On certain questions of the free group automorphisms theory

0.18 MB·English

by  Valeriy G. Bardakov

ON CERTAIN QUESTIONS OF THE FREE GROUP AUTOMORPHISMS
THEORY
VALERIY G. BARDAKOV AND ROMAN MIKHAILOV
Abstract. CertainsubgroupsofthegroupsAut(Fn)ofautomorphismsofafreegroupFn are
considered. Comparing Alexander polynomials of two poly-free groups Cb+4 and P4 we prove
thatthesegroupsarenotisomorphic,despitethefactthattheyhavealotofcommonproperties.
7
This answers the question of Cohen-Pakianathan-Vershinin-Wu from [8]. The questions of
0
0 linearity of subgroupsof Aut(Fn) are considered. As anapplication ofthe properties ofpoison
2 groups in the sense of Formanek and Procesi, we show that the groups of the type Aut(G∗Z)
n for certaingroups G and the subgroupofIA-automorphisms IA(Fn)⊂Aut(Fn) are not linear
a for n≥3. This generalizes the recent result of Pettet that IA(Fn) are not linear for n≥5.
J

6
1

]
R 1. Introduction
G
.
h
1.1. The group of basis conjugating automorphisms. Let F be a free group of rank
t n
a
n ≥ 2 with a free generator set {x ,x ,...,x } and Aut(F ) the group of automorphisms of F .
m 1 2 n n n
Taking the quotient of F by its commutator subgroup F′, we get a natural homomorphism
[ n n

1 ξ : Aut(F ) −→ Aut(F /F′) = GL (Z),
n n n n
v
1 Z
where GL ( ) is the general linear group over the ring of integers. The kernel of this homo-
4 n
4 morphism consists of automorphisms acting trivially modulo the commutator subgroup F′. It
n
1
is called the group of IA-automorphisms and denoted by IA(F ) (see [2, Chapter 1, § 4]). The
0 n
7 group IA(F2) is isomorphic to the group of inner automorphisms Inn(F2), which is isomorphic
0 to the free group F .
2
/
h J. Nilsen (for the case n ≤ 3) and W. Magnus (for all n) have shown that (see [2, Chapter 1,
t
a § 4]) the group IA(Fn) is generated by the following automorphisms
m
: x 7−→ x [x ,x ] if k 6= i,j, x 7−→ x−1x x if i 6= j,
v ε : i i j k ε : i j i j
i ijk (cid:26) xl 7−→ xl if l 6= i, ij (cid:26) xl 7−→ xl if l 6= i.
X
r The subgroup of the group IA(F ) is generated by the automorphisms ε , 1 ≤ i 6= j ≤ n
a n ij
is called the group of basis conjugating automorphisms. We will denote this group by Cb .
n
The group Cb is a subgroup of the group of conjugating automorphisms C . (Recall that any
n n
automorphism of C sends a generator x to an element of the type f−1x f , where f ∈ F ,
n i i π(i) i i n
and π is a permutation from the symmetric group S .) Clearly, if π is the identity permutation,
n
then the described element lies in the group Cb .
n
It is shown by J. McCool [3] that the group of basis conjugating automorphisms Cb is
n
generated by the automorphisms ε , 1 ≤ i 6= j ≤ n, defined above has the following relations
ij
The research of the first author has been supported by the Scientific and Research Council of Turkey
(TU¨BI˙TAK). The research of the second author was partially supported by Russian Foundation Fond of Fun-
damental Research, grant N 05-01-00993,and the Russian Presidential grant MK-1487.2005.1.
1
ON CERTAIN QUESTIONS OF THE FREE GROUP AUTOMORPHISMS THEORY 2
(we denote different indexes by different symbols):
ε ε = ε ε ,
ij kl kl ij
ε ε = ε ε ,
ij kj kj ij
(ε ε )ε = ε (ε ε ).
ij kj ik ik ij kj
It is shown in [6] that the group of conjugating automorphisms Cb , n ≥ 2 decomposes into
n
a semi-direct product:
Cb = D ⋋(D ⋋(...⋋(D ⋋D ))...), (1)
n n−1 n−2 2 1
where the group D , i = 1,2,...,n − 1, is generated by the elements ε ,ε ,...,ε ,
i i+1,1 i+1,2 i+1,i
ε , ε , ...,ε . The elements ε ,ε ,...,ε generate a free group of rank i, but
1,i+1 2,i+1 i,i+1 i+1,1 i+1,2 i+1,i
the elements ε ,ε ,...,ε generate a free abelian group of rank i.
1,i+1 2,i+1 i,i+1
1.2. The braid group as a subgroup of Aut(F ). The braid group B , n ≥ 2, on n strands
n n
is defined by the generators σ ,σ ,...,σ with defining relators
1 2 n−1
σ σ σ = σ σ σ if i = 1,2,...,n−2,
i i+1 i i+1 i i+1
σ σ = σ σ if |i−j| ≥ 2.
i j j i
There exists a natural homomorphism from the braid group B onto the permutation group
n
S , sending the generator σ to the transposition (i,i+1), i = 1,2,...,n−1. The kernel of this
n i
homomorphism is called the pure braid group and denoted by P . The group P is generated by
n n
the elements a , 1 ≤ i < j ≤ n which can be expressed in terms of generators of B as follows:
ij n
a = σ2,
i,i+1 i
a = σ σ ...σ σ2σ−1 ...σ−1 σ−1 , i+1 < j ≤ n.
ij j−1 j−2 i+1 i i+1 j−2 j−1
The pure braid group P is a semi-direct product of the normal subgroup U , which is the
n n
free group with free generators a ,a ,...,a with the group P . Analogously, P is a
1n 2n n−1,n n−1 n−1
semi-directproductofthefreesubgroupU withthefreegeneratorsa ,a ,...,a
n−1 1,n−1 2,n−1 n−2,n−1
with the subgroup P etc. Hence the group P has the following decomposition
n−2 n
P = U ⋋(U ⋋(...⋋(U ⋋U ))...), U ≃ F ,i = 2,3,...,n. (2)
n n n−1 3 2 i i−1
The pure braid group P is defined by relations (for ν = ±1)
n
a−νa aν = (a a )ν a (a a )−ν ,
ik kj ik ij kj kj ij kj
a−νa aν = (a a )ν a (a a )−ν , m < j,
km kj km kj mj kj kj mj
a−νa aν = a−ν,a−ν νa a−ν,a−ν −ν , i < k < m,
im kj im ij mj kj ij mj
(cid:2) (cid:3) (cid:2) (cid:3)
a−νa aν = a , k < i; m < j or m < k,
im kj im kj
where [a,b] = a−1b−1ab is the commutator of elements a and b.
The braid group B can be embedded into the automorphism group Aut(F ). In this em-
n n
bedding the generator σ , i = 1,2,...,n−1, defines the following automorphism
i
x 7−→ x x x−1,
i i i+1 i
σ :  x 7−→ x ,
i i+1 i
 x 7−→ x if l 6= i,i+1.
l l

ON CERTAIN QUESTIONS OF THE FREE GROUP AUTOMORPHISMS THEORY 3
The generator a of the pure braid group P defines the following automorphism
rs n
x 7−→ x if s < i or i < r,
i i

 x 7−→ x x x x−1x−1,
 r r s r s r
a : 
rs 

 x 7−→ [x−1,x−1]x [x−1,x−1]−1 if r < i < s,
i r s i r s


 xs 7−→ xrxsx−r1.



As was shown by E. Artin, the automorphism β from Aut(F ) belongs to the braid group
n
B if and only if β satisfies the following two conditions:
n
1) β(x ) = a−1x a , 1 ≤ i ≤ n,
i i π(i) i
2) β(x x ...x ) = x x ...x ,
1 2 n 1 2 n
where π is a permutation from S , and a ∈ F .
n i n
The set of automorphisms from C , which act trivially on the product x x ...x is exactly
n 1 2 n
the braid group B . It was shown by A. Savushkina [4], that the braid group B intersects
n n
with the subgroup Cb by the pure braid group P . Thus the group P is a subgroup of
n n n
the basis conjugating automorphism group Cb for all n ≥ 2. Furthermore, the concordance
n
of decompositions (1) and (2) of groups Cb and P respectively, takes a place: there are
n n
corresponding embeddings U ≤ D , i = 1,2,...,n−1.
i+1 i
1.3. The group Cb+. Denote by Cb+ the subgroup of the group Cb , n ≥ 2, generated by the
n n n
elements ε , i > j. Groups Cb+ are poly-free groups: it is shown in [14] that there exists the
ij n
splitting exact sequence:
1 → F → Cb+ → Cb+ → 1,
n−1 n n−1
where F is the free subgroup in Cb+, generated by the elements
n−1 n
ε ,ε ,...,ε .
n1 n2 n,n−1
Description of the Lie algebra constructed from lower central filtration of the basis conjugat-
ing automorphism group, is a non-trivial problem; the methods used in the description of Lie
algebras for pure braid groups, do not work in this case. However, in the case of the group Cb+,
n
the situation is different. Lie algebras and cohomology rings of groups Cb+ were described in
n
[8]. It was also shown in [8] that the group Cb+ is isomorphic to the pure braid group P for
n n
n = 2,3. Groups Cb+ and P are quite similar. For instance, the groups Cb+ and P are stably
n n n n
isomorphic, namely, suspensions over their classifying spaces ΣK(Cb+,1) and ΣK(P ,1) are
n n
homotopically equivalent for alln ≥ 1 (see [8]). The conjecture that Cb+ and P areisomorphic
n n
for all n ≥ 1 comes naturally (question 1, [8]). One of the main results of this paper is the
proof that the groups Cb+ and P are non-isomorphic (Theorem 1).
4 4
1.4. The questions of linearity. As it was shown above, for any n ≥ 2, there is the following
chain of subgroups of the automorphism group Aut(F ):
n
P ⊂ Cb ⊂ IA (F ) ⊂ Aut(F ).
n n n n n
It was shown in [15], [16] that the braid groups B are linear. Hence the pure braid groups
n
P are also are linear. The situation with groups Aut(F ) is as follows. The group Aut(F ) is
n n n
not linear for all n ≥ 3 (see [10]), however the group Aut(F ) is linear (such a representation
2
was constructed, for example, in [11]). A certain non-linear subgroup (called poison group) of
Aut(F )wasconstructedin[10]. Thefollowingquestionrisesnaturally(seequestion15.17[12]):
n
arethegroupsof IA-automorphismsIA(F )andthegroupsofbasis conjugatingautomorphisms
n
ON CERTAIN QUESTIONS OF THE FREE GROUP AUTOMORPHISMS THEORY 4
Cb linear for n ≥ 2? Recently A. Pettet showed that the group IA(F ) is not linear for n ≥ 5.
n n
In this paper we show that the groups IA(F ) are not linear for all n ≥ 3 (Theorem 5).
n
Therefore, the complete answer to the question 15.17 from [12] is given.
2. The group Cb+ is not isomorphic to the pure braid group P
4 4
2.1. Fox differential calculus. Recall the definition and main properties of Fox derivatives
[5, chapter 3], [7, chapter 7].
LetF beafreegroupofranknwithfreegeneratorsx ,x ,...,x . Letϕbeanendomorphism
n 1 2 n
of the group F . Denote by Fϕ the image of F under the endomorphism ϕ. Let Z be the ring
n n n
of integers, ZG the integral group ring of G.
For every j = 1,2,...,n define a map
∂
: ZF −→ ZF
n n
∂x
j
by setting
∂x 1 for i = j,
i
1) =
∂x (cid:26) 0 otherwise,
j
∂x−1 −x−1 for i = j,
2) i = i
∂x (cid:26) 0 otherwise,
j
∂(wv) ∂w ∂v
3) = (v)τ +w , w,v ∈ ZF ,
n
∂x ∂x ∂x
j j j
where τ : ZF −→ Z is the trivialization operation, sending all elements of F to the identity,
n n
∂ ∂g
4) a g = a , g ∈ F , a ∈ Z.
g g n g
∂x ∂x
j (cid:16)X (cid:17) X j
Denote by ∆ the augmentation ideal of the ring ZF , i.e. the kernel of the homomorphism
n n
τ. It is easy to see that for every v ∈ ZF , the element v − vτ belongs to ∆ . Furthermore,
n n
there is the following formula
n
∂v
v−vτ = (x −1), (3)
j
∂x
Xj=1 j
which is called the main formula of Fox calculus. Formula (3) implies that the elements {x −
1
1,x −1,...,x −1} determine the basis of the augmentation ideal ∆ .
2 n n
2.2. Groups Cb+ and P . As it was noted in the introduction, the poly-free groups Cb+ and
4 4 n
P have similar properties. Clearly, one has
n
Cb+ ≃ P ≃ Z;
2 2
Cb+ ≃ P ≃ F ⋊Z.
3 3 2
However, the next result shows that the groups Cb+ and P being semidirect products of F
4 4 3
with F ⋊Z, are not isomorphic.
2
Theorem 1. Groups Cb+ and P are not isomorphic.
4 4
Proof. It is well-known (see [9]) that the group P decomposes as a direct product of its cen-
4
ter generated by a single element a a a a a a and a subgroup H = U ⋋ U , U =
12 13 23 14 24 34 4 3 4
ha ,a ,a i, U = ha ,a i. It is easy to check that the center of the group Cb+ is the infinite
14 24 34 3 13 23 4
cyclic group generated by the element ε ε ε and the whole group Cb+ is a direct product
21 31 41 4
of its center and a subgroup G = D+ ⋋D+, where D+ = hε ,ε ,ε i, D+ = hε ,ε i. Since
3 2 3 41 42 43 2 31 32
ON CERTAIN QUESTIONS OF THE FREE GROUP AUTOMORPHISMS THEORY 5
a center is a characteristic subgroup for every group, it is enough to show that the group G is
not isomorphic to the group H. We will compare Alexander polynomials for the groups G and
H.
The presentation of the group Cb implies that the group G is defined by the following set
4
of relations:
ε−1ε ε = ε , ε−1ε ε = ε , ε−1ε ε = ε ε ε−1,
31 41 31 41 31 42 31 42 31 43 31 41 43 41
ε−1ε ε = ε , ε−1ε ε = ε , ε−1ε ε = ε ε ε−1.
32 41 32 41 32 42 32 42 32 43 32 42 43 42
We present these relations in the following form:
r = ε−1ε−1ε ε , r = ε−1ε−1ε ε , r = ε−1ε−1ε−1ε ε ε ,
11 41 31 41 31 21 42 31 42 31 31 43 41 31 43 31 41
r = ε−1ε−1ε ε , r = ε−1ε−1ε ε , r = ε−1ε−1ε−1ε ε ε .
12 41 32 41 32 22 42 32 42 32 32 43 42 32 43 32 42
We define the homomorphism ϕ : G → hti, by setting
ϕ(ε ) = ϕ(ε ) = ϕ(ε ) = ϕ(ε ) = ϕ(ε ) = t
31 32 41 42 43
and extend it to the homomorphism of integral group rings by linearity
ϕ : ZG −→ Zhti.
Let us find Fox derivatives of the relators r , with respect to the symbols ε , ε , ε , ε ,
ij 31 32 41 42
ε and their ϕ-images. We have the following
43
ϕ ϕ ϕ
∂r ∂r ∂r
11 = t−2(t−1), 11 = t−2(1−t), 12 = t−2(t−1),
(cid:18)∂ε (cid:19) (cid:18)∂ε (cid:19) (cid:18)∂ε (cid:19)
31 41 32
ϕ ϕ ϕ
∂r ∂r ∂r
12 = t−2(1−t), 21 = t−2(t−1), 21 = t−2(1−t),
(cid:18)∂ε (cid:19) (cid:18)∂ε (cid:19) (cid:18)∂ε (cid:19)
41 31 42
ϕ ϕ ϕ
∂r ∂r ∂r
22 = t−2(t−1), 22 = t−2(1−t), 31 = t−1(t−1),
(cid:18)∂ε (cid:19) (cid:18)∂ε (cid:19) (cid:18)∂ε (cid:19)
32 42 31
ϕ ϕ ϕ
∂r ∂r ∂r
31 = t−2(t−1), 31 = t−3(1−t2), 32 = t−1(t−1),
(cid:18)∂ε (cid:19) (cid:18)∂ε (cid:19) (cid:18)∂ε (cid:19)
41 43 32
ϕ ϕ
∂r ∂r
32 = t−2(t−1), 32 = t−3(1−t2).
(cid:18)∂ε (cid:19) (cid:18)∂ε (cid:19)
42 43
Derivatives with respect to other generators are zero.
With the obtained values of derivatives we form the Alexander matrix. After elementary
transformations of rows and columns and deleting zero rows and columns, we get the diagonal
matrix
(1−t) diag(−t−2,−t−2,−t−2,−t−2(1+t)).
Since the Alexander polynomial is defined up to the multiplication with a unit of the ring Zhti,
we get
∆ (t) = (1−t)4(1+t).
G
We now consider the group H. Defining relations of the group P imply that the group H
4
can be defined by the following relations
a a a−1 = a−1a a , a−1a a = [a−1,a−1]a [a−1,a−1],
13 14 13 34 14 34 13 24 13 14 34 24 34 14
a−1a a = a ,a a a−1 = a−1a a ,
23 14 23 14 23 24 23 34 24 34
a−1a a = a a a−1, a−1a a = a a a−1.
13 34 13 14 34 14 23 34 23 24 34 24
ON CERTAIN QUESTIONS OF THE FREE GROUP AUTOMORPHISMS THEORY 6
These relations can be presented in the following form
q = a a a−1a−1a−1a , q = a−1a a a a a−1a−1a−1a a a−1a−1,
11 13 14 13 34 14 34 21 13 24 13 14 34 14 34 24 34 14 34 14
q = a−1a−1a a , q = a a a−1a−1a−1a ,
12 14 23 14 23 22 23 24 23 34 24 34
q = a−1a a a a−1a−1, q = a−1a a a a−1a−1.
31 13 34 13 14 34 14 32 23 34 23 24 34 24
As above, we define the homomorphism ϕ from the group H to the infinite cyclic group hti,
by setting
ϕ(a ) = ϕ(a ) = ϕ(a ) = ϕ(a ) = ϕ(a ) = t,
13 23 14 24 34
and extend it by linearity to the homomorphism of group rings
ϕ : ZH −→ Zhti.
WecomputetheFoxderivativesoftherelatorsq ,writtenabove, withrespecttothevariables
ij
a , a , a , a , a and find their images under the homomorphism ϕ. We get the following
13 23 14 24 34
identities (the derivatives of all other relators can be written analogously)
ϕ ϕ ϕ
∂q ∂q ∂q
11 = 1−t, 11 = t−1(t2 −1), 11 = t−1(1−t),
(cid:18)∂a (cid:19) (cid:18)∂a (cid:19) (cid:18)∂a (cid:19)
13 14 34
ϕ ϕ ϕ
∂q ∂q ∂q
21 = t−1(t−1), 21 = −(1−t)2, 21 = t−1(1−t),
(cid:18)∂a (cid:19) (cid:18)∂a (cid:19) (cid:18)∂a (cid:19)
13 14 24
ϕ ϕ ϕ
∂q ∂q ∂q
21 = (t−1)2, 32 = t−1(t−1), 32 = t−1,
(cid:18)∂a (cid:19) (cid:18)∂a (cid:19) (cid:18)∂a (cid:19)
34 23 24
ϕ
∂q
32 = t−1(1−t2).
(cid:18)∂a (cid:19)
34
We form the matrix from the values of the calculated Fox derivatives. After elementary
transformations over rows and columns, we get the diagonal matrix
(1−t) diag(1,−t−2,1,t−2 +t−1 +1).
The Alexander polynomial of the group H is equal to
∆ (t) = (1−t)4(t2 +t+1).
H
Thus the Alexander polynomials of the groups G and H are different, and hence, the groups
G and H are not isomorphic (see, for example [7] for the proof that Alexander invariants, in
particular the Alexander polynomial, are invariants of a group and do not depend on a given
group presentation). (cid:3)
3. Isomorphism problem for a certain class of Lie algebras
3.1. A class of Lie algebras. The last section gives a motivation for finding methods for
proving that two given subgroup of automorphism groups are not isomorphic. In this section
we will consider a similar question.
Consider a homomorphism
φ : F → IA(F ),
n k
where n,k ≥ 2. We can form a natural semi-direct product
G := F ⋊F
φ k n
and ask whether G is isomorphic to G for two different homomorphisms φ ,φ : F →
φ1 φ2 1 2 n
IA(F ).
k
ON CERTAIN QUESTIONS OF THE FREE GROUP AUTOMORPHISMS THEORY 7
For a given group G, denote by L(G) the Lie algebra constructed from the lower central
series filtration:
L(G) := γ (G)/γ (G),
i i+1
Mi≥1
γ (G) = G, γ (G) = [γ (G),G].
1 i+1 i
Here we will consider the class of Lie algebras L(G ) for different homomorphisms φ and
φ
prove that some Lie algebras from this class are not isomorphic.
Since F acts trivially on H (F ), there is the following exact splitting sequence of Lie alge-
n 1 k
bras:
0 → L(F ) → L(G ) → L(F ) → 0
k φ n
by [17].
3.2. Scheuneman’s invariants. Let k be a field and L a finitely-generated Lie algebra of
nilpotence class two. Let us choice a basis of L: {x ,...,x ,y ,...,y } such that {y ,...,y } is
1 n 1 r 1 r
a basis of [L,L]. Let U(L) be a universal enveloping algebra of L. Then U(L) can be identified
with a polynomial algebra over k with non-commutative variables x ,...,x ,y ,...,y and
1 n 1 r
relations
x y = y x , for all i,j;
i j j i
y y = y y , for all i,j;
i j j i
x x −x x = linear combination of y ’s;
i j j i k
Consider an alternative sum
I(x ,...,x ) = (−1)|σ|x ...x , (4)
1 n i1 in
X
σ=(i1,...,in)∈Sn
where |σ| denotes the sign of the permutation σ.
Two polynomials f(z ,...,z ) and g(z ,...,z ) are k-equivalent if
1 m 1 m
f(z ,...,z ) = ag(z′,...,z′ ),
1 m 1 m
where a ∈ k,a 6= 0,
z′ = b z , b ∈ k, det(b ) 6= 0.
j ij j ij ij
X
The following theorem is due to Scheuneman:
Theorem 2. [13] The k-equivalence class of I(x ,...,x ) is an invariant of k-isomorphism of
1 n
the Lie algebra L.
For k-equivalent polynomials f(z ,...,z ) and g(z ,...,z ) and z′ = b z , b ∈ k,
1 m 1 m i ij j ij
det(bij) 6= 0, such that f(z1,...,zm) = ag(z1′,...,zm′ ), P
Hes(f)(z ,...,z ) = am(det(b ))2Hes(g)(z′,...,z′ ),
1 m ij i m
where Hes(f)(z ,...,z ) is the Hessian det( ∂2f ) (see, for example, Lemma 8 [13]).
1 m ∂zi∂zj
ON CERTAIN QUESTIONS OF THE FREE GROUP AUTOMORPHISMS THEORY 8
3.3. Non-isomorphic Lie algebras. Considerthefollowinghomomorphismφ ofafreegroup
α
F with basis {u ,u ,u } to the automorphism group of a free group F with basis {t ,t ,t }:
3 1 2 3 3 1 2 3
t 7−→ t α
φ : u 7→ 1 1 1
α 1 (cid:26) ti 7−→ ti for t = 2,3,
t 7−→ t α
φ : u 7→ 2 2 2
α 2 (cid:26) ti 7−→ ti for t = 1,3,
t 7−→ t α
φ : u 7→ 3 3 3
α 3 (cid:26) ti 7−→ ti for t = 1,2,
where α = (α ,α ,α ),
1 2 3
α ∈ {[t ,t ],[t ,t ],[t ,t ]},
1 1 2 2 3 1 3
α ∈ {[t ,t ],[t ,t ],[t ,t ]},
2 2 1 2 3 1 3
α ∈ {[t ,t ],[t ,t ],[t ,t ]}.
3 1 2 3 2 3 1
Lie algebras L(G ) which correspond to the groups G can be described as Lie algebras with
α α
generators t ,t ,t ,u ,u ,u and defining relations:
1 2 3 1 2 3
[t ,u ] = α , [t ,u ] = α , [t ,u ] = α ,
1 1 1 2 2 2 3 3 3
[t ,u ] = [t ,u ] = [t ,u ] = [t ,u ] = [t ,u ] = [t ,u ] = 0.
2 1 3 1 1 2 3 2 1 3 2 3
Consider the quotient
L := L(G )/[L(G ),L(G ),L(G )].
α φα φα φα φα
Clearly, L is 12-dimensional algebra with basis
α
{t ,t ,t ,u ,u ,u ,y ,y ,y ,y ,y ,y },
1 2 3 1 2 3 1 2 3 4 5 6
where y = [t ,t ],y = [t ,t ],y = [t ,t ],y = [u ,u ],y = [u ,u ],y = [u ,u ]. Consider
1 1 2 2 1 3 3 2 3 4 1 2 5 1 3 6 2 3
the polynomial I(t ,t ,t ,u ,u ,u ). The direct computation gives the following expression
1 2 3 1 2 3
I(t ,t ,t ,u ,u ,u ) =
1 2 3 1 2 3
[t ,t ][t ,u ][u ,u ]−[t ,t ][t ,u ][u ,u ]+[t ,u ][t ,t ][u ,u ]−[t ,u ][t ,u ][t ,u ]
1 2 3 3 1 2 1 3 2 2 1 3 1 1 2 3 2 3 1 1 2 2 3 3
= y y α −y y α +y y α −α α α = P (y ,...,y ).
1 4 3 2 5 2 3 6 1 1 2 3 (α1,α2,α3) 1 6
By Theorem 2, we have
k −equivalence classes of cubic forms P } ⊆ {k −isomorphism classes of G }
(α1,α2,α3) φα
(cid:8)
Consider the following cases:
1) Let a = (α ,α ,α ) with α = [t ,t ], α = [t ,t ], α = [t ,t ]. Then
1 1 2 3 1 2 3 2 1 3 3 1 2
P (y ,...,y ) = y2y −y2y +y2y −y y y ,
a1 1 6 1 4 2 5 3 6 1 2 3
Hes(P )(y ,...,y ) = 64y2y2y2.
a1 1 6 1 2 3
2) Let a = (α ,α ,α ) with α = [t ,t ], α = [t ,t ], α = [t ,t ]. Then
2 1 2 3 1 2 3 2 2 3 3 1 2
P (y ,...,y ) = y2y −y y y +y2y −y y2,
a2 1 6 1 4 2 5 3 3 6 1 3
Hes(P ) = 16y2y4.
a2 1 3
3) Let a = (α ,α ,α ) with α = [t ,t ], α = [t ,t ], α = [t ,t ]. Then
3 1 2 3 1 2 3 2 2 3 3 1 2
P (y ,...,y ) = y y y −y y y +y2y −y3,
a3 1 6 1 4 3 2 5 3 3 5 3
Hes(P ) = 4y6.
a3 3
ON CERTAIN QUESTIONS OF THE FREE GROUP AUTOMORPHISMS THEORY 9
Clearly, all these three cases define non-k-equivalent polynomials. Hence, the corresponding
groups and Lie algebras are not isomorphic.
Proposition 1. The Lie algebras L(G ),L(G ),L(G ) are pairwise non-isomorphic.
a1 a2 a3
Remark. The invariant polynomials for the two-step nilpotent quotients of Lie algebras of the
groups Cb+/Z(Cb+) and P /Z(P ) define cubic forms which lie on Hilbert’s null-cone, hence
4 4 4 4
they do not differ from each other by an invariant of a Hessian type1.
4. On non-linearity of certain automorphism groups
4.1. Poison group. Let G be a group. Consider the group H(G), defined as an HNN-
extension:
H(G) = hG×G,t | t(g,g)t−1 = (g,1), g ∈ Gi.
It is easy to see that the group H(G) can be presented as a semi-direct product
H(G) ≃ (G∗Z)⋊G,
where the action of G on G∗Z = G∗hti is defined by
g : g te1g te2...g tek 7→ g (tg−1)e1g (tg−1)e2...g (tg−1)en, g,g ∈ G,e ∈ Z.
1 2 k 1 2 k i i
For example, if G = Z, we get the following group
H(G) = ha,b | [a,[b,a]] = 1i.
It follows from elementary Tietze moves that
H(Z) = ha,a′,t | [a,a′] = 1, taa′t−1 = ai = ha,a′,t | [a,a′] = 1, a′ = a−1t−1at = [a,t]i =
= ha,t | [a,[a,t]] = 1i,
where a = (a,1),a′ = (1,a) ∈ G×G.
Denote by NAF the class of nilpotent-by-abelian-by-finite groups. The following result due
to Formanek and Procesi presents a way of constructing non-linear groups.
Theorem 3. [10] If G ∈/ NAF, then the group H(G) is non- linear.
The simplest example of a group which does not lie in the class NAF, clearly, is a free
non-cyclic group. Thus, for G = F , the group H(F ), called a poison group, is non-linear. This
2 2
fact plays an important role in the proof of non-linearity of groups Aut(F ) for ≥ 3; the poison
n
group is a subgroup in Aut(F ), n ≥ 3. This statement can be generalized.
n
Theorem 4. Let G ∈/ NAF, then Aut(G∗Z) is non-linear.
Proof. We will realize the group H(G) as a subgroup in Aut(G ∗ Z) and the statement will
follow from Theorem 3. Elements of the subgroup G×G in H(G) we will denote as (g,g′), i.e.
we put dash for elements from the second copy of G. The group G∗Z we will describe in terms
of generators g ∈ G and a free generator t. Consider the homomorphism
f : H(G) → Aut(G∗Z),
given by setting
f : g 7→ ig, g ∈ G, t 7→ it, g′ 7→ sg′, g′ ∈ G,
where ig is the conjugation by g, it is the conjugation by t, sg′ (g′ ∈ G) is the automorphism of
G∗Z acting trivially on G and sending the element t to the element tg′−1. It can be checked
that f is a group homomorphism. Every element of the group H(G) can be written without
1Authors thank V.L. Popov for helping analyzing these cubic forms and for this remark
ON CERTAIN QUESTIONS OF THE FREE GROUP AUTOMORPHISMS THEORY 10
’dash’ elements, since tgg′t−1 = g and, therefore, g′ = [g,t]. Hence, in the case of the existence
of a non-trivial kernel of f, there is an element of G ∗ Z, acting trivially by conjugation, i.e.
lying in the center of G∗Z. However, any non-trivial free product has a trivial center, therefore,
f is a monomorphism. Thus H(G) is a subgroup of Aut(G∗Z). Theorem 3 implies that the
group Aut(G∗Z) is non-linear. (cid:3)
Clearly, one can consider different embeddings of groups H(G) in corresponding automor-
phism groups. Consider the case G = F .
2
Let F be a free group with basis x ,x ,x and a ,a ,a some elements of F , such that the
3 1 2 3 1 2 3 3
subgroup ha ,a ,a i is free of rank 3. Define automorphisms α , i = 1,2,3 of the group F as
1 2 3 i 3
conjugation with a . The following statement can be checked straightforwardly:
i
Proposition 2. Let φ ,φ ∈ Aut(F ) be automorphisms which satisfy the following conditions:
1 2 3
φ (a ) = φ (a ) = a ,
1 1 2 1 1
φ (a ) = φ (a ) = a ,
1 2 2 2 2
φ (a ) = a a , φ (a ) = a a .
1 3 3 1 2 3 3 2
Then the subgroup of Aut(F ), generated by elements α ,α ,α ,φ ,φ is isomorphic to the
3 1 2 3 1 2
poison group H(F ).
2
As an example lets take a = x , with a and a arbitrary elements of hx ,x i, which do not
3 3 1 2 1 2
lie in one cyclic subgroup. Define
x 7→ x , i = 1,2
φ : i i
1 (cid:26) x 7→ x a ,
3 3 1
x 7→ x , i = 1,2
φ : i i
2 (cid:26) x 7→ x a .
3 3 2
The conditions 2 can be checked straightforwardly. Then the subgroup of Aut(F ), generated
3
by the elements α ,φ , i = 1,2,3,j = 1,2 is isomorphic to H(F ). In particular, in the case
i j 2
a ,a ∈ γ (F ), we have the following
1 2 2 2
Theorem 5. The group IA(F ) contains a subgroup isomorphic to H(F ), and hence, IA(F )
3 2 3
is not linear.
Observe also that the poison group H(F ) is residually finite. It follows from Baumslag’s
2
theorem which states that every finitely generated subgroup of an automorphism group of a
residually finite groupis itself residually finite. Also observe that the non-linearity of the poison
group can be used for construction of other non-linear groups given by commutator relations.
For example, the group H(F ) contains the following normal subgroup of index 2:
2
H = hx ,x ,x ,x ,x | [x ,x ] = [x ,x ] = [xx5,x ] = [xx5,x ] = 1,
1 2 3 4 5 1 3 2 4 1 3 2 4
[x x ,x ] = [x x ,x ] = [xx5x ,x ] = [xx5x ,x ] = 1i,
1 3 2 2 4 1 1 3 4 2 4 3
which is of cause non-linear.
5. Questions
(1) Describe the Lie algebra of the group Cb , n ≥ 2.
n
(2) Let L be a free Lie algebra with n generators, n ≥ 3. Does the group Aut(L ) contain
n n
the poison group as a subgroup?
(3) Are the groups Cb+ linear for n ≥ 3?
n