QUASI-COMMUTATIVE POLYNOMIAL ALGEBRAS AND THE POWER PROPERTY OF 2 × 2 QUANTUM MATRICES

VOL. 71 1996 NO. 2

QUASI-COMMUTATIVE POLYNOMIAL ALGEBRAS AND THE POWER PROPERTY OF 2 × 2 QUANTUM MATRICES

BY

PIOTR J E¸ D R Z E J E W I C Z (TORU ´N)

Let K be a field. Recall (e.g. [4], 3.1, [5], 4.2.1) that a quadratic algebra is a graded associative K-algebra

A =

∞

M

k=0

A

,

where A

= K, dim

A

< ∞ and A is generated by A

with the ideal of relations generated by quadratic ones:

A = T (A

)/(R

),

where T (A

) is the tensor algebra of A

and R

⊂ A

. It is convenient to write

A ↔ {A

, R

}.

In this paper we consider quadratic algebras of a special type, with the relations quite similar to ordinary commutativity relations. This approach generalizes different examples of quadratic algebras. In the case of two generators we can unify the definitions of algebras

A

= Khx

, x

i/(x

x

− q

x

x

) and

A

= Khx

, x

i/(x

x

− x

x

− x

)

(notation from [4], 1.2, [5], 4.2.8, 4.4.3). This will allow us to have a com- mon view of some properties of quantum matrices connected with these two algebras. In particular, our Theorem is a generalization of [5], 1.3.8(v) and [3], (ii).

Definition 1. Let V be an m-dimensional linear space over K. Denote by R the subspace of V

spanned by elements x ⊗ y − y ⊗ x for x, y ∈ V . For each P ∈ GL(V ) we define the quadratic algebra

A

= A

[V ] ↔ {V, (I ⊗ P )(R)},

1991 Mathematics Subject Classification: Primary 16W30.

[217]

where I is the identity operator. A quasi-commutative polynomial algebra is a quadratic algebra of the form A

[V ] for some V and P ∈ GL(V ).

Lemma 1. Quadratic algebras A

[V ] and A

[V ] are isomorphic if and only if there exist C ∈ GL(V ) and α ∈ K \ {0} such that P

= α · CP

C

. P r o o f. Any isomorphism A

[V ] → A

[V ] is an extension of a linear automorphism C : V → V such that (C ⊗ C)(I ⊗ P

)(R) = (I ⊗ P

)(R), and this condition is equivalent to (I ⊗ CP

C

)(R) = (I ⊗ P

)(R), which means that α · CP

C

= P

for some α 6= 0.

Now we obtain a linear basis of A

[V ]. Choose a basis x

, . . . , x

of V and its dual basis x

, . . . , x

of V

. We have

A

=

∞

M

k=0

A

= Khx

, . . . , x

i/(x

P (x

) − x

P (x

), 1 ≤ i < j ≤ m).

One easily verifies that in A

the following relations hold:

x

P (x

) . . . P

(x

) = x

P (x

) . . . P

(x

)

for all i

, . . . , i

∈ {1, . . . , m} and σ ∈ S

. This implies that the monomials x

P (x

) . . . P

(x

)

with 1 ≤ i

≤ i

≤ . . . ≤ i

≤ m span A

. Since they are linearly indepen- dent (proof by induction on k), this is a basis of A

. It makes this algebra very similar to the algebra of commutative polynomials. In particular, we have

dim A

= m + k − 1 k

 .

Note that any quadratic algebra with two generators and one non-de- generate relation is quasi-commutative polynomial, but some well known quadratic algebras with more than two generators are not. This is discussed in the following two lemmas.

Lemma 2. Let A ↔ {A

, R

}, where dim A

= 2 and dim R

= 1. If for x, y ∈ V we have R

6= K(x ⊗ y), then A is a quasi-commutative polynomial algebra.

P r o o f. Take a

∈ K such that the generating relation is r = a

x

⊗ x

+ a

x

⊗ x

+ a

x

⊗ x

+ a

x

⊗ x

= x

⊗ (a

x

+ a

x

) − x

⊗ (−a

x

− a

x

).

Since r is not of the form x⊗y, the operator P defined by P (x

) = (−a

x

−

a

x

) and P (x

) = a

x

+ a

x

is non-degenerate and R

= (I ⊗ P )(x

⊗

x

− x

⊗ x

). Hence A = A

.

Lemma 3. The algebra A = Khx

, . . . , x

i/(x

x

− q

x

x

, 1 ≤ i <

j ≤ m) is a quasi-commutative polynomial algebra if and only if there exist q

, . . . , q

such that q

= q

q

for all i < j.

P r o o f. Suppose that A = A

. Let (p

) be the matrix of P with respect to the basis x

, . . . , x

. Take any i, j such that 1 ≤ i < j ≤ m. We have

r

=

m

X

l=1

p

x

⊗ x

−

m

X

k=1

p

x

⊗ x

= x

⊗ P (x

) − x

⊗ P (x

) ∈ R

, x

⊗ x

+ q

x

⊗ x

∈ R

,

so that p

− q

p

= (x

⊗ x

+ q

x

⊗ x

)(r

) = 0.

On the other hand, if there exist q

, . . . , q

such that q

= q

q

for all i < j, then for p

= δ

q

we get A = A

.

As a consequence, note that for m > 2 and q 6= 1 the algebra A

= Khx

, . . . , x

i/(x

x

− q

x

x

, 1 ≤ i < j ≤ m)

is not quasi-commutative polynomial. The algebra from Lemma 3 is con- sidered in [8] and the case of q

= q

q

is connected with the version of the power property given at the end of that paper.

Now recall some general constructions of “quantum endomorphism semi- groups”.

Let A ↔ {V, R

}. Put

E(A) ↔ {V

⊗ V, S

((R

)

⊗ R

)},

where (R

)

⊂ (V ⊗ V )

' (V

⊗ V

) is the annihilator of R

and S

: V

⊗ V

⊗ V ⊗ V → V

⊗ V ⊗ V

⊗ V is the isomorphism interchanging the 2nd and 3rd components (see [4], 4.5b, [5], 4.2.6). The canonical map

V → (V

⊗ V ) ⊗ V : x

7→

m

X

i=1

z

⊗ x

,

where z

= x

⊗x

for 1 ≤ i, k ≤ m, extends to a homomorphism of algebras δ

: A → E(A) ⊗ A.

E(A) is far from any kind of commutativity, it has m

generators and only

· (m

−

) =

relations. To obtain a good analogue of a commutative algebra we have to add the “second half” of the relations.

Let A ↔ {V, R

}, B ↔ {V

, R

}. Put

E(A, B) ↔ {V

⊗ V, S

((R

)

⊗ R

+ R

⊗ (R

)

)}

(compare [4], 6.2, [5], 4.2.7, [6], 1.4). The canonical maps V → (V

⊗ V ) ⊗ V

(as above) and

V

→ (V

⊗ V ) ⊗ V

: x

7→

m

X

k=1

z

⊗ x

extend to homomorphisms of algebras

δ

: A → E(A, B) ⊗ A, δ

: B → E(A, B) ⊗ B.

E(A, B) can be thought of as the “greatest common factor” of E(A) and E(B), both of them being generated by V

⊗ V , the latter via the canonical isomorphism V ⊗ V

' V

⊗ V .

Now, we apply these constructions to quasi-commutative polynomial al- gebras and write down the relations in terms of the basis z

= x

⊗ x

of V

⊗ V , which can be considered as a matrix Z = (z

).

Definition 2. Let P ∈ GL(V ). Put

E

= E

[V

⊗ V ] = E(A

[V ]).

The relations of E

[V

⊗ V ] in the above basis are the following:

z

t

= z

t

, 1 ≤ i ≤ m, 1 ≤ k < l ≤ m,

z

t

− z

t

= z

t

− z

t

, 1 ≤ i < j ≤ m, 1 ≤ k < l ≤ m, where t

are the entries of the matrix T = P

ZP .

It is useful to write these relations in matrix form:

 z

z

z

z



·

 t

−t

−t

t



=  D

0 0 D



for all i < j, k < l and suitable D

(i.e. defined by these relations).

Definition 3. Let P, Q ∈ GL(V ). Put

E

= E

[V

⊗ V ] = E(A

[V ], A

[V

]).

The relations of E

[V

⊗ V ] consist of the ones of E

[V

⊗ V ] (above) and the ones of E

[V ⊗ V

]:

z

s

= z

s

, 1 ≤ i < j ≤ m, 1 ≤ k ≤ m,

z

s

− z

s

= z

s

− z

s

, 1 ≤ i < j ≤ m, 1 ≤ k < l ≤ m, where (s

) = QZQ

, or in matrix form:

 z

−z

−z

z



·  s

s

s

s



=  D

0 0 D

 for all i < j, k < l.

From now on we assume that m = 2 and we consider only 2 × 2 matrices.

Define

 a b c d



=

 d −b

−c a



.

Note that for any 2 × 2 matrix M we have (M

)

= M , tr M

= tr M , M + M

= (tr M ) · I and M · M

= (detM ) · I. Also, (M

)

= (M

)

for any positive integer k. If M is invertible, then (M

)

= (M

)

. If the entries of matrices M and N commute, then (M N )

= N

M

.

Let P, Q ∈ GL

(K). The relations of E

[V

⊗ V ] reduce to one matrix equation

Z(P

ZP )

= DET · I,

where DET = D

. The relations of E

[V

⊗ V ] are the following:

Z(P

ZP )

= DET

· I, Z

QZQ

= DET

· I, where DET

= D

and DET

= D

.

Lemma 4. If tr(QP ) 6= 0, then dim R

= 6 and DET

= DET

. If tr(QP ) = 0, then dim R

= 5.

P r o o f. Let R = K(x

⊗ x

− x

⊗ x

) and R

= K(x

⊗ x

− x

⊗ x

).

Since

(x

⊗ Q

(x

) − x

⊗ Q

(x

))(x

⊗ P (x

) − x

⊗ P (x

)) = tr(QP ), we have (I ⊗ Q

)(R

) ⊂ ((I ⊗ P )(R))

if and only if tr(QP ) = 0. Therefore

dim((I ⊗ P )(R))

⊗ (I ⊗ P )(R) ∩ (I ⊗ Q

)(R

) ⊗ ((I ⊗ Q

)(R

))

=  1 if tr(QP ) = 0, 0 if tr(QP ) 6= 0.

Finally, dim R

= 5 if tr(QP ) = 0 and dim R

= 6 if tr(QP ) 6=

0. Now suppose that tr(QP ) 6= 0. For any 2 × 2 matrices A, B we have tr(AB

) = tr(A · tr B − AB) = tr((tr A) · B − AB) = tr(A

B). Since Z(QZP )

= (QP )

· DET

and Z

QZP = QP · DET

, we get

tr(QP ) · DET

= tr((QP )

) · DET

= tr(Z(QZP )

)

= tr(Z

QZP ) = tr(QP ) · DET

, and hence DET

= DET

.

So, when tr(QP ) 6= 0, the relations of E

[V

⊗ V ] take the form Z(P

ZP )

= Z

QZQ

= DET · I.

Take any p, q ∈ K \ {0}, pq 6= −1. For P =  1 0

0 p



, Q =  1 0 0 q

 ,

we get the quantum matrix

of the algebra M

(2) with the relations ba = pab, dc = pcd, ca = qac, db = qbd,

cb = p

qbc, da = ad + (q − p

)bc,

which is discussed in [6]–[8] and, for p = q, in [1], [2], [4], [5], [9], [10].

The power property, first noticed for M

(2), states that if the entries of the matrix Z satisfy the conditions with parameters p, q, then the entries of Z

satisfy analogous conditions with p

, q

.

Let char K 6= 2 and p, q ∈ K. For P =  1 p

0 1



, Q =  1 q 0 1

 ,

we obtain the algebra M

(2) (considered in [3], [5], [6]) with the following relations:

ac = ca + qc

, dc = cd + pc

,

da = ad + pca − qcd, bc = cb + pqc

+ pca + qcd,

ba = ab + pqcd + pcb + pa

− pad, bd = db + q

cd + qcb − qad + qd

. We observe that the algebra with these relations (for p, q 6= 0) is isomorphic to one with p

= 1, q

= p

q (a

= pa, b

= b, c

= p

c, d

= pb).

In the case of the quantum matrix Z with the relations of M

(2), the entries of its nth power Z

satisfy the relations given by parameters np, nq (see [3]).

This phenomenon is clear from the following theorem.

Theorem. Let dim V = 2, P, Q ∈ GL(V ), P Q = QP , tr(QP ) 6= 0. For any positive integer n the following equalities hold in E

[V

⊗ V ]:

Z

(P

Z

P

)

= (Z

)

Q

Z

Q

= DET

· I.

The proof will follow from Lemmas 5 and 6. The theorem remains true also in the case of tr(QP ) = 0, provided we add the relation DET

= DET

. The power property seems to be possible because these quantum matri- ces have enough commutation relations, namely 6 relations for 4 generators of E

. But it turns out that we need only 3 relations of E

with one addi- tional cubic relation to prove that the nth power satisfies the corresponding relations.

The equality Z(P

ZP )

= DET·I is equivalent to Z(ZP )

= P

·DET.

We have Z

P

= Z(ZP +(ZP )

)P −Z(ZP )

P = ZP ·tr(ZP )−DET·detP , i.e. putting TR

= tr(ZP ), DET

= DET · detP , we get an analogue of the Hamilton–Cayley Formula:

Z

P

− ZP · TR

+ DET

= 0.

For M

(2) this formula was stated in [2], [9], and for M

(2) in [3]. Note that this implies the formula

Z

P

= Z

P

· tr(ZP ) − Z

P

· DET · detP, for n ≥ 2, which will be useful below.

Let us add to E

the cubic relation we need.

Definition 4. Denote by E

= E

[V

⊗ V ] the algebra generated by V

⊗ V with the relations

Z(ZP )

= P

· DET, DET · tr(ZP ) = tr(ZP ) · DET.

Lemma 5. Let dim V = 2 and P ∈ GL(V ). For any positive integer n the following equalities hold in E

[V

⊗ V ]:

Z

(Z

P

)

= (P

)

· DET

and DET · tr(Z

P

) = tr(Z

P

) · DET.

P r o o f. Induction on n. For n = 0 and n = 1 the formulas are obvious.

Take any n ≥ 2. Assume that the formulas hold for n − 1 and n − 2. We have

Z

(Z

P

)

= Z

(Z

P

)

· tr(ZP ) − Z

(Z

P

)

· DET · detP

= Z(P

)

· DET

· tr(ZP ) − Z

(P

)

· DET

· DET · detP

= (Z · tr(ZP ) − Z

P )(P

)

· DET

= Z(ZP )

(P

)

· DET

= (P

)

· DET

.

Since tr(Z

P

) = tr(Z

P

) · tr(ZP ) − tr(Z

P

) · DET · detP , we get DET · tr(Z

P

) = tr(Z

P

) · DET.

Note that applying Lemma 5 to E

[V ⊗ V

], we get (Z

)

((Z

)

(Q

)

)

= ((Q

)

)

· DET

,

and (Z

)

is of course very different from (Z

)

, so this is not what we need.

But we can get what we need by a dual argument.

Definition 5. Denote by E

= E

[V ⊗ V

] the algebra generated by V ⊗ V

with the relations

(Q

Z)

Z = (Q

)

· DET, DET · tr(Q

Z) = tr(Q

Z) · DET.

Lemma 6. Let dim V = 2 and Q ∈ GL(V ). For any positive integer n the following equalities hold in E

[V ⊗ V

]:

(Q

Z

)

Z

= (Q

)

·DET

and DET·tr(Q

Z

) = tr(Q

Z

)·DET.

The proof is analogous to the proof of Lemma 5, but now we use the Hamilton–Cayley Formula for E

[V ⊗ V

]:

Q

Z

− tr(Q

Z) · Q

Z + DET · detQ

= 0.

P r o o f o f t h e T h e o r e m. It is enough to prove that the equalities

DET · tr(ZP ) = tr(ZP ) · DET, DET · tr(Q

Z) = tr(Q

Z) · DET

hold in E

[V

⊗ V ].

We have ZP

Z

= P

· DET and Z

QZ = Q · DET, so P

Q · DET · Z = ZP

Z

QZ = ZP

Q · DET, therefore

DET · ZP = Q

P ZP

Q · DET · P = Q

P ZP P

Q · DET.

This implies

DET · tr(ZP ) = tr(Q

P ZP P

Q) · DET = tr(ZP ) · DET.

Analogously DET · tr(Q

Z) = tr(Q

Z) · DET.

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Faculty of Mathematics and Informatics Nicholas Copernicus University

Chopina 12/18 87-100 Toru´n, Poland

E-mail: pjedrzej@mat.uni.torun.pl

Received 20 February 1996;

revised 15 March 1996

Added in proof. After submitting this paper for publication I found out that the quasi-commutative algebras were considered in the papers: M. A r t i n, J. T a t e and M. V a n d e n B e r g h, Some algebras associated to automorphisms of elliptic curves, The Grothendieck Festschrift, Vol. I, Birkh¨auser, Boston, 1990, 33–85, and M. A r t i n and M. V a n d e n B e r g h, Twisted homogeneous coordinate rings, J. Algebra 133 (1990), 249–271.