ANNALES
DE LA FACULTÉ DES SCIENCES
Mathématiques
STEPANYU. OREVKOV
Automorphism group of the commutator subgroup of the braid group
Tome XXVI, no5 (2017), p. 1137-1161.
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Automorphism group of the commutator subgroup of the braid group
(∗)Stepan Yu. Orevkov(1)
ABSTRACT. — LetBn0 be the commutator subgroup of the braid groupBn. We prove that Aut(Bn0) = Aut(Bn) forn>4. This answers a question asked by Vladimir Lin.
RÉSUMÉ. — SoitBn0 le groupe dérivé du groupe de tressesBn. On montre que Aut(Bn0) = Aut(Bn) pourn>4, ce qui répond à une question posée par Vladimir Lin.
1. Introduction
LetBnbe the braid group withnstrings. It is generated byσ1, . . . , σn−1
(calledstandard orArtingenerators) subject to the relations
σiσj =σjσi for|i−j|>1 ; σiσjσi=σjσiσj for|i−j|= 1. LetB0n be the commutator subgroup ofBn. Vladimir Lin [16] posed a prob- lem to compute the group of automorphisms of B0n. In this paper we solve this problem.
Theorem 1.1. — If n > 4, then the restriction mapping Aut(Bn) → Aut(B0n)is an isomorphism.
Dyer and Grossman [5] proved that Out(Bn) ∼= Z2 for any n. The only nontrivial element of Out(Bn) corresponds to the automorphism Λ defined by σi 7→ σi−1 for all i = 1, . . . , n−1. The center of Bn is gener- ated by ∆2 where ∆ = ∆n =Qn−1
i=1
Qn−i
j=1σj is Garside’s half-twist. Thus
(*)Reçu le 11 novembre 2015, accepté le 5 juillet 2016.
(1) Institut de Mathématiques de Toulouse, Université Paul Sabatier, 118 route de Narbonne, 31062 Toulouse, France — Steklov Mathematical Institute, 8 Gubkina St., 119991 Moscow, Russia — stepan.orevkov@math.univ-toulouse.fr
I am grateful to the referee for many very useful remarks.
Article proposé par Jean-Pierre Otal.
Aut(Bn) ∼= (Bn/h∆2i)o Z2. For an element g of a group, we denote the inner automorphismx7→gxg−1 by ˜g.
Corollary 1.2. — Ifn>4, thenOut(B0n)is isomorphic to the dihedral group Dn(n−1) =Zn(n−1)o Z2. It is generated by Λ and σ˜1 subject to the defining relationsΛ2= ˜σn(n−1)1 = Λ˜σ1Λ˜σ1= id.
Forn= 3, the situation is different. It is proven in [11] thatB03 is a free group of rank two generated byu=σ2σ1−1 andt=σ1−1σ2 (in fact, the free base ofB03 considered in [11] is u, v with v =t−1u). So, its automorphism group is well-known (see [18, §3.5, Theorem N4]). In particular (see [18, Corollary N4]), there is an exact sequence
1 //B03 ι //Aut(B03) π //GL(2,Z) //1
whereι(x) = ˜xandπtakes each automorphism of B03 to the induced auto- morphism of the abelianization ofB03 (which we identify withZ2 by choos- ing the images ofuand tas a base). We have ˜σ1(u) =t−1u, ˜σ2(u) =ut−1,
˜
σ1(t) = ˜σ2(t) =u, Λ(u) =t−1, Λ(t) =u−1whence π(˜σ1) =π(˜σ2) =
1 1
−1 0
, π(Λ) =
0 −1
−1 0
.
Thus, again (as in the case n > 4) the image of Aut(B3) in Out(B03) ∼= GL(2,Z) is isomorphic toD6but this time it is not the whole group Out(B03).
Let Sn be the symmetric group and An its alternating subgroup. Let µ=µn:Bn →Snbe the homomorphism which takesσito the transposition (i, i+ 1) and let µ0 be the restriction of µ toB0n. ThenPn = kerµn is the group ofpure braids. LetJn=B0n∩Pn = kerµ0. Note that the image ofµ0 isAn. The following diagram commutes where the rows are exact sequences and all the unlabeled arrows (except “→1”) are inclusions:
1 //Jn //
B0n µ
0 //
An //
1
1 //Pn //Bn
µ //Sn //1
(1.1)
Recall that a subgroup of a groupGis calledcharacteristicif it is invariant under each automorphism ofG. Lin proved in [15, Theorem D] thatJn is a characteristic subgroup of B0n forn >5 (note that this fact is used in our proof of Theorem 1.1 forn>5). By Theorem 1.1, this result extends to the casen= 4.
Corollary 1.3. — J4 is a characteristic subgroup of B04.
Note that J3 is not a characteristic subgroup of B03. Indeed, let ϕ ∈ Aut(B03) be defined by u 7→ u, t 7→ ut. Then ut ∈ J3 whereas ϕ−1(ut) = t6∈J3.
2. Preliminaries
Let e : Bn → Z be the homomorphism defined by e(σi) = 1 for all i= 1, . . . , n. Then we haveB0n = kere.
2.1. Groups
For a group G, we denote its unit element by 1, the center by Z(G), the commutator subgroup by G0, the second commutator subgroup (G0)0 byG00, and the abelianizationG/G0 byGab. We denotex−1yx byyx (thus x(y˜ x) =y) and we denote the commutator xyx−1y−1 by [x, y]. Forg ∈G, we denote the centralizer ofginGbyZ(g, G). IfH is a subgroup ofG, then, evidently,Z(g, H) =Z(g, G)∩H.
Lemma 2.1. — LetGbe a group generated by a setA. Assume that there exists a homomorphisme:G→Zsuch thate(A) ={1}. Lete¯be the induced homomorphism Gab→Z. Let Γ be the graph such that the set of vertices is Aand two verticesaandb are connected by an edge when[a, b] = 1.
If the graphΓis connected, then (kere)ab∼= ker ¯e.
Proof. — LetK= kere. Let us show thatG0 ⊂K0. SinceK0 is normal in G, andG0is the normal closure of the subgroup generated by [a, b],a, b∈ A, it is enough to show that [a, b]∈K0 for anya, b∈ A.
We define a relation ∼ on A by setting a ∼ b if [a, b] ∈ K0. Since Γ is connected, it remains to note that this relation is transitive. Indeed, if a∼b∼c, then [a, c] = [a, b][bab−2, bcb−2][b, c]∈K0.
ThusG0 ⊂K0 whenceG0 =K0 and we obtain Kab=K/K0 =K/G0 =
ker ¯e.
Remark 2.2. — The fact thatB00n=B0n forn>5 proven by Gorin and Lin [11] (see also [15, Remark 1.10]) is an immediate corollary of Lemma 2.1.
Indeed, if we setG = Bn and A = {σi}n−1i=1, then Γ is connected whence B0n/B00n= (kere)ab∼= ker ¯e={1}. In the same way we obtainG00=G0 when Gis an Artin group of type Dn (n>5),E6,E7,E8,F4, orH4.
2.2. Pure braids
Recall that Pn is generated by the braids σ2ij, 1 6 i < j 6 n, where σij =σji =σj−1. . . σi+1σiσi+1−1 . . . σj−1−1 . For a pure braidX, let us denote the linking number of thei-th andj-th strings by lkij(X). IfX is presented by a diagram with under- and over-crossings, then lkij(X) is the half-sum of the signs of those crossings where thei-th andj-th strings cross. LetAij
be the image ofσij2 inPabn . We have, evidently,
lkγ(i),γ(j)(X) = lki,j(Xγ), for any X∈Pn, γ ∈Bn (2.1) (here γ(i) = µ(γ)(i) which is coherent with the interpretation of Bn as a mapping class group; see Section 4.1).
It is well known thatPabn is freely generated by{Aij}16i<j6n. This fact is usually derived from Artin’s presentation ofPn(see [1, Theorem 1.18]) but it also admits a very simple self-contained proof based on the linking numbers.
Namely, letLbe the free abelian group with a free base{aij}16i<j6n. Then it is immediate to check that the mappingPn →L,X 7→P
i<jlki,j(X)aij is a homomorphism and that the induced homomorphism Pabn → L is the inverse ofL→Pabn ,aij 7→Aij. In particular, we see that the quotient map Pn →Pabn is given byX7→P
i<jlki,j(X)Aij.
Lemma 2.3. — Ifn>5, then the mappingJn→Pabn ,X7→Plkij(X)Aij defines an isomorphismJabn ∼={P
xijAij |P
xij= 0} ⊂Pabn .
Proof. — Follows from Lemma 2.1 with Pn, e|Pn, and {σij2}16i<j6n
standing forG,e, andArespectively.
So, when n > 5, we identify Jabn with its image in Pabn . The following proposition will not be used in the proof of Theorem 1.1.
Proposition 2.4.
(1) Jabn is a free abelian group and rkJabn =
n 2
+
( 1, n∈ {3,4},
−1, otherwise.
(2) E3={u¯t,¯t¯u,¯ u¯3,¯t3} and E4 =E3∪ {¯c2,w¯2,(¯cw)¯ 2} are free bases of Jab3 and Jab4 respectively; u, t, w, c are defined in the beginning of Section 6.
(Here and belowx¯stands for the image ofxunder the quotient mapJn→Jabn.) (3) Letpn:Jabn →Pabn ,n= 3,4, be induced by the compositionJn→
Pn →Pabn . Then impn=n X
xijAij|X
xij = 0o
, kerpn=h¯u3,¯t3i.
Proof. — (1). The result is obvious for n = 2 and it follows from Lem- ma 2.3 forn>5.
Forn = 3, the result follows from the following argument proposed by the referee. We have B03 ∼= π1(Γ) where Γ is the bouquet S1∨S1. Since
|B03/J3|= 3 (see (1.1)), we haveJab3 ∼=H1(˜Γ) where ˜Γ→Γ is a connected 3-fold covering. Then the Euler characteristic of ˜Γ is χ(˜Γ) = 3χ(Γ) = −3 whence rkH1(˜Γ) = 4.
The group Jab4 can be easily computed by the Reidemeister–Schreier method either as kerµ0 using Gorin and Lin’s [11] presentation forB04, or as ker(e|P4) using Artin’s presentation [1] ofP4. Here is the GAP code for the first method:
f:=FreeGroup(4); u:=f.1; v:=f.2; w:=f.3; c:=f.4;
g:=f/[u*c/u/w, u*w/u/w*c/w/w, v*c/v/w*c, v*w/v/w*c*c/w*c/w*c/w*c];
u:=g.1; v:=g.2; w:=g.3; c:=g.4; # group B’(4) according to [11]
s:=SymmetricGroup(4); t1:=(1,2); t2:=(2,3); t3:=(3,4);
U:=t2*t1; V:=t1*t2; W:=t2*t3*t1*t2; C:=t3*t1; # U=mu(u),V=mu(v),...
mu:=GroupHomomorphismByImages(g,s,[u,v,w,c],[U,V,W,C]);
AbelianInvariants(Kernel(mu)); # should be [0,0,0,0,0,0,0]
t
u u
u
t t
u
t
Figure 2.1. The graphs Γ and ˜Γ in the proof of Proposition 2.4 (2) forn= 3. In Figure 2.1 we show the graphs Γ and ˜Γ discussed above.
We see that the loops in ˜Γ represented by the elements ofE3form a base of H1(˜Γ).
(3) forn= 3. The claim about imp3 is evident and a computation of the linking numbers shows thatp3(¯u3) =p3(¯t3) = 0.
(2,3) forn= 4. The claim about imp4 is evident and a computation of the linking numbers shows thatp4(E4\ {t¯3,u¯3}) is a base of imp4. One can check that the homomorphismB04 → B04/K4 ∼=B03 maps J4 to J3. Hence it induces a homomorphism Jab4 → Jab3 which takes ¯u3 and ¯t3 of Jab4 to
¯
u3 and ¯t3 of Jab3 . Hence rk(kerp4) > rkh¯u3,¯t3i = 2. Since rkJab4 = 7 and rk(imp4) = 5, we conclude that kerp4=h¯u3,¯t3i.
Remark 2.5. — Note that the braid closures of bothu3 andt3are Bor- romean links. So, maybe, it could be interesting to study how the considered base ofJab3 is related to Milnor’sµ-invariant.
2.3. Mixed braid groups and the cabling map
Letn>1 andm~ = (m1, . . . , mk),m1+· · ·+mk=n,mi ∈Z,mi>0.
Themixed braid groupBm~ (see [19, 20, 10]) is defined asµ−1(Sm~) where Sm~ is the stabilizer of the following vector under the natural action of Sn
onZn:
1, . . . ,1
| {z }
m1
,2, . . . ,2
| {z }
m2
, . . . , k, . . . , k
| {z }
mk
.
We emphasize two particular cases: B1,...,1 is the pure braid group and Bn−1,1is the Artin group corresponding to the Coxeter group of typeBn−1. We define the cabling map ψ = ψm~ : Bk×(Bm1 × · · · ×Bmk) → Bn by sending (X;X1, . . . , Xk) to the braid obtained by replacing each strand of X by a geometric braid representing Xi embedded into a small tubular neighbourhood of this strand.
Note thatψm~ is not a homomorphism but its restriction toPk×Q
iBmi
is. We haveψ(Pk×Q
iPmi)⊂Pn and ψ(Pk×Q
iBmi)⊂Bm~.
3. Jabn as an An-module and its automorphisms
Letn>5. As we mentioned already, by [15, Theorem D],Jnis a charac- teristic subgroup ofB0n, i.e.,Jn is invariant under any automorphism ofB0n (in fact a stronger statement is proven in [15]).
Lemma 3.1. — Let ϕ∈Aut(B0n) be such that µ0ϕ=µ0. Let ϕ∗ be the automorphism ofJabn induced byϕ|Jn. Thenϕ∗=±id.
Proof. — The exact sequence 1 → Jn → B0n → An → 1 (see (1.1)) defines an action of An on Jabn by conjugation. The condition µ0ϕ = µ0 implies that ϕ∗ is An-equivariant. Let V be a complex vector space with basee1, . . . , en endowed with the natural action ofSninduced by the action on the base. We identifyPabn with its image in the symmetric square Sym2V by the homomorphism Aij 7→ eiej. Then, by Lemma 2.3, we may identify Jabn with{Pcijeiej |Pcij = 0}. These identifications are compatible with the action ofAn.
For a partitionλ= (λ1, . . . , λr), we denote the corresponding irreducible representation of Sn overC (theCSn-module) by Vλ, see, e.g., [7, §4]. For an elementv of aCSn-module, lethviCSn be theCSn-submodule generated
byv. We sete0=e1+· · ·+en,U =he0iCSn =Ce0, andU⊥=he1−e2iCSn. Consider the followingCSn-submodules of Sym2V:
W0=he21iCSn, W1=hwiCSn=Cw where w=P
i<jeiej, W2=h(e1−e2)(e3+· · ·+en)iCSn, W3=h(e1−e2)(e3−e4)iCSn. We have Sym2V = Sym2(U⊕U⊥) = Sym2U ⊕Sym2U⊥⊕(U ⊗U⊥) and U⊥∼=Vn−1,1 (that isVλforλ= (n−1,1)). It is known (see [17, Lemma 2.1]
or [7, Exercise 4.19]) that Sym2Vn−1,1∼=U⊕Vn−1,1⊕Vn−2,2∼=V ⊕Vn−2,2. Thus
Sym2V ∼=V ⊕V ⊕Vn−2,2. (3.1)
LetW =Jabn ⊗C. It is clear that Sym2V =W0⊕W1⊕W. SinceW0∼=V and W1∼=U, we obtainW ∼=U⊥⊕Vn−2,2by cancelling outU⊕V in (3.1). Note that (e1−e2)(e3+· · ·+en) = (e1−e2)(e0−(e1+e2)) = (e1−e2)e0−(e21−e22), hence the mappingei−ej 7→(ei−ej)e0−(e2i−e2j) induces an isomorphism ofCSn-modulesU⊥∼=W2. The identity
(n−2)(e1−e2)e3= (e1−e2)(e3+· · ·+en) +X
i>4
(e1−e2)(e3−ei) (3.2) shows that W2+W3 =h(e1−e2)e3iCSn =W. One easily checks that W2 andW3 are orthogonal to each other with respect to the scalar product on W+W1for which{eiej}i,jis an orthonormal basis. ThereforeW =W2⊕W3
is the decomposition ofW into irreducible factors.
We haveW2∼=Vn−1,1 andW3∼=Vn−2,2. Since the corresponding Young diagrams are not symmetric, W2 and W3 are irreducible as CAn-modules (see [7, §5.1]). Since dimW2 6= dimW3 and ϕ∗ is An-equivariant, Schur’s lemma implies that ϕ∗|Wk, k = 2,3, is multiplication by a constant ck. Moreover, since ϕ∗ is an automorphism of Jabn (a discrete subgroup), we have ck = ±1. If c3 = −c2 = ±1, then (3.2) contradicts the fact that
ϕ∗((e1−e2)e3)∈Jabn .
Letν ∈Aut(S6) be defined by (12)7→(12)(34)(56), (123456)7→(123)(45).
It is well known thatν represents the only nontrivial element of Out(S6).
Lemma 3.2. — Let ϕ∈Aut(B06). Thenµ0ϕ6=νµ0.
Proof. — Given a commutative ringkand akA6-moduleV correspond- ing to a representation ρ : A6 → GL(V, k), we denote the kA6-module corresponding to the representation ρν by ν∗(V). It is clear that ν∗ is a covariant functor which preserves direct sums (hence irreducibility), tensor products, symmetric powers etc.
Suppose thatµ0ϕ= νµ0. As in the proof of Lemma 3.1, we endow Jab6 with the action ofA6. The conditionµ0ϕ=νµ0 implies that ϕinduces an
isomorphism ofA6-modulesJab6 ∼=ν∗(Jab6 ). Let us show that these modules are not isomorphic.
We have Jab6 ⊗C ∼= V5,1 ⊕V4,2 (see the proof of Lemma 3.1). Hence ν∗(Jab6 )⊗C∼= ν∗(V5,1)⊕ν∗(V4,2). We have dimV5,1 = 5 6= 9 = dimV4,2, thus, to complete the proof, it is enough to show thatV5,16∼=ν∗(V5,1) (note thatV4,2∼=ν∗(V4,2)). Indeed,ν exchanges the conjugacy classes of the per- mutations a= (123) andb = (123)(456), hence we haveχ(a) = 2 6=−1 = χ(b) =χν(a) whereχandχνare the characters ofA6corresponding toV5,1
and toν∗(V5,1) respectively.
4. Centralizers of pure braids
Centralizers of braids are computed by González–Meneses and Wiest [10].
For pure braids the answer is much simpler and it can be easily obtained as a specialization of the results of [10].
4.1. Nielsen–Thurston trichotomy
The following definitions and facts we reproduce from [10, Section 2]
where they are taken from different sources, mostly from the book [12] which can be also used as a general introduction to the subject.
Let D be a disk in C that contains Xn = {1, . . . , n}. The elements of Xn will be calledpunctures. It is well known thatBn can be identified with the mapping class group D/D0 where D is the group of diffeomorphisms β:D→Dsuch thatβ|∂D= id∂Dandβ(Xn) =Xn, andD0is the connected component of the identity. Sometimes, by abuse of notation, we shall not distinguish between braids and elements ofD. ForA, B⊂D, we writeA∼B ifβ0(A) =B for some β0∈ D0.
An embedded circle inD\Xn is called an essential curve if it encircles more than one but less than n points of Xn. A multicurve in D\Xn is a disjoint union of embedded circles. It is calledessential if all its components are essential.
Letβ ∈ D. We say that a multicurveCinD\Xnisstabilized orpreserved byβ ifβ(C)∼C (the components ofC may be permuted byβ). The braid represented byβ is calledreducible ifβ stabilizes some essential multicurve.
A braidβis calledperiodicif some power ofβbelongs toZ(Bn). If a braid is neither periodic nor reducible, then it is calledpseudo-Anosov; see [12].
4.2. Canonical reduction systems. Tubular and interior braids
An essential curve C is called a reduction curve for a braid β if it is stabilized by some power ofβ and any other curve stabilized by some power ofβis isotopic inD\Xnto a curve disjoint fromC. An essential multicurve is called acanonical reduction system (CRS) forβ if its components represent all isotopy classes of reduction curves for β (each class being represented once). It is known that there exists a canonical reduction system for any braid and that it is unique up to isotopy, see [3], [12, §7], [10, §2]. If a braid is periodic or pseudo-Anosov, the CRS is empty. The following properties of CRS are immediate consequences of their existence and uniqueness.
Proposition 4.1. — Let C be the CRS forβ ∈ D. ThenC is the CRS forβ−1.
Proposition 4.2. — Let β, γ ⊂ D and let C be the CRS forβ. Then γ−1(C)is the CRS forβγ.
Proposition 4.3. — Let β, γ∈ D represent commuting braids. Then:
(1) γ preserves the CRS ofβ.
(2) Ifγ is pure, then it preserves each reduction curve of β.
Proof. — (1) follows from Proposition 4.2. (2) follows from (1) We say that a braid is in almost regular form if its CRS is a union of round circles (‘almost’ because the definition of regular form in [10] includes some more conditions which we do not need here). By Proposition 4.2 any braid is conjugate to a braid in almost regular form.
Letβ be an element of D which represents a reducible braid in almost regular form and letC be a CRS for β. Without loss of generality we may assume thatβ(C) =C andC is a union of round circles. Let R=R0∪R00 whereR0 is the union of the outermost components ofCandR00is the union of small circles around the points ofXnnot encircled by curves fromR0. Let C1, . . . , Ck be the connected components ofR numbered from left to right.
Recall that the geometric braid (a union of strings in the cylinder [0,1]× D) is obtained from β as follows. Let{βt:D→D}t∈[0,1] be an isotopy such that β0 =β, β1 = idD, andβt|∂D= id∂D for anyt. Then the i-th string of the geometric braid is the graph of the mapping t 7→ βt(i) and the whole geometric braid isS
t {t} ×βt(Xn)
. Similarly, starting from the circlesCi, we define the embedded cylinders (tubes)S
t {t} ×βt(Ci)
, i= 1, . . . , k.
Letmi be the number of punctures encircled byCi. Following [10, §5.1], we define theinterior braid β[i]∈Bmi, i= 1, . . . , k, as the element ofBmi
corresponding to the union of strings contained in the i-th tube, and we
define thetubular braid βˆofβ as the braid obtained by shrinking each tube to a single string. Letm~ = (m1, . . . , mk) and letψm~ be the cabling map (see Section 2.3). Then we haveβ =ψm~( ˆβ;β[1], . . . , β[k]).
Recall that C is a CRS for β. Leta be an open connected subset of D such that∂a⊂C∪∂D. With each suchawe associate the braid which is the union of the strings ofβ starting ataand the strings obtained by shrinking the tubes corresponding to the interior components of ∂a. We denote this braid by β[a]. For example, if a is the exterior component of D\C, then β[a]= ˆβ.
4.3. Periodic and reducible pure braids
The structure of the centralizers of periodic and reducible braids be- comes extremely simple if we restrict our attention to pure braids only. The following fact immediately follows from a result due to Eilenberg [6] and Kerékjártó [13] (see [10, Lemma 3.1]).
Proposition 4.4. — A pure braid is periodic if and only if it is a power of∆2.
The following fact can be considered as a specialization of the results of [10].
Proposition 4.5. — Let β be a pure n-braid.
(1) Ifβ is periodic, then Z(β;Pn) =Pn.
(2) Ifβ is pseudo-Anosov, thenZ(β;Pn) is the free abelian group gen- erated by∆2 and some pseudo-Anosov braid which may or may not coincide withβ.
(3) If β is reducible non-periodic and in almost regular form, then ψm~ maps Z( ˆβ;Pk)×Z(β[1];Pm1)× · · · ×Z(β[k];Pmk) isomorphically ontoZ(β;Pn) (see Section 4.2).
Proof. — (1). Follows from Proposition 4.4.
(2). Follows from [10, Proposition 4.1].
(3). (See also the proof of [10, Proposition 5.17]). By Proposition 4.3 we haveZ(β;Pn)⊂ψm~(Pk×Q
Pmi). The injectivity of the considered mapping and the fact that ψ−1m~ (Z(β;Pn)) is as stated, are immediate consequences from the following observation: if two geometric braids are isotopic, then the braids obtained from them by removal of some strings are isotopic as
well.
Lemma 4.6. — Let m~ = (m1, . . . , mk), m1+· · ·+mk =n, andp∈Z. Thenψm~(∆pk; ∆pm
1, . . . ,∆pm
k) = ∆pn.
Proof. — The result immediately follows from the geometric characteri- zation of ∆ as a braid all whose strings lie on a half-twisted band. Note that the sub-bands of the half-twisted band arising from consecutive strings also
consist of half-twisted bands.
IfX is a periodic pure braid, thenX = ∆2d, d∈Z, by Proposition 4.4.
In this case we setd= degX, the degree ofX. It is clear that lkij(X) =d for anyi < j.
Lemma 4.7. — Let C be the CRS for a reducible pure braid represented by β ∈ D. Let a and b be two neighboring components of D\C and let X =β[a] andY =β[b] be the braids associated with aandb (see the end of Section 4.2). Suppose that each ofX andY is periodic. ThendegX6= degY. Proof. — Suppose that degX = degY =p, i.e.,X = ∆2pk and Y = ∆2pm for some k, m> 2. Let Ci be the component of C that separates aand b.
We may assume thatais exterior toCi. Letcbe the closure ofa∪b. Then we have
β[c] =ψ1,...,1,m,1,...,1(∆2pk ; 1, . . . ,1,∆2pm,1, . . . ,1) = ∆2pk+m−1
by Lemma 4.6. Henceβ[c] preserves any closed curve, in particular a curve which separates some two strings of β[b] and encircles a string of β[c] not belonging toβ[b]. Such a curve is not isotopic to any curve disjoint fromCi. This fact contradicts the condition thatCi is a reduction curve.
Lemma 4.8. — Z(σ12σ−23 ;Jn)∼=Pn−2×Z forn>4.
Proof. — The CRS forσ21σ3−2 consists of two round circles: one of them encircles the punctures 1 and 2, and the other one encircles the punctures 3 and 4. Then Proposition 4.5 (3) implies that ψ = ψm~ : Pn−2 ×(P2 × P2) → Pn, m~ = (2,2,1n−4), is injective and imψ = Z(σ21σ−23 ;Pn). One easily checks that the mapping Pn−2 ×P2 → Z(σ12σ3−2;Jn), (X, σ1k) 7→
ψ(X;σ1k, σ−m1 ),m =e(ψ(X; 1,1)) +k, is an isomorphism. Indeed, any ele- mentY ofZ(σ12σ3−2;Pn) is of the formY =ψ(X;σ1k, σ1−m) and the condition
e(Y) = 0 becomese(ψ(X; 1,1)) +k−m= 0.
Lemma 4.9. — Letβbe a reduciblen-braid in almost regular form. Sup- pose that βˆ ∈ Pk and that the β[i]’s (see Section 4.2) are pairwise non- conjugate. Thenψm~ mapsZ( ˆβ;Pk)×Z(β[1];Bm1)× · · · ×Z(β[k];Bmk)iso- morphically ontoZ(β;Bn)
Proof. — See the proof of Proposition 4.5 (3).
5. Proof of Theorem 1.1 forn>5 5.1. Invariance of the conjugacy class ofσ1σ−13
Suppose thatn>5. Letϕ∈Aut(B0n) be such thatµ0ϕ=µ0andϕ∗= id whereϕ∗ is as in Lemma 3.1. Then we have
lki,j(X) = lki,j(ϕ(X)), X ∈Jn, 16i < j6n. (5.1) Letτ =ψ2,n−2(1;σ(n−2)(n−3)1 ,∆−2). We haveτ ∈Jn.
Lemma 5.1. — Let X be σ12σ3−2 or τ. Let α ∈ D represent ϕ(X). Let C be a simple closed curve preserved byα. Suppose thatC encircles at least two punctures. Then the punctures 1 and 2 are in the same component of D\C.
Proof. — Suppose that 1 and 2 are separated byC. Without loss of gen- erality we may assume that 1 is outsideCand 2 is insideC. Letpbe another puncture insideC. Then we have lk1,p(α) = lk1,2(α) which contradicts (5.1) because lk1,2(X)6= 0 and lk1,p(X) = 0 for anyp6= 2.
Lemma 5.2. — Let α∈ D representϕ(σ12σ3−2). Then the CRS for αis invariant under some element ofDwhich exchanges {1,2} and{3,4}.
Proof. — Follows from Propositions 4.1 and 4.2 because αis conjugate toα−1 and the conjugating element ofDexchanges{1,2} and{3,4}.
Lemma 5.3. — Let α∈ Drepresent ϕ(τ). LetC be a component of the CRS forα. ThenC cannot separate iandj for all36i < j6n.
Proof. — Let β ∈ D represent ϕ(σij2σ1−2). Since α and β commute, β preserves C by Proposition 4.3 (2). Hence C cannot separate i and j by Lemma 5.1 applied toβ(note thatβis conjugate toσ12σ−23 ; see the beginning
of Section 5.2).
Lemma 5.4. — Let α∈ D representϕ(σ12σ−23 ). Suppose that n>6. Let C be a component of the CRS forα. Then:
(1) C cannot separate1 and2. It cannot separate 3 and4.
(2) C cannot separatei andj for56i < j6n.
(3) C cannot separate{1,2,3,4} from {5, . . . , n}.
(4) C cannot encircle5, . . . , n.
Proof. — (1). Follows from Lemma 5.1 and Lemma 5.2.
(2). Letβ∈ Drepresentϕ(σ2ijσ−21 ). Sinceαandβ commute,β preserves C by Proposition 4.3 (2). Hence C cannot separate i and j by Lemma 5.1 applied toβ (see the proof of Lemma 5.3).
(3). Suppose thatC separates 1,2,3,4 from 5,6, . . . , n. Let β ∈ D rep- resent ϕ(σ12σ5−2). Then β is conjugate to α. Let γ ∈ D be a conjugating element. Thenγ(C) is a component of the CRS forβ and it separates the punctures 1,2,5,6 from all the other punctures. Sinceαandβ commute,β preserves C. This is impossible because the geometric intersection number ofCandγ(C) is nonzero.
(4). Combine (1), (3), and Lemma 5.2.
Lemma 5.5. — Let α ∈ D represent ϕ(σ12σ−23 ). Suppose that α is re- ducible non-periodic. Then the CRS forα has exactly two components: one of them encircles1and2, and the other one encircles3 and4.
Proof. — If n> 6, the result follows from Lemma 5.2 and Lemma 5.4.
Suppose thatn= 5 and the CRS is not as stated. By combining Lemma 5.2 with Lemma 4.7, we conclude that the CRS consists of a single circle which encircles 1, 2, 3, 4. The interior braid cannot be periodic by (5.1), hence it is pseudo-Anosov. Therefore,Z(α;P5)∼=Z2 by Proposition 4.5 (2) whence
Z(α;J5) =Z. This contradicts Lemma 4.8.
Lemma 5.6. — ϕ(σ1σ−13 ) is conjugate inBn toσ1σ−13 .
Proof. — Let α ∈ D represent ϕ(σ21σ−23 ). If α is pseudo-Anosov, then Z(α;Pn)∼=Z2 by Proposition 4.5 (2), hence Z(α;Jn) is abelian which con- tradicts Lemma 4.8. If αis periodic, then it is a power of ∆2 by Proposi- tion 4.4. This contradicts (5.1), hence α is reducible non-periodic and its CRS is as stated in Lemma 5.5.
Suppose that ˆα is pseudo-Anosov. Then Z( ˆα;Pn) ∼= Z2 by Proposi- tion 4.5 (2) whence Z(α;Pn) ∼= Z4 by Proposition 4.5 (3) and therefore Z(α;Jn) is abelian which contradicts Lemma 4.8. Thus, ˆα is periodic. By Proposition 4.4 this means that ˆα is a power of ∆2. This fact combined with (5.1) implies ˆα= 1. It follows that ϕ(σ21σ−23 ) is conjugate to σ12kσ−2k3 for some k, and we have k = 1 by (5.1). The uniqueness of roots up to conjugation [9] implies thatϕ(σ1σ−13 ) is conjugate toσ1σ−13 .
Lemma 5.7. — ϕ(τ)is conjugate in Pn toτ.
Proof. — Let α ∈ D represent ϕ(τ). By Proposition 4.3, it cannot be pseudo-Anosov because it commutes withϕ(σ1σ3−1) which is reducible non- periodic by Lemma 5.6. Ifαwere periodic, then it would be a power of ∆2 by Proposition 4.4. This contradicts (5.1), henceαis reducible.
LetC be the CRS for α. By Lemmas 5.1 and 5.3, one of the following three cases occurs.
Case 1. — C is connected, the punctures 1 and 2 are inside C, all the other punctures are outsideC. Then the tubular braid ˆαcannot be pseudo- Anosov becauseαcommutes withϕ(σ1σ3−1), hence it preserves a circle which
separates 3 and 4 from 5, . . . , n. Hence ˆαis periodic which contradicts (5.1) combined with Proposition 4.4. Thus this case is impossible.
Case 2. — C is connected, the punctures 1 and 2 are outsideC, all the other punctures are insideC. This case is also impossible and the proof is almost the same as in Case 1. To show that ˆαcannot be pseudo-Anosov, we note thatαpreserves a curve which encircles only 1 and 2.
Case 3. — C has two components:c1 andc2 which encircle {1,2} and {3, . . . , n}respectively. The interior braid α[2] cannot be pseudo-Anosov by the same reasons as in Case 1, becauseαpreserves a circle separating 3 and 4 from 5, . . . , n. Henceα[2] is periodic. Using (5.1), we conclude thatαis a conjugate of τ. Since the elements of Z(τ;Bn) realize any permutation of {1,2}and{3, . . . , n}, the conjugating element can be chosen inPn.
Lemma 5.8. — There existsγ∈Pn such that
ϕ(σ1σ−1i ) = (σ1σ−1i )γ fori= 3, . . . , n . (5.2) Proof. — Due to Lemma 5.7, without loss of generality we may assume that ϕ(τ) = τ and τ(C) = C where C is the CRS forτ consisting of two round circlesc1 andc2which encircle{1,2}and {3, . . . , n}respectively.
By Lemma 4.9,ψ2,n−2restricts to an isomorphismψ:P2×B2×Bn−2→ Z(τ) :=Z(τ;Bn). Letπ1:Z(τ)→P2 andπ3:Z(τ)→Bn−2 be defined as πi= pri◦ψ−1.
LetH =π−11 (1)∩B0n; note that the elements ofπ−11 (1) correspond to geometric braids whose first two strings are inside the cylinder [0,1]×c1and the other strings are inside the cylinder [0,1]×c2. Thenπ3|H :H →Bn−2is an isomorphism and its inverse is given by Y 7→ψ2,n−2(1, σ1−e(X), Y), that isσi7→σ1−1σi+2,i= 1, . . . , n−3.
Let us show thatϕ(H) = H. Indeed, let X ∈ H. Since X ∈ Z(τ;B0n) andϕ(τ) =τ, we haveϕ(X)∈Z(τ;B0n). The fact that π1(X) = 1 follows from (5.1) applied to a power of X belonging toJn. Hence ϕ(H)⊂H. By the same argumentsϕ−1(H)⊂H.
Thusϕ|H is an automorphism of H and we haveH ∼=Bn−2. Hence, by Dyer and Grossman’s result [5] cited after the statement of Theorem 1.1, there existsγ ∈H such that ˜γϕ|H is either idH or Λ|H. The latter case is impossible by (5.1). Thus there existsγ∈Bn such that (5.2) holds.
It remains to show thatγcan be chosen inPn. By replacingγwithσ1γ if necessary, we may assume that 1 and 2 are fixed byγ. By combining (2.1), (5.1), and (5.2), we conclude thatγ({i, j}) ={i, j} for any i, j∈ {3, . . . , n}
and the result follows.
5.2. Conjugates ofσ1 and simple curves which connect punctures
We fixn>2 and we considerDand the set of puncturesXn={1, . . . , n} ⊂ Das above. LetIbe the set of all smooth simple curves (embedded segments) I⊂Dsuch that∂I⊂Xn andI◦⊂D\Xn. Here we denoteI◦=I\∂I and
∂I ={a, b} wherea andb are the ends ofI. Recall that we write I∼I1 if I1 =α(I) for someα∈ D0 (see Section 4.1), i.e., if I andI1 belong to the same connected component ofI.
LetI ∈ I and let β ∈ Dbe such that β(I) is the straight line segment [1,2]. Then we define the braid σI as σ1β. It is easy to see thatσI depends only on the connected component ofI that containsI. The CRS for σI is a single closed curve which encloses I and separates it from Xn\∂I. By definition, all conjugates of σ1 are obtained in this way. In particular, we haveσi =σ[i,i+1] andσij =σI for an embedded segmentI which connects itoj passing through the upper half-plane.
Lemma 5.9. — For any β∈ D,I∈ I, we haveσββ(I)=σI.
With this notation, a corollary of Lemma 5.8 can be formulated as follows.
Lemma 5.10. — Letn>5 and letϕ∈Aut(Bn0) be as in Section 5.1.
(1) Let I, J ∈ I be such that Card(I∩J) = Card(∂I∩∂J) = 1 (i.e., I∩J is a common endpoint ofI andJ). Then there existI1, J1∈ I such thatI1∪J1 is homeomorphic to I∪J and
ϕ(σIσ−1J ) =σI1σJ−1
1 , ϕ(σ−1I σJ) =σI−1
1 σJ1. (5.3) (2) Let I, J ∈ I, I∩J = ∅. Then the conclusion is the same as in
Part (1).
(3) Let I andJ be as in Part (1) and letK∈ I be such thatK∩(I∪ J) = ∅. Then there exist I1, J1, K1 ∈ I such that I1∪J1∪K1 is homeomorphic toI∪J∪K, and (5.3)holds as well as
ϕ(σKσI−1) =σK1σI−1
1 , ϕ(σKσJ−1) =σK1σJ−1
1 . (5.4)
Proof. — (3). Let γ be as in Lemma 5.8 and let β ∈ D be such that β(K) = [1,2],β(I) = [3,4], andβ(J) = [4,5]. We set K1 =α−1(K), I1 = α−1(I),J1=α−1(J) whereα=β−1γϕ(β). Then we have
ϕ(σKσI−1) =ϕ((σ1σ3−1)β) by definition ofσI andσK
= (σ1σ3−1)γϕ(β) by Lemma 5.8
=σK1σI−1
1 by Lemma 5.9
and, similarly, ϕ(σKσ−1J ) =σK1σJ−1
1. Since σK commutes with σI and σJ, we haveσεIσJ−ε= (σKσ−1I )−ε(σKσJ−1)ε,ε=±1, thus (5.4) implies (5.3).
