Acta Polytechnica


doi:10.14311/AP.2013.53.0399
Acta Polytechnica 53(5):399–404, 2013 © Czech Technical University in Prague, 2013

available online at http://ojs.cvut.cz/ojs/index.php/ap

EXTREMAL VECTORS FOR VERMA TYPE
REPRESENTATION OF B2

Čestmír Burdíka,∗, Ondřej Navrátilb
a Department of Mathematics, Czech Technical University in Prague, Faculty of Nuclear Sciences and Physical

Engineering, Trojanova 13, 120 00 Prague 2, Czech Republic
b Department of Mathematics, Czech Technical University in Prague, Faculty of Transportation Sciences, Na

Florenci 25, 110 00 Prague, Czech Republic
∗ corresponding author: burdices@kmlinux.fjfi.cvut.cz

Abstract. Starting from the Verma modules of the algebra B2 we explicitly construct factor
representations of the algebra B2 which are connected with the unitary representation of the group
SO(3, 2). We find a full set of extremal vectors for representations of this kind. So we can explicitly
resolve the problem of the irreducibility of these representations.
Keywords: Verma modules, height-weight representation, reducibility, extremal vectors.
Submitted: 16 July 2013. Accepted: 28 July 2013.

1. Introduction
Representations of Lie algebras are important in many
physical models. It is therefore useful to study various
methods for constructing them.

The general method of construction of the highest-
weight representation for the semisimple Lie algebra
was developed in [1, 2]. The irreducibility of such rep-
resentations (now called Verma modules) was studied
by Gelfand in [3]. The theory of these representations
is included in Dixmier’s book [4].

In the 1970’s prof. Havlíček with his coworkers dealt
with the construction of realizations of the classical
Lie algebras, see [5]. Our aim in this paper is to
show how one can use realizations of the Lie algebra
to construct so called extremal vectors of the Verma
modules. To work with a specific Lie algebra, we
choose Lie algebra so(3, 2), which plays an important
role in physics, e.g. in AdS/CFT theory, see [6, 7].
In the construction of the Verma modules for B2,

the representations depend on parameters (λ1,λ2).
For connection with irreducible unitary representa-
tions of SO(3, 2) we take λ2 ∈ N0, and in section 3 we
explicitly construct the factor-Verma representation.
Further, we construct a full set of extremal vectors.
These vectors are called subsingular vectors in [8].

In this paper, we use an almost elementary par-
tial differential equation approach to determine the
extremal vectors in any factor-Verma module of B2.
It should be noted that our approach differs from a
similar one used in [9]. First, we identify the factor-
Verma modules with a space of polynomials, and the
action of B2 on the Verma module is identified with
differential operators on the polynomials. Any ex-
tremal vector in the factor-Verma module becomes a
polynomial solution of a system of variable-coefficient
second-order linear partial differential equations.

2. The root system for Lie
algebra B2

In the Lie algebra g = B2 we will take a basis composed
by elements H1, H2, Ek and Fk, where k = 1, . . . , 4,
which fulfill the commutation relations

[H1, E1] = 2E1, [H1, E2] = −E2,
[H1, E3] = E3, [H1, E4] = 0,
[H2, E1] = −2E1, [H2, E2] = 2E2,
[H2, E3] = 0, [H2, E4] = 2E4,
[H1, F1] = −2F1, [H1, F2] = F2,
[H1, F3] = −F3, [H1, F4] = 0,
[H2, F1] = 2F1, [H2, F2] = −2F2,
[H2, F3] = 0, [H2, F4] = −2F4,
[E1, E2] = E3, [E1, E3] = 0,
[E1, E4] = 0, [E2, E3] = 2E4,
[E2, E4] = 0, [E3, E4] = 0,
[F1, F2] = −F3, [F1, F3] = 0,
[F1, F4] = 0, [F2, F3] = −2F4,
[F2, F4] = 0, [F3, F4] = 0,
[E1, F1] = H1, [E1, F2] = 0,
[E1, F3] = −F2, [E1, F4] = 0,
[E2, F1] = 0, [E2, F2] = H2,
[E2, F3] = 2F1, [E2, F4] = −F3,
[E3, F1] = −E2, [E3, F2] = 2E1,
[E3, F3] = 2H1 + H2, [E3, F4] = F2,
[E4, F1] = 0, [E4, F2] = −E3,
[E4, F3] = E2, [E4, F4] = H1 + H2.

We can take as h the Cartan subalgebra with the bases
H1 and H2.
We will denote λ = (λ1,λ2) ∈ h∗, for which we

have
λ(H1) = λ1, λ(H2) = λ2.

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Č. Burdík, O. Navrátil Acta Polytechnica

The root systems g = B2 with respect to these bases
H1 and H2 are R =

{
±αk ; k = 1, 2, 3, 4

}
, where

α1 = (2,−2), α2 = (−1, 2),
α3 = α1 + α2 = (1, 0), α4 = α1 + 2α2 = (0, 2).

If we choose positive roots R+ = {α1, α2, α3, α4
}
,

the basis in root system R is B =
{

α1, α2
}
.

If we define H3 = 2H1 + H2 and H4 = H1 + H2,
the following relations

[Hk, Ek] = 2Ek, [Hk, Fk] = −2Fk, [Ek, Fk] = Hk

are valid for any k = 1, . . . , 4.

3. The extremal vectors for
Verma type representation

We denote by n+, and n− the Lie subalgebras gen-
erated by elements Ek, and Fk, respectively, where
k = 1, . . . , 4, and b+ = h+ n+. Let us further consider
λ = (λ1,λ2) ∈ h∗ the one-dimensional representation
τλ for the Lie algebra b+ such that for any H ∈ h and
E ∈ n+

τλ(H + E)|0〉 = λ(H)|0〉.

The element |0〉 will be called the lowest-weight vector.
Let further be

W(λ) = U(g) ⊗U(b+) C|0〉,

where b+-module C|0〉 is defined by τλ.
It is clear that W (λ) ∼ U(n−)|0〉 and it is the U(g)-

module for the left regular representation, which will
be called the Verma module. 1
It is a well-known fact that every U(g)-submodule

of the module W(λ) is isomorphic to module W(µ),
where

µ = λ −n1α1 −n2α2,

for n1,n2 ∈ N0 = {0, 1, 2, . . .}. For the lowest-weight
vector of the representation W(µ) ⊂ W(λ), |0〉µ, is
fulfilled

H|0〉µ = µ(H)|0〉µ, H ∈ h, E|0〉µ = 0, E ∈ n+.

Such vectors |0〉µ will be called extremal vectors
W(λ).

From the well-known result for the Verma modules
we know that the Verma module W(λ) is irreducible
iff

λ1 /∈ N0, λ2 /∈ N0,
λ1 + λ2 + 1 /∈ N0, 2λ1 + λ2 + 2 /∈ N0.

If λ1 ∈ N0, resp. λ2 ∈ N0, then the extremal vectors
are

Fλ1+11 |0〉 = |0〉µ1, resp. F
λ2+1
2 |0〉 = |0〉µ2,

1In Dixmier’s book the Verma module M(λ) is defined with
respect to τλ−δ, where δ = 12

∑4
k=1αk = (1, 1). So we have

W(λ) = M(λ + δ).

where

µ1 = λ − (λ1 + 1)α1 = (−λ1 − 2, 2λ1 + λ2 + 2),
µ2 = λ − (λ2 + 1)α2 = (λ1 + λ2 + 1,−λ2 − 2). (1)

If W(µ) is a submodule W(λ), we will define the
U(g)-factor-module

W(λ|µ) = W(λ)/W(µ).

Now we can study the reducibility of a representation
like that.
Again, the extremal vector is called any nonzero

vector v ∈ W (λ|µ) for which there exists ν ∈ h∗ such
that

Hkv = νkv, Ekv = 0, k = 1, 2. (2)
It is clear that Ekv = 0 for k = 1, 2, 3, 4.
In this paper, we find all such extremal vectors in

the space W(λ|µ2), where λ2 ∈ N0 and µ2 is given
by (1).

4. Differential equations for
extremal vectors

Let λ2 ∈ N0 and µ2 be given by equation (1). It is
easy to see that the basis in the space W(λ|µ2) is
given by the vectors

|n〉 = |n1,n3,n4,n2〉 = (λ2 −n2)! Fn11 F
n3
3 F

n4
4 F

n2
2 |0〉,

where n1,n3,n4 ∈ N0 and n2 = 0, 1, . . . ,λ2.2
Now by direct calculation we obtain

H1|n〉 = (λ1 − 2n1 + n2 −n3)|n〉,
H2|n〉 = (λ2 + 2n1 − 2n2 − 2n4)|n〉,
E1|n〉 = n1(λ1 −n1 + n2 −n3 + 1)|n1 − 1,n3,n4,n2〉

− (λ2 −n2)n3|n1,n3 − 1,n4,n2 + 1〉
+ n3(n3 − 1)|n1,n3 − 2,n4 + 1,n2〉,

E2|n〉 = n2|n1,n3,n4,n2 − 1〉
+ 2n3|n1 + 1,n3 − 1,n4,n2〉

−n4|n1,n3 + 1,n4 − 1,n2〉. (3)

It is possible to rewrite the action by the second order
differential operators (see [10, 11]) on the polynomial
functions z1, z2, z3 a z4, which are in variable z2 up
to the level λ2. If we put

|n1,n3,n4,n2〉 = (λ2 −n2)! Fn11 F
n3
3 F

n4
4 F

n2
2 |0〉

↔ zn11 z
n2
2 z

n3
3 z

n4
4 ,

we obtain from equations (3) for the action on poly-
nomials f = f(z1,z2,z3,z4)

H1f = λ1f − 2z1f1 + z2f2 −z3f3,
H2f = λ2f + 2z1f1 − 2z2f2 − 2z4f4,
E1f = λ1f1 −z1f11 + z2f12 −z3f13

−λ2z2f3 + z22f23 + z4f33,
E2f = f2 + 2z1f3 −z3f4, (4)

2If λ2 /∈ Z we can use a similar construction with basis |n〉 =
Γ(λ2 −n2 + 1)Fn11 F

n3
3 F

n4
4 F

n2
2 |0〉, where n1,n2,n3,n4 ∈ N0.

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vol. 53 no. 5/2013 Extremal Vectors for Verma Type Representation of B2

where fk =
∂f

∂zk
.

The conditions for extremal vectors (2) are now

λ1f − 2z1f1 + z2f2 −z3f3 = ν1f,
λ2f + 2z1f1 − 2z2f2 − 2z4f4 = ν2f,

λ1f1 −z1f11 + z2f12
−z3f13 −λ2z2f3 + z22f23 + z4f33 = 0,

f2 + 2z1f3 −z3f4 = 0, (5)

where ν1 and ν2 are complex numbers.
The condition on the degree of the polynomial

f(z1,z2,z3,z4) in variable z2 can be rewritten in the
following way

∂λ2+1f

∂zλ2+12
= 0.

5. The extremal vectors
The extremal vectors are in one-to-one correspondence
to polynomial solutions of the systems of equations
(5), which are in variable z2 of maximal degree λ2.
You can find all such solutions in the appendix.

For any λ1 and λ2 there exists a constant solution
f(z1,z2,z3,z4) = 1. But such a solution gives v = |0〉,
which is not interesting.

A further solution exists only in the cases λ1 ∈ N0,
λ1 + λ2 + 1 ∈ N0 or 2λ1 + λ2 + 2 ∈ N0.
For λ1 ∈ N0 there is a function f(z1,z2,z3,z4) =

zλ1+11 , and we obtain the extremal vector

v = Fλ1+11 |0〉.

For λ1 + λ2 + 1 ∈ N0 and 2λ1 + λ2 + 4 ≤ 0 we find
the solution

f(z1,z2,z3,z4) =
(
z4 + z2z3 −z1z22 )

λ1+λ2+2

=
∑

(n1,n3)∈Dλ

(−1)n1 (λ1 + λ2 + 2)!
n1! n3! (λ1 + λ2 −n1 −n3 + 2)!

×zn11 z
2n1+n3
2 z

n3
3 z

λ1+λ2−n1−n3+2
4 ,

where Dλ =
{

(n1,n3) ∈ N20 ; n1 + n3 ≤ λ1 + λ2 + 2
}
.

The extremal vector corresponding to this solution is

v =
∑

(n1,n3)∈Dλ

(−1)n1 (λ2 − 2n1 −n3)!
n1! n3! (λ1 + λ2 −n1 −n3 + 2)!

× Fn11 F
n3
3 F

λ1+λ2−n1−n3+2
4 F

2n1+n3
2 |0〉.

If 2λ1 + λ2 + 2 ∈ N0, we introduce

N = 2λ1 + λ2 + 3, `2 =
[1

2λ2
]
, M =

[1
2N
]
.

Then we can rewrite the solution from the appendix
in the following way:

For λ1 being a half integer, i.e. λ1 = `1 − 12 , where
`1 ∈ Z, we have

f =
M∑

n4=0

min(λ2,N−2n4)∑
n2=0

(−1)n2
cn2,n4
n2! n4!

×zn2+n41 z
n2
2 z

N−n2−2n4
3 z

n4
4 ,

where

cn2,n4 =




∑min(`2,M−n4)
n=[ 12 (n2+1)]

22n+n2+2n4

× `2! M!(2n−n2)! (`2−n)! (M−n−n4)!, λ2 even,∑min(`2,M−n4)
n=[ 12 n2]

22n+n2+2n4

× `2! M!(2n−n2+1)! (`2−n)! (M−n−n4)!, λ2 odd.

For these solutions we obtain the extremal vectors

v =
M∑

n4=0

min(λ2,N−2n4)∑
n2=0

(−1)n2
(λ2 −n2)!
n2! n4!

cn2,n4

× Fn2+n41 F
N−n2−2n4
3 F

n4
4 F

n2
2 |0〉.

If λ1 is an integer we have λ1 ≤−2. The solution
of the differential equations in this case is

f =
M∑

n4=0

N−2n4∑
n2=0

(−1)n2
dn2,n4
n2! n4!

×zn2+n41 z
n2
2 z

N−n2−2n4
3 z

n4
4 ,

where

dn2,n4 =




∑M−n4
n=[ 12 (n2+1)]

2n+n2+2n4 (2`2−1)!!(2`2−2n−1)!!
× M!(2n−n2)! (M−n−n4)!, λ2 even,∑M−n4

n=[ 12 n2]
2n+n2+2n4 (2`2−1)!!(2`2−2n−1)!!

× M!(2n−n2+1)! (M−n−n4)!, λ2 odd,

and the extremal vectors are

v =
M∑

n4=0

N−2n4∑
n2=0

(−1)n2
(λ2 −n2)!
n2! n4!

dn2,n4

× Fn2+n41 F
N−n2−2n4
3 F

n4
4 F

n2
2 |0〉.

6. Appendix: Polynomial
solutions of differential
equations

To obtain extremal vectors we need to find the poly-
nomial solutions

f(z1,z2,z3,z4) =
∑

n1,n2,n3,n4≥0
cn1,n2,n3,n4z

n1
1 z

n2
2 z

n3
3 z

n4
4

of the system of equations (5), which are of less degree
than (λ2 + 1) in the variable z2.
To simplify the solution of the first equations, we

put

f(z1,z2,z3,z4)
= z−ρ21 (4z1z4 + z

2
3 )
ρ2+ρ1/2g(t,x1,x2,x3),

where ρ1 = λ1 −ν1, ρ2 = 12 (λ2 −ν2), x2 = z1, x3 = z2
and

t =
(2z1z2 −z3)2

4z1z4 + z23
, x1 =

2z1z2 −z3
z3

,

401



Č. Burdík, O. Navrátil Acta Polytechnica

or z1 = x2, z2 = x3 and

z3 =
2x2x3
1 + x1

, z4 =
x2x

2
3(x21 − t)

t(1 + x1)2
.

The first order equations are equivalent to the condi-
tions

gx1 = gx2 = gx3 = 0,

and so g(t,x1,x2,x3) = g(t).
The equations of the second order give the system

of three equations

(2λ1ρ1 + 2λ1ρ2 + λ2ρ1 + 2λ2ρ2
−ρ21 − 2ρ1ρ2 − 2ρ

2
2 + 3ρ1 + 4ρ2)g = 0,

(2λ1 + λ2 −ρ1 − 2ρ2 + 3)(1 − t)g′

+ ρ2(λ1 −ρ1 −ρ2 + 1)g = 0,
4t(1 − t)g′′ + 2

(
1 + (2λ1 + 2λ2 + 1)t

)
g′

+ (2λ1ρ1 −ρ21 + 3ρ1 + 2ρ2)g = 0. (6)

As we want to obtain polynomial solutions
f(z1,z2,z3,z4), which are in variable z2 of less or
equal degree λ2 ∈ N0, there must be solution g(t) of
the system (6), which is the polynomial in

√
t of less

or equal degree λ2.
If we exclude derivatives of g from the second and

the third equations,we find that nonzero solutions can
exist only in the following six cases:

(1.) ρ1 = 0, ρ2 = 0;
(2.) ρ1 = 2λ1 + 2, ρ2 = −λ1 − 1;
(3.) ρ1 = 0, ρ2 = λ1 + λ2 + 2;
(4.) ρ1 = 2λ1 + 2, ρ2 = λ2 + 1;
(5.) ρ1 = 2λ1 + λ2 + 3, ρ2 = 0;
(6.) ρ1 = −λ2 − 1, ρ2 = λ1 + λ2 + 2.

Case 1 (ρ1 = ρ2 = 0). A function that corresponds
to the extremal vector is f(z1,z2,z3,z4) = g(t), where
g(t) is the solution of the system

(2λ1 + λ2 + 3)(1 − t)g′ = 0,
2t(1 − t)g′′ +

(
1 + (2λ1 + 2λ2 + 1)t

)
g′ = 0. (7)

For each λ1 and λ2 this system has the solution g(t) =
1 which corresponds to the extremal vector

f(z1,z2,z3,z4) = 1.

But for 2λ1 + λ2 + 3 = 0 we obtained for g(t) the
equation

2t(1 − t)g′′ +
(
1 + (λ2 − 2)t

)
g′ = 0,

which also has a non-constant solution

g(t) = G(
√
t), where G(x) =

∫ (
1−x2

)(λ2−1)/2dx.
However this solution does not give a polynomial
function f(z1,z2,z3,z4) for any λ2.

Case 2 (ρ1 = 2λ1 + 2, ρ2 = −λ1 −1). The function
that corresponds to the extremal vector is in this case

f(z1,z2,z3,z4) = zλ1+11 g(t),

where g(t) is the solution of system (7). As in event 1
we find that the non-constant polynomial solutions

f(z1,z2,z3,z4) = zλ1+11

get only λ1 ∈ N0.

Case 3 (ρ1 = 0, ρ2 = λ1 + λ2 + 2.). The function
for the extremal vectors is

f(z1,z2,z3,z4) =
(

4z1z4 + z23
z1

)λ1+λ2+2
g(t),

where g(t) is the solution of the system

(λ2 + 1)
(
(1 − t)g′ + (λ1 + λ2 + 2)g

)
= 0,

2t(1 − t)g′′ +
(
1 + (2λ1 + 2λ2 + 1)t

)
g′

+(λ1 + λ2 + 2)g = 0. (8)

As we assume that λ2 ∈ N0,for each λ1,λ2 this system
has the solution

g(t) = (1 − t)λ1+λ2+2.

This solution corresponds to the function

f(z1,z2,z3,z4) =
(
z4 + z2z3 −z1z22 )

λ1+λ2+2, (9)

which is a non-constant polynomial for λ1 + λ2 + 1 ∈
N0.
This function is a polynomial in the variable z2 of

degree 2λ1 + 2λ2 + 2. It gives sought solutions for
2λ1 + λ2 + 4 ≤ 0.

Thus, function (9) provides a permissible solution
for the λ2 ∈ N0 only if λ1 ∈ Z, −λ2 − 1 ≤ λ1 ≤
−12λ2 − 2, from which follows λ2 ≥ 2.

Case 4 (ρ1 = 2λ1 + 2, ρ2 = λ2 + 1.). In this case,
the function that can match the extremal vector is

f(z1,z2,z3,z4) = z−λ2−11 (4z1z4 + z
2
3 )
λ1+λ2+2g(t),

where g(t) is the solution of system (8). So

f(z1,z2,z3,z4) = zλ1+11
(
z4 + z2z3 −z1z22 )

λ1+λ2+2.

To give a polynomial solution, which we have found, to
this function there must be λ1 ∈ N0 and λ1 + λ2 + 1 ∈
N0. But in this case, the degree of polynomial f in
the variable z2 is greater than λ2 and, therefore, is
not a permissible solution.

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vol. 53 no. 5/2013 Extremal Vectors for Verma Type Representation of B2

Case 5 (ρ1 = 2λ1 + λ2 + 3, ρ2 = 0.). The function
corresponding to the possible extremal vectors is

f(z1,z2,z3,z4) = (4z1z4 + z23 )
λ1+ 12 λ2+

3
2 g(t),

where function g(t) meets the equation

4t(1 − t)g′′ + 2
(
1 + (2λ1 + 2λ2 + 1)t

)
g′

−λ2(2λ1 + λ2 + 3)g = 0. (10)

This equation has two linearly independent solutions

g1(t) = F
(
−12λ2,−λ1 −

1
2λ2 −

3
2 ;

1
2 ; t
)
,

g2(t) =
√
tF
(1

2 −
1
2λ2,−λ1 −

1
2λ2 − 1;

3
2 ; t
)
,

where F(α,β; γ; t) is the hypergeometric function

F(α,β; γ; t) =
∞∑
n=0

(α)n(β)n
n! (γ)n

tn,

where

(α)n =
Γ(α + n)

Γ(α)
= α(α + 1) . . . (α + n− 1).

These solutions correspond to the functions

f1 =
∞∑
n=0

(−12λ2)n(−λ1 −
1
2λ2 −

3
2 )n

n! ( 12 )n

× (2z1z2 −z3)2n(4z1z4 + z23 )
λ1+ 12 λ2−n+

3
2 ,

f2 =
∞∑
n=0

( 12 −
1
2λ2)n(−λ1 −

1
2λ2 − 1)n

n! ( 32 )n

× (2z1z2 −z3)2n(4z1z4 + z23 )
λ1+ 12 λ2−n+1.

For at least one of these functions to be a nonconstant
polynomial, must be 2λ1+λ2+3 ∈ N, i.e. 2λ1+λ2+2 ∈
N0.
If 2λ1 + λ2 + 3 is even, we get the solution

f1 =
λ1+ 12 λ2+

3
2∑

n=0

(−12λ2)n(−λ1 −
1
2λ2 −

3
2 )n

n! ( 12 )n

× (2z1z2 −z3)2n(4z1z4 + z23 )
λ1+ 12 λ2−n+

3
2 ,

and for 2λ1 + λ2 + 3 odd, we have the solution

f2 =
λ1+ 12 λ2+1∑

n=0

( 12 −
1
2λ2)n(−λ1 −

1
2λ2 − 1)n

n! ( 32 )n

× (2z1z2 −z3)2n+1(4z1z4 + z23 )
λ1+ 12 λ2−n+1.

If 2λ1 + λ2 + 3 is even and λ2 is even, then λ1 is a
half integer, i.e. λ1 = `1 − 12 , where `1 ∈ Z, counts in
f1 only to n ≤ 12λ2, i.e.

f =
min( 12 λ2,λ1+

1
2 λ2+

3
2 )∑

n=0

(−12λ2)n(−λ1 −
1
2λ2 −

3
2 )n

n! ( 12 )n

× (2z1z2 −z3)2n(4z1z4 + z23 )
λ1+ 12 λ2−n+

3
2 ,

and, therefore, f is in the variable z2 of a polynomial
of degree not exceeding λ2.
If 2λ1 + λ2 + 3 is even and λ2 is odd, i.e. λ1 is an

integer, the function

f =
λ1+ 12 λ2+

3
2∑

n=0

(−12λ2)n(−λ1 −
1
2λ2 −

3
2 )n

n! ( 12 )n

× (2z1z2 −z3)2n(4z1z4 + z23 )
λ1+ 12 λ2−n+

3
2 ,

in the variable z2 is a polynomial of degree 2λ1 +λ2 +3.
Thus admissible solutions get only λ1 ≤−2.

If 2λ1 + λ2 + 3 is odd, then solution f2 comes into
play. If 12 (λ2−1) ∈ N0, i.e. for odd λ2 and half integer
λ1 sum in f2 only n ≤ 12 (λ2 − 1), then the solutions
are

f =

min( 12 λ2−
1
2 ,

λ1+ 12 λ2+1)∑
n=0

( 12 −
1
2λ2)n(−λ1 −

1
2λ2 − 1)n

n! ( 32 )n

× (2z1z2 −z3)2n+1(4z1z4 + z23 )
λ1+ 12 λ2−n+1

in the z2 polynomial of degree not exceeding λ2.
But for 2λ1 + λ2 + 3 odd and λ2 even, i.e. λ1 ∈ Z,

the solution is

f =
λ1+ 12 λ2+1∑

n=0

( 12 −
1
2λ2)n(−λ1 −

1
2λ2 − 1)n

n! ( 32 )n

× (2z1z2 −z3)2n+1(4z1z4 + z23 )
λ1+ 12 λ2−n+1.

In the variable z2 it is a polynomial of degree 2λ1 +
λ2 + 3. Therefore we get a permissible solution for
2λ1 + λ2 + 3 ≤ λ2, i.e. λ1 ≤−2.

Case 6 (ρ1 = −λ2 − 1, ρ2 = λ1 + λ2 + 2.). In this
case,

f(z1,z2,z3,z4)

= z−λ1−λ2−21 (4z1z4 + z
2
3 )
λ1+ 12 λ2+

3
2 g(t),

where function g(t) is the solution of equation (10).
For this function f to be polynomial, must be 2λ1 +

λ2 + 3 ∈ N0 and −λ1 − λ2 − 2 ∈ N0. But these
conditions are not fulfilled for any λ2 ∈ N0.

Acknowledgements
The work of Č.B. and O.N. was supported in part by
the GACR-P201/10/1509 grant, and by research plan
MSM6840770039.

References
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of complex semisimple Lie algebras. Yale University, 1966.

[2] D. Verma. Structure of certain induced
representations of complex semisimple lie algebras. Bull
Amer Math Soc 74:160–166, 1968.

403



Č. Burdík, O. Navrátil Acta Polytechnica

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[8] V. Dobrev. Subsingular vectors and conditionally
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[9] X. Xu. Differential equations for singular vectors of
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[10] Č. Burdík. Realization of the real semisimple Lie
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404

http://arxiv.org/abs/math/0305180v1

	Acta Polytechnica 53(5):399–404, 2013
	1 Introduction
	2 The root system for Lie algebra B2
	3 The extremal vectors for Verma type representation
	4 Differential equations for extremal vectors
	5 The extremal vectors
	6 Appendix: Polynomial solutions of differential equations
	Acknowledgements
	References