Transitive factorizations of free partially commutative monoids and Lie algebras

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Transitive factorizations of free partially commutative monoids and Lie algebras

Jean-Gabriel Luque, Gérard Duchamp

To cite this version:

Jean-Gabriel Luque, Gérard Duchamp. Transitive factorizations of free partially commutative monoids

and Lie algebras. Discrete Mathematics, Elsevier, 2002, 246 Issue 1-3, pp.83 - 97. �hal-00086323�

ccsd-00086323, version 1 - 18 Jul 2006

Transitive factorizations of free partially commutative monoids and Lie algebras

G´erard Duchamp and Jean-Gabriel Luque LIFAR, Facult´e des Sciences et des Techniques,

76821 Mont-Saint-Aignan CEDEX, France.

July 18, 2006

Abstract

LetM(A, θ) be a free partially commutative monoid. We give here a necessary and sufficient condition on a subalphabetB⊂Asuch that the right factor of a bisectionM(A, θ) =M(B, θ_B).T be also partially commutative free. This extends strictly the (classical) elimination the- ory on partial commutations and allows to construct new factorizations ofM(A, θ) and associated bases ofL_K(A, θ).

R´esum´e

SoitM(A, θ) un mono¨ıde partiellement commutatif libre. Nous don- nons une condition n´ecessaire et suffisante sur un sous alphabetB ⊂ A pour que le facteur droit d’une bisection de la forme M(A, θ) = M(B, θ_B).T soit partiellement commutatif libre. Ceci nous permet d’´etendre strictement et de fa¸con optimale la th´eorie (classique) de l’´elimination avec commutations partielles et de construire de nouvelles factorisations deM(A, θ) ainsi que les bases deL_K(A, θ) associ´ees.

1 Introduction

A factorization of a monoid is a direct decomposition M =

Y

i∈I

M

where M and the M

are monoids and I is totally ordered. This notion is due

to Sch¨ utzenberger (see [16, 17] where the link with the free Lie algebra

is studied). Then, in his Ph. D. [19], Viennot showed how combinatorial bases of the free Lie algebra could be constructed by composition of bisec- tions (i.e. |I| = 2) obtained by elimination of generators (ideas initiated by Lazard [13] and Shirshov [18]). One of the authors with D. Krob found similar decompositions for the free partially commutative monoid into free factors and studied the link with Lie algebras and groups [6]. This works generalizes the completely free case, but is restricted to the situation where the outgoing factors are also free.

Here, we study the general problem of eliminating generators in these struc- tures and first remark that in any (set theoretical) direct decomposition

M(A, θ) = M(B, θ

).T

(with B ⊂ A, a subalphabet) the complement is a monoid. We get a cri- terion to characterize the case when T is free partially commutative and construct bases of the associated Lie algebras. The case of the group is also mentionned.

2 Definitions and background

We recall that the free partially commutative monoid is defined by genera- tors and relations as

M

(A, θ) = hA|ab = ba, (a, b) ∈ θi

,

where A is an alphabet and θ ⊂ A × A is an antireflexive (i.e. without loops) and symmetric graph on A (θ is called an independence relation).

Thus,

(A, θ) is the quotient A

/

where ≡

is the congruence generated by the set {(ab, ba)|(a, b) ∈ θ}.

Definition 1 If X is a subset of

(A, θ), we set

θ

= {(x

, x

) ∈ X

|Alph(x

) × Alph(x

) ⊂ θ}.

Note that (x

, x

) ∈ θ

implies Alph(x

) ∩ Alph(x

) = ∅, similarly we set θ

= θ

.

As in [7], we denote IA(t) = {z ∈ A|t = zw} and T A(t) = {z ∈ A|t = wz}.

If X is a subset of

(A, θ), we denote hXi the submonoid generated by X.

In [3] and [4], Choffrut introduces the partially commutative codes as some generating sets of free partially commutative submonoids. Let X be a set, we can prove easily that this definition is equivalent to the fact that each trace t ∈ hXi admits a unique decomposition on X up to the commutations (i.e. (X, θ

) is the independence alphabet of hXi the submonoid generated by X).

Example 1 (i) Each subalphabet B of A is a partially commutative code.

(ii) Let (A, θ) = a − b c. The set {c, cb, ca} is a code but not the set {b, a, ca, cb}.

3 Transitive bisections

3.1 Generalities

We recall the definition of a factorization in the sense of Sch¨ utzenberger (cf.

Viennot in [19] and [20]), this notion will be reused extensively at the end of the paper.

Definition 2 (i) Let

be a monoid and (

)

an ordered family of sub- monoids (the total ordering on J will be denoted <). The family (M

)

will be called a factorization of

if and only if every m ∈

=

− {1}

has a unique decomposition

m = m

m

. . . m

with i

> i

> . . . > i

and for each k ∈ [1..n], m

∈

.

(ii) In the case of a free partially commutative monoid, a factorization will be denoted by the sequence of the minimal generating sets of its components.

In the maximal case (each monoid has a unique generator), the factorization is called complete.

Example 2 (Complete factorizations in free and free partially commutative monoids.)

In the free monoid, it exists many complete factorizations. The most famous

of this kind being the Lyndon factorization (defined as the set of primitive

words minimal in their conjugacy classes) is an example of a complete fac-

torization [14, 15, 16]. Hall sets defined in [17] give us a wider example.

The set of Lyndon traces (i.e. the generalization of Lyndon words to the partially commutative case, defined by Lalonde in [11]) endowed with the lexicographic ordering is a complete factorization of the free partially com- mutative monoid.

In the smallest case (|J | = 2), the factorization is called a bisection. Let M be a monoid, then (M

, M

) is a bisection of M if and only if the mapping

M

× M

→ M (m

, m

) → m

m

is one to one.

Remark 1 Not every submonoid is a left (right) factor of a bisection. If M = M

M

is a bisection then M

satisfies (u, uv ∈ M

) ⇒ (v ∈ M

) (see [5]), however, this condition is not sufficient as shown by M

= 2

⊂

= M.

In case M =

(A, θ), one can prove the following property.

Proposition 3 Let (A, θ) be an independence relation and B ⊂ A. Then

(B, θ

) is the left (resp. right) factor of a bisection of

(A, θ).

Proof We treat, here, only the left case, the right case being symmetrical.

It is clear that M = {t ∈

(A, θ)|IA(t) ⊂ A − B } is always a monoid and that we have the (set theoretical) equality

(A, θ) =

(B, θ

).M . It suffices to prove the unicity of the decomposition of a trace. Let w, w

∈

(B, θ

) and t, t

∈ M such that wz = w

z

. Using Levi’s lemma, we find four traces p, q, r, s such that w = ps, t = rq, w

= pr and t

= sq. But, by definition of M, we have uv ∈ M implies u ∈ M , then r, s ∈ M ∩

(B, θ

) = {1}. It follows w = w

and t = t

, which gives the result.

2 In the sequel, we denote Z = A − B.

In the left case, the right submonoid above has

β

(B) = {zw/z ∈ Z, w ∈

(B, θ

), IA(zw) = {z}}

as minimal generating subset.

Remark 2 The monoid hβ

(B )i may not be free partially commutative.

For example, if A = {a, b, c},

θ : a − b c

and B = {c} then a, b, ac, bc ∈ β

(B ) and a.bc = b.ac.

3.2 Transitively factorizing subalphabet

Here we discuss a criterium for the complement hβ

(B)i to be a free partially commutative submonoid.

Definition 4 Let B ⊂ A, we say that B is a transitively factorizing subal- phabet (TFSA) if and only β

(B) is a partially commutative code.

We prove the following theorem.

Theorem 5 Let B ⊂ A. The following assertions are equivalent.

(i) The subalphabet B is a TFSA.

(ii) The subalphabet B satisfies the following condition.

For each z

6= z

∈ Z and w

, w

, w

, w

∈

(A, θ) such that IA(z

w

) = IA(z

w

) = {z

} and IA(z

w

) = IA(z

w

) = {z

} we have

z

w

z

w

= z

w

z

w

⇒ w

= w

, w

= w

.

(iii) For each (z, z

) ∈ Z

∩ θ, the dependence

graph has no partial graph

like

z − b

− . . . − b

− z

. with b

, . . . , b

∈ B .

1The dependence graph is defined byA×A−∆−θwhere ∆ ={(a, a)/a∈A}.

2We repeat here the notion of partial graph. A graphG^′= (S^′, A^′) is a partial graph ofG= (S, A) if and only if S^′ ⊂S and A^′ ⊂A∩S^′×S^′ (G^′ is a subgraph ofGwhen equalityS=S^′ occurs).

Proof It is easy to see that (i)⇒(ii) : by contraposition, if B does not sat- isfy (ii) we can find z

w

, z

w

, z

w

, z

w

∈ β

(B ) such that z

w

.z

w

= z

w

.z

w

with w

6= w

or w

6= w

and this implies obviously that β

(B) is not a partially commutative code.

Let us prove that (ii)⇒(iii). Suppose that

z − b

− . . . − b − n − z

is a partial graph of the dependence graph, then it exists a subgraph of the dependence graph of the form

z − c

− . . . − c

− z

with c

∈ B. Consider the smallest integer k such that (c

, z

) 6∈ θ. Then we have zc

. . . c

.z

c

. . . c

= z

.zc

. . . c

, which proves that B does not satisfy (ii).

Finally, we prove that (iii)⇒ (i). For each z ∈ Z, we define B

the set of letters of B having in the dependence graph a path leading to z and all inner points belonging to B . Clearly the assertion (iii) is equivalent to the fact that (z, z

) ∈ θ

implies ({z} ∪ B

) × ({z

} ∪ B

) ⊆ θ. It follows that β

(B ) × β

(B) ⊂ θ

.

Consider the mapping µ from Z into KhhA, θii defined by µ(z) = β

(B).

As (z, z

) ∈ θ

⇒ [µ(z), µ(z

)] = [β

(B), β

(B)] = 0 and hµ(z), 1i = 0

, we can extend µ as a continuous morphism from KhhZ, θ

ii in KhhA, θii. Let s be the morphism from hβ

(B)i in

(Z, θ

) defined by s(zw) = z for each zw ∈ β

(B). We have

hβ

(B)i = s

(

(Z, θ

)) = P

w∈M(Z,θZ)

s

(w)

= P

w∈M(Z,θZ)

µ(w) = µ(

(Z, θ

)) Let P (θ

) be the polynomial such that

M

(Z, θ

) = 1 P (θ

) . As µ is a continuous morphism, we have

hβ

(B )i = 1

µ(P (θ

)) = 1 P (θ

)

3Here, for a seriesS=P

αuu, we denoteshS, wi=αw.

which is the characteristic series of

(β

(B), θ

).

2 Remark 3 (i) Elimination in [7] deals with the particular case when A − B is totally non commutative. In this case B is a TFSA of A.

(ii) As an example of other case, consider the independence alphabet due to the graph

θ = a − b − c.

The monoid hβ

(c)i is free partially commutative, its alphabet is β

(c) = {b} ∪ {ac

/n ≥ 0}, its independence graph is

θ

=

ac

| . ..

a − b − ac

.. . and

< β

(B) > = 1 1 −

b + P

n≥0

ac

+ P

n≥0

abc

.

4 Factorizations and bases of free partially com- mutative Lie algebra

4.1 Transitive factorizations

We recall some definitions given by Viennot in [19].

Definition 6 Let

be a monoid,

a submonoid of

and

= (

)

a factorization of

. We denote

|

= (

)

where K = {k ∈ J |

⊆

} (in the general case it is not a factorization).

Definition 7 Let ≺ be the partial order on the set of all the factorizations of a monoid

defined by

= (

)

≺

= (

)

(

is finer than

) if and only if J

admits a decomposition J

= P

i∈J

J

as an ordered sum of intervals such that for each i ∈ J, (

)

is a factorization of

.

The following property is straightforward.

Proposition 8 Let

= (

)

be a factorization and

be a factorization such that

≺

then for each i ∈ I ,

|

is a factorization of

.

Definition 9 Let

= (B

, B

) be a bisection and

= (Y

)

a factoriza- tion. We say that Y

is cut by

if and only if

(

) = hB

i ∩ hY

i and

(

) = hB

i ∩ hY

i are not trivial (i.e. not {1}).

We need the following lemma.

Lemma 10 Let

= (B

, B

) be a bisection of

(A, θ) and

= (Y

)

a factorization with n > 1, such that it exists a factorization

= (G

)

with

,

≺

then

≺

if and only if no Y

is cut by

.

Proof We use the decomposition of K as an ordered sum of intervals K = J

+J

= P

i∈[1,n]

I

as in definition 7. The assertion (ii) implies the existence of an integer k ∈ [1, n] such that J

= P

i∈[1,k]

I

and J

= P

i∈[k+1,n]

I

. This allows us to conclude.

2 Note 1 In the preceding lemma, the existence of a common bound

is essential as shown by the following example (with

(A, ∅) = {a, b, c}

and the rational expressions written as in [1])

B

= (a, ba

∪ ca

) and

= (b, a, ab

a

∪ cb

a

)

No factor of

is cut by

and the two factorizations admit no common upper bound.

In the sequel, we use the notion of a composition of factorizations as it is defined by Viennot in [19]. We recall it here.

Definition 11 Let

= (

)

be a factorization of a monoid

and for some k ∈ I ,

= (

)

a factorization of

. The composition of

and

is the factorization

◦

= (

)

where I

= I ∪ I

− {k} is ordered by i < j if and only if

(i) (i, j ∈ I and i <

j) or (i, j ∈ I

and i <

j)) (ii) i ∈ I, i <

k and j ∈ I

(iii) i ∈ I

, j >

k and j ∈ I

and

M^′′_i

=

if i ∈ I

if i ∈ I

Definition 12 A transitive factorization is a factorization which is com- posed of transitive bisections (in finite number).

Lemma 13 Let

= (Y

)

be a transitive factorization and

= (B, β

(B)) be a transitive bisection such that it exists a factorization

finer that

and

. Then it exists at most one Y

cut by

and for a such i we have

(i) The subset T = Y

∩

(B, θ

) is a TFSA of Y

and

(

) is the right monoid of the associated bisection.

(ii) The sequence (Y

, . . . , Y

, T ) is a transitive factorization of

(B, θ

).

(iii) The sequence (β

(T ), Y

, . . . , Y

) is a transitive factor- ization of

(β

(B), θ

)

Sketch of the proof First it suffices to remark that, if i > j are two indices such that Y

and Y

are cut by

then

(

) ⊆

(B, θ

)∩

(β

(B ), θ

(B )) = {1} and this contradicts our hypothesis, hence i = j.

Let us prove assertion (i).

1) First, we remark that

M

(Y

, θ

) =

(

).

(

)

and using the equality

(

) =

(T, θ

) we prove that

(

) = hβ

(Y

)i.

2) We show that if T is not a TFSA of Y

then B is not a TFSA of A and this implies the result.

Let us prove (ii) and (iii) by induction on p. If p = 1 the result is triv- ial. If p > 1 , we can write

under the form

=

◦

◦

where

= (B

, β

(B

)) is a transitive bisection,

= (Y

, . . . , Y

) a transitive factorization of

(B

, θ

) and

= (Y

, . . . , Y

) a transitive factoriza- tion of the monoid

(β

(B

), θ

Z′(B^′)

). If

=

the result is trivial.

If

6=

, we have necessarily B ⊂ B

or B

⊂ B. We suppose that B

⊂ B (the other case is symmetric), and we consider the transitive trisec- tion (B

, β

(B

), β

(B)). Using the induction hypothesis we find that

(Y

, . . . , Y

, T ) and (β

(T ), Y

, . . . , Y

)

are transitive factorizations (respectively of the monoid

(β

, θ

) and

(β

(B), θ

)). And then

(Y

, . . . , Y

, T ) =

◦ (Y

, . . . , Y

, T ) ◦ (B

, β

(B)) is a transitive factorization.

2 Lemma 14 Let

= (B, β

(B)) be a transitive bisection and

= (Y

)

be a transitive factorization such that

≺

. Then the factorizations

|

and

|

are transitive.

Proof We can prove the result by induction on n.

2 Proposition 15 Let

= (Y

)

and

= (Y

)

be two finite transitive factorizations such that it exists a factorization

with

,

≺

then it exists a transitive finite factorization

such that

(i)

,

≺

≺

(ii) For each j ∈ J ,

|

is a transitive finite factoriza- tion.

(iii) For each j ∈ J

,

|

j,θ_Y′

j)

is a transitive finite factoriza- tion.

Sketch of the proof We set J = [1, n], J

= [1, n

] and we prove the result by induction on n. If n = 1 the result is trivial. If n = 2, lemmas 10, 13 and 14 give us the proof. If n ≥ 2, we set

=

◦

◦

where

= (B, β

(B)) is a transitive bisection of

(A, θ),

a transitive factorization of

(B, θ

) and

a transitive factorization of

(β

(B), θ

). Using lemmas 10, 13 and 14 we define a factorization

F^′′

=

If B≺

(Y

, . . . , Y

, T, β

(T ), Y

, . . . , Y

) Otherwise such that

,

≺

≺

,

|

j,θ_Y′

j)

is transitive for each j ∈ [1, n] (in fact

this factorization is trivial for all j ∈ [1, n] but at most one for which it is a

transitive bisection),

|

and

|

are transitive. Using

the induction hypothesis we can construct two factorizations

and

such that

F₁

,

|

≺

≺

|

and

F₂

,

|

βZ(B))

≺

≺

|

βZ(B))

and satisfying (ii) and (iii). We set

=

◦

◦

, then

,

≺

≺

and the induction hypothesis, the construction of

and lemma 14 allow us to conclude.

2 Corollary 16 Let

= (Y

)

≺

be two transitive finite factorizations then for each i ∈ I,

|

is a transitive finite factorization.

Proof It suffices to use proposition 15 with

,

≺

.

2 The following definition is an adaptation to partial commutations of a definition given by Viennot in [19].

Definition 17 A factorization (Y

)

of

(A, θ) has locally the property

if and only if for each finite subalphabet B ⊂ A and n ≥ 0 it exists a factorization (Y

)

with the property

such that there is an strictly increasing mapping φ : I

→ I satisfying

Y

∩ B

= Y

∩ B

and Y

∩ B

= ∅ if j 6∈ φ(I

)

Definition 18 We denote CLT F (A, θ) the set of the complete locally tran- sitive finite factorizations.

Example 3 Consider the following independence graph a − b − c − d.

We construct a complete locally transitive finite factorization

as follow.

We eliminate successively the traces c, ac

, b, d, ac and a. So we have M (A, θ) = c

.(ac

)

.b

.d

.(ac)

.a

.M

where M is a (non-commutative) free monoid. It suffices to take a Lazard

factorization on M to construct a complete locally transitive finite factor-

ization of

(A, θ).

We can remark that one can not obtain this factorization using only tran- sitive bisections with a non commutative right member. Examining all the transitive bisections of this kind

1.

= ({a, c}, β

(a, c)) 5.

= ({a, c, d}, β

(a, c, d)) 2.

= ({b, c}, β

(b, c)) 6.

= ({b, c, d}, β

(b, c, d)) 3.

= ({b, d}, β

(b, d)) 7.

= ({a, b, d}, β

(a, b, d)), 4.

= ({a, b, c}, β

(a, b, c))

we can easily prove that

could not be written like

=

◦

◦

with i ∈ {1, 2, . . . , 7}.

4.2 Transitive elimination in L

( A, θ )

The algebra of trace polynomials KhA, θi = K[

(A, θ)] endowed with the classical Lie bracket is a Lie algebra ([2, 6, 14] when θ = ∅). The free partially commutative Lie algebra is at first defined as the free object with respect to the given commutations [6]. Here we will use its realisation as the sub-Lie algebra of KhA, θi generated by the letters [12]. We will denote it L

(A, θ). One can show that this definition is equivalent to L

(A, θ) = L

(A)/

where L

(A) is the free Lie algebra and I

is the Lie ideal of L

(A) generated by the polynomials [a, b] with (a, b) ∈ θ. The following theorem proves that elimination in L

(A, θ) and transitive factorization of

(A, θ) occur under the same condition.

Theorem 19 Let (B, Z) be a partition of A (i) We have the decomposition

L

(A, θ) = L

(B, θ

) ⊕ J

where J is the Lie ideal generated (as a Lie algebra) by τ

(B ) = {[. . . [z, b

], . . . b

] | zb

. . . b

∈ β

(B )}.

(ii) The subalgebra J is a free partially commutative Lie algebra if B is a TFSA of A.

(iii) Conversely if J is a free partially commutative Lie algebra

with code τ

(B) then B is a TFSA.

Proof (i) We have the classical Lazard bisection L

(A) = L

(B) ⊕ L

(T

(B))

where T

(B) = {[. . . [z, b

], . . .], b − n] | z ∈ Z, b

, . . . , b

∈ B }. Then using the natural mapping L

(A) → L

(A, θ) (as [. . . [z, b

], . . .], b

] maps to 0 if zb

. . . b

6∈ β

(B)) we get the claim.

(ii) The proof goes as in [7], due to the fact that, for a TFSA, (c) below still holds, we sketch the proof.

Define a mapping ∂

from β

(B) to L

(β

(B), θ

) by

∂

=

zwb if zwb ∈ β

(B), 0 otherwise.

a) We prove that if B is TFSA, ∂

can be extended as a derivation of the Lie algebra L

(β

(B), θ

).

b) We define ∂ a mapping from B to Der(L

(β

(B), θ

)) by ∂(b) = ∂

and we extend it as a Lie morphism from L

(B, θ

) into Der(L

(β

(B), θ

)).

c) We prove that the semi-direct product L

(B, θ

) ∝

L

(β

(B), θ

) and the Lie algebra L

(A, θ) are isomorphic using the universal property of the latter. Hence, J is a free partially commutative Lie algebra isomorphic to L

(β

(B ), θ

).

(iii) If the dependence graph admits the following subgraph z − b

− . . . − b

− z

with b

∈ B and z, z

∈ Z we have the identity

[z, [[. . . [z

, b

] . . . , b

], b

]] = [. . . [z

, b

] . . . b

], [z, b

]].

Which implies that τ

(B) is not a code for J .

2

4.3 Construction of bases of L

( A, θ )

In this section, we define a class of bases which contains the bases found

by Duchamp and Krob in [6], [7] and [4] using chromatic partitions and the

partially commutative Lyndon bases found by Lalonde (see Lalonde [11],

Krob and Lalonde [10]).

Definition 20 Let

= (Y

)

be a finite transitive factorization. We denote e

, the set of the n-uplets (

, . . . ,

) of transitive bisections such that

=

◦ . . . ◦

.

Let

be a transitive factorization and

= (

, . . . ,

) ∈ e

, we denote

= (

, . . . ,

).

In the following, if

is the sequence (Y

)

, we denote Cont

= S

i∈J

Y

as in [19].

Definition 21 Let

= (Y

)

be a finite transitive factorization. A bracketing of

along

∈ e

is a mapping Π

from Cont

to L

(A, θ) induc- tively defined as follows. If n = 1, then

is a sequence of length 1 under the form ((B, β

(B ))) and

Π

(w) =

w if w ∈ B,

[. . . [z, b

] . . . b

] if w = zb

. . . b

∈ β

(B ) and z ∈ Z.

If n > 1, let

= (

, . . . ,

) ∈ e

. We set

◦ · · · ◦

= (Y

)

and j ∈ [1, n] such that

= (Y

, β

j−Y_j^′′

(Y

)) (remark that in this case, one has Cont

− Cont

◦ · · · ◦

= β

j−Y_j^′′

(Y

)). And

Π

(w) =

 

 

 



 

 

  Π

n

(w) if w ∈ Cont

◦ · · · ◦

, [. . . [Π

n

(y

), Π

n

(v

)], . . . Π

n

(v

)] if w = y

v

. . . v

, w ∈ β

j−Y_j^′′

(Y

), y

∈ Y

− Y

and v

, . . . , v

∈ Y

. Using theorem 19 in an induction on n we prove the following proposition.

Proposition 22 Let

= (Y

)

be a transitive factorization. For each

∈ e

, we have the following decomposition

L

(A, θ) = M

i∈[1,n−1]

L

(Π

(Y

), θ

) where

θ

= {(Π

(y

), Π

(y

))|(y

, y

) ∈ θ

}.

Definition 23 Let

= (Y

)

be a locally transitive finite factorization, a bracketing of

is a mapping Π from S

i∈J

Y

to L

(A, θ) such that for

each finite subalphabet B ⊂ A and each integer n ≥ 0, it exists a transitive finite factorization

= (Y

)

and

∈ e

such that for each t ∈ Cont

∩ B

, Π(t) = Π

(t).

Lemma 24 Let

= (Y

)

≺

be two finite transitive factorizations. Then, for each

∈ e

, it exists

∈ e

such that for each t ∈ Cont

, Π

(t) = Π

(t).

Proof It is a direct consequence of corollary 16.

2 Theorem 25 Let (A, θ) be an independence alphabet. Each locally finite transitive factorization of

(A, θ) admits a bracketing.

Proof Let

= (Y

)

be a locally finite transitive factorization. Using proposition 15, one can construct a sequence of finite transitive factorizations (F

)

CardB<∞

such that

1. if n ≤ n

and B ⊂ B

then

F_n,B

≺

, 2. for each n ≥ 0 and B ⊂ A

F_n,B

≺

,

3. for each n ≥ 0 and each finite subalphabet B, if we set

= (Y

)

, it exists a strictly increasing mapping φ

from [1, k

] to J verifying

M

(Y

, θ

i

) ∩ B

=

(Y

, θ

) and

j / ∈ φ

([1, k

]) ⇒

(Y

, θ

) ∩ B

= {1}.

By lemma 24, we can define for each n > 0 and each finite subalphabet B of A a sequence

∈

˜

such that for each m < n, B

⊂ B and t ∈ Cont

∩ B

we have Π

t = Π

t.

Thus, we can define Π as the mapping from Cont

into L

(A, θ) such that Πt = Π

|t|,

Alph

t.

2

We have easily the following result.

Proposition 26 Let

= ({l

})

∈ CLT F (A, θ) and Π be a bracketing of

then the family (Π(l

))

is a basis of L

(A, θ) as K-module.

Example 4 We set A = {a, b, c, d} and θ = a − b − c − d. We construct locally (for n ≤ 3) the following basis.

[[a,d],b], [[a,d],d], [[a,d],a], [a,d], [a,[a,c]], a, [a,c], [[a,c],c], [[a,d],c], [b,d], [[b,d[,b[, [[b,d],d], b, c, d.

5 The case of the group

The free partially commutative group [9] can be defined by the presentation

(A, θ) =< A; {ab = ba}

>

.

To extend the elimination process, we need the alphabet of the inverse let- ters. Recall that one can construct the free partially commutative group using ”reduced” traces [4, 8]. If A is an alphabet, we define ˜ A = A ∪ A where A = {a}

is a disjoint copy of A. The set ˜ A is then provided with the involution x → x such that x = x. Thus, θ is extended by

θ ˜ = {(x, y) ∈ A ˜

|{(x, y), (x, y), (x, y), (x, y)} ∩ θ 6= ∅}.

We define the natural mapping s

: ˜ A →

(A, θ) such that s

(a) = a and s

(a) = a

for each letter a ∈ A.

As s

is compatible wih the commutations of ˜ θ (i.e. if (x, y) ∈ θ ˜ then s

(x)s

(y) = s

(y)s

(x)), one has the factorization

A ˜

(A, θ)

M

( ˜ A, θ) ˜ . s

s

The mapping s is onto. For each g ∈

(A, θ), it exists an unique preimage with minimal length in s

(g), this element is called ”reduced expression”

of g (the subset of these traces will be denoted by red( ˜ A, θ)). The links with ˜ the bisections (B, β

(B)) is given by the following.

Lemma 27 Let B ⊂ A, z 6∈ B and w ∈ red( ˜ B, θ ˜

). The following asser-

tions are equivalent.

1. wzw is a reduced trace (i.e. wzw ∈ red( ˜ A, θ)). ˜ 2. zw ∈ β

( ˜ B).

Proof Straightforward, using the criterion given in [8]:

Let t = a

a

. . . a

∈

( ˜ A, θ), ˜ t is not a reduced trace if and only if it exists 1 ≤ i < j ≤ n with a

= a

and such that for each k, i < k < j, (a

, a

) ∈ θ. ˜ 2

We denote β

( ˜ B) the set β

( ˜ B) ∩ red( ˜ A, θ) with the commutation ˜ ˜ θ

provided by the definition 1. One has an analogue of the theorem 19.

Proposition 28 Let (B, Z) be a partition of A.

(i) One has the decomposition as the semi direct product

(A, θ) =

(B, θ

) ⋉ H

where H

is the normal subgroup generated by Z. It is the subgroup generated by

ρ

(B) = {w

zw|zw ∈ β

( ˜ B )}

(ii) The subgroup H

is free partially commutative for the code ρ

(B) and the commutations

θ ˆ

:= {(t, t

) ∈ ρ

(B)

|tt

= t

t and t 6= t

}.

(iii) The natural mapping α :

(β

( ˜ B), θ ˜

Z( ˜B)

) → H

is one to one if and only if B is TFSA.

Proof (i) The decomposition given by (i) is the image of the non commuta- tive Lazard elimination in the free group. The unicity of the decomposition with respect to the semidirect product can be obtain (as in the classical case) by sending all the element of Z to one.

(ii) Let ˆ ρ = {a

}

be an alphabet and ˆ θ be the commutation relation defined by (a

, a

) ∈ θ ˆ if and only if t 6= t

and tt

= t

t.

For each b ∈ B, we define the mapping σ

: ˆ ρ → ρ ˆ by σ

(a

) = a

. Re-

marking that b

tb belongs to ρ

(B ) and then (a

, a

) ∈ θ ˆ implies (σ

(a

), σ

(a

)) ∈

θ, this mapping can be extended in an automorphism ˆ σ

of

(ˆ ρ, θ). Let ˆ σ be

the mapping from B to Aut(

(ˆ ρ, θ)) defined by ˆ σ(b) = σ

. As σ

σ

= σ

σ

when (b, b

) ∈ θ

, σ can be extended as a morphism from

(B, θ

) in Aut(

(ˆ ρ, θ)). Using the same proof than in theorem 19, we find that the ˆ semidirect product

(B, θ

) ∝

(ˆ ρ, θ) and ˆ

(A, θ) are isomorphic.

(iii) Suppose that B is not TFSA then it exists a (z

, z

) ∈ θ

and a minimal path in the non commutation graph

z

− a

− · · · − a

− c − b

− · · · − b

− z

.

Let r

= z

a

· · · a

and r

= z

b

· · · b

. Due to the fact that the chain is of minimal length, one has (r

, r

) ∈ θ ˜

Z( ˜B)

and α(r

)α(r

) = α(r

)α(r

).

But r

c and r

c do not commute and their images α(r

c) = c

α(r

)c and α(r

c) = c

α(r

)c do. This proves that α is not one to one.

The converse follows from the fact that, when B is a TFSA, the commutation graph (β

( ˜ B ), θ ˜

Z( ˜B)

) and (ρ

(B ), θ ˆ

) are obviously isomorphic.

2

Note 2 In general α is into.

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