Varieties of characters and knot symmetries (I)

Transcription

1 Varieties of characters and knot symmetries (I) Joan Porti Universitat Autònoma de Barcelona June 2017 INdAM Meeting Geometric Topology in Cortona

2 Organization Day 1 Character Varieties. Definition and computation Day 2 Distinguished curves for hyperbolic knots (and of hyperbolic 3-manifolds of finite volume). Day 3 Knot symmetries (joint with Luisa Paoluzzi) (Distinguish in the variety of characters whether a knot symmetry has fixed points or not)

3 Motivation hom(π 1 (M 3 ), (P)SL 2 C)/PSL 2 C contains the space of hyperbolic structures on M 3, as PSL 2 (C) = Isom(H 3 ). Used in the proof of the hyperbolic Dehn filling theorem, hyperbolic cone manifolds, also for degeneration of structures, and many other situations... Culler-Shalen theory of surfaces associated to ideal points. A-polynomial. Representations have been used to distinguish knots, in Casson s invariant,... Study the variety of characters as an algebraic object. Dynamics of group actions...

4 Variety of representations Γ = γ 1,..., γ n (r j ) j J finitely generated group {( ) a b SL 2 C = a, b, c, d C, ad bc = 1} c d Def: R(Γ) = hom(γ, SL 2 C) It is an affine algebraic set (zero set of polynomials in C 4n ) R(Γ) SL 2 C (n) SL 2 C C 4n ρ (ρ(γ 1 ),..., ρ(γ n )) The algebraic structure is independent of the presentation. The action by conjugation is algebraic SL 2 C R(Γ) R(Γ) A, ρ γ Aρ(γ)A 1 but the quotient R(Γ)/PSL 2 (C) may be not Hausdorff.

5 Quotient in the algebraic category Γ = γ 1,..., γ n (r j ) j J R(Γ) = hom(γ, SL 2 C) C 4n C-algebra of functions: C[R(Γ)] = C[x 1,..., x 4n ]/I where I = {p C[x 1,..., x 4n ] p(r(γ)) = 0} Look for functions invariant by conjugation: C[R(Γ)] SL 2C For γ Γ, the trace function is τ γ : R(Γ) C ρ tr(ρ(γ)) Thm: (Procesi) γ 1,..., γ N Γ such that C[R(Γ)] SL 2C = τ γ1,..., τ γn Setting y j = τ γj, C[R(Γ)] SL 2C = C[y1,..., y N ]/J hence the zero set of J in C N is the algebraic quotient.

6 trace functions τ γ : R(Γ) C ρ tr(ρ(γ)) Variety of characters characters χ ρ : Γ C γ tr(ρ(γ)) Def: The variety of characters is X (Γ) = {χ ρ ρ R(Γ)} Since C[R(Γ)] SL2C = τ γ1,..., τ γn, X (Γ) has a natural algebraic structure so that C[X (Γ)] = C[R(Γ)] SL2C. For f : R(Γ) C polynomial and SL 2 C-invariant, there exists a unique f : X (Γ) C such that R(Γ) X (Γ) Conjugate representations have the same character: R(Γ) R(Γ)/PSL 2 C X (Γ) ρ [ρ] χ ρ What is the difference between R(Γ)/PSL 2 C and X (Γ)? f f C

7 Irreducible representations Def: ρ R(Γ) is reducible if ρ(γ) has an invariant ( line in ) C 2 equivalently, ρ R(Γ) is reducible if ρ(γ) 0 Lemma: ρ is reducible iff tr ρ([γ, Γ]) = 2 Cor: R red (Γ) and X red (Γ) Zariski closed Lemma: For ρ, ρ R(Γ) irreducible, ρ and ρ are conjugate iff χ ρ = χ ρ, In fact, PSL 2 C R irr (Γ) X irr (Γ) is a principal bundle For reducible characters χ, there is a unique conjugacy class of diagonal representations ρ with χ ρ = χ For every continuous f : R(Γ)/PSL 2 C H with H Hausdorff: R(Γ)/PSL 2 C f H X (Γ) f

8 Free group of rank two Lemma: For A, B SL 2 C: (i) tr(ab) = tr(ba); (ii) tr(a 1 ) = tr(a); (iii) tr(ab) + tr(a 1 B) = tr(a) tr(b). For (iii), A 2 tr(a)a + Id = 0 times A 1 B and take trace. Thm: (Fricke-Klein) Let F 2 = a, b, we have an isomorphism (τ a, τ b, τ ab ): X (F 2 ) = C 3 Setting x = τ a, y = τ b, and z = τ ab, C[X (F 2 )] = C[x, y, z]. Cor: γ F 2, τ γ is a (unique) polynomial on x, y, z

9 Groups of rank two F 2 = a, b, x = τ a, y = τ b, z = τ ab, C[X (F 2 )] = C[x, y, z] τ γ = τ γ 1 τ γµ = τ µγ τ γ τ µ = τ γµ + τ γ 1 µ Powers: τ a 2 = τ 2 a τ 1 = x 2 2 τ a 3 = τ a τ a 2 τ a = x 3 3x τ a n = τ a τ a n 1 τ a n 2 =Chebyshev pol. on x. [a, b] = aba 1 b 1 τ aba 1 b 1 = τ aba 1τ b τ aba 1 b = y 2 τ aba 1 b τ aba 1 b = τ ab τ a 1 b τ a 2 = z τ a 1 b x τ a 1 b = τ a τ b τ ab = xy z τ [a,b] = x 2 + y 2 + z 2 xyz 2 Z 2 = F 2 = a, b [a, b] = 1. We have: X (Z 2 ) = {(x, y, z) C 3 x 2 + y 2 + z 2 xyz 4 = 0}

10 The trefoil knot exterior Γ = a, b aba = bab bab 1 = a 1 ba τ a = τ b Choose x = τ a = τ b and y = τ ab 1 Write τ abab 1 τ ab = 0: τ ab = τ a τ b τ ab 1 = x 2 y τ abab 1 = τ ba τ ab 1 τ b 2 = (x 2 y)y x τ abab 1 τ ab = x 2 y 2x 2 y 2 +y +2 = (y 2)(x 2 y 1) = 0 We have X (Γ) = {(x, y) C 2 (y 2)(x 2 y 1) = 0} y = 2: abelian (and reducible) representations x 2 y 1 = 0: τ ab = x 2 y = 1 and τ aba = τ a τ ab τ b = x 1 x = 0 ρ(ab) 3 = ρ(aba) 2 = Id (set α = ab and β = aba): Γ = α, β, t α 3 = β 2 = t, [α, t] = [β, t] = 1 Irreducible reps map the center to ± Id

11 The figure eight knot exterior (I) Γ = a, b w a = b w with w = ab 1 a 1 b x = τ a = τ b, y = τ ab 1 τ ω τ b 1 wa = (y 2)(y 2 (x 2 1)y + x 2 1) = 0 y 2 = 0 consists of abelian representations y 2 (x 2 1)y + x 2 1 = 0 contains the holonomy of the hyperbolic structure and all deformations corresponding to Dehn filling. Hence X (Γ) = {(x, y) C 2 (y 2)(y 2 (x 2 1)y + x 2 1) = 0} For a 2-bridge knot exterior in general Γ = a, b w a = b w for some w = w(a, b) X (Γ) is a plane curve. For a 2-bridge knot exterior, X (Γ) may have arbitrarily many components (Ohtsuki-Riley-Sakuma)

12 The figure eight knot exterior (II) Γ = a, b w a = b w with w = ab 1 a 1 b ( ) ( ) s 1 s 0 Write ρ(a) = 0 s 1 and ρ(b) = 2 y s 1 s + s 1 = τ a = τ b = x and τ a 1 b = y ρ(w a) ρ(b w) = (y 2 (x 2 1)y + x 2 1) ( 0 ) 1 y 2 0 Component of irreducible representations (Whittemore 1973): y 2 (x 2 1)y + x 2 1 = 0

13 The figure eight knot exterior (III) It is a bundle over S 1 with fibre a punctured torus. Γ = α, β, µ µαµ 1 = αβ, µβµ 1 = βαβ 1 F 2 = α, β Γ Z = µ 1 φ: F 2 F 2, φ(α) = αβ, φ(β) = βαβ, X (F 2 ) = C 3 with coordinates x 1 = τ α, x 2 = τ β, x 3 = τ αβ φ(x 1, x 2, x 3 ) = (x 1, x 2, x 3 ) x 3 = x 1, x 1 + x 2 = x 1 x 2 X (F 2 ) φ = {χ χ φ = χ} ={(x 1, x 2 ) C 2 x 1 + x 2 = x 1 x 2 } res: X (Γ) X (F 2 ) φ is surjective and, for χ X (F 2 ) φ : 2 characters if χ irreducible res 1 (χ) = 1 irred. character if χ reducible non trivial a line of red. ch. if χ trivial X irr (Γ, PSL 2 C) = X (F 2 ) φ

14 The Whitehead link exterior Γ = a, b aw = wa where w = bab 1 a 1 b 1 ab. Coordinates x = τ a, y = τ b, z = τ ab, then X (Γ) = {(x, y, z) C 3 pq = 0} with { p = xy (x 2 + y 2 2)z + xyz 2 z 3 q = x 2 + y 2 + z 2 xyz 4 {q = 0} = X (Z 2 ) abelian/reducible characters {p = 0} Component containing a lift of the holonomy of the complete hyperbolic structure (and Dehn fillings)

15 X (F n ), n 3 F. González-Acuña and J.M. Montesinos-Amilibia (1993), based on H. Vogt (1889) and W. Magnus (1980) F 3 = a, b, c (τ a, τ b, τ c, τ ab, τ bc, τ ca ): X (F 3 ) C 6 is a branched covering There is one more coordinate algebraically dependent: τ abc and τ acb are the solutions of the equation z 2 Pz + Q = 0 with P = τ a τ bc + τ b τ ca + τ c τ ab τ a τ b τ c Q = τ 2 a + τ 2 b + τ 2 c + τ 2 ab + τ 2 bc + τ 2 ca + τ ab τ bc τ ca τ a τ b τ ab τ b τ c τ bc τ c τ a τ ca 4 eg τ abc + τ acb = P and τ abc τ acb = Q. F 4 = a, b, c, d τ abcd is a polynomial on the traces of words on a, b, c and d of length 3 with coefficients in 1 2 Z

16 Γ = γ 1,..., γ n {w j } j J X (Γ), Γ of finite type Thm: (González-Acuña and Montesinos-Amilibia) X (Γ) = {χ X (F n ) χ γi ω j = χ γi, i = 1,..., n, j J} Fico and Montesinos give the explicit description of X (F n ), defined with polynomials with coefficients in Z γ F n, τ γ is a polynomial in the traces of words of length 3 on the generators, with coefficients in 1 2 Z. Can take reduction mod p, p 2 prime, of X (Γ), to compute the variey of characters in an algebraicly closed field of characteristic p.

17 Varieties of characters and knot symmetries (II) Joan Porti Universitat Autònoma de Barcelona June 2017 INdAM Meeting Geometric Topology in Cortona

18 Organization Day 1 Character Varieties. Definition and computation Day 2 Canonical component for hyperbolic 3-manifolds of finite volume (...and further components). Day 3 Knot symmetries (joint with Luisa Paoluzzi) (Action of knot symmetries on X (S 3 K))

19 Manifolds of finite volume M 3 orientable & (complete) hyperbolic. Its volume is finite iff with T 2 i Ends(M 3 ) = (T1 2 T k 2 ) [0, ) = S 1 S 1, and each end is a cusp. By defn, a link L S 3 is hyperbolic if S 3 L is hyperbolic (hence of finite volume). If M 3 is hyperbolic, then hol: π 1 M 3 Isom + (H 3 ) = PSL 2 C lifts to ρ 0 (Culler, Thurston): π 1 (M 3 ) ρ 0 hol SL 2 C PSL 2 C Use that M 3 is parallelizable, and SO(3) PSL 2 C H 3 is the frame bundle on H 3 Results in this talk work also for X (M 3, PSL 2 C)

20 Canonical component Let M 3 be hyperbolic, orientable, of finite volume, and with k > 0 ends (that are cusps), ρ 0 = hol: π 1 M 3 SL 2 C. Thm: (Thurston) χ ρ0 is a smooth point of X (M 3 ), and the component X 0 (M 3 ) that contains χ ρ0 has dimension k. Def: X 0 (M 3 ) is called the canonical component of X (M 3 ) Perhaps not unique? If there are several components, then they are isomorphic, because other lifts of hol are ( 1) ɛ ρ 0 for ɛ: π 1 M 3 Z/2Z Isometries in PSL 2 C preserve orientation, so χ ρ0 χ ρ0. Characters close to χ ρ0 are holonomies of incomplete hyperbolic structures (used for hyperbolic Dehn filling). For a hyperbolic knot exterior S 3 K, X 0 (K) := X 0 (S 3 K) is a curve. Furthermore the A-polynomial is nontrivial

21 Example: cone manifold structures on the fig 8 knot The n-cyclic branched covering of S 3 branched on the figure eight knot is: hyperbolic if n 4 Euclidean if n = 3 spherical if n = 2 Setting α = 2π n, particular case of: There exists a family of cone manifold structures on S 3, with singular locus the figure eight knot, and with cone angle α (0, 4π 3 ) that are hyperbolic if α < 2π 3 Euclidean if α = 2π 3 spherical if 2π 3 < α < 4π 3 As the trace of a rotation of angle α is ±2 cos( α 2 ), look at characters s.t. χ(µ) [ 2, 2] for a meridian µ π 1 (S 3 K)

22 Transition geometry in the character variety K S 3 fig eight knot, µ Γ = π 1 (S 3 K) a meridian. Since χ(µ) = ±2 cos α 2, look at the set {χ X (Γ), χ(µ) [ 2, 2]} χ ρ0 χ ( 1) ɛ ρ 0 χ ρ0 χ ( 1) ɛ ρ τ µ 1 < χ(µ) < 2 are holonomies of hyperbolic cone structures (in magenta the complex conjugate, ie opposite orientation) The blue curve consists of characters in SU(2). Pairs (χ, χ ) with 1 < χ(µ) = χ (µ) < 1 are lifts of spherical holonomies as SO(4) = SU(2) SU(2) χ(µ) = ±1 Euclidean collapse.

23 Back to the canonical component thm Thm: If M 3 hyperbolic of finite volume, with one single end, ρ 0 lift of the holonomy, then χ ρ0 is a smooth point of X (M 3 ), and the component X 0 (M 3 ) that contains χ ρ0 is a curve Proof based on: (a) for each component Y X (M 3 ) containing an irreducible character χ so that χ(π 1 M 3 ) {±2}, dim Y 1 (b) The dimension of the Zariski tangent space at χ 0 is dim Tχ Zar 0 X (Γ) = 1 (a) is a lower bound of the dimension. It is proved in Thurston notes (sufficient for hyperbolic Dehn filling) (b) gives an upper bound of the dimension. (a)+(b) smoothness. (the dimension of the Zariski tangent space is an upper bound for the dimension, with equality precisely at smooth points)

24 Thurston s lower bound (one cusp case) Thm: Let χ ρ X (M 3 ) be irreducible so that χ(π 1 M 3 ) {±2}. For each component Y X (M 3 ) containing χ, dim Y 1 Proof: Chose a simple closed curve α with base point in M 3, s.t. 1. tr(ρ(α)) ±2 2. the restriction ρ α,π1( M is irreducible. 3 ) Set M = M 3 N (α). As α is simple: χ(m ) = 1 2 χ( M ) = 1. M has the homotopy type of a 2-dim CW-complex with 1 zero-cell, r one-cells and (r 2) two-cells: π 1 M has presentation with r generators and r 2 relations dim R(π 1 M ) 6 dim X (π 1 M ) 3 View α π 1 ( M ) and chose β π 1 ( M ) a meridian around α, so that [α, β] bounds a disk in M 3 Claim: For ρ R(M ) in a neighborhood of ρ, the conditions tr(ρ (β)) = tr(ρ ([α, β])) = 2 imply ρ (β) = Id The claim and dim X (π 1 M ) 3 yield the theorem.

25 Zariski tangent space Def: For V = {x C N p 1 (x) = = p r (x) = 0} the Zariski tangent space at x V is T Zar x V = {v C N p 1 (x + εv),..., p r (x + εv) O(ε 2 )} dim Tx Zar V dim(component of V containing x) with equality iff x smooth point of V (and one single comp) Def: Crossed morphisms Z 1 = {θ : Γ sl 2 C θ(γµ) = θ(γ) + ρ(γ)θ(µ)ρ(γ 1 )} Inner morphisms: B 1 = {θ a Z 1 θ a (γ) = ρ(γ)aρ(γ 1 ) a, γ Γ} H 1 (Γ, Ad ρ) = Z 1 /B 1 T Zar Thm: (Weil) ρ R(Γ) θ γ (Id +εθ(γ))ρ(γ) = ρ ε (γ) B 1 = orbit by conjugation Cor: (Weil) For χ ρ irreducible, H 1 (Γ, Ad ρ) = Tχ Zar ρ X (Γ) Thm: (Garland) M 3 hyperbolic, orient., finite vol, and k cusps. For ρ 0 R(M 3 ) lift of the holonomy, H 1 (π 1 M 3, Ad ρ 0 ) = C k Z 1 =

26 Cohomology thm Thm: (Garland) M 3 hyperbolic, orient, finite vol, and k cusps. For ρ 0 R(M 3 ) lift of the holonomy, H 1 (π 1 M 3, Ad ρ 0 ) = C k Use de Rham cohomology: Flat bundle sl 2 C E Ad ρ0 M where E Ad ρ0 = M sl 2 C/ with (x, a) (γx, Ad ρ0(γ) a) Ω p (M, E Ad ρ0 ) = Γ( p T M E Ad ρ0 ) = vector valued p-forms d : Ω p (M, E Ad ρ0 ) Ω p+1 (M, E Ad ρ0 ) yields de Rham cohom, and H dr (M3, E Ad ρ0 ) = H (π 1 M 3, Ad ρ 0 ) Thm: (...? 1960 s) closed L 2 -forms are exact. Set U = M 3 {Compact core} = Ti 2 (0, + ) hence H 1 (M, U, Ad ρ 0 ) 0 H 1 (M, Ad ρ 0 ) H 1 (U, Ad ρ) and dim H 1 (M, Ad ρ 0 ) = i dim 1 2 H1 (T 2 i, Ad ρ 0 ) = i 1 = k.

27 Galois conjugates M 3 hyperbolic, orientable, finite volume, ρ 0 : π 1 M 3 SL 2 C lift of the holonomy. ρ 0 (π 1 M 3 ) SL 2 K, for K a number field (Vinberg). May assume that the finite extension K Q is Galois. Rmk: For each σ Galois(K), χ ρ σ 0 is a smooth point of X (M 3 ) and the unique component that contains χ ρ σ 0 has dimension the number of cusps. Why? View the space of crossed morphisms as a linear space Z 1 = {θ : π 1 M 3 sl 2 C θ(γµ) = θ(γ) + ρ 0 (γ)θ(µ)ρ 0 (γ 1 )} From Γ = π 1 M 3 = γ 1,..., γ n r 1,..., r m, embed Z 1 sl 2 C sl 2 C θ (θ(γ 1 ),..., θ(γ n )) The image is the kernel of a linear map given by r 1,..., r m, with coefficients in K. So dim(z 1 (π 1 M 3, Ad ρ σ 0 )) = dim(z 1 (π 1 M 3, Ad ρ 0 )).

28 Further components Have shown X (M 3 ) has the canonical component and the abelian ones. Look for further components. Thm: (Ohtsuki-Riley-Sakuma) Among 2-bridge knots K and K, the set {f : π 1 (S 3 K) π 1 (S 3 K ) epimorphism}/{inner} may have arbitrarily large cardinality. Cor: (Ohtsuki-Riley-Sakuma) For 2-bridge knots K, X (K) can have arbitrarily many components Rmk: For 2-bridge knots K, components of X (K) are plane curves: As π 1 (S 3 K) = a, b a w(a, b) = w(a, b) b, use coordiates x = τ a = τ b and y = τ ab 1. Hence X (K) C 2 By Thuston s estimate, the dim of components is at least 1. Cannot have dim 2. Thm: (Paoluzzi-P, using an idea of Riley) Given m N, m 2, for a Montesinos knot K, X (K) may have arbitrarily many components of dim m.

29 Varieties of characters and knot symmetries (III) Joan Porti Universitat Autònoma de Barcelona June 2017 INdAM Meeting Geometric Topology in Cortona

30 Recap R(M) = hom(π 1 M, SL 2 C). χ ρ (γ) = tr(ρ(γ)) X (M) = {χ ρ : π 1 M C ρ R(M)} = R(M)//SL 2 C M 3 hyperbolic, orientable and of finite volume. The canonical component X 0 (M 3 ) has dimension the number of cusps. X (M 3 ) may have other irreducible components. Fico-Montesinos: coeffs. of the polynomial equations in 1 2 Z, so that its reduction mod p is the variety of SL 2 F-characters, for F = algebraically closed field of characteristic p 2 For a knot K S 3, X (K) := X (S 3 K)

31 Knot symmetries Let K be a knot in S 3 and ψ : (S 3, K) (S 3, K) a diffeomorphism of finite order p, that preserves the orientation of S 3. Either Fix(ψ) = or Fix(ψ) = S 1 unknotted in S 3 Def: ψ is said to be: a free symmetry if the group ψ = Z/pZ acts freely. a periodic symmetry if Fix(ψ) is a circle disjoint from K. a strong inversion if ψ has order 2 and Fix(ψ) K = 2. a pseudo periodic symmetry otherwise. Assume that ψ has order a prime p 2. Hence ψ is either free or periodic. Goal: show different behavior between periodic and free symmetries in the variety of characters.

32 Example: periodic (non-free) symmetry ψ O = (S 3 K)/ψ Σ=Fix(ψ)/ψ (K Fix(ψ))/ψ = 6 2 2

33 Example: free symmetry L(5, 1) = S 3 /ψ 1 5

34 Components of the variety of characters K S 3 hyperbolic knot ψ : (S 3, K) (S 3, K) symmetry of prime order p 2. X (K) ψ = {χ X (K) χ ψ = χ} Thm: (Paoluzzi-P) If ψ is periodic then X (K) ψ has at least p 1 2 components that are also components of X (K). Rmks: (a) For each prime p > 4 there is a knot K p with a free symmetry ψ of order p so that X (K p ) ψ has at most 20 components (b) When reducing mod p, all the components of the theorem collapse to a single one. (c) If we look at components of X (K), without ψ-invariance, many further components may appear, for ψ either free or periodic.

35 Components of the variety of characters K S 3 hyperbolic knot ψ : (S 3, K) (S 3, K) symmetry of prime order p 2. X (K) ψ = {χ X (K) χ ψ = χ} Thm: (Paoluzzi-P) If ψ is periodic then X (K) ψ has at least p 1 2 components that are also components of X (K). The proof has 3 steps. Set M = S 3 K, O = M/ψ. (1) The restriction map res: X irr (O) X irr (M) ψ is a bijection (2) Find several components for X (O) (3) Use (1) + (2) to find several components for X (M) ψ

36 Step 1: Extending ψ-invariant characters M = S 3 K O = M/ψ π 1 M π 1 O Z/pZ µ 1 ψ ψ : π 1 M π 1 M γ µγµ 1 π 1 O = π 1 M, µ µ p = 1, µγµ 1 = ψ µ, γ π 1 M Lemma The restriction map res: X irr (O) X irr (M) ψ is a bijection Proof Want to extend χ ρ X irr (M) ψ to π 1 O in a unique way. Since χ ρ ψ = χ ρ A SL 2 C s.t. ρ(ψ (γ)) = Aρ(γ)A 1. A is unique up to sign. If furthermore we require A p = Id A exists and unique because p odd Set ρ(µ) = A. (need to show also that restriction of irreducible is irreducible) µ

37 Step 2: Finding components for X (O) O is hyperbolic and hol: π 1 O PSL 2 C lifts to ρ 0 : π 1 O SL 2 C because p is odd. ρ 0 (π 1 O) SL 2 K, for K a number field (Vinberg). May assume that the finite extension K Q is Galois. Since µ p = 1, then tr(ρ 0 (µ)) = 2 cos π p and {τ µ (ρ σ πr 0 ) σ Galois(K)} = { 2 cos p r = 1, 3, 5,..., p 2}. τ µ = 2 cos πr p 1 p distinguishes 2 components Y 1,..., Y p 1 of 2 X (O) that contain different Galois conjugates χ ρ σ 0 of χ ρ0 By Garland s Thm, since O has a cusp, H 1 (π 1 O, Ad ρ 0 ) = C H 1 (π 1 O, Ad ρ σ 0 ) = C, σ Galois(K) each Y r is a curve. Next step: using that res: X irr (O) X irr (M) ψ is a bijection, the curves Y 1,..., Y p 1 yield different components of X (M) ψ. 2

38 Step 3: components for X (M) ψ We know that res: X irr (O) X irr (M) ψ is a bijection. We have Y 1,..., Y p 1 components of X irr (O) that are curves. 2 X (O) Y 1 res X (M) ψ X (M). W i = res(y i ) Y p 1 2 Since Y i = Yi irr (a finite set), W i = res(y i ) (a finite set) and W 1,..., W p 1 are different components of X (M) ψ 2 As dim H 1 (π 1 M, Ad ρ σ 0 ) = dim H1 (π 1 M, Ad ρ 0 ) = 1, W 1,..., W p 1 are also components of X (M). 2

39 Characteristic p Rmk: Let F be an algebraically closed field of characteristic p. As an element of SL 2 F of order p is conjugate to ( ) 1, 0 1 all these components W 1,..., W p 1 2 to a single one mod p. in the theorem collapse

40 Free symmetries: a family of examples Link K 0 A Consider q p -Dehn filling on A and take the covering (S 3, K q ) (L(p, q), K 0) p where p > 4 prime and p, q coprime ψ : (S 3, K q ) (S 3, K q ) free p p symmetry of order p X (6 2 2 ) has two components: X 0(6 2 2 ) and X ab (6 2 2 ). The map X 0 (6 2 2 ) X ( N (A)) is dominant and its generic fibre is finite. γ π 1 N (A) primitive, {τ γ = 2} is a line in X ( N (A)), X (L(p, q) K 0 ) has at most C components, C uniform X (S 3 K q p )ψ has at most C components (C uniform) and ψ has order p (that can be any prime > 4).

41 Thanks for your attention!