TThe
SL(2 , C ) -character variety of aMontesinos knot Haimiao Chen ∗ Beijing Technology and Business University, Beijing, China
Abstract
For each Montesinos knot K , we find a simple method to determine theSL(2 , C )-character variety, and show that it can be decomposed as X ( K ) (cid:116)X ( K ) (cid:116) X ( K ) (cid:116) X (cid:48) ( K ), where X ( K ) consists of trace-free characters, X ( K ) consists of characters of “connected sums” of representations ofthe constituent rational links, X ( K ) is a high-genus algebraic curve, and X (cid:48) ( K ) generically consists of finitely many points. Keywords:
SL(2 , C )-character variety; irreducible representation; Mon-tesinos knot MSC2020:
Throughout this section, let G = SL(2 , C ). Let Γ be a finitely presented group.A G -representation of Γ is a homomorphism ρ : Γ → G . The G -representationvariety of Γ is hom(Γ , G ). Call a representation ρ reducible if elements in Im( ρ )have a common eigenvector, which is equivalent to Im( ρ ) ⊂ a U a − for some a ∈ G , where U denotes the subgroup consisting of upper-triangular matrices;in particular, call ρ abelian if Im( ρ ) is abelian. Call ρ irreducible if ρ is notreducible.The character of ρ is the function χ ρ : Γ → C , g (cid:55)→ tr( ρ ( g )) . As a fact, twoirreducible representations ρ, ρ (cid:48) are conjugate (i.e. there exists a ∈ G such that ρ (cid:48) ( x ) = a ρ ( x ) a − for all x ∈ Γ) if and only if χ ρ = χ ρ (cid:48) . Focusing on irreduciblerepresentations, the (irreducible) G -character variety of Γ is X irr G (Γ) = { χ ρ : ρ ∈ hom(Γ , G ) is irreducible } . When Γ = π ( L ) := π ( S − L ) for a link L , we call X irr G (Γ) the G -charactervariety of L , and so on.There are many reasons for caring about representations. Here are two ofthem. First, in general, having a representation ρ : π ( L ) → GL( d, F ), one candefine the twisted Alexander polynomial of L associated to ρ , which is usefulin many situations (referred to [6] and the references therein). Second, for ∗ Email: [email protected] a r X i v : . [ m a t h . G T ] F e b nots K , a systematic approach to the understanding of the left orderability of π ( K ( p/q )) was proposed by Culler-Dunfield, by using continuous paths ofSL(2 , R )-representations of K .Nowadays a well-known fact is that, when Γ = π ( M ) with M a 3-manifold,the character variety encodes much geometric/topologic information about M .See [14, 15] and the references therein for more details.We focus on computing character varieties of links. The importance forcomputation lies in that knowledge on character variety is still too limited, andconcrete computations may give directions for further researches. Till now,character varieties have been determined for only a few links. Existing re-sults include: Torus knots [10], double twist knots [9], double twist links [12],( − , m + 1 , n )-pretzel links and twisted Whitehead links [16]. For classicalpretzel knots, the author [3, 4] found their character varieties. the irreduciblecharacter variety of each classical pretzel knot can be embedded in C , consist-ing of a finite set of points, some conics, and a high-genus algebraic curve; thedefining equations can be explicitly written down. The result of [3] was recentlyapplied by Khan and Tran [8] to show left orderability for some Dehn surgerieson odd pretzel knots. As a related work, in [11], SL(2 , C )-character varietiesof Montesinos knots of Kinoshita-Terasaka type were studied (without explicitcomputations), to reveal some phenomenons; for instance, for each d in somerange, there is a d -dimensional component of X ( K ).The main result of this paper is Theorem 1.1.
For each Montesinos knot K , the irreducible character variety X irrSL(2 , C ) ( K ) = X ( K ) (cid:116) X ( K ) (cid:116) X ( K ) (cid:116) X (cid:48) ( K ) , where • X ( K ) consists of trace-free characters; • X ( K ) consists of characters of “connected sums” of representations ofrational links N ([ p i /q i ]) ; • X ( K ) is a high-genus algebraic curve; • X (cid:48) ( K ) generically is a finite set. The achievement is at least two-fold. This is the first time to systematicallydeal with a class of knots whose fundamental groups can have arbitrarily manygenerators. The description of X irrSL(2 , C ) ( K ) provides necessary conditions fordeciding whether a given knot is Montesinos, which in general is a difficultproblem [7].The content is organized as follows. Section 2 contains a preliminary onbasic notions. In Section 3, we take a through investigation for representationsof rational tangle, giving some useful formulas. This is our another novel con-tribution. Section 4 is devoted to proving Theorem 1.1. Parts of the job relyessentially on the previous works [2, 5]. To hidden the un-important subtletiesand to make the main idea prominent, many details are omitted, to, but theycan be filled whenever required. 2 Preparation
We adopt notations and conventions following [2, 5].By a “tangle” we simultaneously mean an un-oriented tangle diagram andthe tangle it presents.
Figure 1:
Four tangles in T : (a) [0], (b) [ ∞ ], (c) [1], (d) [ − Figure 2: (a) T + T ; (b) T ∗ T Let T denote the set of tangles T with four ends T nw , T ne , T sw , T se . Thesimplest tangles in T are shown in Figure 1. Defined on T are horizontalcomposition + and vertical composition ∗ , as illustrated in Figure 2.For k (cid:54) = 0, the horizontal composite of | k | copies of [1] (resp. [ − k ] if k > k < | k | copies of [1] (resp.[ − /k ] if k > k < k , . . . , k s ∈ Z , the rationaltangle [[ k ] , . . . , [ k s ]] by definition is (cid:40) [ k ] ∗ [1 /k ] + · · · + [ k s ] , (cid:45) s, [1 /k ] + [ k ] ∗ · · · + [ k s ] , | s. Its fraction is given by the continued fraction [[ k , . . . , k s ]] ∈ Q , which is definedinductively as[[ k ]] = k ; [[ k , . . . , k j ]] = k j + 1 / [[ k , . . . , k j − ]] , j ≥ . Denote [[ k ] , . . . , [ k s ]] as [ p/q ] when its fraction equals p/q .Each T ∈ T gives rise to two links: the numerator N ( T ) and the denom-inator D ( T ), as shown in Figure 3. For p/q ∈ Q , the link N ([ p/q ]) is called a rational link or . A link of the form D ([ p /q ] ∗ · · · ∗ [ p m /q m ]) iscalled a Montesinos link and denoted by M ( p /q , . . . , p m /q m ); in particular, M ( p , . . . , p m ) is a pretzel link . 3 igure 3: (a) the numerator N ( T ); (b) a tangle T ; (c) the denominator D ( T ) Let e = (cid:18) (cid:19) , d ( λ ) = (cid:18) λ λ − (cid:19) , p = (cid:18) (cid:19) , u + κ ( ξ ) = (cid:18) κ ξ κ − (cid:19) , u − κ ( ξ ) = (cid:18) κ ξ κ − (cid:19) . For λ (cid:54) = t − , ± µ (cid:54) = 0, put h λt ( µ ) = 1 λ + 1 (cid:18) λt µ ( t − λ − λ − − λµ − t (cid:19) ;for t (cid:54) = 0, put k t ( α ) = (cid:18) t/ α ( t / − − α ) / (2 t )2 t t/ − α (cid:19) . The following is elementary but useful:
Lemma 2.1.
Suppose a i ∈ SL(2 , C ) with tr( a ) = tr( a ) = t . (a) If a a = d ( λ ) with λ + λ − (cid:54) = t − , ± , then there exists µ (cid:54) = 0 suchthat a = h λt ( µ ) and a = h λt ( − λ − µ ) . (b) If a a = p , then a , a are both upper-triangular. (c) If a a = − p and t (cid:54) = 0 , then a = k t ( α ) and a = k t ( α − t ) for some α . Direct computation showstr (cid:0) h λt ( µ ) − h λt ( ν ) (cid:1) = 2 t + ( λ + λ − + 2 − t )( µν − + µ − ν ) λ + λ − + 2 , (1)tr (cid:0) k t ( α ) − k t ( β ) (cid:1) = ( α − β ) + 2 . (2)For r = η + η − and n ∈ Z , put ω n ( r ) = (cid:40) ( η n − η − n ) / ( η − η − ) , η (cid:54) = ± ,nη n − , η ∈ {± } . For any z ∈ SL(2 , C ) with tr( z ) = r , one has z n = ω n ( r ) z − ω n − ( r ) e . (3)4 Representations of rational tangles
Given a tangle T , let D ( T ) denote the set of directed arcs of T ; each arc givestwo directed arcs. By an (SL(2 , C )-) representation of a tangle T , we mean amap ρ : D ( T ) → SL(2 , C ) such that ρ ( x − ) = ρ ( x ) − for each x ∈ D ( T ), and ρ ( z ) = ρ ( x ) ρ ( y ) ρ ( x ) − for each crossing as illustrated in Figure 4. To presentsuch a representation, it is sufficient to give each arc a direction and label anelement of SL(2 , C ) beside it. Call ρ trace- t if tr( ρ ( x )) = t for each x ∈ D ( T ).In virtue of Wirtinger presentation, a representation ρ of T is the same as ρ : π ( B − T ) → SL(2 , C ), where B is a 3-ball containing T such that ∂ B ∩ T exactly consists of the end points of T , and the basepoint is over T . Figure 4:
A representation satisfies ρ ( z ) = ρ ( x ) ρ ( y ) ρ ( x ) − at each crossing Given a representation ρ of T ∈ T , let x nw = ρ ( T nw ) , x ne = ρ ( T ne ) , x sw = ρ ( T sw ) , x se = ρ ( T se ) . Then x nw x ne = ( x sw ) − ( x se ) − and x sw x nw = ( x se ) − ( x ne ) − . Let z = tr( x nw x ne ) , ˙ z = tr( x sw x nw ) , ` z = tr( x nw x se ) , ´ z = tr( x ne x sw ) . Figure 5:
A representation of the rational tangle [[ k ] , . . . , [ k s ]]: (a) s is odd; (b) s iseven Convention 3.1.
We always present a rational tangle as in Figure 5 (a) or (b),when s is odd or even, respectively. Labeled beside each directed arc is its valueat which a representation takes. 5rom now on till the end of this section, suppose ρ is a representation ofthe rational tangle T = [[ k ] , . . . , [ k s ]]. Let x = x (0) , y = y (0) . Call ( x , y ) the generating pair for ρ . Let t = tr( x ) = tr( y ), and r = tr( xy ). Lemma 3.2. (a) y ∈ C (cid:104) x , x ne , x sw , x se (cid:105) , so ρ is determined by x nw , x ne , x sw , x se . (b) If t (cid:54) = 0 , then ρ is reducible if and only if r ∈ { , t − } . (c) If ρ is irreducible, then x nw x ne = e if and only if z = 2 .Proof. (a) By backward induction on s , this is immediate.(b) ( ⇒ ) is obvious.For ( ⇐ ), suppose r (cid:54) = 2 , t − • If r (cid:54) = t − , ±
2, then up to conjugacy we may assume xy = d ( λ ),and applying Lemma 2.1 (a), it is easy to see that x does not share aneigenvector with xy . • If r = −
2, then xy (cid:54) = − e : otherwise t = tr( x ) = − tr( y − ) = − t , so up toconjugacy we may assume xy = − p ; applying Lemma 2.1 (c), it is easyto see that x does not share an eigenvector with xy .(c) ( ⇒ ) is trivial. For ( ⇐ ), assume z = 2 but x nw x ne (cid:54) = e , then up toconjugacy we may assume p = x nw x ne = ( x sw ) − ( x se ) − . By Lemma 2.1 (b), x nw , x ne , x sw , x se are all upper-triangular, which by (a) would imply that ρ isreducible.We can recursively compute y ( j ) , x ( j ) , j = 1 , . . . , s , writing each one as alinear combination of e , x , y , xy , with coefficients being polynomials in t and r . Thus z, ˙ z, ` z are all polynomials in t and r , (actually with integer coeffi-cients). However, such computations are usually tedious. We shall take a cleverapproach.Introduce ψ n ( λ, γ ) = 2 t + (2 − λ − λ − − t )( γλ n + γ − λ − n )2 − λ − λ − . (4)Let − λ j − λ − j = r j = tr (cid:0) x ( − j y ( j ) (cid:1) . In generic case, up to conjugacy we may assume x = h − λ t (1), y = h − λ t ( λ − ).Then y (1) = h − λ t ( λ k ), x (1) = h − λ t ( λ k − ), and by (1), r = ψ k ( λ , x − = h − λ t (1), y (1) = h − λ t ( λ − ), y (0) = h − λ t ( γ ), x (1) = h − λ t ( γ λ − ). Then r = ψ ( λ , γ ) , tr (cid:0) xx (1) (cid:1) = ψ k − ( λ ,
1) = ψ − ( λ , γ ) , implying ψ n ( λ , γ ) = r ω n +1 ( − r ) − ψ k − ( λ , ω n ( − r ) + 2 t ( λ n − − λ − n − )( λ − ( λ + 1) . y (2) = h − λ t ( γ λ k ), x (2) = h − λ t ( γ λ k − ) − . r = ψ k ( λ , γ ) = r ω k +1 ( − r ) − ψ k − ( λ , ω k ( − r ) + 2 t ( λ k − − λ − k − )( λ − ( λ + 1) . Up to conjugacy, assume x (0) = h − λ t (1), y (2) = h − λ t ( λ − ), y (1) = h − λ t ( γ ), x (2) = h − λ t ( γ λ − ). Then r = ψ ( λ , γ ) , tr( x − x (2) ) = ψ k − ( λ , γ ) = ψ − ( λ , γ ) , implying ψ n ( λ , γ ) = r ω n +1 ( − r ) − ψ k − ( λ , γ ) ω n ( − r ) + 2 t ( λ n − − λ − n − )( λ − ( λ + 1) . In general, there exist γ , γ , . . . ∈ C ∗ such that r j = ψ k j ( λ j − , γ j − ) , (5) ψ n ( λ j , γ j ) = r j − ω n +1 ( − r j ) − ψ k j − ( λ j − , γ j − ) ω n ( − r j )+ 2 t ( λ nj − − λ − n − j )( λ j − ( λ j + 1) . (6)Without knowing the γ j ’s, we can obtain the expressions for z = r s , ˙ z = r s − , ` z = ψ k s − ( λ s − , γ s − ) . When ρ is reducible, traces are not sufficient to determine ρ , and we need tolook closer at it. Then Lemma 3.3 of [5] can be applied, with small modifications.For k ∈ Z and a, b ∈ C × , introduce u k ( a, b ) = (cid:40) [ h ] a b a (1 − b ) , k = 2 h, [ h + 1] a b a − [ h ] a b a b , k = 2 h + 1 ,G k ( a, b ) = (cid:20) − u k ( a, b ) u k ( a, b )1 − u k − ( a, b ) u k − ( a, b ) (cid:21) . Suppose x ( j ) = u + κ ( ξ j ) (cid:15) j , y ( j ) = u + κ ( ζ j ) (cid:15) (cid:48) j , with (cid:15) j , (cid:15) (cid:48) j ∈ {± } . Put η (cid:48) = 1, η (cid:48) = u k ( κ (cid:15) , κ (cid:15) (cid:48) ), η = u k − ( κ (cid:15) , κ (cid:15) (cid:48) ), and recursively compute η j , η (cid:48) j by (cid:20) η (cid:48) j η j (cid:21) = G k j (cid:0) κ (cid:15) (cid:48) j − , κ (cid:15) j − (cid:1) (cid:20) η (cid:48) j − η j − (cid:21) , j ≥ . Then we have x ne = u + κ ((1 − η (cid:48) s ) ξ + η (cid:48) s ζ ) (cid:15) ne , (7) x sw = u + κ ((1 − η (cid:48) s − ) ξ + η (cid:48) s − ζ ) (cid:15) sw , (8) x se = u + κ ((1 − η s ) ξ + η s ζ ) (cid:15) se , (9)where (cid:15) ne , (cid:15) sw , (cid:15) se ∈ {± } are determined by (cid:15) , (cid:15) (cid:48) and T in an obvious way. Remark 3.3.
Note that x ne = ( x (0) ) ± if and only if (cid:15) ne = ± (cid:15) and η (cid:48) s ( ξ − ζ ) =0. If x ne = ( x nw ) − so that ρ gives rise to a representation of N ( T ), then by(15) in [5], η (cid:48) s = 0 is equivalent to ∆ N ( T ) ( κ ) = 0.7 Irreducible trace- t representations of Montesinosknots with t (cid:54) = 0 The case when t = 0 has been well-understood [2], so let us assume t (cid:54) = 0.Consider a Montesinos knot K = M ( p /q , . . . , p m /q m ). Let T i = [ p i /q i ].Let ( x i , y i ) be the generating pair for ρ | T i ; let r i = tr( x i y i ). Let x ne i = ρ ( T ne i ), etc. Clearly, x sw i = ( x nw i +1 ) − and x se i = ( x ne i +1 ) − . Let z i = tr( x nw i x ne i ) , ˙ z i = tr( x nw i x sw i ) , ` z i = tr( x nw i x se i ) . Note that x nw1 x ne1 = · · · = x nw m x ne m := g . Suppose tr( g ) = τ = λ + λ − , sothat z i = τ for all i . Example 4.1.
Let K = 12 a M (7 / , , / Figure 6:
The knot 12 a M (7 / , , / We have x = a − , y = c = x − , y = b − . Using (5), (6), we can compute˙ z = t ( r − + r (3 − r ) , ` z = r − r + 2 + t r (2 − r ) ,z = 2 t − r + ˙ z ` z = 2 t − r + ˙ z ( r − r + 2 + t r (2 − r )) , ˙ z = r , ` z = t − r ,z = tr([ c , x ]) = 2 t − − r ( t − r ) , ˙ z = t (2 − r ) + r − , ` z = t (2 − ˙ z ) + r ( ˙ z − ˙ z − ,z = t ( ˙ z − − r ( ˙ z − ˙ z − z + 1) . Lemma 4.2. (a) If τ = 2 , then g = e . (b) If τ = − , then g (cid:54) = − e . roof. (a) If τ = 2 but g (cid:54) = e , then up to conjugacy we may assume g = p . ByLemma 2.1 (b), x nw i , x ne i are all upper-triangular, so by Lemma 3.2 (a), ρ wouldbe reducible, contradicting the assumption.(b) If g = − e , then t = tr(( x ne i ) − ) = − tr( x nw i ) = − t , which is impossibleunder the assumption t (cid:54) = 0.The discussion is divided separately into three cases: g = e , τ / ∈ { , t − } ,and τ = t − (cid:54) = ±
2, which respectively corresponds to X ( K ), X ( K ) and X (cid:48) ( K ). = e In this case, ρ | T i gives rise to a representation of ρ i : N ( T i ) → SL(2 , C ) andvice-versa. We say that ρ is the connected sum of the ρ i ’s.It may happen that for some i , ρ | T i is abelian or reducible; in the later case∆ N ( T i ) ( t −
2) = 0, by Remark 3.3.Suppose ρ | T i is irreducible (which is equivalent to r i (cid:54) = 2 , t −
2) for all i .With z i = 2 and ˙ z i = tr( x − i x i +1 ), the task reduces to finding a , . . . , a m with a · · · a m = e and tr( a i ) = ˙ z i . Then x i +1 = x i a i , so that x i = x a · · · a i − . Fix x = (cid:18) t − (cid:19) . Then ρ ’s up to conjugacy correspond bijectively toelements of the moduli space (cid:8) ( a , . . . , a m ) ∈ (SL(2 , C )) m : a · · · a m = e , tr( a i ) = ˙ z i (cid:9) / ∼ , where ( a , . . . , a m ) ∼ ( a (cid:48) , . . . , a (cid:48) m ) means the existence of c such that cx = x c and a (cid:48) i = ca i c − for all i .Therefore, X ( K ) can have high-dimensional components, when m ≥ τ (cid:54) = 2 , t − When τ (cid:54) = ± , t −
2, up to conjugacy g = d ( λ ). Then there exists µ i (cid:54) = 0 suchthat x nw i = h λt ( µ i ), x ne i = h λt ( − λ − µ i ). Applying (1) to ( µ, ν ) = ( µ i , µ i +1 ) and( µ, ν ) = ( µ i , − λ − µ i +1 ), we can solve µ i µ i +1 = (1 + λ − ) ˙ z i + (1 + λ )` z i − t (1 − λ )( τ + 2 − t ) . Hence m (cid:89) i =1 ((1 + λ − ) ˙ z i + (1 + λ )` z i − t ) = (1 − λ ) m ( τ + 2 − t ) m . (10)When τ = −
2, by Lemma 4.2 (b), up to conjugacy we can assume g = − p .Then x nw i = k t ( α i ), x ne i = k t ( α i − t ). Applying (2) to ( α, β ) = ( α i , α i +1 ) and( α, β ) = ( α i , α i +1 − t ), we can solve2 t ( α i +1 − α i ) = ˙ z i − ` z i + t . m (cid:88) i =1 (` z i − ˙ z i ) = mt . (11)Conversely, when the equations are satisfied, it is easy to construct the cor-responding representation. Note that if ` z i − ˙ z i (cid:54) = t for some i , then x sw (cid:54) = − ( x se ) − is ensured.Actually, (10) and (11) can be unified as1 λ + 1 (cid:16) m (cid:89) i =1 ((1 + λ − ) ˙ z i + (1 + λ )` z i − t ) − (1 − λ ) m ( τ + 2 − t ) m (cid:17) = 0 . (12)This can be seen by noting that the left-hand-side of (12) can be written as apolynomial in λ + 1, with constant term − (cid:16) mt + m (cid:88) i =1 ( ˙ z i − ` z i ) (cid:17) ( − t ) m − . Therefore, X ( K ) is defined by z i = λ + λ − = τ, ≤ i ≤ m, λ + 1 (cid:16) m (cid:89) i =1 ((1 + λ − ) ˙ z i + (1 + λ )` z i − t ) − (1 − λ ) m ( τ + 2 − t ) m (cid:17) = 0 , which are m + 1 equations in m + 2 variables r , . . . , r m , t, λ ; remember that z i , ˙ z i , ` z i are polynomials in t, r i . Remark 4.3.
Note that (12) will hold if λ is replace by λ − . With some moreefforts, we may rewrite (12) as a polynomial in τ . τ = t − (cid:54) = ± Let t = κ + κ − and λ = κ . Up to conjugacy we may assume g = d ( κ ). Foreach i , either x nw i = x ne i = d ( κ ), or there exists ξ i (cid:54) = 0 such that x nw i = u (cid:15) i κ ( ξ i ), x ne i = u (cid:15) i κ ( − κ − (cid:15) i ξ i ), with (cid:15) i ∈ { + , −} . Set (cid:15) i = 0 and ξ i = 0 if x nw i = x ne i = d ( κ ). The irreducibility of ρ is equivalent to { + , −} ⊆ { (cid:15) , . . . , (cid:15) m } . There arefive possibilities.1) If (cid:15) i = (cid:15) i +1 = 0, then ρ | T i is abelian.2) If (cid:15) i = 0 (cid:54) = (cid:15) i +1 , then ρ | T i is reducible and non-abelian. Since x ne i = x nw i but x se i (cid:54) = x sw i , we see from the final part of Section 3 that κ has finitelymany possible values, and y i (so that ρ | T i ) is determined by ξ i +1 .3) If (cid:15) i (cid:54) = 0 = (cid:15) i +1 , the situation is similar.10) If (cid:15) i (cid:15) i +1 = +, then ρ | T i is reducible and non-abelian. Now ξ i , ξ i +1 isconstrained by a linear equation, with coefficients being polynomials in κ .5) If (cid:15) i (cid:15) i +1 = − , then ρ | T i is irreducible, and z i = κ + κ − , ˙ z i = 2 − ξ i ξ i +1 , ` z i = 2 + κ (cid:15) i ξ i ξ i +1 . Hence ξ i ξ i +1 is equal to a multi-valued function of κ .For generic K , case 2) or 3) occurs at most once. Suppose i is the smallest i with (cid:15) i (cid:54) = 0. Take a further conjugation to make ξ i = 1.Each tuple (cid:15) = ( (cid:15) , . . . , (cid:15) m ) with (cid:15) i ∈ { , + , −} and { + , −} ⊆ { (cid:15) , . . . , (cid:15) m } gives rise to a system of equations, whose solution set X (cid:48) ( K ; (cid:15) ) is finite. All ofthem together contribute X (cid:48) ( K ) = (cid:116) (cid:15) X (cid:48) ( K ; (cid:15) ). References [1] C. Ashley, J.-P. Burelle and S. Lawton, Rank 1 character varieties of finitelypresented groups,
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