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The inverse braid monoid
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Advances in Mathematics 186 (2004) 438–455
http://www.elsevier.com/locate/aim
The inverse braid monoid D. Easdown and T.G. Lavers School of Mathematics and Statistics, University of Sydney, Sydney, NSW 2006, Australia Received 19 July 2001; accepted 10 July 2003 Communicated by Ross Street
Abstract We introduce an inverse monoid which plays a similar role with respect to the symmetric inverse semigroup that the braid group plays with respect to the symmetric group. r 2003 Elsevier Inc. All rights reserved. MSC: 20F36; 20M18 Keywords: Braid groups; Inverse monoids
1. Introduction and preliminaries We describe an inverse monoid which arises naturally when considering usual geometric braids from which strings may have been removed. This monoid contains many copies of the Artin braid groups and may be described by an elegant presentation, from which we can quickly deduce Popova’s presentation [13] (see also [9,11]) of the symmetric inverse monoid. For basic terminology and facts about inverse semigroups we refer the reader to [7,8] or [12]. We denote the symmetric inverse monoid on n letters by In : We denote by group/X jRS
and
monoid/X jRS
group and monoid presentations, respectively, where X is the alphabet of generators and R the set of relations. We will consistently use plain angular brackets to
Corresponding author. E-mail addresses: [email protected] (D. Easdown), [email protected] (T.G. Lavers).
0001-8708/$ - see front matter r 2003 Elsevier Inc. All rights reserved. doi:10.1016/j.aim.2003.07.014
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have the following meaning: if M is a monoid and X is a subset of the group of units of M; then /X S denotes the subgroup of (the group of units of) M generated by X : Sequences are used throughout this paper and we alert the reader to an important convention. It is standard usage that fxi ; y; xj g denotes the empty sequence if either (i) i ¼ j þ 1 and the subscripts are ascending; or (ii) j ¼ i þ 1 and the subscripts are descending. We extend this notation to words over an alphabet fs1 ; y; sn1 g so that the word si ysj denotes the empty word (or group identity element) if either (i) or (ii) occurs. The reader will see from context whether subscripts are intended to be ascending or descending. Artin’s braid group is defined in [1,4,6]. We adapt the treatment given in the first few pages of [6] to give a sequence of definitions, leading below to the definition of IBn ; the inverse braid monoid on n strings. Take the usual coordinate system for R3 (in which for descriptive purposes we think of the z-axis as pointing downwards). Choose z0 oz1 and call the planes z ¼ z0 and z1 upper and lower, respectively. Mark nX1 distinct points P1 ; y; Pn on a line in the upper plane and project them orthogonally onto the lower plane yielding points P01 ; y; P0n : An arc is the image of an embedding from the unit interval ½0; 1 into R3 : A (partial) braid on n strings is a system b ¼ fb1 ; y; bm g of m arcs for some mpn such that (i) there is a rank m partial one–one mapping bF : f1; y; ng-f1; y; ng with domain fi1 ; y; im g such that bj connects Pij to P0ij bF for j ¼ 1; y; m; (ii) each arc intersects each intermediate parallel plane between (and including) the upper and lower planes exactly once; (iii) the union b1 ,?,bm of the arcs intersect each intermediate parallel plane between the upper and lower planes in exactly m distinct points. We may think of the object just defined as a usual (full-string) braid with n m strings removed. When m ¼ 0 we get the empty braid. Two braids b ¼ fb1 ; y; bm g and g ¼ fg1 ; y; gm0 g are called equivalent if (i) bF ¼ gF; so that in particular m ¼ m0 and we may write the domain of bF as fi1 ; y; im g for some i1 o?oim ; and (ii) b and g are homotopy equivalent, by which we mean that there exist continuous maps Fj : ½0; 1 ½0; 1 -R3 for j ¼ 1 to m; such that, for all tA½0; 1 ; Fj ðt; 0Þ ¼ bj ðtÞ and
Fj ðt; 1Þ ¼ gj ðtÞ;
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for all sA½0; 1 ; Fj ð0; sÞ ¼ Pij
and
Fj ð1; sÞ ¼ P0ij bF ;
and, for each sA½0; 1 ; if we define bs ¼ fbs1 ; y; bsm g where bsj ðtÞ ¼ Fj ðs; tÞ
for j ¼ 1 to m;
then bs is itself a braid. If m ¼ n then this reduces to the definition in [6] for equivalence of braids without missing strings. Write b g if b and g are equivalent and ½b for the equivalence class containing b: Define the product or composite b1 b2 of two braids b1 and b2 as follows: (i) translate b2 parallel to itself in space so that the upper plane of b2 and the lower plane of b1 coincide (so that b2 hangs below b1 ); (ii) keeping the plane z ¼ z0 fixed, contract the resulting systems of arcs so that the translated lower plane of b2 moves into the position of the plane z ¼ z1 ; (iii) remove any arcs that do not now join the upper and lower planes. For example, if
then a2 ; ab; ba are obtained as follows:
so that a2
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Clearly if b b0 and g g0 then bg b0 g0 so that the operation ½b ½g ¼ ½bg
on equivalence classes of braids is well defined. It is easy to see that this operation is associative. Put Bn ¼ f½b jb is a braid and bF is a permutation of f1; y; ngg and Gn ¼ group/s1 ; y; sn1 jsi sj ¼ sj si si siþ1 si ¼ siþ1 si siþ1
if ji jj41;
for i ¼ 1 to n 2S:
Then Bn is the usual (Artin) braid group, and we recall Artin’s Theorem: Theorem 1.1 (Artin [1–3], Birman [4], Hansen [6]). The groups Gn and Bn are isomorphic. The relations that appear in the above presentation are referred to as braid relations. The isomorphism in Artin’s Theorem identifies each si with the
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equivalence class of the braid
Now let IBn denote the set of equivalence classes of all braids which clearly forms a monoid equipped with the product given earlier. We will show that IBn is inverse. Note that clearly the map ½b /bF is a well-defined epimorphism from IBn to In : Let ½b AIBn : Define b1 to be the mirror image of b after reflecting through a parallel plane midway between the upper and lower planes. Clearly, if b g then b1 g1 so that we may sensibly define ½b 1 ¼ ½b1 : For any subset X ¼ fi1 ; y; im g of f1; y; ng denote the following braid by eX ; consisting of m straight arcs connecting Pij to P0ij for j ¼ 1 to m:
Then it is easy to see that bb1 eX and b1 b eY where X and Y are the domain and range, respectively, of bF; from which it follows that ½b ½b 1 ½b ¼ ½b
and
½b 1 ½b ½b 1 ¼ ½b 1 :
Thus ½b 1 is an inverse of ½b (which we shall see shortly is unique). In particular IBn is regular. Let ½d AIBn be an idempotent. Then ðdFÞ2 ¼ dF; so that dF is an idempotent in In ; whence dF is the identity map on its domain X : But then ½d lies in a copy of the braid group on jX j strings, so must coincide with its group identity element, that is ½d ¼ ½eX : This proves that the set EðIBn Þ of idempotents of IBn is precisely f½eX jX Df1; y; ngg: Clearly ½eX ½eY ¼ ½eX -Y for any subsets X and Y of f1; y; ng; so that EðIBn Þ forms a commutative subsemigroup of IBn : Thus we have proved
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Theorem 1.2. IBn is an inverse monoid with semilattice of idempotents isomorphic to the semilattice of subsets of f1; y; ng under intersection. 2. Coset decompositions We list some useful (and probably well-known) coset decompositions of certain subgroups of the braid group. Throughout this section nX1 is fixed and (in view of Artin’s Theorem) we may regard Gk as embedded in Gn in the obvious way when kpn: Now fix kpn 1: Define the following subgroups of Gn : H ¼ /s1 ; y; sn2 ; s2n1 S; K ¼ /skþ1 ; ysn1 ; x1 s2k xjxAGk S; L ¼ /s1 ; y; sk1 ; s2k ; skþ1 ; y; sn1 S (so if k ¼ n 1 then H ¼ L). Lemma 2.1. We have the coset decompositions Gn ¼ H,Hsn1 ,Hsn1 sn2 ,?,Hsn1 ys1 [ ¼ si1 ys1 si2 ys2 ysik ysk L: 0pi1 o?oik pn1
Proof. The coset decomposition for H follows from that of L (in the case k ¼ 1 and by observing that the map si /sni induces an anti-isomorphism of Gn ). Put W ¼ fsi1 ys1 ysik ysk j 0pi1 o?oik pn 1g: Suppose 0pi1 o?oik pn 1 and put w ¼ si1 ys1 ysik ysk : Let cAf1; y; n 1g and eAf71g: To prove the lemma it suffices to show that sec wLDWL: If im ¼ m 1 for m ¼ 1; y; k; then w is empty, so sec wL ¼
L sc L
if cak; if c ¼ k;
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sc AW ; and we are done. Suppose then that, for some mAf1; y; kg; is ¼ s 1
for 1pspm 1;
and
im Xm;
so w ¼ sim ysm ysik ysk : The proof now splits into cases. Case (i): If cpm 2 then, by commutativity relations, sec wL ¼ wsec L ¼ wL; and we are done. Case (ii): Suppose c ¼ m 1: If e ¼ 1 then sec wL ¼ sm1 wL and sm1 wAW and we are done. If e ¼ 1 then, by repeated use of the braid relations, sec wL ¼ sm1 s2 m1 wL 1 ¼ sm1 sim ysmþ1 s2 m1 sm ðsm1 sm1 Þsimþ1 ysmþ1 ysik ysk L 1 ¼ sm1 sim ysmþ1 sm ðsm1 s2 m sm1 Þsimþ1 ysmþ1 ysik ysk L
¼y 1 ¼ sm1 sim ysm ysik ysk ðsm1 ysk1 s2k s1 k1 ysm1 ÞL
¼ sm1 wL; and again we are done. Case (iii): If mpcoim then, using the braid relations, sec wL ¼ sim yscþ2 sec scþ1 sc ysm ysik ysk L ¼ sim yscþ2 scþ1 sc secþ1 sc1 ysm ysik ysk L ¼ sim ysm secþ1 simþ1 ysmþ1 ysik ysk L ¼y ¼ sim ysm ysik ysk secþkmþ1 L ¼ wL; at the last step because c þ k m þ 1Xk þ 1; and we are done. Case (iv): Suppose c ¼ im : If e ¼ 1 then sec w ¼ sim 1 ysm simþ1 ysmþ1 ysik ysk AW ; and we are done. If e ¼ 1 then, by the braid relations and the calculation in case (iii), and since sm ysk1 s2k ðsm ysk1 Þ1 AL; sec wL ¼ s2im sim 1 ysm simþ1 ysmþ1 ysik ysk L 1 ¼ sim 1 ysm ðsim ysmþ1 Þs2m s1 mþ1 ysim simþ1 ysmþ1 ysik ysk L
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¼ sim 1 ysm ðsim ysmþ1 Þs2m simþ1 ysmþ1 ysik ysk L ¼ sim 1 ysm ðsim ysmþ1 Þsimþ1 ysmþ1 ysik ysk ðsm ysk1 Þs2k ðsm ysk1 Þ1 L ¼ sim 1 ysm ðsim ysmþ1 Þsimþ1 ysmþ1 ysik ysk L ¼ sim 1 ysm simþ1 ysmþ1 ysik ysk L; and again we are done. Case (v): Suppose c ¼ im þ 1: If im þ 1oimþ1 then sc wAW ; whilst also, by the braid relations and the calculations in cases (ii) and (iii), 1 s1 c wL ¼ sim þ1 sim ysm ysik ysk L 1 1 ¼ sim ysm ðsim þ1 ysmþ1 Þs1 m ðsmþ1 ysim þ1 Þ
simþ1 ysmþ1 ysik ysk L ¼ sim ysm ðsim þ1 ysmþ1 Þs1 m simþ1 ysmþ1 ysik ysk L ¼ sim ysm ðsim þ1 ysmþ1 Þsm simþ1 ysmþ1 ysik ysk L 1 ¼ sim ysm ðsim þ1 ysmþ1 Þsm ðs1 mþ1 ysim þ1 Þ
simþ1 ysmþ1 ysik ysk L ¼ sim þ1 wL; and we are done. On the other hand, if im þ 1 ¼ imþ1 then, by the braid relations and the calculation in case (ii), sec wL ¼ seim þ1 sim ysm simþ1 ysmþ1 ysik ysk L ¼ sim ysm simþ1 ysmþ2 semþ1 simþ2 ysmþ2 ysik ysk L ¼ sim ysm simþ1 ysmþ2 smþ1 simþ2 ysmþ2 ysik ysk L ¼ wL; and again we are done. Case (vi): If c4ik þ 1 then c4k þ 1; so, by commutativity relations, sec wL ¼ wsec L ¼ wL; and we are done. Case (vii): Suppose im þ 1ocpik þ 1; so iN1 þ 1ocpiN þ 1 for some NAfm þ 1; y; kg:
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Note that cXN þ 1: Then sec wL ¼ sim ysm ysiN1 ysN1 sec siN ysN ysik ysk L: We have four subcases: If coiN then, by the calculation in case (iii), sec wL ¼ wL: If c ¼ iN then, by the calculation in case (iv), sec wL ¼ sim ysm ysiN1 ysN1 siN 1 ysN ysik ysk L: If c ¼ iN þ 1 and iN þ 1oiNþ1 then, by the first calculation in case (v), sec wL ¼ sim ysm ysiN1 ysN1 siN þ1 ysN ysik ysk L: If c ¼ iN þ 1 and iN þ 1 ¼ iNþ1 then, by the second calculation in case (v), sec wL ¼ wL: In each of these subcases sec wL ¼ w0 L for some w0 AW ; so we are done. This exhausts all possibilities and the proof of Lemma 2.1 is complete. & Lemma 2.2. We have the coset decomposition L ¼
S
xAGk
xK:
71 Proof. Let xAGk : It suffices to check s72 k xAGk K and si xAGk K for iak: But certainly 1 72 s72 k x ¼ xðx sk xÞAxK;
since x1 s72 k xAK: If iXk þ 1 then 71 s71 i x ¼ xsi AxK; 71 since s71 i AK: If ipk 1 then si xAGk ; and we are done. &
3. A presentation for the inverse braid monoid In this section we provide a monoid presentation for IBn : For the remainder of this paper we will denote the equivalence class of a braid by the braid itself. Put 1 1 1 Mn ¼ monoid/s1 ; y; sn1 ; s1 1 ; y; sn1 ; ejsi si ¼ si si ¼ 1 for all i;
braid relations for s1 ; y; sn1 ; e2 ¼ e ¼ es2n1 ¼ s2n1 e; esi ¼ si e for ipn 2; esn1 e ¼ sn1 esn1 e ¼ esn1 esn1 S: We will prove
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Theorem 3.1. The monoids Mn and IBn are isomorphic. Let C denote the mapping from the set of generators for Mn into IBn given by, for each i;
It is easy to check that the relations in the presentation for Mn hold for the images of the generators, so that C induces a well-defined homomorphism, which we also denote by C; from Mn into IBn : To prove Theorem 3.1 it suffices to show that C is one–one and onto. Proof that C is onto: For the time being, we identify the generators of Mn with 71 their images under C: It suffices to prove that s71 1 y; sn1 ; e generate IBn : Let yAIBn ; and suppose that y possesses k strings. Put ek ¼ ef1;y;kg ; the identity braid with the last n k strings removed. Thus e0 is the empty braid, en ¼ 1; en1 ¼ e; and, by a straightforward calculation,
ek ¼ esn1 yskþ1 esn1 yskþ2 eyesn1 sn2 esn1 e:
Then by stretching near the upper and lower planes (if necessary) we see that y has a decomposition of the form y ¼ abg; where the following picture holds for some
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sequences i1 ; y; ik and j1 ; y; jk :
But b ¼ ek b0 for some b0 ABn ; and, by a simple calculation, a ¼ ðsi1 ys1 si2 ys2 ysik ysk Þek ; g ¼ ek ðsk ysjk sk1 ysjk1 ys2 ysj2 s1 ysj1 Þ; 71 each of which are combinations of s71 1 ; y; sn1 ; e; completing the proof that C is onto. 71 Proof that C is one–one: Consider the alphabet S ¼ fs71 1 ; y; sn ; eg: To prove that C is one–one it suffices to find a set W of words (normal forms) such that, with respect to the congruence generated by the relations given in the presentation for Mn ; each word over S is congruent to a word in W ; and C is one–one on the congruence classes of words in W : Without causing confusion, we will identify words with their congruence classes.
Lemma 3.2. Suppose ipjpn 1: Then esn1 ysj esn1 ysi ¼ esn1 ysi esn1 ysjþ1 : Proof. We use the braid relations and the relations esn1 e ¼ esn1 esn1
and
esk ¼ sk e
for kpn 2:
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An induction begins because, if j ¼ n 1; then esn1 ysj esn1 ysi ¼ esn1 esn1 ysi ¼ esn1 esn2 ysi ¼ esn1 ysi e: If jon 1; then, by an inductive hypothesis, esn1 ysj esn1 ysi ¼ esn1 ysjþ1 esn1 ysjþ2 ðsj sjþ1 sj Þsj1 ysi ¼ esn1 ysjþ1 esn1 ysjþ2 ðsjþ1 sj sjþ1 Þsj1 ysi ¼ ðesn1 ysjþ1 esn1 ysi Þsjþ1 ¼ ðesn1 ysi esn1 ysjþ2 Þsjþ1 ¼ esn1 ysi esn1 ysjþ1 ; verifying the inductive step.
&
Note that e occurs on both sides of any relation in which it appears, so that the group of units of Mn is a copy of the braid group consisting of (congruence classes of) words over S\feg: For 1pmpn put Gm ¼ /s1 ; y; sm1 S; which therefore agrees with our earlier notation. It is convenient also for the statements below to put G0 ¼ |: Let 0pkpn: Put en ¼ dn ¼ 1; the empty word, and for kon; ek ¼ esn1 yskþ1 esn1 yskþ2 eyesn1 sn2 esn1 e; dk ¼ esn1 esn2 sn1 eyeskþ2 ysn1 eskþ1 ysn1 e:
Lemma 3.3. (i) ek ¼ dk : (ii) xek ¼ ek x if xAGk : (iii) yek ¼ ek y ¼ ek if yA/x1 s2k x; skþ1 ; y; sn1 jxAGk S: (iv) e2k ¼ ek : Proof. (i) An induction begins because en1 ¼ en1 sn1 en1 ¼ dn1 : If kon 1 then, by commutativity relations and an inductive hypothesis, ek ¼ esn1 yskþ2 esn1 yskþ3 eyesn1 eskþ1 skþ2 ysn2 sn1 e ¼ ekþ1 skþ1 ysn1 e ¼ dkþ1 skþ1 ysn1 e ¼ dk ; which verifies the inductive step. (ii) This is immediate if k ¼ n: If kon then it follows quickly from commutativity relations since the only generators appearing in ek are e and si where iXk þ 1:
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(iii) Suppose k ¼ n 1 (the statement being vacuous when k ¼ n). For xAGn1 ; using some of the relations, x1 s2n1 xen1 ¼ x1 s2n1 ex ¼ x1 ex ¼ x1 xe ¼ e ¼ en1 ; and similarly en1 x1 s2n1 x ¼ en1 ; which starts an induction. Suppose now kon 1: Then, for xAGk ; by (ii), braid relations and an inductive hypothesis, x1 s2k xek ¼ x1 s2k ek x ¼ x1 esn1 yskþ2 ðs2k skþ1 sk Þs1 k ekþ1 x ¼ x1 esn1 yskþ2 ðskþ1 sk s2kþ1 Þekþ1 s1 k x ¼ x1 esn1 ysk ekþ1 s1 k x ¼ x1 esn1 ysk s1 k ekþ1 x ¼ x1 esn1 yskþ1 ekþ1 x ¼ x1 ek x ¼ x1 xek ¼ ek : Also, by (i) and a similar argument, ek x1 s2k x ¼ dk x1 s2k x ¼ dk : Further, sn1 ek ¼ sn1 dk ¼ sn1 esn1 ey ¼ esn1 ey ¼ dk ¼ ek ; which starts another induction. For k þ 1pjon 1; an inductive hypothesis gives sj ek ¼ esn1 ysjþ2 ðsj sjþ1 sj Þsj1 yskþ1 ekþ1 ¼ esn1 ysjþ2 ðsjþ1 sj sjþ1 Þsj1 yskþ1 ekþ1 ¼ esn1 yskþ1 sjþ1 ekþ1 ¼ esn1 yskþ1 ekþ1 ¼ ek ; and ek sj ¼ dk sj ¼ dk ¼ ek ; by a similar argument, establishing the inductive step. This completes the proof of (iii). (iv) An induction begins because e2n ¼ 12 ¼ 1 ¼ en
and
e2n1 ¼ e2 ¼ e ¼ en1 :
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Suppose kon 1: Then, by (iii) and an inductive hypothesis, e2k ¼ ek esn1 yskþ1 ekþ1 ¼ ek ekþ1 ¼ esn1 yskþ1 e2kþ1 ¼ esn1 yskþ1 ekþ1 ¼ ek ; establishing the inductive step, and completing the proof of Lemma 3.3.
&
Lemma 3.4. Suppose that 1pl1 ol2 o?olr pn and put k ¼ n r: Put v ¼ esn1 ysl1 esn1 ysl2 eyesn1 yslr : Then there exists a sequence 0pj1 oj2 o?ojk pn 1 such that v ¼ ek ðsk ysjk sk1 ysjk1 ys2 ysj2 s1 ysj1 Þ: (Note the convention in the 3rd paragraph of Section 1, and that the subscripts within each of the subwords sn1 ysl1 ; sn1 ysl2 ; y; sn1 yslr of the first expression for v are descending, whilst the subscripts within each of the subwords sk ysjk ; sk1 ysjk1 ; y; s1 ysj1 of the second expression for v are ascending. Also if r ¼ n the second expression for v is intended to denote e0 :) Proof. If r ¼ n then by the convention about descending subscripts v ¼ e0 ; as required. Suppose ron: By the commutativity relations, v ¼ ek w where w ¼ sk ysc1 skþ1 ysc2 ysn2 yscr1 sn1 yscr : If r ¼ 1 then w ¼ sn1 ysc1 ¼ sn1 ysjn1 sn2 ysjn2 ys1 ysj1 where, for j ¼ 1 to n 1; ji ¼
i1 i
if ioc1 ; if iXc1 ;
and in the second expression for w; groups of subscripts are interpreted as formally ascending (so that si ysji is empty when ioc1 ). This starts an induction.
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Now suppose 1oron: If w ¼ 1 then c1 ¼ k þ 1; c2 ¼ k þ 2; y; cr ¼ n so that w ¼ sk ysjk ys1 ysj1 ; where jk ¼ k 1; y; j1 ¼ 0; interpreted as ascending groups of subscripts (so all groups yield empty words), and we are done. Suppose wa1: Then w ¼ sk ysc1 skþ1 ysc2 yskþm1 yscm for some mX1; with descending groups of subscripts (and each group yielding a nonempty word). By commutativity relations, w ¼ sk skþ1 yskþm1 ðsk1 ysc1 sk ysc2 yskþm2 yscm Þ: Put jk ¼ k þ m 1 and apply an inductive hypothesis to conclude w ¼ sk ysjk ðsk1 ysjk1 ys1 ysj1 Þ for some 0pj1 o?ojk1 pk þ m 2; where groups of subscripts are now interpreted as ascending. The lemma follows by induction. & Consider any word w over S: Regarded as an element of Mn we have that w ¼ xex1 eyexk for some kX0 and some x; x1 ; y; xk AGn : Put H ¼ /s1 ; y; sn2 ; s2n2 S; so that e commutes with all elements of H: By the first coset decomposition in Lemma 2.1, we may rewrite w so that x1 ; y; xk lie in the set f1; sn1 ; sn1 sn2 ; y; sn1 ys1 g: Using Lemma 3.2 and the relation e2 ¼ e we may further rewrite w so that it has the form w ¼ xesn1 ysl1 esn1 ysl2 eyesn1 yslr for some (possibly new) r; where 1pl1 ol2 o?olr pn; and for some (possibly new) xAGn : Put k ¼ n r: (For heuristic purposes think of k as the number of strings left in the image of w under F:) Put v ¼ esn1 ysl1 esn1 ysl2 eyesn1 yslr : Our aim is now to control the subword x in our expression w ¼ xv: Put K ¼ /x1 s2k x; skþ1 ; y; sn1 jxAGk S
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and L ¼ /s1 ; y; sk1 ; s2k ; skþ1 ; y; sn1 S: By the second coset decomposition in Lemma 2.1, x ¼ yx0 for some x0 AL and y ¼ si1 ys1 ysik ysk for some sequence 0pi1 o?oik pn 1: By Lemma 2.2, x0 ¼ zx00 for some x00 AK and some zAGk : But, by Lemma 3.4, v can be rewritten to begin with ek ; so that, by Lemma 3.3, w ¼ yx0 v ¼ yzx00 ek v ¼ yzek v ¼ yek zek v: By Lemma 3.4, this shows that every word over S is congruent to a word in the set W ¼ fsi1 ys1 ysik ysk ek xek sk ysjk ys1 ysj1 j kAf0; y; ng; xAGk ;
0pi1 o?oik pn 1 and 0pj1 oyojk pn 1g
(where here the expression for a typical word in W is an element of Gn when k ¼ n; and is intended to mean simply e0 when k ¼ 0). The image of a typical element of W under C is pictured in the earlier proof that C is onto (noting that the image of e0 is the empty braid). Observe that altering any of k; (the congruence class of) x; or i1 ; y; ik ; j1 ; y; jk alters the (equivalence class of the) braid exhibited in that diagram (relying on Artin’s Theorem in the case of x). Thus C is one–one on W and the proof of Theorem 3.1 is complete.
4. The symmetric inverse monoid By adding the relations s21 ¼ ? ¼ s2n1 ¼ 1 to the presentation for the braid group described in Artin’s Theorem we obtain the well-known Moore presentation for the symmetric group [5, p. 63]. Using this in Popova’s description [13] of presentations of the symmetric inverse monoid yields Theorem 4.1 below. We will use Theorem 3.1 to provide an alternative proof of Theorem 4.1. Put Jn ¼ monoid/s1 ; y; sn1 ; ejs21 ¼ ? ¼ s2n1 ¼ 1; braid relations for s1 ; y; sn1 ; e2 ¼ e; esi ¼ si e for ipn 2; esn1 e ¼ sn1 esn1 e ¼ esn1 esn1 S: Theorem 4.1 (Popova [13]). The monoids Jn and In are isomorphic. Recall from [4,6] that the pure braid group Pn consists of all (equivalence classes of) braids on n strings which induce the identity permutation of f1; y; ng: The following
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lemma is immediate from the braid relations and [4, Lemma 1.82]; [6, Lemma 4.2] or [10, Theorem N7]: Lemma 4.2. Pn is generated by the braids ðws2i w1 ÞC for i ¼ 1 to n 1 and as w 71 ranges over all words in s71 1 ; y; sn1 : Proof of Theorem 4.1. In the proof of Theorem 3.1 we saw that C : Mn -IBn is an isomorphism, and in Section 1 we observed that the map F : IBn -In ; b/bC is an epimorphism (noting that we continue to identify braids with their equivalence classes). Put Y ¼ CF so that Y is the homomorphism: Mn -In induced by the map s71 i /
1
y i1
i
iþ1
iþ2 y
n
1
y i1
iþ1
i
iþ2 y
n
e/
1 1
y n1 y n1
;
n :
Therefore Y is an epimorphism and In DMn =ker Y: Let R be the congruence on Mn generated by the relations s21 ¼ ? ¼ s2n1 ¼ 1: Then Jn DMn =R; so it remains to prove that R ¼ ker Y: Since Y takes s2i for each i to the identity permutation of f1; y; ng; it is clear that RDker Y: Suppose conversely that w1 ; w2 AMn and w1 Y ¼ w2 Y: Then w1 C and w2 C must be braids with the same number k of strings emanating from the same vertices on the upper plane, and such that, for i ¼ 1; y; k; the ith strings of each must terminate at the same vertex on the lower plane. Thus w1 C ¼ ab1 g
and
w2 C ¼ ab2 g
for some braids a; b1 ; b2 ; g such that a and g are as depicted in the diagram in the proof that C is onto in Section 3, and further b1 and b2 may be regarded as elements of Bk which induce the same permutation of f1; y; kg: Hence b1 b1 2 APk ; so, by Lemma 4.2, b1 b1 2
¼
m Y
! 2 x1 i sji xi
C
i¼1
for some mX0; letters sji and words xi for i ¼ 1; y; m: Now a ¼ uC; g ¼ vC; b1 ¼ b1 C and b2 ¼ b2 C for some u; v; b1 ; b2 AMn ; so w1 C ¼ ðub1 vÞC ¼
u
m Y i¼1
! 2 x1 i sji xi
!
b2 v C
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whence w1 ¼ u
m Y
! 2 x1 i sji xi
b2 v R ub2 v ¼ w2 :
i¼1
Thus ker YDR; and the proof of Theorem 4.1 is complete.
&
References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13]
E. Artin, Theorie der Zo¨pfe, Abh. Math. Sem. Univ. Hamburg 4 (1926) 47–72. E. Artin, Theory of braids, Ann. of Math. (2) 48 (1947) 101–126. E. Artin, Braids and permutations, Ann. of Math. (2) 48 (1947) 643–649. J. Birman, Braids, links and mapping class groups, Ann. of Math. Stud. 82 (1975). H.S.M. Coxeter, W.O.J. Moser, Generators and relations for discrete groups, in: Ergebnisse der Mathematik und ihrer Grenzgebiete, Vol. 14, Springer, Berlin, 1972. V.L. Hansen, Braids and Coverings: Selected Topics, in: London Mathematical Society Student Texts, Vol. 18, Cambridge University Press, Cambridge, 1989. P.M. Higgins, Techniques of Semigroup Theory, Oxford University Press, Oxford, 1992. J.M. Howie, An Introduction to Semigroup Theory, Academic Press, New York, 1976. S.L. Lipscomb, Symmetric Inverse Semigroups, in: Mathematical Surveys and Monographs, Vol. 46, American Mathematical Society, Providence, RI, 1996. W. Magnus, A. Karrass, D. Solitar, Combinatorial Group Theory, Wiley, New York, 1966. J.C. Meakin, An invitation to inverse semigroup theory, preprint. M. Petrich, Inverse Semigroups, Wiley, New York, 1984. L.M. Popova, Defining relations in some semigroups of partial transformations of a finite set, Uchenye Zap. Leningrad Gos. Ped. Inst. 218 (1961) 191–212 (in Russian).
Advances in Mathematics 186 (2004) 438–455
http://www.elsevier.com/locate/aim
The inverse braid monoid D. Easdown and T.G. Lavers School of Mathematics and Statistics, University of Sydney, Sydney, NSW 2006, Australia Received 19 July 2001; accepted 10 July 2003 Communicated by Ross Street
Abstract We introduce an inverse monoid which plays a similar role with respect to the symmetric inverse semigroup that the braid group plays with respect to the symmetric group. r 2003 Elsevier Inc. All rights reserved. MSC: 20F36; 20M18 Keywords: Braid groups; Inverse monoids
1. Introduction and preliminaries We describe an inverse monoid which arises naturally when considering usual geometric braids from which strings may have been removed. This monoid contains many copies of the Artin braid groups and may be described by an elegant presentation, from which we can quickly deduce Popova’s presentation [13] (see also [9,11]) of the symmetric inverse monoid. For basic terminology and facts about inverse semigroups we refer the reader to [7,8] or [12]. We denote the symmetric inverse monoid on n letters by In : We denote by group/X jRS
and
monoid/X jRS
group and monoid presentations, respectively, where X is the alphabet of generators and R the set of relations. We will consistently use plain angular brackets to
Corresponding author. E-mail addresses: [email protected] (D. Easdown), [email protected] (T.G. Lavers).
0001-8708/$ - see front matter r 2003 Elsevier Inc. All rights reserved. doi:10.1016/j.aim.2003.07.014
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have the following meaning: if M is a monoid and X is a subset of the group of units of M; then /X S denotes the subgroup of (the group of units of) M generated by X : Sequences are used throughout this paper and we alert the reader to an important convention. It is standard usage that fxi ; y; xj g denotes the empty sequence if either (i) i ¼ j þ 1 and the subscripts are ascending; or (ii) j ¼ i þ 1 and the subscripts are descending. We extend this notation to words over an alphabet fs1 ; y; sn1 g so that the word si ysj denotes the empty word (or group identity element) if either (i) or (ii) occurs. The reader will see from context whether subscripts are intended to be ascending or descending. Artin’s braid group is defined in [1,4,6]. We adapt the treatment given in the first few pages of [6] to give a sequence of definitions, leading below to the definition of IBn ; the inverse braid monoid on n strings. Take the usual coordinate system for R3 (in which for descriptive purposes we think of the z-axis as pointing downwards). Choose z0 oz1 and call the planes z ¼ z0 and z1 upper and lower, respectively. Mark nX1 distinct points P1 ; y; Pn on a line in the upper plane and project them orthogonally onto the lower plane yielding points P01 ; y; P0n : An arc is the image of an embedding from the unit interval ½0; 1 into R3 : A (partial) braid on n strings is a system b ¼ fb1 ; y; bm g of m arcs for some mpn such that (i) there is a rank m partial one–one mapping bF : f1; y; ng-f1; y; ng with domain fi1 ; y; im g such that bj connects Pij to P0ij bF for j ¼ 1; y; m; (ii) each arc intersects each intermediate parallel plane between (and including) the upper and lower planes exactly once; (iii) the union b1 ,?,bm of the arcs intersect each intermediate parallel plane between the upper and lower planes in exactly m distinct points. We may think of the object just defined as a usual (full-string) braid with n m strings removed. When m ¼ 0 we get the empty braid. Two braids b ¼ fb1 ; y; bm g and g ¼ fg1 ; y; gm0 g are called equivalent if (i) bF ¼ gF; so that in particular m ¼ m0 and we may write the domain of bF as fi1 ; y; im g for some i1 o?oim ; and (ii) b and g are homotopy equivalent, by which we mean that there exist continuous maps Fj : ½0; 1 ½0; 1 -R3 for j ¼ 1 to m; such that, for all tA½0; 1 ; Fj ðt; 0Þ ¼ bj ðtÞ and
Fj ðt; 1Þ ¼ gj ðtÞ;
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for all sA½0; 1 ; Fj ð0; sÞ ¼ Pij
and
Fj ð1; sÞ ¼ P0ij bF ;
and, for each sA½0; 1 ; if we define bs ¼ fbs1 ; y; bsm g where bsj ðtÞ ¼ Fj ðs; tÞ
for j ¼ 1 to m;
then bs is itself a braid. If m ¼ n then this reduces to the definition in [6] for equivalence of braids without missing strings. Write b g if b and g are equivalent and ½b for the equivalence class containing b: Define the product or composite b1 b2 of two braids b1 and b2 as follows: (i) translate b2 parallel to itself in space so that the upper plane of b2 and the lower plane of b1 coincide (so that b2 hangs below b1 ); (ii) keeping the plane z ¼ z0 fixed, contract the resulting systems of arcs so that the translated lower plane of b2 moves into the position of the plane z ¼ z1 ; (iii) remove any arcs that do not now join the upper and lower planes. For example, if
then a2 ; ab; ba are obtained as follows:
so that a2
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Clearly if b b0 and g g0 then bg b0 g0 so that the operation ½b ½g ¼ ½bg
on equivalence classes of braids is well defined. It is easy to see that this operation is associative. Put Bn ¼ f½b jb is a braid and bF is a permutation of f1; y; ngg and Gn ¼ group/s1 ; y; sn1 jsi sj ¼ sj si si siþ1 si ¼ siþ1 si siþ1
if ji jj41;
for i ¼ 1 to n 2S:
Then Bn is the usual (Artin) braid group, and we recall Artin’s Theorem: Theorem 1.1 (Artin [1–3], Birman [4], Hansen [6]). The groups Gn and Bn are isomorphic. The relations that appear in the above presentation are referred to as braid relations. The isomorphism in Artin’s Theorem identifies each si with the
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equivalence class of the braid
Now let IBn denote the set of equivalence classes of all braids which clearly forms a monoid equipped with the product given earlier. We will show that IBn is inverse. Note that clearly the map ½b /bF is a well-defined epimorphism from IBn to In : Let ½b AIBn : Define b1 to be the mirror image of b after reflecting through a parallel plane midway between the upper and lower planes. Clearly, if b g then b1 g1 so that we may sensibly define ½b 1 ¼ ½b1 : For any subset X ¼ fi1 ; y; im g of f1; y; ng denote the following braid by eX ; consisting of m straight arcs connecting Pij to P0ij for j ¼ 1 to m:
Then it is easy to see that bb1 eX and b1 b eY where X and Y are the domain and range, respectively, of bF; from which it follows that ½b ½b 1 ½b ¼ ½b
and
½b 1 ½b ½b 1 ¼ ½b 1 :
Thus ½b 1 is an inverse of ½b (which we shall see shortly is unique). In particular IBn is regular. Let ½d AIBn be an idempotent. Then ðdFÞ2 ¼ dF; so that dF is an idempotent in In ; whence dF is the identity map on its domain X : But then ½d lies in a copy of the braid group on jX j strings, so must coincide with its group identity element, that is ½d ¼ ½eX : This proves that the set EðIBn Þ of idempotents of IBn is precisely f½eX jX Df1; y; ngg: Clearly ½eX ½eY ¼ ½eX -Y for any subsets X and Y of f1; y; ng; so that EðIBn Þ forms a commutative subsemigroup of IBn : Thus we have proved
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Theorem 1.2. IBn is an inverse monoid with semilattice of idempotents isomorphic to the semilattice of subsets of f1; y; ng under intersection. 2. Coset decompositions We list some useful (and probably well-known) coset decompositions of certain subgroups of the braid group. Throughout this section nX1 is fixed and (in view of Artin’s Theorem) we may regard Gk as embedded in Gn in the obvious way when kpn: Now fix kpn 1: Define the following subgroups of Gn : H ¼ /s1 ; y; sn2 ; s2n1 S; K ¼ /skþ1 ; ysn1 ; x1 s2k xjxAGk S; L ¼ /s1 ; y; sk1 ; s2k ; skþ1 ; y; sn1 S (so if k ¼ n 1 then H ¼ L). Lemma 2.1. We have the coset decompositions Gn ¼ H,Hsn1 ,Hsn1 sn2 ,?,Hsn1 ys1 [ ¼ si1 ys1 si2 ys2 ysik ysk L: 0pi1 o?oik pn1
Proof. The coset decomposition for H follows from that of L (in the case k ¼ 1 and by observing that the map si /sni induces an anti-isomorphism of Gn ). Put W ¼ fsi1 ys1 ysik ysk j 0pi1 o?oik pn 1g: Suppose 0pi1 o?oik pn 1 and put w ¼ si1 ys1 ysik ysk : Let cAf1; y; n 1g and eAf71g: To prove the lemma it suffices to show that sec wLDWL: If im ¼ m 1 for m ¼ 1; y; k; then w is empty, so sec wL ¼
L sc L
if cak; if c ¼ k;
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sc AW ; and we are done. Suppose then that, for some mAf1; y; kg; is ¼ s 1
for 1pspm 1;
and
im Xm;
so w ¼ sim ysm ysik ysk : The proof now splits into cases. Case (i): If cpm 2 then, by commutativity relations, sec wL ¼ wsec L ¼ wL; and we are done. Case (ii): Suppose c ¼ m 1: If e ¼ 1 then sec wL ¼ sm1 wL and sm1 wAW and we are done. If e ¼ 1 then, by repeated use of the braid relations, sec wL ¼ sm1 s2 m1 wL 1 ¼ sm1 sim ysmþ1 s2 m1 sm ðsm1 sm1 Þsimþ1 ysmþ1 ysik ysk L 1 ¼ sm1 sim ysmþ1 sm ðsm1 s2 m sm1 Þsimþ1 ysmþ1 ysik ysk L
¼y 1 ¼ sm1 sim ysm ysik ysk ðsm1 ysk1 s2k s1 k1 ysm1 ÞL
¼ sm1 wL; and again we are done. Case (iii): If mpcoim then, using the braid relations, sec wL ¼ sim yscþ2 sec scþ1 sc ysm ysik ysk L ¼ sim yscþ2 scþ1 sc secþ1 sc1 ysm ysik ysk L ¼ sim ysm secþ1 simþ1 ysmþ1 ysik ysk L ¼y ¼ sim ysm ysik ysk secþkmþ1 L ¼ wL; at the last step because c þ k m þ 1Xk þ 1; and we are done. Case (iv): Suppose c ¼ im : If e ¼ 1 then sec w ¼ sim 1 ysm simþ1 ysmþ1 ysik ysk AW ; and we are done. If e ¼ 1 then, by the braid relations and the calculation in case (iii), and since sm ysk1 s2k ðsm ysk1 Þ1 AL; sec wL ¼ s2im sim 1 ysm simþ1 ysmþ1 ysik ysk L 1 ¼ sim 1 ysm ðsim ysmþ1 Þs2m s1 mþ1 ysim simþ1 ysmþ1 ysik ysk L
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¼ sim 1 ysm ðsim ysmþ1 Þs2m simþ1 ysmþ1 ysik ysk L ¼ sim 1 ysm ðsim ysmþ1 Þsimþ1 ysmþ1 ysik ysk ðsm ysk1 Þs2k ðsm ysk1 Þ1 L ¼ sim 1 ysm ðsim ysmþ1 Þsimþ1 ysmþ1 ysik ysk L ¼ sim 1 ysm simþ1 ysmþ1 ysik ysk L; and again we are done. Case (v): Suppose c ¼ im þ 1: If im þ 1oimþ1 then sc wAW ; whilst also, by the braid relations and the calculations in cases (ii) and (iii), 1 s1 c wL ¼ sim þ1 sim ysm ysik ysk L 1 1 ¼ sim ysm ðsim þ1 ysmþ1 Þs1 m ðsmþ1 ysim þ1 Þ
simþ1 ysmþ1 ysik ysk L ¼ sim ysm ðsim þ1 ysmþ1 Þs1 m simþ1 ysmþ1 ysik ysk L ¼ sim ysm ðsim þ1 ysmþ1 Þsm simþ1 ysmþ1 ysik ysk L 1 ¼ sim ysm ðsim þ1 ysmþ1 Þsm ðs1 mþ1 ysim þ1 Þ
simþ1 ysmþ1 ysik ysk L ¼ sim þ1 wL; and we are done. On the other hand, if im þ 1 ¼ imþ1 then, by the braid relations and the calculation in case (ii), sec wL ¼ seim þ1 sim ysm simþ1 ysmþ1 ysik ysk L ¼ sim ysm simþ1 ysmþ2 semþ1 simþ2 ysmþ2 ysik ysk L ¼ sim ysm simþ1 ysmþ2 smþ1 simþ2 ysmþ2 ysik ysk L ¼ wL; and again we are done. Case (vi): If c4ik þ 1 then c4k þ 1; so, by commutativity relations, sec wL ¼ wsec L ¼ wL; and we are done. Case (vii): Suppose im þ 1ocpik þ 1; so iN1 þ 1ocpiN þ 1 for some NAfm þ 1; y; kg:
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Note that cXN þ 1: Then sec wL ¼ sim ysm ysiN1 ysN1 sec siN ysN ysik ysk L: We have four subcases: If coiN then, by the calculation in case (iii), sec wL ¼ wL: If c ¼ iN then, by the calculation in case (iv), sec wL ¼ sim ysm ysiN1 ysN1 siN 1 ysN ysik ysk L: If c ¼ iN þ 1 and iN þ 1oiNþ1 then, by the first calculation in case (v), sec wL ¼ sim ysm ysiN1 ysN1 siN þ1 ysN ysik ysk L: If c ¼ iN þ 1 and iN þ 1 ¼ iNþ1 then, by the second calculation in case (v), sec wL ¼ wL: In each of these subcases sec wL ¼ w0 L for some w0 AW ; so we are done. This exhausts all possibilities and the proof of Lemma 2.1 is complete. & Lemma 2.2. We have the coset decomposition L ¼
S
xAGk
xK:
71 Proof. Let xAGk : It suffices to check s72 k xAGk K and si xAGk K for iak: But certainly 1 72 s72 k x ¼ xðx sk xÞAxK;
since x1 s72 k xAK: If iXk þ 1 then 71 s71 i x ¼ xsi AxK; 71 since s71 i AK: If ipk 1 then si xAGk ; and we are done. &
3. A presentation for the inverse braid monoid In this section we provide a monoid presentation for IBn : For the remainder of this paper we will denote the equivalence class of a braid by the braid itself. Put 1 1 1 Mn ¼ monoid/s1 ; y; sn1 ; s1 1 ; y; sn1 ; ejsi si ¼ si si ¼ 1 for all i;
braid relations for s1 ; y; sn1 ; e2 ¼ e ¼ es2n1 ¼ s2n1 e; esi ¼ si e for ipn 2; esn1 e ¼ sn1 esn1 e ¼ esn1 esn1 S: We will prove
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Theorem 3.1. The monoids Mn and IBn are isomorphic. Let C denote the mapping from the set of generators for Mn into IBn given by, for each i;
It is easy to check that the relations in the presentation for Mn hold for the images of the generators, so that C induces a well-defined homomorphism, which we also denote by C; from Mn into IBn : To prove Theorem 3.1 it suffices to show that C is one–one and onto. Proof that C is onto: For the time being, we identify the generators of Mn with 71 their images under C: It suffices to prove that s71 1 y; sn1 ; e generate IBn : Let yAIBn ; and suppose that y possesses k strings. Put ek ¼ ef1;y;kg ; the identity braid with the last n k strings removed. Thus e0 is the empty braid, en ¼ 1; en1 ¼ e; and, by a straightforward calculation,
ek ¼ esn1 yskþ1 esn1 yskþ2 eyesn1 sn2 esn1 e:
Then by stretching near the upper and lower planes (if necessary) we see that y has a decomposition of the form y ¼ abg; where the following picture holds for some
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sequences i1 ; y; ik and j1 ; y; jk :
But b ¼ ek b0 for some b0 ABn ; and, by a simple calculation, a ¼ ðsi1 ys1 si2 ys2 ysik ysk Þek ; g ¼ ek ðsk ysjk sk1 ysjk1 ys2 ysj2 s1 ysj1 Þ; 71 each of which are combinations of s71 1 ; y; sn1 ; e; completing the proof that C is onto. 71 Proof that C is one–one: Consider the alphabet S ¼ fs71 1 ; y; sn ; eg: To prove that C is one–one it suffices to find a set W of words (normal forms) such that, with respect to the congruence generated by the relations given in the presentation for Mn ; each word over S is congruent to a word in W ; and C is one–one on the congruence classes of words in W : Without causing confusion, we will identify words with their congruence classes.
Lemma 3.2. Suppose ipjpn 1: Then esn1 ysj esn1 ysi ¼ esn1 ysi esn1 ysjþ1 : Proof. We use the braid relations and the relations esn1 e ¼ esn1 esn1
and
esk ¼ sk e
for kpn 2:
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An induction begins because, if j ¼ n 1; then esn1 ysj esn1 ysi ¼ esn1 esn1 ysi ¼ esn1 esn2 ysi ¼ esn1 ysi e: If jon 1; then, by an inductive hypothesis, esn1 ysj esn1 ysi ¼ esn1 ysjþ1 esn1 ysjþ2 ðsj sjþ1 sj Þsj1 ysi ¼ esn1 ysjþ1 esn1 ysjþ2 ðsjþ1 sj sjþ1 Þsj1 ysi ¼ ðesn1 ysjþ1 esn1 ysi Þsjþ1 ¼ ðesn1 ysi esn1 ysjþ2 Þsjþ1 ¼ esn1 ysi esn1 ysjþ1 ; verifying the inductive step.
&
Note that e occurs on both sides of any relation in which it appears, so that the group of units of Mn is a copy of the braid group consisting of (congruence classes of) words over S\feg: For 1pmpn put Gm ¼ /s1 ; y; sm1 S; which therefore agrees with our earlier notation. It is convenient also for the statements below to put G0 ¼ |: Let 0pkpn: Put en ¼ dn ¼ 1; the empty word, and for kon; ek ¼ esn1 yskþ1 esn1 yskþ2 eyesn1 sn2 esn1 e; dk ¼ esn1 esn2 sn1 eyeskþ2 ysn1 eskþ1 ysn1 e:
Lemma 3.3. (i) ek ¼ dk : (ii) xek ¼ ek x if xAGk : (iii) yek ¼ ek y ¼ ek if yA/x1 s2k x; skþ1 ; y; sn1 jxAGk S: (iv) e2k ¼ ek : Proof. (i) An induction begins because en1 ¼ en1 sn1 en1 ¼ dn1 : If kon 1 then, by commutativity relations and an inductive hypothesis, ek ¼ esn1 yskþ2 esn1 yskþ3 eyesn1 eskþ1 skþ2 ysn2 sn1 e ¼ ekþ1 skþ1 ysn1 e ¼ dkþ1 skþ1 ysn1 e ¼ dk ; which verifies the inductive step. (ii) This is immediate if k ¼ n: If kon then it follows quickly from commutativity relations since the only generators appearing in ek are e and si where iXk þ 1:
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(iii) Suppose k ¼ n 1 (the statement being vacuous when k ¼ n). For xAGn1 ; using some of the relations, x1 s2n1 xen1 ¼ x1 s2n1 ex ¼ x1 ex ¼ x1 xe ¼ e ¼ en1 ; and similarly en1 x1 s2n1 x ¼ en1 ; which starts an induction. Suppose now kon 1: Then, for xAGk ; by (ii), braid relations and an inductive hypothesis, x1 s2k xek ¼ x1 s2k ek x ¼ x1 esn1 yskþ2 ðs2k skþ1 sk Þs1 k ekþ1 x ¼ x1 esn1 yskþ2 ðskþ1 sk s2kþ1 Þekþ1 s1 k x ¼ x1 esn1 ysk ekþ1 s1 k x ¼ x1 esn1 ysk s1 k ekþ1 x ¼ x1 esn1 yskþ1 ekþ1 x ¼ x1 ek x ¼ x1 xek ¼ ek : Also, by (i) and a similar argument, ek x1 s2k x ¼ dk x1 s2k x ¼ dk : Further, sn1 ek ¼ sn1 dk ¼ sn1 esn1 ey ¼ esn1 ey ¼ dk ¼ ek ; which starts another induction. For k þ 1pjon 1; an inductive hypothesis gives sj ek ¼ esn1 ysjþ2 ðsj sjþ1 sj Þsj1 yskþ1 ekþ1 ¼ esn1 ysjþ2 ðsjþ1 sj sjþ1 Þsj1 yskþ1 ekþ1 ¼ esn1 yskþ1 sjþ1 ekþ1 ¼ esn1 yskþ1 ekþ1 ¼ ek ; and ek sj ¼ dk sj ¼ dk ¼ ek ; by a similar argument, establishing the inductive step. This completes the proof of (iii). (iv) An induction begins because e2n ¼ 12 ¼ 1 ¼ en
and
e2n1 ¼ e2 ¼ e ¼ en1 :
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Suppose kon 1: Then, by (iii) and an inductive hypothesis, e2k ¼ ek esn1 yskþ1 ekþ1 ¼ ek ekþ1 ¼ esn1 yskþ1 e2kþ1 ¼ esn1 yskþ1 ekþ1 ¼ ek ; establishing the inductive step, and completing the proof of Lemma 3.3.
&
Lemma 3.4. Suppose that 1pl1 ol2 o?olr pn and put k ¼ n r: Put v ¼ esn1 ysl1 esn1 ysl2 eyesn1 yslr : Then there exists a sequence 0pj1 oj2 o?ojk pn 1 such that v ¼ ek ðsk ysjk sk1 ysjk1 ys2 ysj2 s1 ysj1 Þ: (Note the convention in the 3rd paragraph of Section 1, and that the subscripts within each of the subwords sn1 ysl1 ; sn1 ysl2 ; y; sn1 yslr of the first expression for v are descending, whilst the subscripts within each of the subwords sk ysjk ; sk1 ysjk1 ; y; s1 ysj1 of the second expression for v are ascending. Also if r ¼ n the second expression for v is intended to denote e0 :) Proof. If r ¼ n then by the convention about descending subscripts v ¼ e0 ; as required. Suppose ron: By the commutativity relations, v ¼ ek w where w ¼ sk ysc1 skþ1 ysc2 ysn2 yscr1 sn1 yscr : If r ¼ 1 then w ¼ sn1 ysc1 ¼ sn1 ysjn1 sn2 ysjn2 ys1 ysj1 where, for j ¼ 1 to n 1; ji ¼
i1 i
if ioc1 ; if iXc1 ;
and in the second expression for w; groups of subscripts are interpreted as formally ascending (so that si ysji is empty when ioc1 ). This starts an induction.
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Now suppose 1oron: If w ¼ 1 then c1 ¼ k þ 1; c2 ¼ k þ 2; y; cr ¼ n so that w ¼ sk ysjk ys1 ysj1 ; where jk ¼ k 1; y; j1 ¼ 0; interpreted as ascending groups of subscripts (so all groups yield empty words), and we are done. Suppose wa1: Then w ¼ sk ysc1 skþ1 ysc2 yskþm1 yscm for some mX1; with descending groups of subscripts (and each group yielding a nonempty word). By commutativity relations, w ¼ sk skþ1 yskþm1 ðsk1 ysc1 sk ysc2 yskþm2 yscm Þ: Put jk ¼ k þ m 1 and apply an inductive hypothesis to conclude w ¼ sk ysjk ðsk1 ysjk1 ys1 ysj1 Þ for some 0pj1 o?ojk1 pk þ m 2; where groups of subscripts are now interpreted as ascending. The lemma follows by induction. & Consider any word w over S: Regarded as an element of Mn we have that w ¼ xex1 eyexk for some kX0 and some x; x1 ; y; xk AGn : Put H ¼ /s1 ; y; sn2 ; s2n2 S; so that e commutes with all elements of H: By the first coset decomposition in Lemma 2.1, we may rewrite w so that x1 ; y; xk lie in the set f1; sn1 ; sn1 sn2 ; y; sn1 ys1 g: Using Lemma 3.2 and the relation e2 ¼ e we may further rewrite w so that it has the form w ¼ xesn1 ysl1 esn1 ysl2 eyesn1 yslr for some (possibly new) r; where 1pl1 ol2 o?olr pn; and for some (possibly new) xAGn : Put k ¼ n r: (For heuristic purposes think of k as the number of strings left in the image of w under F:) Put v ¼ esn1 ysl1 esn1 ysl2 eyesn1 yslr : Our aim is now to control the subword x in our expression w ¼ xv: Put K ¼ /x1 s2k x; skþ1 ; y; sn1 jxAGk S
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and L ¼ /s1 ; y; sk1 ; s2k ; skþ1 ; y; sn1 S: By the second coset decomposition in Lemma 2.1, x ¼ yx0 for some x0 AL and y ¼ si1 ys1 ysik ysk for some sequence 0pi1 o?oik pn 1: By Lemma 2.2, x0 ¼ zx00 for some x00 AK and some zAGk : But, by Lemma 3.4, v can be rewritten to begin with ek ; so that, by Lemma 3.3, w ¼ yx0 v ¼ yzx00 ek v ¼ yzek v ¼ yek zek v: By Lemma 3.4, this shows that every word over S is congruent to a word in the set W ¼ fsi1 ys1 ysik ysk ek xek sk ysjk ys1 ysj1 j kAf0; y; ng; xAGk ;
0pi1 o?oik pn 1 and 0pj1 oyojk pn 1g
(where here the expression for a typical word in W is an element of Gn when k ¼ n; and is intended to mean simply e0 when k ¼ 0). The image of a typical element of W under C is pictured in the earlier proof that C is onto (noting that the image of e0 is the empty braid). Observe that altering any of k; (the congruence class of) x; or i1 ; y; ik ; j1 ; y; jk alters the (equivalence class of the) braid exhibited in that diagram (relying on Artin’s Theorem in the case of x). Thus C is one–one on W and the proof of Theorem 3.1 is complete.
4. The symmetric inverse monoid By adding the relations s21 ¼ ? ¼ s2n1 ¼ 1 to the presentation for the braid group described in Artin’s Theorem we obtain the well-known Moore presentation for the symmetric group [5, p. 63]. Using this in Popova’s description [13] of presentations of the symmetric inverse monoid yields Theorem 4.1 below. We will use Theorem 3.1 to provide an alternative proof of Theorem 4.1. Put Jn ¼ monoid/s1 ; y; sn1 ; ejs21 ¼ ? ¼ s2n1 ¼ 1; braid relations for s1 ; y; sn1 ; e2 ¼ e; esi ¼ si e for ipn 2; esn1 e ¼ sn1 esn1 e ¼ esn1 esn1 S: Theorem 4.1 (Popova [13]). The monoids Jn and In are isomorphic. Recall from [4,6] that the pure braid group Pn consists of all (equivalence classes of) braids on n strings which induce the identity permutation of f1; y; ng: The following
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lemma is immediate from the braid relations and [4, Lemma 1.82]; [6, Lemma 4.2] or [10, Theorem N7]: Lemma 4.2. Pn is generated by the braids ðws2i w1 ÞC for i ¼ 1 to n 1 and as w 71 ranges over all words in s71 1 ; y; sn1 : Proof of Theorem 4.1. In the proof of Theorem 3.1 we saw that C : Mn -IBn is an isomorphism, and in Section 1 we observed that the map F : IBn -In ; b/bC is an epimorphism (noting that we continue to identify braids with their equivalence classes). Put Y ¼ CF so that Y is the homomorphism: Mn -In induced by the map s71 i /
1
y i1
i
iþ1
iþ2 y
n
1
y i1
iþ1
i
iþ2 y
n
e/
1 1
y n1 y n1
;
n :
Therefore Y is an epimorphism and In DMn =ker Y: Let R be the congruence on Mn generated by the relations s21 ¼ ? ¼ s2n1 ¼ 1: Then Jn DMn =R; so it remains to prove that R ¼ ker Y: Since Y takes s2i for each i to the identity permutation of f1; y; ng; it is clear that RDker Y: Suppose conversely that w1 ; w2 AMn and w1 Y ¼ w2 Y: Then w1 C and w2 C must be braids with the same number k of strings emanating from the same vertices on the upper plane, and such that, for i ¼ 1; y; k; the ith strings of each must terminate at the same vertex on the lower plane. Thus w1 C ¼ ab1 g
and
w2 C ¼ ab2 g
for some braids a; b1 ; b2 ; g such that a and g are as depicted in the diagram in the proof that C is onto in Section 3, and further b1 and b2 may be regarded as elements of Bk which induce the same permutation of f1; y; kg: Hence b1 b1 2 APk ; so, by Lemma 4.2, b1 b1 2
¼
m Y
! 2 x1 i sji xi
C
i¼1
for some mX0; letters sji and words xi for i ¼ 1; y; m: Now a ¼ uC; g ¼ vC; b1 ¼ b1 C and b2 ¼ b2 C for some u; v; b1 ; b2 AMn ; so w1 C ¼ ðub1 vÞC ¼
u
m Y i¼1
! 2 x1 i sji xi
!
b2 v C
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whence w1 ¼ u
m Y
! 2 x1 i sji xi
b2 v R ub2 v ¼ w2 :
i¼1
Thus ker YDR; and the proof of Theorem 4.1 is complete.
&
References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13]
E. Artin, Theorie der Zo¨pfe, Abh. Math. Sem. Univ. Hamburg 4 (1926) 47–72. E. Artin, Theory of braids, Ann. of Math. (2) 48 (1947) 101–126. E. Artin, Braids and permutations, Ann. of Math. (2) 48 (1947) 643–649. J. Birman, Braids, links and mapping class groups, Ann. of Math. Stud. 82 (1975). H.S.M. Coxeter, W.O.J. Moser, Generators and relations for discrete groups, in: Ergebnisse der Mathematik und ihrer Grenzgebiete, Vol. 14, Springer, Berlin, 1972. V.L. Hansen, Braids and Coverings: Selected Topics, in: London Mathematical Society Student Texts, Vol. 18, Cambridge University Press, Cambridge, 1989. P.M. Higgins, Techniques of Semigroup Theory, Oxford University Press, Oxford, 1992. J.M. Howie, An Introduction to Semigroup Theory, Academic Press, New York, 1976. S.L. Lipscomb, Symmetric Inverse Semigroups, in: Mathematical Surveys and Monographs, Vol. 46, American Mathematical Society, Providence, RI, 1996. W. Magnus, A. Karrass, D. Solitar, Combinatorial Group Theory, Wiley, New York, 1966. J.C. Meakin, An invitation to inverse semigroup theory, preprint. M. Petrich, Inverse Semigroups, Wiley, New York, 1984. L.M. Popova, Defining relations in some semigroups of partial transformations of a finite set, Uchenye Zap. Leningrad Gos. Ped. Inst. 218 (1961) 191–212 (in Russian).