Reidemeister torsion on the variety of characters

Transcription

1 Reidemeister torsion on the variety of characters Joan Porti Universitat Autònoma de Barcelona October 28, 2016 Topology and Geometry of Low-dimensional Manifolds Nara Women s University

2 Overview Goal M oriented hyperbolic 3-manifold, vol(m) <, one cusp ρ: π 1 M SL 2 (C) a representation The Reidemeister torsion τ(m, ρ) is invariant by conjugation of ρ, and defines a function X (M) = hom(π 1 M, SL 2 (C)// SL 2 (C) C that we want to study Plan of the talk: 1. Tools: Reidemeister torsion, variety of characters, aciclicity 2. The torsion function X (M) C 3. Compose the representation with Sym n : SL 2 (C) SL n+1 (C)

3 L(p, q) = S 3 / t. Torsion: recall Lens spaces (z 1, z 2 ) t (e 2πi p z 1, e 2πi p q z 2 ). z2 te2 e1 te1 z1 The lens ẽ 3 = {0 θ 1 2π p } is a fundamental domain for t ẽ 3 = (t 1)ẽ 2 ẽ 2 = (1 + t + + t p 1 )ẽ 1 ẽ 1 = (t r 1)ẽ 0 t ξ C, ξ p = 1 ξ 1 e2 r q 1 mod p ẽ 3 = (ξ 1)ẽ 2 ẽ 2 = 0 ẽ 1 = (ξ r 1)ẽ 0 H (L(p, q), ξ) = 0 Def: (Reidemeister) 1 τ(l(p, q), ξ) := (ξ 1)(ξ r 1) 1 {τ(l(p, q), ξ)} ξ p =1 ξ 1 = {τ(l(p, q ), ξ)} ξ p =1 ξ 1 q = ±q ±1 mod p

4 Torsion of a CW-complex K compact CW-complex, ρ: π 1 K SL n (C). Def: C (K, ρ) := C n ρ π1 K C CW ( K, Z) {ej i} } j i-cells of K {v k } k basis for C n c i = {v k ẽj i} j,k C-basis for C i (K, ρ) i : C i+1 (K, ρ) C i (K, ρ)). If H (K, ρ) = 0, divide each basis into two c i = c i c i, so that card(c i+1 ) = card(c i ) and minor( i, c i+1, c i ) 0 Notice that c n = c n and c 0 = c Def: τ(k, ρ) := n 1 i=0 0. minor( i, c i+1, c i )( 1)i+1 C /{±1} τ(k, ρ) is a combinatorial invariant (by cellular homeos and subdivision) and invariant of the conjugacy class of ρ I use the opposite convention from yesterday 1/τ(K, ρ)!!

5 Variety of representations Algebraic structure: π 1 K = γ 1,..., γ n (r i ) i I finitely generated. hom(π 1 K, SL 2 (C)) SL 2 (C) (n) SL 2 (C) C 4n ρ (ρ(γ 1 ),..., ρ(γ n )) The (r i ) i I yield polynomial equations. {e i j } j i-cells of K {v 1, v 2 } basis for C 2 } c i = {v k ẽj i} j,k C-basis for C i (K, ρ) The coefficients of in the basis c i are polynomial on ρ, thus τ(k, ρ) = n 1 minor( i, c i+1, c i )( 1)i+1 C /{±1} i=0 is a rational function hom(π 1 K, SL 2 (C)) C. If χ(k) is even, then there is no sign indeterminacy, (as dim C 2 is also even, the orderings of {v 1, v 2 } and of the cells e i j do not affect the determinant)

6 Assume Def: Variety of characters M oriented hyperbolic 3-manifold, vol(m) <, with 1 cusp. X (M) = hom(π 1 M, SL 2 (C))// SL 2 (C) Any polynomial/rational function hom(π 1 M, SL 2 (C)) C invariant by conjugation induces a function X (M) C. Distinguished component X 0 (M): component of X (M) that contains the lift of the holonomy π 1 M Isom + (H 3 ) = PSL 2 (C) = SL 2 (C)/{± Id} Such a lift exits (Culler, Thurston) {lifts of hol to SL 2 (C)} {spin structures on M} because PSL 2 (C) = Frame(H 3 ) and SL 2 (C) = Spin(H 3 ) Since M has one cusp, X 0 (M) is a C-curve. (it contains lift of holonomy of Dehn fillings) uestion Before defining X 0(M) C [ρ] τ(m, ρ), does H (M, ρ) = 0?

7 Assume Acyclicity M oriented hyperbolic 3-manifold, vol(m) <, with 1 cusp. Lemma ρ 0 :π 1 M SL 2 (C) lift of hol. H (M, ρ 0 ) = H (M, ρ 0 ) = 0 Proof (Sketch) M = M M, M = T 2 L 2 forms in Ω (M, E ρ ) are exact (Ragunathan, Garland, Matsushima-Murakami, s): H 1 (M, M, ρ) 0 H 1 (M, ρ) H 1 ( M, ρ) Up to conj. ρ 0 (π 1 M) ± ( ) but ρ 0(π 1 M) ( ) H 0 ( M, ρ 0 ) = (C 2 ) ρ 0(π 1 M) = 0 H ( M, ρ 0 ) = 0 (by Poincaré duality+χ(t 2 ) = 0) H 1 (M, ρ 0 ) = 0. (by the exact seq.) Cor: In addition H 0 (M, ρ 0 ) = (C 2 ) ρ 0(π 1 M) = 0. + Euler characteristic & duality homology/cohomology. Acyclicity holds in X 0 (M) except for a finite set, by semicontinuity of (co-)homology.

8 The torsion function Assume M or. hyp 3-manifold, vol(m) <, with 1 cusp. { τ(m, ρ) if H (M, ρ) = 0 Def: T M (ρ) = 0 if H (M, ρ) 0 and [ρ] nontrivial The sign is well defined (since dim C 2 and χ(m) are even, ordering of cells and of basis for C 2 do not change the sign) emark: [ρ] X 0 (M) nontrivial, H 0 (M, ρ) = (C 2 ) ρ(π 1M) = 0 Hence T M : hom(π 1 M, SL 2 (C)) {ρ tr ρ 2} C is algebraic Since M collapses to a 2-dim CW-complex. and H 0 (M, ρ) = H 0 (M, ρ) = 0 for ρ nontrivial, denominators in the def of torsion do not vanish. it defines an algebraic function T M : X 0 (M) {trivial} C Remark: When β 1 (M) = 1, characters in X 0 (M) are nontrivial. hence T M : X 0 (M) C polynomial

9 Example M = S 3 fig-8 knot. π 1 (M) = a, b, m mam 1 = ab, mbm 1 = bab X 0 (M) = {(x, y) C 2 x 2 x 1 = (x 1)y 2 } x = tr a, y = tr m = tr ma = tr mb, tr b = x/(x 1) = y 2 1 x Kitano (1994): T M = 2 2y = 2 2 tr m Can define a twisted Alexander polynomial M ([ρ], t), with M ([ρ], 1) = T M ([ρ]): M (, t) = t 2 2t y + 1 Conj. (Dunfield, Friedl, and Jackson 2011) For M = S 3 hyp knot, deg M ([ρ 0 ], t) = 2 genus(k) and M ([ρ 0 ], t) is monic iff M is fibered.

10 Branched coverings on the Fig-8 knot M n S 3 n-fold cyclic branched covering, branched over the fig-eight knot K, Σ n = K M n lift of branching locus. τ(m n, ρ n ) = τ(m n Σ n, ρ n )τ(σ n, ρ n ) 1 = τ(m n Σ n, ρ n ) 2(1 cosh(λ(σ n)) (ρ n : π 1 M n SL 2 (C) lift of holonomy, λ= complex length ) Recall that M (y, t) = t 2 2yt + 1. Fox formula: τ(m n Σ n, ρ n ) = n 1 k=0 Hence, since λ(σ n ) = 3π/n + O(1/n 3 ): log τ(m n, ρ n ) lim n n = 1 2π z =1 1 = lim n n log (±2, z) = 1 2π (±2 cos(π/n), e 2πi k n ) n 1 log (±2, e 2πi k n ) k=0 z =1 log z 2 4z+1 = log(2 + 3)

11 Assume Dehn fillings M oriented hyperbolic 3-manifold, vol(m) <, with 1 cusp. Choose a frame for H 1 (M, Z) = Z 2. M p/q Dehn filling (with filling meridian ±(p, q) Z 2 ). It is hyperbolic for p 2 + q 2 large, and a lift of its holonomy [ρ p/q π1 M] [ρ 0 ], holonomy of M. τ(m p/q, ρ p/q ) is a topological invariant of the spin mfld M p/q Thm: τ(m p/q ) = T M (ρ p/q M ) 1 2(1 cosh(λ(γ p/q ))) where λ(γ p/q ) C complex length of γ p/q soul of filling torus Proof: Mayer-Vietoris to the pair (M, D 2 S 1 ) & τ(t 2 ) = 1 Cor: τ(m p/q, ρ p/q ) is dense in [ 1 4 T M(ρ 0 ), + ) (Because Re(λ(γ p/q )) 0 as p 2 + q 2 but Im(λ(γ p/q )) is dense in R/2πZ) τ(m p/q, ρ p/q ) distinguishes spin structures (for M = S 3 fig 8, T M (y) = 2 2y and T M (2) T M ( 2)).

12 Representations of SL 2 (C) Sym n : SL 2 (C) SL n+1 (C) irreducible C n+1 = Sym n (C 2 )={homog. polynomials on C 2 of deg n} If C 2 = v 1, v 2, then Sym n (C 2 ) = v n 1, v n 1 1 v 2,, v n 2. Aim Want to define T n+1 M ([ρ]) = τ(m, Symn ρ) For ρ 0 lift of holonomy, H (M, Sym n ρ 0 ) = 0 iff n + 1 even. Can define T n+1 M. For n + 1 odd, H i (M, Sym n ρ 0 )) = C for i = 1, 2, and there are natural choices of basis for H i (M, Sym n ρ 0 ), depending on peripheral elements 1 γ π 1 T 2. { T n+1 M : X 0(M) C for n + 1 even T n+1 M,γ : X 0(M) C for n + 1 odd, 1 γ π 1 T 2 Can study its domain, Dehn fillings, twisted polynomials, etc...

13 Representations of SL 2 (C) (continued) T n+1 {M ([ρ]) = τ(m, Symn ρ), with Sym n : SL 2 (C) SL n+1 (C) T n+1 M : X 0(M) C for n + 1 even T n+1 M,γ : X 0(M) C for n + 1 odd, 1 γ π 1 T 2 M p/q Dehn filing { dense in [ 1 τ(m p/q, Sym n ρ p/q ) (ρ 0), ) for n + 1 even goes to as p 2 + q 2 for n + 1 odd (Menal-Ferrer-P, based on Müller s work) log T lim 2k+2 M ([ρ 0]) log T = lim 2k+1 M,γ ([ρ 0]) = 1 k (2k+2) 2 k (2k+1) 2 4π vol(m) For n + 1 = 3, Sym 2 = Ad (1/(yesterday s torsion)) T n+1 2 n+1 M Conj K S 3 hyperbolic knot. K N = Kashaev invariant, N N K N = e CS+i V 1 2π i τ N 3/2 (1 + O( 1 N )) where CS + i V = (Chern-Simons + i Volume)(S 3 K) τ = τ(s 3 K, Ad hol, meridian)

14 Examples of torsion for higher representations M = S 3 fig-8, y = tr m. n + 1 even: T 2 M = 2 (1 y) T 4 M = ( y 2 2 y 2 ) 2 T 6 M = 2(y 1)( y 8 + 2y 7 13y 6 20y y y 3 33y 2 18y 18 ) T 8 M = (y ( 1)2 2 y 7 4 y 6 21 y y y y 2 18 y 6 ) 2 T 10 M = 2 (y 1) ( y y y y y 8 y y y y 4 74 y y y 6 ) 2. n + 1 odd: l =longitude, m =meridian T 3 M,l = ±(5 2y 2 ), T 3 M,m = ± 1 2 (y 2 1)(y 2 5) = ±(2 x + 1 y 2 )/2 T 5 M,l = ±4(1 6y 2 + y 4 ), Thanks for your attention