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Tresses sur les surfaces et invariants d’entrelacs
Paolo Bellingeri
To cite this version:
Paolo Bellingeri. Tresses sur les surfaces et invariants d’entrelacs. Mathématiques [math]. Université
Joseph-Fourier - Grenoble I, 2003. Français. �tel-00002853�
HAL Id: tel-00002853
https://tel.archives-ouvertes.fr/tel-00002853
Submitted on 21 May 2003
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Remer iements
La liste des remer iements qui suit est sommaire et loin d'être exhaustive. Je m'ex use par
avan e auprès de
eux que j'ai oublié et leur assure que ma pensée va aussi vers eux.
Le premier groupe de remer iements est professionel. Je remer ie d'abord mes dire teurs,
Louis Funar et Vlad Sergies u, pour le soutien, humain et mathématique, pendant tout mon
séjour à l'Institut Fourier, depuis mon arrivée en DEA. Le premier
j'ai suivi à Grenoble était un
ours de Christine Les op sur l'integrale de Kontsevi h. Je suis
heureux de la retrouver à la n et je la remer ie d'avoir a
Je remer ie Thomas Fiedler et Luis Paris d'avoir a
remarques,
ours de Topologie que
epté de faire partie du jury.
epté d'être mes rapporteurs; leur
onseils et suggestions ont été fondamentals et omplémentaires. Je les remer ie
aussi de me faire l'honneur de faire partie du jury. C'est grâ e à Luis Paris que j'ai
onnu
le GDR de tresses dont je fais partie depuis presque quatre ans. Dans
onnu
plusieurs experts de tresses ave
lesquels j'ai dis uté et
e
ontexte j'ai
ollaboré à maintes reprises. Parmi
eux, je remer ie Nuno Fran o, Eddy Godelle, Patri k Dehornoy, Juan González-Meneses et
Bert Wiest.
Je tiens à remer ier l'Institute Simion Stoilow pour l'a
ueil reçu pendant mon séjour à
Bu arest, en parti ulier Barbu Ber eanu et Stefan Papadima pour leurs remarques, suggestions
et en ouragements.
Ma thèse ainsi que mes re her hes sont aussi le fruit d'autres dis ussions, à l'intérieur
ou a l'extérieur de l'Institut Fourier; sans les
iter tous, je tiens à remer ier Roland Ba her,
Mi hael Eisermann, Emmanuel Ferrand, Soa Lambropoulou, Hugh Morton, Marta Rampihini et Mohammed Uludag. Depuis quelques années j'ai le plaisir de parti iper à des projets de
vulgarisation de mathématiques, en Italie et en Fran e. Je remer ie Maria Dedò, Simonetta Di
Sieno, Sylvestre Gallot, Mar el Morales et Cristina Turrini de leur
onan e et enthousiasme.
Je n'oublie pas la superbe équipe de l'Institut Fourier, tout parti ulièrement l'impe
Arlette, Bruno, Corinne, Myriam et Janni k, ainsi que tous les amis thésards et post-do
j'ai ren ontré pendant
able
que
es années. Une pensée parti ulière va sans doute aux nouvelles mamans
Hélène et Sophie et aux in ontournables Constantin, Fran, Guillaume, Matthieu, Stéphane,
Torsten et Xavier.
Le dernier remer iement, le plus important, va à ma femme, Anneli, et à ma famille italopolonaise. Leur soutien et leur patien e
onstituent le pilier
entral de mon travail.
1
Introdu tion
Avant propos
Bn ,
Le groupe de tresses
introduit par Artin en 1926 ([1℄), joue un rle remarquable dans
plusieurs domaines des mathématiques, en parti ulier dans la théorie des n÷uds.
Les Théorèmes d'Alexander et de Markov établissent une
orrespondan e entre n÷uds (et
entrela s) et tresses. Plus pré isément deux entrela s sont isotopes si et seulement si les tresses
qui les représentent sont reliées par une suite de mouvements élémentaires dans la tour
de Markov
Cela implique que la re her he des invariants d'entrela s
vérient les
sur la tour de
[n2C [Bn ℄,
orrespond à la
onstru tion des
onstru tion resta longtemps purement
théorique, jusqu'aux années 80 et à la dé ouverte du polynme de Jones ([58℄). La
les
algèbres de He ke
tra es
'est-à-dire, des familles de fon tionnelles linéaires qui
onditions du théorème de Markov. Cette
algébrique de
[n2Bn.
onstru tion
e polynme est basée sur la dénition (indu tive) d'une tra e de Markov sur
([23℄), qui sont des quotients de dimension nie des algèbres des groupes
de tresses.
Les travaux de Jones
onduisent à la question naturelle suivante:
Peut-on dénir d'autres invariants d'entrela s ave
des
onstru tions analogues au polynme
de Jones (et son extension, le polynme HOMFLY-PT) ?
Le premier résultat de
ubiques
ette thèse (dans l'ordre
hronologique) est une réponse armative à
ette question. Plus pré isément dans le hapitre 5 nous nous intéressons aux algèbres de He ke
, qui sont d'autres quotients des algèbres des groupes de tresses. En suivant l'appro he
de Jones, nous
onstruisons deux nouveaux invariants d'entrela s dans
R 3 . Ces invariants,
férents des invariants HOMFLY-PT et de Kauman, sont ré ursivement
dif-
al ulables et dénis
univoquement par deux relations skein (gure 1).
L'autre sujet de
ette thèse est l'étude des tresses et, plus généralement, des tresses singulières
F on peut dénir le groupe de tresses B (n; F ) ave
Bn ([42℄). Ces groupes sont une généralisation naturelle
du groupe fondamental de la surfa e F et ils sont liés aux Mapping lass groups et à la théorie
des espa es de ongurations ([17℄). Un sous-groupe remarquable de B (n; F ) est le groupe de
tresses pures P (n; F ), qui est le noyau de la proje tion de B (n; F ) dans le groupe symétrique
à n éléments.
sur les surfa es. Étant donnée une surfa e
une
onstru tion analogue à
elle de
Nous exhibons de nouvelles présentations, simples, pour les groupes de tresses et de tresses
pures sur les surfa es. Ces présentations sont des extensions de présentations usuelles de
Bn et
du groupe fondamental de la surfa e. Le nombre de générateurs et relations est inférieur aux
autres présentations
onnues et, à notre
g 1) est nouveau dans la littérature.
onnaissan e, le
as d'une surfa e à bord (de genre
L'intérêt pour les groupes de tresses est également motivé par la re her he d'invariants d'entrela s sur les
3-variétés.
En eet, il existe une généralisation du théorème de Markov pour les
3-variétés ([86℄), qui relie les entrela s de la variété M ave les tresses sur la surfa e F , où
F est la surfa e asso iée à la dé omposition à livre ouvert de M ([73℄). Nous étendons aux
i
3-variétés la
tra e de Markov
onstru tion de Jones et nous obtenons un résultat partiel, en
sur un
ertain quotient de l'algèbre de
B (n; F ).
onstruisant une
Les tresses singulières sont des tresses ayant un nombre ni de points doubles. Les tresses
n brins sur le disque , ave la omposition usuelle de hemins, forment le monoïde
SBn , appelé monoïde de tresses singulières à n brins sur le disque. Le monoïde SB (n; F ) de
tresses singulières à n brins sur une surfa e F , a été introduit dans [48℄ an de dénir des
singulières à
invariants de type ni ([4℄) pour les tresses sur les surfa es. Nous obtenons qu'il se plonge dans
un groupe et que le problème du mot est résoluble dans
la
ara térisation des
entralisateurs de
te hniques de Fenn, Rolfsen et Zhu pour
SB (n; F ). Ces résultats dé
oulent de
e monoïde, que nous obtenons en généralisant des
SBn .
Nous détaillons nos résultats dans les paragraphes suivants.
Tresses sur les surfa es
Dans le premier
tresses
B (n; F ).
Théorème 1.
hapitre nous démontrons des nouvelles présentations pour les groupes de
Soit F une surfa e orientable de genre g 1 et ave p omposantes de bord. Le groupe B(n; F )
admet la présentation suivante.
Générateurs: 1; : : : ; n 1; a1 ; : : : ; ag ; b1; : : : ; bg ; z1 ; : : : ; zp 1 :
Relations:
Relations de tresses, i.e.
(Théorème 1.1.1)
Relations mixtes:
i i+1 i = i+1 i i+1 ;
i j = j i pour ji j j 2 :
(R1) ar i = i ar (1 r g; i 6= 1) ;
br i = i br (1 r g; i 6= 1) ;
(R2) 1 1 ar 1 1 ar = ar 1 1 ar 1 1 (1 r g) ;
1 1 br 1 1 br = br 1 1 br 1 1 (1 r g) ;
(R3) 1 1 as 1 ar = ar 1 1 as 1 (s < r) ;
1 1 bs 1 br = br 1 1 bs 1 (s < r) ;
1 1 as 1 br = br 1 1 as 1 (s < r) ;
1 1 bs 1 ar = ar 1 1 bs 1 (s < r) ;
(R4) 1 1 ar 1 1 br = br 1 1 ar 1 (1 r g) ;
(R5) zj i = i zj (i 6= n 1; j = 1; : : : ; p 1) ;
(R6) 1 1 zi 1 ar = ar 1 1 zi 1 (1 r g; i = 1; : : : ; p 1; n > 1) ;
1 1 zi 1 br = br 1 1 zi 1 (1 r g; i = 1; : : : ; p 1; n > 1) ;
(R7) 1 1 zj 1 zl = zl 1 1 zj 1 (j = 1; : : : ; p 1; j < l) ;
(R8) 1 1 zj 1 1 zj = zj 1 1 zj 1 1 (j = 1; : : : ; p 1) :
ii
Les tresses géométriques
de tresses d'Artin et de
présentations
orrespondant aux générateurs sont les générateurs usuels du groupe
1 (F ).
Nous renvoyons au
hapitre 1 et au Théorème 1.1.2 pour les
orrespondant aux surfa es fermées et aux Théorèmes 1.5.2 and 1.5.3 pour le
as
de surfa es non orientables. La preuve est inspirée d'une preuve de Morita pour la présentation
d'Artin de
Bn ([71℄). Nous obtenons ensuite une présentation pour P (n; F ), dans le
surfa e orientable. Cette présentation est une extension de la présentation
de tresses pures
Pn Bn
as d'une
lassique du groupe
([17℄).
Théorème 2. (Théorème
Soit F une surfa e orientable de genre g 1 ave
P (n; F ) admet la présentation suivante:
Générateurs:
1.6.1)
p>0
omposantes de bord. Le groupe
fAi;j j 1 i 2g + p + n 2; 2g + p j 2g + p + n 1; i < j g:
Relations:
si
ou
(P R1) Ai;j1 Ar;sAi;j = Ar;s (i < j < r < s) (r + 1 < i < j < s);
(i = r + 1 < j < s r < 2 g
r 2g) ;
1
1
(P R2) Ai;j Aj;sAi;j = Ai;s Aj;sAi;s (i < j < s) ;
(P R3) Ai;j1 Ai;sAi;j = Ai;sAj;sAi;s Aj;s1 Ai;s1 (i < j < s) ;
(P R4) Ai;j1 Ar;sAi;j = Ai;sAj;sAi;s1 Aj;s1 Ar;s Aj;sAi;sAj;s1 Ai;s1
(i + 1 < r < j < s)
(i + 1 = r < j < s r < 2g
r > 2g) ;
1
1
(ER1) Ar+1;j Ar;s Ar+1;j = Ar;s Ar+1;sAj;sAr+1;s
r < 2g
;
1
(ER2) Ar 1;j Ar;s Ar 1;j = Ar 1;s Aj;sAr 11;sAr;s Aj;sAr 1;sAj;s1 Ar 11;s
r < 2g
:
ou
si
si
si
paire et
si
si
ou
impaire et
paire
impaire
Le Théorème 1.6.2 fournit un résultat analogue pour les surfa es orientables fermées.
Pn est un produit semi-dire t de Pn 1 et de Fn , le groupe libre de
rang n, où l'a tion induite de Pn 1 sur l'abelianisé de Fn est triviale. On dit alors que Pn est un
T1
d
d
d+1
produit quasi-dire t de Pn 1 et de Fn . Par onséquent,
d=0 I (Pn ) = f0g et I (Pn ) =I (Pn )
k
est un Z-module libre pour tout d 0, où I est la puissan e k -ième de l'idéal d'augmentation
3
de ZPn . Ce résultat est fondamental dans la théorie de Vassilev pour les entrela s dans R
Le groupe de tresses pures
(voir [74℄).
Le groupe
Kn (F ), qui est la lture normale de Pn dans P (n; F ), est étudié dans [48℄. On
Kn (F ) est un produit quasi-dire t iteré de groupes libres de rang inni et on
démontre que
onstruit un invariant universel de type ni pour les tresses sur une surfa e fermée.
Nous introduisons le groupe
où
E est la surfa
Y (n; F ), déni
lture normale dans P (n; F ) de P (n; E ),
F . Nous obtenons que Y (n; F ), qui ontient
omme la
e obtenue en enlevant les anses de
iii
proprement
K (n;TF ),
est un produit quasi-dire t iteré de groupes libres (Proposition 1.6.3).
1 I (Y (n; F ))d = f0g et I (Y (n; F ))d =I (Y (n; F ))d+1 est un Z-module libre
Par onséquent,
d=0
pour tout d 0. D'autre part, lorsque F est une surfa e de genre g 1 à bord, P (n; F ) est un
produit semi-dire t iteré de groupes libres de rang ni, mais il n'est pas quasi-dire t, à
ause
des relations (ER1) et (ER2) dans le Théorème 2 (voir se tion 1.6.3).
multipli atif
Nous remarquons aussi que la relation (R4) dans le Théorème 1 implique qu'il n'existe pas un
invariant universel
de type ni pour les tresses sur les surfa es de genre
1 ([7℄).
Graphes et présentations de tresses
Dans le
hapitre 2 nous poursuivons la re her he de présentations pour les groupes de tresses
sur les surfa es. Sergies u ([84℄) a démontré que l'on peut asso ier à tout graphe à
n sommets
sur le plan ( onnexe, sans bou les ni interse tions) une présentation pour le groupe de tresses
Bn . Ce résultat a été ensuite généralisé pour des autres familles de graphes. Les présentations
ainsi obtenues sont en général très redondantes mais elles permettent de relier les relations
des tresses à la géométrie du graphe. En parti ulier, les présentations par graphes ont été
utilisées dans le problème de
Nous allons don
Bn
Bn ([53℄).
onjugaison pour
des monoides de tresses positives dans
onsidérer le
([18℄) et dans le problème de plongement
as des graphes sur une surfa e
F
et des groupes de tresses
orrespondant. Nous démontrons que l'on peut asso ier à tout graphe à
n
sommets sur la
sphère ( onnexe, sans bou les et interse tions) une présentation pour le groupe de tresses
B (n; S 2 ) (Théorème 2.2.1) et nous déduisons quelques résultats sur les auto2
2
morphismes de B (n; S ) (Corollaire 2.3.1). Nous démontrons aussi que Out(B (n; S )) est
2
isomorphe à Z2 Z2, pour n 4 (Proposition 2.4.2). Les automorphismes 1 ; 2 de B (n; S )
1
dénis par 1 (j ) = j
pour j = 1; : : : ; n
1 et 2 (j ) = j U pour j = 1; : : : ; n 1, où
n
U = (1 n 1 ) est le générateur du entre de B (n; S 2 ), sont des représentants pour les
2
générateurs de Out(B (n; S )).
sur la sphère
Tresses singulières sur les surfa es
Dans le
de
hapitre 3, nous étudions le monoïde de tresses singulières,
B (n; F ), leur inverses, plus des générateurs singuliers 1 ; : : : n
tresses ave
SB (n; F ). Les générateurs
1,
qui
un point double, forment un ensemble de générateurs pour
orrespondent à des
SB (n; F ).
Nous dé-
montrons que les tresses sur les surfa es satisfont une propriété analogue aux tresses singulières
sur le disque ([40℄).
Théorème 3. (Théorème
Pour tout x 2 SB(n; F ), les propriétés suivantes sont équivalentes:
1. j x = xk ,
2. jr x = xkr ; pour quelques r 2 Z n f0g,
3. jr x = xkr ; pour tout r 2 Z,
4. j x = xk ;
3.3.2)
iv
5. jr x = xkr ; pour quelques r 2 Z n f0g.
lasses
L'idée de la preuve est de
de la surfa e
n brins sur la surfa e F omme des mapping
est un ensemble de n points distin ts. En parti ulier, dans
onsidérer les tresses à
F n P , où P
les Théorèmes 3.3.1 and 3.3.2 on traduit les relations du Théorème 3 en termes d'a tion de
tresses sur les
lasses d'isotopies d'ar s (un ar
extremités dans
est un plongement de l'intervalle unitaire ave
P ). Comme appli ation du Théorème 3 et d'une propriété de rédu tion pour
les tresses singulières (Lemme 3.4.2), on déduit des preuves simples pour les résultats suivants:
Théorème 4. (Théorème
3.4.1)
Théorème 5. (Théorème
3.5.2)
Le monoïde SB(n; F ) se plonge dans un groupe.
Le problème du mot pour SB(n; F ) est résoluble.
Algèbres de He ke sur les surfa es
Dans le
hapitre 4 nous rappelons quelques dénitions et
He ke, tra es de Markov et
seront utiles dans le
onstru tions
lassiques (algèbres de
onstru tion algébrique du polynome d'HOMFLY-PT) qui nous
hapitre 5, et nous introduisons les algèbres de He ke sur la surfa e
F
omme le quotient
Hn (q; F ) = C [B (n; F )℄=(j2 + (1 q)j
où
q ; j = 1; : : : ; n 1) ;
j sont les générateurs usuels des groupes des tresses. Nous
as q = 1.
onstruison une tra e de Markov
pour le
Théorème 6. (Théorème
Soit b l'ensemble des lasses0 de onjugaison de 1(F ) et b0 = b f1g. Soit S(C b0 ) l'algèbre
tensorielle symétrique de C b . Pour tout z 2 C , il y a une (unique) famille Tn de fon tionnelles
linéaires
Tn : Hn(1; F ) ! S (C b 0 )
telles que
4.1.1)
Tn(xy) = Tn(yx) 8x; y 2 Hn(1; F ) ;
Tn+1(xn) = zTn(x) 8x 2 Hn(1; F ) ;
Tn+1(n 1A1 nx) = AbTn(x) 8x 2 Hn(1; F ) 8A 2 B (1; F ) ;
Tn(1) = 1 ;
où Ab dénote la lasse de onjugaison de A 2 B(1; F ) = 1(F ).
Nous pensons que
uls sont bien plus
quons que
e résultat s'étend aux algèbres
Hn(q; F ) (voir aussi [78℄). Toutefois, les
al-
ompliqués et l'utilisation d'un ordinateur semble né essaire. Nous remar-
es algébres ont été pré édemment étudiées dans le
as parti ulier
F = S 1 I ([66℄,
[77℄). En suivant l'appro he de Jones, une tra e de Markov ainsi que l' invariant d'entrela s
orrespondant ont été ainsi
onstruits dans le
pour le tore solide avait été pré édemment
as du tore solide
F
I . Le module de skein
al ulé par Turaev ([87℄, [88℄).
v
Invariants d'entrela s satisfaisants une relation skein
Dans le
hapitre 5 on
ubique
onsidère une autre généralisation des algèbres de He ke et on dénit
deux nouveaux invariants polynomiaux qui sont
al ulables ré ursivement et qui sont diérents
de polynmes d'HOMFLY-PT et Kauman. Nous rappelons que le polynme de Jones vérie
la relation skein (d'é heveau) suivante :
!
t 1V
En autres termes, on
!
tV
onsidère trois entrela s ave
plan) sauf au voisinage du
d'un noeud, on peut
al ul ré ursif de
t1=2 )V
1
A
le même diagramme (même proje tion sur le
roisement representé en gure. Etant donné un diagramme planaire
hanger
ertains
représente le diagramme trivial. De
pour un
1=2
= (t
0
roisements pour obtenir un nouveau diagramme qui
ette manière on peut utiliser la relation skein
V . En remplaçant le fa
quadratique
teur
(t
1=2
i-dessus
t1=2 ) par x on obtient l'invariant
HOMFLY-PT. On peut remarquer que la relation qui dénit le polynme de HOMFLY-PT est
. En eet, en rajoutant un
V
0
1
A
roisement positif on obtient la relation skein suivante:
!
= xtV
Le polynme de Kauman est l'autre extension
+ t2 V
0
1
A
onnue du polynme de Jones et il est déni
par les relations skein suivantes sur les diagrammes non orientés.
!
!
!
=z +
+
= a (
Quelques manipulations élémentaires montrent que
0
B
1
0
1
C
A = ( + z) a
On a ré emment démontré que
n'est pas susante pour un
relations skein dont une
ave
1
!!
)
vérie une relation skein
0
1
z
( + 1) a
A
0
1
A + ( )
a
ette relation ne peut pas être
al ul ré ursif de
([31℄). La re
omplète,
1
A
'est-à-dire, elle
her he d'un système
ubiques
ubique, est parti ulièrement intéressante et di ile. En
L.Funar ([8℄) nous avons obtenu deux nouveaux invariants
Théorème 7. (Théorème
ubique:
omplet de
ollaboration
.
Ils existent deux invariants I( ; ) et I (z; Æ) qui sont uniquement dénis par les deux relations
skein en gure 1 ( et par leur valeur sur le noeud trivial qui est traditionellement 1). Ces
invariants prennent valeurs dans
5.1.1)
Z[
; ; (2
2 )=2 ;
(H(
; ))
vi
(
2+2
)=2 ℄
;
= αw
= Aw
+E
-2
+F
+G w
+F
+N w 3
-1
+B w
+L w 2
+I w
3
+ w
-1
-1
+B w
+E
+H w
2
+ βw
+L w 2
+C w
+D
+G w
+H w
+M w 2
+M w 2
+P w 4
+O w 3
Figure 1: Les relations skein.
et respe tivement
où 1 2 f0; 1g
H(
; )
Z[z =2; Æ =2 ℄
;
(P (z; Æ) )
est le nombre de omposantes mod 2 et
:= 8
6
36
8
5 2 + 2 4 4 + 36 4
4 + 38
+8
6
34
3
17 + 8;
3 3 + 17 3 + 8 2 5 + 32 2 2
et respe tivement
P(z; Æ) := z23 + z18Æ 2z16 Æ2 z14Æ3 2z9 Æ4 + 2z7Æ5 + Æ6z5 + Æ7:
I i on denote par (Q) l'idéal engendré par l'élément Q dans l'algèbre respe tive.
A; B; C; ::; P orrespondant à I( ; ) sont donnés i-dessous. Pour obtenir les
(z; Æ) , il sut de faire le hangement de variable w = ( z 4 =(Æz ))1=2 ,
oe ients asso iés à I
7
2
4
= (z + Æ )=(z Æ) et = (Æ z 2 )=z 3 dans le tableau 1.
Les polynmes
vii
2 ))1=2
w = (( 2 + 2 )=(2
2
2
B=(
)
2
D = (1 + 2 + 2 3 )
3)
F = (1 + 2
H = ( 3 2 2 2 + 2)
2 3
2)
L = (2 3 + 3 2
3
N = (1 + 4 + 3 2 2
2
5
2
P = (3
2 3
+4
A=( 2
)
2)
C=( 2
3)
E = (1 + + 2 2
3
2
G=(
2 2 )
4
3
2 2 2
I=(
3 )
4
M =(
2 3 2 + 2)
3 ) O = (1 + 3
3
+3 2 2
4
3)
4)
Tableau 1
La preuve du Théorème est une extension de l'appro he de Jones et elle est detaillée dans
les se tions 1, 2 et 3 du
hapitre 5. La première des relations skeins
onsidérations sur les quotients
l'algèbre de He ke
i-dessus provient de
ubiques des algèbres de groupes de tresses
ubique par analogie ave
les algèbres de He ke
C [Bn ℄.
On dénit
lassiques (voir [23℄):
H (Q; n) = C [Bn ℄=(Q(j ) ; j = 1; : : : ; n 1) ;
où
Q(j ) = j3
j2
Notre but est de
j
2C.
1; ;
onstruir des tra es de Markov sur la tour d'algèbres de He ke
ubiques, qui
dénissent des invariants pour les entrela s. La diéren e entre les algèbres de He ke usuelles
et
elles
don
ubiques est de la même nature que
nis) et
elle entre les groupes de Coxeter sphériques (et
eux hyperboliques (en général innis). En eet, pour
Q(0) 6= 0 on a (voir [28℄):
dimC H (Q; 3) = 24, et H (Q; 3) est isomorphe à l'algèbre du groupe tétraédral < 2; 3; 3 >
d'ordre 24 (i.e.
SL(2; Z3 )).
dimC H (Q; 4) = 648, et H (Q; 4) est isomorphe à l'algèbre du groupe G25 , selon la
las-
si ation de Shepard-Todd ([85℄).
H (Q; 5) est l'algèbre de He ke
onje turé que
y lotomique du groupe
G32 , qui est d'ordre 155520. Il est
ette algèbre est libre de dimension nie,
e qui impliquerait, en utilisant
le théorème de déformation de Tits, qu'elle est isomorphe à l'algèbre de
G32 .
dimC H (Q; n) = 1 pour n 6.
En parti ulier la dénition dire te d'une tra e sur
H (Q; n), n 6 se heurte au problème
la dimension innie.
Pour rester justement dans un
ontexte de dimension nie on introduit les quotients
de
Kn ( ; ),
H (Q; 3). La forme exa te de ette relation est:
2 12 2 + A 12 22 12 + B 1 22 12 + B 12 22 1 +C 12 2 12 + D 1 22 1 + E 1 2 12 +
E 12 2 1 + F 22 12 + F 12 22 + G 2 12 + G 12 2 + H 22 1 + H 1 22 + I 1 2 1 +
L 2 1 + L 1 2 + M 12 + M 22 + N 1 + O 2 + P = 0
où A; B; : : : ; P sont les polynmes du tableau 1.
en rajoutant une relation de plus qui vit dans
Remarque Les algèbres Kn ( ; ) sont de dimension nie pour tout n.
On donne une expli ation intuitive du
simple (pour
Q générique) et se dé
hoix de
ompose
omme
viii
ette relation. L'algèbre
C3
H (Q; 3)
est semi-
M23 M3 , où Mm est l'algèbre des
m m. Si on quotiente par le fa teur C M22 M3 on obtient l'algèbre de He ke
usuelle Hq (3). De même l'algèbre de Birman-Wenzl qui est liée au polynme de Kauman,
2
3
s'obtient en passant au quotient par C M2 . Dans notre as, on prend le quotient par C .
matri es
Notre résultat prin ipal est une
onséquen e immédiate du résultat te hnique
i-dessous:
Il y a exa tement quatre valeurs de (z; zb) pour lesquelles
il existe une (unique) tra e de Markov T sur la tour K( ; ) ave les paramètres (z; zb),
'est-à-dire:
1. T (xy) = T (yx), pour tout x; y 2 Kn( ; ), et tout n.
2. T (xn 1) = zT (x), pour tout x 2 Kn( ; ), et tout n.
3. T (xn 11) = zbT (x), pour tout x 2 Kn( ; ), et tout n.
Le premier ouple (z; zb) est
Théorème 8.
(Théorème 5.1.2)
2 )=(
z = (2
+ 4); zb = (
2+2
)=(
+ 4);
et la tra e asso iée est T ; : Kn( ; ) ! Z[ ; ; 1=( + 4)℄=(H( ; )).
Les trois autres solutions ne sont pas des fon tions rationnelles
et 'est plus onvenable de
2
onsidérer ; et zb omme fon tions de z; Æ, où Æ = z ( z + 1). Plus pré isément on a une
tra e de Markov
T (z; Æ) : K( ; ) ! Z[z1; Æ1 ℄=(P(z; Æ) );
où = (Æ z2)=z3 ; = (z7 + Æ2)=(z4 Æ); zb = z4=Æ:
Idée de la preuve.Kn+1 ( ; )
D'abord tout élément de
peut être é rit
omme
ombinaison
an , où a; b 2 Kn ( ; ) et = f0; 1; 2g. Ce i implique que
Kn ( ; ) s'étend d'une manière unique à une tra e de Markov sur
linéaire d'éléments du type
une tra e de Markov sur
Kn+1 ( ; ). La partie
ompliquée
on erne don
l'existen e d'une telle tra e de Markov.
Notre méthode, fortement inspirée de [11℄, est une amélioration de
elle utilisée dans [43℄.
On dénit un graphe géant dont les sommets sont les éléments du semi-groupe abélien engendré
par le groupe libre à
n
1
générateurs. Les arêtes
orrespondent aux éléments qui dièrent
par exa tement une relation parmi les relations qui dénissent
orientation sur les arêtes, en
arêtes
orrespondant aux
Kn ( ; ).
One donne une
hoisissant un pro essus de rédu tion des mots, sauf pour les
ommutations:
ai j b
! aj ib (j i j j> 1),
qui restent non
orientées.
On prouve que, par rapport à l'ordre partiel ainsi déni, ils existent des éléments minimaux
(peut-être plusieurs) dans
haque
omposante
onnexe du graphe. Ensuite on
onsidère la
suite as endante de graphes qui modélise les fon tionnelles sur la tour d'algèbres
satisfaisant les
onditions 2.) et 3.)
K ( ; )
i-dessus.L'uni ité des éléments minimaux pour la réunion
de graphes est équivalente à un nombre ni d'obstru tions.
Plus pré isément on montre que toute fon tionnelle
K4 ( ; )
admet une extension à tous les
ondition de
Kn ( ; ), n
omme avant qui est bien dénie sur
5.
Si l'on rajoute maintenant la
ommutativité 1.) (pour en faire une tra e de Markov) on montre à nouveau
qu'on peut se ramener à la
En parti ulier
ommutativité dans
K4 ( ; ).
es obstru tions sont en nombre ni,
e qui nous a permis de les traiter à
l'aide d'un ordinateur. Les valeurs des paramètres se trouvent en utilisant la
ix
ommutativité
sur
K3 ( ; ) et ensuite les
al uls expli its montrent que toutes les obstru tions appartiennent
à l'idéal engendré par le polynme
H(
; )
(et respe tivement
Maintenant, étant donnée une tra e de Markov
T
P (z;Æ) ).
on dénit un invariant pour les entrela s à
l'aide de la formule standard:
1
I (x) =
z zb
n
2
e(x)
zb 2
1
T (x);
z
x 2 Bn est une tresse dont la lture est l'entrela s L et e(x) est la somme des exposants de
x. On trouve ainsi les invariants I( ; ) et I (z; Æ) du Théorème 7. Des al uls expli ites montrent
où
que:
Ces invariants distinguent les n÷uds ave
au plus 10
roisements, ayant le même invariant
HOMFLY-PT.
I(
; )
= I(
n÷uds ave
;
)
pour les n÷uds amphi hirals et
au plus 10
roisements, dont la
de Kauman et HOMFLY-PT et le
Tout
2-
I(
; )
déte te la
hiralité de tous les
hiralité n'est pas dete tée par les polynmes
ables de HOMFLY-PT.
omme HOMFLY-PT, Kauman et leurs
2-
ables, les invariants
I(
; )
et
I (z; Æ)
semblent ne pas distinguer les n÷uds mutants.
Il est très di ile, à l'état a tuel, de
omprendre à quel point
es polynmes dièrent des
polynmes usuelles de Kauman et HOMFLY-PT. En parti ulier, on se pose la question si les
indéterminations engendrées par les polynmes
Conje ture.
de Z[
et
P
sont essentielles.
Il y a une tra e de Markov sur H (Q; n) à valeurs dans une extension algébrique
la tra e de Markov sous-ja ente à I( ; ).
; ℄ qui relève
Remarquons que les polynmes
tionelles, don
au plus 8
H
et
P
dénissent des
ourbes algébriques planes non ra-
on ne peut pas expli iter une variable. Dans l'Appendi e du
donne un tableau ave
ave
H
les valeurs des polynmes
roisements.
x
I(
hapitre 5 on
; 0) (K ) et I(0; ) (K ) pour tous les n÷uds
Contents
1 Braids on surfa es
5
1.1
Presentations for surfa e braid groups
1.2
Preliminaries
1.3
1.4
1.5
5
7
1.2.1
Fadell-Neuwirth brations . . . . . . . . . . . . . . . . . . . . . . . . . .
7
1.2.2
Geometri
8
interpretations of generators and relations . . . . . . . . . . .
Outline of the proof of Theorem 1.1.1
1.3.1
The indu tive assertion
1.3.2
The existen e of a se tion
1.3.3
End of the proof
. . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . .
9
9
10
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13
Proof of Theorem 1.1.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14
1.4.1
About the se tion
14
1.4.2
Proof of Lemma 1.4.1
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
Other presentations and remarks
1.5.1
1.6
. . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Braids on
p-pun
. . . . . . . . . . . . . . . . . . . . . . . . . .
tured spheres
. . . . . . . . . . . . . . . . . . . . . . .
14
15
15
1.5.2
Braids on non-orientable surfa es . . . . . . . . . . . . . . . . . . . . . .
16
1.5.3
González-Meneses' presentations
17
1.5.4
Appli ations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Surfa e pure braid groups
1.6.1
. . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Presentations for surfa e pure braid groups
losure of
Pn
in
. . . . . . . . . . . . . . . .
P (n; F )
19
20
20
1.6.2
Remarks on the normal
. . . . . . . . . . . . .
23
1.6.3
Almost-dire t produ ts . . . . . . . . . . . . . . . . . . . . . . . . . . . .
24
2 Braid presentations via graphs
27
2.1
Introdu tion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
27
2.2
Sphere braid groups presentations via graphs
28
2.3
2.4
. . . . . . . . . . . . . . . . . . .
2.2.1
Denitions and Statement of the Main Theorem
2.2.2
Geometri
. . . . . . . . . . . . .
28
interpretation of relations . . . . . . . . . . . . . . . . . . . .
30
Proof of Theorem 2.2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
31
2.3.1
Preliminaries
2.3.2
Indu tive steps
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3.3
Automorphisms and isometries
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The outer automorphisms group of
31
34
. . . . . . . . . . . . . . . . . . . . . . .
36
. . . . . . . . . . . . . . . . . . . .
37
B (n; S 2 )
3 Singular braids
39
3.1
Denitions and results
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
39
3.2
Preliminaries
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
40
1
3.2.1
3.3
3.4
Mapping
lass groups
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
40
3.2.2
Braids and ar s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
42
3.2.3
Isotopy invariants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
43
Statements of Main Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . .
44
3.3.1
Centralisers of
. . . . . . . . . . . . . . . . . . . . . . . . . . .
44
3.3.2
Singular ribbons
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
45
3.3.3
Centralisers on
B (n; F ) .
SB (n; F ) . . . . .
SB (n; F ) embeds in a group
The monoid
. . . . . . . . . . . . . . . . . . . . .
46
. . . . . . . . . . . . . . . . . . . . .
47
3.4.1
Extended singular braids . . . . . . . . . . . . . . . . . . . . . . . . . . .
47
3.4.2
Singular braids embed in extended singular braids
. . . . . . . . . . . .
49
3.5
The word problem is solvable . . . . . . . . . . . . . . . . . . . . . . . . . . . .
50
3.6
Monoid presentations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
51
4 Generalized He ke Algebras
4.1
4.2
4.3
55
Introdu tion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Preliminaries
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
56
4.2.1
Markov tra es . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
56
4.2.2
Algebrai
onstru tion of HOMFLY-PT polynomial . . . . . . . . . . . .
57
Proof of Theorem 4.1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
58
5 Cubi He ke algebras and new invariants for links
5.1
5.2
5.3
5.4
5.5
65
Introdu tion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
65
5.1.1
A short history . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
65
5.1.2
The main result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
66
5.1.3
Cubi
68
He ke algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.1.4
Outline of the proof
5.1.5
Properties of the invariants
Markov tra es on
. . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Kn ( ; )
. . . . . . . . . . . . . . . . . . . . . . . . . . . . .
He ke algebra
H (Q; 3)
69
70
72
A base for the
. . . . . . . . . . . . . . . .
72
5.2.2
The homogeneous quotient of rank 3 . . . . . . . . . . . . . . . . . . . .
72
5.2.3
Uniqueness of Markov tra e on
CPC Obstru tions
ubi
. . . . . . . . . . . . . . . . . . . . . . . . .
5.2.1
. . . . . . . . . . . . . . . . .
73
75
5.3.1
The pentagonal
The
5.3.3
The bi oloured graph
The
Kn ( ; ) .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.3.2
ondition
olored pentagon
. . . . . . . . . . . . . . . . . . . . . . . . . .
ondition: the denition of
(H ): the sub-module H .
n
omputation of obstru tions
n
75
. . . . . . . . . . .
77
. . . . . . . . . . . . .
79
. . . . . . . . . . . . . . . . . . . . . . . . . .
90
5.4.1
Commutativity obstru tions . . . . . . . . . . . . . . . . . . . . . . . . .
90
5.4.2
The CPC obstru tions for n=4
92
The existen e of Markov tra es
5.5.1
5.6
55
Statements
. . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . .
93
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
93
5.5.2
Proof of Theorem 5.5.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . .
93
5.5.3
Proof of Theorem 5.5.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . .
94
5.5.4
Corollaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
95
The invariants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
I(
; )
5.6.1
The denition of
5.6.2
The
5.6.3
Chirality and other properties of
ubi al behaviour
. . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
I(
; )
. . . . . . . . . . . . . . . . . .
96
96
97
97
5.7
I (z; Æ)
5.6.4
The denition of
5.6.5
Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Appendix
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Bibliography
98
99
99
100
3
4
Chapter 1
Braids on surfa es
1.1 Presentations for surfa e braid groups
Let F be an orientable surfa e and let P = fP1 ; : : : ; Pn g be a set of n distin t points of F . A
geometri braid on F based at P is an n-tuple = ( 1; : : : ; n) of paths i : [0; 1℄ ! F su h
that
i (0) = Pi ; i = 1; : : : ; n;
i (1) 2 P ; i = 1; : : : ; n;
1 (t); : : : ; n (t) are distin t points of F
for all
t 2 [0; 1℄.
The usual produ t of paths denes a group stru ture on the set of braids up to homotopies
B (n; F ), does not depend on the hoi e of P and it is
F . On the other hand, let be Fn F = F n n , where
is the big diagonal, i.e. the n-tuples x = (x1 ; : : : xn ) for whi h xi = xj for some i 6= j .
There is a natural a tion of n on Fn F by permuting oordinates. We all the orbit spa e
F^n F = Fn F=n onguration spa e. Then the braid group B (n; F ) is isomorphi to 1 (F^n F ).
We re all that the pure braid group P (n; F ) on n strings on F is the kernel of the natural
proje tion of B (n; F ) in the permutation group n . This group is isomorphi to 1 (Fn F ). The
among braids. This group, denoted
alled the braid group on
rst aim of this
A
p-pun
n
strings on
hapter is to give (new) presentations for braid groups on orientable surfa es.
tured surfa e of genus
losed surfa e of genus
g 1.
g
1 is the surfa e obtained by deleting p points on a
Theorem 1.1.1 Let F be an orientable p-pun tured surfa e of genus g 1, with p 1. The
group B(n; F ) admits the following presentation (see also Se tion 1.2.2):
Generators: 1 ; : : : ; n 1 ; a1 ; : : : ; ag ; b1 ; : : : ; bg ; z1 ; : : : ; zp 1 :
Relations:
Braid relations, i.e.
i i+1 i = i+1 i i+1 ;
i j = j i for ji j j 2 :
5
Mixed relations:
(R1) ar i = i ar (1 r g; i 6= 1) ;
br i = i br (1 r g; i 6= 1) ;
(R2) 1 1 ar 1 1 ar = ar 1 1 ar 1 1 (1 r g) ;
1 1 br 1 1 br = br 1 1 br 1 1 (1 r g) ;
(R3) 1 1 as 1 ar = ar 1 1 as 1 (s < r) ;
1 1 bs 1 br = br 1 1 bs 1 (s < r) ;
1 1 as 1 br = br 1 1 as 1 (s < r) ;
1 1 bs 1 ar = ar 1 1 bs 1 (s < r) ;
(R4) 1 1 ar 1 1 br = br 1 1 ar 1 (1 r g) ;
(R5) zj i = i zj (i 6= 1; j = 1; : : : ; p 1) ;
(R6) 1 1 zi 1 ar = ar 1 1 zi 1 (1 r g; i = 1; : : : ; p 1; n > 1) ;
1 1 zi 1 br = br 1 1 zi 1 (1 r g; i = 1; : : : ; p 1; n > 1) ;
(R7) 1 1 zj 1 zl = zl 1 1 zj 1 (j = 1; : : : ; p 1; j < l) ;
(R8) 1 1 zj 1 1 zj = zj 1 1 zj 1 1 (j = 1; : : : ; p 1) :
Theorem 1.1.2 Let F be a losed orientable surfa e of genus g 1. The group B (n; F )
admits the following presentation:
Generators: 1 ; : : : ; n 1 ; a1 ; : : : ; ag ; b1 ; : : : ; bg :
Relations:
Braid relations as in Theorem 1.1.1.
Mixed relations:
ar i = i ar (1 r g; i 6= 1) ;
br i = i br (1 r g; i 6= 1) ;
(R2) 1 1 ar 1 1 ar = ar 1 1 ar 1 1 (1 r g) ;
1 1 br 1 1 br = br 1 1 br 1 1 (1 r g) ;
(R3) 1 1 as 1 ar = ar 1 1 as 1 (s < r) ;
1 1 bs 1 br = br 1 1 bs 1 (s < r) ;
1 1 as 1 br = br 1 1 as 1 (s < r) ;
1 1 bs 1 ar = ar 1 1 bs 1 (s < r) ;
(R4) 1 1 ar 1 1 br = br 1 1 ar 1 (1 r g) ;
(T R) [a1 ; b1 1 ℄ [ag ; bg 1 ℄ = 1 2 n2 1 2 1 ;
(R1)
where [a; b℄ := aba 1b 1.
We may assume that Theorem 1.1.1 provides also a presentation for
an orientable surfa e with
p
boundary
B (n; F ), when F
is
omponents. We re all that the rst presentations of
6
braid groups on
losed surfa es were found by S ott ([82℄), afterwards revised by Kulikov and
Shimada ([63℄). Re ently González-Meneses redu ed signi antly the number of generators
([46℄). Our presentation has the same number of generators than González-Meneses'one, but
it uses the standard generators of the fundamental group of the surfa e and the number of
relations is smaller. At our knowledge, the
Our proof is inspired by Morita's
ase of pun tured surfa es is new in the literature.
ombinatorial proof for the
lassi al presentation of Artin's
braid group ([71℄). We will explain this approa h while proving Theorem 1.1.1. After that we
will show how to make this te hnique t for obtaining Theorem 1.1.2. We remark that our
argument is quite shorter than previous ones, sin e we do not need a presentation for surfa e
pure braid groups. In Se tion 1.5 we give presentations for braid groups on non orientable
surfa es.
1.2 Preliminaries
1.2.1 Fadell-Neuwirth brations
The main tool one uses is the Fadell-Neuwirth bration, with its generalisation and
ing exa t sequen es. As observed in [34℄, if
not), the map
! Fn 1 F
: Fn F
F
orrespond-
is a surfa e ( losed or pun tured, orientable or
dened by
(x1 ; : : : ; xn ) = (x1 ; : : : ; xn 1 )
is a bration with ber
F nfx1 ; : : : ; xn
1
g. The exa t homotopy sequen e of the bration gives
us the exa t sequen e
2(Fn F ) ! 2(Fn 1 F ) ! 1(F n fx1 ; : : : ; xn 1g)
! P (n; F ) ! P (n 1; F ) ! 1:
Sin e a pun tured surfa e (with at least one pun ture) has the homotopy type of a one
dimensional
omplex, we dedu e
k (Fn F ) = k (Fn 1 F ) = = k (F ); k 3
and
If
F
is an orientable surfa
2 (Fn F ) 2 (Fn 1 F ) 2 (F ) :
2
e and F 6= S , all higher homotopy groups are trivial. Thus, if F
an orientable surfa e dierent from the sphere we
(P BS ) 1
where
is the map that forgets
the last path pointed at
admits a se
losed orientable surfa
se tion is shown in [17℄ in the
xn .
(P BS ) has been ompletely solved in [52℄. It
F has pun tures. On the other hand, when
e of genus g 2, (PBS) splits if and only if n = 2. An expli it
The problem of the existen e of a se tion for
is a
is
on lude that there is an exa t sequen e
! 1(F n fx1 ; : : : ; xn 1g) ! P (n; F ) ! P (n 1; F ) ! 1;
is possible to show that
F
an
tion, when
ase of the torus.
7
1.2.2 Geometri interpretations of generators and relations
F
Let
be an orientable surfa e. Let
Be (n; F )
be the group with the presentation given in
Theorem 1.1.1 or Theorem 1.1.2 respe tively. The geometri
Be (n; F ),
represent
when
F
F
is a
losed surfa e of genus
L
as a polygon
Se tion 1.5.3). We
an
of
L.
sides with the standard identi ation of edges (see also
L,
whi h we draw with the usual over
rossing points. Figure 1.1 presents the generators of
βi
αi
interpretation for generators of
1 is the same as in [46℄, ex ept that we
onsider braids as paths on
and under information at the
realized as braids on
4g
g
αi
βi
αi
βi
bi
ai
Pi
P1
Pn
P1
ai
σi
bi
ai (respe
Note that in the braid
i
( the wall
tively
Pi+1
Pn
Figure 1.1: Generators as braids (for
goes through the wall
Be (n; F )
F
an orientable
losed surfa e).
bi ) the only non trivial string is the rst one, whi h
1 : : : ; n 1 are the lassi al braid
i ). Remark also that
generators on the disk.
βr
αr
βr
αr
βr
βr
αr
αr
11 01
00
11 01
00
Figure 1.2: Geometri
1 1 ar 1 1 br
It is easy to
and
interpretation for relation (R4) in Theorem 1.1.1; homotopy between
(on the left) and
br 1 1 ar 1
(on the right).
B (n; F ). The non trivial strings of ar
onsidered to be disjoint and then (R1) holds in B (n; F ). On the
he k that the relations above hold in
i when i 6= 1, may be
1 1 ar 1 1 is the braid whose the only non trivial string is the se
other hand,
goes through the the wall
Then
1 1 ar 1 1 and ar
is veried. The
ase of
r
and disjoint from the
ar .
(R2)
orresponding non trivial string of
ommute. Similarly we have that
(R3)
ond one, whi h
1 1 br 1 1 and br
ommute and
is similar. Figure 1.2 presents a sket h of a homotopy between
1 1 ar 1 1 br and br 1 1 ar 1 . Thus, (R4) holds in B (n; F ).
Let sr (respe tively tr ) be the rst string of ar (respe tively br ), for r = 1; : : : ; 2g , and
onsider all the paths s1 ; t1 ; : : : ; sg ; tg . We ut L along them and we glue the pie es along the
L. We obtain a new fundamental domain (see Figure 1.3, for the ase of a surfa e of
2), alled L1 , with vertex P1 . On L1 it is lear that [a1 ; b1 1 ℄ [ag ; bg 1 ℄ is equivalent to
edges of
genus
8
α1
t1
β1
s2
α
P1
α1
s1
s1
P1
t
2
P1
t1
β1
s
1
P1
β1
t1
α1
P1
β
2
α
β2
2
α
t
β2
2
s2
P1
P1
2
P1
2
t
s2
P1
Figure 1.3: The fundamental domain
2
L1 .
t1
P1
s1
P1
P1
Pn
tg
Figure 1.4: Braid
[a1 ; b1 1 ℄ [ag ; bg 1 ℄.
the braid of Figure 1.4, equivalent to the braid
in
B (n; F ).
There is an analogous geometri
p-pun
interpretation of generators of
Be (n; F ), for F
an orientable
i ; aj ; bj is the same as above. We only have
zi , where the only non trivial string is the rst one, whi h is a loop around
tured surfa e. The denition of generators
to add generators
the
1 2 : : : n2 1 : : : 2 1 and then (TR) is veried
i-th
pun ture (Figure 1.5). As above, relations
paths (Figure 1.6).
Remark that a loop of the rst string around the
geometri
braid
an be easily
p-th pun
ture
he ked on
orresponding
an be represented by the
orresponding to the element
[a1 ; b1 1 ℄ [ag ; bg 1 ℄1 1 n 1 1 1 2 z1 zp 1 :
Therefore, one has natural morphisms
n : Be (n; F ) ! B (n; F ). We prove that n are a
tually
isomorphisms.
1.3 Outline of the proof of Theorem 1.1.1
1.3.1 The indu tive assertion
F a surfa e of genus g with one pun ture. One applies
n of strands. For n = 1, Be (1; F ) = 1 (F ) = B (1; F ), then 1 is
We outline the ideas of the proof for
an indu tion on the number
an isomorphism.
Consider the subgroup
B 0 (n; F ) = 1 (n 1 ) and the map
: B 0 (n; F ) ! B (n 1; F )
9
α r
βr
αr
βr
αr
Pn
β
r
Pi
Pn
P1
Pi+1
P1
ar
σi
br
P1
Pn
γ
γi
1
γp
z
Figure 1.5: Generators as braids (for
F
i
an orientable surfa e with
Be 0 (n; F ) be the subgroup
a1 ; : : : ; ag ; b1 ; : : : ; bg ; 1 ; : : : ; n 2 ; 1 ; : : : ; n 1 ; !1 ; : : : ; !2g ; where
whi h forgets the last string. Now, let
of
p pun
Be (n; F )
tures).
generated by
1 1
j = n 1 j +1j2 j +1
(n 1 = n2 1 ) ;
n 1
!2r 1 = n 1 1 1 1 ar 1 n 1 r = 1; : : : ; g ;
!2r = n 1 1 1 1 br 1 n 1 r = 1; : : : ; g :
We
onstru t the following diagram:
Be 0 (n; F )
njBe0 (n;F )
#
B 0 (n; F )
~
e
!
B ( n 1; F )
#
n
1
! B (n 1; F )
~ is dened as n 1 1 njB0 (n;F ) . It is well dened, sin e n 1 is an isomorphism by
~(ai ) = ai ; ~(bi ) = bi for i = 1; : : : ; g and
the indu tive assumption, and it is onto. In fa t, ~(j ) = j for j = 1; : : : ; n 2.
The map
1.3.2 The existen e of a se tion
~ has got a natural se tion s : Be (n 1; F ) ! Be 0 (n; F )
j ; s(ai ) = ai ; s(bi ) = bi for j = 1; : : : ; n 2 and i = 1; : : : ; 2g.
The morphism
10
dened as:
s(j ) =
P1
γ
P
2
Figure 1.6: The braids
γ
γ
j
1
1 1 zj 1
and
P1
l
1 1 zj 1 1 .
onsidered disjoint from the non trivial string of
zj .
ommutes with the braid
Remark 1.3.1
of the pun ture.
γ
P
2
1
γ
j
γ
l
1 1 zj 1 an be
1
1
the braid 1 zj 1
The non trivial string of
zl , for j < l. Similarly,
Geometri ally this se tion onsists of adding a straight strand just to the left
G and a subset G of elements of G we set hGi for the subgroup of G generated
by G and hhGii for the subgroup of G normally generated by G . From now on, given a; b two
b
1 ab and b a = bab 1 .
elements of a group G, we set a = b
Given a group
Lemma 1.3.1
Proof:
2
Let G = f1; : : : ; n 1; !1; : : : ; !2g g. Then Ker(~) = hGi.
= 1 n
2 122 n 1 and = 1 1 = n 1 1 2 1
1 2 n 1 . By onstru tion we have hGi Ker(~). The existen e of a se tion s implies
~) = hhGii. In fa t, suppose that there is su h x 2 Ker(~) that x 2= hhGii. Thus,
that Ker (
0 6 1 on generators a1 ; : : : ; ag ; b1 ; : : : ; bg ; 1 ; : : : ; n 2 ; of Be 0 (n; F ) su h that
there is a word x =
~(x0 ) = 1, be ause all other generators of Be 0 (n; F ) are in hGi. This is false, sin e x0 = s(~(x0 )).
e 0 (n; F )
To prove that hGi is normal, we need to show that h ; h 2 hGi for all generators
of B
and for all h 2 G .
We set
i) Let
be the
i = 1; : : : ; n
(
1)
1
= n
1
lassi al braid generator
j , j = 1; : : : ; n 2. It is
h1; : : : ; n 1i, sin e it is already true in
belong to
= ar or
(j = 2; : : : n
1). Note that
ii) Let
ar 1
= !2
b =
1
1
r
r
1
1a = 1 !2 1
1
and
1b = 1 !2
and
!2r
1
1
r
r
r
r
for
for
imply
1 we obtain:
= n 1 2 ar 1 ar 1 ar 1 ar 1 1 1 1 2 1 n 1 1 =
= n 1 2 ar 1 ar 1 1 ar 1 1 ar 1 1 2 1 n 1 1 =
1
= n 1 2 1 ar 1 1 1 ar 1 1 2 1 n 1 1 = !2 1 :
1
r
11
j
and
j i
j = j = j
r = 1; : : : ; g ;
r = 1; : : : ; g :
We show only the rst equation (the other is similar). By iterated appli ation of
ar i
lassi al braid groups
= j !
!i
i = !i (i = 1; : : : ; 2g ).
= br (r = 1; : : : ; g). Commutativity relations
([71℄, [83℄). On the other hand,
j
lear that
[ar ; 1 ar 1 1 ℄ =
Set
a2;s = 1 1 as 1 for s = 1; : : : ; g and respe
tively
b2;s = 1 1 bs1 for s = 1; : : : ; g. In the
same way as above we nd that:
(RC 1) (12 )a = a2 (12 ) (r = 1; : : : ; g) ;
(12 )b = b2 (12 ) (r = 1; : : : ; g) ;
2
(RC 2) a (12 ) = (12 )a2 1 (r = 1; : : : ; g) ;
b ( 2 ) = ( 2 )b2 1 2 (r = 1; : : : ; g ) :
1
1
r
r
;r
;r
r
;r
r
Now, remark that relations
;r
(R3) and (R4) imply the following relations:
(R30 ) ar 1 as 1 1 = 1 as 1 1 ar (r < s) ;
br 1 as 1 1 = 1 as 1 1 br (r < s) ;
ar 1 bs 1 1 = 1 bs 1 1 ar (r < s) ;
br 1 bs 1 1 = 1 bs 1 1 br (r < s) ;
(R40 ) ar 1 1 br 1 1 = 1 br 1 1 ar (1 r g) ;
Relations
(RC 1); (RC 2); (R30 ); (R40 )
ombined with relations
(R2); (R3); (R4) give:
ar a = a
2;s
2;s (s < r ) ;
br a = a
2;s
2;s (s < r ) ;
ar b = b
2;s
2;s (s < r ) ;
br b = b
2;s
2;s (s < r ) ;
2
aa2;rr = a2;r 1 a2;r (1 r g) ;
2
bb2r;r = b2;r 1 b2;r (1 r g) ;
ar a = 12 a
2;r
2;r (1 r g ) ;
2
br b = 1 b
2;r
2;r (1 r g ) ;
2
a
[
a
;
a2;sr = 2;r 1 ℄ (a2;s ) (r < s) ;
2
ba2;sr = [a2;r ;1 ℄ (b2;s ) (r < s) ;
2
ab2r;s = [b2;r ;1 ℄ (a2;s ) (r < s) ;
2
bb2r;s = [b2;r ;1 ℄ (b2;s ) (r < s) ;
ar a = [12 ;a2;r1 ℄ (a ) (s < r ) ;
2;s
2;s
br a = [12 ;b2;r1 ℄ (a ) (s < r ) ;
2;s
2;s
1
2
ar b = [1 ;a2;r ℄ (b ) (s < r ) ;
2;s
2;s
1
2
br b = [1 ;b2;r ℄ (b ) (s < r ) ;
2;s
2;s
a
2
r
b2;r = (a2;r 1 a2;r1 )b2;r [1 2 ; a2;r ℄ (1 r
ar b = 2 b [a 1 ; 2 ℄ (1 r g ) ;
2;r
1 2;r 2;r 1
br a = a 2 (1 r g ) ;
2;r
2;r 1
ab2r;r = a2;r b2;r 12 b2;r1 (1 r g) :
12
g) ;
A
onsequen e of these identities and relation
1; : : : ; g.
(R1) is that !ia ; a !i ; !ib ; b !i 2 hGi, for i; r =
r
r
r
r
Set also f!1; : : : ; !2g ; 1; : : : n 1g in B0(n; F ) for fn(!1); : : : ; n(!2g ); n(1);
: : : n (n 1 )g. Then Ker() is freely generated by f!1 ; : : : ; !2g ; 1 ; : : : ; n 1 g.
Proof:
Lemma 1.3.2
The diagram
P (n; F )
\
#
B 0(n; F )
is
!
P (n 1; F )
\
#
! B (n 1; F )
ommutative and the kernels of horizontal maps are the same. As stated in Se tion 1.2.1,
n fP1 ; : : : ; Pn 1 g; Pn ).
Ker() = 1 (F
ure 1.7 and the non trivial strings of
n fP1 ; : : : ; Pn 1g
h!1; : : : ; !2g ; 1 ; : : : n 1j ; i.
group of
F
based on
If the fundamental domain is
!j ; i are
Pn , it is
onsidered as loops of the fundamental
lear that
Pn
ω2
Pn
n fP1 ; : : : ; Pn 1 g; Pn ) =
ω
3
Pn
ω
4
3
τ1
Pn
Pn
P1
ω4
ω
ω2
Pn
Figure 1.7: Interpretation of
Lemma 1.3.3 njBe0 (n;F )
Proof:
ω4
1 (F
ω
ω1
ω3
hanged as in Fig-
ω1
ω
Pn
1
Pn
2
!j ; i as loops of the fundamental group.
is an isomorphism.
From the previous Lemmas it follows that the map from
isomorphism. The Five Lemma and the indu tive assumption
Ker(~)
to
Ker()
on lude the proof.
is an
1.3.3 End of the proof
In order to show that
n
is an isomorphism, let us remark rst that it is onto. In fa t, from
Lemma 1.3.3 it follows that the image of
surje ts on
n .
Sin e the index of
Be (n; F )
B 0 (n; F )
in
13
Pn and on the other hand Be (n; F )
B (n; F ) is n, it is su ient to show that
ontains
[Be (n; F ) : Be 0 (n;SF )℄ n. Consider the elements j = j n 1 (we set n = 1) in Be (n; F ).
e0
e
We laim that
i i B (n; F ) = B (n; F ). We only have to show that for any (positive or
e
e 0 (n; F )
negative) generator g of B (n; F ) and i = 1; : : : ; n there exists j = 1; : : : ; n and x 2 B
su h that
g is a
If
gi = j x :
lassi al braid, this result is well-known ([24℄). Other
!j . Thus every element of Be (n; F )
1
0
e
e i j 2
= B (n; F ) for i 6= j we are done.
the denition of
Sin
ases
ome almost dire tly from
an be written in the form
i Be 0 (n; F ).
p > 1. This time Be 0 (n; F ) is the subgroup of Be (n; F )
generated by a1 ; : : : ; ag ; b1 ; : : : ; bg ; 1 ; : : : ; n 2 ; 1 ; : : : ; n 1 ; !1 ; : : : ; !2g ; 1 ; : : : ; p 1 where
j ; !r are dened as above and j = n 1 1 1 1 zj 1 n 1 .
The previous proof holds also for
1.4 Proof of Theorem 1.1.2
1.4.1 About the se tion
B 0 (n; F ) = 1 (n 1 ). Let Be 0 (n; F ) be
generated by a1 ; : : : ; ag ; b1 ; : : : ; bg ; 1 ; : : : ; n 2 ; 1 ; : : : ; n 1 ; !1 ; : : : ;
dened as above. Remark that 1 2 hGi sin e from (TR) relation, the
The steps of the proof are the same. We set again
the subgroup of
Be (n; F )
where
are
!2g ;
j ; !r
following relation
1 = [!1 ; !2 1 ℄ [!2g 1 ; !2g1 ℄n 11 2 1 ;
e 0 (n; F ). When F is a losed surfa e the orresponding ~ has no se
holds in B
tion (see Se tion
1.2.1). Nevertheless, we are able to prove the analogous of Lemma 1.3.1 (see se tion 1.4.2).
Lemma 1.4.1
n
1
g.
Let F be a losed surfa e. Then Ker(~) is generated by f!1; : : : ; !2g ; 2; : : : ;
The following Lemma is analogous to Lemma 1.3.2.
Let F be a losed surfa e and set also f!1; : : : ; !2g ; 2; : : : ; n 1g in B0(n; F )
for fn(!1); : : : ; n(!2g ); n(2); : : : ; n(n 1)g. Ker() is freely generated by f!1; : : : ; !2g ; 2; : : : ;
n 1 g.
Lemma 1.4.2
g
j = j n 1 (where n = 1). We may on lude by he king that for any generator
Be (n; F ) (or its inverse) and i = 1; : : : ; n there exists j = 1; : : : ; n and x 2 Be 0 (n; F ) su h
Let
of
that
gi = j x ;
whi h is a sub- ase of previous situation.
1.4.2 Proof of Lemma 1.4.1
To
on lude the proof of Theorem 1.1.2, we give the demonstration of Lemma 1.4.1. Let us
begin with the following Lemma.
14
Lete 0F be a losed surfa e and G = f2; : : : ; n 1; !1; : : : ; !2g g. The subgroup
hGi is normal in B (n; F )
Proof:
Lemma 1.4.3
It su es to
onsider relations in Lemma 1.3.1. Remark that from relations shown in
Lemma 1.3.1, it follows also that the set
jj = 1; : : : n 1; word on f!11 ; : : : ; !2g1 gg ;
is a system of generators for hh1 ; : : : ; n 1 ii hhn 1 ii.
f j
1
In order to prove Lemma 1.4.1, let us
Ker~ tn
onsider the following diagram
~
e
!
B (n 1; F )
! Be 0(n; F )
i
Æ
qn
#
~0 #
Be 0 (n; F )=hhn
Ker~0 !
i0
1
ii
qn is the natural proje tion, ~0 is dened by ~0 Æ qn = ~ and tn is dened
0
by i Æ tn = qn Æ i. Sin e tn is well dened and onto we dedu e that Ker (tn ) = hhn 1 ii.
~0 has a natural se tion s : Be (n 1; F ) ! Be 0 (n; F )=hhn 1 ii dened as s(ai ) = [aj ℄,
Now, s(bi ) = [bj ℄ and s(j ) = [j ℄, where [x℄ is a representative of x 2 Be 0 (n; F ) in Be 0 (n; F )=hhn 1 ii.
~0) = hhKii, where
Thus, using the same argument as in Lemma 1.3.1, we derive that Ker (
K = f[!1 ℄; : : : ; [!2g ℄; [2 ℄; : : : [n 1℄g. From Lemma 1.4.3 it follows that hKi = hhKii. Moreover,
sin e i 2 hhn 1 ii for i = 1; : : : ; n
2, Ker(~0 ) = h[!1 ℄; : : : ; [!2g ℄i.
In this diagram
From the exa t sequen e
1 ! hhn
it follows that the set
tem of generators for
1
f!1; : : : ; !2g g
Ker(~).
h2 ; : : : ; n 1; !1 ; : : : ; !2g i.
ii ! Ker(~) ! Ker(~0) ! 1
and a system of generators for
hhn 1 ii
form a sys-
From the remark in Lemma 1.4.3 it follows that
Ker(~) =
1.5 Other presentations and remarks
1.5.1 Braids on p-pun tured spheres
We re all that the exa t sequen e
1
! 1(F n fP1 ; : : : ; Pn 1 g; Pn ) ! P (n; F ) ! P (n 1; F ) ! 1
holds also when
F = S2
([35℄). Thus, previous arguments may be repeated in the
ase of the
sphere, to obtain a new proof for the well-known presentation of braid groups on the sphere
as quotients of
lassi al braid groups. When
F
to the following result.
15
is a
p-pun
tured sphere, our argument leads
Theorem 1.5.1
Let F be a p-pun tured sphere. The group B(n; F ) admits the following pre-
sentation:
Generators: 1 ; : : : ; n 1 ; z1 ; : : : ; zp 1 :
Relations:
Braid relations, i.e.
Mixed relations:
i i+1 i = i+1 i i+1 ;
i j = j i for ji j j 2 :
(R1) zj i = i zj (i 6= 1; j = 1; : : : ; p 1) ;
(R2) 1 1 zj 1 zl = zl 1 1 zj 1 (j = 1; : : : ; p 1; j < l) ;
(R3) 1 1 zj 1 1 zj = zj 1 1 zj 1 1 (j = 1; : : : ; p 1) ;
We remark that this presentation
oin ides with the presentation shown in [66℄.
1.5.2 Braids on non-orientable surfa es
Previous te hniques
an be used in the
ase of non-orientable surfa es to prove the following
Theorems.
Theorem 1.5.2 Let F be a non-orientable p-pun tured surfa e of genus g 1, with p 1.
The group B(n; F ) admits the following presentation:
Generators: 1 ; : : : ; n 1 ; a1 ; : : : ; ag ; z1 ; : : : ; zp 1 :
Relations:
Braid relations, i.e.
Mixed relations:
(R1)
(R2)
(R3)
(R4)
(R5)
(R6)
(R7)
i i+1 i = i+1 i i+1 ;
i j = j i for ji j j 2 :
ar i = i ar (1 r g; i 6= 1) ;
1 1 ar 1 1 ar = ar 1 1 ar 1 (1 r g) ;
1 1 as 1 ar = ar 1 1 as 1 (s < r) ;
zj i = i zj (i 6= 1; j = 1; : : : ; p 1) ;
1 1 zi 1 ar = ar 1 1 zi 1 (1 r g; i = 1; : : : ; p 1; n > 1) ;
1 1 zj 1 zl = zl 1 1 zj 1 (j = 1; : : : ; p 1; j < l) ;
1 1 zj 1 1 zj = zj 1 1 zj 1 1 (j = 1; : : : ; p 1) :
16
Let F be a losed non-orientable surfa e of genus g 2. The group B(n; F )
admits the following presentation:
Generators: 1 ; : : : ; n 1 ; a1 ; : : : ; ag :
Relations:
Braid relations as in Theorem 1.1.1.
Mixed relations:
Theorem 1.5.3
(R1)
(R2)
(R3)
(T R)
We give only a geometri
ar i = i ar (1 r g; i 6= 1) ;
1 1 ar 1 1 ar = ar 1 1 ar 1 (1 r g) ;
1 1 as 1 ar = ar 1 1 as 1 (s < r) ;
a21 a2g = 1 2 n2 1 2 1 :
dene the path
F
2g sides as in Figure 1.8, and we make an additional
interpretation for the generators. To represent a braid in
onsider the surfa e as a polygon of
e as in the left hand of the Figure 1.8 and
we
ut:
ut the polygon along it. We get
represented as in the right hand side of the same gure, where we
an also see how we
F
hoose
P1 ; : : : ; Pn . We show generators in Figure 1.9. Generators j and zj are as above.
r = 1; : : : ; g, the braid ar onsists on the rst string passing through the r-th wall,
the points
For all
while the other strings are trivial paths. Relations
an be easily veried drawing
orresponding
braids. The relation (TR) in Theorem 1.5.3 is shown in [46℄. We remark that Theorem 1.5.3
provides also a presentation for braid groups on the proje tive plane (see also [90℄).
α g-1
α
αg
αg-1
α
α2
α
α
2
g
P1
Pn
00
11
00
1
0
1
1
0011
00 0
11
0 0
1
0 11
1
0
1
1
α1
e
g
α
αg
e
e
1
α1
Figure 1.8: Representation of a non-orientable surfa e
F.
1.5.3 González-Meneses' presentations
Let
F
be a
losed orientable surfa e of genus
g
1. Using the same arguments outlined in
previous Se tions we may provide an other presentation for
B (n; F ).
Theorem 1.5.4 Let F be a losed orientable surfa e of genus g 1. The group B (n; F )
admits the following presentation:
Generators: 1 ; : : : ; n 1 ; b1 ; : : : ; b2g :
17
αr
αr
P
P
Pn
P1 0
11
0
11
0 00
1
0
1
P
i
1
1
0
10
1
0 0
1
n
11
0
01
01
1
00
0
1
P1 0
1 1
0
01
0 0 1
1
1
0
Pn
P
i+1
e
e
e
e
σ
ar
z
F
j
a non-orientable surfa e).
Relations:
Braid relations as in Theorem 1.1.1.
Mixed relations:
(R1)
(R2)
(R3)
(T R)
A
e
e
i
Figure 1.9: Generators as braids (for
j
br i = i br (1 r 2g; i 6= 1) ;
bs 1 1 br 1 1 = 1 br 1 1 bs (1 s < r 2g) ;
br 1 1 br 1 1 = 1 1 br 1 1 br (1 r 2g) ;
b1 b2 1 : : : b2g 1 b2g1 b1 1 b2 : : : b2g1 1 b2g = 1 2 n2
losed orientable surfa e
F
of genus
g
1
an be easily veried on
2 1 :
is represented as a polygon
where opposite edges are identied. Figure 1.10 gives a geometri
Relations
1
L
of
4g
sides,
interpretation of generators.
orresponding braids.
αi
P1
Pi
Pn
Pi+1
αi
bi
σi
Figure 1.10: Generators as braids (for
The presentation in Theorem 1.5.4 is
F
an orientable
losed surfa e).
lose to González-Meneses' presentation.
Theorem 1.5.5 ([46℄) Let F be a losed orientable surfa e of genus g 1. The group B (n; F )
admits the following presentation:
Generators: 1 ; : : : ; n 1 ; a1 ; : : : ; a2g :
18
Relations:
i i+1 i = i+1 i i+1 ;
i j = j i for ji j j 2 ;
[ar ; A2;s ℄ = 1 (1 r; s 2g; r 6= s) ;
[ar ; i ℄ = 1 (1 r 2g; i 6= 1) ;
[a1 : : : ar ; A2;r ℄ = 12 (1 r 2g) ;
a1 : : : a2g a1 1 : : : a2g1 = 1 2 n2 1 2 1 ;
(1)
(2)
(3)
(4)
(5)
(6)
where A2;r = 1 1(a1 : : : ar
j
1
1
1
1 ar+1 : : : a2g )1 :
Remark that the geometri
interpretation of
aj
is odd and respe tively to
1,
when
j
orresponds to the braid generator
is even. Tedious
aj when
omputations show that relations
bj 's with aj 's) imply relations in Theorem 1.5.5.
in Theorem 1.5.4 (after repla ing generators
In the same way, Theorem 1.5.3
bj
an be also veried dire tly,
he king that the relations in
Theorem 1.5.3 imply all relations of the González-Meneses' presentation for braid groups on
non orientable
losed surfa es in [46℄. However, we remark that the presentation in Theorem
1.5.3 is simpler and with less relations than González-Meneses' one.
On the other hand, it seems di ult to give an algebrai
proof of the equivalen e between
presentation in Theorem 1.1.2 and presentation in Theorem 1.5.5.
1.5.4 Appli ations
We
on lude this Se tion with some remarks. Let
Consider a
onne ted subsurfa e
a boundary
omponent of
F
E F , su
be a surfa e, possibly with boundary.
omponent of
F . We suppose also that E
: B (n; E ) ! B (n; F ) indu ed by the in
2
not ontain a disk D . We may provide
or lies in the interior of
n
is known ([75℄) that the natural map
is inje tive if and only if
F
h that every boundary
F nE
does
E
either is
ontains
lusion
P . It
EF
an analogous
hara terisation about surje tion.
Let F be a surfa e of genus g 1 with p 0 boundary omponents, and
let E be a subsurfa e of F . The natural map n : B(n; E) ! B(n; F ) indu ed by the in lusion
E F is surje tive if and only if F n E is a disjoint union of disks.
Proof:
E
F
k
n : B (n; E ) ! B (n; F )
Proposition 1.5.1
When
is obtained from
removing
disks, the natural map
is onto and Theorems 1.1.1, 1.1.2, 1.5.2 and 1.5.3 give a des ription of
the natural morphism
1
is a surje tion if and only if
p
F nE
Ker( n ). Remark that
: 1 (E; P1 ) ! 1 (F; P1 )
is a disjoint union of disks. Now
onsider a pure braid
2 P (n; F ) as a n-tuple of paths (p1 ; : : : ; pn) and let : P (n; F ) ! 1(F )n
dened by
(p) = (p1 ; : : : ; pn ). The following
P (n; E )
( n )jP (n;E )
#
P (n; F )
ommutative diagram holds
19
! 1(E )n
#
1
! 1(F )n
1
be the map
Sin e
is surje
not surje tive. Thus, sin e
on
( n )jP (n;E ) is not surje tive on P (n; F ) when 1 is
1 (P (n; F )) belongs to P (n; E ), it follows that
n is not surje tive
n
tive ([17℄) we dedu e that
B (n; F ) when
1
is not surje tive.
The fa t that the natural map n : B(n; E) ! B(n; F ) is onto when E
is obtained from F removing k disks an also be obtained from the remark that B(n; E) is a
subgroup of B(n + k; F ) and that the map n orresponds to the usual proje tion B(n + k; F ) !
B (n; F ). The existen e of a braid ombing in B (n + k; F ) ([66℄) implies the laim.
Proposition 1.5.2 Let F be a orientable surfa e of genus g 1, possibly with boundary. Let
Nn (F ) be the normal losure of Bn in B (n; F ). The quotient B (n; F )=Nn (F ) is isomorphi to
H1 (F ), the rst homology group of the surfa e F .
Proof:
j
1
Remark 1.5.1
It is su ient to repla e all
with
in Theorems 1.1.1 and 1.1.2.
1.6 Surfa e pure braid groups
Several presentations for surfa e braid groups are known, when
F
is a
losed surfa e or a holed
disk ([46℄, [52℄, [66℄, [82℄). In Theorem 1.6.1 we provide a presentation for pure braid groups
on orientable surfa es with boundary. This presentation is
of the pure braid group
Pn
braid groups on orientable
lose to the standard presentation
on the disk. We provide also the analogous presentation for pure
losed surfa es.
1.6.1 Presentations for surfa e pure braid groups
Let F be an orientable surfa e of genus g 1 with p > 0 boundary omponents. P (n; F ) admits the following presentation:
Generators:
Theorem 1.6.1
Relations:
fAi;j j 1 i 2g + p + n 2; 2g + p j 2g + p + n 1; i < j g:
if
for even
if
or
or
(P R1) Ai;j1 Ar;s Ai;j = Ar;s (i < j < r < s) (r + 1 < i < j < s);
(i = r + 1 < j < s
r < 2g r 2g) ;
1
1
(P R2) Ai;j Aj;sAi;j = Ai;s Aj;sAi;s (i < j < s) ;
(P R3) Ai;j1 Ai;s Ai;j = Ai;sAj;sAi;s Aj;s1 Ai;s1 (i < j < s) ;
(P R4) Ai;j1 Ar;s Ai;j = Ai;s Aj;sAi;s1 Aj;s1 Ar;s Aj;sAi;sAj;s1 Ai;s1
(i + 1 < r < j < s)
(i + 1 = r < j < s
r < 2g r > 2g) ;
1
1
(ER1) Ar+1;j Ar;s Ar+1;j = Ar;s Ar+1;sAj;sAr+1
;s
r
r < 2g ;
(ER2) Ar 11;j Ar;s Ar 1;j = Ar 1;s Aj;sAr 11;sAr;s Aj;sAr 1;sAj;s1 Ar 11;s
r
r < 2g :
or
if
or
for odd
if even and
if odd and
20
if
or
Proof:
The
hoi e of the notation is motivated by the notation for standard generators of
from [15℄. Let
Pe (n
1; F )
Pn
be the group dened by above presentation. We give in Figure
h = 2g + p 1. In respe t of the
presentation for B (n; F ) given in Theorem 1.1.1, the elements Ai;j are the following braids:
Ai;j = j h i+1 hi2 hi+11 h j 1h, for i 2g + p ;
1.11 a pi ture of
orresponding braids on the surfa e. Let
Ai;j = j 1zi 12g 1 1 j 1h, for 2g < i < 2g + p ;
A2i;j = j 1ag 1i+1 1 1 j 1h, for 1 i g ;
A2i 1;j = j h 1bg 1i+11 1 j 1h, for 1 i g .
The relations (PR1),
arise when we
: : : , (PR4)
orrespond to the
onsider two generators
A2i;j , A2i
lassi al relations for
1;k ,
for
1
Pn . The new relations
i g and j 6= k. They
orre-
spond to two loops based at two dierent points whi h go around the same handle. Relations
(ER1) and (ER2)
an be veried by expli it pi tures or using relations in Theorem 1.1.1. The
g
1
1
A
p-1
A
2g,2g+p
2g+1,2g+p+1
1
1
0
2
1
0
A 1,2g+p
Figure 1.11: Geometri
string of the braid
Ai;j
te hnique to prove that
1
0
A
Ai;j .
interpretation of
(P R1); : : : ; (ER2) is a
We mark again with
n
1
0
2g+p+1, 2g+p+n-1
Ai;j
the only non trivial
omplete system of relations for
P (n; F ) is well
known ([46℄, [52℄, [66℄, [82℄). As shown in [57℄, given an exa t sequen e
1 ! A ! B ! C ! 1;
and presentations hGA ; RA i and hGC ; RC i, we an derive a presentation hGB ; RB i for B , where
GB is the set of generators GA and oset representatives of GC . The relations RB are given by
the union of three sets. The rst orresponds to relations RA , and the se ond one to writing
ea h relation in C in terms of orresponding oset representatives as an element of A. The last
set
orresponds to the fa t that the a tion under
oset representative of
A is an element of A. We an apply this
n = 1. By indu tion, suppose that for
e
n 1, P (n 1; F ) = P (n 1; F ). The set of elements Ai;2g+n+p 1 (i = 1; : : : ; 2g + n + p 2)
is a system of generators for 1 (F n fP1 ; : : : ; Pn 1 g; Pn ). To show that (P R1); : : : ; (ER2)
is a omplete system of relations for P (n; F ) it su es to prove that relations RP (n;F ) are a
onsequen e of relations (P R1); : : : ; (ER2). Sin e 1 (F nfP1 ; : : : ; Pn 1 g; Pn ) is a free group on
generators of
C (and their inverses) on ea
onjugation of ea h
h generator of
result on (PBS) sequen e. The presentation is
the given generators, we just have to
orre t for
he k the se ond and the third set of relations. Consider
Ai;j in P (n 1; F ) the generator Ai;j in P (n; F ).
1
Relations lift dire tly to relations in P (n; F ). The a tion of Ai;j on 1 (F nfP1 ; : : : ; Pn 1 g; Pn )
may be dedu ed from that of Ai;j . In fa t, relations (PR2) and (PR3) imply that
Ai;j Ai;2g+n+p 1 Aj;2g+n+p 1 = Ai;2g+n+p 1Aj;2g+n+p 1Ai;j ;
as
oset representative for the generator
21
for all
i < j < 2g + n + p 1, and from this relation and relations (PR2) we dedu
Ai;j Ai;2g+n+p 1 Ai;j1 = Aj;21g+n+p 1Ai;2g+n+p 1 Aj;2g+n+p
for all
;
i < j < 2g + n + p 1). It follows that
As;j Ai;2g+n+p 1As;j1 2 hA1;2g+n+p 1 ; : : : ; A2g+n+p
for all
1
e that
s < j < 2g + n + p 1.
Thus we have proved that
hA1;2g+n+p 1; : : : ; A2g+n+p
and that also the third set of relations of
In the same way we
RP (n;F ) is a
2;2g+n+p 1
i;
i is a normal subgroup
onsequen e of (P R1); : : : ; (ER2). 2;2g+n+p 1
an prove the following Theorem.
Theorem 1.6.2 Let F be an orientable losed surfa e of genus g 1. P (n; F ) admits the
following presentation:
Generators: fAi;j j 1 i 2g + n 1; 2g + 1 j 2g + n; i < j g:
Relations:
if
for even
if
or
or
(P R1) Ai;j1 Ar;sAi;j = Ar;s (i < j < r < s) (r + 1 < i < j < s);
(i = r + 1 < j < s
r < 2g r > 2g) ;
1
1
(P R2) Ai;j Aj;sAi;j = Ai;s Aj;sAi;s (i < j < s) ;
(P R3) Ai;j1 Ai;sAi;j = Ai;s Aj;sAi;sAj;s1 Ai;s1 (i < j < s) ;
(P R4) Ai;j1 Ar;sAi;j = Ai;sAj;sAi;s1 Aj;s1 Ar;sAj;sAi;s Aj;s1 Ai;s1
(i + 1 < r < j < s)
(i + 1 = r < j < s
r < 2g r > 2g) ;
1
1
(ER1) Ar+1;j Ar;sAr+1;j = Ar;sAr+1;s Aj;sAr+1
;s
r
r < 2g ;
(ER2) Ar 11;j Ar;sAr 1;j = Ar 1;sAj;sAr 11;s Ar;sAj;sAr 1;s Aj;s1 Ar 11;s
r
r < 2g ;
or
if
or
for odd
if
or
if even and
if odd and
(T R)
[A2g;12g+k ; A2g 1;2g+k ℄
2Y
g+n
j =2g+k+1
[A2;12g+k ; A1;2g+k ℄ =
2g+
k 1
Y
l=2g+1
Al;2g+k A2g+k;j k = 1; : : : ; n :
Let E be a holed disk. Theorem 1.6.1 provides a presentation for P (n; E)
([66℄). Let us re all that P (n; E) is a (proper) subgroup of Pn+k , where k is the number of
holes in E.
Remark 1.6.2 We re all that Pn embeds in P (n; F ) ([75℄) and thus Pn is isomorphi to the
subgroup
Remark 1.6.1
hAi;j j 2g + 1 i < j 2g + ni ;
22
when F is a losed surfa e and Pn is isomorphi to
Pn = hAi;j j 2g + p i < j 2g + p + n 1i ;
when F is a surfa e with p > 0 boundary omponents. Consider the sub-surfa e E obtained
removing g handles from F . The group P (n; E) embeds in P (n; F ) ([75℄) and it is isomorphi
to the subgroup
hfAi;j j 2g +1 i < j 2g + ng[fA2k 1;l ; A2k;l1 A2k1 1;l A2k;l j 1 k g ; 2g +1 l 2g + ngi ;
when F is a losed surfa e and respe tively to the subgroup
hfAi;j j 2g + 1 i 2g + p + n 2; 2g + p j 2g + p + n 1; i < j g[
[fA2k 1;l ; A2k;l1 A2k1 1;l A2k;l j 1 k g; 2g + p l 2g + p + n 1gi ;
when F is a surfa e with p > 0 boundary omponents.
Remark 1.6.3 When F is a surfa e with genus, from relation (ER1) we dedu e that generators Ai;j for 2g + p i < j 2g + n + p 1, whi h generate a subgroup isomorphi to Pn,
are redundant. Then Theorem 1.6.1 provides a (homogeneous) presentation for P (n; F ) with
(2g + p 1)n generators.
1.6.2 Remarks on the normal losure of Pn in P (n; F )
Kn (F ),
Pn in P (n; F ) ([45℄).
Lemma 1.6.1 Let : P (n; F ) ! 1 (F )n be the map dened by (p) = (p1 ; : : : ; pn ). Let F be
a losed orientable surfa e possibly with boundary. Let Kn(F ) be the normal losure of Pn in
P (n; F ). Then
Ker() = Kn (F ) :
Proof: We outline the ase of a surfa e F with boundary. The in lusion Kn(F ) Ker()
P (n; F )
b (n; F ) with generators
is obvious. The quotient group
is isomorphi to the group P
Kn (F )
fAi;j j 1 i 2g + p 1; 2g + p j 2g + p + n 1g and relations f[Ai;j ; Ak;l℄ = 1; j 6= lg.
b (n; F ) and (F )n .
The morphism indu es an isomorphism between P
1
As
orollary of previous presentations we give an easy proof of a well-known fa t on
the normal
losure of
Proposition 1.6.1
lassi al pure braid group
Let F be an orientable surfa e possibly with boundary. When F is a torus
[P (n; F ); P (n; F )℄ = Kn (F ) :
Otherwise the stri t in lusion holds:
Proof:
[P (n; F ); P (n; F )℄ Kn (F ) :
Kn (F ) [P (n; F ); P (n; F )℄ follows from relation (ER1). Suppose that
P (n; F )
[P (n; F ); P (n; F )℄ = Kn (F ) = Ker() for g > 1. It follows that
is abelian. This is
Ker()
n
false sin e 1 (F ) is not abelian for g > 1. Let w 2 [P (n; F ); P (n; F )℄. The sum of exponents
Ai;j in w must be zero. The proje tion of (w) on any oordinate is the sub-word of w
The in lusion
onsisting of the generators asso iated to
is zero, if
F
orresponding strand. Sin e the sum of exponents
is a torus this proje tion is trivial and the
23
laim follows.
1.6.3 Almost-dire t produ ts
Let us re all the following denition:
We say that a group G is residually a P -group if for ea h g 2 G, g 6= 1,
there exists a normal subgroup N of G su h that g 2= N and G=N has the property P .
lower entral series fGigi0
G
Gi
G1 = G
1 1
Denition 1.6.1
The
of a group
is the series of groups
dened by
Gi+1 = [G; Gi ℄, where [G; Gi ℄ is the subgroup of G generated by all ommutators hkh k
for h 2 G and k 2 Gi . Set G(i) = Gi =Gi+1 . Magnus proved that free groups, as well
,
,
as
fundamental groups of orientable surfa es, are residually torsion free nilpotents (see [6℄ for an
G is residually torsion free nilpotent then G is biorderable ([81℄). The group G is alled
bi-orderable if there exists a total order < on G su h that for all g; h; k in G the relation g < h
implies that kg < kh and gk < hk .
outline of the proof ). We refer to [67℄ for more results on residual nilpoten e. We re all just
that if
In this se tion we
1.
2.
onsider subgroups of surfa e braid groups having the following properties:
T1
d
d=0 I (G) = f0g;
I (G)d =I (G)d+1 is a
d 0, where IGk
augmentation ideal of the group ring of the group G.
free
Z-module
for all
Free groups have properties 1) and 2) (see [41℄). We stress that
G is residually nilpotent.
means the
T1
d
d=0 I (G)
k-th power of the
= f0g implies that
Let A; C be two groups. If C a ts on A and the indu ed a tion on the
abelianization of A is trivial, we say that A o C is an
of A and C .
Proposition 1.6.2 ([36℄) Let A; C be two groups. If C a ts on A and the indu ed a tion on
the abelianization of A is trivial, then
m
X
I (A o C )m = I (A)k I (C )m k for all m 0
k=0
and
Denition 1.6.2
almost-dire t produ t
(A o C )(m) = A(m) o C(m) :
The pure braid group
Pn
is an
almost-dire t produ t
of free groups ([37℄). In parti ular
Pn
inherits the properties of free groups that we des ribed above. These properties have been
used in [74℄ in order to
The group
onstru t an universal nite type invariant for braids.
Kn (F ), the normal
losure of
lassi al pure braid group
Pn in P (n; F ), is an almost-
dire t produ t of (innitely generated) free groups ([48℄). Moreover, it has been
an universal nite type invariant for braids on surfa es, where the group
onstru ted
Kn (F ) plays the rle
Pn .
Let F be a surfa e with boundary omponents. Consider the sub-surfa e E obtained removing
the handles of F . Let Yn (F ) be the normal losure of P (n; E ) in P (n; F ). Using our presentation for surfa e pure braid groups, we prove that the group Yn (F ), whi h ontains properly
Kn (F ), is an almost-dire t produ t of free groups.
of
Proposition 1.6.3
The group Yn(F ) is an almost-dire t produ t of free groups.
24
Proof:
We sket h a proof for
F
orientable surfa e with one boundary
omponent. Let
1 (F )n
be provided with presentation
hAj;2g+k j = 1; : : : ; 2g; k = 1; : : : ; nj[Aj;2g+k ; Al;2g+q ℄ = 1 for all j; l = 1; : : : ; 2g; 1 k 6= q n i ;
where
Aj;2g+k are the loops dened in Theorem 1.6.1. Let Fg;n be the group with presentation
hA^
j;2g+k j = 1; : : : ; g; k = 1; : : : ; nj[A^
j;2g+k ; A^
l;2g+q ℄ = 1 for all j; l = 1; : : : ; g; 1 k 6= q n i :
Let
One
: 1 (F )n
an pro
! Fg;n be the map dened by (A2i
= A^
i;2g+k and (A2i;2g+k ) = 1.
eed as in Lemma 1.6.1 for showing that Ker ( Æ ) = Yn (F ) . Thus the following
1;2g+k )
ommutative diagram holds:
1
"
1
1
1
where
Fg;1
! Fg;1
"
1
1
"
"
! Fg;n
"
Æ
! Fg;n
"
Æ
1
!1
! Ker() ! P (n; F ) !
P (n 1; F ) ! 1
"
"
"
! Gn
"
! Yn(F ) ! Yn 1(F ) ! 1
"
"
1
1
1
is the free group on g generators and Gn = Yn (F ) \ Ker ( ) is a free group.
The following set is a system of generators for Gn.
f Aj;2g+n 1 j2g < j < 2g + n and 1 j < 2g; j even g
where is a word on fA2k1 1;2g+nj1 k gg.
Proof:
Lemma 1.6.2
Consider the verti al sequen e
! Gn ! Ker() ! Fg;1 ! 1 :
Re all that Ker ( ) = 1 (F n fP1 ; : : : ; Pn 1 g; Pn ). A set of free generators for this group is
given by fAj;2g +n j1 j < 2g + ng. The map Æ sends Aj;2g +n in 1 for 2g < j < 2g + n and
1 j < 2g; j even. On the other hand, Æ (A2k 1;2g+n ) = A^
k;2g+1 for k = 1; : : : ; g .
1
25
implies that Yn 1(F ) a ts by onjugation on
Gn and thus on the abelianization Gn =[Gn ; Gn ℄. The following Lemma on ludes the proof. Re all that the existen e of a se tion for
Lemma 1.6.3
Proof:
The a tion of Yn 1(F ) by onjugation on Gn=[Gn; Gn℄ is trivial.
t 2 fAj;k j2g < k < 2g + n; 2g < j < k and 1 j < 2g ; j even g and f
fAj;2g+nj2g < j < 2g + n and 1 j < 2g ; j even g. We need to verify that every t a
trivially on Gn =[Gn ; Gn ℄. Presentation in Theorem 1.6.1 shows that
Let
(A) tft
for every
t
and
f . Now
1
and
1 j < 2g; j
onsider the a tion of
even
t
on
g,
(B ) tA2s
1;2g+n t
ts
f (mod [Gn ; Gn ℄) ;
again to Theorem 1.6.1 for showing that for every
k
2
1
= hA2s
A2s;2g+n , for s = 1; : : : ; g. We refer on e
t 2 fAj;k j2g < k < 2g + n; 2g < j <
1;2g+n
(1 s g) ;
h 2 Gn . Let be a word on fA2k1 1;2g+n j1 k < gg. From (A) and (B) it follows that,
1t 1 =
for every t 2 fAj;k j2g < k < 2g + n; 2g < j < k and 1 j < 2g; j even g, t f
t t 1 tft 1t 1 t 1 = h tft 1 1 h 1 h f 1 h 1 f 1 , where h is an element of Gn .
where
Remark 1.6.4 We noti e that lassi al te hniques do not apply to the whole group P (n; F ).
The main obstru tion is that, even when the exa t sequen e (PBS) splits, the a tion of P (n; F )
on the abelianisation of 1(F n fx1; : : : ; xn 1g) is not trivial, be ause of relations (ER1) and
(ER2). In parti ular, when F is a surfa e of genus g 1, it is presently unknown whether the
graded group asso iated to the lower entral series of P (n; F ) is torsion free.
Remark 1.6.5 A ording to [51℄ and [54℄, the mapping lass group of a pointed surfa e (see
Denition 3.2.1) is residually nite. As P (n; F ) is a (normal) subgroup of the mapping lass
group of a pointed surfa e ([15℄), it follows that P (n; F ) is residually nite.
26
Chapter 2
Braid presentations via graphs
2.1 Introdu tion
To any planar,
onne ted graph with
n
verti es, without loops or interse tions, it
asso iated a presentation for the braid group
the graph we asso iate the braid
e
Bn (Sergies
whi h is a
an be
e of
e (see Figure 2.1).
u, [83℄ and [84℄). To ea h edge
lo kwise half-twist along
1
0
1
0
1
0
111
000
000
111
e
000
111
000
111
1
0
1
0
111
000
1
0
000
111
000
111
000
111
000
111
000
111
000
111
000
111
000
111
00
11
1
0
00
11
00
11
00
11
00
11
1
000
11
1
0
1
0
1
0
1
0
1
0
1
0
0
1
1
0
1
0
Figure 2.1: Edges and geometri
Sergies u provided a
1
0
e
1
0
braids.
Bn . Afterwards,
omplete set of relations using this set of generators for
Birman, Ko and Lee ([18℄) extended this result to inner- omplete graphs in order to give a new
proof for the
graphs
onjugation problem in
Bn . Re
linearly spanned
ently Han and Ko ([53℄) showed that it is possible
to asso iate braid group presentations to a more general family of graphs (
)
ontaining above graphs. We re all also that these presentations turned out useful in
other related
ontexts (see for instan e [13℄ and [14℄). In this
hapter we provide an analogous
result for sphere braids (Theorem 2.2.1) and we prove some results on automorphisms of
B (n; S 2 ). In parti
to
Z2 Z2.
ular, we prove that the outer automorphisms group of
27
B (n; S 2 ) is isomorphi
2.2 Sphere braid groups presentations via graphs
2.2.1 Denitions and Statement of the Main Theorem
Denition 2.2.1 Let be a graph on an orientable surfa e F . The graph is alled
if is onne ted, nite and it has no loops or interse tions.
Let
of
of
be a normal graph on
the
normal
F . Let S ( ) be the set of verti es of . We asso iate to the edges
braids on F (Figure 2.1) and we dene B ( ; F ) the subgroup
orresponding geometri
B (S ( ); F )
generated by these braids. In the following we will use the same notation for
X = f j edge of g and orresponding
B ( ; F ). It an be easily veried that B ( ; F ) = B (S ( ); F ) if F = D2 or S 2 .
Otherwise B ( ; F ) NS ( ) F , where NS ( ) F is the normal losure of BS ( ) in B (S ( ); F ).
Proposition 1.5.2 shows that the in lusion NS ( ) F B (S ( ); F ) is proper.
2
From now on,
is a normal graph on S .
2
Suppose that
is not a tree. The set S n
is the disjoint union of a nite number of open
2
disks D1 ; : : : ; Dm , m > 1. The boundary of Dj on S is a subgraph (Dj ) of . We hoose a
point O in the interior of (Dj ) and an edge of (Dj ), with verti es v1 and v2 . We suppose
that the triangle O v1 v2 is anti lo k-wise oriented. To the subgraph
(Dj ) we asso iate a
polygon Pj with pj edges as follows.
We suppose that the edges e1 ; : : : ; ep and the verti es x1 ; : : : ; xp of P are anti lo k-wise
oriented. The edge e1 is labelled with the edge (e1 ) = of (Dj ). The verti es x1 ; x2 of e1
are labelled with the verti es v (x1 ) = v1 ; v (x2 ) = v2 . Ea h edge ei+1 of P (we set ep +1 = e1 )
is labelled with (ei+1 ) (Dj ), the rst edge on the left of (ei ) adja ent to v (xi+1 ). The
vertex xi+2 is labelled with the other vertex adja ent to (ei+1 ). If v (xi+1 ) is a uni-valent
vertex then (ei+1 ) = (ei ) and v (xi+2 ) = v (xi ).
elements in the free group generated by the set
braids in
j
j
j
The sequen e (e1 ) : : : (ep ) dened below is the (anti lo k-wise oriented)
pseudo- y le asso iated to Dj .
Denition 2.2.2
The pseudo- y le
j
(e1 ) : : : (ep ) is uniquely dened up to
j
y li
permutation.
Let = (e1 ) : : : (ep) be a pseudo- y le of . If there exist a pair i; j,
su h that (ei) = (ej ), we say that
is the start edge of a reversing if j 6= i 1 (we set e0 = ep).
(ei ) is the end edge of a reversing if j 6= i + 1 (we set ep+1 = e1 ).
Denition 2.2.3
1 i; j p
(ei )
We set
Let
1 : : : p for the pseudo-
be a maximal tree of
(e1 ) : : : (ep ).
. We start from x going through and at the vertex y we
y le
hoose
the rst edge on the left. We iterate this pro ess until meeting an uni-valent vertex, say
orresponding to z and we start again the pro
x after we passed two times through ea h edge of .
where we go ba k through the edge
way we
ome ba k to
Denition 2.2.4
2.2).
a disk
ess. In this
Set Æx; () for the word in X orresponding to the above ir uit (Figure
Æx; () orresponds to a braid where the only non trivial string orresponds
x. The proje tion of this string on the sphere is a simple (oriented) loop bounding
The element
to the vertex
z,
ontaining all other verti es.
28
α
ζ
σ
x
ε
γ
β
δ
Figure 2.2:
Æx; () = 2 2
Æ2 2 2 .
Let be a normal graph with n verti es. The braid group B(n; S2) admits
the presentation hX j R i, where X = f j edge of g and R is the set of following
relations:
Separate relations (SR): if i \ j = ; then ij = j i ;
Adja en y relations (AR): if i; j have a ommon vertex, then ij i = j i j ;
Nodal relations (NR): if f1 ; 2 ; 3g have only one ommon vertex and they are lo kwise ordered (Figure 2.3), then
Theorem 2.2.1
1 2 3 1 = 2 3 1 2 ;
Pseudo- y le relations (PR): if 1 : : : m is a pseudo- y le and 1 is not the start edge
or m the end edge of a reversing (Denition 2.2.3 and Figure 2.4), then
1 2 m
1
= 2 3 m :
Tree relations (TR): Æx; () = 1, for every maximal tree , every vertex x 2 and every edge 2 adja ent to x.
σ
σ2
1
σ
3
Figure 2.3: Nodal relation.
The pseudo- y le 32412 in Figure 2.4 (respe tively the pseudo- y le 41232)
does not verify the hypothesis
of the denition of the relation (PR). Spheri al braids or2
responding to the words 41 and 3412 (respe tively the braids orresponding to the
words 4123 and 12332) do not represent the same element in B(n; S2).
Denition 2.2.5 Let be a normal graph and let 1 : : : p be a pseudo- y le of . We set
R ::: for the set of (PR) relations satised, up to y li permutation, by the pseudo- y le
1 : : : p.
Remark 2.2.1
1
p
29
σ1
σm
σ2
σ2
σ
σ
1
3
σ
4
Figure 2.4: Pseudo- y le relation; on the left
On the right
1 2 32 = 2 32 4 = 32 4 1
Remark 2.2.2
and
1 2 m 1 = 2 m = = m m
3 4 1 2 = 4 1 2 3 .
2.
The denition of pseudo- y le extends naturally to a tree . In parti ular
generated by the set X , where the sequen e
. The set of (TR) relations of implies the
Æx;1 ( ) is the word 1 p in the free group
1 : : : p is the pseudo- y le asso iated to S 2 n
set of relations R1::: .
p
Let be a star. For any lo k-wise ordered subset fi ; : : : ; i j j 2 g
of edges of the following relation holds in the group hX j R i:
Remark 2.2.3
1
j
i1 : : : i i1 = i i1 : : : i :
j
j
j
2.2.2 Geometri interpretation of relations
The natural map
: hX j R
i ! B ( ; S 2) is an homomorphism. It is geometri ally evident
that the relations (AR) and (SR) hold in
B ( ; S 2 ). Let
2.9. Corresponding spheri al braids verify the relation
1 ; 2 ; as in Figure
1 = 1 2
1
1 2 1 = 2 1 1 2 .
ontain a triangle
= 1 2 1 1
and thus
B ( ; S 2 ). The relation 1 2 = 2 follows from the braid relation
Let 1 ; 2 ; 3 be arranged as in Figure 2.5. We add three edges 1 ; 2 ; 3 . The nodal relation
follows from pseudo- y le relations on triangles 1 2 3 , 2 1 3 and 3 1 2 . In fa t, 1 2 3 1 =
2 3 3 1 = 2 3 3 1 = 2 3 1 2 . All other pseudo- y le relations follow from indu tion on
in
the length of the
y le.
σ
1
τ3
τ
σ
2
2
τ1
σ
3
Figure 2.5: Nodal relation holds in
Let
x; ; . The word
Æx; ()
B ( ; S 2 ).
orresponds to the geometri
the only non verti al string is the string asso iated to the vertex
Æx; () = 1 in B ( ; S 2 ).
30
B ( ; S 2 ), where
x going around (with
wise orientation) all other strings (Figure 2.6). This braid is isotopi
then
braid in
lo k-
to the trivial braid and
σ
x
Γ
Γ
Γ
Γ
τ
Figure 2.6: The braid
Æx; () asso
iated to the tree
=
n .
2.3 Proof of Theorem 2.2.1
2.3.1 Preliminaries
The steps of the proof are similar to [84℄. We need some preliminary Lemmas.
σ
m-1
σ2
σm
σ1
Figure 2.7: Proje tion of the graph
on the fa e bounded by
1 ; 2 ; : : : ; m .
Let be a normal graph on the sphere and let 1 : : : m be a pseudo- y le on
. Let hX j R i be the group dened in Theorem 2.2.1.
The groups hX j R i and hX j R n fR ::: gi are isomorphi .
Proof:
Lemma 2.3.1
1
We
an represent the graph
pseudo- y le
1 : : : m
m
on the plane, proje ting
(Figure 2.7). We need to show that the relation
m 2 = m
holds in
on the fa e bounded by the
1
1
hX j R n fR ::: g i.
1
σ
m
∆
m-1
σm
∆
(m)
(1)
∆
σ
(2)
m-1
∆
(m)
∆
σ2
∆
∆’
Figure 2.8: The maximal trees
31
(2)
∆
(1)
σ1
and 0 .
σ2
; 0 in su h that the tree ontains 2 ; : : : ; m , the tree
0 ontains 1 ; : : : ; m 1 , and [ f1 g = 0 [ fm g. The graph n f2 ; : : : ; m g = 0 n
f1 ; : : : ; m 1 g is a set of subtrees (1) ; (2) ; : : : ; (m) of (Figure 2.8). The (TR) relations
0
on and yield the following relation:
Consider two maximal trees
(A) m 2
2 2 3
m
1
= m
1
1 1 1 2 m
1 m
;
(1) (2)
(m)
1 ; : : : ; m are sub-words asso iated to orresponding sub-trees ; ; : : : ; .
Let
n 1 n n+1 q , where n 1 = n et i = j , for i; j > n + 1. We apply the
1 = 1
(PR) relation on the pseudo- y le n n+1
q 1 2 m 1 n 2 n 1 ,
where
6
n n+1 q 1 2 m 1 n 2 = n+1 q 1 2 m 1 n 2 n 1 n ;
and we derive
1 q 1 n+1
n n+1 q 1
We premultiply by
1 1
2
m 1 n 2 = 1 2 m1 n 2n 1 :
and we apply (AR) relations in order to obtain
1 1
n n+1 q 1 q 1 n+1
n
2
m 1 n 2 = 2 m 1 n 2n 1 :
It follows that
1 1 1
1
= 1 n
1 2
m1 n 2n 1n 12 1 1m1 2
1 1
n 1
1 1 :
From iterated appli ations of (AR) and (NR) relations it follows that
m 1 n 2n 1 n 12 1 1m1 = n 11 n 12 1 1 m 1 n 2 n 1 :
Then
1 1 1
holds in
1
=
2 2
m m
1
1 1
m 2 2
1 1
1
m 2 2
hX j R n fR ::: g i. The braids 1 and
2 2 m m 1 = 1 1 2 1
m
ommute, and thus
m:
From equation (A), it follows that
m 2 = m
1
1 :
The Following Lemmas establish that for any graph
angles to
the groups
hX
0 j R 0i
and
0
obtained removing or adding tri-
hX j R i dened in Theorem 2.2.1 are isomorphi .
Let 1; 2 be two adja ent edges of , whi h are not ontained in any pseudoy le. Let 0 = [ be the graph obtained
adding an edge to to form an anti lo k-wise
triangle 12 (Figure 2.9). If B( ; S2) = hX j R i then B( 0; S2) = hX 0 j R 0 i.
Proof: 2
hX 0 j R ; = 12 1 1i
1
0 2
Lemma 2.3.2
By Tietze's transformation we obtain that
for B ( ; S ) = B ( ; S ).
B ( 0 ; S 2 ).
Sin e
is a presentation
fR ; = 1 21 g R 0 , hX 0 j R 0 i is a presentation for
32
τ
σ
σ2
1
Figure 2.9: Adding or removing a triangle.
Let be an edge of 0 bounding
only one pseudo- y le, whi h is an anti lo kwise triangle 12 (Figure 2.9). Let = 20 n . If hX 0 j R 0 i is a presentation for B( 0; S2)
then hX j R i is a presentation for B( ; S ).
Proof: 1
R0
hX 0 j R ; =
Lemma 2.3.3
1 2 1
We need to show that the set of relations
are veried in the group
i. Sergies u showed ( Lemma 1.3 in [84℄) that the relations (SR), (AR) and (NR) for
the graph
0
are a
onsequen e of the relations (SR), (AR) and (NR) for the graph
and
belongs to the pseudo- y le 1 2 . The orresponding
= 1 2 1 1 and 1 2 1 = 2 1 2 . We prove that the
1
relation Æx; () = 1, for any Æ; x; , is a onsequen e of the set of relations fR ; = 1 2 1 g.
If 2
= the laim follows. Suppose that 2 . We have two ases (we refer to the Figure
the relation
= 1 2 1
1.
The edge
pseudo- y le relations derive from
2.10):
α0
α1
τ
β1
α3
β2
τ
σ1
σ
2
β
β3
0
α2
Figure 2.10: Tree relations for
1.
2.
0=
[ are generated by the set fR [ = 1 21 1 g.
Æx; () =
;
0 1 1 2 1 3 ,
where
Æx; () =
.
0 1 2 2 2 3 ,
where
Repla e
= 1 2 1 1 . In the rst
0 1 1 1 2 1
1
2 1 2 1
1
i
i
are the sub-words obtained following the rest of
are the sub-words obtained following the rest of
ase, from (SR) it follows that
1 3
=
0 1 1 1 2 1
2 1 2 3
2
=
0 1 1 1 2 2 2 3
=
2 1
and then
Æx; () = 1 holds in hX 0 j R ; = 1 2 1 1 i. In the se ond
2 = 3 3 3 : : : p p p , where k , for k = 1; : : : ; p, orresponds to an edge of
One dedu es that the relation
ase, let write
1
1
33
1 and 2 and k , for k = 1; : : : ; p, orresponds to a tree disjoint from 1
1 ommute:
The elements k (k = 1; : : : ; p) and 1 2 1
adja ent to
and
2 .
k 1 2 1 1 = 2 1 1 2 k = 1 2 1 1 k :
One derives that
[ 2 ; 1 2 1 1 ℄ = 1. From 1
1
1 2 2 2 1 2 1 3
1
1
0 1 2 1 1 2 1 2 1 2 1 3
=
1 1 ,
it follows that
1
1
0 1 2 1 1 2 2 1 2 1 1 3 =
0 1 2 1 2 2 1 3 :
1 . We remark that belongs also to the
Therefore Æx; () = 1 holds in X 0 R ; = 1 2 1
2 . A ording to Lemma 2.3.1
pseudo- y le P bounding the other onne ted omponent of S
0 1 2 1
1
1
h
we
=
=
j
i
an suppose that the set of pseudo- y le relations
is isomorphi
to
hX
0 jR
; = 1 2 1
1
i.
RP
n
are redundant and thus
hX 0 j R 0 i
2.3.2 Indu tive steps
A is a vertex of valen e greater than two. We all the valen e of ,
), the sum of valen es of all nodes of .
Denition 2.3.1
v(
node
In order to prove Theorem 2.2.1, we pro eed by indu tion on the number of
ponents of
i) Let
S2
n
om-
.
be a tree. We re all that the braid group on the sphere is a quotient of the braid
group on the disk.
Theorem 2.3.1 ([35℄)
onne ted
The group B(n; S2) admits the following presentation
Generators: 1; : : : ; n 1 :
Relations:
i i+1 i = i+1 i i+1 ;
i j = j i for ji j j 2 ;
2
1 2 n 1 2 1 = 1 :
Then, it follows that Theorem 2.2.1 holds when
is a straight line and
that Theorem 2.2.1 holds for all trees with valen e less than
q > 0 and
v( ) = 0.
let
Suppose
be a tree su h
v( ) = q. Let v0 be an uni-valent vertex of . As in Denition 2.2.4, we run on the tree
v0 and hoosing to turn on the right at ea h node. Let v1 be the vertex pre eding
the rst node. Let v2 be the rst uni-valent vertex after v0 (see Figure 2.11).
We repla e the edge between v1 and the rst node with an edge 1 joining v1 to the vertex
v2 (see Figure 2.12). The graph 1 so obtained is su h that v( 1 ) < v( ), and then Theorem
that
starting from
2.2.1 holds for
2
1.
From Lemma 2.3.2 it follows that Theorem 2.2.1 is veried for the graph
obtained adding an edge
2
between
v1
and the other vertex adja ent to
2.13). From Lemma 2.3.3 one dedu es that Theorem 2.2.1 holds for the graph
Iterating the pro ess we derive that the result holds for the initial tree
ii) Suppose that Theorem 2.2.1 holds when the number of
less than
onne ted
p > 1. Let be a normal graph su h that n has p
1 whi h bounds two pseudo- y les 1 ; 2 : : : ; n
S2
remove an edge
en ounter two
ases.
34
(see Figure
3=
.
omponents of
onne ted
and
v2
2
n f1g.
S2 n
is
omponents. We
1 ; 2 ; : : : ; m
of
. We
01
01
01
01
τ
01 v 1
01
10 v 2
01
01v
0
Γ
Figure 2.11: We suppose
01
01
01
embedded in the sphere.
10 v 2
τ1
01
01 v 1
01
01
01v
0
Γ1
Figure 2.12: Repla ing the edge
1.
n
1 .
with
is not the end edge of a reversing. From indu tion hypothesis and Tietze's trans-
formation, we dedu e that hX j R nf1 g ; 1 n 1 = 2 : : : n i is a presentation for
B ( ; S 2 ). A ording to Lemma 2.3.1 we an suppose that pseudo y le relations for the
y le 2 ; : : : ; n ; m ; : : : ; 2 are redundant. Therefore R
ontains the set fR nf1 g ; 1 n 1 = 2 : : : n g and we on lude that hX j R i is a presentation for B ( ; S 2 ).
2. If
n is the end edge of a reversing, there exists l < n su
l + 1 i < j n. It follows that the relation
h that
l = l+1 , and i 6= j
for
l+1 l+2 n 1 2 : : : l
holds in
1
= l+2 n 1 2 : : : l 1 l
hX j R i. Multiplying by 1 1 n 1 l+21 and applying (AR) relations we obtain
2 : : : l 1 l = 1 1 n 1 l+21 l+1 l+2 n 1 2 : : : l
1
=
= l+1 l+2 n 1 n 1 l+21 l+11 2 : : : l 1 ;
what yields
1 = n 1 l+11 2 l 1 l l
Thus the above argument
1
1
2 1l+1 n :
on ludes the proof.
35
01
01
01
10 v 2
τ2
τ1
01
01 v 1
01
01
01v
0
Γ2
Figure 2.13: Adding and removing triangulations.
In order to show Theorem 2.2.1, one 2 an also onsider a normal graph on
the disk and add to Sergies u's relations for B( ; D ) ([84℄) all (TR) relations for . Let
hX j Z i be the group so obtained. Lemmas 2.3.2 and 2.3.3 as well as the indu tive steps hold
true and then hX j Z2 i is isomorphi to B(n; S2). Anyway, our approa h allows to onsider
as embedded on S and to prove in an algebrai way the redundan e of the pseudo- y le
relations.
Remark 2.3.1
2.3.3 Automorphisms and isometries
The presentation of Theorem 2.2.1 is redundant however useful be ause we
relations on the graph
an read the
.
Let F be a sub-set 2of R22 (S2). The symmetry group of F , (F ), is+ the
set of ongruent transformations of R (S ) that leave F invariant. We denote (F ) the
sub-group of (F ) generated by rotations. The symmetry group (F ) is dis rete if the set
f(P )j 2 (F )g is dis rete for any point P 2 F .
Corollary 2.3.1 Every nite group H of O(3) is isomorphi to a subgroup of Aut(B (n; S 2 )),
for some n.
Proof:
( ) = H
B( ; S2)
Denition 2.3.2
Let
be a normal graph on the sphere su h that
2H
. Sin e relations of
of
B(
To every ree tion 2 H we asso iate the morphism that moves the generator 2
1 . This map is an automorphism of B ( ; S 2 ). The subgroup K
of B ( ; S ) in the braid ( )
2
of Aut(B ( ; S )) generated by the set fg
j g generator of H g is isomorphi to H .
hold by rotations we asso iate to every rotation
; S 2 ).
We re all that
B1 is the indu
the
tive limit of the sequen e
orresponding automorphism
B1 B2 : : : . Sergies
u showed
that it is possible to asso iate to every innite graph on the plane, lo ally nite and without
loops or interse tions, a presentation for
B1
([84℄). The following Corollary is the analogous
of Corollary 2.3.1
Let F be a subset of R2 . Let (F ) be dis rete. Then (F ) is isomorphi to
a subgroup of Aut(B1).
Proof:
F R2
(F )
Corollary 2.3.2
For every subset
lo ally nite su h that
su h that
is dis rete, there exists an innite graph
(F ) = ( ) (see for instan
36
e [27℄).
2.4 The outer automorphisms group of B (n; S 2)
Let
G be a group. Let Aut(G) be the automorphism group of G, Inn(G) the inner automorG and Out(G) = Aut(G)=Inn(G) the outer automorphism group of G. We
phism group of
on lude the Chapter with the
omputation of the outer automorphisms group of sphere braid
Out(Bn ) = Z2 and the morphism : Bn ! Bn dened by (j ) = j 1
for j = 1; : : : ; n
1 is the generator of Out(Bn ) ([33℄).
Denition 2.4.1
For an element x 2 BP(n; S2), written in standard generators of Theorem
Q 2.3.1 as x = i j we denote by e(x) = i i mod 2(n 1), the exponent sum of x.
groups. We re all that
i
i
It follows from the presentation in Theorem 2.3.1 that the exponent sum is well-dened
(it is independent on the word
2
hosen in the generators). Let
in
Out(B (n; S 2 ))
and let
. The exponent sum is invariant up to inner auto2
morphisms, and thus we an set e((x)) for e((x)), for x 2 B (n; S ).
2
2
2
Let ZB (n; S ) be the enter of B (n; S ). Let M(n; S ) be the mapping lass group of the
n-pun tured sphere (see next hapter).
B (n; S 2 )
) is isomorphi to Z2, for n 4.
Proposition 2.4.1 The group Out(
ZB (n; S 2 )
B (n; S 2 )
2
2
Proof: The quotient ZB
is isomorphi to M(n; S ) ([15℄) and the group Out(M(n; S ))
(n; S 2 )
is isomorphi to Z2 for n 4 ([56℄).
Aut(B (n; S 2 ))
a
Proposition 2.4.2
Proof:
o-representative of
The group Out(B(n; S2)) is isomorphi to Z2 Z2, for n 4.
ZB (n; S 2 ) is isomorphi to Z2 and it is generated by the element U =
(1 n 1 )n ([15℄). Let id be the identity map in Aut(B (n; S 2 )) and let 1 be the map dened
1 for j = 1; : : : ; n 1. Sin e the relations in Theorem 2.3.1 are symmetri ,
by 1 (j ) = j
1 2 Aut(B (n; S 2 )).
Let 2 be the map dened by 2 (j ) = j U for j = 1; : : : ; n
1. By the denition of
2
2 and the fa t that U has order two and it generates ZB (n; S ), we derive that 2 (U ) =
(U )n(n 1)+1 = U . It follows that 2 Æ 2 = id, and thus 2 2 Aut(B (n; S 2 )). In the same
1
way, one an verify that the map 3 , dened by 3 (j ) = j U for j = 1; : : : ; n
1, is an
2
automorphism of B (n; S ).
2
0
2
2
Any automorphism 2 Aut(B (n; S )) indu es an automorphism of Aut(B (n; S )=ZB (n; S )).
2
2
2
Moreover, if 2 Inn(B (n; S )), then 2 Inn(B (n; S )=ZB (n; S )). Therefore, we obtain a
2
2
2
map
: Out(B (n; S )) ! Out(B (n; S )=ZB (n; S )). Let 1 ; 2 ; 3 be the images of 1 ; 2 ; 3
2
in Out(B (n; S )). Sin e e(1 (j )) =
1 mod 2(n 1), for all j = 1; : : : ; n 1, it follows that
1 is a non trivial element of Out(B (n; S 2 )) for n 3.
2
Sin e U has order two and it generates ZB (n; S ), we dedu e that all automorphisms of
B (n; S 2 ) leave invariant U . If an outer automorphism e belongs to Ker( ), then, up to inner
isomorphism, for j = 1; : : : ; n
1, e(j ) = j or e(j ) = j U . The relations i i+1 i =
i+1 i i+1 imply that either e(j ) = j for all j = 1; : : : ; n 1, or e(j ) = j U for all
j = 1; : : : ; n 1. Thus Ker( ) is generated by 2 .
2
We prove that, for n 4, 2 is not an inner automorphism of B (n; S ), and thus Ker ( )
2
is isomorphi to Z2 . Sin e Out(B (n; S ))=Ker ( ) is trivial or isomorphi to Z2 (Proposition
2
2.4.1), it follows that Out(B (n; S )) has order two or four.
The subgroup
37
Consider the exa t sequen e
1
! P (n; S 2 ) ! B (n; S 2) !
n ! 1 :
Z n = f1g and ZP (n; S 2 ) = ZB (n; S 2 ). On the other hand, the automorphism
2 leaves invariant the group P (n; S 2 ). Moreover, the restri tion of 2 on P (n; S 2 ) is the
2
^2 : n ! n be the
identity map (if x 2 P (n; S ) then e(x) is an even number). Let automorphism indu ed by 2 . This automorphism is the identity on n . The automorphism
2 is not the identity on B (n; S 2 ), sin e U 6= 1.
1 for all
Suppose that 2 is an inner automorphism. Let
2 B (n; S 2 ) su h that 2( ) =
2
2 B (n; S ). Sin e ^2 = id, then ( ) 2 Z n and thus ( ) = 1. It follows that belongs
2
2
2
2
to ZP (n; S ). Sin e ZP (n; S ) = ZB (n; S ), we dedu e that 2 is the identity on B (n; S ),
Remark that
whi h is false.
n = 2k; k > 1, the automorphism of B (n; S 2 )=ZB (n; S 2 ) dened by (j ) = j 1 for
j = 1; : : : ; n 1 an be hoosen as representative of the generator of Out(B (n; S 2 )=ZB (n; S 2 )).
2
Thus
is onto and Out(B (n; S )) is isomorphi to Z2 Z2 .
When n = 2k + 1, we derive that e(2 (j )) = 2k + 1 mod 4k and e(3 (j )) = 2k
1 mod 4k,
2 ; 3 are distin t non trivial elements of Out(B (n; S 2 )).
and then, for k > 1, the elements 1 ; 2
3 and it is isomorphi to Z2 Z2.
It follows that Out(B (n; S )) is generated by 1 ; 2 ; When
The group B(3; S2) has order 12 and it is isomorphi to2 the group 2T , the
semi-dire t 2produ t of Z3 by Z4 ([35℄). It 2 an be veried that Aut(B(3; S )) = B(3; S ) and
Inn(B (3; S )) = Z6 and then Out(B (3; S )) = Z2. In this ase, e(1 (j )) = e(2 (j )) = 3
mod 4, and it is simple to verify that 1 2. We remark also that Out(B(2; S2)) is trivial.
Remark 2.4.1
Let
F
be an orientable surfa e dierent from the sphere, and let
lass group of the surfa e
F
with
n-pun
M(n; F ) be the mapping
tures (see Denition 3.2.1).
The hara terisation of Out(M(n; F )) is ompletely solved (see [56℄ and [70℄).
The braid group B(n; F ) is a subgroup of M(n; F ) ([15℄), but no result is known about
Out(B (n; F )), ex ept when F is the annulus ([30℄).
Remark 2.4.2
38
Chapter 3
Singular braids
3.1 Denitions and results
Singular braids have been introdu ed in [3℄ and [16℄ as an extension of
singular rossing
lassi al braids on
n
strings. String of singular braids are allowed to interse t in a nite number of double points
(
). The isotopy
lasses of these singular braids, with the analogous multipli a-
n strings on D2 , denoted by SBn . The generators
j of Bn and their inverses, plus new monoid gener-
tion, form the monoid of singular braids on
of the monoid are the usual generators
1 ; : : : ; n 1 , where j orresponds to the singular braid with a rossing point involving
j -th string and the (j + 1)-th one. Baez and Birman showed also a omplete system of
relations for this monoid. To the usual braid relations (and the invertibility of j ) we need to
ators
the
add following relations:
ji j j = 1 ;
i j = j i for ji j j 2 ;
i j = j i for ji j j 2 ;
i j i = j i j
i i = i i
for
for
i = 1; : : : ; n 1 :
Singular braids are related to nite type invariants for knots. Several properties of singular
braids have been studied ([5℄, [26℄, [32℄, [39℄, [40℄ and [44℄). In parti ular it has been shown
that
SBn
embeds in a group, that the word problem for
a Markov Theorem for singular braids. An interesting
the embedding of the singular braid monoid
([16℄, [40℄, [92℄). On the other hand one
singular braid monoid on
SBn
SBn
is solvable and that it exists
onje ture for singular braids
on erns
in the group ring of the braid group
an extend the surfa e braid group
B (n; F )
n strings on F , SB (n; F ). This monoid has been introdu
Bn
to the
ed in [48℄,
SB (n; F )
B (n; F ), their inverses and the singular generators 1 ; : : : ; n 1 ,
whi h orrespond to the singular braid generators of SBn . González-Meneses ([49℄) provided
a presentation for SB (n; F ), when F is an orientable losed surfa e. In the last se tion we
give a presentation for the monoid SB (n; F ), when F is an orientable surfa e, possibly with
in order to dene nite type invariants for surfa e braids. A system of generators for
is given by the generators of
boundary. Fenn, Rolfsen and Zhu ([40℄) gave a
group
Bn whi
In parti
hara terisation of those elements of the braid
j . They proved an analogous result for SBn .
ular, it was established that the sub-monoids of those elements ommuting with j on
h
ommute with usual generators
39
one side or with singular generator
to surfa e braid groups
monoids
SB (n; F ):
Theorem 3.1.1
j
in the other side are the same. We extend these results
B (n; F ) (for an orientable surfa e F ) and
orresponding singular braid
For all x 2 SB(n; F ), the following properties are equivalent:
1. j x = xk ,
2. jr x = xkr ; for some r 2 Z n f0g,
3. jr x = xkr ; for all r 2 Z,
4. j x = xk ;
5. jr x = xkr ; for some r 2 N n f0g.
P = fP1 ; : : : ; Png be a set of n distin t points of F . An ar is an embedding a : [0; 1℄ ! F
su h that a(0); a(1) 2 P and a(x) 2
= P for all x 2 (0; 1). As in [40℄ the main idea is to onsider
braids as mapping lasses of the surfa e F nP (Se tion 3.2.1) and to study the a tion of braids
Let
on isotopy
lasses of ar s (Se tions 3.2.2 and 3.2.3 ). In parti ular, in Theorem 3.3.1 and 3.3.2
we identify all solutions
x of j x = xk
by a natural
obje ts having what Fenn, Rolfsen and Zhu
alled
-bands
riterion involving braids as geometri al
(j; k)
. As appli ation of Theorem
3.1.1 and of a redu tion property of singular braids (Lemma 3.4.2), we obtained simple proofs
for the following statements.
Theorem A (Theorem
3.4.1)
Theorem B (Theorem
3.5.2)
SB (n; F ) embeds in a group.
The word problem for SB(n; F ) is solvable.
Remark 3.1.1 Let M be a monoid, with presentation hG j Ri. The monoid M embeds in a
group if and only if it embeds in the group dened by the presentation hG j Ri.
3.2 Preliminaries
3.2.1 Mapping lass groups
We re all some denitions about mapping
are [15℄, [64℄ and [76℄. From now on, let
F
lass groups. The main referen es for this Se tion
be a
ompa t,
onne ted, oriented surfa e and let
P = fP1 ; : : : ; Pn g be a set of n distin t points in the interior of F .
Denition 3.2.1 We denote H(F; P ) the group of orientation-preserving homeomorphisms
h : F ! F , su h that h(P ) = P . The pun tured mapping lass group of F relatively to P is
dened to be the group of isotopy lasses of elements of H(F; P ). The pun tured mapping lass
group does not depend on the hoi e of P . We denote this group M(n; F ).
In the following we onsider a simple losed urve in F n P as an embedding
: S1 ! F n P
whi h does not interse t the boundary of
denote the image of
F n P . The simple
point of
P.
Two simple
t 2 [0; 1℄ su
losed
h that
. The simple
losed
urves
h0
urve
; 0
losed
F.
By abuse of notation, we use the symbol
urve
is essential if it does not bound a dis
is generi , if it does not bound a dis
are isotopi
is the identity and
if there exists a
F
ontaining
ontinuous family
h1 ( ) = 0 . We denote
40
in
'
0.
ht
to
in
0 or 1
2 H(F; P ),
Let
: S 1 ! F nP be a simple losed urve. Choose an embedding A : [0; 1℄
S 1 ! F n P su h that A(1=2; z ) = (z ) for all z 2 S 1 , and we onsider the homeomorphism
T 2 H(F; P ) dened by
Denition 3.2.2
(T Æ A)(t; z ) = A(t; e2it z ); t 2 [0; 1℄; z 2 S 1 ;
and T is the identity on the exterior of the image of A. The Dehn twist along is dened to
be the element 2 M(n; F ) whi h represents T (gure 3.1).
We re all that the denition of
generi
simple
losed
does not depend on the
hoi e of
A and that
two isotopi
urves dene the same Dehn twist.
An ar is an embedding a : [0; 1℄ ! F su h that a(0); a(1) 2 P and a(x) 2= P
for all x 2 (0; 1). A (j; k)-ar is an ar su h that a(0) = Pj and a(1) = Pk .
Denition 3.2.3
a to denote the image of a. Note that two (j; k)-ar s
ontinuous family of (j; k )-ar s. As above
s a and b is denoted by a ' b.
By abuse of notation, we use the symbol
are isotopi
if and only if they
the isotopy of the ar
an be
onne ted by a
Let a : [0; 1℄ ! F be an ar . Choose an embedding A : D2 ! F su h that:
a(t) = A(t 1=2) for all t 2 [0; 1℄,
A(D2 ) \ P = fa(0); a(1)g,
and we onsider the homeomorphism T 2 H(F; P ) dened by
Denition 3.2.4
(T Æ A)(z ) = A(t; e2ijzj z ); z 2 D2 ;
and T is the identity on the exterior of the image of A. The braid twist along a is dened to
be the element 2 M(n; F ) whi h represents T (gure 3.1).
c
γ
α
a
i
j
Figure 3.1: The Dehn twist
Note that the denition of
and the braid twist
does not depend on the
dene the same braid twist.
hoi e of
.
A and that two
isotopi
ar s
The Isotopy Extension Theorem denes a map from B (n; F ) to M(n; F ). It is well-known that
B (n; F ) embeds in M(n; F ) when g > 1. Thus we are allowed to onsider braids as elements
of M(n; F ).
41
β1
α1
β1
α
Pi
P1
α
Pi+1
1
Pn
β2
2
β2
α2
Figure 3.2: The fundamental domain of an orientable
2.
P1 ; : : : ; Pn as in Chapter 1. We x a segment [i; i +1℄
i = 1; : : : ; n 1 as in Figure 3.2.
Let i ; aj ; bj ; zl be the braid generators dened in Chapter 1. The braid generator i orresponds to the braid twist dened by the segment [i; i +1℄. On the other hand, let j be the non
trivial string of aj (respe tively bj ). We onsider two generi simple losed urves 0 ; 1 on the
1
fundamental domain of F as in Figure 3.3 and we hoose an embedding A : [0; 1℄ S ! F nP
We represent
F
losed surfa e of genus
and we
with end points
Pi
and
hoose the points
Pi+1
for
of the annulus su h that
A(3=4; S 1 ) =
A(1=4; S 1 ) =
A(1=2; S 1 ) =
The braid
aj (respe
twist along
i
0,
j.
tively
bj )
orresponds to the homeomorphism
0 1
1 , where
i
(Figure 3.3). This homeomorphism is the identity on the exterior of
to the generator
two generi
1,
zk
simple
βj
we asso iate an element of the pun tured mapping
losed
urves
αj
0; 0
0 1
βj
around the
k-th boundary
c
βj
αj
c1
1
c
lass group dened by
omponent.
αj
βj
αj
is the Dehn
A. Similarly,
c0
0
Figure 3.3: The homeomorphism asso iated with a generator
aj .
3.2.2 Braids and ar s
We adapt some denitions and propositions introdu ed in [40℄. We
a
olle tion
= ( 1; : : : ;
n)
of
n
disjoint strings in
42
F
I
an represent braids as
su h that
i
runs monoti ally
in
t
2I
from the point in
Pi 0
to some point
Pk 1
2 P 1. As above an isotopy is a
deformation through braids with xed ends and two braids are
are isotopi . Similarly to the
lassi al
onsidered equivalent if they
ase, we noti e that two equivalent braids are related
by an horizontal isotopy ([86℄). The produ t of braids
orresponds to
omposing of mapping
H(F; P )) on the left or on
the right respe tively as follows, : F ! F orresponds to a mapping lass F 0 ! F 1
and : F ! F orresponds to a mapping lass F 1 ! F 0. In parti ular braids a t on
the right and on the left on the set of ar s on F up to isotopy in H(F; P ).
lasses of
F . One
Denition 3.2.5
an have a braid
A
ribbon
a ts on
F
(up to isotopy in
is an embedding
R : [0; 1℄ [0; 1℄ ! F [0; 1℄ ;
su h that R(s; t) 2 F t. Let be a braid and let A be a (j k)-ar in F 0. Then the isotopy
orresponding to moves A through a ribbon whi h is proper for , meaning that
R(0; t) and R(1; t) tra e out two strings of the braid, while the rest of the ribbon is disjoint
from ;
R(s; 0) = A and R(s; 1) = A .
Proposition 3.2.1 ([40℄) Let 2 B (n; F ) and let A and B be ar s. Then A = B if and
only if there is a proper ribbon for onne ting A F 0 to B F 1.
Denition 3.2.6 We say that 2 B (n; F ) has a (j; k)-band if there exists a ribbon proper
for and onne ting [j; j + 1℄ 0 to [k; k + 1℄ 1.
Proposition 3.2.2 ([40℄) Let 2 B (n; F ). The braid has a (j; k)-band if and only if
[j; j + 1℄ = [k; k + 1℄. If has a (j; k)-band then j = k .
Proposition 3.2.3 Let 2 B (n; F ). If jr = kr for some integer r, then fj; j + 1g =
fk; k + 1g.
Proof:
r
r 1
The
let
ase
odd is trivial, sin e it su es to
onsider the asso iated permutation. Thus,
r be even. Then, k
2 P (n; F ). Let j;j+1 : P (n; F ) ! P (2; F ) be the map that forgets
j -th one and the (j + 1)-th one. Suppose that fj g 2= fk; k + 1g. It
all strands ex ept the
follows
This is false, sin
1 = j;j +1( kr 1 ) = j;j +1(jr ) = 1r :
2
e Pn embeds in P (n; F ) ([75℄) and P2 = h1 i = Z.
3.2.3 Isotopy invariants
a and b be two simple
a and b is
Let
of
losed
urves in
I (a; b) =
inf
F . Following [38℄ and [76℄, the index of interse
fja0 \ b0j ; a0 ' a; b0 ' bg :
Adapting above denition to ar s, the index of interse tion of the ar s
I (a; b) =
inf
fjint(a0 ) \ int(b0)j ; a0 ' a; b0 ' bg :
43
a and b is
tion
We note that
Let
set
Ea
a.
I (a; b) =
inf
fja0 \ bj ; a0 ' ag
and that if
a ' b then I (a; b) = 0.
a be an ar and let the braid twist asso iated to a. The square of is a Dehn twist. We
ab for a generi simple losed urve su h that Tba = 2 where Tba is the Dehn twist along ab.
b bounds a disk D (a
b) in F ontaining an ar a0 isotopi to
h generi simple losed urves a
Proposition 3.2.4 ([76℄)
Let a and b be two simple losed urves and n any integer Then
I (Tan (b); b) = jnjI (a; b)2 ;
where Tan is the n-th power of the Dehn twist along a.
Lemma 3.2.1 If A is a (l; k)-ar , with fk; lg \ fj; j + 1g = ; su h that A j = A, then
I (A; [j; j + 1℄) = 0, i.e. A and [j; j + 1℄ are disjoint up to isotopy.
Proof:
A j = A
Ab j = Ab
T[j;j +1℄(Ab) =
The hypothesis
\
implies that
and therefore
Ab. It follows that I (T[j;j +1℄(Ab); Ab) = 0. From the formula
b [j;\
I (A;
j + 1℄) = 0, whi h implies I (A; [j; j + 1℄) = 0.
Lemma 3.2.2
Proof:
\
in Proposition 3.2.4 we dedu e
If A is a (j; j + 1)-ar , su h that A jr = A for some r, then A ' [j; j + 1℄.
b [j;\
I (A;
j + 1℄) = 0.
0
0
0
b
b
b
b
\
\
Let A ' A su h that jA \ [j; j + 1℄j = ;. The disks D (A ); D ([j; j + 1℄) annot be disjoint
b0 ) D ([j;\
(they ontains Pj ). Thus, either D (A
j + 1℄) or D([j;\
j + 1℄) D(Ab0 ). We dedu e
b0 ' [j;\
that A
j + 1℄ and therefore [j; j + 1℄ and A are isotopi .
Using the same argument as in Lemma 3.2.1 we
The following
an suppose that
orollary is the analogous for braid twists of the well-known result on Dehn
twists (see e.g. Theorem 7.5 in [55℄).
Let a be an (i; j)-ar in F and let Ta be the braid twist along a. Let b be a
(k; l)-ar , where fi; j g\fk; lg = ;. The braid twist Ta and Tb ommute if and only if I (a; b) = 0.
Proof:
a
b
Ta
Tb
1
Corollary 3.2.1
It is evident that if
and
are disjoint (up to isotopy)
and
ommute. Re all
that as in the ase of Dehn twist Ta Tb Ta
= TT (b) . Commutation hypothesis implies that
TT (b) = Tb . Now, onsider braids asso iated to TT (b) and Tb . Thus, with the previous notation
b TT (b) = b Tb = b and Lemma 3.2.2 implies Ta (b) isotopi to b, i.e. b Ta = b. From Lemma
3.2.1 we dedu e that a and b are disjoint (up to isotopy).
a
a
a
a
3.3 Statements of Main Theorems
3.3.1 Centralisers of B (n; F )
Let us state the rst Theorem on
Theorem 3.3.1
For ea h
entralisers of
B (n; F ).
2 B (n; F ), the following properties are equivalent:
44
1. j = k ,
2. jr = kr ; for any integer r,
3. jr = kr ; for some nonzero integer r,
4. has a (j, k)-band,
5. [j; j + 1℄ = [k; k + 1℄.
Proof:
(5) ) (4) ) (1)
(2)
From Proposition 3.2.2 it follows that
) (3). It remains to show that (3) ) (5).
Proposition 3.2.3 this equation is possible only if
Suppose that for some
r,
jr
fj; j + 1g = fk; k + 1g. From the remark
= kr and that kr has a proper (k; k)-band, we on lude that there is a proper
1 r from [k; k +1℄ 0 to [k; k +1℄ 1. Dene A = [k; k +1℄ = [k; k +1℄ 1 .
ribbon R for
j
We may assume (up to isotopy) that R(; 1=3) = A 1=3 and R(; 2=3) = A (2=3). Then
r
r
there is a ribbon for j onne ting A to A. From Proposition 3.2.1 we dedu e that A j = A.
By Lemma 3.2.2, A = [j; j + 1℄ and the laim is proved.
that
1 r
j
(1) )
= kr . By
, and it is obvious that
Denitions 3.2.5 and 3.2.6 an be extended naturally to mapping lass groups.
The Theorem 3.3.1 holds true for any 2 M(n; F ).
Remark 3.3.1
3.3.2 Singular ribbons
SB (n; F ). Two singular braids x1 and x2 are equivalent if
Ht of F [0; 1℄ su h that H0 = idF [0;1℄ and H1 (x1 ) = x2 . Let B be a
ball of radius entered at the singularity p. Denote st = Ht (x1 ), pt = Ht (p) and Bt = Ht (B ).
We an suppose without loss of generality, that Bt is the ball of radius entered at pt and
that Bt \ st is as in Figure 3.4.
Theorem 3.3.1
an be extended to
there exists an isotopy
B
p
Figure 3.4: The neighbourhood of the singularity
p.
We stress that equivalent singular braids do not need to be related by a level preserving isotopy
H
in
F [0; 1℄.
A
is a map R : I I ! F I su h that R embeds I t into
F t, ex ept for nitely many points t, for whi h the image is a single point in F t. One
also assumes, at these singular points, that there is a tangent plane in F t for the singular
ribbon.
proper
f0; 1g I
Denition 3.3.1
singular ribbon
As for braids, we say that a singular ribbon is
for a singular braid if it sends
along two of its strings and the image is disjoint from the other strings of the singular braid. An
45
isotopy of a singular braid
In
an be extended to an isotopy of any of its proper singular ribbons.
ontrast to the ordinary situation, it is not always possible to nd a singular ribbon proper
for a given singular braid and a given ar
A = R(I; 0). Anyway, we have the following.
Proposition 3.3.1 ([40℄) If a singular ribbon R is proper for the singular braid x and R(I; 0)
and R(I; 1) are isotopi as ar s to [j; j + 1℄ 0 and [k; k + 1℄ 1 respe tively, then j x = xk
in SB(n; F ).
Denition 3.3.2 A singular braid has a (j; k)-band if it has a proper (singular) ribbon onne ting [j; j + 1℄ 0 to [k; k + 1℄ 1.
Remark 3.3.2 For a singular braid x, having a (j; k)-band is a su ient ondition for satisfying j x = xk .
Lemma 3.3.1 Let 2 B (n; F ) and y 2 SB (n; F ), su h that both i y and y have (singular)
(j; k)-bands. Then i y also has a (singular) (j; k)-band.
Proof:
A = [j; j + 1℄ The proof is an immediate extension of Lemma 6.4 in [40℄. Let
. Sin e
y has a (j; k)-band, we have a proper ribbon R su h that R(I; 0) and R(I; 1) are isotopi as
ar s to [j; j + 1℄ 0 and [k; k + 1℄ 1. After an isotopy we may suppose R as the omposition
of two ribbons R1 and R2 for
and y , su h that R1 (I; 1) = R2 (I; 0) = A. The hypothesis
that i y has a (j; k )-band implies that A i = A. Considerations on asso iated permutation
show that either fj; j + 1g = fi; i + 1g ( ase 1) or fj; j + 1g \ fi; i + 1g = ; ( ase 2). In
the rst ase, A is an (i; i + 1)-ar . Sin e A i = A, Lemma 3.2.2 implies that A = [i; i + 1℄
and then
has a (j; i)-band. On the other hand R2 is a proper band onne ting [i; i + 1℄ 0
to [k; k + 1℄ 1 and then y has a (i; k )-band. Combining these bands with the obvious singular
(i; i)-band for i provides a (j; k)-band for i y.
If fj; j + 1g and fi; i + 1g are disjoint sets, A = R1 (I; 1) = R2 (I; 0) is disjoint from
[i; i + 1℄ (Lemma 3.2.1). Thus we may insert i between and y so that the singular strands
are disjoint from the band, and we on lude that i y has a (j; k )-band.
an ellation property
Proposition 3.3.2 ([49℄) Left and right an ellation hold in SB (n; F ), that is for all x; y; z 2
SB (n; F ) the equation xy = xz (respe tively yx = zx) implies y = z .
Similarly to singular braids, singular surfa e braids have the
3.3.3 Centralisers on SB (n; F )
The following Theorem is an extension of Theorem 3.3.1 to
Theorem 3.3.2
SB (n; F ).
For ea h x 2 SB(n; F ), the following properties are equivalent:
1. j x = xk ,
2. jr x = xkr ; for some nonzero integer r,
3. jr x = xkr ; for any r,
4. j x = xk ;
46
.
5. jr x = xkr ; for some positive integer r,
6. x has a (possibly singular) (j, k)-band.
Proof:
(6) ) (1) ) (3) ) (2)
It is
lear that
and
(6) ) (4) ) (5).
(4) ) (6). Sin e the order of singularities on a string is an isotopy invariant, we dedu e
that fj; j +1g x = fk; k +1g and that the j -th and (j +1)-th strings are disjoint from the
p0 ; : : : ; pm be the ordered set of singularities of j x on the j th string.
j x = l0 1 l1 m l m+1 , where l is the singular generator orresponding to
pq and q+1 2 SB (n; F ). On the other hand we write xk = 1 l1 2 m l m+1 l +1 .
other strings. Let
Let
m
q
m
m
The isotopy in reases of one the index of all singular generators. The trivial singular
(lq ; lq+1 )-band for q+1 . We ombine these bands with the
obvious singular (lq ; lq )-bands for l in order to obtain a singular (j; k )-band for x. The
ase (5) ) (6) is analogous.
braid near
l
q
provides a
q
(2) ) (6). We outline the proof that is the same as in [40℄. We pro
eed by indu tion on
x. Assume x = i y, where is a surfa e braid. The
2
r
2
r
1 2r y = y 2r . Sin e
1 2r is a
hypothesis implies that j x = xk and then
i
i k
j
j
2
r
pure surfa e braid, the i in i yk
orresponds under some homomorphism, to the i in
1 2r y . Hen e the image, under that homeomorphism, of the trivial singular band
i
j
1 2r . Therefore, ommutes with
1 2r .
near the rst i provides a band for
i
j
j
1
2
r
2
r
1
2
r
It follows that i
j y = i yk . By Proposition 3.3.2, we have
j y = yk2r ,
2r y = y 2r . We dedu e that y has a (j; k )-band and from the existen e of a
i.e. j
k
(i; i)-band for 1 j2r we dedu e 1 j2r i y = i yk2r . It follows that also i y has a
(j; k)-band. Lemma 3.3.1 on ludes the proof.
the number of singular generators in
The monoid SB (n; F ) embeds in a group
3.4
3.4.1 Extended singular braids
E we denote by (E ) the free monoid generated by E . Let GB(n;F )
1
be the set of generators of B (n; F ) and G
B (n;F ) the set of their inverses. Let T be the set of
singular generators j . We an asso iate to SB (n; F ) the group dened as follows:
From now on given a set
Generators: GB(n;F ); GB1(n;F ); T
for
j = 1; : : : ; n 1 :
Isotopy relations (IR):
and the additional set
SB (n; F )
relations obtained by substituting j for j .
the relations for
T
1
of singular generators
j ;
(see Se tion 3.6) and the additional
Birth-death relations (BDR): j j = j j = 1 :
F , SG(n; F ). Let be the natural homomorphism from SB (n; F ) to SG(n; F ). The aim of this Se tion is to show that is inje tive.
There is a geometri interpretation for SG(n; F ) analogous to the one proposed by Fenn,
Keyman and Rourke for the group SGn asso iated to SBn ([39℄). Now, we have two types of
We
all this group the singular braid group on
47
singular points, that we label respe tively with a bla k point,
rossing,
j . We
extended singular braids1
an open blob,
orresponding to
Given two words
;
(GB(n;F ) [ T 1 ) , we
I
write
a geometri
braids that
singular
and with
if they are equivalent under
relations of type IR. In other words, there exists an isotopy from
the isotopy preserves balls of radius
j
all these singular braids having two type of singular
.
in
orresponding to
to
. We
an suppose that
entered at the labeled singularities. Also we
interpretation of birth-death relations, that
V
oin ide ex ept an open set
rossings of dierent types. We
an give
orrespond to two extended singular
where we allow the birth or the death of a pair of
V
an assume that
is disjoint from the other strings.
Two words in (GB1(n;F ) [ T ) whi h represent the same element in SG(n; F )
have the same number of singular generators.
Proof:
SG(n; F )
1
Lemma 3.4.1
From the presentation of
SG(n; F ) to the integers whi
it follows that there is a group homomorphism from
h maps
GB(n;F )
in
0, j
in
1 and j
in
1.
Consider
two
words
A = 1 j 2 : : : m j m+1 and B = 01 i 02 : : : 0m i 0m+1 ,
where i; 0i 2 (GB1(n;F )) and i ; j 2 (T ). If A = B in SG(n; F ), then there exists
k 2 f1; : : : ; mg su h that
Lemma 3.4.2
1
l
1
m
m
l
1 j1 2 : : : m jm m+1
=
0
0
0
1
1 j1 2 : : : m jm m+1
=
0
0
0
0
0
0
1 i1 2 : : : k ik k+1 : : : m im m+1 ;
0
1 0 ::: 0 k+1
m im m+1
j11
ik B
1 i1 2 : : : k ik
hold1 in SG(n; F ), where j in A is repla ed by j or
i respe tively.
Proof:
A
B
1
1
and in is repla ed by i or
k
k
If the words
and
represent the same element of
A
sequen e of isotopies and birth-deaths relations relaying
point
orresponding to
j1
in
A. We en
ounter two
SG(n; F ), then there is a nite
B . Let p0 be the singularity
to
ases.
p0 during the sequen e does not mat h a death and thus j1 is sent in
B . The isotopies and birth-death relations an move the singularity p0 but
we an suppose that they do not modify the interior of a ball B (p0 ) of radius entered
at p0 . Let us onsider the singular braid A1 obtained by modifying A only inside B (p0 ),
where we substitute the singularity p0 with the positive rossing j1 . It follows that A1
is equivalent in SG(n; F ) to the singular braid B1 , whi h orresponds to the singular
braid B ex ept that i is repla ed by i .
1. Suppose that
some
i
k
in
k
k
p0 mat hes a death at the step d1 of the sequen e. It follows that at a
b1 < d1 , there is a birth s(b1 ) s(b1 ) (or s(b1 ) s(b1 ) ) on the j -th and the j +1-th string.
0
Restart from the step b1 and set p1 and p1 the opposite singularities orresponding to
s(b1 ) and s(b1 ) . We iterate the pro ess following p1 and so on. Sin e the sequen e is
nite there exists a birth bq su h that the orresponding singularity pq is sent by the
sequen e in a singular point of B , orresponding to some singular generator i . As above
onsider the singular braid A1 obtained by repla ing j1 with the positive rossing j1
0
and the singular braid B1 obtained by repla ing i with i . Dene B (pi ) and B (pi )
2. Suppose that
step
k
k
48
k
the balls of radius
pi
entered at
and
p0i . During the sequen
e of isotopies and birth-
A to B we an suppose that ex ept in steps of type dl or bl the balls
B (p0i ) remain balls of radius entered at the orresponding singularities.
Repla e A with A1 . At ea h step bi we modify singular braids only inside a small open
0
0
set V B (pi ) [ B (pi ) repla ing pi with a positive rossing s(b ) and pi with a negative
1
rossing s(b ) . This orresponds to substitute the birth bi with an horizontal isotopy. It
follows that, at ea h step di (re all that bi < di ) the subword s(d ) s(d ) (or s(d ) s(d ) )
1
1
1
1
is repla ed by s(d ) s(d ) (or s(d ) s(d ) ). Thus we substitute s(d ) s(d ) (or s(d ) s(d ) )
deaths" from
B (pi )
and
i
i
i
i
i
i
i
with the empty word, whi h
i
i
i
i
i
i
i
orresponds to an horizontal isotopy. This pro edure gives
A1 to B1 .
a sequen e of isotopy and birth-death relations from
3.4.2 Singular braids embed in extended singular braids
Theorem 3.4.1
Proof:
The monoid SB(n; F ) embeds in a group.
A; B in (GB1(n;F ) [ T ) . Let A = B in SG(n; F ). We want to show
that A = B in SB (n; F ). We pro eed by indu tion on the number m of singular generators
`
m
m
in A and B . We set SB (n; F ) =
m SB (n; F ) , where SB (n; F ) means the set of singular
0
braids with m singularities. If m = 0 the statement is true, be ause B (n; F ) = SB (n; F )
embeds in SG(n; F ). In fa t, there is a retra tion morphism r : SG(n; F ) ! B (n; F ), whi h
sends ea h braid generator to itself and singular generators i to identity. The omposition
jB(n;F ) Æ r is the identity on B (n; F ). Now, suppose that the statement is true for m 1
0
0
0
0
singular generators, and set A =
1 j1 2 : : : m j m+1 and B = 1 i1 2 : : : m i m+1 ,
1 ) . We an suppose 1 to be the empty word. Lemma 3.4.2
0
where i ; i are words in (G
B (n;F )
Consider two words
m
m
implies that the equalities
j1
2 : : : m jm m+1
0
0
0
0
0
0
0
0
0
1 i1 2 : : : k ik k+1 : : : m im m+1
=
0
0
0
j11 2 : : : m j m+1 = 1 i1 2 : : : k i 1 k+1 : : :
hold in SG(n; F ) for some k 2 f1; : : : ; mg. Thus we derive
m
0
j21 1 i1
m im m+1 ;
k
0
0
2 ::: k
=
0
0
0 2
1 i1 2 : : : k ik
SG(n; F )
in
and by indu tion on the number of singular generators this relation holds true also in
SB (n; F ).
From Theorem 3.3.2 it follows:
j1
0
0
0
1 i1 2 : : : k
=
0
0
0
1 i1 2 : : : k ik
in
SB (n; F ):
(3.1)
and thus
2 j2 3 : : : m jm m+1
whi h is also true in
=
0
0
SB (n; F ), by indu
j1 2 j2 : : :
0 0
0
0
1 i1 2 : : : k k+1 : : : m im m+1
m jm m+1
in
SG(n; F )
tion. We dedu e that
0
= j1 1 i1
49
0
0 0
0
0
2 : : : k k+1 : : : m im m+1
holds in
SB (n; F ), and (3.1) implies
j1 2 j2 : : :
in
m jm m+1
0
0
0
0
0
0
1 i1 2 : : : k ik k+1 : : : m im m+1
=
SB (n; F ).
Re ently González-Meneses showed us another proof of Theorem 3.4.1, when
F is a losed orientable surfa e ([50℄).
Remark 3.4.1
3.5
The word problem is solvable
Theorem 3.5.1
Proof:
The word problem for B(n; F ) is solvable.
P (n; F ). Let F be a
surfa e of genus g with p > 0 boundary omponents and let fAi;j j 1 i 2g + p + n
2; 2g +
p j 2g + p + n 1; i < j g be the set of generators of P (n; F ) dened in Theorem 1.6.1.
The algorithm is similar to the lassi al braid ombing. When F has boundary, the (PBS)
It su es to prove the word problem for the pure braid group
exa t sequen e splits (Se tion 1.2.1) and then
P (n; F ) = 1 (F n fx1 ; : : : ; xn
g; xn ) o 1(F n fx1 ; : : : ; xn 2g; xn 1 ) o o 1(F; x1 ) ;
where the fundamental group 1 (F nfx1 ; : : : ; xj 1 g; xj ) is freely generated by the set fAi;j j i <
j g. We use relations in Theorem 1.6.1 to move all letters in fAi;n1 j i < ng on the right hand
side to obtain a word
Xn
1 n
all the subwords of the form
and we obtain a word
1
equivalent to
xx
1
or
00 = 0 0
1
n
1 x,
x
0
n
. Let
in
n.
the redu ed word obtained removing
The algorithm will end in
equivalent to
, where
0
j (j = 1; : : : ; n)
n
1
steps
is a redu ed
fAi;j j i < j g. Sin e fAi;j j i < j g is a free system of generators for the free group
1 (F n fx1 ; : : : ; xj 1 g; xj ), the word 00 is unique. The ase of losed surfa es is similar ([46℄).
word on
Consider
two words
A; B in (GB1(n;F ) [ T ) . Let A = 1 j 2 : : : m j m+1
0
0
0
and B = 1i 2 : : : mi m+1, where i; 0i 2 (GB1(n;F ) ) and i ; j 2 (T ), for 1 i; l m + 1. A and B represent the same element in SB (n; F ) if and only if there exists
k 2 f1; : : : ; mg su h that
Lemma 3.5.1
0
1
1
m
l
1 j1 2 : : : m jm m+1
=
0
0
0
1
1 j1 2 : : : m jm m+1
=
0
0
0
0
m
l
0
0
1 i1 2 : : : k ik k+1 : : : m im m+1 ;
0
1 0 ::: 0 k+1
m im m+1 ;
j11
ik B
1 i1 2 : : : k ik
hold1 in SB(n; F ), where j in A is repla ed by j or
i respe tively.
Proof:
1
1
and in is repla ed by i or
k
k
The if part follows from Lemma 3.4.2. Conversely, we suppose
1
to be the empty
word and we pro eed as in Theorem 3.4.1. The equalities
j1
2 : : : m jm m+1
=
j11
2 : : : m jm m+1
=
0
0
0
0
0
0
1 i1 2 : : : k ik k+1 : : : m im m+1
0
0
0
1 i1 2 : : : k ik
50
0
1 0 ::: 0 k+1
m im m+1
;
in
SB (n; F ) imply that
0
0
j21 1 i1
0
2 ::: k
=
0
0
0 2
1 i1 2 : : : k ik
SG0 (n; F )0 and, by Theorem 3.4.1, also in SB (n; F ).
1 i1 2 : : : k has a (j1 ; ik )-band, and thus
holds in
that
0
2 : : : m jm m+1
Then we
0
0
0 0
0
0
1 i1 2 : : : k k+1 : : : m im m+1
:
on lude as in Theorem 3.4.1.
Theorem 3.5.2
Proof:
=
From Theorem 3.3.2 we derive
The word problem for SB(n; F ) is solvable.
(GB1(n;F ) [ T ) . We pro eed by indu tion on the num0
ber m of singularities in A. For m = 0 we have SB (n; F ) = B (n; F ) and for B (n; F )
the word problem is solvable (Theorem 3.5.1). Let A =
1 j1 2 : : : m j m+1 and B =
0
0
0
0
(1)
(2)
1
1 i1 2 : : : m i m+1 . Let j = j and j = j . Set the words
A and B
Let
be two words in
m
m
A(1s) =
Br(s) =
r = 1; : : : ; m. Then all the A(1s)
0
0
0 (s) 0
0
0
1 i1 2 : : : r ir r+1 : : : m im m+1 ;
Br(s)
m 1 singular generators. By
(s)
(s)
the indu tion hypothesis there exists an algorithm de iding whether A1 = Br
in SB (n; F )
or not. From Lemma 3.5.1, A = B in SB (n; F ) if and only if there exists r = 1; : : : ; m su h
(1)
(1)
(2)
(2)
that A1 = Br
and A1 = Br .
for
s = 1; 2
(s)
1 j1 2 : : : m jm m+1 ;
and
and
have
3.6 Monoid presentations
In following Theorems we provide presentations for singular braid monoids on orientable surfa es. Relations
an easily veried on
system of relations one
orresponding braids. To prove that it is a
omplete
an repeat arguments in [49℄.
Let F be an orientable surfa e with p > 0 boundary omponents. The monoid
SB (n; F ) admits the following presentation:
Generators: 11 ; : : : ; n1 1 ; a1 1 ; : : : ; ag 1 ; b1 1 ; : : : ; bg 1 ; z11 ; : : : ; zp11; 1 ; : : : ; n 1 :
Relations:
Group relations:
Theorem 3.6.1
i 1 i = i i 1 = 1 (1 i n 1) ;
ar 1 ar = ar ar 1 = br 1 br = br br 1 = 1 (1 r g) ;
zj 1 zj = zj zj 1 = 1 (1 j p 1) :
Braid relations, i.e:
(R1) i i+1 i = i+1 i i+1 ;
(R2) i j = j i
ji j j 2 :
for
51
Mixed relations:
ar i = i ar (1 r g; i 6= 1) ;
br i = i br (1 r g; i 6= 1) ;
(R4) 1 1 ar 1 1 ar = ar 1 1 ar 1 1 (1 r g) ;
1 1 br 1 1 br = br 1 1 br 1 1 (1 r g) ;
(R5) 1 1 as 1 ar = ar 1 1 as 1 (1 s < r g) ;
1 1 bs 1 br = br 1 1 bs1 (1 s < r g) ;
1 1 as 1 br = br 1 1 as1 (1 s < r g) ;
1 1 bs 1 ar = ar 1 1 bs1 (1 s < r g) ;
(R6) 1 1 ar 1 1 br = br 1 1 ar 1 (1 r g) ;
(R7) zj i = i zj (i 6= 1; 1 j p 1) ;
(R8) 1 1 zi 1 ar = ar 1 1 zi 1 (1 r g; 1 i p 1; n > 1) ;
1 1 zi 1 br = br 1 1 zi 1 (1 r g; 1 i p 1; n > 1) ;
(R9) 1 1 zj 1 zl = zl 1 1 zj 1 (1 j < l p 1) ;
(R10) 1 1 zj 1 1 zj = zj 1 1 zj 1 1 (1 j p 1) :
(R3)
Singular relations:
i j i = j i j for ji j j = 1 ;
i j = j i for ji j j 2 ;
i j = j i for ji j j 2 ;
i i = i i (1 i n 1) ;
(ai;r ai+1;r )i = i (ai;r ai+1;r ) (1 i n 1; 1 r g) ;
(bi;r bi+1;r )i = i (bi;r bi+1;r ) (1 i n 1; 1 r g) ;
(R16) i aj;r = aj;r i (j 6= i; i + 1; 1 r g) ;
i bj;r = bj;r i (j 6= i; i + 1; 1 r g) ;
(R17) (zi;r zi+1;r )i = i (zi;r zi+1;r ) (1 i n 1; 1 r p 1) ;
(R18) i zj;r = zj;r i (j 6= i; i + 1; 1 r p 1) ;
(R11)
(R12)
(R13)
(R14)
(R15)
where ai;r = i 11 1 1ar 1 1 i 11, bi;r = i 11 1 1br1 1 i 11 and zi;r = i 11 1 1
zr 1 i 1 .
Theorem 3.6.2 Let F be a losed orientable surfa e of genus g 1. The monoid SB (n; F )
admits the following presentation:
Generators: 11 ; : : : ; n1 1 ; a1 1 ; : : : ; ag 1 ; b1 1 ; : : : ; bg 1 ; 1 ; : : : ; n 1 :
Relations:
Group relations:
i 1 i = i i 1 = 1 (1 i n 1) ;
ar 1 ar = ar ar 1 = br 1 br = br br 1 = 1 (1 r g) :
52
Braid relations, i.e.
Mixed relations:
(R1) i i+1 i = i+1 i i+1 ;
(R2) i j = j i
ji j j 2 :
for
ar i = i ar (1 r g; i 6= 1) ;
br i = i br (1 r g; i 6= 1) ;
(R4) 1 1 ar 1 1 ar = ar 1 1 ar 1 1 (1 r g) ;
1 1 br 1 1 br = br 1 1 br 1 1 (1 r g) ;
(R5) 1 1 as 1 ar = ar 1 1 as 1 (1 s < r g) ;
1 1 bs 1 br = br 1 1 bs 1 (1 s < r g) ;
1 1 as 1 br = br 1 1 as 1 (1 s < r g) ;
1 1 bs 1 ar = ar 1 1 bs 1 (1 s < r g) ;
(R6) 1 1 ar 1 1 br = br 1 1 ar 1 (1 r g) ;
(T R) [a1 ; b1 1 ℄ [ag ; bg 1 ℄ = 1 2 n2 1 2 :1 :
(R3)
Singular relations:
i j i = j i j for ji j j = 1 ;
i j = j i for ji j j 2 ;
i j = j i for ji j j 2 ;
i i = i i (1 i n 1) ;
(ai;r ai+1;r )i = i (ai;r ai+1;r ) (1 i n 1; 1 r g) ;
(bi;r bi+1;r )i = i (bi;r bi+1;r ) (1 i n 1; 1 r g) ;
(R12) i aj;r = aj;r i (j 6= i; i + 1; 1 r g) ;
i bj;r = bj;r i (j 6= i; i + 1; 1 r g) ;
(R7)
(R8)
(R9)
(R10)
(R11)
where [a; b℄ := aba 1b 1 and ai;r , bi;r are dened as in previous Theorem.
53
54
Chapter 4
Generalized He ke Algebras
4.1 Introdu tion
Any braid
x
in
Bn
yields an oriented link
xb
by
losing up the strands of the braid as in
losure
gure 4.1. The up to down orientation of braid strands indu es an orientation for the
Alexander was the rst to observe that any oriented link
an be identied with the
losure.
of
a braid ([2℄). On the other hand, Markov provided moves relating two braids with the same
losure.
Two losed braids xb; yb are equivalent links if and only if one
an relate the braids x; y in [n2Bn by a sequen e of the following elementary moves:
: z 2 Bn is repla ed by z 1 2 Bn, for some 2 Bn.
(or its inverse, namely a
): z 2 Bn is repla ed by zn1 2
Bn+1 .
Proposition 4.1.1 (Markov)
Conjugation
Stabilization
destabilization
Markov Tra es
To obtain invariants of links, we may pro eed
onstru ting suitable fun tionals on
C [Bn ℄,
alled
. This pro ess is well-known for He ke algebras, whi h are nite dimensional
quotients of
C [Bn ℄.
Denition 4.1.1
The He ke algebra Hn(q) (of type A) is the quotient
Hn (q) = C [Bn ℄=(j2 + (1 q)j
q ; j = 1; : : : ; n 1) :
β
Figure 4.1: The
losure of the braid
55
.
Famous
onstru tion of Jones polynomial using He ke algebras posed the problem of similar
C [Bn ℄. We would extend Jones' approa h to the ase of
3-manifolds. The rst task is to dene and verify the existen e of Markov tra es on
C [B (n; F )℄, or suitable quotients. We propose to onsider the following quotient of C [B (n; F )℄.
onstru tions on other quotients of
links in
Let F be a surfa e with at least one boundary omponent. The surfa e He ke
algebra Hn(q; F ) is the quotient
Denition 4.1.2
Hn (q; F ) = C [B (n; F )℄=(j2 + (1 q)j
where j is the lassi generator of Bn.
q ; j = 1; : : : ; n 1) ;
H1 (q; F ) is the group algebra of B (1; F ) = 1 (F ) We remark that the
natural embedding Bn ! B (n; F ) indu es an embedding of the usual He ke algebra Hn (q ) into
Hn (q; F ). On the other hand, when F has boundary, B (n; F ) embeds naturally in B (n +1; F ),
and thus we an dene the tower [n1 Hn (q; F ).
b
We su eeded nding a Markov tra e on [n1 Hn (q; F ) for the spe ialization q = 1. Let b0 = b
be the set of onjugation lasses of 1 (F ) and f1g. Let S (C b 0 ) be the symmetri
b0 .
algebra of the ve tor spa e C The He ke algebra
For any z 2 C , there exists an unique family Tn of linear fun tionals
Theorem 4.1.1
Tn : Hn(1; F ) ! S (C b 0 )
su h that
1. Tn(xy) = Tn(yx) 8x; y 2 Hn(1; F ) ;
2. Tn+1(xn) = zTn(x) 8x 2 Hn(1; F ) ;
3. Tn+1(n 1A1 nx) = AbTn(x) 8x 2 Hn(1; F ) 8A 2 B(1; F ) ;
4. Tn(1) = 1 ;
where Ab denotes the onjugation lass of the element A 2 B(1; F ) = 1(F ).
and
M
rd skein module
We noti e that when
proved that the
3
of links in
orresponding link invariant have been
[77℄).
M = F I , where F is a holed disk, it has been
M is isomorphi to S (C b 0 ) ([78℄). A Markov tra e
is a handlebody, i.e.
onstru ted in the
ase of the solid torus ([65℄ and
3-manifolds were dened by Reshetikhin and Turaev
3
ase when the manifold is S one obtains the olored Jones polynomial at roots
Quantum invariants for links in arbitrary
in [80℄. In the
of unity. In general is presently unknown whether these invariants
ome from a polynomial
evaluated at roots of unity.
4.2 Preliminaries
4.2.1 Markov tra es
Let us re all some
C [Bn ℄
lassi
results. The natural embedding
! C [Bn+1 ℄. Let Tn : C [Bn ℄ ! C
Bn
! Bn+1 indu es the inje tion
be a family of linear fun tionals fullling following
onditions.
56
Tn(xy) = Tn(yx); 8x; y 2 Bn;
Tn+1(xn) = zTn(x) 8x 2 Bn;
Tn+1(xn 1 ) = zbTn(x) 8x 2 Bn,
b 2 C . Su h a family is alled a Markov tra e, and usually one drops the subs ript
for some z; z
Q
n from T . For an element x 2 Bn written in terms of the standard generators as x = i i we
i
denote by
e(x) the exponent sum of x. Sin
e relations in
Bn
are homogeneous, the exponent
sum is well-dened.
Corollary 4.2.1 Let T be a Markov tra e. The fun tion F asso iating to the losed braid x^
(for x 2 Bn) the value
is a link invariant.
F (x) = z
e(x)+n
2
zb
e(x)
n
2
T (x) ;
In order to nd Markov tra es we fo us on He ke algebras
image of
j
in
Hn (q). V. Jones and A. O
Hn (q).
We set again
j
for the
neanu showed that:
For any z 2 C there exists an unique Markov tra e T on the He ke
algebras Hn(q) verifying:
Tn(xy) = Tn(yx); 8x; y 2 Hn(q);
Tn+1(xn) = zTn(x) 8x 2 Hn(q);
Tn(1) = 1.
Proposition 4.2.1
The main idea in the proof is that every element in
ombination of words
of Hn (q)
Hn (q)
an be written in terms of a linear
xn y with x; y 2 Hn(q) and words from Hn (q). Moreover, every element
Hn(q) ([58℄):
an be written in terms of the standard basis of
(i1 i1
r1 )(i2 i2 r2 ) (ip ip rp ) ;
1 i1 < i2 < < ip n 1 and rj 2 f0; 1; : : : ; ij 1g. The natural in lusion of Bn
Bn+1 indu es the in lusion on orresponding He ke algebras. The above indu tive basis
1
allows to onstru t a fun tional T on H := [n=1 Hn (q ) with values in C . One proves easily
where
into
that
T
is a tra e. We remark that the denition of a Markov tra e
on erns also the behaviour
ondition T (an ) = z T (a) and the
j2 = (q 1)j + q imply the relation T (an 1 ) = zbT (a), where zb = z qq+1 .
of the tra e with respe t to the other stabilization. The
quadrati
relation
4.2.2 Algebrai
onstru tion of HOMFLY-PT polynomial
A Markov tra e gives rise to a link invariant and we derive the following
Corollary 4.2.2
x 2 Bn . Then
orollary.
Set = 1 qzq+z . Consider the link L presented as the losure of the braid
is a link invariant.
XL (q; ) =
p(1 q)
(1 q)
57
n 1
p
( )e(x) T (x) ;
This is
alled HOMFLY-PT polynomial of
pp
L and it is
ommon to
p
p1q ;
t = q; x = q
and to denote it
Let
L+ ; L ; L0
hange variables as follows:
PL (t; x) = XL (q; ) :
be oriented link diagrams that are identi al, ex ept in one
rossing as in the
gure below.
L+
; L
; L0
:
skein relation
Theorem 4.2.1 PL (t; x) is the unique Laurent polynomial in t and x verifying the skein
relation
0
1
!
!
Thus we
an express HOMFLY-PT polynomial with the usual
t 1P
tP
= xP and taking the value 1 for the unknot.
The Jones polynomial follows from the spe ialisation
p
PL (
.
A
1=t;
p 1(pt (1=pt))).
4.3 Proof of Theorem 4.1.1
ondition F with boundary implies that B (n; F ) embeds naturally
B (n + 1; F ). Therefore, there is no ambiguity to onsider an element of B (n; F ) as an
element of B (m; F ), for m > n.
B (n; F )
b (n; F ) the quotient
Let us denote B
2
hhj ; j = 1; : : : ; n 1ii . The algebra Hn(1; F ) an be onn
sidered as the group algebra of 1 (F ) o n , where n a ts by permutation of the oordinates
n
in 1 (F ) .
First, we re all that the
in
Proposition 4.3.1
Proof:
Let
The group Bb (n; F ) is isomorphi to the semi-dire t produ t 1(F )n o n.
: P (n; F )
! 1(F )n
be the map whi h forgets about the braiding and keeps
only the fundamental group information of ea h strand (see also Lemma 1.6.1). The following
diagram holds:
1
! P (n; F ) ! B (n; F ) ! n ! 1
1
where
: B (n; F ) ! n
! 1
is the
whi h indu es a surje tion
b
of
p
?
?
(F )n
anoni
Bb (n; F )
!
id
?
?
Bb (n; F )
b
#
! n ! 1 ;
B (n; F ) onto
n ,
on n . Set sj for the transposition (j; j + 1) in n .
proje tion of
58
the symmetri
group
b is provided with the se tion s : n ! Bb (n; F ) whi h sends ea h generator sj
b (n; F ). Thus the lower exa t sequen e splits and the laim
in the equivalen e lass of j on B
The morphism
follows.
We
have
proved
that
the
map
indu es an isomorphism : Bb (n; F ) !
1 (F )n o n . The inverse : 1 (F )n o n ! Bb (n; F ) is dened as follows:
(p1 ; 1; ; 1) = p1, where p1 2 1(F ) = B (1; F )
(1; : : : ; 1; pi ; 1; : : : ; 1) = i 1 1p11 i 1 , where pi 2 1(F ) = B (1; F ), for i =
2; : : : ; n 1.
(1; : : : ; 1; ) = s(), where 2 n.
We an suppose that s() is written in the
form :
Remark 4.3.1
normal
(si1 si1
r1 )(si2 si2 r2 ) (sip sip rp ) ;
where 1 i1 < i2 < < ip n 1 and rj 2 f0; 1; : : : ; ij
normal form for the element 2 Bb (n; F ).
Hn(1; F ). The algebra H1 (1; F ) is the group algebra of
B (1; F ), the group freely generated by fa1 ; : : : ; ag ; b1 ; : : : ; bg ; z1 ; : : : ; zp 1 g.
We set again
j
for the image of
j
. The word Æ ( ) is the
1g
in
Every element x 2 Hn+1(1; F ) an be written as linear ombination of words,
ea h of one of the following types:
1. wn 1;
2. wn 1nvn 1;
3. wn 1n 1A1 n,
where wn 1; vn 1 are some words in Hn(1; F ) and A some word in H1(1; F ).
Proof:
Hn(1; F )
Bb (n; F )
x
b
Theorem 4.3.1
Sin e
B (n + 1; F )
is the group algebra of
, it su es to prove that ea h word
in
an be written as a word of one of following types:
1.
wn
2.
wn 1 n vn
3.
wn 1 n 1 A1 n ,
1;
Bb (n; F ) and A an element of B (1; F ). We pro eed
by indu tion on n. The laim is true for n = 1. Braid relations in Theorem 1.1.1 and the
2
relation j = 1 imply A1 B1 = 1 B1 A, for all A; B in B (1; F ). It yields that any element
b (2; F ) an be rewritten as a word A 1 B 2 , for A; B in B (1; F ) and ; = f0; 1g. Let
of B
1 2
1
1
x 2 Bb (n + 1; F ). There exists a word z equivalent to x, where n appears at most twi e. Let
n x0 n x00 n , x0 ; x00 2 Bb (n; F ) be a sub-word of x. We distinguish four ases
where
wn 1 ; vn
1;
1
are some elements of
59
1.
x0 = yn 1 1 A1 n 1 ; x00 = y0 n
1; F ) ; A; A0 2 B (1; F ).
1
1A0 1 n 1; for some y; y0 2 Bb (n
n yn 1 1 A1 n 1 n y0 n 1 1 A0 1 n 1 n =
= yn n 1 1 A1 n 1 n y0 n 1 1 A0 1 n 1 n =
= yy0n 1 1 A0 1 n 1 n n 1 1 A1 n 1 n2 = wn v ;
where
2.
w; v 2 Bb (n; F ).
x0 = yn
B (1; F ).
1
1 A1 n 1; x00 = y0n 1y00 ; for some y; y0 ; y00 2 Bb (n 1; F ); A 2
n yn 1 1 A1 n 1 n y0 n 1 y00 n =
= yn n 1 1 A1 n 1 n y0 n 1 y00 n =
= yy0 n 1 y00 n n 1 1 A1 n 1 n2 = wn v ;
where
3.
w; v 2 Bb (n; F ).
x0 = yn 1y0 ; x00 = y00 n
B (1; F ).
1
1 A1 n 1 ; for some y; y0 ; y00 2 Bb (n 1; F ), A 2
n yn 1 y0 n n 1 1 A1 n 1 n =
= n n n 1 1 A1 n 1 n yn 1 n n 1 y0 = wn v ;
where
4.
w; v 2 Bb (n; F ).
x0 = yn 1 y0 ; x00 = un 1 u0 for some y; y0 ; u; u0 2 Bb (n 1; F ).
nyn 1 y0 nun 1 u0 n = yn 1 n n 1 y0 un 1 n u0 ;
(a) if
y0 u = an 2 b for some a; b 2 Bb (n 2; F ),
yn 1 n n 1 y0 un 1 n u0 = yn 1 an n 1 n 2 n 1 n bu0 =
= yn 1 an 2 n 1 n n 1 n 2 bu0 = wn v ;
w; v 2 Bb (n; F ).
0
if y u = an 2 1 A1 n
where
(b)
2
for some
a 2 Bb (n 2; F ); A 2 B (1; F ),
yn 1 n n 1 y0 un 1 n u0 =
= yn 1 an n 1 n 2 1 A1 n 2 n 1 n u0 =
wn 1 A1 n ;
w 2 Bb (n; F ); A 2 B (1; F ).
Equivalen es with = are justied be ause [g; k 1 A1 k ℄ = 1
A 2 B (1; F ). Let x = wn un v, w; u; v 2 Bb (n; F ). We pro eed as in 4.
b (n 1; F ),
1. if u = an 1 b for some a; b 2 B
where
where
w0 ; v0 2 Bb (n; F ).
wan n 1 n bv = w0 n v0 ;
60
for any
g
2 Bb (k; F ),
2. if
u = an
where
3. if
2
1A1 n 2 for some a 2 Bb (n 1; F ); A 2 B (1; F ),
wan n 1 1 A1 n 1 n u = w0 n 1 A1 n ;
w0 2 Bb (n; F ).
u = a for some a 2 Bb (n 1; F ),
wan n v = w0 ;
where
w0 2 Bb (n 1; F ).
Let
A 2 B (1; F ). We set A(i) = i
1
1A1 i
1,
for
i = 1; : : : ; n, (A(1) = A).
Every element of Hn+1(1; F ) an be written uniquely as a linear ombination
of words ea h of one of the following types:
1. wn 1;
2. wn 1n j for j = 1; : : : ; n 1;
3. wn 1n iA(i) for i = 1; : : : ; n,
where wn 1 is a word in Hn(1; F ) and A a word in H1(1; F ).
Proof:
x 2 Bb (n + 1; F )
Theorem 4.3.2
First, we prove that ea h word
an be written as a word of one of
following types:
1.
wn
2.
wn 1 n j
3.
wn 1 n i A(i)
where
1;
wn
1
j = 1; : : : ; n 1;
for
i = 1; : : : ; n,
Bb (n; F ) and A an element of B (1; F ). From previous Theorem it
b (n; F )
laim for words of the type wn 1 n vn 1 , where wn 1 ; vn 1 are in B
is an element of
su es to show the
and
for
A in B (1; F ).
We reason by indu tion. We en ounter three possibilities:
2 Bb (n 1; F ) then x = wn 1vn 1n = wn0 1n, where wn0 1 2 Bb (n; F );
b (n 1; F ), then
if vn 1 = un 2 n 1 j , where un 2 2 B
x = wn 1 un 2 n n 1 j = wn0 1 n j ;
1. if
2.
vn
1
3.
wn0
2 Bb (n; F );
b (n 1; F ) and A 2 B (1; F ), then
if vn 1 = un 2 n 1 1 A1 n 1 , where un 2 2 B
0
(
n
x = wn 1 un 2 n n 1 1 A1 n 1 = wn 1 n A ) , where wn0 1 2 Bb (n; F ).
where
1
61
(n)
wb(x) = A(1)
1 An , where A1 ; : : : ; An 2
B (1; F ) and is an element of Bb (n; F ) in the generators f1 ; : : : n g. We an suppose written
in the normal form (see Remark 4.3.1). We reason by indu tion on n. We have to he k only the
(1)
(n)
(i)
b (n; F ). Suppose that w
ase w (x) = wn 1 n i A , where wn 1 2 B
n 1 = A1 A n 1 ,
b (1; F ) and is an element of , written in its normal form.
where A1 ; : : : ; An 1 belong to B
n
A tually the word
w(x)
an be rewritten as a word
Thus,
(n)
(1)
(n)
(i)
(i)
w(x) = A(1)
1 An 1 n i A = A1 An 1 n i A i n n i =
(n 1) (n)
(1)
(n 1) (n) 0
= A(1)
1 An 1 A n i = A1 An 1 An ;
(n )
(n) and 0 = is an element of written in its normal form.
where An = A
n
i
n
b (x) is the normal form Æ The word w
(x) of x (Remark 4.3.1).
T on [n1Hn(1; F ) using Theorem 4.3.2. Let x be an
Bb (n + 1; F ). By Theorem 4.3.2, the element x is equivalent to a word in
Bb (n; F ) or a word wn v, with w; v 2 Bb (n; F ) or a word un 1 A1 n , with u 2 Bb (n; F )
and A 2 B (1; F ). Assume that the T is dened on Hn (1; F ). Dene now T (x) = z T (wv ) if
x = wn v and T (x) = AbT (u) if x = un 1 A1 n . The map T is well dened and it
extends by linearity to Hn+1 (1; F ).
In order to prove the existen e of T , it remains to prove that T (xy ) = T (yx), for all
x; y 2 [n1 Hn(1; F ). Before ontinuing with proof, we note that the uniqueness of T follows immediately sin e for any x 2 Hn (1; F ), T (x) an be omputed indu tively using rules
We
an
onstru t indu tively the tra e
arbitrary word of
1), 2), 3), and 4) of Theorem 4.1.1.
T (xy) = T (yx), for all x; y 2 [n1Hn(1; F ). We suppose that
x; y 2 Bb (n; F ) and we prove that T (xy) = T (yx), for all x; y 2
Bb (n + 1; F ). We an suppose that y = wn v, with w; v 2 Bb (n; F ) or y = un 1 A1 n ,
b (n; F ) and A 2 B (1; F ).
with u 2 B
We pro eed with
he king that
the assumption holds for all
x 2 Bb (n; F ) the laim follows by the denition of T .
b (n; F ) and A 2 B (1; F ). Then,
For instan e, let y = un 1 A1 n , with u 2 B
When
T (xy) = AbT (xu) = AbT (ux) = T (yx) :
0
b (n; F ) and A 2 B (1; F ) we en ounter two possiWhen x = an 1 A 1 n , with a 2 B
bilities:
1. If
2.
y = un 1 A1 n , with u 2 Bb (n; F ) and A 2 B (1; F ), then
0 AT (au) =
T (xy) = T (aun 1 A0A1 n) = Ad
d0 T (ua) = T (ua AA0 ) = T (yx) :
= AA
n
1
1
n
b (n; F ), then
If y = wn v , with w; v 2 B
T (xy) = T (awn 1 A1 n 1v) = zT (awn 1 1A1 n 1v) =
= z T (wn 1 1 A1 n 1 va) = T (wn 1 1 A1 n va) = T (yx) :
62
Finally suppose that
x = an b. Without loss of generality, we
a and b to be the
an suppose
empty word.
y = un 1 A1 n , for some u 2 Bb (n; F ); A 2 B (1; F ).
0
0
0 b (n 1; F ) and A0 2 B (1; F ).
Suppose u = u n 1 1 A 1 n 1 , for some y 2 B
Let
1.
T (ny) = T (u0 nn 1 1 A01 n 1nn 1 1 A1 n) =
= T (u0 n 1 1 A1 n 1 nn 1 1 A0 1 n 1 n n) =
d0 T (u0 ) :
= z T (u0 n 1 1 AA0 1 n 1 ) = z AA
T (yn) = T (u0n 1 1A0 1 n 1n 1A1 nn) =
= z T (u0 n 1 1 A0 1 n 1 n 1 1 A1 n 1 ) =
d0 T (u0 ) :
0 AT (u0 ) = z AA
= z T (u0 n 1 1 A0 A1 n 1 ) = z Ad
2. Suppose
u = u0 n 1 u00 , for some u0 ; u00 2 Bb (n 1; F ).
T (yn) = T (u0 n 1u00 n 1A1 nn) =
= z T (u0 n 1 u00 n 1 1 A1 n 1 ) =
= z T (u0 n 2 1 A1 n 1 u00 ) =
= z 2 T (u0 n 2 1 A1 n 2 u00 ) ;
T (ny) = T (u0 nn 1u00 n 1A1 n) =
= z T (u0 n 1 1 A1 n 1 n 1 u00 ) =
= z 2 T (u0 n 2 1 A1 n 2 u00 ) :
y = wn v, for some w; v 2 Bb (n; F ).
0
00
0
00
1. Suppose w = w n 1 w , v = v n 1 v
Let
for some
w0 ; w00 ; v0 ; v00 2 Bb (n 1; F ).
T (yn) = T (w0 n 1w00nv0 n 1v00 n) =
= T (w0 n 1 w00 v0 n 1 n n 1 v00 ) = z 2 T (w0 w00 v0 v00 ) = T (n y) :
w = w0 n 1 w00 , v = v0 n
and A 2 B (1; F ).
2. Suppose
1
1A1 n
0
1 , for some w ; w
00 ; v0 2 Bb (n 1; F )
T (yn) = T (w0 n 1w00nv0 n 1 1 A1 n 1n) = zAbT (w0 w00 v0) ;
T (ny) = T (w0 nn 1nw00v0 n 1 1 A1 n 1 ) =
= z T (w0 w00 v0 n 1 1 A1 n 1 ) = z AbT (w0 w00 v0 ) :
w = w0 n 1 1 A1 n 1 , v = v0 n 1 v00 , for some v0 ; v00 ; w0 2 Bb (n 1; F )
and A 2 B (1; F ). As in previous point
3. Suppose
T (yn) = T (w0 n 1 1A1 n 1nv0 n 1v00n) = zAbT (w0 v0v00) ;
T (ny) = T (w0 nn 1 1A1 n 1nv0 n 1v00 ) = zAbT (w0 v0v00) :
63
w = w0 n 1 1 A1 n
Bb (n 1; F ) ; A; A0 2 B (1; F ).
4. Suppose
1,
v = v0 n
1
1 A01 n
1,
for some
v; w 2
T (yn) = T (0wn 1 1 A1 n 1nv0n 1 1A01 n 1n)
= T (w0 v0 n 1 1 A1 n 1 n n 1 1 A0 1 n 1 n ) =
= A0 Ab T (w0 v0 ) ;
T (n y) = T (nw0 n 1 1 A1 n 1 nv0 n 1 1A01 n 1)
= T (w0 v0 n n 1 1 A1 n 1 n n 1 1 A0 1 n 1 ) =
= AbA0 T (w0 v0 ) :
64
Chapter 5
Cubi He ke algebras and new
invariants for links
This
hapter is a joint work with L.Funar ([8℄).
5.1 Introdu tion
5.1.1 A short history
John Conway showed that the Alexander polynomial of a knot, when suitably normalized,
satises the following skein relation:
r
Given a knot diagram one
!
!
r
an always
= (t
1
t1=2 )r A
hange some of the
diagram represents the unknot. Therefore one
putation of
1=2
0
rossings su h that the modied
an use the skein relation for a re ursive
r, although this algorithm is rather time
om-
onsuming (exponential).
In the mid eighties Jones dis overed another invariant verifying a dierent but quite similar
skein relation, namely:
!
t 1V
!
tV
= (t
1=2
t1=2 )V
0
1
A
whi h was further generalized to HOMFLY-PT invariant (Chapter 4) by repla ing the fa tor
(t1=2 t
1=2 ) with
a new variable
x. The latter one was shown to spe
ialize to both Alexander
and Jones polynomials. The Kauman polynomial is another extension of Jones polynomial
whi h satises a skein relation in the realm of unoriented diagrams. Spe i ally the formulas
!
!
+
!
=z = a (
65
!!
+
)
dene a regular isotopy invariant of links, whi h
an be renormalized (by using the writhe
of the oriented diagram) in order to be ome a link invariant. Remark that some elementary
manipulations show that
0
1
B
C
A
veries a
ubi al skein relation:
1
0
1
= ( + z) a
1
0
z
( + 1) a
A
1
0
1
A + ( )
a
A
It has been re ently proved ([31℄, see also problem 1.59 [59℄) that this relation alone is not
su ient for a re ursive
alled
omputation of
(whenever this is possible the skein relations are
omplete).
These invariants were generalized to quantum invariants asso iated to Lie (super Lie, et )
algebras and their representations. Turaev ([89℄) identied the HOMFLY-PT and Kauman
polynomials with the invariants obtained from the series
Kuperberg ([61℄) dened the
An
and
Bn ; Cn ; Dn
respe tively.
G2 quantum invariant of knots by means of skein relations making
use of trivalent graphs diagrams and exploited further these ideas for spiders of rank 2 Lie
algebras. The skein relations satised by the quantum invariants
oming from simple Lie
algebras were approa hed also via weight systems and the Kontsevi h integral in ([68, 69℄) for
the
lassi al series and in ([9, 10℄) for the
Noti e that any link invariant
ase of
g2 .
oming from some R-matrix
R
veries a skein relation of
the type
n
X
j =0
whi h
*
aj
9
>
>
>
=
+
j twists
>
>
>
;
=0
an be derived from the polynomial equation satised by the matrix
R.
Let us mention that the skein relations are somewhat related to the representation theory
of the Hopf algebra asso iated to
relations are
Lie algebras or by
G2
R. In parti
ular there are no other invariants whose skein
ompletely known and one expe ts that the invariants obtained from other super
abling the previous ones satisfy skein relations of degree at least 4 (as the
invariant does).
This makes the sear h for an expli it set of
one relation is
omplete skein relations, in whi h at least
ubi al, parti ularly di ult and interesting. This problem was rst
in [43℄ and solved in a parti ular
ase. In this
onstru ting a deformation of the previously
hapter we
onsidered
ubi He ke algebras
omplete the previous results by
onsidered quotients (of the
,
see Se tion 5.1.3) and of the Markov tra es supported by these algebras. In parti ular the link
invariants obtained this way will be re ursively
Kauman and the
2-
omputable and dierent from HOMFLY-PT,
abling of HOMFLY-PT.
5.1.2 The main result
In this hapter we will dene two link invariants by means of (a
omplete set of ) skein relations.
More pre isely we will prove the following Theorem (see Se tion 5.5):
There exist two link invariants I( ; ) and I (z; Æ) whi h
are (uniquely) determined by the two skein relations shown in Figure 5.1 and their value for
the unknot (whi h, traditionally, is 1). These invariants take values in
Theorem 5.1.1 (Main Theorem)
66
= αw
= Aw
+E
-2
+E
2
+ βw
-1
+B w
+G w
+F
+L w 2
+I w
+H w
-1
-1
+B w
+F
3
+ w
+L w 2
+N w 3
+C w
+D
+G w
+H w
+M w 2
+M w 2
+P w 4
+O w 3
Figure 5.1: The skein relations
Z[
2 )=2 ;
; ; (2
(H(
and respe tively
where H(
:= 8
2+2
)=2 ℄
;
Z[z =2; Æ =2 ℄
;
(P (z; Æ) )
is the number of omponents mod 2 and
1 2 f0; 1g
; )
; ))
(
6
36
8
5 2 + 2 4 4 + 36 4
4 + 38
+8
6
34
17 3 + 8;
3 3 + 17 3 + 8 2 5 + 32 2 2
and respe tively
P(z; Æ) := z23 + z18Æ 2z16 Æ2 z14Æ3 2z9 Æ4 + 2z7Æ5 + Æ6z5 + Æ7:
Here (Q) denotes the ideal generated by the element Q in the algebra under onsideration.
The polynomials
order to obtain those
= (z 7 + Æ2 )=(z 4 Æ)
A; B; C:::; P
I( ; ) are given in the table below. In
(
z;
Æ
)
4
1=2 and repla e
orresponding to I
it su es to set w = ( z =(Æz ))
2
3
and
= (Æ z )=z in the other entries of table 1.
orresponding to
67
2 ))1=2
w = (( 2 + 2 )=(2
2
B=( 2
)
D = (1 + 2 + 2 2 3 )
3)
F = (1 + 2
3
H=(
2 2 2 + 2)
3
2 3
2)
L = (2
+3 2
3
N = (1 + 4 + 3 2 2
2
5
2
P = (3
2 3
+4
4
3)
A=( 2
)
2
2)
C=(
3)
E = (1 + + 2 2
G=( 3 2 2 2 )
3 2 2 2
I=( 4
3 )
4
M =(
2 3 2 + 2)
3 ) O = (1 + 3
3
+3 2 2
4)
Table 1
5.1.3 Cubi He ke algebras
Denition 5.1.1
where Q(j ) = j3
Our aim is to
The ubi He ke algebra H (Q; n) is the quotient
H (Q; n) = C [Bn ℄=(Q(j ) ; j = 1; : : : ; n 1) ;
j2
j
2 C.
1; ;
onstru t Markov tra es on the tower of
tra es dene link invariants (Se tion 4.2.1). The
the (generi )
in [21, 22℄ in
y lotomi
ubi
ubi
He ke algebras sin e Markov
He ke algebras are parti ular
ases of
He ke algebras introdu ed by Broué and Malle (see [20℄) and studied
onne tion with braid group representations. Let us stress that, for
Q(0) =
6 0,
the following results are known (see also [28, 20, 21, 22℄ and [29℄ p.148-149):
dimC H (Q; 3) = 24, and H (Q; 3) is isomorphi
hedral group
to the group algebra of the binary tetra-
< 2; 3; 3 > of order 24 (equivalently, SL(2; Z3)).
dimC H (Q; 4) = 648,
and
H (Q; 4)
is the group algebra of
G25
in the Shepard-Todd
lassi ation (see [85℄).
dimC H (Q; 5) is the
is
y lotomi
He ke algebra of group
G32 ,
whose order is
155520.
It
onje tured that this algebra is free of nite dimension whi h would imply (by using
the Tits deformation theorem) that it is isomorphi
to the group algebra of
G32 .
dimC H (Q; n) = 1 for n 6.
n
6 is highly a nontrivial matter, in
parti ular it would involve the expli it solution of the
onjuga y problem in these algebras
Thus a dire t denition of the tra e on
H (Q; n)
for
whi h seems out of rea h.
In order to deal with nite dimensional algebras one introdu es smaller quotients
by adding one more relation living in
H (Q; 3). The exa
Kn ( ; )
t form of this relation is
2 12 2 +A 12 22 12 + B 1 22 12 +B 12 22 1 +C 12 2 12 + D 1 22 1 +E 1 2 12 + E 12 2 1 +
F 22 12 + F 12 22 + G 2 12 + G 12 2 + H 22 1 + H 1 22 + I 1 2 1 + L 2 1 + L 1 2 +
M 12 + M 22 + N 1 + O 2 + P = 0
where
A; B; : : : ; P
are the polynomials from table 1.
68
Remark 5.1.1
The algebras Kn(
; ) are nite dimensional for any n.
Let us explain the heuristi s behind that
hoi e for the additional relation. The algebra
H (Q; 3) is semisimple (for generi Q) and de omposes as C 3 M23 M3 ; where Mm is the
algebra of m m matri es. As explained in Se tion 5.2 the usual quadrati He ke algebra
Hq (3) arises when the fa tor C M22 M3 is killed. It is known that Jones and HOMFLYPT polynomials an be derived by the unique Markov tra e on the tower Hq (n). In a similar
way the Birman-Wenzl algebra, whi h yields the Kauman polynomial ([43℄) is obtained when
we quotient by
C
The geometri
Figure 1 is the
M22 . In our situation the extra relation kills exa tly the fa tor C 3 .
interpretation of these relations is now obvious: the rst skein relation in
ubi al relation
relation denes the algebras
Our main theorem is a
orresponding to the quotients
Kn ( ; ).
H (Q; n) and the se
ond skein
onsequen e of the more te hni al result below (see Se tions
5.2,5.3,5.4).
For exa tly four values of the (z; zb) there exists an unique Markov tra es T
) with parameters (z; zb) i.e. verifying:
Theorem 5.1.2
on Kn( ;
1. T (xy) = T (yx);
2. T (xn 1) = zT (x);
3. T (xn 11) = zbT (x):
The rst ouple (z; zb) is
2 )=(
z = (2
+ 4); zb = (
2+2
)=(
+ 4);
and the orresponding tra e is T ; : Kn( ; ) ! Z[ ; ; 1=( + 4)℄=(H( ; )).
The other three solutions are not rational
fun tions on the parameters and we prefer to give
2
; and zb as fun tions of z; Æ (Æ = z ( z + 1)). More pre isely we set
T (z; Æ) : K( ; ) ! Z[z1; Æ1 ℄=(P(z; Æ) );
where
= (Æ
z 2 )=z 3 ;
= (z 7 + Æ2 )=(z 4 Æ) zb = z 4 =Æ:
5.1.4 Outline of the proof
As explained in Chapter 4, one
an
Hn+1(q). We will prove by re
of (Kn ( ; ); Kn ( ; ))-bimodules
algebra
Kn ( ; ) Kn ( ; )
Kn 1 ( ; ) Kn (
onstru t indu tively a basis for the (quadrati ) He ke
urren e on
n (Lemma 5.3.1)
; ) Kn ( ; )
that there is a surje tion
Kn 1 ( ; ) Kn (
; )
! Kn+1( ; )
x y z u v ! x + yn z + un2 v.
Sin e there is a system of generators for Kn+1 ( ; ) onstru ted out of one for Kn ( ; ), the
extension of a Markov tra e on Kn ( ; ), if ever exists, it must be unique.
given by
However the previous morphism is not inje tive and the most di ult step is to prove that the
anoni al extension is a well-dened linear fun tional and it satises the tra e
69
ommutativity.
A = Z[ ; ; z; zb℄=H ,
rewriting
system
Let
A[Fn ℄ where Fn
where
H
is an ideal of
Z[
; ; z; zb℄.
Our approa h is to
onsider a
(we refer to [25℄ for a survey on the subje t) for the tower of the ring algebra
is the free monoid in the generators
f1; : : : ; n 1g. The method of proof is
greatly inspired from [11℄.
A[Fn ℄ and whi
First, one denes a graph whi h verti es are the elements of
h edges
spond to elements diering by exa tly one relation (from the set of relations dening
One denes a redu tion pro ess for elements in
A[Fn ℄ introdu
orre-
Kn ( ; )).
ing the following orientations
on some edges. The arrows show the orientation, if exa tly one monomial is
hanged using
one of the following rules
aj3 b ! aj2 b + aj b + ab;
aj +1 j j +1 b ! aj j +1 j b;
aj +1 j2 j +1 b ! aSj b;
aj +1 j2 j2+1 b ! aCj b;
aj2+1 j2 j +1 b ! aDj b;
P
Sj ; Cj and Dj are of the form i Pi ja jb+1 j , Pi are polynomials in ; and ai ; bi ; i 2
f0; 1; 2g. An element z is irredu ible (or minimal) if there is not an element u su h that z ! u.
where
Several edges remain unoriented. They
i
i
i
orrespond to a
hange in a monomial of type
ai j b ! aj i b whenever j i j j> 1.
The reason for introdu ing the extra relations (whi h obviously hold in
the existen e of des ending paths to some minimal points even if
H (Q; n)) is to insure
losed oriented loops may
be found in the graph. Our aim is to show that this rewriting system is
onne ted
onuent, i.e. every
omponent has at most one minimal element, up to unoriented equivalen e (see
Se tion 5.3.1). The existen e is proved in Lemma 5.3.4 and we
he k the uniqueness by means
of so- alled Pentagon Lemma. Anyway, to show that the rewriting system is
enlarge our graph to a tower of graphs modeling not one algebra
on the whole tower
[1
n=2 Kn ( ; )
satisfying a re urren e
onuent, we shall
Kn ( ; ) but the fun
tionals
ondition whi h permits to redu e
further the minimal elements. We will nd that the Pentagon Lemma applies ex ept a nite
number of
5.3.2)
K4 ( ; ). Here the Colored Pentagon Lemma (see Se
an be applied and the problem is redu ed to some algebrai
wish to
a
ongurations that lie in
he k the
ommutativity
onstraint appears in
tion
omputations. When we
ondition for the fun tional to be a tually a Markov tra e,
K4 ( ; ) on the variables ; ; z; zb. Then there are only two types of
obstru tions to the existen e of a Markov tra es:
CPC obstru tions (Colored Pentagon Condition), dening an ideal
ommutativity obstru tions, implying a
These nitely many obstru tions have been
in the prin ipal ideal generated by
H(
; )
onstraint on
; ; z; zb (Se
he ked by using the
(respe tively
P (z;Æ) ).
H
in
Z[
; ; z; zb℄.
tion 5.4).
omputer and all of them lie
5.1.5 Properties of the invariants
In the next Se tion we will
tra es
T(
; )
and
ompute these obstru tions and derive the existen e of the two
T (z; Æ) . As explained in Chapter 4, given a Markov tra e T , we get a link
invariant by setting:
70
where
x
2 n
1
I (xb) =
z zb
2
1
e(x)
zb 2
z
T (x);
L and e(x)
I (z; Æ) . We nd that:
is a braid representative of the link
Therefore we nd two invariants
n
I(
; )
and
they distinguish all knots with number
is the exponent sum of
x.
rossing at most 10 that have the same HOMFLY-
PT polynomial (and then they are independent from HOMFLY-PT). However, like
HOMFLY-PT and Kauman polynomials, they seem to not distinguish among mutants
knots (in parti ular they don't separate Kinoshita-Terasaka and Conway knots).
I(
; )
= I(
with number
;
)
for amphi heiral knots, and
I(
; )
dete ts the
hirality of all the knots
rossing at most 10, where HOMFLY-PT and Kauman polynomials fail.
The invariant I( ; ) is independent from the 2- abling of HOMFLY-PT.
I( ; ) and I (z; Æ) have a ubi al behaviour.
ubi al behaviour
Denition 5.1.2 A Laurent polynomial j 2Z j aj is a (n; k)-polynomial (for n; k 2 N ) if
j = 0 for j 6= hn + k , for all h 2 Z.
P
Remark 5.1.2 P The HOMFLY-PT polynomial an be written as k2z Rk (l)mk and respe tively as k2z Sk (m)lk , where Rk (l) and Sk(m) are (2; k)-Laurent polynomials with
R2k+1 (l) = S2k+1 (m) = 0.
The Kauman polynomial an be written as Pk2z Uk (l)mk (respe tively as Pk2z Tk (m)lk ),
where Uk (l) and Tk (m) are (2; k + 1)-Laurent polynomials.
Let us explain briey what we meant by
.
P
In this respe t the HOMFLY-PT and Kauman polynomials have a quadrati
Proposition 5.1.1 I( ; )
some l 2 f0; 1; 2g so that
I(
;
behaviour.
and I (z; Æ) have a ubi al behaviour, i.e. for ea h link L there exists
) (L) =
P
Pk (
P k2N
k2N Qk (
)
)
k
k
=
where Pk ; Qk ; Mk ; Nk are (3; k + l)-polynomials, and
I (z; Æ) (L) =
X
Hk (Æ)z k =
k2Z
where Hk; Gk are (3; k)-Laurent polynomials.
71
P
Mk ( ) k
;
) k
Pk2N
k2N Nk (
X
k2Z
Gk (z )Æk ;
5.2 Markov tra es on Kn( ;
)
5.2.1 A base for the ubi He ke algebra H (Q; 3)
Usually
ubi
He ke algebras are dened as the quotients
H (Q; n) = C [Bn ℄=(Q(j ) ; j = 1; : : : ; n 1)
Q(j ), ubi polynomial
; ; ; 2 C . One may onsider
1 Q; 1) (the towers of ubi He ke
of the group algebra of the braid group by the ideal generated by
with parameters
=1
in the
Q(j ) = j3
j2
j
polynomial sin e H (Q; 1) and H (
and
ubi
, i.e.
algebras) are isomorphi . In [43℄ it was shown that:
For all ubi polynomials Q with Q(0) 6= 0 one has dimC H (Q; 3) = 24.
A onvenient base of the ve2 tor spa e H (Q; 3) is2
2
Proposition 5.2.1
e1 = 1; e2 = 1 ; e3 = 1 ; e4 = 2 ; e5 = 2 ; e6 = 1 2 ; e7 = 2 1 ; e8 = 1 2 ; e9 =
2 12 ; e10 = 1 22 ; e11 = 22 1 ; e12 = 12 22 ; e13 = 22 12 ; e14 = 1 2 1 ; e15 = 12 2 1 ; e16 =
1 2 12 ; e17 = 1 22 12 ; e18 = 12 2 12 ; e19 = 12 22 1 ; e20 = 1 22 1 ; e21 = 12 22 12 ; e22 =
2 12 2 ; e23 = 2 12 2 1 = 1 2 12 2 ; e24 = 2 12 2 12 = 1 2 12 2 1 = 12 2 12 2 .
We refer also to [43℄ for the following identities:
j +1 j2 j +1 j = j j +1 j2 j +1 ;
j2+1 j2 j +1 = j j2+1j2 + (j +1 j2 j +1 j j2+1 j ) + (j2 j +1 j j2+1 );
j +1 j2 j2+1 = j2 j2+1 j + (j +1 j2 j +1 j j2+1 j ) + (j +1 j2 j2+1 j ):
5.2.2 The homogeneous quotient of rank 3
P (1) of H (Q; 1) is homogeneous if any identity F (i ; i+1 ; :::; j ) = 0, whi h
P (1) remains valid under the translation of indi es i.e. also F (i+k ; i+k+1 ; :::; j +k )
= 0, for k 2 Z; k 1 i. One onsiders the Markov tra es supported by the quotients
Kn ( ; ) = H (Q; n)=In , where In is the (two-sided) ideal generated by:
The quotient
holds in
2
2
j j2 1 j +( 2 )j2 1 j2 j2 1 +( 2
)j 1j2 j2 1 + ( 2
)j2 1j2 j 1 +( 2
2 ) 2 2 + (1+2 + 2 2
3 ) 2 +(1+ + 2 2
3 ) 2 +(1+ +
j 1 j j 1
j 1 j j 1
j 1 j j 1
2 2
3 ) 2 +(1+2
3 ) 2 2 + (1+2
3 ) 2 2 +( 3 2 2 2 ) 2 +
j
j
1
j j 1
j 1
j j 1
j 1 j
( 3 2 2 2 )j2 1 j + ( 3 2 2 2 + 2 )j2 j 1 + ( 3 2 2 2 + 2 )j 1 j2 +
3 2 2 2
2 3
2 ) + (2 3 + 3 2
( 4
3 )j 1 j j 1 + (2 3 + 3 2
j j 1
2 3
2 ) + ( 4 2
2
2
2
4
3
+ )j 1 + (
2
3 2 + 2 )j2 + (1 + 4 +
j 1 j
4
3 )
2 2
3
4 ) +3 2
5 2
3 2 2 3
3 2 + 4 3;
j 1 + (1 + 3 + 3
j
where
j = 1; : : : ; n 1. Then K1 ( ; ) is a homogeneous quotient of H (Q; 1).
is a semisimple algebra whi h de omposes generi3 ally as C 3 M23 M3 , where Mn is the algebra of n n matri es. The morphism into C is obtained via the
abelianization map and that into M2 is2 part of the proje tion onto the quadrati He3ke algebra
dened by a divisor of Q (whi h is C M2). One identies then K3( ; ) = M2 M3.
Remark 5.2.1 H (Q; 3)
In fa t it su es to show that the ideal
of
R0 ; R1 ; R2 , where
I3 is a ve
72
tor spa e of dimension
3. Let R be the span
2
2
R0 := 2 12 2 + ( 2
) 12 22 12 + ( 2
)1 22 12 + ( 2
)12 22 1 +( 2
2 ) 2 2 + (1 + 2
3 ) 2 + (1 +
3 ) 2 + (1 +
+ 2 2
+ 2 2
+
1 2 1
1 2 1
1 2 1
2 2
3 ) 2 +(1 + 2
3 ) 2 2 + (1 + 2
3 ) 2 2 + ( 3 2
2 ) 2
2
2 1
1 2 1
2 1
1 2
+( 3 2 2 2 )12 2 + ( 3 2 2 2 + 2 )22 1 +( 3 2 2 2 + 2 )1 22 + ( 4
3 2 2 2
2 ) + (2 3 +3 2
2 3
2 ) +
3 )1 2 1 +(2 3 +3 2 2 3
2 1
1 2
4
2
2
2
4
2
2
2
2
2
3
4
3 )
(
2 3
+ )1 + (
2 3
+ )2 +(1 + 4 + 3
1
3
4 ) + 3 2
5 2
+(1 + 3 + 3 2 2
3 2 + 4 3;
2
R1 := 1 R0 = 1 2 12 2
12 22 12 + (1 + )1 22 12 + (1 + )12 22 1 +(1 + )12 2 12
2
2
2
(
2 )1 2 1 +(
2 )1 2 12 + ( 2 2 )12 2 1 + ( 2 )22 12 +( 2 )12 22 +
2
2
2
2
2
2
2
(
)2 1 + (
)12 2 + ( 2
)22 1 + ( 2
)1 22 + ( 3 + +
3 ) + (1 +
3 ) + (1 + 2
3 ) 2 +
3 2 )1 2 1 + (1 + + 2 2
+ 2 2
2 1
1 2
1
3
2
3
2
2
3
2
4
2
(1 + 2
)2 + (
2 2 + )1 +(
2 2 )2 +
2 3 + 2;
R2 := 1 R1 = 12 2 12 2 + 12 22 12
1 22 12
12 22 1
12 2 12 + 2 1 22 1 + ( 2 +
2
2
2
2
2
2
2
)1 2 1 + ( + )1 2 1 + ( )2 1 + ( )1 2 + (1 + )2 12 + (1 + )12 2 + (1 +
)22 1 + (1 + )1 22 + ( 3
+ 1)1 2 1 + ( 2 2 )2 1 + ( 2 2 )1 2 +
2
2
2
2
2
2
2 + 2 ) + 1 + 2
3:
(
)1 + (
)2 + (
+
)1 + (
2
Lemma 5.2.1
Proof:
As ve tor spa es R = I3 in H (Q; 3).
Remark that
1 R0 = R0 1 = R1 ; 1 R1 = R1 1 = R2 ; 1 R2 = R2 1 = R0 + R1 + R2 :
and after some messy
omputations ( omputer aided) we obtain that
2 R0 = R0 2 = R1 ; 2 R1 = R1 2 = R2 ; 2 R2 = R2 2 = R0 + R1 + R2 :
From these relations we nd that
xR0 y 2 R for all x; y 2 H (Q; 3), hen e I3 R. The other
in lusion is trivial.
5.2.3 Uniqueness of Markov tra e on Kn ( ; )
From now on we will work with the group ring
Z[
; ℄ [B1 ℄ instead of C [ B1 ℄.
Let z; zb 2 C ; ; 2 C , R be a Z [ ; ; z; zb℄ module and H an ideal of R.
i) T is an admissible fun tional on K1( ; ) (taking values in R=H ) if the following onditions
are fullled:
T (xny) = zT (xy) for any x; y 2 Kn( ; );
Denition 5.2.1
T (xn 1 y) = zbT (xy) for any x; y 2 Kn( ; ):
ii) An admissible fun tional T is a Markov tra e if
T (ab) = T (ba)
for any a; b 2 Kn(
73
; ):
Markov tra es on the quadrati He ke algebras (see [58℄) have the following
multipli ative property: T (xn) = T (x)T (n), for x 2 H (Q; n), whi h implies that: T (xy) =
T (x)T (y), for x 2 H (Q; n); y 2< 1; n ; n+1; :::; n+k >.
Remark 5.2.2
However we
annot expe t that this property will extend to higher level algebras and Markov
tra es on them.
The Markov tra e T is multipli ative if T (xnk ) = T (x)T (nk ) holds when
x 2 H (Q; n); k 2 Z.
Remark 5.2.3 In the ase of ubi He ke algebras the Markov tra es are multipli ative. In
fa t using the identity n2 = n + + n 1 we derive then the multipli ativity for k2 = 2,
and by re urren e for all k. In parti ular if T is a Markov tra e it follows that T (anb) =
tT (ab) a; b 2 Bn ; where t = z + + zb.
Denition 5.2.2
One
an state now the unique extension property of Markov tra es.
For xed (z; t) 2 (C )2 there exists at most one Markov tra e on Kn(
with parameters (z; t).
Proof:
Ln
Proposition 5.2.2
Dene re ursively the modules
; )
by
L2 = H (Q; 2),
L3 = C < 1i 2j 1k ; i; j; k 2 f0; 1; 2g >,
Ln+1 = C < an" b j a; b 2 basis of Ln ; " 2 f1; 2g > Ln .
We need the following result
Lemma 5.2.2
Proof:
Under the natural proje tion on Kn(
; ), Ln
surje ts onto Kn(
; ).
n = 2 it is lear. For n = 3 we know that 2 12 2 ; 1 2 12 2 ; 12 2 12 2 2 (L3 ).
Consider now w 2 Kn+1 ( ; ) represented by a word in the i 's having only positive
exponents. We assume that the degree of the word in the variable n is minimal among all
linear ombinations of words (with positive exponents) representing w .
For
If the degree is less or equal to 1 there is nothing to prove.
w = un2 v, u; v 2 Kn ( ; ) so using the indu tion hypothesis
"
we are done, or else w = un zn v , where u; z; v 2 Kn ( ; ). Therefore z = xn 1 y where
x; y 2 Kn 1 ( ; ) by the indu tion and " 2 f0; 1; 2g. If " = 0 then w an be redu ed to uzn2 v.
If " = 1 then w = un xn 1 yn v = uxn 1 n n 1 yv hen e the degree of w an be lowered
2
by 1, whi h ontradi ts our assumption. If " = 2 then w = uxn n 1 n yv . One derives
2
i
j
k
n n 1 n 2 C < n 1 n n 1 ; i; j; k 2 f0; 1; 2g >,
0 2 0
hen e we redu ed the problem to the ase when w is a word of type u n v .
If the degree of w is at least 3 we will ontradi t the minimality. In fa t w ontains either
0
a b
00
a subword w = n un , u 2 Kn ( ; ) and a + b 3, or else a subword w = n un vn ,
u; v 2 Kn ( ; ).
"
In the rst ase using the indu tion we an write u = xn 1 y , x; y 2 Kn 2 ( ; ).
0
a+b
a+b 1 xy + a+b 2 xy + a+b 3 xy , hen e the degree of w
If " = 0 then w = n xy = n
n
n
If the degree is 2 then either
an be lowered by 1.
If
" = 1 then w0 = na 1 xn n 1 n ynb 1 = na 1 xn 1 n n 1 ynb
an be redu ed by one unit.
74
1,
and again its degree
" = 2 then a or b equals 2. Set a = 2. We an write
0
w = xn2 n2 1 n ynb 1 = xn 1 n2 n2 1 ynb 1 + (n n2 1 n
+ (n2 1 n n 1 n2 )ynb 1 .
If
n 1 n2 n 1 )ynb 1 +
w.
u = xn" 1 y, v = rnÆ 1s with x; y; r; s 2 Kn 1 ( ; ).
00 ontains the subword
If " or Æ equals 1 then, after some obvious ommutation the word w
n n 1 n whi h an be repla ed by n 1 n n 1 hen e lowering its degree.
00 = xn 2 n yr2 n s. We use the homogeneity to repla e
If " = Æ = 2 then w
n 1
n 1
n n2 1 n by a sum of elements of type ni 1 nj nk 1 . Ea h term of the expression of w00
whi h omes from a fa tor having j < 2 has the degree less than it had before. The remaining
i
2 k
2
2
terms are xn 1 n n 1 yrn 1 n s, so they ontains a subword n un whose degree we already
still
ontradi ting the minimality of the degree of
In the se ond
know that it
ase we
an write also
an be redu ed as above. This proves our
T
T (xn" y) = T (n" )T (yx)
holds, and Kn ( ; ) it is an algebra hen e yx 2 Kn ( ; ) . Therefore the extension of T , by
Now the Markov tra es
re ursion, from
proposition.
2
on
H (Q; 1)
laim.
are multipli ative hen e
Kn ( ; ) to Kn+1 ( ; ) if ever exists it is unique. This ends the proof of our
5.3 CPC Obstru tions
5.3.1 The pentagonal ondition
The following Lemma is also a
Lemma 5.3.1
There is a surje tion of (Kn(
Kn ( ; ) Kn ( ; )
given by
onsequen e of the previous one:
Kn 1 ( ;
; ); Kn ( ; ))-bimodules
K
(
;
)
Kn ( ; ) K 1 ( ; ) Kn ( ; )
) n
n
! Kn+1( ; )
x y z u v ! x + yn z + un2 v.
hoi e of T (1) 2 R. Look now at the
K ( ; ). We wish to use the following transformations on the words (one way):
3
(C0)(j+1) aj +1 b ! aEj +1 b;
(C1)(j) aj +1 j j +1 b ! aj j +1 j b;
2
(C2)(j) aj +1 j j +1 b ! aSj b;
2 2
(C12)(j) aj +1 j j +1 b ! aCj b;
2
2
(C21)(j) aj +1 j j +1 b ! aDj b,
2
2
where Ej +1 = j +1 + j +1 +1, Sj = j +1 j j +1
R(0; j ) , Cj = j2 j2+1 j + (j +1j2 j +1
j j2+1 j )+ (j +1 j2 j2+1 j ) and Dj = j j2+1 j2 + (j2+1 j2 j2+1 j j2+1j )+ (j2 j +1
j j2+1 ), j = 0; : : : ; n 2. Our aim is to redu e the degree of n 1 as mu h as possible in
Kn ( ; ). A ording
to the previous Lemma every word in Kn ( ; ) is equivalent to a sum
P
"
of words of type
i xi n 1 yi . Unfortunately we are for ed to use also the transformations
i j $ j i for j i j j> 1,
In parti ular admissible fun tionals are unique up to the
algebra
i
whi h have to be used in both dire tions.
Assume this is the redu tion pro ess we want to
P
"
i xi n 1 yi
i
with
xi ; yi
2 Kn 1 ( ; ). Of
normal form
arry out. We eventually obtain a sum
ourse this
for the word we started
with is not unique sin e we may perform again permutations of its letters in ea h term. But if
any two su h normal forms are equivalent under permutations of its letters (i.e. of the letters
75
vn-1
v2
v1
vn
e
Figure 5.2: The pentagon
i j with j i j j> 1) then we will get an almost
ondition
anoni al des ription of the basis of
Kn ( ; ).
Indeed the last assumption is equivalent to say that the surje tion of the previous Lemma is
an isomorphism. Unfortunately this is not the
to the uniqueness of this almost
ase. One
an however obtain the obstru tions
anoni al form as follows.
We return now to the module of admissible fun tionals. The last group of relations enables
us to make a further redu tion, namely
an 1 b ! z ab,
an 1 b ! t ab.
This way we
an redu e a word to a linear
that we are using a re urren e on
uniquely redu ed to an element of
su es to
Kn 1 ( ; ). Assume
of Kn 1 ( ; ) an be
ombination of words lying in
n. This means that ea h element
R (the value of the fun tional on
the element). Thus it
he k the obstru tions dire tly on the values in order to obtain that the fun tional
is well-dened. One formalizes this at follows. Let
be a semi-oriented graph. This means
that some of its edges are oriented and the remaining ones are unoriented. A path
a semi-oriented path if either
the
vj
v1 v2 :::vn is
! vj+1 or else vj vj+1 is unoriented, for all j . If all edges of
hain are unoriented we say that its endpoints are unoriented equivalents.
is an open pentagon onguration in (o.p. .) if v2 ! v1, v2v3:::vn 1
is an unoriented path and vn 1 ! vn.
pentagon ondition
Denition 5.3.2 veries the pentagon ondition (PC) if for any open pentagon onguration v1v2:::vn there exist semi-oriented paths v1x1x2:::xme and vny1y2:::ype having the same
endpoint (see Figure 5.2).
Denition 5.3.1 v1 v2 :::vn
We state rst the
Set now
for semi-oriented graphs:
x y if there exists a semi-oriented path from y to x in
a partial order relation. A ne essary and su ient
loops exist in
to
x.
. One says that
. Of
ourse
ondition is that no
x is minimal if y x implies that y
is not always
losed semi-oriented
is unoriented equivalent
Suppose that the (PC) holds. If a onne ted omponent C of the graph has
a minimal element mC then it is unique up to unoriented equivalen e.
Lemma 5.3.2
76
x
y
z1
u1
z2
u2
e
Figure 5.3: Proof of Pentagon Lemma
Proof:
Consider two minimal elements x and y whi h lie in C . Then there exists some path
xx0 x1 :::xn y joining them. Sin e x is minimal the losest oriented edge (if ever exists ) is ingoing,
and the same is true for y . If this path is not unoriented again from minimality there are at
least two oriented edges. Therefore open pentagon
ongurations (i.e. those
where (PC) applies) exist. We apply then (PC) iteratively whenever su h
ongurations exist
xz1 z2 :::zp e and yu1 u2 :::us e
or has appeared. When this pro ess stops we nd two semi-oriented
e. So e x and e y. Again from minimality these paths must be
having the same endpoint
unoriented so
ongurations
x and y are unoriented equivalent (see Figure 5.3).
A priori one annot say too mu h about the existen e of su h minimal elements. If had been a partial order with des ending hain ondition then the existen e of
minimal elements would be standard. However in the present ase even if is not a partial
order the existen e of minimal elements an be established.
Remark 5.3.1
5.3.2 The olored pentagon ondition: the denition of
Suppose now we have a sequen e of disjoint graphs
subset of verti
ea h
0
es Vn
onne ted
whi h are minimal elements in their
there exists a distinguished
onne ted
omponents. Suppose that
omponent admits at least one minimal element. Ea h su h vertex from
has exa tly one outgoing edge going to a vertex of
n
n . In every n
n
for the union of all
j,
n 1.
We
Vn0
olor these new edges in red. Set
j n and with the red edges added in ea
h rank
j.
is if any onne ted omponent of n has an unique minimal
element (with respe t to n) in 0 up to unoriented equivalen e.
olored
Denition 5.3.4 We say that n veries the olored pentagon ondition (CPC) if for any
open pentagon onguration v1v2:::vn in n there exist bi oloured semi-oriented paths (in n)
from v1 0and vn having the same endpoint. In addition
if xy is an unoriented edge in n with
x; y 2 Vn then there exist semi-oriented paths in n starting from x and y with red edges and
having the same endpoint (see the Figure 5.4).
Denition 5.3.3 n
oherent
We state now the
version of the Pentagon Lemma for this type of graphs.
77
v2
vn-1
0
1
11
00
1
0
1 00
0
11
1 0
0
1
1
0
111
000
111
000
1111
0000
111
000
1111
0000
111
000
v1 0
1
1
0
0
1
1
0
1 0
0
1
1
0
v
1
0
0n
1
Γ
1
0
Γn-1
1
0
1
0
Γ0
1
0
e
Figure 5.4: The
Lemma 5.3.3
is oherent.
Proof:
Suppose that
n 1
Vn0
0
Vn-1
V00
V00
olored Pentagon Condition
is oherent and the (CPC) ondition is fullled. Then
Z[ ; ; z; zb℄Fn ,
0
Z[
where
F
n.
n
Its verti es are the elements of the ring algebra
n letters
f1 ; 2; :::; n g.PThe verti es of
P
v = i i xi and w = i i yi , i ; i 2
by an oriented edge if exa tly one monomial of v is
is the free monoid in the
Z[ ;
Fn , are
will be the elements of
; ; z; zb℄ and xi ; yi 2
n
The proof is similar to that of Pentagon Lemma.
Now we are ready dene our graph
; z; zb℄.
related
Two verti es
hanged following one of the rules
aj3 b ! aEj b;
(C1)(j) aj +1 j j +1 b ! aj j +1 j b;
2
(C2)(j) aj +1 j j +1 b ! aSj b;
2 2
(C12)(j) aj +1 j j +1 b ! aCj b;
2
2
(C21)(j) aj +1 j j +1 b ! aDj b;
where Ej ; Sj ; Cj ; Dj as above. An unoriented edge between v and w
in a monomial of v of type
(Pij ) ai j b ! aj i b whenever j i
j j> 1.
(C0)(j)
orresponds to a
hange
Remark that the use of (C12) and (C21) is somewhat ambiguous sin e we may always use (C2)
for a subword. Their role is to break in some sense the
now the following sets of words in the
i 's:
losed oriented loops in
n.
Consider
W0 = f1g,
Wn+1 = Wn [ Wn n+1 Zn [ Wn n2 +1 Zn .
Zn = fni0 ni1 1 :::ni p ; i1 ; i2 ; :::; ip 2 f1; 2g; p = 0; n 1g.
0
b℄-module generated
Let Vn be the set of verti es orresponding to elements of the Z[ ; ; z; z
by Wn .
p
Lemma 5.3.4
unique.
Proof:
the
Ea h onne ted omponent of n has a minimal element in Vn0, not ne essarily
We prove our laim by indu tion on n. For n = 0 it is obvious. Let now w be a word in
i 's having only positive exponents. If its degree in n is zero or one we apply the indu tion
78
hypothesis and we are done. If the degree is 2 and it
apply the indu tion hypothesis. One
ontains the subword
n2
we are able to
an also suppose that no exponents greater than 2 o
ur
w = xn yn z with x; y; z 2 Fn 1 . The
"
indu tion applied to y implies that w xn an 1 bz with a; b 2 Fn 1 . Then several transforms
of type (Pnj ) and (C") will do the job. Consider now that the degree is stri tly greater than
by using (C0) several times. If the degree is 2 then
2. So we have a subword of type
n xn
with
3 +
4
or else one of the type
n xn yn .
The se ond
ase redu e to the rst one as above. Next say that
Pnj ) leads us to
x
an" 1b, a; b 2 Fn
n n" 1 n .
2.
" = 1 we apply two
times (C1) and we are done. Otherwise we shall apply (C
) and then (C1) if
6= or both
(C12) and (C21) and then (C1) if
= = 2. This proves that every vertex des ends to Vn0 .
Several appli ations of (
onsider the word
But these verti es have not outgoing edges as
If
an be easily seen. When we use the unoriented
edges some new verti es have to be added. But it is easy to see that these also does not have
outgoing edges. Sin e any vertex has a semi-oriented path ending in
Vn0 we are done.
The moves (C12) and (C21) are really ne essary for the on lusion of2 2the
previous Lemma to remain valid. For instan e look at the ase = = 0. From j+1 only (C2) an be applied and in the linear ombination we obtain the fa tor j2+1j2j+1j .j+1If
we ontinue, then at ea h stage we shall nd one of these two monomials.2 When
all possible
2
redu tions are performed at the se ond stage we re over the word j+1j j+1 so we have a
losed oriented loop in the graph. This onne ted omponent has no minimal element unless
we enlarge the graph by using the extra transformations (C12) and (C21). For general ;
a similar argument holds, and it an be he ked by a2 omputer
program. If one does not use
(C12) or (C21) then the redu tion pro ess for j+1j j2+1 yields at the sixth stage a sum of
words generating an oriented loop.
Remark 5.3.2
5.3.3 The bi oloured graph n (H ): the sub-module H
We are able now to dene the bi oloured graph
P
k
i;k (i;k) x(i;k) n y(i;k) ;
v=
w = i;k (i;k) uk x(i;k) y(i;k) , whi h is a vertex
Finally 0 (H ) is the graph having the verti es
Ea h minimal vertex
P
are
onne ted by an unoriented edge i the
R=H , H
being a
ertain submodule of
R.
of
n(H ). The red edges are dened as follows.
where k = 0; 1; 2; is joined by a red edge to
n 1 , where we set u0 = 1; u1 = z; u2 = t.
orresponding to the module
R and two verti
orresponding elements lie in the same
The submodule
H
es
oset of
is ne essary be ause going on
dierent des ending paths we might obtain dierent elements. Then, we have to nd whether
there exists
H
so that
We will test the
(H ) is
n
onditions of
n = 1; 2 there are no
oherent.
oheren e of ea h
onditions on
n
by re urren e on
n.
Noti e that for
H . Our strategy is to make use of the Colored Pentagon
Lemma in the following way. For those
dire tly we shall
(H )
ongurations that we
he k that the (CPC) (whi h is weaker sin e it
is veried.
annot prove the (PC) holds
on erns all the tower
n (H ))
[w0 ; w1 ; :::; wn ℄. This means
that w1 ! w0 , w1 ; :::; wn 1 are unoriented equivalent and wn 1 ! wn . We say that this o.p. .
is irredu ible if none of the verti es w1 ; w2 ; :::; wn 1 has an outgoing edge.
Consider an open pentagon
onguration (abbreviated o.p. . )
79
2
2
A σ j+1 σ j σ j+1 B
(C2)(j)
(C12)(j)
A σ j+1 S j B
A Cj B
?
Figure 5.5: The o.p. . for
Aj2+1 j2 j +1B
i) In order to verify (PC) it su es to restri t to irredu ible ongurations.
ii) It su es to verify (PC) only for monomials from F .
iii) Suppose wj0 = Awj B, for j = 0; : : : ; n (so A; B are nnot tou hed by any transform) in
the o.p. ..
If (PC) holds for [w0; w1; :::; wn℄ it also holds for [w00 ; w10 ; :::; wn0 ℄.
iv) Suppose that (PC) holds for [w0; w1; :::; wn ℄ and for [y0; y1; :::; ym ℄. Then for all A; B; C
the (PC) is valid also for
[Aw0 By1 C; Aw1 By1 C; :::; Awn 1 By1 C; Awn 1 By2C; Awn 1 By3C; :::
::; Awn 1 Bym 1 C; Awn 1 Bym C ℄.
In fa t when we x the endpoints of the o.p. . we an mix the unoriented edges of ea h
subja ent o.p. . in any order we want. Let (ik ; jk ) 2 f0; 1; :::; ng f0; 1; :::; mg; k = 1; p su h
that i0 = 0 < i1 i2 ::: ip; jp = m > jp 1 ::: 0; and ik+1 ik + jk+1 jk = 1 for all
k. Then the o.p. . Awi Byj C; Awi Byj C; :::; Awi Byj C fullls the (PC).
Proof:
Lemma 5.3.5
0
0
1
1
i) We may always de ompose a
p
p
onguration into irredu ible ones and iterate the
onstru tion.
ii) The redu tion transforms on dierent monomials
ommute with ea h other so we are
done.
iii) Obvious.
iv) The redu tions of
xn
1
Thus the top line of a o.p. .
letters giving in order
w0
1
of
B
or
w = Aw0 B . Then ea
ommute again with ea h other.
orresponds to a word
w2 ; w3 ; :::; wn
whi h fulll the following two
i) Set
y1
and
w = w1 has no proper subwords
onditions:
h of the
onsidered permutations a ts only on the letters of
w0 . Thus the transform w" of w0
is equivalent to
an study the (PC) for irredu ible
The rst step is to
and a sequen e of permutations of its
1 . We may suppose that
ii) The redu tion transforms performed at
Now we
w1
he k if the (PC)
w1
and
w0 .
w2 a
ts a tually on
tou hed by redu tions and on the subword
are as in Se tion 5.3.1.
w0
and
ongurations as in Lemma 5.3.5.
j2+1 j2 j +1
80
one
A,
w".
(n = 2) and
A and B are subwords not
an apply (Cij) or (C2). Cj and
ondition holds when the top line is trivial
there are two or more outgoing edges. For instan e, see Figure 5.5.
Sj
2
Lemma 5.3.6
Proof:
If the top line is trivial then the (PC) holds.
By Lemma 5.3.5 we have a nite number of
ab , where ab and b are subwords
2
j +1 j2 j +1 ; j +1j2 j2+1 g, j = 1; : : : ; n
form
ases to test. These are the words of the
belong to the set
2.
fj3+1; j+1j j+1; j+1j2 j+1;
The number of
ases to study
an be easily
redu ed, sin e
If b is the identity, the (PC) trivially holds.
By homogeneity of the redu tions (C ")(j) it su es to onsider j = 1.
For a word w = w1; : : : ; wl its symmetri is the word w = wl ; : : : ; w1 . If the (PC) holds
for
ab
, (PC) holds also for the symmetri
word
(ab ) (this result follows from the form
of redu tions).
Several
ases, as
The non trivial
A tually, we have to
1
2
j"+1
j2 j"+1
j3+1j j +1 ,
ases appear when a (Cij)-move (and then a (C2)-move)
he k only
"i = 2; 3)
(
an be easily tested at hand.
are
A; B
depi ted in Figure 5.5 (
j2+1 j2 j +1 , sin e j +1 j2 j2+1 is its symmetri
an be applied.
and the
ases
onsequen es of these ones. Then, we start from the situation
are empty words). If we apply (Cij) whether is possible on
j +1 Sj , after a long and messy
Cj .
omputation we nd the same minimal element asso iated to
Remark 5.3.3 Using a omputer program one an get the oriented graph asso iated to the
word j2+1j2j+1 (Figure 5.6). The verti es are of the type P j wj , j polynomials in ; and
wj words in j ; j +1. An oriented edge between an outgoing vertex a and an ingoing vertex
b indi ates that the redu tion pro edure applied to a yields b. When there are no subwords
j2+1 j2 j +1 or j +1 j2 j2+1 the edges are spotted. As we already noti ed in Remark 5.3.2, if we
apply six times the pro edure without (Cij) we nd a loop.
Let us study the
w1 and wn
1
applied, we
ase when the top line is non trivial. By Lemma 5.3.6 we
an suppose that
have ea h one exa t one outgoing edge. In parti ular, when a (Cij)-move
hoose always the edge (Cij) in the redu tion pro ess.
Now the top line is determined by the sequen e of transpositions of the letters of
an be
w. Let l be
w. Otherwise this is the same to giving a permutation 2 Sl with a pres ribed
Tj for the transposition whi h inter hanges the letters
on the positions j and j + 1. Noti e that for a xed w not all are suitable. In fa t only a
subset of the group of permutations, whi h we all permitted, may work. Say P (w ) is the set
of permitted permutations. If ew : f1; 2; :::; l g ! f1; 2; :::; n
1g is the evaluation map
ew (j ) =index of the letter lying in position j on w
then Tj is permitted (where 2 P (w )) i
j e(w) (j ) e(w) (j + 1) j> 1.
0
Say that two permitted permutations and are equivalent if for the o.p. . orresponding
0
to and the (PC) is valid or not for both in same time.
the length of
de omposition into transpositions. Set
i) Suppose that 1Tj Ti2 2 P (w), j i
these two permutations are equivalent.
Lemma 5.3.7
81
j j> 1.
Then 1TiTj 2 2 P (w) and
2
σ j+1
σ 2j σ j+1
C2
Cij
C2
Cij
C2
Cij
Cij
C2
Cij
C2
CijC2
C2
C2
C2 Cij
Cij
C2
Cij
Cij
C2
Unique minimal element in Γj+1
C2 C2 Cij
Cij
Cij
C2 Cij Cij
Cij
C2
C2
C2
C2
Cij
Cij
C2
Figure 5.6: The graph underlying to
j2+1 j2 j +1
ii) Suppose that 1Ti+1TiTi+12 2 P (w). Then 1TiTi+1Ti2 2 P (w) and these two permutations are2equivalent. The onverse is still true.
iii) If 1Ti 2 2 P (w) then 12 is permitted and equivalent to the previous one.
Proof:
j e (w) (j ) e (w)(j + 1) j> 1
The existen e in the rst
ase is equivalent to
and
j e(w) (i) e(w) (i + 1) j> 1, so it is symmetri . In the se ond ase also it is equivalent to
j e (w) (j + "1) e (w)(j + "2 ) j> 1 for all "j 2 f0; 1; 2g, so it is again symmetri . The
equivalen e is trivial.
2
2
2
2
One uses a graphi al representation for the de omposition of
into transpositions similar
to the braid pi tures (see Figure 5.7), where we spe ify on the top and bottom lines of the
diagram the values of the evaluation maps.
This gure en odes all information about the o.p. . be ause the two words
w and (w) have
unique redu tion. For the moment one draws only those traje tories of the six (to ten) elements whi h enter in the two blo ks whi h redu es. Suppose for instan e that the two redu tion
w = xiiiy and (w) = x0 jjjy0 . Say that i = j . The traje tories of the
may be disjointed sin e the transposition a ting on the ouple ii is trivial in fa t. So the
moves are two (C0). So
i0 s
possible traje tories t into 4
ases whi h may be seen in Figure 5.8.a,b, ,d.
Suppose now we have two traje tories of
that
i and j 6= i whi
h interse ts. First of way we derive
j i j j> 1. Orient all the ar s from the top to the bottom.
i) Suppose that the ar s labeled i and j have algebrai interse tion number 0.
Then we an repla e the diagram by an equivalent one where the ar s are disjoint.
Lemma 5.3.8
82
e w (j); j=1, 2, ..., n -1
e
σ (w)
k
(j); j= 1, 2, ..., n -1
Figure 5.7: The
k
omplete diagram asso iated to an o.p. .
i i i
a
c
i i i
i i i
i i i
i i i
i i i
b
i i i
d
i i i
Figure 5.8: The essential traje tories for (C0)(i)-(C0)(i)
83
i
j
i
j
i
j
i
j
Figure 5.9: Disjointing traje tories
Figure 5.10: Non minimal biangle pro edure
ii) Suppose that the ar s labeled i and j have algebrai interse tion number 1. Then we an
repla e the diagram by an equivalent one where the ar s have exa tly one interse tion point.
Proof:
We
We
onsider the diagram is that from Figure 5.9.
an assume that the biangle in the middle is minimal, hen e it does not
biangle. In fa t we
an apply repeatedly the disjointedness pro edure only for minimal bian-
gles. Su h biangle have two walls: one
no other ar
Let
ontain any other
oming from
i and the other from j . From minimality
ross twi e the same wall (see Figure 5.10).
L and R su h that: the set of ar s labeled by something not ommuting
ontained in L, and those labeled by some k not ommuting with i are ontained in
onsider the region
with j is
R. Then the situation is that from Figure 5.11.
Thus all ar s whi h
j . The same
ross the biangle are labeled by some
k whi
h
i and
i and j
ommutes with both
ommutation transforms may be performed whenever we make the ar s
disjoint.
A similar reasoning permits to say that the diagrams from Figure 5.12 are equivalent.
k
i
j
L
R
i
j
k
Figure 5.11: The regions
84
R and L
i
j
k
i
k i j
j
k
k i j
Figure 5.12: Equivalent diagrams
j j j
i i i
j j j
i i i
Figure 5.13: The diagram for (C0)(i)-(C0)(j) when
When the triangle in the middle is not tou hed by any ar
lemma 5.3.8 ii). If it is minimal, any ar
with
j.
whi h
j i j j> 1
then it is a simple
onsequen e of
ross it is labeled by something whi h
ommutes
Remark now the similitude of Figures 5.9 and 5.12 with the Reidemester's moves on link
diagrams. So we
tangent (in a
Now we
an a tually isotopy our ar s leaving the endpoints xed and keeping the
C 1 -approximation of ar
an
s) away from the horizontal.
ontinue our dis ussion on the traje tories of
i0 s
and
j 0 s.
If
j i j j= 1 the
traje tories are disjoint so there are as in Figure 5.13.
If
i and j
ommute then there are essentially sixteen diagrams (up to isotopy) whi h
an be
seen in Figure 5.14.
In order to represent graphi ally the possible diagrams for the (C1), (C2), (C12), (C21) moves
we shall gure the traje tories of a
ouple of neighbor points having the same label as a single
thi ker traje tory. This may be done sin e every ar
5.15) between the traje tories of the the two
traje tories of
i and i + 1 are disjoint.
rossing the dashed region (see Figure
i0 s has a label
i
ommuting with . In addition the
j 6= i 1; i; i + 1; i + 2 the twelve diagrams
from above appear appropriately labeled. For j = i
1; i; i + 2 some diagrams are not realized
be ause the ar s labeled by i
1 and i does not interse t, so several ases have to be left. For
j = i + 1 another diagram have to be onsidered, that from Figure 5.16.
Suppose we are in the
ase (C1)(i)-(C0)(j). For
The same situation we en ounter when we des ribe the possible traje tories for the
ouple of
redu tion transforms (C2)-(C0), (C12)-(C0), (C21)-(C0). A simple analysis shows that in the
remaining
ases the only new diagrams are those from Figure 5.17.
The other ones are obtained from the previous twelve using the suitable labeling, and taking
into a
ount the
onstraints of disjointedness imposed by the labels. We say now that a
85
i i i
j j j
j j
i i i
j j j
j j j
j j
j
i i i
i i i
j j j
i i i
i i i
j j j
i i i
i i j j j
i
j
i i i
j j
j j
j
j
i i i
i i
j j j i
i i i
j j j
i i i
j j j
j
j j j
i i i
j j j
i i i
j
i i i
j
i i j j j i
ii
j i i i
j j j
i i i
i i i
j j j
j j j
j j
i i i
j j j
j j
i j j j i i
j j j
i i i
i i i
j j j i
i
j j j i i
i i i
j j j
i j j j i i
Figure 5.14: The 16 diagrams for (C0)(i) -(C0)(j) in the
i i i
i i
j j j
i i i
j j j i i
ommuting
ii
i
i
Figure 5.15: The graphi al representation of the dashed region
i +1 i i +1
i +1 i +1 i +1
Figure 5.16: The new diagram for (C1)(i)-(C0)(i+1)
86
j j
ase
i+1 i i+1 i i+1
i+1 i
i+1 i i+1
i+1 i i+1
i+1 i
i i+1
i+1 i i+1
i+1 i i+1
i+1 i i+1
Figure 5.17: The new diagrams for (Cx)(i)-(Cy)(i)
x; y 6= 0
Σj
α λ
j
i i i
j
v
u
i
Σi
j
i
j j j
i
Figure 5.18: The whole gure of a non-intera tive diagram without
intera tive
diagram is
if there is some marked ar
rossings
relating the top and bottom blo ks where
the redu tion transforms a t. Our task will be to eliminate the non-intera tive diagrams where
the (PC) trivially holds.
Lemma 5.3.9
Proof:
We
The usual (PC) is valid in n for non-intera tive diagrams.
onsider rst the
ase where no
rossings of the essential ar s exist. The typi al
ase is that from Figure 5.13. We draw now all traje tories as in Figure 5.18. We have the
dashed regions
Everything
that
U
U
and
V
whi h are bounded by the
rossing the regions
and
V
U
and
V
i0 s ar
ommutes with
s and respe tively
i and j
j 0 s ar
respe tively. We
s.
laim rst
are tangent to the end lines from left and right respe tively. If not there exists
lying to the left of U . Assume that this ar is the rst from the left having
this property. In parti ular ommutes with every label
whi h stands to the left of . Thus
we may perform these ommutation transforms at any moment, to get on the rst position.
Sin e does not ross U we may leave it on the the rst position repla ing the o.p. . by an
some ar
labeled
equivalent one. Thus the new
onguration
orresponds to a word whi h is not minimal with
respe t to the redu tion pro edure (see Lemma 5.3.5 and the subsequent
Let now
i
be the
onvex hull of the three points labeled
lying on the bottom line. Similarly set
ar
whi h arrive on
i
must
ross
move these endpoints using the
U
j
for the
i
omments).
oming from essential ar s and
j 0 s on the top line. Every
some k ommuting with i. We an
onvex hull of the
hen e is labeled by
ommutation rules from the left or the right a
following prin iple: if the start point of the ar
labeled
k
ording to the
is in the left of the blo k of
i0 's on
the top line, then we move to the left. Otherwise we move to the right. The only problem
whi h we
labeled
an have is in the following
ase: the start point of some
l, both arrive on i, but this time the endpoint of l
argument shows that these two ar s
ross ea h other. Therefore
87
k
is in the left of the ar
is in the left of
k
and
l
are
k. A topologi
al
ommuting and
Σj
α λ
j
i i i
j
v
u
i
j
Σi
i
i
j j j
k
l
Figure 5.19: The simpli ation of a non-intera tive diagram without
rossings of essential ar s
i i i
j j j
Figure 5.20: The standard non-intera tive diagram
we
an perform our transforms as it was said (see Figure 5.19).
Finally we re over a diagram whi h this time has
rossings but is equivalent to the standard
one of Figure 5.20.
Without loss of generality we
an set
= = 0 in the redu
the notation. Suppose now that the redu tion transforms
tion transforms in order to simplify
AiiiB
! AB and CjjjD ! CD
are also performed. We may use the simpli ation transforms ( ommutations whi h are still
j are ollapsed) for above for ea h word: to AB in the part of j 0 s
0
and to CD in the part of i s. Due to the parti ular form of the standard diagram we shall get
0
0
0 0
(see the Figure 5.18) the words UjjjV and U iiiV respe tively, with UV = U V . So again
valid even if the
i
or the
the use of a redu tion transform will get the same word. Thus the (PC) is satised for these
ongurations. It is almost the same reasoning for the other non-intera tive diagrams without
rossings.
It remains the
ase when
rossings of essential ar s appear. But the
ommutation transforms
j 0 s on the top line will be all
on the same part with respe t to the iii blo k. In other words we make j and the blo k iii
disjoint. The same is true for the bottom line. The worst ase is again when iii is in the left
of j on the top line and down the situation is reversed. But again i and j ommutes with
everything starts or arrives on the onvex hulls of iii [ j and jjj [ i . So we an rearrange
may be also be performed in su h way that the starting points of
them to obtain the same order in the top and bottom lines. This ends the proof of the Lemma.
88
i i i
i
i i i
i
a
b
i i i
c
i i i
i i
i i
i +1 i i +1
i +1 i
i i i
i i +1
i +1 i i +1
e
i +1 i i +1
i +1 i i +1 i +1 i +1
i i i
d
i +1 i +1i +1
i +1 i
f
Figure 5.21: The normal forms of intera tive
So it remains to look at the intera tive
i +1 i i +1
ongurations
ongurations. The same reasoning permits us to
restri t to the normal forms drawn in Figure 5.21.a-f.
Some of the traje tories may be thi k traje tories. The
ases
a,b, ,d and f
are trivially veried
K3 ( ; ) is involved.
"
w=
i+1 i" i+1 xi2+1 whi h is unoriented equivalent to w0 = i+1 i" xi3+1 . Here x ommutes with
i+1 and so we may suppose it lies in Fi 1 . Therefore x ! x0 ij1 1 ij2 2 :::ij p , with x0 2 Fi 2 .
So again we an restri t to the ase x0 = 1. Consider the ase " = 2 (the others are trivial).
j
j2
Set q = i 2 :::i p . We have the following situation
be ause only the
Let us
onsisten y of relations dening
he k a sub ase of
d,
orresponding to (C )-C(0). The monomial has the form
p
p
w0
w
=
Z
Z
~
Z
Sj ij1 1 i2+1 q
where
Sj ; Ej
i+1 i2 Ej ij1 1 q
as above. From Lemmas 3.5 and 3.6 it follows that (PC) holds for
i+1 i2 i3+1 ij1 1 q
=
Z
Z
~
Z
Sj i2+1 ij1 1 q
Sin e
i+1 i2 Ej ij1 1 q
Sj ij1 1 i2+1 q is unoriented equivalent to Sj i2+1 ij1 1 q we have done. All other
e are similar.
In the
ase
e
the situation is dierent. Using the
ases but
ommutation rules, as above we must
ij1 1 . So we must he k the ongurations
w = xi+1 i" i+1 i 1 iÆ i+1 ij2 2 :::ij p ,
where x 2 Fi 1 . At this point one annot prove that the (PC)
preserve the term
p
holds. In fa t it does not
hold sin e the surje tion of Lemma 5.3.1 has a nontrivial kernel in rank
89
n = 3.
Fortunately
we proved that the
ongurations that don't verify (CPC)
obstru tions. Therefore one
an dene
H
as the ideal
ome from a nite number of
ontaining all these obstru tions, and
see whether it is nontrivial.
Lemma 5.3.10
Proof:
It su es to onsider x = 1; p = 1.
One observes that any admissible fun tional T on K1 ( ; ) satises:
T (xuv) = T (u)T (xv) for x; v 2 H (Q; m) and u 2< 1; m ; m+1 ; :::; m+k >.
In fa t for k = 0 this is the multipli ativity of T . For k > 0 then in the redu tion pro ess
" where T (u) = T ( " ). When redu ing again one derives T (xuv ) =
one repla es u by
m
m
"
T (m )T (xv).
Further the (CPC) is equivalent to the existen e of an admissible fun tional.
We have therefore to
he k the o.p. .
3 2 1 3 2Æ 3
and
orresponding to following
ouples
3 2 3 1 2Æ 3 ; ; ; ; Æ; = 1 or 2
Then the only possible obstru tions to the existen e of Markov tra e
ome out from these
ouples. In Se tion 5.5 we study these obstru tions and we nd the ideal
H
in
R
ontaining
them.
5.4 The omputation of obstru tions
5.4.1 Commutativity obstru tions
We are now
on erned with the
At the rst stage (i.e.
ommutativity
onstraints:
T (ab) = T (ba)
for all
a; b:
K3 ( ; )) we obtain the identities
T (2 12 2 ) = T (12 22 ); T (1 2 12 2 ) = T (2 1 2 12 ):
Thus the following equations should be satised:
T (R0 ) = T (R1 ) = 0:
i.e.
( 3 +3 +4)t2 +(3 2 7 2 6 +2 4 )t+ (3 2 5 2 3 2 +4 3 )+ (2 3 + 2 6 2
2 + 2 )z 2 = ( 2
10 )zt + ( 3 3 + 7 2 2 + 9 + 4 3 2 4 )z + (3 3 + 7 2 2 3
2
3
4
2
2
2
2
2
2 )t +(4+5
2 )t+ ( 2 3 + )+(2 +5
2 )zt+( +2 3 5 2 6 )z +
(4 + 2 2 +
2 3 )z 2 = 0
These yield the following values for the parameters:
either
z=
2+2
+4
; t=
2+2
+4
;
or else
t=
2 z
2z 2 +
;
2+ z
where
z veries (
90
+ 1)z 3 + ( +
2 )z 2 + 2
z + 1 = 0:
One
he ks then the
ommutativity
onstraints by indu tion on
n. It su
es to
onsider
b 2 f1 ; :::; n g and a lying in a system of generators of Kn+1 ( ; ), let us say Wn (Se tion
3.2). For b = i , i < n it is obvious. It remains to he k whenever T (an ) = T (n a). We have
three
ases
a 2 Kn ( ; ).
ii) a = xn y , x; y 2 Kn ( ; ).
2
iii) a = xn y , x; y 2 Kn ( ; ).
i)
whi h will be dis ussed in
ombination with the six sub ases
x 2 Kn 1 ( ; ), and y 2 Kn 1 ( ; ).
2) x 2 Kn 1 ( ; ), and y = un 1 v , u; v 2 Kn 1 ( ; ).
2
3) x 2 Kn 1 ( ; ), and y = un 1 v , u; v 2 Kn 1 ( ; ).
4) x = rn 1 s, r; s 2 Kn 1 ( ; ), y = un 1 v , u; v 2 Kn 1 ( ; ).
2
5) x = rn 1 s, r; s 2 Kn 1 ( ; ), y = un 1 v , u; v 2 Kn 1 ( ; ).
2
2
6) x = rn 1 s, r; s 2 Kn 1 ( ; ), y = un 1 v , u; v 2 Kn 1 ( ; ).
1)
Now (*,i), (1,ii) and (1,iii) are trivial.
T (nxnun 1v) = tzT (xuv) = T (xnun 1vn).
T (nxn2 un 1v) = ( t + z + 1)T (xun 1v) = ( t + z + 1)zT (xuv)
= T (xun 1 n n2 1 v) = T (xn2 un 1 vn ).
2
2
2 2
2
(3,ii) T (n xn un 1 v ) = t T (xuv ) = T (n n 1 )T (xuv ) = T (n n 1 n )T (xuv ) =
= T (xun n2 1 n v) = T (xn un2 1 vn ).
2 2
2
(3,iii) T (n xn un 1 v ) = ( t + z + 1)T (xun 1 v ) = ( t + z + 1)tT (xuv )
2
2
2
2
= T (xuv)T (n n 1 n ) = T (xn un 1 vn ).
(2,ii)
(2,iii)
For the other
ases, we need also to know the form of
su.
Set
su = pn" 2 q
with
p; q
2
Kn 2 ( ; ) where " = 0; 1 or 2. We an show by a dire t omputation that the equalities hold
also for (4; ii); (4; iii); (6; ii); and (6; iii). Using Maple we have found that in the ases (5; ii)
2
and (5; iii) for su = pn 2 q there are only two new equations, whi h are not onsequen es
of the identities T (R0 ) = T (R1 ) = 0). Spe i ally we have three obstru tions in ea h ase,
2
namely the polynomial oe ients of T (rpn 2 qv ); T (rpn 2 qv ) and T (rpqv ).
from
(5; ii) we have
T (rpn2 2qv) yields the equation L := 3
4 +5 2 5 2 +2 4
7 + 3 + (13 3 2 10 2 4 + 13 2
7 3 3 7 2 2
6 3 2 4 +3 +
6
3
5
2
2
2
3
2
4
2
2 )t+( 6
6 +3 +5
)t + ( 16
5 2 2 2 +3 5 +2 5
3
3
4
2
6
4
4
2
5
13 +11
2
)z +( 2 +15 +2
11 3 3 +15 3 +6 )zt+
3 5+6 4 3 3 3 2+2 2 4 9 5
( 3
9 2
10 4 )z 2 = 0;
the
oe ient of
T (rpn 2qv) vanishes is equivalent to M :=
4 +6 2
2 5
2 3 +7 4 3 + 11 3 2 + 6 7 2 4 5 3 5 + 2 7 +( 21 3
2 2 6 +2 2 +
2
3
4
2
2
3
4
5
5
2
2
4
14
13
7 +10
2
+2 )t + ( 7
+6 + 10 3 + 4 +
2 5 5 3 3 )t2 +( 3 6 +2 3 6 +5 +11 2 2 +16 5 2 +8 3 +25 4
11 4 4
4 4 10 3 3 )z +(11 4 3 14 2 +10 3 2 +4 3 15 5 27 4 2 3 5 )zt
+(4 2 4 2 3 + 4 5 +19 5 3 4 +4 2 3 4 2 +21 3 6 5 3 +9 6 )z 2 = 0;
the
oe ient of
91
the
from
T (rpqv) from whi h one derives N := 12
2 3+
8 6 2 6 2
+3 2 + 11 3 4 4 5
6 4 2 7 3 + ( 21 3 3 + 7 4 + 5 3 +10 4
7
2
2
2
2
2
17
+ 12 5 )t +( 4 4 + 10 3 2 3 + 6 + 5 2 6 2 4
3 3 )t2 +(3 + 3 3 + 2 2 7 + 16 3 2 2 6 7 4 13 5 +5 2
13 3
4
3
2
3
5
3
4
2
3
2
6
5
+25
)z + (
12
+ 10 + 13
2
+2
24 4
2
3
3
3
5
2
4
2
2
3
6
2
5
5 )zt + (5 + 4
+ 14
+8
+7
+
+5
2
6
4
4
2
7
)z = 0:
oe ient of
(5; iii) one obtains the obstru
the
the
the
T (rpn2 2qv) yields L = 0;
oe ient of T (rpn 2 qv ) yields
M = 0;
oe ient of T (rpqv ) yields
N = 0:
oe ient of
As pointed out in Se tion 5.3 the
oheren e of
(H ) depends on the following
n
ouples:
3 2 1 3 2Æ 3 et 3 2 3 1 2Æ 3 ; ; ; ; Æ; = 1 or 2
a word w = w1 ; : : : ; wl its symmetri is the word w = wl ; : : : ; w1 .
T (w) = T (w ) holds one
we must
5
2
6
tions:
5.4.2 The CPC obstru tions for n=4
Re all that for
2
an redu e ourselves to the study of 24
ouples. The
Sin e
ouples that
he k are the following:
(1:i) : 3 2 Pi 22 3
and
(2:i) : 3 2 Pi 2 32
and
(3:i) : 3 22 Pi 2 32
and
3 2 Pi0 22 3 ;
3 2 Pi0 2 32 ;
3 22 Pi0 2 32 ;
(4:i) : 32 22 Pi 22 3 and 32 22 Pi0 22 3 ;
(5:i) : 32 2 Pi 22 32 and 32 2 Pi0 22 32 ;
(6:i) : 32 22 Pi 2 3 and 32 22 Pi0 2 3 ;
P1 = 1 3 ; P2 = 12 3 ; P3 = 1 32 ; P4 = 12 32 ; P10 = 3 1 ; P20 = 3 12 ; P30 =
32 1 ; P40 = 32 12 :
From now on we denote the orresponding ouples by the respe tive label (i; j ). For general
; the omputation is very long and and we needed a omputer program. For more
where
information about the
ode, see Remark 5.6.2.
15 dierent obstru tions from these CPC obstru tions, and the following identities among the obstru tions: (5:2) =
(3:2); (6:2) = (1:2); (1:4) = (1:2). Thus, we must
onsider the ouples (1; 2); (2; 4); (3; 2); (3; 3); (3; 4); (4; 1); (4; 2); (4; 3); (4; 4); (5; 3); (5; 4);
(6; 4).
One nds
The exa t form of the obstru tions will be given in the next Se tion.
92
5.5 The existen e of Markov tra es
5.5.1 Statements
Theorem 5.5.1
There exists an unique Markov tra e
T(
; )
with6 parameters
z = (2
5 2
4 4
4
8
8
Z[
and
; ; (4 + ) 1 ℄
(H( ; ) )
2 )=(
+ 4)
zb = ( 2 + 2
3
3
3
34
+17 +8 2 5 + 32 2 2
+36
Æ = z 2 ( z + 1), so that the obstru
ome Laurent polynomials in z and Æ .
It is
be
+2
: K ( ; ) !
onvenient now to put
)=(
36
+ 4)
4 +38
Set = (z7 + Æ2 )=(z4Æ), = (Æ z2)=z3 and zb =
There exists an unique Markov tra e with parameters (z; zb)
where P(z; Æ) = z23 + z18Æ
2z 16 Æ2
z 14 Æ3
; )
; ) :=
17 3 +8:
+8
tions below in the se ond
Theorem 5.5.2
T (z; Æ) : K(
, where6 H(
ase
z 2 =( z + 1) = z 4 =Æ.
1 1
! Z[z (z;; ÆÆ) ℄
P
(
)
2z 9 Æ4 + 2z 7 Æ5 + Æ6 z 5 + Æ7
.
5.5.2 Proof of Theorem 5.5.1
The parameters
z; t have to satisfy the
ondition
T (R0 ) = T (R1 ) = 0
Consider rst the simple solutions
T(
; )
z = (2
2 )=(
+ 4) and t = ( 2 + 2 )=(
for the admissible fun tional asso iated to these values of the parameters. Noti e that
in this
ase
zb = t. Set u := 1=(
+ 4), z0 := 2
2
The ommutativity obstru tions
The equations en ountered above for
(5; ii) amount to
u2 H( ; ) = 0,
u2( + 2)H( ; ) = 0,
2 )H
u2 (
( ; ) = 0:
CPC obstru tions
+ 4). We set
(1.2):
u3 (
2 )H
( ; ) W;
(2.4):
u3 (
2 )( 2 +
)H(
(3.2):
u3 (
2 2+2+
+
(3.3):
u3 (
+ 2)H(
; ) W;
3 )H
( ; ) W;
; ) W;
93
and
t0 :=
2+2
=: zb0 .
(3.4):
u3
2 )H
( ; ) W;
(
(4.1):
u3 (
(4.2):
u3 (
3+2+2
(4.3):
u3 (
3
(4.4):
trivial,
2
3 2
2
2
W =( +2
)(
2
3 )H
( ; ) W;
; ) W;
2 + 4 )H
)H(
( ; ) W;
u3 ( + 2 2 )(
(6.4):
where
2 2
u3 (
; ) W;
2 2
3 )H
)H(
( ; ) W;
u3 ( 2 + 2 + 2
(5.3):
(5.4):
2 )( 2 +
2 )H
( ; ) W;
2 +4+
+2 +
2) = 3 + 8
3+6
.
5.5.3 Proof of Theorem 5.5.2
There are three more solutions of
(
+ 1)z 3 + (
+
2 )z 2 + 2
rational fun tions on
z and
T (R0 ) = T (R1 ) = 0, given by t = 2
z + 1 = 0. In this
.
The ommutativity obstru tions
ZB1=(z7 (z + 1)4 ) = 0;
ZB2=(z9 (z + 1)5 ) = 0;
ZB3=(z7 (z + 1)5 ) = 0:
The CPC obstru tions
(1.2):
ZB4B5 B6 =(z 13 (z + 1)8 );
(2.4):
ZB4B6 B7 =(z 15 (z + 1)9 );
(3.2):
ZB4 B8 =(z 15 (z + 1)9 );
(3.3):
ZB4B9 =(z 11 (z + 1)7 );
(3.4):
ZB4 B5 B6 =(z 13 (z + 1)8 );
(4.1):
ZB4 B6 B7 =(z 15 (z + 1)9 );
(4.2):
ZB4 B5 B10 =(z 17 (z + 1)10 );
(4.3):
ZB4 B5 B11 =(z 17 (z + 1)10 );
(4.4):
trivial,
(5.3):
ZB4B12 =(z 13 (z + 1)8 );
(5.4):
ZB4B5 B13 =(z 19 (z + 1)11 );
94
z 2z 2 +
2+ z
, where
z veries
ase the obstru tions are better expressed as
ZB4B5 B6 B14 =(z 17 (z + 1)10 );
(6.4):
where
Z; B1 ; : : : ; B14
are the following polynomials in
z;
:
Z = 1 + 7z + 21z2 2 + z3 + 35z3 3 + 35z4 4 + 21z5 5 + 7z6 6 + z7 7 + z9 6 + 8z8 5 +
23z 7
4 + 32z 6 3 + 23z 5 2 + 8z 4
B1 = 3z 3 + z 4 + 1 + z ;
B5 = 1 + z 3 + z 2
B2 = 5z 3 + 10z 4 + 6z 5
B3 = + 2z
2 + 4z 3
2z 6 + z 9
2 + z 6 3 + 4z 6 + 2z 7
+ 5z 4
2 + z5 3 + z2 3
B4 = (z + z 2 + 1 + z z 2 )(z + 1 + 2z 3 )(z 4
2z 2 + 3z 3 + z 3 + z 4 + z 4 );
B6 = z 3
2 + 2z
3 + 1 + 2z
z9
3
5z 8
2
6z 7 ;
+ 1 + 3z + 3z 2
2 + z3 3;
2z 5 ;
2
z3
2 + z 2 2 + 1 + 2z
;
+ 2z 2
2 + z3;
B7 = 1 + 4z + 6z 2
2 + 2z 3 + 4z 3 3 + z 4 4 + z 6 3 + 4z 5 2 + 5z 4
B8 = z 2
2
3+
+ 2z
z 2z 2 +
2z 2
+ z 6 );
z3 ;
B9 = 1 + 6z + 16z 2 2 + 3z 3 + 25z 3 3 + 25z 4 4 + 16z 5
13z 7 4 + 24z 6 3 + 24z 5 2 + 13z 4 + z 7 + z 6 + z 9 ;
5
+ 6z 6
6
+ z7
7
+ 3z 8
5
+
B10 = 1 + 6z + 16z2 2 + 3z3 + 25z3 3 + 25z4 4 + 16z5 5 + 6z6 6 + z7 7 + z9 6 +
7z 8
5 + 20z 7 4 + 31z 6 3 + 28z 5 2 + 14z 4
+ z6 + z9 + z9
3 + 2z 8 2 + 2z 7
;
B11 = 6z + 16z2 2 + 3z3 + 10z8 2 + 5z8 5 + z7 7 + z9 6 + 12z7 + 12z7 4 + 19z6 3 +
20z 5
2 + 12z 4
+ 6z 6
6 + 3z 9 3 + 5z 6 + z 9 + 1 + 25z 3 3 + 25z 4 4 + 16z 5 5 ;
B12 = 2 + 4z5 3 2z5 + 2z4 5 + 8z 2 + 12z2 3 2z2 + 8z3 4 + 3z4 2 2z3 + z6 4 ;
B13 = 1 + 8z + 29z2 2 + 63z3 3 + 80z6 3 + 29z7 7 + 13z9 6 + 17z9 3 + 91z4 4 +
57z 5 2 + 23z 4 + 4z 3 + 6z 6 + 4z 9 + 91z 5
22z 7 + z 12 + z 9 9 z 12 6 + z 10 4 + 2z 10
5 + 63z 6 6
7 + 8z 8 8
+ 39z 8 5 + 70z 7 4 + 30z 8
3z 11 5 + 3z 11 2 + 7z 10 ;
B14 = 2 + 8z + 12z2 2 + 4z3 + 8z3 3 + 2z4 4 + z6 3 + 6z5 2 + 9z4 + 2z6 :
Noti e that
Z (z; ) = P (z; Æ) (z; Æ).
5.5.4 Corollaries
Corollary 5.5.1
There exists an unique Markov tra e
with parameters z =
T : K(0; 2) ! (86 Z17[℄ 3 + 1) ;
2 , t = and zb = ,
95
2
+
respe tively
T : K ( 2; 0) ! (86 Z17[℄ 3 + 1) ;
with parameters z = , t = 2 and zb =
2 .
We have a similar situation for the other three solutions. In fa t for
)2 ,
(t
be ause
where
zb = t
t satises
.
Corollary 5.5.2
(t3
t2
4
+5
2t + 1
2
3)
= 0. In parti
= 0, we derive z =
b3
ular z
zb2 + 1 = 0
There exists an unique Markov tra e
with parameters z =
and respe tively
1
3
T : K (0; +2 1 ) ! (9 2Z[6+ ℄ 3 + 1) ;
23 + 1
2
2 , zb = and t =
,
3
1
T : K( +2 1 ; 0) ! (9 2Z[6+ ℄ 3 + 1) ;
with parameters z = , zb =
2
23 + 1
2
and t =
.
5.6 The invariants
5.6.1 The denition of I(
; )
2 )=(
z = (2
+ 4), t = ( 2 + 2 )=( + 4), u := 1=(
2
z0 := 2
and t0 :=
+ 2 =: zb0 (noti e that in this ase zb = t).
Denition 5.6.1 Let us set for an oriented link L
As in Se tion 5.2 we set
2
I(
;
1
) (L) = z zb
n
2
1
e(x)
zb 2
z
T(
; ; z0=2 ; zb0=2 ℄
;
(H( ; ) )
to L. Here 1 is the number of
; ) (x) 2
where x 2 Bn is any braid whose losure is isotopi
mod 2.
Lemma 5.6.1 I( ; ) is well-dened.
Proof: j 1 = j2 j
+ 4),
Z[
omponents
, we an suppose that x is a positive braid. All the elements
x are polynomials in the variables z; t of degree at most n 1. The
0
substitutions z = uz0 and t = ut0 imply that, if T( ; ) (x) and T( ; ) (x) are representatives of
0 T( ; ) (x) = un 1 G( ; )H( ; ) , where G( ; ) is a polynomial
the tra e of x, then T( ; ) (x)
in ; . It follows
( )
1
1 2 zb0 2 e
I( ; ) (L) =
T( ; )(x);
z0 zb0
z0
Sin e
in
0 (H )
asso iated to
e x
n
where
Te(
; ) (x) := u
n+1 T
( ; ) (x)
96
2 (ZH[ ; ℄) :
( ; )
5.6.2 The ubi al behaviour
Proposition 5.6.1
For any link K there exists some l 2 f0; 1; 2g su h that
I(
; ) (K ) =
P
where Pk ; Qk ; Mk ; Nk are (3; k + l)-polynomials.
Proof:
Mk ; Nk
(3; k + l)
+
+
We will show that
k
k
Pk ( )
)
P k2N
k2N Qk (
are
=
P
Mk ( )
)
Pk2N
k2N Nk (
k
k
-polynomials. The other
ase is analogous.
x 2 Bn , where Bn is the set of positive braids and n is su h that x 2= Bn+ 1 .
Then e(x) = jxj where jxj means the length of x. In the pro ess omputing the value of
the tra e on the word x we make two types of redu tions: either one uses the relations in
2
some Kn ( ; ), or else one repla es al b by zab (respe tively al b by tab), where a; b are
subwords, and this way one lowers the rank n. Using the relations the word y is repla ed by
P P
k
s ( k2N Dk;s ( ) )ys where the ys are a nite number of words in Bn and the oe ients
Dk;s( ) are (3; k + e(x) ls )-polynomials where ls = jys j. In the se ond ase the word w is
0
00
0
repla ed by zw + tw where jw j = jw j
1 and jw00 j = jwj 2. When we introdu e the z and
t as fun tions on and one nds that
X
T( ; )(x) = us Vk ( ) k ;
Suppose rst that
k
k2N
where
0 sk n 1 and Vk ( ) are (3; k + e(x))-polynomials. In parti
Te(
Now
us
n+1
k
where
=
P
k2N Yk (
)
k
; ) (x) =
where
Te(
X
k2N
us
k
n+1 V
k(
; ) (x) =
Lk ( ) are (3; k + e(x))-polynomials.
Taking into a
ubi
) k:
Yk ( ) are (3; k)-polynomials. Thus it follows
X
k2N
Lk ( ) k ;
The same is true for non ne essarily positive
exponents (using the
ular
x
2 Bn,
by getting rid of the negative
relation).
ount the normalization fa tor in front of the tra e we obtain the
Corollary 5.6.1 I( ; 0) (K ) =
a3i ; b3i 2 Z[ 12 ℄.
P
i2N a3i
3i
laim.
and, respe tively, I(0; )(K ) = Pi2Zb3i 3i, where
5.6.3 Chirality and other properties of I(
; )
Set x 2 Bn for the word one obtains from x when ea h j is repla ed by j .
Then T( ; )(x) = T( ; )(x ) holds true. Consequently for amphi heiral K , I( ; )(K ) =
I( ; ) (K ) is fullled.
Proof: Q(j )
R0
Q(j )
R0
1
Lemma 5.6.2
Let
substitutions
!
(respe tively
,
Q(j ) = j 3 Q(j ) = 0.
H(
;
)
!
) denotes the image of
and
Using a
l
! l
for
(respe tively
l = 1; : : : ; n 1. It is easy to he k that
R0 = R1 = 0. Sin e H( ; ) =
omputer we veried that
we are done.
The following properties have been
he ked with a
97
) after the
omputer program:
I(
; )
is independent from HOMFLY-PT and in parti ular it distinguishes knots that
have the same HOMFLY-PT polynomial. The knots 5.1 and 10.132 have the same
HOMFLY-PT polynomial but dierent
the other three
; )
; 0)
I(0;
and
)
invariants. This holds true for
rossing 10
(8:8; 10:129); (8:16; 10:156); (10:25; 10:56).
ouples of prime knots with number
fails to distinguish, i.e.
I(
I(
dete ts the
hirality of those knots with
that HOMFLY-PT
rossing number at most 10, where
HOMFLY-PT fails i.e. the knots 9.42, 10.48, 10.71, 10.91, 10.104 and 10.125).
The Kauman polynomial does not dete t the
Therefore
The
2-
kindly
I(
; )
hirality of
9:42
and
10:71
(see [79℄).
is independent from the Kauman polynomial.
abling of HOMFLY-PT does not dete t the
ommuni ated by H. Morton). Therefore
I(
; )
hirality of
10:71
(this result was
is independent from the
2-
abling
of HOMFLY-PT.
I(
; )
does not distinguish a well-known pair of mutant knots, the Conway knot
the Kinoshita-Terasaka knot
(KT ).
(C ) and
5.6.4 The denition of I (z; Æ)
Denition 5.6.2
Let us set for an oriented link L
I (z; Æ) (L) =
1
z zb
n
2
1
e(x)
zb 2
z
T (z; Æ) (x) 2 Z[z
=2; Æ=2 ℄
(
P(z; Æ))
;
where x 2 Bn is any braid whose losure is isotopi to L and ; ; t; zb as in Theorem 5.5.2.
Here 1 is the number of omponents mod 2, 2 f1; 2g.
Remark 5.6.1 This invariant doesn't dete t the amphi heirality of knots. Also I (z; Æ) does not
distinguish the Conway knot and the Kinoshita-Terasaka knot.
Proposition 5.6.2
I (z; Æ) (K ) =
X
Hk (Æ)z k =
k2Z
where Hk; Gk are (3; k)-Laurent polynomials.
Proof:
X
Gk (z )Æk ;
k2Z
The proof is analogous to the proof of Proposition 5.6.1.
For evaluating obstru tions and tra es of braids we used a Delphi ode. The
input is a word (or a linear ombinations) and we restri ted to words representing 5-braids. One
transforms rst the word to a sum of positive words, by using the ubi relations. Furthermore
the transformations Ci and Cij are used in order to redu e the shape of the word as mu h
as possible. When it gets stalked one allows permutations of the letters. The nal result is
the value of the tra e on the braid element. The program is available on
.
Remark 5.6.2
www-fourier.ujf-
grenoble.fr/
bellinge.html
98
5.6.5 Comments
As explained in Se tion 5.1.3, there are three essentially distin t link invariants whi h
from Markov tra es on the
nomial
ubi
Q one has a Markov tra
He ke algebras. For ea h quadrati
fa tor
Pi of the
ubi
ome
poly-
H (Pi ; n), yielding a reparameterized
(z;Æ) ).
Kauman polynomial and I( ; ) (or I
e whi h fa tors through
HOMFLY-PT invariant. The two others are the
It would be very interesting to nd whether there exists some relation among them. First of
way one expe ts there exists a lift of the invariant we des ribed to a genuine two-parameter
invariant.
There exists a Markov tra e on H (Q; n) taking values in an algebrai extension of Z[ ; ℄ lifting the Markov tra e underlying I( ; ). In other words the non-determina y
H( ; ) in I( ; ) an be removed.
Conje ture 5.6.1
Noti e that the polynomials
rational. In parti ular one
H and P
dene irredu ible planar algebrai
urves whi h are non-
annot express expli itly the invariants as one variable polynomial.
How far are these invariants from the usual Kauman and HOMFLY-PT polynomials is hard
to determine in the present state. One might expe t they give rise to some ni e weight systems
for parti ular values of the parameters, whi h should be
ompared with those
oming from
Lie algebras.
5.7 Appendix
The values of the polynomials for
with no more than 8
bold entry in the table is the
zero
oe ients of
3k
I(
; 0) (K )
rossings. The se ond
and
I (6:2) = [ 5
The entry A" in the last
oe ient of
3k
0
and
(respe tively
respe tively, for
19
4
3
I(0; ) (K )
are displayed below for all knots
olumn is a braid representative for the knot. A
0 ).
The other entries are the non
k 2 Z. For example,
1 6
℄; I (6:2) = [ 16
2
3 + 19
olumn means that the knot is amphi heiral.
99
2 3 ℄:
3:1
4:1
5:1
5:2
6:1
6:2
6:3
7:1
7:2
7:3
7:4
7:5
7:6
7:7
8:1
8:2
8:3
8:4
8:5
8:6
8:7
8:8
8:9
8:10
8:11
8:12
8:13
8:14
8:15
8:16
8:17
8:18
8:19
8:20
8:21
13
1 2 1 1 2 1
15
12 22 1 1 2
1 1 2 1 1 3 2 1 3 2
1 1 2 1 1 23
1 1 22 1 2 2
17
1 1 33 2 12 3 1 2
12 2 1 1 24
12 2 32 1 1 2 3 1 2
14 2 1 1 22
1 2 1 1 2 3 23 3
1 3 1 2 3 1 2 1 1 2 3 1 2
1 1 2 3 2 1 1 1 42 3 2 4 1
1 1 25 1 1 2
1 2 2 1 1 42 3 4 1 2 1 3
13 3 2 1 3 2 1 2 1
13 2 1 13 2 1
1 1 2 1 1 3 1 23 32
14 2 2 1 2 1
1 1 2 12 3 1 22 3 2
1 1 2 1 3 23
1 1 22 1 2 23
1 1 22 3 1 2 32 1 1 2
1 2 1 3 4 1 3 4 1 2 1 3 1 2 1
12 2 3 1 2 1 1 3 2 2
12 22 1 1 3 1 2 3 1 2
12 2 1 1 32 22 3
12 2 1 12 2 1 1 2 1
1 1 2 1 1 22 1 2 2
1 2 1 1 2 1 1 2 1 1 2 1
1 2 1 2 1 22 1
13 2 1 3 2
1 2 2 12 23
1=4
8 10 1
0 7=8 1=8
2 17=8 1=4
8 -16 10 1
-5
19=4 1=2
-3
1=2
0
5=8 9=16
-3
11=2 21=8
-1
7=4 19=16
0
17=8 9=4
0
9=8 9=8
-4
37=8 1=2
8 -20 21=2
16 43 37 12 1
4 59=8 23=8 1=4
8 -8 1
8 8 3=4
1 3 19=8 1=4
5 21=2 21=4 1=2
3 9=4 1=4
3 17=4 1=2
-7
9 1
1 2 1=4
8 21 147=8 6 1=2
24 44 21 2
8 12 21=4 1=2
6 85=8 21=4 1=2
0
17=8 9=4
-3 3=2 1=4
-11
19=2 1
8 -16 10 1
0 3=8 1=16
5 9=2 1=2
1
1 1=8
-1
100
1
1=16
1=4
1=8
1=4
1=8
1
8 2
8 10
24 4
8 2
16 19
-3 1=2
56 8
64
64
64
24
24
A
1
2
A
64 -6
48 -4
+ 128 -78 8
4
20
-19 37=2
2
2
64 144 -88 9
24 36 4
8 -8 1
8 -24 19 2
24 36 4
A
1
1=4
16 -25 3
16 -21 5=2
-7 9
1
8 -8 1
64 136 -79 8
24 44 21 2
8 -28 39=2 2
8 18 2
64 32 4
-7 1
-11 19=2
1
8 -16 10 1
64 64 1
8 0
8 0
A
A
A
A
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