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ADVANCES IN MATHEMATICS
Vol.41, No.2
April, 2012
Simplicity of a Class of
Quantum Torus Lie Algebras
LUO Liuhong1,∗ , LU Caihui2
(1. College of Science, Beijing Forestry University, Beijing, 100083, P. R. China; 2. School of
Mathematical Sciences, Capital Normal University, Beijing, 100037, P. R. China)
±1 ±1
Abstract: Let Cq := Cq [t±1
1 , t2 , t3 ] be the quantum torus associative algebra and Lq =
Cq /C be the quantum torus Lie algebra. In this article we define a system of exponential
equations associated to q = (qij )3i,j=1 and call it characteristic equations of Cq , and we prove
that Lq is not simple if and only if the characteristic equations have nonzero solution in Z3 . For
|q| = 0, we show that there exists a maximal Abelian subalgebra I in Cq such that I strictly
contains the center Z(Cq ). We also point out that such a subalgebra does not exist in Cq for
|q| 6= 0.
Key words: Lie algebra; quantum torus; simplicity
MR(2000) Subject Classification: 08A35 / CLC number: O152.5
Document code: A
Article ID: 1000-0917(2012)02-0139-11
0 Introduction
Let q = (qij )ni,j=1 be an n × n matrix over C satisfying
qii = 1,
−1
qij = qji
,
i, j = 1, 2, · · · , n,
(0.1)
where n is a positive integer. We define a q-quantum torus algebra Cq which is a unital
associative algebra over C generated by t1 , t2 , · · · , tn and subjects to the defining relations
ti tj = qij tj ti ,
ti t−1
= t−1
i
i ti = 1,
i 6= j, i, j = 1, 2, · · · , n.
(0.2)
±1
±1
We write it as Cq := Cq [t±1
1 , t2 , · · · , tn ]. If all qij = 1, then Cq is a Laurent polynomial algebra
and an abelian associative algebra. Based on this reason, we always assume there exists at least
one pair i 6= j such that qij 6= 1.
We define a q-quantum torus Lie algebra Lq = Cq /C be the associated Lie algebra. Let
D be a subalgebra of DerLq , then we can define the holomorph of Lq by D:
Lq [D] = D
M
Lq .
Lq [D] is a Lie algebra associated with quantum torus Lie algebra. We can easily obtain various
Lie algebras owing to different choices of D. For simplicity, we call these Lie algebras as the Lie
algebras constructed from q-quantum torus.
Received date: 2010-01-11. Revised date: 2010-11-03.
E-mail: ∗ [email protected]
140
Æ
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41
In the classification of the extended affine Lie algebras[1, 2] , the quantum torus algebra
±1
±1
Cq [t±1
1 , t2 , · · · , tn ] and its skew derivation algebra Derskew (Cq ) play important roles. In recent
years, there are many research results about this kind of algebras. For example, when the order
of matrix q is n, Zhao[7] computed explicitly all 2-cocycles of the quantum differential operator
N
algebra Fq [D] = Fq
F [D], where F is a field of characteristic 0. The author also showed that
the second cohomology group of Fq [D] has dimension n if D = 0; dimension 1 if dimD = 1;
and dimension 0 if dimD > 1, etc. When n = 2, Jiang and Meng[4, 5] gave the derivation
algebra of the Virasoro-like algebra and the structure of automorphism group. The structure
and automorphism group of a class of derivation Lie algebra over Lq were discussed by Chen,
etc.[3] and Zheng, Tan[8] . Besides, Lin and Tan[6] investigated a class of irreducible modules of
skew derivation Lie algebra over quantum torus.
1 q12
When n = 2, q =
is a 2 × 2 matrix, and q12 q21 = 1, |q| = 0. The only factor
q21 1
which can influence the structure of the quantum torus Lie algebra is q12 . On the other hand,
q12 just has two different choices: to be a root of unity or to be not. In this case, it is not so
difficult to conclude that Lq is simple when q12 is a non-root of unity; and Lq is not simple when
q12 is a root of unity.
But when n ≥ 3, the results are not so easy to get. This is because on the one hand q
may or may not be invertible, and on the other hand {qij |i, j = 1, 2, · · · , n} has three possible
choices: (i) all qij are roots of unity; (ii) all qij are not roots of unity; (iii) there exist roots and
non-roots of unity in qij at the same time. It is no doubt that these differences will influence
the structure of Cq and Lq , for example, their center, simplicity and the existence of maximal
abelian subalgebra, etc. All of them are very interesting, meaningful and worth further research.
In this paper, we mainly focus on the simplicity of Lq = Cq /C when n = 3 (or equivalently,
the discussion of the center of Lq ). We provide a sufficient and necessary condition which
ensures that Lq is not simple(Theorem 2.1) by defining characteristic equations(Definition 2.1),
and furthermore we guess that the conclusion probably holds for any n. In Section 3, we specialize
in the situation of |q| = 0, and obtain a maximal abelian subalgebra(denoted by I) which strictly
contains the center Z(Cq ). And then, we give the structure of I by Equation (3.4) and Theorem
3.8, and give an example to demonstrate that such abelian subalgebra does not exist when |q| 6= 0.
In addition, we also discuss the different impact on the simplicity of Lq which results from the
different choices of qij in Section 2 and Section 3.
The terms and symbols in this paper are the same as [3].
1 Some Known Results
αn
1 α2
For α = (α1 , α2 , · · · , αn ) ∈ Zn , where Z is a set of all integers, we write tα := tα
1 t2 · · · tn .
For α, β ∈ Zn , we define two functions σ(α, β) and f (α, β) over Zn × Zn by
tα tβ = σ(α, β)tα+β ,
tα tβ = f (α, β)tβ tα .
(1.1)
Simplicity of a Class of Quantum Torus Lie Algebras
2
141
It is easy to get
α β
Y
σ(α, β) =
qiji j ,
f (α, β) =
1≤j<i≤n
n
Y
α β
qiji j ,
(1.2)
i,j=1
and
f (α, β) = σ(α, β)σ(β, α)−1 ,
Lemma 1.1[1]
f (−α, β) = f (α, β)−1 .
(1.3)
For α, β, α′ , β ′ ∈ Zn , we have
(i) f (α + α′ , β) = f (α, β)f (α′ , β);
(ii) f (α, β + β ′ ) = f (α, β)f (α, β ′ );
(iii) f (α, β) = f (β, α)−1 ;
(iv) f (α, kα) = 1, ∀k ∈ Z.
Lemma 1.2[1]
(i) The associative algebra Cq is simple if and only if Z(Cq ) = C, where
Z(Cq ) is the center of Cq ;
(ii) The Lie algebra Lq is simple if and only if Cq is a simple associative algebra.
We define the radical of f as
radf = {α ∈ Zn | f (α, Zn ) = 1}.
(1.4)
Applying (1.2) and Lemma 1.1, we deduce the following fundamental results easily:
(a) 0 = (0, 0, · · · , 0) ∈ radf ;
(b) For any α, β ∈ radf and any k, l ∈ Z, we have kα + lβ ∈ radf .
Therefore, radf is an abelian group.
Obviously, for any α ∈ radf , we have tα ∈ Z(Cq ). Then,
X
Ctα ⊆ Z(Cq ).
(1.5)
α∈radf
To prove
P
α∈radf
Ctα ⊇ Z(Cq ), we introduce a kind of derivation ∂i , i = 1, 2, · · · , n of Cq
for any tα ∈ Cq by
∂i (tα ) = αi tα .
(1.6)
It is not difficult to prove that ∂i is the derivation of associative algebra Cq , and the transformation induced by ∂i on the Lie algebra Lq is also a derivation. We still write it as ∂i . It is known
that Z(Lq ) is invariant under the actions of ∂i , i = 1, 2, · · · , n.
P
(j)
(j)
(j)
(j)
Lemma 1.3 Let z = 1≤j≤s,α(j) ∈Zn λj tα ∈ Z(Cq ), α(j) = (α1 , α2 , · · · , αn ), then
(j)
tα
∈ Z(Cq ), and α(j) ∈ radf, j = 1, 2, · · · , s.
Proof We use mathematical induction on the term s to prove this lemma.
(1)
When s = 2, let z = λ1 tα
(2)
+ λ2 tα
∈ Z(Cq ), where α(1) 6= α(2) , so there must exist at
(1)
(2)
least one different component. Without loss of generality, assume that α1 6= α1 , then we have
(
(1)
(2)
z = λ1 tα + λ2 tα ,
(1)
(2)
(1)
(2)
∂1 (z) = α1 λ1 tα + α1 λ2 tα .
142
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#
41
1
1
(2)
α(1)
, λ2 tα in terms of z, ∂1 (z) ∈
(1)
(2) is invertible, we can solve λ1 t
α1
α1
(1)
(2)
Z(Cq ) to obtain tα , tα ∈ Z(Cq ).
Since the matrix
We suppose that the statement is always true for any term less than s. Let us see what will
happen to s.
(1)
Let z = λ1 tα
(1)
(1)
(2)
+ λ2 tα
(s)
+ · · · + λs tα
∈ Z(Cq ), and α(i) 6= α(j) if i 6= j. Fix α(1) =
(1)
(1)
(α1 , α2 , · · · , αn ), then there at least exists one component in α(1) , say α1 , such that there
(j)
(1)
exist some α(j) (j ≥ 2) of which the first component α1 6= α1 . Thus we can group the expression
of z by α(j) s’ first components. If the first components are equal, then they are in the same
group. So we obtain r groups distinguished by r different numbers and denote these numbers by
a1 , a2 , · · · , ar . We have to remember that ai 6= aj if i 6= j. The next step is to sum up the terms
P
P
(j)
in the same group, then we get z = rk=1 zk where zk = j,α(j) =a λj tα .
1
k
Let ∂1 act on z, ∂1 (z), ∂12 (z), · · · , ∂1r−2 (z), we obtain r equations

 

z
1
1
···
1
 ∂1 (z)  

a2
···
ar 

  a1


=

..

···
··· ···
···



.
r−1
r−1
r−1
r−1
a
a
·
·
·
a
r
1
2
∂
(z)
1
in matrix form

z1
z2 

..  .
. 
zr
(1.7)
Since the determinant of the matrix is a nonzero Vandermonde determinant, we can solve zk in
terms of ∂1i (z) ∈ Z(Cq ) to obtain zk ∈ Z(Cq ), k = 1, 2, · · · , r. On the other hand, the lengths of
the expressions of z1 , z2 , · · · , zr are all less than s, then by induction hypothesis on s, we know
(1)
(2)
(s)
λ1 tα , λ2 tα , · · · , λs tα
(1)
(2)
(s)
∈ Z(Cq ). In other words, tα , tα , · · · , tα
∈ Z(Cq ). So we have
α
proved that any t appeared in z is indeed in Z(Cq ).
The proof of the above lemma shows that tα which appears in the expression of z is commutative with any tβ . In other words, f (α, β) = 1, ∀β ∈ Z3 . It implies that α ∈ radf . Thus,
P
z ∈ α∈radf Ctα . This proves
X
Z(Cq ) =
Ctα .
(1.8)
α∈radf
By using Lemma 1.2 and Lemma 1.3, we know that Z(Cq ) can be described clearly by the
structure of radf . Thus we get that the Lie algebra Lq is simple if and only if Z(Lq ) = 0 and if
and only if radf = {0}.
In the following, we shall discuss how to use the properties of radf to describe the simplicity
of the q-quantum torus Lie algebra when n = 3.
2 Simplicity of the Quantum Torus Lie Algebra Lq (n = 3)
When n = 3, the determinant of q = (qij )3i,j=1 is
|q| = q12 q23 q31 +
(q12 q23 q31 − 1)2
1
−2=
.
q12 q23 q31
q12 q23 q31
(2.1)
Then we have
|q| = 0 ⇔ q12 q23 q31 = 1.
(2.2)
Simplicity of a Class of Quantum Torus Lie Algebras
2
By using (0.1), it is easy to obtain the three equivalent forms of (2.2):

q21 = q23 q31 ,
q32 = q31 q12 ,

q13 = q12 q23 .
143
(2.3)
For α = (α1 , α2 , α3 ), β = (β1 , β2 , β3 ) ∈ Z3 , we deduce from (1.2) that
f (α, β) =
3
Y
α βj
qiji
α1 β2 −α2 β1 α2 β3 −α3 β2 α3 β1 −α1 β3
= q12
q23
q31
.
(2.4)
i,j=1
If |q| = 0, by using (2.3) it can be further simplified and then we get the following three
equivalent forms:
α (β2 +β3 )−(α2 +α3 )β1 (α1 +α2 )β3 −α3 (β1 +β2 )
q23
(α1 +α3 )β2 −α2 (β1 +β3 ) α3 (β1 +β2 )−(α1 +α2 )β3
q12
q31
α2 (β1 +β3 )−(α1 +α3 )β2 (α2 +α3 )β1 −α1 (β2 +β3 )
q23
q31
.
f (α, β) = q121
=
=
(2.5)
(2.6)
(2.7)
Definition 2.1 Suppose that q = (qij )3i,j=1 , we define the characteristic equations of
the quantum torus algebra Cq to be the following exponential equations:
y
xi
qik
= qkjj ,
1 ≤ i 6= j 6= k ≤ 3, xi , yj ∈ Z.
(2.8)
α = (α1 , α2 , α3 ) ∈ Z3 is said to be a solution of the characteristic equations (2.8), if its components α1 , α2 , α3 satisfy the three equations
 α1
q12
q α2
 23
α3
q31
in (2.8) simultaneously. Namely,
α3
= q23
, (E1)
α1
= q31
, (E2)
α2
= q12
. (E3)
Theorem 2.1 Whether |q| = 0 or |q| 6= 0, we have radf 6= {0} if and only if there exists
non-zero α = (α1 , α2 , α3 ) ∈ Z3 such that the characteristic equations (E1), (E2), (E3) hold at
the same time.
Proof ⇒ Setting 0 6= α = (α1 , α2 , α3 ) ∈ radf , for any β = (β1 , β2 , β3 ) ∈ Z3 , we have
α1 β2 −α2 β1 α2 β3 −α3 β2 α3 β1 −α1 β3
f (α, β) = q12
q23
q31
= 1.
(2.9)
α1 −α3
α1
α3
In (2.9), fix β = (0, 1, 0), then q12
q23 = 1 ⇒ q12
= q23
; (E1) holds. Fix β = (0, 0, 1), then
α2 −α1
q23
q31
α2
α1
α3 −α2
α3
α2
= 1 ⇒ q23
= q31
; (E2) holds. Fix β = (1, 0, 0), then q31
q12 = 1 ⇒ q31
= q12
; (E3)
holds. Hence, the necessary condition follows.
⇐
If there exists 0 6= α = (α1 , α2 , α3 ) ∈ Z3 such that (E1), (E2), (E3) hold simultaneously,
α1
α3
α1 −α3
then by using (E1), we have q12
= q23
⇒ q12
q23 = 1. For β1 = (0, 1, 0), applying (2.4) we can
obtain f (α, β1 ) = 1.
Similarly, using (E2) and (E3) , for β2 = (0, 0, 1) and β3 = (1, 0, 0) separately, we shall get
f (α, β2 ) = 1 and f (α, β3 ) = 1.
144
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41
However, β1 = (0, 1, 0), β2 = (0, 0, 1), β3 = (1, 0, 0) constitute a Z-base of Z3 . Thus, for any
β = (k3 , k1 , k2 ) ∈ Z3 , we always have a unique expression as β = k1 β1 + k2 β2 + k3 β3 .
By using Lemma 1.1, we see
f (α, β) = f (α, k1 β1 + k2 β2 + k3 β3 )
= f (α, k1 β1 )f (α, k2 β2 )f (α, k3 β3 )
= f (α, β1 )k1 f (α, β2 )k2 f (α, β3 )k3 = 1.
So, 0 6= α ∈ radf , then radf 6= {0}. Hence, the sufficient condition follows.
Lemma 2.2 Suppose |q| = 0 and α = (α1 , α2 , α3 ) ∈ radf .
(i) If there exists at least one non-root of unity in {qij |1 ≤ i 6= j ≤ 3}, then α1 + α2 + α3 = 0.
(ii) If all {qij |1 ≤ i 6= j ≤ 3} are roots of unity, then α1 + α2 + α3 = km, where m is the
m
smallest positive integer such that all qij
= 1, 1 ≤ i 6= j ≤ 3 and k is some integer.
Proof Suppose α = (α1 , α2 , α3 ) ∈ radf . Based on Theorem 2.1, we know that (E1), (E2),
(E3) hold simultaneously. That is to say,
α1
α3
q12
= q23
,
α2
α1
q23
= q31
,
α3
α2
q31
= q12
.
Using the three equations mentioned above, we deduce that
α1 α2 α3
α3 α3 α3
q12
q12 q12 = q23
q31 q12 = (q12 q23 q31 )α3 .
Notice that |q| = 0, we know from (2.2) that q12 q23 q31 = 1. Thus, we have
α1 +α2 +α3
q12
= 1.
(2.10)
α1 +α2 +α3
q23
= 1;
(2.11)
α1 +α2 +α3
q31
= 1.
(2.12)
Similarly,
(i) If there exists at least one non-root of unity in q12 , q23 and q31 , then it follows from
(2.10)–(2.12) that α1 + α2 + α3 = 0.
(ii) If all q12 , q23 , q31 are roots of unity, then there exists a smallest positive integer m such
m
that qij
= 1, 1 ≤ i 6= j ≤ 3. Thus from (2.10), (2.11) and (2.12), we have m|α1 + α2 + α3 .
Therefore, α1 + α2 + α3 = km, where k is some integer.
Lemma 2.3 Suppose |q| = 0. If there exist two roots of unity in {qij |1 ≤ i 6= j ≤ 3}, then
the third one must be a root of unity.
Proof Without loss of generality, let q12 , q23 be the two roots of unity, and m is the smallest
m
m
positive integer such that q12
= q23
= 1. Since |q| = 0, we have q12 q23 q31 = 1. Then,
m m m
(q12 q23 q31 )m = q12
q23 q31 = 1.
m
m
m
Replacing q12
= q23
= 1 in it, we have q31
= 1. Thus, q31 is a root of unity too.
Theorem 2.4 Whether |q| = 0 or |q| 6= 0, if there exist at least two roots of unity in
{qij | 1 ≤ i 6= j ≤ 3}, then radf 6= {0}. In other words, Lq is not a simple Lie algebra.
Simplicity of a Class of Quantum Torus Lie Algebras
2
145
Proof We shall discuss the argument in two cases.
Case 1
Suppose that q12 , q23 , q31 are all roots of unity and the corresponding orders in
the multiplicative group C∗ are m1 , m2 and m3 , respectively. Let m be the minimal common
m
multiple of m1 , m2 , m3 . Then qij
= 1, 1 ≤ i 6= j ≤ 3. Fix α = (α1 , α2 , α3 ) such that α1 =
0, α2 = m, α3 = 0, then α 6= 0.
Case 2
Suppose that there are two roots of unity, say q12 and q23 . Let m be the smallest
m
m
positive integer such that q12
= q23
= 1, and q31 is not a root of unity. From Lemma 2.3, we
know that |q| 6= 0. We also set α = (α1 , α2 , α3 ) and α1 = 0, α2 = m, α3 = 0, then α 6= 0.
In these two cases, it is not difficult to verify that the α satisfies (E1), (E2), (E3) at the
same time.
α1
α3
α1
α3
0
0
(i) Since q12
= q12
= 1, q23
= q23
= 1, then q12
= q23
; (E1) holds.
α2
α1
α2
α1
m
0
(ii) Since q23
= q23
= 1, q31
= q31
= 1, then q23
= q31
; (E2) holds.
α3
α2
α3
α2
0
m
(iii) Since q31
= q31
= 1, q12
= q12
= 1, then q31
= q12
; (E3) holds.
Thus we obtain that 0 6= α ∈ radf by Theorem 2.1. Hence, Lq is not simple.
Corollary 2.5 If Lq is a simple Lie algebra, then there exist at least two non-roots of
unity in {qij |1 ≤ i 6= j ≤ 3}.
However, whether |q| = 
0 or |q| 6= 0, the
 inverse proposition of Theorem 2.4 is not true.
1
1
i
2
Example 1 Let q =  −i 1 − 2i , namely, q12 = i, q23 = − 2i , q31 = 2. By simple
2 2i 1
calculation, we get q12 q23 q31 = i × (− 2i ) × 2 = 1, and |q| = 0. Setting α = (4, −4, 0), for any
β = (β1 , β2 , β3 ) ∈ Z3 , based on (1.2) we have
α1 β2 −α2 β1 α2 β3 −α3 β2 α3 β1 −α1 β3
f (α, β) = q12
q23
q31
−4β3 −0×β2
= i4β2 +4β1 (− 2i )
20×β1 −4β3
= 1 × (− 2i × 2)−4β3 = 1.
Thus, 0 6= α = (4, −4, 0) ∈ radf . So Lq is not a simple Lie algebra. But q23 = − 2i , q31 = 2 are
non-roots of unity. This example demonstrates that the inverse proposition of Theorem 2.4 is
not true when |q| = 0.

1
1
i
4
Example 2 Let q =  −i 1 − 2i , namely, q12 = i, q23 = − 2i , q31 = 4. Since
4 2i 1
q12 q23 q31 = i × (− 2i ) × 4 = 2 6= 1, then |q| 6= 0. Setting α = (4, −8, 0), for any β = (β1 , β2 , β3 ) ∈

Z3 , based on (1.2) we have
α1 β2 −α2 β1 α2 β3 −α3 β2 α3 β1 −α1 β3
f (α, β) = q12
q23
q31
= i4β2 +8β1 (− 2i )−8β3 −0×β2 40×β1 −4β3
= 1 × ((− 2i )2 × 4)−4β3 = 1.
Thus, 0 6= α = (4, −8, 0) ∈ radf . So Lq is not a simple Lie algebra. But q23 = − 2i , q31 = 4 are
non-roots of unity. This example demonstrates that the inverse proposition of Theorem 2.4 is
not true when |q| 6= 0.
146
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#
41
3 Further Results on |q| = 0
We assume |q| = 0 throughout this section; it implies that q12 q23 q31 = 1. If there exists at
least one non-root of unity in {qij | 1 ≤ i 6= j ≤ 3}, then denote
N0 = {α = (α1 , α2 , α3 ) ∈ Z3 | α1 + α2 + α3 = 0}.
If {qij | 1 ≤ i 6= j ≤ 3} are all roots of unity, then there exists m which is the smallest positive
m
integer such that qij
= 1, 1 ≤ i 6= j ≤ 3. Denote
Nm = {α = (α1 , α2 , α3 ) ∈ Z3 | α1 + α2 + α3 ∈ mZ}.
For simplicity, we write the above two subsets uniformly as N , namely, N = N0 if there is at
least one non-root of unity, or N = Nm otherwise.
Lemma 3.1 Suppose α = (α1 , α2 , α3 ) ∈ N , then for any β = (β1 , β2 , β3 ) ∈ Z3 we have
α1 −α3 β1 +β2 +β3
f (α, β) = (q12
q23 )
=
=
α2 −α1 β1 +β2 +β3
(q23
q31 )
α3 −α2 β1 +β2 +β3
(q31 q12 )
.
(3.1)
(3.2)
(3.3)
Proof Applying the definition of N , α1 + α2 + α3 = km, k ∈ Z. By using (2.5) we obtain
α (β2 +β3 )−(α2 +α3 )β1 (α1 +α2 )β3 −α3 (β1 +β2 )
q23
α1 (β2 +β3 )−(km−α1 )β1 (km−α3 )β3 −α3 (β1 +β2 )
q23
q12
α1 (β1 +β2 +β3 ) m −kβ1 −α3 (β1 +β2 +β3 ) m kβ3
q23
(q23 )
q12
(q12 )
α1 −α3 β1 +β2 +β3
.
(q12
q23 )
f (α, β) = q121
=
=
=
Then (3.1) follows. Similarly we can deduce (3.2) and (3.3) based on (2.6) and (2.7).
Theorem 3.2 radf ⊆ N .
Proof Let α = (α1 , α2 , α3 ) ∈ radf . When |q| = 0, it is easy to get from (2.10), (2.11),
(2.12) that
α1 +α2 +α3
qij
= 1,
1 ≤ i 6= j ≤ 3.
If there exists one non-root of unity in {qij | 1 ≤ i 6= j ≤ 3}, then α1 + α2 + α3 = 0. If all
{qij | 1 ≤ i 6= j ≤ 3} are roots of unity, then α1 + α2 + α3 = km, where m is the smallest positive
m
integer such that all qij
= 1, 1 ≤ i 6= j ≤ 3. Thus we know from the definition of N that α ∈ N .
Thus, radf ⊆ N .
Under the condition |q| = 0, we can improve the results of Theorem 2.1 and Theorem 2.4
as follows.
Theorem 3.3 radf 6= 0 if and only if there exists at least one of the characteristic equations (E1), (E2), (E3) having nonzero solution in Z2 . Namely, there exists (0, 0) 6= (x, y) ∈ Z2
y
x
such that some qik
= qkj
, 1 ≤ i 6= j 6= k ≤ 3 holds.
Proof ⇒ Let 0 6= α = (α1 , α2 , α3 ) ∈ radf , then based on Theorem 2.1 we know that α
satisfies (E1), (E2), (E3), and furthermore at least one is nonzero solution because α 6= 0.
Simplicity of a Class of Quantum Torus Lie Algebras
2
⇐
147
y
x
Without loss of generality, we let q12
= q23
, where (0, 0) 6= (x, y) ∈ Z2 . We set
α1 = x, α3 = y, α2 = m − (x + y), where m = 0 if there exists some non-root of unity in
m
{qij | 1 ≤ i 6= j ≤ 3}, or m is the smallest positive integer such that qij
= 1, 1 ≤ i 6= j ≤ 3. Then
α1
α3
α1 −α3
setting α = (α1 , α2 , α3 ), we have α1 + α2 + α3 = m. By using q12
= q23
, we obtain q12
q23 = 1.
For any β = (β1 , β2 , β3 ) ∈ Z3 , since α ∈ N , using Lemma 3.1 we get
α1 −α3 β1 +β2 +β3
f (α, β) = (q12
q23 )
= 1.
Thus, α ∈ radf and α 6= 0. Hence, radf 6= {0}.
Theorem 3.4 If |q| = 0, the quantum torus Lie algebra Lq is not simple if and only if
there exists at least one of the characteristic equations (E1), (E2), (E3) having nonzero solution
in Z2 .
Corollary 3.5 If there exists one root of unity in {qij | 1 ≤ i 6= j ≤ 3}, then Lq is not
simple.
m
Proof Let q12 be the only root of unity, and m satisfy q12
= 1. Then let α1 = m, α2 =
α1
α3
−m, α3 = 0, we know q12
= q23
= 1. By using Theorem 3.3, we know radf 6= {0}. Thus, Lq is
not simple.
1
2
However, the inverse proposition
ofCorollary 3.5 is not true.

1 21 4
Example 3 Let q =  2 1 8 , namely, q12 = 12 , q23 = 8, q31 = 14 . Since q12 q23 q31 =
1
1
1
4
8
1
× 8 × 4 = 1, then |q| = 0. Setting α = (−3, 2, 1), for any β = (β1 , β2 , β3 ) ∈ Z3 , it is obvious
that α1 + α2 + α3 = 0. Based on (3.1) we have
f (α, β) =
α1 −α3 β1 +β2 +β3
(q12
q23 )
=
1 −3
2
×8
−1
β1 +β2 +β3
= 1.
Thus, 0 6= α ∈ radf . Hence, Lq is not a simple Lie algebra, but q12 , q23 , q31 are all not roots of
unity. This example demonstrates that the inverse proposition of Corollary 3.5 is not true.
Next, we define a subalgebra of the associative algebra Cq as follows:
X
I :=
Ctα .
(3.4)
α∈N
For any α, β ∈ N , let α1 + α2 + α3 = km, β1 + β2 + β3 = k ′ m, k, k ′ ∈ Z, then by simple
P3
computation, it is easy to get i=1 (αi + βi ) = (k + k ′ )m. It means α + β ∈ N . Thus, from
tα tβ = σ(α, β)tα+β , we know tα tβ ∈ I. Therefore, I is a subalgebra of Cq .
The subalgebra I can be defined in the same way when |q| = 0 or |q| 6= 0, but the properties
are totally different in the two different cases.
Proposition 3.6 I is not an abelian subalgebra when |q| 6= 0.
Example 4
Example 2 in Section 2 shows q12 = i, q23 = − 2i , q31 = 4 and |q| 6= 0. Here
we fix α = (1, −1, 0), β = (0, 1, −1), then we can deduce that tα , tβ ∈ I. However,
α1 β2 −α2 β1 α2 β3 −α3 β2 α3 β1 −α1 β3
f (α, β) = q12
q23
q31
= i1 (− 2i )1 41 6= 1.
148
Æ
#
41
It means that tα and tβ are not commutative. Therefore, I is not an abelian subalgebra.
Now, let us turn to the situation |q| = 0, and we will see a very different but interesting
result.
Define g(α, β) = σ(α, β) − σ(β, α), then we have the Lie bracket [tα , tβ ] = g(α, β)tα+β .
Lemma 3.7 For any tα , tβ ∈ Cq , g(α, β) = 0 if and only if f (α, β) = 1.
Proof It follows directly from (1.3) that g(α, β) = 0 if and only if f (α, β) = 1.
Theorem 3.8 I is a maximal abelian subalgebra of Cq which strictly contains the center
Z(Cq ).
Proof First, we know from Theorem 3.2 that I ⊇ Z(Cq ). Next we manage to prove that
I contains Z(Cq ) strictly. Suppose that I = Z(Cq ), then N = radf . Setting α = (1, −1, 0) ∈
N = radf , based on Theorem 2.1 we know that α satisfies the characteristic equations at the
same time. Then from (E1), we get
α1
α3
1
0
q12
= q23
⇒ q12
= q23
= 1 ⇒ q12 = 1.
Similarly, if we take α = (0, 1, −1) and α = (1, 0, −1), then we can deduce that q23 = q31 = 1
respectively. These results contradict the hypothesis at the beginning of this article that there
exists at least one pair i 6= j such that qij 6= 1. Thus, N ⊃ radf and I ⊃ Z(Cq ).
The next step is to prove I is an abelian subalgebra. For any α, β ∈ N , we have α1 +α2 +α3 =
km, β1 + β2 + β3 = k ′ m, k, k ′ ∈ Z. By using Lemma 3.1, it is easy to see that
α1 −α3 k m
α1 −α3 β1 +β2 +β3
= (q12
q23 )
f (α, β) = (q12
q23 )
mα1 −mα3 k′
= (q12 q23
) = 1.
′
Then, tα tβ = tβ tα , ∀α, β ∈ I. So I is an abelian subalgebra.
At last we will show that I is a maximal abelian subalgebra of Cq . Let I ′ be any abelian
P
(j)
subalgebra of Cq which contains I. For any x ∈ I ′ , let x = 1≤j≤s,β (j) ∈Z3 λj tβ , where s is a
positive integer. Then, for any tα ∈ I, we have
[tα , x] =
s
X
λj [tα , tβ
(j)
]=
j=1
Thus all g(α, β
λj g(α, β (j) )tα+β
(j)
= 0.
j=1
But β (1) , β (2) , · · · , β (s) are different, then tα+β
(j)
s
X
(1)
, tα+β
(2)
, · · · , tα+β
(s)
are linearly independent.
) = 0, j = 1, 2, · · · , s. From Lemma 3.7, we know that f (α, β (j) ) = 1, j =
(j)
1, 2, · · · , s. This implies that tα and tβ , j = 1, 2, · · · , s are commutative. Since tα is an
arbitrary element in I, we conclude that tβ
(j)
(j)
(j)
is commutative with any element in I.
(j)
(j)
(j)
(j)
For any j, let β (j) = (β1 , β2 , β3 ) ∈ Z3 , and suppose that β1 + β2 + β3
= km + d
where 0 ≤ d < m. We take that δ = (0, 1, −1), η = (1, −1, 0), γ = (1, 0, −1), then δ, η, γ ∈ N .
Hence, tδ , tη , tγ ∈ I. So we know that tβ
f (γ, β (j) ) = 1,
(j)
is commutative with tδ , tη and tγ . Thus, we have
f (η, β (j) ) = 1,
f (δ, β (j) ) = 1.
Since δ = (δ1 , δ2 , δ3 ) = (0, 1, −1) ∈ N , applying Lemma 3.1 we see
(j)
δ2 −δ1 β1
f (δ, β (j) ) = (q23
q31 )
(j)
(j)
+β2 +β3
km+d
d
= q23
= q23
= 1.
Simplicity of a Class of Quantum Torus Lie Algebras
2
149
d
d
Similarly, by simple calculation, we have q12
= q31
= 1.
If there exists a non-root of unity in q12 , q23 , q31 , then d = 0. If q12 , q23 , q31 are all roots of
(j)
(j)
(j)
unity, then m|d. But d < m, hence d = 0 and β1 + β2 + β3 = km.
(j)
(j)
(j)
In any case, we can get β1 + β2 + β3 ∈ mZ, thus β (j) ∈ N . So, tβ
′
(j)
∈ I, j = 1, 2, · · · , s.
′
Then, x ∈ I. On the other hand, x is arbitrary in I . Then we deduce I ⊆ I. Therefore, I ′ = I.
Corollary 3.9 I/C is a maximal abelian subalgebra of the quantum torus Lie algebra Lq
and it strictly contains the center Z(Lq ).
References
[1] Allison, B., Azam, S., Berman, S., Gao Y. and Pianzola, A., Extended affine Lie algebras and their root
systems, Mem. Amer. Math. Soc., 1997, 126(603): 1-122.
[2] Berman, S., Gao Y. and Krylyuk, Y.S., Quantum tori and the structure of elliptic quasi-simple Lie algebras,
J. Funct. Anal., 1996, 135(2): 339-389.
[3] Chen Y.M., Xue M., Lin W.Q. and Tan S.B., Structure and automorphism group of a class of derivation Lie
algebras over quantum torus, Chinese. Ann. of Math.(Ser. A), 2005, 26(6): 755-764.
[4] Jiang C.B. and Meng D.J., The derivation algebra of the associative algebra Cq [x, y, x−1 , y −1 ], Comm. in
Algebra, 1998, 26(6): 1723-1736.
[5] Jiang C.B. and Meng D.J., The Automorphism group of the derivation algebra of the Virasoro-like algebra,
Advances in Mathematics(China), 1998, 27(2): 175-183.
[6] Lin W.Q. and Tan S.B., Representations of the Lie algebra of skew derivations on quantum torus, Advances
in Mathematics(China), 2005, 34(4): 477-487.
[7] Zhao K.M., Weyl type algebras from quantum tori, Comm. Contemp. Math., 2006, 8(2): 135-165.
[8] Zheng Z.J. and Tan S.B., Automorphism group for a class of Lie algebras over quantum torus, Acta. Math.
Sinica(Chinese Series), 2007, 503: 591-600.
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