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Presentations of Schur covers of braid groups

  • Toshiyuki Akita , Rikako Kawasaki EMAIL logo and Takao Satoh
Published/Copyright: August 16, 2023

Abstract

In this paper, we consider several basic facts of Schur covers of the symmetric groups and braid groups. In particular, we give explicit presentations of Schur covers of braid groups.

1 Introduction

For a group 𝐺, consider a central group extension

0H2(G)EϖG1

of 𝐺 by the integral second homology H2(G) of 𝐺. If the cohomology class uH2(G,H2(G)) corresponding to this central extension is mapped to idH2(G) by the homomorphism κ:H2(G,H2(G))HomZ(H2(G),H2(G)) coming from the universal coefficient theorem, the group 𝐸 is called a Schur cover of 𝐺. A concept of Schur covers was originally introduced by Schur in a series of his works [9, 10] in the study of projective representations of finite groups. In particular, he determined the Schur cover of the symmetric groups and gave a finite presentation for it. (See Theorem 2.2.)

For n1, let Bn be the braid group of 𝑛-strands. The group Bn was studied in his original works by Artin [1] who gave its first finite presentation. Let Sn be the symmetric group of degree 𝑛. There is a natural surjective homomorphism from Bn to Sn. In other words, Bn is an extension of Sn. From the viewpoint of Artin groups, Sn is the Coxeter group of Bn. Now the study of the braid groups has achieved good progress and has a broad range of remarkable results by many authors. For the study of the Schur covers of Bn, it turns out that there is a unique non-trivial central extension of Bn by H2(Bn) up to isomorphism from the universal coefficient theorem. (For details, see Section 2.2.) Namely, the Schur cover of Bn is essentially unique; denote it by Cn. On the other hand, Huebschmann [5] studied the homomorphism ϖ:CnBn from a crossed module viewpoint and showed that it is a free Bn-crossed module generated by a single generator of Bn. Thus the group Cn has two interesting features. One is the non-trivial central extension, and the other is being a free crossed module.

Hitherto, no explicit presentation of the Schur cover of Bn has been obtained. In the main theorem of this paper, we give it as follows.

Theorem 1

For any n4, a presentation for the Schur cover Cn of the braid group Bn is given by generators σ1*,,σn1*,z* subject to relations

(z*)2=1,σi*σi+1*σi*=σi+1*σi*σi+1*(1in2),σi*σj*=z*σj*σi*(i<j1),σi*z*=z*σi*(1in1).

Here ϖ:CnBn satisfies ϖ(σi*)=σi (1in1), ϖ(z*)=1.

In order to show this, we use 2-cocycles of Bn and combinatorial group theory. In Section A, as an appendix, we give explicit constructions of generators of H2(Sn,Z/2Z)(Z/2Z)2 for n4. These cocycles are used to construct those of Bn. This paper is based on the master’s thesis [8] of the second author at Hokkaido University in 2022.

2 Schur covers

First of all, we review and fix notation for Artin’s braid groups and related groups. For n1, let Bn be the braid group of 𝑛 strands. The groups Bn were introduced by Artin in 1925, and Artin gave the finite presentations for them in 1947. More precisely, Artin [1] obtained the first finite presentation of Bn as follows:

(2.1)Bn=σ1,,σn1σi+1σiσi+1σi1σi+11σi1(1in2),[σi,σj](1i<jn1,|ij|2).

By the universal mapping property, there is unique homomorphism onto Sn. It is also well known that Sn has the following finite presentation:

(2.2)Sn=s1,,sn1ai:=si+1sisi+1si1si+11si1=1(1in2),bij:=[si,sj]=1(1i<jn1,|ij|2),ci:=si2=1(1in1).

Thus we see that there is the standard surjective homomorphism π:BnSn such that π(σi)=si for 1in1.

Here, for a group 𝐺, we review the correspondence between central extensions and 2-cocycles of 𝐺. Let 𝐴 be an abelian group and

0AiEϖG1

a central extension of 𝐺 by 𝐴. Namely, i(A) is contained in the center of 𝐸. Choose a set-theoretic section s:GE of 𝜛 such that s(1)=1. Further, let fE:G×GA be the map defined by (g,h)s(g)s(h)(s(gh))1. Then fE is a normalized 2-cocycle of 𝐺 to 𝐴.

Conversely, if a normalized 2-cocycle f:G×GA of 𝐺 is given, then 𝑓 defines a central extension of 𝐺 by 𝐴 as follows. Let Ef be the set A×G equipped with the product defined by (a,g)(b,h)=(a+b+f(g,h),gh). From the 2-cocycle condition of 𝑓, Ef becomes a group. Let E(G,A) be the set of equivalence classes of central extensions of 𝐺 by 𝐴. It is well known that the correspondence [E][fE] defines a bijection E(G,A)H2(G,A), whose inverse is given by [f][Ef].

We review the Schur covers of the symmetric groups and the braid groups. Here, for the simplicity, for a group 𝐺, we will abbreviate the integral homology group Hp(G,Z) of 𝐺 as Hp(G). For any p1 and any trivial 𝐺-module 𝑀, let us consider the universal coefficient theorem

0ExtZ(Hp1(G),M)Hp(G,M)κHomZ(Hp(G),M)0.
Definition 2.1

For groups 𝐺 and 𝐸, let

(2.3)0H2(G)EϖG1

be a central extension and uH2(G,H2(G)) the cohomology class corresponding to this central extension. The central extension 𝐸 is called a Schur cover of 𝐺 if κ(u)=idH2(G).

We remark that a Schur cover is not unique in general. If 𝐺 is a perfect group, there exists a unique Schur cover up to isomorphism. Then (2.3) is called the universal central extension of 𝐺.

2.1 Schur covers of the symmetric groups

Here we review Schur covers of the symmetric group Sn. For any n4, it is known that

H1(Sn)Z/2Z,H2(Sn)Z/2Z.

Hence a Schur cover 𝐸 of Sn satisfies

(2.4)0Z/2ZESn1.

By using the universal coefficient theorem and

HomZ(H2(Sn),Z/2Z)Z/2Z,ExtZ1(H1(Sn),Z/2Z)Z/2Z,

we have H2(Sn,Z/2Z)Z/2ZZ/2Z. Hence there exist four non-equivalent central extensions which satisfy (2.4). As HomZ(H2(Sn),H2(Sn))Z/2Z consists of two elements, which are the identity map idH2(Sn) and the zero map, there exist two elements of H2(Sn,H2(Sn)) which are mapped to idH2(Sn). Each of them induces the extension (2.4) and the following commutative diagram:

Therefore, there exist two Schur covers of Sn. The 2-cocycles which correspond to the two Schur covers are given by the following theorem.

Theorem 2.2

Theorem 2.2 (Schur [9, 10] (see also [2]))

Consider Z/2Z as a multiplicative group, and let 𝑧 be its generator.

  1. For any α,βZ/2Z={1,z}, there exists the 2-cocycle

    τ[α,β]:Sn×SnZ/2Z

    such that

    τ[α,β](si,si)=α,τ[α,β](sk,sl)={1(k<l),β(l<k).

    Then we have H2(Sn,Z/2Z)={[τ[α,β]]α,β{1,z}}.

  2. Let

    0Z/2ZιE[α,β]ϖSn1

    be the central extension corresponding to [τ[α,β]]. Then Eτ[α,β] has a presentation whose generators are t1,,tn1,z subject to relations

    ti2=α,z2=1,zti=tiz,tjtj+1tj=tj+1tjtj+1,tktl=βtltk(1in1, 1jn2, 1k<l1n2).

    Here ι(z)=z, ϖ(ti)=si and ϖ(z)=1.

  3. If β=z, then for any αZ/2Z, the central extension E[α,β] is a Schur cover of Sn.

In Appendix A, we give explicit calculations to obtain the 2-cocycles τ[α,β] by using combinatorial group theory since there is no explicit calculations in [9].

2.2 Schur covers of braid groups

Here we consider a Schur cover of the braid group Bn. For any n4, it is well known that

H1(Bn)Z,H2(Bn)Z/2Z.

A Schur cover of Bn satisfies

(2.5)0Z/2ZEϖBn1.

By using the universal coefficient theorem and

Hom(H2(Bn),Z/2Z)Z/2Z,ExtZ1(H1(Bn),Z/2Z)=0,

we have H2(Bn,Z/2Z)Z/2Z. In this case, there are two non-equivalent central extensions of (2.5). Furthermore, since

κ:H2(Bn,Z/2Z)Hom(H2(Bn),Z/2Z)

is an isomorphism, the non-trivial central extension of Bn by Z/2Z is the unique Schur cover. For the standard surjection π:BnSn, consider the four 2-cocycles μ[α,β]:=τ[α,β](π×π):Bn×BnZ/2Z of Bn, and the following commutative diagram induced from the universal coefficient theorem:

For any α,βZ/2Z={1,z}, let

0Z/2ZC[α,β]Bn1

be a central extension corresponding to μ[α,β]. We obtain the following proposition.

Proposition 2.3

In the notation above,

  1. if β=1, then the central extension is trivial;

  2. C[1,z] and C[z,z] are equivalent central extensions of Bn, and each of them is a Schur cover of Bn.

3 Presentations of extensions of groups

In this section, according to [6], we review a method to obtain a presentation for an extension of presented groups. Let

1KιGqN1

be a group extension, and assume that presentations for 𝐾 and 𝑁 are given by

K=YS,N=XR

respectively. Set X*:={x*xX} and Y*:={y*yY}. Let 𝐹 be the free group on X*Y*, and set

S*:={(y1*)e1(yk*)ekFy1e1ykekS}.

For any xX, take xq1(x), and set X:={xGxX}. Let φ:FG be the surjective homomorphism defined by x*x and y*ι(y) for any x*X*, y*Y*. For any r=x1f1xlflR, set r*:=(x1*)f1(xl*)flF. Then

φ(r*)=(x1)f1(xl)flKer(q)=Im(ι).

Hence we can write φ(r*)=yr,1e1yr,kek for some yr,iY and ei=±1. Set vr:=(yr,1*)e1(yr,k*)ekF, and R*:={r*vr1FrR}. Since Im(ι) is the normal subgroup of 𝐺, we see that φ((x*)1y*x*)Im(ι) for any x*X* and y*Y*. Thus we can write φ((x*)1y*x*)=ι(yx,y,1a1yx,y,kak) for some yx,y,iY and ai=±1. Set

wx,y:=(yx,y,1*)a1(yx,y,k*)akF,T*:={(x*)1y*x*wx,y1Fx*X*andy*Y*}.
Theorem 3.1

Theorem 3.1 (Presentations of extensions of groups)

In the notation above, we have G=X*Y*R*S*T*.

For a proof, see [6].

4 Presentations of Schur covers of braid groups

In this section, we prove our main theorem. By Theorem 2.2, for any

α,βZ/2Z={1,z},

the 2-cocycle μ[α,β]:Bn×BnZ/2Z defined in Section 2.2 satisfies the following equations:

{μ[α,β](σi,σi)=α,μ[α,β](σk,σl)=1(k<l),μ[α,β](σl,σk)=β(k<l).

Let

1Z/2ZιC[α,β]ϖBn1

be the central extension corresponding to μ[α,β]. The group C[α,β] can be identified with Z/2Z×Bn as a set. Under this identification, the group operation on C[α,β] is represented as

(x,σ)(y,τ)=(xyμ[α,β](σ,τ),στ),

and the homomorphisms ι:Z/2ZC[α,β], ϖ:C[α,β]Bn are represented as ι(s)=(s,1), ϖ(x,σ)=σ. Furthermore, there exists a standard set-theoretic section BnC[α,β], σ(1,σ). Consider the presentation zz2 of Z/2Z and Artin’s presentation of Bn as mentioned in (2.1). For simplicity, we write the above presentations as Z/2=YS, Bn=XR. Let F(X) be the free group on 𝑋, and φ:F(X)Bn the standard surjection. Since φ|X:XBn is injective, we regard 𝑋 as a subset of Bn through 𝜑, and we write φ(x) as 𝑥 for any xX by abuse of language. In a similar way, we consider YZ/2Z.

We use Theorem 3.1 to obtain a presentation of C[α,β]. First, set

X*:={σ1*,,σn1*}andY*:={z*}.

Let 𝐹 be the free group F(X*Y*) on X*Y*. Here S*={(z*)2}F. Set X:={(1,σi)1in1}. The standard homomorphism ψ:FC[α,β] is defined by z*z and σi*(1,σi). For each i,j such that 1i<jn1, define rijR and rij*F by

rij:={σiσi+1σi(σi+1σiσi+1)1(i=j1),σiσj(σjσi)1(i<j1),rij*:={σi*σi+1*σi*(σi+1*σi*σi+1*)1(i=j1),σi*σj*(σj*σi*)1(i<j1).

Then we have

φ(rij*)={(1,σi)(1,σi+1)(1,σi)(1,σi+1)1(1,σi)1(1,σi+1)1(1in2),(1,σi)(1,σj)(1,σi)1(1,σj)1(i<j1).

Consider the element (1,σi)(1,σj)(1,σi)1(1,σj)1 for i<j1. By using the equations

(1,σi)(1,σj)=(11μ[α,β](σi,σj),σiσj)=(1,σiσj),(1,σj)(1,σi)=(11μ[α,β](σj,σi),σjσi)=(β,σiσj)=(β,1)(1,σiσj),

we obtain

(1,σi)(1,σj)(1,σi)1(1,σj)1=(1,σiσj)(1,σiσj)1(β,1)1=(β,1)1=(β,1).

Next, we consider the element

(1,σi)(1,σi+1)(1,σi)(1,σi+1)1(1,σi)1(1,σi+1)1for 1in2.

Similarly, we have

(1,σi)(1,σi+1)(1,σi)=(11μ[α,β](σi,σi+1)(1,σi)=(1,σiσi+1)(1,σi)=(11μ[α,β](σiσi+1,σi),σiσi+1σi),
(1,σi+1)(1,σi)(1,σi+1)=(1,σi+1)(11μ(σiσi+1),σiσi+1)=(1,σi+1)(1,σiσi+1)=(11μ[α,β](σi+1,σiσi+1),σi+1σiσi+1)=(μ[α,β](σi+1,σiσi+1),σi+1σiσi+1).

Lemma 4.1

For any α,βZ/2Z={1,z} and any 1<i<n1,

μ[α,β](σiσi+1,σi)=μ[α,β](σi+1,σiσi+1).

Proof

We use the same notation as in Theorem 2.2. Consider the central extension

1Z/2ZιE[α,β]ϖSn1

corresponding to the 2-cocycle τ[α,β]:Sn×SnZ/2Z. The group E[α,β] can be identified with Z/2Z×Sn as a set. Thus the operation on E[α,β] is given by

(x,si)(y,sj)=(xyτ[α,β](si,sj),sisj).

In this case, it follows from Schur [9] that

(1,si)(1,si+1)(1,si)=(1,si+1)(1,si)(1,si+1).

Furthermore, we see τ[α,β](sisi+1,si)=τ[α,β](si+1,sisi+1) from

(LHS)=(τ[α,β](si,si+1),sisi+1)(1,si)=(1,sisi+1)(1,si)=(τ[α,β](sisi+1,si),sisi+1si),
(RHS)=(1,si+1)(τ[α,β](si,si+1),sisi+1)=(τ[α,β](si+1,sisi+1),si+1sisi+1).
Since μ[α,β]=τ[α,β](π×π), we obtain the required result. ∎

From Lemma 4.1,

(1,σi)(1,σi+1)(1,σi)(1,σi+1)1(1,σi)1(1,σi+1)1=(μ[α,β](σiσi+1,σi),σiσi+1σi)(μ[α,β](σi+1,σiσi+1),σi+1σiσi+1)1=(1,1).

Set

R*:={σi*σi+1*σi*(σi+1*σi*σi+1*)11in2}{σi*σj*(σj*σi*)1(z*)1i<j1}.

We have φ((σi*)1z*σi*)Im(ι) for any σi*X* and z*Y*, and

φ((σi*)1z*σi*)=(1,σi)1(z,1)(1,σi)=(z,σi1)(z,1)(1,σi)=(11,σi1)(1,σi)=(z,1)=ι(z).

Hence the set T* of Theorem 3.1 is

T*={(σi*)1z*σi*(z*)1Fσi*X*,z*Y*}.

From the above calculations and Theorem 3.1, we obtain the following theorem.

Theorem 4.2

For any n4 and α,βZ/2Z={1,z}, a presentation for C[α,β] is given by generators σ1*,,σn1*,z* subject to relations

(z*)2=1,σi*σi+1*σi*=σi+1*σi*σi+1*(1in2),σi*σj*={σj*σi*(β=1,i<j1),z*σj*σi*(β=z,i<j1),σi*z*=z*σi*(1in1).

Here ϖ(σi*)=σi (1in1), ϖ(z*)=1, ι(z)=z*.

From Proposition 2.3, we have the following corollary.

Corollary 4.3

For any n4, a presentation for the Schur cover Cn of the braid group Bn is given by generators σ1*,,σn1*,z* subject to relations

(z*)2=1,σi*σi+1*σi*=σi+1*σi*σi+1*(1in2),σi*σj*=z*σj*σi*(i<j1),σi*z*=z*σi*(1in1).

Here ϖ:CnBn satisfies ϖ(σi*)=σi (1in1), ϖ(z*)=1.

A Appendix: Constructions of generators of H2(Sn,Z/2Z)

In this section, by using the presentation of Sn and combinatorial group theory, we give explicit constructions of generators of H2(Sn,Z/2Z)(Z/2Z)2 for n4. Here we do not show H2(Sn)Z/2Z, but the order of H2(Sn) is at most 2. Hence, by combining the fact that H2(Sn)Z/2Z, we obtain an explicit generator of H2(Sn). By using this, we construct generators of H2(Sn,Z/2Z).

A.1 A generator of H2(Sn,Z)

Recall the presentation (2.2) of the symmetric group Sn. Let 𝐹 be the free group on {s1,,sn1}, and φ:FSn the standard surjection defined by the correspondence si(ii+1). The kernel R:=Ker(φ) of 𝜑 is the normal closure of all of ais, bijs and cis in 𝐹. The homological five-term exact sequence of the group extension

(A.1)1RFSn1

is

0=H2(F,Z)H2(Sn,Z)H1(R,Z)Fι*H1(F,Z)φ*H1(Sn,Z)0.

Here ι:RF is the inclusion map, and the maps ι* and φ* are the induced homomorphisms by 𝜄 and 𝜑 respectively. Since 𝐹 is a free group, we have H2(F,Z)=0, and hence H2(Sn,Z)Ker(ι*). Since the signature map sgn:Sn{±1} induces the abelianization of Sn, we identify H1(Sn,Z) with {±1} via the map sgn. Consider the natural identifications

H1(R,Z)F=R/[F,R]andH1(F,Z)=F/[F,F].

For any σF, set σ:=σ(mod[F,F]). Then it follows that F/[F,F] is the free abelian group of rank n1 with basis s1,,sn1. The map φ* maps each si to −1, and it is easily seen that Ker(φ*) is the free abelian group of rank n1 with basis s2(s1)1,s3(s2)1,,sn1(sn2)1,cn1.

Next, we find a minimal generating set of R/[F,R]. For any rR, denote by r′′ the coset class of 𝑟 in R/[F,R].

Lemma A.1

For n4, R/[F,R] is generated by a1′′,,an2′′,cn1′′,b13′′ as an abelian group.

Proof

Notice that, for any rR and xF, we have xrx1r(mod[F,R]). Thus R/[F,R] is generated by the coset classes of all the relators

ai(1in2),bij(1i<jn1,|ij|2),ci(1in1)

as an abelian group. We reduce these generators. Observe that

siaisi1=ai1(si+1sisi+12si1si+11)si2,

and hence

ciai2ci+1(mod[F,R])for any 1in2.

Therefore, we can remove the generators c1′′,,cn2′′ from the generating set.

Next, for any 1i<jn2 such that |ij|2, observe

sjsj+1bijsj+11sj1=[sjsj+1sisj+11sj1,sjsj+1sjsj+11sj1]=[[sjsj+1,si]si,aj1sj+1].

By using commutator formulae

[x,yz]=[x,y][x,z][[z,x],y],[xy,z]=[x,[y,z]][y,z][x,z],

since [sjsj+1,si], aj1R, we have

[[sjsj+1,si]si,aj1sj+1][si,sj+1](mod[F,R]),

and hence bij′′=bi,j+1′′R/[F,R]. Similarly, we can see

bi1,j′′=bi,j′′R/[F,R]

for any 2i<jn1 such that |ij|2. Thus we obtain

bij′′=b13′′R/[F,R]

for any |ij|2, and the required result. ∎

Since ι* satisfies

ι*(a1′′)=s2(s1)1,,ι*(an2′′)=sn1(sn2)1,ι*(cn1′′)=cn1,

it turns out that the subgroup of R/[F,R] generated by a1′′,,an2′′,cn1′′ is a free abelian group of rank n1 and that

R/[F,R]a1′′,,an2′′,cn1′′b13′′.

We see that (bij′′)2=1 is induced from the observation

sibijsi1=si2(sjsi2sj1)bij1R.

By considering Hopf’s formula, we see

H2(Sn,Z)(R[F,F])/[F,R]=b13′′.

Since H2(Sn)Z/2Z, we see that b13′′ is a non-trivial 2-torsion element, and it generates H2(Sn).

A.2 Generators of H2(Sn,Z/2Z)

Here we give explicit constructions of generators of H2(Sn,Z/2Z)(Z/2Z)2 for n4. The Z/2Z-coefficient cohomological five-term exact sequence of the group extension (A.1) is

0H1(Sn,Z/2Z)φ*H1(F,Z/2Z)ι*H1(R,Z/2Z)FtgH2(Sn,Z/2Z)H2(F,Z/2Z)=0.

Here the maps ι* and φ* are the induced homomorphisms by 𝜄 and 𝜑 respectively. Since 𝐹 is a free group, we have H2(F,Z/2Z)=0, and hence

H2(Sn,Z/2Z)Coker(ι*).

Notice that

H1(F,Z/2Z)HomZ(F/[F,F],Z/2Z),H1(R,Z/2Z)FHomZ(R/[F,R],Z/2Z).

Consider F/[F,F] and R/[F,R] as additive groups. For any 1in1, let si*:F/[F,F]Z/2Z be the homomorphism defined by

k1s1++kn1sn1kī,

where k̄Z/2Z means the coset class of kZ. Then s1*,,sn1* form a basis of H1(F,Z/2Z) as a Z/2Z-vector space. Recall that

R/[F,R]a1′′,,an2′′,cn1′′b13′′.

Any element xR/[F,R] can be uniquely written as

x=l1a1′′++ln2an2′′+ln1cn1′′+l̄b13′′

for some liZ, l̄Z/2Z. Define ai*, cn1* and b13*HomZ(R/[F,R],Z/2Z) by

ai*(x):=lī(1in2),cn1*(x):=ln1̄,b13*(x):=l̄.

Then a1*,,an2*,cn1*,b13* is a basis of HomZ(R/[F,R],Z/2Z) as a Z/2Z-vector space. We have

ι*(si*)={a1*(i=1),ai+1*ai*(2in2),an2*(i=n1)

and see Im(ι*)=a1*,,an2*. Hence we obtain that H2(Sn,Z/2Z) is generated by tg(cn1*) and tg(b13*). Furthermore, since H2(Sn,Z/2Z)(Z/2Z)2, the elements tg(cn1*), tg(b13*) form a basis of H2(Sn,Z/2Z) as a Z/2Z-vector space.

Next we review the transgression map tg:H1(R,Z/2Z)FH2(Sn,Z/2Z). For any [f]H1(R,Z/2Z)F, we examine tg([f]). Let TF be a set of left coset representatives of 𝑅 in 𝐹. We assume that 𝑇 contains 1, s1,,sn1 and sisj for any 1i<jn1, and fix 𝑇. In the following, we define elements mσZ/2Z for any σF. First, for any τT, since

[τf]=[f]H1(R,Z/2Z)F,

there exists some mτZ/2Z such that

(τff)(x)=xmτmτ

for any xR. Since 𝐹 acts on Z/2Z trivially, we may choose mτ=0 for any τT. For any σF, there exist unique elements τT and rR such that σ=τr. Set mσ:=mτ+τ(f(r′′))=f(r′′). By using mσs for all σF, define the map αf:F×FZ/2Z by

αf(σ,ρ):=mσρσmρmσ=mσρmρmσ.

The map αf satisfies the following.

  • For any σ,ρ,μG,

    αf(ρ,μ)αf(σρ,μ)+αf(σ,ρμ)αf(σ,ρ)=0.
  • For any σ,ρG and any r,lR,

    αf(σr,ρl)=αf(σ,ρ).
Thus we see that αf induces the 2-cocycle αf̄:Sn×SnZ/2Z by

αf̄(φ(σ),φ(ρ)):=αf(σ,ρ)

and have tg([f])=[αf̄].

We give a few examples of calculations of images of αf̄ for f=cn1* and b13*. First, from the 2-cocycle condition of αf̄, we have

αf̄(φ(σ),1)=αf̄(1,φ(σ))=αf̄(1,1)=m1=0

for any σF. Notice that msi=0 for any 1in1 since siT. Hence we have

αf̄(φ(sisi+1),φ(si))=msaisi+1simsisi+1=msi+1sisi+1msisi+1=αf̄(φ(si+1),φ(sisi+1)).

Next we consider αf̄(φ(si),φ(sj)). For any 1i,jn1, we can write sisjF as τr for τT and rR as follows:

sisj={sjsi(si1sj1[si,sj]sjsi)(j<i),1si2(i=j),sisj1(i<j).

Thus, for the case where f=cn1*, since

ci′′=(ai′′)2(ai+1′′)2(an2′′)2cn1′′,

we have

msisj={1(i=j),0(ij),

and hence

αcn1*̄(φ(si),φ(sj))={1(i=j),0(ij).

On the other hand, for the case where f=b13*, we have

msisj={1(i<j),0(ij),

and hence

αb13*̄(φ(si),φ(sj))={1(j<i),0(ij).

From these facts, it is easily seen that the 2-cocycle of Sn appearing in Theorem 2.2 exists.

  1. Communicated by: Christopher W. Parker

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Received: 2023-02-01
Revised: 2023-07-01
Published Online: 2023-08-16
Published in Print: 2024-03-01

© 2023 Walter de Gruyter GmbH, Berlin/Boston

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