = 0. Moreover, by Proposition 3.7, on this component we have 2 −1 − s2 + s1s2z w11 = 2 . s1 + s1s2 − s2z Here we use the same notations as in the previous subsections.

For α ∈ (0,αW2n−1 ), by the Schlafli formula we have

π π 2 −1 − s2 + s1s2z Vol E 2 −1 (α)= 2 log |w | dω = 2 log dω W n 11 s + s s2 − s z Zα Zα 1 1 2 2 2 iω/2 −1−s2+s1s2z where s1 = s2 = s = e and z (with 2 ≥ 1) satisfy s1+s1s2−s2z

(xy − vz)Sn−1(v ) − (xy − 2 z)Sn−2(v)=0. We refer the reader to [DMM] for the volume formula of hyperbolic cone-manifolds of links using the Schlafli formula. 2 −2 −1−s2+s1s2z z−(s +1) −1 Note that 2 = 2 ≥ 1 is equivalent to Im z ≥ 0. Since x = y = s+s s1+s1s2−s2z z−(s +1) 2 2 2 2 2 2 and v = x + y + z − xyz − 2=2x + z − x z − 2, the proof of Theorem 1.3 for W2n−1 is complete. The proof for W 2n is similar.

Acknowledgements The author has been partially supported by a grant from the Simons Foundation (#354595 to Anh Tran).

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Department of Mathematical Sciences, The University of Texas at Dallas, Richard- son, TX 75080, USA E-mail address: [email protected]