The Grothendieck-Teichm¨ uller group and the outer automorphism groups of the profinite braid groups (joint

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The Grothendieck-Teichm¨ uller group and the outer automorphism groups of the profinite braid groups (joint

work with Hiroaki Nakamura)

Arata Minamide

RIMS, Kyoto University

June 28, 2021

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Discrete case

n >3: an integer

Bn: the (Artin) braid group onn strings Example: (n= 4)

◦ =

Arata Minamide (RIMS, Kyoto University) The Grothendieck-Teichm¨uller group June 28, 2021 2 / 18

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Discrete case

n >3: an integer

◦ =

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Discrete case

n >3: an integer

◦ =

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Discrete case

n >3: an integer

◦ =

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Note: We have B_n ∼=

⟨

σ₁, σ₂, . . . , σ_n₋₁

σ_i·σ_i+1·σ_i = σ_i+1·σ_i·σ_i+1; σ_i·σ_j = σ_j·σ_i (|i−j| ≥ 2)

⟩

ι∈Aut(Bn): the involutive automorphism ofBn determined by the formula σi 7→ σ_i⁻¹ (i= 1,2, . . . , n−1)

Theorem (Dyer-Grossman)

The natural surjection Aut(B_n)↠Out(B_n) induces

⟨ι⟩ →^∼ Out(B_n).

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⟨

σ₁, σ₂, . . . , σ_n₋₁

⟩

ι∈Aut(B_n): the involutive automorphism of B_n determined by the formula σi 7→ σ_i⁻¹ (i= 1,2, . . . , n−1)

⟨ι⟩ →^∼ Out(B_n).

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⟨

σ₁, σ₂, . . . , σ_n₋₁

⟩

ι∈Aut(B_n): the involutive automorphism of B_n determined by the formula σi 7→ σ_i⁻¹ (i= 1,2, . . . , n−1)

⟨ι⟩ →^∼ Out(B_n).

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Profinite case

Bbn: the profinite completion of Bn

S_n: the symmetric group onn letters ↷

(A¹_Q)_n ^def= {(x₁, . . . , x_n)∈(A¹_Q)ⁿ |x_i ̸=x_j (i̸=j)} The structure morphism (A¹_Q)_n/S_n→Spec(Q) induces

G_Q ^def= Gal(Q/Q) ,→ Out(π₁(((A¹_Q)_n/S_n)×_QQ)) ∼= Out(Bb_n).

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Profinite case

Bbn: the profinite completion of Bn

S_n: the symmetric group onn letters ↷

(A¹_Q)_n ^def= {(x₁, . . . , x_n)∈(A¹_Q)ⁿ |x_i ̸=x_j (i̸=j)}

The structure morphism (A¹_Q)_n/S_n→Spec(Q) induces

G_Q ^def= Gal(Q/Q) ,→ Out(π₁(((A¹_Q)_n/S_n)×_QQ)) ∼= Out(Bb_n).

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Drinfeld and Ihara defined a certain subgroup dGT ⊆ Aut(Z[∗Z), called the (profinite) Grothendieck-Teichm¨uller group, such that there exists a commutative diagram

G_Q Out(Bbn)

GTd

∃ ∃

Open problem: Is G_Q ,→ GTd anisomorphism?

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G_Q Out(Bbn)

GTd

∃ ∃

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G_Q Out(Bbn)

GTd

∃ ∃

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Theorem (M.-Nakamura) Write

Z_n ^def= Ker(Zb^×↠(bZ/n(n−1)Zb)^×).

Then we have a natural homomorphism Zn → Out(Bbn).

Moreover, this homomorphism and GTd→Out(Bb_n) induce Zn×GTd →^∼ Out(Bbn).

Note: If G_Q →^∼ GT, then we haved Z_n×G_Q →^∼ Out(Bb_n).

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Definition of Zn →Aut(Bbn) (↠ Out(Bbn))

Note: The center Cn ⊆ Bn is an infinite cyclic group (∼= Z) generated by ζ_n ^def= (σ₁· · ·σ_n₋₁)ⁿ.

Let ν ∈ Z_n

=⇒ ν = 1 +n(n−1)e (e∈Zb) Set ϕ_ν(σ_i) ^def= σ_i·ζ_n^e ∈ Bb_n (i= 1, . . . , n−1)

=⇒ {ϕ_ν(σ_i)}i=1,... ,n−1 satisfy the “braid relations”.

=⇒ We obtain a homomorphism ϕν :Bbn → Bbn.

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Let ν ∈ Z_n =⇒ ν = 1 +n(n−1)e (e∈Zb)

Set ϕ_ν(σ_i) ^def= σ_i·ζ_n^e ∈ Bb_n (i= 1, . . . , n−1)

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Let ν ∈ Z_n =⇒ ν = 1 +n(n−1)e (e∈Zb) Set ϕ_ν(σ_i) ^def= σ_i·ζ_n^e ∈ Bb_n (i= 1, . . . , n−1)

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Lemma 1 It holds that

ϕ₁ = id; ϕ_ν₁_·_ν₂ = ϕ_ν₁ ◦ϕ_ν₂ (ν₁, ν₂ ∈ Z_n).

Proof.

This follows from the formula

ϕ_ν(ζ_n) = ϕ_ν((σ₁· · ·σ_n₋₁)ⁿ) = (σ₁· · ·σ_n₋₁)ⁿ·ζ_nⁿ⁽ⁿ⁻^1)e = ζ_n^ν

=⇒ ϕν :Bbn → Bbn is a bijection(cf. ϕν◦ϕ_ν−1 = id).

=⇒ We obtain ahomomorphism

ϕ:Zn → Aut(Bbn) ν 7→ ϕν

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ϕ₁ = id; ϕ_ν₁_·_ν₂ = ϕ_ν₁ ◦ϕ_ν₂ (ν₁, ν₂ ∈ Z_n).

Proof.

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ϕ₁ = id; ϕ_ν₁_·_ν₂ = ϕ_ν₁ ◦ϕ_ν₂ (ν₁, ν₂ ∈ Z_n).

Proof.

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ϕ₁ = id; ϕ_ν₁_·_ν₂ = ϕ_ν₁ ◦ϕ_ν₂ (ν₁, ν₂ ∈ Z_n).

Proof.

b

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Outline of the proof of Theorem Step1 Write Bn

def= Bn/Cn. We show that the composite GTd → Out(Bb_n) → Out(Bbn) is an isomorphism. (Note that we have Bbn→∼ Bbn/Cbn.)

Corollary of Step1

Γ1,2: the pure mapping class group of torus w/ 2 marked pts Corollary (M.-Nakamura)

We have a natural isomorphism

GT [d →^∼ Out(Bb4)] →^∼ Out(bΓ1,2).

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def= Bn/Cn. We show that the composite GTd → Out(Bb_n) → Out(Bbn)

is an isomorphism. (Note that we have Bbn→∼ Bbn/Cbn.) Corollary of Step1

GT [d →^∼ Out(Bb4)] →^∼ Out(bΓ1,2).

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Γ1,2: the pure mapping class group of torus w/ 2 marked pts

Corollary (M.-Nakamura) We have a natural isomorphism

GT [d →^∼ Out(Bb4)] →^∼ Out(bΓ1,2).

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Step2 We show that there is a central extension

1 Zn ^ϕ Aut(Bbn) Aut(Bbn) 1.

Then we have

1 Zn Out(Bbn) Out(Bbn) 1.

GTd

splitting

∼ Step1

Therefore, we conclude that Zn×dGT →^∼ Out(Bbn).

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Then we have

GTd

splitting

∼ Step1

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Then we have

GTd

splitting

∼ Step1

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Then we have

GTd

splitting

∼ Step1

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Then we have

GTd

splitting

∼ Step1

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Then we have

GTd

splitting

∼ Step1

×d →^∼ b

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Details of Step1 Idea Observe that Pn

def= Ker(Bn↠S_n) may be identified with Γ_0,n+1: the pure mapping class group of sphere w/n+ 1marked pts.

Note: bΓ_0,n+1 may be identified with the ´etale π₁ of

the(n−2)-nd config. sp. of P¹_Q\ {0,1,∞} · · · anabelian variety! [cf. combinatorial anabelian geometry].

Theorem (Hoshi-M.-Mochizuki) We have a natural isomorphism

Sn+1×dGT →^∼ Out(Γb0,n+1).

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Note: bΓ0,n+1 may be identified with the ´etale π1 of the(n−2)-nd config. sp. of P¹_Q\ {0,1,∞}

· · · anabelian variety!

[cf. combinatorial anabelian geometry]. Theorem (Hoshi-M.-Mochizuki)

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Note: bΓ0,n+1 may be identified with the ´etale π1 of

the(n−2)-nd config. sp. of P¹_Q\ {0,1,∞} · · · anabelian variety!

[cf. combinatorial anabelian geometry]. Theorem (Hoshi-M.-Mochizuki)

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[cf. combinatorial anabelian geometry].

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[cf. combinatorial anabelian geometry].

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Denote by

Γ_0,[n+1]: the mapping class group of sphere w/ n+ 1marked pts

∼=

⟨

ω1, ω2, . . . , ωn

“braid relations”; (ω1· · ·ωn)ⁿ⁺¹= 1; ω₁· · ·ω_n−1·ω²_n·ω_n−1· · ·ω₁= 1

⟩ .

Note: We have a commutative diagram

1 Pbn Bbn Sn 1

1 bΓ_0,n+1 Γb_0,[n+1] S_n+1 1.

≀

— where Bbn→bΓ_0,[n+1] is defined to be σi7→ωi.

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Denote by

∼=

⟨

ω1, ω2, . . . , ωn

“braid relations”; (ω1· · ·ωn)ⁿ⁺¹= 1;

ω₁· · ·ω_n₋₁·ω²_n·ω_n₋₁· · ·ω₁= 1

⟩ .

1 Pbn Bbn Sn 1

1 bΓ_0,n+1 Γb_0,[n+1] S_n+1 1.

≀

— where Bbn→bΓ_0,[n+1] is defined to be σi7→ωi.

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Denote by

∼=

⟨

ω1, ω2, . . . , ωn

“braid relations”; (ω1· · ·ωn)ⁿ⁺¹= 1;

ω₁· · ·ω_n₋₁·ω²_n·ω_n₋₁· · ·ω₁= 1

⟩ .

1 Pbn Bbn S_n 1

1 bΓ_0,n+1 Γb_0,[n+1] S_n+1 1.

≀

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Lemma 2 Let

1 ∆ Π G 1

be an exact sequence of finitely generated profinite groups. Write ρ:G→Out(∆) for the outer rep’n assoc. to the exact sequence.

Suppose that

∆ and G arecenter-free.

∆⊆Π is a characteristic subgroup. Then we have an exact sequence

1 Z_Out(∆)(Im(ρ)) Out(Π) Out(G)

— where Z_Out(∆)(Im(ρ)) is the centralizer of Im(ρ) in Out(∆).

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Lemma 2 Let

1 ∆ Π G 1

Suppose that

∆⊆Π is a characteristic subgroup.

Then we have an exact sequence

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Lemma 2 Let

1 ∆ Π G 1

Suppose that

∆⊆Π is a characteristic subgroup.

Then we have an exact sequence

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We would like to apply Lemma 2 to the exact sequence

1 Pbn Bbn Sn 1

Γb0,n+1

≀

to obtain an exact sequence

1 Z_Out(b_Γ

0,n+1)(Sn) Out(Bbn) Out(Sn)

Z_GT_d_×_S

n+1(S_n) {1}

≀

[HMM] (n̸=6)

≀

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1 Pbn Bbn Sn 1

Γb0,n+1

≀

to obtain an exact sequence 1 Z_Out(b_Γ

Z_GT_d_×_S

n+1(S_n) {1}

GTd

≀

[HMM] (n̸=6)

≀

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1 Pbn Bbn Sn 1

Γb0,n+1

≀

Z_GT_d_×_S

n+1(S_n) {1}

≀

[HMM] (n̸=6)

≀

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1 Pbn Bbn Sn 1

Γb0,n+1

≀

Z_GT_d_×_S

n+1(S_n) {1}

GTd

≀

[HMM] (n̸=6)

≀

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1 Pbn Bbn Sn 1

Γb0,n+1

≀

Z_GT_d_×_S

n+1(S_n) {1}

≀

[HMM] (n̸=6)

≀

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Note: Pbn [∼=bΓ0,n+1] and Sn are center-free.

Thus, to apply Lemma 2, it suﬃces to check the following:

Proposition

Pbn⊆Bbn is a characteristic subgroup.

Lemma 3

Let Gbe a residually finite gp (i.e., G ,→G);b N ⊆Ga finite index normal subgp. Suppose that Ker(G↠^∀ Q^def= G/N) coincides with N. Then Ker(Gb↠^∀ Q) coincides with Nb.

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Proposition

Lemma 3

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Proposition

Lemma 3

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Theorem (E. Artin) Let

φ:Bn ↠ Sn

be a surjective homomorphism.

Then, up to some autom. ∈Aut(Sn), φ is the “standard surjection” (i.e.,φ(σi) = (i, i+ 1)∈Sn), with the following two exceptions for n= 4:

(a) φ(σ₁) = (1,2,3,4), φ(σ₂) = (2,1,3,4), φ(σ₃) = (1,2,3,4); (b) φ(σ1) = (1,2,3,4), φ(σ2) = (2,1,3,4), φ(σ3) = (4,3,2,1).

In particular, if n≥5, then Ker(Bn

↠∀ S_n) =Pn

=⇒ Ker(Bbn

↠∀ Sn) =Pbn (cf. Lemma 3)

=⇒ Pbn⊆Bbn is characteristic!

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φ:Bn ↠ Sn

be a surjective homomorphism. Then, up to some autom. ∈Aut(Sn), φ is the “standard surjection” (i.e.,φ(σi) = (i, i+ 1)∈Sn), with the following two exceptions for n= 4:

(a) φ(σ₁) = (1,2,3,4), φ(σ₂) = (2,1,3,4), φ(σ₃) = (1,2,3,4);

(b) φ(σ1) = (1,2,3,4), φ(σ2) = (2,1,3,4), φ(σ3) = (4,3,2,1).

↠∀ S_n) =Pn

=⇒ Ker(Bbn

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φ:Bn ↠ Sn

(a) φ(σ₁) = (1,2,3,4), φ(σ₂) = (2,1,3,4), φ(σ₃) = (1,2,3,4);

(b) φ(σ1) = (1,2,3,4), φ(σ2) = (2,1,3,4), φ(σ3) = (4,3,2,1).

In particular, if n≥5, then

Ker(Bn

↠∀ S_n) =Pn

=⇒ Ker(Bbn

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φ:Bn ↠ Sn

(a) φ(σ₁) = (1,2,3,4), φ(σ₂) = (2,1,3,4), φ(σ₃) = (1,2,3,4);

(b) φ(σ1) = (1,2,3,4), φ(σ2) = (2,1,3,4), φ(σ3) = (4,3,2,1).

↠∀ S_n) =Pn

=⇒ Ker(Bbn

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φ:Bn ↠ Sn

(a) φ(σ₁) = (1,2,3,4), φ(σ₂) = (2,1,3,4), φ(σ₃) = (1,2,3,4);

(b) φ(σ1) = (1,2,3,4), φ(σ2) = (2,1,3,4), φ(σ3) = (4,3,2,1).

↠∀ S_n) =Pn

=⇒ Ker(Bb ↠^∀ S ) =Pb (cf. Lemma 3)

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φ:Bn ↠ Sn

(a) φ(σ₁) = (1,2,3,4), φ(σ₂) = (2,1,3,4), φ(σ₃) = (1,2,3,4);

(b) φ(σ1) = (1,2,3,4), φ(σ2) = (2,1,3,4), φ(σ3) = (4,3,2,1).

↠∀ S_n) =Pn

=⇒ Ker(Bbn

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Remark: In the proof of [DG, Theorem 11] claiming that P4⊆ B4 is characteristic, there is an inaccurate argument. They forgot to treat thecase (b). (Moreover, the argument which was applied to “eliminate the case (a)” does not function properly for the case (b).)

=⇒ We need another argument to prove that Pb4 ⊆Bb4 is characteristic. Ingredients The following “anabelian results”:

a special property of finite subgroups of free profinite products (cf. Herfort-Ribes);

a special property of free profinite groups (cf. Lubotzky-van den Dries).

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=⇒ We need another argument to prove that Pb4 ⊆Bb4 is characteristic.

Ingredients The following “anabelian results”:

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Details of Step2

We consider the following sequence:

1 Z_n ^ϕ Aut(Bb_n) ^p¹ Aut(Bbn) 1.

Exactness at the middle

Let α ∈ Ker(p1).

Note: α(ζ_n) = ζ_n^ν (ν ∈ Zb^×); α(σ_i) = σ_i·ζ_n^eⁱ (e_i ∈ Zb)

=⇒ All e_i are the same constant e ∈ Zb (cf. the “braid relations”).

=⇒ ν = 1 +n(n−1)e ∈ Zn (cf. ζn = (σ1· · ·σn−1)ⁿ).

=⇒ α = ϕν ∈ Im(ϕ).

Remark: Using this argument, we can also prove the “centrality”.

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Exactness at the middle Let α ∈ Ker(p1).

=⇒ ν = 1 +n(n−1)e ∈ Zn (cf. ζn = (σ1· · ·σn−1)ⁿ).

=⇒ α = ϕν ∈ Im(ϕ).

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=⇒ ν = 1 +n(n−1)e ∈ Zn (cf. ζn = (σ1· · ·σn−1)ⁿ).

=⇒ α = ϕν ∈ Im(ϕ).

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=⇒ ν = 1 +n(n−1)e ∈ Zn (cf. ζn = (σ1· · ·σn−1)ⁿ).

=⇒ α = ϕν ∈ Im(ϕ).

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=⇒ ν = 1 +n(n−1)e ∈ Zn (cf. ζn = (σ1· · ·σn−1)ⁿ).

=⇒ α = ϕν ∈ Im(ϕ).