Twisted de Rham Complex on Line and Singular Vectors in sl 2 ^ Verma Modules
SSymmetry, Integrability and Geometry: Methods and Applications SIGMA (2019), 075, 26 pages Twisted de Rham Complex on Lineand Singular Vectors in (cid:99) sl 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: [email protected], [email protected]
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, 2019https://doi.org/10.3842/SIGMA.2019.075
Abstract.
We consider two complexes. The first complex is the twisted de Rham complexof scalar meromorphic differential forms on projective line, holomorphic on the complementto a finite set of points. The second complex is the chain complex of the Lie algebra of sl -valued algebraic functions on the same complement, with coefficients in a tensor prod-uct of contragradient Verma modules over the affine Lie algebra (cid:99) sl . In [Schechtman V.,Varchenko A., Mosc. Math. J. (2017), 787–802] a construction of a monomorphism of thefirst 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–Fuchssingular vectors) is reflected in the relations between the cohomology classes of the de Rhamcomplex. In this paper we prove these results. Key words: twisted de Rham complex; logarithmic differential forms; (cid:99) sl -modules; Lie alge-bra chain complexes Dedicated to Dmitry Borisovich Fuchson the occasion of his 80-th birthday
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 finiteset of points. The second complex is the chain complex of the Lie algebra of sl -valued algebraicfunctions on the same complement, with coefficients in a tensor product of contragradient Vermamodules over the affine Lie algebra (cid:99) sl . In [9] a construction of a monomorphism of the firstcomplex to the second was suggested. That construction gives a relation between the singularvectors 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. Thepaper [9] provides supporting evidence to the results formulated in [9], but not the proofs. Thegoal of this paper is to give the proofs to the results formulated in [9], namely, the proofs thatthe construction in [9] indeed gives a homorphism of complexes and relates the resonances inthe de Rham complex and the (cid:99) sl singular vectors. a r X i v : . [ m a t h . AG ] S e p A. Slinkin and A. VarchenkoThe construction in [9] has two motivations.The first motivation was to generalize the principal construction of [8]. In [8], the tensorproducts of contragradient Verma modules over a semisimple Lie algebra were identified withthe spaces of the top degree logarithmic differential forms over certain configuration spaces. Alsothe logarithmic parts of the de Rham complexes over the configuration spaces were identifiedwith 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 Liealgebras is reflected in the relations between the cohomology classes of logarithmic differentialforms. 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 thiscorrespondence is developed for another non-trivial class of singular vectors, namely for (a partof) Malikov–Feigin–Fuchs singular vectors, cf. [6].The paper has the following structure. In Section 2 we introduce the de Rham complex ofa master function and resonance relations. In Section 3 we discuss (cid:99) sl Verma modules, the Kac–Kazhdan reducibility conditions. We formulate Theorem 3.2 which describes certain relations ina 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 deRham complex of the master function to the chain complex of the Lie algebra of sl -valuedalgebraic functions. Theorem 4.1 says that the map is a monomorphism of complexes. Theproof of Theorem 4.1 is the second new result of this paper. Section 5 is devoted to the proofof Theorem 3.2. The proof is straightforward but rather nontrivial and lengthy. Consider C with coordinate t . Define the master function by the formulaΦ( t ) = n (cid:89) i =1 ( t − z i ) − m i /κ , where z , . . . , z n , m , . . . , m n , κ ∈ C are parameters. Fix these parameters and assume that z , . . . , z n are distinct. Set z n +1 = ∞ , m n +1 = m + · · · + m n − . Denote U = C − { z , . . . , z n } .Consider the twisted de Rham complex associated with Φ,0 −→ Ω ( U ) ∂ −→ Ω ( U ) −→ . (2.1)Here Ω p ( U ) is the space of rational differential p -forms on C regular on U . The differential ∂ isgiven by the formula ∂ = d + α ∧ · , (2.2)where d is the standard de Rham differential and the second summand is the left exteriormultiplication by the form α = − κ n (cid:88) i =1 m i d tt − z i = dΦΦ . wisted de Rham Complex on Line and Singular Vectors in (cid:99) sl Verma Modules 3Formula (2.2) is motivated by the computationd(Φ ω ) = Φd ω + dΦ ∧ ω = Φ(d ω + α ∧ ω ) . The complex Ω • ( U ) is the complex of global algebraic sections of the de Rham complex of (cid:0) O an U , ∂ (cid:1) , where ∂ = d + α ∧ · is considered as the integrable connection on the sheaf O an U ofholomorphic functions on U .If the monodromy of Φ is non-trivial, that is, if at least one of the numbers m /κ, . . . , m n /κ is not an integer, then H (Ω • ( U )) = 0 , dim H (Ω • ( U )) = n − , see for example [7]. • ( U ) The functions1( t − z i ) a for a ∈ Z > and t a for a ∈ Z ≥ form a basis of Ω ( U ). The differential formsd t ( t − z i ) a for a ∈ Z > and t a d t for a ∈ Z ≥ form a basis of Ω ( U ). The differential ∂ is given by the formulas κ∂ (cid:18) t − z i ) a (cid:19) = − ( m i + aκ ) d t ( t − z i ) a +1 + a (cid:88) k =1 (cid:88) j (cid:54) = i m j ( z j − z i ) k d t ( t − z i ) a +1 − k − (cid:88) j (cid:54) = i m j ( z j − z i ) a d tt − z j , (2.3) κ∂ (cid:0) t a (cid:1) = aκ − n (cid:88) j =1 m j t a − dt − a − (cid:88) k =1 n (cid:88) j =1 m j z kj t a − − k d t − n (cid:88) j =1 m j z aj d tt − z j . (2.4) The equations(i) m i + ( a − κ = 0 for some a ∈ Z > , i ∈ { , . . . , n } ,(ii) m n +1 + 2 − aκ = 0 for some a ∈ Z > ,(iii) κ = 0,are called the resonance relations for the parameters m , . . . , m n +1 , κ of the de Rham complex.If κ = 0, then the twisted de Rham complex is not defined. If the resonance relation m i + aκ = 0 is satisfied for some a , then the first term in the right-hand side of (2.3) equalszero. Similarly, if the resonance relation m n +1 + 2 − aκ = 0 is satisfied for some a , then the firstterm in the right-hand side of (2.4) equals zero. A. Slinkin and A. Varchenko Let Ω ( U ) ⊂ Ω ( U ) be the subspace generated over C by function 1. Let Ω ( U ) ⊂ Ω ( U ) bethe subspace generated over C by the differential forms ω j = d tt − z j , j = 1 , . . . , n. These subspaces form the logarithmic subcomplex (Ω • log ( U ) , ∂ ) of the de Rham complex(Ω • ( U ) , ∂ ). We have ∂ : 1 (cid:55)→ α. For generic m , . . . , m n , κ , the embedding (Ω • log ( U ) , ∂ ) (cid:44) → (Ω • ( U ) , ∂ ) is a quasi-isomorphism,the logarithmic forms ω , . . . , ω n generate the space H (Ω • ( U )), and the cohomological relation n (cid:80) i =1 m i ω i ∼ ω , . . . , ω n ,see [9, Corollary 6.4]. For example, if m n +1 + 2 − κ = 0, then n (cid:80) j =1 z j m j ω j ∼
0, and if m n +1 +2 − κ = 0, then n (cid:88) j =1 z j m j ω j − κ n (cid:88) j =1 z j m j (cid:32) n (cid:88) i =1 z i m i ω i (cid:33) ∼ . (cid:99) sl -modules (cid:99) sl Let sl be the Lie algebra of complex (2 × e , f , h be standardgenerators subject to the relations[ e, f ] = h, [ h, e ] = 2 e, [ h, f ] = − f. Let (cid:99) sl be the affine Lie algebra (cid:99) sl = sl (cid:2) T, T − (cid:3) ⊕ C c with the bracket (cid:2) aT i , bT j (cid:3) = [ a, b ] T i + j + i (cid:104) a, b (cid:105) δ i + j, c, where c is central element, (cid:104) a, b (cid:105) = tr( ab ). Set e = e, f = f, h = h,e = f T, f = eT − , h = c − h. These are the standard Chevalley generators defining (cid:99) sl as the Kac–Moody algebra correspon-ding to the Cartan matrix (cid:0) − − (cid:1) . π The Lie algebra (cid:99) sl has an automorphism π , π : c (cid:55)→ c, eT i (cid:55)→ f T i , f T i (cid:55)→ eT i , hT i (cid:55)→ − hT i . wisted de Rham Complex on Line and Singular Vectors in (cid:99) sl Verma Modules 5
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 (cid:99) sl Verma module with generating vector v . The Vermamodule is generated by v subject to the relations e v = 0 , e v = 0 , h v = mv, h v = ( k − m ) v. Let ˆ n − ⊂ (cid:99) sl be the Lie subalgebra generated by f , f and U ˆ n − its enveloping algebra. Themap U ˆ n − → V ( m, k − m ), F (cid:55)→ F v , is an isomorphism of U ˆ n − -modules.The space V ( m, k − m ) has a Z ≥ -grading: a vector f i · · · f i p v with i j ∈ { , } has deg-ree ( p , p ), if p i is the number of i ’s in the sequence i , . . . , i p . For γ ∈ Z ≥ , 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 singularvector if e ω = e ω = 0. The Verma module V ( m, k − m ) is reducible, if and only if it containsa singular vector. 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 relationsholds:(a) m − l + 1 + ( a − κ = 0,(b) m + l + 1 − aκ = 0,(c) κ = 0,where l, a ∈ Z > . 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 ofdegree ( la, l ( a − l ( a − , 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 > and κ ∈ C , the monomials F ( l, a, κ ) = f l +( a − κ f l +( a − κ f l +( a − κ · · · f l − ( a − κ f l − ( a − κ ,F ( l, a, κ ) = f l +( a − κ f l +( a − κ f l +( a − κ · · · f l − ( a − κ f l − ( a − κ are well-defined as elements of U ˆ n − . If m = l − − ( a − κ , then F ( l, a, κ ) v ∈ V ( m, k − m ) isa singular vector of degree ( la, l ( a − and if m = − l − aκ , then F ( l, a, κ ) v ∈ V ( m, k − m ) is a singular vector of degree ( l ( a − , la ) . An explanation of the meaning of complex powers in these formulas see in [6].For example for m = − κ , we have F (1 , , κ ) v = f v = eT v, and for m = − κ , we have F (1 , , κ ) v = f κ f f − κ v = f (cid:16) eT (cid:17) v + (1 + κ ) hT eT v − (1 + κ ) κ eT v. A. Slinkin and A. Varchenko
The
Shapovalov form on an (cid:99) sl Verma module V with generating vector v is the unique symmetricbilinear form S ( · , · ) on V such that S ( v, v ) = 1 , S ( f i x, y ) = S ( x, e i y ) for i = 1 , x, y ∈ V. For γ ∈ Z ≥ , let V ∗ γ be the vector space dual to V γ . Define V ∗ = ⊕ γ V ∗ γ . The space V ∗ is an (cid:99) sl -module with the (cid:99) sl -action defined by the formulas: (cid:104) f i φ, x (cid:105) = (cid:104) φ, e i x (cid:105) , (cid:104) e i φ, x (cid:105) = (cid:104) φ, f i x (cid:105) , where φ ∈ V ∗ , x ∈ V , i = 1 ,
2. The (cid:99) sl -module V ∗ is called the contragradient Verma module.The Shapovalov form S considered as a map S : V −→ V ∗ is a morphism of (cid:99) sl -modules. V and V ∗ Let V be an (cid:99) sl Verma module V . For every γ = ( p , p ) ∈ Z ≥ with p (cid:54) = p , we fix a basis inthe homogeneous component V γ ⊂ V .For p > p , we fix the basis (cid:26) fT i · · · fT i a hT j · · · hT j b eT k · · · eT k c v (cid:27) , where0 ≤ i a ≤ i a − ≤ · · · ≤ i , ≤ j b ≤ j b − ≤ · · · ≤ j , ≤ k c ≤ k c − ≤ · · · ≤ k ; a (cid:88) s =1 i s + b (cid:88) s =1 j s + c (cid:88) s =1 k s + a − c = p , a (cid:88) s =1 i s + b (cid:88) s =1 j s + c (cid:88) s =1 k s = p . (3.1)For p < p , we fix the basis (cid:26) eT k · · · eT k c hT j · · · hT j b fT i · · · fT i a v (cid:27) , with the indices satisfying (3.1). Notice that for any x ∈ sl the elements xT i and xT j commute.These collections of vectors are bases by the Poincar´e–Birkhoff–Witt theorem.For any γ , we fix a basis in the γ -homogeneous component V ∗ γ ⊂ V ∗ as the basis dual of thebasis in V γ specified above. If { w i } is a basis in V γ , then we denote by { ( w i ) ∗ } the dual basis in V ∗ γ . Theorem 3.2 ([9, Theorem 5.12]) . For m, k ∈ C and a ∈ Z > , the following identities hold inthe contragradient Verma module V ( m, k − m ) ∗ , fT a − ( v ) ∗ = ( m + ( a − k + 2)) (cid:18) fT a − v (cid:19) ∗ + a − (cid:88) (cid:96) =1 (cid:20) hT (cid:96) (cid:18) fT a − − (cid:96) v (cid:19) ∗ + 2 eT (cid:96) (cid:88) i + j = a − − (cid:96)i ≥ j ≥ (cid:18) fT i fT j v (cid:19) ∗ (cid:21) , (3.2) eT a ( v ) ∗ = ( a ( k + 2) − m − (cid:16) eT a v (cid:17) ∗ wisted de Rham Complex on Line and Singular Vectors in (cid:99) sl Verma Modules 7+ a − (cid:88) (cid:96) =0 (cid:20) − hT (cid:96) +1 (cid:16) eT a − (cid:96) − v (cid:17) ∗ + 2 fT (cid:96) (cid:88) i + j = a − (cid:96)i ≥ j ≥ (cid:16) eT i eT j v (cid:17) ∗ (cid:21) , (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 − k + 2)and a ( k + 2) − m −
2. The vanishing of these factors corresponds to the resonance conditions m i + ( a − κ = 0 and m n +1 + 2 − aκ = 0 for the de Rham complex in Section 2.3, if we recallthat κ = k + 2. Remark.
Theorem 3.2 says that the action of the element fT a − of degree ( a, a −
1) on thecovector ( v ) ∗ can be expressed in terms of the actions of the elements hT l and eT l of smallerdegree on some other covectors. Similarly the action of the element eT a of degree ( a − , a ) onthe covector ( v ) ∗ can be expressed in terms of the actions of the elements hT l , fT l of smallerdegree on some other covectors. Let S : V ( m, k − m ) → V ( m, k − m ) ∗ be the Shapovalov form. Denote X a ( m, k − m ) = S − (cid:18) ( m + ( a − k + 2)) (cid:18) fT a − v (cid:19) ∗ (cid:19) ,Y a ( m, k − m ) = S − (cid:16) ( m + 2 − a ( k + 2)) (cid:16) eT a v (cid:17) ∗ (cid:17) . For generic values of m and k , the Shapovalov form S is non-degenerate and X a and Y a are welldefined elements of V ( m, k − m ). The chosen basis in V ( m, k − m ) allows us to compare thesevectors for different values of k , m . The vectors X a ( m, k − m ), Y a ( m, k − m ) are holomorphicfunctions of k , m for generic k , m .Recall the resonance lines in the ( m, k )-plane, given by the equations m − l + 1 + ( a − k + 2) = 0 , m + l + 1 − a ( k + 2) = 0 , k + 2 = 0 , for some a, l ∈ Z > , see Section 3.4. Theorem 3.3 ([9, Theorem 6.2]) . For a ∈ Z > let ( m , k ) be a point of the line m + ( a − k +2) = 0 , which does not belong to other resonance lines. Then the function X a ( m, k − m ) can beanalytically continued to the point ( m , k ) , and X a ( m , k − m ) is a (nonzero) singular vectorof V ( m , k − m ) , hence it is proportional to the Malikov–Feigin–Fuchs vector F (1 , a, k + 2) .Similarly, for a ∈ Z > let ( m , k ) be a point of the line m + 2 − a ( k + 2) = 0 , which does notbelong to other resonance lines. Then the function Y a ( m, k − m ) can be analytically continued tothe point ( m , k ) , and Y a ( m , k − m ) is a ( nonzero ) singular vector of V ( m , k − m ) , henceit is proportional to the Malikov–Feigin–Fuchs vector F (1 , a, k + 2) . A. Slinkin and A. Varchenko sl ( U ) Recall that { z , . . . , z n , z n +1 = ∞} are pairwise distinct points of the complex projective line P and U = P − { z , . . . , z n , z n +1 } . Fix local coordinates t − z , . . . , t − z n , /t on P at these points,respectively. Let sl ( U ) be the Lie algebra of sl -valued rational functions on P regular on U ,with the pointwise bracket. Thus, an element of sl ( U ) has the form e ⊗ u + h ⊗ u + f ⊗ u with u i ∈ Ω ( U ), and the bracket is defined by the formula [ x ⊗ u , y ⊗ u ] = [ x, y ] ⊗ ( u u ). sl ( U )-modules We say that an (cid:99) sl -module W has the finiteness property, if for any w ∈ W and x ∈ sl , we have xT j · w = 0 for all j (cid:29)
0. For example, the contragradient Verma module has the finitenessproperty.Let W , . . . , W n +1 be (cid:99) sl -modules with the finiteness property. Then the Lie algebra sl ( U )acts on W ⊗ · · · ⊗ W n +1 by the formula x ⊗ u · ( w ⊗ · · · ⊗ w n +1 ) = (cid:0) [ x ⊗ u ( t )] ( z ) w (cid:1) ⊗ w ⊗ · · · ⊗ w n +1 + · · · + w ⊗ · · · ⊗ w n − ⊗ (cid:0) [ x ⊗ u ( t )] ( z n ) w n (cid:1) ⊗ w n +1 + w ⊗ · · · ⊗ w n ⊗ (cid:0) π ([ x ⊗ u ( t )] ( ∞ ) ) w n +1 (cid:1) , where for x ⊗ u ∈ sl ( U ) the symbol [ x ⊗ u ( t )] ( z j ) denotes the Laurent expansion of x ⊗ u at t = z j and [ x ⊗ u ( t )] ( ∞ ) denotes the Laurent expansion at t = ∞ ; the symbol π in the last termdenotes the (cid:99) sl -automorphism defined in Section 3.2.The finiteness property of the tensor factors ensures that the actions of the Laurent seriesare well-defined.The (cid:99) sl -action gives us a map µ : sl ( U ) ⊗ (cid:0) ⊗ n +1 j =1 W j (cid:1) → ⊗ n +1 j =1 W j . (4.1) 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 C p ( g , W ) = ∧ p g ⊗ W, d( g p ∧ · · · ∧ g ⊗ w ) = p (cid:88) i =1 ( − i − g p ∧ · · · ∧ (cid:98) g i ∧ · · · ∧ g ⊗ g i w + (cid:88) ≤ i 2. For j = 1 , . . . , n + 1,let V j be the (cid:99) sl Verma module V ( m j , k − m j ) and V ∗ j the corresponding contragradient Vermamodule. Consider the chain complex C • (cid:0) sl ( U ) , ⊗ n +1 j =1 V ∗ j (cid:1) and its last two terms → sl ( U ) ⊗ (cid:0) ⊗ n +1 j =1 V ∗ j ) d −→ ⊗ n +1 j =1 V ∗ j → , where d = µ , see formula (4.1).wisted de Rham Complex on Line and Singular Vectors in (cid:99) sl Verma Modules 9We assign degree 0 to the term ⊗ n +1 j =1 V ∗ j of this complex and assign degree 1 to the differen-tial d, so that the whole complex sits in the non-positive area. Consider the twisted de Rham complex in (2.1) corresponding to κ = k + 2 with degrees shiftedby 1, namely, the complex Ω • ( U )[1],0 → Ω ( U ) ∂ −→ Ω ( U ) → , where the shift [1] means that we assign degree p − p ( U ). Define a linear map η : Ω ( U ) −→ ⊗ n +1 j =1 V ∗ j by the formulasd t ( t − z m ) a (cid:55)→ − κ ( v ) ∗ ⊗ · · · ⊗ (cid:18) fT a − v m (cid:19) ∗ ⊗ · · · ⊗ ( v n +1 ) ∗ , (4.2) t a − d t (cid:55)→ κ ( v ) ∗ ⊗ · · · ⊗ ( v n ) ∗ ⊗ (cid:16) eT a v n +1 (cid:17) ∗ , (4.3)for a > 0. Define a linear map η : Ω ( U ) −→ sl ( U ) ⊗ (cid:0) ⊗ n +1 j =1 V ∗ j (cid:1) by the formulas1( t − z m ) a (cid:55)→ f ( t − z m ) a ⊗ ( v ) ∗ ⊗ · · · ⊗ ( v n +1 ) ∗ − a (cid:88) l =1 (cid:20) e ( t − z m ) l ⊗ ( v ) ∗ ⊗ · · · ⊗ (cid:88) i + j = a − li ≥ j ≥ (cid:18) fT i fT j v m (cid:19) ∗ ⊗ · · · ⊗ ( v n +1 ) ∗ + h ( t − z m ) l ⊗ ( v ) ∗ ⊗ · · · ⊗ (cid:18) fT a − l v m (cid:19) ∗ ⊗ · · · ⊗ ( v n +1 ) ∗ (cid:21) , (4.4)for a > t a (cid:55)→ f t a ⊗ ( v ) ∗ ⊗ · · · ⊗ ( v n +1 ) ∗ − a − (cid:88) l =0 (cid:20) et l ⊗ ( v ) ∗ ⊗ · · · ⊗ ( v n ) ∗ ⊗ (cid:88) i + j = a − l,i ≥ j ≥ (cid:16) eT i eT j v n +1 (cid:17) ∗ + ht l +1 ⊗ ( v ) ∗ ⊗ · · · ⊗ ( v n ) ∗ ⊗ (cid:16) eT a − l − v n +1 (cid:17) ∗ (cid:21) , (4.5)for a ≥ Theorem 4.1 ([9, Theorem 5.12]) . Formulas (4.2) – (4.5) define a homomorphism of complexes η : Ω • ( U )[1] → C • (cid:0) sl ( U ); ⊗ n +1 j =1 V ∗ j (cid:1) , namely we have d η = η ∂. The homomorphism is injective. Proof . First we calculate η ( ∂ (( t − z p ) − a )),1( t − z p ) a ∂ (cid:55)→ − κ ( m p + aκ ) d t ( t − z p ) a +1 + 1 κ a (cid:88) k =1 (cid:88) j (cid:54) = p m j ( z j − z p ) k d t ( t − z p ) a +1 − k − κ (cid:88) j (cid:54) = p m j ( z j − z p ) a d tt − z jη (cid:55)→ ( m p + κa )( v ) ∗ ⊗ · · · ⊗ (cid:18) fT a v p (cid:19) ∗ ⊗ · · · ⊗ ( v n +1 ) ∗ − a (cid:88) k =1 (cid:88) j (cid:54) = p m j ( z j − z p ) k ( v ) ∗ ⊗ · · · ⊗ (cid:18) fT a − k v p (cid:19) ∗ ⊗ · · · ⊗ ( v n +1 ) ∗ + (cid:88) j (cid:54) = p m j ( z j − z p ) a ( v ) ∗ ⊗ · · · ⊗ ( f v j ) ∗ ⊗ · · · ⊗ ( v n +1 ) ∗ . Then we calculate d (cid:0) η (( t − z p ) − a ) (cid:1) ,1( t − z p ) a η (cid:55)→ f ( t − z p ) a ⊗ ( v ) ∗ ⊗ · · · ⊗ ( v n +1 ) ∗ − a (cid:88) l =1 (cid:20) h ( t − z p ) l ⊗ ( v ) ∗ ⊗ · · · ⊗ (cid:18) fT a − l v p (cid:19) ∗ ⊗ · · · ⊗ ( v n +1 ) ∗ + e ( t − z p ) l ⊗ (cid:88) i + j = a − li ≥ j ≥ ( v ) ∗ ⊗ · · · ⊗ (cid:18) fT i fT j v p (cid:19) ∗ ⊗ · · · ⊗ ( v n +1 ) ∗ (cid:21) µ (cid:55)→ ( v ) ∗ ⊗ · · · ⊗ (cid:20) ( m p + a ( k + 2)) (cid:18) fT a v p (cid:19) ∗ + a (cid:88) l =1 (cid:20) hT l (cid:18) fT a − l v p (cid:19) ∗ + 2 eT l (cid:88) i + j = a − li ≥ j ≥ (cid:18) fT i fT j v p (cid:19) ∗ (cid:21)(cid:21) ⊗ · · · ⊗ ( v n +1 ) ∗ + (cid:88) j (cid:54) = p m j ( z j − z p ) a ( v ) ∗ ⊗ · · · ⊗ ( f v j ) ∗ ⊗ · · · ⊗ ( v n +1 ) ∗ − a (cid:88) l =1 (cid:20) ( v ) ∗ ⊗ · · · ⊗ hT l (cid:18) fT a − l v p (cid:19) ∗ ⊗ · · · ⊗ ( v n +1 ) ∗ + ( v ) ∗ ⊗ · · · ⊗ (cid:88) i + j = a − li ≥ j ≥ eT l (cid:18) fT i fT j v p (cid:19) ∗ ⊗ · · · ⊗ ( v n +1 ) ∗ + (cid:88) j (cid:54) = p m j ( z j − z p ) l ( v ) ∗ ⊗ · · · ⊗ (cid:18) fT a − l v p (cid:19) ∗ ⊗ · · · ⊗ ( v n +1 ) ∗ (cid:21) = ( κa + m p )( v ) ∗ ⊗ · · · ⊗ (cid:18) fT a v p (cid:19) ∗ ⊗ · · · ⊗ ( v n +1 ) ∗ − a (cid:88) l =1 (cid:88) j (cid:54) = p m j ( z j − z p ) l ( v ) ∗ ⊗ · · · ⊗ (cid:18) fT a − l v p (cid:19) ∗ ⊗ · · · ⊗ ( v ∗ n +1) + (cid:88) j (cid:54) = p m j ( z j − z p ) a ( v ) ∗ ⊗ · · · ⊗ ( f v j ) ∗ ⊗ · · · ⊗ ( v n +1 ) ∗ . wisted de Rham Complex on Line and Singular Vectors in (cid:99) sl Verma Modules 11In this calculation we use formula (3.2) to express the action of fT a on ( v p ) ∗ . These formulasshow that d (cid:0) η (( t − z p ) − a ) (cid:1) = η (cid:0) ∂ (( t − z p ) − a ) (cid:1) .Now we calculate η ( ∂ ( t a )), t a ∂ (cid:55)→ κ (cid:18) aκ − n (cid:88) j =1 m j (cid:19) t a − d t − κ a − (cid:88) s =1 n (cid:88) j =1 m j z sj t a − s − d t − κ n (cid:88) j =1 m j z aj d tt − z jη (cid:55)→ (cid:18) aκ − n (cid:88) j =1 m j (cid:19) ( v ) ∗ ⊗ · · · ⊗ ( v n ) ∗ ⊗ (cid:16) eT a v n +1 (cid:17) ∗ − a − (cid:88) s =1 n (cid:88) j =1 m j z sj ( v ) ∗ ⊗ · · · ⊗ ( v n ) ∗ ⊗ (cid:16) eT a − s v n +1 (cid:17) ∗ + n (cid:88) j =1 m j z aj ( v ) ∗ ⊗ · · · ⊗ ( f v j ) ∗ ⊗ · · · ⊗ ( v n +1 ) ∗ . Then we calculate d (cid:0) η ( t a ) (cid:1) , t a η (cid:55)→ f t a ⊗ ( v ) ∗ ⊗ · · · ⊗ ( v n +1 ) ∗ − a − (cid:88) l =0 (cid:20) ht l +1 ⊗ ( v ) ∗ ⊗ · · · ⊗ ( v n ) ∗ ⊗ (cid:16) eT a − l − v n +1 (cid:17) ∗ + et l ⊗ ( v ) ∗ ⊗ · · · ⊗ ( v n ) ∗ ⊗ (cid:88) i + j = a − li ≥ j ≥ (cid:16) eT i eT j v n +1 (cid:17) ∗ (cid:21) µ (cid:55)→ n (cid:88) j =1 m j z aj ( v ) ∗ ⊗ · · · ⊗ ( f v j ) ∗ ⊗ · · · ⊗ ( v n +1 ) ∗ + ( v ) ∗ ⊗ · · · ⊗ ( v n ) ∗ ⊗ (cid:20) ( − m n +1 − a ( k + 2)) (cid:16) eT a v n +1 (cid:17) ∗ + a − (cid:88) l =0 (cid:20) − hT l +1 (cid:16) eT a − l − v n +1 (cid:17) ∗ + 2 fT l (cid:88) i + j = a − li ≥ j ≥ (cid:16) eT i eT j v n +1 (cid:17) ∗ (cid:21)(cid:21) − ( v ) ∗ ⊗ · · · ⊗ ( v n ) ∗ ⊗ a − (cid:88) l =0 (cid:20) fT l (cid:88) i + j = a − li ≥ j ≥ (cid:16) eT i eT j v n +1 (cid:17) ∗ − hT l +1 (cid:16) eT a − l − v n +1 (cid:17) ∗ (cid:21) − a − (cid:88) s =1 n (cid:88) j =1 m j z sj ( v ) ∗ ⊗ · · · ⊗ ( v n ) ∗ ⊗ (cid:16) eT a − s v n +1 (cid:17) ∗ = (cid:18) aκ − n (cid:88) j =1 m j (cid:19) ( v ) ∗ ⊗ · · · ⊗ ( v n ) ∗ ⊗ (cid:16) eT a v n +1 (cid:17) ∗ − a − (cid:88) s =1 n (cid:88) j =1 m j z sj ( v ) ∗ ⊗ · · · ⊗ ( v n ) ∗ ⊗ (cid:16) eT a − s v n +1 (cid:17) ∗ + n (cid:88) j =1 m j z aj ( v ) ∗ ⊗ · · · ⊗ ( f v j ) ∗ ⊗ · · · ⊗ ( v n +1 ) ∗ . In this calculation we use formula (3.3) to express the action of eT a on ( v n +1 ) ∗ . Notice also thatcalculating the action on V ∗ n +1 we use the automorphism π , see Section 3.2.These formulas showthat d (cid:0) η ( t a ) (cid:1) = η ( ∂ ( t a )).Clearly the maps η , η are injective. Theorem 4.1 is proved. (cid:4) Under the monomorphism η of Theorem 4.1 the image of the logarithmic subcomplex (Ω • log ( U ) , ∂ )is the chain complex C • (cid:0) n − , ⊗ n +1 j =1 V ∗ j (cid:1) of the nilpotent subalgebra n − ⊂ sl generated by f . Moreprecisely, we have η : 1 (cid:55)→ f ⊗ ( v ) ∗ ⊗ · · · ⊗ ( v n +1 ) ∗ , d tx − t j (cid:55)→ − κ ( v ) ∗ ⊗ · · · ⊗ ( f v j ) ∗ ⊗ · · · ⊗ ( v n +1 ) ∗ ,j = 1 , . . . , n , and µ : f ⊗ ( v ) ∗ ⊗ · · · ⊗ ( v n +1 ) ∗ (cid:55)→ n (cid:88) j =1 m j ( v ) ∗ ⊗ · · · ⊗ ( f v j ) ∗ ⊗ · · · ⊗ ( v n +1 ) ∗ . Far-reaching generalizations of this identification of the logarithmic subcomplex with thechain complex of the nilpotent Lie algebra n − see in [8]. The Lie algebra (cid:99) sl has an automorphism ρ , corresponding to the involution of the Dynkindiagram: ρ ( e i ) = e − i , ρ ( f i ) = f − i , ρ ( h i ) = h − i , i = 1 , . We have ρ = id. In other words, ρ acts by the formulas e ↔ f T, f ↔ eT − , h ↔ c − h. Lemma 5.1. For i ∈ Z > , we have ρ : fT i (cid:55)→ eT i +1 , eT i (cid:55)→ fT i − , hT i (cid:55)→ − hT i . Proof . We have fT = 12 (cid:20) f, hT (cid:21) = 12 (cid:104) f, (cid:104) eT , f (cid:105)(cid:105) δ −→ (cid:104) eT , (cid:104) f, eT (cid:105)(cid:105) = 12 (cid:20) eT , − hT (cid:21) = eT ,fT i = 12 (cid:20) fT i − , (cid:104) eT , f (cid:105)(cid:21) δ −→ (cid:104) eT i , (cid:104) f, eT (cid:105)(cid:105) = 12 (cid:20) eT i , − hT (cid:21) = eT i +1 . Similarly we prove that ρ (cid:0) eT i (cid:1) = fT i − , ρ (cid:0) hT i (cid:1) = − hT i . (cid:4) For m ∈ C , let σ m : (cid:99) sl → End( V ( m, k − m )) be the Verma module structure. Let σ m ◦ ρ : (cid:99) sl → End( V ( m, k − m )) be the twisted module structure.Clearly the (cid:99) sl -modules ( σ m ◦ ρ, V ( m, k − m )) and ( σ m − k , V ( m − k, m )) are isomorphic. If v m ∈ V ( m, k − m ) and v k − m ∈ V ( k − m, m ) are generating vectors, then an isomorphism χ : ( σ m ◦ ρ, V ( m, k − m )) → ( σ m − k , V ( m − k, m )) is defined by the formula, f i l · · · f i v k − m (cid:55)→ f − i l · · · f − i v m , for any i , . . . , i l ∈ { , } . The isomorphism χ restricts to isomorphisms of the graded compo-nents, V ( k − m, m ) ( p ,p ) → V ( m, k − m ) ( p ,p ) .wisted de Rham Complex on Line and Singular Vectors in (cid:99) sl Verma Modules 13In Section 3.6 we fixed bases of the homogeneous components V ( p ,p ) with p (cid:54) = p of anyVerma module V . By Lemma 5.1, under the isomorphism χ the chosen basis of V ( k − m, m ) ( p ,p ) is mapped to the chosen basis of V ( m, k − m ) ( p ,p ) up to multiplication of the basis vectorsby ± 1. This ± ρ (cid:0) hT i (cid:1) = − hT i . In particular, we have χ : fT i v k − m (cid:55)→ eT i +1 v m , fT i fT j v k − m (cid:55)→ eT i +1 eT j +1 v m . Let σ ∗ m : (cid:99) sl → End( V ( m, k − m ) ∗ ) be the contragradient Verma module structure. Let σ ∗ m ◦ ρ : (cid:99) sl → End( V ( m, k − m ) ∗ ) be the twisted module structure. The isomorphism χ inducesan 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 ∗ ( p ,p ) with p (cid:54) = p of any con-tragradient Verma module V ∗ . Under the isomorphism χ ∗ , the chosen basis of V ( k − m, m ) ∗ ( p ,p ) is mapped to the chosen basis of V ( m, k − m ) ∗ ( p ,p ) up to multiplication of the basis vectorsby ± 1. In particular, we have χ ∗ : (cid:18) fT i v k − m (cid:19) ∗ (cid:55)→ (cid:16) eT i +1 v m (cid:17) ∗ , (cid:18) fT i fT j v k − m (cid:19) ∗ (cid:55)→ (cid:16) eT i +1 eT j +1 v m (cid:17) ∗ . 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 fT a − ( v k − m ) ∗ = ( − m − a ( k + 2)) (cid:18) fT a − v k − m (cid:19) ∗ + a − (cid:88) (cid:96) =1 (cid:20) hT (cid:96) (cid:18) fT a − − (cid:96) v k − m (cid:19) ∗ + 2 eT (cid:96) (cid:88) i + j = a − − (cid:96)i ≥ j ≥ (cid:18) fT i fT j v k − m (cid:19) ∗ (cid:21) . The isomorphism χ ∗ sends this relation to the relation in V ( m, k − m ) ∗ , eT a ( v m ) ∗ = ( − m − a ( k + 2)) (cid:16) eT a v m (cid:17) ∗ + a − (cid:88) (cid:96) =1 (cid:20) − hT (cid:96) (cid:16) eT a − (cid:96) v m (cid:17) ∗ + 2 fT (cid:96) − (cid:88) i + j = a − − (cid:96)i ≥ j ≥ (cid:16) eT i +1 eT j +1 v m (cid:17) ∗ (cid:21) , which is exactly the relation in formula (3.3). Thus formula (3.2) implies formula (3.3). Let V = V ( m, k − m ) and V ∗ = V ( m, k − m ) ∗ . Lemma 5.2. For x ∈ V , φ ∈ V ∗ , k ∈ Z ≥ , we have (cid:28) fT k ϕ, x (cid:29) = (cid:10) ϕ, eT k x (cid:11) , (cid:68) eT k ϕ, x (cid:69) = (cid:10) ϕ, f T k x (cid:11) , (cid:28) hT k ϕ, x (cid:29) = (cid:10) ϕ, hT k x (cid:11) . Proof . The proof is by induction. We prove the first equality, the others are proved similarly.We have [ f , f ] = hT , hence [ f , [ f , f ]] = fT . Similarly [ e , [ e , e ]] = 2 eT . So for k = 1, wehave (cid:28) fT ϕ, x (cid:29) = (cid:28) 12 [ f , [ f , f ]] ϕ, x (cid:29) = (cid:28) ϕ, 12 [[ e , e ] , e ] x (cid:29) = (cid:104) ϕ, eT x (cid:105) . (cid:2) f , fT k − (cid:3) = hT k , hence (cid:2) f , (cid:2) f , fT k − (cid:3)(cid:3) = fT k . Similarly, (cid:2)(cid:2) eT k − , f T (cid:3) , e (cid:3) = 2 eT k . Then (cid:28) fT k ϕ, x (cid:29) = (cid:28) (cid:20) f , (cid:20) f , fT k − (cid:21)(cid:21) ϕ, x (cid:29) = (cid:10) ϕ, (cid:2)(cid:2) eT k − , e (cid:3) , e (cid:3) x (cid:11) = (cid:10) ϕ, eT k x (cid:11) . (cid:4) We reformulate formula (3.2) as( m + ( a − k + 2)) (cid:18) fT a − v (cid:19) ∗ = fT a − ( v ) ∗ − a − (cid:88) (cid:96) =1 (cid:20) hT (cid:96) (cid:18) fT a − − (cid:96) v (cid:19) ∗ + 2 eT (cid:96) (cid:88) i + j = a − − (cid:96)i ≥ j ≥ (cid:18) fT i fT j v (cid:19) ∗ (cid:21) , (5.1)and will prove it in this form.Each term in (5.1) is an element of the homogeneous component V ∗ ( a,a − . In Section 3.6 wespecified a basis of the dual component V ( a,a − . We will calculate the value of the right-handside in (5.1) on an arbitrary basis vector and will obtain the value of the left-hand side on thatvector.The basis in V ( a,a − consists of the vectors fT i · · · fT i r hT j · · · hT j s eT l · · · eT l r − v, where0 ≤ i r ≤ i r − ≤ · · · ≤ i , ≤ j s ≤ j s − ≤ · · · ≤ j , ≤ l r − ≤ l r − ≤ · · · ≤ l ; r (cid:88) u =1 i u + s (cid:88) u =1 j u + r − (cid:88) u =1 l u = a − . We partition the basis in four groups. Group O consists of the single basis vector fT a − v . Group Iconsists of all basis vectors with r = 1, but different from fT a − v . Group II consists of all basisvectors with r = 2. Group III consists of all basis vectors with r ≥ fT a − v equals m +( a − k + 2). Hence we need to show that the value of the right-hand side on the basisvector fT a − v equals m + ( a − k + 2). Similarly the value of the left-hand side on any basisvector of Groups I–III equals zero. Hence we need to prove that the value of the right-handside on any basis vector of Groups I–III equals zero. These four statements are the content ofPropositions 5.3, 5.4, 5.7, and 5.9 below. These propositions prove Theorem 3.2. Proposition 5.3. The value of the right-hand side of (5.1) on the basis vector fT a − v equals m + ( a − k + 2) . Proof . By Lemma 5.2 we have (cid:28) fT a − ( v ) ∗ , fT a − v (cid:29) = (cid:28) ( v ) ∗ , eT a − fT a − v (cid:29) = (cid:28) ( v ) ∗ , (cid:20) h + ( a − c + fT a − eT a − (cid:21) v (cid:29) = m + ( a − k, since eT a − v is of degree ( − a, − a + 1), hence zero.wisted de Rham Complex on Line and Singular Vectors in (cid:99) sl Verma Modules 15By Lemma 5.2, for (cid:96) ∈ { , . . . , a − } we have (cid:28) hT (cid:96) (cid:18) fT a − − (cid:96) v (cid:19) ∗ , fT a − v (cid:29) = (cid:28) (cid:18) fT a − − (cid:96) v (cid:19) ∗ , hT (cid:96) fT a − v (cid:29) = (cid:28) (cid:18) fT a − − (cid:96) v (cid:19) ∗ , − fT a − − (cid:96) v (cid:29) = − . By Lemma 5.2 for (cid:96) ∈ { , . . . , a − } we have (cid:28) eT (cid:96) (cid:88) i + j = a − − (cid:96)i ≥ j ≥ (cid:18) fT i fT j v (cid:19) ∗ , fT a − v (cid:29) = (cid:28) (cid:88) i + j = a − − (cid:96)i ≥ j ≥ (cid:18) fT i fT j v (cid:19) ∗ , f T (cid:96) fT a − v (cid:29) = (cid:28) (cid:88) i + j = a − − (cid:96)i ≥ j ≥ (cid:18) fT i fT j v (cid:19) ∗ , fT a − f T (cid:96) v (cid:29) = 0 , since f T (cid:96) v is of degree ( − (cid:96) + 1 , − (cid:96) ) ≤ (0 , − (cid:28) fT a − ( v ) ∗ − a − (cid:88) (cid:96) =1 (cid:20) hT (cid:96) (cid:18) fT a − − (cid:96) v (cid:19) ∗ + 2 eT (cid:96) (cid:88) i + j = a − − (cid:96)i ≥ j ≥ (cid:18) fT i fT j v (cid:19) ∗ (cid:21) , fT a − v (cid:29) = m + ( a − k − ( − a − 1) = m + ( a − k + 2) . Proposition 5.3 is proved. (cid:4) 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 = fT a − − n hT j · · · hT j s v, where n ∈ { , . . . , a − } , j + · · · + j s = n, j i ≥ . Lemma 5.5. In the notation above, if s = 1 , then (cid:28) fT a − ( v ) ∗ , w (cid:29) = 2 nk, (cid:28) hT (cid:96) (cid:18) fT a − − (cid:96) v (cid:19) ∗ , w (cid:29) = nk, if (cid:96) = n , − , if (cid:96) > a − − n , , if (cid:96) ≤ a − − n , (5.2) (cid:28) eT (cid:96) (cid:88) i + j = a − − (cid:96)i ≥ j ≥ (cid:18) fT i fT j v (cid:19) ∗ , w (cid:29) = (cid:40) , if (cid:96) ≤ n , , if (cid:96) > n . Note that the first line in (5.2) is not mutually exclusive with the second and third linesin (5.2). Proof . We have w = fT a − − n hT n v . Then (cid:28) fT a − ( v ) ∗ , fT a − − n hT n v (cid:29) = (cid:28) ( v ) ∗ , eT a − fT a − − n hT n v (cid:29) (cid:28) ( v ) ∗ , (cid:20) hT n + fT a − − n eT a − (cid:21) hT n v (cid:29) . Note that eT a − hT n v is of degree ( − a + n, − a + 1 + n ) ≤ ( − , eT a − hT n v = 0. Hence (cid:28) ( v ) ∗ , hT n hT n v (cid:29) = (cid:28) ( v ) ∗ , (cid:20) nk + hT n hT n (cid:21) v (cid:29) = 2 nk. We have (cid:28) hT (cid:96) (cid:18) fT a − − (cid:96) v (cid:19) ∗ , fT a − − n hT n v (cid:29) = (cid:28) (cid:18) fT a − − (cid:96) v (cid:19) ∗ , hT (cid:96) fT a − − n hT n v (cid:29) ,hT (cid:96) fT a − − n hT n v = (cid:20) − f T (cid:96) + n − a +1 + fT a − − n hT (cid:96) (cid:21) hT n v = − f T (cid:96) + n − a +1 hT n v + fT a − − n hT (cid:96) hT n v. Note that the second summand is nonzero if and only if (cid:96) = n . In that case we have (cid:28) (cid:18) fT a − − n v (cid:19) ∗ , fT a − − n hT n hT n v (cid:29) = 2 nk. For the first summand, if (cid:96) + n − a + 1 ≤ 0, then − f T (cid:96) + n − a +1 hT n v is a basis vector and sopairing with (cid:16) fT a − − (cid:96) v (cid:17) ∗ gives zero. If (cid:96) + n − a + 1 > 0, then (cid:28) (cid:18) fT a − − (cid:96) v (cid:19) ∗ , − f T (cid:96) + n − a +1 hT n v (cid:29) = (cid:28) (cid:18) fT a − − (cid:96) v (cid:19) ∗ , − (cid:20) fT a − − (cid:96) + hT n f T (cid:96) + n − a +1 (cid:21) v (cid:29) = − , where we used f T (cid:96) + n − a +1 v = 0.Finally, (cid:28) eT (cid:96) (cid:88) i + j = a − − (cid:96)i ≥ j ≥ (cid:18) fT i fT j v (cid:19) ∗ , fT a − − n hT n v (cid:29) = (cid:28) (cid:88) i + j = a − − (cid:96)i ≥ j ≥ (cid:18) fT i fT j v (cid:19) ∗ , f T (cid:96) fT a − − n hT n v (cid:29) ,f T (cid:96) fT a − − n hT n v = fT a − − n f T (cid:96) hT n v = fT a − − n (cid:20) f T (cid:96) − n + hT n f T (cid:96) (cid:21) v = 2 fT a − − n fT n − (cid:96) v, since f T (cid:96) v = 0. Note that ( a − − n ) + ( n − (cid:96) ) = a − − (cid:96) , hence if i = a − − n and j = n − (cid:96) (or vice versa depending on what is greater) we have (cid:28) (cid:88) i + j = a − − (cid:96)i ≥ j ≥ (cid:18) fT i fT j v (cid:19) ∗ , fT a − − n fT n − (cid:96) v (cid:29) = 2 , whenever n − (cid:96) ≥ (cid:4) For s = 1 Proposition 5.4 follows from Lemma 5.5: (cid:28) fT a − ( v ) ∗ − a − (cid:88) (cid:96) =1 (cid:20) hT (cid:96) (cid:18) fT a − − (cid:96) v (cid:19) ∗ + 2 eT (cid:96) (cid:88) i + j = a − − (cid:96)i ≥ j ≥ (cid:18) fT i fT j v (cid:19) ∗ (cid:21) , fT a − − n hT n v (cid:29) = 2 nk − nk + 4 n − · n = 0 . wisted de Rham Complex on Line and Singular Vectors in (cid:99) sl Verma Modules 17 Lemma 5.6. For s ≥ , we have (cid:28) fT a − ( v ) ∗ , w (cid:29) = 0 , (5.3) a − (cid:88) (cid:96) =1 (cid:28) hT (cid:96) (cid:18) fT a − − (cid:96) v (cid:19) ∗ , w (cid:29) = 0 , (5.4) a − (cid:88) (cid:96) =1 (cid:28) eT (cid:96) (cid:88) i + j = a − − (cid:96)i ≥ j ≥ (cid:18) fT i fT j v (cid:19) ∗ , w (cid:29) = 0 . (5.5) Proof . Recall that w = fT a − − n hT j · · · hT js v with j + · · · + j s = n . We have (cid:28) fT a − ( v ) ∗ , w (cid:29) = (cid:10) ( v ) ∗ , eT a − w (cid:11) ,eT a − w = eT a − fT a − − n hT j · · · hT j s v = (cid:20) hT n + fT a − − n eT a − (cid:21) hT j · · · hT j s v. We have hT n hT j · · · hT js v = 0, since hT n commutes with all hT ji . Indeed, we have n > j i since j + · · · + j s = n , j i ≥ 1, and s ≥ fT a − − n eT a − hT j · · · hT js v = 0 since eT a − hT j · · · hT js v is of degree ( − a + n, − a +1 + n ) ≤ ( − , s . For s = 2 we have (cid:28) hT (cid:96) (cid:18) fT a − − (cid:96) v (cid:19) ∗ , w (cid:29) = (cid:28) (cid:18) fT a − − (cid:96) v (cid:19) ∗ , hT (cid:96) w (cid:29) ,hT (cid:96) w = hT (cid:96) fT a − − n hT j hT j v = (cid:20) − f T (cid:96) − a +1+ n + fT a − − n hT (cid:96) (cid:21) hT j hT j v = − f T (cid:96) − a +1+ n hT j hT j v + fT a − − n hT (cid:96) hT j hT j v. Note that for fT a − − n hT (cid:96) hT j hT j v to give a nonzero pairing with (cid:0) fT a − − (cid:96) v (cid:1) ∗ we need (cid:96) = n , whichimplies that hT (cid:96) commutes with hT j and hT j ( (cid:96) > j i since j + j = n = (cid:96) and j i ≥ fT a − − n hT (cid:96) hT j hT j v gives zero for all (cid:96) .Also note that whenever (cid:96) ≤ a − − n , f T (cid:96) − a +1+ n hT j hT j v is a basis vector and so pairingwith (cid:0) fT a − − (cid:96) v (cid:1) ∗ gives zero. If (cid:96) > a − − n , then − f T (cid:96) − a +1+ n hT j hT j v = − (cid:20) f T (cid:96) − a +1+ n − j + hT j f T (cid:96) − a +1+ n (cid:21) hT j v = − f T (cid:96) − a +1+ n − j hT j v − hT j f T (cid:96) − a +1+ n hT j v. (5.6)If (cid:96) ≤ a − − n + j , the first summand gives zero when pairing with (cid:0) fT a − − (cid:96) v (cid:1) ∗ , since forsuch (cid:96)f T (cid:96) − a +1+ n − j hT j v is a basis vector. For (cid:96) > a − − n + j , we have − f T (cid:96) − a +1+ n − j hT j v = − (cid:20) fT a − − (cid:96) v + hT j f T (cid:96) − a +1+ n − j (cid:21) v = − fT a − − (cid:96) v, since f T (cid:96) − a +1+ n − j v is of degree ( − (cid:96) + a − − n + j + 1 , − (cid:96) + a − − n + j ) ≤ (0 , − (cid:96) ∈ { a − − n + j + 1 , . . . , a − } we get (cid:28) (cid:18) fT a − − (cid:96) v (cid:19) ∗ , − f T (cid:96) − a +1+ n − j hT j v (cid:29) = − 88 A. Slinkin and A. Varchenkoand zero for other values of (cid:96) . The total number of elements in the set { a − − n + j +1 , . . . , a − } equals j .Whereas, for the second summand in (5.6) we have − hT j f T (cid:96) − a +1+ n hT j v = − hT j (cid:20) f T (cid:96) − a +1+ n − j + hT j f T (cid:96) − a +1+ n (cid:21) v = − hT j f T (cid:96) − a +1+ n − j v, since f T (cid:96) − a +1+ n v is of degree ( − (cid:96) + a − − n + 1 , − (cid:96) + a − − n ) ≤ (0 , − (cid:96) > a − − n + j , then f T (cid:96) − a +1+ n − j v is of degree ( − (cid:96) + a − − n + j +1 , − (cid:96) + a − − n + j ) ≤ (0 , − (cid:96) ≤ a − − n + j , then − hT j f T (cid:96) − a +1+ n − j v = − (cid:20) − fT a − − (cid:96) + f T (cid:96) − a +1+ n − j hT j (cid:21) = 8 fT a − − (cid:96) − f T (cid:96) − a +1+ n − j hT j . The second summand gives zero when pairing with (cid:0) fT a − − (cid:96) v (cid:1) ∗ . So for (cid:96) ∈ { a − − n + 1 , . . . , a − − n + j } we get (cid:28) (cid:18) fT a − − (cid:96) v (cid:19) ∗ , − hT j f T (cid:96) − a +1+ n hT j v (cid:29) = 8and zero for other values of (cid:96) . The total number of elements in the set { a − − n + 1 , . . . , a − − n + j } equals j .Therefore, a − (cid:88) (cid:96) =1 (cid:28) hT (cid:96) (cid:18) fT a − − (cid:96) v (cid:19) ∗ , fT a − − n hT j hT j v (cid:29) = − j + 8 j = 0and so for s = 2 we proved (5.4).Now suppose that (5.4) holds for all natural numbers up to s . Then hT (cid:96) fT a − − n hT j . . . hT j s +1 v = (cid:20) − f T (cid:96) − a +1+ n + fT a − − n hT (cid:96) (cid:21) hT j · · · hT j s +1 v. Note that for fT a − − n hT (cid:96) hT j · · · hT js +1 v to give a nonzero pairing with (cid:0) fT a − − (cid:96) v (cid:1) ∗ we need (cid:96) = n .That assumption implies that hT (cid:96) commutes with hT ji for all i ∈ { , . . . , s + 1 } since (cid:96) > j i as j + · · · + j s +1 = n = (cid:96) and j i ≥ 1. Hence fT a − − n hT (cid:96) hT j · · · hT js +1 v gives zero for all (cid:96) .Also note that whenever (cid:96) ≤ a − − n , the vector f T (cid:96) − a +1+ n hT j · · · hT js +1 v is a basis vectorand so pairing with (cid:0) fT a − − (cid:96) v (cid:1) ∗ gives zero.If (cid:96) > a − − n , then − f T (cid:96) − a +1+ n hT j · · · hT j s +1 v = − (cid:20) f T (cid:96) − a +1+ n − j + hT j f T (cid:96) − a +1+ n (cid:21) hT j · · · hT j s +1 v = − fT a − − ( n − j ) − (cid:96) hT j · · · hT j s +1 v − hT j f T − a +1+ n + (cid:96) hT j · · · hT j s +1 v. (5.7)Note that by induction hypothesis we have0 = a − (cid:88) (cid:96) =1 (cid:28) hT (cid:96) (cid:18) fT a − − (cid:96) v (cid:19) ∗ , fT a − − ( n − j ) hT j · · · hT j s +1 v (cid:29) wisted de Rham Complex on Line and Singular Vectors in (cid:99) sl Verma Modules 19= a − (cid:88) (cid:96) =1 (cid:28) (cid:18) fT a − − (cid:96) v (cid:19) ∗ , hT (cid:96) fT a − − ( n − j ) hT j · · · hT j s +1 v (cid:29) . So we add this zero term multiplied by − . 7) to get − a − (cid:88) (cid:96) =1 (cid:28) (cid:18) fT a − − (cid:96) v (cid:19) ∗ , fT a − − ( n − j ) − (cid:96) hT j · · · hT j s +1 v (cid:29) − a − (cid:88) (cid:96) =1 (cid:28) (cid:18) fT a − − (cid:96) v (cid:19) ∗ , hT (cid:96) fT a − − ( n − j ) hT j · · · hT j s +1 v (cid:29) (5.8)= − a − (cid:88) (cid:96) =1 (cid:28) (cid:18) fT a − − (cid:96) v (cid:19) ∗ , (cid:20) fT a − − ( n − j ) − (cid:96) + hT (cid:96) fT a − − ( n − j ) (cid:21) hT j . . . hT j s +1 v (cid:29) = − a − (cid:88) (cid:96) =1 (cid:28) (cid:18) fT a − − (cid:96) v (cid:19) ∗ , fT a − − ( n − j ) hT (cid:96) hT j · · · hT j s +1 v (cid:29) , where in the last step we use commutation relations.Note that for fT a − − ( n − j hT (cid:96) hT j · · · hT js +1 v to give a nonzero pairing with (cid:0) fT a − − (cid:96) v (cid:1) ∗ we need (cid:96) = n − j . That assumption implies that hT (cid:96) commutes with hT ji for all i ∈ { , . . . , s + 1 } since (cid:96) > j i as j + · · · + j s +1 = n − j = (cid:96) and j i ≥ 1. Hence fT a − − ( n − j hT (cid:96) hT j · · · hT js +1 v gives zerofor all (cid:96) .For the second summand in (5.7) we have − a − (cid:88) (cid:96) =1 (cid:28) (cid:18) fT a − − (cid:96) v (cid:19) ∗ , hT j f T − a +1+ n + (cid:96) hT j · · · hT j s +1 v (cid:29) = − a − (cid:88) (cid:96) =1 (cid:28) (cid:18) fT a − − (cid:96) v (cid:19) ∗ , hT j (cid:20) fT a − − ( n − j ) − (cid:96) + hT j f T − a +1+ n + (cid:96) (cid:21) hT j · · · hT j s +1 v (cid:29) = − a − (cid:88) (cid:96) =1 (cid:28) (cid:18) fT a − − (cid:96) v (cid:19) ∗ , hT j fT a − − ( n − j ) − (cid:96) hT j · · · hT j s +1 v (cid:29) (5.9) − a − (cid:88) (cid:96) =1 (cid:28) (cid:18) fT a − − (cid:96) v (cid:19) ∗ , hT j hT j f T − a +1+ n + (cid:96) hT j · · · hT j s +1 v (cid:29) . (5.10)In (5.9) we note that hT j fT a − − ( n − j ) − (cid:96) hT j · · · hT j s +1 v = (cid:20) − fT a − − ( n − j − j ) − (cid:96) + fT a − − ( n − j ) − (cid:96) hT j (cid:21) hT j · · · hT j s +1 v = − fT a − − ( n − j − j ) − (cid:96) hT j · · · hT j s +1 v + fT a − − ( n − j ) − (cid:96) hT j hT j · · · hT j s +1 v. Note that in both terms the number of h ’s is less than or equal to s , so we use the exact samereasoning as in (5.8) to show that a − (cid:88) (cid:96) =1 (cid:28) (cid:18) fT a − − (cid:96) v (cid:19) ∗ , fT a − − ( n − j − j ) − (cid:96) hT j . . . hT j s +1 v (cid:29) = 0 , a − (cid:88) (cid:96) =1 (cid:28) (cid:18) fT a − − (cid:96) v (cid:19) ∗ , fT a − − ( n − j ) − (cid:96) hT j hT j · · · hT j s +1 v (cid:29) = 0 , a − (cid:88) (cid:96) =1 (cid:28) (cid:18) fT a − − (cid:96) v (cid:19) ∗ , hT j hT j f T − a +1+ n + (cid:96) hT j · · · hT j s +1 v (cid:29) the factor f T − a +1+ n + (cid:96) can be pulled to the right by using the same argument (first commute f T − a +1+ n + (cid:96) with hT j and then pull f T − a +1+ n − j + (cid:96) to the left). Ultimately, we get a − (cid:88) (cid:96) =1 (cid:28) (cid:18) fT a − − (cid:96) v (cid:19) ∗ , hT j hT j hT j · · · hT j s +1 f T − a +1+ n + (cid:96) v (cid:29) = 0 , since (cid:96) > a − − n and so f T − a +1+ n + (cid:96) v = 0. Therefore, a − (cid:88) (cid:96) =1 (cid:28) hT (cid:96) (cid:18) fT a − − (cid:96) v (cid:19) ∗ , fT a − − n hT j · · · hT j s v (cid:29) = 0 , and formula (5.4) is proved.We prove formula (5.5) by induction on s . For s = 2, we have (cid:28) eT (cid:96) (cid:88) i + j = a − − (cid:96)i ≥ j ≥ (cid:18) fT i fT j v (cid:19) ∗ , w (cid:29) = (cid:28) (cid:88) i + j = a − − (cid:96)i ≥ j ≥ (cid:18) fT i fT j v (cid:19) ∗ , f T (cid:96) w (cid:29) ,f T (cid:96) w = f T (cid:96) fT a − − n hT j hT j v = fT a − − n f T (cid:96) hT j hT j v. Note that f T (cid:96) hT j hT j v is of degree ( n − (cid:96) + 1 , n − (cid:96) ), hence nonzero only if (cid:96) ≤ n . For such (cid:96) wehave fT a − − n (cid:20) f T (cid:96) − j + hT j f t (cid:96) (cid:21) hT j v = 2 fT a − − n f T (cid:96) − j hT j v + fT a − − n hT j f T (cid:96) hT j v = 2 fT a − − n f T (cid:96) − j hT j v + 2 fT a − − n hT j f T (cid:96) − j v. (5.11)If (cid:96) ≤ j , then the first summand in (5.11) gives zero when pairing with any vector with two f ’s. If (cid:96) > j , then2 fT a − − n f T (cid:96) − j hT j v = 2 fT a − − n (cid:20) f T (cid:96) − j − j + hT j f T (cid:96) − j (cid:21) v = 4 fT a − − n fT n − (cid:96) v. If (cid:96) > j , then the second summand in (5.11) is zero simply because f T (cid:96) − j v = 0. If (cid:96) ≤ j ,then 2 fT a − − n hT j f T (cid:96) − j v = 2 fT a − − n (cid:20) − f T (cid:96) − j − j + f T (cid:96) − j hT j (cid:21) v = − fT a − − n fT n − (cid:96) v, since fT a − − n f T (cid:96) − j hT j v is a basis vector, hence pairing with a vector consisting of two f ’s giveszero. Therefore, a − (cid:88) (cid:96) =1 (cid:28) (cid:88) i + j = a − − (cid:96)i ≥ j ≥ (cid:18) fT i fT j v (cid:19) ∗ , f T (cid:96) w (cid:29) wisted de Rham Complex on Line and Singular Vectors in (cid:99) sl Verma Modules 21= a − (cid:88) (cid:96) =1 (cid:28) (cid:88) i + j = a − − (cid:96)i ≥ j ≥ (cid:18) fT i fT j v (cid:19) ∗ , fT a − − n fT n − (cid:96) v (cid:29) (5.12)+ a − (cid:88) (cid:96) =1 (cid:28) (cid:88) i + j = a − − (cid:96)i ≥ j ≥ (cid:18) fT i fT j v (cid:19) ∗ , − fT a − − n fT n − (cid:96) v (cid:29) . (5.13)Note that in the expression in (5.12) for each (cid:96) ∈ { j + 1 , . . . , n } there exists exactly one pair ofindices ( i, j ) = (max { a − − n, n − (cid:96) } , min { a − − n, n − (cid:96) } ) that gives 4 when pairing. All otherpairs ( i, j ) give zero. Similarly, the expression in (5.13) equals − (cid:96) ∈ { , . . . , j } andexactly one corresponding pair ( i, j ), and zero otherwise. Also note that the number of elementsin each set { j + 1 , . . . , n } and { , . . . , j } equals j . Hence we get4 j − j = 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 f T (cid:96) fT a − − n hT j · · · hT j s +1 v = fT a − − n f T (cid:96) hT j · · · hT j s +1 v = fT a − − n (cid:20) f T (cid:96) − j + hT j f T (cid:96) (cid:21) hT j · · · hT j s +1 v = 2 fT a − − n f T (cid:96) − j hT j · · · hT j s +1 v + fT a − − n hT j f T (cid:96) hT j · · · hT j s +1 v. (5.14)Note that if (cid:96) ≤ j , then the first summand in (5.14) is a basis vector and hence its pairingwith a vector consisting of two f ’s gives zero. If (cid:96) > j we have a − (cid:88) (cid:96) =1 (cid:28) (cid:88) i + j = a − − (cid:96)i ≥ j ≥ (cid:18) fT i fT j v (cid:19) ∗ , fT a − − n f T (cid:96) − j hT j · · · hT j s +1 v (cid:29) = a − (cid:88) (cid:96) = j +1 (cid:28) (cid:88) i + j = a − − (cid:96)i ≥ j ≥ (cid:18) fT i fT j v (cid:19) ∗ , fT a − − n f T (cid:96) − j hT j · · · hT j s +1 v (cid:29) = a − (cid:88) (cid:96) = j +1 (cid:28) eT (cid:96) − j (cid:88) i + j = a − − (cid:96)i ≥ j ≥ (cid:18) fT i fT j v (cid:19) ∗ , fT a − − n hT j · · · hT j s +1 v (cid:29) = a − − j (cid:88) k =1 (cid:28) eT k (cid:88) i + j = a − − j − ki ≥ j ≥ (cid:18) fT i fT j v (cid:19) ∗ , fT ( a − − j ) − ( n − j ) hT j · · · hT j s +1 v (cid:29) = 0by induction hypothesis. For the second summand in (5.14) we have fT a − − n hT j f T (cid:96) hT j · · · hT j s +1 v = fT a − − n hT j (cid:20) f T (cid:96) − j + hT j f T (cid:96) (cid:21) hT j · · · hT j s +1 v = 2 fT a − − n hT j f T (cid:96) − j hT j · · · hT j s +1 v + fT a − − n hT j hT j f T (cid:96) hT j · · · hT j s +1 v. (5.15)Note that fT a − − n hT j f T (cid:96) − j hT j · · · hT j s +1 v = fT a − − n (cid:20) − f T (cid:96) − j − j + f T (cid:96) − j hT j (cid:21) hT j · · · hT j s +1 v − fT a − − n f T (cid:96) − j − j hT j · · · hT j s +1 v + fT a − − n f T (cid:96) − j hT j hT j · · · hT j s +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 − (cid:88) (cid:96) =1 (cid:28) (cid:88) i + j = a − − (cid:96)i ≥ j ≥ (cid:18) fT i fT j v (cid:19) ∗ , fT a − − n hT j f T (cid:96) − j hT j · · · hT j s +1 v (cid:29) = 0 . Now in the second summand in (5.15), fT a − − n hT j hT j f T (cid:96) hT j · · · hT j s +1 v, we pull f T (cid:96) to the right and at each step we use induction hypothesis to argue that we keepgetting zeros. Ultimately, we get a vector fT a − − n hT j · · · hT j s +1 f T (cid:96) v, which is zero, since f T (cid:96) has grading ( − (cid:96) + 1 , − (cid:96) ) ≤ (0 , − 1) and so f T (cid:96) v = 0. Therefore, a − (cid:88) (cid:96) =1 (cid:28) eT (cid:96) (cid:88) i + j = a − − (cid:96)i ≥ j ≥ (cid:18) fT i fT j v (cid:19) ∗ , fT a − − n hT j · · · hT j s v (cid:29) = 0 . Formula (5.5) and Lemma 5.6 are proved. (cid:4) Proposition 5.4 is proved. (cid:4) 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 = fT i fT i hT j · · · hT j s eT l v. Lemma 5.8. We have (cid:28) fT a − ( v ) ∗ , w (cid:29) = 2 s +1 ( m − lk ) , (5.16) (cid:28) hT (cid:96) (cid:18) fT a − − (cid:96) v (cid:19) ∗ , w (cid:29) = s +2 ( m − lk ) , if i = i = a − − (cid:96), s +1 ( m − lk ) , if i (cid:54) = i and i or i = a − − (cid:96), , otherwise , (5.17) (cid:28) eT (cid:96) (cid:88) i + j = a − − (cid:96)i ≥ j ≥ (cid:18) fT i fT j v (cid:19) ∗ , w (cid:29) = (cid:40) − s ( m − lk ) , if (cid:96) = a − − i − i , , otherwise . (5.18)wisted de Rham Complex on Line and Singular Vectors in (cid:99) sl Verma Modules 23 Proof . We have (cid:28) fT a − ( v ) ∗ , w (cid:29) = (cid:10) ( v ) ∗ , eT a − w (cid:11) = (cid:28) ( v ) ∗ , (cid:20) hT a − − i + fT i eT a − (cid:21) fT i hT j · · · hT j s eT l v (cid:29) = (cid:28) ( v ) ∗ , hT a − − i fT i hT j · · · hT j s eT l v (cid:29) + (cid:28) ( v ) ∗ , fT i eT a − fT i hT j · · · hT j s eT l v (cid:29) . (5.19)Note that eT a − fT i hT j · · · hT js eT l is of degree ( − i − , − i ) ≤ ( − , eT a − fT i hT j · · · hT js eT l v = 0. In the first summand in (5.19) we pull hT a − − i to the right to get (cid:28) ( v ) ∗ , − f T a − − i − i hT j · · · hT j s eT l v (cid:29) = · · · = (cid:28) ( v ) ∗ , − s +1 f T a − − i − i − j −···− a s eT l v (cid:29) = (cid:28) ( v ) ∗ , − s +1 f T l eT l v (cid:29) = (cid:10) ( v ) ∗ , s +1 ( h − lc ) v (cid:11) = 2 s +1 ( m − lk ) , where at each step we do not write monomials of negative degree, since they give zero whenapplied to v . This proves formula (5.16).We have (cid:28) hT (cid:96) (cid:18) fT a − − (cid:96) v (cid:19) ∗ , w (cid:29) = (cid:28) (cid:18) fT a − − (cid:96) v (cid:19) ∗ , hT (cid:96) w (cid:29) ,hT (cid:96) w = (cid:20) − f T (cid:96) − i + fT i hT (cid:96) (cid:21) fT i hT j · · · hT j s eT l v = − f T (cid:96) − i fT i hT j · · · hT j s eT l v + fT i (cid:20) − f T (cid:96) − i + fT i hT (cid:96) (cid:21) hT j · · · hT j s eT l v = (cid:20) − fT i f T (cid:96) − i − fT i f T (cid:96) − i + fT i fT i hT (cid:96) (cid:21) hT j · · · hT j s eT l v. Note that the vector fT i fT i hT (cid:96) hT j · · · hT js eT l v after pulling hT (cid:96) to the right either becomesa zero vector or a vector with two f ’s, which of course gives zero when pairing with a basisvector with one f . Also, note that the only possibility for the vector fT i f T (cid:96) − i hT j · · · hT js eT l v togive a nonzero pairing with (cid:0) fT a − − (cid:96) v (cid:1) ∗ is when i = a − − (cid:96) . Similarly fT i f T (cid:96) − i hT j · · · hT js eT l v gives a nonzero number only if i = a − − (cid:96) . First consider the case i = i = a − − (cid:96) . Wehave − fT a − − (cid:96) f T (cid:96) − a +1 hT j · · · hT j s eT l v = − fT a − − (cid:96) (cid:20) f T (cid:96) − a +1 − j + hT j f T (cid:96) − a +1 (cid:21) hT j · · · hT j s eT l v. Note that f T (cid:96) − a +1 hT j · · · hT js eT l is of degree ( − j , − j ) ≤ ( − , − f T (cid:96) − a +1 hT j · · · hT j s eT l v = 0 . So we get − fT a − − (cid:96) f T (cid:96) − a +1 − j hT j · · · hT j s eT l v · · · = − s +2 fT a − − (cid:96) f T (cid:96) − a +1 − j −···− j s eT l v = − s +2 fT a − − (cid:96) f T l eT l v = 2 s +2 fT a − − (cid:96) ( h − lc ) v = 2 s +2 ( m − lk ) , where at each step we don’t write monomials of negative degree, since they give zero whenapplied to v .4 A. Slinkin and A. VarchenkoFor i = a − − (cid:96) (cid:54) = i and i = a − − (cid:96) (cid:54) = i we have − fT a − − (cid:96) f T (cid:96) − a +1 hT j . . . hT j s eT l v = 2 s +1 ( m − lk ) , where we performed the exact same computation as above. Therefore, formula (5.17) is proved.We have (cid:28) eT (cid:96) (cid:88) i + j = a − − (cid:96)i ≥ j ≥ (cid:18) fT i fT j v (cid:19) ∗ , w (cid:29) = (cid:28) (cid:88) i + j = a − − (cid:96)i ≥ j ≥ (cid:18) fT i fT j v (cid:19) ∗ , f T (cid:96) w (cid:29) ,f T (cid:96) w = fT i fT i f T (cid:96) hT j · · · hT j s eT l v = fT i fT i (cid:20) f T (cid:96) − j + hT j f T (cid:96) (cid:21) hT j · · · hT j s eT l v. The only nonzero pairing happens when (cid:96) is such that i + i = a − − (cid:96) . In that case f T (cid:96) hT j · · · hT js eT l has degree ( − j , − j ) ≤ ( − , − f T (cid:96) hT j · · · hT js eT l v = 0. Therefore wehave 2 fT i fT i f T (cid:96) − j hT j · · · hT j s eT l v = 2 s fT i fT i f T (cid:96) − j −···− j s eT l v = 2 s fT i fT i f T l eT l v, where we pulled f T (cid:96) − j to the right and did not write monomials of negative degree, since theygive zero when applied to v . Hence we get2 s fT i fT i ( − h + lc ) v. Therefore, (cid:28) eT (cid:96) (cid:88) i + j = a − − (cid:96)i ≥ j ≥ (cid:18) fT i fT j v (cid:19) ∗ , w (cid:29) = (cid:88) i + j = a − − (cid:96)i ≥ j ≥ (cid:28) (cid:18) fT i fT j v (cid:19) ∗ , s fT i fT i ( − h + lc ) v (cid:29) = − s ( m − lk ) , since for i = i , j = i we get − s ( m − lk ) and zero for other pairs ( i, j ). Formula (5.18) andLemma 5.8 are proved. (cid:4) By Lemma 5.8, we have (cid:28) fT a − ( v ) ∗ − a − (cid:88) (cid:96) =1 (cid:20) hT (cid:96) (cid:18) fT a − − (cid:96) v (cid:19) ∗ + 2 eT (cid:96) (cid:88) i + j = a − − (cid:96)i ≥ j ≥ (cid:18) fT i fT j v (cid:19) ∗ (cid:21) , w (cid:29) = 2 s +1 ( m − lk ) − · s +1 ( m − lk ) + 2 · s ( m − lk ) = 0 . Note that (cid:28) a − (cid:88) (cid:96) =1 hT (cid:96) (cid:18) fT a − − (cid:96) v (cid:19) ∗ , w (cid:29) = 2 · s +1 ( m − lk )in both cases i = i and i (cid:54) = i . Also note that (cid:28) a − (cid:88) (cid:96) =1 eT (cid:96) (cid:88) i + j = a − − (cid:96)i ≥ j ≥ (cid:18) fT i fT j v (cid:19) ∗ , w (cid:29) (cid:54) = 0only if (cid:96) is such that i + i = a − − (cid:96) . Therefore, Proposition 5.7 is proved. (cid:4) wisted de Rham Complex on Line and Singular Vectors in (cid:99) sl Verma Modules 25 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 = fT i · · · fT i r hT j · · · hT j s eT l · · · eT l r − v, where r ≥ Lemma 5.10. For every (cid:96) ∈ { , . . . , a − } , we have (cid:28) fT a − ( v ) ∗ , w (cid:29) = 0 , (5.20) (cid:28) hT (cid:96) (cid:18) fT a − − (cid:96) v (cid:19) ∗ , w (cid:29) = 0 , (5.21) (cid:28) eT (cid:96) (cid:88) i + j = a − − (cid:96)i ≥ j ≥ (cid:18) fT i fT j v (cid:19) ∗ , w (cid:29) = 0 . (5.22) Proof . We have (cid:28) fT a − ( v ) ∗ , w (cid:29) = (cid:10) ( v ) ∗ , eT a − w (cid:11) = (cid:28) ( v ) ∗ , (cid:20) hT a − − i + fT i eT a − (cid:21) fT i · · · fT i r hT j · · · hT j s eT l · · · eT l r − v (cid:29) . Note that eT a − fT i · · · fT ir hT j · · · hT js eT l · · · eT lr − v is of degree ( − i − , − i ) ≤ ( − , (cid:28) ( v ) ∗ , (cid:20) − f T a − − i − i + fT i hT a − − i (cid:21) fT i · · · fT i r hT j · · · hT j s eT l · · · eT l r − v (cid:29) . As above, note that hT a − − i fT i · · · fT ir hT j · · · hT js eT l · · · eT lr − v is of degree ( − i − , − i ) ≤ ( − , (cid:28) ( v ) ∗ , − fT i · · · fT i r f T a − − i − i hT j · · · hT j s eT l · · · eT l r − v (cid:29) = 0 , since f T a − − i − i hT j · · · hT js eT l · · · eT lr − v is of degree ( − i −· · ·− i r − r +2 , − i −· · ·− i r ) ≤ ( − , r ≥ 3, hence zero. Formula (5.20) is proved.We have (cid:28) hT (cid:96) (cid:18) fT a − − (cid:96) v (cid:19) ∗ , w (cid:29) = (cid:28) (cid:18) fT a − − (cid:96) v (cid:19) ∗ , hT (cid:96) w (cid:29) ,hT (cid:96) w = (cid:20) − f T (cid:96) − i + fT i hT (cid:96) (cid:21) fT i · · · fT i r hT j · · · hT j s eT l · · · eT l r − v = − fT i · · · fT i r f T (cid:96) − i hT j · · · hT j s eT l · · · eT l r − v (5.23)+ fT i hT (cid:96) fT i · · · fT i r hT j · · · hT j s eT l · · · eT l r − v. (5.24)6 A. Slinkin and A. VarchenkoObserve that for (cid:96) ≤ i in (5.23) we have a basis vector, hence it gives zero when pairing with (cid:0) fT a − − (cid:96) v (cid:1) ∗ . If (cid:96) > i , then we pull f T (cid:96) − i to the right and notice that no matter how f T (cid:96) − i interacts with h ’s and e ’s, it does not affect the number of f ’s, which is greater or equal thantwo. Hence, the vector in (5.23) gives zero when pairing with (cid:0) fT a − − (cid:96) v (cid:1) ∗ .In (5.24) note that hT (cid:96) fT i = − f T (cid:96) − i + fT i hT (cid:96) so that either hT (cid:96) is pulled to the right not affecting the number of f ’s or it gives f T (cid:96) − i , forwhich we apply the same argument as above after pulling it to the right to argue that the pairingof the vector in (5.24) with (cid:0) fT a − − (cid:96) v (cid:1) ∗ is zero. Formula (5.21) is proved.We have (cid:28) eT (cid:96) (cid:88) i + j = a − − (cid:96)i ≥ j ≥ (cid:18) fT i fT j v (cid:19) ∗ , w (cid:29) = (cid:28) (cid:88) i + j = a − − (cid:96)i ≥ j ≥ (cid:18) fT i fT j v (cid:19) ∗ , f T (cid:96) w (cid:29) ,f T (cid:96) w = f T (cid:96) fT i · · · fT i r hT j · · · hT j s eT l · · · eT l r − v = fT i · · · fT i r f T (cid:96) hT j · · · hT j s eT l · · · eT l r − v. As in formula (5.21), no matter how f T (cid:96) interacts with h ’s and e ’s, the number of f ’s remainsunchanged, i.e., we have more than or equal to three f ’s, so that pairing with (cid:0) fT i fT j v (cid:1) ∗ is zero.Formula (5.22) is proved. (cid:4) Proposition 5.9 follows from Lemma 5.10. (cid:4) Theorem 3.2 is proved. Acknowledgements The authors thank V. Schechtman for useful discussions. The second author was supported inpart by NSF grant DMS-1665239. 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