A.K.Aringazin, D.A.Kirukhin and R.M.Santilli- Isotopic Generalization of the Legendre, Jacobi, and Bessel Functions

8/3/2019 A.K.Aringazin, D.A.Kirukhin and R.M.Santilli- Isotopic Generalization of the Legendre, Jacobi, and Bessel Functions

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ISOTOPIC GENERALIZATION OF

THE LEGENDRE, JACOBI, AND

BESSEL FUNCTIONS

A.K.Aringazin1,2, D.A.Kirukhin1,

andR.M.Santilli2

1Department of Theoretical PhysicsKaraganda State University

Karaganda 470074, Kazakhstan

and2Istituto per la Ricerca di Base

Castello Principe PignatelliI-86075 Monteroduni (IS)

Molise, Italy

Received June 25, 1995Algebras, Groups & Geometries 12 (1995) 255-305.

Abstract

In this paper, we consider the Lie-isotopic generalizations of theLegendre, Jacobi, and Bessel functions.

Permanent address.

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Contents

1 Introduction 3

2 The group SU(2) 42.1 Parametrizations . . . . . . . . . . . . . . . . . . . . . . . . . 42.2 IsoEuler angles for matrix product . . . . . . . . . . . . . . . 62.3 Relation to the group of rotations . . . . . . . . . . . . . . . . 8

3 Irreps ofSU(2) 9

4 Matrix elements of the irreps ofSU(2) and isoLegendre poly-nomials 10

5 Basic properties of the isoLegendre polynomials 13

5.1 Symmetry relations . . . . . . . . . . . . . . . . . . . . . . . . 135.2 Counter relations . . . . . . . . . . . . . . . . . . . . . . . . . 145.3 Relation to classical orthogonal polynomials . . . . . . . . . . 15

5.3.1 IsoJacobi polynomials . . . . . . . . . . . . . . . . . . 155.3.2 isoLegendre polynomials . . . . . . . . . . . . . . . . . 165.3.3 Adjoint isoLegendre functions . . . . . . . . . . . . . . 16

6 Functional relations for isoLegendre functions 16

6.1 Theorem of composition . . . . . . . . . . . . . . . . . . . . . 16

6.1.1 Theorem of composition for isoLegendre polynomials . 196.2 Multiplication rules . . . . . . . . . . . . . . . . . . . . . . . . 206.3 Orthogonality relations . . . . . . . . . . . . . . . . . . . . . . 22

7 Recurrency relations for isoLegendre functions 23

8 The group QU(2) 288.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288.2 Parametrizations . . . . . . . . . . . . . . . . . . . . . . . . . 288.3 Relation to the group SH(3) . . . . . . . . . . . . . . . . . . 30

9 Irreps ofQU(2) 30

9.1 Description of the irreps . . . . . . . . . . . . . . . . . . . . . 309.2 Representations T(g0) . . . . . . . . . . . . . . . . . . . . . . 31

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10 Matrix elements of the irreps of QU(2) and isoJacobi func-

tions 32

10.1 The matrix elements . . . . . . . . . . . . . . . . . . . . . . . 32

11 IsoJacobi function Blmn(z) 33

12 IsoJacobi function Bl00 3412.1 Symmetry relations for Blmn(z) and Bl(z) . . . . . . . . . . . 35

13 Functional relations for Blmn(z) 35

14 Recurrency relations for Blmn 37

15 The group M(2) 3915.1 D efinitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3915.2 Parametrizations . . . . . . . . . . . . . . . . . . . . . . . . . 40

16 Irreps of M(2) 4116.1 Description of the irreps . . . . . . . . . . . . . . . . . . . . . 4116.2 Infinitesimal operators . . . . . . . . . . . . . . . . . . . . . . 4316.3 T he irreps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

17 Matrix elements of the irreps of M(2) and isoBessel func-tions 44

17.1 M atrix elements . . . . . . . . . . . . . . . . . . . . . . . . . 4417.2 IsoBessel functions with opposite sign indeces . . . . . . . . . 4517.3 Expansion series for IsoBessel functions . . . . . . . . . . . . 46

18 Functional relations for isoBessel function 47

18.1 Theorem of composition . . . . . . . . . . . . . . . . . . . . . 4718.2 Theorem of multiplication . . . . . . . . . . . . . . . . . . . . 48

19 Recurrency relations for Jn(z) 49

20 Relations between IsoBessel functions and Plmn(z) 5020.1 IsoEuclidean plane and sphere . . . . . . . . . . . . . . . . . . 5020.2 IsoBessel and isoJacobi functions . . . . . . . . . . . . . . . . 51

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1 Introduction

In this paper, we consider the Lie-isotopic generalization of the Legendre,Jacobi, and Bessel functions.

We describe in detail the group of rotations of three-dimensional isoEu-clidean space, and the group locally isomorphic to it, SU(2), consisting ofisounitary isounimodular 2 2 matrices. Also, we study the group QU(2)of quasiunitary matrices and the group M(2) of isometric transformationsof isoEuclidean plane.

These studies are of interest both in mathematical and physical pointsof view. We refer the interested reader to monographs [3] for comprehensivereview on the Lie-isotopic formalism and its applications.

The isotopic generalizations of the groups SO(3), SU(2), and M(2) areof continuing interest in the literature. From physical point of view, ourinterest is that the Lie-isotopic generalizations of the Legendre functions aswell as the other special functions, such as Jacobi and Bessel functions, canbe used in formulating the nonpotential scattering theory [1, 2, 6, 7] whenone considers non-zero isoangular momenta.

The paper is organized as follows.Sections 2-7 are devoted to representations of the group SU(2) and isoLe-

gendre functions. Namely, in Sec.2, we consider the group SU(2). In Sec.3,we consider unitary irreducible representations (irreps) of the group SU(2).In Sec.4, we present matrix elements of the unitary irreps of SU(2), andisoLegendre functions Pl

mn(z). In Sec.5, we present basic properties of the

isoLegendre functions. In Sec.6, we present functional relations satisfied bythe isoLegendre functions. In Sec.7, we present recurrency relations satisfiedby the isoLegendre functions.

Sections 8-14 are devoted to representations of the group QU(2) andisoJacobi functions.

Sections 15-20 are devoted to representations of the group M(2) andisoBessel functions.

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2 The group SU(2)

In this Section, we consider representations of the group SU(2), elements ofwhich are isounitary isounimodular 2 2 matrices, and its relation to thegroup SO(3) of rotations of three dimensional isoEuclidean space.

2.1 Parametrizations

Denote SU(2) the set of isounitary isounimodular 2 2 matrices, namely,of the matrices

u = I

. (1)

If u1 SU(2) and u2 SU(2) then

(u1 u2) = u2 u1 = u12 u11 = (u1 u2)1 (2)

and det(u1 u2) = 1. Therefore, u1 u2 SU(2). Also, it is easy to showthat u11 SU(2). We arrive at the conclusion that SU(2) is a group.

Let u SU(2). Since

u = I

(3)

and

u1 = I

, (4)

then = and = .Thus, any matrix u SU(2) has the form

u = I

, (5)

where I = diag(g111 , g122 ), det I = . Since det u = 1 then

|||| + |||| = 1, (6)

and vice versa, if u is a matrix of the form (5) and Eq.(6) holds, thenu SU(2).

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From the above consideration, it follows that the elements of SU(2)

can be uniquely determined by two complex numbers (, ) obeying eq.(6). These complex numbers can be presented by three real parameters,for example, by ||, Arg, and Arg. I f = 0, one can use anotherparametrization, namely Euler angles, , , and , which are related to ||,arg , and arg according to the following relations:

|| = g1/211 cos[1/2/2] isocos[/2],

Arg = +

2, Arg =

+ 2

, (7)

where

1/2

, 1/2

, 1/2

. (8)The values of the Euler angles are not determined by (7) uniquely, so thatwe must put additionally

0 < 2, 0 < < , 2 < 2. (9)

From (7) it follows that || = g1/222 sin(1/2/2) and that the matrix u =u(, , ) has the following form:

u =

gg

1/211 cos/2e

i(+)/2 i2g1/211 sin /2e

i()/2

i2g1/222 sin /2e

i()/2 gg1/211 cos /2e

i(+)/2

. (10)

From (5) and (10) we have

g111/2 cos[1/2] = 2||2 1,

exp[i1/2/2] = i ||| , (11)

exp i1/2/2 =1/2 exp{i1/2/2}

|| .

Also, from (10) we have the following factorization

u(,,

) =

exp{i1/2/2

}0

0 exp{i1/2/2} g11 0

0 g22

g1/211 cos /2 igg

1/222 sin/2

g1/222 sin /2 ig

1/211 cos /2

g11 00 g22

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exp{i1/2/2

}0

0 exp{i1/2/2} u(, 0, 0)u(0, , 0)u(0, 0, ). (12)

Diagonal matrices exp[i1/2/2] 0

0 exp[i1/2/2]

(13)

form a one-parameter subgroup of SU(2). Thus every matrix u SU(2) liesin the left and right conjugacy class in respect to ths subgroup containingthe matrix of the form

g1/211 cos[

1/2/2] ig1/222 sin[

1/2/2]

ig1/222 sin[

1/2/2] g1/211 cos[

1/2/2]

. (14)

Note that the matrices represented by (14) form a one-parameter subgroupof SU(2).

2.2 IsoEuler angles for matrix product

Let u = u1u2 is a product of two matrices u1, u2 SU(2). Denote thecorresponding isoEuler angles by (, , ), (1, 1, 1), and (2, 2, 2).To express the isoEuler angles of u via the isoEuler angles of u1 and u2, we

consider the case when 1 = 1 = 2 = 0. For this case we have

u =

g1/211 cos 1/2 ig

1/222 sin 1/2

ig1/222 sin 1/2 g

1/211 cos 1/2

g11 00 g22

g1/211 cos 2/2exp{i2/2} i2 sin 2/2exp{i2/2}i2 sin 2/2exp{i2/2} g1/211 cos 2/2exp{i2/2}

. (15)

Using (11) we have from (15) in sequence

cos[1/2] = cos[11/2] cos[2

1/2]g1/211

sin[11/2] sin[21/2]gg1/211 cos[21/2], (16)

exp{i1/2} = sin[11/2]g

1/211 cos[2

1/2]

sin[1/2]

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+(g11)

1 cos[11/2] sin[2

1/2]g1/211 cos[2

1/2]

sin[1/2]

+i sin[2

1/2]sin[21/2]

sin[1/2], (17)

exp{i1/2( + )/2} = g111 cos[2

1/2/2]2 exp{i1/22/2}cos[11/2/2]

g1/222 sin[1

1/2/2] sin[21/2/2] exp{i1/22/2}

g1/211 cos[

1/2/2]. (18)

It is more convenient to use the following expressions:

g22g11

1/2tan[1/2]

=sin[2

1/2] sin[21/2]

g111 cos[11/2] sin[21/2]cos[21/2] + g

1/211 sin[1

1/2] cos[21/2],

(19)g22g11

1/2tan[1/2]

=sin[1

1/2] sin[21/2]

sin[11/2]g111 cos[2

1/2]cos[21/2] + g1/211 cos[1

1/2]sin[21/2].

(20)

Due to the results of this particular case can easily turn to the generalcase. Indeed, according to (12) we have

u(1, 1, 1)u(2, 2, 2))

= 5u(1, 0, 0)u(0, 1, 0)u(0, 0, 1)u(2, 0, 0)u(0, 2, 0)u(0, 0, 2). (21)

Note thatu(0, 0, 1)u(2, 0, 0) = u(2 + 1, 0, 0). (22)

We observe that the result of the product u(, , ) u(1, 0, 0) gives thematrix u( + 1, , ). Similarly, the result of the product u(0, 0, 1) u(, , ) gives the matrix u(, , + 1. From these observations it follows

that the formulas (16)-(18) are valid in general case with the replacements2 2 + 1, 1, and 2.

Namely, in an explicit form

cos[1/2] = cos[11/2]cos[2

1/2]

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gg

1/222 sin[1

1/2]sin[21/2] cos[(2 + 1)

1/2], (23)

exp{iD} = g1/211 sin[1

1/2] cos[21/2]

sin[1/2]

+g111 cos[1

1/2] sin[21/2] cos[(2 + 1)

1/2]

sin[1/2]

+ig1/222 sin[2

1/2] sin[(2 + 1)1/2]

sin[1/2], (24)

exp{i1/2( 1 + 2)/2}

=g111 cos[1

1/2/2] cos[21/2/2]exp{i1/2(2 + 1)/2}

g

1/2

11 cos[1/2

/2]

g1/222 sin[1

1/2/2] sin[21/2/2] exp{i1/2(2 + 1)/2}

g1/211 cos[

1/2/2]. (25)

2.3 Relation to the group of rotations

Let us define the relation between the groups SU(2) and SO(3). To thisend, we identify the vector x(x1, x2, x3) of three dimensional isoEuclideanspace with the complex 2 matrix of the form

hx =

x3 x1 + ix2

x1

ix2

x3

, (26)

where x = xI = x1. The set of the matrices of the form (26) consistsof isoHermitean matrices g with Trg = 0. Namely, with every matrix u SU(2) we relate the transformation T(u),

T(u)hx = uhx u. (27)Since for the isounitary matrices we have u = u1, the traces of hx andT(u)hx coincide so that the trace of T(u)hx is zero. Also, we have

(T(u)h(x)) = (uhxu) = uhxu = uhxu = T(u)hx, (28)so that the matrix T(u)hx is indeed isoHermitean. On the other hand, for

isoHermitean matrices we have the following representation:

T(u)hx =

y3 y1 + iy2

y1 iy2 y3

1y3 y1 + iy21y1 iy2 y3

= hy,

(29)

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where y(y1, y2, y3) is a vector in three dimensional isoEuclidean space.

From (27) it can be seen that the components of y are linear combinationsof the components of x so that T(u) is a linear transformation of the threedimensional isoEuclidean space E3. From the local isomorphism between thegroups SU(2) and SO(3) it follows that rotations ofE3 can be parametrizedby the isoEuler angles (, , ). Here, the angle varies from 0 to 2 sinceu and u correspond to the same rotation.

Due to (12) the matrices u(, 0, 0) and u(0, 0, ) can be presented asexp{i1/2t/2} 0

0 exp{i1/2t/2}

3(t), (30)

where 3(t) is the rotation by the angle

t around the axis Ox3, and u(0,

, 0)has the form

g1/211 cos[t

1/2/2] ig1/222 sin[t

1/2/2]

i(g22)1/2 sin[t1/2/2] g

1/211 cos[t

1/2/2]

1(t), (31)

which is the rotation around the axis Ox1. From this observation, we havethe following decomposition for arbitrary rotation g of E3:

g(, , ) = g(, 0, 0)g(0, , 0)g(, 0, 0)

=

g1/211 cos g1/222 sin 0

g1/222 sin g

1/211 cos 0

0 0 0

g11 0 00 g22 0

0 0 g33

1 0 0

0 g1/211 cos g1/222 sin

0 (g22)1/2 sin (g11)

1/2 cos

g11 0 00 g22 0

0 0 g33

1 0 0

0 g1/211 cos g1/222 sin

0 (g22)1/2 sin (g11)

1/2 cos

. (32)

3 Irreps of SU(2)

Recall that with any isounimodular complex 2 2 matrix g we associate thelinear transformation,

w1 = 2z1 +

2z2 = 2(z1 + z2), (33)

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w2 = z1 + 2z2 =

2(z1 + z2),

of two dimensional linear complex space. Such a transformation can bepresented by the operator,

T(g)f(z1, z2) = f(z1 + z2; z1 + z2), (34)

acting on the space of functions of two complex variables. Evidently,

T(g1g2) = T(g1) + T(g2),

so that T(g) is a representation of the group SL(2, C). Similarly to thetheorem from Ref.[4] we have the following

Proposition 1. Every irreducible isounitary representationT(u) of

SU(2)

is equivalent to one of the representations Tl(u), where l = 0, 1/2, 1, . . ..The prove of the Proposition 1 is analogous to that of given in Ref.[4],

and we do not present it here.From Proposition 1 it follows that in the space of subgroup SU(2) there

exists the orthogonal normalized basis, fl, , fl, such that the operatorsT(u) are represented in this basis by the same matrices as the operatorsTl(u) in the basis {k(x)}, where

k(x) = s+1/2 x

lk

(l k)!(l + k)! , (35)

l k l, s = 1, . . . , n .We call such a basis isocanonical. It is easy to verify that isocanonical basisis determined uniquely up to scalar factor , with || = 1. More precisely,isocanonical basis consists of normalized eigenvectors of the operator T(h),where

h =

exp{i1/2t/2} 0

0 exp{i1/2t/2}

. (36)

4 Matrix elements of the irreps ofSU(2) and isoLe-gendre polynomials

In this Section, we calculate matrix elements of the irreps Tl(u) of SU(2),and express the matrix elements tlmn(g) through the isoEuler angles (, ,) of the matrix g.

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The representations Tl(g) of SL(2, C) are given by

Tl(g)(x) = (x + 1)2l

x + 1

x + 1, (37)

where (x) is polynomial of degree 2l on x, and g SL(2, C).Using the isocanonical basis of Sec. 3 and the formula aij = (ej ,ei), where

{ei} is orthonormalized basis, we write down the matrix element,

tlmn(g) = (Tl(g)n,m) =

Tl(g)x

ln,xlm

(l m)!(l + m)!(l n)!(l + n)!3/2 , (38)

wheren(x) =

xln3/2(l n)!(l + n)! , (39)

l n l, (l n)! = s(l n)(l n + 1) . . . , s = 1, 2, . . .On the other hand,

T(g)xln = (x + 1)ln(x + 1)l+n, (40)

so that (38) yields

tlmn =(x +

1)ln(x + 1)l+n ,xlm1(l m)!(l + m)!(l n)!(l + n)! 3/2. (41)

Taking into account that (xlk, xlm) = 0 at k = m and (xlm, xlm) =(l m)!(l + m)!2s+1, we have finally from (41)

tlmn(g) =

(l m)!(l + m)!(l n)!(l + n)!

2+l

N

j=M

Clmjln Cjl+n

lmjjm+jnl+nj

=

(l m)!(l + m)!(l n)!(l + n)!24slmmnl+n

N

j=M

4s1

j!(l m j)!(l + n j)!(m n +j)!

j

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= (l m)!(l + m)!(l n)!(l + n)!lmmnl+n

N

j=M

(9/11s)1/2j

j!(l m j)!(l + n j)!(m n +j)!

j, (42)

where M = max(0, n m), N = min(l m, l + n). We should to note thatthe matrix element (42) in fact does not depend on because of isounimod-ularity of g implying = 1.

We are in a position to express tlmn(g) in terms of the isoEuler angles.Due to (32),

Tl[g(, , )] = Tl[(g(, 0, 0)]Tl[g(0, , 0)]T[g(0, 0, )], (43)

so that finding the general matrixTl(g) reduces to finding of the matricesTl[g(, 0, 0)], Tl[g(0, , 0)], and Tl[g(0, 0, )].

The matrix g(, 0, 0) is diagonal,

g(, 0, 0) =

exp{i3/2/2} 0

0 exp{i3/2/2}

. (44)

For this matrix, we have (see, for example, Ref.[4] for the ordinary case)

Tl[g(, 0, 0)]3/2xln = exp i2n1/23/2xln. (45)

Hence, the matrix of the operator Tl[g(, 0, 0)] is diagonal too, with thenonzero elements being exp[i5/2], l n l. The matrix of theoperator Tl[g(0, 0, )] has similar form.

Let us denote matrix element of the operator Tl[g(0, , 0)] as tlmn().

Then, according to diagonality of the matrices of the operators Tl[g(, 0, 0)]and Tl[g(0, 0, )], we obtain

tlmn = tlmn[g(, 0, 0)]t

lmn()t

lnn[g(0, 0, )exp{i2(m + n)}tlmn().

(46)It remains to obtain tlmn(). The matrix g(0, , 0) has the form

g(0, , 0) =

g1/211 cos /2 ig

1/222 sin /2

ig1/222 sin /2 g

1/211 cos /2

, (47)

where 0 Re < .In the same manner as in Ref.[4] we then have

tlmn() = imn53s+2j

(l m)!(l n)!(l + m)!(l + n)!

g11g22

1/2cotanm+n[1/2/2]

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l

j=max(m,n)

(l +j)!i2j

(l j)!(j m)!(j n)! g1/222 sin[

1/2

/2]. (48)

Parameter varies within the range 0 Re < so that, in this range,different values of correspond to different values of z = g

1/211 cos[

1/2].So, tlmn() can be viewed as a function on isocos . In accordance to this, weput

tlmn() = Plmn(g

1/211 cos[

1/2]. (49)

Then, (46) can be rewritten as

tlmn() = exp i1/2(m + n)Plmn(z). (50)

With the use of (50), Eq.(48) leads to the following definition of the isoLe-gendre polynomials:

Plmn = imn52s+3j+(m+n)/2

(l m)!(l n)!(l + m)!(l + n)!

1 + z

1 z(m+n)/2

l

j=max(m,n)

(l +j)!i2j

(l j)!(j m)!(j n)!

1 z2

j. (51)

The factor ((1 + z)/(1 z))(m+n)/2 is twovalued since m and n areboth integer or half-integer. Single valued definition in (51) comes when

taking into account that 0 Re < and z maps this range to the planez cutted along the real axis, (; 1) and (1; ). In the cutted plane thefactor is single valued.

5 Basic properties of the isoLegendre polynomials

In this Section, we study the basic relations obeyed by the isoLegendrepolynomials.

5.1 Symmetry relations

We will show that Plmn(z) is invariant under the changing of signs of the

indeces m and n. For this purpose, we use the relation

g()g() = g()g(), (52)

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where we have denoted for brevity

g() =

g1/211 cos[t

1/2/2] ig1/222 sin[t

1/2/2]

i(g22)1/2 sin[t1/2/2] g

1/211 cos[t

1/2/2]

. (53)

From (52)-(53) it follows that

Tl()Tl() = Tl()Tl(). (54)

Recall that the matrix elements of Tl() are just Plmn(z). Also, it is known

that tlmn() = 0 at m+n = 0, and tlm,m() = i2l. Replacing the operatorsin (54) by their matrix elements we obtain

Plm,n(z) = Plm,n(z), (55)

from which we havePlmn(z) = P

lm,n(z). (56)

According to the explicit representation (51), we then also obtain

Plmn(z) = Plnm(z). (57)

The relations (55), (56) and (57) are the basic symmetry relations forthe isoLegendre polunomials.

The relations (56) and (57) means, particularly, that Plmn(z) depends on

m and n through the combinations |m + n| and |m n|.Also, it is straightforward to verify that the following relation holds,

Plmn(z) = i2(lmn)Plm,n(z). (58)

5.2 Counter relations

The function Plmn(z) is defined in complex plane cutted along the lines(; 1) and (1; ). On the upper and lower neighbours of these linesPlmn(z) takes different values. From (51) it follows that for z > 1 we have

Plmn(z + i0) = 1

mnPlmn(z i0). (59)

Similarly, for z < 1,

Plmn(z + i0) = 1

m+nPlmn(z i0). (60)

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5.3 Relation to classical orthogonal polynomials

In Sec. 4, we have defined the isoLegendre function Plmn(z), and obtainedone of the representaions of it. Now, we relate Plmn(z) to some of classicalorthogonal polynomials - isoJacobi, adjoint isoLegendre, and isoLegendrepolynomials.

This relations allows, particularly, to establish properties of the polyno-mials by the use of the properties of the isoLegendre function.

5.3.1 IsoJacobi polynomials

IsoJacobi polynomials are defined by

P,k (z) = (1)k2kk! (1 z)1/2(1 + z)1/2

dk

dzk[(1 z2)k(1 + z)1/2(1 + z)1/2]5ks. (61)

Comparing (61) with the following representation of the isoLegendre func-tion,

Plmn(z) =nml

2l

(l + m)!

(l n)!(l + n)(l n)!

(1 + z)(m+n)/2(1 z)(nm)/2 dlm

dzlm[(1 z)ln(1 + z)l+n]2s+2l, (62)

we obtain

P,k (z) = 2mn

(l n)!(l + n)!(l m)!(l + m)!

(1 z)(nm)/2(1 z)(n+m)/2Plmn(z)2+2mn, (63)where

l = k + +

2, m =

+

2, n =

2

. (64)

From (30) we see that = mn and = m + n are integer numbers. Thus,Plmn(z) is reduced to isoJacobi polymomials, for which and are integer

numbers.

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5.3.2 isoLegendre polynomials

IsoLegendre polynomials are defined by

Pl(z) = l

2ll!

dl

dzl[(1 z2)l](s+l). (65)

This implies that Pl(z) = P00l (z). Comparing (65) and (62) we obtain

Pl(z) = Pl00(z). (66)

5.3.3 Adjoint isoLegendre functions

The adjoint isoLegendre function Pml (z), where m

0 (l, m are integer), is

defined by

Pml (z) =(1)l

2l+ml!(1 z2)m/2 d

l

dzl[(1 z2)l](s+2l+m/2), (67)

that is

Pml (z) =2m(l + m)!

l!(1 z2)m/2Pm,ml+m (z)m/2. (68)

Comparing (68) with (62) leads to the following relation:

Pml (z) = im

(l + m)!

(l m)! Plm0(z)

2m. (69)

Let us rewrite (69) by taking into account (56),

Pml (z) = im

(l + m)!

(l m)! Plm0(z)

2m, m 0. (70)

6 Functional relations for isoLegendre functions

In this Section, we derive basic theorems of composition and multiplicationof Plmn(z), and the condition of its orthogonality.

6.1 Theorem of composition

Many important properties ofPlmn(z) are related to the theorem of compo-sition. To derive the rule, let us use the relation

Tl(g1g2) = Tl(g1)T

l(g2), (71)

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from which it follows that

tl(g1g2) =l

k=l

tl(g1)tl(g2), (72)

and it can be rewritten as

tlmn(g1g2) = exp{i1/2(m + n)}Plmn(z). (73)

Fortlmn(g1 = P

lmk(z), t

lkn(g2 = exp{i3/22}Plkn(z), (74)

where , , are isoEuler angles of the matrix g1g2. These angles are

expressed through the angles 1,2, 2 due to the following formulas:

cos[1/2] = cos[11/2]cos[2

1/2]

g1/222 sin[11/2]sin[21/2]cos[21/2], (75)

exp{i1/2} = sin[11/2]g

1/211 cos[2

1/2]

sin[1/2]

+g111 cos[1

1/2] sin[21/2]cos[2]

1/2]

sin[1/2]

+ig1/2

22sin[2

1/2]sin[2 + 1]1/2]

sin[1/2] , (76)

exp{i( + )/2} = g111 cos[1/2] cos[2/2] exp{i2/2}exp{2/2}

g1/211 cos[/2]

,

g1/222 sin[1/2] sin[2/2] exp{i2/2}

g1/211 cos[/2]

, (77)

where 0 Re < , 0 Re < 2, and 2 Re < 2.Inserting equations (73) and (74) into (72), we obtain

exp{

i1/2(m + n)}

Pl

mn

(z)

=l

k=l

exp{i3/22}Plmk(z1)Plkn(z2)2. (78)

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(a) Let 2 = 0, then if Re(1 + 2) <

= 1 + 2 and = = 0.

Accordingly, (78) takes the form

Plmn(z1 + z2) =l

k=l

Plmk(z1)Plkn(z2)

Plmn[g1/211 cos[(1 + 2)

1/2] (79)

=l

k=l

Plmk(g1/211 cos[1

1/22]Plkn(g1/211 cos[2

1/22]).

(b) Let 2 = 0, then ifRe(1 + 2) > = 2 1 2, = = .Therefore,

Plmn(z1 + z2) = 2mnl

k=l

Plmk(g1/211 cos[1

1/22])Plkn(g1/211 cos[2

1/22]). (80)

(c) Let 2 = , then if Re 1 Re 2, = 1 2, = 0, = .Therefore,

Plmn(z1 + z2)

=l

k=l

(2nk)Plmk(g1/211 cos[11/22])Plkn(g1/211 cos[21/22]). (81)

(d) In particular, at 1 = 2 = , we have

lk=l

(1nk)Plmk(g1/211 cos[11/22])

Plkn(g1/211 cos[21/22]) = mn. (82)(e) At = 2 , the formulas (75)-(77) take the following forms:

cos[1/2] = cos[11/2]cos[2

1/2], (83)

exp{

i1/2}

=sin[1

1/2]g1/211 cos[2

1/2] + isin[21/2]

sin[1/2], (84)

exp{i1/2( + )/2} = cos[(1 + 2)1/2/2] + i cos[(1 2)1/2/2]

cos[21/2],

(85)

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respectively.

Instead of (84) and (85), it is more convenient to define

gg1/222 tan[

1/2] =sin[{21/2]

sin[1

1/2] cos[21/2], (86)

g1/222 tan[

1/2] =sin[{11/2]

cos[11/2] sin[21/2]. (87)

Thenexp{i1/2(m + n)}Plmn(g1/211 cos[21/2])

= ikl

k=l

Plmk(g1/211 cos[1

1/2])Plkn(g1/211 cos[2

1/2]). (88)

6.1.1 Theorem of composition for isoLegendre polynomials

Consider particular cases of the function Plmn(z), namely, the IsoLegendrepolynomials and adjoint isoLegendre polynomials. The polynomials are de-fined due to

Pl(z) = Pl00(z), P

ml = i

m

(l + m)!

(l m)! Plm0(z)

2m. (89)

Taking into account the formulas from Sec 6.1 and using (89) we get

exp{i3/2m}Plm(g1/211 cos[1/2 ])

= im

(l + m)!

(l m)!l

k=l

ik

(l k)!(l + k)!

(90)

exp{i3/2m2}Plmk(g1/211 cos[11/2])Pkl (g1/211 cos[21/2])5+mk,where , , 2, and 1, 2 are related to each other as in Sec. 6.1.

If we put m = n = 0, we obtain, particularly,

Pl(g111 cos[11/2]) cos[2

1/2]

g122 sin([11/2]sin([21/2]g1/211 cos[21/2] sin[21/2]) (91)

= (1)l

k=l

ik

(l k)!(l + k)!

exp{i3/2m2}

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Pkl (g

1/211 cos[1

1/2])Pkl (g1/211 cos[2

1/2])4.

Due to the symmetry Plm0(z) = Plm0(z) and (89), the relation (91) can

be reduced:

Pml (z) = 1

(l + m)!

(l m)! Pml (z)

2m. (92)

Thus, from (91) it follows that the polynomials - Pl(z) obey the followingtheorem of composition:

Pl(g111 cos[11/2]) cos[2

1/2]

g122 sin([11/2] sin[21/2]g1/211 cos[21/2] sin[21/2]) (93)

= (1) lk=l

ik

(l k)!(l + k)!

exp{i3/2m2}

Pkl (g1/211 cos[11/2])Pkl (g1/211 cos[21/2])3.

6.2 Multiplication rules

Let in the composition rule

exp{i3/2(m + n)}Pl(g1/211 cos[1/2])

= exp

{i3/2k2

}Plmk(g

1/211 cos[1

1/2])Plkn(g1/211 cos[2

1/2])2. (94)

If2 is a real angle, then this formula can be viewed as a Fourier expansionof the function

exp{i3/2(m + n)}Pl(g1/211 cos[1/2]).Therefore,

Plmk(g1/211 cos[1

1/2])Plkn(g1/211 cos[2

1/2])

=3

2

exp{i3/2(k2 m n)}Plmn(g1/211 )d(21/2). (95)

Putting m = n = 0 in this formula, we get

3

2

exp{i3/2k2}Plmn(g1/211 cos[1/2)d(21/2)

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=(l k)!(l + k)! Pkl (g1/211 cos[11/2])Pkl (g1/211 cos[21/2])3k. (96)

Since Plmn(g1/211 cos[

1/2]) is an even function in respect to 2, the aboveequality can be rewritten

2

Plmn(g1/211 cos[

1/2])g1/211 cos[k2

1/2]d(21/2)

= Pkl (g1/211 cos[1

1/2])Pkl (g1/211 cos[2

1/2]). (97)

If we now let additionally k = 0, we obtin the further reduction

Plmn(g1/211 cos[

1/2)d(21/2)

= Pkl (g1/211 cos[1

1/2])Pkl (g1/211 cos[2

1/2]). (98)

Let us rewrite eq. (98) in a more convenient form. Assuming 1, 2, 2 tobe real numbers such that 0 1 < and 0 1 + 2 < , we redefine thevariable

cos[1/2] = cos[11/2]cos[2

1/2]

g122 sin[11/2]sin[21/2]cos[21/2]2. (99)

Introduce the notationTn(x) = g

1/211 cos[ng

1/211 arccosx].

This function defines Chebyshev-I polynomial. From the last equation itfollows that

g1/211 cos[k

3/22]

= Tkg111 (cos[1

1/2]g111 cos[21/2] cos[1/2])

g122 sin[11/2]g122 sin[2

1/2]. (100)

In turn, from the condition (100) it follows

d2 =

g1/222 sin[

1/2]dg1/211 (cos[

1/2] cos[(1 + 2)1/2])g1/211 (cos[(1 2) cos[1/2].

(101)

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Since when varying 2 from to the variable varies in the range from

|1 + 2| to |1 2|, the above made redefinition transfroms the integral tothe form

2

2

1+2|1+2|

Pkl (g1/211 cos[

1/2])

Tk g111 cos[1

1/2]g111 cos[21/2] g1/211 cos[1/2]

g122 sin[11/2]g122 sin[2

1/2]

g1/222 sin[

1/2]d(1/2)

g

1/2

11 (cos[1/2

] cos[(1 + 2)1/2

])g

1/2

11 (cos[(1 2)1/2

]cos[1/2

])

= Pkl (g1/211 cos[1

1/2])Pkl (g1/211 cos[2

1/2]). (102)

The expression in the denominator has a simple geometrical meaning: itis equal to the square of the spherical triangle with the sides 1, 2 and ,divided to 42.

6.3 Orthogonality relations

In this Section, we apply theorems of orthogonality and completeness ofthe system of matrix elements of pairwise nonequivalent irreducible isouni-tary representations of compact group to the group SU(2). Since dimen-sion of the representation Tl(u) of the group SU(2) is 2l + 1, the functions

2l + 1tlmn(u) form complete orthogonal normalized system in respect toinvariant measure du on this group. In other words, the functions tlmn(u)fulfill the relations

SU(2)

tlmn(u)2tkpq(u)du =

2

(2l + 1)lkmpnq. (103)

Inserting expression for the matrix elements

tlmn(, , ) = exp

{i3/2(m + n)

}Pl(g

1/211 cos[

1/2]) (104)

into (103) and using the fact that the measure du on the group SU(2) isgiven by

du =4

16gg

1/222 sin[

1/2]ddd, (105)

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we turn to the following specific cases.

(a) If l = k or m = p or n = q, then2

2

20

0

Plmn(g1/211 cos[1

1/2])Pkmn(g1/211 cos[1

1/2])g1/222 sin[

1/2]

6 exp{i3/2(p m)} exp{i3/2(q n)}d(1/2)d(1/2)d(1/2).(106)

(b) Let p = m and q = n, then, at l = k,

0

Plmn(g1/211 cos[1

1/2])Pkmn(g1/211 cos[1

1/2])g1/211 cos[]

3d(1/2) = 0.

(107)Analogously, from (103) it follows

0

|Plmn(g1/211 cos[11/2])|2g1/211 sin[1/2]d(1/2) =2

2l + 1. (108)

Further, putting x = g1/211 cos[

1/2] we get the orthogonality relations forPlmn(x):

1

1

Plmn(x)Pkmn(x)d(x) =

2

2l + 1

lk. (109)

7 Recurrency relations for isoLegendre functions

In this Section, we derive the formulas relating the functions Plmn(z), indecesof which are differ from each other by one, that is, recurrency relations, whichcan be viewed as an infinitesimal form of the theorem of composition. Theserelations are then follow from the composition rules at infinitesimal 2.

To obtain the reccurency rules, we diffirentiate the equation below on 2and put 2 = 0:

P

l

mn(z) =(l

m)!(l + m)!

(l n)!(l + n)!

7

2

20

d(g1/211 cos[

1/2

2]exp

i1/2

2

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+ig1/2

22 sin[

1/2

2 ]exp i1/2

2 )ln

(ig1/2

22 sin[

1/2

2 ]exp

i1/2

2

g1/211 cos[1/2

2]exp

i1/22

)ln exp i1/2. (110)

First, we find

d

d[Plmn(g 111/2 cos[1/2])]|=0

=5

4

(l m)!(l + m)!(l n)!(l + n)! (111)

20

d(ln)exp i3/2(n + 1) +(l + n)exp i3/2(n 1) exp3/2m.

It is obvious that the r.h.s. of this equation is zero unless m = n 1. Atm = n + 1, from (111) we get

d

d[Pn+1n cos[

1/2

2]|=0 =

i

23/2

(l n)(l + n + 1). (112)

Similarly,

d

d

[Pn1n (g1/211 cos[

1/2

2

]=0 =i

2

3/2(l + n)(l n + 1). (113)Now, we are ready to derive the recurrency relations.Using the factorization

Plmn[g1/211 cos[(1 + 2)

1/2]] =l

k=l

Plmk(g1/211 cos[1

1/2])

Plkn(g1/211 cos[2

1/2])

obtained in Sec. GS6, putting 2 = 0 and replacing g1/211 cos[2

1/2]) by z,we obtain the recurrency relation in the form

1 z2dP

lmn(z)

dz= i

23/2[

(l n)(l + n + 1) (114)

Plm,n+1(z) +

(l + n)(l n + 1)Plm,n1(z)].

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To derive the second recurrency relation, we use the particular case of

the theorem of composition which corresponds to 2 = 2 . Namely, wedifferentiate the formula

exp i3/2(m + n)Plmn(g1/211 cos[11/2]) =l

k=l

ikPlmk(g1/211 cos[1

1/2]

Plkn(g1/211 cos[2

1/2])

and put 2 = 0. After strightforward computations, we have

i3

m dd2

+ n dd2

|2=0

Plmk(g1/211 cos[11/2] dPlmk(g

1/211 cos[1

1/2])

d1

d

2|2=0 (115)

12

3/2

(l + n)(l n + 1)Plm,n1(g1/211 cos[11/2](l n)(l + n + 1)Plm,n+1(g1/211 cos[21/2])

It remains to find d/d2 and d/d2. To this end, we differentiate the

equality cos[1/2

] = cos[11/2

] cos[21/2

]. Since, at2 = 0, we have = 1, = 0, and = 2 =

2 , it follows that d/d2|2=0 = 0. Similarly,

d

d2|2=0 =

g1/211

cos[11/2](116)

andd

d2|2=0 = (

g11g22

)1/2ctan[11/2]

i

m nz1 z2

Plmn(z)

= 12

(l + n)(l n + 1)Plm,n1(z)

(l n)(l + n + 1)Plm,n+1(z)

.

(117)

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From the reccurency relations obtained above it is straightforward to

write down the following recurrency relations:

1 z2 dP

lmn(z)

dz+

nz m1 z2 P

lmn(z) = i5/2

(l n)(l + n + 1)Plm,n+1(z)

(118)and, analogously,

1 z2 dP

lmn(z)

dz nz m

1 z2 Plmn(z) = i5/2

(l + n)(l n + 1)Plm,n+1(z).

(119)Due to the symmetry, we have from (118) and (119)

1 z2 dPlmn(z)

dz+ mz n

1 z2 Plmn(z) = i5/2

(l m)(l + m + 1)Plm,n+1(z)

(120)and

1 z2 dP

lmn(z)

dz mz n

1 z2 Plmn(z) = i5/2

(l + m)(l m + 1)Plm,n+1(z).

(121)Adding (118) to (119), we obtain the recurrency relations for three Ps:

2

n mz1 z2

Plmn(z) = i

3/2

(l + n)(l n + 1)Plm,n1(z)

(l n)(l + n + 1)Plm,n+1(z)

, (122)

1 z2Plmn(z) = i5/2

(l + n)(l n + 1)Plm,n1(z)

+

(l n)(l + n + 1)Plm,n+1(z)

. (123)

Putting m = 0 in (118) and (119), and using

Pl0n(z) = in2

(l n)(l + n)

Pnl (z) (124)

we obtain, finally, the recurrency rules for the adjoint isoLegendre polyno-mials,

1 z2 dPnl (z)

dz+ 2

nz

1 z2 Pnl (z) = Pn+1l (z) (125)

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and

1 z2dP

nl (z)

dz 2 n

1 z2 Pnl (z) = 3(l + n)(l n + 1)Pn1l (z). (126)

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8 The group QU(2)

In this Section, we consider the group QU(2) consisting of isounimodu-lar isoquasiunitary matrices representations of which lead to isoJacobi andisoLegendre functions.

8.1 Definitions

The representations ofQU(2) are in many ways similar to that of the groupSU(2). However, in contrast to SU(2), the group QU(2) is not compact,thus having continuous series of isounitary representations.

Similarly to the description of the group SU(2), we describe the group

QU(2) as a set of isounimodular isoquasiunitary 2 2 matrices

g0 =

, (127)

where and are given by (7), satisfying

g0sg0 = s, (128)

where

s = 1 0

0 1 , g0 =

det g0 = 1, ||

2 ||2 = 1.(129)


The matrices g0 above have been defined by the complex numbers and. However, in various aspects it is suitable to define them by the isoEulerangles. Constraints on the isoEuler angles following from the requirementthat g0 QU(2) are

g1/211 cos[

1/2

2]exp{i3/2( + )/2}

= g1/211 cos[1/2/2]exp{i3/2( + )/2} (130)

and

g1/211 sin[

1/2

2]exp{i3/2( )/2}

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=

g1/2

22 sin[

1/2

2 ]exp{i3/2

( )/2}, (131)which we rewrite in the following form:

cos[i1/2

2] =

cos[

1/2

2]exp{i3/2( + )/2} (132)

and

sin[i1/2

2] =

sin[

i1/2

2]exp{i3/2( + )/2}. (133)

The angles + and + are real. So, if g0 = g0(, , )

QU(2) then cos[1/2

/2] is an imaginary number, i.e. = i is real.Taking into account the constraints (132) and (133), we obtain the fol-

lowing ranges for the parameters:

0 < 2, 0 < , 2 < 2. (134)

In terms of these parameters, the matrix g0 is

g0 =

g1/211 cos[ i1/22 ]e i

3/2(+)2 i2g1/222 sin[i1/2]e

i3/2()2

i2g1/222 sin[ i1/2

2 ]ei3/2()

2 g1/211 cos[

1/2

2 ]ei3/2(+)

2

.

(135)Thus, we see that the group QU(2) is one of the real types of subgroupsof SL(2, C). In the following, we use the parameters (134) instead of theisoEuler angles (, , ).

Let us find the transformation laws for these parameters under the mul-tiplying of two elements ofQU(2). We introduce the notation g01 = (0, 1, 0)and g02 = (2, 2, 0) so that g01g02 = (, , ). Using the formulas (16)-(18) we find

cos[i1/2] == cos[i11/2] cos[i2

1/2]g1/211 (136)

sin[i11/2]sin[i21/2]g1/211 cos[21/2],

exp{i1/2( + )/2} = 2(g1

11

cos[i11/2]cos[i2

1/2]exp{

i1/22/2}cos[i1/2]

(137)

+g122 sin[i1

1/2]sin[i21/2]exp{i1/22/2})

cos[i1/2]

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and

exp{i1/2} = 1/2( sin[i11/2]cos[i21/2]g1/222 sin[i

1/2](138)

+1/2 cos[i1

1/2]sin[i21/2]g

1/211 cos[2

1/2]

g1/222 sin[i

1/2]

ig122 sin[i2

1/2] sin[21/2]

g1/222 sin[i

1/2]).

It is easy to check that the element g0(, , ) is an inverse of g0( , , ).

8.3 Relation to the group SH(3)

Let us define the group SH(3) as the group of isolinear transformation ofthree dimensional isoEuclidean space E3 acting transitively on (iso)hyperboloidsand (iso)conics. This transformation is an isohyperbolic one.

The relation between the groups QU(2) and SH(3) is similar to thatbetween SU(2) and S0(3). Namely, to every point x(x1, x2, x3) E3 weassociate the quasiuntary matrix

hx =

x1 x2 + ix3

x2 ix3 x1

. (139)

Then,T(g0)hx = g0h

xg0. (140)

Accordingly,

T(g0)hx =

1y1 y2 + iy3

y2 iy3 y1

, (141)

where x = g1/211 x, y = g

1/222 y, and y(y1, y2, y3)inE3.

9 Irreps of QU(2)

9.1 Description of the irreps

Denote = (l, ), where l is complex number and = 0, 1/2. With every we associate the space D of functions (z) of complex variable z = x + iysuch that:

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(1) (z) is ofC class on x and y at every point z = x + iy except for

z = 0;(2) for any a > 0 the following equation is satisfied:

(az) = a2(z). (142)

(3) (z) is an even (odd) function at = 0(1/2),

(z) = (1)2(z). (143)

For subsequent purposes, we realize the space D on a circle. Namely,

with every function (z) we associate the function f such that, at = 0,

f(exp{i1/2}) = (exp{i1/2}) (144)and, at = 1/2,

f(exp{i1/2}) = exp{i1/2}(exp{i1/2}). (145)

Thus, the space D can be represented as the space D of functions on circle.

9.2 Representations T(g0)

To every element

g0 =

of the group QU(2) we associate the operator in the space D,

T(g0)(z) = (z +

z). (146)

Clearly, function T(g0)(z) has the same homogeneity degree as the func-tion (z), and so the operator T(g0) is an automorphism of the space D.Also, it is easy to verify that

T(g01)T(g01) = T(g01g02). (147)

Action of the operator T(g0) can then be strightforwardly derived.Namely, for = (l, 0) we have

T(g)f(exp{i1/2})

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= |exp{i1/2} + |2l2l+1f exp{i1/2

}+

exp{i1/2} +

, (148)

and for = (l, 1/2)T(g)f(exp{i1/2})

= |exp{i1/2}+|2l12l+1(exp{i1/2}+)f exp{i1/2} +

exp{i1/2} +

.

(149)

10 Matrix elements of the irreps of QU(2) and iso-

Jacobi functions

10.1 The matrix elements

Let us choose the basis exp{im3/2} in space D, and define the matrixelemets of T(h), where

h =

exp{it1/2/2} 0

0 exp{it1/2/2}

. (150)

In the same manner as for g of QU(2) we can represent

h =

ei1/2

/2 00 ei

1/2/2

g1/211 cos[ i1/2

2 ] ig1/222 sin[ i1/2

2 ]ig1/222 sin[ i

1/2

2 ] g1/211 cos[

i1/2

2

(151)

e i1/22 0

0 ei1/2

2

,

where , , and are isoEuler angles of g0. From (153) we define T(g),namely,

g1/211 cos[

i1/2

2 ] ig1/2

22 sin[i1/2

2 ]

ig1/222 sin[ i1/2

2 ] g1/211 cos[

i1/2

2 ]

. (152)

Then, straighforward calculations yield [4]

tmn =4+l

2

20

d (g1/211 cos[

i1/2

2]ig1/222 sin[

i1/2

2]exp{i1/2})l+n+

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(g1/2

11 cos[

i1/2

2 ] (153)

ig1/222 sin[i1/2

2]exp{i1/2})ln exp{i(m n)3/2}.

Introduce the function Blmn(isocosh) defining

Blmn(g1/211 cos[i

1/2]) =4+l

2

20

d (g1/211 cos[

i1/2

2]ig1/222 sin[

i1/2

2]

exp{i1/2})l+n(g1/211 cos[i1/2

2] (154)

ig1/222 sin[i1/2

2]exp{i1/2})ln exp{i(m n)3/2}.

Comparing (153) and (154) we have

tmn = Blmn(g

1/211 cos[i

1/2]) = (l, ), (155)

wherem = m + , n = n + , 0 < , (156)

l is a complex number, m and n are simultaneously integer or half-integernumbers. From the expansion (153) it follows that

T(g0) =

T(

h)

T(g)

T(

h). (157)

So we can write

tmn(, , ) = exp{i3/2(m + n)}Blmn(g1/211 cos[i1/2]), (158)

where m,n, and are defined according to (156). Since Blmn(z) playsthe same role for QU(2) as the function Plmn(z) for SU(2), we call B

lmn(z)

isoJacobi function of the variable z = g1/211 cos[i

1/2].

11 IsoJacobi function Blmn(z)

Integral representation of the isoJacobi function Blmn(z) can be readily de-rived (see [4] for the usual case),

Blmn(g1/211 cos[i

1/2]) =

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4+l

2

exp

{(l

n + 1/2)3/2t

}2g1/211 (cos[i1/2] cos[it1/2])

zmn

+ (g1/2

11 cos[i1/2

/2]

(159)

z+ig1/222 sin[i1/2/2])2n + zmn (g1/211 cos[i1/2/2]

zigg1/222 sin[i1/2/2])2n

dt,

where

z =exp{1/2} g1/211 cos[i1/2]

g1/222 sin[i

1/2]

i exp

{it1/2/2

}2g

1/211 (cos[i

1/2]

cos[it1/2])

g1/222 sin[i1/2] . (160)

As one can see, the representation (159) is simplified when n = m andalso when n = 0.

When n = m we have directly from (159)

Blnn(z) =5/2

0

g1/211 cos[(l n + 121/2]g

1/211 cos[(2n

3/2)1/2]dtg111 cos

[ i1/2

2 ] + g122 sin

[ i1/2

2 ].

(161)When n = 0 we have

Blm0(z) =

52+m

2

exp

{(l + 12)

3/2t

}(zm+ + z

m )dt

2(g1/211 (cos[i1/2] cos[it1/2])) . (162)

Particularly, when in addition m = 0 we have

Bl00(z) =3/2

0

cos[i(l + 1/2)t]dtcos2[ i

1/2

2 ] cos2[ it1/2

2 ]. (163)

12 IsoJacobi function Bl00

Let us put m = n = 0 in (155). Then

t

00(g0) = Bl00(z). (164)

We call Bl00(z) isoJacobi function with index l and denote it simply Bl(z),namely,

Bl(z) = t00(0, , 0, ) = B

l00(z), (165)

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where = (l, 0) and z = g1/211 cos[i

1/2].The following integral representations for the isoJacobi function Bl(z)

can be written:

Bl(z) =l+1

2

0

(g1/211 cos[i

1/2] g1/222 sin[i1/2]g1/211 cos[i1/2])ld,(166)

Bl(z) =l+1

2i

(g1/211 cos[i

1/2] i z2 + 1

22zg1/222 sin[i

1/2])ldz

z, (167)

Bl(z) =l

2g

1/222 sin[

1/2l]

0

(g1/211 cos[i(l +

12)t

1/2]dt

cos[it1/2])l + cos[i1/2]

. (168)

From (166) it can be seen that when l is integer the isoJacobi function Bl(z)coincides with the isoLegendre polynomial,

Bl(z) = Pl(z), (169)

which has been considered in Secs. 2-7.

12.1 Symmetry relations for Blmn(z) and Bl(z)

Similarly to the isoLegendre polynomials Plmn(z), the isoJacobi functionsBlmn(z) satisfy the following symmetry relations:

Blmn(z) = Blmn(z) (170)

andBl(z) = Bl1(z). (171)

13 Functional relations for Blmn(z)

Functional relations for isoJacobi functions Blmn(z) can be derived in a sim-ilar fashion as it for isoLegendre functions Plmn(z). Particularly, we have

exp{

i3/2(m + n)}

Bl

mn

(z) =

k=

exp

i3/2k2Bl

mk

(z1)Bl

kn

(z2),

(172)

where z = g1/211 cos[i

1/2], z1 = g1/211 cos[i1

1/2], z2 = g1/211 cos[i2

1/2],and , 1, 2, , and are defined due to eqs. (136)-(138).

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So, as a consequence of (172) we have the following particular cases.

(a) Let 2 = 0, then = 1 + 2, = = 0, and we have

Blmn(g1/211 cos[i(1 + 2)

1/2]) =

k=

Blmk(g1/211 cos[i1

1/2]) (173)

Blkn(g1/211 cos[i21/2]).(b) Let 2 = , then 1 2, = 1 2, = 0, = , and we have

Blmn(g1/211 cos[i(1 2)1/2]) =

k=

Blmk(g1/211 cos[i1

1/2])2 (174)

Blkn(g1/211 cos[i21/2]).(c) Particularly, when in addition 1 = 2, we have

k=

Blmk(g1/211 cos[i1

1/2])2Blkn(g1/211 cos[i2

1/2]) = Blmn(1) (175)

= mn mn1.Theorem of composition for isoLegendre function.

Let us define isoLegendre function and adjoint isoLegendre function asfollows

Bl(z) = Bl00(z) (176)

and

Bml (z) =(l + m + 1)

(l + 1)Bm0l (z), B

lm0(z) =

(l + 1)

(l m + 1) B0ml (z).

(177)Putting m = n = 0 in (172) and using (176) and (177) we obtain

Bl(z) =(l k + 1)(l + k + 1)

3ei3/2k2Bkl (z1)B

kl (z2), (178)

where

g1/211 cos[i

1/2] = g111 cos[i11/2]g111 cos[i2

1/2]+ (179)

g122 sin[i11/2]2 sin[i2

1/2]g111 cos[i21/2].

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The composition formula for the adjoint isoLegendre function follows

from (172) with n = 0, namely, we have

Bl(z) =(l + m + 1)

(l + k + 1)3ei

3/2k2Bkl (z1)Bkl (z2), (180)

where

(l + m + 1) = (l + m)(l + m)and(l + m + 1) =

0

elm2xl+m1dx.

(181)Multiplication formula.

Multiplying both sides of the equation (172) by exp{i3/2k2} we obtain

Blmk(z1)Blmk(z2) =

2

2

20

d2 ei3/2(k2mn)Blmn(z). (182)

Putting m = n = 0 in (182) and using the symmetry relations we get

Bkl (z1)Bkl (z2) =

2

2

20

ei3/2k2 (183)

Blmn(z1z2 + z3z4z5)d2,

where z1 = g1/211 cos[i1

1/2], z2 = g1/211 cos[i2

1/2], z3 = g1/211 sin[i1

1/2],

z4

= g1/2

11sin[i

21/2], and z

5= g

1/2

11cos[i

21/2]. Particularly,

Bl(z1)Bl(z2) =

2

20

Blmn(z1z2 + z3z4z5)d2. (184)

14 Recurrency relations for Blmn

Recurrency relations for Blmn can be derived in the same manner as it forPlmn. So, we do not represent the calculations here, and write down the finalresults.

z2 1dBlmn(z)

dz=

(l + n)

2Bl

m,n1(z) +

(l

n)

2Bl

m,n+1(z), (185)

m nzz2 1

dBlmn(z)

dz= (l + n)

2Blm,n+l(z) +

(l n)2

Blm,n+1(z). (186)

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From (185) and (186) we have

z2 1 dB

lmn(z)

dz m nz

z2 1 dBlmn(z)

dz= (l n)Blm,n+1(z) (187)

and

z2 1 dB

lmn(z)

dz+

m nzz2 1

dBlmn(z)

dz= (l + n)Blm,n+1(z). (188)

Using the symmetry relations we have

z2 1dBlmn(z)

dz

+n mzz2 1

dBlmn(z)

dz

= (l + m +1)Blm,n+1(z) (189)

and

z2 1 dB

lmn(z)

dz+

m nzz2 1

dBlmn(z)

dz= (lm+1)Blm,n+1(z). (190)

Also,

(l n)Blm,n+1(z) (l + n)Blm,n1(z) =2(m nz)

z2 1 2Blmn(z), (191)

(l+m+1)Blm+1,n

(z)

(l

m+1)Blm1,n

(z) =2(n

mz)

z2 12Bl

mn(z). (192)

The differential equation satisfied by isoJacobi function is

z2 1 d

2Blmn(z)

dz22z dB

lmn(z)

dzm

2 + n2 2mnz2 1 B

lmn(z) = l(l+1)B

lmn(z).

(193)The differential equation satisfied by adjoint isoLegendre function is

z2 1 d

2Bml (z)

dz22z dB

ml (z)

dz m

22

z2 1 Bml (z) = l(l +1)B

ml (z), (194)

and the equation satisfied by isoLegendre function is

z2 1 d

2Bl(z)

dz2 2zdBl(z)

dz= l(l + 1)Bl(z). (195)

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15 The group M(2)

In this Section, we consider linear transformations of isoEuclidean plane.

15.1 Definitions

The motion of isoEuclidean plane E2 is similar to that of the ordinary Eu-clidean plane E2 so the definition of the group M(2) is similat to that ofM(2).

Choosing local coordinates (x, y) on E2, we write the motion g : (x, y) (x, y) in th following form:

x = xg1/2

11cos[1/2]

yg

1/2

22sin[1/2] + a, (196)

y = xg1/222 sin[

1/2] yg1/211 cos[1/2] + b,where

x = g1/211 x, y = g

1/222 y, (197)

so that, in an explicit form,

x = x(g1/211 g22)cos[

1/2] y(g1/211 g22)sin[1/2] + a, (198)

y = xg3/211 sin[

1/2] y(g1/211 g22) cos[1/2] + b.Here, a, b, and parametrize the motion g so that every element g M(2)can be defined by the three parameters having the following ranges:

< a < , < b < , 0 < 2. (199)Another realization of M(2) comes with the identification of g(a,b,)

with the matrix

T(g) =

g

1/211 cos[

1/2] g1/222 sin[1/2] ag1/222 sin[

1/2] g1/211 cos[

1/2] b0 0 1

. (200)

It can be easily verified that

T(g1)T(g2) = T(g1g1),

so that T(g) is a representation of M(2). This representation is an exactone, i.e. T(g1) = T(g2) if g1 = g2. Thus, we conclude that the group M(2)is realized as group of 3 3 real matrices (201).

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The group M(2) can be realized also as the group of 2

2 complex

matrices. Namely, by the identification of g(a,b,) with the matrix

Q(g) =

exp{i3/2} z

0 1)

, (201)

wherez = a + ib. (202)

It is easy to verify that Q(g1)Q(g2) = Q(g1g1) and Q(g1) = Q(g2) ifg1 = g2.


For the parametrization above, let us find the composition law. Let g1 =g(a1, b1, 1) and g2 = g(a2, b2, 2). Then

T(g1g2) = (203) g

1/211 cos[1 + 2] g1/222 sin[1 + 2] a1 + a2g1/211 cos[1] b2g1/222 sin[1]

g1/222 sin[1 + 2] g

1/211 cos[1 + 2] b1 + a2g

1/222 sin[1] + b2g

1/211 cos[1]

0 0 1

,

so that the law is

a = a1 + a2g1/211 cos[1

1/2] b2g1/222 sin[11/2], (204)

b = b1 + a2g1/222 sin[1

1/2] + b2g1/211 cos[1

1/2], (205)

= 1 + 2. (206)

Denoting x = (a1, b1) and y = (a1, b1) we rewrite the formulas (204)-(206)as follows:

g(x, )g(y, ) = g(x + y, + ). (207)

From this equation it follows that if g = g(x, ) then

g1 = g(x, 2 ). (208)Another useful parametrization can be represented by isoEuler angles.

On the plane, we parametrize the vector x = (a, b) by isopolar angles a =rg

1/211 cos[

1/2] and b = g1/222 sin[

1/2]. The set of parameters for g isthen (r, , ), with the rabges

0 r < ,quad0 < 2,quad0 < 2. (209)

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Decomposition for element of M(2) reads

g(r, , ) = g(0, , 0)g(r, 0, 0)g(0, 0, ). (210)Transformations corresponding to g(0, 0, ) and g(0, 0, ) are rotationswhile g(r, 0, 0) defines a parallel transport along the axis Ox. For g1 =g(r, 0, 1) and g2 = g(r2, 0, 0), we have from eqs.(204)-(206)

g1g2 = g(r, , )

, where

r =

r21 + r

22 + 2r1r2g

1/211 cos[

1/2] (211)

and

r2 = xb21x + yb22y + zb

23z; r

21 = x1b

21x1 + y1b

22y1 + z1b

23z1, (212)

r22 = x2b21x2 + y2b

22y2 + z2b

23z2,

exp i3/2 =r1 + r2exp{i3/21

} r, (213)

= 1. (214)

To find the parameters of the composition g1g2 for g1 = g(r1, 1, 1) andg2 = g(r2, 2, 2), one should replace 1 by 1 + 2 1, by 1,and by 2 in (211)-(214).

From decomposition (210) and equation

g(0, 0, 1 + 2 1) = g(0, 0, 1 1)g(0, 0, 2) (215)we get

g1g2 = g(0, 0, 1)g(r1, 0, 0, )g(0, 0, 1 + 2 1)g(r2, 0, 0, )g(0, 0, 2 + 2)3.(216)

16 Irreps ofM(2)

16.1 Description of the irreps

Denote the space of smooth functions f(x) on circle x1b21x1 + x1b21x1 =

1

by D. To every element g(a, ) M(2) we associate the operator Tc(g)acting on f(x),

Tc(g)f(x) = ec(a,x)f(x). (217)

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Here, c is fixed complex number, x is vector to which the vector x is

transformed by rotation on angle , and (a, x) = a11x1g1/211 + a2x2g1/211 .Let us show that Tc(g) is the representation of M(2). For g1 = g(a, ) andg2 = g(b, ) we have

Tc(g1)Tc(g2)f(x) = Tc(g1)ec(b,x)f(x)e

c(a,x)ec(b,x f(x). (218)

Since (b, x) = (b, x) the following equation is valid:

Tc(g1)Tc(g2)f(x) = ec(a+b,x)f(x). (219)

On the other hand, owing to (207)

g1g2 = g(a, )g(b, ) = g(a + b, + ), (220)

so thatTc(g1g2)f(x) = e

c(a+b,x)f(x). (221)

Thus, Tc(g1g2) = Tc(g1Tcg2), i.e. Tc(g) is representation of M(2).Parametrical equations of the circle, x1b

21x2 + x2b

21x2 =

1, have theform

x1 = g111 cos[

1/2], x2 = 1/2 sin[1/2], 0 < 2, (222)

so that one can think of functions f(x) D as functions depending on ,

f(x) = f(). (223)

The operator can be rewritten as

Tc(g)f() = exp{c2rg1/211 cos[( )1/2]}f( ), (224)where

a = (rg1/211 cos[

1/2], rg1/222 sin[

1/2]), g = g(a, ).

By introducing scalar product,

(f1,f2) =2

2

2

0

f1()

f2()d, (225)

we make the space D to be isoHilbert space E. Then, Tc(g) is isounitaryin respect to the scalar product (225) if and only if c = i is an imaginarynumber.

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16.2 Infinitesimal operators

The operator Tc(w1(t)), where

w1(t) =

1 0 t1

0 1 00 0 1

)

, (226)

w1 2, transforms function f() to

Tc(w1(t))f() = exp{c2trg1/211 cos[1/2]}f(), (227)

so that

A1 = dTc(w1(t))

dt|t=0 = cg1/211 cos[1/2], (228)

i.e. A1 acts as a multiplication operator.Similarly, one can prove that the infinitesimal operator A2 corresponding

to the subgroup 2 represented by the matrices

w2(t) =

1 0 00 t1 00 0 1

)

(229)

is given by

A2 = C(g11g1/2

22 ) sin[1/2

]. (230)Also, for the subgroup 3 consisting of the matrices

w3(t) =

g

1/211 cos[t

1/2] g1/222 sin[t1/2] 0g1/222 sin[t

1/2] g1/211 cos[t

1/2] 00 0 1

)

(231)

we have

A3 = dd

. (232)

16.3 The irrepsThe prove of irreducibility of the representation T(g) of the group M(2) canbe given in the same way as it of T(g), and we do not present it here.

Below, we consider two choices of c.

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(a) c

= 0. We have

Tc(w3())f() = f( ). (233)

(b) c = 0. We have

Tc(g)f() = f( ), (234)

where g = (x,). This representation is reducible since it can be decomposedinto direct sum of the one-dimensional representations

T0n(g) = ei3/2n. (235)

Note that Tc(g) with c = 0 and T0n(g), where n is integer number, constituteall possible irreps of M(2).

17 Matrix elements of the irreps ofM(2) and isoBesselfunctions

17.1 Matrix elements

In the space E, we choose the orthonormal basis {exp(i5/2n)} consistingof eigenfunctions of the operator Tc(w), w 3. The matrix elements arewritten in this basis as

tcmn(g) = (Tc(g)ein3/2 ,eim

3/2). (236)

Taking into account definition (225) and eq.(224) we get

tcmn(g) =exp{in3/2}

23

20

d ec2rg

1/211 cos[()

1/2]ei(nm)3/2

.

(237)Let r = = 0, i.e. g defines rotation on isoangle . Due to orthogonality

of the functions exp{in5/2}, we have

tcmn(g) tcmn() = exp{in3/2})mn. (238)

Thus, the rotation is represented by a diagonal matrix Tc(), with non-zeroelements being exp{in5/2}, < n < .

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Let = = 0. In this case, g defines transplacement on r along the Ox

axis so that (237) takes the form

tcmn(g) tcmn(r) =2

2

20

d exp{C2rg1/211 cos[1/2]+i(nm)3/2}.(239)

Replacing by /2 , we then have

tcmn(r) =2

2inm

20

d exp{c2rg1/222 sin[1/2] i(n + m)3/2}.(240)

Let us denote

Jn(x) = 2

2inm

20

d exp{2g1/222 sin[1/2] in3/2}, (241)

and refer to Jn(x) as isoBessel function.Using this definition we have from (240), in a compact writting,

tcmn(r) = inmJnm(ic2r). (242)

Now, to obtain tcmn(g) in a eneral case it is suffice to make the replace-ment = 2 in the integral (237). Namely, using (241) we obtain

tcmn(g) = inmexp{in3/2 + i(n + m)3/2}Jnm(ic2r). (243)

Indeed, from (243) it follows that

Tc(g) = Tc()Tc(r)Tc( ). (244)

Since the matrices Tc(g) and Tc( ) are both diagonal, with the non-zero elements exp{i3/2n} and exp{i3/2n( )} respectively, whiletcmn(r) = i

nmJnm(ic2r) we come to (243).If g is an identity transformation, g = g(0, 0, 0), then Tc(g) is the isounit

matrix. Consequently, we have the following relations:

Jnm(0) = , J(0) = 1, Jn(0) = 0, (n = 0).

17.2 IsoBessel functions with opposite sign indeces

In this section, we find the relation between the isoBessel functions withopposite sign indeces.

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In the space of functions f(), introduce the operator Q acting according

toQf() = f(). (245)

This operator commutes with operator Tc(g) Tc(r), where g = g(r, 0, 0).Indeed,

Tc(r)Qf() = Tc(r)f() = exp{c2rg1/211 cos[1/2]}f().Consequently,

QTc(r) = Tc(r)Q. (246)

Operator Q acts by changing the basis element, exp{in5/2} to exp{in5/2},so the matrix has the form (qmn), where qm,m =

1 and qmn = 0 form + n = 0. Thus, from (246) we obtain

tcm,n(r) = tcm,n(r). (247)

Then, taking into account (242) we get

im+nJn+m(i2cr) = imnJnm(i2cr). (248)Putting in (248) m = 0 and z = i2cr, we finally have

Jn(z) = 1nJn(z). (249)

17.3 Expansion series for IsoBessel functions

Our aim is to derive the expansion series for isoBessel function in x. Tothis end, we use integral representation (241). Expanding the exponentexp{i2xg1/2 sin[2]} and integrating over all the terms we obtain

Jn(x) =k=0

akxk(g

k+1/211 g

k22), (250)

where

ak =3ks

2k! 2

0d exp{i5/2n}(i(g11g1/222 sin[2])k. (251)

Here, s = 1, 2, 3, . . . On the other and, owing to the Euler formula,

(ig1/222 sin[

2])k =exp{i3/2

ei3/2}2k2k (252)

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=

km=0

(

1)m2(1k)mCmk exp{

i(k

2m)3/2}2k .

Inserting this formula into (251), one can observe that ak is non-zero iff(k n) is an even number, i.e. k n = 2m, m 0. Ifk = n + 2m then

ak =(1)m

2km!(k m)!k+2s =(1)n3m2s

2n+2mm!(n + m)!. (253)

So, we finally have

Jn(x) = (g5/22s11 g

22s22 )(x/2)

nk

m=0nmx2m

22mm!(n + m)!. (254)

18 Functional relations for isoBessel function

18.1 Theorem of composition

Theorem of composition for isoBessel function can be derived in the samemanner as it for isoLegendre function Plmn. One should use the equalityTc(g1g2) = Tc(g1)Tc(g2, that is

tcmn(g1g2) =

k=

tcmk(g1)tckn(g2). (255)

Let us put g1 = g(r1, 0, 0) and g2 = g(r2, 2, 0). Then the parameters r,, and corresponding to the composition g = g1g2 can be expressed viaparameters r1, r2, and 2 as

r =

r21 + r

22 + 2r1r2g

1/211 cos[2

1/2], (256)

ei3/2 = r1 + r2e

i3/22, (257)

= 0, (258)

where r21, r22, and r

2 are defined due to (212).Inserting the matrix elements (243) into (255) and putting m = 0 and

R = i1 we have after some algebra

ei5/2nJn(r) =

k=

ei5/2kJnk(r1)Jk(r2), (259)

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where r, r1, r2, , and 2 are defined according to (256)-(257).

The formula (258) represents the theorem of composition of isoBesselfunctions.

Particularly, at n = 0 we have from (258)

J0(r) =

k=

(1k

)ei5/2k2 Jk(r1)

3Jk(r2). (260)

Below, we consider some useful particular cases of the theorem.(a) At 2 = 0, we have r = r1 + r2 and = 0, so that

Jn(r1 + r2) =

k=

Jnk(r1)Jk(r2). (261)

(b) At 2 = and r1 r2, we have = 0 and r = r1 r2, so that

Jn(r1 r2) =

k=

(1)Jnk(r1)1kJk(r2). (262)

(c) For = /2, we have

(r1 + ir2r1 ir2 )

n2

n2+1Jn(

r21 + r

22) =

k=

ikJnk(r1)1+kJk(r2). (263)

(d) For r = r1 = r2 we have

k=

Jn+k(r)Jk(r) = Jn(0) =1, n = 0

0, n = 0 (264)

18.2 Theorem of multiplication

Multiplying both sides of equation (259) by exp{i3/2m2}/2 and inte-grating over 2 in the range (0, 2), we have

2

2

2

0

ei(nm2Jn(r)d2 = Jnm(r1)Jm(r2), (265)

where r, r1, r2, , and 2 are defined according to (256)-(258). Here, wehave used the fact that exp{i3/2n2} are orthogonal so that all the termsare zero except for those with k = m.

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The equation (265) represents the theorem of product for isoBessel func-

tions.Let us consider specific case of the theorem characterized by r1 = r2 = R.

From this condition it follows that r = 2R2gg1/211 cos[

22 ] and = 2, so

that

Jnm(r1)Jm(r2) =2

2

0

ei(n2m)Jn(2R2g1/211 cos[

22

])d. (266)

Replacing the variable in the above integral by r, we note that when 2varies from 0 to the variable r varies from r1 + r2 to |r1 r2|, while when2 varies from to 2 the variable r varies from |r1 r2| to r1 + r2. Inaddition,

dr

d2=

4r21r

22 (r2 r21 r22)

2r, (267)

where minus and plus signs correspond to 0 2 and 2 2respectively. Thus, we have

Jnm(r1)Jm(r2) =22

r1+r2|r1r2|

ei(nm2Jn(r)rdr4r21r

22 (r2 r21 r22)2

, (268)

where and 2 are related to r according to (256)-(258).At m = n = 0 the formula (268) takes the most simple form,

J0(r1)J0(r2) =22

r1+r2/r1r2/

J0(r)rdr4r21r

22 (r2 r21 r22)

. (269)

19 Recurrency relations for Jn(z)

As it for isoLegendre functions Plmn(z), recurrency relations for isoBesselfunctions follow from the composition theorem. Namely, we should first putr2 in this theorem to be infinitesimal.

Let us find derivatives of the isoBessel function on x at the point x = 0.

Differentiating (241) we have

Jn(0) =i2

2

20

exp i2ng1/222 sin[]d (270)

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=

2

420 [exp i

2

(n 1) exp i

2

(n + 1)]d

.

This integral is non-zero only when n = 1. Also,

J1(0) = J1(0) =

1

2. (271)

Differentiating both sides of (263) on r2 and putting r2 = 0 we find

2Jn(x) = Jn1(x) Jn+1(x). (272)

Here, we used (271) and replace r by x.Similarly, from (265) we find

2n

xJn(x) = Jn1(x) + Jn+1(x). (273)

Combining (272) and (273) we finally obtain

Jn1(x) =n

xJn(x) + J

n(x), (274)

Jn+1(x) =n

xJn(x) Jn(x) (275)

. These formulas can be presented also in the following form:

Jn1(x) = (n

x +d

dx )Jn(x), (276)

Jn+1(x) = (n

x d

dx)Jn(x). (277)

20 Relations between IsoBessel functions and Plmn(z)

20.1 IsoEuclidean plane and sphere

Two-dimensional sphere can be mapped to isoEuclidean plane in a standardway. Namely, this can be done in taking the limit R for the radius ofthe sphere. Accordingly, M(2) can be considered as some limit of SO(3).More precisely, replacing , , , 1, 2, and 2 by , r/R, , r1/R, r2/R,and 1 in (135) defining multiplications in SO(3) we should retain leadingterms in the limit R . Simple calculations show that the result isexactly the formulas (211)-(214) defining multiplications in M(2).

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20.2 IsoBessel and isoJacobi functions

The relation between the groups M(2) and SO(3) makes it possible to relatematrix elements of its irreducible isounitary representations. Thus, isoBesselfunctions, as matrix elements of representations TI(g) of M(2), can be

derived from Plmn, which are matrix elements of representations Tl(g) ofSO(3). The limiting procedure is R and l .

To obtain concrete formulas we note first that Plmn has the integralrepresentation,

Plmn(g1/211 cos[]) = (

7

2)

(l m)!(l + m)!(l n)!(l + n)!

20

d (g1/211 cos[

2]ei/2

(278)+ig

1/222 sin[

2]ei/2)(g

1/222 sin[

2]ei/2 + ig

1/211 cos[

2]ei/2)eim.

Putting = r/l and taking the limit l we find

liml

Plmn(g1/211 cos[

r

l]) =

2(l+1)

2

20

(1 +ir5/2

2lexp i)ln (279)

(1 + ir5/2

2lexp i)exp i3/2(m n)d(1/2).

Note that at m = n = 0 the above relation takes the following simple

form:liml

Pl(g1/211 cos[

rl

1/2]) = J0(r), (280)

so that J0(r) appears as the limit from the isoLegendre polynomial.

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References

[1] R.Mignani, Lett. Nuovo Cim. 39, 406 (1984); 43, 355 (1985); HadronicJ. 9, 103 (1986).

[2] R.M.Santilli, Hadronic J. Suppl.4B, issue no.2 (1989).

[3] R.M.Santilli, Elements of hadronic mechanics, Vol. 1, 2 (NaukovaDumka, Kiev, 1994).

[4] N.Ya.Vilenkin, Special functions and theory of representations of groups.(Nauka, Moscow, 1978).

[5] A.G.Sitenko, Scattering theory (Naukova Dumka, Kiev 1975).

[6] A.K.Aringazin, D.A.Kirukhin, and R.M.Santilli, Nonpotential two-body

elastic scattering problem, Hadronic J 18 (1995).

[7] A.K.Aringazin, D.A.Kirukhin, R.M.Santilli, Nonpotential elastic scat-tering of spinning particles, Hadronic J 18 (1995).

A.K.Aringazin, D.A.Kirukhin and R.M.Santilli- Isotopic Generalization of the Legendre, Jacobi, and Bessel Functions

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