Free field realizations of certain modules for affine Lie algebra slˆ(n,C)
For the affine Lie algebra slˆ(n,C) we study a realization in terms of infinite sums of partial differential operators of a family of representations introduced in [BBFK]. These representations generalize a construction of Imaginary Verma modules [F1]. The realization constructed in the paper exten...
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irk-123456789-1547682019-06-17T01:30:41Z Free field realizations of certain modules for affine Lie algebra slˆ(n,C) Martins, R.A. For the affine Lie algebra slˆ(n,C) we study a realization in terms of infinite sums of partial differential operators of a family of representations introduced in [BBFK]. These representations generalize a construction of Imaginary Verma modules [F1]. The realization constructed in the paper extends the free field realization of Imaginary Verma modules constructed by B.Cox [С1]. 2011 Article Free field realizations of certain modules for affine Lie algebra slˆ(n,C) / R.A. Martins // Algebra and Discrete Mathematics. — 2011. — Vol. 12, № 1. — С. 28–52. — Бібліогр.: 15 назв. — англ. 1726-3255 2000 Mathematics Subject Classification:17B67, 81R10 http://dspace.nbuv.gov.ua/handle/123456789/154768 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України |
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For the affine Lie algebra slˆ(n,C) we study a realization in terms of infinite sums of partial differential operators of a family of representations introduced in [BBFK]. These representations generalize a construction of Imaginary Verma modules [F1]. The realization constructed in the paper extends the free field realization of Imaginary Verma modules constructed by B.Cox [С1]. |
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Martins, R.A. Free field realizations of certain modules for affine Lie algebra slˆ(n,C) Algebra and Discrete Mathematics |
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Martins, R.A. |
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Martins, R.A. |
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Free field realizations of certain modules for affine Lie algebra slˆ(n,C) |
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Free field realizations of certain modules for affine Lie algebra slˆ(n,C) |
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Free field realizations of certain modules for affine Lie algebra slˆ(n,C) |
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Free field realizations of certain modules for affine Lie algebra slˆ(n,C) |
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Free field realizations of certain modules for affine Lie algebra slˆ(n,C) |
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free field realizations of certain modules for affine lie algebra slˆ(n,c) |
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Інститут прикладної математики і механіки НАН України |
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Free field realizations of certain modules for affine Lie algebra slˆ(n,C) / R.A. Martins // Algebra and Discrete Mathematics. — 2011. — Vol. 12, № 1. — С. 28–52. — Бібліогр.: 15 назв. — англ. |
series |
Algebra and Discrete Mathematics |
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AT martinsra freefieldrealizationsofcertainmodulesforaffineliealgebraslˆnc |
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2025-07-14T06:52:37Z |
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2025-07-14T06:52:37Z |
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Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 12 (2011). Number 1. pp. 28 – 52
c© Journal “Algebra and Discrete Mathematics”
Free field realizations of certain modules
for affine Lie algebra ŝl(n,C)
Renato A. Martins
Communicated by V. V. Kirichenko
Abstract. For the affine Lie algebra ŝl(n,C) we study a
realization in terms of infinite sums of partial differential operators
of a family of representations introduced in [BBFK]. These repre-
sentations generalize a construction of Imaginary Verma modules
[F1]. The realization constructed in the paper extends the free field
realization of Imaginary Verma modules constructed by B.Cox [C1].
1. Introduction
Representation theory of affine Lie algebras is a very rich subject with
many applications in Mathematics and Physics. Representations of affine
Lie algebras have some very distinct features which have no analogs in
the finite dimensional case. For example, there exist modules for affine
Lie algebras containing both finite and infinite-dimensional weight spaces.
The simplest example of such modules is given by the so-called Imaginary
Verma modules [F1]. These representations correspond to nonstandard
partitions of the root system which are not equivalent under the Weyl
group to the standard partition into positive and negative roots (for
details see [DFG]). For affine Lie algebras, there are always only finitely
many equivalence classes of such nonstandard partitions (see [F2]). These
partitions give rise to Verma-type modules, which were first studied and
classified by Jakobsen and Kac [JK], and by Futorny [F2, F3] (see also
[C2, F1, FS, FK]).
In a recent paper [BBFK] different Borel-type subalgebras that do not
correspond to partitions of the root system of affine Lie algebra g were
considered. They correspond to functions ϕ : N → {±} on the set N of
positive integers, and give rise to a class of modules called ϕ-Imaginary
2000 Mathematics Subject Classification: 17B67, 81R10.
R. A. Martins 29
Verma modules. These modules can be viewed as induced from ϕ-highest
weight modules over the Heisenberg subalgebra of g. If ϕ(n) = + for
all n ∈ N, then ϕ-Imaginary Verma modules are just Imaginary Verma
modules. It was shown in [BBFK] that ϕ-Imaginary Verma module is
irreducible if and only if its central charge is nonzero.
In the case of Imaginary Verma modules for affine Lie algebra ŝl(n,C)
their free field realization was constructed by B.Cox in [C1]. This realiza-
tion generalizes the classical Wakimoto construction in terms of infinite
sums of partial differential operators. Later this has been generalized by
B.Cox and V.Futorny [CF] for other Verma type modules.
The purpose of this paper is to extend the construction of [C1] to all
ϕ-Imaginary Verma modules over the affine Lie algebra ŝl(n,C).
The structure of the paper is as follows. In Section 3, we recall a
construction of J-Imaginary Verma modules for affine Lie algebras follow-
ing [BBFK]. In Section 4, we consider the case n = 2 and construct a
realization ρJ of J-Imaginary Verma modules for affine sl(2). Finally, in
Section 5, we construct a representation ρJ : ŝl(n,C) → gl(C[x, y]), for all
J, using ϕJ and two auxiliary anti-automorphisms ρ1 : ŝl(n,C) → ŝl(n,C)
and ρ2 : gl(C[x, y]) → gl(C[x, y]). Note that generically (when the central
charge of the module is not zero) J-Imaginary Verma module is irreducible.
Hence, our construction provides free field realization of a large family of
irreducible modules for affine ŝl(n,C).
2. Preliminaries
Let sl(n,C) be a complex Lie algebra of n×n-matrices with trace zero
with the Killing form (X|Y ) = tr(XY ). Denote by Eij the matrix units,
i, j = 1, . . . , n and set Hi = Eii − Ei+1,i+1, i = 1, . . . , n − 1. Then Eij ,
i 6= j and Hi’s form a basis of sl(n,C).
The affine Lie algebra ŝl(n,C) is the universal central extension with
the 1-dimensional center Cc of the loop algebra sl(n,C)⊗ C[t, t−1]. Set
X(m) := tm ⊗ X, for all X,Y ∈ sl(n,C) and m ∈ Z. Then ŝl(n,C) is
generated by Ei(m), Fi(m) and Hi(m), with m ∈ Z and 1 6 i 6 n, and
the central element c.
Fix a symbol A and a sequence {A(m)}m∈Z. Define:
A+(z) :=
∑
m∈Z>0
A(m)z−m, (2.1)
A−(z) :=
∑
m∈Z60
A(m)z−m, (2.2)
A(z) := A+(z) +A−(z) =
∑
m∈Z
A(m)z−m, (2.3)
30 Free field realizations of certain modules
Ȧ(z) := Ȧ+(z) + Ȧ−(z) =
∑
m∈Z
mA(m)z−m, (2.4)
where, for example, Ḣ+
i (z) =
∑
m∈Z>0mHi(m)z−m.
Then we have the following defining relations in ŝl(n,C):
(R1) [Hi(z), Hj(w)] = −(Hi|Hj)cδ̇(z − w),
(R2) [Hi(z), Ej(w)] = Cij
∑
nEj(z)z
nw−n = CijEj(z)δ(z − w),
(R3) [Hi(z), Fj(w)] = −Cij
∑
n Fj(z)z
nw−n = −CijFj(z)δ(z − w),
(R4) [Ei(z), Fj(w)] = δij(Hi(z)δ(z − w)− cδ̇(z − w)),
(R5) [Ei(z), Ej(w)] = [Fi(z), Fj(w)] = 0 if Cij 6= −1,
(R6) [Fi(z1), Fi(z2), Fj(w)] = [Ei(z1), Ei(z2), Ej(w)] = 0 if Cij = −1,
where C = (Cij) denotes the Cartan matrix of type An and [X,Y, Z] :=
[X, [Y, Z]].
ElementsHi, i = 1, . . . , n−1 and c span a Cartan subalgebra of ŝl(n,C),
while Hi(m), m ∈ Z, 1 6 i 6 n, and c span a Heisenberg subalgebra of
ŝl(n,C). The central element c acts a scalar on any irreducible module V .
This scalar is called the central charge of V .
Now fix γ ∈ C∗ and for all 1 6 i 6 n, fix λi ∈ C. Let 2c = γ2. Then:
C[x] := C[xij(m) | i, j,m ∈ Z, 1 6 i, j 6 n] (2.5)
C[y] := C[yi(m) | i,m ∈ N
∗, 1 6 i 6 n] (2.6)
are the algebras generated over C by xij(m) and yi(m), respectively. In
C[x], define operators:
aij := −xij(m); (2.7)
a∗ij(m) :=
∂
∂xij(−m)
. (2.8)
where [aij(m), a∗kl(p)] = δikδjlδm,−p. Fix an arbitrary J ⊆ N
∗. In C[y]
R. A. Martins 31
define operators bi(m), with m ∈ Z and 1 6 i 6 n as follows:
bi(m) :=
−γ−1λi if m = 0;
−γm
∂
∂yi(m)
if m ∈ N
∗ \ J;
−γ−1yi(m) if m ∈ J;
γm
∂
∂yi(−m)
if −m ∈ J;
−γ−1yi(−m) if −m ∈ N
∗ \ J.
(2.9)
where we have [bi(m), bj(p)] = mδijδm,−p. Let
C[x, y] := C[x]⊗C C[y].
3. J-Imaginary Verma modules for g
Fix J ⊆ N
∗ and let ψJ : Z → {0, 1} be the function given by:
ψJ(m) :=
{
1 if m ∈ J or −m ∈ N
∗ \ J;
0 if m ∈ N
∗ \ J or −m ∈ J.
(3.1)
Let g = ŝl(n,C) and consider SψJ
= {α + nδ | α ∈ ∆̇+, n ∈ Z} ∪
{nδ | n ∈ N, ϕJ(n) = 1} ∪ {−mδ | m ∈ N, ϕJ(m) = 0}. Then the spaces
gSψJ
=
⊕
α∈SψJ
gα and g−SψJ
=
⊕
α∈−SψJ
gα are subalgebras of g such
that g = g−SψJ
⊕ h⊕ gSψJ
of g. Let λ ∈ h∗ and suppose λ(c) = a, a ∈ C.
Let bψJ
= h ⊕ gSψJ
be the Borel subalgebra corresponding to SψJ
, and
note that bψJ
⊃ Cc⊕ L+ψJ
. Let Cvλ be a module of dimension one for bψJ
such that for all h ∈ h, gSψJvλ = 0 and hvλ = λ(h)vλ.
Remark 3.1. We call the g-module
MψJ
(λ) := MbψJ
(λ) = U(g)⊗U(bψJ )
Cvλ
a J-Imaginary Verma module. Identifying 1 ⊗ vλ with vλ, the U(L)-
submodule of MψJ
(λ) generated by vλ is isomorphic to MψJ
(a). If ψJ(n) = 1
for all n, then MψJ
(λ) coincides with the imaginary Verma module MS(λ)
where S = ∆+.
Here we present some basic properties of MψJ
(λ).
Proposition 3.2. [BBFK, Proposition 3.4] Let λ ∈ h∗ be such that
λ(c) = a. If a 6= 0, then the following statements are true for MψJ
(λ).
• MψJ
(λ) is a free U(g−SψJ
)-module of rank 1.
32 Free field realizations of certain modules
• MψJ
(λ) has a unique maximal submodule and hence a unique irre-
ducible quotient.
• supp
(
MψJ
(λ)
)
=
⋃
β∈Q̇+
{λ− β + nδ | n ∈ Z} where Q̇+ is the free
abelian monoid generated by all the simple roots in ∆̇+.
• If ψJ(k) 6= ψJ(ℓ) for some k, ℓ ∈ N, then dimMψJ
(λ)µ = ∞ for any
µ ∈ supp
(
MψJ
(λ)
)
.
We have the following irreducibility criterion for modules MψJ
(λ).
Theorem 3.3. [BBFK, Theorem 3.5] Let λ ∈ h∗, λ(c) = a. Then MψJ
(λ)
is irreducible if and only if a 6= 0.
4. The case n = 2
We denote E(m) instead of E1(m), am instead of a11(m), etc.
Definition 4.1. Let ϕJ : ŝl(2,C) → gl(C[x, y]) be the function given by:
ϕJ(E(m)) := −
∂
∂xm
;
ϕJ(F (m)) :=
∑
l,j xlxj−l−m
∂
∂xj
+
∑
j∈J xj−m
∂
∂yj
+ 2K
∑
i∈J jyjx−j−m +
Kmx−m − 2Jx−m;
ϕJ(H(m)) := −2
∑
j xj−m
∂
∂xj
−ψJ(m)
∂
∂ym
+ψJ(−m)2mKy−m+2δm,0J.
Lemma 4.2. We have the following:
(a) [ϕJ(H(m)), ϕJ(E(n))] = 2ϕJ(E(m+ n));
(b) [ϕJ(H(m)), ϕJ(H(n))] = 2δm,−nmK;
(c) [ϕJ(H(m)), ϕJ(F (n))] = −2ϕJ(F (m+ n));
(d) [ϕJ(E(m)), ϕJ(F (n))] = ϕJ(H(m+ n)) +mδm,−nK.
Proof. We have:
(a) [ϕJ(H(m)), ϕJ(E(n))]
= [−2
∑
j
xj−m
∂
∂xj
− ψJ(m)
∂
∂ym
+ ψJ(−m)2mKy−m + 2δm,0J,−
∂
∂xn
]
= [−2
∑
j
xj−m
∂
∂xj
,−
∂
∂xn
] = 2
∂
∂xm+n
= 2ϕJ(E(m+ n)).
R. A. Martins 33
(b) [ϕJ(H(m)), ϕJ(H(n))]
= [−2
∑
j
xj−m
∂
∂xj
− ψJ(m)
∂
∂ym
+ ψJ(−m)2mKy−m + 2δm,0J,
− 2
∑
j
xj−n
∂
∂xj
− ψJ(n)
∂
∂yn
+ ψJ(−n)2nKy−n + 2δn,0J ]
= 4
∑
j
xj−m
∂
∂xj+n
− 4
∑
j
xj−n
∂
∂xj+m
− δm,−nψJ(m)ψJ(−n)2nK
+ 2δm,−nψJ(n)ψJ(−m)Km+ 0 = 2δm,−nmK(ψJ(m) + ψJ(−m))
= 2δm,−nmK.
(c) [ϕJ(H(m)), ϕJ(F (n))]
=
[
− 2
∑
j
xj−m
∂
∂xj
− ψJ(m)
∂
∂ym
+ ψJ(−m)2mKy−m + 2δm,0J,
∑
l,i
xlxi−l−n
∂
∂xi
+
∑
i∈J
xi−n
∂
∂yi
+ 2K
∑
i∈J
iyix−i−n +Knx−n − 2Jx−n
]
= −2
(∑
j
xj−m
(∑
i
xi−j−n
∂
∂xi
+
∑
l
xl
∂
∂xj+l+n
)
−
∑
i,l
xlxi−l−n
∂
∂xi+m
)
− 2
∑
j
xj−mψJ(j + n)
∂
∂yj+n
− 4K
∑
j
xj−mψJ(−j − n)(−j − n)y−j−n
− 2Knx−n−m + 4JX−n−m − 2ψJ(m)KψJ(m)mx−m−n
− ψJ(−m)2KmψJ(−m)x−m−n
= −4
∑
j,i
xjxi−j−m−n
∂
∂xi
+ 2
∑
l,i
xlxi−l−m−n
∂
∂xi
− 2
∑
j
ψJ(j)xj−m−n
∂
∂yj
− 4K
∑
j
ψJ(−j)(−j)y−jxj−m−n − 2Knx−m−n
+ 4Jx−m−n − 2Kmx−m−n
34 Free field realizations of certain modules
= −2
∑
j,i
xjxi−j−m−n
∂
∂xi
− 2
∑
j∈J
xj−m−n
∂
∂xj
− 4K
∑
j∈J
jyjx−j−m−n − 2K(m+ n)x−m−n + 4Jx−m−n
= −2ϕJ(F (m+ n)).
(d) [ϕJ(E(m)), ϕJ(F (n))]
=
[
−
∂
∂xm
,
∑
l,j
xlxj−l−n
∂
∂xj
+
∑
j∈J
xj−n
∂
∂yj
+ 2K
∑
j∈J
jyjx−j−n +Knx−n − 2Jx−n
]
= −
(∑
j
xj−m−n
∂
∂xj
+
∑
l
xl
∂
∂xl+m+n
+ ψJ(m+ n)
∂
∂ym+n
− 2KψJ(−m− n)(m+ n)Y−m−n − δm,−n(Km+ 2J)
)
= −2
∑
j
Xj−m−n
∂
∂xj
− ψJ(m+ n)
∂
∂ym+n
+ 2ψJ(−m− n)(m+ n)Ky−m−n + 2δm,−nJ +Kmδm,−n
= ϕJ(H(m+ n)) +mδm,−nK.
Now let ρ1 : ŝl(2,C) → ŝl(2,C) be the function given by
ρ1(E(m)) = −F (−m), ρ1(F (m)) = −E(−m), ρ1(H(m)) = H(−m)
and ρ1(K) = K.
Let ρ2 : gl(C[x, y]) → gl(C[x, y]) be the function given by
ρ2(x−m) =
∂
∂xm
, ρ2(
∂
∂xm
) = x−m, ρ2(yk) = −
∂
∂yk
and ρ2(
∂
∂yk
) = −yk.
Note that ρ1 and ρ2 are anti-automorphisms of ŝl(2,C) and gl(C[x, y]),
respectively.
Definition 4.3. Let ρJ := ρ2 ◦ ϕJ ◦ ρ1. We have:
(a) ρJ(F (m)) = ρ2 ◦ ϕJ ◦ ρ1(F (m))
= ρ2 ◦ ϕJ(−E(−m)) = ρ2
( ∂
∂x−m
)
= xm.
(b) ρJ(E(m)) = ρ2 ◦ ϕJ ◦ ρ1(E(m)) = ρ2 ◦ ϕJ(−F (−m))
R. A. Martins 35
= ρ2
(
−
∑
l,j
xlxj−l+m
∂
∂xj
−
∑
j∈J
xj+m
∂
∂yj
− 2K
∑
j∈J
jyjx−j+m −K(−m)xm + 2Jxm
)
= −
∑
l,j
x−j
∂
∂x−l
∂
∂x−j+l−m
−
∑
j∈J
(−yj)
∂
∂x−j−m
− 2K
∑
j∈J
j(−
∂
∂yj
)
∂
∂xj−m
−K(−m)
∂
∂x−m
+ 2J
∂
∂x−m
= −
∑
l,j
xj
∂
∂xl
∂
∂xj−l−m
+
∑
j∈J
yj
∂
∂x−j−m
+ 2K
∑
j∈J
j
∂
∂yj
∂
∂xj−m
+ (Km+ 2J)
∂
∂x−m
= −
∑
l,j
xj+l+m
∂
∂xl
∂
∂xj
+
∑
j∈J
yj
∂
∂x−j−m
+ 2K
∑
j∈J
j
∂
∂yj
∂
∂xj−m
+ (Km+ 2J)
∂
∂x−m
.
(c) ρJ(H(m)) = ρ2 ◦ ϕJ ◦ ρ1(H(m)) = ρ2 ◦ ϕJ(H(−m))
= ρ2
(
− 2
∑
j
xj+m
∂
∂xj
− ψJ(−m)
∂
∂y−m
+ ψJ(m)2(−m)Kym + 2δm,0J
)
= −2
∑
j
x−j
∂
∂x−j−m
+ ψJ(−m)y−m + ψJ(m)2mK
∂
∂ym
+ 2δm,0J
= −2
∑
j
xj+m
∂
∂xj
+ ψJ(−m)y−m + ψJ(m)2mK
∂
∂ym
+ 2δm,0J.
The function ρJ satisfy (R1) - (R6) because ϕJ satisfy these relations
and ρ1 and ρ2 are anti-automorphisms. Hence, it defines a representation
of ŝl(2,C). Now we can start to study a candidate ρJ when n is arbitrary.
5. Free field realization of ŝl(n,C)
Let ρJ be the function given by:
(a) ρJ(Fr)(w) := −ar,r+1(w) +
r−1∑
j=1
aj,r+1(w)a
∗
jr(w);
36 Free field realizations of certain modules
(b) ρJ(Hr)(w) := 2ar,r+1(w)a
∗
r,r+1(w)
+
r−1∑
i=1
(
ai,r+1(w)a
∗
i,r+1(w)− air(w)a
∗
ir(w)
)
+
n∑
j=r+2
(
arj(w)a
∗
rj(w)− ar+1,j(w)a
∗
r+1,j(w)
)
− γbr(w) +
γ
2
(
b+r−1(w) + b+r+1(w)
)
;
(c) ρJ(Er)(w) := ar,r+1(w)a
∗
r,r+1(w)
2
−
n∑
j=r+2
ar+1,j(w)a
∗
r+1,j(w) +
r−1∑
j=1
ajr(w)a
∗
j,r+1(w)
+
n∑
j=r+2
(
arj(w)a
∗
rj(w)− ar+1,j(w)a
∗
r+1,j(w)
)
a∗r,r+1(w)
− γa∗r,r+1(w)br(w) +
γ
2
a∗r,r+1(w)
(
b+r−1(w) + b+r+1(w)
)
−
γ
2
a∗r,r+1(w).
Theorem 5.1. The function ρJ : ŝl(n,C) → gl(C[x, y]) is a representation
of ŝl(n,C).
To prove that ρJ defines a representation of ŝl(n,C) we need only verify
the relations (R1)− (R6) but to do it, we need the following lemmas:
Lemma 5.2. We have the following:
[aij(z), a
∗
kl(w)] = δikδjlδ(z − w); (5.1)
[aij(z)a
∗
kl(z), amn(w)a
∗
pq(w)]
= δpiδqjamn(w)a
∗
kl(z)δ(z − w)− δkmδlnaij(z)a
∗
pq(w)δ(w − z); (5.2)
[aij(z), ȧ
∗
kl(w)] = δikδjlδ̇(z − w); (5.3)
ȧ∗ij(z)δ(w − z) = a∗ij(z)δ̇(w − z)− a∗ij(w)δ̇(w − z); (5.4)
[bi(z), b
±
j (w)] = [b∓i (z), b
±
j (w)] = δij δ̇
∓(w − z); (5.5)
[bi(z), bj(w)] = [b+i (z), b
−
j (w)] + [b−i (z), b
+
j (w)] = δij δ̇(w − z); (5.6)
[ar,r+1(z)a
∗
r,r+1(z), ar,r+1(w)a
∗
r,r+1(w)a
∗
r,r+1(w)]
= ar,r+1(w)a
∗
r,r+1(z)a
∗
r,r+1(w)δ(w − z); (5.7)
[arj(z)a
∗
rj(z), f(w)a
∗
kl(w)] = f(w)δrkδjla
∗
rj(z)δ(w − z), (5.8)
where f(w) comutes with arj(z), a
∗
rj(z) and a∗kl(w).
R. A. Martins 37
Proof. We have:
(5.1) [aij(z), a
∗
kl(w)] =
∑
m,n∈Z
[aij(m), a∗kl(n)]z
−mw−n
=
∑
m,n∈Z
δikδjlδm,−nz
−mw−n =
∑
n∈Z
δikδjlz
nw−n
= δikδjlδ(z − w).
(5.2) [aij(z)a
∗
kl(z), amn(w)a
∗
pq(w)]
= aij(z)[a
∗
kl(z), amn(w)]a
∗
pq(w) + aij(z)amn(w)a
∗
kl(z)a
∗
pq(w)
− amn(w)[a
∗
pq(w), aij(z)]a
∗
kl(z)− amn(w)aij(z)a
∗
pq(w)a
∗
kl(z)
= δpiδqjamn(w)a
∗
kl(z)δ(z − w)− δkmδlnaij(z)a
∗
pq(w)δ(w − z).
(5.3) [aij(z), ȧ
∗
kl(w)] =
∑
m,n∈Z
[aij(m), na∗kl(n)]z
−mw−n
=
∑
m,n∈Z
nδikδjlδm,−nz
−mw−n =
∑
n∈Z
δikδjlnz
nw−n
= δikδjlδ̇(z − w).
(5.4) ȧ∗ij(z)δ(w − z)
= (
∑
m∈Z
(−m)a∗ij(m)z−m)(
∑
n∈Z
z−nwn)
=
∑
m,n∈Z
(−m)a∗ij(m)z−m−nwn
=
∑
m,n∈Z
−(m+ n)a∗ij(m)z−m−nw(m+n)−m
+
∑
m,n∈Z
na∗ij(m)z−m−nwn
=
∑
m,n∈Z
−na∗ij(m)z−nwn−m +
∑
m,n∈Z
na∗ij(m)z−mwnz−n
= −(
∑
m∈Z
a∗ij(m)w−m)δ̇(w − z) +
∑
m∈Z
a∗ij(m)z−mδ̇(w − z)
= a∗ij(z)δ̇(w − z)− a∗ij(w)δ̇(w − z).
(5.5) [bi(z), b
−
j (w)]
=
∑
m∈Z
n∈Z∗
−
[bi(m), bj(n)]z
−mw−n =
∑
m∈Z
n∈Z∗
−
mδijδm,−nz
−mw−n
=
∑
m∈Z∗+
n∈Z∗
−
mδijδm,−nz
−mw−n = [b+i (z), b
−
j (w)]
38 Free field realizations of certain modules
=
∑
n∈Z−
(−n)δijznw−n = δij δ̇
+(w − z).
and similarly for the other case.
(5.6) [bi(z), bj(w)] = [b+i (z), b
−
j (w)] + [b−i (z), b
+
j (w)]
=
∑
m,n∈Z
mδijδm,−nz
−mw−n = δij
∑
n∈Z
(−n)znw−n
= δij δ̇(w − z).
(5.7) [ar,r+1(z)a
∗
r,r+1(z), ar,r+1(w)a
∗
r,r+1(w)a
∗
r,r+1(w)]
= ar,r+1(z)[a
∗
r,r+1(z), ar,r+1(w)]a
∗
r,r+1(w)a
∗
r,r+1(w)
− ar,r+1(w)a
∗
r,r+1(w)[a
∗
r,r+1(w), ar,r+1(z)]a
∗
r,r+1(z)
+ ar,r+1(z)ar,r+1(w)a
∗
r,r+1(z)a
∗
r,r+1(w)a
∗
r,r+1(w)
− ar,r+1(w)a
∗
r,r+1(w)ar,r+1(z)a
∗
r,r+1(w)a
∗
r,r+1(z)
= −ar,r+1(z)δ(w − z)a∗r,r+1(w)a
∗
r,r+1(w)
+ ar,r+1(w)a
∗
r,r+1(w)δ(w − z)a∗r,r+1(z)
+ ar,r+1(w)[ar,r+1(z), a
∗
r,r+1(w)]a
∗
r,r+1(z)a
∗
r,r+1(w)
= ar,r+1(w)a
∗
r,r+1(z)a
∗
r,r+1(w)δ(w − z).
(5.8) [arj(z)a
∗
rj(z), f(w)a
∗
kl(w)]
= f(w)(arj(z)a
∗
rj(z)a
∗
kl(w)− a∗kl(w)arj(z)a
∗
rj(z))
= f(w)[arj(z), a
∗
kl(w)]a
∗
rj(z) = f(w)δrkδjla
∗
rj(z)δ(w − z).
Lemma 5.3. We have the following:
n∑
j=r+2
s−1∑
k=1
[aks(w)a
∗
k,s+1(w), arj(z)a
∗
rj(z)]
= −δs,r+1ar,r+1(w)a
∗
r,r+2(z)δ(w − z); (5.9)
r−1∑
j=1
n∑
k=s+2
[as+1,k(w)a
∗
sk(w), ajr(z)a
∗
j,r+1(z)] = 0; (5.10)
n∑
j=r+2
s−1∑
k=1
[aks(w)a
∗
k,s+1(w), ar+1,j(z)a
∗
r+1,j(z)] = 0; (5.11)
n∑
j=r+2
n∑
k=s+2
[as+1,k(w)a
∗
sk(w), arj(z)a
∗
rj(z)− ar+1,j(z)a
∗
r+1,j(z)]
= −2δrsδ(w − z)
n∑
j=r+2
ar+1,j(w)a
∗
rj(z)
R. A. Martins 39
+ δr,s+1δ(w − z)
n∑
j=r+2
arj(z)a
∗
r−1,j(w)
+ δs,r+1δ(w − z)
n∑
j=r+3
ar+2,j(w)a
∗
r+1,j(z); (5.12)
r−1∑
j=1
[a∗s,s+1(w), ajr(z)a
∗
j,r+1(z)] = −δr,s+1a
∗
r−1,r+1(z)δ(w − z); (5.13)
n∑
j=r+2
[a∗s,s+1(w), ar+1,j(z)a
∗
rj(z)] = −δr+1,sa
∗
r,r+2(z)δ(w − z). (5.14)
Now we are able to verify (R1) - (R6).
Lemma 5.4. (R1) [ρJ(Hr)(z), ρJ(Hs)(w)] = −(Hr|Hs)cδ̇(z − w).
Proof. Observe that if |r−s|> 1, then [ρJ(Hr)(z), ρJ(Hs)(w)] = 0 because
all summations are equal to zero. So, writing in terms of δs,f(r), we have
the following:
[ρJ(Hr)(z), ρJ(Hs)(w)] = δrs([ρJ(Hr)(z), ρJ(Hr)(w)])
+ δr,s+1([ρJ(Hr)(z), ρJ(Hr−1)(w)])
+ δr,s−1([ρJ(Hr)(z), ρJ(Hr+1)(w)])
= δrs
(
[2ar,r+1(z)a
∗
r,r+1(z), ρJ(Hr)(w)]
+
r−1∑
i=1
[ai,r+1(z)a
∗
r,r+1(z)− air(z)a
∗
ir(z),
ρJ(Hr)(w)] +
n∑
j=r+2
[arj(z)a
∗
rj(z)− ar+1,j(z)a
∗
r+1,j(z), ρJ(Hr)(w)]
+ [−γbr(z) +
γ
2
(b+r−1(z) + b+r+1(z)), ρJ(Hr)(w)]
)
+ δr,s+1
(
[2ar,r+1(z)a
∗
r,r+1(z), ρJ(Hr−1)(w)]
+
r−1∑
i=1
[ai,r+1(z)a
∗
i,r+1(z)− air(z)a
∗
ir(z), ρJ(Hr−1)(w)]
+
n∑
j=r+2
[arj(z)a
∗
rj(z)− ar+1,j(z)a
∗
r+1,j(z), ρJ(Hr−1)(w)]
+ [−γbr(z) +
γ
2
(b+r−1(z) + b+r+1(z)), ρJ(Hr−1)(w)]
)
+ δr,s−1
(
[ρJ(Hr)(z), ρJ(Hr+1)(w)]
)
40 Free field realizations of certain modules
= δrs
(
4[ar,r+1(z)a
∗
r,r+1(z), ar,r+1(w)a
∗
r,r+1(w)]
+
r−1∑
i=1
[ai,r+1(z)a
∗
r,r+1(z), ai,r+1(w)a
∗
i,r+1(w)]
+
r−1∑
i=1
[air(z)a
∗
ir(z), air(w)a
∗
ir(w)] + 0 + γ2[br(z), br(w)]
+
γ2
4
[b+r−1(z), b
+
r−1(w)] +
γ2
4
[b+r+1(z), b
+
r+1(w)]
)
+ δr,s+1
(
[2ar,r+1(z)a
∗
r,r+1(z),
n∑
j=r+1
−a(r−1)+1,j(w)a
∗
(r−1)+1,j(w)]
+
r−1∑
i=1
[−air(z)a
∗
ir(z),
r−2∑
j=1
−ajr(w)a
∗
jr(w)]
+
n∑
j=r+2
[arj(z)a
∗
rj(z),
n∑
i=r+1
−ari(w)a
∗
ri(w)]
+ [−γbr(z),
γ
2
b+(r−1)+1(w)] + [
γ
2
b+r−1(z),−γbr−1(w)]
)
+ δr,s−1
(
[ρJ(Hr)(z), ρJ(Hr+1)(w)]
)
= δrs(0 + 0 + 0 + 0 + 2cδ̇(w − z) + 0 + 0)
+ δr,s+1(−2[ar,r+1(z)a
∗
r,r+1(z), ar,r+1(w)a
∗
r,r+1(w)]
−
r−2∑
i=1
[air(z)a
∗
ir(z), air(w)a
∗
ir(w)]
+
n∑
j=r+2
[arj(z)a
∗
rj(z),−arj(w)a
∗
rj(w)]
− cδ̇−(w − z) + cδ̇−(z − w)) + δr,s−1([ρJ(Hr)(z), ρJ(Hr+1)(w)])
+ δrs(2cδ̇(w − z)) + δr,s+1(0 + 0 + 0 +−cδ̇−(w − z)− cδ̇+(w − z))
+ δr,s−1([ρJ(Hr)(z), ρJ(Hr+1)(w)])
= δrs(2cδ̇(w − z)) + δr,s+1(−cδ̇(w − z)) + δr,s−1(−cδ̇(w − z)),
where in the last equality we have:
[ρJ(Hr)(z), ρJ(Hr+1)(w)] =
− [ρJ(Hr+1)(w), ρJ(Hr)(z)] = cδ̇(z − w) = −cδ̇(w − z).
Then the relation (R1) is verified.
R. A. Martins 41
Lemma 5.5. (R2)[ρJ(Hr)(z), ρJ(Es)(w)] = CrsρJ(Es)(z)δ(z − w).
Proof. We have the following:
(a) 2[ar,r+1(z)a
∗
r,r+1(z), ρJ(Es)(w)]
= δsr(2[ar,r+1(z)a
∗
r,r+1(z), ar,r+1(w)a
∗
r,r+1(w)a
∗
r,r+1(w)]
+ 2[ar,r+1(z)a
∗
r,r+1(z),
(
n∑
j=r+2
arj(w)a
∗
rj(w)− ar+1,j(w)a
∗
r+1,j(w))a
∗
r,r+1(w)]
+ 2[ar,r+1(z)a
∗
r,r+1(z),
− γa∗r,r+1(w)(
1
2
+ br(w)−
1
2
b+r−1(w)−
1
2
b+r+1(w))])
+ δs,r+1(2[ar,r+1(z)a
∗
r,r+1(z), ar,r+1(w)a
∗
r,r+2(w)])
+ δs,r−1(−2a∗r−1,r+1(z)ar,r+1(w)δ(w − z))
(5.8)
(5.7)
= δsr(2ar,r+1(w)a
∗
r,r+1(z)a
∗
r,r+1(w)δ(w − z)
+ 2
n∑
j=r+2
(arj(w)a
∗
rj(w)− ar+1,j(w)a
∗
r+1,j(w))a
∗
r,r+1(z)δ(w − z)
− 2γa∗r,r+1br(w)δ(w − z) + γa∗r,r+1(z)(b
+
r−1(w) + b+r+1(w))δ(w − z)
− γ2a∗r,r+1(z)δ̇(w − z)) + δs,r+1(2ar,r+1(z)a
∗
r,r+2(w)δ(w − z))
+ δs,r−1(−2a∗r−1,r+1(z)ar,r+1(w)δ(w − z)).
(b)
r−1∑
i=1
[ai,r+1(z)a
∗
i,r+1(z)− air(z)a
∗
ir(z), ρJ(Es)(w)]
= δsr(
r−1∑
i=1
[ai,r+1(z)a
∗
i,r+1(z)− air(z)a
∗
ir(z),
r−1∑
j=1
ajr(w)a
∗
j,r+1(w)])
+ δs,r+1(
r−1∑
i=1
[ai,r+1(z)a
∗
i,r+1(z), ai,r+1(w)a
∗
i,r+2(w)])
+ δs,r−1(−(ar−1,r(w)a
∗
r−1,r(z)a
∗
r−1,r(w) +
n∑
j=r+1
(ar−1,j(w)a
∗
r−1,j(w)
− arj(w)a
∗
rj(w))a
∗
r−1,r(z) +
r−2∑
j=1
aj,r−1(w)a
∗
jr(z)
− ar,r+1(w)a
∗
r−1,r+1(z)− γa∗r−1(w)br−1(w)
+
γ
2
(a∗r−1,r(w)b
+
r−2(w)− a∗r−1,r(w)b
+
r (w))δ(w − z)
42 Free field realizations of certain modules
+
γ2
2
a∗r−1,r(z)δ̇(w − z))
(5.2)
= δsr(2
r−1∑
i=1
air(w)a
∗
i,r+1(w)δ(w − z))
+ δs,r+1(−
r−1∑
i=1
ai,r+1(z)a
∗
i,r+2(w)δ(w − z))
+ δs,r−1(−(ar−1,r(w)a
∗
r−1,r(z)a
∗
r−1,r(w) +
n∑
j=r+1
(ar−1,j(w)a
∗
r−1,j(w)
− arj(w)a
∗
rj(w))a
∗
r−1,r(z) +
r−2∑
j=1
aj,r−1(w)a
∗
jr(z)
− ar,r+1(w)a
∗
r−1,r+1(z)− γa∗r−1(w)br−1(w)
+
γ
2
(a∗r−1,r(w)b
+
r−2(w)− a∗r−1,r(w)b
+
r (w))δ(w − z)
+
γ2
2
a∗r−1,r(z)δ̇(w − z)).
(c)
n∑
j=r+2
[arj(z)a
∗
rj(z)− ar+1,j(z)a
∗
r+1,j(z), ρJ(Es)(w)]
= δsr(
n∑
j=r+2
[−ar+1,j(z)a
∗
r+1,j(z),−ar+1,j(w)a
∗
r+1,j(w)]
+
n∑
j=r+2
[arj(z)a
∗
rj(z), arj(w)a
∗
rj(w)]
+
n∑
j=r+2
[arj(z)a
∗
rj(z)− ar+1,j(z)a
∗
r+1,j(z),
(arj(w)a
∗
rj(w)− ar+1,j(w)a
∗
r+1,j(w))a
∗
r,r+1(w)])
+ δs,r+1(−(ar+1,r+2(w)a
∗
r+1,r+2(z)a
∗
r+1,r+2(w)
+
n∑
j=r+3
(ar+1,j(w)a
∗
r+1,j(w)
− ar+2,j(w)a
∗
r+2,j(w))a
∗
r+1,r+2(z)− ar,r+1(w)a
∗
r,r+2(z)
−
n∑
j=r+3
ar+2,j(w)a
∗
r+1,j(w)
− γa∗r+1,r+2(w)br+1(w) +
γ
2
a∗r+1,r+2(w)(b
+
r (w) + b+r+2(w)))δ(w − z)
R. A. Martins 43
+
γ2
2
a∗r+1,r+2(z)δ̇(w − z)) + δs,r−1(
n∑
j=r+2
arj(w)a
∗
r−1,j(w)δ(w − z))
(5.2)
(5.7)
= δsr(0 + 0 + 2
n∑
j=r+2
ar+1,j(w)a
∗
rj(z)δ(w − z))
+ δs,r+1(−(ar+1,r+2(w)a
∗
r+1,r+2(z)a
∗
r+1,r+2(w)
+
n∑
j=r+3
(ar+1,j(w)a
∗
r+1,j(w)
− ar+2,j(w)a
∗
r+2,j(w))a
∗
r+1,r+2(z)− ar,r+1(w)a
∗
r,r+2(z)
−
n∑
j=r+3
ar+2,j(w)a
∗
r+1,j(w)
− γa∗r+1,r+2(w)br+1(w) +
γ
2
a∗r+1,r+2(w)(b
+
r (w) + b+r+2(w)))δ(w − z)
+
γ2
2
a∗r+1,r+2(z)δ̇(w − z)) + δs,r−1(
n∑
j=r+2
arj(w)a
∗
r−1,j(w)δ(w − z)).
Finally, we have:
(d) [−γbr(z) +
γ
2
(b+r−1(z) + b+r+1(z)), ρJ(Es)(w)]
= δsr(γ
2[br(z), a
∗
r,r+1(w)br(w)])
+ δs,r+1(−
γ2
2
a∗r+1,r+2(w)([br(z), b
+
r (w)] + [b+r+1(z), br+1(w)]))
+ δs,r−1(
γ2
2
a∗r−1,r(w)([br(z), b
+
r (w)] + [b+r−1(z), br−1(w)]))
= δsr(γ
2a∗r,r+1(w)δ̇(w − z)) + δs,r+1(−
γ2
2
a∗r+1,r+2(w)δ̇(w − z))
+ δs,r−1(
γ2
2
a∗r−1,r(w)δ̇(w − z)).
Adding the four equations up we get:
[ρJ(Hr)(z), ρJ(Es)(w)]
= δsr2ρJ(Er)(z)δ(w − z) + δs,r+1(−ρJ(Er+1)(z)δ(w − z))
+ δs,r−1(−ρJ(Er−1)(z)δ(w − z)),
proving (R2).
Lemma 5.6. (R3)[ρJ(Hr)(z), ρJ(Fs)(w)] = −CrsρJ(Fs)(z)δ(z − w).
44 Free field realizations of certain modules
Proof. We have:
(a) 2[ar,r+1(z)a
∗
r,r+1(z), ρJ(Fs)(w)]
= δsr(2ar,r+2(z)δ(z − w))
+ δs,r+1(2[ar,r+1(z)a
∗
r,r+1(z), ar,r+2(w)a
∗
r,r+1(w)]) + δs,r−10
= δsr(2ar,r+2(z)δ(z − w)) + δs,r+1(2ar,r+2(w)a
∗
r,r+1(z)δ(w − z)).
(b)
r−1∑
i=1
[ai,r+1(z)a
∗
i,r+1(z)− air(z)a
∗
ir(z), ρJ(Fs)(w)]
= δsr(
r−1∑
i=1
[ai,r+1(z)a
∗
i,r+1(z), ai,r+1(w)a
∗
ir(w)]
−
r−1∑
i=1
[air(z)a
∗
ir(z), ai,r+1(w)a
∗
ir(w)])
+ δs,r+1(
r∑
j=1
[ai,r+1(z)a
∗
i,r+1(z), aj,r+2(w)a
∗
j,r+1(w)])
+ δs,r−1([ar−1,r(z)a
∗
r−1,r(z), ar−1,r(w)]
−
r−2∑
i=1
air(z)[a
∗
ir(z), air(w)]a
∗
i,r−1(w))
= δsr(−2
r−1∑
i=1
ai,r+1(z)a
∗
ir(w)δ(z − w))
+ δs,r+1(
r−1∑
i=1
ai,r+2(w)a
∗
i,r+1(z)δ(z − w))
+ δs,r−1((−ar−1,r(z) +
r−2∑
i=1
air(z)a
∗
i,r−1(w))δ(z − w)).
(c)
n∑
j=r+2
[arj(z)a
∗
rj(z)− ar+1,j(z)a
∗
r+1,j(z), ρJ(Es)(w)]
= δsr0 + δs,r+1(
n∑
j=r+2
r∑
k=1
[arj(z)a
∗
rj(z), ak,r+2(w)a
∗
k,r+1(w)]
+
n∑
j=r+2
[ar+1,j(z)a
∗
r+1,j(z), ar+1,r+2(w)]) + δs,r−10
= δs,r+1(−ar,r+2(z)a
∗
r,r+1(w)δ(z − w)− ar+1,r+2(z)δ(z − w)).
(d) [−γbr(z) +
γ
2
(b+r−1(z) + b+r+1(z)), ρJ(Fs)(w)] = 0.
R. A. Martins 45
Adding the four equations up, we have the result.
Lemma 5.7. (R4) [ρJ(Er)(z), ρJ(Fs)(w)] = δsr(ρJ(Hr)(z)δ(z − w) −
cδ̇(z − w)).
Proof. Consider the case |s− r| > 1. We have [ρ(Er)(z),−ar,r+1(w)] = 0
and
∑r−1
j=1
∑s−1
k=1[akr(z)a
∗
k,r+1(z), aj,s+1(w)a
∗
js(w)] = 0, because r 6= s.
Finally:
s−1∑
j=1
[
n∑
k=r+2
(ark(z)a
∗
rk(z)− ar+1,k(z)a
∗
r+1,k(z))a
∗
r,r+1(z)
+
n∑
k=r+2
ar+1,k(z)a
∗
rk, aj,s+1(w)a
∗
js(w)]
=
s−1∑
j=1
n∑
k=r+2
[ark(z)a
∗
rk(z)− ar+1,k(z)a
∗
r+1,k(z), aj,s+1(w)a
∗
js(w)]a
∗
r,r+1(z)
−
s−1∑
j=1
n∑
k=r+2
[ar+1,k(z)a
∗
rk(z), aj,s+1(w)a
∗
js(w)]
=
s−1∑
j=1
n∑
k=r+2
(aj,s+1(w)a
∗
rk(z)δskδrj − ark(z)a
∗
js(w)δs+1,kδjr
− aj,s+1(w)a
∗
r+1,k(z)δr+1,jδks
+ ar+1,k(z)a
∗
js(w)δr+1,jδk,s+1)a
∗
r,r+1(z)δ(z − w)
+
s−1∑
j=1
n∑
k=r+2
(ar+1,k(z)a
∗
js(w)δrjδk,s+1
− aj,s+1(w)a
∗
rk(z)δr+1,jδks)δ(z − w) = ⊛
If s + 1 > n then aj,s+1(w) = 0 = δk,s+1 and ⊛ = 0. Now suppose
s+ 1 6 n. If s− 1 < r then j 6 s− 1 < r, δjr = δj,r+1 = 0 and ⊛ = 0. If
s−1 > r we have s−1 > r because |s−r|> 1. Then s > r+2 and for this
reason the terms j = r, r + 1 and k = s, s+ 1 appears in the expression.
Then:
⊛ =(ar,s+1(w)a
∗
rs(z)− ar,s+1(z)a
∗
rs(w)− ar+1,s+1(w)a
∗
r+1,s(z)
+ ar+1,s+1(z)a
∗
r+1,s(w))a
∗
r,r+1(z)δ(z − w)
+ (ar+1,s+1(z)a
∗
rs(w)− ar+1,s+1(w)a
∗
rs(z))δ(z − w) = 0.
So we only have nontrivial expressions when r = s, r = s − 1 or
r = s+ 1. Then:
[ρJ(Er)(z), ρJ(Fs)(w)] = δrs([ρJ(Er(z)),−ar,r+1(w)]
46 Free field realizations of certain modules
+ [ρJ(Er(z)),
r−1∑
j=1
(aj,r+1(w)a
∗
jr(w)])
+ δr,s−1([ρJ(Es−1(z)),−as,s+1(w)]
+ [ρJ(Es−1(z)),
s−1∑
j=1
(aj,s+1(w)a
∗
js(w)])
+ δr,s+1([ρJ(Es+1)(z),−as,s+1(w)]
+ [ρJ(Es+1)(z),
s−1∑
j=1
aj,s+1(w)a
∗
js(w)])
= δrs
(
(2ar,r+1(z)a
∗
r,r+1(z)
+
n∑
j=r+2
(a∗rj(z)arj(z)− a∗r+1,j(z)ar+1,j(z))− γbr(z)
+
γ
2
(b+r−1(z) + b+r+1(z)))δ(z − w)−
γ2
2
δ̇(z − w)
+
r−1∑
j=1
[ajr(z)a
∗
j,r+1(z), aj,r+1(w)a
∗
jr(w)]
)
+ δr,s+1(0 + 0)
+ δr,s−1(−
n∑
j=s+1
[as,s+1(w), asj(z)a
∗
s−1,s(z)a
∗
sj(z)]
+ as,s+1(z)a
∗
s−1,s(z)δ(w − z))
= δrs((2ar,r+1(z)a
∗
r,r+1(z)
+
n∑
j=r+2
(a∗rj(z)arj(z)− a∗r+1,j(z)ar+1,j(z))− γbr(z)
+
γ
2
(b+r−1(z) + b+r+1(z)))δ(z − w)
−
γ2
2
δ̇(z − w) +
r−1∑
j=1
(aj,r+1(z)a
∗
j,r+1(z)
− ajr(z)a
∗
jr(z))δ(z − w)) + δr,s−1(−as,s+1(z)a
∗
s−1,s(z)δ(w − z)
+ as,s+1(z)a
∗
s−1,s(z)δ(w − z))
= δsr(ρJ(Hr)(z)δ(z − w)− cδ̇(z − w)).
Then the lemma is verified.
Lemma 5.8. We have the following:
(R5|R6) [ρJ(Fr)(z), ρJ(Fs)(w)] = [ρJ(Er)(z), ρJ(Es)(w)] = 0 if Crs 6= −1;
R. A. Martins 47
[ρJ(Fr)(z1), ρJ(Fr)(z2), ρJ(Fs)(w)]
= [ρJ(Er)(z1), ρJ(Er)(z2), ρJ(Es)(w)] = 0 if Crs = −1.
Proof. We have:
[ρJ(Fr)(z), ρJ(Fs)(w)]
= [−ar,r+1(z), ρJ(Fs)(w)] + [
r−1∑
j=1
aj,r+1(z)a
∗
jr(z), ρJ(Fs)(w)]
= δs,r+1ar,r+2(w)δ(w − z) + δr,s+1ar−1,r+1(w)δ(w − z)
− δr,s+1
s−1∑
j=1
aj,r+1(w)a
∗
js(w)δ(w − z)
+ δs,r+1
r−1∑
j=1
aj,r+2(w)a
∗
jr(w)δ(w − z)
= (δs,r+1(ar,r+2(w) +
r−1∑
j=1
aj,r+2(w)a
∗
jr(w)) + δr,s+1(ar−1,r+1(w)
−
r−2∑
j=1
aj,r+1(w)a
∗
j,r−1(w)))δ(w − z).
Then [ρJ(Fr)(z), ρJ(Fs)(w)] = 0 if |r − s|6= 1 (or equivalently,
if Crs 6= −1). Now:
[ρJ(Er)(z), ρJ(Es)(w)] = [ar,r+1(z)a
∗
r,r+1(z)
2, ρJ(Es)(w)]
+ [
n∑
j=r+2
(arj(z)a
∗
rj(z)− ar+1,j(z)a
∗
r+1,j(z))a
∗
r,r+1(z), ρJ(Es)(w)]
+ [
r−1∑
j=1
ajr(z)a
∗
j,r+1(z)−
n∑
j=r+2
ar+1,j(z)a
∗
rj(z), ρJ(Es)(w)]
+ [−γa∗r,r+1(z)(br(z)−
1
2
a∗r,r+1(z)(b
+
r−1(z) + b+r+1(z)))
−
γ2
2
ȧ∗r,r+1(z), ρJ(Es)(w)]
= (−δrs
n∑
j=r+2
(arj(w)a
∗
rj(w)− ar+1,j(w)a
∗
r+1,j(w))a
∗
r,r+1(z)
2δ(w − z)
− δr,s+1ar,r+1(w)a
∗
r−1,r(w)a
∗
r,r+1(z)
2δ(w − z)
+ 2δr,s−1ar,r+1(z)a
∗
r,r+1(z)a
∗
r,r+2(w)δ(w − z)
− 2δr,s+1ar,r+1(z)a
∗
r,r+1(z)a
∗
r−1,r+1(w)δ(w − z)
48 Free field realizations of certain modules
− δrsγa
∗
r,r+1(z)
2(br(w)−
1
2
(b+r−1(w) + b+r+1(w)))δ(w − z)
− δrs
γ2
2
a∗r,r+1(z)
2δ̇(w − z)) + (δrs
n∑
j=r+2
(arj(w)a
∗
rj(w)
− ar+1,j(w)a
∗
r+1,j(w))a
∗
r,r+1(z)
2δ(w − z)
+ δs,r+1as,s+1(w)a
∗
s,s+1(z)
2a∗s,s+1(w)δ(w − z)
+ δs,r+1
n∑
k=r+3
(ar+1,k(w)a
∗
r+1,k(w)
− ar+2,k(w)a
∗
r+2,j(w))a
∗
r,r+1(z)a
∗
r+1,r+2(z)δ(w − z)
− δr,s+1
n∑
j=r+2
(arj(z)a
∗
rj(z)− ar+1,j(z)a
∗
r+1,j(z))
a∗r−1,r(w)a
∗
r,r+1(w)δ(w − z)
+
n∑
j=r+2
(arj(z)a
∗
rj(z)− ar+1,j(z)a
∗
r+1,j(z))
(δr,s+1a
∗
r,r+2(w)− δr,s+1a
∗
r−1,r+1(w))δ(w − z)
− δs,r+1(ar,r+1(w)a
∗
r,r+2(z)
+
n∑
j=r+3
ar+2,k(w)a
∗
r+1,j(z))a
∗
r,r+1(z)δ(w − z)
+ 2δrs
n∑
j=r+2
ar+1,j(w)a
∗
rj(z)a
∗
r,r+1(z)δ(w − z)
− δr,s+1
n∑
j=r+2
arj(z)a
∗
r−1,j(w)a
∗
r,r+1(z)δ(w − z)
+ a∗r,r+1(z)a
∗
r+1,r+2(z)δs,r+1(γbs(w)δ(w − z)
−
γ
2
(b+s−1(w) + b+s+1(w))δ(w − z) +
γ2
2
δ̇(w − z)))
+ ((−2δs+1,ras,s+1(z)a
∗
s,s+1(z)a
∗
s,s+2(w)
+ 2δr+1,sas,s+1(z)a
∗
s,s+1(z)a
∗
s−1,s+1(w)
− δr,s+1(
n∑
j=s+2
(asj(z)a
∗
sj(z)− as+1,j(z)a
∗
s+1,j(z))a
∗
s,s+2(w)
+ as,s+1(w)a
∗
s,s+1(z)a
∗
s,s+2(z))−
n∑
j=s+3
as+2,k(w)a
∗
s+1,j(z)a
∗
s,s+1(z)
R. A. Martins 49
+ δs,r+1
n∑
j=s+2
asj(z)a
∗
s−1,j(w)a
∗
s,s+1(z)
+ δs,r+1
n∑
j=s+2
(asj(z)a
∗
sj(z)− as+1,j(z)a
∗
s+1,j(z))a
∗
s−1,s+1(w)
+ δs+1,r(−
r−2∑
j=1
aj,r−1(w)a
∗
j,r+1(z) +
n∑
j=r+2
ar+1,j(z)a
∗
r−1,j(w))
+ δs,r+1(
r−1∑
j=1
ajr(z)a
∗
j,r+2(w)−
n∑
j=r+3
ar+2,j(w)a
∗
rj(z)))δ(w − z)
+ (δr+1,sa
∗
s−1,s+1(z)− δr,s+1a
∗
r−1,r+1(z))((γbs(z)
−
γ
2
(b+s−1(z) + b+s+1(z)))δ(w − z)
+
γ
2
δ̇(w − z))−
γ2
2
δr,s+1ȧr−1,r+1(z)δ(w − z)) + (δsrγa
∗
s,s+1(z)
2(bs(w)
−
1
2
(b+s−1(w) + b+s+1(w)))δ(w − z) + δsr
γ2
2
a∗s,s+1(z)
2δ̇(w − z)
− a∗s,s+1(w)a
∗
s+1,s+2(w)δr,s+1(γbr(z)δ(z − w)
−
γ
2
(b+r−1(z) + b+r+1(z))δ(z − w) +
γ2
2
δ̇(z − w))
+ (δs,r+1a
∗
s−1,s+1(z)− δs+1,ra
∗
r−1,r+1(z))((γbr(z)−
γ
2
(b+r−1(z)
+ b+r+1(z)))δ(w − z) +
γ2
2
δ̇(w − z))−
γ2
2
δr,s+1ȧr−1,r+1(z)δ(w − z)
+ γ2a∗r,r+1(z)a
∗
s,s+1(w)(δrs −
1
2
(δr,s+1 + δs,r+1))δ̇(w − z)).
After doing the calculations above, the term in δrs become equal to
zero. So we have only terms in δr,s+1 and δr,s−1. Then:
[ρJ(Er)(z), ρJ(Es)(w)] = 0 if |r − s|6= 1,
and (R5) is verified. To finish we have:
[ρJ(Fr)(z1), ρJ(Fr)(z2), ρJ(Fs)(w)] := [ρJ(Fr)(z1), [ρJ(Fr)(z2), ρJ(Fs)(w)]]
= [−ar,r+1(z1), (ar−1,r+1(w)−
r−2∑
j=1
aj,r+1(w)a
∗
j,r−1(w))δ(w − z2)]
+ [
r−1∑
j=1
aj,r+1(z1)a
∗
jr(z1), ar−1,r+1(w)
50 Free field realizations of certain modules
−
r−2∑
l=1
al,r+1(w)a
∗
l,r−1(w))δ(w − z2)] = 0 + 0 = 0.
Similarly for s = r + 1. If s = r − 1 we have:
(a) [ar,r+1(z1)a
∗
r,r+1(z1)
2, [ρJ(Er)(z2), ρJ(Er−1)(w)]]
= −ar,r+1(w)a
∗
r,r+1(z1)
2a∗r−1,r+1(w)δ̇(z1 − z2)δ(w − z2)
+
n∑
j=r+2
(arj(z2)a
∗
rj(z2)− ar+1,j(z2)a
∗
r+1,j(z2))
× a∗r,r+1(z1)
2a∗r−1,r(w)δ(z1 − z2)δ(w − z2)
+
n∑
j=r+2
arj(z2)a
∗
r−1,j(w)a
∗
r,r+1(z1)
2δ(z1 − z2)δ(w − z2)
+ a∗r,r+1(z1)
2a∗r−1,r(w)δ(z1 − z2)δ(w − z2)(γbr(z2)
−
γ
2
(b∗r−1(z2) + b+r+1(z2)))
−
γ2
2
a∗r,r+1(z1)
2a∗r−1,r(w)δ̇(w − z1)δ(z2 − w).
(b) [
n∑
j=r+2
(arj(z1)a
∗
rj(z1)− ar+1,j(z1)a
∗
r+1,j(z1))a
∗
r,r+1(z1),
[ρJ(Er)(z2), ρJ(Er−1)(w)]]
= −
n∑
j=r+2
(arj(z1)a
∗
rj(z1)− ar+1,j(z1)a
∗
r+1,j(z1))
× a∗r,r+1(z1)
2a∗r−1,r(w)δ(z1 − z2)δ(w − z2)
−
n∑
j=r+2
(arj(z1)a
∗
rj(z1)− ar+1,j(z1)a
∗
r+1,j(z1))
× a∗r,r+1(z1)a
∗
r−1,r+1(w)δ(z1 − z2)δ(w − z2)
− 2
n∑
j=r+2
ar+1,j(z1)a
∗
rj(w)a
∗
r,r+1(z1)a
∗
r−1,r(w)δ(z1 − z2)δ(w − z2)
−
n∑
j=r+2
arj(z2)a
∗
r−1,j(w)a
∗
r,r+1(z1)
2δ(z1 − z2)δ(w − z2)
−
n∑
j=r+2
ar+1,j(z1)a
∗
r−1,j(w)a
∗
r,r+1(z1)δ(w − z2)δ(z1 − z2).
R. A. Martins 51
(c) [
r−1∑
j=1
ajr(z1)a
∗
j,r+1(z1)
−
n∑
j=r+2
ar+1,j(z1)a
∗
rj(z1), [ρJ(Er)(z2), ρJ(Er−1)(w)]]
= (a∗r−1,r+1(z1)a
∗
r,r+1(w)
n∑
j=r+2
(a∗rj(z2)arj(z2)− a∗r+1,j(z2)ar+1,j(z2))
− a∗r−1,r+1(z1)a
∗
r,r+1(w)(γbr(z1)−
γ
2
(b+r−1(z1) + b+r+1(z1)))
+
γ2
2
a∗r−1,r+1(z1)ȧ
∗
r,r+1(w)
+ 2
n∑
j=r+2
ar+1,j(z1)a
∗
rj(z1)a
∗
r−1,r(w)a
∗
r,r+1(w)
+
n∑
j=r+2
ar+1,j(z1)a
∗
r−1,j(w)a
∗
r,r+1(z2))δ(z1 − z2)δ(w − z2).
(d) [−γa∗r,r+1(z1)(br(z1)−
1
2
a∗r,r+1(z1)(b
+
r−1(z1) + b+r+1(z1)))
−
γ2
2
ȧ∗r,r+1(z1), [ρJ(Er)(z2), ρJ(Er−1)(w)]]
= −a∗r,r+1(z2)
2a∗r−1,r(w)δ(z1 − z2)δ(w − z2)(γbr(z1)
−
γ
2
(b+r−1(z1) + b+r+1(z1)))
−
γ2
2
a∗r,r+1(z2)
2a∗r−1,r(w)δ̇(z1 − w)δ(w − z2)
+ a∗r,r+1(z2)a
∗
r−1,r+1(w)δ(z1 − z2)δ(w − z2)(γbr(z1)
−
γ
2
(b+r−1(z1) + b+r+1(z1)))
−
γ2
2
a∗r,r+1(z2)a
∗
r−1,r+1(w)δ̇(z1 − z2)δ(w − z1)
− γ2a∗r,r+1(z1)a
∗
r−1,r(w)a
∗
r,r+1(w)δ(z2 − w)δ̇(z2 − z1)
+
γ2
2
a∗r,r+1(z1)a
∗
r−1,r+1(z2)δ̇(z1 − z2)δ(w − z2).
Adding the four summations up we have:
[ρJ(Er)(z1), ρJ(Er)(z2), ρJ(Er−1)(w)] = 0
Similarly: [ρJ(Er)(z1), ρJ(Er)(z2), ρJ(Er+1)(w)] = 0 and then (R6) is
proved.
52 Free field realizations of certain modules
6. Acknowledgment
This work is part of the Ph.D. Thesis of the author, who was supported
by FAPESP (Process number: 2008/06860-3). The author is grateful to
his supervisor V. Futorny and to B. Cox for stimulating discussions.
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Contact information
R. A. Martins Instituto de Matemática e Estat́istica, Universi-
dade de São Paulo, Rua do Matão, 1010- Cidade
Universitária, São Paulo SP, Brasil
E-Mail: renatoam@ime.usp.br
Received by the editors: 15.07.2011
and in final form 15.07.2011.
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