First and second cohomologies of grading-restricted vertex algebras

Transcription

1 First and second cohomologies of grading-restricted vertex algebras Yi-Zhi Huang Abstract Let V be a grading-restricted vertex algebra and W a V -module. We show that for any m Z +, the first cohomology Hm(V, 1 W ) of V with coefficients in W introduced by the author is linearly isomorphic to the space of derivations from V to W. In particular, Hm(V, 1 W ) for m N are equal (and can be denoted using the same notation H 1 (V, W )). We also show that the second cohomology H 1 (V, W ) of V with coefficients in W introduced by the author corresponds bijectively to the set of equivalence classes of square-zero extensions of V by W. In the case that W = V, we show that the second cohomology H 1 (V, V ) corresponds bijectively to the set of equivalence classes of first order deformations of V. 1 Introduction The present paper is a sequel to the paper [H]. We discuss the first and second cohomologies of grading-restricted vertex algebras introduced by the author in that paper. Let V be a grading-restricted vertex algebra and W a V -module. Recall from [H] that for each m Z + and n N, we have an n-th cohomology Hm(V, n W ) of V with coefficients in W. For each n N, We also have an n-th cohomology H (V, n W ) of V with coefficients in W which is isomorphic to the inverse limit of the inverse system {Hm(V, n W )} m Z+. We also have an additional second cohomology H 1 (V, W ) of V with coefficients in W. In the present paper, we discuss only Hm(V, 1 W ) for m Z + and H 1 (V, W ). 1

2 Let V be a grading-restricted vertex algebra and W a V -module. grading-preserving linear map f : V W is called a derivation if A f(y V (u, z)v) = Y W W V (f(u), z)v + Y W (u, z)f(v) = e zl( 1) Y W (v, z)f(u) + Y W (u, z)f(v) for u, v V. We use Der (V, W ) to denote the space of all such derivations. We have the following result for the first cohomologies of V with coefficients in W : Theorem 1.1. Let V be a grading-restricted vertex algebra and W a V - module. Then H 1 m(v, W ) is linearly isomorphic to the space of derivations from V to W for any m Z +, that is, H 1 m(v, W ) is linearly isomorphic to Der (V, W ) for any m Z +. In particular, H 1 m(v, W ) for m N are isomorphic (and can be denoted using the same notation H 1 (V, W )). Definition 1.. Let V be a grading-restricted vertex algebra. A square-zero ideal of V is an ideal W of V such that for any u, v W, Y V (u, x)v = 0. Definition 1.3. Let V be a grading-restricted vertex algebra and W a Z- graded V -module. A square-zero extension (Λ, f, g) of V by W is a gradingrestricted vertex algebra Λ together with a surjective homomorphism f : Λ V of grading-restricted vertex algebras such that ker f is a squarezero ideal of Λ (and therefore a V -module) and an injective homomorphism g of V -modules from W to Λ such that g(w ) = ker f. Two square-zero extensions (Λ 1, f 1, g 1 ) and (Λ, f, g ) of V by W are equivalent if there exists an isomorphism of grading-restricted vertex algebras h : Λ 1 Λ such that the diagram 0 W Λ 1 V 0 g 1 f 1 1 W h 1 V is commutative. 0 W Λ V 0, g f The notion of square-zero extension of V by W is an analogue of the notion of square-zero extension of an associative algebra by a bimodule. (see, for example, Section 9.3 of [W]).

3 We have the following result for the second cohomology H 1 (V, W ) of V with coefficients in W : Theorem 1.4. Let V be a grading-restricted vertex algebra and W a V - module. Then the set of the equivalence classes of square-zero extensions of V by W corresponds bijectively to H 1 (V, W ). Definition 1.5. Let t be a complex variable. A family of grading-restricted vertex algebras up to the first order in t is a Z-graded vector space V, a family Y t : V V V ((x)) for t C of linear maps of the form Y t = Y 0 +tψ where Y 0 and Ψ are linear maps from V V to V ((x)) independent of t, and an element 1 V satisfying all the axioms for grading-restricted vertex algebras up to the first order in t. More precisely, the triple (V, Y t, 1) satisfies the grading restriction condition, lower-truncation condition for vertex operators, L(0)- bracket formula and the following conditions: 1. Identity property up to the first order in t: Y t (1, x) = 1 V + O(t ).. Creation property up to the first order in t: For u V, Y t (u, x)1 V [[x]] and lim x 0 Y t (u, x)1 = v + O(t ). 3. Duality up to the first order in t: For v 1, v, v 3 V and v V, the coefficients of t 0 and t 1 terms of v, Y t (v 1, z 1 )Y t (v, z )v 3 v, Y t (v, z )Y t (v 1, z 1 )v 3 v, Y t (Y t (v 1, z 1 z )v, z )v 3 are absolutely convergent in the regions z 1 > z > 0, z > z 1 > 0 and z > z 1 z > 0, respectively, to common rational functions in z 1 and z with the only possible poles at z 1, z = 0 and z 1 = z. Definition 1.6. Let (V, Y V, 1) be a grading-restricted vertex algebra. A first order deformation of V is a family Y t : V V V ((x)) for t C of linear maps of the form Y t = Y V + tψ where Ψ : V V V ((x)) v 1 v Ψ(v 1, x)v is a linear map such that (V, Y t, 1) for t C is a family of grading-restricted vertex algebras up to the first order in t. Two first order deformations Y (1) t 3

4 and Y () t, t C, of (V, Y V, 1) are equivalent if there exists a family f t : V V, t C of linear maps of the form f t = 1 V + tg where g : V V is a linear map preserving the gradings of V such that for v 1, v V. We have: f t (Y (1) t (v 1, x)v ) Y () t (f t (v 1 ), x)f t (v ) t V ((x)) (1.1) Theorem 1.7. The set of equivalence classes of first order deformations of a grading-restricted vertex algebra is in bijection with the set of equivalence classes of square-zero extensions of V by V. From Theorems 1.4 and 1.7, we obtain immediately the following result for the second cohomology H 1 (V, V ) of V with coefficients in V : Theorem 1.8. Let V be a grading-restricted vertex algebra. Then the set of the equivalence classes of first order deformations of V correspond bijectively to H 1 (V, V ). We prove Theorems 1.1, 1.4 and 1.7 in Sections, 3 and 4, respectively. Acknowledgments grant PHY The author is grateful for partial support from NSF First cohomologies and spaces of derivations We prove Theorem 1.1 in the present section. First, we need the following: Lemma.1. Let f : V W be a derivation. Then f(1) = 0. Proof. By definition, f(1) = f(y V (1, z)1) = lim f(y V (1, z)1) z 0 = lim e zl( 1) Y W (1, z)f(1) + lim Y W (1, z)f(1) z 0 z 0 = f(1). 4

5 So f(1) = 0. Let Φ : V W z1 be an element of Cm(V, 1 W ) satisfying δmφ 1 = 0. Since Φ satisfies the L(0)-conjugation property, for v V (n) and z C, z L(0) (Φ(v))(0) = (Φ(z L(0) v))(0) = z n (Φ(v))(0). Thus (Φ(v))(0) W (n). So (Φ(v))(0) is a grading-preserving linear map from V to W. Since δ 1 mφ = 0, R( w, Y W (v 1, z 1 )(Φ(v ))(z ) ) R( w, (Φ(Y V (v 1, z 1 z )v ))(z ) ) = 0 +R( w, Y W (v, z )(Φ(v 1 ))(z 1 ) ) for v 1, v V and w W. By L( 1)-derivative property for Φ and the vertex operator map Y W, R( w, Y W (v, z )(Φ(v 1 ))(z 1 ) ) = R( w, e z 1L( 1) Y W (v, z 1 + z )(Φ(v 1 ))(0) ). Thus we have R( w, Y W (v 1, z 1 )(Φ(v ))(z ) ) R( w, (Φ(Y V (v 1, z 1 z )v ))(z ) ) = 0. Let z = 0. We obtain +R( w, e z 1L( 1) Y W (v, z 1 + z )(Φ(v 1 ))(0) ) R( w, Y W (v 1, z 1 )(Φ(v ))(0) ) R( w, (Φ(Y V (v 1, z 1 )v ))(0) ) = 0. +R( w, e z 1L( 1) Y W (v, z 1 )(Φ(v 1 ))(0) ) Since w is arbitrary, we obtain (Φ(Y V (v 1, z 1 )v ))(0) = e z 1L( 1) Y W (v, z 1 )(Φ(v 1 ))(0) + Y W (v 1, z 1 )(Φ(v ))(0) = Y W W V ((Φ(v 1 ))(0), z 1 )(Φ(v ))(0) + Y W (v 1, z 1 )(Φ(v ))(0) 5

6 for v 1, v V. This means that (Φ( ))(0) : V W is a derivation from V to W. Note that δm(c 0 m(v, 0 W )) = 0. So we obtain a linear map from H 1 (V, W ) to the space of derivations from V to W. Conversely, given any derivation f from V to W, let Φ f : V W z1 be given by (Φ f (v))(z 1 ) = f(y V (v, z 1 )1) = YW W V (f(v), z 1 )1 for v V, where we have used Lemma.1. By Theorem 5.6. in [FHL], the map from V to W z1 given by v Y W W V ((Φ(v))(0), z 1)1 is composable with m vertex operators for any m N. Thus Φ f C 1 m(v, W ) for any m N. For v 1, v V and w W, ((δ 1 mφ f )(v 1 v ))(z 1, z ) = R( w, Y W (v 1, z 1 )Y W W V (f(v ), z )1 ) R( w, Y W W V (f(y V (v 1, z 1 z )v ), z )1) ) +R( w, Y W (v, z )Y W W V (f(v 1 ), z 1 )1 ) = R( w, Y W (v 1, z 1 )Y W W V (f(v ), z )1 ) R( w, Y W W V (Y W W V (f(v 1 ), z 1 z )v ), z )1) ) R( w, Y W W V (Y W (v 1, z 1 z )f(v ), z )1) ) +R( w, Y W (v, z )Y W W V (f(v 1 ), z 1 )1 ) = R( w, Y W (v 1, z 1 )Y W W V (f(v ), z )1 ) R( w, e z L W ( 1) Y W W V (f(v 1 ), z 1 z )v ) R( w, e z L W ( 1) Y W (v 1, z 1 z )f(v ) ) +R( w, Y W (v, z )Y W W V (f(v 1 ), z 1 )1 ) = R( w, Y W (v 1, z 1 )Y W W V (f(v ), z )1 ) R( w, Y W W V (f(v 1 ), z 1 )e z L V ( 1) v ) R( w, Y W (v 1, z 1 )e z L W ( 1) f(v ) ) +R( w, Y W (v, z )Y W W V (f(v 1 ), z 1 )1 ) = R( w, Y W (v 1, z 1 )Y W W V (f(v ), z )1 ) R( w, Y W W V (f(v 1 ), z 1 )Y W (v, z )1 ) R( w, Y W (v 1, z 1 )Y W W V (f(v ), z )1 ) +R( w, Y W (v, z )Y W W V (f(v 1 ), z 1 )1 ) = R( w, Y W W V (f(v 1 ), z 1 )Y V (v, z )1 ) +R( w, Y W (v, z )Y W W V (f(v 1 ), z 1 )1 ). 6

7 (.1) From Theorem 5.6. in [FHL], we know that the right-hand side of (.1) is 0. So we obtain a linear map f Φ f from the space Der (V, W ) to H 1 m(v, W ) = C 1 m(v, W ). Clearly these two maps are inverse to each other and thus Der (V, W ) and H 1 m(v, W ) are isomorphic. 3 Second cohomologies and square-zero extensions In this section, we prove Theorem 1.4. Let (Λ, f, g) be a square-zero extension of V by W. Then there is an injective linear map Γ : V Λ such that the linear map h : V W Λ given by h(v, w) = Γ(v) + g(w) is a linear isomorphism. By definition, the restriction of h to W is the isomorphism g from W to ker f. Then the gradingrestricted vertex algebra structure and the V -module structure on Λ give a grading-restricted vertex algebra structure and a V -module structure on V W such that the embedding i : W V W and the projection p 1 : V W V are homomorphisms of grading-restricted vertex algebras. Moreover, ker p 1 is a square-zero ideal of V W, i is an injective homomorphism such that i (W ) = ker p 1 and the diagram 0 W i V W p 1 V 0 1 W h 1 V 0 W g Λ f V 0 (3.1) of V -modules is commutative. So we obtain a square-zero extension (V W, p 1, i ) equivalent to (Λ, f, g). We need only consider square-zero extension of V by W of the particular form (V W, p 1, i ). Note that the difference between two such square-zero extensions are in the vertex operator maps. So we use (V W, Y V W, p 1, i ) to denote such a square-zero extension. We now write down the vertex operator map for V W explicitly. Since (V W, Y V W, p 1, i ) is a square-zero extension of V, there exists Ψ(u, x)v 7

8 W ((x)) for u, v V such that Y V W ((v 1, 0), x)(v, 0) = (Y V (v 1, x)v, Ψ(v 1, x)v ), Y V W ((v 1, 0), x)(0, w) = (0, Y V (v 1, x)w ), Y V W ((0, w 1 ), x)(v, 0) = (0, Y W W V (w, x)v ), Y V W ((0, w 1 ), x)(0, w ) = 0 for v 1, v V and w 1, w W. Thus we have Y V W ((v 1, w 1 ), x)(v, w ) = (Y V (v 1, x)v, Y W (v 1, x)w + Y W W V (w 1, x)v + Ψ(v 1, x)v ) (3.) for v 1, v V and w 1, w W. The vacuum of V W is (1, 0). Since Y V W ((v, w), x)(1, 0) = e xl V W ( 1) (v, w) for v W and w W, we have = (e xl V ( 1) v, e xl W ( 1) w) = (Y V (v, x)1, Y W W V (w, x)1) Ψ(v, x)1 = 0 (3.3) for v V. We identify (V W ) with V W. For v 1, v V and w W, (0, w ), Y V W ((v 1, 0), z 1 )Y V W ((v, 0), z )(1, 0) = w, Ψ(v 1, z 1 )Y V (v, z )1 + Y W (v 1, z 1 )Ψ(v, z )1 = w, Ψ(v 1, z 1 )Y V (v, z )1, (0, w ), Y V W ((v, 0), z )Y V W ((v 1, 0), z 1 )(1, 0) = w, Ψ(v, z )Y V (v 1, z 1 )1 + Y W (v, z )Ψ(v 1, z 1 )1 = w, Ψ(v, z )Y V (v 1, z 1 )1, (0, w ), Y V W (Y V W ((v 1, 0), z 1 z )(v, 0), z )(1, 0) = w, Y W W V (Ψ(v 1, z 1 z )v, z )1 + Ψ(Y V (v 1, z 1 z )v, z )1 = w, Y W W V (Ψ(v 1, z 1 z )v, z )1 are absolutely convergent in the region z 1 > z > 0, z > z 1 > 0, z > z 1 z > 0, respectively, to one rational function in z 1 and z with 8

9 the only possible poles at z 1, z = 0 and z 1 = z. Using our notation in [H], we denote this rational function by R( w, Ψ(v 1, z 1 )Y V (v, z )1 ) or or R( w, Ψ(v, z )Y V (v 1, z 1 )1 ) R( w, Y W W V (Ψ(v 1, z 1 z )v, z )1 ). Then we obtain an element, denoted by E(Ψ(v 1, z 1 )Y V (v, z )1) or or E(Ψ(v, z )Y V (v 1, z 1 )1) E(Y W W V (Ψ(v 1, z 1 z )v, z )1), of W z1,z given by w, E(Ψ(v 1, z 1 )Y V (v, z )1) = R( w, Ψ(v 1, z 1 )Y V (v, z )1 ) or w, E(Ψ(v, z )Y V (v 1, z 1 )1) = R( w, Ψ(v, z )Y V (v 1, z 1 )1 ) or w, E(Y W W V (Ψ(v 1, z 1 z )v, z )1) = R( w, Y W W V (Ψ(v 1, z 1 z )v, z )1 ). By definition, we have for v 1, v V. Let E(Ψ(v 1, z 1 )Y V (v, z )1) = E(Ψ(v, z )Y V (v 1, z 1 )1) = E(Y W W V (Ψ(v 1, z 1 z )v, z )1) Φ : V V W z1,z 9

10 be the linear map given by (Φ(v 1 v ))(z 1, z ) = E(Ψ(v 1, z 1 )Y V (v, z )1) = E(Ψ(v, z )Y V (v 1, z 1 )1) = E(Y W W V (Ψ(v 1, z 1 z )v, z )1) (3.4) for v 1, v V and (z 1, z ) F C. We first prove that Φ Ĉ 1 (V, W ). By the L( 1)-derivative property and the L(0)-bracket formula for V W, we have d dx Y V W ((v, 0), x) = Y V W ((L V ( 1)v, 0), x) (3.5) = [L V +W ( 1), Y V W ((v, 0), x)], (3.6) [L V +W (0), Y V W ((v, 0), x)] = Y V W ((L V (0)v, 0), x) + x d dx Y V W ((v, 0), x) (3.7) for v V. By (3.5), (3.6), (3.7), (3.) and the L( 1)-derivative property and the L(0)-bracket formula for V, we obtain d dx Ψ(v, x) = Ψ(L V ( 1)v, x) (3.8) = L W ( 1)Ψ(v, x) Ψ(v, x)l V ( 1), (3.9) L W (0)Ψ(v, x) Ψ(v, x)l V (0) = Ψ(L V (0)v, x) + x d Ψ(v, x) dx (3.10) for v V. From (3.10), we obtain z L W (0) Ψ(v, x) = Ψ(z L V (0) v, zx)z L V (0) (3.11) for v V. For v 1, v V and w W, by (3.8) and the L( 1)-derivative property for V, we obtain z 1 w, (Φ(v 1 v ))(z 1, z ) = z 1 w, E(Ψ(v 1, z 1 )Y V (v, z )1) 10

11 = R( w, Ψ(v 1, z 1 )Y V (v, z )1 ) z ( 1 ) = R w, Ψ(v 1, z 1 )Y V (v, z )1 z 1 = R( w, Ψ(L V ( 1)v 1, z 1 )Y V (v, z )1 ) = w, E(Ψ(L V ( 1)v 1, z 1 )Y V (v, z )1) = w, (Φ(L V ( 1)v 1 v ))(z 1, z ) (3.1) and w, (Φ(v 1 v ))(z 1, z ) z = w, E(Ψ(v 1, z 1 )Y V (v, z )1) z = R( w, Ψ(v 1, z 1 )Y V (v, z )1 ) z ( = R w, Ψ(v 1, z 1 ) ) Y V (v, z )1 z = R( w, Ψ(v 1, z 1 )Y V (L V ( 1)v, z )1 ) = w, E(Ψ(v 1, z 1 )Y V (L V ( 1)v, z )1) = w, (Φ(v 1 L V ( 1)v ))(z 1, z ). (3.13) Using (3.8), (3.9) and the L( 1)-derivative property for V, we obtain ( + ) w, (Φ(v 1 v ))(z 1, z ) z z ( = + ) w, E(Ψ(v 1, z 1 )Y V (v, z )1) z z ( = + ) R( w, Ψ(v 1, z 1 )Y V (v, z )1) ) z z ( ) = R w, Ψ(v 1, z 1 )Y V (v, z )1 z ( 1 +R w, Ψ(v 1, z 1 ) ) Y V (v, z )1 z = R( w, Ψ(L V ( 1)v 1, z 1 )Y V (v, z )1 ) +R( w, Ψ(v 1, z 1 )Y V (L V ( 1)v, z )1 ) 11

12 = R( w, L W ( 1)Ψ(v 1, z 1 )Y V (v, z )1 ) = R( L W (1)w, Ψ(v 1, z 1 )Y V (v, z )1 ) = L W (1)w, E(Ψ(v 1, z 1 )Y V (v, z )1) = w, L W ( 1)E(Ψ(v 1, z 1 )Y V (v, z )1) = w, L W ( 1)(Φ(v 1 v ))(z 1, z ) (3.14) for v 1, v V and w W, From (3.1), (3.13) and (3.14), we see that Φ satisfies the L( 1)-derivative property. Also for v 1, v V and w W, by (3.11) and the L(0)-bracket formula for V, we have w, z L W (0) (Φ(v 1 v ))(z 1, z ) = w, z L W (0) E(Ψ(v 1, z 1 )Y V (v, z )1) = z L W (0) w, E(Ψ(v 1, z 1 )Y V (v, z )1) = R( z L W (0) w, Ψ(v 1, z 1 )Y V (v, z )1 ) = R( w, z L W (0) Ψ(v 1, z 1 )Y V (v, z )1 ) = R( w, Ψ(z L V (0) v 1, zz 1 )Y V (z L V (0) v, zz )1 ) = w, E(Ψ(z L V (0) v 1, zz 1 )Y V (z L V (0) v, zz )1) = w, (Φ(z L V (0) v 1 z L V (0) v ))(zz 1, zz ), that is, Φ satisfies the L(0)-conjugation property. Since V W is a grading-restricted vertex algebra, for v 1, v, v 3 V and w W, the series and (0, w ), Y V W ((v 1, 0), z 1 )Y V W ((v, 0), z )Y V W ((v 3, 0), z 3 )(1, 0) (0, w ), Y V W (Y V W ((v 1, 0), z 1 z )(v, 0), z )Y V W ((v 3, 0), z 3 )(1, 0) are absolutely convergent in the regions given by z 1 > z > z 3 > 0 and by z > z 1 z, z 3 > 0 and z z 3 > z 1 z, respectively, to a same rational function with the only possible poles at z 1 = z, z 1 = z 3, z = z 3. But by (3.) and (3.3), these series are equal to w, Ψ(v 1, z 1 )Y V (v, z )Y V (v 3, z 3 )1 + w, Y W (v 1, z 1 )Ψ(v, z )Y V (v 3, z 3 )1 1

13 and w, Ψ(Y V (v 1, z 1 z )v, z )Y V (v 3, z 3 )1 + w, Y W W V (Ψ(v 1, z 1 z )v, z )Y V (v 3, z 3 )1, respectively, and are absolutely convergent to a same rational function which in our convention is equal to R( w, Ψ(v 1, z 1 )Y V (v, z )Y V (v 3, z 3 )1 + w, Y W (v 1, z 1 )Ψ(v, z )Y V (v 3, z 3 )1 ) and R( w, Ψ(Y V (v 1, z 1 z )v, z )Y V (v 3, z 3 )1 + w, Y W W V (Ψ(v 1, z 1 z )v, z )Y V (v 3, z 3 )1 ). In particular, we have proved that Φ Ĉ 1 (V, W ). Since by (3.4), (Φ(v 1 v ))(z 1, z ) = E(Ψ(v 1, z 1 )Y V (v, z )1) for v 1, v V and (z 1, z ) F C, that is, for v 1, v V, We obtain = E(Ψ(v, z )Y V (v 1, z 1 )1) = (Φ(v v 1 ))(z, z 1 ) = (σ 1 (Φ(v v 1 )))(z 1, z ) Φ(v 1 v ) σ 1 (Φ(v v 1 )) = 0 σ J ;1 ( 1) σ σ(φ(v 1 v )) = Φ(v 1 v ) σ 1 (Φ(v v 1 )) = 0 for v 1, v V. So Φ C (V, W ). Next we show that δ 1 (Φ) = 0. For v 1, v, v 3 V, w W, w, ((δ 1 (Φ))(v 1 v v 3 ))(z 1, z, z 3 ) 13

14 = R( w, (E (1) W (v 1; Φ(v v 3 )))(z 1, z, z 3 ) + w, (Φ(v 1 E () (v v 3 ; 1)))(z 1, z, z 3 ) ) R( w, (Φ(E () (v 1 v ; 1) v 3 ))(z 1, z, z 3 ) + w, (E W ;(1) W V (Φ(v 1 v ); v 3 ))(z 1, z, z 3 ) ) = R( w, Y W (v 1, z 1 )Ψ(v, z )Y V (v 3, z 3 )1 + w, Ψ(v 1, z 1 )Y V (v, z )Y V (v 3, z 3 )1 ) R( w, Ψ(Y V (v 1, z 1 z )v, z )Y V (v 3, z 3 )1 + w, Y W W V (Ψ(v 1, z 1 z )v, z )Y V (v 3, z 3 )1 ). (3.15) Since V W is a grading-restricted vertex algebra, we have the associativity property R( (0, w ), Y V W ((v 1, 0), z 1 )Y V W ((v, 0), z )Y V W ((v 3, 0), z 3 )(1, 0) ) = R( (0, w ), Y V W (Y V W ((v 1, 0), z 1 z )(v, 0), z ) which, by (3.) and (3.3), is equivalent to R( w, Ψ(v 1, z 1 )Y V (v, z )Y V (v 3, z 3 )1 + w, Y W (v 1, z 1 )Ψ(v, z )Y V (v 3, z 3 )1 ) = R( w, Ψ(Y V (v 1, z 1 z )v, z )Y V (v 3, z 3 )1 Y V W ((v 3, 0), z 3 )(1, 0) ), + w, Y W W V (Ψ(v 1, z 1 z )v, z )Y V (v 3, z 3 )1 ), as we have noticed above. So the right-hand side of (3.15) is 0. Thus Φ + δc 1 (V, 1 W ) is an element of H 1 (V, W ). Conversely, given any element of H 1 (V, W ), let Φ C 1 (V, W ) be a rep- resentative of this element. Then for any v 1, v V, there exists N such that for w W, w, (Φ(v 1 v ))(z, 0) is a rational function of z with the only possible pole at z = 0 of order less than or equal to N. For v 1, v V, let Ψ(v 1, x)v W ((x)) be given by w, Ψ(v 1, x)v x=z = w, (Φ(v 1 v ))(z, 0). for z C. For v 1, v V, define Y V W (v 1, x)v using (3.). So we obtain a vertex operator map Y V W. Reversing the proof above, we see that V W equipped with the vertex operator map Y V +W and the vacuum (1, 0) is a 14

15 grading-restricted vertex algebra and together with the projection p 1 : V W V and the embedding i : W V W, V W is a square-zero extension of V by W. Next we prove that two elements of ker δ 1 obtained this way differ by an element of δ 1 C 1 (V, W ) if and only if the corresponding square-zero extensions of V by W are equivalent. Let Φ 1, Φ ker δ 1 be two such elements obtained from square-zero extensions (V W, Y (1) V W, p 1, i ) and (V W, Y () V W, p 1, i ). Assume that Φ 1 = Φ + δ 1 (Γ) where Γ C 1 (V, W ). Since we have w, ((δ 1 (Γ))(v 1 v ))(z 1, z ) = R( w, Y W (v 1, z 1 )(Γ(v ))(z ) ) R( w, (Γ(Y V (v 1, z 1 z )v ))(z ) ) +R( w, Y W (v, z )(Γ(v 1 ))(z 1 ) ), R( w, Ψ 1 (v 1, z 1 )Y V (v, z )1 ) = w, (Φ 1 (v 1 v ))(z 1, z ) = w, (Φ (v 1 v ))(z 1, z ) + w, (δ 1 (Γ))(z 1, z ) = R( w, Ψ (v 1, z 1 )Y V (v, z )1 +R( w, Y W (v 1, z 1 )(Γ(v ))(z ) ) R( w, (Γ(Y V (v 1, z 1 z )v ))(z ) ) +R( w, Y W (v, z )(Γ(v 1 ))(z 1 ) ) = R( w, Ψ (v 1, z 1 )Y V (v, z )1 +R( w, Y W (v 1, z 1 )(Γ(v ))(z ) ) R( w, (Γ(Y V (v 1, z 1 z )v ))(z ) ) +R( w, e (z 1 z )L W ( 1) Y W (v, z 1 )(Γ(v 1 ))(z ) ). (3.16) Let z go to zero on both sides of (3.16). We obtain w, Ψ 1 (v 1, z 1 )v = w, Ψ (v 1, z 1 )v + w, Y W (v 1, z 1 )(Γ(v ))(0) w, (Γ(Y V (v 1, z 1 )v ))(0) 15

16 + w, e z 1L W ( 1) Y W (v, z 1 )(Γ(v 1 ))(0) = w, Ψ (v 1, z 1 )v + w, Y W (v 1, z 1 )(Γ(v ))(0) w, (Γ(Y V (v 1, z 1 )v ))(0) + w, YW W V ((Γ(v 1 ))(0), z 1 )v. Then Ψ 1 (v 1, x)v = Ψ (v 1, x)v + Y W (v 1, x)(γ(v ))(0) (Γ(Y V (v 1, x)v ))(0) + Y W W V ((Γ(v 1 ))(0), x)v. (3.17) For v 1, v V and w 1, w W, by (3.) and (3.17), we have Y (1) V W ((v 1, w 1 ), x)(v, w ) = (Y V (v 1, x)v, Y W (v 1, x)w + Y W W V (w 1, x)v + Ψ 1 (v 1, x)v ) = (Y V (v 1, x)v, Y W (v 1, x)w + Y W W V (w 1, x)v + Ψ (v 1, x)v ) +(Y V (v 1, x)v, Y W (v 1, x)(γ(v ))(0)) (Y V (v 1, x)v, (Γ(Y V (v 1, x)v ))(0)) +(Y V (v 1, x)v, Y W W V ((Γ(v 1 ))(0), x)v ) = Y () V W ((v 1, w 1 + (Γ(v 1 ))(0)), x)(v, w + (Γ(v ))(0)) (Y V (v 1, x)v, (Γ(Y V (v 1, x)v ))(0)). (3.18) We now define a linear map h : V W V W by e(v, w) = (v, w + (Γ(v))(0)) for v V and w W. Then h is a linear isomorphism and (3.18) can be rewritten as h(y (1) V W ((v 1, w 1 ), x)(v, w )) = Y () V W (h(v 1, w 1 ), x)h(v, w ). (3.19) for v 1, v V and w 1, w W. Thus h is an isomorphism of gradingrestricted vertex algebras from (V W, Y (1) () V W, (1, 0)) to (V W, Y V W, (1, 0)) such that the diagram 0 W i V W p 1 V 0 1 W h 1 V (3.0) 0 W i V W p 1 V 0 16

17 is commutative. Thus the two square-zero extensions of V by W are equivalent. Conversely, let (V W, Y (1) V +W, p 1, i ) and (V W, Y () V +W, p 1, i ) be two equivalent square-zero extensions of V by W. So there exists an isomorphism h : V W V W of grading-restricted vertex algebras such that (3.0) is commutative. We have the following lemma which is also needed in the next section: Lemma 3.1. There exists a linear map g : V V such that for v V and w W. h(v, w) = (v, w + g(v)) Proof. Let h(v, w) = (f(v, w), g(v, w)) for v V and w W. Then by (3.0), we have f(v, w) = v and g(0, w) = w. Since h is linear, we have g(v, w) = g(v, 0)+g(0, w) = w+g(v, 0). So h(v, w) = (v, w+g(v, 0)). Taking g(v) to be g(v, 0), we see that the conclusion holds. Let (Γ(v))(z 1 ) = e z 1L W ( 1) g(v) W. Then Γ : V W z1 is an element of C(V, 1 W ). By definition, we have g(v) = (Γ(v))(0) and h(v, w) = (v, w + (Γ(v))(0)) for v V and w W. Let Φ 1 and Φ be elements of ker δ 1 obtained from (V W, Y (1) V W, p 1, i ) and (V W, Y () V +W, p 1, i ), respectively, and Ψ 1 and Ψ the corresponding maps from V V to W ((x)). Then since h is a homomorphism of grading-restricted vertex algebras, (3.19) holds for v 1, v V and w 1, w W. Thus the two sides of (3.18) are equal for v 1, v V and w 1, w W. So the two expressions in the middle of (3.18) are equal for v 1, v V and w 1, w W. Thus we have (3.17) for v 1, v V. Formula (3.17) implies that the two sides of (3.16) are equal for v 1, v V. Thus the middle expressions in (3.16) are all equal for v 1, v V. In particular, we obtain Φ 1 = Φ + Γ. So Φ 1 and Φ differ by an element of δ 1 C 1 (V, W ). 4 Square-zero extensions and first order deformations In this section, we prove Theorem

18 Let Y t : V V V ((x)), t U, be a first order deformation of V. By definition, there exists such that Ψ : V V V ((x)) v 1 v Ψ(v 1, x)v Y t (v 1, x)v = Y V (v 1, x)v + tψ(v 1, x)v for v 1, v V and (V, Y t, 1) is a family of grading-restricted vertex algebras up to the first order in t. The identity property for (V, Y t, 1) up to the first order in t gives Y V (1, x)v + tψ(1, x)v = v + O(t ) for v V. So we obtain Ψ(1, x)v = 0 (4.1) for v V. The creation property for (V, Y t, 1) up to the first order in t gives for v V. Then we have lim (Y V (v, x) + tψ(v, x))1 = v + O(t ) x 0 lim Ψ(v, x)1 = 0 (4.) x 0 for v V. The duality property up to the first order in t can be written explicitly as follows: For v 1, v, v 3 V and v V, v, (Y V (v 1, z 1 )Ψ(v, z ) + Ψ(v 1, z 1 )Y V (v, z ))v 3 (4.3) v, (Y V (v, z )Ψ(v 1, z 1 ) + Ψ(v, z )Y V (v 1, z 1 ))v 3 (4.4) v, (Y V (Ψ(v 1, z 1 z )v, z ) + Ψ(Y V (v 1, z 1 z )v, z ))v 3 (4.5) are absolutely convergent in the regions z 1 > z > 0, z > z 1 > 0 and z > z 1 z > 0, respectively, to a common rational function in z 1 and z with the only possible poles at z 1, z = 0 and z 1 = z. Let Y V V : (V V ) (V V ) (V V )[[x, x 1 ]] (u 1, v 1 ) (u, v ) Y V V ((u 1, v 1 ), x)(u, v ) 18

19 be given by Y V V ((u 1, v 1 ), x)(u, v ) = (Y V (u 1, x)u, Y V (u 1, x)v + Y V (v 1, x)u + Ψ(u 1, x)u ) (4.6) for u 1, u, v 1, v V. By (4.6) and (4.1), Y V V ((1, 0), x)(u, v) = (Y V (1, x)u, Y V (1, x)v + Y V (0, x)u + Ψ(1, x)u) = (u, v) for u, v V, that is, (V V, Y V V, (1, 0)) has the identity property. By (4.6) and (4.), lim Y V V ((u, v), x)(1, 0) x 0 = (lim Y V (u, x)1, lim Y V (u, x)0 + lim Y V (v, x)1 + lim Ψ(u, x)1) x 0 x 0 x 0 x 0 = (u, v) for u, v V, that is, (V V, Y V V, (1, 0)) has the creation property. By (4.6), we have (u, v ), Y V V ((u 1, v 1 ), z 1 )Y V V ((u, v ), z )(u 3, v 3 ) = (u, v ), Y V V ((u 1, v 1 ), z 1 ) (Y V (u, z )u 3, Y V (u, z )v 3 + Y V (v, z )u 3 + Ψ(u, z )u 3 ) = u, Y V (u 1, z 1 )Y V (u, z )u 3 + v, Y V (u 1, z 1 )Y V (u, z )v 3 + v, Y V (u 1, z 1 )Y V (v, z )u 3 + v, Y V (u 1, z 1 )Ψ(u, z )u 3 + v, Y V (v 1, z 1 )Y V (u, z )u 3 + v, Ψ(u 1, z 1 )Y V (u, z )u 3. (4.7) By the properties of V and the absolute convergence of (4.3), we see that the left-hand side of (4.7) is absolutely convergent when z 1 > z > 0. Similarly, by (4.6), we have (u, v ), Y V V ((u, v ), z )Y V V ((u 1, v 1 ), z 1 )(u 3, v 3 ) = u, Y V (u, z )Y V (u 1, z 1 )u 3 + v, Y V (u, z )Y V (u 1, z 1 )v 3 + v, Y V (u, z )Y V (v 1, z 1 )u 3 + v, Y V (u, z )Ψ(u 1, z 1 )u 3 + v, Y V (v, z )Y V (u 1, z 1 )u 3 + v, Ψ(u, z )Y V (u 1, z 1 )u 3 (4.8) and the left-hand side of (4.8) is absolutely convergent when z > z 1 > 0. Moreover, since (4.3) and (4.4) converges absolutely when z 1 > z > 0 19

20 and when z > z 1 > 0, respectively, to a common rational function with the only possible poles at z 1, z, z 1 z = 0, the left-hand side of (4.7) and left-hand side of (4.8) also converges absolutely when z 1 > z > 0 and when z > z 1 > 0, respectively, to a common rational function with the only possible pole at z 1 z = 0. By (4.6) again, we have (u, v ), Y V V (Y V V ((u 1, v 1 ), z 1 z )(u, v ), z )(u 3, v 3 ) = (u, v ), Y V V ((Y V (u 1, z 1 z )u, Y V (u 1, z 1 z )v + Y V (v 1, z 1 z )u +Ψ(u 1, z 1 z )u ), z )(u 3, v 3 ) = u, Y V (Y V (u 1, z 1 z )u, z )u 3 + v, Y V (Y V (u 1, z 1 z )u, z )v 3 + v, Y V (Y V (u 1, z 1 z )v, z )u 3 + v, Y V (Y V (v 1, z 1 z )u, z )u 3 + v, Y V (Ψ(u 1, z 1 z )u, z )u 3 + v, Ψ(Y V (u 1, z 1 z )u, z )u 3. (4.9) By the properties of V and the absolute convergence of (4.5) and (4.9), we see that the left-hand side of (4.9) is absolutely convergent when z > z 1 z > 0. Moreover, since (4.3) and (4.5) converges absolutely when z 1 > z > 0 and when z > z 1 z > 0, respectively, to a common rational function with the only possible poles at z 1, z, z 1 z = 0, the left-hand side of (4.7) and left-hand side of (4.9) also converges absolutely when z 1 > z > 0 and when z > z 1 z > 0, respectively, to a common rational function with the only possible poles at z 1, z, z 1 z = 0. So (V V, Y V V, (1, 0)) has the duality property. Note that the L( 1)-derivative property is in fact a consequence of the other axioms for vertex algebras. Thus (V V, Y V V, (1, 0)) is a gradingrestricted vertex algebra. By definition, p 1 (Y V V ((u 1, v 1 ), x)(u, v )) = p 1 (Y V (u 1, x)u, Y V (u 1, x)v + Y V (v 1, x)u + Ψ(u 1, x)u ) = Y V (u 1, x)u = Y V (p 1 (u 1, v 1 ), x)p 1 (u, v ) for u 1, u, v 1, v V. Also ker p 1 = 0 V 0

21 and Y V V ((0, v 1 ), x)(0, v ) = (0, 0) for v 1.v V. So p 1 is a surjective homomorphism of grading-restricted vertex algebras and ker p 1 is a square-zero ideal of V V. We use YV V V to denote the vertex operator map for V V when V V is viewed as a V -module. Then by definition, i (Y V (v 1, x)v ) = (0, Y V (v 1, x)v ) = Y V V V (v 1, x)(0, v ) = Y V V V (v 1, x)i (v ) for v 1, v V. So i is an injective homomorphism of V -modules. Clearly, we have i (V ) = ker p 1. Thus (V V, Y V V, p 1, i ) is a square-zero extension of V by V. Conversely, let (V V, Y V V, p 1, i ) be a square-zero extension of V by V. Then there exists such that Ψ : V V V ((x)) v 1 v Ψ(v 1, x)v Y V V ((u 1, 0), x)(u, 0) = (Y V (u 1, x)u, Ψ(u 1, x)u ) for u 1, u V. The identity property and the creation property of the grading-restricted vertex algebra (V V, Y V V, (1, 0)) give (4.1) and (4.). The duality property for (V V, Y V V, (1, 0)) gives (4.3), (4.4) and (4.5). For t C, define Y t (v 1, x)v = Y V (v 1, x)v + tψ(v 1, x)v for v 1, v V. Then (4.1) and (4.) imply that Y t satisfies the identity property and the creation property up to the first order in t and (4.3), (4.4) and (4.5) imply that Y t satisfies the duality property up to the first order in t. Thus (V, Y t, 1) is a grading-restricted vertex algebras up to the first order in t, that is, Y t is a first-order deformation of (V, Y V, 1). Now we prove that two first-order deformations of V are equivalent if and only if the corresponding square-zero extensions of V by V are equivalent. 1

22 Consider two equivalent first-order deformations of V given by Y (1) t : V V V ((x)) and Y () t : V V V ((x)) for t C. Then there exist a family f t : V V, t C, of linear maps of the form f t = 1 V + tg where g : V V is a linear map preserving the grading of V such that (1.1) holds for v 1, v V. By definition, there exist linear maps and Ψ 1 : V V V ((x)) v 1 v Ψ 1 (v 1, x)v Ψ : V V V ((x)) v 1 v Ψ (v 1, x)v such that Y (1) t = Y V + tψ 1 and Y () t = Y V + tψ. By (1.1), we have Ψ 1 (v 1, x)v Ψ (v 1, x)v = g(y V (v 1, x)v ) + Y V (g(v 1 ), x)v + Y V (v 1, x)g(v ) (4.10) for v 1, v V. Let (V V, Y (1) V V, p 1, i ) and (V V, Y () V V, p 1, i ) be the square-zero extensions of V by V constructed from Y (1) t and Y () t. Let h : V V V V be defined by h(v 1, v ) = (v 1, v + g(v 1 )) for v 1, v V. Clearly, h is a linear isomorphism. For u 1, u, v 1, v V, by definition and (4.10), h(y (1) V V ((u 1, v 1 ), x)(u, v )) = h(y V (u 1, x)u, Y V (u 1, x)v + Y V (v 1, x)u + Ψ 1 (u 1, x)u ) = (Y V (u 1, x)u, Y V (u 1, x)v + Y V (v 1, x)u +Ψ 1 (u 1, x)u + g(y V (u 1, x)u )) = (Y V (u 1, x)u, Y V (u 1, x)v + Y V (v 1, x)u +Ψ (u 1, x)u + Y V (g(u 1 ), x)u + Y V (u 1, x)g(u ))) = (Y V (u 1, x)u, Y V (u 1, x)(v + g(u )) = Y () V V (h(u 1, v 1 ), x)h(u, v ). +Y V ((v 1 + g(u 1 )), x)u + Ψ (u 1, x)u )

23 So h is in fact an isomorphism from the algebra (V V, Y (1) V V, (1, 0)) to the algebra (V V, Y () V V, (1, 0)). Now it is clear that the following diagram is commutative: 0 V i V V p 1 V 0 1 W h 1 V 0 V i V V p 1 V 0, So these two first order deformations are equivalent. Conversely, let (V V, Y (1) V V, p 1, i ) and (V V, Y () V V, p 1, i ) be two equivalent square-zero extensions of V by V. Let Ψ 1, Ψ : V V V ((x)) be given by Y (1) V V ((u 1, 0), x)(u, 0)) = (Y V (u 1, x)u, Ψ 1 (u 1, x)u ), Y () V V ((u 1, 0), x)(u, 0)) = (Y V (u 1, x)u, Ψ (u 1, x)u ) for u 1, u V. Then Y (1) t, Y () t : V V V ((x)) given by Y (1) t (v 1, x)v = Y V (v 1, x)v + tψ 1 (v 1, x)v, Y () t (v 1, x)v = Y V (v 1, x)v + tψ (v 1, x)v for v 1, v V are first-order deformations of (V, Y V, 1) by the proof above. Let h : V V V V be an equivalence from (V V, Y (1) V V, p 1, i ) to (V V, Y () V V, p 1, i ). Then by Lemma 3.1, there exists a linear map g : V V such that h(v 1, v ) = (v 1, v + g(v 1 )) for v 1, v V. Using the fact that h is an isomorphism of grading-restricted vertex algebras from (V V, Y (1) () V V, (1, 0)) to (V V, Y V V, (1, 0)), we obtain (4.10) which implies (1.1). Thus the two first-order deformations Y (1) t and are equivalent. Y () t References [FHL] I. Frenkel, Y. Huang and J. Lepowsky, On axiomatic approaches to vertex operator algebras and modules, Memoirs American Math. Soc. 104,

24 [H] [W] Y.-Z. Huang, A cohomology theory of grading-restricted vertex algebras, to appear; arxiv: C. Weibel, An introduction to homological algebras, Cambridge Studies in Adv. Math., Vol. 38, Cambridge University Press, Cambridge, Beijing International Center for Mathematical Research, Peking University, Beijing , China, Kavali Institute For Theoretical Physics China, CAS, Beijing , China and Department of Mathematics, Rutgers University, 110 Frelinghuysen Rd., Piscataway, NJ (permanent address) address: yzhuang@math.rutgers.edu 4