The volume of hyperbolic cone-manifolds of the knot with Conway's notation C(2n,3)
NNovember 6, 2018 3:18 WSPC/INSTRUCTION FILE Conway2n3xj
Journal of Knot Theory and Its Ramificationsc (cid:13)
World Scientific Publishing Company
The volume of hyperbolic cone-manifolds of the knot withConway’s notation C (2 n, Ji-Young Ham ∗ Department of Science, Hongik University, 94 Wausan-ro, Mapo-gu, Seoul, 04066,Department of Mathematical Sciences, Seoul National University, 1 Gwanak-ro, Gwanak-gu,Seoul 08826,[email protected]
Joongul Lee † Department of Mathematics Education, Hongik University, 94 Wausan-ro, Mapo-gu, Seoul,04066,[email protected]
ABSTRACTLet C (2 n,
3) be the family of two bridge knots of slope (4 n +1) / (6 n +1). We calculatethe volumes of the C (2 n,
3) cone-manifolds using the Schl¨afli formula. We present theconcrete and explicit formula of them. We apply the general instructions of Hilden,Lozano, and Montesinos-Amilibia and extend the Ham, Mednykh, and Petrov’s methods.As an application, we give the volumes of the cyclic coverings over those knots. Forthe fundamental group of C (2 n, Keywords : hyperbolic orbifold, hyperbolic cone-manifold, volume, C (2 n,
1. Introduction
Let us denote C (2 n,
3) by T n and the exterior of T n by X n . By Mostow-Prasadrigidity, X n has a unique hyperbolic structure. Let ρ ∞ be the holonomy representa-tion from π ( X n ) to P SL (2 , C ) and denote ρ ∞ ( π ( X n )) by Γ, a Kleinian group. X n is a ( P SL (2 , C ) , H )-manifold and can be identified with H / Γ. Thurston’sorbifold theorem guarantees an orbifold, X n ( α ), with T n as the singular locus ∗ This work was supported by Basic Science Research Program through the National ResearchFoundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (No.NRF-2008-341-C00004 and NRF-R01-2008-000-10052-0). † This work was supported by 2015 Hongik University Research Fund.1 a r X i v : . [ m a t h . G T ] M a r ovember 6, 2018 3:18 WSPC/INSTRUCTION FILE Conway2n3xj Ji-Young Ham, Joongul Lee and the cone-angle α = 2 π/k for some nonzero integer k can be identified with H / Γ (cid:48) for some Γ (cid:48) ∈ P SL (2 , C ); the hyperbolic structure of X n is deformed to thehyperbolic structure of X n ( α ). For the intermediate angles whose multiples arenot 2 π and not bigger than π , Kojima [11] showed that the hyperbolic structureof X n ( α ) can be obtained uniquely by deforming nearby orbifold structures. Notethat there exists an angle α ∈ [ π , π ) for each T n such that X n ( α ) is hyperbolicfor α ∈ (0 , α ), Euclidean for α = α , and spherical for α ∈ ( α , π ] [18,8,11,19]. Forfurther knowledge of cone manifolds you can consult [1,7].Explicit volume formulae for hyperbolic cone-manifolds of knots and links areknown a little. The volume formulae for hyperbolic cone-manifolds of the knot4 [8,11,12,15], the knot 5 [13], the link 5 [16], the link 6 [17], and the link 6 [2]have been computed. In [9] a method of calculating the volumes of two-bridgeknot cone-manifolds were introduced but without explicit formulae. In [7], explicitvolume formulae for hyperbolic twist knot cone-manifolds are computed. Similarmethods are used for computing Chern-Simons invariants of twist knot orbifoldsin [6].The main purpose of the paper is to find an explicit and efficient volume formulafor hyperbolic cone-manifolds of the knot T n . The following theorem gives thevolume formula for X n ( α ). Note that if 2 n of T n is replaced by an odd integer, then T n becomes a link with two components. Also, note that the volume of hyperboliccone-manifolds of the knot with Conway’s notation C ( − n, −
3) is the same as thatof the knot with Conway’s notation C (2 n, T n has to be hyperbolic, we exclude the case when n = 0. Theorem 1.1.
Let X n ( α ) , ≤ α < α be the hyperbolic cone-manifold withunderlying space S and with singular set T n of cone-angle α . Then the volume of X n ( α ) is given by the following formula Vol ( X n ( α )) = (cid:90) πα log (cid:12)(cid:12)(cid:12)(cid:12) M − + xM + x (cid:12)(cid:12)(cid:12)(cid:12) dα, where x with Im( x ) ≤ is a zero of the Riley-Mednykh polynomial P n = P n ( x, M ) which is given recursively by P n = (cid:40) QP n − − M P n − if n > ,QP n +1) − M P n +2) if n < − , with initial conditions P − = M x + (cid:0) M − M + 1 (cid:1) x + M ,P = M − for n < P ( x, M ) = 1 for n > ,P = − M x + (cid:0) − M + M − M (cid:1) x + (cid:0) − M + M − M + M − (cid:1) x + M , ovember 6, 2018 3:18 WSPC/INSTRUCTION FILE Conway2n3xj Volumes of C (2 n,
3) 3 and M = e iα and Q = − M x + (cid:0) − M + 2 M − M (cid:1) x + (cid:0) − M + 2 M − M + 2 M − (cid:1) x +2 M . From Theorem 1.1, the following corollary can be obtained. The following corol-lary gives the hyperbolic volume of the k -fold cyclic covering over T n , M k ( X n ),for k ≥ Corollary 1.2.
The volume of M k ( X n ) is given by the following formula Vol ( M k ( X n ))) = k (cid:90) π πk log (cid:12)(cid:12)(cid:12)(cid:12) M − + xM + x (cid:12)(cid:12)(cid:12)(cid:12) dα, where x with Im( x ) ≤ is a zero of the Riley-Mednykh polynomial P n = P n ( x, M ) and M = e i α .
2. Two bridge knots with Conway’s notation C (2 n, Fig. 1. A two bridge knot with Conway’s notation C (2 n,
3) for n > n <
A knot is a two bridge knot with Conway’s notation C (2 n,
3) if it has a regulartwo-dimensional projection of the form in Figure 1. For example, Figure 2 is knot C (4 , n right-handed horizontal cross-ings. Recall that we denote it by T n . In [20, Proposition 1], the fundamental groupof two-bridge knots is presented. We will use the fundamental group of X n in [10].In [10], the fundamental group of X n is calculated with 3 right-handed verticalcrossings as positive crossings instead of three left-handed vertical crossings. Thefollowing proposition is tailored to our purpose. The reduced word relation of theone in the following proposition can also be obtained by reading off the fundamen-tal group from the Schubert normal form of T n with slope n +16 n +1 for n > T n with slope n +16 n +1 for n < Ji-Young Ham, Joongul Lee
Fig. 2. The knot C (4 ,
3) (7 in the Rolfsen’s knot table) the Hoste-Shanahan’s question whether their presentation is actually derived fromSchubert’s canonical 2-bridge diagram or not. Proposition 2.1. π ( X n ) = (cid:10) s, t | swt − w − = 1 (cid:11) , where w = ( ts − tst − s ) n . (cid:0) P SL (2 , C ) , H (cid:1) structure of X n ( α ) Let R = Hom( π ( X n ) , SL(2 , C )). Given a set of generators, s, t , of the fundamentalgroup for π ( X n ), we define a set R ( π ( X n )) ⊂ SL(2 , C ) ⊂ C to be the set ofall points ( ρ ( s ) , ρ ( t )), where ρ is a representaion of π ( X n ) into SL(2 , C ). Sincethe defining relation of π ( X n ) gives the defining equation of R ( π ( X n )) [21], R ( π ( X n )) is an affine algebraic set in C . R ( π ( X n )) is well-defined up to iso-morphisms which arise from changing the set of generators. We say elements in R which differ by conjugations in SL(2 , C ) are equivalent . A point on the variety givesthe (cid:0) P SL (2 , C ) , H (cid:1) structure of X n ( α ).Let ρ ( s ) = (cid:20) M M − (cid:21) , ρ ( t ) = (cid:20) M − M − M − − x M − (cid:21) . Then ρ becomes a representation if and only if M and x satisfies a polynomialequation [21,14]. We call the defining polynomial of the algebraic set { ( M, x ) } asthe Riley-Mednykh polynomial . The Riley-Mednykh polynomial
Since we are interested in the excellent component (the geometric component) of R ( π ( X n )), in this subsection we set M = e iα . Given the fundamental group ofovember 6, 2018 3:18 WSPC/INSTRUCTION FILE Conway2n3xj Volumes of C (2 n,
3) 5 X n , π ( X n ) = (cid:10) s, t | swt − w − = 1 (cid:11) , where w = ( ts − tst − s ) n , let S = ρ ( s ) , T = ρ ( t ) and W = ρ ( w ). Then the trace of S and the trace of T are both 2 cos α . Lemma 3.1.
For c ∈ SL(2 , C ) which satisfies cS = T − c and c = − I , SW T − W − = − ( SW c ) Proof. ( SW c ) = SW cSW c = SW T − c ( T S − T ST − S ) n c = SW T − ( S − T S − T − ST − ) n c = − SW T − W − . From the structure of the algebraic set of R ( π ( X n )) with coordinates ρ ( s )and ρ ( t ) we have the defining equation of R ( π ( X n )). By plugging in e iα into M of that equation, we have the following theorem. Theorem 3.2. ρ is a representation of π ( X n ) if and only if x is a root of thefollowing Riley-Mednykh polynomial P n = P n ( x, M ) which is given recursively by P n = (cid:40) QP n − − M P n − if n > ,QP n +1) − M P n +2) if n < − , with initial conditions P − = M x + (cid:0) M − M + 1 (cid:1) x + M ,P = M − for n < P ( x, M ) = 1 for n > ,P = − M x + (cid:0) − M + M − M (cid:1) x + (cid:0) − M + M − M + M − (cid:1) x + M , and Q = − M x + (cid:0) − M + 2 M − M (cid:1) x + (cid:0) − M + 2 M − M + 2 M − (cid:1) x +2 M . Proof.
Note that
SW T − W − = I , which gives the defining equations of R ( π ( X n )), is equivalent to ( SW c ) = − I in SL(2 , C ) by Lemma 3.1 and( SW c ) = − I in SL(2 , C ) is equivalent to tr( SW c ) = 0.We can find two c ’s in SL(2 , C ) which satisfies cS = T − c and c = − I by directcomputations. The existence and the uniqueness of the isometry (the involution)which is represented by c are shown in [4, p. 46]. Since two c ’s give the same elementin PSL(2 , C ), we use one of them. Hence, we may assume c = (cid:34) − M √− M − M − M x √− M − M − M xM (cid:35) , ovember 6, 2018 3:18 WSPC/INSTRUCTION FILE Conway2n3xj Ji-Young Ham, Joongul Lee S = (cid:34) e iα e − iα (cid:35) , T = (cid:34) e iα − e iα − e − iα − x e − iα (cid:35) , and let U = T S − T ST − S .Recall that P n is the defining polynomial of the algebraic set { ( M, x ) } and thedefining polynomial of R ( π ( X n )) corresponding to our choice of ρ ( s ) and ρ ( t ).We will show that P n is equal to tr( SW c ) / tr( Sc ) times M n for n > M n − for n <
0. One can easily see tr(
SU c ) = P tr( Sc ) /M , tr( SW c ) = P − tr( Sc ) /M and tr( U ) = Q/M = tr( U − ). We set P as the statement of the theorem. Now,we only need the following recurrence relations.tr( SW c ) = tr( SU n c ) = tr( SU n − cU − ) = tr( SU n − c )tr( U − ) − tr( SU n − cU )= tr( SU n − c )tr( U − ) − tr( SU n − c ) = 0 if n > SW c ) = tr( SU n c ) = tr( SU n +1 cU ) = tr( SU n +1 c )tr( U ) − tr( SU n +1 cU − )= tr( SU n +1 c )tr( U ) − tr( SU n +2 c ) = 0 if n < − , where the third equality comes from the Cayley-Hamilton theorem. Since tr( Sc ),tr( SU c ), and tr( SU − c ) have tr( Sc ) = √− M − M − M xM as a common factor, allof tr( SW c )’s have tr( Sc ) as a common factor. We left the common factor out oftr( SW c ), multiplied it by a power of M n if n > M n − for n < P n , the Riley-Mednykh polynomial.
4. Longitude
Let l = ww ∗ s − n , where w ∗ is the word obtained by reversing w . Let L = ρ ( l ) .Then l is the longitude which is null-homologus in X n . Recall ρ ( w ) = U n . We canwrite ρ ( w ∗ ) = (cid:101) U n . It is easy to see that U and (cid:101) U can be written as U = (cid:18) u u u u (cid:19) and (cid:101) U = (cid:18) ˜ u ˜ u ˜ u ˜ u (cid:19) where ˜ u ij is obtained by u ij by replacing M with M − . Similar computationwas introduced in [10]. Definition 4.1.
The complex length of the longitude l is the complex number γ α modulo 4 π Z satisfying tr( ρ ( l )) = 2 cosh γ α . ovember 6, 2018 3:18 WSPC/INSTRUCTION FILE Conway2n3xj Volumes of C (2 n,
3) 7
Note that l α = | Re ( γ α ) | is the real length of the longitude of the cone-manifold X n ( α ).The following lemma was introduced in [10] with slightly different coordinates. Lemma 4.2. [10] u L + ˜ u M − n = 0 . Theorem 4.3. L = − M − n − M − + xM + x . Proof.
By directly computing u L + ˜ u M − n = 0 in Lemma 4.2, the theoremfollows.
5. Proof of Theorem 1.1
For n ≥ M = e i α , P n ( x, M ) have 3 n component zeros, and for n < − − (3 n + 1) component zeros. For each n , there exists an angle α ∈ [ π , π ) such that T n ( α ) is hyperbolic for α ∈ (0 , α ), Euclidean for α = α , and spherical for α ∈ ( α , π ] [18,8,11,19]. From the following Equality (5.1), when | L | = 1, which happenswhen α = α , Im( x ) = 0. Hence, when α increases from 0 to α , two complexnumbers x and x approach a same real number. In other words, P n ( x, e iα ) hasa multiple root. Denote by D ( X n ( α )) the discriminant of P n ( x, M ) over x . Then α will be one of the zeros of D ( X n ( α )).From Theorem 4.3, we have the following equality, | L | = (cid:12)(cid:12)(cid:12)(cid:12) M − + xM + x (cid:12)(cid:12)(cid:12)(cid:12) = (cos α + Re( x )) + (Im( x ) − sin α ) (cos α + Re( x )) + (Im( x ) + sin α ) . (5.1)For the volume, we choose L with | L | ≥ x ) ≤ ρ ( X n ) = ρ ( X n (0)) for each component with | L | ≥ α of D ( X n ( α )) with α ∈ [ π , π ) on it. The component which gives the maximal volumeis the excellent component [3,5]. On the geometric component we have the volumeof a hyperbolic cone-manifold X n ( α ) for 0 ≤ α < α :Vol( X n ( α )) = − (cid:90) αα l α dα = − (cid:90) αα log | L | dα = − (cid:90) απ log | L | dα = (cid:90) πα log | L | dα = (cid:90) πα log (cid:12)(cid:12)(cid:12)(cid:12) M − + xM + x (cid:12)(cid:12)(cid:12)(cid:12) dα, ovember 6, 2018 3:18 WSPC/INSTRUCTION FILE Conway2n3xj Ji-Young Ham, Joongul Lee where the first equality comes from the Schl¨afli formula for cone-manifolds (Theo-rem 3.20 of [1]), the second equality comes from the fact that l α = | Re ( γ α ) | is thereal length of the longitude of X n ( α ), the third equality comes from the fact thatlog | L | = 0 for α < α ≤ π by Equality 5.1 since all the characters are real (theproof of Proposition 6.4 of [19]) for α < α ≤ π , and α ∈ [ π , π ) is a zero of thediscriminant D ( X n ( α )). Acknowledgments
The authors would like to thank Alexander Mednykh and Hyuk Kim for theirvarious helps and anonymous referees.
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