Quotient groups of IA-automorphisms of a free group of rank 3

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by  V. Metaftsis

Quotient groups of IA-automorphisms of a free group of rank 3
7
V. Metaftsis, A.I. Papistas and H. Sevaslidou
1
0
2

n Abstract
a
J We prove that, for any positive integer c, the quotient groupγc(M3)/γc+1(M3) of the

0 lower central series of the McCool group M3 is isomorphic to two copies of the quotient
1 groupγc(F3)/γc+1(F3)ofthelowercentralseriesofafreegroupF3ofrank3asZ-modules.

Furthermore, we give a necessary and sufficient condition whether the associated graded
]
R Lie algebra gr(M3) of M3 is naturally embedded into the Johnson Lie algebra L(IA(F3))
G of the IA-automorphisms of F3.
.
h
t 1 Introduction and Notation
a
m
Let G be a group. For a positive integer c, let γ (G) be the c-th term of the lower central
[ c
1 series of G. We point out that γ2(G) = G′; that is, the derived group of G. We write
IA(G) for the kernel of the natural group homomorphism from Aut(G) into Aut(G/G′) and
v
8 we call its elements IA-automorphisms of G. For a positive integer c ≥ 2, the natural group
7
epimorphismfromG onto G/γ (G) inducesagroup homomorphismπ fromAut(G) of Ginto
4 c c
2 Aut(G/γc(G)) of G/γc(G). Write IcA(G) = Kerπc. Note that I2A(G) = IA(G). It is proved
0 by Andreadakis [1, Theorem1.2], that if Gis residually nilpotent (that is, γ (G) = {1}),
Tc≥1 c
.
1 then I A(G) = {1}.
Tc≥2 c
0
Throughout this paper, by “Lie algebra”, we mean Lie algebra over the ring of integers
7
Z. Let G be a group. Write gr (G) = γ (G)/γ (G) for c ≥ 1 and denote by (a,b) the
1 c c c+1
: commutator (a,b) = a−1b−1ab with a,b ∈ G. The (restricted) direct sum of the quotients
v
Xi grc(G) is the associated graded Lie algebra of G, gr(G) = Lc≥1grc(G). The Lie bracket mul-
tiplication in L (G) is [aγ (G),bγ (G)] = (a,b)γ (G), where aγ (G) and bγ (G)
c+1 d+1 c+d+1 c+1 d+1
r
a are the images of the elements a ∈ γc(G) and b ∈ γd(G) in the quotient groups grc(G) and
gr (G), respectively, and (a,b)γ (G) is the image of the group commutator (a,b) in the
d c+d+1
quotient group gr (G). Multiplication is then extended to gr(G) by linearity.
c+d
For a positive integer n, with n ≥ 2, we write F for a free group of rank n with a free
n
generating set {x ,...,x }. For c ≥ 2, we write F = F /γ (F ). Thus, F is the
1 n n,c−1 n c n n,c−1
free nilpotent group of rank n and class c − 1. The natural group epimorphism from F
n
onto F induces a group homomorphism π : Aut(F ) −→ Aut(F ). We write
n,c−1 n,c−1 n n,c−1
I A(F ) = Kerπ . It is well known that, for t,s ≥ 2, (I A(F ),I A(F )) ⊆ I A(F ).
c n n,c−1 t n s n t+s−1 n
Since F is residually nilpotent, we have I A(F ) = {1}. Since F /gr (F ) ∼= F
n Tc≥2 c n n,c c n n,c−1
and gr (F ) is a fully invariant subgroup of F , the natural group epimorphism from F
c n n,c n,c
onto F induces a group homomorphism ψ : Aut(F ) −→ Aut(F ). It is
n,c−1 c,c−1 n,c n,c−1
well known that ψ is onto. For c ≥ 2, we define A∗(F ) = Imπ ∩ Kerψ . For
c,c−1 c n n,c c,c−1
t ∈ {2,...,c}, the natural group epimorphism from F onto F /γ (F ) induces a group
n,c n,c t n,c
homomorphism σ : Aut(F ) → Aut(F /γ (F )). Write I A(F ) = Kerσ , and, for
c,t n,c n,c t n,c t n,c c,t
t = 2, IA(F ) = I A(F ). We note that, for c≥ 2, F /γ (F )∼= F . Thus, for c≥ 2,
n,c 2 n,c n,c c n,c n,c−1
1
A∗(F ) = Imπ ∩I A(F ). It is easily proved that A∗(F ) ∼= I A(F )/I A(F ) as free
c n n,c c n,c c n c n c+1 n
abelian groups (see, also, [1, Section 4, p. 246]). Furthermore, for n,c ≥ 2, rank(A∗(F )) ≤
c n
n rank(gr (F )) = n µ(d)nc/d, where µ is the M¨obius function. We point out that
c n c Pd|c
rank(gr (F )) = 1 µ(d)nc/d for all n,c≥ 2 (see, for example, [8]).
c n c Pd|c
Fromnowon,forr ≥ 2,wewriteLr(IA(F )) = I A(F )/I A(F ). Form the(restricted)
n r n r+1 n
direct sum of the free abelian groups Lr(IA(F )), and denoted by
n
L(IA(Fn)) = MLr(IA(Fn)).
r≥2
It has the structure of a graded Lie algebra with Lr(IA(F )) as component of degree r−1 in
n
the grading and Lie multiplication given by
[φI A(F ),ψI A(F )] = (φ−1ψ−1φψ)I A(F ),
j+1 n κ+1 n j+κ n
for all φ ∈ I A(F ), ψ ∈ I A(F ) (j,κ ≥ 2). Multiplication is then extended to L(IA(F ))
j n κ n n
by linearity. The above Lie algebra is usually called the Johnson Lie algebra of IA(F ). We
n
point out that, for a positive integer c, γ (IA(F )) ⊆ I A(F ).
c n c+1 n
Let H be a finitely generated subgroup of IA(F ) with H/H′ torsion-free. For a positive
n
q
integer q, let L (H) = γ (H)(I A(F ))/I A(F ). Form the (restricted) direct sum of
1 q q+2 n q+2 n
q
abelian groups L (H) = L (H). It is easily verified that L (H) is a Lie subalgebra of
1 Lq≥1 1 1
L(IA(F )). Furthermore, if {y H′,...,y H′} is a Z-basis for H/H′, then L (H) is generated
n 1 m 1
as Lie algebra by the set {y (I A(F )),...,y (I A(F ))}. By a natural embedding of gr(H)
1 3 n m 3 n
into L(IA(F )), we mean that there exists a Lie algebra isomorphism φ from gr(H) onto
n
L (H) satisfying the conditions φ(y H′) = y (I A(F )), i = 1,...,m. In this case, we also
1 i i 3 n
say that gr(H) is naturally isomorphic to L (H).
1
For n ≥ 2, itwas shownby Magnus [7], usingwork of Nielsen [12], thatIA(F ) hasa finite
n
generating set {χ ,χ : 1 ≤ i,j,k ≤ n;i 6= j,k;j < k}, where χ maps x 7→ x (x ,x ) and
ij ijk ij i i i j
χ maps x 7→ x (x−1,x−1), with both χ and χ fixing the remaining basis elements. Let
ijk i i j k ij ijk
M be the subgroup of IA(F ) generated by the subset S = {χ : 1≤ i,j ≤ n; i6= j}. Then,
n n ij
M is called the McCool group or the basis conjugating automorphisms group. It is easily
n
verified that the following relations are satisfied by the elements of S, provided that, in each
case, the subscripts i,j,k,q occurring are distinct: (χ ,χ ) = (χ ,χ ) = (χ χ ,χ ) = 1.
ij kj ij kq ij kj ik
It has been proved in [9] that M has a presentation hS |Zi, where Z is the set of all possible
n
relations of the above forms. Since γ (M ) ⊆ γ (IA(F )) ⊆ I A(F ) for all c ≥ 1, and since
c n c n c+1 n
F is residually nilpotent, we have γ (M ) = {1} and so, M is residually nilpotent.
n Tc≥1 c n n
In the present paper, we show the following result.
Theorem 1 1. For a positive integer c,
γ (M )/γ (M ) ∼= γ (F )/γ (F )⊕γ (F )/γ (F )
c 3 c+1 3 c 3 c+1 3 c 3 c+1 3
as free abelian groups.
2. Let H be the subgroup of M generated byχ ,χ ,χ . Then, L (M ) isadditively equal
3 21 12 23 1 3
to the direct sum of the Lie subalgebras L (H) and L (Inn(F )), where Inn(F ) denotes
1 1 3 3
the group of inner automorphisms of F . Furthermore, gr(M ) is naturally isomorphic
3 3
to L (M ) as Lie algebras if and only if gr(H) is naturally isomorphic to L (H) as Lie
1 3 1
algebras.
2
In [10, Theorem 1], it is shown that M is a Magnus group. The proof of it was long
3
and tedious. In Section 2, we present a rather simple proof avoiding many of the technical
results. The new approach gives us the description of each quotient group γ (M )/γ (M )
c 3 c+1 3
as in Theorem 1 (1). By a result of Sjogren [14] (see, also, [5, Corollary 1.9]), M satisfies the
3
Subgroup Dimension Problem. That is, each γ (M ) is equal to the c-th dimension subgroup
c 3
ofM . Furthermore,thenewapproachhelpsustogiveanecessaryandsufficientconditionfor
3
a natural embedding of gr(M ) into L(IA(F )). For n ≥ 2, let Inn(F ) denote the subgroup
3 3 n
of IA(F ) consisting of all inner automorphisms of F . In Section 3, by using an observation
n n
of Andreadakis [1, Section 6, p. 249], we show that gr(Inn(F )) is naturally embedded into
n
L(IA(F )). Hence, γ (Inn(F ))/γ (Inn(F )) is isomorphic to a subgroup of Lc+1(IA(F ))
n c n c+1 n n
for all c ≥ 1. Since Inn(F ) ∼= F , we obtain
n n
1
Xµ(d)nc/d ≤ rank(Lc+1(IA(Fn)))
c
d|c
for all n,c, with n ≥ 2. For c = 1 and n ≥ 2, we have rank(L2(IA(F ))) = n2(n−1) (see, [1,
n 2
Theorem 5.1]). For n = 2 we have IA(F ) = Inn(F ), by a result of Nielsen [11] and by a
2 2
result of Andreadakis [1, Theorem 6.1], we have rank(Lc+1(IA(F ))) = 1 µ(d)2c/d. For
2 c Pd|c
n = 3, by Theorem 1(2), we may give a lower bound of rank(Lc+1(IA(F ))) in terms of the
3
rank of Lc(H) for all c. In fact, we observe that
1
1 1
Xµ(d)2c/d+ Xµ(d)3c/d ≤ rank(Lc+1(IA(F3)))
c c
d|c d|c
(see, Remark 2 below). For n ≥ 4 and c≥ 2, Satoh [15, Corollary 3.3] provides a lower bound
for rank(Lc+1(IA(F ))).
n
2 The associated Lie algebra of M
3
2.1 Lazard elimination
For a free Z-module A, let L(A) be the free Lie algebra on A, that is, the free Lie algebra
on A, where A is an arbitrary Z-basis of A. Thus, we may write L(A) = L(A). For a
positive integer c, let Lc(A) denote the cth homogeneous component of L(A). It is well-
known that L(A) = Lc(A). For Z-submodules A and B of any Lie algebra, let [A,B]
Lc≥1
be the Z-submodule spanned by [a,b] where a ∈ A and b ∈ B. Furthermore, B ≀A denotes
the Z-submodule defined by B ≀A= B +[B,A]+[B,A,A]+···.
Throughout this paper, we use the left-normed convention for Lie commutators. The
following result is a version of Lazard’s ”Elimination Theorem” (see [2, Chapter 2, Section
2.9, Proposition 10]). In the form written here it is a special case of [3, Lemma 2.2] (see, also,
[10, Section 2.2]).
Lemma 1 Let U and V be free Z-modules, and consider the free Lie algebra L(U⊕V). Then,
U and V ≀ U freely generate Lie subalgebras L(U) and L(V ≀ U), and there is a Z-module
decomposition L(U⊕V)= L(U)⊕L(V ≀U). Furthermore, V ≀U = V ⊕[V,U]⊕[V,U,U]⊕···
and, for each n ≥ 0, there is a Z-module isomorphism [V,U,...,U]∼= V ⊗U ⊗···⊗U under
| {nz } | {nz }
which [v,u ,...,u ] corresponds to v⊗u ⊗···⊗u for all v ∈ V and all u ,...u ∈ U.
1 n 1 n 1 n
3
As a consequence of Lemma 1, we have the following result.
Corollary 1 For free Z-modules U and V, we write L(U ⊕ V) for the free Lie algebra
on U ⊕ V. Then, there is a Z-module decomposition L(U ⊕ V) = L(U) ⊕ L(V) ⊕ L(W),
where W = W ⊕W ⊕··· such that, for all m ≥ 2, W is the direct sum of submodules
2 3 m
[V,U,U,...,U,V,...,V] with a+b = m−2 and a,b ≥ 0. Each [V,U,U,...,U,V,...,V] is
| {az } | {bz } | {az } | {bz }
isomorphic to V ⊗U ⊗U ⊗···⊗U ⊗V ⊗···⊗V as Z-module. Furthermore, L(W) is the
| {az } | {bz }
ideal of L(U ⊕V) generated by the module [V,U].
2.2 A decomposition of a free Lie algebra
Let X be the free Z-module with a Z-basis {x ,...,x } and L = L(X) the free Lie algebra
1 6
on X. For i = 1,2,3, let v = x +x . Furthermore, we write
2i 2i−1 2i
U = Z-span{x ,x ,x } and V = Z-span{v ,v ,v }.
1 3 5 2 4 6
Since X = U ⊕V, we have L = L(U ⊕V) and so, L is free on X = {x ,x ,x ,v ,v ,v }. Let
1 3 5 2 4 6
J be the subset of L,
J = {[v ,x ],[v ,x ],[v ,x ],[v ,v ]−[v ,x ],[v ,v ]−[v ,x ],[v ,v ]−[v ,x ],
2 1 4 3 6 5 4 2 4 1 2 4 2 3 4 6 4 5
[v ,x ],[v ,x ],[v ,x ]}.
6 1 6 3 2 5
The aim of this section is to show the following result.
Proposition 1 With the above notation, let L = L(U⊕V) be the free Lie algebra on U⊕V.
Let J be the ideal of L generated by the set J. Then, L = L(U)⊕L(V)⊕J. Moreover, J is
a free Lie algebra.
For non negative integers a and b, we write [V,U, U, V] for [V,U,U,...,U,V,...,V].
a b
| {az } | {bz }
By Lemma 1 and Corollary 1, we have
L = L(U ⊕V)
= L(U)⊕L(V ≀U)
= L(U)⊕L(V)⊕L(W),
where W = W ⊕W ⊕··· such that, for all m ≥ 2,
2 3
Wm = M [V,U, aU, bV].
a+b=m−2
Furthermore, L(V ≀U) and L(W) are the ideals in L generated by the modules V ≀U and
[V,U], respectively. In particular, L(W) is the ideal in L generated by the natural Z-basis
[V,U] = {[v ,x ],[v ,x ],[v ,x ],[v ,x ],[v ,x ],[v ,x ],[v ,x ],[v ,x ],[v ,x ]}
2 1 4 3 6 5 4 1 2 3 4 5 6 1 6 3 2 5
of [V,U]. Let X be the natural Z-basis of V ≀U. That is,
V,U
XV,U = V ∪([[V, aU]),
a≥1
4
where [V, U] is the natural Z-basis of the module [V, U]. Let ψ be the Z-linear mapping
a a 2
from [V,U] into L(V ≀U) with
ψ ([v ,x ])= [v ,v ]−[v ,x ], ψ ([v ,x ]) = [v ,v ]−[v ,x ], ψ ([v ,x ]) = [v ,v ]−[v ,x ]
2 4 1 4 2 4 1 2 2 3 2 4 2 3 2 4 5 4 6 4 5
andψ fixestheremainingelementsof[V,U]. Itisclearenoughthatψ isaZ-linearmonomor-
2 2
phism of [V,U] into L(V ≀ U). For a ≥ 3, let ψ be the mapping from [V,U, U] into
a (a−2)
L(V ≀U) satisfying the conditions ψ ([v,u,u ,...,u ]) = [ψ ([v,u]),u ,...,u ] for all v ∈ V
a 3 a 2 3 a
and u,u ,...,u ∈U. We define a map
3 a
Ψ :X → L(V ≀U)
V,U
by Ψ(v) = v for all v ∈ V and, for a ≥ 2, Ψ(v) = ψ (v) for all v ∈ [V,U, U]. Since
a (a−2)
L(V ≀U) is free on X , we obtain Ψ is a Lie algebra homomorphism. By applying Lemma
V,U
2.1 in [4], we see that Ψ is a Lie algebra automorphism of L(V ≀U). Since L(W) is a free Lie
subalgebra of L(V ≀U) and Ψ is an automorphism, we have Ψ(L(W)) is a free Lie subalgebra
of L(V ≀U). In fact,
Ψ(L(W)) =L(Ψ(W)),
that is, Ψ(L(W)) is a free Lie algebra on Ψ(W).
Lemma 2 With the above notation, L(Ψ(W)) is an ideal in L.
Proof. Since Ψ is an automorphism of L(V ≀U), we obtain Ψ(L(W)) = L(Ψ(W)) is an ideal
in L(V ≀U). We point out that
L(V ≀U) = Ψ(L(V ≀U))
(By Corollary 1) = Ψ(L(V)⊕L(W))
(Ψ automorphism) = Ψ(L(V))⊕Ψ(L(W))
= L(V)⊕L(Ψ(W))
and so,
L = L(U)⊕L(V)⊕L(Ψ(W)).
To show that L(Ψ(W)) is an ideal in L, it is enough to show that [w,u] ∈ L(Ψ(W)) for all
w ∈ L(Ψ(W)) and u ∈ L. Since L(Ψ(W)) is an ideal in L(V ≀U) and Lemma 1, it is enough
to show that [w,u] ∈ L(Ψ(W)) for all w ∈ L(Ψ(W)) and u ∈ L(U). Furthermore, we may
show that [w,x ,...,x ] ∈ L(Ψ(W)) for all w ∈ L(Ψ(W)) and x ,...,x ∈ {x ,x ,x }.
i1 ik i1 ik 1 3 5
Write
C = Ψ([V,U])∪( [ [Ψ([V,U]), aU, bV]). (1)
a+b≥1
a,b≥0
Since C is a Z-basis for Ψ(W), we have L(Ψ(W)) = L(C). Thus, we need to show that
[w,x ,...,x ] ∈ L(Ψ(W)) for all w ∈ C and x ,...,x ∈ {x ,x ,x }. Since C is a free
i1 ik i1 ik 1 3 5
generating set of L(Ψ(W)), the equation (1) and the linearity of the Lie bracket, we may
assume that w ∈ [Ψ([V,U]), U, V] with a+b ≥ 1. Clearly, we may assume that b ≥ 1.
a b
Since L(Ψ(W)) is ideal in L(V ≀U) and, by the Jacobi identity, we may further assumethat w
has a form [v,y ,...,y , v , v , v ] with y ,...,y ∈ U, µ,ν,ρ ≥ 0 and µ+ν +ρ≥ 1.
j1 ja µ 2 ν 4 ρ 6 j1 ja
By using the Jacobi identity in the expression [w,x ,...,x ], and replacing [v ,x ],[v ,x ]
i1 ik 4 1 2 3
and[v ,x ]by[v ,v ]−ψ ([v ,x ]),[v ,v ]−ψ ([v ,x ])and[v ,v ]−ψ ([v ,x ]),respectively,
4 5 4 2 2 4 1 2 4 2 2 3 4 6 2 4 5
as many times as it is needed and since L(Ψ(W)) is an ideal in L(V ≀U), we may show that
[w,x ,...,x ] ∈ L(Ψ(W)). Therefore, L(Ψ(W)) is an ideal in L. (cid:3)
i1 ik
5
Example 1 In the present example, we explain the procedure described in the above proof.
Let w = [ψ ([v ,x ]),x ,v ,v ]. Then,
2 4 1 3 2 4
[w,x ] = [ψ ([v ,x ]),x ,v ,v ,x ]
1 2 4 1 3 2 4 1
= [ψ ([v ,x ]),x ,v ,x ,v ]+[ψ ([v ,x ]),x ,v ,[v ,x ]]
2 4 1 3 2 1 4 2 4 1 3 2 4 1
= [ψ ([v ,x ]),x ,v ,x ,v ]+
2 4 1 3 2 1 4
[ψ ([v ,x ]),x ,v ,[v ,v ]]−[ψ ([v ,x ]),x ,v ,ψ ([v ,x ])]
2 4 1 3 2 4 2 2 4 1 3 2 2 4 1
= [ψ ([v ,x ]),x ,x ,v ,v ]+[ψ ([v ,x ]),x ,[v ,x ],v ]+
2 4 1 3 1 2 4 2 4 1 3 2 1 4
[ψ ([v ,x ]),x ,v ,[v ,v ]]−[ψ ([v ,x ]),x ,v ,ψ ([v ,x ])] ∈ L(Ψ(W)).
2 4 1 3 2 4 2 2 4 1 3 2 2 4 1
Proof of Proposition 1. Since ψ ([V,U]) ⊆ J and J is ideal in L, we get L(Ψ(W)) ⊆ J. Since
2
J ⊆ L(Ψ(W)), by Lemma 2, we obtain J ⊆ L(Ψ(W)). Therefore, J = L(Ψ(W)). That is, J
is a free Lie algebra. Furthermore, L = L(U)⊕L(V)⊕J. (cid:3)
For c ≥ 2, let Jc = J∩Lc. Since J is homogeneous, we get J = Jc. From the above
Lc≥2
proof, we have the following result.
Corollary 2 With the above notation, let Ψ be the Lie algebra automorphism of L(V ≀ U)
defined naturally on V ≀U by means of ψ . Then, J = L(Ψ(W)). Furthermore, for c ≥ 2,
2
Lc = Lc(U)⊕Lc(V)⊕Jc.
2.3 A description of gr(M )
3
Ouraim in this section is toshow thefollowing result. For its proof,we usesimilar arguments
as in [10, Section 6].
Theorem 2 Let M be the McCool subgroup of the IA-automorphisms IA(F ) of F . Then,
3 3 3
∼
gr(M ) = L/J as Lie algebras. In particular, gr(M ) is isomorphic as a Lie algebra to an
3 3
(external) direct sum of two copies of a free Lie algebra of rank 3. Furthermore, for each c,
γ (M )/γ (M )∼= γ (F )/γ (F )⊕γ (F )/γ (F )
c 3 c+1 3 c 3 c+1 3 c 3 c+1 3
as free abelian groups.
Following the notation of the previous subsection, let us denote by F the free group
generated by {x ,...,x }. It is well known that L (F) is a free Lie algebra of rank 6; freely
1 6
generated by the set {x F′ : i = 1,...,6}. The free Lie algebras L and L (F) are isomorphic
i
and from now on we identify the two Lie algebras. Furthermore, Lc = γ (F)/γ (F) for all
c c+1
c ≥ 1. Define
r = (x ,x ), r = (x ,x ), r = (x ,x )
1 1 2 2 3 4 3 5 6
r = (x x ,x ), r = (x x ,x ), r = (x x ,x )
4 1 2 5 5 3 4 6 6 1 2 4
r = (x x ,x ), r = (x x ,x ), r = (x x ,x ),
7 3 4 2 8 5 6 3 9 5 6 1
and R= {r ,...,r }.
1 9
LetN = RF bethenormalclosureofRinF. For apositiveinteger d,letN = N∩γ (F).
d d
We point out that for d ≤ 2, N = N. Further, for d ≥ 2, N = N ∩γ (F). Define
d d+1 d d+1
I (N) = N γ (F)/γ (F). It is easily verified that I (N) ∼= N /N as Z-modules. The
d d d+1 d+1 d d d+1
following result was shown in [10, Section 6].
Proposition 2 For a positive integer c, N is generated by the set {(r±1,g , ...,g ) : r ∈
c+2 1 s
R,s ≥ c,g ,...,g ∈ F \{1}}. Furthermore, I (N) = Jc+2 for all c≥ 1.
1 s c+2
6
Since F is residually nilpotent, we have N = {1}. Also N ⊆ F′, we get I (N) = 0.
Td≥2 d 1
Moreover, I (N) ∼= N /N as Z-modules for all d ≥ 2, and by Proposition 2 we have,
d d d+1
N 6= N for all d ≥ 2. Define
d d+1
I(N) =MNdγd+1(F)/γd+1(F) = MId(N).
d≥2 d≥2
Since N is a normal subgroup of F, we have I(N) is an ideal of L (see [6]).
Corollary 3 I(N)= J.
Proof. Since J = Jd and I (N) = J2, we have from Proposition 2 that I(N)= J. (cid:3)
Ld≥2 2
Proof of Theorem 2. Since M /M′ ∼= F/NF′ = F/F′, we have L (M ) is generated as a
3 3 3
Lie algebra by the set {α : i = 1,...,6} with α = x M′. Since L is a free Lie algebra of
i i i 3
rank 6 with a free generating set {x ,...,x }, the map ζ from L into L (M ) satisfying the
1 6 3
conditions ζ(x ) = α , i = 1,...,6, extends uniquely to a Lie algebra homomorphism. Since
i i
L (M ) is generated as a Lie algebra by the set {α : i = 1,...,6}, we have ζ is onto. Hence,
3 i
L/Kerζ ∼= L (M ) as Lie algebras. By definition, J ⊆ Kerζ, and so ζ induces a Lie algebra
3
epimorphism ζ from L/J onto L (M ). In particular, ζ(x +J) = α , i = 1,...,6. Note that
3 i i
ζ induces ζ , say, a Z-linear mapping from (Lc+J)/J onto γ (M )/γ (M ). For c≥ 2,
c c 3 c+1 3
γ (M )/γ (M ) ∼= γ (F)γ (F)N/γ (F)N ∼= γ (F)/(γ (F)∩γ (F)N).
c 3 c+1 3 c c+1 c+1 c c c+1
Since γ (F) ⊆ γ (F), we have by the modular law,
c+1 c
γ (F)/(γ (F)∩γ (F)N) = γ (F)/γ (F)N .
c c c+1 c c+1 c
But, by Proposition 2, for c≥ 3,
γ (F)/γ (F)N ∼= (γ (F)/γ (F))/I (N) ∼= Lc/Jc.
c c+1 c c c+1 c
Since I (N) = J2, we obtain, for c ≥ 2,
2
γ (F)/γ (F)N ∼= (γ (F)/γ (F))/I (N) ∼= Lc/Jc.
c c+1 c c c+1 c
Therefore, for c≥ 1,
γ (M )/γ (M ) ∼=Lc/Jc ∼= (Lc)∗,
c 3 c+1 3
by Corollary 2, where (Lc)∗ = Lc(U) ⊕ Lc(V). Since both L(U) and L(V) are free Lie
algebras of rank 3, we have L(U) ∼= L(V) ∼= gr(F ) in a natural way and so, for c ≥ 1,
3
Lc(U) ∼=Lc(V)∼= γ (F )/γ (F ) as free abelian groups. Hence, for c ≥ 1,
c 3 c+1 3
γ (M )/γ (M )∼= γ (F )/γ (F )⊕γ (F )/γ (F )
c 3 c+1 3 c 3 c+1 3 c 3 c+1 3
as free abelian groups and so, rank(γ (M )/γ (M )) = rank(Lc)∗. Since J = Jc, we
c 3 c+1 3 Lc≥2
have (Lc+J)/J ∼= Lc/(Lc∩J)= Lc/Jc ∼= (Lc)∗ and so, we obtain Kerζ is torsion-free. Since
c
rank(γ (M )/γ (M )) = rank(Lc)∗,wehaveKerζ = {1}andso,ζ isisomorphism. Sinceζ
c 3 c+1 3 c c
is epimorphism and each ζ is isomorphism, we have ζ is isomorphism. Hence, L/J ∼= L (M )
c 3
as Lie algebras. (cid:3)
7
3 Embeddings
In this section, we shall give a criterion for the natural embeddingof gr(M ) into L (IA(F )).
3 1 3
We shall prove the following result.
Lemma 3 Let H be a finitely generated subgroup of IA(F ), n ≥ 2, with H/H′ torsion-free,
n
and let {y H′,...,y H′} be a Z-basis for H/H′. Then, gr(H) is naturally isomorphic to
1 m
L (H) if and only if γ (H) = H ∩(I A(F )) for all c.
1 c c+1 n
Proof. We assume that γ (H) = H ∩ (I A(F )) for all c. For c ≥ 1, let ψ be the
c c+1 n c
natural Z-module epimorphism from gr (H) onto Lc(H). Since γ (H) ∩ (I A(F )) =
c 1 c c+2 n
H ∩(I A(F )) = γ (H) for all c, we get ψ is isomorphism for all c ≥ 1. Since gr(H)
c+2 n c+1 c
is the (restricted) direct sum of the quotients gr (H), there exists a group homomorphism ψ
c
from gr(H) into L (IA(F )) such that each ψ is the restriction of ψ on gr (H). It is easily
1 n c c
shown that ψ is a Lie algebra homomorphism. Since ψ(y H′) = y (I A(F )), i = 1,...,n,
i i 3 n
we get ψ is a Lie algebra epimorphism. Furthermore, since each ψ is a Z-module isomor-
c
phism, we obtain ψ is a Lie algebra isomorphism. Conversely, let φ be a Lie algebra isomor-
phism from gr(H) onto L (H) satisfying the conditions φ(y H′) = y (I A(F )), i= 1,...,m.
1 i i 3 n
Then, φ induces a Z-module isomorphism φ from gr (H) onto Lc(H) for all c. In par-
c c 1
ticular, φ ((y ,...,y )γ (H)) = (y ,...,y )(I A(F )) for all i ,...,i ∈ {1,...,m}.
c i1 ic c+1 i1 ic c+2 n 1 c
Furthermore, gr (H) is Z-module isomorphic to γ (H)/(γ (H)∩(I A(F ))). Since gr (H)
c c c c+2 n c
is polycyclic and so, it is a hopfian group, we have γ (H) = γ (H) ∩(I A(F )) for all
c+1 c c+2 n
c. We claim that γ (H) = H ∩(I A(F )) for all c. Since γ (H) ⊆ I A(F ), it is enough
c c+1 n c c+1 n
to show that H ∩ (I A(F )) ⊆ γ (H). To get a contradiction, let α ∈ H ∩ (I A(F ))
c+1 n c c+1 n
and α ∈/ γ (H). Since γ (H) ⊆ I A(F ) for all c and I A(F ) = {1}, we get H is
c c c+1 n Tt≥2 t n
residually nilpotent. Thus, there exists a unique d ∈ N such that α ∈ γ (H) \ γ (H).
d d+1
Therefore, α ∈/ H ∩ (I A(F )). Since α ∈ γ (H) \ γ (H) and α ∈/ γ (H), we have
d+2 n d d+1 c
γ (H) ⊆ γ (H) and so, d + 1 ≤ c. Let k be a non-negative integer such that c =
c d+1
d + 1 + k. Since H ∩ (I A(F )) = H ∩ (I A(F )), we have α ∈ I A(F ). But
c+1 n d+2+k n d+2+k n
I A(F ) ⊆ I A(F ) and so, α ∈ I A(F ), which is a contradiction. Therefore,
d+2+k n d+2 n d+2 n
γ (H) = H ∩(I A(F )) for all c. (cid:3)
c c+1 n
Remark 1 It is known that IA(F )/γ (IA(F )), with n ≥ 2, is torsion-free and its rank is
n 2 n
n2(n−1)
. Furthermore, γ (IA(F )) = I A(F ) (see, for example, [13]). It was conjectured by
2 2 n 3 n
Andreadakis [1] that γ (IA(F )) = I A(F ) for all c. By Lemma 3, gr(IA(F )) is naturally
c n c+1 n n
isomorphictoL (IA(F ))ifandonlyifAndreadakisconjectureisvalid. Now, ifH isafinitely
1 n
generated subgroup of IA(F ) with H/H′ torsion free, then the statement gr(H) is naturally
n
isomorphic to L (H) seems to be an “Andreadakis Conjecture” for H.
1
3.1 The associated Lie algebra of Inn(F )
n
In this section, we show that the associated Lie algebra gr(Inn(F )) of the inner automor-
n
phisms Inn(F ) of F , with n ≥ 2, is naturally embedded into L(IA(F )). Throughout this
n n n
section, we write E =Inn(F ). Recall that, for g ∈ F , τ (x) = gxg−1 for all x ∈ F . Thus,
n n n g n
E = {τ :g ∈ F }. The following result has been proved in [1, Section 6].
n g n
Lemma 4 For a positive integer c, γ (E )= E ∩I A(F ).
c n n c+1 n
8
Using the above we may show the following.
Proposition 3 Letnbepositive integer, withn ≥ 2. Then, gr(E )isnaturally embedded into
n
L(IA(F )). In particular, for all c, gr (E ) is isomorphic to a Z-submodule of Lc+1(IA(F )).
n c n n
Proof. Since F is centerless, we have F ∼= E in a natural way and so, E is finitely
n n n n
generated. Moreover, gr(E )isafreeLiealgebraofrankn. Namely,gr(E )isfreelygenerated
n n
by the set {τ E′ : i = 1,...,n}. Let φ be the mapping from {τ E′ : i = 1,...,n} to
xi n xi n
L (E ) satisfying the conditions φ(τ E′) = τ I A(F ), i = 1,...,n. Since gr(E ) is free on
1 n xi n xi 3 n n
{τ E′ :i = 1,...,n}, φ is extended to a Lie algebra epimorphism. By Lemma 3, it is enough
xi n
to show thatγ (E )= E ∩I A(F )for all c, which is valid by Lemma4. Therefore, gr(E )
c n n c+1 n n
is naturally embedded into L(IA(F )). Since
n
Lc(E ) ∼= γ (E )/(γ (E )∩(I A(F )))
1 n c n c n c+2 n
= γ (E )/(E ∩I A(F )))
c n n c+2 n
= gr (E )
c n
for all c, we have gr (E ) is isomorphic to a Z-submodule of Lc+1(IA(F )) for all c. (cid:3)
c n n
Corollary 4 For positive integers n and c, with n ≥ 2,
1
Xµ(d)nc/d ≤ rank(Lc+1(IA(Fn)))
c
d|c
where µ is the Mo¨bius function.
Proof. Since F ∼= E as groups and gr(F ) is a free Lie algebra, we get gr(F ) ∼= gr(E )
n n n n n
as Lie algebras. Hence, for c ≥ 1, gr (F ) ∼= gr (E ) as Z-modules. Since rank(gr (F )) =
c n c n c n
1 µ(d)nc/d, we obtain, by Proposition 3, the required result. (cid:3)
c Pd|c
3.2 The Lie algebra L (M )
1 3
Write b = χ , b = χ , b = χ , u = χ χ , u = χ χ and u = χ χ . Then, M is
1 21 2 12 3 23 2 31 21 4 32 12 6 23 13 3
generated by the set {b ,b ,b ,u ,u ,u } and its presentation is given by
1 2 3 2 4 6
M = hb ,b ,b ,u ,u ,u : (u ,b ),(u ,u b−1),(u ,b ),
3 1 2 3 2 4 6 2 1 2 4 2 2 3
(u ,u b−1),(u ,b ),(u ,b−1u ),(u ,b ),(u ,b ),(u ,b )i.
4 2 1 4 2 4 3 6 6 1 6 2 6 3
We point out that u = τ , u = τ and u = τ . Let H and E be the subgroups of
2 x−1 4 x−1 6 x−1
1 2 3
M generated by the sets {b ,b ,b } and {u ,u ,u }, respectively. We point out that E =
3 1 2 3 2 4 6
Inn(F ). One can easily see that H is a free group of rank 3. Thus, gr(H) ∼= gr(E) ∼= gr(F )
3 3
as Lie algebras.
Proposition 4 L (M ) is additively equal to the direct sum of the Lie subalgebras L (H)
1 3 1
and L (E).
1
9
∼
Proof. By the proof of Proposition 3, gr(E) = L (E) and so, L (E) is a free Lie algebra of
1 1
rank 3. Since b ∈/ I A(F ), we have L (H) is a non-trivial subalgebra of L(IA(F )). In fact,
1 3 3 1 3
L (H) is generated by the set {b (I A(F )) : i = 1,2,3}. Since (τ ,φ) = τ for all
1 i 3 3 g g−1φ−1(g)
φ∈ Aut(F ) and g ∈ F , we have L (E) is an ideal in L(IA(F )) and so, L (H)+L (E) is a
3 3 1 3 1 1
Lie subalgebra of L (M ). Let w ∈ L (H)∩L (E). Since both L (H) and L (E) are graded
1 3 1 1 1 1
Lie algebras, we may assume that w ∈ Ld(H)∩Ld(E) for some d. Thus, there are u∈ γ (H)
1 1 d
and v ∈ γ (E) such that w = u(I A(F )) = v(I A(F )). To get a contradiction, we
d d+2 3 d+2 3
assume that u,v ∈/ I A(F ). Therefore, v ∈ γ (E) \ γ (E) and so, there exists ω ∈
d+2 3 d d+1
γ (F )\γ (F ) such that v = τ ρ, where ρ ∈ γ (E). Since γ (E) ⊆ I A(F ), we get
d 3 d+1 3 ω d+1 d+1 d+2 3
u−1τ ∈ I A(F ). Sinceu−1 fixesx ,wehavex−1(u−1τ (x )) = (x ,u−1(ω−1)) ∈γ (F ).
ω d+2 3 3 3 ω 3 3 d+2 3
Since u−1(x )= x y , y ∈ γ (F ), j = 1,2, and γ (F ) is a fully invariant subgroup of F ,
j j j j d+1 3 d 3 3
we have u−1(ω−1)= ω−1ω , with ω ∈ γ (F ) and so, (w,x )∈ γ (F ). Since gr(F ) is a
1 1 d+1 3 3 d+2 3 3
free Lie algebra of rank 3 with a free generating set {x F′ : i = 1,2,3} and γ (F )/γ (F )
i 3 d 3 d+1 3
is the d-th homogeneous component of gr(F ), we obtain (w,x ) ∈ γ (F )\γ (F ), which
3 3 d+1 3 d+2 3
is a contradiction. Therefore, L (H)∩L (E) = {0}. By the proof of Theorem 2, we obtain
1 1
L (M ) = L (H)⊕L (E). (cid:3)
1 3 1 1
Remark 2 By Proposition 4, for all c, we obtain Lc(M ) = Lc(H)⊕Lc(E). Since Lc(E) ∼=
1 3 1 1 1
γ (E)/γ (E) ∼= γ (F )/γ (F ), we have rank(Lc(E)) = 1 µ(d)3c/d. Thus, for any c,
c c+1 c 3 c+1 3 1 c Pd|c
1
rank(Lc1(H))+ c Xµ(d)3c/d ≤ rank(Lc+1(IA(F3))).
d|c
Let H be the subgroup of H generated by the set {χ ,χ }. We point out that H is a free
1 21 23 1
groupofrank2. Then,gr(H )isafreeLiealgebraofrank2. Sincebothχ andχ fixx and
1 21 23 1
x , it may be shown that γ (H ) = H ∩I A(F ) for all c. Therefore, γ (H )∩I A(F )=
3 c 1 1 c+1 3 c 1 c+2 3
γ (H ) for all c and so, Lc(H ) ∼= gr (H ) for all c ≥ 1. Since gr(H ) is a free Lie algebra
c+1 1 1 1 c 1 1
on H = {χ H′,χ H′}, the mapping ψ from H into L (H ) satisfying the conditions
1 21 1 23 1 1 1 1
ψ(aH′) = a(I A(F )) with a ∈ {χ ,χ } can be extended to a Lie algebra homomorphism
1 3 3 21 23
ψ. Since L (H ) is generated as Lie algebra by the set {χ (I A(F )),χ (I A(F ))}, we have
1 1 21 3 3 23 3 3
ψ is onto. By Lemma 3, ψ is a natural Lie algebra isomorphism from gr(H ) onto L (H ).
1 1 1
Since Lc(H ) ∼= gr (H ) for all c≥ 1, we have rank(Lc(H )) = 1 µ(d)2c/d. Since L (H )
1 1 c 1 1 1 c Pd|c 1 1
is a Lie subalgebra of L (H), we obtain Lc(H ) ≤ Lc(H) for all c. Therefore,
1 1 1 1
1
c Xµ(d)2c/d ≤ rank(Lc1(H))
d|c
for all c and so,
1 1
Xµ(d)2c/d+ Xµ(d)3c/d ≤ rank(Lc+1(IA(F3)))
c c
d|c d|c
for all c.
In our next result, we give a necessary and sufficient condition for a natural embedding
of gr(M ) into L(IA(F )).
3 3
Proposition 5 Let H be the subgroup of M generated by χ ,χ ,χ . Then, gr(M ) is
3 21 12 23 3
naturally isomorphic to L (M ) as Lie algebras if and only if gr(H) is naturally isomorphic
1 3
to L (H) as Lie algebras.
1
10