q-Integration on Quantum Spaces

arXiv:hep-th/0206083v3 27 Nov 2003

q-Integration on Quantum Spaces Hartmut Wachter∗ Sektion Physik, Ludwig-Maximilians-Universit¨at, Theresienstr. 37, D-80333 M¨ unchen, Germany July 30, 2013

Abstract In this article we present explicit formulae for q-integration on quantum spaces which could be of particular importance in physics, i.e., q-deformed Minkowski space and q-deformed Euclidean space in three or four dimensions. Furthermore, our formulae can be regarded as a generalization of Jackson’s q-integral to three and four dimensions and provide a new possibility for an integration over the whole space being invariant under translations and rotations.

1

Introduction

The history of the natural sciencies shows us that revolutionary changes in our understanding of physical phenomena have often been accompanied by new developments in mathematics. Newtonian mechanics is one famous example for such a situation in physics, as its general formulation would not have been able without the ideas of differential calculus. Due to this fact one may believe that a new theory giving a more detailed description of nature has to be based on a modified version of traditional mathematics. Quantum spaces which are defined as co-module algebras of quantum groups and which can be interpreted as deformations of ordinary co-ordinate algebras [1] could provide a proper framework for developing such a modification [2], [3]. For our purposes it is sufficient to consider a quantum space as an algebra Aq of formal power series in the non-commuting co-ordinates X1 , X2 , . . . , Xn Aq = C [[X1 , . . . Xn ]] /I ∗

e-mail:[email protected]

1

(1)

where I denotes the ideal generated by the relations of the non-commuting co-ordinates. The algebra Aq satisfies the Poincar´e-Birkhoff-Witt property, i.e. the dimension of the subspace of homogenous polynomials should be the same as for commuting co-ordinates. This property is the deeper reason why the monomials of normal ordering X1 X2 . . . Xn constitute a basis of Aq . In particular, we can establish a vector space isomorphism between Aq and the commutative algebra A generated by ordinary co-ordinates x1 , x2 , . . . , xn : W W(xi11

. . . xinn )

:

A −→ Aq ,

=

X1i1

(2)

. . . Xnin .

This vector space isomorphism can be extended to an algebra isomorphism introducing a non-commutative product in A, the so-called ⋆-product [4], [5]. This product is defined by the relation W(f ⋆ g) = W(f ) · W(g)

(3)

where f and g are formal power series in A. In [6] we have calculated the ⋆-product for quantum spaces which could be of particular importance in physics, i.e. q-deformed Minkowski space and q-deformed Euclidean space in three or four dimensions. Additionally, for each of these quantum spaces exists a symmetry algebra [7], [8] and a covariant differential calculus [9], which can provide an action upon the quantum spaces under consideration. By means of the relation W(h ⊲ f ) := h ⊲ W(f ),

h ∈ H, f ∈ A,

(4)

we are also able to introduce an action upon the corresponding commutative algebra. In our previous work [10] we have presented explicit formulae for such representations. In the following it is our aim to derive some sort of q-integration on commutative algebras which are, via the algebra isomorphism W, related to q-deformed Minkowski space or q-deformed Euclidean space in three and four dimensions. In some sense our considerations can be supposed to be a generalization of the celebrated Jackson-integral [11] to higher dimensions. To this end we will start with introducing new elements which are formally inverse to the partial derivatives of the given covariant differential calculi. Such an extension of the algebra of partial derivatives will lead to additional commutation relations. Finally, the representations of the partial derivatives

2

given in [10] will aid us in identifying this new elements with particular solutions to some q-difference equations. In this way, we can interpret our results as a method to discretise classical integrals of more than one dimension. Furthermore, we will see that we have to distinguish between left and right integrals. For both types formulae for integration by parts can be derived. It is also possible to define volume integrals which are invariant under translations or the action of symmetry generators if surface terms are neglected. Suprisingly, these integrals obey a rather simple structure, as the single integrations for the different directions become independent from each other if we integrate over the entire space.

2

q-Deformed Euclidean space in three dimensions

From [12] we know that the partial derivatives ∂ + , ∂ 3 , ∂ − satisfy the same relations as the quantum space coordinates X + , X 3 , X − . Thus we have ∂ 3∂ + = q2∂ + ∂ 3,

∂ −∂ 3 = q2∂ 3 ∂ −,

∂ − ∂ + = ∂ + ∂ − + λ(∂ 3 )2 ,

(5)

where λ = q − q −1 and q > 1. Now, we would like to extend the algebra of partial derivatives by inverse elements, such that ∂ + (∂ + )−1 = (∂ + )−1 ∂ + = 1, 3

3 −1

−

− −1

∂ (∂ )

∂ (∂ )

(6)

3 −1 3

= (∂ )

∂ = 1,

− −1 −

= (∂ )

∂ = 1.

It can easily be shown that these relations imply the formulae (∂ 3 )−1 ∂ + = q −2 ∂ + (∂ 3 )−1 , − −1 3

(∂ )

∂

= q

−2 3

− −1

∂ (∂ )

∂ 3 (∂ + )−1 = q −2 (∂ + )−1 ∂ 3 , 3 −1

−

,

∂ (∂ )

=q

−2

(7)

3 −1 −

(∂ )

∂ ,

(∂ − )−1 ∂ + = ∂ + (∂ − )−1 − q −4 λ(∂ 3 )2 (∂ − )−2 , ∂ − (∂ + )−1 = (∂ + )−1 ∂ − − q −4 λ(∂ + )−2 (∂ 3 )2 . In addition, we can also find the following identities: (∂ 3 )−1 (∂ + )−1 = q 2 (∂ + )−1 (∂ 3 )−1 , − −1

(∂ )

3 −1

(∂ )

2

3 −1

= q (∂ )

(∂ − )−1 (∂ + )−1 =

∞ X

i

− −1

(∂ )

λ [[i]]q4 !

i=0



(8)

, −1 i

 !2 q4

· (∂ + )−(i+1) (∂ 3 )2i (∂ − )−(i+1) . 3

From the commutation relations between symmetry generators and partial derivatives [12] we find the expressions L+ (∂ + )−1 = (∂ + )−1 L+ , +

3 −1

L (∂ )

3 −1

= (∂ )

(9)

+

L +q

3 −2 − 21

−1 +

∂ (∂ )

τ

,

L+ (∂ − )−1 = (∂ − )−1 L+ + q −1 ∂ 3 (∂ − )−2 τ

− 21

,

L− (∂ − )−1 = (∂ − )−1 L− , −

3 −1

−

+ −1

L (∂ ) L (∂ )

3 −1

= (∂ )

−3

−

−4

L −q

+ −1

= (∂ )

(10)

−

L −q

3 −2 − − 21

(∂ )

∂ τ

,

+ −2 3 − 21

(∂ )

∂ τ

,

1

1

τ − 2 (∂ + )−1 = q −2 (∂ + )−1 τ − 2 , τ

− 12

τ

(∂ − )−1 = q 2 (∂ − )−1 τ

− 12

(∂ 3 )−1 = (∂ 3 )−1 τ

1

− 21

− 21

(11)

,

, 1

Λ 2 (∂ A )−1 = q 2 (∂ A )−1 Λ 2 ,

A = ±, 3.

(12)

Applying the substitutions ∂ A → ∂ˆA ,

(∂ A )−1 → (∂ˆA )−1 ,

A = ±, 3

(13)

to all expressions presented so far we get the corresponding relations of the second differential calculus (generated by the conjugated partial derivatives ∂ˆA [12]). In [10] it was shown that according to   A A ∂ A ⊲ F = (∂(i=0) ) + (∂(i>0) ) F (14)

the representations of our partial derivatives can be divided up into a classical part and corrections vanishing in the undeformed limit q → 1. Thus, seeking a solution to the equation ∂A ⊲ F = f

(15)

for given f it is reasonable to consider the following expression: F

= =

∂A


−1

A (∂(i=0) )

⊲f =

1 f A A (∂(i=0) ) + (∂(i>0) ) 1



f A A 1 + (∂(i=0) )−1 (∂(i>0) ) 4

(16)

= =

1 1+ ∞ X k=0

A A (∂(i=0) )−1 (∂(i>0) )

·

1 A (∂(i=0) )

f

h ik A A A (−1)k (∂(i=0) )−1 (∂(i>0) ) (∂(i=0) )−1 f.

A A To apply this formula, we need to identify the contributions ∂(i=0) and ∂(i>0) the representations of our partial derivatives consist of. However, their explicit form can be read off quite easily from the results in [10]. Hence we have + )f (∂(i=0)

= −qDq−4 f (q 2 x3 ),

3 (∂(i=0) )f

= Dq32 f (q 2 x+ ),

− )f (∂(i=0)

= −q −1 Dq+4 f

+ )f (∂(i>0)

= −qλx+ (Dq32 )2 f,

3 (∂(i>0) )f

= 0,

− )f (∂(i>0)

= 0.

(17)

and (18)

A Furthermore, it is easily seen that the inverse operators (∂(i=0) )−1 are given by + )−1 f (∂(i=0)

= −q −1 (Dq−4 )−1 f (q −2 x3 ),

3 (∂(i=0) )−1 f

= (Dq32 )−1 f (q −2 x+ ),

− )−1 f (∂(i=0)

= −q(Dq+4 )−1 f,

(19)

where DqAa and (DqAa )−1 denote Jackson derivatives and Jackson integrals, respectively. Inserting these expressions into formula (16) we can represent the inverse counterparts of the partial derivatives in the form (∂ − )−1 L f

= −q(Dq+4 )−1 f,

(∂ 3 )−1 L f

= (Dq32 )−1 f (q −2 x+ ),

(∂ + )−1 L f

= −q

−1

(20)

∞ ik h X 3 2 + − −1 k 2k(k+1) x (Dq4 ) (Dq2 ) (−λ) q

k=0 − −1 · (Dq4 ) f (q −2(k+1) x3 ).

5

Repeating the identical steps as before, we can also find solutions to the equations ∂ˆA ˜⊲ F = f, F ⊳˜ ∂ A = f, F ⊳ ∂ˆA = f,

(21)

where the explicit form of the action of our partial derivatives is again taken from [10]. However, in [10] it was also shown that the different representations of our partial derivatives can be transformed into each other by a straightforward application of so-called crossing-symmetries. Due to this fact the solutions to equation (21) are easily obtained from (20), if we apply the transformations ± → ∓ q → 1/q

(∂ ± )−1 ⊲ f

←→

± → ∓ q → 1/q

(∂ 3 )−1 ⊲ f

←→

(∂ˆ∓ )−1 ˜⊲f,

(22)

(∂ˆ3 )−1 ˜⊲f,

which concretely mean, that the expressions on the right and left hand side are related to each other by the substitutions1 x± → x∓ ,

q ±1 → q ∓1 ,

n ˆ ± → −ˆ n∓

(23)

(Dq±a )−1 → (Dq∓−a )−1 .

Dq±a → Dq∓−a , In the same way we have f˜ ⊳ (∂ ± )−1 3 −1

f˜ ⊳ (∂ ) f ⊳ (∂ˆ± )−1 f ⊳ (∂ˆ3 )−1

+↔−

←→ −q 6 (∂ˆ∓ )−1 ˜⊲ f, +↔− ←→ −q 6 (∂ˆ3 )−1 ˜⊲ f,

(24)

+↔−

(25)

←→ −q −6 (∂ ∓ )−1 ⊲ f,

+↔−

←→ −q −6 (∂ 3 )−1 ⊲ f,

which symbolizes a transition described by the substitutions x± → x∓ ,

Dq±a → Dq∓a ,

(Dq±a )−1 → (Dq∓a )−1 ,

n ˆ± → n ˆ∓.

(26)

Next, we would like to have a closer look at the question in which sense the operator (∂ A )−1 inverse to the operator ∂ A is. First of all, we require 1

For notation see appendix A, please.

6

that all lower limits of Jackson integrals appearing in the representations of (∂ A )−1 are set equal to zero. In this case one can easily verify the identities   A A A −1 x Dqa (Dqa ) f 0 = f, (27) xA (DqAa )−1 DqAa f 0

leading to

A ∂(i=0)



¯

xA A (∂(i=0) )−1 f 0

=



f |x0

A

= f,

¯

xA A −1 A (∂(i=0) ) ∂(i=0) f 0

=

(28) ¯ A

f |x0 .

It is important to realize, that in the above formulae the coordinates giving the upper limits of integration have to be labeled by different indices than A the corresponding operators (∂(i=0) )−1 . Let us now introduce the operator xA¯ A A −1 A ˆ C f = − (∂(i=0) ) (∂(i>0) )f .

(29)

0

With the identities (14), (16) and (28) one can then compute that xA¯ (∂ A )−1 ∂ A f 0 ∞ ∞ xA¯ xA¯ X X ¯ A k A k A ˆ ˆ (C ) f (C ) (f − f (x = 0) − = 0

k=0

¯

= f − f (xA = 0) −

∞ X k=1

0

k=1

¯ (Cˆ A )k f (xA = 0)

(30)

¯ xA

0

where the limits of the integration intervals always refer to the Jackson integrals in the expressions of (∂ A )−1 or Cˆ A . In addition to this one readily checks that   A¯ A A −1 x = f, (31) ∂ (∂ ) f 0

as the representations of (∂ A )−1 have been determined in such a way that they give solutions to equation (15). As a next step we wish to present formulae for integration by parts. Before doing this let us collect some notation that will be used in the following.

7

In the case of right and left derivatives we use the abbreviations ∂LA f

(32)

A ∂R f

≡ ∂ A ⊲ f, ≡ f ⊳˜ ∂ A ,

∂ˆLA f ∂ˆA f

≡ ∂ˆA ˜⊲ f, ≡ f ⊳ ∂ˆA .

(33)

R

Likewise for right and left integrals we abbreviate (34)

(∂ A )−1 L f

≡ (∂ A )−1 ⊲ f, ≡ f ⊳˜ (∂ A )−1 ,

(∂ˆA )−1 R f −1 A (∂ˆ ) f

≡ (∂ˆA )−1 ˜⊲ f, ≡ f ⊳ (∂ˆA )−1 .

(35)

(∂ A )−1 L f

R

Furthermore, we set ¯

A f kx0

∞ xA¯ X ¯ A k A ˆ ≡ f − f (x = 0) − (C ) f (x = 0) . ¯ A

0

k=1

(36)

Now, we are in a position to write down rules for integration by parts. In complete analogy to the classical case these rules can be derived from the Leibniz rules of the corresponding partial derivatives [10]. In this way we get − a (∂ − )−1 L (∂L f ) ⋆ g |x+ =0

(37)

1/2 −1/2 τ f ) ⋆ ∂L− g |ax+ =0 , = f ⋆ g kax+ =0 − (∂ − )−1 L (Λ

a 3 3 (∂ f ) ⋆ g (∂ 3 )−1 L x =0 L

a 1/2 f ) ⋆ ∂L3 g ax3 =0 = f ⋆ g x3 =0 − (∂ 3 )−1 L (Λ 1/2 + L f ) ⋆ ∂L− g ax3 =0 , − λλ+ (∂ 3 )−1 L (Λ + a (∂ + )−1 L (∂L f ) ⋆ g |x− =0

= f⋆

g kax− =0

−

1/2 1/2 τ f) ⋆ (∂ + )−1 L (Λ

∂L+ g |ax− =0

1/2 1/2 + τ L f ) ⋆ ∂L3 g bx− =a − qλλ+ (∂ + )−1 L (Λ
− b 1/2 1/2 + 2 g f ) ⋆ ∂ (Λ τ L − q 2 λ2 λ+ (∂ + )−1 L x− =a L 8

(38)

(39)

and a ˆ+ (∂ˆ+ )−1 L (∂L f ) ⋆ g |x− =0 −1/2 −1/2 τ f ) ⋆ ∂ˆL+ g |ax− =0 , = f ⋆ g kax− =0 − (∂ˆ+ )−1 L (Λ

a ˆ3 (∂ˆ3 )−1 L (∂L f ) ⋆ g x3 =0

a −1/2 ˆ3 g a3 (Λ f ) ⋆ ∂ = f ⋆ g x3 =0 − (∂ˆ3 )−1 L x =0  L  Λ−1/2 L− f ⋆ ∂ˆL+ g ax3 =0 , − λλ+ (∂ˆ3 )−1 L

a ˆ− (∂ˆ− )−1 L (∂L f ) ⋆ g |x+ =0 −1/2 1/2 τ f ) ⋆ ∂ˆL− g |ax+ =0 = f ⋆ g kax+ =0 − (∂ˆ− )−1 L (Λ − q −1 λλ+ (∂ˆ− )−1 (Λ−1/2 τ 1/2 L− f ) ⋆ ∂ˆ3 g |a+ L

L

(40)

(41)

(42)

x =0

−1/2 1/2 τ (L− )2 f ) ⋆ ∂ˆL+ g |ax+ =0 . − q −2 λ2 λ+ (∂ˆ− )−1 L (Λ

In the case of right integrals, however, we have to consider Leibniz rules for right derivatives [10]. Hence, we have − a (∂ − )−1 R f ⋆ ∂R g |x+ =0

= f⋆

g kax+ =0

−

(43)

− −1/2 1/2 τ g) |ax+ =0 , (∂ − )−1 R (∂R f ) ⋆ (Λ

3 a (∂ 3 )−1 R f ⋆ ∂R g x3 =0

3 −1/2 g) ax3 =0 = f ⋆ g ax3 =0 − (∂ 3 )−1 R (∂R f ) ⋆ (Λ −1/2 1/2 + − τ L g) ax3 =0 , + λλ+ (∂ 3 )−1 R (∂R f ) ⋆ (Λ + a (∂ + )−1 R f ⋆ ∂R g |x− =0

=

and

(44)

(45)

−1/2 −1/2 + f ⋆ g kax− =0 − (∂ + )−1 τ g) |ax− =0 R (∂R f ) ⋆ (Λ 3 −1/2 + L g) |ax− =0 + q −1 λλ+ (∂ + )−1 R (∂R f ) ⋆ (Λ
2 − −1/2 1/2 τ L+ g) |ax− =0 − λ2 λ+ (∂ + )−1 R (∂R f ) ⋆ (Λ

ˆ+ a (∂ˆ+ )−1 R f ⋆ ∂R g |x− =0 1/2 1/2 ˆ+ τ g) |ax− =0 , = f ⋆ g kax− =0 − (∂ˆ+ )−1 R (∂R f ) ⋆ (Λ

(46)

ˆ3 a (∂ˆ3 )−1 R f ⋆ ∂R g x3 =0

ˆ3 f ) ⋆ (Λ1/2 g) a3 ( ∂ = f ⋆ g ax3 =0 − (∂ˆ3 )−1 R x =0 R

(47)

9

ˆ+ f ) ⋆ (Λ1/2 τ 1/2 L− g) a3 , ( ∂ + λλ+ (∂ˆ3 )−1 R x =0 R

ˆ− a (∂ˆ− )−1 R f ⋆ ∂R g |x+ =0 1/2 −1/2 ˆ− τ g) |ax+ =0 = f ⋆ g kax+ =0 − (∂ˆ− )−1 R (∂R f ) ⋆ (Λ 1/2 − ˆ3 L g) |ax+ =0 + qλλ+ (∂ˆ− )−1 R (∂R f ) ⋆ (Λ
2 1/2 1/2 ˆ+ τ L− g) |ax+ =0 . − λ2 λ+ (∂ˆ− )−1 R (∂R f ) ⋆ (Λ

(48)

We want to close this section by discussing a method for an integration over the entire space. For this purpose we could define Z dq V f ≡ (∂ + )−1 (∂ 3 )−1 (∂ − )−1 ⊲ f (49) L

for a 3-dimensional volume integral. But this is not the only possibility for defining a global integration, as the operators (∂ A )−1 , A = ±, 3, do not mutually commute. However, for functions vanishing at infinity one can show by inserting the representations of (20) that the following indentities hold: (∂ + )−1 (∂ 3 )−1 (∂ − )−1 ⊲ f = q

−4

= q

−2

= q

−2

= q

−2

= q

−2

− −1

(∂ )

3 −1

(∂ )

3 −1

(∂ )

(∂ )

− −1

(∂ )

+ −1

(∂ )

− −1

(∂ )

+ −1

(∂ )

3 −1

⊲ f + S.T.

− −1

⊲ f + S.T.

3 −1

⊲ f + S.T.

3 −1

⊲ f + S.T.

(∂ ) (∂ )

(∂ )

− −1

(∂ )

⊲ f + S.T.

+ −1

(∂ )

+ −1

(50)

+ −1

(∂ ) (∂ )

where S.T. stands for neglected surface terms. And in the same manner we can also find as an explicit formula for calculating volume integrals (∂ + )−1 (∂ 3 )−1 (∂ − )−1 ⊲ f = =

3 − + )−1 ⊲ f + O.T. )−1 (∂(i=0) )−1 (∂(i=0) (∂(i=0) q −4 (Dq−4 )−1 (Dq32 )−1 (Dq+4 )−1 f (q −2 x+ , q −2 x3 )

(51) + S.T.

if again surface terms are dropped. Using the crossing-symmetries (22), (24) and (25) it immediately follows that ± → ∓ q → 1/q (∂ + )−1 (∂ 3 )−1 (∂ − )−1 ⊲ f ←→ (∂ˆ− )−1 (∂ˆ3 )−1 (∂ˆ+ )−1 ˜⊲ f

10

(52)

and f˜ ⊳ (∂ + )−1 (∂ 3 )−1 (∂ − )−1 f ⊳ (∂¯− )−1 (∂¯3 )−1 (∂¯+ )−1

+↔−

←→ (∂¯− )−1 (∂¯3 )−1 (∂¯+ )−1 ˜⊲ f,

+↔−

+ −1

←→ (∂ )

3 −1

(∂ )

− −1

(∂ )

(53)

⊲ f,

where (∂¯A )−1 = −q 6 (∂ˆA )−1 . From a direct calculation using the representations in [10] and appying them to the last expressions of (51) one can also verify that our volume integrals are invariant under both translations and rotations. This means, in explicit form, that we have Z Z Z (54) dq V ∂ A ⊲ f = ε(∂ A ) dq V f = 0, dq V f = ∂A ⊲ L L L Z Z Z ± ± ± dq V L ⊲ f = ε(L ) dq V f = 0, dq V f = L ⊲ L Z Z ZL ZL dq V f, dq V τ ⊲ f = ε(τ ) dq V f = dq V f = τ⊲ L

L

L

L

where we have used as a shorthand notation Z dq V f ≡ (∂ + )−1 (∂ 3 )−1 (∂ − )−1 ⊲ f, L Z dq V f ≡ f ˜⊳ (∂ + )−1 (∂ 3 )−1 (∂ − )−1 , R Z dq V f ≡ (∂¯− )−1 (∂¯3 )−1 (∂¯+ )−1 ˜⊲ f, ZL dq V f ≡ f ⊳ (∂¯− )−1 (∂¯3 )−1 (∂¯+ )−1 .

(55)

R

In the same way one can check invariance under right representations, hence  Z Z Z (56) dq V f ˜⊳ ∂ A = ε(∂ A ) dq V f = 0, ⊳ ∂A = dq V f ˜ L L L  Z Z Z ± ± ± dq V f ⊳ L = ε(L ) dq V f = 0, dq V f ⊳ L = L L  LZ Z Z Z dq V f. dq V f ⊳ τ = ε(τ ) dq V f = dq V f ⊳ τ = L

L

L

L

Because of the crossing-symmetries these indentities carry over to all of the other volume integrals. However, one has to notice that in the case of

11

conjugated integrals translation invariance now reads Z Z Z A A A ¯ ¯ ¯ dq V f = ∂ ⊲ dq V ∂ ⊲ f = ε(∂ ) dq V f = 0, L L  Z Z ZL dq V f ⊳ ∂¯A = ε(∂¯A ) dq V f = 0, dq V f ⊳ ∂¯A = L Z ZL ZL A A A dq V f = ⊲ ∂¯ ˜ dq V f = 0, dq V ∂¯ ˜⊲ f = ε(∂¯ ) R R  Z Z ZR dq V f ⊳ ∂¯A = ε(∂¯A ) dq V f = 0. dq V f ⊳ ∂¯A = R

R

3

(57)

R

q-Deformed Euclidean space in four dimensions

The 4-dimensional Euclidean space can be treated in very much the same way as the 3-dimensional one. Therefore we will restrict ourselves to stating the results only. Again, we start with the commutation relations [15] ∂ 1 ∂ 2 = q∂ 2 ∂ 1 ,

∂ 1 ∂ 3 = q∂ 3 ∂ 1 , ∂2∂3 = ∂3∂2,

∂ 2 ∂ 4 = q∂ 4 ∂ 2 ,

∂ 3 ∂ 4 = q∂ 3 ∂ 2 ,

(58)

∂ 4 ∂ 1 = ∂ 1 ∂ 4 + λ∂ 2 ∂ 3 ,

where λ = q − q −1 with q > 1, and introduce elements ∂ i by
−1
−1 i ∂i ∂i = ∂i ∂ = 1, i = 1, . . . , 4.


−1

, i = 1, . . . , 4,

Now, the additional commutation relations become (∂ 2 )−1 ∂ 1 = q∂ 1 (∂ 2 )−1 , 3 −1 1

(∂ )

∂

4 −1 2

(∂ )

∂

1

3 −1

2

4 −1

3

4 −1

= q∂ (∂ ) = q∂ (∂ )

4 −1 3

= q∂ (∂ )

4 −1 1

1

(∂ )

∂

(∂ )

∂

4

1 −1

∂ (∂ )

3 −1 2

(∂ )

∂

4 −1

= ∂ (∂ )

∂ 2 (∂ 1 )−1 = q(∂ 1 )−1 ∂ 2 ,

,

2

1 −1

4

2 −1

4

3 −1

∂ (∂ )

,

∂ (∂ )

,

∂ (∂ ) 2

2 3

2

= ∂ (∂ )

= q(∂ )

3

,

∂ ,

2 −1 4

= q(∂ ) = q(∂ )

4 −2

∂ ,

3 −1 4

∂ ,

,

1 −2 2 3

∂ − q λ(∂ )

3 −1

(59)

1 −1 3

− q λ∂ ∂ (∂ )

1 −1 4

= (∂ )

3

2 −1

∂ (∂ )

∂ ∂ , = (∂ 2 )−1 ∂ 3 .

Furthermore, we obtain identities of the form (∂ 2 )−1 (∂ 1 )−1 = q −1 (∂ 1 )−1 (∂ 2 )−1 , 3 −1

(∂ )

4 −1

(∂ )

1 −1

(∂ )

2 −1

(∂ )

= q

−1

= q

−1

1 −1

(∂ )

2 −1

(∂ ) 12

3 −1

,

4 −1

,

(∂ )

(∂ )

(60)

(∂ 4 )−1 (∂ 3 )−1 = q −1 (∂ 3 )−1 (∂ 4 )−1 , (∂ 3 )−1 (∂ 2 )−1 = (∂ 2 )−1 (∂ 3 )−1 , (∂ 4 )−1 (∂ 1 )−1 =

∞ X

λi [[i]]q−2 !

i=0



−1 i



q −2

!2

· (∂ 1 )−(i+1) (∂ 2 )i (∂ 3 )i (∂ 4 )−(i+1) . Next, the commutation relations with the symmetry generators read

−1 +
−2 2 1 −1 = q −1 ∂ 1 L+ L1 + q −1 ∂ 1 ∂ , 1 ∂

−1 −1 2 L+ = q ∂2 L+ 1 ∂ 1,


−2 4 −1 −1 3 −1 L+ = q −1 ∂ 3 L+ ∂3 ∂ , 1 ∂ 1 −q

−1 −1 4 L+ = q ∂4 L+ 1 ∂ 1, 1 L+ 2 ∂

L+ 2 ∂


2 −1

3 L+ 2 ∂

4 L+ 2 ∂ 1 L− 1 ∂

L− 1 − L1 L− 1

∂

∂4

1 L− 2 ∂

L− 2 ∂


−1


−1


−1


−1


−1


2 −1


3 −1

4 L− 2 ∂

K1 ∂ 1 K1 ∂


−1


2 −1

∂3

L− 2 ∂


−1


−1


−1


2 −1

K1 ∂ 3 K1 ∂ 4


−1


−1

= q −1 ∂ 1


−1

−1 L+ ∂1 2 +q


2 −1

−1 = q −1 ∂ L+ ∂ 2 −q
−1 + = q ∂3 L2 ,
−1 = q ∂4 L+ 2,


−2


2 −2


−1 − L1 , = q −1 ∂ 1

−2 2 −1 − L1 + q 3 ∂ 1 ∂ 2 , = q ∂
−1 3 −1 − = q ∂ L1 ,

−2 −1 − = q ∂4 L1 − q 3 ∂ 3 ∂ 4 , = q −1 ∂ 1


−1

L− 2,

∂3,

(61)

(62)

∂4,

(63)

(64)


2 −1

= q −1 ∂ L− 2,

−2 −1 3 1 = q ∂3 L− ∂3 , 2 +q ∂

3 2 4 −2 4 −1 − , = q ∂ L2 − q ∂ ∂

= q ∂1


−1

K1 ,
−1 K1 , = q −1 ∂ 2
−1 = q ∂3 K1 ,
−1 = q −1 ∂ 4 K1 , 13

(65)

K2 ∂ 1 K2 ∂ K2 ∂ K2 ∂


−1


2 −1


3 −1


4 −1

= q ∂1 = q ∂ = q

−1

= q

−1

Applying the substitutions ∂ i → ∂ˆi ,


−1


2 −1 ∂

∂

K2 ,

(66)

K2 ,


3 −1


4 −1

K2 , K2 .

(∂ i )−1 → (∂ˆi )−1 ,

i = 1, . . . , 4,

(67)

we get the corresponding relations for the second differential calculus spanned by the conjugated derivatives. With the representations in [10] the same reasonings we have already applied to the 3-dimensional Euclidean space lead immediately to the expressions (∂ˆ1 )−1 ⊲ f

= q

(∂ˆ2 )−1 ⊲ f

=

(∂ˆ3 )−1 ⊲ f

=

(∂ˆ4 )−1 ⊲ f

=

∞ X

h ik (−λ)k q −k(k+1) x1 (Dq4−2 )−1 Dq2−2 Dq3−2

(68)

k=0 · (Dq4−2 )−1 f (q k+1 x2 , q k+1 x3 ), (Dq3−2 )−1 f (qx1 ), (Dq2−2 )−1 f (qx1 ), q −1 (Dq1−2 )−1 f.

In complete analogy to the 3-dimensional case we have the transformation i −1

(∂ )

i → i′ q → 1/q

′

˜ ⊲ f ←→ (∂ˆi )−1 ⊲ f,

i′ = 5 − i,

(69)

whose explicit form is given by the substitutions x± → x∓ ,

q ±1 → q ∓1 ,

i′ q −a

Dqi a → D

,

′

n ˆ i → −ˆ ni i′ q −a

(Dqi a )−1 → (D

(70)

)−1 .

For the relationship between right and left integrals we now have j↔j ′

′

f ⊳ (∂ i )−1 ←→ −q 4 (∂ˆi )−1 ⊲ f, j↔j ′

(71)

′

f˜ ⊳ (∂ˆi )−1 ←→ −q −4 (∂ i )−1 ˜⊲ f, which symbolizes that we have to apply the substitutions ′

xj → xj ,

′

Dqja → Dqja ,

′

(Dqja )−1 → (Dqja )−1 , 14

′

n ˆj → n ˆj .

(72)

Next, we would like to present the formulae for integration by parts. For left integrals belonging to the differential calculus with unhatted derivatives we find a 1 (73) (∂ 1 )−1 L (∂L f ) ⋆ g x4 =0 a

a 1/2 1/2 1/2 K1 K2 f ) ⋆ ∂L1 g 4 , = f ⋆ g x4 =0 − (∂ 1 )−1 L (Λ x =0

a 2 3 (∂ f ) ⋆ g (∂ 2 )−1 L L x =0 a

a −1/2 1/2 2 −1

K2 f ) ⋆ ∂L2 g 3 = f ⋆ g x3 =0 − (∂ )L (Λ1/2 K1 x =0 a 1/2 1/2 1/2 + 1 2 −1 − qλ (∂ )L (Λ K1 K2 L1 f ) ⋆ ∂L g 3 ,

(74)

x =0

a 3 (∂ 3 )−1 L (∂L f ) ⋆ g x2 =0 a

a 3 1/2 1/2 −1/2 f ) ⋆ ∂ g (Λ K K = f ⋆ g x2 =0 − (∂ 3 )−1 3 L 1 2 L x =0 a 1/2 1/2 1/2 + 1 3 −1 − qλ (∂ )L (Λ K1 K2 L2 f ) ⋆ ∂L g 2 ,

(75)

x =0

b 4 (∂ 4 )−1 L (∂L f ) ⋆ g x1 =a b

1/2 −1/2 −1/2 4 (Λ K K f ) ⋆ ∂ g = f ⋆ g ax1 =0 − (∂ 4 )−1 L 1 1 2 L x =a b 1/2 1/2 −1/2 + K1 K2 L1 f ) ⋆ ∂L3 g 1 + qλ (∂ 4 )−1 L (Λ x =a b −1/2 1/2 1/2 2 K1 K2 L+ + qλ (∂ 4 )−1 2 f ) ⋆ ∂L g 1 L (Λ x =a b 1/2 1/2 1/2 + + 2 2 4 −1 + q λ (∂ )L (Λ K1 K2 L1 L2 f ) ⋆ ∂L1 g 1 .

(76)

x =a

For the second differential calculus with hatted derivatives one can compute likewise a ˆ1 f ) ⋆ g (77) ( ∂ (∂ˆ1 )−1 L L x4 =0

a −1/2 −1/2 −1/2 K1 K2 f ) ⋆ ∂ˆL1 g 4 = f ⋆ g ax4 =0 − (∂ˆ1 )−1 L (Λ x =0 a −1/2 1/2 −1/2 − 2 −1 1 −1 ˆ ˆ K1 K2 L1 f ) ⋆ ∂L g 4 + q λ (∂ )L (Λ xa =0 −1/2 −1/2 1/2 − −1 1 −1 3 ˆ ˆ K1 + q λ (∂ )L (Λ K2 L2 f ) ⋆ ∂L g 4 x =0 a −1/2 1/2 −1/2 − − −2 2 ˆ1 −1 K1 K2 L1 L2 f ) ⋆ ∂ˆL4 g 4 , + q λ (∂ )L (Λ x =0

15

b ˆ2 f ) ⋆ g ( ∂ (∂ˆ2 )−1 L L x3 =a a

a −1/2 1/2 −1/2 ˆ2 g (Λ K K f ) ⋆ ∂ = f ⋆ g x3 =0 − (∂ˆ2 )−1 L 1 2 L x3 =0 a 1/2 1/2 −1/2 ˆ4 , K1 K2 L− − q −1 λ (∂ˆ2 )−1 2 f ) ⋆ ∂L g 3 L (Λ

(78)

x =0

ˆ3 f ) ⋆ g ( ∂ (∂ˆ3 )−1 L L a

−1/2 −1/2 1/2 ˆ3 g (Λ K K f ) ⋆ ∂ = f ⋆ g ax2 =0 − (∂ˆ3 )−1 L 1 2 L x2 =0 a 1/2 1/2 −1/2 ˆ4 , K1 K2 L− − q −1 λ (∂ˆ3 )−1 1 f ) ⋆ ∂L g 2 L (Λ

(79)

x =0

b ˆ4 f ) ⋆ g ( ∂ (∂ˆ4 )−1 L L x1 =a a

a −1/2 1/2 1/2 4 ˆ4 g (Λ K K f ) ⋆ ∂ = f ⋆ g x1 =0 − (∂ˆ )−1 L 1 2 L 1

x =0

(80)

.

Now, we come to the corresponding formulae for right integrals which in the case of the differential calculus with unhated derivatives read 1 a f ⋆ ∂ g (∂ 1 )−1 (81) R R x4 =0 a

a 1 −1/2 −1/2 −1/2 K1 K2 g) 4 , = f ⋆ g x4 =0 − (∂ 1 )−1 R (∂R f ) ⋆ (Λ x =0

2 a (∂ 2 )−1 R f ⋆ ∂R g x3 =0 a

a −1/2 1/2 −1/2 2 −1 2

K1 K2 g) 3 = f ⋆ g x3 =0 − (∂ )R (∂R f ) ⋆ (Λ x =0 a 1/2 −1/2 1 −1/2 , K1 K2 L+ + λ (∂ 2 )−1 1 g) 3 R (∂R f ) ⋆ (Λ

(82)

x =0

3 a (∂ 3 )−1 R f ⋆ ∂R g x2 =0 a

3 −1/2 −1/2 1/2 (∂ f ) ⋆ (Λ K K g) = f ⋆ g ax2 =0 − (∂ 3 )−1 2 R 1 2 R x =0 a −1/2 −1/2 1/2 + 3 −1 1 K1 K2 L2 g) 2 , + λ (∂ )R (∂R f ) ⋆ (Λ

(83)

x =0

4 a (∂ 4 )−1 R f ⋆ ∂R g x1 =0 a

4 −1/2 1/2 1/2 (∂ f ) ⋆ (Λ K K g) = f ⋆ g ax1 =0 − (∂ 4 )−1 1 R 1 2 R x =0 a −1/2 1/2 1/2 + 4 −1 3 K1 K2 L1 g) 1 − λ (∂ )R (∂R f ) ⋆ (Λ x =0

16

(84)

a 2 −1/2 1/2 1/2 + (∂ f ) ⋆ (Λ K K L g) − λ (∂ 4 )−1 1 R 1 2 2 R

+λ

2

x =0 a −1/2 1/2 1/2 + + 4 −1 1 K1 K2 L1 L2 g) 1 . (∂ )R (∂R f ) ⋆ (Λ x =0

In the same manner we get for the second differential calculus with unhated derivatives a 1 1 −1 ˆ ˆ (85) (∂ )R f ⋆ ∂R g 4 x =0 b

1/2 1/2 1/2 ˆ1 K1 K2 g) 4 = f ⋆ g ax4 =0 − (∂ˆ1 )−1 R (∂R f ) ⋆ (Λ x =a b 1/2 1/2 1/2 ˆ2 K1 K2 L− − λ (∂ˆ1 )−1 1 g) 4 R (∂R f ) ⋆ (Λ x =a b 1/2 1/2 − −1 3 1/2 1 − λ (∂ˆ )R (∂ˆR f ) ⋆ (Λ K1 K2 L2 g) 4 x =a b ˆ4 f ) ⋆ (Λ1/2 K 1/2 K 1/2 L− L− g) , ( ∂ + λ2 (∂ˆ1 )−1 R 1 2 1 2 R 4 x =a

a ˆ2 g f ⋆ ∂ (∂ˆ2 )−1 R R x3 =0 a

a ˆ2 f ) ⋆ (Λ1/2 K −1/2 K 1/2 g) ( ∂ = f ⋆ g x3 =0 − (∂ˆ2 )−1 R 1 2 R x3 =0 a 1/2 −1/2 1/2 − ˆ4 K1 K2 L2 g) 3 , + λ (∂ˆ2 )−1 R (∂R f ) ⋆ (Λ

(86)

x =0

a ˆ3 g f ⋆ ∂ (∂ˆ3 )−1 R R x2 =0 a

a ˆ3 f ) ⋆ (Λ1/2 K −1/2 K −1/2 g) ( ∂ = f ⋆ g x2 =0 − (∂ˆ3 )−1 R 1 2 R x2 =0 a 1/2 −1/2 4 1/2 − 3 −1 L1 g) 2 , + λ (∂ˆ )R (∂ˆR f ) ⋆ (Λ K1 K2

(87)

x =0

b ˆ4 g f ⋆ ∂ (∂ˆ4 )−1 R R x1 =a a

a ˆ4 f ) ⋆ (Λ1/2 K −1/2 K −1/2 g) ( ∂ = f ⋆ g x1 =0 − (∂ˆ4 )−1 R 1 2 R 1

x =0

(88) .

Finally, we would like to present the relations concerning our method for integration over the whole space. If surface terms are neglected, we can again state the identities (∂ 1 )−1 (∂ 2 )−1 (∂ 3 )−1 (∂ 4 )−1 = (∂ 1 )−1 (∂ 3 )−1 (∂ 2 )−1 (∂ 4 )−1 = 1 −1

q(∂ )

2 −1

(∂ )

4 −1

(∂ )

3 −1

(∂ )

1 −1

= q(∂ ) 17

3 −1

(∂ )

4 −1

(∂ )

2 −1

(∂ )

=

(89)

q(∂ 2 )−1 (∂ 1 )−1 (∂ 3 )−1 (∂ 4 )−1 = q(∂ 3 )−1 (∂ 1 )−1 (∂ 2 )−1 (∂ 4 )−1 = q 3 (∂ 2 )−1 (∂ 4 )−1 (∂ 3 )−1 (∂ 1 )−1 = q 3 (∂ 3 )−1 (∂ 4 )−1 (∂ 2 )−1 (∂ 1 )−1 = q 3 (∂ 4 )−1 (∂ 2 )−1 (∂ 1 )−1 (∂ 3 )−1 = q 3 (∂ 4 )−1 (∂ 3 )−1 (∂ 1 )−1 (∂ 2 )−1 = q 4 (∂ 4 )−1 (∂ 2 )−1 (∂ 3 )−1 (∂ 1 )−1 = q 4 (∂ 4 )−1 (∂ 3 )−1 (∂ 2 )−1 (∂ 1 )−1 = = q 2 × remaining combinations. Let us note that the quantities (∂ˆi )−1 , i = 1, . . . , 4, obey the same relations. Since the results of the various volume integrals in (89) differ by normalisation factors only, we can restrict attention to one of the above expressions. Thus, for performing the integration over the whole space it is sufficient to consider the following formula: (∂ˆ1 )−1 (∂ˆ2 )−1 (∂ˆ3 )−1 (∂ˆ4 )−1 ⊲ f = q

4

(90)

(Dq4−2 )−1 (Dq3−2 )−1 (Dq2−2 )−1 (Dq1−2 )−1 f (q 2 x1 , qx2 , qx3 ).

It is quite clear that the transformations (69) and (71) carry over into our formulae for volume integrals. Hence, (∂ˆ1 )−1 (∂ˆ2 )−1 (∂ˆ3 )−1 (∂ˆ4 )−1 ⊲ f i → i′ q → 1/q

←→

(91)

(∂ 4 )−1 (∂ 3 )−1 (∂ 2 )−1 (∂ 1 )−1 ˜⊲ f

and f ˜⊳ (∂¯1 )−1 (∂¯2 )−1 (∂¯3 )−1 (∂¯4 )−1

(92)

j↔j ′

←→ (∂ 4 )−1 (∂ 3 )−1 (∂ 2 )−1 (∂ 1 )−1 ˜⊲ f, f ⊳ (∂ 4 )−1 (∂ 3 )−1 (∂ 2 )−1 (∂ 1 )−1

(93)

j↔j ′

←→ (∂¯1 )−1 (∂¯2 )−1 (∂¯3 )−1 (∂¯4 )−1 ⊲ f, where (∂ i )−1 = −q 4 (∂ˆi )−1 . With the same reasonings applied to q-deformed Euclidean space in three dimensions we can immediately verify rotation and translation invariance of our 4-dimensional volume integrals. Using the notation Z dq V f ≡ (∂ 4 )−1 (∂ 3 )−1 (∂ 2 )−1 (∂ 1 )−1 ˜⊲ f, (94) L Z dq V f ≡ f ⊳ (∂ 4 )−1 (∂ 3 )−1 (∂ 2 )−1 (∂ 1 )−1 , R

18

Z

ZL

dq V f

≡ (∂¯1 )−1 (∂¯2 )−1 (∂¯3 )−1 (∂¯4 )−1 ⊲ f,

dq V f

≡ f ˜⊳ (∂¯1 )−1 (∂¯2 )−1 (∂¯3 )−1 (∂¯4 )−1 ,

R

we can explicitely Z dq V f ⊲ ∂i ˜ L/R Z ± dq V f Lj ⊲ L/R Z dq V f Kj ⊲ L/R

∂¯i ⊲

Z

write Z Z dq V f = 0, dq V ∂ i ˜⊲ f = ε(∂ i ) = L/R L/R Z Z ± ± dq V f = 0, dq V Lj ⊲ f = ε(Lj ) = L/R L/R Z Z Z dq V f = dq V Kj ⊲ f = ε(Kj ) =

dq V f

Z

=

L/R

L± j

⊲

Kj ⊲

Z

dq V f

Z

=

L/R

Z

dq V f

Z

=

L/R

and Z

dq V f

L/R

Z

dq V f L/R

Z

dq V f L/R

Z Z

dq V f L/R

Z

dq V f L/R

! !

dq V f

L/R

!

!

L/R

dq V L/R

Z

⊳ Kj

Z

⊳ ∂¯i =

⊳

L± j

⊳ Kj

= =

⊲f =

dq V f = 0,

ε(L± j )

Z Z Z

dq V f ˜⊳ ∂ i = ε(∂ i ) L/R

L/R

Z

Z Z

L/R

L/R

dq V f ⊳ Kj = ε(Kj )

Z

L/R

Z

19

dq V f, L/R

dq V f = 0,

dq V f ⊳ Kj = ε(Kj ) L/R

Z

(98)

L/R

Z

=

dq V f = 0, dq V f =

Z

dq V f ⊳ L/R

(97)

L/R

L/R

ε(L± j )

L± j

dq V f

L/R

dq V f = 0, Z

dq V f ⊳ ∂¯i = ε(∂¯i )

dq V f = 0, Z dq V f =

L/R

± dq V f ⊳ L± j = ε(Lj )

L/R

(96)

L/R

dq V Kj ⊲ f = ε(Kj )

Z

=

L± j

Z

L/R

⊳ L± = j

!

!

˜ ⊳ ∂i =

dq V ∂¯i ⊲ f = ε(∂¯i )

dq V f,

L/R

L/R

L/R

(95)

dq V f = 0, L/R

dq V f = L/R

Z

dq V f. L/R

4

q-Deformed Minkowski-space

In principle all considerations of the previous two sections pertain equally to q-deformed Minkowski space [12] [16], [17], [18]2 , apart from the fact that the results now entail a more involved structure. The partial derivatives of q-deformed Minkowski space satisfy the relations ∂µ∂0 = ∂0∂µ, − 3/0

∂ ∂

= q ∂

+ 3/0

∂ ∂ − +

µ = {0, +, −, 3/0},

(99)

2 3/0 −

= q

+ −

∂ ∂ −∂ ∂

∂ ,

−2 3/0 +

∂

= λ(∂

∂ ,

3/0 3/0

∂

+ ∂ 0 ∂ 3/0 ),

with λ = q − q −1 and q > 1. As usual, we introduce inverse elements (∂ µ )−1 , µ = ±, 0, 3/0, by (∂ µ )−1 ∂ µ = ∂ µ (∂ µ )−1 = 1,

µ = 0, +, −, 3/0.

(100)

From these requirements we get the relations ∂ 0 (∂ µ )−1 = (∂ µ )−1 ∂ 0 , (∂

3/0 −1 ±

)

∂

± −1 3/0

(∂ ) −

∂

+ −1

∂ (∂ )

= q

∓2 ±

= q

±2 3/0

∂ (∂

∂

µ = 0, +, −, 3/0,

3/0 −1

,

± −1

,

)

(∂ )

+ −1 −

= (∂ )

∂ −q

−2

(101)

λ(∂ + )−2 (q −2 ∂ 3/0 + ∂ 0 )∂ 3/0 ,

(∂ − )−1 ∂ + = ∂ + (∂ − )−1 − q −2 λ∂ 3/0 (q −2 ∂ 3/0 + ∂ 0 )(∂ − )−2 . Furthermore, we have (∂ 0 )−1 (∂ µ )−1 = (∂ µ )−1 (∂ 0 )−1 , (∂

3/0 −1

)

± −1

(∂ )

(∂ − )−1 (∂ + )−1

= q

±2

+ −1

(∂ )

(∂

µ = 0, +, −, 3/0,

3/0 −1

)

(102)

,

 !2 −1 i (λλ−1 = q2 + ) [[i]]q 2 ! i q−2 i=0   X k X i 6 k i(i+2k) (q 4j λ+ )p · (−q ) q k q2 ∞ X



p=0

j+k=i

+ p−(i+1)

· (∂ )

(∂

3/0 2j

) (Sq )k,p (∂ 0 , ∂ 3/0 )(∂ − )p−(i+1) ,

where Sk,p is a polynomial of degree 2(k − p) with its explicit form presented in appendix A. 2

For a different version of q-deformed Minkowski space see also [19].

20

Next, we come to the commutation relations involving the Lorentz generators [20], [17], [21]. Explicitely, they are given by T + (∂ 0 )−1 = (∂ 0 )−1 T + ,

(103)

T + (∂ 3/0 )−1 = (∂ 3/0 )−1 T + −

1/2 q 1/2 λ+ (∂ 3/0 )−2 ∂ + ,

T + (∂ + )−1 = q 2 (∂ + )−1 T + , 1/2

T + (∂ − )−1 = q −2 (∂ − )−1 T + − q −1/2 λ+ (∂ − )−2 (∂ 3/0 + q −2 ∂ 0 ), T − (∂ 0 )−1 = (∂ 0 )−1 T − −

T (∂ −

3/0 −1

)

− −1

T (∂ )

= (∂ = q

3/0 −1

)

−2

T

(104) −

− −1

(∂ )

−

1/2 q −1/2 λ+ (∂ 3/0 )−2 ∂ − ,

T −, 1/2

T − (∂ + )−1 = q 2 (∂ + )−1 T − − q 1/2 λ+ (∂ + )−2 (∂ 3/0 + q 2 ∂ 0 ), τ 3 (∂ 0 )−1 = (∂ 0 )−1 τ 3 , 3

τ (∂ 3

3/0 −1

= (∂

+ −1

4

)

τ (∂ )

(105)

3/0 −1 3

)

τ ,

+ −1 3

= q (∂ )

τ ,

τ 3 (∂ − )−1 = q −4 (∂ − )−1 τ 3 , T 2 (∂ 3/0 )−1 = q(∂ 3/0 )−1 T 2 , 2

+ −1

T (∂ )

= q

−1

+ −1

(∂ )

(106)

2

T , −1/2

T 2 (∂ − )−1 = q(∂ − )−1 T 2 − q −3/2 λ+ ∂ 3/0 (∂ − )−2 τ 1 ,
−1 2
−1 1 )T , )τ + (σ 2 ⊲ ∂ 3 T 2 (∂ 3 )−1 = (T 2 ⊲ ∂ 3

S 1 (∂ 3/0 )−1 = q −1 (∂ 3/0 )−1 S 1 , 1

− −1

−1

S (∂ ) = q (∂ ) S ,
−1 −1/2 S1 ∂+ = q(∂ + )−1 S 1 + q −1/2 λ+ (∂ + )−2 ∂ 3/0 σ 2 ,
−1 2
−1 1 )S , )σ + (τ 1 ⊲ ∂ 3 S 1 (∂ 3 )−1 = (S 1 ⊲ ∂ 3

τ 1 (∂ 3/0 )−1 = q −1 (∂ 3/0 )−1 τ 1 , 1

− −1

1

+ −1

τ (∂ )

τ (∂ )

(107)

− −1 1

(108)

− −1 1

= q(∂ ) = q

−1

τ , −1/2 2

+ −1 1

(∂ )

τ + q −1/2 λ+

λ (∂ + )−2 ∂ 3/0 T 2 ,

τ 1 (∂ 3 )−1 = (τ 1 ⊲ (∂ 3 )−1 )τ 1 + λ2 (S 1 ⊲ (∂ 3 )−1 )T 2 , σ 2 (∂ 3/0 )−1 = q(∂ 3/0 )−1 σ 2 ,

(109) 21

σ 2 (∂ + )−1 = q(∂ + )−1 σ 2 , −1/2 2

σ 2 (∂ − )−1 = q −1 (∂ − )−1 σ 2 − q 1/2 λ+

λ (∂ − )−2 ∂ 3/0 S 1 ,

σ 2 (∂ 3 )−1 = (σ 2 ⊲ (∂ 3 )−1 )σ 2 + λ2 (T 2 ⊲ (∂ 3 )−1 )S 1 , where we have to insert the actions ∞ X (k,k+1) T 2 ⊲ (∂ 3 )−1 = q −3/2 (q 2 α0 )k (K−1 )1,q2 k=0

X

·

j−1/2 λ+

0≤i+j≤k

(110)

  k (∂ + )j+1 (aq (q 2j ∂ 3/0 ))i i

3/0 −2(k+1) − j ) (∂ ) , · (Sq )k−i,j (∂ 0 , ∂ 3/0 )(∂ 0 + 2q 2j+1 λ−1 + ∂ 1

3 −1

S ⊲ (∂ )

= −q

−3/2

∞ X (k,k+1) (q −2 α0 )k (K−1 )1,q−2 k=0

X

·

j−1/2 λ+

0≤i+j≤k

(111)

  k (∂ + )j (aq (q 2j ∂ 3/0 ))i i

3/0 −2(k+1) − j+1 ) (∂ ) , · (Sq )k−i,j (∂ 0 , ∂ 3/0 )(∂ 0 + 2q 2j+1 λ−1 + ∂

τ 1 ⊲ (∂ 3 )−1 = q

∞ X

(k,k)

(q 2 α0 )k (K−1 )1,q2

k=0

X

·

λj+

0≤i+j≤k

(112)

  k (∂ + )j (aq (q 2j ∂ 3/0 ))i i

3/0 −2k−1 − j ) (∂ ) , · (Sq )k−i,j (∂ 0 , ∂ 3/0 )(∂ 0 + 2q 2j+1 λ−1 + ∂ 2

3 −1

σ ⊲ (∂ )

= q

−1

∞ X (k,k) (q −2 α0 )k (K−1 )1,q−2 k=0

·

X

0≤i+j≤k

λj+

(113)

  k (∂ + )j (aq−1 (q 2j ∂ 3/0 ))i i

3/0 −2k−1 − j ) (∂ ) · (Sq )k−i,j (∂ 0 , ∂ 3/0 )(∂ 0 + 2q 2j−1 λ−1 + ∂

with

λ2 . (114) λ2+ Note that these expressions have been formulated by using the abbreviations and conventions listed in appendix A. Applying the substitutions ∂ µ → ∂ˆµ , (∂ µ )−1 → (∂ˆµ )−1 , µ = ±, 0, 3/0, (115) α0 = −

22

to the formulae (99)-(113) yields the corresponding identities for the differential calculus of the hatted derivatives. As in the Euclidean case the representations of the partial derivatives of q-deformed Minkowski space split into two parts [10]. Hence   µ µ ) F, µ = ±, 0, 3/0. (116) ) + (∂ˆ(i>0) ∂ˆµ ⊲ F = (∂ˆ(i=0)

Now, we can follow the same lines as in the previous sections. Thus, solutions to the difference equations ∂ˆµ ⊲ F = f,

µ = ±, 0, 3/0

(117)

are given by F

= (∂ˆµ )−1 ⊲ f ∞ ik h X µ µ µ ) (∂ˆ(i=0) )−1 (∂ˆ(i>0) )−1 f, (−1)k (∂ˆ(i=0) =

(118) µ = ±, 0, 3/0.

k=0

The operators we have to insert in (118) for evaluating can in explicit terms be written as 3/0

+ 3/0 3 − , x , x ) = (Dq3− q2 )−1 f (q −2 x+ ), (∂ˆ(i=0) )−1 L f (x , x

(119)

− + 3/0 3 − , x , x ) = −q(Dq+2 )−1 f, )−1 (∂ˆ(i=0) L f (x , x + + 3/0 3 − , x , x ) = −q −1 (Dq−2 )−1 f (q −2 q+ x3 ), )−1 (∂ˆ(i=0) L f (x , x 3/0

0 + 3/0 3 − (∂ˆ(i=0) , x , x ) = (Dq2 )−1 f (q− x3 ) )−1 L f (x , x

and 3/0

(∂ˆ(i>0) )L f (x+ , x3/0 , x3 , x− ) =

∞ X

αl+

l=1

X

M−

0≤i+j≤l


l

i,j

(120)

(x)(T 3 )ij f,

− )L f (x+ , x3/0 , x3 , x− ) (121) (∂ˆ(i>0) ∞ o λ X l X n + k α+ (M )i,j (x)(T1− )ij f + q −1 (M − )ki,j (x)(T2− )ij f , = λ+ l=0

0≤i+j≤l

+ )L f (x+ , x3 , x3/0 , x− ) (∂ˆ(i>0)

(122) 23

3/0

= −qλx+ (Dq2 Dq32 f )−1 f (q+ x3 ) ∞ o X n X (M − )ki,j (x)(T1+ )ij f + λ(M + )ki,j (x)(T2+ )ij f αk+ −q 0≤i+j≤k

k=1

λ −q λ+ λ − λ+

X

k+l α+

0≤k+l 0, q > 1 and xA < 0,we set 0 (DqAa )−1 f xA

= (1 − q a )

xA (DqAa )−1 f −∞ = (1 − q a )

0 (DqA−a )−1 f A x

xA (DqA−a )−1 f

−∞

∞ X

k=1 ∞ X

(q −ak xA )f (q −ak xA ),

(170)

(q ak xA )f (q ak xA ),

k=0

= −(1 − q −a )

∞ X (q −ak xA )f (q −ak xA ), k=0

∞ X = −(1 − q −a ) (q ak xA )f (q ak xA ). k=0

Note that the formulae (168) and (170) also yield expressions for qintegrals over any other interval [13]. 6. Additionally, we need operators of the following form n ˆ A ≡ xA

∂ . ∂xA

(171)

7. Calculations for q-deformed Minkowski space show that it is reasonable to give the following repeatedly appearing polynomials a name of their own:  pi−j−1 j p1  P i−j Q  P P aq−1 (x0 , q 2pl x3/0 ), . . . 0 3/0 (Sq )i,j (x , x ) ≡ p1 =0 p2 =0 pi−j =0 l=0   1, if j = i   ≡ −qx3/0 (qx3/0 + λ+ x0 ), (172) aq x0 , x3/0

k k M ± i,j (x) ≡ M ± i,j (x0 , x+ , x3/0 , x− )    i k j = λ+ aq±1 (q 2j x3/0 ) i
j · x+ x− (Sq )k−j,j (x0 , x3/0 ), +− k,l )i,j,u (x0 , x+ , x3/0 , x− ) (M +− )k,l i,j,u (x) ≡ (M    k l u + − u = λ (x x ) i j + l−j  k−i  · aq (q 2u x3/0 ) aq−1 (q 2u x3/0 )

· (Sq )i+j,u (x0 , x3/0 ). 34

(k ,...,k )

8. In [10] we have introduced the quantities (Kα )a11,...,al l , ki ∈ N, ai , α ∈ R. We also refer to [10] for a review of their explicit calculation. These quantities can be used to define new operators by setting ( (k ,...,k ) (Kn )a11,...,al l xn−k1 −...−kl , ,...,kl ) n x = Da(k11,...,a (173) l 0, if n < k1 + . . . + kl , Especially, in the case of q-deformed Minkowski space we need the operators (k,l)

3 k,l (D1,q ) ≡ D1,q2 ,

(174)

(k,l)

3 k,l (D2,q ) ≡ Dy− /x3 ,q2 y− /x3 , (k,l,i,j)

3 k,l (D3,q )i,j ≡ Dy+ /x3 ,q2 y+ /x3 ,y− /x3 ,q2 y− /x3 ,

where y± = y± (x0 , x3/0 ) = x0 +

2q ±1 3/0 x . λ+

(175)

Notice that these operators have to act upon the coordinate x3 , only. 9. In Sect. 4 the representations of (∂ˆµ )−1 , µ = ±, 0, 3/0, have been formulated by using the operators (O3/0 )k f (O1− )k f (O2− )k f (O10 )k f

3 k,k+1 = (D2,q ) f (q 2 x+ ),

(176)

3 k+1,k+1 = x− (D2,q ) f (q 2 x+ ),

(177)

= x

3/0

3 k,k+1 ) f (q 2 x+ ), Dq+2 (D2,q

3/0

3 k,k ) f = Dq2 (D1,q

(178) 3/0

3 k,k+1 2 + 3/0 + ) f, Dq2 Dq2 (D1,q − q 3 λ−1 + λ x x

(O20 )k f

k,k+1 − 3 Dq2 f (q 2 x3/0 ), = x− (D1,q −1 )

(O30 )k f

3 k,k+1 ) f, = qx+ Dq+2 (D2,q

(O40 )k f

k,k − 3 Dq2 f, = x3/0 Dq+2 (D1,q −1 )

(Q01 )k,l f

3 k+1,k = (x0 + q −1 λx3 )(D3,q )l,l+1 f 3 k,k+1 + 3/0 + )l,l+1 f Dq2 (D3,q + qλ−1 + λ(q + λ+ )x x

35

(179)

3 k+1,k+1 2 + 3/0 0 )l,l+1 f, (x + q −1 λx3/0 )Dq+2 (D3,q − q 3 λ−1 + λ x x

(Q02 )k,l f

3 k+1,k+1 = (D3,q )l,l+1 f,

(Q03 )k,l f

2 + 3/0 + 3 k+1,k+1 3 k+1,k Dq2 (D3,q )l+1,l+1 f, = q −1 (D3,q )l+1,l+1 f − q 2 λ−1 + λ x x

(O1+ )k f (O2+ )k f (Q+ 1 )k,l f

3 k,k − = (D1,q Dq2 f, −1 )

=

(180)

3/0 3 k,k+1 ) f, x+ Dq2 (D1,q

3 k,k+1 = (q + λ+ )x+ (D3,q )l,l+1 f

(181)

3 k+1,k+1 )l,l+1 f, − q 2 λx+ (x0 + q −1 λx3 )(D3,q

(Q+ 2 )k,l f

3 k+1,k+1 = x+ (D3,q )l+1,l+1 f.

Acknowledgement First of all I want to express my gratitude to Julius Wess for his efforts, suggestions and discussions. Also I would like to thank Michael Wohlgenannt, Fabian Bachmeier, Christian Blohmann, and Marcus Dietz for useful discussions and their steady support.

References [1] N.Yu. Reshetikhin, L.A. Takhtadzhyan, L.D. Faddeev, Quantization of Lie Groups and Lie Algebras, Leningrad Math. J. 1 (1990) 193. [2] J. Wess, q-deformed Heisenberg Algebras, in H. Gausterer, H. Grosse and L.Pittner, eds., Proceedings of the 38. Internationale Universit¨ atswochen f¨ ur Kern- und Teilchen physik, no. 543 in Lect. Notes in Phys., Springer -Verlag, Schladming (2000), math-phy 9910013. [3] S. Majid, Braided geometry: A new approach to q-deformations, in R. Coquereaux et al., eds., First Caribbean Spring School of Mathematics and Theoretical Physics, World Sci., Guadeloupe (1993). [4] J.E. Moyal, Quantum mechanics as a statistical theory, Proc. Camb. Phil. Soc. 45 (1949) 99. [5] J. Madore, S. Schraml, P. Schupp, J. Wess, Gauge Theory on Noncommutative Spaces, Eur. Phys. J. C16 (2000) 161, hep-th/0103120. [6] H. Wachter, M. Wohlgenannt, *-Products on Quantum Spaces, Eur. Phys. J. C23 (2002) 761-767, hep-th/0103120. 36

[7] V.G. Drinfeld, Hopf algebras and the quantum Yang-Baxter equation, Soviet Math. Dokl. 32 (1985) 254-258. [8] M. Jimbo, A q-analogue of U(g) and the Yang-Baxter equation, Lett. Math. Phys. 10 (1985) 63-69. [9] J. Wess, B. Zumino, Covariant differential calculus on the quantum hyperplane, Nucl. Phys. B. Suppl. 18 (1991), 302-312. [10] C. Bauer, H. Wachter, Operator representations on quantum spaces, , Eur. Phys. J. C 31 (2003) 261, math-ph/0201023. [11] F.H. Jackson, q-Integration, Proc. Durham Phil. Soc. 7 (1927), 182-189. [12] A. Lorek, W. Weich, J. Wess, Non Commutative Euclidean and Minkowski Structures, Z. Phys. C76 (1997) 375, q-alg/9702025. [13] A. Klimyk, K. Schm¨ udgen, Quantum Groups and their Representations, Springer Verlag, Berlin (1997). [14] S. Majid, Foundations of Quantum Group Theory, University Press, Cambridge (1995). [15] H. Ocambo, SO q (4) quantum mechanics, Z. Phys. C70 (1996) 525. [16] U. Carow-Watamura, M. Schlieker, M. Scholl, S. Watamura, Z. Phys. C48 (1990) 159. [17] W.B. Schmidtke, J. Wess, B. Zumino, A q-deformed Lorentz Algebra in Minkowski phase space, Z. Phys. C52 (1991) 471. [18] S. Majid, Examples of braided groups and braided matrices, J. Math. Phys. 32 (1991) 3246-3253. [19] V. K. Dobrev, New q-Minkowski space-time and q-Maxwell equations hierarchy from q-conformal invariance, Phys. Lett. 341B (1994) 133138 & 346B (1995) 427. [20] O. Ogievetsky, W.B. Schmittke, J. Wess, B. Zumino, q-deformed Poincar´e Algebra, Commun. Math. Phys. 150 (1992) 495. [21] M. Rohregger, J. Wess, q-Deformed Lorentz Algebra in Minkowski phase space, Eur. Phys. J.7 (1999) 177. [22] H. Steinacker, Integration on quantum Euclidean Space and sphere, J. Math. Phys. 37 (1996) 4738. 37

[23] C. Chryssomalakos, B. Zumino, Translations integrals and Fourier transforms in the quantum plane, preprint LBL-34803, UCCB-PTH93/30, in Salamfestschrift, edited by A. Ali, J. Ellis, and S. RandjbarDaemi, World Sci., Singapore, 1993. [24] G. Fiore, The SO q (N)-symmmetric harmonic oszillator on the quantum Euclidean space..., Int. J. Mod. Phys., A8 (1993) 4679. [25] A. Hebecker, W. Weich, Free particle in q-deformed configuration space, Lett. Math. Phys. 26 (1992) 245. [26] M.A. Dietz, Symmetrische Formen auf Quantenalgebren, diplomatheses, Universit¨ at Hamburg, Fakult¨at f¨ ur Physik (2001). [27] A. Kempf, S. Majid, Algebraic q-integration and Fourier theory on quantum and braided spaces, J. Math. Phys. 35 (1994) 6802.

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