An algebraic characterization of simple closed curves on surfaces with boundary

0.4 MB·English

by  Moira Chas

An algebrai

hara
terization of simple
losed
urves
on surfa
es with boundary
9
0
0
2 Moira Chas and Fabiana Krongold

n
a
J
Abstra
t We
hara
terize in terms of the Goldman Lie algebra whi
h
onjuga
y
0
3
lasses in the fundamental group of a surfa
e with non empty boundary are
represented by simple
losed
urves.

]
T Weprove the following: A non power
onjuga
y
lass X
ontains an embedded
G representative if and only if the Goldman Lie bra
ket of X with the third power
. of X is zero.
h
t The proof uses
ombinatorial group theory and Chas'
ombinatorial des
rip-
a
tion of the bra
ket re
ast here in terms of an exposition of the Cohen-Lustig
m
algorithm. Using results of Ivanov, Korkmaz and Luo there are
orollaries
har-
[
a
terizingwhi
hpermutationsof
onjuga
y
lassesarerelatedtodi(cid:27)eomorphisms

4
of the surfa
es.
v
4 This result
ompletes the solution of a problem posed by Turaev in the eight-
4 ies.
9
Our main theorem
ounts the minimal possible number of self-interse
tion
3
. points of representatives of a
onjuga
y
lass X in terms of the bra
ket of X with
1
0 the third power of X.
8
0
:
v
i
X 1 Introdu
tion
r
a Z
During the middle eighties Goldman [10℄ de(cid:28)ned a Lie algebra stru
ture on the -
module generated by non-trivial
onjuga
y
lasses and explained it as a universal
Poisson stru
ture (cid:28)tting with
al
ulations of S
ott Wolpert [28℄ using the symple
-
ti
stru
ture on Tei
hZmüller spa
e. In the late eighties Turaev added a Lie
obra
ket
stru
ture [27℄ on this -module and showed the two operations, bra
ket and
obra
ket,
de(cid:28)ned a Lie bialgebra in the sense of Drinfeld. Turaev formulated a question whether
the vanishing of the
obra
ket
hara
terized embedded non-power
onjuga
y
lasses.
Turaev was motivated by the group theory statement:
s
π −→ F ×F
g g g
Every surje
tion
ontains in its kernel an embedded
onjuga
y
lass.
2000 Mathemati
s Subje
t Classi(cid:28)
ation: Primary 57M99, Se
ondary 17B62.
Key words and phrases: surfa
es,
onjuga
y
lasses, Lie algebras, interse
tion number, hyperboli
geometry, embedded
urves.
1
π g F
g g
Here is the fundamental group of a
losed surfa
e of genus , is the free group
g
on generators and the surje
tion is the data of a Heegaard de
omposition of a simply

onne
ted three manifold. This statement was shown to be equivalent to the Poin
arè

onje
ture ([16℄ and [26℄.)
CounterexamplestoTuraev's
onje
tureweregivenin[5℄andeviden
e wasobtained
for a bra
ket
hara
terization of embedded
onjuga
y
lasses. Before the
ounterex-
amples were found, attempts to answer Turaev's question led to the String Topology
paper [3℄.
In this paper we also report on the empiri
al eviden
e that the vanishing of the

obra
ket of non-trivial powers of a
onjuga
y
lass may also
hara
terize embedded

lasses.
Themaintheoremofthispaper, whi
h holdsforsurfa
es withnon-emptyboundary,
is:
V
Theorem 4.1: Let be a non-power
onjuga
y
lass of the fundamental group of
p q p
an oriented surfa
e. Let and be distin
t positive integers with one of them,
q
or at least three. The number of terms (
ounted with multipli
ity) of the Goldman
hVp,Vqi 2 · p · q
bra
ket is times the minimal possible number of self-interse
tions of
V
representatives of the
onjuga
y
lass .
Our proof of the main theorem really happens in the analysis of Proposition 2.19.
This result distinguishes the
onjuga
y
lasses of various sets of linear words. The
analog of this proposition for
losed surfa
es is work in progress and may yield an anal-
ogous
hara
terization of embedded
onjuga
y
lasses there. Beyond Proposition 2.19,
the proof uses the
ombinatorial presentation of the Lie bialgebra from [5℄. However,
for the purposes of a simpler and more self-
ontained exposition we adapt propositions
of Cohen-Lustig [7℄ to arrive in this paper at a modi(cid:28)ed
ombinatorial presentation of
the Goldman Lie algebra.
The surje
tion above
an be realized by a simpli
ial map from the surfa
e to the
Cartesian square of a trivalent graph. Dis
overing some stru
ture analogous to the
Goldman bra
ket for the latter spa
e and obtaining a deeper understanding of the
on-

omitant algebra presented here
ould possibly lead to a topologi
al or algebrai
proof
of the group theoreti
statement above whi
h is now known to be true by Perelman's
work ([21℄, [22℄ and [23℄. See also Morgan-Tian [13℄).
The paper
loses with some further algebrai
questions and problems whi
h are
related to the ideas here.
In moredetail,thepaperisorganizedasfollows: (cid:28)rst we develope the
ombinatorial
aspe
ts of our work (Se
tion 2) and then, the topologi
-geometri
aspe
ts (Se
tion 3).
More pre
isely, in Se
tion 2 we de(cid:28)ne linear and
y
li
words, an equivalen
e relation
on the set of
y
li
permutations of pairs of words and an order on the set of half
in(cid:28)nite words. These
on
epts were already introdu
ed in [2℄ and [7℄. In this paper, for
2
ompleteness, we give proofs of all the
ombinatorial results we state. Using all these
elements, we de(cid:28)ne a
ertain bilinear map, the
ombinatorial Goldman Lie bra
ket.
Also,weprovethat
ertainpairsof
y
li
words
annotbe
onjugate(Proposition2.19).
The argument of Proposition 2.19 (espe
ially Figure 3) is the heart of the proof of the
main result of this paper.
In Se
tion3, we de(cid:28)ne boundary expansions andwe showthat theGoldmanbra
ket
is the bilinear map we de(cid:28)ned
ombinatorially in Se
tion 2.
Even if for Se
tion 2 the de(cid:28)nitions and results of Se
tion 3 are not needed, we
will make referen
e to Se
tion 3 when exposing Se
tion 2 in order to
larify and give
meaning to an otherwise
hallenging
ombinatorial dis
ussion.
Finally,inSe
tion4, we prove ourmainresult, Theorem4.1, sayingthatthebra
ket
"
ounts" the number of self-interse
tion points of a
onjuga
y
lass. We
on
lude with
Se
tion 5 by stating some questions and
onje
tures relative to these problems.
This work bene(cid:28)ted from dis
ussions with Laura Ciobanu and Miguel Paternain.
TheauthorsareindebtedtoDennisSullivanforhishelpinthehistori
alremarksandto
the referee for a
areful reading and ex
ellent suggestions. The resear
h was supported
by the NSF grant 1034525 - 1- 2977. The Instituto Argentino de Matemati
as, with
its kind hospitality, also
ontributed to the
ompletion of this work.
2 Cy
li
words, linking pairs and a bilinear map
Z V
In this se
tion we introdu
[e, ]t:heVf⊗reeV −-→moVdule of
y
li
words of[
,e]rtain alphabet
and de(cid:28)ne a bilinear map . From the de(cid:28)nition of , it will follow
[, ]
straightforwardly that is antisymmetri
. In fa
t, this bilinear map satis(cid:28)es the
Ja
obi identity. We will give an indire
t proof of this fa
t, by showing that the the

ombinatorialbilinearmapandtheGoldmanLiebra
ketof
urvesonasurfa
e
oin
ide.
In [11℄ a purely
ombinatorial proof of the Ja
obi identity will be exhibited.
[, ]
The bilinear map
oin
ides with the one de(cid:28)ned in [5℄, but the presentation we
give here, is easier to pro
ess for the reader familiar with Hyperboli
Geometry.
2.1 Linear and
y
li
words
In this subse
tion we introdu
e de(cid:28)nitions and re
all basi
well known results about
linear and
y
li
words.
q q 2q
Let be a positive integer. A -alphabet or, brie(cid:29)y, an alphabet is a set of
{a ,a ,...,a ,a ,a ,...,a }
1 2 q 1 2 q
sqymbols, A , endowe2qd with a (cid:28)xed linear order. We denote a
-alphabetby q. (There areapproximately q di(cid:27)erent dis
ussions depending on the
A A (cid:0) (cid:1) A
q q q
order of .) The elements of are letters. A linear word in is a non-empty (cid:28)nite
3
v v ...v v A i ∈ {0,1,...,n−1}.
0 1 n−1 i q
sequen
e ofsymbols su
h that belongsto forea
h
v v = v
For ea
h letter , .
V = v v ...v n V
0 1 n−1
Let V be a liVne=arvwordv. We..s.avy that , the numbeVr of letters of is
n−1 n−2 0
thve le6=ngvth of . By die∈(cid:28)n{it0i,o1n,,...,n−1} V . The linear word v is fr6=eelvy redu
eVd
i i+1 n−1 0
if for ea
h . If is freely redu
ed and then
is
y
li
ally redu
ed.
V = v v ...v
0 1 n−1
Notation 2.1. When dealing with letters denoting linear wordVs n we
wVi=ll valwva.y.s.v
onsider subindi
es of leitters mod thevlength of , that isv . VThus if
0 1 n−1 i h
is a linear word and is an integer, denotes the letter of where
h n i−h h < n
is the only non-negative integer su
h that divides and .
Consider(tVh,eWeq)uivalen
e reVlation on the set of linear woWrds, geVne=raWtedvvby the pvairs
of the formA su
h that is a
y
li
permutation of or where is
q
a letter in . The equivalen
e
lasses under this equivalen
e relation are
alled
y
li
words. (Observe that these are the
onjuga
y
lasses of the free group generated by
a ,a ,...,a
1 2 q
). Thus, every
y
li
woVrd
an be labeled by a unique redu
ed oriented
ring of symbols. (see Figure 1). If is a (not ne
essarily redu
ed) linear word, we
V V
denote the equivalen
e
lass of by . Observe that the de(cid:28)nition of
y
li
word we
are giving here does not
oin
ide witbh the one given in [5℄ whi
h allowed unredu
ed
rings of symbols.
a
3
a a
5 4
x
a a
6 7
a a
4 a 4
1
A
7
Figure 1: A ring in the letters of
V = v v ...v j
0 1 n−1
v vIf ...v v ...v is a linear wVord and is an integer theVn the linear word
j j+1 n−1 0 j−1 j
V is denoted by . Noti
e that the length of equals the length
j
of .
V V V
A linearword is a linear representative of a
y
li
word if is
y
li
allyredu
ed
V
aVnd belongs to the equivalen
e
lass . ObservVe that all the linear representatives of
are exa
tly all the
y
li
permutations of . Thus, in parti
ular, all the linear
V
representatives of a
y
li
word have the same length.
V V k Vk
If is a
y
li
word, \a linear representative and an integer, we de(cid:28)ne as
Vk k (V−k) k
if is positive and as if is negative. (These are the basi
operations on

conjuga
y
lasses in groups and are well de(cid:28)ned by thVerse pres
rViprtions). A linre≥ar2(resp.

y
li
) word is primitive ifVit
annot be writtenVas (resp. ) for some and
some redu
ed linear word (resp.
y
li
word ). The length of a
y
li
word is the
length of any linear representative. (Re
all that, by de(cid:28)nition, a linear representative
of a
y
li
word is
y
li
ally redu
ed.)
4
We start by gathering together some elementary well known results we will need
throughout these pages.
V V = V i V
i
Lemma 2.2. (1) If is a liinear word and for somVe integer then either is
not a primitive word or is a multiple of the length of .
V W
(2) If is a
y
li
ally reduk
ed linear wVor=d Wthekn there exists a primitive linear word
and a positive integer su
h that .
V W V = W
(3) If anid are
yV
li=
aWlly redu
ed words su
h that then there exists an
i
integer su
h that . b b
V k l (V )l =
k
(4) (IfVl)is a
y
li
ally redu
ed linear word and and are positive integers then
k
.
2.2 An equivalen
e relation on the set of
y
li
permutations
of pairs of words
V W n m
Let and be two
y
li
ally redu
ed words of length and respe
tively. In this
{0,1,...,n−
Subse
tion we de(cid:28)ne an equivalen
e relationon the set of ordered pairs in
1} × {0,1,...,m − 1}
. This de(cid:28)nition is purely
ombinatorial but has the following
geometri
interpretation: This equivalen
e relation is su
h that there exists a one
(i,j)
to one
orresponden
e between pairs of integers equivalent to and fundamental
domainstraversedsimultaneouslybythetwVoaxesWofthetwohyper(bio,lji)
transformations
asso
iated in Subse
tion 3.1 to the words and and the pair V . MoWre pre
(iis,ejly),
in the
ontext of Subse
tion 3.1, given two
y
li
allyredu
ed wordVs andW and ,
i j
one
an determine whi
h fundamental domains are traversed by]and "starting"
D (i,j)
1
at . There is a natural bije
tion between the elemVents of andWthe pair of
i j
fundamental domains that are traversed by the axis of and the axis of .
V = v v ...v W = w w w ...w
0 1 n−1 0 1 2 m−1
De(cid:28)nition 2.3. LAet I(V,W)and be two
y{
(lji,
kal)ly:
q
0red≤u
jed<wonrds in0 ≤. kDe<nomte}by R(tVhe,Wse)t of ordered pairs of integers I(V,W)
and and by the equivalen
e relation on
generated by the pairs
(j,k) ∼ (j +1,k+1) v = w
j k
(1) if .
(j +1,k) ∼ (j,k +1) v = w
j k
(2) if .
]
(i,j) (i,j)
Denote by the equivalen
e
lass of .
I(V,W)
We use with the entries of pairs in the same
onvention we use with
subindi
es in words: namely, integers are taken mod the length of the
orrespond-
ing word.
Next we prove an auxiliaryresult whi
h willbe used in the proof of Proposition 2.5.
5
V W (i,j),(k,h) (l,f)
Lemma 2.4. LI(eVt,Wa)nd (i,bje)two
(yk
,lhi
)ally redu
ed words. Let and(k,h)
be elements of . If and are related as in De(cid:28)nition 2.3(1) then
(l,f)
and are not related as in De(cid:28)nition 2.3(2).
(i,j) (k,h)
Proof. Sin
e and are related as in De(cid:28)nition 2.3(1) there are two possibili-
ties.
(k,h) = (i+1,j +1) v = w
i j
(a) and
(k,h) = (i−1,j −1) v = w
i−1 j−1
(b) and
(k,h) (l,f)
Suppose that and are related as in De(cid:28)nition 2.3(2). Then there are two
possibilities.
(l,f) = (k +1,h−1) v = w
k h−1
(i) and
(l,f) = (k −1,h+1) v = w
k−1 h
(ii) and
(a) (i) v = w = w = v = v . V
i j h−1 k i+1
Assume that and hold. Then Then is not
redu
ed, a
ontradi
tion. The other three
ases follow by similar arguments.
R(VIn,Wth)e next proposition we give a
hara
terization of the equivalen
e
lasses of
.
V W C
ProposiRti(oVn,W2.)5. Let and be two
y
li
ally redu
ed words.(iI,fj) is an equivalen
e

lass of then there exists an ordered pair of integers su
h that exa
tly
one of the following holds.
C = {(i,j),(i+1,j+1),...(i+c,j+c)} c
(1) for some non-negative integer . Moreover
v 6= w
i+c j+c
.
C = {(i,j),(i+ 1,j − 1),...(i + c,j − c)} c
(2) for some positive integer . Moreover
v 6= w
i+c j−c−1
.
v = w s
i+s j+s
(3) for every integer .
v = w s
i+s j−s
(4) for every integer .
(1) (2) V W n
Moreover, in
ases and , if the length of and the length of are equal to
n c < n
for some positive integer then .
C C (1)
Proof. If
ontains exa
tly one element, then is of type and the results holds.
C C
Thus we
an assume now that
ontainsmorethan one element. Then
ontains two
elements related as in De(cid:28)nition 2.3(1) or two elements related as in De(cid:28)nition 2.3(2).
6
C
We assume that
ontains two elements related as in De(cid:28)nition 2.3(1). The other

ase
an be studied in an analogous way.
(h,k) C C
Let be an element in . By Lemma 2.4, there are no pairs of elements in
T (h+t,k+t)
related as in De(cid:28)nition 2.3(2). Denote by the set of integers su
h that
C C (h+t,k+t)
belongs to . Note that every element of has the form for some integer
t T T T
in . By de(cid:28)nition, zero belongs to so is a non-empty set. It is not hard to
T T T
see that has an upper bound if and only if has a lower bound. Assume that
u l T
has an upper bound. Let (resp. ) denote the maximum (res. minimum) of . Set
i = h − l j = k − l C = {(i,j),(i + 1,j + 1),...,(i + u + l,j + u + l)}
and . Then
(i+u+l+1,j+u+l+1) = (h+u+1,k+u+1) ∈/ C v 6= w
h+u h+u
. This implies that .
C (1)
Thus is of type .
T C (3)
On the other hand, if does not have an upper bound then is of type .
V W n
Assume now that the length of and the length of are equal to for some
n T n T
positive integer . Observe that if
ontains at least integers, then
ontains all
the integers. This
ompletes the proof.
C
In the
onditions of Proposition 2.5, to ea
h equivalen
e
lass we asso
iate a
C
non-negative integer whi
h will be
alled the negative length of . The negative length
c (2) 0
is de(cid:28)ned as the number if holds and otherwise.
V W
Let and be two
y
li
ally redu
ed linear words. In Subse
tion 2.4, in order to
(i,j)
de(cid:28)Rn(eVt,hWe)bilinearmap, wV[eWneed to
hoose a representative of an equivalen
e
lass
i j
of and
onsider . ByLemma2.7, thisisindependent oftherepresentative
(i,j)
. Nevertheless, in order to fa
ilitate the proof of Theorem 3.16, we will use a
(rei,pjr)esenIt(aVti,vWe )with
ertain properties, wvhi
6=hwis de(cid:28)nved6=aswfollows. A pair of integers
i j i j−1
in is said to be extremal if and .
V W C
Corollary 2.6. LetR(Van,Wd ) be two
y
li
ally redu
ed linear woCrds and let be an
equivalen
e
lass of as in Lemma 2.5(1) or (2). Then
ontains a unique
extremal pair.
The following lemma follows straightforwardly from Proposition 2.5.
V W A
LI(eVm,Wm)a 2.7. Let and be two
yA
li
ally redu
edAl(iin,eja)r=woV[rdWs. Let bAe a map from
q i j
to the set of
y
li
worRds(Vin,W) de(cid:28)ned by . Then is
onstant
over ea
h equivalen
e
lass of .
c (i,j) n m
MoreovVer, if Wis the negative lengVt[hWof n+, amnd−2cand are, respe
tively, the
i j g
lengths of and then the length of is .
A = {a,b,a,b}
2
EVxampWle 2.8. Consider the ordered alphabet R(V,.WF)or ea
h of the words
and below, we list the set of equivalen
e
lasses of . The pairs in bold
7
are the extremal pairs (note that
ertain equivalen
e
lasses do not
ontain extremal
pairs).
V = ab W = aab {{(0,0),(1,1),(0,1),(1,2)},{(1,0)},{(0,2)}}
(1) , ,
V = W = aabb {{(0,0),(1,1),(2,2),(3,3)},{(0,1),(1,2)},{(1,0),(2,1)},
(2) ,
{(2,3),(3,0)},{(3,2),(0,3)},{(0,2)},{(2,0)},{(1,3)},{(3,1)}}
V = aab W = a {{(0,0),(1,0),(2,0)}}
(3) , , .
2.3 Linking pairs and sign
A
q
The set of redu
ed half in(cid:28)nite words in the ordered alphabet D is in
orresponden
e
with a subset of points in the boundary of the Poin
aré disk (see Lemma 3.5). In
thisASubse
tion, we will study following [2℄ an order on the set of half in(cid:28)nite words
q
iDn , whi
h re(cid:29)e
ts the
y
li
order of the
VorrespWonding points in the bounda−ry1 o1f
. 0Using this order, fRor(Vev,eWry).pair of words and , we asso
iate an integer, ,
or to ea
h
lass in We will see in Subse
tion 3.3 that this
ombinatorial
de(cid:28)nition is related to whether two axis of
ertain hyperboli
transformation interse
t
and if they do interse
t, whi
h is the sign of that interse
tion.
A T = t t t ...
q 0 1 2
WUe=nouwudeu(cid:28)n.e..an order in the set of half in(cid:28)nite words in . LeTt < U
0 1 2
and be two distin
t half in(cid:28)nite words. We say that if one of
the following holds.
t < u
0 0
(i)
j t = u i
i i
(ii) there exists a non-negative integer su
h that for every su
h that
0 ≤ i ≤ j t u
j+1 j+1
and
omeAs before in the netw alphabet obtained by
y
li
ally
q j
permuting the order of in su
h a way that is the (cid:28)rst element.
T U T < U U < T T = U
Note that for every pair of words and , either or or . In other
words, the order is linear.
T ,T ,...T T < T <
1 2 k 1 2
··· A<(cid:28)Tnite sTequ<en
·e··o<f hTalf<in(cid:28)Tnite words T ,T ,i.s.l.inTearly ordered if
k k 2 1 1 2 k
or T ,T ,...T. The sequen
e is
y
li
ally ordered if a
1 2 k

y
li
permutation of is linearly ordered.
A
q
Remark 2.9. The alphabet is initially endowed with a linear order. Any
y
li
permutation of thAis linear order yields the same
y
li
order on the set of in(cid:28)nite words
q
on the letters of , and this
y
li
Aorder is what determines all our further results.
q
The reason why one initiallyendows with a linear order and not with a
y
li
order
is that linear orders are easier to handle.
8
V = v v v ...v V∞
0 1 2 n−1
VV.G.i.ven aV(cid:28)−n∞ite lineVarVw..o.rd we de(cid:28)ne the in(cid:28)nite word as
and and .
A = {a,b,a,b} V = aabb
2
Example 2.10. Consider the ordered alphabet and . In the
above order we have
V∞ < V∞ < V∞ < V∞ < V∞ < V∞ < V∞ < V∞
0 1 3 2 2 1 3 0
V = v v v ...v W = w w w ...w
0 1 2 n−1 0 1 2 m−1
ALet and (i,j) ∈ I(V,bWe)two
y
li
ally redu
ed words
q
in . To ea
h ordered pair of integers we asso
iate an element of
{1,−1,0}
as follows
1 V∞,W∞,V−∞,W−∞
i j i j
s (i,j) =  −1 if V∞,W−∞,V−∞,W∞ is
y
li
ally ordered
i j i j
V,W  if is
y
li
ally ordered
0
otherwise.

Aproofofavariantofthefollowinglemma,usinghyperboli
geometry
anbe found
in [7℄. Here, for
ompleteness, we give the argument.
V W s
Lemma 2.11. Let and be two
y
lRi
(aVll,yWre)d.u
ed linear words. The fun
tion V,W
is
onstant on ea
h equivalen
e
lass of
Proof. Itsis enough to prove that if two pairs are related as in De(cid:28)nition 2.3(1) or
(2) then V,W has the same value for these (tiw,oj)pairs.(i+W1e,jst+ud1y) theI(V(cid:28)r,sWt)
ase. The
se
ond follows similarly. Consider two pairs and in for whi
h
v = w
i j
de(cid:28)nition De(cid:28)nition 2.3(1) holds. Thus . Assume that
V∞ < W∞ < V∞ < W∞.
i j i j
V
(The other possibilities follows by analogous arguments). Sin
e is
y
li
ally redu
ed,
v 6= v w = v < v ≤ w
i i−1 j i i−1 j−1
. By de(cid:28)nition of order we have, .
v = w V∞ < W∞ V∞ = v v V∞ W∞ =
v vAsWsu∞me V(cid:28)r∞st t<haWt∞i+1 j+1. Sin
e i j and i i i+1 i+2 and j
i i+1 i+2 i+1 j+1
, . We
onsider the following two
ases
∞ ∞ ∞ ∞
v = w V < W V W
i−1 j−1 i j i j
(1) ∞ . S∞in
e ∞ , and∞ , both start with the same lett∞er,
V = v v V < v v W = W V∞ W∞ V
i+1 ∞i i−1 i−1 i i−1 i−1 i+1. The words i+1 and j+1 (resp. i+1
W
and j+1∞) sta∞rt with the same l∞etter.∞Then the
y
li
order of the sequen
es
V∞,W∞V ,W V∞ ,W∞ V ,W
i j i j i+1 j+1 i+1 j+1
and
oin
ide.
v 6= w v = w < v < w
i−1 j−1 i j i−1 j−1
(2) . Hen
e, . This inequality implies that
∞ ∞ ∞
V = v v V < w w W .
i+1 i i−1 i−1 j j−1 j−1
V∞ < W∞ V∞ W∞
Sin
e i+1 j+1 and both words i+1 and j+1 start with the same letter, this

ase is done.
9
v 6= w v w
i+1 j+1 i+1 j+1
Finally assume that A . Then v
omes before in thVe∞alphabet
q i i+1
obtained by
y
li
ally permuting so that is th∞e (cid:28)rst elem∞ent. Sin
e starts
v W∞ w V W v
with i+1, j+1∞starts w∞ith j+1 and both words i+1 and j+1 start with i, if we
V < W

an show that i+1 j+1, the result follows. We
onsider the following
ases
∞ ∞
v = w (1) V < W
(1) i−1 j−1. As in of the previous
ase, i+1 j+1.
v 6= w v = w < v < w
i−1 j−1 i j i−1 j−1
(2) . Hen
e, . Therefore,
∞ ∞ ∞ ∞
v v V = V < W = w w W .
i i−1 i−1 i+1 j+1 j j−1 j−1
V W (0,0)
Example 2.12. If and be as in Example 2.8(1) then the
lass that
ontains
−1
has sign . The other two
lasses have sign zero.
V = W = aabb
If assin(E1,x3a)m=pl1e 2s.8(2(3),b1y) t=he−
1al
ulations in Example 2.10 one
an
dedu
e the following: V,V , V,V and the sign of the
lasses that do
(1,3) (3,1)
not
ontain or is zero.
V W
Finally, if and are as in Example 2.8(3), the sign of the unique equivalen
e

lass is one.
V W
De(cid:28)nition 2.13(.i,Lje)t Ia(nVd,W) be two
y
lsi
al(liy,jr)ed6=u
e0d linear words. An ordered
pair of integers i]n su
h that V,W is a linking pair. The set
(i,j) (i,j) LP(V,W)
of equivalen
eL
Pla(sVse,sW) su
h that is a linking pair is denoted by .
Observe that is well de(cid:28)ned be
ause of Lemma 2.11.
V W C
LeRm(mV,aW2).14.CLet and be two
y
li
ally redu
eCd linear words and let be a
lass
Cin∈/ LP(V,.WI)f is as in Lemma 2.5(3) or (4) then is not a linking pair. In symbols,
.
∞
C V∞ = W∞ V∞ = W
i j i j
(Pir,ojo)f∈. ICf is as in Lemma 2s.5(3()i,ojr) =(4)0 then or for ea
h
. Thus, by de(cid:28)nition V,W .
V W
Remark 2.15. For ea
h pair of
y
li
ally redu
ed words and , it is noLtPh(aVrd,Wto)
prove that there is a one to one
orresponden
e between the linking pairs
introdu
ed in [7℄ and the linked pairs of [5℄.
Z
2.4 A bilinear map on free -module of
y
li
words
V Z A
q
Let denote the free -module generated by the set of
y
li
words in .
Remark 2.16. By de(cid:28)Vnition, the
y
li
word
onVsisting of theZempty word is an
element of the base of . Had we
hosen to take as the free -module generated
10