Hyperbolically twisted Alexander polynomials of knots Nathan M. - PowerPoint PPT Presentation

Hyperbolically twisted Alexander polynomials of knots Nathan M. Dunfield University of Illinois Stefan Friedl Köln Nicholas Jackson Warwick Gauge Theory Seminar, April 6, 2012 This talk available at http://dunfield.info/ Math blog: http://ldtopology.wordpress.com/

Hyperbolically twisted Alexander Setup: polynomials of knots • Knot: K = S 1 ֓ S 3 ◦ (K) • Exterior: M = S 3 − N Nathan M. Dunfield University of Illinois Stefan Friedl Köln Nicholas Jackson A basic and fundamental invariant of K its Warwick Alexander polynomial (1923): ∆ K (t) = ∆ M (t) ∈ Z [t, t − 1 ] Gauge Theory Seminar, April 6, 2012 This talk available at http://dunfield.info/ Math blog: http://ldtopology.wordpress.com/

Universal cyclic cover: corresponds to the kernel of the unique epimorphism π 1 (M) → Z . Setup: • Knot: K = S 1 ֓ S 3 ◦ (K) • Exterior: M = S 3 − N � M M S A basic and fundamental invariant of K its Alexander polynomial (1923): ∆ K (t) = ∆ M (t) ∈ Z [t, t − 1 ] S

Universal cyclic cover: corresponds to the kernel of the unique epimorphism π 1 (M) → Z . A M = H 1 ( � M ; Q ) is a module over Λ = Q [t ± 1 ] , where � t � is the covering group. � M As Λ is a PID, �� n � M � Λ A M = p k (t) k = 0 Define S � n p k (t) ∈ Q [t, t − 1 ] ∆ M (t) = k = 0 Figure-8 knot: ∆ M = t − 3 + t − 1 S

A M = H 1 ( � M ; Q ) is a module over Λ = Q [t ± 1 ] , Genus: where � t � is the covering group. � � g = min genus of S with ∂S = K As Λ is a PID, � � = min genus of S gen. H 2 (M, ∂M ; Z ) �� n � � Λ A M = p k (t) k = 0 Fundamental fact: Define 2 g ≥ deg ( ∆ M ) � n p k (t) ∈ Q [t, t − 1 ] ∆ M (t) = k = 0 Note deg ( ∆ M ) = dim Q (A M ) . Proof: As A M is generated by H 1 (S ; Q ) ≅ Q 2 g , the inequality fol- Figure-8 knot: lows. ∆ M = t − 3 + t − 1

∆ (t) determines g for all alternating knots and all fibered knots. Kinoshita-Terasaka knot: ∆ (t) = 1 but g = 2 . Genus: � � g = min genus of S with ∂S = K � � = min genus of S gen. H 2 (M, ∂M ; Z ) Fundamental fact: 2 g ≥ deg ( ∆ M ) Note deg ( ∆ M ) = dim Q (A M ) . Proof: As A M is generated by H 1 (S ; Q ) ≅ Q 2 g , the inequality fol- Idea: Improve ∆ M by looking at H 1 ( � M ; V ρ ) for lows. the system of local coefficients coming from a representation α : π 1 (M) → GL (V) . [Lin 1990; Wada 1994,...] Twisted Alexander polynomial: τ M,α ∈ F [t ± 1 ]

∆ (t) determines g for all alternating knots and all fibered knots. Technically, it’s best to define τ M,α as a torsion, a la Reidemeister/Milnor/Turaev. Kinoshita-Terasaka knot: ∆ (t) = 1 but g = 2 . Genus bound: When α is irreducible and non- trivial: 1 2 g − 1 ≥ dim V deg (τ M,α ) ( ⋆ ) Proof: deg (τ M,α ) = dim H 1 ( � M ; V α ) � � � χ(S) � ≤ dim H 1 (S ; V α ) = ( dim V) · Thm (Friedl-Vidussi, using Agol and Wise) If M is hyperbolic, then there exists some α where (⋆) is sharp. Idea: Improve ∆ M by looking at H 1 ( � M ; V ρ ) for the system of local coefficients coming from a Idea: By Wise, π 1 (M) is virtually special, hence representation α : π 1 (M) → GL (V) . [Lin 1990; RFRS. By Agol, there exists a finite cover of M Wada 1994,...] where the lift of S is a limit of fiberations. Use α associated to this cover. Twisted Alexander polynomial: τ M,α ∈ F [t ± 1 ]

Technically, it’s best to define τ M,α as a torsion, a la Reidemeister/Milnor/Turaev. Assumption: M is hyperbolic, i.e. ◦ = H 3 � for a lattice Γ ≤ Isom + H 3 M Γ Genus bound: When α is irreducible and non- trivial: Thus have a faithful representation 1 2 g − 1 ≥ dim V deg (τ M,α ) ( ⋆ ) where V = C 2 . α : π 1 (M) → SL 2 C ≤ GL (V) Proof: Hyperbolic Alexander polynomial: deg (τ M,α ) = dim H 1 ( � M ; V α ) � � � t ± 1 � � χ(S) � coming from H 1 ( � ≤ dim H 1 (S ; V α ) = ( dim V) · M ; V α ) . τ M (t) ∈ C Examples: Thm (Friedl-Vidussi, using Agol and Wise) • Figure-8: τ M = t − 4 + t − 1 If M is hyperbolic, then there exists some α where • Kinoshita-Terasaka: (⋆) is sharp. τ M ≈ ( 4 . 417926 + 0 . 376029 i)(t 3 + t − 3 ) − ( 22 . 941644 + 4 . 845091 i)(t 2 + t − 2 ) Idea: By Wise, π 1 (M) is virtually special, hence RFRS. By Agol, there exists a finite cover of M + ( 61 . 964430 + 24 . 097441 i)(t + t − 1 ) where the lift of S is a limit of fiberations. Use α − ( − 82 . 695420 + 43 . 485388 i) associated to this cover.

Assumption: M is hyperbolic, i.e. ◦ = H 3 � for a lattice Γ ≤ Isom + H 3 Basic Properties: M Γ • τ M is an unambiguous element of C [t ± 1 ] Thus have a faithful representation with τ M (t) = τ M (t − 1 ) . where V = C 2 . α : π 1 (M) → SL 2 C ≤ GL (V) � � tr ( Γ ) • The coefficients of τ M lie in Q and Hyperbolic Alexander polynomial: are often algebraic integers. � t ± 1 � coming from H 1 ( � • τ M (ζ) ≠ 0 for any root of unity ζ . M ; V α ) . τ M (t) ∈ C • τ M = τ M (t) Examples: • M amphichiral ⇒ τ M (t) ∈ R [t ± 1 ] . • Figure-8: τ M = t − 4 + t − 1 • Genus bound: • Kinoshita-Terasaka: 4 g − 2 ≥ deg τ M (t) τ M ≈ ( 4 . 417926 + 0 . 376029 i)(t 3 + t − 3 ) − ( 22 . 941644 + 4 . 845091 i)(t 2 + t − 2 ) For the KT knot, g = 2 and deg τ M (t) = 6 so this is sharp, unlike with ∆ M . + ( 61 . 964430 + 24 . 097441 i)(t + t − 1 ) − ( − 82 . 695420 + 43 . 485388 i)

Knots by the numbers: 313,231 number of prime knots with at most 15 crossings. [HTW 98] Basic Properties: 22 number which are non-hyperbolic. • τ M is an unambiguous element of C [t ± 1 ] number where 2 g > deg ( ∆ M ) . 8,834 with τ M (t) = τ M (t − 1 ) . � � tr ( Γ ) • The coefficients of τ M lie in Q and 7,972 number of non-fibered knots are often algebraic integers. where ∆ M is monic. • τ M (ζ) ≠ 0 for any root of unity ζ . number where 4 g − 2 > deg (τ M ) . 0 • τ M = τ M (t) 0 number of non-fibered knots • M amphichiral ⇒ τ M (t) ∈ R [t ± 1 ] . where τ M is monic. • Genus bound: Conj. τ M determines the genus and fibering for 4 g − 2 ≥ deg τ M (t) any hyperbolic knot in S 3 . For the KT knot, g = 2 and deg τ M (t) = 6 so this is sharp, unlike with ∆ M . Computing τ M : Approximate π 1 (M) → SL 2 C to 250 digits by solving the gluing equations asso- ciated to some ideal triangulation of M to high precision.

Knots by the numbers: 313,231 number of prime knots with at most 15 crossings. [HTW 98] Basic Properties: 22 number which are non-hyperbolic. • τ M is an unambiguous element of C [t ± 1 ] number where 2 g > deg ( ∆ M ) . 8,834 with τ M (t) = τ M (t − 1 ) . � � tr ( Γ ) • The coefficients of τ M lie in Q and 7,972 number of non-fibered knots are often algebraic integers. where ∆ M is monic. • τ M (ζ) ≠ 0 for any root of unity ζ . number where 4 g − 2 > deg (τ M ) . 0 • τ M = τ M (t) 0 number of non-fibered knots • M amphichiral ⇒ τ M (t) ∈ R [t ± 1 ] . where τ M is monic. • Genus bound: Conj. τ M determines the genus and fibering for 4 g − 2 ≥ deg τ M (t) any hyperbolic knot in S 3 . For the KT knot, g = 2 and deg τ M (t) = 6 so this is sharp, unlike with ∆ M . Computing τ M : Approximate π 1 (M) → SL 2 C to 250 digits by solving the gluing equations asso- ciated to some ideal triangulation of M to high precision.

Knots by the numbers: 313,231 number of prime knots with at most 15 crossings. [HTW 98] 22 number which are non-hyperbolic. Genus and fibering for most of these knots was previously unknown; Haken-style normal surface number where 2 g > deg ( ∆ M ) . 8,834 algorithms are impractical in this range, various 7,972 number of non-fibered knots tricks were used. where ∆ M is monic. Q. How can we prove this conjecture? number where 4 g − 2 > deg (τ M ) . 0 Not known to be true for infinitely many non- 0 number of non-fibered knots where τ M is monic. fibered knots! Conj. τ M determines the genus and fibering for If conjecture and GRH are true, then knot genus any hyperbolic knot in S 3 . is in NP ∩ co-NP . Computing τ M : Approximate π 1 (M) → SL 2 C to 250 digits by solving the gluing equations asso- ciated to some ideal triangulation of M to high precision.

Approach 1: Deform the representation Genus and fibering for most of these knots was Can consider other reps to SL 2 C , understand how previously unknown; Haken-style normal surface τ M,α varies as you move around the character va- algorithms are impractical in this range, various riety: tricks were used. Example: m 037 , X 0 = C \ {− 2 , 0 , 2 } Q. How can we prove this conjecture? � t + t − 1 � τ X 0 (t) = (u + 2 ) 4 16 u 2 � � u 4 + 4 u 3 − 8 u 2 + 16 u + 16 + (u + 2 ) Not known to be true for infinitely many non- 8 (u − 2 )u 2 fibered knots! Can sometimes connect this universal polynomial If conjecture and GRH are true, then knot genus to ∆ M . is in NP ∩ co-NP . Ideal points corresponding to S : not helpful.