We calculate the twisted Alexander polynomials of $(-2,3,2n+1)$-pretzel knots
associated to their holonomy representations. As a corollary, we obtain new
supporting evidences of Dunfield, Friedl and Jackson's conjecture, that is, the
twisted Alexander polynomials of hyperbolic knots associated to their holonomy
representations determine the genus and fiberedness of the knots.Comment: 8 pages, 2 figure

Aso, Airi

アソ, アイリ

阿蘇, 愛理

English

arXiv

Institutional Repositories DataBase (IRDB)

Twisted Alexander Polynomials of (−2, 3, 2n + 1)-Pretzel Knots

首都大学東京修士（理学）We calculate the twisted Alexander polynomials of (−2, 3, 2n + 1)-pretzel knots associated to their holonomy representations.首都大学東京, 2018-03-25, 修士（理学）thesi

22774

22775

Tokyo Metropolitan University Institutional Repository Miyako-Dori / 首都大学東京機関リポジトリ

TWISTED ALEXANDER POLYNOMIALS OF (−2, 3, 2n+ 1)-PRETZEL KNOTSAIRI ASODEPARTMENT OF MATHEMATICS AND INFORMATION SCIENCES,TOKYO METROPOLITAN UNIVERSITYAbstract. We calculate the twisted Alexander polynomials of (−2, 3, 2n + 1)-pretzel knotsassociated to their holonomy representations.1. IntroductionThe notion of twisted Alexander polynomials was introduced by Wada [W] and Lin [L] indepen-dently in 1990s. The definition of Lin is for knots in S3 and the definition of Wada is for finitelypresented groups. The twisted Alexander polynomial is a generalization of the Alexander polyno-mial and is defined for the pair of a group and its representations. By Kitano and Morifuji [KM],it is known that Wada’s twisted Alexander polynomials of the knot groups for any nonabelianrepresentations into SL2(F) over a field F are polynomials. In this paper, by using the followingdefinition due to Wada, we compute the twisted Alexander polynomials of (−2, 3, 2n + 1)-pretzelknots Kn depicted in Figure 1 associated to their holonomy representations ρm : G(Kn)→ SL2(C)given in following section.Figure 1. (−2, 3, 2n+ 1)-pretzel knotDefinition 1.1. Let G(K) = π1(S3 \K) be the knot group of a knot K presented byG(K) = ⟨x1, · · · , xn r1, · · · , rn−1⟩.Let Γ denote the free group generated by x1, · · · , xn and φ : ZΓ → ZG(K) the natural ringhomomorphism. Let ρ : G(K) → GLd(C) be a d-dimensional linear representation of G(K) andΦ : ZΓ→Md(C[t, t−1]) the ring homomorphism defind byΦ = (ρ˜⊗ α˜) ◦ φ,where α˜ : ZG(K) → Z⟨t, t−1⟩ and ρ˜ are respective ring homomorphisms induced by the abelian-ization α : G(K)→ ⟨t⟩ and ρ. We putAi,j = Φ(∂ri∂xj),Key words and phrases. twisted Alexander polynomials, pretzel knot, holonomy representation.12 A. ASOwhere∂∂xjdenotes the Fox derivative (or free derivative) with respect to xj , that is, a mapZΓ→ ZΓ satisfying the conditions∂∂xjxi = δij ,∂∂xjgg′ =∂∂xjg +∂∂xjg′,where δij denotes the Kronecker symbol and g, g′ ∈ Γ. Then, the twisted Alexander polynomial ofK is defined by∆K,ρ =detAρ,kdetΦ(xk − 1) ,where Aρ,k is the 2(n − 1) × 2(n − 1) matrix obtained from Aρ = (Ai,j) by removing the k-thcolumn, i.e.Aρ,k =⎛⎜⎝ A1,1 · · · A1,k−1 A1,k+1 · · · A1,n... ... ... ...An−1,1 · · · An−1,k−1 An−1,k+1 · · · An−1,n⎞⎟⎠ .If K is hyperbolic, i.e. the complement S3 \ K admits a complete hyperbolic metric of finitevolume, the most important representation is its holonomy representation into SL2(C) which isa lift of the representation into the group of orientation-preserving isometries of the hyperbolic3-space H3. In fact, the twisted Alexander polynomials of some hyperbolic knots associated totheir holonomy representations are computed by Dunfield, Friedl and Jackson [DFJ]. Recently,the twisted Alexander polynomials of some infinite families of knots, twist knots and genus onetwo-bridge knots associated to their holonomy representations, are computed by Morifuji [Mo1]and Tran [T1] and genus one two-bridge knots associated to the adjoint representations of theirholonomy representations is also computed by Tran [T2].(−2, 3, 2n+1)-pretzel knot is an infinite family of knots which contains the Fintushel-Stern knoti.e. (−2, 3, 7)-pretzel knot. It plays an important role in studying of exceptional surgeries of knots[Ma]. The A-polynomials of (−2, 3, 2n+1)-pretzel knot are computed by Tamura-Yokota [TY] andGaroufalidis-Mattman [GM].Acknowledgement: The author would like to thank professor Yoshiyuki Yokota for supervising andgiving helpful comments. She also would like to thank professor Teruhiko Soma and professorManabu Akaho for giving valuable comments.2. Holonomy representationsIn this section, we give a presentation of knot group G(Kn) and its holonomy representationρm : G(Kn)→ SL2(C), where m represents the eigenvalue of the meridian of Kn.Let L be the link depicted in Figure 2 and E = S3 \ L. Then, the Wirtinger presentation (see[CF]) of π1(E) is given by⟨a, b, x {axba(xb)−1}−1x = xb{axba(xb)−1}−1(axb)−1xb, [x, axba(xb)−1] = 1⟩,where a, b and x is Wirtinger generators assigned to the corresponding pass depicted in Figure 2.Note that En := S3 \Kn is obtained from L by (− 1n )-surgery along the trivial component, that is,removing the tubular neighborhood of the trivial component and re-gluing the solid torus again.Therefore, by the van Kampen theorem, we haveπ1(En) = ⟨a, b, x {axba(xb)−1}−1x = xb{axba(xb)−1}−1(axb)−1xb, x = {axba(xb)−1}n⟩.Figure 2. Link L3Proposition 2.1. For a non-zero complex number m, there exists a representation ρm : π1(En)→SL2(C) such thatρm(a) =⎛⎝ m −(m2 − s) (s2n+1 + 1)m(s+ 1)0 m−1⎞⎠ , ρm(b) = 1sα⎛⎜⎝ β − (sα−mβ)(msα− β)mββm(msα− β) + sαm⎞⎟⎠ ,andρm(x) =⎛⎝ sn 0sn − s−ns2n+1 + 1s−n⎞⎠ ,where s is a solution to0 = m8(s− 1)(s+ 1)2(s2n − s2)s2n+2(1)−m6{s6n+3+(2s6 + s5 − 4s4 + s3 + s2 − s− 1)s4n+1−(s6 + s5 − s4 − s3 + 4s2 − s− 2)s2n+2 + s6}+m4{(s2+1)s6n+2 + (s6 + 2s5 − 3s4 − 2s3 + 6s2 − 4s− 2)s4n+3−(2s6 + 4s5 − 6s4 + 2s3 + 3s2 − 2s− 1)s2n + (s2 + 1)s5}−m2{s6n+3+(2s6 + s5 − 4s4 + s3 + s2 − s− 1)s4n+1−(s6 + s5 − s4 − s3 + 4s2 − s− 2)s2n+2 + s6}+(s− 1)(s+1)2(s2n − s2)s2n+2and α,β are given byα = (s2 − 1)s2n{−m6(s− 1)s2(s2n+1 + 1) +m4(s2n+2(s4 − 2s2 + 3s− 1) + s4 − 3s3 + 2s2 − 1)−m2s(s2n(2s3 − s2 + 1)− s(s3 − s+ 2)) + s2(s2n − s2)},β = m7s2n+2(s2 − 1)(s3 + 1)−m5s3{s4n(s3 − s2 + 1) + s2n−2(s− 1)(s3 + s+ 1)(s3 + s2 + 1)− (s3 − s+ 1)}+m3s2(s3 + 1)(s2n − 1)(s2n + s2)−ms3(s2n − s2)(s2n + s).In what follows, for simplicity, we denote the right hand side of (1) by r0.Proof. For simplicity, put A = ρm(a), B = ρm(b), X = ρm(x). By the aid of Mathematica, wehaveAXBA(XB)−1 =⎛⎝ s 0s2 − 1s(s2n+1 + 1)1s⎞⎠+ r1⎛⎜⎝1m3s(s2n+1 + 1)α2− 1m3s(s+ 1)α2s+ 1m3s2(s2n+1 + 1)2α2− 1m3s2(s2n+1 + 1)α2⎞⎟⎠ ,wherer1 = −α2ms(m2s2n+2 −m2 − s2n+1 + s) + αβ(m2 − 1)(m2 + 1)s2n+1(s+ 1)+β2ms2n(m2s2n+1 −m2s− s2n+2 + 1) ≡ 0 mod r0.Therefore, by (1), we have X = {AXBA(XB)−1}n, that is, ρm(x) = ρm({axba(xb)−1}n).On the other hand, we can observeAXB{AXBA(XB)−1} ≡ XBX−1{AXBA(XB)−1}XB mod r0and so AXB{AXBA(XB)−1} = XBX−1{AXBA(XB)−1}XB by (1). Further more, we obtainXB{AXBA(XB)−1}−1(AXB)−1XB = XB(AXB{AXBA(XB)−1})−1XB= XB(XBX−1{AXBA(XB)−1}XB)−1XB= {AXBA(XB)−1}−1Xthat is, ρm({axba(xb)−1}−1x) = ρm (xb{axba(xb)−1}−1(axb)−1xb). This completes the proof.!4 A. ASORemark 2.2. Since the representation ρm comes from the holonomy representation obtained fromthe ideal triangulation of E given in [TY], the holonomy representation ρm of G(Kn) is given bythe solution to (1) which maximizes the hyperbolic volume of S3 \Kn.3. Calculation of the twisted Alexander polynomialThe following is the main result of this paper.Theorem 3.1. The twisted Alexander polynomial of Kn associated to ρm is given by∆Kn,ρm(t) = 1 +2n−1∑i=0λi(ti+3 + t4n−i+3) + t4n+6,whereλi =⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩(1 +m2)(Hsi/2+1β − s(si/2+1 − s−(i/2+1))(η1 + η2))Hmβif 0 ≤ i ≤ 2n− 2 and i is even,s(i−1)/2 − s−(i−1)/2s− s−1 if 0 ≤ i ≤ 2n− 2 and i is odd,sn−1 − s−(n−1)s− s−1 −(s2 − 1)η1Hsnβif i = 2n− 1and we putH = 1−m2s+m2s2n+1 − s2n+2,η1 = mα−ms2n+1α+ s2nβ +m2s2nβ,η2 = −msα+ms2n+1α− s2nβ − s2n+1β.To prove Theorem 3.1, it suﬃces to showProposition 3.2. For simplicity, we put S = sn and T = tn. The twisted Alexander polynomial∆Kn,ρm(t) is given byS − T 2s− t2sS(ms−mST 2 + (1 +m2)(1− s2)StT 2m(1− s2)t2 +(1 +m2)(1− sSt2T 2)(η1 + η2)Hmt3β)+1− ST 21− st2sS((1 +m2)(1− s2)S −mSt+mstT 2m(1− s2)t3 −(1 +m2)(sS − t2T 2)(η1 + η2)Hmt3β)+1t6+ T 4 +(1− s2)(1 + t2)T 2η1HSt4β.By multiplying t6 and rearranging with respect to t , we obtain the formula of Theorem 3.1,when we useS − T 2s− t2 =Ssn−1∑i=0(t2s)i,ST 2 − 1st2 − 1 =n−1∑i=0(st2)i.4. Proof of Proposition 3.2Recall thatπ1(En) = ⟨a, b, x {axba(xb)−1}−1x = xb{axba(xb)−1}−1(axb)−1xb, x = {axba(xb)−1}n⟩= ⟨a, c (acac−1)n−1 = c(acac−1)−1(ac)−1c⟩.Then the twisted Alexander polynomial of Kn is given by∆Kn,ρm(t) =detΦ(∂∂a(acac−1)n−1 − ∂∂ac(acac−1)−1(ac)−1c)detΦ(c− 1) ,5whereΦ(∂∂a(acac−1)n−1 − ∂∂ac(acac−1)−1(ac)−1c)=n−1∑i=1t2(i−1)ρm({axba(xb)−1}i−1){ρm(1) + t2(n+1)ρm(axb)}+ t4n+1ρm(xbxba−1)(2)+t2n−1ρm(xb{axba(xb)−1}−1)+ t−3ρm(xb{axba(xb)−1}(axb)−1) .For simplicity, we putγ1 = sα−mβ , γ2 = msα− β , γ3 = m2s(sS2 + 1)α.By the aid of Mathematica, the first term of the right hand side of (2) is given byn−1∑i=1t2(i−1)(AXBA(XB)−1)i−1(E + t2(n+1)AXB)=⎛⎜⎜⎝(ST 2 − st2)(St2βT 2 +mα)mst2(st2 − 1)α −T 2(ST 2 − st2)(γ1η2 + (mα− β)γ3)m2s(s+ 1)S (st2 − 1)αβmC1α− St2T 2C2βmsS(sS2 + 1)t2(s− t2)(st2 − 1)αC3t4T 4 + C4t2T 4 + C5t6T 2 + C6t4T 2 + C7(s+ 1)S2t2(s− t2)(st2 − 1)γ3β⎞⎟⎟⎠ ,whereC1 = −t4s(s2 − 1)S − T 2{t2(S2 − s4)− s(S2 − s2)},C2 = −t2(t2 − 1)s(s+ 1)S + T 2{t2(S2 + s3) + s(S2 − s)},C3 = (s3 + S2)γ1η2 − {s3(msα+ β)− S2(mα− β)}γ3,C4 = −s(s+ S2)γ1η2 + s{s(msα+ β)− S2(mα− β)}γ3,C5 = −s(s+ 1)S{γ1η2 + (η1 + η2 − (1 +m2S2 − sS2)β)γ3},C6 = s(s+ 1)S{sαη2 −m(s+ 1)S2βγ2},C7 = s(s+ 1)S(st2 − 1)(St2 − sT 2)βγ3.Similarly, the second term of the right hand side of (2) is given byXBXBA−1 =⎛⎜⎜⎝S2D1γ3αmsD1D2 − (sS2 + 1)(sS2D1 +mγ3α)β2(s+ 1)γ3αβ2(s+ 1)D2(sS2 + 1)γ3αmsS2D1D2 + s(sS2 + 1)(m2sα2 − S2β2)D2S2(sS2 + 1)γ3αβ2−m⎞⎟⎟⎠ ,whereD1 = −(s+ 1)αγ2 +m(η1 + γ2 +mS2γ1)β,D2 = −αη2 +mS2(η1 +mS2γ1 + γ2)β,the third term of the right hand side of (2) is given byXB{AXBA(XB)−1}−1=⎛⎜⎜⎝SE1ms (sS2 + 1)αβ−Sγ1γ2mαβ(s+ 1)E2msS (sS2 + 1)2 αβE3mS (sS2 + 1)αβ⎞⎟⎟⎠ ,whereE1 = (s2 − 1)αγ2 +m(η1 +mS2γ1 − sγ2)β,E2 = (s− 1)αη2 +mS2(η1 +mS2γ1 − sγ2)β,E3 = −sαη2 +m(s+ 1)S2βγ2,and the fourth term of the right hand side of (2) is given byXB(AXBAXBA(XB)−1)−1 =⎛⎜⎜⎝mF3γ23β2F4m(s+ 1)γ3αβ2m(s2 − 1)F1F2S2(sS2 + 1)γ23β2mF5S2γ23β2⎞⎟⎟⎠ ,6 A. ASOwhereF1 = m(s+ 1)S2(η1 +mS2γ1)β − η2α,F2 = m(s+ 1)S2(sS2 + 1)β2 − sF1,F3 = −{mβ(η1 +mS2γ1) + sγ1γ2 − γ2α}F2 +ms(s+ 1)S2(sS2 + 1)γ1γ2β2,F4 = (s2 − 1){m(η1 +mS2γ1)β − γ2α}F2+γ3{mγ2α− (m2η1 + s2η2 +m3S2γ1 − s2(S2 − 1)γ2)β −msγ1γ2}α,F5 = (s− 1)(sF1 −mγ3α)F2 −m2S2(sS2 + 1)γ3αβ2.Therefore, the determinant of the right hand side of (2) is written as∑i,j Ui,jtiT jm3S2t6(s− t2)(st2 − 1)β2ι ,whereU0,0 = U4,0 = U6,0 = U2,4 = U10,4 = U6,8 = U8,8 = U12,8 = −m3sS2β2ι,U2,0 = U10,8 = m3(s2 + 1)S2β2ι,HU3,0 ≡ HU9,8 ≡ −m2(m2 + 1)sS2β(Hsβ − (s2 − 1)(η1 + η2))ι mod r0,U5,0 ≡ U7,8 ≡ m2(m2 + 1)sS2β2ι mod r0,HU1,2 ≡ HU11,6 ≡ m2(m2 + 1)(s− 1)sSβη2ι mod r0,HU2,2 = HU6,2 = HU8,2 = HU4,6 ≡ HU6,6 = HU10,6 ≡ m3(s2 − 1)sSβη1ι mod r1,HU3,2 ≡ HU9,6 ≡ m2(m2 + 1)(s− 1)Sβ{HsS2β − s(sS2 + 1)η1 − (s2S2 + s2 + 1)η2}ι mod r0,H2U4,2 ≡ H2U8,6≡ m(s− 1)sS{H2m3αβ +H(m2 + 1)(m2s+ s+ 1)βη2 − (m2 + 1)2(s2 − 1)η2(η1 + η2)}ιmod r0,HU5,2 ≡ HU7,6 ≡ −m2(m2 + 1)(s− 1)sSβη2ι mod r0,HU7,2 ≡ HU5,6 ≡ m2(m2 + 1)(s− 1)sSβ(HS2β − (sS2 + 1)η1 − (sS2 − 1)η2)ι mod r1,H2U3,4 ≡ H2U9,4 ≡ −m2(m2 + 1)(s− 1)2s(s+ 1)η1η2ι mod r0,H2U4,4 = H2U8,4≡ m{H2m2(s2 − s+ 1)S2β2 + (m2 + 1)2(s− 1)2sη2(−HS2β + (sS2 + 1)η1 + sS2η2)}ιmod r1,H2U5,4 ≡ H2U7,4≡ −(m2 + 1)(s− 1)s{(s− 1)η2(m3Hα+ (m2 + 1)η2) +m2S2Hβ(Hβ − (s+ 1)(η1 + η2))}ιmod r0,H2U6,4 ≡ −2ms(HmSβ − (m2 + 1)(s− 1)η2)(HmSβ + (m2 + 1)(s− 1)η2)ι mod r0,where we put ι = m2s2(s+ 1)S(sS2 + 1)3α3β, and the other Ui,j ’s are 0.On the other hand, by the aid of Mathematica,detΦ(c− 1) = det(t2n+1ρm(xb)−(1 00 1))=mSHβ +mSHt2T 4β − (m2 + 1)(s− 1)tT 2η2mSHβ− (S2 − 1)tT 2mS(sS2 + 1)Hαβr1=mSHβ +mSHt2T 4β − (m2 + 1)(s− 1)tT 2η2mSHβ.Consequently, we have∆Kn,ρm(t) =∑i,j Vi,jtiT jHm2St6(s− t2)(st2 − 1)β ,(3)7whereV0,0 = V4,0 = V6,0 = V4,4 = V6,4 = V10,4 = −Hm2sSβ,V2,0 = V8,4 = Hm2(s2 + 1)Sβ,V3,0 = V7,4 = m(m2 + 1)sS{(s2 − 1)(η1 + η2)−Hsβ},V5,0 = V5,4 = Hm(m2 + 1)sSβ,V2,2 = V8,2 = m2s(s2 − 1)η1,V3,2 = V7,2 = m(m2 + 1)(s− 1)s{(s+ 1)η1 + η2}V4,2 = V6,2 = (s− 1)s{(m2 + 1)η2 +Hm3α},V5,2 = −2m(m2 + 1)(s− 1)sη2,and the other Vi,j ’s are 0. By the aid of Mathematica, the diﬀerence between the right hand sideof (3) and the formula in Proposition 3.2 is equal tosζ1 + tζ2 − 2t2ζ1 + t3ζ2 + st4ζ1Hm2St3(s+ 1)(s− t2)(st2 − 1)βT2,whereζ1 = m(m2 + 1)s(s+ 1)(HS2β − s(S2 − 1)η1 − (sS2 − 1)η2),ζ2 = Hm2s(mα−ms2α+ sβ + S2β)− (s2 − 1)(m2η1 +m2s3η1 + sη2 +m2sη2).Note that ζ1 = 0 by the definition of H, η1 and η2 and thatζ2 = m{(m2(s2 − s+ 1)− s)(s3S2 + 1)−Hs(s− 1)}r0 = 0.This completes the proof of Proposition 3.2.References[CF] R. H. Crowell and R. H. Fox, Introduction to knot theory, Springer-Verlag(1963).[CGLS] M. Culler, C. M. Gordon, J. Luecke and P. B. Shalen, Dehn Surgery on Knots, Ann. Math, 125 (1987),237–300.[DFJ] N. Dunfield, S. Friedl and N. Jackson, Twisted Alexander polynomials of hyperbolic knots, Exp. Math. 21(2012), 329–352.[GKM] H. Goda, T. Kitano and T. Morifuji, Twisted Alexander polynomials (in Japanese), Sugaku-Memoirs 5(2005).[GM] S. Garoufalidis and T. W. Mattman, The A-polynomial of the (−2, 3, 3 + 2n) pretzel knots, New York J.Math. 17 (2011) 269–279.[KM] T. Kitano and T. Morifuji, Divisibility of twisted Alexander polynomials and fibered knots, Ann. ScuolaNorm. Sup. Pisa Cl. Sci. (5) Vol. IV (2005), 179–186.[L] X. S. Lin, Representations of knot groups and twisted Alexander polynomials, Acta Math. Sin. 17 (2001),361–380.[Ma] Thomas W. Mattman, The Culler-Shalem seminorms of the (−2, 3, n) pretzel knot, J. Knot Theory Ram-ifications, 11 (2002), 1251.[Mo1] T. Morifuji, Twisted Alexander polynomials of twist knots for nonabelian representations, Bull. Sci. Math.132 (2008), no. 5, 439–453.[Mo2] T. Morifuji, Representations of knot groups into SL(2,C) and twisted Alexander polynomials, Handbookof Group Actions (Vol. I), Advanced Lectures in Mathematics 31 (2015), 527–576.[T1] A. Tran, Twisted Alexander polynomials of genus one two-bridge knots, preprint 2015, arXiv:1506.05035.[T2] A. Tran, Adjoint twisted Alexander polynomials of genus one two-bridge knots, J. Knot Theory Ramifica-tions 25 (2016), 1650065.[TY] N. Tamura and Y. Yokota, A formula for the A-polynomials of (−2, 3, 2n + 1)-pretzel knots, Tokyo J.Math. 27 (2004), 263–273.[W] M. Wada, Twisted Alexander polynomial for finitely presentable groups, Topology 33 (1994), 241–256.E-mail address: aso-airi@ed.tmu.ac.jp

Twisted Alexander Polynomials of $(-2,3,2n+1)$ -Pretzel Knots

Abstract

Similar works

Full text

Available Versions

Institutional Repositories DataBase (IRDB)

Tokyo Metropolitan University Institutional Repository Miyako-Dori / 首都大学東京機関リポジトリ

Twisted Alexander Polynomials of (−2,3,2n+1)(-2,3,2n+1)(−2,3,2n+1)-Pretzel Knots

Abstract

Similar works

Full text

Available Versions

Institutional Repositories DataBase (IRDB)

Tokyo Metropolitan University Institutional Repository Miyako-Dori / 首都大学東京機関リポジトリ

Twisted Alexander Polynomials of $(-2,3,2n+1)$ -Pretzel Knots