AN INTRODUCTION TO VERTEX ALGEBRAS
DANIELE VALERI
Abstract. These are my notes from the course "An introduction to vertex algebras" given by prof. Victor Kac at the University of Rome "La Sapienza" during the months of december 2008 and january 2009. The exposition essentially follows the main reference [Kac96].
Contents 0. A nice picture 1 1. Wightman's axioms of QFT1 2. Calculus of formal distributions3 3. Some equivalent de nitions of a Vertex Algebra 10 4. An example: the free boson 22 5. Poisson Vertex Algebras 24 6. Representation theory 26 7. On W-algebras 30 References 30
0. A nice picture There are 4 basic frameworks of physical theories quantization ( ClFTquasiclassicalo QFT G W limit chiralization chiralization quasiclassical limit ClM o QM 6 quantization (Cl=classical, Q=quantum, M= mechanics, FT= eld theory) and the corresponding algebraic structures are
PVA o 'VA G W
PA o AssA 5 (P=Poisson, V=Vertex, Ass=Associative, A=algebras).
1. Wightman's axioms of QFT
Let M be the Minkovski space time, that is M = Rd endowed with the norm 2 2 2 2 kxk = x0 − x1 − ... − xd−1. A Poincaré group is a group of symmetries of M, including translations, that preserve the norm. A quantum eld theory (QFT) is the following data 1 · space of states, a Hilbert space H; · vacuum vector, a vector |0i ∈ H; · unitary representation π of P in H, P 3 g −→π π(g) unitary operator in H; α ·F = {Φ }α∈I collection of quantum elds. De nition 1.1. A quantum eld is a distribution on M with values in operators on H. This means that a quantum eld is a continuous linear functional on the space of test functions on M with values in the space of linear operators on H. One requires that these data satisfy the following Wightman's axioms: (vacuum) π(g)|0i = |0i for all g ∈ P ; (covariance) π(g)Φα(ϕ)π(g)−1 = Φα(π(g)(ϕ)); α1 α2 αs (completeness) Φ (ϕ1)Φ (ϕ2) ... Φ (ϕs) span a dense set in H; (locality) Φα(ϕ)Φβ(ψ) = Φβ(ψ)Φα(ϕ) if supp ϕ and supp ψ are separated by the light cone. We have a special case when d = 2. In this case the light cone is given by the equations x − x = 0 0 1 . x0 + x1 = 0 We can consider left chiral elds (respectively right) that are those elds for which α ∂x0+x1 Φ = 0 (respectively α ). Taking all left (or right) chiral elds of a -dimensional QFT produces a ∂x0−x1 Φ = 0 2 left (or right) vertex (= chiral) algebra. The data of this algebra is given by the quadruple (V, |0i,T, F), where · the vector space V is the space of states; ·| 0i ∈ V is the vacuum vector; · T is the (only one) operator on V ; α ·F = {Φ }α∈I is the collection of quantum elds. In this case we can give a more formal de nition of a quantum eld. De nition 1.2. A quantum eld is a series of the form
X n Φ(z) = Φnz , n∈Z where Φn ∈ End V , such that Φnv = 0 for n << 0 and for any v ∈ V . By de nition we get that Φ(z)v ∈ V ((z)) for any v ∈ V . This algebra satis es the following axioms (vacuum) T |0i = 0; (covariance) d ; [T, Φ(z)] = dz Φ(z) (completeness) α1 α2 αs span ; Φn1Φn2 ... Φns|0i V (locality) (z − w)N Φα(z)Φβ(w) = (z − w)N Φβ(w)Φα(z) for N >> 0. Exercise 1. Prove equivalence with Wightman's axioms.
Solution. In this case P is a one-dimensional abelian Lie group (the group of translations in the x0 direction). The representation of the corresponding one-dimensional abelian Lie algebra is given by a vector space V and T ∈ End V . Hence, the vacuum axiom becomes etT |0i = |0i, for every t ∈ R. This is equivalent to the fact that T |0i = 0. Moreover, the covariance axiom in this case reads etT Φ(z)e−tT = , for every . If we apply d to both sides and we set we get d . Φ(z + t) t ∈ R dt t = 0 [T, Φ(z)] = dz Φ(z) Equivalence of completeness axioms is clear. For the equivalence of locality axioms see [Kac96]. Example 1.3. Let V be a unital commutative associative algebra, then we de ne a vertex algebra structure by |0i = 1, T = 0 and F = {La}, where La(b) = ab.
Exercise 2. Given a unital algebra V , the property that LbLa = LaLb is equivalent to commutativity and associativity.
Solution. Since V is unital, (LaLb)1 = (LbLa)1 implies commutativity. Moreover, from LaLb = LbLa, we get a(bc) = b(ac), (1.1) for all a, b, c ∈ V . Using commutativity and (1.1), we have a(bc) = a(cb) = c(ab) = (ab)c. On the other hand, let us assume that V is associative and commutative. Then, for a, b ∈ V , we have
La(Lb(c)) = a(bc) = (ab)c = (ba)c = Lb(La(c)) , 2 for all . Hence, . c ∈ V LaLb = LbLa Lemma 1.4. For any vertex algebra and for any of its quantum elds Φα(z) we have that Φα(z)|0i ∈ V [[z]]. Proof. Let us write (is a usual convention that will be clear later)
α X α −n−1 Φ (z) = Φ(n)z . n∈Z By covariance axiom we get α α [T, Φ(n)] = −nΦ(n−1) and we know, by de nition of quantum eld, that α for . We want to prove that Φ(n)v = 0 n >> 0 α for . Take the minimal non-negative such that α and apply . We get Φ(n)|0i = 0 n ∈ Z+ N Φ(N)|0i = 0 T α α α α 0 = T Φ(N)|0i = [T, Φ(N)]|0i + Φ(N)T |0i = −NΦ(N−1)|0i by vacuum axiom. So if , then α , but this contradict our choice of unless . N 6= 0 Φ(N−1)|0i = 0 N N = 0 Due to the above lemma, we have a map
α α Φ (z) −→ Φ (z)|0i|z=0 ∈ V. This map is called eld-state correspondence.
2. Calculus of formal distributions A distribution is a linear function on the space of test functions with values in a vector space U (with some continuity properties). We choose the space of test functions to be C[z, z−1]. Let us x some notation: given a vector space U we denote ·U [z] the space of polynomials with coe cients in U; ·U [z, z−1] the space of Laurent polynomials with coe cients in U; ·U [[z]] the space of power series with coe cients in U; ·U ((z)) the space of Laurent series with coe cients in U; ·U [[z, z−1]] the space of in nite series in both directions with coe cients in U. If U is an algebra, then U[[z, z−1]] is a module over C[z, z−1], while the remaining are all algebras. A U-valued formal distribution in z is a linear function on C[z, z−1] with values in U. Proposition 2.1. The space of U-valued formal distribution is canonically identi ed with U[[z, z−1]], that is it consists of series of the form X n unz , n∈Z where un ∈ U. Before proving the proposition we need the notion of residue: given a U-valued formal distribution P n Φ(z) = unz , its residue is the linear function de ned by
Resz Φ(z) = u−1. The basic property of the residue is that
Resz ∂zΦ(z) = 0. Proof (of Proposition 2.1). Let Φ(z) be a U-valued formal distribution and set n for Φ(n) = Φ(z)(z ), n ∈ Z, P −n−1 −1 then Φ(z) = Φ(n)z . On the other hand, each Φ(z) ∈ U[[z, z ]] de nes a formal distribution by the formula
Φ(z)(ϕ) = Resz Φ(z)ϕ(z). Indeed X −n−1 m X m−n−1 ( Φ(n)z )(z ) = Resz Φ(n)z = Φ(m). n∈Z n∈Z In particular, quantum elds are End V -valued formal distributions. 3 Remark 2.2. The same is true for many variables. For example for two variables a U-valued formal distribution is a series of the type
X −m−1 −n−1 Φ(z, w) = a(m,n)z w . m,n∈Z Example 2.3. An important example of formal distribution is the formal δ-function, de ned by X w n δ(z, w) = z−1 ∈ [[z, z−1, w, w−1]]. z C n∈Z Consider now the algebras w A = [z, z−1, w, w−1][[ ]], z,w C z z A = [z, z−1, w, w−1][[ ]]. w,z C w Clearly −1 −1 . Denote by the algebra of rational functions of the form Az,w, Aw,z ⊂ C[[z, z , w, w ]] R P (z, w) , zmwn(z − w)k where P (z, w) ∈ C[z, w], then we have the following homomorphisms of algebras:
iz,w : R −→ Az,w f −→ expansion in the domain |z| > |w|,
iw,z : R −→ Aw,z f −→ expansion in the domain |w| > |z|. Example 2.4. The expansions of the function (z − w)−1 are of great interest: n 1 expansion of 1 1 −1 X w iz,w = · w = z ∈ Az,w; z − w z 1 − z z n∈Z+ n 1 expansion of 1 1 −1 X z iw,z = − · z = w ∈ Aw,z. z − w w 1 − w w n∈Z+ Thus we can also de ne the δ-function in terms of the expansions of (z − w)−1, namely 1 1 δ(z, w) = i − i . z,w z − w w,z z − w It follows also an important formula for the derivatives of the δ-function: 1 X n 1 1 ∂k δ(z, w) = wn−kz−n−1 = i − i . (2.1) k! w k z,w (z − w)k+1 w,z (z − w)k+1 n∈Z Proposition 2.5. The formal δ-function has the following properties: a) δ(z, w) = δ(w, z); b) ∂zδ(z, w) = −∂wδ(z, w); c) 1 1 ∂n−mδ(z, w) if m ≤ n, (z − w)m ∂n δ(z, w) = (n−m)! w n! w 0 if m > n; d) if a(z) ∈ U[[z, z−1]], then
a(z)δ(z, w) = a(w)δ(z, w) and Resz a(z)δ(z, w) = a(w); e) λ(z−w) n n e ∂wδ(z, w) = (λ + ∂w) δ(z, w). Proof. a), b) and c) follow easily from (2.1). By c) for m = n = 0 we know that zδ(z, w) = wδ(z, w), then znδ(z, w) = wnδ(z, w). By linearity it follows that a(z)δ(z, w) = a(w)δ(z, w). If we now take the residue in this last relation, we get
Resz a(z)δ(z, w) = a(w) Resz δ(z, w) = a(w), proving d). λ(z−w) −λ(z−w) Finally, let us prove e). We note that e ∂we = λ + ∂w. Indeed, take any function f(w), then λ(z−w) −λ(z−w) λ(z−w) −λ(z−w) −λ(z−w) e ∂we f(w) = e (λe f(w) + e ∂wf(w)) = (λ + ∂w)f(w). 4 Then λ(z−w) n −λ(z−w) n e ∂we = (λ + ∂w) . Apply now to δ(z, w) and get λ(z−w) n −λ(z−w) λ(z−w) n n e ∂we δ(z, w) = e ∂wδ(z, w) = (λ + ∂w) δ(z, w), −λ(z−w) −λ(w−w) since by d), we know that e δ(z, w) = e δ(z, w) = δ(z, w). In particular, from the above proof, it follows that
Resz ϕ(z)δ(z, w) = ϕ(w) for any test function ϕ(z) ∈ C[z, z−1]. This analogy with the Dirac δ-function in distribution theory clari es why δ(z, w) is called formal δ-function. De nition 2.6. A formal distribution a(z, w) ∈ U[[z, z−1, w, w−1]] is called local if (z − w)N a(z, w) = 0 for some . N ∈ Z+ By Proposition 2.5(c), it follows that the formal δ-function and its derivatives are local. Moreover the following theorem states that any local distribution is a linear combination of them.
Theorem 2.7 (Decomposition Theorem). A local formal distribution a(z, w) ∈ U[[z, z−1, w, w−1]] has a unique decomposition X ∂j δ(z, w) a(z, w) = cj(w) w , j! j∈Z+ where the sum is nite and cj(w) ∈ U[[w, w−1]]. The coe cients cj(w) are given by the formula j j c (w) = Resz(z − w) a(z, w). (2.2)
∂j δ(z,w) Proof. Let P j w , where j are given by formula (2.2). Then the second b(z, w) = a(z, w)− c (w) j! c (w) sum is nite since (z − w)ja(z, w) = 0 for j >> 0. Note that X ∂j−nδ(z, w) Res (z − w)nb(z, w) = Res (z − w)na(z, w) − Res cj(w) w = z z (j − n)! j≥n X ∂j−n Res δ(z, w) = cn(w) − cj(w) w = cn(w) − cn(w) = 0. (j − n)! j≥n So, n for any . Hence Resz(z − w) b(z, w) = 0 n ∈ Z+ X j b(z, w) = z bj(w), j∈Z+ −1 with bj(w) ∈ U[[w, w ]]. Indeed, since Resz b(z, w) = 0, we get b−1(w) = 0. In the same way, we have
0 = Resz(z − w)b(z, w) = Resz zb(z, w) = b−2(w). Iterating, it follows that b(z, w) ∈ U[[w, w−1]][[z]]. But we also know that (z − w)N b(z, w) = 0, since cj(w) = 0 for j ≥ N. Hence b(z, w) = 0 and the decomposition holds for coe cients cj(w) given by (2.2). Let us prove uniqueness. We claim that if
N X ∂j δ(z, w) cj(w) w = 0, j! j=0
j −1 j j0 where c (w) ∈ U[[w, w ]], then all c (w) = 0. If not, let j0 be the maximal index for which c 6= 0 and multiply both sides for (z − w)j0 . Then by Proposition 2.5(c) we get cj0 δ(z, w) = 0. Taking the residue j in this relation we have c 0 (w) = 0, that contradicts our assumption. Corollary 2.8. a(z, w) is local if and only if it admits such a decomposition. Exercise 3. Given a(z, w) ∈ U[[z, z−1, w, w−1]] we may construct a linear operator −1 −1 Da(z,w) : C[z, z ] −→ U[[w, w ]]
ϕ(z) −→ Resz ϕ(z)a(z, w). Prove that j a) D j = c(w)∂ , in particular D = 1; c(w)∂wδ(z,w) w δ(z,w) 5 b) a(z, w) is local if and only if Da(z,w) is a dierential operator of nite order; c) if is local, then is local too and ∗ . a(z, w) a(w, z) Da(z,w) = Da(w,z) Here ∗ is the formal adjoint de ned by ∂∗ = −∂ and f ∗ = f if f is a formal series. Solution. Part a) follows from a straightforward computation:
j j j D j ϕ(z) = Res ϕ(z)c(w)∂ δ(z, w) = c(w)∂ Res ϕ(z)δ(z, w) = c(w)∂ ϕ(w). c(w)∂wδ(z,w) z w w w We know that is local if and only if P j j and the sum is nite. Then, a(z, w) a(z, w) = c (w)∂wδ(z, w)/j! by part a), it follows that P j j is a dierential operator of nite order. This proves Da(z,w) = c (w)∂w/j! b). To prove the last assertion we need Proposition 2.5(b). We have
j j j D j ϕ(z) = Res ϕ(z)c(z)∂ δ(w, z) = Res c(z)ϕ(z)∂ δ(z, w) = Res c(z)ϕ(z)(−∂ ) δ(z, w) = c(z)∂z δ(w,z) z z z z w j j = (−∂w) Resz ϕ(z)c(z)δ(z, w) = (−∂w)c(w)ϕ(w). Since a(z, w) is local, by linearity we get j ∗ j X ∂ X (−∂w) D∗ = cj(w) w = cj(w) = D . a(z,w) j! j! a(w,z) Exercise 4. If a(z, w) is local, then any of its derivatives (by z or by w) is local. Solution. By symmetry it su ces to prove the result only for one derivative. Suppose (z−w)na(z, w) = 0, then n+1 n+1 n (z − w) ∂wa(z, w) = ∂w((z − w) a(z, w)) + (n + 1)(z − w) a(z, w) = 0. Let U = g be a Lie algebra and let a(z), b(z) ∈ g[[z, z−1]]. De nition 2.9. The pair a(z), b(z) is called local if [a(z), b(w)] ∈ g[[z, z−1, w, w−1]] is local. Clearly, this means that n for some . Applying the Decomposition (z − w) [a(z), b(w)] = 0 n ∈ Z+ Theorem (Theorem 2.7) we get for a local pair a(z), b(z) N X ∂j δ(z, w) [a(z), b(w)] = cj(w) w , j! j=0 j j where c (w) = Resz(z − w) [a(z), b(w)]. Let us write X −n−1 X −n−1 and j X j −n−1 a(z) = a(n)z , b(z) = b(n)z c (z) = c(n)z . n∈Z n∈Z n∈Z Comparing the coe cient of z−m−1w−n−1 in [a(z), b(w)], we get X m [a , b ] = cj . (m) (n) j (m+n−j) j∈Z+ This equality is known as the commutator formula and it is useful in many computations.
Example 2.10. Let g be a nite dimensional Lie algebra with a symmetric invariant bilinear form (· | ·), that is for any , , . De ne −1 , where is a central element. ([a, b] | c) = (a | [b, c]) a b c ∈ g bg = g[t, t ] + CK K We de ne a Lie bracket on by bg m n m+n [at , bt ] = [a, b]t + mδm,−n(a | b)K, (2.3) for any a, b ∈ g and m, n ∈ Z. Exercise 5. Check Lie algebra axioms for . bg Solution. The fact that [atm, atm] = 0 and linearity of (2.3) implies antisymmetry. We are left to prove the Jacobi identity. For a, b, c ∈ g and n, m, k ∈ Z, we have m n k m n+k m+n+k [at , [bt , ct ]] = [at , [b, c]t + nδn,−k(a | b)K] = [a, [b, c]]t + mδm,−n−k(a | [b, c])K. Similarly we get n k m m+n+k [bt , [ct , at ]] = [b, [c, a]]t + nδn,−m−k(b | [c, a])K, k m n m+n+k [ct , [at , bt ]] = [c, [a, b]]t + kδk,−m−n(c | [a, b])K. 6 Thus we are left to prove Jacobi identity only for n + m + k = 0. In this case it su ces to prove that m(a | [b, c]) + n(b | [c, a]) + k(c | [a, b]) = 0. But this is clear by the condition n + m + k = 0 and the invariance of the scalar product. De nition 2.11. is called a nization of . bg (g, (· | ·)) To an element we associate the following -valued formal distribution, called current, a ∈ g bg X a(z) = (atn)z−n−1. n∈Z Then (2.3) is equivalent, for a, b ∈ g, to the relations
[a(z), b(w)] = [a, b](w)δ(z, w) + (a | b)K∂wδ(z, w), [K, a(z)] = 0. To prove this fact we can use the commutator formula: here we have c0(w) = [a, b](w) and c1(w) = (a|b)K. Another way is to think of the current as a(z) = aδ(t, z). Then we get
[a(z), b(w)] = [aδ(t, z), bδ(t, w)]=[a, b]δ(w, t)δ(z, t)+(a | b)K Resz ∂zδ(w, t)δ(z, t) =
= [a, b]δ(t, w)δ(z, w) + (a | b)K∂wδ(z, w) = [a, b](w)δ(z, w) + (a | b)K∂wδ(z, w). Example 2.12. The Virasoro algebra is the Lie algebra with basis and the following {Ln, n ∈ Z,C} commutation relations m3 − m [L ,L ] = (m − n)L + δ C, (2.4) m n m+n 12 m.−n where C is a central element. Let X −n−2 X −n−1 L(z) = Lnz = L(n)z , n∈Z n∈Z where L(n) = Ln−1. Exercise 6. The relations (2.4) are equivalent to C ∂3 δ(z, w) [L(z),L(w)] = (∂ L(w))δ(z, w) + 2L(w)∂ δ(z, w) + w . w w 2 3! Solution. We use again commutator formula. We have 0 , 1 and 3 C . c (w) = ∂wL(w) c (w) = 2L(w) c (w) = 2 Then 0 , 1 and 3 C . Hence, cn = −nL(n−1) c(n) = 2L(n) c(n) = δn,−1 2 m(m − 1)(m − 2) C [L ,L ]=(−m − n)L +2mL + δ = (m) (n) (m+n−1) (m+n−1) 3! m+n−3,−1 2 m(m − 1)(m − 2) =(m − n)L + δ C. (m+n−1) m−1,−n+1 12 But , then rescaling and we get (2.4). [L(m),L(n)] = [Lm−1,Ln−1] m 7→ m + 1 n 7→ n + 1 It is clear from examples that the pairs for and for the Virasoro algebra (a(z), b(z))a,b∈g bg (L(z),L(w)) are local. Example 2.13. A special case of is obtained when , with . Then P n −n−1 bg g = Ca (a|a) = 1 a(z) = (at )z and [a(z), a(w)] = ∂wδ(z, w)K. The formal distribution a(z) is called a free boson. Example 2.14. Another important example of formal distribution is the free fermion. Let us call it ϕ(z), then it satis es the relation [ϕ(z), ϕ(w)] = δ(z, w). To explain how this distribution is de ned we need the notion of Lie superalgebra. A super space is a vector space
V = V0¯ ⊕ V1¯, where ¯, ¯ . If , we set and call this element even. Similarly, if , we 0 1 ∈ Z 2Z a ∈ V0¯ p(a) = 0 a ∈ V1¯ set p(a) = 1 and call this element odd.A bilinear form (· | ·) on it, is a bilinear form on V such that (V0¯,V1¯) = 0. It is called symmetric (respectively antisymmetric) if it is symmetric when restricted to V0¯ and skewsymmetric when restricted to V1¯ (respectively if it is skewsymmetric when restricted to V0¯ and symmetric when restricted to V1¯). Note that, given a symmetric bilinear form on a super vector space V , it becomes antisymmetric on the super vector space with reverse parity ΠV .A superalgebra is a super space with a product such that , where , . A = A0¯ ⊕ A1¯ AαAβ ⊂ Aα+β α β ∈ Z 2Z 7 For example, if V is a superspace, then End V = End V0¯ ⊕End V1¯ is an associative superalgebra, where a 0 0 c End V¯ = ∈ End V and End V¯ = ∈ End V . 0 0 b 1 d 0 The commutator in an associative superalgebra is de ned as
[a, b] = ab − p(a, b)ba,
where p(a, b) = (−1)p(a)p(b). This bracket satis es the axioms of a Lie superalgebra, namely (antisymmetry) [a, b] = −p(a, b)[b, a]; (Jacobi identity) [a, [b, c]] = [[a, b], c] + p(a, b)[b, [a, c]]. Come back to example. We need Cliord a nization: take a superspace A with a skewsymmetric bilinear form h·, ·i and let Ab = A[t, t−1] + CK, K central, with the bracket de ned by m n [at , bt ] = ha, biδm,−n−1K. Currents are de ned by P(atn)z−n−1, for a ∈ A and their bracket is [a(z), b(w)] = ha, biδ(z, w). Now, if we take A = Cϕ, with ϕ odd and hϕ, ϕi = 1 then we get the formal distribution of free fermion. Rewrite the decomposition for the bracket of a local pair a(z), b(z) ∈ g[z, z−1] in the following way (just rename the coe cients cj(w))
N X ∂j δ(z, w) [a(z), b(w)] = a(w) b(w) w , (2.5) (j) j! j=0 where j a(w)(j)b(w) = Resz(z − w) [a(z), b(w)]. (2.6) This expression is called operator product expansion (OPE) for a local pair a(z), b(z). In particular for each j ∈ Z we get a j-th product given by (2.6). This is just the singular part of the OPE of the product of two formal distributions. We want to get the "complete" OPE: we need the notion of normally ordered P n product. Let a(z) = anz be a formal distribution, we de ne
X n a(z)+ = anz (creation part), n∈Z+ X n a(z)− = anz (annihilation part). n<0 De nition 2.15. If U is an associative algebra, the complete normally ordered product of a(z), b(z) ∈ U[[z, z−1]] is
: a(z)b(w) := a(z)+b(w) + b(w)a(z)−.
Let us note that (∂za(z))± = ∂z(a(z)±). Lemma 2.16 (Formal Cauchy Formulas). The following relations hold 1 1 a(w) = Res a(z)i , a(w) = − Res a(z)i (2.7) + z z,w z − w − z w,z z − w and, dierentiating k times by z, get
k ∂wa(w)± 1 = ± Resz a(z)iz,w . k! w,z (z − w)k+1
P n Proof. Let a(w) = anw , then 1 X X w k X X Res a(z)i = Res a zn · z−1 = Res a zn−k−1wk = a wn = a(w) . z z,w z − w z n z z n n + n∈Z k∈Z+ n∈Z n∈Z+ k∈Z+ This proves the rst of the formulas (2.7). The other is proved in the same way. By dierentiating these formulas we get the lemma. 8 Exercise 7. For a local pair of U-valued formal distributions, U associative algebra, the OPE formula is equivalent to the following two equalities: X 1 a(z)b(w) = a(w) b(w)i + : a(z)b(w):, (j) z,w (z − w)j+1 j∈Z+ X 1 b(w)a(z) = a(w) b(w)i + : a(z)b(w): . (j) w,z (z − w)j+1 j∈Z+ These relations are the so called complete OPE. Solution. By (2.1), we can rewrite (2.5) in the following way X 1 1 [a(z), b(w)] = a(w) b(w) i − i . (j) z,w (z − w)j+1 w,z (z − w)j+1 j∈Z+ If we separate positive and negative powers in z in this expression we obtain X 1 [a(z) , b(w)] = a(w) b(w)i − (j) z,w (z − w)j+1 j∈Z+ X 1 [a(z) , b(w)] = − a(w) b(w)i . + (j) w,z (z − w)j+1 j∈Z+ The proof now follows since we have
a(z)b(w) = a(z)+b(w) + a(z)−b(w) = [a(z)−, b(w)]+ : a(z)b(w):,
b(w)a(z) = b(w)a(z)+ + b(w)a(z)− = [b(w), a(z)+]+ : a(z)b(w): . Physicists usually write this OPE as
X a(w)(j)b(w) a(z)b(w) ∼ , (z − w)j+1 j∈Z+ writing only what we have called singular part of the OPE. Look now at the regular part of the OPE and take the limit z → w. Then we get
: a(z)b(z) := a(z)+b(z) + b(z)a(z)−. This expression is called normal ordered product of a(z) and b(z). It usually diverges, but we will prove that converges if our formal distributions are quantum elds. First recall the de nition of a quantum eld.
P −n−1 De nition 2.17. A quantum eld is an End V -valued formal distribution a(z) = a(n)z such that for any v ∈ V , a(z)v ∈ V ((z)).
This means that a(n)v = 0 for n >> 0. Proposition 2.18. If a(z) and b(z) are quantum elds, then : a(z)b(z): is a well de ned quantum eld.
Proof. We have to check that : a(z)b(z): v = (a(z)+b(z) + b(z)a(z)−)v ∈ V ((z)). We know that b(z)v ∈ V ((z)) and a(z)+ ∈ End V [[z]], then a(z)+b(z) ∈ V ((z)). For the second term we have ! X i X i b(z)a(z)−v = b(z) ai(v)z = b(z)ai(v)z ∈ V ((z)). nite nite Expanding in Taylor series the regular part of the OPE we get X : ∂ja(w)b(w): : a(z)b(w) := (z − w)j. j! j∈Z+ This expression holds in the domain |z − w| < |z|. We de ne : ∂ja(z)b(z): a(z) b(z) = , (−j−1) j! for , so j ∈ Z+ : a(z)b(z) := a(z)(−1)b(z). 9 Proposition 2.19. For all j ∈ Z and a(z), b(z) quantum elds, we have j j a(w)(j)b(w) = Resz a(z)b(w)iz,w(z − w) − b(w)a(z)iw,z(z − w) . Proof. For it is (2.6). For it follows from Cauchy formulas multiplied by . j ∈ Z+ j < 0 b(w) Exercise 8. Show that , for is a quantum eld if are. a(z)(j)b(z) j ∈ Z+ a(z), b(z) n Solution. Remember that a(w)(n)b(w) = Resz[a(z), b(w)](z − w) . Expanding we get X n a(w) b(w) = Res (−1)k [a , b ]zk−m−1wn−k−l−1 = (n) z k (m) (l) k,l,m X n X X = (−1)k [a , b ]wn−k−l−1 = (−1)k[a , b ] w−l−1. k (k) (l) (k) (l−k+n) k,l l∈Z k∈Z+
When apply to v, we can choose N such that b(l−n) kills the nite set of vectors v, a(0)v, . . . , a(n)v, for l > N. Then the commutators terms kill v. Let V be a vector (super)space and let a(z), b(z) be End V -valued quantum elds, then for each j ∈ Z we have de ned the j-th product, which is again a eld, a(w)(j)b(w).
Proposition 2.20. ∂w is a derivation of all these products. Proof. For we have j. Then we get j ∈ Z+ a(w)(j)b(w) = Resz[a(z), b(w)](z − w) j j−1 ∂w a(w)(j)b(w) = Resz[a(z), ∂wb(w)](z − w) − Resz[a(z), b(w)]j(z − w) =
= a(w)(j)∂wb(w) + ∂wa(w)(j)b(w), j−1 j since − Resz[a(z), b(w)]j(z − w) = Resz[∂za(z), b(w)](z − w) . On the other hand, for j > 0 we have : ∂j a(w)b(w): ∂j a(w) ∂j a(w) a(w) b(w) = w = w + b(w) + b(w) w − . (−j−1) j! j! j! Then we get ∂j+1a(w) ∂j a(w) ∂j a(w) ∂j+1a(w) ∂ a(w) b(w) = w + b(w) + w + ∂ b(w) + ∂ b(w) w − + b(w) w − = w (−j−1) j! j! w w j! j!
= ∂wa(w)(−j−1)b(w) + a(w)(−j−1)∂wb(w). 3. Some equivalent definitions of a Vertex Algebra
j De nition 3.1. A vertex algebra is the data of (V, |0i,T, F = {a (z)}j∈J ) where V is a vector (su- per)space (space of state), |0i is a (even) vector (vacuum vector), T ∈ End V is an (even) operator (translation operator) and F is a collection of End V - valued quantum elds with the following axioms: (vacuum) T |0i = 0; j j (translation covariance) [T, a (z)] = ∂za (z); (completeness) all vectors aj1 aj2 . . . ajs |0i span V ; (n1) (n2) (ns) (locality) Nij i j for some . (z − w) [a (z), a (z)] = 0 Nij ∈ Z+ We have already proved in Section1 the following lemma:
Lemma 3.2. aj(z)|0i ∈ V [[z]]. The proof of this lemma uses only the rst two axioms.
De nition 3.3. An End V - valued quantum elds a(z) is called translation covariant if [T, a(z)] = ∂za(z).
By Lemma 3.2 it follows that a(z)|0i ∈ V [[z]]. Moreover, let Fe denote the space of all translation covariant End V -valued quantum elds, then we have the map s Fe −→ V
a(z) −→ a(z)|0i|z=0 . This map is called eld-state correspondence. 10 Theorem 3.4 (Extension Theorem). Let (V, |0i,T, F) be a vertex algebra and let F denote the space of all translation covariant End V -valued quantum elds which are local to each of the quantum elds from F, then a) all axioms of vertex algebra for (V, |0i,T, F) still hold; b) the map s : F −→ V is bijective.
We need several lemmas before proving the theorem. First note that s(a(z)) = a(−1)|0i. Lemma 3.5. contains 1 , it is -invariant and it is closed under all -th products, . Fe V ∂z n n ∈ Z
Proof. Clearly 1V ∈ Fe, in fact [T, 1V ] = 0 = ∂z1V . Moreover, dierentiating by z the equality [T, a(z)] = ∂za(z), we get [T, ∂za(z)] = ∂z(∂za(z)). Finally, we need to prove that if [T, a(z)] = ∂za(z) and [T, b(z)] = , then the same holds for , . Indeed, we have ∂zb(z) a(z)(j)b(z) j ∈ Z j j [T, a(w)(j)b(w)] = T, Resz a(z)b(w)iz,w(z − w) − p(a, b)b(w)a(z)iw,z(z − w) = j j = Resz [T, a(z)b(w)]iz,w(z − w) − p(a, b)[T, b(w)a(z)]iw,z(z − w) = j j = Resz ([T, a(z)]b(w) + a(z)[T, b(w)])iz,w(z − w) −p(a, b)([T, b(w)]a(z) + b(w)[T, a(z)])iw,z(z − w) = j j = Resz (∂za(z)b(w) + a(z)∂wb(w))iz,w(z − w) − p(a, b)(∂wb(w)a(z) + b(w)∂za(z))iw,z(z − w) =
= ∂wa(w)(j)b(w) + a(w)(j)∂wb(w) = ∂w(a(w)(j)b(w)). Lemma 3.6. Let a(z), b(z) ∈ Fe and s(a(z)) = a, s(b(z)) = b, then: a) s(∂za(z)) = T a = a(−2)|0i; b) s(a(z)(n)b(z)) = a(n)b.
Proof. We have a(z)|0i = a(−1)|0i + za(−2)|0i + .... Apply T to both sides and get
LHS = T a(z)|0i = [T, a(z)]|0i + a(z)T |0i = ∂za(z),
RHS = T a(−1)|0i + z(...) = [T, a(−1)]|0i + z(...) = a(−2)|0i + z(...). So and this proves a). For b), we have s(∂za(z)) = ∂za(z)|z=0 = a(−2)|0i a(w) b(w)|0i = Res a(z)b(w)|0ii (z − w)n − Res b(w)a(z)|0ii (z − w)n . (n) |w=0 z z,w |w=0 z w,z |w=0 n n Now, we note that b(w)|0i ∈ V [[w]]. Then the rst term of the sum is Resz a(z)bz , while iw,z(z − w) ∈ C[[w, w−1]][[z]] and a(z)|0i ∈ V [[z]], so the second term is 0. This means that n a(w)(n)b(w) = Resz a(z)bz = a(n)b. Exercise 9. Prove that the dierential equation d , where and dz f(z) = R(z)f(z) f(z) ∈ U[[z]] R(z) ∈ End U[[z]], has a unique solution with given f(0). Solution. Write X n X n f(z) = anz , an ∈ U , and R(z) = Rnz ,Rn ∈ End U . n∈Z+ n∈Z+ Then d X X f(z) = na zn−1 = (n + 1)a zn . dz n n+1 n>0 n∈Z+ and n ! X X n R(z)f(z) = Rn−kak z . n∈Z+ k=0 It follows that f(z) satis es the dierential equation if and only if its coe cients satisfy the recursion a0 = f(0) 1 Pn an+1 = n+1 k=0 Rn−kak n ≥ 0 which is uniquely solvable for xed f(0). Lemma 3.7. Let a(z) ∈ Fe, then a) a(z)|0i = ezT a, where a = s(a(z)); wT −wT b) e a(z)e = iz,wa(z + w); 11 wT −wT c) e a(z)±e = iz,wa(z + w)±. Proof. We prove only part a). Let us note that both sides are equal when z = 0 and satisfy the same dierential equation, in fact d a(z)|0i = ∂ a(z)|0i = [T, a(z)]|0i = T a(z)|0i. dz z Clearly, on the other hand d ezT a = T ezT a. dz Exercise 10. Prove b) and c) using the fact that both sides satisfy the dierential equation d f(w) = ad T f(w), dw where f(w) ∈ (End V [[z, z−1]])[[w]] and f(0) = a(z). Solution. Let us prove part b) (the proof of part c) is the same). Clearly both sides of the dierential equation are in (End V [[z, z−1]])[[w]] and satisfy the same initial condition. If we dierentiate both sides we get d LHS = ewT a(z)e−wT = T ewT a(z)e−wT − ewT a(z)e−wT T = ad T ewT a(z)e−wT , dw while the other side is d RHS = i a(z + w) = i ∂ a(z + w) = i ad T a(z + w) = ad T i a(z + w). dw z,w z,w z z,w z,w
Lemma 3.8 (Uniqueness Lemma). Let F 0 ⊂ Fe be a collection of translation covariant quantum elds such that s : F 0 −→ V is surjective, let a(z) ∈ Fe be such that s(a(z)) = 0 and all pairs (a(z), b(z)), where b(z) ∈ F 0, are local, then a(z) = 0. Proof. By locality we have n n , . By part a) of Lemma (z − w) b(w)a(z)|0i = ±(z − w) a(z)b(w)|0i n ∈ Z+ 3.7 we have (z − w)nb(w)a(z)|0i = (z − w)nb(w)ezT s(a(z)) = 0. Then we have obtained 0 = ±(z − w)na(z)ewT b, where b = s(b(w)). If we put w = 0 we get zna(z)b = 0, then a(z)b = 0. From the arbitrariness of b it follows that a(z) = 0. Lemma 3.9 (Dong's Lemma). Let a(z), b(z) and c(z) be quantum elds which are pairwise local, then the pair is local for any . (a(z)(n)b(z), c(z)) n ∈ Z Proof. For the proof see [Kac96, Lemma 3.2].
0 Proof of Theorem 3.4. Let F be the minimal subspace of the space of all quantum elds containing 1V 0 and F, ∂z-invariant and closed under all n-th products. Then by Lemma 3.5 F ⊂ Fe and by Lemma 3.9 we get the following chain F ⊂ F 0 ⊂ F ⊂ Fe. Moreover, by Dong's Lemma, F 0 contains the following quantum elds
j1 js−1 js 1 a(z) = a (z)(n1)(... (a (z)(ns−1)(a (z)(ns) V )) ... ). But, by Lemma 3.6, s(a(z)) = aj1 . . . ajs−1 ajs |0i. Hence s(F 0) = V by completeness axiom, while (n1) (ns−1) (ns) from Lemma 3.8 it follows that s : F → V is injective. This means that F 0 = F and s is bijective that 0 proves b). Part a) is now clear since local axioms hold because all elds from F are pairwise local. Corollary 3.10. We have a) |0i(z) = 1V ; b) T a(z) = ∂za(z); c) (a(n)b)(z) = a(z)(n)b(z). 12 By Extension Theorem s is an isomorphism, so we have s−1 : V −→ F a −→ a(z) where
−1 j1 js j1 js s (a . . . a |0i) = a (z) (... (a (z) 1V ) ... ). (3.1) (n1) (ns) (n1) (ns) In fact both sides lie in F and they have the same image under s by lemma 3.6. Hence they are equal by Uniqueness Lemma. The map s−1 is called state- eld correspondence. Proposition 3.11 (Skewsymmetry). Let a, b ∈ V , a(z) = s−1(a), b(z) = s−1(b), then a(z)b = p(a, b)ezT b(−z)a. Proof. By locality (z − w)na(z)b(w)|0i = p(a, b)(z − w)nb(w)a(z)|0i. By part a) of Lemma 3.7 we have n wT n zT −zT zT n zT (z − w) a(z)e b = p(a, b)(z − w) e e b(w)e a = p(a, b)(z − w) e iw,zb(w − z)a. n n n zT If n >> 0, then (z − w) b(w − z)a ∈ V [[z − w]]. If we put w = 0, then z a(z)b = p(a, b)z e b(−z)a.
j1 js In most important examples V is spanned by vectors a = a . . . a |0i, where n1, . . . , ns ∈ (−n1−1) (−ns−1) , then (3.1) becomes Z+ ∂n1 ∂ns s−1(a) =: z aj1 (z) ... z ajs (z):, n1! ns! where : · : is taken from right to left, that means : a(z)b(z)c(z) :=: a(z): b(z)c(z) :: . The Extension Theorem (Theorem 3.4) gives the second equivalent de nition of vertex algebra.
De nition 3.12. A vertex algebra is the data of (V, |0i, T, s), where s is a map from V to a space of End V -valued quantum elds, s(a) = a(z), such that the following axioms hold (vacuum) , ; T |0i = 0 a(z)|0i|z=0 = a (translation covariance) [T, a(z)] = ∂za(z); (locality) (z − w)Nab [a(z), b(w)] = 0.
The completeness axiom is automatically satis ed because a(−1)|0i = a span V . Often a(z) is denoted Y (a, z) and called vertex operator. The set of quantum elds of a vertex algebra V contains 1V = |0i(z), is ∂z-invariant and is closed under all n-th products, that is
a(z)(n)b(z) = (a(n)b)(z). The last equality is called n-th product formula. The commutator formula in a vertex algebra reads
X ∂j δ(z, w) [a(z), b(w)] = (a b)(w) w . (j) j! j∈Z+ This formula follows from the Decomposition Theorem and the n-th product identity. If we compare the coe cients of z−m−1 we get the following equivalent formulas, for n, m ∈ Z, to the commutator formula: X m [a , b(w)] = (a b)(w)wm−j; (m) j (j) j∈Z+ (3.2) X m [a , b ] = (a b) . (m) (n) j (j) (m+n−j) j∈Z+ Exercise 11. Given f(z) ∈ C((z)) and a ∈ V , let X af = Resz f(z)a(z) = fna(n). n≥−N n n In particular, if f = z , then af = Resz z a(z) = a(n). Prove that X [af , bg] = a(j)b f(j)g . j! j∈Z+ 13 Solution. Let P n and P m be elements in then by (3.2) we get f = n≥−N fnz g = m≥−M gmz C((z)) X X X n [af , bg] = fngm[a(n), b(m)] = fngm a(j)b . j (n+m−j) n≥−N j∈Z+ n≥−N m≥−M m≥−M On the other hand, since f (j)g X n = f g zn+m−j, j! j n m n≥−N m≥−M one gets, for any , j ∈ Z+ X n n+m−j−k−1 a(j)b f(j)g = Resz f(z) a(j)b f(j)g (z) = Resz fngm a(j)b (k) z = j! j! j k∈Z+ n≥−N m≥−M X = fngm a(j)b (n+m−j) . n≥−N m≥−M Example 3.13. A ne Lie algebras: −1 , P n −n−1. We know that bg = g[t, t ] + CK a(z) = (at )z
[a(z), b(w)] = [a, b](w)δ(z, w) + (a | b)∂wδ(z, w)K. Then we get 0 [af , bg] = [a, b]fg + (a | b) Resz f (z)g(z)K. P −n−2 Example 3.14. Virasoro algebra: L(z) = Lnz with the commutator given by C [L(z),L(w)] = ∂ L(w)δ(z, w) + 2L(w)∂ δ(z, w) + ∂3 δ(z, w). w w 12 w Then we get
1 000 [L ,L ] = L 0 0 + Res f (z)g(z). f g f g−fg 12 z Example 3.15. Take A to be a vector superspace with a skew-symmetric bilinear form h·, ·i and set −1 . De ne m n , then for P n −n−1, , the Ab = A[t, t ] + C1 [at , bt ] = ha, biδm,−n−11 a(z) = (at )z a ∈ A commutator is given by [a(z), b(w)] = ha, biδ(z, w)1. So we get
[af , bg] = ha, bi Resz fg. Now we state a famous identity, called Borcherds identity, that is a generalization of the n-th product identity and the commutator formula. Theorem 3.16 (Borcherds identity).
X ∂j δ(z, w) a(z)b(w)i (z − w)n − p(a, b)b(w)a(z)i (z − w)n = (a b)(w) w . z,w w,z (n+j) j! j∈Z+ Proof. LHS is a local formal distribution in z and w by the locality axiom, then, by Decomposition Theorem, we have: X ∂j δ(z, w) LHS = cj(w) w , j! j where
j j n n c (w) = Resz (z − w) (a(z)b(w)iz,w(z − w) − p(a, b)b(w)a(z)iw,z(z − w) ) = j+n j+n = Resz a(z)b(w)iz,w(z − w) − p(a, b)b(w)a(z)iw,z(z − w) = a(w)(n+j)b(w) = (a(n+j)b)(w). 14 Comparing coe cients of z−m−1w−k we get, for m, n, k ∈ Z and a, b, c ∈ V , the original Borcherds identity: X m X n (a b) c = (−1)j a (b c)− (3.3) j (n+j) (m+k−j) j (m+n−j) (k+j) j∈Z+ j∈Z+ n − (−1) p(a, b)b(n+k−j)(a(m+j)c) . Thanks to Borcherds identity we obtain the third de nition of vertex algebra. De nition 3.17. A vertex algebra is the data of with the following axioms (V, |0i, a(n)b | n ∈ Z) (a, b, c ∈ V ): (vacuum) , for every , and , for every ; |0i(n)a = δn,−1a n ∈ Z a(n)|0i = δn,−1a n ≥ −1 P m P j n n (Borcherds identity) (a(n+j)b)(m+k−j)c= (−1) a(m+n−j)(b(k+j)c)−(−1) p(a, b)b(n+k−j)(a(m+j)c) . j∈Z+ j j∈Z+ j Let us note that Borcherds identity implies locality axiom. Exercise 12. Derive the second de nition of vertex algebra from the third one. Solution. From the vacuum axiom it follows that and . De ne on by |0i(z) = idV a(z)|0i|z=0 = a T V T a = a(−2)|0i. Then, clearly, T |0i = 0. Put c = |0i, m = 0 and k = −2 in the identity (3.3) and get
(a(n)b)(−2)|0i = a(n)(b(−2)|0i) − na(n−1)(b(−1)|0i). Then
[T, a(n)]b = T a(n)b − a(n)T b = (a(n)b)(−2)|0i − a(n)(b(−2)|0i) = −na(n−1)(b(−1)|0i) = −na(n−1)b, then , since is arbitrary. [T, a(z)] = ∂za(z) b Exercise 13. Let V be a commutative associative algebra with 1 and let T be a derivation of V , then (V, |0i = 1, T, a → eT za) is a vertex algebra. Moreover, any vertex algebra for wich all a(z) do not involve negative powers of z is one of these. Proof. Let us check the vacuum axiom. Since T is a derivation we have T (1) = T (1·1) = T (1)·1+1·T (1) = , then . Moreover zT zT zT . For the translation 2T (1) T |0i = T (1) = 0 e a|0i|z=0 = e a · 1|z=0 = e a|z=0 = a covariance axiom we have X zk X zk [T, ezT a]b = T (ezT a · b) − ezT a · T b = (T ezT a)b = T k+1(a) · b = ∂ T k(a) · b = (∂ ezT a)b. k! z k! z k∈Z+ k∈Z+ The locality axiom holds since V is commutative. Clearly, a(z) does not involve negative powers of z, the converse follows from the Extension Theorem. j De nition 3.18. A regular formal distribution Lie (super)algebra is the data of (g,T, F = {a (z)}j∈J ), where is a (even) derivation of , the coe cients of j P j −n−1 span , i j are all T g a (z) = a(n)z g (a (z), a (z)) local pairs and T a(z) = ∂za(z). Example 3.19. The algebra with and is a regular formal distribution bg F = {a(z)}a∈g ∪ {K} T = −∂t Lie algebra.
Exercise 14. Prove that ∂za(z) = −∂ta(z). Solution. It is a straightforward computation:
X n −n−1 X n −n−2 X n−1 −n−1 ∂za(z) = ∂z (at )z = − (n + 1)(at )z = − n(at )z = −∂ta(z). n∈Z n∈Z n∈Z
Example 3.20. The Virasoro algebra with F = {L(z),C} and T = L−1 (note that [L−1,L(z)] = ∂zL(z)) is a regular formal distribution Lie algebra.
j Let (V = U(g), |0i = 1,T, F = {a (z)}j∈J ). Is this a vertex algebra, according to the rst de nition? All axioms hold, but we also need j for and this is false, because formal distributions, a(n)1 = 0 n >> 0 in general, are not quantum elds. We need to take U(g) , where is -invariant. Let V = U(g)g− g− T j g− = span{a(n), n ∈ Z+, j ∈ J}. 15 This is a subalgebra of g, called the annihilation subalgebra, since we have X m [a , b ] = (a b) (m) (n) j (j) (m+n−j) j∈Z+ and is T -invariant since T am = −ma(m−1). Example 3.21. In the previous examples we have M bg− = g[t], V ir− = CLn, Ab− = A[t]. n≥−1 Theorem 3.22. U(g) is a vertex algebra. U(g)g−, im(1),T, F Proof. We prove by induction on s that a (aj1 . . . ajs 1) = 0, for N >> 0. (N) (n1) (ns)
For s = 0 this is true since g− is in the kernel of the projection to the quotient. For s > 0 we have a (aj1 . . . ajs 1) = [a , aj1 ]aj2 . . . ajs 1 + aj1 a aj2 . . . ajs 1. (N) (n1) (ns) (N) (n1) (n2) (ns) (n1) (N) (n2) (ns) The second summand is 0 for N >> 0 by induction hypothesis. For the rst recall that j1 X N j [a(N), a ] = c = 0 (n1) j (N+n1−j) j∈Z+ for N >> 0. Let us de ne U(g) . V (g, F) = U(g)g−, |0i = im(1),T, F Example 3.23. The vertex algebra V k(g) = V (g, F) , b V (bg, F)(K − k) k ∈ C, is called universal a ne vertex algebra of level k. Example 3.24. The vertex algebra c V = V (V ir, F) V (V ir, F)(C − c), c ∈ C, is called universal Virasoro vertex algebra of charge c. Example 3.25. The vertex algebra F (A) = V (A,b F) V (A,b F)(1 − im(1)), is called fermions vertex algebra based on A. Choose a basis {aj} of g, then by Poincaré-Birkho-Witt theorem the elements a = aj1 aj2 . . . ajs |0i, (−n1−1) (−n2−1) (−ns−1) with , span . Since −1 −1 , we have −1 −1 , so the state- eld ni ∈ Z+ V bg = g[t ]t + K + g[t] V = U(g[t ]t )|0i correspondence looks ∂n1 aj1 (z) . . . ∂ns ajs (z) a(z) =: z z : . n1! . . . ns! Exercise 15. Write state- eld correspondence for the Virasoro case and for F (A). Solution. The vertex algebra c is spanned by elements , with . Hence, the V L−n1−2 ··· L−ns−2|0i ni ∈ Z+ state- eld correspondence is
n1 ns ∂z L(z) . . . ∂z L(z) (L−n1−2 ··· L−ns−2|0i)(z) =: : . n1! . . . ns!
j1 js j Similarly, F (A) is spanned by a ··· a |0i, with ni ∈ + and {a } is a basis of A. We get (−n1−1) (−ns−1) Z the following state- eld correspondence
n1 j1 ns js j1 js ∂z a (z) . . . ∂z a (z) (a(−n −1) ··· a(−n −1)|0i)(z) =: : . 1 s n1! . . . ns! 16 We present now another construction of V (g, F). Assume that g admits a descending ltration
... ⊃ g−1 ⊃ g0 ⊃ g1 ... completed such that T gj ⊂ gj−1. In our example the ltration was given by powers of t. Let U(g) consist P of all series uj such that all but nitely many uj lie in U(g)gN , for each N. Let F be the minimal collection of formal distribution containing 1, T -invariant and closed under all n-th products and denote by F its closure. In particular we have
X X completed : a(z)b(z):(n)= a(j)b(n−j−1) + b(n−j−1)a(j) ∈ U . j<0 j∈Z+
Then (F, |0i = 1, T, a(z)(n)b(z)) is a vertex algebra isomorphic to V (g, F) and the isomorphism is given by the map F −→ V (g, F) de ned by representation of U(g)completed in V (g, F). De nition 3.26. Let V be a vertex algebra, then an element L ∈ V is called an energy-momentum state and the corresponding eld L(z) is called an energy-momentum eld if a) 1 3 1 ; [L(z),L(w)] = ∂wL(w)δ(z, w) + 2L(w)∂wδ(z, w) + 12 ∂wδ(z, w)C V ,C ∈ C b) L−1 = T ; c) L0 is a diagonalizable operator on V . Recall that part a) is equivalent to the relation m3 − m [L ,L ] = (m − n)L + δ C1 , m n m+n 12 m,−n V P −n−2 for L(z) = Lnz . If L0a = ∆aa, then ∆a is called the energy or the conformal weight of a. By the commutator formula we get X ∂j δ(z, w) [L(z), a(w)] = (L a)(w) w = (T a)(w)δ(z, w) + (L a)(w)∂ δ(z, w) + ... = j−1 j! 0 w j∈Z+ = ∂wa(w)δ(z, w) + (L0a)(w)∂wδ(z, w) + .... By taking the coe cients of z−m−2, we get the equivalent formula X m + 1 [L , a(z)] = (L a)(z)zm+1−j. (3.4) m j j−1 j∈Z+
In particular [L0, a(z)] = z∂za(z) + (L0a)(z). De nition 3.27. A diagonalizable operator H on V is called an energy operator if
[H, a(z)] = (z∂z + ∆a)a(z), (3.5)
where L0a = ∆aa.
The operator L0 coming from energy-momentum eld is a special case of energy operator.
Remark 3.28. We always have translation covariance, that is [T, a(z)] = ∂za(z). Equation (3.5) means scale covariance. The most general covariance is given by equation (3.4), that is covariance with respect to the whole conformal group. Let us write
X −n−∆a a(z) = anz , (3.6) n∈−∆a+Z if H(a) = ∆aa. Then equation (3.5) is equivalent to
[H, an] = −nan or [H, a(n)] = (∆a − n − 1)a(n). (3.7) Proposition 3.29. We have a) ∆|0i = 0; b) ; ∆a(n)b = ∆a + ∆b − n − 1 c) ∆T a = ∆a + 1; d) writing a(z) and b(z) in the form (3.6), we get X m − 1 + ∆a [a , b ] = (a b) . m n j (j) m+n j∈Z+ 17 Proof. For a) we have 0 = [H, 1V ] = [H, |0i(z)] = (z∂z + ∆|0i)1V , then ∆|0i = 0. Moreover
H(a(n)b) = [H, a(n)]b + a(n)H(b) = (∆a − n − 1)a(n)b + ∆ba(n)b,
that proves b). Now, T a = a−2|0i, then
∆T a = ∆a + ∆|0i − (−2) − 1 = ∆a + 1 and so c) is proved. The part d) follows from the commutator formula if we replace by a(m) am+1−∆a and by . b(n) bn+1−∆b Remark 3.30. Note that a ∈ V and the corresponding eld a(z) has conformal weight with respect to energy-momentum eld if
[L(z), a(w)] = ∂wa(w)δ(z, w) + ∆a(w)∂wδ(z, w) + ...,
but [L(z),L(w)] = ∂wL(w)δ(z, w) + 2L(w)∂wδ(z, w), then ∆L = 2. Remark 3.31. Conformal weight is a good book-keeping device X ∂j δ(z, w) [a(z), b(w)] = (a b)(w) w . (j) j! c Example 3.32. V 3 L = L(−1)|0i = L−2|0i. The corresponding eld is the Virasoro eld. Hence the only thing to check is that is a diagonalizable operator. The vectors , with L0 L−j1−1 ...L−js−1|0i c j1 ≥ ... ≥ js ≥ 1, form a basis of V , since
V ir = V ir+ ⊕ V ir−, where P . Each of this vector is an eigenvector of : V ir− = n≥−1 Ln H = L0 (3.8) L0(L−j1−1 ...L−js−1|0i) = (j1 + . . . js + s)L−j1−1 ...L−js−1|0i. The vectors have energy . L−j1−1 ...L−js−1|0i j1 + . . . js + s
Exercise 16. Compute the dimension of the N-th eigenspace of L0. Solution. From (3.8) it follows that the -th eigenspace of is spanned by vectors N L0 L−j1 ...L−js |0i where
j1 ≥ · · · ≥ js ≥ 2 and N = j1 + ··· + js . So its dimension is the number of partitions of N in a sum of integers ≥ 2. Let p(N) denote the number of partitions of N. Then the number of partitions of N in a sum of integers ≥ 2 is p(N) − p(N − 1). Indeed, any partition containing 1 can be written as 1 plus a partition of N − 1. De nition 3.33. Given a formal distribution a(z) ∈ U[[z, z−1]], we de ne its formal Fourier transform to be λ λz Fz a(z) = Resz e a(z). P −n−1 In particular, if a(z) = a(n)z , then X X λm X λn F λa(z) = Res a zm−n−1 = a . z z (n) m! (n) n! m∈Z+ n∈Z n∈Z+ So λ −1 is a linear map. Fz : U[[z, z ]] −→ U[[λ]] Proposition 3.34. λ satis es the following properties: Fz a) λ λ ; Fz ∂za(z) = −λFz a(z) b) λ zT λ+T , if ; Fz (e a(z)) = Fz a(z) a(z) ∈ U((z)) c) λ −λ ; Fz a(−z) = −Fz a(z) d) λ n λw n. Fz ∂wδ(z, w) = e λ
Proof. Since Resz ∂z(·) = 0, we get λ λz λz λz λ Fz ∂za(z) = Resz e ∂za(z) = − Res(∂ze )a(z) = −λ Resz e a(z) = −λFz a(z), that proves a), while b) is clear. Let us prove d);
λ n λz n n λz n λw n λw Fz ∂wδ(z, w) = Resz e ∂wδ(z, w) = ∂w Res e δ(z, w) = ∂we = λ e . Exercise 17. Prove part c). 18 Solution. De ne the even formal distribution b(z) = eλza(−z) + e−λza(z).
Then b(z) has no odd powers, this means that Resz b(z) = 0. So λz −λz Resz e a(−z) + Resz e a(z) = 0. De nition 3.35. Let V be a vertex algebra, with a −→ a(z) the state- eld correspondence, de ne the λ-bracket on V by λ [aλb] = Fz a(z)b. If we apply λ to using the commutator formula we get Fz [a(z), b(w)]c X ∂j δ(z, w) X ∂j δ(z, w) F λ[a(z), b(w)]c = F λ (a b)(w)c w = (a b)(w)cF λ w = z z (j) j! (j) z j! j∈Z+ j∈Z+ X λj = eλw (a b)c = eλw[a b](w)c. (j) j! λ j∈Z+ Since , applying λ we get [a(z), b(w)]c = a(z)b(w)c − p(a, b)b(w)a(z)c Fz λw [aλb(w)c] = p(a, b)b(w)[aλc] + e [aλb](w)c. (3.9) Proposition 3.36. The λ-bracket on V has the following properties:
(sesquilinearity) [T aλb] = −λ[aλb] and [aλT b] = (T + λ)[aλb]; (skewsymmetry) [aλb] = −p(a, b)[b−λ−T a]; (Jacobi identity) [aλ[bµc]] = p(a, b)[bµ[aλc]] + [[aλb]λ+µc]. Proof. We prove only the Jacobi identity. Let us apply µ to equation (3.9), then we get Fw µ λw (λ+µ)w [aλ[bµc] = p(a, b)[bµ[aλc]] + Fw e [aλb](w)c = p(a, b)[bµ[aλc]] + Resw e [aλb](w)c =
= p(a, b)[bµ[aλc]] + [[aλb]λ+µc]. Exercise 18. Prove the remaining parts of the proposition. Solution. Let us prove the sesquilinearity property. In the rst case we have λ λ λ [T aλb] = Fz (T a)(z)b = Fz ∂za(z)b = −λFz a(z)b = −λ[aλb]. For the second case we have λ λ λ λ λ [aλT b] = Fz a(z)T b = Fz [a(z),T ]b + Fz T (a(z)b) = −Fz ∂za(z)b + ∂zFz a(z)b = (λ + ∂)[aλb]. To prove the skewsymmetry property we have only to apply λ to the relation Fz λ zT a(z)b = p(a, b)Fz e b(−z)a. Then we get λ+T −λ−T [aλb] = p(a, b)Fz b(−z)a = −p(a, b)Fz b(z)a = −p(a, b)[b−λ−T a], where we have used properties b) and c) of the formal Fourier transform. De nition 3.37. A C[T ]-module V endowed with the λ-bracket V ⊗V −→ V [λ] such that sesquilinearity, skewsymmetry and Jacobi identity hold is called a Lie conformal (super)algebra.
Theorem 3.38. Any simple and nitely generated as C[T ]-module Lie conformal algebra is one of the following i) , where , , or Curg = C[T ] ⊗ g [aλb] = [a, b] a, b ∈ g ii) , where . V ir = C[T ]L [LλL] = (T + 2λ)L We want to arrive to another equivalent de nition of vertex algebra. If we write X λn [a b] = (a b) λ (n) n! n∈Z+ we need only to consider in addition the product, called normally ordered product,
: ab := a(−1)b. 19 Let us recall that T n : ab := a b, n ∈ . n! (−n−1) Z+ If we compare the coe cients of both sides of equation (3.9) we get Z λ [aλ : bc :] = p(a, b): b[aλc] : + : [aλb]c : + [[aλb]µc]dµ. (3.10) 0 Let us prove this formula: we need only to compute the constant term of X λmwm X eλw[a b](w)c = [a b] cw−n−1, λ m! λ (n) m∈Z+ n that is equal to X λm X λm+1 Z λ [aλb](m−1)c =: [aλb]c : + [aλb](m)c =: [aλb]c : + [[aλb]µc]dµ. m! (m + 1)! 0 m∈Z+ m∈Z+ Equation (3.10) is called non-commutative Wick formula. Proposition 3.39. The normally ordered product is quasi-commutative, that is Z 0 : ab := p(a, b): ba : + [aλb]dλ, −T and is quasi-associative, that is, X X :: ab : c :=: a : bc :: + a(−j−2)(b(j)c) + p(a, b) b(−j−2)(a(j)c). j∈Z+ j∈Z+ Exercise 19. Prove the proposition: deduce quasi-commutativity from skewsymmetry of V and quasi- associativity from the −1-product formula : a(z)b(z) := (a(−1)b)(z), by taking the costant term in z. Solution. The skewsymmetry in V tells us that b(z)a = −p(a, b)eT z(a(−z)b). Taking the coe cient of z−n−1 we get the relation X T j b a = −p(a, b) (−1)j+n (a b) (n) j! (n+j) j∈Z+ and for n = −1 we get X T j X T j+1 b a = p(a, b) (−1)j (a b) = p(a, b)a b + p(a, b) (−1)j+1 (a b) (−1) j! (j−1) (−1) (j + 1)! (j) j∈Z+ j∈Z+ that means Z 0 : ba := p(a, b): ab : −p(a, b) [aλb]dλ. −T For the quasi-associativity we have that (a(−1)b)(−1) is the constant term of : a(z)b(z) := (a(−1)b)(z), but : a(z)b(z) := a(z)+b(z) + p(a, b)b(z)a(z)−, then X X −m−n−2 X X −m−n−2 : a(z)b(z) : = a(n)b(m)z + p(a, b) b(m)a(n)z = m∈Z n≤−1 m∈Z n∈Z+ X X −m−1 X X −m−1 = a(n)b(m−n−1)z + p(a, b) b(m−n−1)a(n)z . m∈Z n≤−1 m∈Z n∈Z+ Taking the constant term we get X X X (a(−1)b)(−1) = a(n)b(−n−2) + p(a, b) b(−n−2)a(n) = a(−1)b(−1) + a(n)b(−n−2)+ n≤−1 n∈Z+ n≤−2 X X X + p(a, b) b(−n−2)a(n) = a(−1)b(−1) + a(−n−2)b(n) + p(a, b) b(−n−2)a(n). n∈Z+ n∈Z+ n∈Z+ Finally, applyng to c we get X X (a(−1)b)(−1)c = a(−1)(b(−1)c) + a(−n−2)(b(n)c) + p(a, b) b(−n−2)(a(n)c). n∈Z+ n∈Z+ 20 We are ready now for the fourth de nition of vertex algebra.
De nition 3.40. A vertex algebra is the data of (V, |0i,T, [aλb], : ab :), where (V,T, [aλb]) is a Lie conformal algebra, (V, |0i,T, : ab :) is a quasi-commutative, quasi-associative unital dierential algebra and the λ-bracket and the normally ordered product are related by the non-commutative Wick formula. For the equivalence with the other de nitions we remind to [BK03].
A homomorphism of vertex algebras ϕ : V1 −→ V2 is a linear map preserving all products such that ϕ(|0i1) = |0i2.A subalgebra U ⊂ V is a subspace containing |0i and closed under all products. In particular TU ⊂ U, since T a = a(−2)|0i. An ideal I ⊂ V is a subspace, not containing |0i, such that and for all and . Indeed, from the skewsymmetry, it follows that every TI ⊂ I a(n)I ⊂ I a ∈ V n ∈ Z ideal is a two-sided ideal. Example 3.41. Let be a vertex algebra, then an j -invariant and -invariant subspace (V, |0i,T, F) a(n) T U not containing |0i is an ideal since all elds are n-th products of aj(z), hence their coe cients are combination of products of j 's. a(n)
k bg Example 3.42. Let V (g) = U(bg) (U(g)g[t] + U(g)(K − k)) = Ind k, then the ideals are the b b bg[t]+CK C -invariant subspaces which are -invariant. bg T c Example 3.43. V is a Virasoro module with T = L−1, so the ideals are the Virasoro-submodules. Proposition 3.44. V c contains a unique maximal ideal. c Proof. L0 is diagonalizable on V with positive eigenvalues and the only eigenvector of eigenvalue 0 is . So any ideal is -invariant and -graded. Then the sum of proper ideal is again a proper ideal |0i L0 Z+ and the claim follows.
The corresponding simple vertex algebra is denoted Vc. Here are some methods of constructing new vertex algebras: a) subalgebras and quotient algebras; b) xed point set of a group of automorphisms Γ of V ; if |Γ| < ∞ this is called an orbifold vertex algebra; c) tensor product of vertex algebras;
i Exercise 20. The tensor product of vertex algebras (Vi, |0ii, T, a (z)), i = 1, 2, is the following vertex algebra: 1 1 X −m−n−2 (V = V1 ⊗V2, |0i = |0i1 ⊗|0i2,T = T1 ⊗ V2 + V1 ⊗T2, a⊗b(z) = a(z)⊗b(z) = (a(m) ⊗b(n))z ). Solution. We should check the axioms of vertex algebra for the tensor product. For the vacuum axioms we have 1 1 T |0i = (T1 ⊗ V2 + V1 ⊗ T2)(|0i1 ⊗ |0i2) = 0 ⊗ |0i2 + |0i1 ⊗ 0 = 0 and a ⊗ b(z)|0i = a ⊗ b(z)(|0i ⊗ |0i ) = a(z)|0i ⊗ b(z)|0i = a ⊗ b. |z=0 1 2 |z=0 1 2|z=0 For the translation covariance we have 1 1 [T, a ⊗ b(z)] = [T1 ⊗ V2 , a ⊗ b(z)] + [ V1 ⊗ T2, a ⊗ b(z)] = [T1, a(z)] ⊗ b(z) + a(z) ⊗ [T2, b(z)] =
= ∂za(z) ⊗ b(z) + a(z) ⊗ ∂zb(z) = ∂z(a(z) ⊗ b(z)).
Finally, let us prove locality. If we apply [a1 ⊗ b1(z), a2 ⊗ b2(z)] to a monomial v1 ⊗ v2 we get
[a1 ⊗ b1(z), a2 ⊗ b2(z)](v1 ⊗ v2) =
= (a1 ⊗ b1(z))(a2 ⊗ b2(z))(v1 ⊗ v2) − (a2 ⊗ b2(z))(a1 ⊗ b1(z))(v1 ⊗ v2) =
= a1 ⊗ b1(z)(a2(z)v1 ⊗ b2(z)v2) − a2 ⊗ b2(z)(a1(z)v1 ⊗ b1(z)v2) =
= a1(z)a2(z)v1 ⊗ b1(z)b2(z)v2 − a2(z)a1(z)v1 ⊗ b2(z)b1(z)v2 =
= a1(z)a2(z)v1 ⊗ b1(z)b2(z)v2 − a2(z)a1(z)v1 ⊗ b1(z)b2(z)v2
+ a2(z)a1(z)v1 ⊗ b1(z)b2(z)v2 − a2(z)a1(z)v1 ⊗ b2(z)b1(z)v2 =
= [a1(z), a2(z)]v1 ⊗ b1(z)b2(z)v2 + a2(z)a1(z)v1 ⊗ [b1(z), b2(z)]v2. n Then multiplying for (z − w) for su ciently large n gives zero. d) common kernel of one or several derivation of a vertex algebra, that is a linear operator D such that D(a(n)b) = (Da)(n)b + a(n)(Db); 21 e) coset vertex algebra: given a subspace U ⊂ V , its centralizer C(U) = {b ∈ V | [a(z), b(w)] = 0∀a ∈ V } is a subalgebra of V .
Exercise 21. For any a ∈ V , a(0) is a derivation.
Solution. Just use the commutator formula with m = 0.
4. An example: the free boson The vector space is , the vacuum vector is and P −n−1 , V = C[xn, n ∈ Z+] |0i = 1 F = {α(z) = n∈ αnz } where Z ∂ ∂x n > 0 αn = n −nx−n n ≤ 0.
Since [αm, αn] = mδm,−n, it follows that [α(z), α(w)] = ∂wδ(z, w), so we get locality axiom. For α = α−1|0i we have [αλα] = λ|0i.
Exercise 22. [T, αn] = −nαn−1, write T . Solution. We claim that P . Indeed, T = n≥2 α−nαn−1
[α−nαn−1, αm] = [α−n, αm]αn−1 + α−n[αn−1, αm] = −nδn,mαn−1 + (n − 1)δn−1,−mα−n. So, if m > 0, then the second term vanishes and we get X [T, αm] = − nδn,mαn−1 = −mαm−1, n≥2 while, if m < 0 the rst term vanishes and we get X [T, αm] = (n − 1)δn−1,−mα−n = −mαm−1. n≥2 Example 4.1. We have X :: αα : α : − : α : αα ::= α(−j−2)(α(j)α) = 2α(−2)|0i. j∈Z+
The corresponding eld is 2∂zα(z) 6= 0. Let 1 , then L = 2 : αα : 1 1 1 1 Z λ [αλL] = [αλ : αα :] = :[αλα]α : + : α[αλα] : + [[αλα]µα]dµ = λα, 2 2 2 2 0
so [Lλα] = (λ + T )α. Moreover 1 1 1 1 Z λ [LλL] = [Lλ : αα :] = :[Lλα]α : + : α[Lλα] : + [[Lλα]µα]dµ = 2 2 2 2 0 1 1 1 Z λ = :(λ + T )αα : + : α(λ + T )α : + [(λ + T )αµα]dµ = 2 2 2 0 1 1 1 Z λ = : T αα : + : αT α : +λ : αα : + (λ − µ)[αµα]dµ = 2 2 2 0 1 1 Z λ λ3 = (T + 2λ)( : αα :) + (λµ − µ2)dµ|0i = (T + 2λ)L + |0i. 2 2 0 24 Thus L de nes a Virasoro algebra with central charge 1 and α is an eigenvector of conformal weight 1. 2 Recall that if a has conformal weight ∆, then [Lλa] = (T + ∆λ)a + o(λ ).
De nition 4.2. a is called primary if [Lλa] = (T + ∆λ)a. Exercise 23. is primary if and only if , for all . a Lna = δn,0∆a n ∈ Z+ 22 Solution. By de nition, it follows that X X λm X λm [L a] = F λL(z)a = Res eλzL(z)a = Res (L )azn+m = L a. λ z z z m! n m! −m−1 m∈Z+ n∈Z m∈Z+ m Comparing the coe cient of λ we have that [Lλa] = (T + ∆λ)a if and only if Lna = δn,0a for all . n ∈ Z+ Let us denote this algebra B. Proposition 4.3. is a vertex algebra with energy-momentum vector 1 , being a primary B L = 2 : αα : α element of conformal weight 1. Proof. We have . Moreover is diagonalizable with eigenvalues in and L−1 = T L0 Z+ X Bn = span(α−ns . . . α−n1 |0i | 0 ≤ n1 ≤ ... ≤ ns, ni = n). has dimension , the number of partition of . Moreover , then is a simple vertex Bn p(n) n B0 = C|0i B algebra. B is called vertex algebra of free boson. We note that V k(g) is a generalization of B: the commutator on is de ned as bg [am, bn] = [a, b]m+n + mδm,−nK. Then we have V 1(Cα) = B. There is a similar construction of an energy-momentum vector L, called the Sugawara construction.
Theorem 4.4. Let g be a simple nite dimensional Lie algebra with a non-degenerate bilinear form P (· | ·). Let {ai} and {bi} be dual basis of g with respect to this form. Let Ω = aibi be the Casimir element and let ˇ be the eigenvalue of on . Let k , , where via , 2h Ω g V = V (g) k ∈ C g ⊂ V a −→ a−1|0i and let 1 P , then T = 2 : aibi : ˇ a) [aλT ] = (k + h)λa, a ∈ g; b) L = 1 T is an energy-momentum eld with c = k dim g , provided that k 6= hˇ; k+hˇ k+hˇ c) each a ∈ g is primary of conformal weight 1. Exercise 24. Prove the theorem. Solution. By non-commutative Wick formula (3.10) we get
1 X 1 X 1 X Z λ [a T ] = :[a a ]b : + : a [a b ] : + [[a a ] b ]dµ = λ 2 λ i i 2 i λ i 2 λ i µ i i i i 0 1 X λk X 1 X λk X = :[a, a ]b : + (a | a )b + : a [a, b ] : + (a | b )a + 2 i i 2 i i 2 i i 2 i i i i i i 1 X Z λ + ([[a, a ], b ] + µk([a, a ] | b )) dµ. 2 i i i i i 0 We recall that, for any a ∈ g, we have X X a = (a | bi)ai = (a | ai)bi, i i thus we obtain X X X X :[a, ai]bi := ([a, ai] | bj): ajbi := − ([a, bj] | ai): ajbi :=− : aj[a, bj]: . i i,j i,j j P Furthermore, we not that ([a, ai] | ai)bi = tr ad a = 0. Indeed, using the dual basis, we also have P P tr ad a = ([a, bi] | ai) = − ([a, ai] | bi) = − tr ad a. Then we get λ X λ X λ [a T ] = λka + [[a, a ], b ] = λka + [a , [b , a]] = λka + Ω(a) = (k + hˇ)λa, λ 2 i i 2 i i 2 i i P where we used Jacobi identity and the fact that [ai, bi] = 0. Part c) follows easily, since we have [aλL] = λa, then by skewsymmetry we get [Lλa] = (∂ + λ)a, for all a ∈ g. 23 Finally, we prove part b). Using again non-commutative Wick formula, we get ! 1 X 1 X Z λ [L L] = [L : a b :] = :[L a ]b : + : a [L b ] : + [[L a ] b ]dµ = λ ˇ λ i i ˇ λ i i i λ i λ i µ i 2(k + h) i 2(k + h) i 0 ! 1 X Z λ = :(λ + ∂)a · b : + : a (λ + ∂)b : + (λ − µ)([a , b ] + µk(a | b ))dµ = ˇ i i i i i i i i 2(k + h) i 0 k dim g Z λ k dim g = (∂ + 2λ)L + (λ − µ)µdµ = (∂ + 2λ)L + λ3. 2(k + hˇ) 0 12(k + hˇ) 5. Poisson Vertex Algebras We want to de ne the notion of quasiclassical limit of a family of vertex algebras. Let
(V~, |0i~,T~, [aλb]~, : ab :~) be a family of vertex algebras, meaning that is a free module over such that all axioms hold and V~ C[[~]] [V V ] ⊂ V [λ]. The quasiclassical limit is V = V . We have that V 3 1 = im(|0i ), T induces an ~λ ~ ~ ~ ~ ~V~ ~ ~ operator ∂ on V and : ab : induces a product on V. ˜ We de ne a Poisson λ-bracket {aλb} on V as follows: let a˜ and b be preimages of a, b ∈ V V in V and let ~ [˜a ˜b] λ V {aλb} = ∈ ~V = V. ~ Proposition 5.1. (V, 1, ∂, ab, {aλb}) is a Poisson vertex algebra, that is the following axioms hold: a) (V, 1, ∂, ab) is a unital commutative associative dierential algebra; b) (V, ∂, {aλb}) is a Lie conformal algebra; c) {·λ·} and · are related by the left Leibniz rule: {aλbc} = {aλb}c + b{aλc}. Proof. By quasi-commutativity we have in V~ Z 0 : ab :~ − : ba :~= [aλb]~dλ. −T ~ But , so in the limit , we have . Similarly, quasi-associativity implies [aλb]~ ∈ ~V~ ~ → 0 ab = ba associativity when ~ → 0. This proves part a). Part b) is obvious since all axioms of Lie conformal algebras are homogeneous. For c) if we divide by ~ the non-commutative Wick formula and take the limit we get the Leibniz rule.
Exercise 25. Deduce from {aλb} = −{b−λ−∂ a}, that
{bcλa} = {bλ+∂ a}→c + {cλ+∂ a}→b. This is called right Leibniz rule. Solution. Using our assumption and left Leibniz rule we get
{bcλa} = −{a−λ−∂ bc} = −{a−λ−∂ b}c − {a−λ−∂ c}b = {bλ+∂ a}→c + {cλ+∂ a}→b. Example 5.2. Let be a regular formal distibution Lie algebra, in particular is a (g, F,T ) g = bg0 Virasoro algebra and is the annihilator subalgebra (for example and P ), g− bg− = g0[t] V ir− = n≥−1 CLn then we have the associative vertex algebra U(g) . Let us consider with bracket V (g, F) = U(g)g− g~ [a, b] = ~[a, b], so ~ V (g , F) = U [[ ]](g~) ~ ~ C ~ UC[[~]](g~)g~− over C[[~]] is a family of vertex algebras. The quasiclassical limit of the family is denoted by V(g, F) and is constructed as follows in the case : choose a basis i of so that P i n −n−1 g = bg0 {a }i∈J g0 F = { (a t )z }i∈J and let u = aj |0i in V (g , F) and u be its image in V(g, F), then ej (−1) ~ ~ j (n) V(g, F) = C[uj | j ∈ J, n ∈ Z+] with (n) (n+1) and P k k, where i j P k k and extend by sesquilinearity ∂uj = uj {uiλuj} = k∈J cija [a , a ] = k cija and both Leibniz rules. 24 A special case is when . We set , where P −n−1 with g0 = C B~ = V~(g~, {α(z)}) α(z) = αnz . The quasiclassical limit is a Poisson vertex algebra called Gardner-Fadeev- [αn, αm]~ = ~mδm,−n B Zakharov Poisson vertex algebra. Exercise 26. ∼ (n) with (n) (n+1) and . B = C[u | n ∈ Z+] ∂u = u {uλu} = λ Proof. Since is one dimensional, by previous example we have ∼ (n) with (n) g0 B = C[u | n ∈ Z+] ∂u = u(n+1), where u is the image in B of u = α |0i ∈ B . Moreover, [u u] = λ, then we get {u u} = λ. e −1 ~ eλe ~ λ Example 5.3. The Virasoro Poisson vertex algebra is (n) with (n) (n+1) and V = C[u | n ∈ Z+] ∂u = u λ3 . {uλu} = (∂ + 2λ)u + 12 c Let M be a nite dimensional manifold and V be the algebra of C∞ functions on M. Introduce in V a Poisson bracket {·, ·} such that (1) V is a unital commutative associative algebra; (2) Lie algebra axioms hold; (3) Leibniz rule holds: {a, bc} = {a, b}c + b{a, c}. Also choose a function h ∈ V, called Hamiltonian function, then a system of Hamiltonian equations on {ui}i∈I (here ui's are "topological" generators of V) is: du i = {h, u }, i ∈ I (5.1) dt i and . A function is called a conservation law if df on the trajectories. This can be ui = ui(x, t) f dt = 0 expressed in terms of the Poisson bracket as . Indeed, by the Leibniz rule, df , . {h, f} = 0 dt = {h, f} f ∈ V The system (5.1) is called integrable if we have "su ciently many" commuting conservation laws. In the case of an in nite dimensional Hamiltonian system, the manifold M is replaced by Mf = P tn (n) Map(I,M). If ui are coordinates in M, then on M we let ui(t) = u . So, coordinates in M f n∈Z+ n! i f are (n), . We take for the space of functions on the algebra (n) . Thus we ui i ∈ I Mf V = C[ui |∈ I, n ∈ Z+] have a dierential algebra, since (n) (n+1). One consider usually local Poisson brackets ∂ui = ui
{ui(x), uj(y)} = Bij(u(y), ∂y)δ(x, y), (5.2) where R . M f(x)δ(x, y) = f(y) Example 5.4. We can just de ne {u(x), u(y)} = δ(x, y). By bilinearity and Leibniz rule we can extend this bracket to any P,Q ∈ V: X ∂P (x) ∂Q(y) {P (x),Q(y)} = ∂α∂β{u (x), u (y)}. (5.3) (α) (β) x y i j α,β,i,j ∂ui ∂uj
This is called Poisson bracket if it is skewsymmetric. This happens if and only if Bij is a skewsymmetric dierential operator and satis es the Jacobi identity. Hamiltonian functions and integrals of motion are not elements of V, but they are local functionals: Z IP = P (x)dx, P ∈ V. M Taking an algebraic point of view, local functionals are elements of V ∂V. Given h ∈ V we denote its R image in V ∂V by h. Using bilinearity and integration by part we get ZZ X ∂P (x) ∂Q(y) ZZ X δP δQ {I ,I } = ∂α∂β{u (x), u (y)}dxdy = {u (x), u (y)}dxdy, P Q (α) (β) x y i j i j δui δuj α,β,i,j ∂ui ∂uj i,j where δP X ∂P = (−∂)n . (n) δui ∂u n∈Z+ i Finally, using (5.2) we can write Z δP δQ {R P, R Q} = B . δu δu This is a Poisson bracket if V ∂V is a Lie algebra with the induced bracket. R A Hamiltonian functional is h ∈ V ∂V and the Hamiltonian equation is duj R δh = { h, uj} = B ; dt δuj 25 R f is a conserved quantity (integral of motion) if {R h, R f} = 0. The Hamiltonian equation is called integrable if there exist in nitely many linearly independent commuting integrals of motion. Taking the formal Fourier transform of both sides of (5.2) we get: Z λ(x−y) {uiλuj} = e {ui(x), uj(y)}dx. Applying the formal Fourier transform to (5.3) gives X ∂Q ∂P {P Q} = (∂ + λ)n{u u } (−∂ − λ)m . λ (n) iλ+∂ j → (m) i,j∈I ∂uj ∂ui m,n∈Z+ Proposition 5.5. We have
(1) this λ-bracket on V de nes the structure of a Poisson vertex algebra, where Bij(∂) = {uj∂ ui}; R (2) an Hamiltonian function is h ∈ V ∂V and the corresponding Hamiltonian system is
dui R = { hλui} ; dt λ=0 R (3) the integrals of motion are elements f ∈ V ∂V such that R R in V { fλ h} λ=0 = 0 ∂V. Lemma 5.6. If is a Poisson vertex algebra, consider , then V {P,Q} = {PλQ}|λ=0 (1) ∂V is a two-sided ideal for this bracket, hence we have an induced bracket on V ∂V; (2) V ∂V is a Lie algebra and V is a left V ∂V-module. Proof. Part a) follows from the sequilinearity axioms, while part b) follows easily because of the axiom of Lie conformal algebras. Example 5.7. Let (n) with , then V = C[u , n ∈ Z+] {uλu} = λ R u, R u2, R u3 − (u0)2,... form an in nite sequence of commuting elements in V ∂V and du = {R u3 − (u0)2, u} = 3uu0 + u000. dt In particular, in V ∂V, we have Z δf δh {R h, R f} = ∂ . δu δu 6. Representation theory De nition 6.1. A representation of a vertex algebra V in a vector (super)space M is a linear map from to a space of -valued quantum elds, M P M −n−1, such that V End M a −→ a (z) = an z M (1) |0i (z) = 1M ; j M M n M M n P M ∂wδ(z,w) (2) a (z)b (z)iz,w(z − w) − p(a, b)b (w)a (z)iw,z(z − w) = (a(n+j)b) (w) . j∈Z+ j! If we put n = 0, we get the commutator formula X ∂j δ(z, w) [aM (z), bM (w)] = (a b)M (w) w . (6.1) (j) j! j∈Z M M M From (6.1) and a (w)(n)b (w) = (a(n)b) (w) follows Borcherds identity. Example 6.2. If we take M = V , then M is a representation by a −→ a(z). Example 6.3. Take V = V k(g) (or, more generally, V (g, F)), then any representation should satisfy the commutator formula. If a(z) = P(atn)z−n−1 and b(z) = P(btn)z−n−1, then (6.1) becomes M M M [a (z), b (w)] = [a, b] (w)δ(z, w) + k(a|b)1M ∂wδ(z, w). This formula means that we have a -module with and M , for , and bg M K = k an (m) = 0 m ∈ M a ∈ g . So any k -module is an extension to the whole vertex algebra of a restricted -module n >> 0 V (g) bg M with K = k. 26 k Recall that we had another construction of V (g) as a subspace of the completed U(bg) (K − k). So, if we have a restricted -module with , we can extend it to the completion of U(g) , bg K = k1M b (K − k) which contains V k(g). In conclusion, any k -module is obtained from a restricted -module with 1 . V (g) bg M K = k M What are the restricted -modules with ? bg K = k Example 6.4. Let U be an arbitrary g-module, then Indbg U can be extended to g[t] by letting g[t]+CK n , for and . This is a highest weight module if and only if is a highest (gt )|U = 0 n >> 0 K|U = k U weight g-module.
k k Let I be the unique maximal ideal of V (g), then Vk(g) = V (g) I is a simple vertex algebra. What are its representations? They are the V k(g)-modules M such that IM M = 0. So we need to study I: if a ∈ I, then aM (z) = 0. Example 6.5. Consider the case , then is an irreducible -module with and k ∈ Z+ Vk(g) bg K = k1 for and . Let be the Chevalley generators of and let , where an|0i = 0 a ∈ V n ∈ Z+ ei, fi, hi g (θ|θ) = 2 θ is the highest root of g. Then g|0i = 0, so fi|0i = 0, hi|0i = 0 and ei|0i = 0, for i = 1, . . . , r = rank(g). Also and one set , −1 and . Then and . Hence g ⊂ bg e0 = e−θt f0 = eθt h0 = k − θ e0|0i = 0 h0|0i = k|0i is an irreducible highest weight module over with highest weight vector of weight such Vk(g) bg |0i Λ0 that Λ0(hi) = kδ0,i. We denote this module by L(Λ0). Proposition 6.6. is a left -module generated by k+1 . I bg f0 |0i Hence, for , a k -module is actually a -module if and only if M k+1 . Thus k ∈ Z+ V (g) M Vk(g) (eθ (z)) = 0 we need to nd all restricted -modules with such that k+1 in . bg M K = k (eθ(z)) = 0 M k Recall that Vk(g) = V (g) I, where I is the maximal submodule. But G, the group with Lie algebra g, acts by of such that is xed and is an invariant subalgebra. Hence acts by automorphisms ... bg K g[t] G of k , hence of , so σ k+1 on for each . V (g) Vk(g) eθ (z) = 0 M σ ∈ G What does the relation a(z)k+1 = 0, for a ∈ g, mean? If a(z) = P(atn)z−n−1, then a(z)k+1 = 0 if and only if X n n (at 1 ) ... (at k+1 ) = 0 on M, ∀N ∈ Z. (6.2) n1+...+nk+1=N We can ask which of the irreducible -modules , with , are -modules. We know that bg L(Λ) Λ(K) = k Vkg L(Λ0) is. Apply both sides of (6.2) to vΛ and take N = −k − 1, then we have a term −1 k (at ) vΛ + (...)vΛ, where in the parenthesis one of the factors is n with . Take , then n for at n ∈ Z+ a = eθ (eθt )vΛ = 0 , that means −1 k+1 . So k+1 . If we now take and , , in n ∈ Z+ (eθt ) vΛ = 0 f0 vΛ = 0 N = 0 a = ealpha α > 0 (6.2) we get k+1 k+1 e−α vΛ + ... (e−αt)vΛ ... = e−α vΛ = 0. Then k+1 . This two conditions are satis ed if and only if , for all . fi vΛ = 0 ∆(hi) ∈ Z+ i Then only the integrable -modules , with , can be -modules. bg L(Λ) Λ(K) = k Vk(g) Theorem 6.7. Let , then all irreducible -modules are . Moreover, every k ∈ Z+ Vk(g) {L(∆)} ∆(K)=k ∆(hi)∈Z+ Vk(g)-module is a direct sum of irreducible modules. It remains to prove that:
i) any irreducible Vk(g)-module is one of the L(∆); ii) any integrable L(∆) is indeed a Vk(g)-module; iii) complete reducibility. c In the case of Virasoro vertex algebras V and Vc we have that any restricted module M over the Virasoro algebra with C = c is a V c-module.
c c Theorem 6.8. Vc = V , that is V is an irreducible module over the Virasoro algebra, if and only if c is not of the form 6(p − q)2 c = 1 − , (6.3) p,q pq where p, q ∈ Z, p, q ≥ 2 and (p, q) = 1. 27 Theorem 6.9. If c is of the form (6.3), then Vc has nitely many irreducible modules, all of them highest weight modules, and any Vc-module is a direct sum of irreducibles.
If we have C = c1, Lnvc,h = 0, for n > 0, and L0vc,h = hvc,h, then we call this module L(c, h). Let H ∈ End V be an energy operator, that we remember is a diagonalizable operator, such that any H-eigenvector a has eigenvalue ∆a. Setting
X −n−∆a a(z) = anz , n∈−∆a+Z we have . Assume that . [H, an] = −nan spec H ⊂ Z+ De nition 6.10. A positive energy V-module is a graded representation such that M M = ⊕j∈Z+ Mj M (6.4) an Mj ⊂ Mj−n. The condition means that M for and M . For example, if is an energy an M0 = 0 n > 0 a0 M0 ⊂ M0 L ∈ V momentum state, then H = L0 is an energy operator and if we have a highest weight representation M of k or c, then the eigenspace decomposition with respect to M V (g) V L0 M M = M˜ j j∈h+Z+ M is such that (6.4) holds (respect to the usual decomposition we have Mj = M˜ j−h). Write a (z) = P M −n−∆a , then Borcherds identity reads an z X n X m + ∆a − 1 (−1)j aM bM − (−1)nbM aM = (a b)M . j m+n−j k+j−n k−j m+j j (n+j) m+k j∈Z+ j∈Z+
Pick v ∈ M0 and apply both sides of Borcherds identity to v. For m = 1, n = −1 and k = −1 we get X ∆a aM bM (v) = (a b)(v). (6.5) 0 0 j (j−1) j∈Z+
Exercise 27. Let m = 1, n = −2 and k = −1, then for v ∈ M0, Borcherds identity is X ∆a (a b)(v) = 0. (6.6) j (j−2) j∈Z+ Solution. The left hand side of Borcherds identity applied to v is zero since both k + j − n and m + j are greater than zero. The right hand side is X ∆a (a b)(v), j (j−2) j∈Z+ thus proving identity (6.6). De ne the product ∗ on V by RHS of (6.5): X ∆a a ∗ b = a b. j (j−1) j∈Z+ Exercise 28. Prove that X ∆a a b = ((T + H)a) ∗ b. j (j−2) j∈Z+
Solution. Recall that (T a)(n) = −na(n−1) and (Ha)(n) = (∆a − n − 1)a(n). Then, using the de nition of the product ∗ on V we get: X ∆a X ∆a ∆a ((T + H)a) ∗ b = a b + (∆ − j) − (j + 1) a b . j (j−2) j a j + 1 (j−1) j∈Z+ j∈Z+
The claim follows since ∆a ∆a−j ∆a . j j+1 (j + 1) = j+1 (j + 1) Thus the map , de ned by M is a homomorphism such that the kernel (V, ∗) −→ End M0 a −→ a0 |M0 contains all elements of the form ((T + H)a) ∗ b. 28 Theorem 6.11 (Zhu). Let J be the span of all elements of the form ((T + H)a) ∗ b in V , then J is a two-sided ideal of the algebra (V, ∗) such that Zhu(V ) = (V J, ∗) is a unital associative algebra with 1 = im(|0i).
Theorem 6.12. Given a positive energy V -module M we have a Zhu(V )-module M0 given by a −→ M . Thus we get a functor from the category of positive energy -modules to the category of - a0 V Zhu(V ) |M0 modules. Conversely, given a Zhu(V )-module M0, one can construct the corresponding almost irreducible positive energy V -module M with given M0. This gives an equivalence of categories. If M is a positive energy V -module, then it is called almost irreducible if every nonzero submodule has a nonzero intersection with M0 Proof. Letting n = 0 in Borcherds identity we get the commutator formula: X m + ∆a − 1 [aM , bM ] = (a b)M , m k j (j) m+k j∈Z with . So we get a -graded, with , Lie algebra m, k ∈ Z Z deg an = n M g = gj. j∈Z Let , given a representation of the Zhu algebra in , we get in particular a represen- g∈Z+ = ⊕j∈Z+ gj M0 tation of the Lie algebra in , hence of . g0 M g∈Z+ Exercise 29. Prove that a ∗ b − b ∗ a = [a , b ] . |M0 0 0 |M0 Consider g . Indg = Mf = ⊕j∈ + Mfj ∈Z+ Z Theorem 6.13. Let M , then the whole Borcherds M = f (maximal g-submodule intersecting M0 trivially) identity holds, hence M is a V -module.
Example 6.14. V = V k(g) is spanned by elements (a1t−k1 ) ... (ast−ks )|0i = a1 . . . as |0i, where (−k1) (−ks) i a ∈ g and ki ≥ 1. But vectors of the form a(−2)b + ... ∈ J, hence mod J we can remove factors with , then 1 s . We have a surjective homomorphism ki > 1 V = a(−1) . . . a(−1)|0i + J U(g) −→ Zhu(V k(g)) 1 s 1 s V a . . . a −→ a(−1) . . . a(−1)|0i ∈ J This gives an isomorphism U(g) −→∼ Zhu(V k(g)).
Example 6.15. If V = V c, then Zhu(V c) =∼ C[x]. Example 6.16. We have that . If , then the element Zhu(Vk(g)) = U(g) (maximal ideal) k ∈ Z+ −1 k+1 k is zero in and it goes to k+1 under the previous isomorphism. (eθt ) |0i ∈ V (g) Vk(g) eθ ∈ U(g) So we have the following theorem:
Theorem 6.17. If k ∈ , then Zhu(V (g)) =∼ U(g) k+1 . Z+ k (eθ ) Let k+1 and let be the Lie group corresponding to , so contains all the elements J = (eθ ) G g J σ k+1, . Since is simple, the span of σ is , hence there is a basis of elements 1 d of (eθ ) σ ∈ G g {eθ } g a . . . a g such that alla ai are nilpotent elements of g and (ai)k+1 ∈ J. By Poincaré-Birkho-Witt theorem, 1 k1 d k elments (a ) ... (a ) d span U(g), but these elements with ki ≤ k span U(g) J, then Zhu(Vk) is a nite dimensional semisimple associative algebras, so it is isomorphic to a direct sum of matrix algebras. There is a conjecture that states that if Zhu(V ) is a nite dimensional semisimple associative algebra, then any representation of is sum of irreducibles. The case is still open. V k 6∈ Z+ c ∼ c Example 6.18. Since Vc = V I, then Zhu(Vc) = C[x] (f(x)). But we know that Vc = V unless 2 6(p−q) , , . If , then c is not simple and is a polynomial of c 6= cp,q = 1 − pq p, q ≥ 2 (p, q) = 1 c = cp,q V f(x) degree (p−1)(q−1) . If and , then , so and . If and , 2 p = 3 q = 2 deg f(x) = 1 f(x) = x cp,q = 0 p = 4 q = 3 then deg f(x) = 3 and cp,q = 1/2, this is called Ising model. Exercise 30. Prove that the energy-momentum element L ∈ V goes to a central element in Zhu(V ) = V (). 29 7. On W-algebras
Take k ∈ C, g a simple Lie (super)algebra with a non-degenerate invariant bilinear form (·|·) and f a nilpotent element of g, then we associate to them a so called W -algebra, Wk(g, f). Let us see how this is constructed.
Include f in a sl2-triple (e, h = 2x, f), that is [e, f] = h, [h, e] = 2e and [h, f] = −2f. With respect to ad x, g decomposes as M g = gi, g = g− ⊕ g0 ⊕ g+. 1 i∈ 2 Z We have a non-degenerate pairing between and , so ∗ . In particular ∗ . Let be g− g+ g− = g+ f ∈ g+ G+ the Lie group corresponding to : it acts on and ∗ and we have ∗ . also acts g+ g+ g+ G+ · f = O ⊂ g+ G+ on and ∗ ∗ is the moment map for this action. We consider g µ : g −→ g+ µ−1(O) G+ ∗ which is called the classical Hamiltonian reduction of the Hamiltonian action of G+ on g . The corre- sponding Poisson algebra of functions is called the classical W-algebra. If you take f = 0 you get nite vertex algebras. Some famous examples are c ∼ k Virasoro V = W (sl2, f); N5 =∼ Wk(osp(1|2), f); N2 =∼ Wk(sl(2|1), f); N3 =∼ Wk(osp(3|2), f); N4 =∼ Wk(sl(2|2), f), with f lowest root vector. References [BK03] Bojko Bakalov, Victor Kac, Field algebras, IMRN 3, 2003, 123-159. [Kac96] Victor Kac, Vertex Algebras for Beginners, University Lecture Series, AMS 10, 1996 (2nd ed. 1998).
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