Twisted de Rham Complex on Line and Singular Vectors in sl₂ˆ Verma Modules
We consider two complexes. The first complex is the twisted de Rham complex of scalar meromorphic differential forms on the projective line, holomorphic on the complement to a finite set of points. The second complex is the chain complex of the Lie algebra of sl₂-valued algebraic functions on the sa...
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| Cite this: | Twisted de Rham Complex on Line and Singular Vectors in sl₂ˆ Verma Modules / A. Slinkin, A. Varchenko // Symmetry, Integrability and Geometry: Methods and Applications. — 2019. — Т. 15. — Бібліогр.: 9 назв. — англ. |
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| citation_txt | Twisted de Rham Complex on Line and Singular Vectors in sl₂ˆ Verma Modules / A. Slinkin, A. Varchenko // Symmetry, Integrability and Geometry: Methods and Applications. — 2019. — Т. 15. — Бібліогр.: 9 назв. — англ. |
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| description | We consider two complexes. The first complex is the twisted de Rham complex of scalar meromorphic differential forms on the projective line, holomorphic on the complement to a finite set of points. The second complex is the chain complex of the Lie algebra of sl₂-valued algebraic functions on the same complement, with coefficients in a tensor product of contragradient Verma modules over the affine Lie algebra sl₂ˆ. In [Schechtman V., Varchenko A., Mosc. Math. J. 17 (2017), 787-802] a construction of a monomorphism of the first complex to the second was suggested, and it was indicated that under this monomorphism, the existence of singular vectors in the Verma modules (the Malikov-Feigin-Fuchs singular vectors) is reflected in the relations between the cohomology classes of the de Rham complex. In this paper, we prove these results.
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Symmetry, Integrability and Geometry: Methods and Applications SIGMA 15 (2019), 075, 26 pages
Twisted de Rham Complex on Line
and Singular Vectors in ŝl2 Verma Modules
Alexey SLINKIN † and Alexander VARCHENKO †‡
† Department of Mathematics, University of North Carolina at Chapel Hill,
Chapel Hill, NC 27599-3250, USA
E-mail: slinalex@live.unc.edu, anv@email.unc.edu
URL: http://varchenko.web.unc.edu/
‡ Faculty of Mathematics and Mechanics, Lomonosov Moscow State University,
Leninskiye Gory 1, 119991 Moscow GSP-1, Russia
Received May 30, 2019, in final form September 21, 2019; Published online September 26, 2019
https://doi.org/10.3842/SIGMA.2019.075
Abstract. We consider two complexes. The first complex is the twisted de Rham complex
of scalar meromorphic differential forms on projective line, holomorphic on the complement
to a finite set of points. The second complex is the chain complex of the Lie algebra of
sl2-valued algebraic functions on the same complement, with coefficients in a tensor prod-
uct of contragradient Verma modules over the affine Lie algebra ŝl2. In [Schechtman V.,
Varchenko A., Mosc. Math. J. 17 (2017), 787–802] a construction of a monomorphism of the
first complex to the second was suggested and it was indicated that under this monomor-
phism the existence of singular vectors in the Verma modules (the Malikov–Feigin–Fuchs
singular vectors) is reflected in the relations between the cohomology classes of the de Rham
complex. In this paper we prove these results.
Key words: twisted de Rham complex; logarithmic differential forms; ŝl2-modules; Lie alge-
bra chain complexes
2010 Mathematics Subject Classification: 17B56; 17B67; 33C80
Dedicated to Dmitry Borisovich Fuchs
on the occasion of his 80-th birthday
1 Introduction
We consider two complexes. The first complex is the twisted de Rham complex of scalar mero-
morphic differential forms on projective line, that are holomorphic on the complement to a finite
set of points. The second complex is the chain complex of the Lie algebra of sl2-valued algebraic
functions on the same complement, with coefficients in a tensor product of contragradient Verma
modules over the affine Lie algebra ŝl2. In [9] a construction of a monomorphism of the first
complex to the second was suggested. That construction gives a relation between the singular
vectors in the Verma modules and resonance relations in the de Rham complex.
That construction of the homomorphism was invented in the middle of 90s, while the paper [9]
was prepared for publication 20 years later, when the proofs were forgotten, if they existed. The
paper [9] provides supporting evidence to the results formulated in [9], but not the proofs. The
goal of this paper is to give the proofs to the results formulated in [9], namely, the proofs that
the construction in [9] indeed gives a homorphism of complexes and relates the resonances in
the de Rham complex and the ŝl2 singular vectors.
This paper is a contribution to the Special Issue on Algebra, Topology, and Dynamics in Interaction in honor
of Dmitry Fuchs. The full collection is available at https://www.emis.de/journals/SIGMA/Fuchs.html
mailto:slinalex@live.unc.edu
mailto:anv@email.unc.edu
http://varchenko.web.unc.edu/
https://doi.org/10.3842/SIGMA.2019.075
https://www.emis.de/journals/SIGMA/Fuchs.html
2 A. Slinkin and A. Varchenko
The construction in [9] has two motivations.
The first motivation was to generalize the principal construction of [8]. In [8], the tensor
products of contragradient Verma modules over a semisimple Lie algebra were identified with
the spaces of the top degree logarithmic differential forms over certain configuration spaces. Also
the logarithmic parts of the de Rham complexes over the configuration spaces were identified
with some standard Lie algebra chain complexes having coefficients in these tensor products, cf.
in [4, 5] a D-module explanation of this correspondence.
The second idea was that the appearance of singular vectors in Verma modules over affine Lie
algebras is reflected in the relations between the cohomology classes of logarithmic differential
forms. This was proved in an important particular case in [1, 2], and in [7] a one-to-one cor-
respondence was established “on the level of parameters”. In [9] and in the present paper this
correspondence is developed for another non-trivial class of singular vectors, namely for (a part
of) Malikov–Feigin–Fuchs singular vectors, cf. [6].
The paper has the following structure. In Section 2 we introduce the de Rham complex of
a master function and resonance relations. In Section 3 we discuss ŝl2 Verma modules, the Kac–
Kazhdan reducibility conditions. We formulate Theorem 3.2 which describes certain relations in
a contragradient Verma module. The proof of Theorem 3.2 is the main new result of this paper.
In Theorem 3.3 we describe the connection between the relations, described in Theorem 3.2,
and the Malikov–Feigin–Fuchs singular vectors. In Section 4 we construct a map of the de
Rham complex of the master function to the chain complex of the Lie algebra of sl2-valued
algebraic functions. Theorem 4.1 says that the map is a monomorphism of complexes. The
proof of Theorem 4.1 is the second new result of this paper. Section 5 is devoted to the proof
of Theorem 3.2. The proof is straightforward but rather nontrivial and lengthy.
2 The de Rham complex of master function
2.1 Twisted de Rham complex
Consider C with coordinate t. Define the master function by the formula
Φ(t) =
n∏
i=1
(t− zi)−mi/κ,
where z1, . . . , zn,m1, . . . ,mn, κ ∈ C are parameters. Fix these parameters and assume that
z1, . . . , zn are distinct. Set
zn+1 =∞, mn+1 = m1 + · · ·+mn − 2.
Denote U = C− {z1, . . . , zn}.
Consider the twisted de Rham complex associated with Φ,
0 −→ Ω0(U)
∂−→ Ω1(U) −→ 0. (2.1)
Here Ωp(U) is the space of rational differential p-forms on C regular on U . The differential ∂ is
given by the formula
∂ = d + α ∧ ·, (2.2)
where d is the standard de Rham differential and the second summand is the left exterior
multiplication by the form
α = −1
κ
n∑
i=1
mi
dt
t− zi
=
dΦ
Φ
.
Twisted de Rham Complex on Line and Singular Vectors in ŝl2 Verma Modules 3
Formula (2.2) is motivated by the computation
d(Φω) = Φdω + dΦ ∧ ω = Φ(dω + α ∧ ω).
The complex Ω•(U) is the complex of global algebraic sections of the de Rham complex of(
Oan
U , ∂
)
, where ∂ = d + α ∧ · is considered as the integrable connection on the sheaf Oan
U of
holomorphic functions on U .
If the monodromy of Φ is non-trivial, that is, if at least one of the numbers m1/κ, . . . ,mn/κ
is not an integer, then
H0(Ω•(U)) = 0, dimH1(Ω•(U)) = n− 1,
see for example [7].
2.2 Basis of Ω•(U)
The functions
1
(t− zi)a
for a ∈ Z>0 and ta for a ∈ Z≥0
form a basis of Ω0(U). The differential forms
dt
(t− zi)a
for a ∈ Z>0 and tadt for a ∈ Z≥0
form a basis of Ω1(U). The differential ∂ is given by the formulas
κ∂
(
1
(t− zi)a
)
= −(mi + aκ)
dt
(t− zi)a+1
+
a∑
k=1
∑
j 6=i
mj
(zj − zi)k
dt
(t− zi)a+1−k
−
∑
j 6=i
mj
(zj − zi)a
dt
t− zj
, (2.3)
κ∂
(
ta
)
=
aκ− n∑
j=1
mj
ta−1dt−
a−1∑
k=1
n∑
j=1
mjz
k
j t
a−1−kdt−
n∑
j=1
mjz
a
j
dt
t− zj
. (2.4)
2.3 Resonances
The equations
(i) mi + (a− 1)κ = 0 for some a ∈ Z>0, i ∈ {1, . . . , n},
(ii) mn+1 + 2− aκ = 0 for some a ∈ Z>0,
(iii) κ = 0,
are called the resonance relations for the parameters m1, . . . ,mn+1, κ of the de Rham complex.
If κ = 0, then the twisted de Rham complex is not defined. If the resonance relation
mi + aκ = 0 is satisfied for some a, then the first term in the right-hand side of (2.3) equals
zero. Similarly, if the resonance relation mn+1 + 2− aκ = 0 is satisfied for some a, then the first
term in the right-hand side of (2.4) equals zero.
4 A. Slinkin and A. Varchenko
2.4 Logarithmic subcomplex
Let Ω0
log(U) ⊂ Ω0(U) be the subspace generated over C by function 1. Let Ω1
log(U) ⊂ Ω1(U) be
the subspace generated over C by the differential forms
ωj =
dt
t− zj
, j = 1, . . . , n.
These subspaces form the logarithmic subcomplex (Ω•log(U), ∂) of the de Rham complex
(Ω•(U), ∂). We have
∂ : 1 7→ α.
For generic m1, . . . ,mn, κ, the embedding (Ω•log(U), ∂) ↪→ (Ω•(U), ∂) is a quasi-isomorphism,
the logarithmic forms ω1, . . . , ωn generate the space H1(Ω•(U)), and the cohomological relation
n∑
i=1
miωi ∼ 0 is the only one, see for example [7].
Each resonance relation implies a new cohomological relation between the forms ω1, . . . , ωn,
see [9, Corollary 6.4]. For example, if mn+1 + 2 − κ = 0, then
n∑
j=1
zjmjωj ∼ 0, and if mn+1 +
2− 2κ = 0, then
n∑
j=1
z2jmjωj −
1
κ
n∑
j=1
zjmj
( n∑
i=1
zimiωi
)
∼ 0.
3 ŝl2-modules
3.1 Lie algebra ŝl2
Let sl2 be the Lie algebra of complex (2× 2)-matrices with zero trace. Let e, f , h be standard
generators subject to the relations
[e, f ] = h, [h, e] = 2e, [h, f ] = −2f.
Let ŝl2 be the affine Lie algebra ŝl2 = sl2
[
T, T−1
]
⊕ Cc with the bracket[
aT i, bT j
]
= [a, b]T i+j + i〈a, b〉δi+j,0c,
where c is central element, 〈a, b〉 = tr(ab). Set
e1 = e, f1 = f, h1 = h,
e2 = fT, f2 = eT−1, h2 = c− h.
These are the standard Chevalley generators defining ŝl2 as the Kac–Moody algebra correspon-
ding to the Cartan matrix
(
2 −2
−2 2
)
.
3.2 Automorphism π
The Lie algebra ŝl2 has an automorphism π,
π : c 7→ c, eT i 7→ fT i, fT i 7→ eT i, hT i 7→ −hT i.
Twisted de Rham Complex on Line and Singular Vectors in ŝl2 Verma Modules 5
3.3 Verma modules
We fix k ∈ C and assume that the central element c acts on all our representations by multipli-
cation by k.
For m ∈ C, let V (m, k −m) be the ŝl2 Verma module with generating vector v. The Verma
module is generated by v subject to the relations
e1v = 0, e2v = 0, h1v = mv, h2v = (k −m)v.
Let n̂− ⊂ ŝl2 be the Lie subalgebra generated by f1, f2 and U n̂− its enveloping algebra. The
map U n̂− → V (m, k −m), F 7→ Fv, is an isomorphism of U n̂−-modules.
The space V (m, k − m) has a Z2
≥0-grading: a vector fi1 · · · fipv with ij ∈ {1, 2} has deg-
ree (p1, p2), if pi is the number of i’s in the sequence i1, . . . , ip. For γ ∈ Z2
≥0, denote by
V (m, k −m)γ ⊂ V (m, k −m) the corresponding γ-homogeneous component.
A homogeneous nonzero vector ω in V (m, k−m), non-proportional to v, is called a singular
vector if e1ω = e2ω = 0. The Verma module V (m, k −m) is reducible, if and only if it contains
a singular vector.
3.4 Reducibility conditions
See Kac–Kazhdan [3]. Set
κ = k + 2.
The Verma module V (m, k−m) is reducible if and only if at least one of the following relations
holds:
(a) m− l + 1 + (a− 1)κ = 0,
(b) m+ l + 1− aκ = 0,
(c) κ = 0,
where l, a ∈ Z>0. If (m,κ) satisfies exactly one of the conditions (a), (b), then V (m, k − m)
contains a unique proper submodule, and this submodule is generated by a singular vector of
degree (la, l(a− 1)) for condition (a) and of degree (l(a− 1), la) for condition (b).
These singular vectors are highly nontrivial and are given by the following theorem.
Theorem 3.1 (Malikov–Feigin–Fuchs, [6]). For a, l ∈ Z>0 and κ ∈ C, the monomials
F12(l, a, κ) = f
l+(a−1)κ
1 f
l+(a−2)κ
2 f
l+(a−3)κ
1 · · · f l−(a−2)κ2 f
l−(a−1)κ
1 ,
F21(l, a, κ) = f
l+(a−1)κ
2 f
l+(a−2)κ
1 f
l+(a−3)κ
2 · · · f l−(a−2)κ1 f
l−(a−1)κ
2
are well-defined as elements of U n̂−. If m = l− 1− (a− 1)κ, then F12(l, a, κ)v ∈ V (m, k−m) is
a singular vector of degree (la, l(a−1)) and if m = −l−1 +aκ, then F21(l, a, κ)v ∈ V (m, k−m)
is a singular vector of degree (l(a− 1), la).
An explanation of the meaning of complex powers in these formulas see in [6].
For example for m = −2 + κ, we have
F21(1, 1, κ)v = f2v =
e
T
v,
and for m = −2 + 2κ, we have
F21(1, 2, κ)v = f1+κ2 f1f
1−κ
2 v = f
( e
T
)2
v + (1 + κ)
h
T
e
T
v − (1 + κ)κ
e
T 2
v.
6 A. Slinkin and A. Varchenko
3.5 Shapovalov form
The Shapovalov form on an ŝl2 Verma module V with generating vector v is the unique symmetric
bilinear form S(·, ·) on V such that
S(v, v) = 1, S(fix, y) = S(x, eiy) for i = 1, 2; x, y ∈ V.
For γ ∈ Z2
≥0, let V ∗γ be the vector space dual to Vγ . Define V ∗ = ⊕γV ∗γ . The space V ∗ is an
ŝl2-module with the ŝl2-action defined by the formulas:
〈fiφ, x〉 = 〈φ, eix〉, 〈eiφ, x〉 = 〈φ, fix〉,
where φ ∈ V ∗, x ∈ V , i = 1, 2. The ŝl2-module V ∗ is called the contragradient Verma module.
The Shapovalov form S considered as a map S : V −→ V ∗ is a morphism of ŝl2-modules.
3.6 Bases in V and V ∗
Let V be an ŝl2 Verma module V . For every γ = (p1, p2) ∈ Z2
≥0 with p1 6= p2, we fix a basis in
the homogeneous component Vγ ⊂ V .
For p1 > p2, we fix the basis{
f
T i1
· · · f
T ia
h
T j1
· · · h
T jb
e
T k1
· · · e
T kc
v
}
,
where
0 ≤ ia ≤ ia−1 ≤ · · · ≤ i1, 1 ≤ jb ≤ jb−1 ≤ · · · ≤ j1, 1 ≤ kc ≤ kc−1 ≤ · · · ≤ k1;
a∑
s=1
is +
b∑
s=1
js +
c∑
s=1
ks + a− c = p1,
a∑
s=1
is +
b∑
s=1
js +
c∑
s=1
ks = p2. (3.1)
For p1 < p2, we fix the basis{
e
T k1
· · · e
T kc
h
T j1
· · · h
T jb
f
T i1
· · · f
T ia
v
}
,
with the indices satisfying (3.1). Notice that for any x ∈ sl2 the elements x
T i and x
T j commute.
These collections of vectors are bases by the Poincaré–Birkhoff–Witt theorem.
For any γ, we fix a basis in the γ-homogeneous component V ∗γ ⊂ V ∗ as the basis dual of the
basis in Vγ specified above. If {wi} is a basis in Vγ , then we denote by {(wi)∗} the dual basis in V ∗γ .
3.7 Main formula
Theorem 3.2 ([9, Theorem 5.12]). For m, k ∈ C and a ∈ Z>0, the following identities hold in
the contragradient Verma module V (m, k −m)∗,
f
T a−1
(v)∗ = (m+ (a− 1)(k + 2))
(
f
T a−1
v
)∗
+
a−1∑
`=1
[
h
T `
(
f
T a−1−`
v
)∗
+ 2
e
T `
∑
i+j=a−1−`
i≥j≥0
(
f
T i
f
T j
v
)∗ ]
, (3.2)
e
T a
(v)∗ = (a(k + 2)−m− 2)
( e
T a
v
)∗
Twisted de Rham Complex on Line and Singular Vectors in ŝl2 Verma Modules 7
+
a−2∑
`=0
[
− h
T `+1
( e
T a−`−1
v
)∗
+ 2
f
T `
∑
i+j=a−`
i≥j≥1
( e
T i
e
T j
v
)∗ ]
, (3.3)
where v is the generating vector of the Verma module V (m, k −m).
Theorem 3.2 was announced in [9]. The proof of Theorem 3.2 is the main result of this paper.
The theorem is proved in Section 5.
Remark. The right-hand sides of formulas (3.2) and (3.3) have the factors m+ (a− 1)(k + 2)
and a(k + 2) −m − 2. The vanishing of these factors corresponds to the resonance conditions
mi + (a− 1)κ = 0 and mn+1 + 2− aκ = 0 for the de Rham complex in Section 2.3, if we recall
that κ = k + 2.
Remark. Theorem 3.2 says that the action of the element f
Ta−1 of degree (a, a − 1) on the
covector (v)∗ can be expressed in terms of the actions of the elements h
T l and e
T l of smaller
degree on some other covectors. Similarly the action of the element e
Ta of degree (a − 1, a) on
the covector (v)∗ can be expressed in terms of the actions of the elements h
T l ,
f
T l of smaller
degree on some other covectors.
3.8 Relation to Malikov–Feigin–Fuchs vectors
Let
S : V (m, k −m)→ V (m, k −m)∗
be the Shapovalov form. Denote
Xa(m, k −m) = S−1
(
(m+ (a− 1)(k + 2))
(
f
T a−1
v
)∗)
,
Ya(m, k −m) = S−1
(
(m+ 2− a(k + 2))
( e
T a
v
)∗)
.
For generic values of m and k, the Shapovalov form S is non-degenerate and Xa and Ya are well
defined elements of V (m, k −m). The chosen basis in V (m, k −m) allows us to compare these
vectors for different values of k, m. The vectors Xa(m, k −m), Ya(m, k −m) are holomorphic
functions of k, m for generic k, m.
Recall the resonance lines in the (m, k)-plane, given by the equations
m− l + 1 + (a− 1)(k + 2) = 0, m+ l + 1− a(k + 2) = 0, k + 2 = 0,
for some a, l ∈ Z>0, see Section 3.4.
Theorem 3.3 ([9, Theorem 6.2]). For a ∈ Z>0 let (m0, k0) be a point of the line m+(a−1)(k+
2) = 0, which does not belong to other resonance lines. Then the function Xa(m, k −m) can be
analytically continued to the point (m0, k0), and Xa(m0, k0−m0) is a (nonzero) singular vector
of V (m0, k0−m0), hence it is proportional to the Malikov–Feigin–Fuchs vector F12(1, a, k0 + 2).
Similarly, for a ∈ Z>0 let (m0, k0) be a point of the line m+ 2− a(k+ 2) = 0, which does not
belong to other resonance lines. Then the function Ya(m, k−m) can be analytically continued to
the point (m0, k0), and Ya(m0, k0−m0) is a (nonzero) singular vector of V (m0, k0−m0), hence
it is proportional to the Malikov–Feigin–Fuchs vector F21(1, a, k0 + 2).
8 A. Slinkin and A. Varchenko
4 Homomorphism of complexes
4.1 Lie algebra sl2(U)
Recall that {z1, . . . , zn, zn+1 =∞} are pairwise distinct points of the complex projective line P1
and U = P1−{z1, . . . , zn, zn+1}. Fix local coordinates t−z1, . . . , t−zn, 1/t on P1 at these points,
respectively. Let sl2(U) be the Lie algebra of sl2-valued rational functions on P1 regular on U ,
with the pointwise bracket. Thus, an element of sl2(U) has the form e ⊗ u1 + h ⊗ u2 + f ⊗ u3
with ui ∈ Ω0(U), and the bracket is defined by the formula [x⊗ u1, y ⊗ u2] = [x, y]⊗ (u1u2).
4.2 sl2(U)-modules
We say that an ŝl2-module W has the finiteness property, if for any w ∈W and x ∈ sl2, we have
xT j · w = 0 for all j � 0. For example, the contragradient Verma module has the finiteness
property.
Let W1, . . . ,Wn+1 be ŝl2-modules with the finiteness property. Then the Lie algebra sl2(U)
acts on W1 ⊗ · · · ⊗Wn+1 by the formula
x⊗ u · (w1 ⊗ · · · ⊗ wn+1) =
(
[x⊗ u(t)](z1)w1
)
⊗ w2 ⊗ · · · ⊗ wn+1 + · · ·
+ w1 ⊗ · · · ⊗ wn−1 ⊗
(
[x⊗ u(t)](zn)wn
)
⊗ wn+1
+ w1 ⊗ · · · ⊗ wn ⊗
(
π([x⊗ u(t)](∞))wn+1
)
,
where for x ⊗ u ∈ sl2(U) the symbol [x ⊗ u(t)](zj) denotes the Laurent expansion of x ⊗ u at
t = zj and [x⊗ u(t)](∞) denotes the Laurent expansion at t =∞; the symbol π in the last term
denotes the ŝl2-automorphism defined in Section 3.2.
The finiteness property of the tensor factors ensures that the actions of the Laurent series
are well-defined.
The ŝl2-action gives us a map
µ : sl2(U)⊗
(
⊗n+1
j=1 Wj
)
→ ⊗n+1
j=1Wj . (4.1)
4.3 Chain complex
For a Lie algebra g and a g-module W we denote by C•(g,W ) the standard chain complex of g
with coefficients in W , where
Cp(g,W ) = ∧pg⊗W,
d(gp ∧ · · · ∧ g1 ⊗ w) =
p∑
i=1
(−1)i−1gp ∧ · · · ∧ ĝi ∧ · · · ∧ g1 ⊗ giw
+
∑
1≤i<j≤p
(−1)i+jgp ∧ · · · ∧ ĝj ∧ · · · ∧ ĝi ∧ · · · ∧ g1 ⊗ [gj , gi]w.
4.4 Two complexes
4.4.1
Let m1, . . . ,mn, k ∈ C, k + 2 6= 0. Define mn+1 = m1 + · · · + mn − 2. For j = 1, . . . , n + 1,
let Vj be the ŝl2 Verma module V (mj , k−mj) and V ∗j the corresponding contragradient Verma
module. Consider the chain complex C•
(
sl2(U),⊗n+1
j=1V
∗
j
)
and its last two terms
→ sl2(U)⊗
(
⊗n+1
j=1V
∗
j )
d−→ ⊗n+1
j=1V
∗
j → 0,
where d = µ, see formula (4.1).
Twisted de Rham Complex on Line and Singular Vectors in ŝl2 Verma Modules 9
We assign degree 0 to the term ⊗n+1
j=1V
∗
j of this complex and assign degree 1 to the differen-
tial d, so that the whole complex sits in the non-positive area.
4.4.2
Consider the twisted de Rham complex in (2.1) corresponding to κ = k+ 2 with degrees shifted
by 1, namely, the complex Ω•(U)[1],
0→ Ω0(U)
∂−→ Ω1(U)→ 0,
where the shift [1] means that we assign degree p− 1 to the term Ωp(U).
4.5 Construction
Define a linear map
η1 : Ω1(U) −→ ⊗n+1
j=1V
∗
j
by the formulas
dt
(t− zm)a
7→ −κ(v1)
∗ ⊗ · · · ⊗
(
f
T a−1
vm
)∗
⊗ · · · ⊗ (vn+1)
∗, (4.2)
ta−1dt 7→ κ(v1)
∗ ⊗ · · · ⊗ (vn)∗ ⊗
( e
T a
vn+1
)∗
, (4.3)
for a > 0. Define a linear map
η0 : Ω0(U) −→ sl2(U)⊗
(
⊗n+1
j=1V
∗
j
)
by the formulas
1
(t− zm)a
7→ f
(t− zm)a
⊗ (v1)
∗ ⊗ · · · ⊗ (vn+1)
∗
−
a∑
l=1
[
e
(t− zm)l
⊗ (v1)
∗ ⊗ · · · ⊗ 2
∑
i+j=a−l
i≥j≥0
(
f
T i
f
T j
vm
)∗
⊗ · · · ⊗ (vn+1)
∗
+
h
(t− zm)l
⊗ (v1)
∗ ⊗ · · · ⊗
(
f
T a−l
vm
)∗
⊗ · · · ⊗ (vn+1)
∗
]
, (4.4)
for a > 0;
ta 7→ fta ⊗ (v1)
∗ ⊗ · · · ⊗ (vn+1)
∗
−
a−2∑
l=0
[
etl ⊗ (v1)
∗ ⊗ · · · ⊗ (vn)∗ ⊗ 2
∑
i+j=a−l,
i≥j≥1
( e
T i
e
T j
vn+1
)∗
+ htl+1 ⊗ (v1)
∗ ⊗ · · · ⊗ (vn)∗ ⊗
( e
T a−l−1
vn+1
)∗ ]
, (4.5)
for a ≥ 0.
Theorem 4.1 ([9, Theorem 5.12]). Formulas (4.2)–(4.5) define a homomorphism of complexes
η : Ω•(U)[1]→ C•
(
sl2(U);⊗n+1
j=1V
∗
j
)
, namely we have
dη0 = η1∂.
The homomorphism is injective.
10 A. Slinkin and A. Varchenko
Theorem 4.1 was announced in [9]. Here is a proof of the theorem.
Proof. First we calculate η1(∂((t− zp)−a)),
1
(t− zp)a
∂7→ −1
κ
(mp + aκ)
dt
(t− zp)a+1
+
1
κ
a∑
k=1
∑
j 6=p
mj
(zj − zp)k
dt
(t− zp)a+1−k
− 1
κ
∑
j 6=p
mj
(zj − zp)a
dt
t− zj
η17→ (mp + κa)(v1)
∗ ⊗ · · · ⊗
(
f
T a
vp
)∗
⊗ · · · ⊗ (vn+1)
∗
−
a∑
k=1
∑
j 6=p
mj
(zj − zp)k
(v1)
∗ ⊗ · · · ⊗
(
f
T a−k
vp
)∗
⊗ · · · ⊗ (vn+1)
∗
+
∑
j 6=p
mj
(zj − zp)a
(v1)
∗ ⊗ · · · ⊗ (fvj)
∗ ⊗ · · · ⊗ (vn+1)
∗.
Then we calculate d
(
η0((t− zp)−a)
)
,
1
(t− zp)a
η07→ f
(t− zp)a
⊗ (v1)
∗ ⊗ · · · ⊗ (vn+1)
∗
−
a∑
l=1
[
h
(t− zp)l
⊗ (v1)
∗ ⊗ · · · ⊗
(
f
T a−l
vp
)∗
⊗ · · · ⊗ (vn+1)
∗
+
e
(t− zp)l
⊗ 2
∑
i+j=a−l
i≥j≥0
(v1)
∗ ⊗ · · · ⊗
(
f
T i
f
T j
vp
)∗
⊗ · · · ⊗ (vn+1)
∗
]
µ7→ (v1)
∗ ⊗ · · · ⊗
[
(mp + a(k + 2))
(
f
T a
vp
)∗
+
a∑
l=1
[
h
T l
(
f
T a−l
vp
)∗
+ 2
e
T l
∑
i+j=a−l
i≥j≥0
(
f
T i
f
T j
vp
)∗ ]]
⊗ · · · ⊗ (vn+1)
∗
+
∑
j 6=p
mj
(zj − zp)a
(v1)
∗ ⊗ · · · ⊗ (fvj)
∗ ⊗ · · · ⊗ (vn+1)
∗
−
a∑
l=1
[
(v1)
∗ ⊗ · · · ⊗ h
T l
(
f
T a−l
vp
)∗
⊗ · · · ⊗ (vn+1)
∗
+ (v1)
∗ ⊗ · · · ⊗
∑
i+j=a−l
i≥j≥0
2
e
T l
(
f
T i
f
T j
vp
)∗
⊗ · · · ⊗ (vn+1)
∗
+
∑
j 6=p
mj
(zj − zp)l
(v1)
∗ ⊗ · · · ⊗
(
f
T a−l
vp
)∗
⊗ · · · ⊗ (vn+1)
∗
]
= (κa+mp)(v1)
∗ ⊗ · · · ⊗
(
f
T a
vp
)∗
⊗ · · · ⊗ (vn+1)
∗
−
a∑
l=1
∑
j 6=p
mj
(zj − zp)l
(v1)
∗ ⊗ · · · ⊗
(
f
T a−l
vp
)∗
⊗ · · · ⊗ (v∗n+1)
+
∑
j 6=p
mj
(zj − zp)a
(v1)
∗ ⊗ · · · ⊗ (fvj)
∗ ⊗ · · · ⊗ (vn+1)
∗.
Twisted de Rham Complex on Line and Singular Vectors in ŝl2 Verma Modules 11
In this calculation we use formula (3.2) to express the action of f
Ta on (vp)
∗. These formulas
show that d
(
η0((t− zp)−a)
)
= η1
(
∂((t− zp)−a)
)
.
Now we calculate η1(∂(ta)),
ta
∂7→ 1
κ
(
aκ−
n∑
j=1
mj
)
ta−1dt− 1
κ
a−1∑
s=1
n∑
j=1
mjz
s
j t
a−s−1dt− 1
κ
n∑
j=1
mjz
a
j
dt
t− zj
η17→
(
aκ−
n∑
j=1
mj
)
(v1)
∗ ⊗ · · · ⊗ (vn)∗ ⊗
( e
T a
vn+1
)∗
−
a−1∑
s=1
n∑
j=1
mjz
s
j (v1)
∗ ⊗ · · · ⊗ (vn)∗ ⊗
( e
T a−s
vn+1
)∗
+
n∑
j=1
mjz
a
j (v1)
∗ ⊗ · · · ⊗ (fvj)
∗ ⊗ · · · ⊗ (vn+1)
∗.
Then we calculate d
(
η0(ta)
)
,
ta
η07→ fta ⊗ (v1)
∗ ⊗ · · · ⊗ (vn+1)
∗ −
a−2∑
l=0
[
htl+1 ⊗ (v1)
∗ ⊗ · · · ⊗ (vn)∗ ⊗
( e
T a−l−1
vn+1
)∗
+ etl ⊗ (v1)
∗ ⊗ · · · ⊗ (vn)∗ ⊗ 2
∑
i+j=a−l
i≥j≥1
( e
T i
e
T j
vn+1
)∗ ]
µ7→
n∑
j=1
mjz
a
j (v1)
∗ ⊗ · · · ⊗ (fvj)
∗ ⊗ · · · ⊗ (vn+1)
∗
+ (v1)
∗ ⊗ · · · ⊗ (vn)∗ ⊗
[
(−mn+1 − 2 + a(k + 2))
( e
T a
vn+1
)∗
+
a−2∑
l=0
[
− h
T l+1
( e
T a−l−1
vn+1
)∗
+ 2
f
T l
∑
i+j=a−l
i≥j≥1
( e
T i
e
T j
vn+1
)∗ ]]
− (v1)
∗ ⊗ · · · ⊗ (vn)∗ ⊗
a−2∑
l=0
[
2
f
T l
∑
i+j=a−l
i≥j≥1
( e
T i
e
T j
vn+1
)∗
− h
T l+1
( e
T a−l−1
vn+1
)∗ ]
−
a−1∑
s=1
n∑
j=1
mjz
s
j (v1)
∗ ⊗ · · · ⊗ (vn)∗ ⊗
( e
T a−s
vn+1
)∗
=
(
aκ−
n∑
j=1
mj
)
(v1)
∗ ⊗ · · · ⊗ (vn)∗ ⊗
( e
T a
vn+1
)∗
−
a−1∑
s=1
n∑
j=1
mjz
s
j (v1)
∗ ⊗ · · · ⊗ (vn)∗ ⊗
( e
T a−s
vn+1
)∗
+
n∑
j=1
mjz
a
j (v1)
∗ ⊗ · · · ⊗ (fvj)
∗ ⊗ · · · ⊗ (vn+1)
∗.
In this calculation we use formula (3.3) to express the action of e
Ta on (vn+1)
∗. Notice also that
calculating the action on V ∗n+1 we use the automorphism π, see Section 3.2.These formulas show
that d
(
η0(ta)
)
= η1(∂(ta)).
Clearly the maps η1, η2 are injective. Theorem 4.1 is proved. �
12 A. Slinkin and A. Varchenko
4.6 Image of logarithmic subcomplex
Under the monomorphism η of Theorem 4.1 the image of the logarithmic subcomplex (Ω•log(U), ∂)
is the chain complex C•
(
n−,⊗n+1
j=1V
∗
j
)
of the nilpotent subalgebra n− ⊂ sl2 generated by f . More
precisely, we have
η : 1 7→ f ⊗ (v1)
∗ ⊗ · · · ⊗ (vn+1)
∗,
dt
x− tj
7→ −κ(v1)
∗ ⊗ · · · ⊗ (fvj)
∗ ⊗ · · · ⊗ (vn+1)
∗,
j = 1, . . . , n, and
µ : f ⊗ (v1)
∗ ⊗ · · · ⊗ (vn+1)
∗ 7→
n∑
j=1
mj(v1)
∗ ⊗ · · · ⊗ (fvj)
∗ ⊗ · · · ⊗ (vn+1)
∗.
Far-reaching generalizations of this identification of the logarithmic subcomplex with the
chain complex of the nilpotent Lie algebra n− see in [8].
5 Proof of Theorem 3.2
5.1 Formula (3.3) follows from formula (3.2)
The Lie algebra ŝl2 has an automorphism ρ, corresponding to the involution of the Dynkin
diagram:
ρ(ei) = e3−i, ρ(fi) = f3−i, ρ(hi) = h3−i, i = 1, 2.
We have ρ2 = id. In other words, ρ acts by the formulas
e↔ fT, f ↔ eT−1, h↔ c− h.
Lemma 5.1. For i ∈ Z>0, we have
ρ :
f
T i
7→ e
T i+1
,
e
T i
7→ f
T i−1
,
h
T i
7→ − h
T i
.
Proof. We have
f
T
=
1
2
[
f,
h
T
]
=
1
2
[
f,
[ e
T
, f
]]
δ−→ 1
2
[ e
T
,
[
f,
e
T
]]
=
1
2
[
e
T
,− h
T
]
=
e
T 2
,
f
T i
=
1
2
[
f
T i−1
,
[ e
T
, f
]]
δ−→ 1
2
[ e
T i
,
[
f,
e
T
]]
=
1
2
[
e
T i
,− h
T
]
=
e
T i+1
.
Similarly we prove that ρ
(
e
T i
)
= f
T i−1 , ρ
(
h
T i
)
= − h
T i . �
Form ∈ C, let σm : ŝl2 → End(V (m, k−m)) be the Verma module structure. Let σm◦ρ : ŝl2 →
End(V (m, k −m)) be the twisted module structure.
Clearly the ŝl2-modules (σm ◦ ρ, V (m, k −m)) and (σm−k, V (m − k,m)) are isomorphic. If
vm ∈ V (m, k − m) and vk−m ∈ V (k − m,m) are generating vectors, then an isomorphism
χ : (σm ◦ ρ, V (m, k −m))→ (σm−k, V (m− k,m)) is defined by the formula,
fil · · · fi1vk−m 7→ f3−il · · · f3−i1vm,
for any i1, . . . , il ∈ {1, 2}. The isomorphism χ restricts to isomorphisms of the graded compo-
nents, V (k −m,m)(p1,p2) → V (m, k −m)(p2,p1).
Twisted de Rham Complex on Line and Singular Vectors in ŝl2 Verma Modules 13
In Section 3.6 we fixed bases of the homogeneous components V(p1,p2) with p1 6= p2 of any
Verma module V . By Lemma 5.1, under the isomorphism χ the chosen basis of V (k−m,m)(p1,p2)
is mapped to the chosen basis of V (m, k − m)(p2,p1) up to multiplication of the basis vectors
by ±1. This ±1 appears due to the formula ρ
(
h
T i
)
= − h
T i . In particular, we have
χ :
f
T i
vk−m 7→
e
T i+1
vm,
f
T i
f
T j
vk−m 7→
e
T i+1
e
T j+1
vm.
Let σ∗m : ŝl2 → End(V (m, k − m)∗) be the contragradient Verma module structure. Let
σ∗m ◦ ρ : ŝl2 → End(V (m, k−m)∗) be the twisted module structure. The isomorphism χ induces
an isomorphism of modules χ∗ : (σ∗m ◦ ρ, V (m, k −m)∗)→ (σ∗m−k, V (m− k,m)∗).
In Section 3.6 we fixed bases in the homogeneous components V ∗(p1,p2) with p1 6= p2 of any con-
tragradient Verma module V ∗. Under the isomorphism χ∗, the chosen basis of V (k−m,m)∗(p1,p2)
is mapped to the chosen basis of V (m, k − m)∗(p2,p1) up to multiplication of the basis vectors
by ±1. In particular, we have
χ∗ :
(
f
T i
vk−m
)∗
7→
( e
T i+1
vm
)∗
,
(
f
T i
f
T j
vk−m
)∗
7→
( e
T i+1
e
T j+1
vm
)∗
.
Assume that the relation in formula (3.2) holds in every contragradient Verma module V ∗.
Then in V (k −m,m)∗ it takes the form
f
T a−1
(vk−m)∗ = (−m− 2 + a(k + 2))
(
f
T a−1
vk−m
)∗
+
a−1∑
`=1
[
h
T `
(
f
T a−1−`
vk−m
)∗
+ 2
e
T `
∑
i+j=a−1−`
i≥j≥0
(
f
T i
f
T j
vk−m
)∗ ]
.
The isomorphism χ∗ sends this relation to the relation in V (m, k −m)∗,
e
T a
(vm)∗ = (−m− 2 + a(k + 2))
( e
T a
vm
)∗
+
a−1∑
`=1
[
− h
T `
( e
T a−`
vm
)∗
+ 2
f
T `−1
∑
i+j=a−1−`
i≥j≥0
( e
T i+1
e
T j+1
vm
)∗ ]
,
which is exactly the relation in formula (3.3). Thus formula (3.2) implies formula (3.3).
5.2 Auxiliary lemma
Let
V = V (m, k −m) and V ∗ = V (m, k −m)∗.
Lemma 5.2. For x ∈ V , φ ∈ V ∗, k ∈ Z≥0, we have〈
f
T k
ϕ, x
〉
=
〈
ϕ, eT kx
〉
,
〈 e
T k
ϕ, x
〉
=
〈
ϕ, fT kx
〉
,
〈
h
T k
ϕ, x
〉
=
〈
ϕ, hT kx
〉
.
Proof. The proof is by induction. We prove the first equality, the others are proved similarly.
We have [f2, f1] = h
T , hence [f1, [f2, f1]] = 2f
T . Similarly [e1, [e2, e1]] = 2eT . So for k = 1, we
have 〈
f
T
ϕ, x
〉
=
〈
1
2
[f1, [f2, f1]]ϕ, x
〉
=
〈
ϕ,
1
2
[[e1, e2], e1]x
〉
= 〈ϕ, eTx〉.
14 A. Slinkin and A. Varchenko
We have
[
f2,
f
Tk−1
]
= h
Tk , hence
[
f1,
[
f2,
f
Tk−1
]]
= 2f
Tk . Similarly,
[[
eT k−1, fT
]
, e
]
= 2eT k. Then〈
f
T k
ϕ, x
〉
=
〈
1
2
[
f1,
[
f2,
f
T k−1
]]
ϕ, x
〉
=
〈
ϕ,
[[
eT k−1, e2
]
, e1
]
x
〉
=
〈
ϕ, eT kx
〉
. �
5.3 The structure of the proof of formula (3.2)
We reformulate formula (3.2) as
(m+ (a− 1)(k + 2))
(
f
T a−1
v
)∗
=
f
T a−1
(v)∗ −
a−1∑
`=1
[
h
T `
(
f
T a−1−`
v
)∗
+ 2
e
T `
∑
i+j=a−1−`
i≥j≥0
(
f
T i
f
T j
v
)∗ ]
, (5.1)
and will prove it in this form.
Each term in (5.1) is an element of the homogeneous component V ∗(a,a−1). In Section 3.6 we
specified a basis of the dual component V(a,a−1). We will calculate the value of the right-hand
side in (5.1) on an arbitrary basis vector and will obtain the value of the left-hand side on that
vector.
The basis in V(a,a−1) consists of the vectors
f
T i1
· · · f
T ir
h
T j1
· · · h
T js
e
T l1
· · · e
T lr−1
v,
where
0 ≤ ir ≤ ir−1 ≤ · · · ≤ i1, 1 ≤ js ≤ js−1 ≤ · · · ≤ j1, 1 ≤ lr−1 ≤ lr−2 ≤ · · · ≤ l1;
r∑
u=1
iu +
s∑
u=1
ju +
r−1∑
u=1
lu = a− 1.
We partition the basis in four groups. Group O consists of the single basis vector f
Ta−1 v. Group I
consists of all basis vectors with r = 1, but different from f
Ta−1 v. Group II consists of all basis
vectors with r = 2. Group III consists of all basis vectors with r ≥ 3.
Notice that the value of the left-hand side of (5.1) on the basis vector f
Ta−1 v equals m +
(a − 1)(k + 2). Hence we need to show that the value of the right-hand side on the basis
vector f
Ta−1 v equals m + (a − 1)(k + 2). Similarly the value of the left-hand side on any basis
vector of Groups I–III equals zero. Hence we need to prove that the value of the right-hand
side on any basis vector of Groups I–III equals zero. These four statements are the content of
Propositions 5.3, 5.4, 5.7, and 5.9 below. These propositions prove Theorem 3.2.
5.4 Group O
Proposition 5.3. The value of the right-hand side of (5.1) on the basis vector f
Ta−1 v equals
m+ (a− 1)(k + 2).
Proof. By Lemma 5.2 we have〈
f
T a−1
(v)∗,
f
T a−1
v
〉
=
〈
(v)∗, eT a−1
f
T a−1
v
〉
=
〈
(v)∗,
[
h+ (a− 1)c+
f
T a−1
eT a−1
]
v
〉
= m+ (a− 1)k,
since eT a−1v is of degree (−a,−a+ 1), hence zero.
Twisted de Rham Complex on Line and Singular Vectors in ŝl2 Verma Modules 15
By Lemma 5.2, for ` ∈ {1, . . . , a− 1} we have〈
h
T `
(
f
T a−1−`
v
)∗
,
f
T a−1
v
〉
=
〈(
f
T a−1−`
v
)∗
, hT `
f
T a−1
v
〉
=
〈(
f
T a−1−`
v
)∗
,−2
f
T a−1−`
v
〉
= −2.
By Lemma 5.2 for ` ∈ {1, . . . , a− 1} we have〈
e
T `
∑
i+j=a−1−`
i≥j≥0
(
f
T i
f
T j
v
)∗
,
f
T a−1
v
〉
=
〈 ∑
i+j=a−1−`
i≥j≥0
(
f
T i
f
T j
v
)∗
, fT `
f
T a−1
v
〉
=
〈 ∑
i+j=a−1−`
i≥j≥0
(
f
T i
f
T j
v
)∗
,
f
T a−1
fT `v
〉
= 0,
since fT `v is of degree (−`+ 1,−`) ≤ (0,−1), hence zero. Therefore,〈
f
T a−1
(v)∗ −
a−1∑
`=1
[
h
T `
(
f
T a−1−`
v
)∗
+ 2
e
T `
∑
i+j=a−1−`
i≥j≥0
(
f
T i
f
T j
v
)∗ ]
,
f
T a−1
v
〉
= m+ (a− 1)k − (−2)(a− 1) = m+ (a− 1)(k + 2).
Proposition 5.3 is proved. �
5.5 Group I
Proposition 5.4. The value of the right-hand side of (5.1) on any basis vector of Group I
equals zero.
Proof. Group I consists of basis vectors of the form
w =
f
T a−1−n
h
T j1
· · · h
T js
v, where n ∈ {1, . . . , a− 1}, j1 + · · ·+ js = n, ji ≥ 1.
Lemma 5.5. In the notation above, if s = 1, then〈
f
T a−1
(v)∗, w
〉
= 2nk,
〈
h
T `
(
f
T a−1−`
v
)∗
, w
〉
=
2nk, if ` = n,
−4, if ` > a− 1− n,
0, if ` ≤ a− 1− n,
(5.2)
〈
e
T `
∑
i+j=a−1−`
i≥j≥0
(
f
T i
f
T j
v
)∗
, w
〉
=
{
2, if ` ≤ n,
0, if ` > n.
Note that the first line in (5.2) is not mutually exclusive with the second and third lines
in (5.2).
Proof. We have w = f
Ta−1−n
h
Tn v. Then〈
f
T a−1
(v)∗,
f
T a−1−n
h
Tn
v
〉
=
〈
(v)∗, eT a−1
f
T a−1−n
h
Tn
v
〉
16 A. Slinkin and A. Varchenko
=
〈
(v)∗,
[
hTn +
f
T a−1−n
eT a−1
]
h
Tn
v
〉
.
Note that eT a−1 h
Tn v is of degree (−a+ n,−a+ 1 + n) ≤ (−1, 0), so that eT a−1 h
Tn v = 0. Hence〈
(v)∗, hTn
h
Tn
v
〉
=
〈
(v)∗,
[
2nk +
h
Tn
hTn
]
v
〉
= 2nk.
We have〈
h
T `
(
f
T a−1−`
v
)∗
,
f
T a−1−n
h
Tn
v
〉
=
〈(
f
T a−1−`
v
)∗
, hT `
f
T a−1−n
h
Tn
v
〉
,
hT `
f
T a−1−n
h
Tn
v =
[
−2fT `+n−a+1 +
f
T a−1−n
hT `
]
h
Tn
v
= −2fT `+n−a+1 h
Tn
v +
f
T a−1−n
hT `
h
Tn
v.
Note that the second summand is nonzero if and only if ` = n. In that case we have〈(
f
T a−1−n
v
)∗
,
f
T a−1−n
hTn
h
Tn
v
〉
= 2nk.
For the first summand, if ` + n − a + 1 ≤ 0, then −2fT `+n−a+1 h
Tn v is a basis vector and so
pairing with
(
f
Ta−1−` v
)∗
gives zero. If `+ n− a+ 1 > 0, then〈(
f
T a−1−`
v
)∗
,−2fT `+n−a+1 h
Tn
v
〉
=
〈(
f
T a−1−`
v
)∗
,−2
[
2
f
T a−1−`
+
h
Tn
fT `+n−a+1
]
v
〉
= −4,
where we used fT `+n−a+1v = 0.
Finally,〈
e
T `
∑
i+j=a−1−`
i≥j≥0
(
f
T i
f
T j
v
)∗
,
f
T a−1−n
h
Tn
v
〉
=
〈 ∑
i+j=a−1−`
i≥j≥0
(
f
T i
f
T j
v
)∗
, fT `
f
T a−1−n
h
Tn
v
〉
,
fT `
f
T a−1−n
h
Tn
v =
f
T a−1−n
fT `
h
Tn
v =
f
T a−1−n
[
2fT `−n +
h
Tn
fT `
]
v = 2
f
T a−1−n
f
Tn−`
v,
since fT `v = 0. Note that (a− 1− n) + (n− `) = a− 1− `, hence if i = a− 1− n and j = n− `
(or vice versa depending on what is greater) we have〈 ∑
i+j=a−1−`
i≥j≥0
(
f
T i
f
T j
v
)∗
, 2
f
T a−1−n
f
Tn−`
v
〉
= 2,
whenever n− ` ≥ 0 and zero otherwise. The lemma is proved. �
For s = 1 Proposition 5.4 follows from Lemma 5.5:〈
f
T a−1
(v)∗ −
a−1∑
`=1
[
h
T `
(
f
T a−1−`
v
)∗
+ 2
e
T `
∑
i+j=a−1−`
i≥j≥0
(
f
T i
f
T j
v
)∗ ]
,
f
T a−1−n
h
Tn
v
〉
= 2nk − 2nk + 4n− 2 · 2n = 0.
Twisted de Rham Complex on Line and Singular Vectors in ŝl2 Verma Modules 17
Lemma 5.6. For s ≥ 2, we have〈
f
T a−1
(v)∗, w
〉
= 0, (5.3)
a−1∑
`=1
〈
h
T `
(
f
T a−1−`
v
)∗
, w
〉
= 0, (5.4)
a−1∑
`=1
〈
e
T `
∑
i+j=a−1−`
i≥j≥0
(
f
T i
f
T j
v
)∗
, w
〉
= 0. (5.5)
Proof. Recall that w = f
Ta−1−n
h
T j1
· · · h
T js v with j1 + · · ·+ js = n. We have〈
f
T a−1
(v)∗, w
〉
=
〈
(v)∗, eT a−1w
〉
,
eT a−1w = eT a−1
f
T a−1−n
h
T j1
· · · h
T js
v =
[
hTn +
f
T a−1−n
eT a−1
]
h
T j1
· · · h
T js
v.
We have hTn h
T j1
· · · h
T js v = 0, since hTn commutes with all h
T ji
. Indeed, we have n > ji since
j1 + · · ·+ js = n, ji ≥ 1, and s ≥ 2.
We also have f
Ta−1−n eT
a−1 h
T j1
· · · h
T js v = 0 since eT a−1 h
T j1
· · · h
T js v is of degree (−a+n,−a+
1 + n) ≤ (−1, 0), hence zero. This proves (5.3).
We prove (5.4) by induction on s. For s = 2 we have〈
h
T `
(
f
T a−1−`
v
)∗
, w
〉
=
〈(
f
T a−1−`
v
)∗
, hT `w
〉
,
hT `w = hT `
f
T a−1−n
h
T j1
h
T j2
v =
[
−2fT `−a+1+n +
f
T a−1−n
hT `
]
h
T j1
h
T j2
v
= −2fT `−a+1+n h
T j1
h
T j2
v +
f
T a−1−n
hT `
h
T j1
h
T j2
v.
Note that for f
Ta−1−nhT
` h
T j1
h
T j2
v to give a nonzero pairing with
( f
Ta−1−` v
)∗
we need ` = n, which
implies that hT ` commutes with h
T j1
and h
T j2
(` > ji since j1 + j2 = n = ` and ji ≥ 1), so that
f
Ta−1−nhT
` h
T j1
h
T j2
v gives zero for all `.
Also note that whenever ` ≤ a − 1 − n, fT `−a+1+n h
T j1
h
T j2
v is a basis vector and so pairing
with
( f
Ta−1−` v
)∗
gives zero. If ` > a− 1− n, then
−2fT `−a+1+n h
T j1
h
T j2
v = −2
[
2fT `−a+1+n−j1 +
h
T j1
fT `−a+1+n
]
h
T j2
v
= −4fT `−a+1+n−j1 h
T j2
v − 2
h
T j1
fT `−a+1+n h
T j2
v. (5.6)
If ` ≤ a − 1 − n + j1, the first summand gives zero when pairing with
( f
Ta−1−` v
)∗
, since for
such `fT `−a+1+n−j1 h
T j2
v is a basis vector. For ` > a− 1− n+ j1, we have
−4fT `−a+1+n−j1 h
T j2
v = −4
[
2
f
T a−1−`
v +
h
T j2
fT `−a+1+n−j1
]
v = −8
f
T a−1−`
v,
since fT `−a+1+n−j1v is of degree (−`+ a− 1− n+ j1 + 1,−`+ a− 1− n+ j1) ≤ (0,−1), hence
must be equal to zero. So for ` ∈ {a− 1− n+ j1 + 1, . . . , a− 1} we get〈(
f
T a−1−`
v
)∗
,−4fT `−a+1+n−j1 h
T j2
v
〉
= −8
18 A. Slinkin and A. Varchenko
and zero for other values of `. The total number of elements in the set {a−1−n+j1+1, . . . , a−1}
equals j2.
Whereas, for the second summand in (5.6) we have
−2
h
T j1
fT `−a+1+n h
T j2
v = −2
h
T j1
[
2fT `−a+1+n−j2 +
h
T j2
fT `−a+1+n
]
v
= −4
h
T j1
fT `−a+1+n−j2v,
since fT `−a+1+nv is of degree (−` + a − 1 − n + 1,−` + a − 1 − n) ≤ (0,−1), hence must be
equal to zero.
If ` > a−1−n+j2, then fT `−a+1+n−j2v is of degree (−`+a−1−n+j2+1,−`+a−1−n+j2) ≤
(0,−1), hence must be equal to zero. If ` ≤ a− 1− n+ j2, then
−4
h
T j1
fT `−a+1+n−j2v = −4
[
−2
f
T a−1−`
+ fT `−a+1+n−j2 h
T j1
]
= 8
f
T a−1−`
− 4fT `−a+1+n−j2 h
T j1
.
The second summand gives zero when pairing with
( f
Ta−1−` v
)∗
. So for ` ∈ {a−1−n+1, . . . , a−
1− n+ j2} we get〈(
f
T a−1−`
v
)∗
,−2
h
T j1
fT `−a+1+n h
T j2
v
〉
= 8
and zero for other values of `. The total number of elements in the set {a− 1− n+ 1, . . . , a−
1− n+ j2} equals j2.
Therefore,
a−1∑
`=1
〈
h
T `
(
f
T a−1−`
v
)∗
,
f
T a−1−n
h
T j1
h
T j2
v
〉
= −8j2 + 8j2 = 0
and so for s = 2 we proved (5.4).
Now suppose that (5.4) holds for all natural numbers up to s. Then
hT `
f
T a−1−n
h
T j1
. . .
h
T js+1
v =
[
−2fT `−a+1+n +
f
T a−1−n
hT `
]
h
T j1
· · · h
T js+1
v.
Note that for f
Ta−1−nhT
` h
T j1
· · · h
T js+1
v to give a nonzero pairing with
( f
Ta−1−` v
)∗
we need ` = n.
That assumption implies that hT ` commutes with h
T ji
for all i ∈ {1, . . . , s + 1} since ` > ji as
j1 + · · ·+ js+1 = n = ` and ji ≥ 1. Hence f
Ta−1−nhT
` h
T j1
· · · h
T js+1
v gives zero for all `.
Also note that whenever ` ≤ a− 1− n, the vector fT `−a+1+n h
T j1
· · · h
T js+1
v is a basis vector
and so pairing with
( f
Ta−1−` v
)∗
gives zero.
If ` > a− 1− n, then
−2fT `−a+1+n h
T j1
· · · h
T js+1
v = −2
[
2fT `−a+1+n−j1 +
h
T j1
fT `−a+1+n
]
h
T j2
· · · h
T js+1
v
= −4
f
T a−1−(n−j1)−`
h
T j2
· · · h
T js+1
v − 2
h
T j1
fT−a+1+n+` h
T j2
· · · h
T js+1
v. (5.7)
Note that by induction hypothesis we have
0 =
a−1∑
`=1
〈
h
T `
(
f
T a−1−`
v
)∗
,
f
T a−1−(n−j1)
h
T j2
· · · h
T js+1
v
〉
Twisted de Rham Complex on Line and Singular Vectors in ŝl2 Verma Modules 19
=
a−1∑
`=1
〈(
f
T a−1−`
v
)∗
, hT `
f
T a−1−(n−j1)
h
T j2
· · · h
T js+1
v
〉
.
So we add this zero term multiplied by −2 to the first summand in (5.7) to get
−2
a−1∑
`=1
〈(
f
T a−1−`
v
)∗
, 2
f
T a−1−(n−j1)−`
h
T j2
· · · h
T js+1
v
〉
− 2
a−1∑
`=1
〈(
f
T a−1−`
v
)∗
, hT `
f
T a−1−(n−j1)
h
T j2
· · · h
T js+1
v
〉
(5.8)
= −2
a−1∑
`=1
〈(
f
T a−1−`
v
)∗
,
[
2
f
T a−1−(n−j1)−`
+ hT `
f
T a−1−(n−j1)
]
h
T j2
. . .
h
T js+1
v
〉
= −2
a−1∑
`=1
〈(
f
T a−1−`
v
)∗
,
f
T a−1−(n−j1)
hT `
h
T j2
· · · h
T js+1
v
〉
,
where in the last step we use commutation relations.
Note that for f
Ta−1−(n−j1)
hT ` h
T j2
· · · h
T js+1
v to give a nonzero pairing with
( f
Ta−1−` v
)∗
we need
` = n− j1. That assumption implies that hT ` commutes with h
T ji
for all i ∈ {2, . . . , s+ 1} since
` > ji as j2 + · · ·+ js+1 = n− j1 = ` and ji ≥ 1. Hence f
Ta−1−(n−j1)
hT ` h
T j2
· · · h
T js+1
v gives zero
for all `.
For the second summand in (5.7) we have
−2
a−1∑
`=1
〈(
f
T a−1−`
v
)∗
,
h
T j1
fT−a+1+n+` h
T j2
· · · h
T js+1
v
〉
= −2
a−1∑
`=1
〈(
f
T a−1−`
v
)∗
,
h
T j1
[
2
f
T a−1−(n−j2)−`
+
h
T j2
fT−a+1+n+`
]
h
T j3
· · · h
T js+1
v
〉
= −4
a−1∑
`=1
〈(
f
T a−1−`
v
)∗
,
h
T j1
f
T a−1−(n−j2)−`
h
T j3
· · · h
T js+1
v
〉
(5.9)
− 2
a−1∑
`=1
〈(
f
T a−1−`
v
)∗
,
h
T j1
h
T j2
fT−a+1+n+` h
T j3
· · · h
T js+1
v
〉
. (5.10)
In (5.9) we note that
h
T j1
f
T a−1−(n−j2)−`
h
T j3
· · · h
T js+1
v
=
[
−2
f
T a−1−(n−j1−j2)−`
+
f
T a−1−(n−j2)−`
h
T j1
]
h
T j3
· · · h
T js+1
v
= −2
f
T a−1−(n−j1−j2)−`
h
T j3
· · · h
T js+1
v +
f
T a−1−(n−j2)−`
h
T j1
h
T j3
· · · h
T js+1
v.
Note that in both terms the number of h’s is less than or equal to s, so we use the exact same
reasoning as in (5.8) to show that
a−1∑
`=1
〈(
f
T a−1−`
v
)∗
,
f
T a−1−(n−j1−j2)−`
h
T j3
. . .
h
T js+1
v
〉
= 0,
a−1∑
`=1
〈(
f
T a−1−`
v
)∗
,
f
T a−1−(n−j2)−`
h
T j1
h
T j3
· · · h
T js+1
v
〉
= 0,
20 A. Slinkin and A. Varchenko
which implies that the expression in (5.9) equals zero. Similarly, one shows that in (5.10),
a−1∑
`=1
〈(
f
T a−1−`
v
)∗
,
h
T j1
h
T j2
fT−a+1+n+` h
T j3
· · · h
T js+1
v
〉
the factor fT−a+1+n+` can be pulled to the right by using the same argument (first commute
fT−a+1+n+` with h
T j3
and then pull fT−a+1+n−j3+` to the left). Ultimately, we get
a−1∑
`=1
〈(
f
T a−1−`
v
)∗
,
h
T j1
h
T j2
h
T j3
· · · h
T js+1
fT−a+1+n+`v
〉
= 0,
since ` > a− 1− n and so fT−a+1+n+`v = 0. Therefore,
a−1∑
`=1
〈
h
T `
(
f
T a−1−`
v
)∗
,
f
T a−1−n
h
T j1
· · · h
T js
v
〉
= 0,
and formula (5.4) is proved.
We prove formula (5.5) by induction on s. For s = 2, we have〈
e
T `
∑
i+j=a−1−`
i≥j≥0
(
f
T i
f
T j
v
)∗
, w
〉
=
〈 ∑
i+j=a−1−`
i≥j≥0
(
f
T i
f
T j
v
)∗
, fT `w
〉
,
fT `w = fT `
f
T a−1−n
h
T j1
h
T j2
v =
f
T a−1−n
fT `
h
T j1
h
T j2
v.
Note that fT ` h
T j1
h
T j2
v is of degree (n− `+ 1, n− `), hence nonzero only if ` ≤ n. For such ` we
have
f
T a−1−n
[
2fT `−j1 +
h
T j1
ft`
]
h
T j2
v = 2
f
T a−1−n
fT `−j1
h
T j2
v +
f
T a−1−n
h
T j1
fT `
h
T j2
v
= 2
f
T a−1−n
fT `−j1
h
T j2
v + 2
f
T a−1−n
h
T j1
fT `−j2v. (5.11)
If ` ≤ j1, then the first summand in (5.11) gives zero when pairing with any vector with two
f ’s. If ` > j1, then
2
f
T a−1−n
fT `−j1
h
T j2
v = 2
f
T a−1−n
[
2fT `−j1−j2 +
h
T j2
fT `−j1
]
v = 4
f
T a−1−n
f
Tn−`
v.
If ` > j2, then the second summand in (5.11) is zero simply because fT `−j2v = 0. If ` ≤ j2,
then
2
f
T a−1−n
h
T j1
fT `−j2v = 2
f
T a−1−n
[
−2fT `−j1−j2 + fT `−j2
h
T j1
]
v = −4
f
T a−1−n
f
Tn−`
v,
since f
Ta−1−n fT
`−j2 h
T j1
v is a basis vector, hence pairing with a vector consisting of two f ’s gives
zero. Therefore,
a−1∑
`=1
〈 ∑
i+j=a−1−`
i≥j≥0
(
f
T i
f
T j
v
)∗
, fT `w
〉
Twisted de Rham Complex on Line and Singular Vectors in ŝl2 Verma Modules 21
=
a−1∑
`=1
〈 ∑
i+j=a−1−`
i≥j≥0
(
f
T i
f
T j
v
)∗
, 4
f
T a−1−n
f
Tn−`
v
〉
(5.12)
+
a−1∑
`=1
〈 ∑
i+j=a−1−`
i≥j≥0
(
f
T i
f
T j
v
)∗
,−4
f
T a−1−n
f
Tn−`
v
〉
. (5.13)
Note that in the expression in (5.12) for each ` ∈ {j1 + 1, . . . , n} there exists exactly one pair of
indices (i, j) = (max{a−1−n, n− `},min{a−1−n, n− `}) that gives 4 when pairing. All other
pairs (i, j) give zero. Similarly, the expression in (5.13) equals −4 for each ` ∈ {1, . . . , j2} and
exactly one corresponding pair (i, j), and zero otherwise. Also note that the number of elements
in each set {j1 + 1, . . . , n} and {1, . . . , j2} equals j2. Hence we get
4j2 − 4j2 = 0.
Therefore, formula (5.5) is proved for s = 2.
Now suppose that formula (5.5) holds for all natural numbers up to s. Then
fT `
f
T a−1−n
h
T j1
· · · h
T js+1
v =
f
T a−1−n
fT `
h
T j1
· · · h
T js+1
v
=
f
T a−1−n
[
2fT `−j1 +
h
T j1
fT `
]
h
T j2
· · · h
T js+1
v
= 2
f
T a−1−n
fT `−j1
h
T j2
· · · h
T js+1
v +
f
T a−1−n
h
T j1
fT `
h
T j2
· · · h
T js+1
v. (5.14)
Note that if ` ≤ j1, then the first summand in (5.14) is a basis vector and hence its pairing
with a vector consisting of two f ’s gives zero. If ` > j1 we have
a−1∑
`=1
〈 ∑
i+j=a−1−`
i≥j≥0
(
f
T i
f
T j
v
)∗
,
f
T a−1−n
fT `−j1
h
T j2
· · · h
T js+1
v
〉
=
a−1∑
`=j1+1
〈 ∑
i+j=a−1−`
i≥j≥0
(
f
T i
f
T j
v
)∗
,
f
T a−1−n
fT `−j1
h
T j2
· · · h
T js+1
v
〉
=
a−1∑
`=j1+1
〈
e
T `−j1
∑
i+j=a−1−`
i≥j≥0
(
f
T i
f
T j
v
)∗
,
f
T a−1−n
h
T j2
· · · h
T js+1
v
〉
=
a−1−j1∑
k=1
〈
e
T k
∑
i+j=a−1−j1−k
i≥j≥0
(
f
T i
f
T j
v
)∗
,
f
T (a−1−j1)−(n−j1)
h
T j2
· · · h
T js+1
v
〉
= 0
by induction hypothesis. For the second summand in (5.14) we have
f
T a−1−n
h
T j1
fT `
h
T j2
· · · h
T js+1
v =
f
T a−1−n
h
T j1
[
2fT `−j2 +
h
T j2
fT `
]
h
T j3
· · · h
T js+1
v
= 2
f
T a−1−n
h
T j1
fT `−j2
h
T j3
· · · h
T js+1
v +
f
T a−1−n
h
T j1
h
T j2
fT `
h
T j3
· · · h
T js+1
v. (5.15)
Note that
f
T a−1−n
h
T j1
fT `−j2
h
T j3
· · · h
T js+1
v =
f
T a−1−n
[
−2fT `−j1−j2 + fT `−j2
h
T j1
]
h
T j3
· · · h
T js+1
v
22 A. Slinkin and A. Varchenko
= −2
f
T a−1−n
fT `−j1−j2
h
T j3
· · · h
T js+1
v +
f
T a−1−n
fT `−j2
h
T j1
h
T j3
· · · h
T js+1
v,
where in each vector the number of h’s is less than or equal to s. Repeating the argument above,
we see that by induction hypothesis we get
a−1∑
`=1
〈 ∑
i+j=a−1−`
i≥j≥0
(
f
T i
f
T j
v
)∗
,
f
T a−1−n
h
T j1
fT `−j2
h
T j3
· · · h
T js+1
v
〉
= 0.
Now in the second summand in (5.15),
f
T a−1−n
h
T j1
h
T j2
fT `
h
T j3
· · · h
T js+1
v,
we pull fT ` to the right and at each step we use induction hypothesis to argue that we keep
getting zeros. Ultimately, we get a vector
f
T a−1−n
h
T j1
· · · h
T js+1
fT `v,
which is zero, since fT ` has grading (−`+ 1,−`) ≤ (0,−1) and so fT `v = 0. Therefore,
a−1∑
`=1
〈
e
T `
∑
i+j=a−1−`
i≥j≥0
(
f
T i
f
T j
v
)∗
,
f
T a−1−n
h
T j1
· · · h
T js
v
〉
= 0.
Formula (5.5) and Lemma 5.6 are proved. �
Proposition 5.4 is proved. �
5.6 Group II
Proposition 5.7. The value on the right-hand side of (5.1) on any basis vector from Group II
equals zero.
Proof. Group II consists of vectors
w =
f
T i1
f
T i2
h
T j1
· · · h
T js
e
T l
v.
Lemma 5.8. We have〈
f
T a−1
(v)∗, w
〉
= 2s+1(m− lk), (5.16)
〈
h
T `
(
f
T a−1−`
v
)∗
, w
〉
=
2s+2(m− lk), if i1 = i2 = a− 1− `,
2s+1(m− lk), if i1 6= i2 and i1 or i2 = a− 1− `,
0, otherwise,
(5.17)
〈
e
T `
∑
i+j=a−1−`
i≥j≥0
(
f
T i
f
T j
v
)∗
, w
〉
=
{
−2s(m− lk), if ` = a− 1− i1 − i2,
0, otherwise.
(5.18)
Twisted de Rham Complex on Line and Singular Vectors in ŝl2 Verma Modules 23
Proof. We have〈
f
T a−1
(v)∗, w
〉
=
〈
(v)∗, eT a−1w
〉
=
〈
(v)∗,
[
hT a−1−i1 +
f
T i1
eT a−1
]
f
T i2
h
T j1
· · · h
T js
e
T l
v
〉
=
〈
(v)∗, hT a−1−i1
f
T i2
h
T j1
· · · h
T js
e
T l
v
〉
+
〈
(v)∗,
f
T i1
eT a−1
f
T i2
h
T j1
· · · h
T js
e
T l
v
〉
. (5.19)
Note that eT a−1 f
T i2
h
T j1
· · · h
T js
e
T l is of degree (−i1 − 1,−i1) ≤ (−1, 0), hence
eT a−1 f
T i2
h
T j1
· · · h
T js
e
T l v = 0. In the first summand in (5.19) we pull hT a−1−i1 to the right to get〈
(v)∗,−2fT a−1−i1−i2
h
T j1
· · · h
T js
e
T l
v
〉
= · · · =
〈
(v)∗,−2s+1fT a−1−i1−i2−j1−···−as
e
T l
v
〉
=
〈
(v)∗,−2s+1fT l
e
T l
v
〉
=
〈
(v)∗, 2s+1(h− lc)v
〉
= 2s+1(m− lk),
where at each step we do not write monomials of negative degree, since they give zero when
applied to v. This proves formula (5.16).
We have〈
h
T `
(
f
T a−1−`
v
)∗
, w
〉
=
〈(
f
T a−1−`
v
)∗
, hT `w
〉
,
hT `w =
[
−2fT `−i1 +
f
T i1
hT `
]
f
T i2
h
T j1
· · · h
T js
e
T l
v
= −2fT `−i1
f
T i2
h
T j1
· · · h
T js
e
T l
v +
f
T i1
[
−2fT `−i2 +
f
T i2
hT `
]
h
T j1
· · · h
T js
e
T l
v
=
[
−2
f
T i2
fT `−i1 − 2
f
T i1
fT `−i2 +
f
T i1
f
T i2
hT `
]
h
T j1
· · · h
T js
e
T l
v.
Note that the vector f
T i1
f
T i2
hT ` h
T j1
· · · h
T js
e
T l v after pulling hT ` to the right either becomes
a zero vector or a vector with two f ’s, which of course gives zero when pairing with a basis
vector with one f . Also, note that the only possibility for the vector f
T i2
fT `−i1 h
T j1
· · · h
T js
e
T l v to
give a nonzero pairing with
( f
Ta−1−` v
)∗
is when i2 = a−1− `. Similarly f
T i1
fT `−i2 h
T j1
· · · h
T js
e
T l v
gives a nonzero number only if i1 = a − 1 − `. First consider the case i1 = i2 = a − 1 − `. We
have
−4
f
T a−1−`
fT 2`−a+1 h
T j1
· · · h
T js
e
T l
v
= −4
f
T a−1−`
[
2fT 2`−a+1−j1 +
h
T j1
fT 2`−a+1
]
h
T j2
· · · h
T js
e
T l
v.
Note that fT 2`−a+1 h
T j2
· · · h
T js
e
T l is of degree (−j1,−j1) ≤ (−1,−1), hence
fT 2`−a+1 h
T j2
· · · h
T js
e
T l
v = 0.
So we get
−8
f
T a−1−`
fT 2`−a+1−j1 h
T j2
· · · h
T js
e
T l
v · · · = −2s+2 f
T a−1−`
fT 2`−a+1−j1−···−js e
T l
v
= −2s+2 f
T a−1−`
fT l
e
T l
v = 2s+2 f
T a−1−`
(h− lc)v = 2s+2(m− lk),
where at each step we don’t write monomials of negative degree, since they give zero when
applied to v.
24 A. Slinkin and A. Varchenko
For i1 = a− 1− ` 6= i2 and i2 = a− 1− ` 6= i1 we have
−2
f
T a−1−`
fT 2`−a+1 h
T j1
. . .
h
T js
e
T l
v = 2s+1(m− lk),
where we performed the exact same computation as above. Therefore, formula (5.17) is proved.
We have〈
e
T `
∑
i+j=a−1−`
i≥j≥0
(
f
T i
f
T j
v
)∗
, w
〉
=
〈 ∑
i+j=a−1−`
i≥j≥0
(
f
T i
f
T j
v
)∗
, fT `w
〉
,
fT `w =
f
T i1
f
T i2
fT `
h
T j1
· · · h
T js
e
T l
v =
f
T i1
f
T i2
[
2fT `−j1 +
h
T j1
fT `
]
h
T j2
· · · h
T js
e
T l
v.
The only nonzero pairing happens when ` is such that i1 + i2 = a − 1 − `. In that case
fT ` h
T j2
· · · h
T js
e
T l has degree (−j1,−j1) ≤ (−1,−1), hence fT ` h
T j2
· · · h
T js
e
T l v = 0. Therefore we
have
2
f
T i1
f
T i2
fT `−j1
h
T j2
· · · h
T js
e
T l
v = 2s
f
T i1
f
T i2
fT `−j1−···−js
e
T l
v = 2s
f
T i1
f
T i2
fT l
e
T l
v,
where we pulled fT `−j1 to the right and did not write monomials of negative degree, since they
give zero when applied to v. Hence we get
2s
f
T i1
f
T i2
(−h+ lc)v.
Therefore,〈
e
T `
∑
i+j=a−1−`
i≥j≥0
(
f
T i
f
T j
v
)∗
, w
〉
=
∑
i+j=a−1−`
i≥j≥0
〈(
f
T i
f
T j
v
)∗
, 2s
f
T i1
f
T i2
(−h+ lc)v
〉
= −2s(m− lk),
since for i = i1, j = i2 we get −2s(m − lk) and zero for other pairs (i, j). Formula (5.18) and
Lemma 5.8 are proved. �
By Lemma 5.8, we have〈
f
T a−1
(v)∗ −
a−1∑
`=1
[
h
T `
(
f
T a−1−`
v
)∗
+ 2
e
T `
∑
i+j=a−1−`
i≥j≥0
(
f
T i
f
T j
v
)∗ ]
, w
〉
= 2s+1(m− lk)− 2 · 2s+1(m− lk) + 2 · 2s(m− lk) = 0.
Note that〈 a−1∑
`=1
h
T `
(
f
T a−1−`
v
)∗
, w
〉
= 2 · 2s+1(m− lk)
in both cases i1 = i2 and i1 6= i2. Also note that〈 a−1∑
`=1
e
T `
∑
i+j=a−1−`
i≥j≥0
(
f
T i
f
T j
v
)∗
, w
〉
6= 0
only if ` is such that i1 + i2 = a− 1− `. Therefore, Proposition 5.7 is proved. �
Twisted de Rham Complex on Line and Singular Vectors in ŝl2 Verma Modules 25
5.7 Group III
Proposition 5.9. The value of the right-hand side of (5.1) on any basis vector of Group III
equals zero.
Proof. A vector in Group III has the form
w =
f
T i1
· · · f
T ir
h
T j1
· · · h
T js
e
T l1
· · · e
T lr−1
v,
where r ≥ 3.
Lemma 5.10. For every ` ∈ {1, . . . , a− 1}, we have〈
f
T a−1
(v)∗, w
〉
= 0, (5.20)〈
h
T `
(
f
T a−1−`
v
)∗
, w
〉
= 0, (5.21)〈
e
T `
∑
i+j=a−1−`
i≥j≥0
(
f
T i
f
T j
v
)∗
, w
〉
= 0. (5.22)
Proof. We have〈
f
T a−1
(v)∗, w
〉
=
〈
(v)∗, eT a−1w
〉
=
〈
(v)∗,
[
hT a−1−i1 +
f
T i1
eT a−1
]
f
T i2
· · · f
T ir
h
T j1
· · · h
T js
e
T l1
· · · e
T lr−1
v
〉
.
Note that eT a−1 f
T i2
· · · f
T ir
h
T j1
· · · h
T js
e
T l1
· · · e
T lr−1
v is of degree (−i1 − 1,−i1) ≤ (−1, 0), hence
zero. So we have〈
(v)∗,
[
−2fT a−1−i1−i2 +
f
T i2
hT a−1−i1
]
f
T i3
· · · f
T ir
h
T j1
· · · h
T js
e
T l1
· · · e
T lr−1
v
〉
.
As above, note that hT a−1−i1 f
T i3
· · · f
T ir
h
T j1
· · · h
T js
e
T l1
· · · e
T lr−1
v is of degree (−i2 − 1,−i2) ≤
(−1, 0), hence zero. Therefore we obtain〈
(v)∗,−2
f
T i3
· · · f
T ir
fT a−1−i1−i2
h
T j1
· · · h
T js
e
T l1
· · · e
T lr−1
v
〉
= 0,
since fT a−1−i1−i2 h
T j1
· · · h
T js
e
T l1
· · · e
T lr−1
v is of degree (−i3−· · ·−ir−r+2,−i2−· · ·−ir) ≤ (−1, 0)
for r ≥ 3, hence zero. Formula (5.20) is proved.
We have〈
h
T `
(
f
T a−1−`
v
)∗
, w
〉
=
〈(
f
T a−1−`
v
)∗
, hT `w
〉
,
hT `w =
[
−2fT `−i1 +
f
T i1
hT `
]
f
T i2
· · · f
T ir
h
T j1
· · · h
T js
e
T l1
· · · e
T lr−1
v
= −2
f
T i2
· · · f
T ir
fT `−i1
h
T j1
· · · h
T js
e
T l1
· · · e
T lr−1
v (5.23)
+
f
T i1
hT `
f
T i2
· · · f
T ir
h
T j1
· · · h
T js
e
T l1
· · · e
T lr−1
v. (5.24)
26 A. Slinkin and A. Varchenko
Observe that for ` ≤ i1 in (5.23) we have a basis vector, hence it gives zero when pairing with( f
Ta−1−` v
)∗
. If ` > i1, then we pull fT `−i1 to the right and notice that no matter how fT `−i1
interacts with h’s and e’s, it does not affect the number of f ’s, which is greater or equal than
two. Hence, the vector in (5.23) gives zero when pairing with
( f
Ta−1−` v
)∗
.
In (5.24) note that
hT `
f
T i2
= −2fT `−i2 +
f
T i2
hT `
so that either hT ` is pulled to the right not affecting the number of f ’s or it gives fT `−i2 , for
which we apply the same argument as above after pulling it to the right to argue that the pairing
of the vector in (5.24) with
( f
Ta−1−` v
)∗
is zero. Formula (5.21) is proved.
We have〈
e
T `
∑
i+j=a−1−`
i≥j≥0
(
f
T i
f
T j
v
)∗
, w
〉
=
〈 ∑
i+j=a−1−`
i≥j≥0
(
f
T i
f
T j
v
)∗
, fT `w
〉
,
fT `w = fT `
f
T i1
· · · f
T ir
h
T j1
· · · h
T js
e
T l1
· · · e
T lr−1
v
=
f
T i1
· · · f
T ir
fT `
h
T j1
· · · h
T js
e
T l1
· · · e
T lr−1
v.
As in formula (5.21), no matter how fT ` interacts with h’s and e’s, the number of f ’s remains
unchanged, i.e., we have more than or equal to three f ’s, so that pairing with
( f
T i
f
T j v
)∗
is zero.
Formula (5.22) is proved. �
Proposition 5.9 follows from Lemma 5.10. �
Theorem 3.2 is proved.
Acknowledgements
The authors thank V. Schechtman for useful discussions. The second author was supported in
part by NSF grant DMS-1665239.
References
[1] Feigin B., Schechtman V., Varchenko A., On algebraic equations satisfied by hypergeometric correlators in
WZW models. I, Comm. Math. Phys. 163 (1994), 173–184.
[2] Feigin B., Schechtman V., Varchenko A., On algebraic equations satisfied by hypergeometric correlators in
WZW models. II, Comm. Math. Phys. 170 (1995), 219–247, arXiv:hep-th/9407010.
[3] Kac V.G., Kazhdan D.A., Structure of representations with highest weight of infinite-dimensional Lie alge-
bras, Adv. Math. 34 (1979), 97–108.
[4] Khoroshkin S., Schechtman V., Factorizable D-modules, Math. Res. Lett. 4 (1997), 239–257,
arXiv:q-alg/9611018.
[5] Khoroshkin S., Varchenko A., Quiver D-modules and homology of local systems over an arrangement of
hyperplanes, IInt. Math. Res. Pap. 2006 (2006), 69590, 116 pages, arXiv:math.QA/0510451.
[6] Malikov F., Feigin B., Fuks D., Singular vectors in Verma modules over Kac–Moody algebras, Funct. Anal.
Appl. 20 (1986), 103–113.
[7] Schechtman V., Terao H., Varchenko A., Local systems over complements of hyperplanes and the Kac–
Kazhdan conditions for singular vectors, J. Pure Appl. Algebra 100 (1995), 93–102, arXiv:hep-th/9411083.
[8] Schechtman V., Varchenko A., Arrangements of hyperplanes and Lie algebra homology, Invent. Math. 106
(1991), 139–194.
[9] Schechtman V., Varchenko A., Rational differential forms on the line and singular vectors in Verma modules
over ŝl2, Mosc. Math. J. 17 (2017), 787–802, arXiv:1511.09014.
https://doi.org/10.1007/BF02101739
https://doi.org/10.1007/BF02099447
https://arxiv.org/abs/hep-th/9407010
https://doi.org/10.1016/0001-8708(79)90066-5
https://doi.org/10.4310/MRL.1997.v4.n2.a6
https://arxiv.org/abs/q-alg/9611018
https://doi.org/10.1155/IMRP/2006/69590
https://arxiv.org/abs/math.QA/0510451
https://doi.org/10.1007/BF01077264
https://doi.org/10.1007/BF01077264
https://doi.org/10.1016/0022-4049(95)00014-N
https://arxiv.org/abs/hep-th/9411083
https://doi.org/10.1007/BF01243909
https://doi.org/10.17323/1609-4514-2016-16-4-787-802
https://arxiv.org/abs/1511.09014
1 Introduction
2 The de Rham complex of master function
2.1 Twisted de Rham complex
2.2 Basis of (U)
2.3 Resonances
2.4 Logarithmic subcomplex
3 sl2"0362sl2-modules
3.1 Lie algebra sl2"0362sl2
3.2 Automorphism
3.3 Verma modules
3.4 Reducibility conditions
3.5 Shapovalov form
3.6 Bases in V and V*
3.7 Main formula
3.8 Relation to Malikov–Feigin–Fuchs vectors
4 Homomorphism of complexes
4.1 Lie algebra sl2(U)
4.2 sl2(U)-modules
4.3 Chain complex
4.4 Two complexes
4.4.1
4.4.2
4.5 Construction
4.6 Image of logarithmic subcomplex
5 Proof of Theorem 3.2
5.1 Formula (3.3) follows from formula (3.2)
5.2 Auxiliary lemma
5.3 The structure of the proof of formula (3.2)
5.4 Group O
5.5 Group I
5.6 Group II
5.7 Group III
References
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| id | nasplib_isofts_kiev_ua-123456789-210220 |
| institution | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| issn | 1815-0659 |
| language | English |
| last_indexed | 2025-12-07T21:24:47Z |
| publishDate | 2019 |
| publisher | Інститут математики НАН України |
| record_format | dspace |
| spelling | Slinkin, A. Varchenko, A. 2025-12-04T13:00:07Z 2019 Twisted de Rham Complex on Line and Singular Vectors in sl₂ˆ Verma Modules / A. Slinkin, A. Varchenko // Symmetry, Integrability and Geometry: Methods and Applications. — 2019. — Т. 15. — Бібліогр.: 9 назв. — англ. 1815-0659 2010 Mathematics Subject Classification: 17B56; 17B67; 33C80 arXiv: 1812.09791 https://nasplib.isofts.kiev.ua/handle/123456789/210220 https://doi.org/10.3842/SIGMA.2019.075 We consider two complexes. The first complex is the twisted de Rham complex of scalar meromorphic differential forms on the projective line, holomorphic on the complement to a finite set of points. The second complex is the chain complex of the Lie algebra of sl₂-valued algebraic functions on the same complement, with coefficients in a tensor product of contragradient Verma modules over the affine Lie algebra sl₂ˆ. In [Schechtman V., Varchenko A., Mosc. Math. J. 17 (2017), 787-802] a construction of a monomorphism of the first complex to the second was suggested, and it was indicated that under this monomorphism, the existence of singular vectors in the Verma modules (the Malikov-Feigin-Fuchs singular vectors) is reflected in the relations between the cohomology classes of the de Rham complex. In this paper, we prove these results. The authors thank V. Schechtman for useful discussions. The second author was supported in part by NSF grant DMS-1665239. en Інститут математики НАН України Symmetry, Integrability and Geometry: Methods and Applications Twisted de Rham Complex on Line and Singular Vectors in sl₂ˆ Verma Modules Article published earlier |
| spellingShingle | Twisted de Rham Complex on Line and Singular Vectors in sl₂ˆ Verma Modules Slinkin, A. Varchenko, A. |
| title | Twisted de Rham Complex on Line and Singular Vectors in sl₂ˆ Verma Modules |
| title_full | Twisted de Rham Complex on Line and Singular Vectors in sl₂ˆ Verma Modules |
| title_fullStr | Twisted de Rham Complex on Line and Singular Vectors in sl₂ˆ Verma Modules |
| title_full_unstemmed | Twisted de Rham Complex on Line and Singular Vectors in sl₂ˆ Verma Modules |
| title_short | Twisted de Rham Complex on Line and Singular Vectors in sl₂ˆ Verma Modules |
| title_sort | twisted de rham complex on line and singular vectors in sl₂ˆ verma modules |
| url | https://nasplib.isofts.kiev.ua/handle/123456789/210220 |
| work_keys_str_mv | AT slinkina twistedderhamcomplexonlineandsingularvectorsinsl2ˆvermamodules AT varchenkoa twistedderhamcomplexonlineandsingularvectorsinsl2ˆvermamodules |