VERTEX ALGEBRAS AND STRONGLY HOMOTOPY LIE ALGEBRAS

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1 University of Kentucky UKnowledge University of Kentucky Doctoral Dissertations Graduate School 2006 VERTEX ALGEBRAS AND STRONGLY HOMOTOPY LIE ALGEBRAS Daniel F. Pinzon University of Kentucky, Click here to let us know how access to this document benefits you. Recommended Citation Pinzon, Daniel F., "VERTEX ALGEBRAS AND STRONGLY HOMOTOPY LIE ALGEBRAS" (2006). University of Kentucky Doctoral Dissertations This Dissertation is brought to you for free and open access by the Graduate School at UKnowledge. It has been accepted for inclusion in University of Kentucky Doctoral Dissertations by an authorized administrator of UKnowledge. For more information, please contact

2 ABSTRACT OF DISSERTATION Daniel F. Pinzon The Graduate School University of Kentucky May 3, 2006

3 VERTEX ALGEBRAS AND STRONGLY HOMOTOPY LIE ALGEBRAS ABSTRACT OF DISSERTATION A dissertation submitted in partial fulfillment of the requirements for the degree of Doctor of Philosophy in the College of Arts and Sciences at the University of Kentucky By Daniel F. Pinzon Lexington, Kentucky Director: Dr. Vassily Gorbounov, Department of Mathematics Lexington, Kentucky May 3, 2006 Copyright c Daniel F. Pinzon May 3, 2006

4 ABSTRACT OF DISSERTATION VERTEX ALGEBRAS AND STRONGLY HOMOTOPY LIE ALGEBRAS Vertex algebras and strongly homotopy Lie algebras (SHLA) are extensively used in qunatum field theory and string theory. Recently, it was shown that a Courant algebroid can be naturally lifted to a SHLA. The 0-product in the de Rham chiral algebra has an identical formula to the Courant bracket of vector fields and 1-forms. We show that in general, a vertex algebra has an SHLA structure and that the de Rham chiral algebra has a non-zero l 4 homotopy. Keywords: vertex algebra, vertex operator algebra, strongly homotopy lie algebra, courant, dorfman, chiral de Rham Daniel F. Pinzon May 3, 2006

5 VERTEX ALGEBRAS AND STRONGLY HOMOTOPY LIE ALGEBRAS By Daniel F. Pinzon Dr. Vassily Gorbounov Director of Dissertation Dr. Serge Ochanine Director of Graduate Studies May 3, 2006

6 RULES FOR THE USE OF DISSERTATIONS Unpublished dissertations submitted for the Doctor s degree and deposited in the University of Kentucky Library are as a rule open for inspection, but are to be used only with due regard to the rights of the authors. Bibliographical references may be noted, but quotations or summaries of parts may be published only with the permission of the author, and with the usual scholarly acknowledgments. Extensive copying or publication of the dissertation in whole or in part requires also the consent of the Dean of the Graduate School of the University of Kentucky.

7 DISSERTATION Daniel F. Pinzon The Graduate School University of Kentucky May 3, 2006

8 VERTEX ALGEBRAS AND STRONGLY HOMOTOPY LIE ALGEBRAS DISSERTATION A dissertation submitted in partial fulfillment of the requirements of the degree of Doctor of Philosophy at the University of Kentucky By Daniel F. Pinzon Lexington, Kentucky Director: Dr. Vassily Gorbounov, Department of Mathematics Lexington, Kentucky May 3, 2006 Copyright c Daniel F. Pinzon May 3, 2006

9 ACKNOWLEDGMENTS The author would like to thank his advisor, Vassily Gorbounov. His patience is very much appreciated and made this work possible. He would also like to thank the members of his committee Dr. Serge Ochanine, Dr. Uwe Nagel, Dr. Al Shapere and Dr. Cidambi Srinivasan. Also, the author s family and friends must be thanked. Much support came from his family, his parents Humberto and Elvira Pinzon, his brother Steven and most of all his wife Katherine, to whom he dedicates this work. Without her, this year would have been unbearable. Thanks to all those around him who gave him the necessary breaks he needed, especially Scott G. and Pippet, Scott M. and Traci, Josh, Pam, Bryson, Summer and James. iv

10 TABLE OF CONTENTS Acknowledgments iv 1 Introduction 1 2 Background Vertex algebras and the Courant bracket Strongly Homotopy Lie Algebras and Courant Algebroids SHLA and Vertex Algebras Lie Algebra Structure on a Vertex Algebra Lifting of a Vertex Algebra to a SHLA References 25 Vita 26 v

11 1 Introduction Vertex algebras were introduced about 20 years ago by Borcherds. They provide a rigorous definition of the chiral part of 2-dimensional conformal field theory, intensively studied by physicists. Since then, they have had important applications to string theory, conformal field theory, and mathematics, by providing tools to study the most interesting representations of infinite dimensional Lie algebras. More recently, a mathematical approximation to the algebraic structure of quantum field theory was given by [MSV] called the Chiral de Rham complex. In particular, it preserves features of the quantum field theory such as the elliptic genus. The Chiral de Rham complex is a sheaf Ω ch X of vertex algebras with the Zarisky topology on a manifold X, i.e. for each open U X then Ω ch X assigns it a vertex algebra Ω ch X (U) and the restriction maps are morphisms of vertex algebras. It comes equipped with a chiral de Rham differential Q 0 = d ch DR and grading by fermionic charge given by J 0. This gives us a cohomology. If X is a Calabi-Yau manifold, i.e. a complex Riemannian manifold with compatible symplectic structure and with vanishing first Chern class, then the sheaf Ω ch X has a structure of a topological vertex algebra. Much of point particle physics can be described in terms of Lie algebras and their representations. Closed string field theory, on the other hand, leads to a generalization of Lie algebra which arose naturally within mathematics in the study of deformations of algebraic structures. For a topological space X, the homotopy groups π (ΩX) form a graded Lie algebra which can be extended non-trivially (though non-canonically) to a strongly homotopy Lie algebra which reflects more accurately the homotopy type of X. 1

12 Guided by the questions given in [LZ] as to what type of structure and higher homotopies does this have. As a first approximation to this we note that L 1 = [Q 0, G] and we see that L 1 is equivalent to the derivation of the vertex algebra. It is known that V/ V is a Lie algebra with bracket the skew-symmetrization of the 0-product and following [BFLS] and [RW] we examine the resolution of V/ V and inductively produce the SHLA structure. In the background chapter, we give the definition of a vertex algebra and a topological vertex algebra. We introduce the Boson-Fermion topological vertex algebra example where functions will be represented by elements in the vertex algebra of conformal weight 0 and vector fields and 1-forms will be represented as elements of conformal weight 1. In this way we show how the skew symmetrization of the 0-product is a representation of the Courant bracket on vector fields and 1-forms. The next section introduces strongly homotopy Lie algebras and following [RW] how a Courant algebroid can be naturally lifted to a SHLA. The next chapter shows how the 0-product and its skew symmetrization are Lie brackets up to derivation of elements. This makes it a good candidate as the l 2 homotopy on the SHLA. The main result of this work is to show that a vertex algebra has a SHLA structure with the skew-symmetrization of the 0-product as the l 2 homotopy. In the example of the de Rham chiral algebra, the bracket induces a different SHLA structure than that of the Courant algebroid. In particular, the l 4 term is nonzero. 2

13 2 Background 2.1 Vertex algebras and the Courant bracket The definition that follows of a vertex algebra comes from [GIII]. The axioms are applied to the states which makes the definition more algebraic at the expense of the more physically intuitive axioms applied to the fields as given in [Kac], but the two are equivalent. Definition A Z 0 -graded Vertex Algebra is a Z 0 -graded k-module V = V i i 0 equipped with the following data: (parity) p : V Z/2 that splits V = V even V odd (vacuum vector) distinguished even vector 1 V 0 : V V such that V i V i+1 a family of k-bilinear operations (n) : V V V ; (x, y) x (n) y (n Z) such that V i(n) V j V i+j n 1 These should satisfy the following axioms (x, y, z V ; n, m Z): (i) 1 n x = δ n, 1 x; x 1 1 = x; x n 1 = 0 if n 0 (ii) 1 = 0; ( x) n y = nx n 1 y (iii) x n y = ( 1) n+1 ( 1) i (i) (y n+i x) where (i) = i /i! (iv) (a m b) n c = i=0 ( m ( 1) i i i=0 ) (a m i (b n+i c) ( 1) m b m+n i (a i c)) 3

14 Axioms (iii) and (iv) show how an operation in a vertex algebra fails to be commutative or associative. Therefore, we must always keep track of parentheses in our operations. As a matter of convention, we omit the parentheses when they are nested in the rightmost possible position such as below a m (b n (c p 1)) = a m b n c p 1 We notice that we can view a n for some n and some element a V as an endomorphism on V. We will use the (0)-product and extensively in this paper and so we note the following useful property. Property The endomorphisms a 0 for any a V and act as derivations on a vertex algebra. Proof. By axiom (iv) we see that ( ) 0 (a 0 b) n c = ( 1) (a0 i(b i n+ic) ( 1) 0 b 0+n i(a i c) ) i i=0 = a 0 (b n c) b n (a 0 c) so the (0)-product is a derivation on any multiplication n Z Also, since (a m b) V then by axiom (i) and (ii), (a m b) = ( (a m b)) 1 1 = (a m b) 2 1. Now we apply (iv) and note that all positive products on 1 are zero. ( ) m (a m b) 2 1 = ( 1) i (a m i (b 2+i 1) ( 1) m b m 2 i (a i 1)) i i=0 = a m b 2 1 ma m 1 b 1 1 = a m b + ( a) m b The following two examples will be used to define a topological vertex algebra, all of which are defined in [MSV]. 4

15 Example Boson Vertex Algebra Fix an integer N > 0. Let H N be the Lie algebra which as a vector space has the base a i n, b i n i = 1,.., N; n Z and C, all these elements being even, with brackets [a i m, b j n] = δ ij δ m+1, n C The Boson vertex algebra B N as a vector space is generated by the words in the symbols a n, b n applied to the vacuum subject to the commutator relations above and the following property a i n1 = b i n1 = 0 if n 0; C1 = 1 where b j 11 V 0 and a i 11 V 1 Example Fermion Vertex Algebra Let CL N be the Lie algebra which as a vector space has the base φ i n, ψn i i = 1,.., N; n Z and C, all these elements being odd with the (anti-commutating) bracket [φ i m, ψn] j = δ ij δ m+1, n C where φ i 11 V 0 and ψ j 11 V 1 We form the Fermion vertex algebra F N in the same way as the Boson vertex algebra. Definition A Topological Vertex Algebra of rank N is a vertex algebra equipped with: (i) an even element L V 2 called the Virosoro element (ii) an even element J V 1 5

16 (iii) Q, G V odd where Q V 1, G V 2 L must have the following properties: [L n, L m ] = (n m)l n+m 1 [L 0, a] = a; a V If a V n, L 1 a = na The following products must hold J 0 J = 0; J 1 J = d; J k J = 0 for k 2 G k G = Q k Q = 0 for k 0 L 0 J = J 2 1; L 1 J = J 1 1; L 2 J = N; L k J = 0 for k 3 L 0 G = G 2 1; L 1 G = 2G 1 1; L k G = 0 for k 2 L 0 Q = Q 2 1; L 1 Q = Q 1 1; L k Q = 0 for k 2 J 0 G = G 1 1; J 1 G = G 1 1; J k G = 0 for k 2 Q 0 G = L 1 1; Q 1 G = J 1 1; Q 2 G = N; Q k G = 0 for k 2 Example de Rham chiral Algebra of rank N We combine the two examples of vertex algebras given above and define the elements L, J, Q, G by L = N b i 2a i + φ i 2ψ i J = N φ i 1ψ i ; Q = N φ i 1a i ; G = N b i 2ψ i 6

17 Polynomials In this section we create polynomials in terms of b i in our de Rham chiral algebra. We use the (-1)-product as multiplication and note that since b i V 0 then the functions we form are of conformal weight zero. We explore how the elements a i affect our functions, we form the equivalent of vector fields and 1-forms and find that the (0)-product represents the Dorfman bracket on vector fields and 1-forms. Proposition Let f = f(b 1, b 2,..., b k ) = f(b) be a polynomial in the topological vertex algebra given in example above. Then a i 0 acting on f behaves like a partial derivative with respect to b i and the vertex algebra differential acting on f behaves like the derham differential df: (i) a i 0f = f b i = f 0a i (ii) ( f) 1 1 = i ( ) f ( b i b 11) i 1 Proof. Let (b (i) ) n b i 1b i 1...b i 11. Then a typical term in f(b) ignoring the coefficient is (b (1) ) n 1 1(b (2) ) n (b (k) ) n k 11. The proof of (i) uses the commutating relations on the endomorphisms a i m, b j n given in So we see that the endomorphism a i 0 commutes with all b j 1 for i j. So, we only need to calculate the following a i 0((b (i) ) n ) = 1 1 (b (i) ) n 1 + b i 1a i 0(b (i) ) n 1 Proceeding inductively, we see that we end up with n terms of (b (i) ) n 1. The second equality on (i) is shown using the commutative property (iii) of a vertex algebra. 7

18 To prove (ii), we recall that ( x) n y = nx n 1 y and that acts as a derivation on all multiplications of V. So ((b (i) ) n ) = ( b i ) 1 b i 1...b i 11 + b i 1( b i ) 1...b i = b i 2b i 1...b i 11 + b i 1b i 2...b i and since all the endomorphisms commute, we can move b 2 to the end. = n((b (i) ) n 1 ) 1 b i 21 = n((b (i) ) n 1 ) 1 ( b i 11) In general, we have ((b (1) ) n 1 1(b (2) ) n (b (k) ) n k 11) = = ( (b (1) ) n 1 ) 1 (b (2) ) n (b (k) ) n k 11 + (b (1) ) n 1 1( (b (2) ) n 2 ) 1...(b (k) ) n k = i b i ((b(1) ) n 1 1(b (2) ) n (b (k) ) n k ) 1 ( b i 1)1 Theorem The Dorfman bracket of vector fields and 1-forms has an identical formula as the 0-product of factors of the form f(b)a 1 1+α(b)b 2 1 on the Boson-Fermion topological vertex algebra. Dorfman Bracket of Vector fields and 1-forms [f + α, g + β] D = [f, g] + i f (dβ) + d(i f β) i g (dα) where f, g are vector fields and α, β are 1-forms. N N N f = f i i, g = g i i α = α i db i, β = N β i db i (i) [f, g] = i,j f i i g j j i,j g j j f i i (ii) i f (dβ) + d(i f β) = i,j f i i β j db j + i β i df i (iii) i g dα = i,j g i i α j db j + i,k g i k α i db k 8

19 (0)-product on the de Rham Chiral Algebra ( f i a i + ) ( α i b i 21 g j a j + ) β j b j 21 i i j j 0 (i) i,j f i 1 i g j a j 1 g j 1 j f i a i 1 (ii) i,j f i 1 i β j 1b j 21 + i,k β i 1 k f i 1b k 21 (iii) i,j g j 1 j α i 1b i 21 + i g i 1 k α i 1b k 21 Proof. [f + α, g + β] D = [f, g] + i f (dβ) + d(i f β) i g (dα) so we note: (i) [f, g] = f i i g j j g j j f i i i,j i,j ( ) (ii) i f (dβ) = f i i dβ j db j i j ( ) = f i i ( k β j db k )db j i j k = i,j,kf i k β j (δ i,k db j δ i,j db k ) = i,j f i i β j db j i,k f i k β i db k ( ) (iii) d(i f β) = d f i i i j β j db j = d( i f i β i ) = i df i β i + i f i dβ i (iv) i g dα = i,j g i i α j db j + i,k g i k α i db k We want to compare this with the vertex algebra calculation: ( f i a i + ) α i b i 21 g j a j + β j b j 2 1 i i j j 0 9

20 We note that this algebra is neither commutative or associative. So we recall that the 0-product is a derivation and the associative rule is: ( ) m (a m b) n c = ( 1) k (a m k (b n+k c) ( 1) m b m+n k a k c) k k=0 Let s look at the terms: (i) ( ) f i a i i 0 ( ) g j a j = f 1 ka i i kg 1a j j + a i 1 kfkg i 1a j j j i,j,k = i,j f i 1a i 0g j 1a j + a i 1f i 0g j 1a j = i,j f i 1a j 1 i g j + a i 1g j 1( j f i ) (ii) ( ) f i a i i 0 j ( ) β j b j 21 = f 1 ka i i kβ 1b j j 21 + a i 1 kfkβ i 1b j j 21 = i,j,k i,j f i 1a i 0β j 1b j 21 + f i 2a i 1β j 1b j 21 = i,j f i 1 i β j 1b j 21 + i β i 1f i 21 (iii) ( ) α i b i 21 i 0 j ( ) g j a j = b i 2 kαkg i j a j α 2 kb i i kg j a j i,j,k = b i 2α0g i j a j α 2b i i 0g j a j i,j,k i,j g j j α i b i 21 + i g i k α i 1b k 21 (iv) ( ) α i b i 21 i 0 j ( ) β j b j 21 = 0 Corollary The skew-symmetrization of the Dorfman bracket of vector fields and 1- forms, called the Courant bracket, has an identical formula as the skew-symmetrization of the 0-product of factors of the form f(b)a α(b)b 2 1 in the B-F topological vertex algebra. 10

21 2.2 Strongly Homotopy Lie Algebras and Courant Algebroids In this section we review what was done in [RW]. We first would like to define the Koszul sign ɛ of a permutation σ on graded indeterminates x 1,..., x n by x 1... x n = ɛ(σ; x 1,..., x n ) x σ(1)... x σ(n) that is, following the convention of supermathematics, that a minus sign is introduced when two odd elements are permuted. For example, x y = ( 1) x y y x. We say that σ S n is an (j,n-j)-unshuffle, 0 j n, if σ(1) < < σ(j) and σ(j + 1) < < σ(n) Definition An L(m)-structure on a graded vector space L is a system of linear maps l k with deg(l k ) = k 2 which are antisymmetric in the sense that l k (x 1,..., x n ) = ( 1) σ ɛ(σ; x 1,..., x n )l k (x σ(1),..., x σ(n) ) (1) for all σ S n and x 1,..., x n L i.e. when two elements are permuted, a minus sign is introduced when at least one them is even. Moreover, the following generalized form of the Jacobi identity is satisfied for any n m: i+j=n+1( 1) i(j 1) σ ( 1) σ ɛ(σ)( 1) i(j 1) l j (l i (x σ(1),..., x σ(i) ), x σ(i+1),..., x σ(n) ) = 0 (2) where the summation is taken over all (i,n-i)-unshuffles with i 1 11

22 A short-hand notation that we will use for (2) is the following: i+j=n+1 ( 1) i(j 1) l j l i Definition We have the following definition for the degree of an n-tuple (x 1,..., x n ) n X where X is a graded vector space. degree(x 1,..., x n ) i x i Example An L(1)-algebra structure on L consists of a degree -1 endomorphism l 1 and the Jacobi identity reduces to l 2 1 = 0, so an L(1)-algebra is a differential space. Example An L(2)-algebra has an L(1)-structure plus a bilinear map that satisfies the antisymmetry condition (1) and the condition l 1 (l 2 (x, y)) = l 2 (l 1 (x), y) + ( 1) x l 2 (x, l 1 (y)) So, an L(2)-algebra is an antisymmetric, non-associative, non-unital differential graded algebra. Example For an L(3) algebra, l 3 is the contracting homotopy for the classical Jacobi identity: ( 1) x 1 x 3 l 2 (l 2 (x 1, x 2 ), x 3 ) + ( 1) x 2 x 3 l 2 (l 2 (x 3, x 1 ), x 2 ) + ( 1) x 1 x 2 l 2 (l 2 (x 2, x 3 ), x 1 ) = ( 1) { x 1 x 3 +1 l 1 (l 3 (x 1, x 2, x 3 )) + l 3 (l 1 (x 1 ), x 2, x 3 ) + ( 1) x1 l 3 (x 1, l 1 (x 2 ), x 3 )+ ( 1) x 1 + x 2 l 3 (x 1, x 2, l 2 (x 3 )) } 12

23 Definition An algebra with an L( )-structure is called a strongly homotopy Lie algebra. It is sometimes called an sh-lie algebra and by the acronym SHLA. As an example of an SHLA, we will introduce Courant algebroids and show how they can be naturally lifted to an SHLA. This result was shown in [Royt]. Definition Given a bilinear, skew-symmetric operation [, ] on a vector space V, its Jacobiator J is the trilinear operator on V: J(x, y, z) = [[x, y], z] + [[y, z], z] + [[z, x], y] for x, y, z V Definition A Courant algebroid is a vector bundle E M equipped with a nondegenerate symmetric bilinear form, on the bundle, a skew symmetric bracket [, ] on Γ(E), and a bundle map ρ : E T M such that the following properties are satisfied: (i) For any e 1, e 2, e 3 Γ(E), J(e 1, e 2, e 3 ) = DT (e 1, e 2, e 3 ) (ii) For any e 1, e 2, e 3 Γ(E), ρ[e 1, e 2 ] = [ρe 1, ρe 2 ] (iii) For any e 1, e 2, e 3 Γ(E), and f C (M), [e 1, fe 2 ] = f[e 1, e 2 ] + (ρ(e 1 )f)e 2 e 1, e 2 Df (iv) ρ D = 0, that is for any f, g C (M), Df, Dg = 0 (v) For any e, h 1, h 2 Γ(E), ρ(e) h 1, h 2 = [e, h 1 ] + D e, h 1, h 2 + h 1, [e, h 2 ] + D e, h 2 13

24 where T (e 1, e 2, e 3 ) = 1 3 [e 1, e 2 ], e 3 + cyclic permutations and D : C (M) Γ(E) is the map defined by D = 1 2 β 1 ρ d 0, where β is the isomorphism between E and E given by the bilinear form and d 0 is the derham differential. Theorem A Courant algebroid structure on a vector bundle E M gives rise naturally to a SHLA structure on the total space X given by the resolution of the coker D with l 1 = d, the derivation of the complex below,... 0 kerd i C (M) D Γ(E) cokerd 0 and the higher structure maps given by the following formulas where e X 0 = Γ(E); f, g X 1 = C (M) and c X 2 = kerd l 2 (e 1, e 2 ) = [e 1, e 2 ] in degree 0 l 2 (e, f) = e, Df in degree 1 l 2 = 0 in degree 2 l 3 (e 1, e 2, e 3 ) = T (e 1, e 2, e 3 ) in degree 0 l 3 = 0 in degree 1 l n = 0 for n 4 Proof. (Sketch) We will sketch the proof below. We see that the Courant algebroid is a good candidate to be extended to a SHLA from the first axiom, which states that the Courant bracket satisfies the Jacobi identity up to a D-exact term. Once we have a resolution of a Lie algebra then we see that we have a differential space, i.e. an L(1)-algebra. So we define l 1 to be the differential. We now need to define l 2 on elements of degree 0 (that is elements from 14

25 X 0 = Γ(E)). The natural choice is the Courant bracket. We notice that SHLA identity (2) is trivial since l 1 (X 0 ) = 0 by construction. Once we have these defined, we can define l 2 on higher degree by induction on the degree as well as higher homotopies l n by induction on n using the SHLA identity (2). The skew-symmetric identity is satisfied automatically using this inductive construction. [Let n=2, degree =1] As stated above, identity (2) for n=2 states: l 1 (l 2 (e, f)) = l 2 (l 1 (e), f) + ( 1) e l 2 (e, l 1 (f)) We observe that the first term is zero since l 1 (e) = 0 and l 2 has degree 0 so l 1 = D on an element of degree 1, so we have D(l 2 (e, f)) = [e, Df] = D e, Df where we used the lemma above for the last equality. So we set l 2 (e, f) = e, Df Similarly, l 2 can be found for higher degrees. [Let n=3, degree = 0] We notice that the LHS of the SHLA identity (2) reduces to the Jacobi identity on elements of degree 0 and from the first axiom of a Courant algebroid we see that (2) reduces to J(e 1, e 2, e 3 ) = DT (e 1, e 2, e 3 ) = l 1 (l 3 (e 1, e 2, e 3 )) = Dl 3 (e 1, e 2, e 3 ) so we define l 3 = T. 15

26 Again, l 3 can be defined on higher degree by induction on the degree. By a dimensional argument, we note that l 4 need not be zero on elements of degree 0. The second axiom gives l 1 l 4 = il 4 = l + 4 = l 2 l 3 l 3 l 2. However, due to the structure of a Courant algebroid, l 2 l 3 l 3 l 2 = 0 For higher homotopies l n for n 5 on elements of degree 0 we note that l n ( n X 0 ) X n 2 = 0 by construction. 16

27 3 SHLA and Vertex Algebras 3.1 Lie Algebra Structure on a Vertex Algebra The first step to finding a natural SHLA structure on V is to find a Lie algebra so that we can take its free resolution. The following will explore the 0-product and examine the well known result of how it fails to be a Lie bracket on V. Proposition Let V be a vertex algebra. The 0-product and it s skew-symmetrization are both brackets that form a Lie algebra structure on V/ V Proof. We recall the commutative axiom for multiplication by (0) x 0 y = y 0 x + ( 1) n+1 ( 1) i (i) (y n+i x) and note that it fails skew symmetry by the derivation of an element in V. Before we look at the Jacobi identity let s note that (x 0 (y n z)) = x 0 (y n z) + x 0 (y n z) = x 0 (y n z) Let s now look at the Jacobi identity: [x, [y, z]] + [y, [z, x]] + [z, [x, y]] = x 0 (y 0 z) + y 0 (z 0 x) + z 0 (x 0 y) We see that from the associative axiom, the 0-product is a derivation. so x 0 (y 0 z) = (x 0 y) 0 z + y 0 (x 0 z) ( ) ( ) = z 0 (x 0 y) y 0 (z 0 x) ( 1) i (i) (z i (x 0 y)) y 0 ( 1) i (i) (z i x) ( ) = z 0 (x 0 y) y 0 (z 0 x) ( 1) i (i) (z i (x 0 y) + y 0 (z i x)) since (x 0 y) = ( x) 0 y +x 0 ( y) = x 0 ( y). Again, we note that the summation term is zero in V/ V and adding this to the other two terms in the Jacobi identity shows that it is satisfied. 17

28 Now we would like to look at the bracket defined by skew symmetrization of (0) : [x, y] = 1 (x 2 0y y 0 x) = 1(x 2 0y + x 0 y+summation term ) This clearly satisfies the Jacobi identity for the same reasons as above. 3.2 Lifting of a Vertex Algebra to a SHLA The following theorem is the new result of this work and closely follows [RW]. We define a complex and a bracket and the higher homotopies are found by induction. Theorem Let V = V i ; X 0 = X 1 = V ; X 2 = ker. A vertex algebra V gives rise i 0 naturally to a SHLA structure on the total space X of the resolution of v/ V below with l 1 as the differential. 0 ker i V V 0 where is from the definition of the vertex algebra and i is the inclusion operator. We note that just as in the Courant algebroid construction, l k = 0 for k 5. In the following we want to define what the l k are in our complex. This is done inductively on k and the degree of the elements using the SHLA identities as in [S]. Step k=1 Note that the first SHLA identity is l 2 1 = 0 and is clearly satisfied. 18

29 Step k=2 The second identity has two parts 1. l 2 (x, y) = ( 1) x y l 2 (y, x) 2. l 1 (l 2 (x, y)) = l 2 (l 1 (x), y) + ( 1) x l 2 (x, l 1 (y)) 0 where x = n if x V n We can define l 2 on elements of X 0 by l 2 = 1(x 2 0y y 0 x) We see that 1. is satisfied by definition. For 2., we see that the RHS is zero since l 1 (X 0 ) = 0. From this definition we can define l 2 for X i for higher degree by induction. We will use the fact that ( x) 0 y = 0 [degree 1] (i)let x X 1, y X 0 then l 2 (l 1 (x), y) + ( 1) x l 2 (x, l 1 (y)) = 1(( x) 2 0y y 0 x) = 1 (y 2 0x) So, we can define l 2 (x, y) = 1(y 2 0x) (ii) Let x X 0, y X 1 then in a likewise manner, we can define l 2 (x, y) = 1(x 2 0y) [degree 2] (i) Let x, y X 1 then l 2 (l 1 (x), y) + ( 1) x l 2 (x, l 1 (y)) = 1(( x) 2 0y + ( y) 0 x) = 0 (ii) Let x X 2, y X 0 then l 2 (l 1 (x), y) + ( 1) x l 2 (x, l 1 (y)) = l 2 (l 1 (x), y) = 1y 2 0x where we note that l 1 on an element of degree 2 is defined by the inclusion. Since l 1 l 2 = l 2 and so we can define it. The other case is similarly found. Since there are no elements of degree 3, we are done. The above results are summarized below. 19

30 l 2 (x, y) = 1(x 2 0y y 0 x) in degree 0 l 2 (x, y) = 1( y x 2 0y x y 0 x) in degree 1 l 2 (x, y) = 1((1 x ) y x 4 0y (1 y ) x y 0 x) in degree 2 l 2 (x, y) = 0 otherwise where x = i if x X i Step k=3 Now let s look at the third SHLA identity: ( 1) x z [[x, y], z] + ( 1) y z [[z, x], y] + ( 1) x y [[y, z], x] = = ( 1) { x z +1 l 1 (l 3 (x, y, z)) + l 3 (l 1 (x), y, z) + ( 1) x l 3 (x, l 1 (y), z)+ ( 1) x + y l 3 (x, y, l 2 (z)) } Again we do this by induction on the degree of the entries. First we ll create some notation Let K x,y = 1 2 (x 0y + y 0 x) = K y,x We note that 1 (x 2 0y y 0 x) = x 0 y + K x,y = x 0 y + ( 1) i (i) (x i y) 2 We observe that [[x, y], z] = 1 4 ((x 0y y 0 x) 0 z z 0 (x 0 y y 0 x)) So, the Jacobiator J(x, y, z) = [[x, y], z] + [[x, y], z] + [[x, y], z] = 1 4 ( (x 0y) 0 z + (y 0 z) 0 x + (z 0 x) 0 y z 0 (x 0 y) x 0 (y 0 z) y 0 (z 0 x) (y 0 x) 0 z (z 0 y) 0 x (x 0 z) 0 y + z 0 (y 0 x) + x 0 (z 0 y) + y 0 (x 0 z)) We notice that most of the terms cancel using the fact that the 0-product is a derivation. 20

31 We are left with only the first three terms. With a little manipulation using the notation defined above we have J(x, y, z) = 1((x 4 0y) 0 z + c.p.) = 1(2K 4 y,x 0 z 2K x,y0 z) [degree 0] Let x, y, z X 0 and note l 1 (X 0 ) = 0 We note that LHS of the second identity is J(x, y, z) so LHS is a cycle and we can easily define l 3 l 3 = 1 ( 1) i (i 1) (x i (y 0 z) y i (x 0 z)) 4 i [degree 1] Let x X 1, y, z X 0 then the LHS is the same as above but RHS = l 1 (l 3 (x, y, z)) l 3 (l 1 (x), y, z). Since l 3 (x, y, z) X 2 then l 1 (l 3 (x, y, z)) = i(l 3 (x, y, z)) = l 3 (x, y, z) and so l 3 is defined. [degree 2] We note then that the degree(l 3 (x, y, z)) 3 so we define l 3 = 0 Step k=4 Let s now look at the fourth SHLA identity: ( 1) i(j 1) l j l i = l 1 l 4 l 2 l 3 + l 3 l 2 l 4 l 1 = 0 i+j=n+1 We see that the last term is zero since l 1 (X 0 ) = 0 and l 1 l 4 = il 4 = l 4. So l 4 is defined and we have: l 4 = l 2 l 3 l 3 l 2 We now want to show that l 4 is not zero, so we look at the following example. Example Non-zero example of l 4 (w, x, y, z) 21

32 Let V be the B-F topological vertex algebra and the vector fields w, x, y, z V 1 X 0 be defined 3 as x = a 1, y = f(b 2 )a 2, z = a 3,w = g i (b)a i. We first note that since the 0-product is a commutator of vector fields, x 0 y = y 0 z = x 0 z = 0 [Calculation of l 2 l 3 ] We see that l 2 l 3 = l 2 (l 3 (x, y, z), w) + l 2 (l 3 (w, y, z), x) + l 2 (l 3 (w, x, z), y) + l 2 (l 3 (w, x, y), z) Since l 3 (x, y, z) X 1 then by the way l 2 is defined above (i) l 2 (l 3 (x, y, z), w) = 1w 2 0l 3 (x, y, z) ( ) = 1w (i 1) 8 0 (y i (x 0 z) x i (y 0 z)) i = 1 (i 1) 8 i w 0 (y i (x 0 z) x i (y 0 z)) = 0 Since (w 0 a) = w 0 a and x 0 z = y 0 z = 0 (ii) l 2 (l 3 (w, y, z), x) = 1 8 x 0(y 1 (w 0 z)) Since the the first summation term is an element of V 1 i and the second summation term is zero. The next two follow the same reasoning except that we note that x 1 (w 0 z) = a 1 1(w 0 z) 1 1 = (w 0 z) 1 a 1 11 = 0. (iii) l 2 (l 3 (w, x, z), y) = 1 8 y 0(x 1 (w 0 z)) = 0 (iv) l 2 (l 3 (w, x, y), z) = 1 8 z 0(x 1 (w 0 y)) = 0 [Calculation of l 3 l 2 ] We calculate that l 3 l 2 = l 3 (l 2 (w, x), y, z) l 3 (l 2 (w, y), x, z) + l 3 (l 2 (w, z), x, y) +l 3 (l 2 (x, y), w, z) l 3 (l 2 (x, z), w, y) + l 3 (l 2 (y, z), w, x) 22

33 (i) l 3 (l 2 (w, x), y, z) = ( 1) i 4 = ( 1) i 4 (i 1) i (i 1) i (y i ([w, x] 0 z)) since y 0 z = 0 (y i ([w, x] 0 z) [w, x] i (y 0 z)) = 1 8 y 1((w 0 x x 0 w) 0 z) since the summation term is an element of V 1 i (ii) l 3 (l 2 (w, y), x, z) = 1 8 x 1((w 0 y y 0 w) 0 z) = 1 8 a1 1((w 0 y y 0 w) 0 z) = 1 8 ((w 0y y 0 w) 0 z) 1 a 1 11 = 0 Since a 1 1 commutes with all factors by the Boson commutation relation. The same argument applies for the following. (iii) l 3 (l 2 (w, z), x, y) = 1 8 (x 1((w 0 z z 0 w) 0 y)) = 0 (iv) l 3 (l 2 (x, y), w, z) = 0 (v) l 3 (l 2 (x, z), w, y) = 0 (vi) l 3 (l 2 (y, z), w, x) = 0 Combining the above we have: l 4 = 1 8 ( x 0(y 1 (w 0 z)) y 1 ((w 0 x x 0 w) 0 z)) We now use the fact that the (0)-product is a derivation to rewrite the first term. x 0 (y 1 (w 0 z)) = (x 0 y) 1 (w 0 z) y 1 (x 0 (w 0 z)) = y 1 ((x 0 w) 0 z) So we have noting that the 0-product is a commutator of vector fields: l 4 (w, x, y, z) = 1y 8 1(x 0 (w 0 z)) ( = 1 f(b 2 ) ) ( ( ) (( 3 ) g i (b) 8 b 2 1 b 1 0 b i ( = f(b 2 ) ) ( 3 ) 2 g i b 2 b 3 b 1 b i ( ) )) b 3

34 We now need to find what the 1-product does to vector fields. We will first calculate it in general and then apply it to our case recalling the associative relation of a vertex algebra and that b i = ai in the B-F topological vertex algebra. (f(b) i 1a i ) 1 (g(b) j 1a j ) i,j = i,j=1;k=0 ( ) 1 (f ( 1) k i k 1 k(a i 1+k(g j a j )) + a i k(fk(g i j a j )) ) We see that the first summation term is always zero because a i 1+k will commute with (gj a j ) and annihilate the vacuum vector since it has no (-(2+k))-products. Also, the second term is zero except when k = 0 since f i k will commute and annihilate the vacuum because there are no (-(k+1))-products for k 1. We now recall that a i 0f = b i f = f 0a i = i,j a i 0(f i 0(g j a j )) = a i 0(g j 1(f i 0a j )) = ( ( a i 0 g j 1 )) b f i j i,j = g j f i b i b + 2 f i j gj b j b i i,j Applying that result to our case we have l 4 = ( 3 ) ( 3 ) 2 g i f(b 2 ) 2 g i 2 f(b 2 ) + b 2 b 3 b 1 b j b 3 b 1 b j b 2 j 3 ( ) ( ) ( ) ( ) 3 g i f(b 2 ) 2 g i 2 f(b 2 ) = + b 1 b 2 b 3 b 2 b 3 b 1 b 2 b 2 which is not zero in general. 24

35 REFERENCES [BFLS] G. Barnich, R. Fulp, T. Lada, and J. Stasheff, The sh Lie structure of Poisson brackets in field theory, hep- th/ , [F] Frenkel, E. and Ben-Zvi, D., Vertex Algebras and Algebraic Curves, Mathematical Surveys and Monographs, Vol. 88, Amer. Math. Soc., Providence, [GII] Gorbounov, V., Malikov, F. and Schechtman, V., Gerbes of chiral differential operators, II, Vertex Algebroids, Invent. Math. 155 (2004), no. 3, [GIII] Gorbounov,V.,Malikov, F., Schechtman,V.: Gerbes of chiral differential operators. III. In: The orbit method in Geometry and Physics. In honor of A.A. Kirillov. Progress in Mathematics 213, Boston, Basel, Berlin: Birkhuser 2003 [Kac] Kac, V., Vertex Algebras for Beginners, University Lecture Series 10, Amer. Math. Soc., [LZ] B. Lian, G. Zuckerman, New perspectives on the BRST-algebraic structure of string theory, Commun. Math. Phys. 154 (1993), ; hep-th/ [MSV] Malikov, F., Schechtman, V., Vaintrob, A.: Chiral de Rham complex. Commun. Math. Phys. 204, (1999) [RW] Roytenberg, D. and Weinstein, A. Courant Algebroids and Strongly Homotopy Lie Algebras, Lett. Math. Phys. 46 (1998),

36 VITA (i) Background (a) Date of Birth: December 29, 1976 (b) Place of Birth: Vineland, NJ (ii) Academic Degrees (a) B.A., Physics(Mathematics Concentration), Cornell University, (b) M.A., Mathematics, University of Kentucky, 2004 (iii) Professional Experience (a) Teaching Assistant, Mathematics Department, University of Kentucky, Fall Spring (b) Primary Instructor, University of Kentucky, Fall 2003-Spring (iv) Publications (a) The shortest enclosure of two connected regions in a corner, Rocky Mountain J. Math., 31 (2001), no. 2,