On the twisted Alexander polynomial for representations into SL_2(C)
aa r X i v : . [ m a t h . G T ] S e p ON THE TWISTED ALEXANDER POLYNOMIAL FORREPRESENTATIONS INTO SL ( C ) ANH T. TRAN
Abstract.
We study the twisted Alexander polynomial ∆
K,ρ of a knot K associated toa non-abelian representation ρ of the knot group into SL ( C ). It is known for every knot K that if K is fibered, then for every non-abelian representation, ∆ K,ρ is monic and hasdegree 4 g ( K ) − g ( K ) is the genus of K . Kim and Morifuji recently proved theconverse for 2-bridge knots. In fact they proved a stronger result: if a 2-bridge knot K isnon-fibered, then all but finitely many non-abelian representations on some componenthave ∆ K,ρ non-monic and degree 4 g ( K ) −
2. In this paper, we consider two special familiesof non-fibered 2-bridge knots including twist knots. For these families, we calculate thenumber of non-abelian representations where ∆
K,ρ is monic and calculate the number ofnon-abelian representations where the degree of ∆
K,ρ is less than 4 g ( K ) − Main results
The twisted Alexander polynomial was introduced by Lin [Li] for knots in S and byWada [Wa] for finitely presentable groups. It is a generalization of the classical Alexanderpolynomial and gives a powerful tool in low dimensional topology. One of the mostimportant applications is the determination of fiberedness [FV2] and genus (the Thurstonnorm) [FV3] of knots by the collection of the twisted Alexander polynomials correspondingto all finite-dimensional representations. For literature on other applications and relatedtopics, we refer to the survey paper by Friedl and Vidussi [FV1].Following [KmM], we study the twisted Alexander polynomial ∆ K,ρ of a knot K associ-ated to a representation ρ of the knot group into SL ( C ). See Section 2 for a definiton of∆ K,ρ . In general, ∆
K,ρ is a rational function in one variable. However if the representation ρ is non-abelian, then ∆ K,ρ is a Laurent polynomial and has degree less than or equalto 4 g ( K ) − g ( K ) is the genus of K . The twisted Alexander polynomials of twoconjugate representations are the same, hence we can always consider representations upto conjugation. It is known for every knot K that if K is fibered, then for every non-abelian representation, ∆ K,ρ is monic [GKM] and has degree 4 g ( K ) − K is non-fibered, then all but finitely many non-abelianrepresentations on some component have ∆ K,ρ non-monic and degree 4 g ( K ) − b (6 n + 1 ,
3) and the twist knots. For these two families, we calculate the number ofnon-abelian representations where ∆
K,ρ is monic and calculate the number of non-abelianrepresentations where the degree of ∆
K,ρ is less than 4 g ( K ) − Mathematics Subject Classification . 57M27.
Key words and phrases. twisted Alexander polynomial, SL ( C )-representation, 2-bridge knot, twistknot, fiberedness, genus. Let b ( p, m ) be the 2-bridge knot associated to a pair of relatively prime odd integers p > m > K m ( m >
0) the m -twist knot (see Figures 1 and 2 inSection 4). Then we have the following. Theorem 1.
For K = b (6 n + 1 , , the number of (conjugacy classes of ) non-abelianrepresentations where the degree of ∆ K,ρ is less than g ( K ) − is equal to n , and thenumber of non-abelian representations where ∆ K,ρ is monic is also equal to n . Theorem 2. (i) For K = K n , the number of non-abelian representations where thedegree of ∆ K,ρ is less than g ( K ) − is equal to − n , and the number of non-abelianrepresentations where ∆ K,ρ is monic is equal to n − − a n − b n , where a n = ( , n ≡ , otherwise ,b n = ( , n ≡ , otherwise . (ii) For K = K n − , the number of non-abelian representations where the degree of ∆ K,ρ is less than g ( K ) − is equal to − n , and the number of non-abelian representationswhere ∆ K,ρ is monic is equal to n − − c n − d n − e n , where c n = ( , n ≡ , otherwise ,d n = ( , n ≡ , otherwise ,e n = ( , n ≡ , otherwise . The paper is organized as follows. We review the definition of the twisted Alexanderpolynomial and some related work on fibering and genus of knots in Section 2. We thenprove Theorem 1 in Section 3 and Theorem 2 in Section 4.We would like the referee for comments and suggestions that greatly improve the ex-position of the paper. We would also like to thank T. Morifuji for comments.2.
Twisted Alexander polynomials
Let K be a knot in S and G K = π ( S \ K ) its knot group. We choose and fix aWirtinger presentation G K = h a , . . . , a ℓ | r , . . . , r ℓ − i . Then the abelianization homomorphism f : G K → H ( S \ K, Z ) ∼ = Z = h t i is given by f ( a ) = · · · = f ( a ℓ ) = t . Here we specify a generator t of H ( S \ K ; Z ) and denote thesum in Z multiplicatively. Let us consider a linear representation ρ : G K → SL ( C ).These maps naturally induce two ring homomorphisms e ρ : Z [ G K ] → M (2 , C ) and e f : Z [ G K ] → Z [ t ± ], where Z [ G K ] is the group ring of G K and M (2 , C ) is the matrix algebraof degree 2 over C . Then e ρ ⊗ e f defines a ring homomorphism Z [ G K ] → M (2 , C [ t ± ]).Let F ℓ denote the free group on generators a , . . . , a ℓ and Φ : Z [ F ℓ ] → M (2 , C [ t ± ]) the WISTED ALEXANDER POLYNOMIALS AND SL ( C )-REPRESENTATIONS 3 composition of the surjection Z [ F q ] → Z [ G K ] induced by the presentation of G K and themap e ρ ⊗ e f : Z [ G K ] → M (2 , C [ t ± ]).Let us consider the ( ℓ − × ℓ matrix M whose ( i, j )-component is the 2 × (cid:18) ∂r i ∂a j (cid:19) ∈ M (cid:0) , Z [ t ± ] (cid:1) , where ∂/∂a denotes the free differential calculus. For 1 ≤ j ≤ ℓ , let us denote by M j the( ℓ − × ( ℓ −
1) matrix obtained from M by removing the j th column. We regard M j asa 2( ℓ − × ℓ −
1) matrix with coefficients in C [ t ± ]. Then Wada’s twisted Alexanderpolynomial of a knot K associated to a representation ρ : G K → SL ( C ) is defined to bea rational function ∆ K,ρ ( t ) = det M j det Φ(1 − a j )and moreover well-defined up to a factor t k ( k ∈ Z ), see [Wa]. It is known that if tworepresentations ρ, ρ ′ are conjugate, then ∆ K,ρ = ∆
K,ρ ′ holds. Hence, in this paper we canalways consider representations up to conjugation.A representation ρ : G K → SL ( C ) is called non-abelian if ρ ( G K ) is a non-abeliansubgroup of SL ( C ). Suppose ρ is a non-abelian representation. It is known that thetwisted Alexander polynomial ∆ K,ρ ( t ) is always a Laurent polynomial for any knot K [KtM], and is a monic polynomial if K is a fibered knot. It is also known that theconverse holds for 2-bridge knots [KmM]. Moreover, if K is a knot of genus g ( K ) thendeg(∆ K,ρ ( t )) ≤ g ( K ) − K is fibered [KtM].We say that the twisted Alexander polynomial ∆ K,ρ ( t ) determines the fiberedness of K if it is a monic polynomial if and only if K is fibered. We also say ∆ K,ρ ( t ) determinesthe knot genus g ( K ) if deg(∆ K,ρ ( t )) = 4 g ( K ) − Proof of Theorem 1
The following result is well-known.
Lemma 3.1.
The knot b ( p, is non-fibered if and only if p ≡ .Proof. The proof is based on the calculation of the Alexander polynomial. Since b ( p,
3) isan alternating knot, it is fibered if and only if its Alexander polynomial is monic, i.e. hasleading coefficient 1, see [Cr, Mu]. The leading coefficient of the Alexander polynomial of b ( p,
3) is 2 if p ≡ p ≡ − (cid:3) The standard presentation of the knot group of K = b ( p, m ) is G K = h a, b | wa = bw i where w = a ε b ε . . . a ε p − b ε p − and ε j = ( − ⌊ jq/p ⌋ , see e.g. [BZ]. Here a and b are 2standard generators of a 2-bridge knot group.For a representation ρ : G K → SL ( C ), let x = tr ρ ( a ) = tr ρ ( b ) and z = tr ρ ( ab ). Let d = p − . By [Le], the representation ρ is non-abelian if and only if R w ( x, z ) = 0 where R w ( x, z ) = tr ρ ( w ) − tr ρ ( w ′ ) + · · · + ( − d − tr ρ ( w ( d − ) + ( − d . Here if u is a word in the letters a, b , then u ′ denotes the word obtained from u by deletingthe two letters at the two ends. Then w ( d − denotes the element obtained from w byapplying the deleting operation d − ANH T. TRAN
Let { S j ( z ) } j be the sequence of Chebyshev polynomials defined by S ( z ) = 1 , S ( z ) = z ,and S j +1 ( z ) = zS j ( z ) − S j − ( z ) for all integers j . By [NT, Proposition 3.1], we have thefollowing description of non-abelian representations of b (6 n + 1 , Lemma 3.2.
For the 2-bridge knot b (6 n + 1 , , one has w = ( ab ) n ( a − b − ) n ( ab ) n and R w ( x, z ) = S n ( z ) − S n − ( z ) − x ( z − S n − ( z )( S n ( z ) − S n − ( z )) . Let r = waw − b − . Then ∂r∂a = w (cid:0) − a ) w − ∂w∂a (cid:1) where ∂w∂a = (cid:0) ab ) n ( a − b − ) n (cid:1) (cid:0) · · · + ( ab ) n − (cid:1) − ( ab ) n (cid:0) · · · + ( a − b − ) n − (cid:1) a − . Suppose ρ : G K → SL ( C ) is a non-abelian representation. Then the twisted Alexanderpolynomial of K associated to ρ is∆ K,ρ ( t ) = det Φ (cid:18) ∂r∂a (cid:19) (cid:14) det Φ(1 − b ) = det Φ (cid:18) ∂r∂a (cid:19) (cid:14) (1 − tx + t ) . We have det Φ (cid:18) ∂w∂a (cid:19) = t n | I + ( I − tA ) t − n ( AB ) − n ( BA ) n ( AB ) − n (cid:2) (cid:0) I + ( AB ) n ( A − B − ) n (cid:1) (cid:0) I + · · · + ( t AB ) n − (cid:1) − ( t AB ) n (cid:0) I + · · · + ( t − A − B − ) n − (cid:1) t − A − (cid:3) | where A = ρ ( a ) and B = ρ ( b ). Here | M | denotes the determinant of M .We will need the following lemma whose proof is elementary. Lemma 3.3.
Let L ( t ) = P sj = r M j t j be a Laurent polynomial in t whose coefficients are × complex matrices. Suppose | M r | and | M s | are non-zero. Then the highest and lowestdegree terms of | L ( t ) | (in t ) are | M s | t s and | M r | t r respectively. The following lemma follows easily from Lemma 3.3.
Lemma 3.4.
For the two-bridge knot K = b (6 n +1 , , the highest and lowest degree termsof det Φ (cid:0) ∂w∂a (cid:1) are | I + A ( AB ) − n ( BA ) n A − | t n and | I + ( AB ) n ( BA ) − n | t respectively,provided that | I + A ( AB ) − n ( BA ) n A − | and | I + ( AB ) n ( BA ) − n | are non-zero. The following two lemmas are standard, see e.g. [MT].
Lemma 3.5.
One has S j ( z ) − zS j ( z ) S j − ( z ) + S j − ( z ) = 1 . Lemma 3.6.
Suppose the sequence { M j } j ∈ Z of × matrices satisfies the recurrencerelation M j +1 = zM j − M j − for all integers j . Then M j = S i − ( z ) M − S i − ( z ) M . Lemma 3.7.
One has | I + A ( AB ) − n ( BA ) n A − | = | I + ( AB ) n ( BA ) − n | = 4 + ( z − z + 2 − x ) S n − ( z ) . WISTED ALEXANDER POLYNOMIALS AND SL ( C )-REPRESENTATIONS 5 Proof.
It is easy to see that | I + A ( AB ) − n ( BA ) n A − | = | I +( AB ) n ( BA ) − n | . By applyingLemma 3.6 twice, once with M j = ( AB ) j and once with M j = ( A − B − ) j , we havetr( AB ) n ( BA ) − n = tr( AB ) n ( A − B − ) n = S n − ( z ) tr ABA − B − + S n − ( z ) tr I − S n − ( z ) S n − ( z )(tr AB + tr A − B − ) . From the identity tr CD = tr C tr D − tr CD − for all matrices C, D in SL ( C ), it iseasy to see that tr ABA − B − = z − zx + 2 x − z − z + 2 − x ). Hence, byLemma 3.5, we obtaintr( AB ) n ( BA ) − n = (2 + ( z − z + 2 − x )) S n − ( z ) + 2 S n − ( z ) − zS n − ( z ) S n − ( z )= 2 + ( z − z + 2 − x ) S n − ( z ) . The lemma follows, since det( I + C ) = 1 + det C + tr C . (cid:3) Genus.
We first note that the genus of K = b (6 n + 1 ,
3) is n . Lemmas 3.2, 3.4 and3.7 imply that the number of (conjugacy classes of) non-abelian representations wherethe degree of ∆ K,ρ is less than 4 g ( K ) − x, z ) ∈ C of the following system 4 + ( z − z + 2 − x ) S n − ( z ) = 0 , (3.1) S n ( z ) − S n − ( z ) − x ( z − S n − ( z )( S n ( z ) − S n − ( z )) = 0 . (3.2)Suppose eq. (3.1) holds. Then x ( z − S n − ( z ) = 4 + ( z − S n − ( z ). Eq. (3.2) isthen equivalent to the following(3.3) S n ( z ) − S n − ( z ) − (4 + ( z − S n − ( z ))( S n ( z ) − S n − ( z )) = 0 . Claim 3.8.
The equation (3.3) has n distinct roots, all of which are different from ± .Proof. Let h ( z ) denote the left hand side of eq. (3.3). Since S j (2) = j + 1 and S j ( −
2) =( − j ( j + 1) for all integers j , we have h (2) = − h ( −
2) = ( − n +1 (2 n + 3).Suppose z = ±
2. We may write z = α + α − where α = ±
1. Since S j ( z ) = α j +1 − α − ( j +1) α − α − for all integers j , we have h ( z ) = α − n (2 α +1) α n + α +2 α +1 . It follows that h ( z ) is a polynomialin z of degree n . We want to show that h ( z ) does not have multiple roots. This holdstrue if we can show that the polynomial h ( α ) = 2 α n +1 + α n + α + 2 does not havemultiple roots α such that | α | ≥
1. (Note that if α is a root of h , then so is α − ).Suppose h has a multiple root α ∈ C such that | α | ≥
1. Then h ( α ) = h ′ ( α ) = 0, orequivalently (2 α + 1) α n + α + 2 = 2((2 n + 1) α + n ) α n − + 1 = 0 . It follows that α +22 α +1 = α n +1) α + n ) , i.e. α + ( + n ) α + 1 = 0. Hence α + α − = − ( + n ) < − . Then we must have α ∈ R and α < −
2, since | α | ≥
1. Hence h ( α ) = (2 α + 1) α n + α + 2 <
0, a contradiction. The claim follows. (cid:3)
For each solution z of eq. (3.3), we write z = α + α − for some α ∈ C ∗ . Then fromthe proof of Claim (3.8), we have α = ± α − n (2 α +1) α n + α +2 α +1 = 0. It follows that α n = − α +22 α +1 = 1. Eq. (3.1) is then equivalent to the following x = 4 + ( z − S n − ( z )( z − S n − ( z ) = ( α + 1) ( α n + 1) α ( α n − = ( α − α , ANH T. TRAN i.e. 9 x = z −
2. Hence Claim 3.8 implies that the system (3.1)+(3.2) has exactly 2 n complex solutions ( x, z ). This means that the number of non-abelian representationswhere the degree of ∆ K,ρ is less than 4 g ( K ) − n .3.0.2. Fiberedness.
The number of non-abelian representations where ∆
K,ρ is monic isequal to the number of solutions ( x, z ) ∈ C of the following system4 + ( z − z + 2 − x ) S n − ( z ) = 1 , (3.4) S n ( z ) − S n − ( z ) − x ( z − S n − ( z )( S n ( z ) − S n − ( z )) = 0 . (3.5)Suppose eq. (3.4) holds. Then x ( z − S n − ( z ) = 3 + ( z − S n − ( z ), and eq. (3.5) isequivalent to the following(3.6) S n ( z ) − S n − ( z ) − (3 + ( z − S n − ( z ))( S n ( z ) − S n − ( z )) = 0 . Claim 3.9.
The equation (3.6) has n distinct roots, all of which are different from ± .Proof. Let h ( z ) denote the left hand side of eq. (3.6). It is easy to see that h (2) = − h ( −
2) = 2( − n +1 . Suppose z = ±
2. We may write z = α + α − where α = ± h ( z ) = α − n ( α n + 1), and the claim follows easily. (cid:3) For each solution z of eq. (3.6), we write z = α + α − for some α ∈ C ∗ . Then from theproof of Claim (3.9), we have α = ± α n = −
1. Eq. (3.4) is then equivalent to thefollowing x = 3 + ( z − S n − ( z )( z − S n − ( z ) = ( α + 1) ( α n + α n + 1) α ( α n − = ( α + 1) α , i.e. 4 x = z + 2. Hence Claim 3.9 implies that the system (3.4)+(3.5) has exactly 2 n complex solutions ( x, z ). This means that the number of non-abelian representationswhere ∆ K,ρ is monic is equal to 2 n .This completes the proof of Theorem 1. Remark 3.10.
One can easily show that all the solutions ( x, z ) of the systems (3.1)+(3.2)and (3.4)+(3.5) satisfying the condition x = 2. It follows that the twisted Alexanderpolynomial associated to any parabolic SL ( C )-representation of the knot group of the2-bridge knot b (6 n + 1 ,
3) determines the fiberedness and genus of b (6 n + 1 , b (6 n + 1 , Proof of Theorem 2
We first note that the m -twist knot K m , m >
0, is non-fibered if and only if m > K is the trefoil knot, and K is the figure-8 knot.)4.1. K n , n > . From [Tr] (and also [NT]), the knot group of K = K n is h a, b | wa = bw i where a, b are meridians depicted in Figure 1 and w = ( ba − ) n b ( ab − ) n . Note that this isnot the standard presentation of a 2-bridge knot.For a representation ρ : G K → SL ( C ), let x = tr ρ ( a ) = tr ρ ( b ) and y = tr ρ ( ab − ).From [Tr], we have the following Lemma 4.1.
For the twist knot K = K n , a representation ρ : G K → SL ( C ) is non-abelian if and only if ( x, y ) ∈ C is a root of the polynomial R even ( x, y ) = ( y + 1) S n − ( y ) − S n ( y ) − S n − ( y ) S n ( y ) + x S n − ( y )( S n ( y ) − S n − ( y )) . WISTED ALEXANDER POLYNOMIALS AND SL ( C )-REPRESENTATIONS 7 a b2n crossings ... Figure 1. K n , n > r = waw − b − . Then ∂r∂a = w (cid:0) − a ) w − ∂w∂a (cid:1) where ∂w∂a = − (1 + · · · + ( ba − ) n − ) ba − + ( ba − ) n b (1 + · · · + ( ab − ) n − ) . Suppose ρ : G K → SL ( C ) is a non-abelian representation. Then the twisted Alexanderpolynomial of K associated to ρ is∆ K,ρ ( t ) = det Φ (cid:18) ∂r∂a (cid:19) (cid:14) det Φ(1 − b ) = det Φ (cid:18) ∂r∂a (cid:19) (cid:14) (1 − tx + t ) . We havedet Φ (cid:18) ∂w∂a (cid:19) = t | I + ( I − tA ) t − ( BA − ) n B − ( AB − ) n (cid:2) − (1 + · · · + ( BA − ) n − ) BA − + ( BA − ) n tB (1 + · · · + ( AB − ) n − ) (cid:3) | . where A = ρ ( a ) and B = ρ ( b ).Let { T j ( z ) } j be the sequence of Chebyshev polynomials defined by T ( z ) = 2 , T ( z ) = z ,and T j +1 ( z ) = zT j ( z ) − T j − ( z ) for all integers j .The following lemma follows easily from Lemma 3.3. Lemma 4.2.
For the twist knot K = K n , the highest and lowest degree terms of det Φ (cid:0) ∂w∂a (cid:1) are | · · · + ( AB − ) n − | t = T n ( y ) − y − t and | − (1 + · · · + ( BA − ) n − ) | t = T n ( y ) − y − t respectively, provided that T n ( y ) − y − is non-zero. Genus.
We first note that the genus of K n is 1. Lemmas 4.1 and 4.2 imply thatthe number of non-abelian representations where the degree of ∆ K,ρ is less than 4 g ( K ) − x, y ) ∈ C of the following system( T n ( y ) − / ( y −
2) = 0 , (4.1) ( y + 1) S n − ( y ) − S n ( y ) − S n − ( y ) S n ( y ) + x S n − ( y )( S n ( y ) − S n − ( y )) = 0 . (4.2)Suppose eq. (4.1) holds. Then y = 2 and T n ( y ) = 2. We write y = β + β − where β = 1.Then T n ( y ) = β n + β − n = 2 is equivalent to β n = 1, or equivalent β = e i πkn for some1 ≤ k ≤ n . It follows that eq. (4.1) has ⌊ n/ ⌋ distinct solutions given by y = 2 cos kπn where 1 ≤ k ≤ n .If y = 2 cos kπn for some 1 ≤ k < n then y = β + β − where β = e i kπn . It follows that S n − ( y ) = 0 and S n ( y ) = 1. Hence R even , the left hand side of eq. (4.2), is equal to − ANH T. TRAN If y = − n must be even), it is easy to see that R even = − (2 n + 1)( nx +2 n + 1), since S j ( −
2) = ( − j ( j + 1) for all integers j . Then eq. (4.2) is equivalent to x = − (2 + n ).Hence the system (4.1)+(4.2) has exactly 1 + ( − n solutions.4.1.2. Fiberedness.
The number of non-abelian representations where ∆
K,ρ is monic isequal to the number of solutions ( x, y ) ∈ C of the following system( T n ( y ) − / ( y −
2) = 1 , (4.3) ( y + 1) S n − ( y ) − S n ( y ) − S n − ( y ) S n ( y ) + x S n − ( y )( S n ( y ) − S n − ( y )) = 0 . (4.4)Suppose eq. (4.3) holds. Then y = 2 and T n ( y ) = y . We write y = β + β − where β = 1. Then T n ( y ) = y is equivalent to β n +1 = 1 or β n − = 1. It follows that thedistinct solutions of eq. (4.3) are y = 2 cos kπn +1 where 1 ≤ k < n +12 , y = 2 cos kπn − where1 ≤ k < n − , and (if n is odd) y = − y = − n must be odd), it is easy to see that eq. (4.4) is equivalent to x = − (2 + n ).Suppose y = 2 cos kπn +1 where 1 ≤ k < n +12 . Then y = β + β − where β = e i kπn +1 . Itfollows that S n − ( y ) = − S n ( y ) = 0. Hence R even = − ( x + 1), and eq. (4.4) isequivalent to x = − y = 2 cos kπn − where 1 ≤ k < n − . Then y = β + β − where β = e i kπn − . Itfollows that S n − ( y ) = 1 and S n ( y ) = y . Hence R even = − ( y + y −
1) + x ( y − y = 1, i.e. k = n − , then R even = −
1. If y = 1 then eq. (4.4) is equivalent to x = y + y − y − .If y = − √ (i.e. k = n − ) or y = − √ (i.e. k = n − ) then y + y − x = 0.Hence the system (4.3)+(4.4) has exactly 2 n − − a n − b n solutions, where a n , b n areas defined in Theorem 2(i).4.2. K n − , n > . From [Tr] (and also [NT]), the knot group of K = K n − is h a, b | wa = bw i where a, b are meridians depicted in Figure 2 and w = ( ab − ) n b ( ba − ) n . Notethat this is not the standard presentation of a 2-bridge knot. b a ... Figure 2. K n − , n > ρ : G K → SL ( C ), let x = tr ρ ( a ) = tr ρ ( b ) and y = tr ρ ( ab − ).From [Tr], we have the following WISTED ALEXANDER POLYNOMIALS AND SL ( C )-REPRESENTATIONS 9 Lemma 4.3.
For the twist knot K = K n − , a representation ρ : G K → SL ( C ) isnon-abelian if and only if ( x, y ) ∈ C is a root of the polynomial R odd ( x, y ) = − ( y + 1) S n − ( y ) + S n − ( y ) + 2 S n − ( y ) S n − ( y ) + x S n − ( y )( S n − ( y ) − S n − ( y )) . Let r = waw − b − . Then ∂r∂a = w (cid:0) − a ) w − ∂w∂a (cid:1) where ∂w∂a = (1 + · · · + ( ab − ) n − ) − ( ab − ) n b (1 + · · · + ( ba − ) n − ) ba − . Suppose ρ : G K → SL ( C ) is a non-abelian representation. Then the twisted Alexanderpolynomial of K associated to ρ is∆ K,ρ ( t ) = det Φ (cid:18) ∂r∂a (cid:19) (cid:14) det Φ(1 − b ) = det Φ (cid:18) ∂r∂a (cid:19) (cid:14) (1 − tx + t ) . We havedet Φ (cid:18) ∂w∂a (cid:19) = t | I + ( I − tA ) t − ( AB − ) n B − ( BA − ) n (cid:2) (1 + · · · + ( AB − ) n − ) − ( AB − ) n tB (1 + · · · + ( BA − ) n − ) BA − (cid:3) | . where A = ρ ( a ) and B = ρ ( b ). The following lemma follows easily from Lemma 3.3. Lemma 4.4.
For the twist knot K = K n − , the highest and lowest degree terms of det Φ (cid:0) ∂w∂a (cid:1) are | · · · + ( BA − ) n − | t = T n ( y ) − y − t and | − (1 + · · · + ( AB − ) n − ) | t = T n ( y ) − y − t respectively, provided that T n ( y ) − y − is non-zero. Applying Lemmas 4.3 and 4.4, the proof of Theorem 2(ii) is similar to that of Theorem2(i). This completes the proof of Theorem 2.
Remark 4.5.
From the proof of Theorem 2, one can easily see that the twisted Alexanderpolynomial associated to any parabolic SL ( C )-representation of the knot group of the m -twist knot K m , m >
0, determines the fiberedness and genus of K m . This gives anotherproof of [MT, Theorem 1.1] and hence of a conjecture of Dunfield, Friedl and Jackson[DFJ] for K m . References [BZ] G. Burde and H. Zieschang,
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