EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 16, No. 4, 2023, 2348-2367
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

The SL2(R) Group Representations on Spaces of
Holomorphic Functions on the Unit Disc

Amjad Saleh Alghamdi

Department of Mathematics, Jamoum University Collage, Umm Al-Qura University,
Saudi Arabia

Abstract. We can realise the representations of the group SL2(R) on the unit disc. This is
due to the isomorphism between the group SL2(R) and the group SU(1, 1). The discrete series
representations for the group SL2(R) given by

πn(g)φ(z) = φ

(
dz − b

a− cz

)
(a− cz)−n, n ∈ Z. (1)

are on the Bergman space where n ≥ 2 [5, 6, 10]. Lang [13, IX] studied the discrete series on the
group SL(R) in the upper half-plane and on the unit disc. For n = 1, the SL2(R) representation is
called the mock discrete series. The representation space of the mock discrete series is the Hardy
space [5, 6, 10]. In this paper we describe the SL2(R) representation on the Dirichlet space.

2020 Mathematics Subject Classifications: 32A10, 31C25
Key Words and Phrases: SL2(R) group, representations, Dirichlet space,SU(1, 1) group

1. Introduction

The Lie group SL2(R) consists of 2 × 2 matrices with real entries and a determinant
equal to one

SL2(R) =
{(

a b
c d

)
: ad− bc = 1, a, b, c, d ∈ R

}
.

It acts on the upper half-plane by Möbius transformation

g · z = az + b

cz + d
,

where g ∈ SL2(R) and z ∈ {z ∈ C : Imz > 0}.
The group SL2(R) contains the following three subgroups:

K =

{(
cos θ sin θ
− sin θ cos θ

)
: θ ∈ R

}
,

DOI: https://doi.org/10.29020/nybg.ejpam.v16i4.4923

Email address: asmghamdi@uqu.edu.sa (A. S. Alghamdi)

https://www.ejpam.com 2348 © 2023 EJPAM All rights reserved.



A. S. Alghamdi / Eur. J. Pure Appl. Math, 16 (4) (2023), 2348-2367 2349

A =

{(
α 0
0 α−1

)
: α > 0

}
,

N =

{(
1 x
0 1

)
: x ∈ R

}
.

The Lie algebra sl2(R) is the set of all 2× 2 real matrices of trace zero. It is a three-
dimensional Lie algebra so we can choose a basis{Z,A,B} of sl2(R) by setting

Z =

(
0 1
−1 0

)
, A =

1

2

(
−1 0
0 1

)
and B =

1

2

(
0 1
1 0

)
. (2)

Note that

[Z,A] = 2B, [Z,B] = −2A, [A,B] = −1

2
Z. (3)

2. The Group SU(1, 1)

The Cayley transform of the upper-half plane to the unit disc D is defined by

w =
z − i

z + i
, (4)

where x ∈ D and z ∈ {z ∈ C, Imz > 0}.
By the transformation (4) we can transfer the action of the group SL2(R) from the upper
half-plane to the action of the group SU(1, 1) on the unit disc, where

SU(1, 1) =

{(
α β
β α

)
: α, β ∈ C, |α|2 − |β|2 = 1

}
.

Furthermore, the matrix

(
a b
c d

)
∈ SL2(R) can be an element of the group SU(1, 1) by

the following identity:

1√
2

(
1 −i
−i 1

)(
a b
c d

)
1√
2

(
1 i
i 1

)
=

(
α β
β α

)
. (5)

Next, any g ∈ SU(1, 1) has a unique decomposition of the form(
α β
β α

)
= |α|

(
1 βα−1

βα−1 1

)( α
|α| 0

0 α
|α|

)

=
1√

1− |u|2

(
1 u
u 1

)(
eiθ 0
0 e−iθ

)
,

(6)

where θ = argα, u = βα−1 and |u| < 1 ( since |α|2 − |β|2 = 1). Let u = reiϕ, then
the identity (6) describes an element g ∈ SU(1, 1) by a triplet of numbers (r, ϕ, θ) where
0 ≤ r < 1 and −π < ϕ, θ ≤ π. The connection with the (α, β) coordinates is as follows:

α =
eiθ√

1− |r|2
, β =

rei(θ−ϕ)√
1− |r|2

,



A. S. Alghamdi / Eur. J. Pure Appl. Math, 16 (4) (2023), 2348-2367 2350

r =

∣∣∣∣βα
∣∣∣∣ , ϕ = − arg

β

α
, θ = argα.

Moreover, the decomposition (6) can be rewritten with the same variables as(
α β
β α

)
=

(
ei

ϕ
2 0

0 e−iϕ
2

) 1√
1−|r|2

r√
1−|r|2

r√
1−|r|2

1√
1−|r|2

(ei(θ−ϕ
2
) 0

0 e−i(θ−ϕ
2
)

)
(7)

The last presentation is a decomposition of the group SU(1, 1) as the product KAK of its
subgroups, which is called the Cartan decomposition. The base of the Lie algebra sl2(R)
consists of the following three matrices:

Z̃ =

(
i 0
0 −i

)
, Ã =

(
0 − i

2
i
2 0

)
and B̃ =

(
0 1

2
1
2 0

)
. (8)

The matrices Z̃, Ã and B̃ satisfy the commutation relation (3). Also, the exponential map
of each matrix generates a one-dimensional subgroup of the SU(1, 1) group, that is

eθZ̃ =

(
eiθ 0
0 e−iθ

)
, (9)

eθà =

(
cosh θ

2 −i sinh θ
2

i sinh θ
2 cosh θ

2

)
, (10)

eθB̃ =

(
cosh θ

2 sinh θ
2

sinh θ
2 cosh θ

2

)
. (11)

3. Induced Representation on the Unit Disc

In this section, we induce a representation of the group SU(1, 1) from the subgroup K.
Mainly, we use the references [8, 11].

The one-dimensional compact subgroup K is defined as follows:

K =

{(
eiθ 0
0 e−iθ

)
, −π < θ ≤ π.

}
(12)

Using the decomposition (6) of any element g ∈ SU(1, 1), we can identify the homogeneous
space X = SU(1, 1)/K with the open unit disc D. Let the section s : D → SU(1, 1) be
defined as follows:

s : u 7→ 1√
1− |u|2

(
1 u
ū 1

)
. (13)

There is a natural projection map p : SU(1, 1) → D, which assigns to an element of SU(1, 1)
its equivalence class in SU(1, 1)/K:

p :

(
α β
β α

)
7→ β

α
. (14)



A. S. Alghamdi / Eur. J. Pure Appl. Math, 16 (4) (2023), 2348-2367 2351

Mapping r : SU(1, 1) → K associates f to the natural projection p, and the section s is
defined as follows:

r :

(
α β
β α

)
7→

(
α
|α| 0

0 α
|α|

)
(15)

For the homogeneous space SU(1, 1)/K defines a left action denoted by ”·” as follows:

g : u 7→ g · u = p(g ∗ s(u)), (16)

where ∗ is the multiplication of the group SU(1, 1).
The invariant measure dµ(u) on D comes from the decomposition dg = dµ(u)dk, where

dg and dk are the Haar measures on G = SU(1, 1) and K respectively. The measure dµ(u)
is given by

dµ(u) =
du ∧ dū

(1− |u|2)2
. (17)

Let χn : T → C be a character of the subgroup K ≃ T defined as follows:

χn(w) = wn, n ∈ Z. (18)

This character induces a representation of SU(1, 1) constructed in the Hilbert space Lχn
2 (SU(1, 1)),

consisting of the functions Fn : SU(1, 1) → C with the property

Fn

[(
α β
β α

)]
= χn

(
α

|α|

)
F

(
β

α

)
, (19)

where F ∈ L2(D). Then, the norm of the function Fn is defined as follows:

∥Fn∥2 =
∫
D
|F (u)|2 du ∧ dū

(1− |u|2)2
. (20)

The space Lχn
2 (SU(1, 1)) is invariant under the left shift of the SU(1, 1) group. The

restriction of the left shift on Lχn
2 (SU(1, 1)) is the left regular representation of SU(1, 1),

which can be written as follows:

[Λ(g)Fn](g
′) = Fn(g

−1 ∗ g′), (21)

where ∗ is a matrix multiplication.
The lifting map Lχn : L2(D) → Lχn

2 (SU(1, 1)) for the subgroup K and its character
χn is defined as follows:

[Lχnf ]

(
α β
β α

)
= χn

(
r

(
α β
β α

))
f

(
p

(
α β
β α

))
=

(
ᾱ

|α|

)n

f

(
β

α

)
.

(22)

The pulling map is given by the following:

P : Lχn
2 (SU(1, 1)) → L2(D),



A. S. Alghamdi / Eur. J. Pure Appl. Math, 16 (4) (2023), 2348-2367 2352

P(F (w, w̄)) = F (s(w)),

such that P ◦ Lχn = I and Lχn ◦ P = I.
Therefore, the representation πn : L2(D) → L2(D), which is induced by the character

χn is given by the following:[
πn

(
α β
β α

)]
= P ◦ Λ

(
α β
β α

)
◦ Lχn .

By simple calculation, we get:

[πn(g)f ](w, w̄) =
(α− βw)n

|α− βw|n
f

(
αw − β

α− βw
,
αw − β

α− βw

)
=

(
α− βw

α− βw̄

)n
2

f

(
αw − β

α− βw
,
αw − β

α− βw

)
.

(23)

[9] For n ∈ Z, an n-peeling is an isometry Pn : L2(D, dw) → L2(D, (1−|w|2)n−2dw∧dw̄)
defined as follows:

Pn : f(w) 7→ [Pnf ](w) =
f(w)

(1− |w|2)
n
2

, w = u+ iv. (24)

The representation (23) is intertwined π̆n◦Pn = Pn◦πn by the n-peeling with the following
representation:

[π̆n(g)f ](w) = (α− βw)−nf

(
αw − β

α− βw

)
, (25)

which is unitary in L2(D, (1 − |w|2)n−2dw ∧ dw̄). The demonstration of the intertwining
properties is based on the following analogue of identity for the unit disc :

1−
∣∣∣∣αw − β

α− βw

∣∣∣∣ = 1− |w|2

|α− βw|2
.

The matrix

(
α β
β α

)
∈ SU(1, 1) is transformed to

(
a b
c d

)
∈ SL2(R) by the identity (5).

Therefore, the representation ρ̆Kn can be transformed to a holomorphic representation of
the group SL2(R):

[π̆n(g)f ](z) = (d− bz)−nf

(
az − c

d− bz

)
, (26)

which is unitary on the upper half-plane where z = x + iy ∈ R2
+ with the measure

dµ(g) = dxdy
y2

.



A. S. Alghamdi / Eur. J. Pure Appl. Math, 16 (4) (2023), 2348-2367 2353

4. Actions of Ladder Operators

In this section, we study the left and right actions of the ladder operators for the
representation πn given by (23). First, the derived representations are given as follows:

[Ef ](w,w) =
d

dt
πn(e

tZ̃)f(w,w)|t=0

= [−inI − 2iw∂w + 2iw∂w̄]f(w,w),
(27)

[A1f ](w,w) =
d

dt
πn(e

tÃ)f(w,w)|t=0

=

[
ni

4
(w + w)I +

i

2
(1 + w2)∂w − i

2
(1 + w2)∂w̄

]
f(w,w),

(28)

[B1f ](w,w) =
d

dt
πn(e

tB̃)f(w,w)|t=0

=

[
n

4
(w − w̄)I +

1

2
(w2 − 1)∂w +

1

2
(w2 − 1)∂w̄

]
f(w,w).

(29)

The ladder operators are defined as

L+ = B1 − iA1 =
n

2
wI + w2∂w − ∂w,

L− = B1 + iA1 =
−n
2
wI + w̄2∂w − ∂w,

and satisfy the following relations:

[E,L±] = ±2iL±, [L+, L−] = −iE. (30)

The Casimir operator is given by

dπn(C) = E2 − 2[L+L− + L−L+]

= (ww − 1)[n2I + 2nw∂w − 2nw̄∂w + 4(ww − 1)∂w∂w].
(31)

The Casimir operator in the polar coordinate w = reiθ is as follows:

dπn(C) = (r2 − 1)(n2I − 2in∂θ)− (r2 − 1)2(∂2r + r−1∂r + r−2∂2θ ). (32)

Lemma 1. The operator (27) has two eigenfunctions:

(i) For m ̸= 2, 4, 6, 8......,

f−m
2
,n(w,w) = w−m

2 (1− ww̄)
1±

√
1−µ
2 F

(
1

2
[1 + n−m±

√
1− µ],

1

2
[1− n±

√
1− µ], 1− m

2
, ww

)
, (33)



A. S. Alghamdi / Eur. J. Pure Appl. Math, 16 (4) (2023), 2348-2367 2354

(ii) For m ̸= −2,−4,−6,−8......,

f̃−m
2
,n(w,w) = w

m
2 (1− ww)

1±
√

1−µ
2 F

(
1

2
[1− n+m±

√
1− µ],

1

2
[1 + n±

√
1− µ], 1 +

m

2
, ww

)
, (34)

where F is a hypergeometric function.

Proof.
To find the eigenfunction of the subgroup K, we will solve the following partial differ-

ential equation by using the method of characteristics:

[Ef ](w,w) = [−inI − 2iw∂w + 2iw∂w̄]f(w, w̄) = 0.

We can write the characteristics for this equation as follows:

df

inf
=

dw

−2iw
=

dw

2iw
.

dw

−2iw
=

dw̄

2iw̄
⇒ C1 = ww.

We need to obtain another integral curve that involves f . This is possible from the
following equation:

df

inf
=

dw

−2iw
, we get C2 = w

n
2 f.

Then, the general solution of (4) is of the form C2 = ϕ(C1), that is

f(w,w) = w−n
2 ϕ(ww̄).

Now, for m ∈ Z the eigenfunction is given by

f−m
2
(w,w) = w−m

2 ϕ(ww), (35)

which satisfies
[Efm](w,w) = i(m− n)fm(w,w). (36)

Therefore, the eigenvalue of the operator E is m− n.
Next,let w = reiθ. Then the eigenfunction (35) will be given by

f−m
2
(r, θ) = (reiθ)−

m
2 ϕ(r2).

The Casimir operator (32) is applied to f−m
2
(r, θ)

[dπn(C)f−m
2
,n](r, θ) = (reiθ)−

m
2
[
(r2 − 1)(n2 − nm)ϕ(r2)

− 2(r2 − 1)2((−m+ 2)ϕ′(r2) + 2r2ϕ
′′
(r2))

]
. (37)



A. S. Alghamdi / Eur. J. Pure Appl. Math, 16 (4) (2023), 2348-2367 2355

To find the value of ϕ in (35), we need to solve the differential equation

[dπn(C)f ](r, θ) = µf(r, θ).

That is,

[(r2 − 1)(n2 − nm)− µ]ϕ(r2)− 2(r2 − 1)2(−m+ 2)ϕ′(r2)

− 4(r2 − 1)2r2ϕ
′′
(r2) = 0. (38)

Let x = r2. Then we get

[(x− 1)(n2 − nm)− µ]ϕ(x)− 2(x− 1)2(−m+ 2)ϕ′(x)

− 4(x− 1)2x2ϕ
′′
(x) = 0. (39)

Now, let
ϕ(x) = xα(1− x)βψ(x).

Then by substitute ϕ(x) in (39), we get

α =
m

2
or 0, β =

1±
√
1− µ

2
.

Hence, we have two solutions:

(i) ϕ(x) = (1− x)
1±

√
1−µ
2 ψ(x),

(ii) ϕ(x) = x
m
2 (1− x)

1±
√
1−µ
2 ψ(x).

By substituting the first solution in the differential equation(39), we get

x(1− x)ψ
′′
(x) +

(
1− m

2
− (1±

√
1− µ+ 1− m

2
)x
)
ψ

′
(x)

+

[
µ

2
+

(
−1∓

√
1− µ

2

)(
1− m

2

)
+

1

4

(
n2 − nm

)]
ψ(x) = 0. (40)

This is a hypergeometric differential equation that takes the following form:

x(1− x)ψ
′′
(x) + [c− (a+ b+ 1)x]ψ

′
(x)− abψ(x) = 0.

By simple calculation, we get

a =
1

2
[1 + n−m±

√
1− µ],

b =
1

2
[1− n±

√
1− µ],

c = 1− m

2
.



A. S. Alghamdi / Eur. J. Pure Appl. Math, 16 (4) (2023), 2348-2367 2356

Then, ψ(x) = F (a, b, c, x), and the solution of (38) is

ϕ(r2) = (1− r2)
1±

√
1−µ
2 F (a, b, c, r2).

The hypergeometric function is given by

F (a, b, c, r2) = 1 +
∞∑
k=1

(a)k(b)k
(c)k

(r2)k

k!
.

Finally, the eigenfunction is given by

f−m
2
,n(w,w) = w−m

2 (1− ww)
1±

√
1−µ
2 F

(1
2
[1 + n−m±

√
1− µ],

1

2
[1− n±

√
1− µ], 1− m

2
, ww

)
,

where m ̸= 2, 4, 6, 8.......
Following the same calculation for the second solution, we get the eigenfunction

f̃−m
2
,n(w,w) = w

m
2 (1− ww̄)

1±
√
1−µ
2 F

(1
2
[1− n+m±

√
1− µ],

1

2
[1 + n±

√
1− µ], 1 +

m

2
, ww

)
,

where m ̸= −2,−4,−6,−8.......

The commutator relation [E,L±] = ±2iL±, implies that

[Ef−m
2
,n(w, w̄)] = e(m−n)iθf−m

2
,n(w, w̄).

Hence E = (m − n)i. Additionally, by using the relation [L+, L−] = −iE, and (31), we
get the following identities:

4L+L− = E2 − 2iE − dπn(C),

4L−L+ = E2 + 2iE − dπn(C).

Then, for dπn(C) = µI,

L+L− = −1

4
[(m− n− 1)2 + µ− 1], (41)

L−L+ = −1

4
[(m− n+ 1)2 + µ− 1]. (42)

Now, since L∗
+ = −L−, we have

∥L−∥ = ∥L∗
−L−∥

1
2 = ∥ − L+L−∥

1
2

=
1

2
[(m− n− 1)2 + µ− 1]

1
2 .

(43)



A. S. Alghamdi / Eur. J. Pure Appl. Math, 16 (4) (2023), 2348-2367 2357

Similarly,

∥L+∥ =
1

2
[(m− n+ 1)2 + µ− 1]

1
2 . (44)

Let 1− µ = (n− 1)2, where n is an integer. The functions (33) are given by

f−m
2
,n(w,w) = w

−m
2 (1− ww)

n
2 . (45)

The functions f−m
2
,n(w,w) = w

−m
2 (1−ww)

n
2 , are L2 summable for n > 1 and m ≤ 0.

Proof. Let w = reiθ, then f−m
2
,n(re

iθ, re−iθ) = (reiθ)
−m
2 (1 − r2)

n
2 . The measure is

dµ = rdr∧dθ
(1−r2)2

. Then,

∥f−m
2
,n∥2 =

∫ 2π

0

∫ 1

0
|f−m

2
,n(re

iθ, re−iθ)|2 rdr ∧ dθ
(1− r2)2

=

∫ 2π

0

∫ 1

0

∣∣∣∣(reiθ)−m
2 (1− r2)

n
2

∣∣∣∣2 rdr ∧ dθ(1− r2)2

=

∫ 2π

0

∫ 1

0
r−m(1− r2)n−2rdrdθ

≤ π

∫ 1

0
(1− r2)n−22rdr, for m ≤ 0

= −π (1− r2)n−1

n− 1

∣∣∣∣1
0

=
π

n− 1
<∞.

Hence, f−m
2
,n are L2 summable if n > 1 and m ≤ 0.

By simple calculation, we get

[L+f−m
2
,n](w,w) =

(
n− m

2

)
f−m

2
+1,n(w,w), (46)

[L−f−m
2
,n](w,w) =

m

2
f−m

2
−1,n(w,w). (47)

At m = 0, we have the function f0,n(w,w) = (1 − ww)
n
2 . Then, L−f0,n(w,w) = 0,

which means that f0,n(w,w) is the vacuum of the operator L−. This is represented by the
following diagram: Next, let 1− µ = (n+ 1)2. Then, the functions (34) are given by

0 f0,n f1,n f2,n · · ·
L−

L+

L−

L+

L−

L+

L−

f̃−m
2
,n(w,w) = w

m
2 (1− ww)

−n
2 , (48)



A. S. Alghamdi / Eur. J. Pure Appl. Math, 16 (4) (2023), 2348-2367 2358

which are L2 summable for n < −1 and m ≥ 0; that is,

∥f̃−m
2
,n∥2 =

∫
D

∣∣∣∣f̃−m
2
,n(w,w)

∣∣∣∣2 dw ∧ dw
(1− |w|2)

<∞, n < −1.

[L+f̃−m
2
,n](w,w) =

−m
2
f̃−(m

2
+1),n(w,w), (49)

[L−f̃−m
2
,n](w,w) =

(
m

2
− n

)
f̃−(m

2
−1),n(w,w). (50)

Atm = 0, we get the function f̃0,n(w,w) = (1−ww)
−n
2 . We can then see that [L+f̃0,n](w,w) =

0, which means that f̃0,n is the vacuum of the operator L+. This is represented by the
following diagram: The Lie derivatives of the representation πn are

· · · f̃−2,n f̃−1,n f̃0,n 0
L+

L−

L+

L−

L+

L−

L+

LZ̃ = −∂θ, (51)

LÃ =
−r
2

sin(ϕ− 2θ)∂θ −
1

2
(1− uū)

[
e2iθ∂u + e−2iθ∂ū

]
, (52)

LB̃ =
r

2
cos(ϕ− 2θ)∂θ −

i

2
(1− uū)

[
e2iθ∂u − e−2iθ∂ū

]
. (53)

The right ladder operators are then represented by

L+ = LÃ+iB̃ = e−2iθ

[
i

2
u∂θ − (1− uū)∂ū

]
, (54)

L− = LÃ−iB̃ = −e2iθ
[
i

2
ū∂θ + (1− uū)∂u

]
. (55)

Proof. The Lie derivative LX for an element X of the Lie algebra su(1, 1) is given by

[LXF ](g) =
d

dt
F (g exp tX)|t=0, (56)

for any differentiable function F on SU(1, 1) and g =

(
α β̄
β ᾱ

)
.

We know that the space Lχn
2 (SU(1, 1)) consists of the functions Fn : SU(1, 1) → C

with the property

Fn

[(
α β
β α

)]
= χn

(
α

|α|

)
F

(
β

α
,
β

α

)
,

where F ∈ L2(D).



A. S. Alghamdi / Eur. J. Pure Appl. Math, 16 (4) (2023), 2348-2367 2359

Hence, for v = α
|α| = eiθ and u = β

α = reiϕ, we have

[LXFn](g) =
d

dt
Fn(g exp tX)

∣∣∣∣
t=0

=
d

dt
χn(v(t))F (u(t), ū(t))

∣∣∣∣
t=0

=
∂χn

∂v

dv(t)

dt

∣∣∣∣
t=0

+
∂F

∂u

du(t)

dt

∣∣∣∣
t=0

+
∂F

∂ū

dū(t)

dt

∣∣∣∣
t=0

.

(57)

From section 2 we have Z̃, Ã and B̃ ∈ su(1, 1) given by (8). Then, the Lie deriva-
tives corresponding to the subgroups exp tZ̃,(9), exp tÃ,(10) and exp tB̃(11) are obtained
through the differentiation of the right action of these subgroups as follows:

[LZ̃Fn](g) =
d

dt
Fn(g exp tZ̃)

∣∣∣∣
t=0

=
d

dt
Fn

(
αeit β̄e−it

βeit ᾱe−it

) ∣∣∣∣
t=0

=
d

dt
χn

(
αeit

|αeit|

)
F

(
β̄e−it

ᾱe−it
,
βe−it

αe−it

) ∣∣∣∣
t=0

=
d

dt
χn(e

i(θ+t))F (u, ū)

∣∣∣∣
t=0

= −∂F
∂θ
,

where α = eiθ√
1−|r|2

and β = rei(θ−ϕ)√
1−|r|2

.

Similarly, it is easy to determine that

[LÃFn](g) =
d

dt
Fn(g exp tÃ)

∣∣∣∣
t=0

=
−r
2

sin(ϕ− 2θ)
∂F

∂θ
− 1

2
(1− uū)

[
e2iθ

∂F

∂u
+ e−2iθ ∂F

∂ū

]
.

[LB̃Fn](g) =
d

dt
Fn(g exp tB̃)

∣∣∣∣
t=0

=
r

2
cos(ϕ− 2θ)

∂F

∂θ
− i

2
(1− uū)

[
e2iθ

∂F

∂u
− e−2iθ ∂F

∂ū

]
.

The function f−m
2
,n given by (45) is an eigenfunction with an eigenvalue in for the

operator LZ̃ . That is, for
Fn(g) = eintf−m

2
,n(w,w),



A. S. Alghamdi / Eur. J. Pure Appl. Math, 16 (4) (2023), 2348-2367 2360

we have
LZ̃eintf−m

2
,n(w,w) = ineintf−m

2
,n(w,w). (58)

Moreover, f̃−m
2
,n (48) is an eigenfunction with an eigenvalue in for the operator LZ̃ .

Lemma 2. We have
LñiB̃ : f−m

2
,n → f−m

2
,n±2,

and
LñiB̃ : f̃−m

2
,n → f̃−m

2
,n±2.

Proof. From the commutator relations [LZ̃ ,LñiB̃] = ±2iLñiB̃, for the eigenfunction
f−m

2
,n given by (45), we can see that

[LZ̃LñiB̃]eintf−m
2
,n = LñiB̃(LZ̃eintf−m

2
,n)± 2iLñiB̃eintf−m

2
,n

= LñiB̃(nieintf−m
2
,n)± 2iLñiB̃eintf−m

2
,n

= (n± 2)iLñiB̃eintf−m
2
,n.

(59)

Similarly,for the eigenfunction f̃−m
2
,n (48), we have

[LZ̃LñiB̃]eintf̃−m
2
,n = (n± 2)iLñiB̃eintf̃−m

2
,n.

The vacuum f0,n(w,w) = (1 − ww)
n
2 is annihilated by the operator LÃ+iB̃. That

is, [LÃ+iB̃eintf0,n](w,w) = 0. Then, all the vectors fj,n = (L+)
jf0,n are vacuums of the

operator LÃ+iB̃ due to the commutation of the left and right actions:

LÃ+iB̃fj,n = LÃ+iB̃(L+)
jf0,n

= (L+)
jLÃ+iB̃f0,n = 0.

(60)

For each vacuum f0,n, the collection of vectors fj,n = (L+)
jf0,n forms an orthogonal basis

of an irreducible component with the respective ladder operators (46) and (47). The left
and the right actions for the eigenfunctions fm,n (45) jointly create the two-dimensional
lattice structure that can be seen in the following diagram:



A. S. Alghamdi / Eur. J. Pure Appl. Math, 16 (4) (2023), 2348-2367 2361

0 f0,2 f0,4 f0,6 · · ·

f1,2 f1,4 f1,6 · · ·0

f2,2 f2,4 f2,6 · · ·0

000

...
...

...

L+

L−

L+

L−

L+

L−

L−

L+

L−

L+

L−

L+

L−

L−

L+

L−

L+

L−

L+

L−

L−

L− L− L−

L−L+

L−L+

L−L+

L−L+

L−L+

L−L+

L−L+

L−L+

L−L+

Figure 1: The left and the right actions of the ladder operators for f−m
2

,n
.

Furthermore, the function f̃0,n(w,w) = (1−ww)
−n
2 is a vacuum of the operator LÃ−iB̃.

That is, [LÃ−iB̃eintf̃0,n](w,w) = 0. Then, all the vectors f̃k,n = (L−)
kf̃0,n are vacuums of

the operator LÃ−iB̃ due to the commutation of the left and right actions:

LÃ−iB̃ f̃k,n = LÃ−iB̃(L−)
kf0,n

= (L−)
kLÃ−iB̃ f̃0,n = 0.

(61)

For each f̃0,n, the collection of vectors f̃k,n = (L−)
kf0,n forms an orthogonal basis of an

irreducible component with the respective ladder operators (49) and (50). The left and the
right actions for the functions f̃m,n (48) jointly create the two-dimensional lattice structure
that can be seen in the following diagram:



A. S. Alghamdi / Eur. J. Pure Appl. Math, 16 (4) (2023), 2348-2367 2362

· · · f̃0,6 f̃0,4 f̃0,2 0

0

f̃−1,6

0

f̃−1,4

0

f̃−1,2 0· · ·

f̃−2,6 f̃−2,4 f̃−2,2· · · 0

...
...

...

L+

L−

L+

L−

L+

L−

L+

L+

L−L+

L+

L−L+

L+

L−L+

L−

L+ L+

L−

L+

L−

L+

L−

L+ L+

L−

L+

L−

L+

L−L+ L−L+ L−L+

L−L+ L−L+ L−L+

Figure 2: The left and the right actions of the ladder operators for f̃−m
2

,n
.

5. Representation on the Dirichlet Space

The Dirichlet space, the Hardy space and the Bergman space are the three classical
spaces of holomorphic functions in the unit disc. In the present section, we find the su(1, 1)
module (which is the space of the derived representation) on the Dirichlet space.

[4] The Dirichlet space D on the unit disc D = {w : |w| < 1} consists of the holomorphic
functions f(w) on D, for which the following semi-norm is finite:

D(f) :=

(
1

π

∫
D
|f ′(w)|2dxdy

) 1
2

, w = x+ iy. (62)

For g =

(
α β
β α

)
∈ SU(1, 1), the SU(1, 1) representation on the Dirichlet space is

defined as follows:

[π̆0(g)f ](w) = f

(
αw − β

α− βz

)
. (63)

The semi-norm D(f) is not a norm because D(f) = 0 whenever f is a constant. Then, π̆0
is a non-unitary representation.

Lemma 3. The Dirichlet space has two su(1, 1) vector module:

• the lowest weight vector module V0+2m = {w0,m : m = 0, 1, 2, 3.....}, with the follow-
ing ladder operators

L+w0,m = imw0,m+1, m ∈ Z+ − {0},
L−w0,m = imw0,m−1, m ∈ Z+ − {0},



A. S. Alghamdi / Eur. J. Pure Appl. Math, 16 (4) (2023), 2348-2367 2363

L+w0,0 = 0,

• the highest weight vector module V̄0+2m = {w0,m : m = 0, 1, 2, 3.....},with the follow-
ing ladder operators

L+w0,m = imw0,m+1, m ∈ Z+ − {0},
L−w0,m = imw0,m−1, m ∈ Z+ − {0},
L−w0,0 = 0,

Proof.
The representation π̆0 (63) is the SU(1, 1) representation π̆n, for n = 0. The represen-

tation π̆n is defined as follows:

[π̆n(g)f ](w) = f

(
αw − β

α− βw

)
(α− βw)−n, (64)

where n ∈ Z. The derived representations for the basis {Z̃, Ã, B̃} (8) are as follows:

E = dπ̆Z̃n =
d

dt
π̆n(e

tZ̃)f(w)|t=0 = [−inI − 2iw∂w]f(w),

A1 = dπ̆Ãn =
d

dt
π̆n(e

tÃ)f(w)|t=0 =
i

2
[nwI + (1 + w2)∂w]f(w),

B1 = dπ̆B̃n =
d

dt
π̆n(e

tB̃)f(w)|t=0 =
1

2
[nwI + (w2 − 1)∂w]f(w).

The commutator relations are

[E,A1] = 2B1, [E,B1] = −2A1, [A1, B1] = −1

2
E.

The ladder operators are defined as

L+ = A1 + iB1 = inwI + iw2∂w, L− = A1 − iB1 = i∂w,

and
[E,L+] = −2iL+, [E,L−] = 2iL− and [L+, L−] = iE. (65)

The Casimir operator is

dπ̆n(C) = dπ̆n(Z̃
2 − 4Ã2 − 4B̃2) = −n2 + 2n. (66)

The representation π̆n on L2(D) is irreducible, and Vn+2m is the one-dimensional subspace
generated by wn,m [13]. Indeed,

π̆n

((
eiθ 0
0 e−iθ

))
(wn,m) = e−iθ(n+2m)wn,m.



A. S. Alghamdi / Eur. J. Pure Appl. Math, 16 (4) (2023), 2348-2367 2364

Hence, Vn+2m is an eigenspace of K with an eigenvalue e−iθ(n+2m), which is the character
of the subgroup K. Then

π̆n(exp tZ̃) = e−i(n+2m)tI on Vn+2m,

and the derived representation is given by

E = −i(n+ 2m)I on Vn+2m.

From the commutator relation (65), we have

E(L+wn,m) = L+(Ewn,m)− 2iL+wn,m

= L+(−i(n+ 2m))− 2iL+ = −i(n+ 2m+ 2)L+,

E(L−wn,m) = L−(Ewn,m) + 2iL−wn,m

= L−(−i(n+ 2m)) + 2iL+ = −i(n+ 2m− 2)L−.

Therefore, the ladder operator L± acts as follows:

L+ : Vn+2m → Vn+2m+2, L− : Vn+2m → Vn+2m−2.

· · · Vn+2m−2 Vn+2m Vn+2m+2 · · ·
L+

L−

L+

L−

L+

L−

L+

L−

Vn+2m = {wn,m : m = 0, 1, 2, 3.....} is the lowest weight module and is given as follows:

Ewn,m = −(n+ 2m)iwn,m,

L+wn,m = A1wn,m + iB1wn,m = (n+m)iwn,m+1, m ∈ Z+ − {0},
L−wn,m = A1wn,m − iB1wn,m = miwn,m−1, m ∈ Z+ − {0}
L−wn,0 = 0,

dπ̆n(C)w = (−n2 + 2n)w, w ∈ Vn+2m.

The vector wn,0 is called the lowest weight vector.

0 wn,0 wn,1 wn,2 · · ·
L−

L+

L−

L+

L−

L+

L−

V̄n+2m = {wn,m : m = 0, 1, 2, 3.....} is the highest weight module and is given as follows:

Ewn,m = −(n− 2m)iwn,m,

L−wn,m = A1wn,m + iB1wn,m = i(n+m)wn,m−1, m ∈ Z+ − {0},
L+wn,m = A1wn,m − iB1wn,m = imwn,m+1, m ∈ Z+ − {0}



A. S. Alghamdi / Eur. J. Pure Appl. Math, 16 (4) (2023), 2348-2367 2365

· · · wn,2 wn,1 wn,0 0
L+

L+

L−

L+

L−

L+

L−

L+wn,0 = 0,

dπ̆n(C)w = (−n2 + 2n)w, w ∈ Vn−2m.

The vector wn,0 is called the highest weight vector.
The vector module Vn+2m is unitarisable if and only if n > 0, and V̄n+2m is unitarisable

if and only if n < 0 [7, p.96].
Next, for the Dirichlet space the su(1, 1) vector module is V0+2m, which is given as

follows:

Ew0,m = −2imw0,m,

L+w0,m = imw0,m+1, m ∈ Z+ − {0},
L−w0,m = imw0,m−1, m ∈ Z+ − {0},
L+w0,0 = 0,

dπ̆0(C) = 0.

This is shown in the following figure:

0 [w0,0] w0,1 w0,2 · · ·
L− L−

L+

L−

L+

L−

In addition, w0,0 is the highest weight vector for the vector module V̄0+2m which is
given by

Ew0,m = −2imw0,m,

L+w0,m = imw0,m+1, m ∈ Z+ − {0},
L−w0,m = imw0,m−1, m ∈ Z+ − {0},
L−w0,0 = 0,

dπ̆0(C) = 0.

and presented in the following figure:

· · · w0,2 w0,1 [w0,0] 0
L+

L+

L−

L+

L−

L+



REFERENCES 2366

6. Conclusion

This paper considers the discrete series representation of the SL2(R) group πn defined
by (1) on spaces of holomorphic functions on the unit disc. The group SL2(R) is more
convenient for complex analysis in the upper half-plane. However, the group SU(1, 1) of
2 × 2 matrices, with complex entries and a determinant equal to one, is well suited in
unit disc D. The Cayley transform (4) defines an isomorphism of the group SL2(R) with
the group SU(1, 1). We present the action of the ladder operator of representation on the
su(1, 1) Lie algebra. This action can be realised on the Bergman space for n ≥ 2, on
the Hardy space for n = 1 and on the Dirichlet space for n = 0. The vector module
of the representation on the Dirichlet space has been described. It is worth to studying
the su(1, 1) vector module on other spaces of holomorphic function, for instance the Poly-
Bergman space [16].

Acknowledgements

I would like to thank Dr. Vladimir Kisil for all his guidance, valuable advises and
comments. Also, my deep gratitude to my family for their continuous love, help and
support. I would like to thank my husband for his patience and kindness. Thanks for my
kids who have made me stronger and filled me with happiness.

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