Weights of Markov traces for Alexander polynomials of mixed links
aa r X i v : . [ m a t h . R T ] D ec Weights of Markov traces forAlexander polynomials of mixed links
Hitoshi Yamanaka
Abstract
Using the Fourier expansion of Markov traces for Ariki-Koike alge-bras over Q ( q, u , . . . , u e ), we give a direct definition of the Alexanderpolynomials for mixed links. We observe that under the correspondingspecialization of a Markov parameter, the Fourier coefficients of Markovtraces take quite simple form.As a consequence, we show that the Alexander polynomial of a mixedlink is essentially equal to the Alexander polynomial of the link obtainedby resolving the twisted parts. In [L1] Lambropoulou initiated the mixed link theory which is the link the-ory in the solid torus. A mixed link is an embedding of a disjoint union offinitely many circles into the solid torus Y . Since the 3-sphere has the canonicalgenus 1 Heegaard decomposition, it can be considered as an embedding into thecomplement S \ Int Y .We say that two mixed links are equivalent if they are joined by an ambientisotopy of S \ Int Y .Lambropoulou showed that there is a variant of the usual braid theory, thatis, every mixed link is equivalent to the closure of a mixed braid and one canconsider analogues of the Markov Moves. More precisely, a mixed braid with n -strands is an n -tuple ( p , . . . , p n ) of embeddings of the closed interval [0 , I \ Cyn ), where 1 yn := (cid:26) ( x, y, z ) ∈ I (cid:12)(cid:12)(cid:12)(cid:12)(cid:18) x − (cid:19) + (cid:18) y − n + 1) (cid:19) ≤ (cid:18) n + 1) (cid:19) (cid:27) ,such that • the curves p i ( I ) and p j ( I ) do not intersect if i = j , • there exists a permutation σ of { , . . . , n } such that p i (0) = (cid:18) , in + 1 , (cid:19) , p i (1) = (cid:18) , σ ( i ) n + 1 , (cid:19) for all i ∈ { , . . . , n } , • the point p i ( t ) lies in the interior of cl( I \ Cyn ) if t ∈ (0 , • the 3rd coordinate of p i ( t ) is increasing with respect to t for all i ∈{ , . . . , n } .Two mixed braids with n -strands are said to be equivalent if they are joined byan ambient isotopy of cl( I \ Cyn ) which fixes the boundary.Then the set of equivalence classes is in bijection with the affine braid group B aff ,n with generators t , t , · · · , t n − satisfying fundamental relations t t t t = t t t t ,t i t j = t j t i ( | i − j | ≥ ,t i t i +1 t i = t i +1 t i t i +1 (1 ≤ i ≤ n − B aff ,n is the semi-direct product of the braid group B n = h t , . . . , t n − i and the free subgroup P n generated by t ′ , t ′ . . . , t ′ n − where t ′ i = t i · · · t t t − · · · t − i ( i = 0 , , . . . , n − B n is embedded in B aff ,n as a subgroup.Moreover, in this setting we have the following analogues of the Markovmoves:(1) α ←→ βαβ − ( α, β ∈ B aff ,n ), 22) α ←→ αt ± n ( α ∈ B aff ,n ).One of the main interest in this paper is to construct an analogue of theAlexander polynomial explicitly.In the usual link theory, Jones [J] discovered a way to construct the HOM-FLYPT polynomial using the Markov traces of Iwahori-Hecke algebras of type A . He gave two methods to derive the Alexander polynomials. The first one isto use the Skein relation for HOMFLYPT polynomials and the second is to usethe “Fourier expansion” of the Markov traces of Iwahori-Hecke algebras of type A , namely, the expression as the linear combination of irreducible characters.Note that the second method is more direct than the first one.As for the mixed links, it is possible to give an analogue of the Alexanderpolynomial using the analogue of HOMFLYPT polynomials and their Skeinrelations given in [L2]. We point out that one can also define the Alexanderpolynomial of a mixed link directly following Jones’s second argument. Forthis purpose a result of Geck-Iancu-Malle [GIM] is quite helpful. In [GIM],they determined the rational polynomials appearing in the coefficients in theexpression of the Markov traces as the linear combination of the irreduciblecharacters of the Ariki-Koike algebra of type G ( e, , n ).Then our second observation is that when we consider the specialization ofparameter, the Fourier coefficients take quite simple form. As a consequence, weshow that the Alexander polynomial of a mixed link is essentially the same asthe Alexander polynomial of the link obtained by resolving the twisting parts.This paper is organized as follows. In Section 2, we recall the definition ofthe Ariki-Koike algebras and its irreducible ordinary representations. In Section3, we recall the definition of the Markov traces for Ariki-Koike algebras andits Fourier expansion due to Geck-Iancu-Malle. In Section 4, we collect somefundamental definitions and results in Lambropoulou’s mixed link theory. InSection 5, we propose a definition of Alexander polynomial for a mixed link.In Section 6, we calculate the Fourier coefficient of Markov traces and prove arelation between the Alexander polynomials for mixed links and the one for theusual links. Acknowledgment.
The work was supported by Grant-in-Aid for JPSP Fellows23 · In this section we recall the definition of the Ariki-Koike algebra and its ordi-nary finite dimensional irreducible representations. For the details we refer theoriginal paper [AK].Let e, n be two positive integers. We denote by k the field of rational func-tions over Q with ( e + 1)-indeterminates q, u , · · · , u e .3 efinition 2.1. The
Ariki-Koike algebra of type G ( e, , n ) is the associative k -algebra H e,n with generators T , T , · · · , T n − satisfying fundamental relations( T − u )( T − u ) · · · ( T − u e ) = 0 , ( T i − q )( T i + 1) = 0 (1 ≤ i ≤ n − ,T T T T = T T T T ,T i T j = T j T i ( | i − j | ≥ ,T i T i +1 T i = T i +1 T i T i +1 (1 ≤ i ≤ n − Remark 2.2.
When e = 1 and u = 1, the corresponding Ariki-Koike algebraof type G (1 , , n ) is just the Iwahori-Hecke algebra of type A n − . Similarly,when e = 2 and u = −
1, we have the Iwahori-Hecke algebra of type B n withunequal parameters.For i ∈ { , , · · · n − } we put L i = T i · · · T T T − · · · T − i . We also definean element T w ( w ∈ S n ) as follows: let w = s i · · · s i ℓ be a reduced expressionof w where s i is the permutation which transposes i and i + 1. Then we put T w = T i · · · T i ℓ . By [AK, Theorem 3.10] the set { L e L e · · · L e n − n − T w | ≤ e i ≤ e − ≤ i ≤ n − , w ∈ S n } forms a k -basis of H e,n .Note that for each positive integer e we have the following inductive system: H e, ⊂ H e, ⊂ · · · ⊂ H e,n ⊂ · · · .We denote by H e the inductive limit of the inductive system, i.e., the union ∞ [ n =1 H e,n .Next, let us recall the definition of multi-Young tableaux and related notions. Definition 2.3. An e -Young diagram of total size n is an e -tuple λ =( λ , · · · , λ e ) consisting of sequences λ i = ( λ i, , · · · , λ i,p ( i ) ) of positive integerswhich satisfies the following two conditions:(1) X ≤ i ≤ n ≤ j ≤ p ( i ) λ i,j = n ,(2) λ i, ≥ λ i, ≥ · · · ≥ λ i,p ( i ) for all i ∈ { , · · · , e } . Remark 2.4.
In the definition of the multi-Young diagram, we allow the casethat λ i = ∅ for some i .Obviuously we can regard an e -Young diagram as an e -tuple of Young dia-grams. Example 2.5.
The 4-Young diagram λ = ((4 , , , ∅ , (3 , , , ∅ ) of total size13 is identified with the following 4-tuple of Young diagrams:4 , ∅ , , ∅ Let a, b be boxes in an e -Young diagram λ . Then the content c ( a ; λ ) of a is the difference(the column number of a ) − (the row number of a )and the axial distance r ( a, b ) from a to b is the difference c ( b ; λ ) − c ( a ; λ ). Definition 2.6.
Let λ be an e -Young diagram. A standard e -tableau T of shape λ is a pair of an e -Young diagram and an ordering on the boxes by { , , · · · , n } which satisfies the following condition: in each Young diagram thewritten numbers are increasing from left to right and from top to bottom. Example 2.7.
Let λ be the 4-Young diagram presented in Example 2.5. Thenin the following two figures the left one is a standard 4-tableau of shape λ butthe right one is not. , ∅ , , ∅ , , ∅ , , ∅ For an e -Young diagram λ , we denote by Std( λ ) the set of standard e -tableaux of shape λ . For an integer k and an indeterminate y we define aLaurent polynomial ∆( k, y ) and a matrix M ( k, y ) as follows:∆( k, y ) = 1 − q k y , M ( k, y ) = 1∆( k, y ) (cid:20) q − k + 1 , y ) q ∆( k − , y ) − q k y ( q − (cid:21) . Finally, for a standard e -tableau T we define the number τ ( i ) so that i is writtenin the τ ( i )-th Young diagram of T . For an e -Young diagram λ of total size n we denote by V ( λ ) the finite dimensional vector space V ( λ ) = M T ∈ Std( λ ) k t where t is the symbol corresponding to a standard e -tableau T .Following [AK] we define a representation of H e,n on V ( λ ) as follows:(1) T t = u τ (1) t ,(2) For i ∈ { , · · · , n − } we define T i t as follows:(2-1) We define T i t = q t , if i and i + 1 are placed as i i + 152-2) We define T i t = − t , if i and i + 1 are placed as ii + 1(2-3) In the other case we define T i t by the following: T i h t , t ′ i = h t , t ′ i M r ( i + 1 , i ) , u τ ( i ) u τ ( i +1) ! .Here t ′ is the symbol corresponding to the standard e -Young tableau T ′ obtained by transposing i and i + 1 in T . Remark 2.8.
To unify the above definitions (1),(2-1),(2-2) and (2-3) we intro-duce the following notation. For the permutation s i = ( i, i + 1) and a standard e -tableau T we define a vector s i t in V ( λ ) by s i t = (cid:26) t ′ (if T ′ is standard)0 (else) . Here, T ′ is the e -Young tableau obtained by transposing i and i + 1 in T .Under this notation we have T i t = ( q − u τ ( i ) u τ ( i ) − q r ( i +1 ,i ) u τ ( i +1) t + q ( u τ ( i ) − q r ( i +1 ,i ) − u τ ( i +1) ) u τ ( i +1) − q r ( i +1 ,i ) u τ ( i ) ( s i t ) . for all i ∈ { , . . . , n − } and a standard e -tableau T .The set { V ( λ ) | λ : e -Young diagram of total size n } gives a complete list of finite dimensional irreducible representations over k upto equivalence.For an e -Young diagram of total size n we denote by χ λ the irreduciblecharacter corresponding to λ . In this section we recall a result of Lambropoulou [L2] and of Geck-Iancu-Malle[GIM] concerning the Markov traces of the Ariki-Koike algebras.Let z, y , · · · , y e − be elements of k . A k -linear map τ : H e −→ k is calledthe Markov trace associated to z, y , · · · , y e − if it satisfies the following prop-erties:(1) τ (1) = 1 , τ ( hh ′ ) = τ ( h ′ h ) ( h, h ′ ∈ H e,n ) , (3) τ ( hT i ) = zτ ( h ) ( h ∈ H e,n , ≤ i ≤ n ) , (4) τ ( hL ji ) = y j τ ( h ) ( h ∈ H e,n , ≤ i ≤ n − , ≤ j ≤ e − . By [L2] for fixed e, z, y , · · · , y e − the Markov trace of H e exists uniquely.Since the Markov trace τ satisfies (2), the restriction τ | H e,n is written as the k -linear combination of irreducible characters. Geck-Iancu-Malle [GIM] deter-mined the coefficients in this expression.To state their result we prepare some notations. Let λ = ( λ , · · · , λ e ) bean e -Young diagram of total size n . By adding some zeros we regard λ and λ p ( p ∈ { , · · · , e } ) as the sequences λ = ( λ , , · · · , λ ,n , λ ,n +1 ) ,λ p = ( λ p, , · · · , λ p,n )of length n + 1 and n respectively. We define finite sets A , · · · , A e as follows: A = { α ,i := λ ,i + n − i + 1 | ≤ i ≤ n + 1 } , A p = { α p,i := λ p,i + n − i | ≤ i ≤ n } (2 ≤ p ≤ n ) . Finally, for a non-negative integer d we denote by σ d the d -th fundamental sym-metric polynomial with respect to u , · · · , u e . Here, we understand σ = 1.Under these notations, we define b D λ ( q, u , · · · , u e ) and R λ ( z, y , · · · , y e − ) asfollows: b D λ ( q, u , · · · , u e )= ( − e )( n ) + n ( e − Y ≤ k ≤ l ≤ e Y ( α,α ′ ) ∈ A k × A l α>α ′ (if k = l ) ( q α u k − q α ′ u l ) e Y k =1 u nk q f ( n,e ) e Y k,l =1 Y α ∈ A k α Y h =1 ( q h u k − u l ) Y ≤ k 1) + 12 (cid:19) + (cid:18) e ( n − 2) + 12 (cid:19) + · · · + (cid:18) e + 12 (cid:19) .Finally, we put D λ ( q, u , u , · · · , u e ) = b D λ ( q, q − u , u , · · · , u e ).7 emma 3.1. ([GIM]) We have τ | H e,n = X λ ( − en e Y k =1 u | λ k |− nk ! D λ ( q, u , u , · · · , u e ) R λ ( z, y , · · · , y e − ) χ λ where λ runs through all e -Young diagrams of total size n and | λ k | stands forthe number of boxes in λ k . Remark 3.2. We see easily by induction on n that f ( n, e ) = 112 en ( n − en − e + 3).In the rest of this paper, we set C ( e, k, x ) = e − Y i =1 ( q ( e − i ) c ( x ) u e − i − k i X j =1 ( − i − j y j σ i − j ) + ( − e − Y ≤ l ≤ el = k u l . In this section we collect some fundamental notions and facts from Lambropoulou’smixed link theory. For the details we refer [L1] and [L2]. Definition 4.1. (1) A mixed link is an embedding of a disjoint union offinitely many circles into the solid torus.(2) Two mixed links are said to be equivalent if these are joined each otherby an ambient isotropy of the solid torus.By considering the canonical genus 1 Heegaard decomposition, we regardthe solid torus as the complement of the interior of a solid torus: Definition 4.2. (1) A a mixed braid with n -strands is an n -tuple ( p , . . . , p n )of embeddings of the closed interval [0 , 1] into the closure cl( I \ Cyn ), where Cyn := (cid:26) ( x, y, z ) ∈ I (cid:12)(cid:12)(cid:12)(cid:12)(cid:18) x − (cid:19) + (cid:18) y − n + 1) (cid:19) ≤ (cid:18) n + 1) (cid:19) (cid:27) ,such that 8 the curves p i ( I ) and p j ( I ) do not intersect if i = j , • there exists a permutation σ of { , . . . , n } such that p i (0) = (cid:18) , in + 1 , (cid:19) , p i (1) = (cid:18) , σ ( i ) n + 1 , (cid:19) for all i ∈ { , . . . , n } , • the point p i ( t ) lies in the interior of cl( I \ Cyn ) if t ∈ (0 , • the 3rd coordinate of p i ( t ) is increasing with respect to t for all i ∈{ , . . . , n } .(2) Two mixed braids with n -strands are said to be equivalent if they arejoined by an ambient isotopy of cl( I \ Cyn ) which fixes the boundary.As in the usual braid theory the set of equivalence classes of mixed linkswith n -strands is equipped with the natural group structure.Let B aff ,n be the n -th affine braid group, namely, the group with generators t , t , · · · , t n − satisfying fundamental relations t t t t = t t t t ,t i t j = t j t i ( | i − j | ≥ ,t i t i +1 t i = t i +1 t i t i +1 (1 ≤ i ≤ n − Lemma 4.3. The group of equivelence classes of mixed braids with n -strandsis isomorphic to B aff ,n as groups. Here, t corresponds toand t i (1 ≤ i ≤ n − 1) corresponds to i i + 19s in the usual braid theory, we can consider the closure of a mixed braid. Lemma 4.4. Every mixed link is equivalent to the closure of a mixed braid.For a mixed braid α we denote by b α the closure of α . The analogue of theMarkov Moves and Alexander’s theorem are given by the following lemma. Lemma 4.5. The closures of two mixed braids are equivalent as mixed linksif and only if the corresponding mixed braids are joined by a sequence of thefollowing two transformations:(1) α ←→ βαβ − ( α, β ∈ B aff ,n )(2) α ←→ αt ± n ( α ∈ B aff ,n ).Let us recall the construction of the analogue of the HOMFLYPT polynomialin mixed link theory. We first introduce new variable t as t = 1 − q + zqz .Let π n : B aff ,n −→ H × e,n be the group homomorphism defined by π ( t i ) = T i (0 ≤ i ≤ n − Definition 4.6. Let b α be a mixed link obtained as the closure of a mixed braid α with n -strands. Then we define the HOMFLYPT polynomial of type G ( e, of b α by X b α ( q, t ) = " − − tqt (1 − q ) n − ( t ) wr ( b α ) τ ( π n ( α )).Here wr ( b α ) stands for the writhe number of the mixed link b α . Lemma 4.7. HOMFLYPT polynomials of type G ( e, 1) satisfy the followingSkein relations: ( qt ) − X L + − ( qt ) X L − = ( q − q − ) X L , X M e = a e − X M e − + · · · + a X M + a X M .10ere, a i is defined by( T − u ) · · · ( T − u e ) = T e − a e − T e − − · · · − a T − a and L + , L − , L , M e , . . . , M , M are given by the following local mixed link di-agrams: L + L − L e e − M e M e − M M In this section we define the Alexander polynomials for mixed links.If we want to define the Alexander polynomial for a mixed link, we mustconsider the specialization t → q − . However, a priori, the specialization doesnot make sense since the term 1 − qt appears in X ( q, t ). To solve this problemwe can use the Skein relations for X ( q, t ). As explained in the previous sectionthe link polynomials X ( q, t ) satisfy the Skein relation. Conversely by givingsome initial conditions, we can define X ( q, t ) by using the Skein relations. Inparticular, we can define the Alexander polynomial for a mixed link. However,this definition is indirect and is not explicit.In fact, thanks to the result of Geck-Iancu-Malle explained in Section 3, wecan define the Alexander polynomial for a mixed link explicitly.To see this let us focus on the term R λ .After the change of variables z = − − q − tq , R λ can be written as b g λ / (1 − tq ) n where b g λ is given by e Y k =1 Y x ∈ λ k − (1 − q )(1 − q c ( x ) ) Y ≤ l ≤ el = k ( q c ( x ) u k − u l ) + (1 − tq )(1 − q ) C ( e, k, x ) ! .Since at least one component of λ , say λ k , is not empty, when x is the (1,1)-component of λ k , the corresponding term is given by (1 − tq )(1 − q ) C ( e, k, x ).This allows us to evaluate at t = q − . 11 efinition 5.1. Let b α be a mixed link obtained as the closure of a mixed braid α . We define Alexander polynomial of type G ( e, for b α by∆ G ( e, ( b α ) = X b α ( q, q − ).The following is a direct consequence of Lemma 4.7. Corollary 5.2. Alexander polynomials of type G ( e, 1) satisfy the followingSkein relations: ∆ G ( e, L + − ∆ G ( e, L − = ( q − q − )∆ G ( e, L ,∆ G ( e, M e = a e − ∆ G ( e, M e − + · · · + a ∆ G ( e, M + a ∆ G ( e, M .Here, the local mixed link diagrams L + , L − , L , M e , M e − , . . . , M , M are as inLemma 4.7.Finally we discuss simplification of the Alexander polynomials of type G ( e, λ has at least two non-empty components. Then the sameconsideration shows that after the specialization t = q − , the corresponding R λ is zero. Similarly, if λ has a component which has at least two diagonalboxes, the corresponding R λ is also zero. This shows that when we considerthe Alexander polynomial of type G ( e, e -Youngdiagrams which have the form ( ∅ , · · · , ∅ , λ ( a ) , ∅ , · · · , ∅ ), where λ ( a ) is the Youngdiagram ( a + 1 , , · · · , 1) of size n .Summarizing the above argument we have the following lemma. Lemma 5.3. Let α be a mixed braid. Then we have∆ G ( e, ( b α )= ( q − − ( n − q − n − wr ( b α ) − X ≤ p ≤ e ≤ a ≤ n − ( − en (cid:16) Y ≤ i ≤ ei = p u − ni (cid:17) D λ ( a ) p g λ ( a ) p χ λ ( a ) p .Here λ ( a ) p = ( ∅ , · · · , ∅ , p ˘ λ ( a ) , ∅ , · · · , ∅ ) ,g λ ( a ) p = ( − n − (1 − q ) C ( e, p ) Y x ∈ λ ( a ) x =(1 , (1 − q )(1 − q c ( x ) ) Y ≤ l ≤ el = p ( q c ( x ) u k − u l ) ! ,C ( e, p ) = e − Y i =1 ( u e − i − k i X j =1 ( − i − j y j σ i − j ) + ( − e − Y ≤ l ≤ el = k u l . Quantum calculus In this section we calculate D λ ( a ) p g λ ( a ) p .We first consider the case of p ∈ { , · · · , e } . In this case α i,j are given by α ,i = n − i + 1 (1 ≤ i ≤ n + 1) , α l,i = n − i ( l = 1 , p, ≤ i ≤ n ), α p,i = a + n ( i = 1) n − i + 1 (2 ≤ j ≤ b + 1) n − i ( b + 2 ≤ n ) . Here b = n − a − Lemma 6.1. For l = 1 , p and i, j ≥ 2, we have the following.(1) j − Y j ′ =1 ( q α ,j ′ − q α ,j ) = q ( j − n − j +1) j − Y h =1 ( q h − . (2) j − Y j ′ =1 ( q α l,j ′ − q α l,j ) = q ( j − n − j ) j − Y h =1 ( q h − . (3) i − Y i ′ =1 ( q α p,i ′ − q α p,i )= q ( i − n − i ) q a + i − q i − b − − i − Y h =1 ( q h − 1) ( b + 3 ≤ i ≤ n ) ,q ( i − n − i ) q a + i − q − i − Y h =1 ( q h − 1) ( i = b + 2) ,q ( i − n − i )+ i − q a + i − − q i − − i − Y h =1 ( q h − 1) (2 ≤ i ≤ b + 1) . Proof. We only prove the 1st identity in (3).(LHS) = i − Y i ′ =1 ( q α p,i ′ − q n − i )= q ( i − n − i ) i − Y i ′ =1 ( q α p,i ′ − ( n − i ) − q ( i − n − i ) ( q a + i − b +1 Y i ′ =2 ( q ( n − i ′ +1) − ( n − i ) − i − Y i ′ = b +2 ( q ( n − i ′ ) − ( n − i ) − q ( i − n − i ) ( q a + i − b +1 Y i ′ =2 ( q i − i ′ +1 − i − Y i ′ = b +2 ( q i − i ′ − q ( i − n − i ) q a + i − q i − b − − i − Y h =1 ( q h − . D λ ( a ) p into the following three factors:(1) ( − e )( n ) + n ( e − q − n Y ≤ l ≤ e u nl q f ( n,e ) Y ≤ l ≤ e ( q − u − u l ) n Y ≤ k We have D λ ( a ) p g λ ( a ) p = ( − a + n C ( e, p ) q n ( n − n − (1 − q ) n − q n . Proof. By Lemma 6.1 (3) we have Y α,α ′ ∈ A p ( q α − q α ′ )= n Y i =2 i Y i ′ =1 ( q α p,i ′ − q α p,i )= q P ni =2 ( n − i )( i − P b +1 i =2 ( i − n Y i =2 i − Y h =1 ( q h − b +1 Y i =2 q a + i − − q i − − × q n +1 − q − × n Y i = b +3 q a + i − q i − b − − . On the other hand, since b Y j =1 (1 − q − j ) = q − P bj =1 j b Y j =1 ( q j − a Y i =1 (1 − q i ) = ( − a a Y i =1 ( q i − b +1 Y i =2 q a + i − − q i − − b Y j =1 (1 − q − j ) = q − P bj =1 j n − Y i = a +1 ( q i − n Y i = b +3 q a + i − q i − b − − a Y i =1 (1 − q i ) = ( − a ( q − a + n Y i = n +2 ( q i − D λ ( a ) p g λ ( a ) p = ( − a + n − C ( e, p ) q P ni =2 ( n − i )( i − (1 − q ) n q n − n Y i =2 i − Y h =1 ( q h − a + n Y i = a +1 ( q i − Y α ∈ A p α Y h =1 ( q h − 1) .Since Y α ∈ A p α Y h =1 ( q h − 1) = n Y i =1 i − Y h =1 ( q h − a + n Y i = a +1 ( q i − n − b − Y h =1 ( q h − n Y i =2 i − Y h =1 ( q h − a + n Y i = a +1 ( q i − , we obtain the desired formula. Lemma 6.3. We have D λ ( a ) p g λ ( a ) p = ( − np u p − u n Y j =1 1 Y h = j − n ( q h u p − u ) Y ≤ l ≤ el = p u p − u l n Y i =1 0 Y h = i − n ( q h u p − u l ) . Proof. We first focus on the factors of D λ ( a ) p which have the form q ∗ u − q ∗ u p .It is given by 16 ( α,α ′ ) ∈ A × A p ( q α − u − q α ′ u p ) Y α ∈ A p α Y h =1 ( q h u p − q − u ) .Now the numerator is equal to n +1 Y i =1 n Y j =1 ( q − i − u − q α p,j u p )= q n P n +1 i =1 ( i − n +1 Y i =1 n Y j =1 ( q α p,j u p − u )= q n ( n +1)( n − n +1 Y i =1 n Y j =1 ( q ( j − − ( i − u p − u ) n +1 Y i =1 ( q ( a + n ) − ( i − u p − u ) n +1 Y i =1 ( q ( n − b − − ( i − u p − u )= q n ( n +1)( n − n +1 Y i =1 n Y j =1 ( q j − i +1 u p − u ) n +1 Y i =1 ( q a + n − i +2 u p − u ) n +1 Y i =1 ( q n − i − b +1 u p − u )and the denominator is equal to Y α ∈ A p α Y h =1 ( q h u p − q − u ) = n Y j =1 j − Y h =1 ( q h u p − q − u ) a + n Y h =1 ( q h u p − q − u ) n − b − Y i =1 ( q h u p − q − u )= q − n ( n +1)2 n Y j =1 j − Y h =1 ( q h +1 u p − u ) a + n Y h =1 ( q h +1 u p − u ) n − b − Y i =1 ( q h +1 u p − u ) . Thus the factor is given by q n ( n +1)( n − n +1 Y i =1 n Y j =1 ( q j − i +1 u p − u ) n +1 Y i =1 ( q a + n − i +2 u p − u ) n − b − Y i =1 ( q h +1 u p − u ) n +1 Y i =1 ( q n − i − b +1 u p − u ) n Y j =1 j − Y h =1 ( q h +1 u p − u ) a + n Y h =1 ( q h +1 u p − u ) .17ow the following three identities hold: n +1 Y i =1 n Y j =1 ( q j − i +1 u p − u ) j − Y h =1 ( q h +1 u p − u ) = n Y j =1 1 Y h = j − n ( q h u p − u ) , n +1 Y i =1 ( q a + n − i +2 u p − u ) a + n Y h =1 ( q h +1 u p − u ) = ( qu p − u ) " a Y i =1 ( q i u p − u ) − , n − b − Y i =1 ( q h +1 u p − u ) n +1 Y i =1 ( q n − i − b +1 u p − u ) = 1( u p − u )( qu p − u ) " b Y j =1 ( q − j u p − u ) − .Thus by combining with the corresponding factors in g λ ( a ) p , we have q n ( n +1)( n − u p − u n Y j =1 1 Y h = j − n ( q h u p − u ) . Next we focus on the factors in D λ ( a ) p which have the form q ∗ u p − q ∗ u l ( l =1 , p ). It is given by Y ≤ k We have D λ ( a ) p g λ ( a ) p = ( − a + en C ( e, p ) (1 − q ) n − q n Y ≤ l ≤ el = p u nl Y ≤ l ≤ el = p u p − u l . Proof. The proof is divided into five parts. We use Lemma 6.2 and Lemma 6.3.(1) The factors which have the form q ∗ − q ∗ . It is given by(1 − q ) n Y α,α ′ ∈ A α>α ′ ( q α − q α ′ ) Y ≤ l ≤ el = p Y α,α ′ ∈ A l α>α ′ ( q α − q α ′ )(1 − q n ) Y α ∈ A α Y h =1 ( q h − Y ≤ l ≤ el = p Y α ∈ A l α Y h =1 ( q h − . By Lemma 6.1 (1) and (2) we have the following two identities: Y α,α ′ ∈ A α>α ′ ( q α − q α ′ ) = q n ( n +1)( n − Y α ∈ A α Y h =1 ( q h − Y α,α ′ ∈ A l α>α ′ ( q α − q α ′ ) = q n ( n − n − Y α ∈ A l α Y h =1 ( q h − q n ( n − en − n − e +5) (1 − q ) n − q n . (2) The factors which have the form q ∗ u k − q ∗ u l . We divide into four cases.(2-1) The factors which have the form q ∗ u p − q ∗ u .This is given by 1 u p − u n Y i =1 1 Y h = i − n ( q h u p − u )( q − u − u p ) n Y α ∈ A α Y h =1 ( q h − u − u p ) . Now the denominator can be calculated as follows:( q − u − u p ) n n +1 Y i =1 n − i +1 Y h =1 ( q h − u − u p )= n Y i =1 n − i +1 Y h =0 ( q h − u − u p ) 19 ( − P ni =1 ( n − i +2) q P ni =1 ( ( n − i )( n − i +1)2 − n Y i =1 n − i +1 Y h =0 ( q − h u p − u )= ( − n ( n +3)2 q n − n n Y i =1 1 Y h = i − n ( q h u p − u ) . Thus this part is given by( − n ( n − q − n − n u p − u . (2-2) The factors which have the form q ∗ u p − q ∗ u l ( l ≥ , l = p ).This is given by Y ≤ l ≤ el = p u p − u l n Y i =1 0 Y h = i − n ( q h u p − u l ) Y p Lemma 6.5. For i, j, l ≥ j − Y j ′ =1 ( q α l,j ′ − q α l,j ) = q ( j − n − j ) j − Y h =1 ( q h − i − Y i ′ =1 ( q α ,i ′ − q α ,i )= q ( i − n − i +1) q a + i − q i − b − − i − Y h =1 ( q h − 1) ( b + 3 ≤ i ≤ n + 1) q ( i − n − i +1) q a + i − q − i − Y h =1 ( q h − 1) ( i = b + 2) q ( i − n − i +1)+ i − q a + i − − q i − − i − Y h =1 ( q h − 1) (2 ≤ i ≤ b + 1) . Lemma 6.6. We have D λ ( a ) p g λ ( a ) p = ( − a + n C ( e, q n ( n +1)( n − (1 − q ) n − q n . Lemma 6.7. We have D λ ( a )1 g λ ( a )1 = q n ( n +1)( n − e − Y ≤ l ≤ e u − u l n Y i =1 − Y h = i − n − ( q h u − u l ) ! . Lemma 6.8. We have D λ ( a )1 g λ ( a )1 = ( − a + en C ( e, 1) (1 − q ) n − q n Y ≤ l ≤ e u − u l . The following is a direct consequence of Lemma 5.3, 6.4 and 6.8. Theorem 6.9. For any mixed braid α with n -strands we have∆ G ( e, ( b α )= ( − n − q − n − wr ( b α ) − − q − q n X ≤ a ≤ n − ≤ p ≤ e ( − a C ( e, p ) Y ≤ l ≤ el = p u p − u l ! χ λ ( a ) p ( π n ( α )).23rom this formula we find the relationship between the Alexander polynomialof a mixed link and of the link which is obtained by resolving the twisted parts. Theorem 6.10. Let α be a mixed braid, α be the braid obtained by avoidingthe powers of t appearing in α , and ∆( α ) be the usual Alexander polynomialof α . Then we have∆ G ( e, ( b α ) = ∆( c α ) X ≤ p ≤ e C ( e, p ) u wr ( α ) p Y ≤ l ≤ el = p u p − u l ! . Here, wr ( α ) is the sum of exponents of t appearing in α . Proof. We first remark that by taking e = 1 we can recover the following formula∆( c α ) = ( − n − q − n − wr ( b α ) − − q − q n X ≤ a ≤ n − ( − a χ λ (1) p ( π n ( α ))of Jones. Since χ λ ( a ) p ( α ) = u wr ( α ) p χ λ (1) p ( α ), we have∆ G ( e, ( b α )= ( − n − q − n − wr ( b α ) − − q − q n × X ≤ a ≤ n − ≤ p ≤ e ( − a C ( e, p ) Y ≤ l ≤ el = p u p − u l ! χ λ ( a ) p ( π n ( α ))= ( − n − q − n − wr ( b α ) − − q − q n × X ≤ p ≤ e C ( e, p ) u wr ( α ) p Y ≤ l ≤ el = p u p − u l ! X ≤ a ≤ n − ( − a χ λ (1) p ( π n ( α ))= ∆( α ) X ≤ p ≤ e C ( e, p ) u wr ( α ) p Y ≤ l ≤ el = p u p − u l ! . as desired. 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