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PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 137, Number 10, October 2009, Pages 3533–3541 S 0002-9939(09)09906-7 Article electronically published on May 19, 2009

THE COMPLEX VOLUMES OF TWIST

JINSEOK CHO, JUN MURAKAMI, AND YOSHIYUKI YOKOTA

(Communicated by Daniel Ruberman)

Abstract. For a given hyperbolic , the third author defined a function whose imaginary part gives the of the . We show that, fora , the function actually gives the complex volume of the knot complement using Zickert’s and Neumann’s theory of the extended Bloch groups and the complex volumes.

1. Introduction In [1], Kashaev defined an invariant of a and conjectured the following: log ∣⟨𝐿⟩ ∣ vol(𝐿)=2𝜋 lim 𝑁 , 𝑁→∞ 𝑁 where 𝐿 is a , vol(𝐿) is the hyperbolic volume of the link comple- ment, ⟨𝐿⟩𝑁 is the Kashaev invariant of 𝐿,and𝑁 is the positive-integer variable. Afterwards, in [3], the complexified was proposed as follows and confirmed in some cases with numerical computations: log⟨𝐿⟩ 𝑖(vol(𝐿)+𝑖 cs(𝐿)) ≡ 2𝜋 lim 𝑁 (mod 𝜋2), 𝑁→∞ 𝑁 where cs(𝐿) is the Cheeger-Chern-Simons invariant of the hyperbolic link comple- ment. We call vol(𝐿)+𝑖 cs(𝐿) the complex volume of 𝐿. On the other hand, the third author developed a special ideal triangulation of a hyperbolic knot complement in [6] and showed that this triangulation has a close relation with the Kashaev invariant. He also defined a function 𝑉 from this triangulation and showed that, after evaluating 𝑉 with the geometric solution of the hyperbolicity equations, the imaginary part of 𝑉 becomes the volume of the knot complement. Therefore, it is natural to expect that, after evaluating, the real part of 𝑉 be- comes the Cheeger-Chern-Simons invariant of the knot complement modulo 𝜋2.In this article, we show such a coincidence for the twist knots using Zickert’s and Neu- mann’s theory of the extended Bloch groups and the complex volumes in [7] and [4].

Received by the editors December 29, 2008. 2000 Mathematics Subject Classification. Primary 57M27; Secondary 51M25, 58J28. Key words and phrases. Twist knot, volume conjecture, complex volume. This work was carried out while the first authorwas visiting Waseda University with the support of the Korea National Foundation Grant funded by the Korean Government (MOEHRD) (KRF-2007-612-C00037).

⃝c 2009 American Mathematical Society Reverts to public domain 28 years from publication 3533

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2. Preliminaries

For a positive integer 𝑛,let𝑇𝑛 be the twist knot with 𝑛 + 4 crossings as in Figure 1. We label some of the crossings with integers 0, 1,...,𝑛+ 1 as in Figure 1. Also consider a (1,1)-tangle diagram whose closure is 𝑇𝑛. Following the method in [6], we assign a complex variable 𝑧𝑘 to the edge of 𝑇𝑛 that connects the over-crossing point of the 𝑘th crossing and the under-crossing point of the (𝑘+1)th crossing. (See Figure 1. The arcs around the crossing points represent the surviving tetrahedra

after the collapsing process, which will be briefly mentioned after Figure 2.)

³ ° °   °                        0  ...... ......        𝑧0     ......   ...... .. 1 ...  ......       𝑧1       ......   ...... .. 2  ..  ......  ..  𝑧   .  2  .  . .

  𝑧𝑛−1  ......   ...... .. 𝑛 ...  ......       𝑧𝑛       ...... 𝑛 +1                           

   

² ± ²

Figure 1

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  B0 B𝑘                          D   C D   C 0   0 𝑘   𝑘                 E0   F0 E𝑘   F𝑘               A A0 𝑘  

Figure 2

Let 𝐼 = {1, 2,...,𝑛}. We put an octahedron A𝑘B𝑘C𝑘D𝑘E𝑘F𝑘 at the 𝑘th crossing of the tangle diagram for 𝑘 ∈ 𝐼 ∪{0,𝑛 +1} so that A𝑘B𝑘 is perpendicular to the plane; A𝑘 and B𝑘 touch the under-crossing point and the over-crossing point respectively; and C𝑘,D𝑘,E𝑘,F𝑘 are to be the northeast, northwest, southwest, and southeast respectively. (See Figure 2.) Note that the octahedron A𝑘B𝑘C𝑘D𝑘E𝑘F𝑘 splits into 4 tetrahedra: in our case, A𝑘B𝑘C𝑘D𝑘,A𝑘B𝑘D𝑘E𝑘,A𝑘B𝑘E𝑘F𝑘,and A𝑘B𝑘F𝑘C𝑘. We identify the edges A0D0 with A0F0,B0C0 with B0E0,A𝑘C𝑘 with A𝑘E𝑘,and B𝑘D𝑘 with B𝑘F𝑘 respectively for 𝑘 ∈ 𝐼 ∪{𝑛 +1}. To obtain the ideal triangu- lation of the knot complement, we collapse some of the tetrahedra and glue the surviving tetrahedra together following the knot diagram. (For details, see [2] or [6].) In our case, we can obtain the ideal triangulation of the knot complement by gluing the 3𝑛 + 2tetrahedra A 0B0E0F0,A𝑘B𝑘C𝑘D𝑘,A𝑘B𝑘D𝑘E𝑘,A𝑘B𝑘E𝑘F𝑘,and A𝑛+1B𝑛+1C𝑛+1D𝑛+1 for 𝑘 ∈ 𝐼. To apply the method in [6] to this triangulation, we parameterize each ideal tetra- hedra with complex numbers as follows. Assign 𝑧0 to the edge A0B0 of A0B0E0F0, 𝑧𝑘 1 𝑧𝑘−1 to the edge A𝑘B𝑘 of A𝑘B𝑘C𝑘D𝑘, to the edge A𝑘B𝑘 of A𝑘B𝑘D𝑘E𝑘, to 𝑧𝑘−1 𝑧𝑘 theedgeA𝑘B𝑘 of A𝑘B𝑘E𝑘F𝑘,and𝑧𝑛 to the edge A𝑛+1B𝑛+1 of A𝑛+1B𝑛+1C𝑛+1D𝑛+1 respectively for 𝑘 ∈ 𝐼. Then, from [5] or [6], the hyperbolicity equations of the knot complement become

𝑧0 1 (2.1) 1 − =(1− 𝑧0)(1 − ), 𝑧1 𝑧0 𝑧𝑘 1 𝑧𝑘−1 (1 − )(1 − )=(1− 𝑧𝑘)(1 − )for𝑘 =1, 2,...,𝑛− 1, 𝑧𝑘+1 𝑧𝑘 𝑧𝑘 1 𝑧𝑛−1 1 − =(1− 𝑧𝑛)(1 − ). 𝑧𝑛 𝑧𝑛

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Note that the solution of these equations which induces the complete hyperbolic structure of the knot complement exists uniquely by virtue of the Lemma 2.3 of [6]. In this article, the unique solution is called the geometric solution. Now let 𝑟𝑘 be the even integer satisfying the following equations for 𝑘 ∈ 𝐼 ∪{0}:

𝑧0 1 (2.2) 𝑟0𝜋𝑖 =log(1− ) − log(1 − 𝑧0) − log(1 − ), 𝑧1 𝑧0 𝑧𝑘 1 𝑧𝑘−1 𝑟𝑘𝜋𝑖 =log(1− ) + log(1 − ) − log(1 − 𝑧𝑘) − log(1 − ) 𝑧𝑘+1 𝑧𝑘 𝑧𝑘 for 𝑘 =1, 2,...,𝑛− 1, 1 𝑧𝑛−1 𝑟𝑛𝜋𝑖 =log(1− ) − log(1 − 𝑧𝑛) − log(1 − ). 𝑧𝑛 𝑧𝑛 Then we can determine the 𝑉 function, following [6], as ( ) ∑ 𝜋2 1 (2.3) 𝑉 (𝑧0,𝑧1,...,𝑧𝑛)=− 𝑟𝑘𝜋𝑖log 𝑧𝑘 + − Li2( ) 6 𝑧0 𝑘∈𝐼∪{0} {( ) ( ) ( )} ∑ 2 2 2 𝜋 𝜋 𝑧𝑘−1 1 𝜋 + Li2(𝑧𝑘−1) − + − Li2( ) + Li2( ) − 6 6 𝑧𝑘 𝑧𝑘 6 (𝑘∈𝐼 ) 𝜋2 + Li (𝑧 ) − . 2 𝑛 6 With these notations, we state the main theorem of this article.

Theorem 2.1. Let 𝑧0,𝑧1,...,𝑧𝑛 be the geometric solution of (2.1), which induces the complete hyperbolic structure of the 𝑇𝑛 complement. Then

2 𝑉 (𝑧0,𝑧1,...,𝑧𝑛) ≡ 𝑖(vol(𝑇𝑛)+𝑖 cs(𝑇𝑛)) (mod 𝜋 ). We will prove this theorem in Section 3. The main technique of the proof is Zickert’s article [7]. To apply it, we need the vertex ordering of each tetrahedron. We equip the vertex ordering according to the order of the vertices as follows: B0A0F0E0,A1B1D1C1,A𝑘B𝑘C𝑘D𝑘 for 𝑘 =2, 3,...,𝑛+1,A𝑘B𝑘E𝑘D𝑘 for 𝑘 ∈ 𝐼,andA𝑘B𝑘E𝑘F𝑘 for 𝑘 ∈ 𝐼. For example, the vertices A𝑘,B𝑘,E𝑘,D𝑘 of the tetrahedron A𝑘B𝑘E𝑘D𝑘 have the order 0, 1, 2, 3 respectively. Note that the vertex ordering of each tetrahedron induces orientations of edges and the tetrahedron. The induced orientation can be different from the original orientation induced by the knot complement. In our case, such a situation arises for A1B1D1C1 and A𝑘B𝑘E𝑘D𝑘 for 𝑘 ∈ 𝐼. To apply the main technique, we need the following lemma, which is induced directly from the gluing pattern of the triangulation.

Lemma 2.2. The ideal triangulation of the 𝑇𝑛 complement given as above is in concordance with the ordering of each edge which is induced by the vertex ordering defined above. In other words, if two edges are glued together by the triangulation, then the orientations of two edges induced by each vertex ordering coincide.

Now we will determine the element of the extended Bloch group ℬˆ(ℂ) corre- sponding to each tetrahedron using the method of [7]. At first, let 𝐽 = {1, 2, 3}

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(𝑗) and define a sign function 𝜎𝑘 as

{ −1if𝑗 =2or(𝑘, 𝑗)=(1, 1), (2.4) 𝜎(𝑗) = 𝑘 1otherwise,

for 𝑘 ∈ 𝐼 ∪{0,𝑛+1} and 𝑗 ∈ 𝐽 except (𝑘, 𝑗)=(0, 1), (0, 2), (𝑛 +1, 2), (𝑛 +1, 3). Also let

(3) (3) (3) (3) (1) (1) (1) (1) (1) (1) (1) (1) (2.5) 𝜎0 [𝑧0 ; 𝑝0 ,𝑞0 ],𝜎1 [𝑧1 ; 𝑝1 ,𝑞1 ],𝜎𝑘 [𝑧𝑘 ; 𝑝𝑘 ,𝑞𝑘 ]for𝑘 ∕=1, (2) (2) (2) (2) (3) (3) (3) (3) (1) (1) (1) (1) 𝜎𝑘 [𝑧𝑘 ; 𝑝𝑘 ,𝑞𝑘 ],𝜎𝑘 [𝑧𝑘 ; 𝑝𝑘 ,𝑞𝑘 ], and 𝜎𝑛+1[𝑧𝑛+1; 𝑝𝑛+1,𝑞𝑛+1]

ˆ be the elements of ℬ(ℂ) corresponding to the tetrahedra B0A0F0E0,A1B1D1C1, A𝑘B𝑘C𝑘D𝑘,A𝑘B𝑘E𝑘D𝑘,A𝑘B𝑘E𝑘F𝑘,andA𝑛+1B𝑛+1C𝑛+1D𝑛+1 for 𝑘 ∈ 𝐼 respec- tively. Then, by the definition above and the parameters of the ideal tetrahedra,

(3) (1) 1 (1) (2.6) 𝑧0 = 𝑧0,𝑧1 = ,𝑧𝑘 = 𝑧𝑘−1 for 𝑘 ∕=1, 𝑧0 (2) 𝑧𝑘−1 (3) 1 (1) 𝑧𝑘 = ,𝑧𝑘 = , and 𝑧𝑛+1 = 𝑧𝑛, 𝑧𝑘 𝑧𝑘

(𝑗) (𝑗) for 𝑘 ∈ 𝐼. To determine the integers 𝑝𝑘 and 𝑞𝑘 for 𝑘 ∈ 𝐼 ∪{0,𝑛+1} and 𝑗 ∈ 𝐽, we use the following definition from (3.5) of [7]. For a tetrahedron with a vertex ordering,

(2.7) 𝑝𝜋𝑖 +log𝑧 =log𝑐(𝑔03) + log 𝑐(𝑔12) − log 𝑐(𝑔02) − log 𝑐(𝑔13),

(2.8) 𝑞𝜋𝑖 − log(1 − 𝑧) = log 𝑐(𝑔02) + log 𝑐(𝑔13) − log 𝑐(𝑔01) − log 𝑐(𝑔23),

where [𝑧; 𝑝, 𝑞] is the element of ℬˆ(ℂ) corresponding to the ideal tetrahedron and 𝑐(𝑔𝑎𝑏) is the complex number assigned to the edge connecting the 𝑎th and 𝑏th vertex. Zickert gave an explicit method for calculating the complex numbers in [7], but, in this article, we do not need the exact numbers. What we need is only the fact that if two edges are identified by the gluing pattern of the ideal triangulation, then the numbers assigned to these identified edges are the same. Using this fact, the following lemma is directly deduced from the gluing pattern of the triangulation.

Lemma 2.3. Let 𝛼𝑘, 𝛽𝑘, 𝛾𝑘, 𝛾𝑘+1, 𝑚𝑘,and𝛿 be the complex numbers assigned to the edges B𝑘E𝑘, B𝑘F𝑘, C𝑘D𝑘, E𝑘F𝑘, A𝑘B𝑘,andD𝑘E𝑘 respectively for 𝑘 ∈ 𝐼. Also let 𝛼0 and 𝛽0 be the complex numbers assigned to the edges B0F0 and B0E0 respectively. Then we can determine all the assigned complex numbers of each edge using 𝛼𝑘, 𝛽𝑘, 𝛾𝑘, 𝑚𝑘, 𝛿, 𝛼0, 𝛽0,and𝛾𝑛+1 for 𝑘 ∈ 𝐼 as in Figure 3.

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B0 B𝑘 B𝑛+1      𝛽   𝛽 𝛿 = 𝛼 𝛿   𝑘  𝑘 𝑛−1      𝛽   𝛼 𝛼𝑘  𝛽 0 0  𝑘  D𝑛+1    D𝑘   C C 𝛾𝑘  𝑘 𝛾𝑛+1 𝑛+1       𝛽0  𝛿  𝑚𝑘  𝛼𝑛       𝛽𝑘−1 𝛼𝑘−1 𝛽𝑛 𝛼𝑛      E F E F  0  𝛿  0 𝑘 𝛾𝑘+1  𝑘              𝛾1 𝛾1 𝛼𝑘−1 𝛼𝑘−1      A A A0 𝑘 𝑛+1

Figure 3

The following lemma is a direct consequence of (2.5), Lemma 2.2, and [4] and [7].

Lemma 2.4. Let 𝑧0,𝑧1,...,𝑧𝑛 be the geometric solution of (2.1) and let (𝑗) (𝑗) (𝑗) (𝑗) ˆ 𝜎𝑘 [𝑧𝑘 ; 𝑝𝑘 ,𝑞𝑘 ] be the element of the extended Bloch group ℬ(ℂ) defined in (2.4), (2.5),and(2.6).Then ∑ (𝑗) ˆ (𝑗) (𝑗) (𝑗) 2 (2.9) 𝑖(vol(𝑇𝑛)+𝑖 cs(𝑇𝑛)) ≡ 𝜎𝑘 𝐿[𝑧𝑘 ; 𝑝𝑘 ,𝑞𝑘 ](mod𝜋 ), 𝑘∈𝐼∪{0,𝑛+1} 𝑗∈𝐽 ˆ 1 𝜋𝑖 𝜋2 where 𝐿[𝑧; 𝑝, 𝑞]=Li2(𝑧)+ 2 log 𝑧 log(1 − 𝑧)+ 2 (𝑞 log 𝑧 + 𝑝 log(1 − 𝑧)) − 6 is the extended dilogarithm function defined in ℬˆ(ℂ). To prove the main theorem, we prove the coincidence of the right side of (2.3) and the right side of (2.9) in the next section.

3. Proof of the Theorem 2.1 (𝑗) To simplify the calculation, we define𝑝 ˆ𝑘 by (𝑗) (𝑗) (𝑗) 𝑝ˆ𝑘 𝜋𝑖 = 𝑝𝑘 𝜋𝑖 +log𝑧𝑘 for 𝑘 ∈ 𝐼 ∪{0,𝑛+1} and 𝑗 ∈ 𝐽. Then, the right side of (2.9) becomes { } ∑ 𝜋2 𝜎(𝑗) Li (𝑧(𝑗)) − 𝑘 2 𝑘 6 𝑘∈𝐼∪{0,𝑛+1} 𝑗∈𝐽 𝜋𝑖 ∑ + 𝜎(𝑗){𝑞(𝑗) log 𝑧(𝑗) +ˆ𝑝(𝑗) log(1 − 𝑧(𝑗))} 2 𝑘 𝑘 𝑘 𝑘 𝑘 𝑘∈𝐼∪{0,𝑛+1} 𝑗∈𝐽 ∑ = 𝑉 (𝑧0,𝑧1,...,𝑧𝑛)+ 𝑟𝑘𝜋𝑖log 𝑧𝑘 𝑘∈𝐼∪{0} 𝜋𝑖 ∑ + 𝜎(𝑗){𝑞(𝑗) log 𝑧(𝑗) +ˆ𝑝(𝑗) log(1 − 𝑧(𝑗))}. 2 𝑘 𝑘 𝑘 𝑘 𝑘 𝑘∈𝐼∪{0,𝑛+1} 𝑗∈𝐽

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Therefore, it is enough to show that ∑ (𝑗) (𝑗) (𝑗) (𝑗) (𝑗) (3.1) 𝜎𝑘 𝜋𝑖{𝑞𝑘 log 𝑧𝑘 +ˆ𝑝𝑘 log(1 − 𝑧𝑘 )} 𝑘∈𝐼∪{0,𝑛+1} 𝑗∈𝐽 ⎧ ⎫ ⎨ ∑ ⎬ +2 𝑟 𝜋𝑖log 𝑧 ≡ 0(mod2𝜋2). ⎩ 𝑘 𝑘⎭ 𝑘∈𝐼∪{0}

Lemma 3.1.

(2) (3) (1) 𝑟0𝜋𝑖 =(𝑞1 − 𝑞0 − 𝑞1 )𝜋𝑖, (3) (2) (2) (1) 𝑟𝑘𝜋𝑖 =(𝑞𝑘 − 𝑞𝑘 + 𝑞𝑘+1 − 𝑞𝑘+1)𝜋𝑖 for 𝑘 =1, 2,...,𝑛− 1, (3) (1) (2) 𝑟𝑛𝜋𝑖 =(𝑞𝑛 − 𝑞𝑛+1 − 𝑞𝑛 )𝜋𝑖. Proof. From Lemma 2.3 and (2.8), we obtain

(1) 1 𝑞1 𝜋𝑖 =log(1− ) + log 𝛽0 +log𝛽1 − log 𝛾1 − log 𝑚1, 𝑧0 (1) 𝑞𝑘+1𝜋𝑖 =log(1− 𝑧𝑘) + log 𝛼𝑘 +log𝛽𝑘+1 − log 𝛾𝑘+1 − log 𝑚𝑘+1 for 𝑘 =1, 2,...,𝑛− 1, (1) 𝑞𝑛+1𝜋𝑖 =log(1− 𝑧𝑛) + log 𝛼𝑛−1 − log 𝛾𝑛+1, (2) 𝑧𝑘−1 𝑞𝑘 𝜋𝑖 =log(1− ) + log 𝛼𝑘−1 +log𝛽𝑘 − log 𝛿 − log 𝑚𝑘 for 𝑘 ∈ 𝐼, 𝑧𝑘 (3) 𝑞0 𝜋𝑖 =log(1− 𝑧0) + log 𝛼0 +log𝛾1 − log 𝛽0 − log 𝛿, (3) 1 𝑞𝑘 𝜋𝑖 =log(1− ) + log 𝛼𝑘−1 +log𝛽𝑘 − log 𝛾𝑘+1 − log 𝑚𝑘 for 𝑘 ∈ 𝐼. 𝑧𝑘

Applying these equations and 𝛿 = 𝛼𝑛−1 to (2.2), we obtain the result. □

Lemma 3.2. ∑ ∑ (𝑗) (𝑗) (𝑗) (𝑗) (𝑗) (𝑗) 𝜎𝑘 𝑝ˆ𝑘 𝜋𝑖log(1 − 𝑧𝑘 ) ≡ 𝜎𝑘 𝑞𝑘 𝜋𝑖log 𝑧𝑘 𝑘∈𝐼∪{0,𝑛+1} 𝑘∈𝐼∪{0,𝑛+1} 𝑗∈𝐽 𝑗∑∈𝐽 2 ≡− 𝑟𝑘𝜋𝑖log 𝑧𝑘 (mod 2𝜋 ). 𝑘∈𝐼∪{0}

Proof. From (2.7) and Lemma 2.3, we know that

(1) 𝑝ˆ1 𝜋𝑖 =log𝛼0 − log 𝛽0, (1) 𝑝ˆ𝑘+1𝜋𝑖 = − log 𝛼𝑘 +log𝛽𝑘, (2) 𝑝ˆ𝑘 𝜋𝑖 =log𝛼𝑘 − log 𝛽𝑘 − log 𝛼𝑘−1 +log𝛽𝑘−1, (3) 𝑝ˆ0 𝜋𝑖 = − log 𝛼0 +log𝛽0, (3) 𝑝ˆ𝑘 𝜋𝑖 =log𝛼𝑘 − log 𝛽𝑘,

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for 𝑘 ∈ 𝐼. Using these equations, (2.2), and Lemma 3.1, we can finish the proof as follows:

∑ (𝑗) (𝑗) (𝑗) 𝜎𝑘 𝑝ˆ𝑘 𝜋𝑖log(1 − 𝑧𝑘 ) 𝑘∈𝐼∪{0,𝑛+1} 𝑗∈𝐽 ∑ (1) 1 (1) = −𝑝ˆ1 𝜋𝑖log(1 − )+ 𝑝ˆ𝑘+1𝜋𝑖log(1 − 𝑧𝑘) 𝑧0 𝑘∈𝐼 ∑ ∑ (2) 𝑧𝑘−1 (3) (3) 1 − 𝑝ˆ𝑘 𝜋𝑖log(1 − )+ˆ𝑝0 𝜋𝑖log(1 − 𝑧0)+ 𝑝ˆ𝑘 𝜋𝑖log(1 − ) 𝑧𝑘 𝑧𝑘 𝑘∈𝐼 { 𝑘∈𝐼 } 1 𝑧0 =(− log 𝛼0 +log𝛽0) log(1 − ) − log(1 − ) + log(1 − 𝑧0) 𝑧0 𝑧1 𝑛∑−1 { 𝑧𝑘−1 + (log 𝛼𝑘 − log 𝛽𝑘) − log(1 − 𝑧𝑘) − log(1 − ) 𝑧𝑘 𝑘=1 } 𝑧 1 +log(1− 𝑘 ) + log(1 − ) 𝑧 𝑧 { 𝑘+1 𝑘 } 𝑧 1 +(log 𝛼 − log 𝛽 ) − log(1 − 𝑧 ) − log(1 − 𝑛−1 ) + log(1 − ) 𝑛 𝑛 𝑛 𝑧 𝑧 ∑ 𝑛 𝑛 (2) (3) (1) ≡− 𝑟𝑘𝜋𝑖log 𝑧𝑘 = −(𝑞1 − 𝑞0 − 𝑞1 )𝜋𝑖log 𝑧0 𝑘∈𝐼∪{0} 𝑛∑−1 (3) (2) (2) (1) (3) (1) (2) − (𝑞𝑘 − 𝑞𝑘 + 𝑞𝑘+1 − 𝑞𝑘+1)𝜋𝑖log 𝑧𝑘 − (𝑞𝑛 − 𝑞𝑛+1 − 𝑞𝑛 )𝜋𝑖log 𝑧𝑛 ∑𝑘=1 (𝑗) (𝑗) (𝑗) 2 ≡ 𝜎𝑘 𝑞𝑘 𝜋𝑖log 𝑧𝑘 (mod 2𝜋 ). 𝑘∈𝐼∪{0,𝑛+1} 𝑗∈𝐽

Applying Lemma 3.2to (3.1), the proof of the theorem is finished. □

References

1. R. Kashaev, The hyperbolic volume of knots from the quantum dilogarithm, Lett. Math. Phys. 39 (1997), no. 3, 269-275. MR1434238 (98b:57012) 2. H. Murakami, Kashaev’s invariant and the volume of a hyperbolic knot after Y. Yokota, Physics and combinatorics 1999 (Nagoya), 244-272, World Sci. Publ., River Edge, NJ, 2001. MR1865040 (2002k:57031) 3. H. Murakami, J. Murakami, M. Okamoto, T. Takata, and Y. Yokota, Kashaev’s conjecture and the Chern-Simons invariants of knots and links, Experiment. Math. 11 (2002), no. 3, 427-435. MR1959752 (2004a:57016) 4. W. Neumann, Extended Bloch group and the Cheeger-Chern-Simons class,Geom.Topol.8 (2004), 413-474. MR2033484 (2005e:57042) 5. Y. Yokota, From the to the A-polynomial of hyperbolic knots,Interdiscip. Inform. Sci. 9 (2003), no. 1, 11-21. MR2023102 (2004j:57014) 6. Y. Yokota, On the volume conjecture for hyperbolic knots, arXiv:math/0009165. 7. C. Zickert, The Chern-Simons invariant of a representation, arXiv:0710.2049.

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use THE COMPLEX VOLUMES OF TWIST KNOTS 3541

Department of Mathematical Sciences,Seoul National University,Seoul 151-742, Korea E-mail address: [email protected]

Department of Mathematics,Faculty of S cience and Engineering,Waseda Univer- sity,3-4-1 Okubo,Shinjuku-ku,Tokyo 169-8555,Japan E-mail address: [email protected]

Department of Mathematics,Tokyo Metropolitan University,Tokyo 192-0397,Japan E-mail address: [email protected]

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