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(1)

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

(2)

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

(3)

Discrete case

n >3: an integer

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

=

(4)

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

(5)

Discrete case

n >3: an integer

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

=

(6)

Note: We have Bn =

σ1, σ2, . . . , σn1

σi·σi+1·σi = σi+1·σi·σi+1; σi·σj = σj·σi (|i−j| ≥ 2)

ι∈Aut(Bn): the involutive automorphism ofBn determined by the formula σi 7→ σi1 (i= 1,2, . . . , n1)

Theorem (Dyer-Grossman)

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

⟨ι⟩ Out(Bn).

Arata Minamide (RIMS, Kyoto University) The Grothendieck-Teichm¨uller group June 28, 2021 3 / 18

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Note: We have Bn =

σ1, σ2, . . . , σn1

σi·σi+1·σi = σi+1·σi·σi+1; σi·σj = σj·σi (|i−j| ≥ 2)

ι∈Aut(Bn): the involutive automorphism of Bn determined by the formula σi 7→ σi1 (i= 1,2, . . . , n1)

Theorem (Dyer-Grossman)

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

⟨ι⟩ Out(Bn).

(8)

Note: We have Bn =

σ1, σ2, . . . , σn1

σi·σi+1·σi = σi+1·σi·σi+1; σi·σj = σj·σi (|i−j| ≥ 2)

ι∈Aut(Bn): the involutive automorphism of Bn determined by the formula σi 7→ σi1 (i= 1,2, . . . , n1)

Theorem (Dyer-Grossman)

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

⟨ι⟩ Out(Bn).

Arata Minamide (RIMS, Kyoto University) The Grothendieck-Teichm¨uller group June 28, 2021 3 / 18

(9)

Profinite case

Bbn: the profinite completion of Bn

Sn: the symmetric group onn letters ↷

(A1Q)n def= {(x1, . . . , xn)(A1Q)n |xi ̸=xj (i̸=j)} The structure morphism (A1Q)n/SnSpec(Q) induces

GQ def= Gal(Q/Q) ,→ Out(π1(((A1Q)n/Sn)×QQ)) = Out(Bbn).

(10)

Profinite case

Bbn: the profinite completion of Bn

Sn: the symmetric group onn letters ↷

(A1Q)n def= {(x1, . . . , xn)(A1Q)n |xi ̸=xj (i̸=j)}

The structure morphism (A1Q)n/SnSpec(Q) induces

GQ def= Gal(Q/Q) ,→ Out(π1(((A1Q)n/Sn)×QQ)) = Out(Bbn).

Arata Minamide (RIMS, Kyoto University) The Grothendieck-Teichm¨uller group June 28, 2021 4 / 18

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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

GQ Out(Bbn)

GTd

Open problem: Is GQ ,→ GTd anisomorphism?

(12)

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

GQ Out(Bbn)

GTd

Open problem: Is GQ ,→ GTd anisomorphism?

Arata Minamide (RIMS, Kyoto University) The Grothendieck-Teichm¨uller group June 28, 2021 5 / 18

(13)

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

GQ Out(Bbn)

GTd

Open problem: Is GQ ,→ GTd anisomorphism?

(14)

Theorem (M.-Nakamura) Write

Zn def= Ker(Zb×↠(bZ/n(n−1)Zb)×).

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

Moreover, this homomorphism and GTdOut(Bbn) induce Zn×GTd Out(Bbn).

Note: If GQ GT, then we haved Zn×GQ Out(Bbn).

Arata Minamide (RIMS, Kyoto University) The Grothendieck-Teichm¨uller group June 28, 2021 6 / 18

(15)

Theorem (M.-Nakamura) Write

Zn def= Ker(Zb×↠(bZ/n(n−1)Zb)×).

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

Moreover, this homomorphism and GTdOut(Bbn) induce Zn×GTd Out(Bbn).

Note: If GQ GT, then we haved Zn×GQ Out(Bbn).

(16)

Theorem (M.-Nakamura) Write

Zn def= Ker(Zb×↠(bZ/n(n−1)Zb)×).

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

Moreover, this homomorphism and GTdOut(Bbn) induce Zn×GTd Out(Bbn).

Note: If GQ GT, then we haved Zn×GQ Out(Bbn).

Arata Minamide (RIMS, Kyoto University) The Grothendieck-Teichm¨uller group June 28, 2021 6 / 18

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

Note: The center Cn Bn is an infinite cyclic group (= Z) generated by ζn def= (σ1· · ·σn1)n.

Let ν Zn

= ν = 1 +n(n−1)e (eZb) Set ϕνi) def= σi·ζne Bbn (i= 1, . . . , n1)

=⇒ {ϕνi)}i=1,... ,n1 satisfy the “braid relations”.

= We obtain a homomorphism ϕν :Bbn Bbn.

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

Note: The center Cn Bn is an infinite cyclic group (= Z) generated by ζn def= (σ1· · ·σn1)n.

Let ν Zn = ν = 1 +n(n−1)e (eZb)

Set ϕνi) def= σi·ζne Bbn (i= 1, . . . , n1)

=⇒ {ϕνi)}i=1,... ,n1 satisfy the “braid relations”.

= We obtain a homomorphism ϕν :Bbn Bbn.

Arata Minamide (RIMS, Kyoto University) The Grothendieck-Teichm¨uller group June 28, 2021 7 / 18

(19)

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

Note: The center Cn Bn is an infinite cyclic group (= Z) generated by ζn def= (σ1· · ·σn1)n.

Let ν Zn = ν = 1 +n(n−1)e (eZb) Set ϕνi) def= σi·ζne Bbn (i= 1, . . . , n1)

=⇒ {ϕνi)}i=1,... ,n1 satisfy the “braid relations”.

= We obtain a homomorphism ϕν :Bbn Bbn.

(20)

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

Note: The center Cn Bn is an infinite cyclic group (= Z) generated by ζn def= (σ1· · ·σn1)n.

Let ν Zn = ν = 1 +n(n−1)e (eZb) Set ϕνi) def= σi·ζne Bbn (i= 1, . . . , n1)

=⇒ {ϕνi)}i=1,... ,n1 satisfy the “braid relations”.

= We obtain a homomorphism ϕν :Bbn Bbn.

Arata Minamide (RIMS, Kyoto University) The Grothendieck-Teichm¨uller group June 28, 2021 7 / 18

(21)

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

Note: The center Cn Bn is an infinite cyclic group (= Z) generated by ζn def= (σ1· · ·σn1)n.

Let ν Zn = ν = 1 +n(n−1)e (eZb) Set ϕνi) def= σi·ζne Bbn (i= 1, . . . , n1)

=⇒ {ϕνi)}i=1,... ,n1 satisfy the “braid relations”.

= We obtain a homomorphism ϕν :Bbn Bbn.

(22)

Lemma 1 It holds that

ϕ1 = id; ϕν1·ν2 = ϕν1 ◦ϕν21, ν2 Zn).

Proof.

This follows from the formula

ϕνn) = ϕν((σ1· · ·σn1)n) = (σ1· · ·σn1)n·ζnn(n1)e = ζnν

= ϕν :Bbn Bbn is a bijection(cf. ϕν◦ϕν1 = id).

= We obtain ahomomorphism

ϕ:Zn Aut(Bbn) ν 7→ ϕν

Arata Minamide (RIMS, Kyoto University) The Grothendieck-Teichm¨uller group June 28, 2021 8 / 18

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Lemma 1 It holds that

ϕ1 = id; ϕν1·ν2 = ϕν1 ◦ϕν21, ν2 Zn).

Proof.

This follows from the formula

ϕνn) = ϕν((σ1· · ·σn1)n) = (σ1· · ·σn1)n·ζnn(n1)e = ζnν

= ϕν :Bbn Bbn is a bijection(cf. ϕν◦ϕν1 = id).

= We obtain ahomomorphism

ϕ:Zn Aut(Bbn) ν 7→ ϕν

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Lemma 1 It holds that

ϕ1 = id; ϕν1·ν2 = ϕν1 ◦ϕν21, ν2 Zn).

Proof.

This follows from the formula

ϕνn) = ϕν((σ1· · ·σn1)n) = (σ1· · ·σn1)n·ζnn(n1)e = ζnν

= ϕν :Bbn Bbn is a bijection(cf. ϕν◦ϕν1 = id).

= We obtain ahomomorphism

ϕ:Zn Aut(Bbn) ν 7→ ϕν

Arata Minamide (RIMS, Kyoto University) The Grothendieck-Teichm¨uller group June 28, 2021 8 / 18

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Lemma 1 It holds that

ϕ1 = id; ϕν1·ν2 = ϕν1 ◦ϕν21, ν2 Zn).

Proof.

This follows from the formula

ϕνn) = ϕν((σ1· · ·σn1)n) = (σ1· · ·σn1)n·ζnn(n1)e = ζnν

= ϕν :Bbn Bbn is a bijection(cf. ϕν◦ϕν1 = id).

= We obtain ahomomorphism

b

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Outline of the proof of Theorem Step1 Write Bn

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

Arata Minamide (RIMS, Kyoto University) The Grothendieck-Teichm¨uller group June 28, 2021 9 / 18

(27)

Outline of the proof of Theorem Step1 Write Bn

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

(28)

Outline of the proof of Theorem Step1 Write Bn

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

Arata Minamide (RIMS, Kyoto University) The Grothendieck-Teichm¨uller group June 28, 2021 9 / 18

(29)

Outline of the proof of Theorem Step1 Write Bn

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

(30)

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).

Arata Minamide (RIMS, Kyoto University) The Grothendieck-Teichm¨uller group June 28, 2021 10 / 18

(31)

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).

(32)

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).

Arata Minamide (RIMS, Kyoto University) The Grothendieck-Teichm¨uller group June 28, 2021 10 / 18

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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).

(34)

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).

Arata Minamide (RIMS, Kyoto University) The Grothendieck-Teichm¨uller group June 28, 2021 10 / 18

(35)

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

×d b

(36)

Details of Step1 Idea Observe that Pn

def= Ker(Bn↠Sn) may be identified with Γ0,n+1: the pure mapping class group of sphere w/n+ 1marked pts.

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

the(n2)-nd config. sp. of P1Q\ {0,1,∞} · · · anabelian variety! [cf. combinatorial anabelian geometry].

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

Sn+1×dGT Out(Γb0,n+1).

Arata Minamide (RIMS, Kyoto University) The Grothendieck-Teichm¨uller group June 28, 2021 11 / 18

(37)

Details of Step1 Idea Observe that Pn

def= Ker(Bn↠Sn) may be identified with Γ0,n+1: the pure mapping class group of sphere w/n+ 1marked pts.

Note: bΓ0,n+1 may be identified with the ´etale π1 of the(n2)-nd config. sp. of P1Q\ {0,1,∞}

· · · anabelian variety!

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

We have a natural isomorphism

Sn+1×dGT Out(Γb0,n+1).

(38)

Details of Step1 Idea Observe that Pn

def= Ker(Bn↠Sn) may be identified with Γ0,n+1: the pure mapping class group of sphere w/n+ 1marked pts.

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

the(n2)-nd config. sp. of P1Q\ {0,1,∞} · · · anabelian variety!

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

We have a natural isomorphism

Sn+1×dGT Out(Γb0,n+1).

Arata Minamide (RIMS, Kyoto University) The Grothendieck-Teichm¨uller group June 28, 2021 11 / 18

(39)

Details of Step1 Idea Observe that Pn

def= Ker(Bn↠Sn) may be identified with Γ0,n+1: the pure mapping class group of sphere w/n+ 1marked pts.

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

the(n2)-nd config. sp. of P1Q\ {0,1,∞} · · · anabelian variety!

[cf. combinatorial anabelian geometry].

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

Sn+1×dGT Out(Γb0,n+1).

(40)

Details of Step1 Idea Observe that Pn

def= Ker(Bn↠Sn) may be identified with Γ0,n+1: the pure mapping class group of sphere w/n+ 1marked pts.

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

the(n2)-nd config. sp. of P1Q\ {0,1,∞} · · · anabelian variety!

[cf. combinatorial anabelian geometry].

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

Sn+1×dGT Out(Γb0,n+1).

Arata Minamide (RIMS, Kyoto University) The Grothendieck-Teichm¨uller group June 28, 2021 11 / 18

(41)

Denote by

Γ0,[n+1]: the mapping class group of sphere w/ n+ 1marked pts

=

ω1, ω2, . . . , ωn

“braid relations”; (ω1· · ·ωn)n+1= 1; ω1· · ·ωn−1·ω2n·ωn−1· · ·ω1= 1

.

Note: We have a commutative diagram

1 Pbn Bbn Sn 1

1 bΓ0,n+1 Γb0,[n+1] Sn+1 1.

— where Bbn0,[n+1] is defined to be σi7→ωi.

(42)

Denote by

Γ0,[n+1]: the mapping class group of sphere w/ n+ 1marked pts

=

ω1, ω2, . . . , ωn

“braid relations”; (ω1· · ·ωn)n+1= 1;

ω1· · ·ωn1·ω2n·ωn1· · ·ω1= 1

.

Note: We have a commutative diagram

1 Pbn Bbn Sn 1

1 bΓ0,n+1 Γb0,[n+1] Sn+1 1.

— where Bbn0,[n+1] is defined to be σi7→ωi.

Arata Minamide (RIMS, Kyoto University) The Grothendieck-Teichm¨uller group June 28, 2021 12 / 18

(43)

Denote by

Γ0,[n+1]: the mapping class group of sphere w/ n+ 1marked pts

=

ω1, ω2, . . . , ωn

“braid relations”; (ω1· · ·ωn)n+1= 1;

ω1· · ·ωn1·ω2n·ωn1· · ·ω1= 1

.

Note: We have a commutative diagram

1 Pbn Bbn Sn 1

1 bΓ0,n+1 Γb0,[n+1] Sn+1 1.

(44)

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 ZOut(∆)(Im(ρ)) Out(Π) Out(G)

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

Arata Minamide (RIMS, Kyoto University) The Grothendieck-Teichm¨uller group June 28, 2021 13 / 18

(45)

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 ZOut(∆)(Im(ρ)) Out(Π) Out(G)

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

(46)

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 ZOut(∆)(Im(ρ)) Out(Π) Out(G)

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

Arata Minamide (RIMS, Kyoto University) The Grothendieck-Teichm¨uller group June 28, 2021 13 / 18

(47)

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 ZOut(bΓ

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

ZGTd×S

n+1(Sn) {1}

[HMM] (n̸=6)

(48)

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 ZOut(bΓ

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

ZGTd×S

n+1(Sn) {1}

GTd

[HMM] (n̸=6)

Arata Minamide (RIMS, Kyoto University) The Grothendieck-Teichm¨uller group June 28, 2021 14 / 18

(49)

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 ZOut(bΓ

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

ZGTd×S

n+1(Sn) {1}

[HMM] (n̸=6)

(50)

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 ZOut(bΓ

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

ZGTd×S

n+1(Sn) {1}

GTd

[HMM] (n̸=6)

Arata Minamide (RIMS, Kyoto University) The Grothendieck-Teichm¨uller group June 28, 2021 14 / 18

(51)

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 ZOut(bΓ

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

ZGTd×S

n+1(Sn) {1}

[HMM] (n̸=6)

(52)

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↠ Qdef= G/N) coincides with N. Then Ker(Gb↠ Q) coincides with Nb.

Arata Minamide (RIMS, Kyoto University) The Grothendieck-Teichm¨uller group June 28, 2021 15 / 18

(53)

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↠ Qdef= G/N) coincides with N. Then Ker(Gb↠ Q) coincides with Nb.

(54)

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↠ Qdef= G/N) coincides with N. Then Ker(Gb↠ Q) coincides with Nb.

Arata Minamide (RIMS, Kyoto University) The Grothendieck-Teichm¨uller group June 28, 2021 15 / 18

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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

Sn) =Pn

= Ker(Bbn

Sn) =Pbn (cf. Lemma 3)

= Pbn⊆Bbn is characteristic!

(56)

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

Sn) =Pn

= Ker(Bbn

Sn) =Pbn (cf. Lemma 3)

= Pbn⊆Bbn is characteristic!

Arata Minamide (RIMS, Kyoto University) The Grothendieck-Teichm¨uller group June 28, 2021 16 / 18

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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

Sn) =Pn

= Ker(Bbn

Sn) =Pbn (cf. Lemma 3)

= Pbn⊆Bbn is characteristic!

(58)

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

Sn) =Pn

= Ker(Bbn

Sn) =Pbn (cf. Lemma 3)

= Pbn⊆Bbn is characteristic!

Arata Minamide (RIMS, Kyoto University) The Grothendieck-Teichm¨uller group June 28, 2021 16 / 18

(59)

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

Sn) =Pn

= Ker(Bb ↠ S ) =Pb (cf. Lemma 3)

= Pbn⊆Bbn is characteristic!

(60)

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

Sn) =Pn

= Ker(Bbn

Sn) =Pbn (cf. Lemma 3)

= Pbn⊆Bbn is characteristic!

Arata Minamide (RIMS, Kyoto University) The Grothendieck-Teichm¨uller group June 28, 2021 16 / 18

(61)

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).

(62)

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).

Arata Minamide (RIMS, Kyoto University) The Grothendieck-Teichm¨uller group June 28, 2021 17 / 18

(63)

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).

(64)

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).

Arata Minamide (RIMS, Kyoto University) The Grothendieck-Teichm¨uller group June 28, 2021 17 / 18

(65)

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).

(66)

Details of Step2

We consider the following sequence:

1 Zn ϕ Aut(Bbn) p1 Aut(Bbn) 1.

Exactness at the middle

Let α Ker(p1).

Note: α(ζn) = ζnν Zb×); α(σi) = σi·ζnei (ei Zb)

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

= ν = 1 +n(n−1)e Zn (cf. ζn = (σ1· · ·σn1)n).

= α = ϕν Im(ϕ).

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

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

(67)

Details of Step2

We consider the following sequence:

1 Zn ϕ Aut(Bbn) p1 Aut(Bbn) 1.

Exactness at the middle Let α Ker(p1).

Note: α(ζn) = ζnν Zb×); α(σi) = σi·ζnei (ei Zb)

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

= ν = 1 +n(n−1)e Zn (cf. ζn = (σ1· · ·σn1)n).

= α = ϕν Im(ϕ).

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

(68)

Details of Step2

We consider the following sequence:

1 Zn ϕ Aut(Bbn) p1 Aut(Bbn) 1.

Exactness at the middle Let α Ker(p1).

Note: α(ζn) = ζnν Zb×); α(σi) = σi·ζnei (ei Zb)

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

= ν = 1 +n(n−1)e Zn (cf. ζn = (σ1· · ·σn1)n).

= α = ϕν Im(ϕ).

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

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

(69)

Details of Step2

We consider the following sequence:

1 Zn ϕ Aut(Bbn) p1 Aut(Bbn) 1.

Exactness at the middle Let α Ker(p1).

Note: α(ζn) = ζnν Zb×); α(σi) = σi·ζnei (ei Zb)

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

= ν = 1 +n(n−1)e Zn (cf. ζn = (σ1· · ·σn1)n).

= α = ϕν Im(ϕ).

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

(70)

Details of Step2

We consider the following sequence:

1 Zn ϕ Aut(Bbn) p1 Aut(Bbn) 1.

Exactness at the middle Let α Ker(p1).

Note: α(ζn) = ζnν Zb×); α(σi) = σi·ζnei (ei Zb)

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

= ν = 1 +n(n−1)e Zn (cf. ζn = (σ1· · ·σn1)n).

= α = ϕν Im(ϕ).

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

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

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