Lifts of holonomy representations and the volume of a link. complement. Hiroshi Goda. (Tokyo University of Agriculture and Technology)

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1 Lifts of holonomy representations and the volume of a link complement Hiroshi Goda (Tokyo University of Agriculture and Technology) Intelligence of Low-dimensional Topology at RIMS May 26,

2 $ 1. Introduction Mostow rigidity a complete hyperbolic 3-mfd M is geometrically determined by its fundamental group π 1 (M). the hyperbolic volume vol(m) should theoretically be computable from a presentation of π 1 (M). π 1 (M) vol(m) A n H 1 (M) Alexander polynomial 2

3 $ 2. Background Reidemeister torsion twisted by the ajoint representation associated with an SL(2, C)-rep. [Porti] A relation between the non-acyclic R-torsion and a zero of the acyclic R-torsion (Yamaguchi [Y, 08]) (knot case) Twisted Alexander invariants and non-abelian R-torsion (Dubois & Yamaguchi [DY]) (link case) 3

4 A volume formula of a closed hyperbolic 3-manifold using the Ray-Singer analytic torsion (by Müller [ 12]) [Müller 93] A volume formula of a cusped hyperbolic 3-manifold using R-torsion (by Menal-Ferrer & Porti [MFP, 14]) [Y] A volume formula of a hyperbolic knot complement using the twisted Alexander invariant [G] [DY] Link case 4

5 3. The Alexander polynomial K: a knot in S 3 E(K) := S 3 N(K) π 1 (E(K)) := G(K) = x 1,..., x k r 1,..., r k 1 : Wirtinger pre. α : G(K) H 1 (E(K); Z) = Z = t : epimorphism µ(meridian) t That is, α(x 1 ) = α(x 2 ) = = α(x k ) = t. α induces the ring homomorphism between group rings over Z: α : ZG(K) Z[t ±1 ] F k = x 1, x 2,, x k : the free group of rank k ϕ : F k G(K): epi. extend by linearity 5 ϕ : ZFk ZG

6 Φ : α ϕ : ZF k Z[t ±1 ]: ring homo. ( ri ) M := Φ ( M x k 1,k (Z[t ±1 ])) : the Alexander matrix, j where x i x j = δ ij, Example. x 1 i = δ x ij x 1 (uv) i, = u + u v j x j x j x j (Fox s free differential calculus) x (xy 1 x 1 yxy 1 xyx 1 y 1 ) = x x + x x (y 1 x 1 yxy 1 xyx 1 y 1 ) = 1 + x ( y 1 x = 1 + xy 1( x 1 + y 1 x (x 1 yxy 1 xyx 1 y 1 ) ) + x 1 x (yxy 1 xyx 1 y 1 ) ) x = 1 xy 1 x 1 + xy 1 x 1 x (yxy 1 xyx 1 y 1 ) = = 6

7 1 xy 1 x 1 + xy 1 x 1 y + xy 1 x 1 yxy 1 xy 1 x 1 yxy 1 xyx 1 α 1 t t = 1 t + 3 t M l : the sub matrix of M deleting l-column Definition. The Alexander polynomial : K (t) = det M l. Example. K = 4 1 (figure 8 knot) G(K) = x, y xy 1 x 1 yxy 1 xyx 1 y 1 41 K (t) = det ( 1 t + 3 t) = 1 t + 3 t 7

8 4. The twisted Alexander invariant ρ : G(K) SL(n, C): rep. extend by linearity ρ : ZG(K) M(n, C) ρ α : ZG(K) M(n, C[t ±1 ]): tensor prod. ring homo. Φ : ( ρ α) ϕ : ZF k M(n, C[t ±1 ]): ring homo. ( ri ) M := Φ ( M x n(k 1),nk (C[t ±1 ])) : j The Alexander matrix associated with ρ M l : the sub matrix of M deleting l -column 8

9 Definition. The twisted Alexander invariant : K,ρ (t) = det M l det Φ(x l 1) Example. K = 4 1 (figure 8 knot) ρ : G(K) SL(2, C) s.t. ρ(x) = 1 1 =: X, ρ(y) = =: Y, where u 2 +u+1 = 0. u 1 r x = 1 xy 1 x 1 + xy 1 x 1 y + xy 1 x 1 yxy 1 xy 1 x 1 yxy 1 xyx 1 ( r ) Φ = I 1 x t XY 1 X 1 + XY 1 X 1 Y + XY 1 X 1 Y XY 1 txy 1 X 1 Y XY 1 XY X 1 Φ(y 1) = ty I ( ) r det Φ x K,ρ (t) = det Φ(y 1) = 1/t2 (t 1) 2 (t 2 4t + 1). = t 2 4t + 1 (t 1) 2 9

10 K: a hyperbolic knot in the 3-sphere. ρ ± n : G(K) Hol. PSL(2, C) ± SL(2, C) irr. SL(n, C) σn K,ρ ± n (t): the twisted Alexander invariant Set A ± K,ρ ± (t) K,2k (t) := 2k K,ρ ± (t) and A K,2k+1(t) := K,ρ 2k+1 (t) K,ρ3 (t) 2 ( K,ρ + 2k+1 (t) = K,ρ 2k+1 (t)) Theorem [G]. lim k log A K,2k+1 (1) (2k + 1) 2 = lim k log A ± K,2k (1) (2k) 2 = Vol(K) 4π 10

11 $ 5. Lifts of the holonomy representation M: an oriented, complete, hyperbolic 3-manifolds of finite volume with M = T1 2 T b 2. Hol M : π 1 (M, p) Isom + H 3 = PSL(2, C)(= SL(2, C)/{±1}) η : a lift of Hol M to SL(2, C). Thus we have a map: Hol (M,η) : π 1 (M, p) SL(2, C). Definition. (1) η is positive on T 2 i if for all γ π 1(T 2 i, p), we have trace Hol (M,η) (γ) = +2. (2) A lift η is acyclic if η is non-positive on each T 2 i. 11

12 Proposition [MFP]. M γ : the mfd obtained by a Dehn filling along γ. A lift η of Hol M to SL(2, C) extends to a lift Hol Mγ to SL(2, C) trace Hol (M,η) (γ) = 2. Proposition [MFP]. Assume that, for each boundary component Tl 2, the map ι : H 1 (Tl 2 ( ) ; Z/2Z) H 1(M; Z/2Z) induced by the inclusion ι has non trivial kernel, then all lifts of Hol M are non-positive on each T 2 l, i.e., all lifts are acyclic. Example. a knot complement. ( a Seifert surface), in general any 3-manifold with M = T 2 (i.e., b = 1) [eg. Hempel 6.8]. 12

13 $ 6. Results of Dubois & Yamaguchi M: an oriented, complete, hyperbolic 3-manifolds of finite volume with M = T1 2 T b 2. They investigated a relation between the R-tosion and the twisted Alexander invariants under the following condition: the hom. i : H 1 ( M; Z) H 1 (M; Z) is onto ( ) and its restriction (i T 2) has rank one for all l. l ( ) = ( ) = ( ) M is the complement of a algebraically split link L ( ) (homologically trivial link) in a homology 3-sphere. i.e., L = K 1 K b such that lk(k i, K j ) = 0 for all i, j. 13

14 Example. (algebraically split link) Every knot. Figure 8 (4 ) 1 Whitehead link Borromean link 14

15 Suppose φ : π 1 (M) Z n epi. & (a (l) 1,..., a(l) n ) Z n >0 such that φ(π 1 (T 2 l )) = ta(l) 1 1 t a(l) n n for all Tl 2. Theorem [DY]. Suppose ( ). Then, we have: lim t 1,...,t n 1 M,φ Ad ρ2 (t 1,..., t n ) bl=1 (t a(l) 1 1 t a(l) n n 1) where λ l is a longitudinal basis on T 2 l. = ( 1) b T(M; Ad ρ 2 ; {λ l }), 15

16 $ 7. A volume formula of a link complement Proposition.[MFP 12] Let M be a hyperbolic mfd with b cusps. Then, dim C H 0 (M; ρ n ) = 0, dim C H 1 (M; ρ n ) = a, dim C H 2 (M; ρ n ) = a, where a = b if n is odd, and a = of cusps for which the lift of the holonomy is positive if n is even, i.e., η is acyclic dim C H i (M; ρ n ) = 0 when n is even. 16

17 Proposition.[MFP 14] Let G l (< π 1 (M)) be some fixed realization of π 1 (T 2 l ) as a subgroup of π 1(M). For each T 2 l choose a non-trivial cycle θ l H 1 (T 2 l ; Z), and non-trivial vector w l V n fixed by ρ n (G l ). 1. A basis of H 1 (M; ρ n ) is given by i l ([w l θ l ]), (l = 1,..., b). 2. A basis of H 2 (M; ρ n ) is given by i l ([w l T 2 l ]), (l = 1,..., b). Set T 2k+1 (M, η) := T(M; ρ 2k+1; {θ l }), T(M; ρ 3 ; {θ l }) T 2k (M, η) := T(M; ρ 2k; {θ l }) T(M; ρ 2 ; {θ l }). 17

18 Theorem [MFP 14]. (1) For any η, lim k (2) If η is acyclic, then lim k log T 2k+1 (M, η) (2k + 1) 2 = Vol(M) 4π log T 2k (M, η) (2k) 2 = Vol(M) 4π Applying the idea of the proof of Dubois & Yamaguchi s Theorem, we have: Theorem[G]. Let L be an algebraically split link. Then, lim k log A L,2k+1 (1,..., 1) (2k + 1) 2 = lim k log A L,2k (1,..., 1) (2k) 2 = Vol(L) 4π 18

19 8. The irreducible representation σ n of SL(2,C) V n the vector space of 2-variables homogeneous polynomials on C with degree n 1, i.e., V n = {a 1 x n 1 + a 2 x n 2 y + + a n y n 1 a 1,..., a n C} = span C x n 1, x n 2 y, x n 3 y 2,..., xy n 2, y n 1 The action of A SL(2,C) is expressed as: A p x = p ( A 1 x ) for p y y x V n y (V n, σ n ) the rep. given by this action of SL(2, C) where σ n means the homomorphism from SL(2, C) to GL(V n ). 19

20 (The fact is, σ n (A) SL(n, C).) Theorem. Every irreducible n-dim. representation of SL(2, C) is unique and equivalent to (V n, σ n ). 20

21 Example. Set X = 1 1 ( SL(2, C) ). X 1 = X 1 x = y p ( X 1 x ) = p x y y y 1 1 x = x y 0 1 y y (x y) 2 = 1x 2 2xy + 1y 2, (x y)y = 1xy y 2, y 2 = 1y 2 T σ 3 (X) =

22 σ 4 (X) = In the same way, we have: (x y) 3 = x 3 3x 2 y + 3xy 2 y 3, (x y) 2 y = x 2 y 2xy 2 +y 3, (x y)y 2 = xy 2 y 3, y 3 = y T 22

23 σ n (X) = 1 n 1 C 1 n 1 C 2 ( 1) n 2 n 1C n 2 ( 1) n n 2 C 1 ( 1) n 3 n 2C n 3 ( 1) n ( 1) n 4 n 3C n 4 ( 1) n T Example. K = 4 1 G(K) = x, y xy 1 x 1 yxy 1 xyx 1 y 1. ρ(x) = 1 1 = X, ρ(y) = 1 0 = Y 0 1 u 1 : holonomy rep. where (u 2 + u + 1 = 0). 23

24 σ n (Y ) = u u 2 2u u n 2 n 2C 1 u n u n 1 n 1C 1 u n 2 n 1C 2 u n 3 n 1C n 2 u 1 T 24

25 9. Experimentation using a computer K = 4 1, Vol(K) = Vol(K) = 4π lim k log A K,2k (1) (2k) 2 K,ρ2 (t) = 1/t2 (t 1) 2 (t 2 4t + 1) (t 1) 2. = t 2 4t + 1 K,ρ4 (t) = 1 t 4(t2 4t + 1) 2. = (t 2 4t + 1) 2 A K,4 (t) = K,ρ 4 (t) K,ρ2 (t) = (t2 4t + 1) 2 t 2 4t + 1 = t2 4t + 1 4π log A K,4(1) 4 2 = π log

26 Vol(K) = 4π lim k log A K,2k+1 (1) (2k + 1) 2 K,ρ3 (t) = 1/t 3 (t 1)(t 2 5t + 1). = (t 1)(t 2 5t + 1) K,ρ5 (t) = 1 t 5(t 1)(t4 9t t 2 9t + 1). = (t 1)(t 4 9t t 2 9t + 1) A K,5 (t) = K,ρ 5 (t) K,ρ3 (t) = t4 9t t 2 9t + 1 t 2 5t + 1 4π log A K,5(1) 4π log =

27 n(even) 4π log A+ K,n (1) 4π log A K,n (1) n 2 n 2 n(odd) 4π log A K,n(1) n

28 10. On Evaluation by Root of unity (in progress) Theorem [DY 12]. M q : q-fold cyclic covering of M, and s = t q Corollary. ˆρ ˆα M q (s) = lim k q 1 k=0 log A K,2k ( 1) (2k) 2 ( ) Suppose t = 1, q = 2, then s = 1. ˆρ ˆα ρ α M (e2πk 1/q t) = Vol(K) 4π. M (1) = ρ α (1) ρ α ( 1) 2 M M 2Vol(M) Vol(M) Vol(M) 28

29 - i 3-rd root of unity OK 4-th root of unity OK 5-th root of unity??? 29

30 Thank you very much for your attention! 30