Example of non-linearizable quasi-cyclic subgroup of automorphism group of polynomial algebra

0.1 MB·English

by  Valeriy G. Bardakov

Example of non-linearizable quasi-cyclic subgroup of
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automorphism group of polynomial algebra
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n
a Valeriy G. Bardakov
J
Sobolev Institute of Mathematics and Novosibirsk State University,
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Novosibirsk 630090, Russia

] Laboratory of Quantum Topology, Chelyabinsk State University,
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G Brat’ev Kashirinykh street 129, Chelyabinsk 454001, Russia.
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h [email protected]
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a
m
[ Mikhail V. Neshchadim

1 Sobolev Institute of Mathematics and Novosibirsk State University,
v
6 Novosibirsk 630090, Russia
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6 [email protected]
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Abstract
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5
1 It is well known that every finite subgroup of automorphism group of polyno-
v: mial algebra of rank 2 over the field of zero characteristic is conjugated with a
i subgroup of linear automorphisms. We prove that it is not true for an arbitrary
X
torsion subgroup. We construct an example of abelian p-group of automorphism
r
a of polynomial algebra of rank 2 over the field of complex numbers, which is not
conjugated with a subgroup of linear automorphisms.
Key words and phrases: algebra of polynomials, polynomial automorphisms,
quasi-cyclic group.
0.1 Introduction
In the present paper we consider the polynomial algebra P = k[x ,x ,...,x ], and free
n 1 2 n
associativealgebraA = khx ,x ,...,x iwithfreegeneratorsx ,x ,...,x overthefield
n 1 2 n 1 2 n
k. We assume that these algebras contain unit elements. We will use symbols AutP
n
This research was partially supported by the Laboratory of Quantum Topology of Chelyabinsk
State University (Russian Federation government grant 14.Z50.31.0020), RFBR-14-01-00014, RFBR-
13-01-00513and Indo-Russian RFBR-13-01-92697.
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and AutA , to denote groups of k-automorphisms of these algebras, i. e. automorphisms
n
which fix elements of k. It is well known [1], that the group AutP is isomorphic to the
2
group AutA . Let P = k[[x ,x ,...,x ]] be the algebra of the formal power series in
2 n 1 2 n
commutative variables. The following inclusions evidently holds
e
P ⊆ P , AutP ⊆ AutP .
n n n n
In the theory of automorphisms of algebra P the following problem is well known
e n e
(see the survey [2]): Is it true that every automorphism from AutP of finite order is
n
conjugated with linear automorphism in AutP ? Corresponding question for involutions
n
was formulated in [3, Question 14.68]. The case of involutions is interesting because of
its connection with the Cancelation Problem (see for example [4]).
For n = 2 and the field k = C of complex numbers the answer is positive, and it
follows from the result of van der Kulk [5] that the group AutP is a free product with
2
amalgamation. For subgroups of AutP more general result is known: every finite sub-
2
group is conjugated with the subgroup of linear automorphisms. P.M. Cohn [7] studied
algebras over the finite field and proved that every finite subgroup of automorphism
group of A is conjugate with subgroup of linear automorphisms if the characteristic of
2
the field does not divide the order of the subgroup. As reviewer noted this result was
firstly proved in [8]. Unfortunately, this publication is not available for wide range of
readers.
On the other hand, we know examples of finite subgroups of AutP , which are not
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conjugatedwithsubgroupsoflinear automorphisms(forexample, dihedral groupoforder
8m for m ≥ 3) [6]. H. Kraft and G. Schwarz [2, p. 61] have formulated the following
question: is there exists a commutative (reductive) subgroup of AutP , which is not
n
conjugated with subgroup of linear automorphisms?
M. A. Shevelin [9] constructed an example of infinite torsion subgroup of automor-
phism group of free Lie algebra of rank 3 over the complex field, which is not conjugated
with subgroup of linear automorphisms. Using this idea we prove the following assertion.
Theorem 0.1 Let p be a prime number. There are an uncountable set of pairwise non-
conjugated subgroups of the group AutP , which are isomorphic to the quasi-cyclic p–
2
group C and which are not conjugated with subgroups of linear automorphisms.
p∞
We are greatfull to all the participants of the scientific seminar “Evariste Galois”
(Novosibirsk State University) for the useful discussions. Also, we thank anonymous
reviewer for helpful notes.
0.2 Quasi-cyclic subgroups
Throughout the paper we use the field of complex numbers C as the main field. Every
automorphism ψ ∈ AutP is completely determined by its action on the generators,
n
2
therefore we identify automorphism ψ with the vector (xψ,xψ,...,xψ), i. e.
1 2 n
ψ = (xψ,xψ,...,xψ).
1 2 n
Let p be a fixed prime number. Recall (see, for example, [10, 1.2.5, 2.4.10]) that a
quasi-cyclic p–group Cp∞ is the set of all complex solutions of the equations xpn = 1,
n = 1,2,... with natural multiplication. Group Cp∞ is an infinite abelian p–group, such
that every proper subgroup of it is cyclic. Let us denote
Gp = {(αx1,αx2)|α ∈ Cp∞}.
Itis obvious thatG isa subgroupoflinear automorphisms inAutP anditis isomorphic
p 2
to Cp∞.
We will prove that there exits other embedding of Cp∞ into the group AutP2. To do
it consider a formal power series
∞
a w , a ∈ C, w = xpk+1
k k k k 2
k=0
X
in the algebra P . Then the map a : P −→ P , which is defined by the equality
2 2 2
∞
e e e
a = x + a w , x
1 k k 2
!
k=0
X
is an automorphism of P . If there are only a finite number of non-zero coefficients a
2 k
here, then a is an automorphism of the algebra P .
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Consider the subgrouep
Ga = a−1G a
p p
of the group AutP . Let us show that Ga lies in the group AutP , i. e. it is a subgroup
2 p 2
of automorphism group of P2. Really, let an automorphism ϕ = (αx1,αx2), α ∈ Cp∞
belongs to the groeup G . Conjugating it by automorphism a we have an equality
p
∞
a−1ϕa = αx +α a (1−αpk)w , αx .
1 k k 2
!
k=0
X
Here and further we assume that automorphisms act from the left to the right. Since
α ∈ Cp∞ we see that for sufficiently large number k0 all the coefficients 1 − αpk, k =
k ,k +1,... are equal to zero and hence a−1ϕa ∈ AutP .
0 0 2
By construction, the group Ga is conjugated with the group G in the automor-
p p
phism group AutP of the formal power series P . Let us find out the question about
2 2
conjugation of Ga with some subgroup of linear automorphisms in the group AutP .
p 2
e e
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Proposition 0.2 If an automorphism
∞
a = x + a w , x
1 k k 2
!
k=0
X
is defined by infinite number of non-zero coefficients a , then Ga is not conjugated with
k p
any subgroup of linear automorphisms.
Proof. Let H be a subgroup of linear automorphisms, which is conjugated with Ga,
p
i. e.
θ−1Gaθ = H,
p
forsomeθ ∈ AutP . Withno lossof generality weassume that H belongsto thegroupof
2
diagonal automorphisms, therefore the last equality means that for every automorphism
ϕ = (αx1,αx2) there exists an automorphism h = (α1x1,α2x2) ∈ H, α1,α2 ∈ Cp∞, such
that the following equality holds
ϕaθ = θh. (1)
Suppose that θ = (f ,f ), where the polynomials f = f (x ,x ) and f = f (x ,x )
1 2 1 1 1 2 2 2 1 2
have degree m an m correspondingly. Since these polynomials generate the algebra P ,
1 2 2
then m > 0 and m > 0. It is not difficult to see that the equality (1) is equivalent to
1 2
the system
∞
αf +α a (1−αpk)fpk+1 = f (α x ,α x ), αf (x ,x ) = f (α x ,α x ).
1 k 2 1 1 1 2 2 2 1 2 2 1 1 2 2
k=0
X
If we write the first equality in the following form
∞
α a (1−αpk)fpk+1 = f (α x ,α x )−αf ,
k 2 1 1 1 2 2 1
k=0
X
then we see that the right hand side of it has the degree, which is less or equal to m .
1
But taking different α, we can get an arbitrary large degree of the left hand side of this
equality. Hence, the equality (1) is impossible. The proposition is proved.
0.3 On conjugation of quasi-cyclic subgroups
In this section we will prove the following proposition.
Proposition 0.3 There exist an uncountable set of subgroups Ga, which are pairwise
p
non-conjugated in AutP .
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Let
∞ ∞
a = x + a w ,x , b = x + b w ,x ,
1 k k 2 1 k k 2
! !
k=0 k=0
X X
a , b ∈ C, k = 0,1,2,... be two automorphisms of P , which are not automorphisms of
k k 2
P , i. e. each of them contains an infinite number of non-zero coefficients. Let us find
2
out are the subgroups Ga and Gb conjugated in AuteP .
p p 2
Let θ be such an automorphism of P , that the following equality holds
2
−1 −1 −1
θ (a ϕa)θ = b ϕb for all ϕ ∈ G .
p
This equality is evidently equivalent to the following equality
−1 −1
(a ϕa)θ = θ(b ϕb). (2)
Suppose that the automorphism θ acts on the generators of P by the following way
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xθ = f (x ,x ), xθ = f (x ,x ).
1 1 1 2 2 2 1 2
Then the equality (2) is equivalent to the system
∞ θ
αx +α a (1−αpk)w = fb−1ϕb, (αx )θ = fb−1ϕb,
1 k k 1 2 2
!
k=0
X
which can be rewritten by the following way
αf +α ∞ a (1−αpk)fpk+1 = f αx +α ∞ b (1−αpk)w , αx ,
1 k=0 k 2 1 1 k=0 k k 2
 (cid:16) (cid:17) (3)
P P
 αf = f αx +α ∞ b (1−αpk)w , αx ,
2 2 1 k=0 k k 2
Bythe choice o(cid:16)f α, the foPllowing equalities holds 1(cid:17)−αpk = 0 for enough large k. Hence,
the system of equations (3) is the system of equalities in the algebra P .
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If the polynomial f contains the variable x , then it is decomposed by the powers of
2 1
x
1
s
f = c (x )xj, c (x ) ∈ C[x ].
2 j 2 1 j 2 2
j=0
X
Then
∞ s ∞ j
f αx +α b (1−αpk)w , αx = c (αx ) αx +α b (1−αpk)w
2 1 k k 2 j 2 1 k k
! !
k=0 j=0 k=0
X X X
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and since α ∈ Cp∞ is arbitrary, then the degree of the right hand side with respect to x2
can be arbitrarily large. Hence, f does not contain x , i. e. f = f (x ) ∈ C[x ]. Since
2 1 2 2 2 2
θ is an automorphism, then
f = βx +β , β ∈ C∗, β ∈ C,
2 2 0 0
and the first equality of (3) has the following form
αf (x ,x )+α ∞ a (1−αpk)(βx +β )pk+1 =
1 1 2 k=0 k 2 0
(4)
P
= f αx +α ∞ b (1−αpk)w , αx .
1 1 k=0 k k 2
(cid:16) (cid:17)
P
Since θ = (f ,βx +β ) we see that
1 2 0
f = γx +g(x ), γ ∈ C∗, g ∈ C[x ].
1 1 2 2
Then the equality (4) has the form
α(γx +g(x ))+α ∞ a (1−αpk)(βx +β )pk+1 =
1 2 k=0 k 2 0
P
= γ αx +α ∞ b (1−αpk)w +g(αx ).
1 k=0 k k 2
(cid:16) (cid:17)
P
or equivalently
∞
αg(x )−g(αx )+α (1−αpk)[a (βx +β )pk+1 −b xpk+1] = 0.
2 2 k 2 0 k 2
k=0
X
Since the degree of g(x ) is bounded and there exist an infinite number of non-zero
2
coefficients a and b , then the following equalities hold for some number k which is
k k 0
large enough
a βpk+1 = b , k ≥ k .
k k 0
If it is not true, then there is no automorphism θ of P such that
2
−1 −1 −1
θ (a ϕa)θ = b ϕb.
Thus we have proved that if the sequences {a } and {b } satisfy the inequality
k k
a βpk+1 6= b
k k
for an infinite set of indexes k and for an arbitrary β ∈ C∗, then the subgroups Ga and
p
Gb are not conjugated in AutP .
p 2
Tofinish theproofofthepropositionwehave toprovethatthereexist anuncountable
set of such sequences.
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Denote by
Ω = {λ = (λ ,λ ,λ ,...)|λ ∈ {0,1}}
1 2 3 i
the set of all infinite binary sequences. The set Ω evidently contains an uncountable set
of different sequences. Furthermore, the following lemma is true
Lemma 0.4 There exists an uncountable subset Ω ⊂ Ω, which posses the following
0
property: for any two sequences λ,µ ∈ Ω inequality λ 6= µ holds for an infinite
0 k k
number of indexes k ∈ N.
Proof. If we define the following relation on the set Ω
λ ∼ µ ⇔ λ and µ different only on finite number of places,
then it is not difficult to see that this relation is equivalence relation and hence, divides
theset Ωonto equivalence classes. Moreover, every equivalence classcontains onlycount-
able number of elements and then the number of equivalence classes Ω/∼ is uncountable.
The transversal set with respect to relation ∼ is the necessary subset Ω .
0
It follows from this lemma, that there is an uncountable set of subgroups Ga, which
p
are pairwise non-conjugated in AutP . It completes the proof of Proposition 2.
2
Now the main theorem follows from the propositions 1 and 2.
We have found a sufficient condition when subgroups Ga and Gb are not conjugated
p p
in AutP .
2
Question 0.5 Find necessary and sufficient conditions when subgroups Ga and Gb are
p p
conjugated in AutP .
2
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