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 january2009. The exposition essentially follows the main reference [Kac96].

Contents

0. A nice picture 11. Wightman's axioms of QFT 12. Calculus of formal distributions 33. Some equivalent denitions of a Vertex Algebra 104. An example: the free boson 225. Poisson Vertex Algebras 246. Representation theory 267. On W-algebras 30References 30

0. A nice picture

There are 4 basic frameworks of physical theories

ClFT

quantization((QFT

quasiclassicallimit

oo

ClM

chiralization

GG

quantization

66 QM

chiralization

WW

quasiclassicallimitoo

(Cl=classical, Q=quantum, M= mechanics, FT=eld theory) and the corresponding algebraic structuresare

PVA

''VA

oo

PA

GG

55 AssA

WW

oo

(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

‖x‖2 = x20 − x2

1 − . . .− x2d−1.

A Poincaré group is a group of symmetries of M , including translations, that preserve the norm. Aquantum eld theory (QFT) is the following data

1

· space of states, a Hilbert space H;· vacuum vector, a vector |0〉 ∈ H;· unitary representation π of P in H, P 3 g π−→ π(g) unitary operator in H;· F = Φαα∈I collection of quantum elds.

Denition 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 Mwith values in the space of linear operators on H.

One requires that these data satisfy the following Wightman's axioms:

(vacuum) π(g)|0〉 = |0〉 for all g ∈ P ;(covariance) π(g)Φα(ϕ)π(g)−1 = Φα(π(g)(ϕ));

(completeness) Φα1(ϕ1)Φα2(ϕ2) . . .Φαs(ϕs) span a dense set in H;(locality) Φα(ϕ)Φβ(ψ) = Φβ(ψ)Φα(ϕ) if suppϕ and suppψ are separated by the light cone.

We have a special case when d = 2. In this case the light cone is given by the equationsx0 − x1 = 0x0 + x1 = 0

.

We can consider left chiral elds (respectively right) that are those elds for which ∂x0+x1Φα = 0

(respectively ∂x0−x1Φα = 0). Taking all left (or right) chiral elds of a 2-dimensional QFT produces a

left (or right) vertex (= chiral) algebra. The data of this algebra is given by the quadruple (V, |0〉, T,F),where

· the vector space V is the space of states;· |0〉 ∈ V is the vacuum vector;· T is the (only one) operator on V ;· F = Φαα∈I is the collection of quantum elds.

In this case we can give a more formal denition of a quantum eld.

Denition 1.2. A quantum eld is a series of the form

Φ(z) =∑n∈Z

Φnzn,

where Φn ∈ EndV , such that Φnv = 0 for n << 0 and for any v ∈ V .

By denition we get that Φ(z)v ∈ V ((z)) for any v ∈ V . This algebra satises the following axioms

(vacuum) T |0〉 = 0;(covariance) [T,Φ(z)] = d

dzΦ(z);(completeness) Φα1

n1Φα2n2 . . .Φ

αsns|0〉 span V ;

(locality) (z − w)NΦα(z)Φβ(w) = (z − w)NΦβ(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 avector space V and T ∈ EndV . Hence, the vacuum axiom becomes etT |0〉 = |0〉, for every t ∈ R. This isequivalent to the fact that T |0〉 = 0. Moreover, the covariance axiom in this case reads etTΦ(z)e−tT =Φ(z + t), for every t ∈ R. If we apply d

dt to both sides and we set t = 0 we get [T,Φ(z)] = ddzΦ(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 dene a vertex algebrastructure by |0〉 = 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 commutativityand 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 haveLa(Lb(c)) = a(bc) = (ab)c = (ba)c = Lb(La(c)) ,

2

for all c ∈ V . Hence, LaLb = LbLa.

Lemma 1.4. For any vertex algebra and for any of its quantum elds Φα(z) we have that Φα(z)|0〉 ∈V [[z]].

Proof. Let us write (is a usual convention that will be clear later)

Φα(z) =∑n∈Z

Φα(n)z−n−1.

By covariance axiom we get

[T,Φα(n)] = −nΦα(n−1)

and we know, by denition of quantum eld, that Φα(n)v = 0 for n >> 0. We want to prove that

Φα(n)|0〉 = 0 for n ∈ Z+. Take the minimal non-negative N such that Φα(N)|0〉 = 0 and apply T . We get

0 = TΦα(N)|0〉 = [T,Φα(N)]|0〉+ Φα(N)T |0〉 = −NΦα(N−1)|0〉

by vacuum axiom. So if N 6= 0, then Φα(N−1)|0〉 = 0, but this contradict our choice of N unless N = 0.

Due to the above lemma, we have a map

Φα(z) −→ Φα(z)|0〉|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 (withsome continuity properties). We choose the space of test functions to be C[z, z−1]. Let us x somenotation: given a vector space U we denote

· U [z] the space of polynomials with coecients in U ;· U [z, z−1] the space of Laurent polynomials with coecients in U ;· U [[z]] the space of power series with coecients in U ;· U((z)) the space of Laurent series with coecients in U ;· U [[z, z−1]] the space of innite series in both directions with coecients 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 identied with U [[z, z−1]],that is it consists of series of the form ∑

n∈Zunz

n,

where un ∈ U .

Before proving the proposition we need the notion of residue: given a U-valued formal distributionΦ(z) =

∑unz

n, its residue is the linear function dened 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

Φ(n) = Φ(z)(zn), for n ∈ Z,

then Φ(z) =∑

Φ(n)z−n−1. On the other hand, each Φ(z) ∈ U [[z, z−1]] denes a formal distribution by

the formula

Φ(z)(ϕ) = Resz Φ(z)ϕ(z).

Indeed

(∑n∈Z

Φ(n)z−n−1)(zm) = Resz

∑n∈Z

Φ(n)zm−n−1 = Φ(m).

In particular, quantum elds are EndV -valued formal distributions.3

Remark 2.2. The same is true for many variables. For example for two variables a U-valued formaldistribution is a series of the type

Φ(z, w) =∑m,n∈Z

a(m,n)z−m−1w−n−1.

Example 2.3. An important example of formal distribution is the formal δ-function, dened by

δ(z, w) = z−1∑n∈Z

(wz

)n∈ C[[z, z−1, w, w−1]].

Consider now the algebras

Az,w = C[z, z−1, w, w−1][[w

z]],

Aw,z = C[z, z−1, w, w−1][[z

w]].

Clearly Az,w,Aw,z ⊂ C[[z, z−1, w, w−1]]. Denote by R the algebra of rational functions of the form

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,wf −→ expansion in the domain |z| > |w|,

iw,z : R −→ Aw,zf −→ expansion in the domain |w| > |z|.

Example 2.4. The expansions of the function (z − w)−1 are of great interest:

iz,w1

z − w= expansion of

1

z· 1

1− wz

= z−1∑n∈Z+

(wz

)n∈ Az,w;

iw,z1

z − w= −expansion of

1

w· 1

1− zw

= w−1∑n∈Z+

( zw

)n∈ Aw,z.

Thus we can also dene the δ-function in terms of the expansions of (z − w)−1, namely

δ(z, w) = iz,w1

z − w− iw,z

1

z − w.

It follows also an important formula for the derivatives of the δ-function:

1

k!∂kwδ(z, w) =

∑n∈Z

(n

k

)wn−kz−n−1 = iz,w

1

(z − w)k+1− iw,z

1

(z − w)k+1. (2.1)

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)

(z − w)m1

n!∂nwδ(z, w) =

1(n−m)!∂

n−mw δ(z, w) if m ≤ n,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) eλ(z−w)∂nwδ(z, w) = (λ+ ∂w)nδ(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 theresidue in this last relation, we get

Resz a(z)δ(z, w) = a(w) Resz δ(z, w) = a(w),

proving d).Finally, let us prove e). We note that eλ(z−w)∂we

−λ(z−w) = λ+ ∂w. Indeed, take any function f(w),then

eλ(z−w)∂we−λ(z−w)f(w) = eλ(z−w)(λe−λ(z−w)f(w) + e−λ(z−w)∂wf(w)) = (λ+ ∂w)f(w).

4

Theneλ(z−w)∂nwe

−λ(z−w) = (λ+ ∂w)n.

Apply now to δ(z, w) and get

eλ(z−w)∂nwe−λ(z−w)δ(z, w) = eλ(z−w)∂nwδ(z, w) = (λ+ ∂w)nδ(z, w),

since by d), we know that e−λ(z−w)δ(z, w) = e−λ(w−w)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 theoryclaries why δ(z, w) is called formal δ-function.

Denition 2.6. A formal distribution a(z, w) ∈ U [[z, z−1, w, w−1]] is called local if (z−w)Na(z, w) = 0for some N ∈ Z+.

By Proposition 2.5(c), it follows that the formal δ-function and its derivatives are local. Moreover thefollowing 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 aunique decomposition

a(z, w) =∑j∈Z+

cj(w)∂jwδ(z, w)

j!,

where the sum is nite and cj(w) ∈ U [[w,w−1]]. The coecients cj(w) are given by the formula

cj(w) = Resz(z − w)ja(z, w). (2.2)

Proof. Let b(z, w) = a(z, w)−∑cj(w)

∂jwδ(z,w)j! , where cj(w) are given by formula (2.2). Then the second

sum is nite since (z − w)ja(z, w) = 0 for j >> 0. Note that

Resz(z − w)nb(z, w) = Resz(z − w)na(z, w)− Res∑j≥n

cj(w)∂j−nw δ(z, w)

(j − n)!=

= cn(w)−∑j≥n

cj(w)∂j−nw Res δ(z, w)

(j − n)!= cn(w)− cn(w) = 0.

So, Resz(z − w)nb(z, w) = 0 for any n ∈ Z+. Hence

b(z, w) =∑j∈Z+

zjbj(w),

with bj(w) ∈ U [[w,w−1]]. 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)Nb(z, w) = 0, sincecj(w) = 0 for j ≥ N . Hence b(z, w) = 0 and the decomposition holds for coecients cj(w) given by(2.2).

Let us prove uniqueness. We claim that if

N∑j=0

cj(w)∂jwδ(z, w)

j!= 0,

where cj(w) ∈ U [[w,w−1]], then all cj(w) = 0. If not, let j0 be the maximal index for which cj0 6= 0 andmultiply both sides for (z − w)j0 . Then by Proposition 2.5(c) we get cj0δ(z, w) = 0. Taking the residuein this relation we have cj0(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

Da(z,w) : C[z, z−1] −→ U [[w,w−1]]

ϕ(z) −→ Resz ϕ(z)a(z, w).

Prove that

a) Dc(w)∂jwδ(z,w) = c(w)∂jw, in particular Dδ(z,w) = 1;

5

b) a(z, w) is local if and only if Da(z,w) is a dierential operator of nite order;c) if a(z, w) is local, then a(w, z) is local too and Da(z,w) = D∗a(w,z).

Here ∗ is the formal adjoint dened by ∂∗ = −∂ and f∗ = f if f is a formal series.

Solution. Part a) follows from a straightforward computation:

Dc(w)∂jwδ(z,w)ϕ(z) = Resz ϕ(z)c(w)∂jwδ(z, w) = c(w)∂jw Resϕ(z)δ(z, w) = c(w)∂jwϕ(w).

We know that a(z, w) is local if and only if a(z, w) =∑cj(w)∂jwδ(z, w)/j! and the sum is nite. Then,

by part a), it follows that Da(z,w) =∑cj(w)∂jw/j! is a dierential operator of nite order. This proves

b). To prove the last assertion we need Proposition 2.5(b). We have

Dc(z)∂jzδ(w,z)ϕ(z) = Resϕ(z)c(z)∂jzδ(w, z) = Resz c(z)ϕ(z)∂jzδ(z, w) = Resz c(z)ϕ(z)(−∂w)jδ(z, w) =

= (−∂w)j Resz ϕ(z)c(z)δ(z, w) = (−∂jw)c(w)ϕ(w).

Since a(z, w) is local, by linearity we get

D∗a(z,w) =

(∑cj(w)

∂jwj!

)∗=∑ (−∂w)j

j!cj(w) = Da(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 suces to prove the result only for one derivative. Suppose (z−w)na(z, w) = 0,then

(z − w)n+1∂wa(z, w) = ∂w((z − w)n+1a(z, w)) + (n+ 1)(z − w)na(z, w) = 0.

Let U = g be a Lie algebra and let a(z), b(z) ∈ g[[z, z−1]].

Denition 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 (z − w)n[a(z), b(w)] = 0 for some n ∈ Z+. Applying the DecompositionTheorem (Theorem 2.7) we get for a local pair a(z), b(z)

[a(z), b(w)] =

N∑j=0

cj(w)∂jwδ(z, w)

j!,

where cj(w) = Resz(z − w)j [a(z), b(w)]. Let us write

a(z) =∑n∈Z

a(n)z−n−1, b(z) =

∑n∈Z

b(n)z−n−1 and cj(z) =

∑n∈Z

cj(n)z−n−1.

Comparing the coecient of z−m−1w−n−1 in [a(z), b(w)], we get

[a(m), b(n)] =∑j∈Z+

(m

j

)cj(m+n−j).

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 ([a, b] | c) = (a | [b, c]) for any a, b, c ∈ g. Dene g = g[t, t−1] +CK, where K is a central element.We dene a Lie bracket on g by

[atm, btn] = [a, b]tm+n +mδm,−n(a | b)K, (2.3)

for any a, b ∈ g and m, n ∈ Z.

Exercise 5. Check Lie algebra axioms for g.

Solution. The fact that [atm, atm] = 0 and linearity of (2.3) implies antisymmetry. We are left to provethe Jacobi identity. For a, b, c ∈ g and n, m, k ∈ Z, we have

[atm, [btn, ctk]] = [atm, [b, c]tn+k + nδn,−k(a | b)K] = [a, [b, c]]tm+n+k +mδm,−n−k(a | [b, c])K.Similarly we get

[btn, [ctk, atm]] = [b, [c, a]]tm+n+k + nδn,−m−k(b | [c, a])K,

[ctk, [atm, btn]] = [c, [a, b]]tm+n+k + 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 suces to prove thatm(a | [b, c]) + n(b | [c, a]) + k(c | [a, b]) = 0. But this is clear by the condition n + m + k = 0 and theinvariance of the scalar product.

Denition 2.11. g is called anization of (g, (· | ·)).

To an element a ∈ g we associate the following g-valued formal distribution, called current,

a(z) =∑n∈Z

(atn)z−n−1.

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 Ln, n ∈ Z, C and the followingcommutation relations

[Lm, Ln] = (m− n)Lm+n +m3 −m

12δm.−nC, (2.4)

where C is a central element. Let

L(z) =∑n∈Z

Lnz−n−2 =

∑n∈Z

L(n)z−n−1,

where L(n) = Ln−1.

Exercise 6. The relations (2.4) are equivalent to

[L(z), L(w)] = (∂wL(w))δ(z, w) + 2L(w)∂wδ(z, w) +C

2

∂3wδ(z, w)

3!.

Solution. We use again commutator formula. We have c0(w) = ∂wL(w), c1(w) = 2L(w) and c3(w) = C2 .

Then c0n = −nL(n−1), c1(n) = 2L(n) and c

3(n) = δn,−1

C2 . Hence,

[L(m), L(n)]=(−m− n)L(m+n−1)+2mL(m+n−1)+m(m− 1)(m− 2)

3!δm+n−3,−1

C

2=

=(m− n)L(m+n−1) + δm−1,−n+1m(m− 1)(m− 2)

12C.

But [L(m), L(n)] = [Lm−1, Ln−1], then rescaling m 7→ m+ 1 and n 7→ n+ 1 we get (2.4).

It is clear from examples that the pairs (a(z), b(z))a,b∈g for g and (L(z), L(w)) for the Virasoro algebraare local.

Example 2.13. A special case of g is obtained when g = Ca, with (a|a) = 1. Then a(z) =∑

(atn)z−n−1

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 itϕ(z), then it satises the relation [ϕ(z), ϕ(w)] = δ(z, w).

To explain how this distribution is dened we need the notion of Lie superalgebra. A super space isa vector space

V = V0 ⊕ V1,

where 0, 1 ∈ Z/2Z. If a ∈ V0, we set p(a) = 0 and call this element even. Similarly, if a ∈ V1, weset 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 toV0 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 vectorspace V , it becomes antisymmetric on the super vector space with reverse parity ΠV . A superalgebra isa super space A = A0 ⊕A1 with a product such that AαAβ ⊂ Aα+β , where α, β ∈ Z/2Z.

7

For example, if V is a superspace, then EndV = EndV0⊕EndV1 is an associative superalgebra, where

EndV0 =

(a 00 b

)∈ EndV

and EndV1 =

(0 cd 0

)∈ EndV

.

The commutator in an associative superalgebra is dened as

[a, b] = ab− p(a, b)ba,

where p(a, b) = (−1)p(a)p(b). This bracket satises 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 anization: take a superspace A with a skewsymmetric

bilinear form 〈·, ·〉 and let A = A[t, t−1] + CK, K central, with the bracket dened by

[atm, btn] = 〈a, b〉δm,−n−1K.

Currents are dened by∑

(atn)z−n−1, for a ∈ A and their bracket is [a(z), b(w)] = 〈a, b〉δ(z, w). Now, ifwe take A = Cϕ, with ϕ odd and 〈ϕ,ϕ〉 = 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 coecients cj(w))

[a(z), b(w)] =

N∑j=0

a(w)(j)b(w)∂jwδ(z, w)

j!, (2.5)

where

a(w)(j)b(w) = Resz(z − w)j [a(z), b(w)]. (2.6)

This expression is called operator product expansion (OPE) for a local pair a(z), b(z). In particular foreach j ∈ Z we get a j-th product given by (2.6). This is just the singular part of the OPE of the productof two formal distributions. We want to get the "complete" OPE: we need the notion of normally orderedproduct. Let a(z) =

∑anz

n be a formal distribution, we dene

a(z)+ =∑n∈Z+

anzn (creation part),

a(z)− =∑n<0

anzn (annihilation part).

Denition 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

a(w)+ = Resz a(z)iz,w1

z − w, a(w)− = −Resz a(z)iw,z

1

z − w(2.7)

and, dierentiating k times by z, get

∂kwa(w)±k!

= ±Resz a(z)iz,ww,z

1

(z − w)k+1.

Proof. Let a(w) =∑anw

n, then

Resz a(z)iz,w1

z − w= Resz

∑n∈Z

anzn · z−1

∑k∈Z+

(wz

)k= Resz∑n∈Zk∈Z+

anzn−k−1wk =

∑n∈Z+

anwn = a(w)+.

This proves the rst of the formulas (2.7). The other is proved in the same way. By dierentiating theseformulas we get the lemma.

8

Exercise 7. For a local pair of U-valued formal distributions, U associative algebra, the OPE formulais equivalent to the following two equalities:

a(z)b(w) =∑j∈Z+

a(w)(j)b(w)iz,w1

(z − w)j+1+ : a(z)b(w) :,

b(w)a(z) =∑j∈Z+

a(w)(j)b(w)iw,z1

(z − w)j+1+ : a(z)b(w) : .

These relations are the so called complete OPE.

Solution. By (2.1), we can rewrite (2.5) in the following way

[a(z), b(w)] =∑j∈Z+

a(w)(j)b(w)

(iz,w

1

(z − w)j+1− iw,z

1

(z − w)j+1

).

If we separate positive and negative powers in z in this expression we obtain

[a(z)−, b(w)] =∑j∈Z+

a(w)(j)b(w)iz,w1

(z − w)j+1

[a(z)+, b(w)] = −∑j∈Z+

a(w)(j)b(w)iw,z1

(z − w)j+1.

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

a(z)b(w) ∼∑j∈Z+

a(w)(j)b(w)

(z − w)j+1,

writing only what we have called singular part of the OPE. Look now at the regular part of the OPEand 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 provethat converges if our formal distributions are quantum elds. First recall the denition of a quantumeld.

Denition 2.17. A quantum eld is an EndV -valued formal distribution a(z) =∑a(n)z

−n−1 suchthat 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 dened 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 thatb(z)v ∈ V ((z)) and a(z)+ ∈ EndV [[z]], then a(z)+b(z) ∈ V ((z)). For the second term we have

b(z)a(z)−v = b(z)

(∑nite

ai(v)zi

)=∑nite

b(z)ai(v)zi ∈ V ((z)).

Expanding in Taylor series the regular part of the OPE we get

: a(z)b(w) :=∑j∈Z+

: ∂ja(w)b(w) :

j!(z − w)j .

This expression holds in the domain |z − w| < |z|. We dene

a(z)(−j−1)b(z) =: ∂ja(z)b(z) :

j!,

for j ∈ Z+, so: 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

a(w)(j)b(w) = Resz(a(z)b(w)iz,w(z − w)j − b(w)a(z)iw,z(z − w)j

).

Proof. For j ∈ Z+ it is (2.6). For j < 0 it follows from Cauchy formulas multiplied by b(w).

Exercise 8. Show that a(z)(j)b(z), for j ∈ Z+ is a quantum eld if a(z), b(z) are.

Solution. Remember that a(w)(n)b(w) = Resz[a(z), b(w)](z − w)n. Expanding we get

a(w)(n)b(w) = Resz∑k,l,m

(−1)k(n

k

)[a(m), b(l)]z

k−m−1wn−k−l−1 =

=∑k,l

(−1)k(n

k

)[a(k), b(l)]w

n−k−l−1 =∑l∈Z

∑k∈Z+

(−1)k[a(k), b(l−k+n)]

w−l−1.

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, forl > N . Then the commutators terms kill v.

Let V be a vector (super)space and let a(z), b(z) be EndV -valued quantum elds, then for each j ∈ Zwe have dened 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 j ∈ Z+ we have a(w)(j)b(w) = Resz[a(z), b(w)](z − w)j . Then we get

∂w(a(w)(j)b(w)

)= Resz[a(z), ∂wb(w)](z − w)j − Resz[a(z), b(w)]j(z − w)j−1 =

= a(w)(j)∂wb(w) + ∂wa(w)(j)b(w),

since −Resz[a(z), b(w)]j(z − w)j−1 = Resz[∂za(z), b(w)](z − w)j . On the other hand, for j > 0 we have

a(w)(−j−1)b(w) =: ∂jwa(w)b(w) :

j!=∂jwa(w)+

j!b(w) + b(w)

∂jwa(w)−j!

.

Then we get

∂w(a(w)(−j−1)b(w)

)=∂j+1w a(w)+

j!b(w) +

∂jwa(w)+

j!∂wb(w) + ∂wb(w)

∂jwa(w)−j!

+ b(w)∂j+1w a(w)−

j!=

= ∂wa(w)(−j−1)b(w) + a(w)(−j−1)∂wb(w).

3. Some equivalent definitions of a Vertex Algebra

Denition 3.1. A vertex algebra is the data of (V, |0〉, T,F = aj(z)j∈J) where V is a vector (su-per)space (space of state), |0〉 is a (even) vector (vacuum vector), T ∈ EndV is an (even) operator(translation operator) and F is a collection of EndV - valued quantum elds with the following axioms:

(vacuum) T |0〉 = 0;(translation covariance) [T, aj(z)] = ∂za

j(z);

(completeness) all vectors aj1(n1)aj2(n2) . . . a

js(ns)|0〉 span V ;

(locality) (z − w)Nij [ai(z), aj(z)] = 0 for some Nij ∈ Z+.

We have already proved in Section 1 the following lemma:

Lemma 3.2. aj(z)|0〉 ∈ V [[z]].

The proof of this lemma uses only the rst two axioms.

Denition 3.3. An EndV - valued quantum elds a(z) is called translation covariant if [T, a(z)] =∂za(z).

By Lemma 3.2 it follows that a(z)|0〉 ∈ V [[z]]. Moreover, let F denote the space of all translationcovariant EndV -valued quantum elds, then we have the map

F s−→ V

a(z) −→ a(z)|0〉|z=0.

This map is called eld-state correspondence.10

Theorem 3.4 (Extension Theorem). Let (V, |0〉, T,F) be a vertex algebra and let F denote the space ofall translation covariant EndV -valued quantum elds which are local to each of the quantum elds fromF , thena) all axioms of vertex algebra for (V, |0〉, 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)|0〉.

Lemma 3.5. F contains 1V , it is ∂z-invariant and it is closed under all n-th products, n ∈ Z.

Proof. Clearly 1V ∈ F , 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)] =∂zb(z), then the same holds for a(z)(j)b(z), j ∈ Z. Indeed, we have

[T, a(w)(j)b(w)] =[T,Resz

(a(z)b(w)iz,w(z − w)j − p(a, b)b(w)a(z)iw,z(z − w)j

)]=

= Resz([T, a(z)b(w)]iz,w(z − w)j − p(a, b)[T, b(w)a(z)]iw,z(z − w)j

)=

= Resz(([T, a(z)]b(w) + a(z)[T, b(w)])iz,w(z − w)j−p(a, b)([T, b(w)]a(z) + b(w)[T, a(z)])iw,z(z − w)j

)=

= Resz((∂za(z)b(w) + a(z)∂wb(w))iz,w(z − w)j − p(a, b)(∂wb(w)a(z) + b(w)∂za(z))iw,z(z − w)j

)=

= ∂wa(w)(j)b(w) + a(w)(j)∂wb(w) = ∂w(a(w)(j)b(w)).

Lemma 3.6. Let a(z), b(z) ∈ F and s(a(z)) = a, s(b(z)) = b, then:

a) s(∂za(z)) = Ta = a(−2)|0〉;b) s(a(z)(n)b(z)) = a(n)b.

Proof. We have a(z)|0〉 = a(−1)|0〉+ za(−2)|0〉+ . . .. Apply T to both sides and get

LHS = Ta(z)|0〉 = [T, a(z)]|0〉+ a(z)T |0〉 = ∂za(z),

RHS = Ta(−1)|0〉+ z(. . .) = [T, a(−1)]|0〉+ z(. . .) = a(−2)|0〉+ z(. . .).

So s(∂za(z)) = ∂za(z)|z=0= a(−2)|0〉 and this proves a). For b), we have

a(w)(n)b(w)|0〉|w=0= Resz a(z)b(w)|0〉iz,w(z − w)n|w=0

− Resz b(w)a(z)|0〉iw,z(z − w)n|w=0.

Now, we note that b(w)|0〉 ∈ V [[w]]. Then the rst term of the sum is Resz a(z)bzn, while iw,z(z−w)n ∈C[[w,w−1]][[z]] and a(z)|0〉 ∈ V [[z]], so the second term is 0. This means that

a(w)(n)b(w) = Resz a(z)bzn = a(n)b.

Exercise 9. Prove that the dierential equation ddz f(z) = R(z)f(z), where f(z) ∈ U [[z]] and R(z) ∈

EndU [[z]], has a unique solution with given f(0).

Solution. Write

f(z) =∑n∈Z+

anzn, an ∈ U , and R(z) =

∑n∈Z+

Rnzn, Rn ∈ EndU .

Thend

dzf(z) =

∑n>0

nanzn−1 =

∑n∈Z+

(n+ 1)an+1zn .

and

R(z)f(z) =∑n∈Z+

(n∑k=0

Rn−kak

)zn .

It follows that f(z) satises the dierential equation if and only if its coecients satisfy the recursiona0 = f(0)an+1 = 1

n+1

∑nk=0Rn−kak n ≥ 0

which is uniquely solvable for xed f(0).

Lemma 3.7. Let a(z) ∈ F , thena) a(z)|0〉 = ezTa, where a = s(a(z));b) ewTa(z)e−wT = iz,wa(z + w);

11

c) ewTa(z)±e−wT = iz,wa(z + w)±.

Proof. We prove only part a). Let us note that both sides are equal when z = 0 and satisfy the samedierential equation, in fact

d

dza(z)|0〉 = ∂za(z)|0〉 = [T, a(z)]|0〉 = Ta(z)|0〉.

Clearly, on the other handd

dzezTa = TezTa.

Exercise 10. Prove b) and c) using the fact that both sides satisfy the dierential equation

d

dwf(w) = adTf(w),

where f(w) ∈ (EndV [[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 dierentialequation are in (EndV [[z, z−1]])[[w]] and satisfy the same initial condition. If we dierentiate both sideswe get

LHS =d

dwewTa(z)e−wT = TewTa(z)e−wT − ewTa(z)e−wTT = adTewTa(z)e−wT ,

while the other side is

RHS =d

dwiz,wa(z + w) = iz,w∂za(z + w) = iz,w adTa(z + w) = adTiz,wa(z + w).

Lemma 3.8 (Uniqueness Lemma). Let F ′ ⊂ F be a collection of translation covariant quantum elds

such that s : F ′ −→ V is surjective, let a(z) ∈ F be such that s(a(z)) = 0 and all pairs (a(z), b(z)), whereb(z) ∈ F ′, are local, then a(z) = 0.

Proof. By locality we have (z −w)nb(w)a(z)|0〉 = ±(z −w)na(z)b(w)|0〉, n ∈ Z+. By part a) of Lemma3.7 we have

(z − w)nb(w)a(z)|0〉 = (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, thenthe pair (a(z)(n)b(z), c(z)) is local for any n ∈ Z.

Proof. For the proof see [Kac96, Lemma 3.2].

Proof of Theorem 3.4. Let F ′ be the minimal subspace of the space of all quantum elds containing 1Vand F , ∂z-invariant and closed under all n-th products. Then by Lemma 3.5 F ′ ⊂ F and by Lemma 3.9we get the following chain

F ⊂ F ′ ⊂ F ⊂ F .Moreover, by Dong's Lemma, F ′ contains the following quantum elds

a(z) = aj1(z)(n1)(. . . (ajs−1(z)(ns−1)(a

js(z)(ns)1V )) . . . ).

But, by Lemma 3.6, s(a(z)) = aj1(n1) . . . ajs−1

(ns−1)ajs(ns)|0〉. Hence s(F ′) = V by completeness axiom, while

from Lemma 3.8 it follows that s : F → V is injective. This means that F ′ = F and s is bijective thatproves b). Part a) is now clear since local axioms hold because all elds from F ′ are pairwise local.

Corollary 3.10. We have

a) |0〉(z) = 1V ;b) Ta(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 −→ Fa −→ a(z)

where

s−1(aj1(n1) . . . ajs(ns)|0〉) = aj1(z)(n1)(. . . (a

js(z)(ns)1V ) . . . ). (3.1)

In fact both sides lie in F and they have the same image under s by lemma 3.6. Hence they are equalby 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)|0〉 = p(a, b)(z − w)nb(w)a(z)|0〉. By part a) of Lemma 3.7 we have

(z − w)na(z)ewT b = p(a, b)(z − w)nezT e−zT b(w)ezTa = p(a, b)(z − w)nezT iw,zb(w − z)a.

If n >> 0, then (z−w)nb(w−z)a ∈ V [[z−w]]. If we put w = 0, then zna(z)b = p(a, b)znezT b(−z)a.

In most important examples V is spanned by vectors a = aj1(−n1−1) . . . ajs(−ns−1)|0〉, where n1, . . . , ns ∈

Z+, then (3.1) becomes

s−1(a) =:∂n1z

n1!aj1(z) . . .

∂nszns!

ajs(z) :,

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 denition of vertex algebra.

Denition 3.12. A vertex algebra is the data of (V, |0〉, T, s), where s is a map from V to a space ofEndV -valued quantum elds, s(a) = a(z), such that the following axioms hold

(vacuum) T |0〉 = 0, a(z)|0〉|z=0= a;

(translation covariance) [T, a(z)] = ∂za(z);(locality) (z − w)Nab [a(z), b(w)] = 0.

The completeness axiom is automatically satised because a(−1)|0〉 = a span V . Often a(z) is denotedY (a, z) and called vertex operator. The set of quantum elds of a vertex algebra V contains 1V = |0〉(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

[a(z), b(w)] =∑j∈Z+

(a(j)b)(w)∂jwδ(z, w)

j!.

This formula follows from the Decomposition Theorem and the n-th product identity. If we compare thecoecients of z−m−1 we get the following equivalent formulas, for n,m ∈ Z, to the commutator formula:

[a(m), b(w)] =∑j∈Z+

(m

j

)(a(j)b)(w)wm−j ;

[a(m), b(n)] =∑j∈Z+

(m

j

)(a(j)b)(m+n−j).

(3.2)

Exercise 11. Given f(z) ∈ C((z)) and a ∈ V , let

af = Resz f(z)a(z) =∑n≥−N

fna(n).

In particular, if f = zn, then af = Resz zna(z) = a(n). Prove that

[af , bg] =∑j∈Z+

(a(j)b

)f(j)gj!

.

13

Solution. Let f =∑n≥−N fnz

n and g =∑m≥−M gmz

m be elements in C((z)) then by (3.2) we get

[af , bg] =∑n≥−Nm≥−M

fngm[a(n), b(m)] =∑j∈Z+

∑n≥−Nm≥−M

(n

j

)fngm

(a(j)b

)(n+m−j)

.

On the other hand, since

f (j)g

j!=

∑n≥−Nm≥−M

(n

j

)fngmz

n+m−j ,

one gets, for any j ∈ Z+,(a(j)b

)f(j)gj!

= Resz f(z)(a(j)b

)f(j)gj!

(z) = Resz∑k∈Z+

n≥−Nm≥−M

(n

j

)fngm

(a(j)b

)(k)zn+m−j−k−1 =

=∑n≥−Nm≥−M

fngm(a(j)b

)(n+m−j) .

Example 3.13. Ane Lie algebras: g = g[t, t−1] + CK, a(z) =∑

(atn)z−n−1. We know that

[a(z), b(w)] = [a, b](w)δ(z, w) + (a | b)∂wδ(z, w)K.

Then we get

[af , bg] = [a, b]fg + (a | b) Resz f′(z)g(z)K.

Example 3.14. Virasoro algebra: L(z) =∑Lnz

−n−2 with the commutator given by

[L(z), L(w)] = ∂wL(w)δ(z, w) + 2L(w)∂wδ(z, w) +C

12∂3wδ(z, w).

Then we get

[Lf , Lg] = Lf ′g−fg′ +1

12Resz f

′′′(z)g(z).

Example 3.15. Take A to be a vector superspace with a skew-symmetric bilinear form 〈·, ·〉 and set

A = A[t, t−1] + C1. Dene [atm, btn] = 〈a, b〉δm,−n−11, then for a(z) =∑

(atn)z−n−1, a ∈ A, thecommutator is given by

[a(z), b(w)] = 〈a, b〉δ(z, w)1.

So we get

[af , bg] = 〈a, b〉Resz fg.

Now we state a famous identity, called Borcherds identity, that is a generalization of the n-th productidentity and the commutator formula.

Theorem 3.16 (Borcherds identity).

a(z)b(w)iz,w(z − w)n − p(a, b)b(w)a(z)iw,z(z − w)n =∑j∈Z+

(a(n+j)b)(w)∂jwδ(z, w)

j!.

Proof. LHS is a local formal distribution in z and w by the locality axiom, then, by DecompositionTheorem, we have:

LHS =∑j

cj(w)∂jwδ(z, w)

j!,

where

cj(w) = Resz((z − w)j(a(z)b(w)iz,w(z − w)n − p(a, b)b(w)a(z)iw,z(z − w)n)

)=

= Resz(a(z)b(w)iz,w(z − w)j+n − p(a, b)b(w)a(z)iw,z(z − w)j+n

)= a(w)(n+j)b(w) = (a(n+j)b)(w).

14

Comparing coecients of z−m−1w−k we get, for m,n, k ∈ Z and a, b, c ∈ V , the original Borcherdsidentity: ∑

j∈Z+

(m

j

)(a(n+j)b)(m+k−j)c =

∑j∈Z+

(−1)j(n

j

)(a(m+n−j)(b(k+j)c)− (3.3)

− (−1)np(a, b)b(n+k−j)(a(m+j)c)).

Thanks to Borcherds identity we obtain the third denition of vertex algebra.

Denition 3.17. A vertex algebra is the data of (V, |0〉, a(n)b | n ∈ Z) with the following axioms(a, b, c ∈ V ):

(vacuum) |0〉(n)a = δn,−1a, for every n ∈ Z, and a(n)|0〉 = δn,−1a, for every n ≥ −1;

(Borcherds identity)∑j∈Z+

(mj

)(a(n+j)b)(m+k−j)c=

∑j∈Z+

(−1)j(nj

)(a(m+n−j)(b(k+j)c)−(−1)np(a, b)b(n+k−j)(a(m+j)c)

).

Let us note that Borcherds identity implies locality axiom.

Exercise 12. Derive the second denition of vertex algebra from the third one.

Solution. From the vacuum axiom it follows that |0〉(z) = idV and a(z)|0〉|z=0= a. Dene T on V by

Ta = a(−2)|0〉. Then, clearly, T |0〉 = 0. Put c = |0〉, m = 0 and k = −2 in the identity (3.3) and get

(a(n)b)(−2)|0〉 = a(n)(b(−2)|0〉)− na(n−1)(b(−1)|0〉).

Then

[T, a(n)]b = Ta(n)b− a(n)Tb = (a(n)b)(−2)|0〉 − a(n)(b(−2)|0〉) = −na(n−1)(b(−1)|0〉) = −na(n−1)b,

then [T, a(z)] = ∂za(z), since b is arbitrary.

Exercise 13. Let V be a commutative associative algebra with 1 and let T be a derivation of V , then(V, |0〉 = 1, T, a→ eTza) is a vertex algebra. Moreover, any vertex algebra for wich all a(z) do not involvenegative 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) =2T (1), then T |0〉 = T (1) = 0. Moreover ezTa|0〉|z=0

= ezTa · 1|z=0= ezTa|z=0

= a. For the translationcovariance axiom we have

[T, ezTa]b = T (ezTa · b)− ezTa · Tb = (TezTa)b =∑k∈Z+

T k+1(a)zk

k!· b = ∂z

∑k∈Z+

T k(a)zk

k!· b = (∂ze

zTa)b.

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.

Denition 3.18. A regular formal distribution Lie (super)algebra is the data of (g, T,F = aj(z)j∈J),

where T is a (even) derivation of g, the coecients of aj(z) =∑aj(n)z

−n−1 span g, (ai(z), aj(z)) are all

local pairs and Ta(z) = ∂za(z).

Example 3.19. The algebra g with F = a(z)a∈g ∪ K and T = −∂t is a regular formal distributionLie algebra.

Exercise 14. Prove that ∂za(z) = −∂ta(z).

Solution. It is a straightforward computation:

∂za(z) = ∂z∑n∈Z

(atn)z−n−1 = −∑n∈Z

(n+ 1)(atn)z−n−2 = −∑n∈Z

n(atn−1)z−n−1 = −∂ta(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.

Let (V = U(g), |0〉 = 1, T,F = aj(z)j∈J). Is this a vertex algebra, according to the rst denition?

All axioms hold, but we also need aj(n)1 = 0 for n >> 0 and this is false, because formal distributions,

in general, are not quantum elds. We need to take V = U(g)/U(g)g−, where g− is T -invariant. Let

g− = spanaj(n), n ∈ Z+, j ∈ J.15

This is a subalgebra of g, called the annihilation subalgebra, since we have

[a(m), b(n)] =∑j∈Z+

(m

j

)(a(j)b)(m+n−j)

and is T -invariant since Tam = −ma(m−1).

Example 3.21. In the previous examples we have

g− = g[t], V ir− =⊕n≥−1

CLn, A− = A[t].

Theorem 3.22.(U(g)

/U(g)g−, im(1), T,F

)is a vertex algebra.

Proof. We prove by induction on s that

a(N)(aj1(n1) . . . a

js(ns)

1) = 0, for N >> 0.

For s = 0 this is true since g− is in the kernel of the projection to the quotient. For s > 0 we have

a(N)(aj1(n1) . . . a

js(ns)

1) = [a(N), aj1(n1)]a

j2(n2) . . . a

js(ns)

1 + aj1(n1)a(N)aj2(n2) . . . a

js(ns)

1.

The second summand is 0 for N >> 0 by induction hypothesis. For the rst recall that

[a(N), aj1(n1)] =

∑j∈Z+

(N

j

)cj(N+n1−j) = 0

for N >> 0.

Let us dene V (g,F) =(U(g)

/U(g)g−, |0〉 = im(1), T,F

).

Example 3.23. The vertex algebra

V k(g) = V (g,F)/V (g,F)(K − k),

k ∈ C, is called universal ane vertex algebra of level k.

Example 3.24. The vertex algebra

V c = 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,F)/V (A,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(−n1−1)aj2(−n2−1) . . . a

js(−ns−1)|0〉,

with ni ∈ Z+, span V . Since g = g[t−1]t−1 + K + g[t], we have V = U(g[t−1]t−1)|0〉, so the state-eldcorrespondence looks

a(z) =:∂n1z aj1(z) . . . ∂nsz ajs(z)

n1! . . . ns!: .

Exercise 15. Write state-eld correspondence for the Virasoro case and for F (A).

Solution. The vertex algebra V c is spanned by elements L−n1−2 · · ·L−ns−2|0〉, with ni ∈ Z+. Hence, thestate-eld correspondence is

(L−n1−2 · · ·L−ns−2|0〉)(z) =:∂n1z L(z) . . . ∂nsz L(z)

n1! . . . ns!: .

Similarly, F (A) is spanned by aj1(−n1−1) · · · ajs(−ns−1)|0〉, with ni ∈ Z+ and aj is a basis of A. We get

the following state-eld correspondence

(aj1(−n1−1) · · · ajs(−ns−1)|0〉)(z) =:

∂n1z aj1(z) . . . ∂nsz ajs(z)

n1! . . . ns!: .

16

We present now another construction of V (g,F). Assume that g admits a descending ltration

. . . ⊃ g−1 ⊃ g0 ⊃ g1 . . .

such that Tgj ⊂ gj−1. In our example the ltration was given by powers of t. Let U(g)completed consistof 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 denoteby F its closure. In particular we have

: a(z)b(z) :(n)=∑j<0

a(j)b(n−j−1) +∑j∈Z+

b(n−j−1)a(j) ∈ Ucompleted.

Then (F , |0〉 = 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) dened by representation of U(g)completed in V (g,F).

Denition 3.26. Let V be a vertex algebra, then an element L ∈ V is called an energy-momentumstate and the corresponding eld L(z) is called an energy-momentum eld if

a) [L(z), L(w)] = ∂wL(w)δ(z, w) + 2L(w)∂wδ(z, w) + 112∂

3wδ(z, w)C1V , C ∈ C;

b) L−1 = T ;c) L0 is a diagonalizable operator on V .

Recall that part a) is equivalent to the relation

[Lm, Ln] = (m− n)Lm+n +m3 −m

12δm,−nC1V ,

for L(z) =∑Lnz

−n−2. If L0a = ∆aa, then ∆a is called the energy or the conformal weight of a. Bythe commutator formula we get

[L(z), a(w)] =∑j∈Z+

(Lj−1a)(w)∂jwδ(z, w)

j!= (Ta)(w)δ(z, w) + (L0a)(w)∂wδ(z, w) + . . . =

= ∂wa(w)δ(z, w) + (L0a)(w)∂wδ(z, w) + . . . .

By taking the coecients of z−m−2, we get the equivalent formula

[Lm, a(z)] =∑j∈Z+

(m+ 1

j

)(Lj−1a)(z)zm+1−j . (3.4)

In particular [L0, a(z)] = z∂za(z) + (L0a)(z).

Denition 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) meansscale covariance. The most general covariance is given by equation (3.4), that is covariance with respectto the whole conformal group.

Let us write

a(z) =∑

n∈−∆a+Zanz−n−∆a , (3.6)

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) ∆|0〉 = 0;b) ∆a(n)b = ∆a + ∆b − n− 1;

c) ∆Ta = ∆a + 1;d) writing a(z) and b(z) in the form (3.6), we get

[am, bn] =∑j∈Z+

(m− 1 + ∆a

j

)(a(j)b)m+n.

17

Proof. For a) we have 0 = [H, 1V ] = [H, |0〉(z)] = (z∂z + ∆|0〉)1V , then ∆|0〉 = 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, Ta = a−2|0〉, then∆Ta = ∆a + ∆|0〉 − (−2)− 1 = ∆a + 1

and so c) is proved. The part d) follows from the commutator formula if we replace a(m) by am+1−∆a

and b(n) by bn+1−∆b.

Remark 3.30. Note that a ∈ V and the corresponding eld a(z) has conformal weight with respect toenergy-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

[a(z), b(w)] =∑

(a(j)b)(w)∂jwδ(z, w)

j!.

Example 3.32. V c 3 L = L(−1)|0〉 = L−2|0〉. The corresponding eld is the Virasoro eld. Hencethe only thing to check is that L0 is a diagonalizable operator. The vectors L−j1−1 . . . L−js−1|0〉, withj1 ≥ . . . ≥ js ≥ 1, form a basis of V c, since

V ir = V ir+ ⊕ V ir−,where V ir− =

∑n≥−1 Ln. Each of this vector is an eigenvector of H = L0:

L0(L−j1−1 . . . L−js−1|0〉) = (j1 + . . . js + s)L−j1−1 . . . L−js−1|0〉. (3.8)

The vectors L−j1−1 . . . L−js−1|0〉 have energy j1 + . . . js + s.

Exercise 16. Compute the dimension of the N -th eigenspace of L0.

Solution. From (3.8) it follows that the N -th eigenspace of L0 is spanned by vectors L−j1 . . . L−js |0〉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 numberof 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.

Denition 3.33. Given a formal distribution a(z) ∈ U [[z, z−1]], we dene its formal Fourier transformto be

Fλz a(z) = Resz eλza(z).

In particular, if a(z) =∑a(n)z−n−1, then

Fλz a(z) = Resz∑m∈Z+

∑n∈Z

a(n)λm

m!zm−n−1 =

∑n∈Z+

a(n)λn

n!.

So Fλz : U [[z, z−1]] −→ U [[λ]] is a linear map.

Proposition 3.34. Fλz satises the following properties:

a) Fλz ∂za(z) = −λFλz a(z);b) Fλz (ezTa(z)) = Fλ+T

z a(z), if a(z) ∈ U((z));c) Fλz a(−z) = −F−λz a(z);d) Fλz ∂

nwδ(z, w) = eλwλn.

Proof. Since Resz ∂z(·) = 0, we get

Fλz ∂za(z) = Resz eλz∂za(z) = −Res(∂ze

λz)a(z) = −λResz eλza(z) = −λFλz a(z),

that proves a), while b) is clear. Let us prove d);

Fλz ∂nwδ(z, w) = Resz e

λz∂nwδ(z, w) = ∂nw Res eλzδ(z, w) = ∂nweλw = λneλw.

Exercise 17. Prove part c).18

Solution. Dene 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

Resz eλza(−z) + Resz e

−λza(z) = 0.

Denition 3.35. Let V be a vertex algebra, with a −→ a(z) the state-eld correspondence, dene theλ-bracket on V by

[aλb] = Fλz a(z)b.

If we apply Fλz to [a(z), b(w)]c using the commutator formula we get

Fλz [a(z), b(w)]c = Fλz∑j∈Z+

(a(j)b)(w)c∂jwδ(z, w)

j!=∑j∈Z+

(a(j)b)(w)cFλz∂jwδ(z, w)

j!=

= eλw∑j∈Z+

(a(j)b)cλj

j!= eλw[aλb](w)c.

Since [a(z), b(w)]c = a(z)b(w)c− p(a, b)b(w)a(z)c, applying Fλz we get

[aλb(w)c] = p(a, b)b(w)[aλc] + eλw[aλb](w)c. (3.9)

Proposition 3.36. The λ-bracket on V has the following properties:

(sesquilinearity) [Taλb] = −λ[aλb] and [aλTb] = (T + λ)[aλb];(skewsymmetry) [aλb] = −p(a, b)[b−λ−Ta];(Jacobi identity) [aλ[bµc]] = p(a, b)[bµ[aλc]] + [[aλb]λ+µc].

Proof. We prove only the Jacobi identity. Let us apply Fµw to equation (3.9), then we get

[aλ[bµc] = p(a, b)[bµ[aλc]] + Fµweλw[aλb](w)c = p(a, b)[bµ[aλc]] + Resw e

(λ+µ)w[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

[Taλb] = Fλz (Ta)(z)b = Fλz ∂za(z)b = −λFλz a(z)b = −λ[aλb].

For the second case we have

[aλTb] = Fλz a(z)Tb = Fλz [a(z), T ]b+ Fλz T (a(z)b) = −Fλz ∂za(z)b+ ∂zFλz a(z)b = (λ+ ∂)[aλb].

To prove the skewsymmetry property we have only to apply Fλz to the relation

a(z)b = p(a, b)Fλz ezT b(−z)a.

Then we get

[aλb] = p(a, b)Fλ+Tz b(−z)a = −p(a, b)F−λ−Tz b(z)a = −p(a, b)[b−λ−Ta],

where we have used properties b) and c) of the formal Fourier transform.

Denition 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 thefollowing

i) Curg = C[T ]⊗ g, where [aλb] = [a, b], a, b ∈ g, orii) V ir = C[T ]L, where [LλL] = (T + 2λ)L.

We want to arrive to another equivalent denition of vertex algebra. If we write

[aλb] =∑n∈Z+

(a(n)b)λn

n!

we need only to consider in addition the product, called normally ordered product,

: ab := a(−1)b.19

Let us recall that

:Tn

n!ab := a(−n−1)b, n ∈ Z+.

If we compare the coecients of both sides of equation (3.9) we get

[aλ : bc :] = p(a, b) : b[aλc] : + : [aλb]c : +

∫ λ

0

[[aλb]µc]dµ. (3.10)

Let us prove this formula: we need only to compute the constant term of

eλw[aλb](w)c =∑m∈Z+

λmwm

m!

∑n

[aλb](n)cw−n−1,

that is equal to∑m∈Z+

λm

mc =: [aλb]c : +

∑m∈Z+

λm+1

(m+ 1)c =: [aλb]c : +

∫ λ

0

[[aλb]µc]dµ.

Equation (3.10) is called non-commutative Wick formula.

Proposition 3.39. The normally ordered product is quasi-commutative, that is

: ab := p(a, b) : ba : +

∫ 0

−T[aλb]dλ,

and is quasi-associative, that is,

:: ab : c :=: a : bc :: +∑j∈Z+

a(−j−2)(b(j)c) + p(a, b)∑j∈Z+

b(−j−2)(a(j)c).

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)eTz(a(−z)b).Taking the coecient of z−n−1 we get the relation

b(n)a = −p(a, b)∑j∈Z+

(−1)j+nT j

j!(a(n+j)b)

and for n = −1 we get

b(−1)a = p(a, b)∑j∈Z+

(−1)jT j

j!(a(j−1)b) = p(a, b)a(−1)b+ p(a, b)

∑j∈Z+

(−1)j+1 T j+1

(j + 1)!(a(j)b)

that means

: ba := p(a, b) : ab : −p(a, b)∫ 0

−T[aλb]dλ.

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

: a(z)b(z) : =∑m∈Z

∑n≤−1

a(n)b(m)z−m−n−2 + p(a, b)

∑m∈Z

∑n∈Z+

b(m)a(n)z−m−n−2 =

=∑m∈Z

∑n≤−1

a(n)b(m−n−1)z−m−1 + p(a, b)

∑m∈Z

∑n∈Z+

b(m−n−1)a(n)z−m−1.

Taking the constant term we get

(a(−1)b)(−1) =∑n≤−1

a(n)b(−n−2) + p(a, b)∑n∈Z+

b(−n−2)a(n) = a(−1)b(−1) +∑n≤−2

a(n)b(−n−2)+

+ p(a, b)∑n∈Z+

b(−n−2)a(n) = a(−1)b(−1) +∑n∈Z+

a(−n−2)b(n) + p(a, b)∑n∈Z+

b(−n−2)a(n).

Finally, applyng to c we get

(a(−1)b)(−1)c = a(−1)(b(−1)c) +∑n∈Z+

a(−n−2)(b(n)c) + p(a, b)∑n∈Z+

b(−n−2)(a(n)c).

20

We are ready now for the fourth denition of vertex algebra.

Denition 3.40. A vertex algebra is the data of (V, |0〉, T, [aλb], : ab :), where (V, T, [aλb]) is a Lieconformal algebra, (V, |0〉, T, : ab :) is a quasi-commutative, quasi-associative unital dierential algebraand the λ-bracket and the normally ordered product are related by the non-commutative Wick formula.

For the equivalence with the other denitions we remind to [BK03].A homomorphism of vertex algebras ϕ : V1 −→ V2 is a linear map preserving all products such that

ϕ(|0〉1) = |0〉2. A subalgebra U ⊂ V is a subspace containing |0〉 and closed under all products. Inparticular TU ⊂ U , since Ta = a(−2)|0〉. An ideal I ⊂ V is a subspace, not containing |0〉, such thatTI ⊂ I and a(n)I ⊂ I for all a ∈ V and n ∈ Z. Indeed, from the skewsymmetry, it follows that everyideal is a two-sided ideal.

Example 3.41. Let (V, |0〉, T,F) be a vertex algebra, then an aj(n)-invariant and T -invariant subspace

U not containing |0〉 is an ideal since all elds are n-th products of aj(z), hence their coecients are

combination of products of aj(n)'s.

Example 3.42. Let V k(g) = U(g)/(U(g)g[t] + U(g)(K − k)) = Indg

g[t]+CK Ck, then the ideals are the

g-invariant subspaces which are T -invariant.

Example 3.43. V c 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.

Proof. L0 is diagonalizable on V c with positive eigenvalues and the only eigenvector of eigenvalue 0 is|0〉. So any ideal is L0-invariant and Z+-graded. Then the sum of proper ideal is again a proper idealand the claim follows.

The corresponding simple vertex algebra is denoted Vc. Here are some methods of constructing newvertex 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;

Exercise 20. The tensor product of vertex algebras (Vi, |0〉i, T, ai(z)), i = 1, 2, is the following vertexalgebra:

(V = V1⊗V2, |0〉 = |0〉1⊗|0〉2, T = T1⊗1V2+1V1

⊗T2, a⊗b(z) = a(z)⊗b(z) =∑

(a(m)⊗b(n))z−m−n−2).

Solution. We should check the axioms of vertex algebra for the tensor product. For the vacuum axiomswe have

T |0〉 = (T1 ⊗ 1V2 + 1V1 ⊗ T2)(|0〉1 ⊗ |0〉2) = 0⊗ |0〉2 + |0〉1 ⊗ 0 = 0

anda⊗ b(z)|0〉|z=0

= a⊗ b(z)(|0〉1 ⊗ |0〉2)|z=0= a(z)|0〉1 ⊗ b(z)|0〉2|z=0

= a⊗ b.For the translation covariance we have

[T, a⊗ b(z)] = [T1 ⊗ 1V2, a⊗ b(z)] + [1V1

⊗ 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.

Then multiplying for (z − w)n for suciently large n gives zero.

d) common kernel of one or several derivation of a vertex algebra, that is a linear operator D such thatD(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 V = C[xn, n ∈ Z+], the vacuum vector is |0〉 = 1 and F = α(z) =∑n∈Z αnz

−n−1,where

αn =

∂∂xn

n > 0

−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|0〉 we have [αλα] = λ|0〉.

Exercise 22. [T, αn] = −nαn−1, write T .

Solution. We claim that T =∑n≥2 α−nαn−1. Indeed,

[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

[T, αm] = −∑n≥2

nδn,mαn−1 = −mαm−1,

while, if m < 0 the rst term vanishes and we get

[T, αm] =∑n≥2

(n− 1)δn−1,−mα−n = −mαm−1.

Example 4.1. We have

:: αα : α : − : α : αα ::=∑j∈Z+

α(−j−2)(α(j)α) = 2α(−2)|0〉.

The corresponding eld is 2∂zα(z) 6= 0.

Let L = 12 : αα :, then

[αλL] =1

2[αλ : αα :] =

1

2: [αλα]α : +

1

2: α[αλα] : +

1

2

∫ λ

0

[[αλα]µα]dµ = λα,

so [Lλα] = (λ+ T )α. Moreover

[LλL] =1

2[Lλ : αα :] =

1

2: [Lλα]α : +

1

2: α[Lλα] : +

1

2

∫ λ

0

[[Lλα]µα]dµ =

=1

2: (λ+ T )αα : +

1

2: α(λ+ T )α : +

1

2

∫ λ

0

[(λ+ T )αµα]dµ =

=1

2: Tαα : +

1

2: αTα : +λ : αα : +

1

2

∫ λ

0

(λ− µ)[αµα]dµ =

= (T + 2λ)(1

2: αα :) +

1

2

∫ λ

0

(λµ− µ2)dµ|0〉 = (T + 2λ)L+λ3

24|0〉.

Thus L denes a Virasoro algebra with central charge 1 and α is an eigenvector of conformal weight 1.Recall that if a has conformal weight ∆, then [Lλa] = (T + ∆λ)a+ o(λ2).

Denition 4.2. a is called primary if [Lλa] = (T + ∆λ)a.

Exercise 23. a is primary if and only if Lna = δn,0∆a, for all n ∈ Z+.22

Solution. By denition, it follows that

[Lλa] = Fλz L(z)a = Resz eλzL(z)a = Resz

∑m∈Z+

∑n∈Z

λm

m!(Ln)azn+m =

∑m∈Z+

λm

m!L−m−1a.

Comparing the coecient of λm we have that [Lλa] = (T + ∆λ)a if and only if Lna = δn,0a for alln ∈ Z+.

Let us denote this algebra B.

Proposition 4.3. B is a vertex algebra with energy-momentum vector L = 12 : αα :, α being a primary

element of conformal weight 1.

Proof. We have L−1 = T . Moreover L0 is diagonalizable with eigenvalues in Z+ and

Bn = span(α−ns . . . α−n1|0〉 | 0 ≤ n1 ≤ . . . ≤ ns,

∑ni = n).

Bn has dimension p(n), the number of partition of n. Moreover B0 = C|0〉, then B is a simple vertexalgebra.

B is called vertex algebra of free boson. We note that V k(g) is a generalization of B: the commutatoron g is dened as

[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 theSugawara construction.

Theorem 4.4. Let g be a simple nite dimensional Lie algebra with a non-degenerate bilinear form(· | ·). Let ai and bi be dual basis of g with respect to this form. Let Ω =

∑aibi be the Casimir

element and let 2h be the eigenvalue of Ω on g. Let V = V k(g), k ∈ C, where g ⊂ V via a −→ a−1|0〉,and let T = 1

2

∑: aibi :, then

a) [aλT ] = (k + h)λa, a ∈ g;

b) L = 1k+h

T is an energy-momentum eld with c = k dim g

k+h, provided that k 6= h;

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

[aλT ] =1

2

∑i

: [aλai]bi : +1

2

∑i

: ai[aλbi] : +1

2

∑i

∫ λ

0

[[aλai]µbi]dµ =

=1

2

∑i

: [a, ai]bi : +λk

2

∑i

(a | ai)bi +1

2

∑i

: ai[a, bi] : +λk

2

∑i

(a | bi)ai+

+1

2

∑i

∫ λ

0

([[a, ai], bi] + µk([a, ai] | bi)) dµ.

We recall that, for any a ∈ g, we have

a =∑i

(a | bi)ai =∑i

(a | ai)bi,

thus we obtain∑i

: [a, ai]bi :=∑i,j

([a, ai] | bj) : ajbi := −∑i,j

([a, bj ] | ai) : ajbi :=−∑j

: aj [a, bj ] : .

Furthermore, we not that∑

([a, ai] | ai)bi = tr ad a = 0. Indeed, using the dual basis, we also havetr ad a =

∑([a, bi] | ai) = −

∑([a, ai] | bi) = − tr ad a. Then we get

[aλT ] = λka+λ

2

∑i

[[a, ai], bi] = λka+λ

2

∑i

[ai, [bi, a]] = λka+λ

2Ω(a) = (k + h)λa,

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

[LλL] =1

2(k + h)

∑i

[Lλ : aibi :] =1

2(k + h)

∑i

(: [Lλai]bi : + : ai[Lλbi] : +

∫ λ

0

[[Lλai]µbi]dµ

)=

=1

2(k + h)

∑i

(: (λ+ ∂)ai · bi : + : ai(λ+ ∂)bi : +

∫ λ

0

(λ− µ)([ai, bi] + µk(ai | bi))dµ

)=

= (∂ + 2λ)L+k dim g

2(k + h)

∫ λ

0

(λ− µ)µdµ = (∂ + 2λ)L+k dim g

12(k + h)λ3.

5. Poisson Vertex Algebras

We want to dene the notion of quasiclassical limit of a family of vertex algebras. Let

(V~, |0〉~, T~, [aλb]~, : ab :~)

be a family of vertex algebras, meaning that V~ is a free module over C[[~]] such that all axioms hold and[V~λV~] ⊂ ~V~[λ]. The quasiclassical limit is V = V~

/~V~. We have that V 3 1 = im(|0〉~), T~ induces an

operator ∂ on V and : ab : induces a product on V.We dene a Poisson λ-bracket aλb on V as follows: let a and b be preimages of a, b ∈ V

/V~ in V

and let

aλb =[aλb]

~∈ V/~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 = aλbc+ baλc.

Proof. By quasi-commutativity we have in V~

: ab :~ − : ba :~=

∫ 0

−T~

[aλb]~dλ.

But [aλb]~ ∈ ~V~, so in the limit ~ → 0, we have ab = ba. Similarly, quasi-associativity impliesassociativity when ~ → 0. This proves part a). Part b) is obvious since all axioms of Lie conformalalgebras are homogeneous. For c) if we divide by ~ the non-commutative Wick formula and take thelimit we get the Leibniz rule.

Exercise 25. Deduce from aλb = −b−λ−∂a, thatbcλa = 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−∂bc− a−λ−∂cb = bλ+∂a→c+ cλ+∂a→b.

Example 5.2. Let (g,F , T ) be a regular formal distibution Lie algebra, in particular g = g0 is aVirasoro algebra and g− is the annihilator subalgebra (for example g− = g0[t] and V ir− =

∑n≥−1 CLn),

then we have the associative vertex algebra V (g,F) = U(g)/U(g)g−. Let us consider g~ with bracket

[a, b]~ = ~[a, b], so

V~(g~,F) = UC[[~]](g~)/UC[[~]](g~)g~−

over C[[~]] is a family of vertex algebras. The quasiclassical limit of the family is denoted by V(g,F) andis constructed as follows in the case g = g0: choose a basis aii∈J of g0 so that F =

∑(aitn)z−n−1i∈J

and let uj = aj(−1)|0〉 in V~(g~,F) and uj be its image in V(g,F), then

V(g,F) = C[u(n)j | j ∈ J, n ∈ Z+]

with ∂u(n)j = u

(n+1)j and uiλuj =

∑k∈J c

kija

k, where [ai, aj ] =∑k c

kija

k and extend by sesquilinearityand both Leibniz rules.

24

A special case is when g0 = C. We set B~ = V~(g~, α(z)), where α(z) =∑αnz

−n−1 with[αn, αm]~ = ~mδm,−n. The quasiclassical limit B is a Poisson vertex algebra called Gardner-Fadeev-Zakharov Poisson vertex algebra.

Exercise 26. B ∼= C[u(n) | n ∈ Z+] with ∂u(n) = u(n+1) and uλu = λ.

Proof. Since g0 is one dimensional, by previous example we have B ∼= C[u(n) | n ∈ Z+] with ∂u(n) =u(n+1), where u is the image in B of u = α−1|0〉 ∈ B~. Moreover, [uλu] = ~λ, then we get uλu = λ.

Example 5.3. The Virasoro Poisson vertex algebra is V = C[u(n) | n ∈ Z+] with ∂u(n) = u(n+1) and

uλu = (∂ + 2λ)u+ λ3

12 c.

Let M be a nite dimensional manifold and V be the algebra of C∞ functions on M . Introduce in Va Poisson bracket ·, · such that

(1) V is a unital commutative associative algebra;(2) Lie algebra axioms hold;(3) Leibniz rule holds: a, bc = a, bc+ ba, c.

Also choose a function h ∈ V, called Hamiltonian function, then a system of Hamiltonian equations onuii∈I (here ui's are "topological" generators of V) is:

duidt

= h, ui, i ∈ I (5.1)

and ui = ui(x, t). A function f is called a conservation law if dfdt = 0 on the trajectories. This can be

expressed in terms of the Poisson bracket as h, f = 0. Indeed, by the Leibniz rule, dfdt = h, f, f ∈ V.The system (5.1) is called integrable if we have "suciently many" commuting conservation laws.

In the case of an innite dimensional Hamiltonian system, the manifold M is replaced by M =

Map(I,M). If ui are coordinates in M , then on M we let ui(t) =∑n∈Z+

tn

n!u(n)i . So, coordinates in M

are u(n)i , i ∈ I. We take for the space of functions on M the algebra V = C[u

(n)i |∈ I, n ∈ Z+]. Thus we

have a dierential algebra, since ∂u(n)i = u

(n+1)i . One consider usually local Poisson brackets

ui(x), uj(y) = Bij(u(y), ∂y)δ(x, y), (5.2)

where∫Mf(x)δ(x, y) = f(y).

Example 5.4. We can just dene u(x), u(y) = δ(x, y).

By bilinearity and Leibniz rule we can extend this bracket to any P,Q ∈ V:

P (x), Q(y) =∑α,β,i,j

∂P (x)

∂u(α)i

∂Q(y)

∂u(β)j

∂αx ∂βy ui(x), uj(y). (5.3)

This is called Poisson bracket if it is skewsymmetric. This happens if and only if Bij is a skewsymmetricdierential operator and satises the Jacobi identity.

Hamiltonian functions and integrals of motion are not elements of V, but they are local functionals:

IP =

∫M

P (x)dx, P ∈ V.

Taking an algebraic point of view, local functionals are elements of V/∂V. Given h ∈ V we denote its

image in V/∂V by

∫h. Using bilinearity and integration by part we get

IP , IQ =

∫ ∫ ∑α,β,i,j

∂P (x)

∂u(α)i

∂Q(y)

∂u(β)j

∂αx ∂βy ui(x), uj(y)dxdy =

∫ ∫ ∑i,j

δP

δui

δQ

δujui(x), uj(y)dxdy,

whereδP

δui=∑n∈Z+

(−∂)n∂P

∂u(n)i

.

Finally, using (5.2) we can write

∫P,∫Q =

∫ (BδP

δu

)δQ

δu.

This is a Poisson bracket if V/∂V is a Lie algebra with the induced bracket.

A Hamiltonian functional is∫h ∈ V

/∂V and the Hamiltonian equation is

dujdt

= ∫h, uj = B

δh

δuj;

25

∫f is a conserved quantity (integral of motion) if

∫h,∫f = 0. The Hamiltonian equation is called

integrable if there exist innitely many linearly independent commuting integrals of motion.Taking the formal Fourier transform of both sides of (5.2) we get:

uiλuj =

∫eλ(x−y)ui(x), uj(y)dx.

Applying the formal Fourier transform to (5.3) gives

PλQ =∑i,j∈I

m,n∈Z+

∂Q

∂u(n)j

(∂ + λ)nuiλ+∂uj→(−∂ − λ)m∂P

∂u(m)i

.

Proposition 5.5. We have

(1) this λ-bracket on V denes the structure of a Poisson vertex algebra, where Bij(∂) = uj∂ui;(2) an Hamiltonian function is

∫h ∈ V

/∂V and the corresponding Hamiltonian system is

duidt

= ∫hλui

∣∣λ=0

;

(3) the integrals of motion are elements∫f ∈ V

/∂V such that

∫fλ∫h∣∣λ=0

= 0 in V/∂V.

Lemma 5.6. If V is a Poisson vertex algebra, consider P,Q = PλQ|λ=0, then

(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 axiomof Lie conformal algebras.

Example 5.7. Let V = C[u(n), n ∈ Z+] with uλu = λ, then∫u,

∫u2,

∫u3 − (u′)2, . . .

form an innite sequence of commuting elements in V/∂V and

du

dt= ∫u3 − (u′)2, u = 3uu′ + u′′′.

In particular, in V/∂V, we have

∫h,∫f =

∫δf

δu∂δh

δu.

6. Representation theory

Denition 6.1. A representation of a vertex algebra V in a vector (super)space M is a linear map fromV to a space of EndM -valued quantum elds, a −→ aM (z) =

∑aMn z

−n−1, such that

(1) |0〉M (z) = 1M ;

(2) aM (z)bM (z)iz,w(z − w)n − p(a, b)bM (w)aM (z)iw,z(z − w)n =∑j∈Z+

(a(n+j)b)M (w)

∂jwδ(z,w)j! .

If we put n = 0, we get the commutator formula

[aM (z), bM (w)] =∑j∈Z

(a(j)b)M (w)

∂jwδ(z, w)

j!. (6.1)

From (6.1) and aM (w)(n)bM (w) = (a(n)b)

M (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 satisfythe commutator formula. If a(z) =

∑(atn)z−n−1 and b(z) =

∑(btn)z−n−1, then (6.1) becomes

[aM (z), bM (w)] = [a, b]M (w)δ(z, w) + k(a|b)1M∂wδ(z, w).

This formula means that we have a g-module M with K = k and aMn (m) = 0, for m ∈ M , a ∈ g andn >> 0. So any V k(g)-module is an extension to the whole vertex algebra of a restricted g-module Mwith K = k.

26

Recall that we had another construction of V k(g) as a subspace of the completed U(g)/(K − k). So,

if we have a restricted g-module with K = k1M , we can extend it to the completion of U(g)/(K − k),

which contains V k(g).In conclusion, any V k(g)-module is obtained from a restricted g-module M with K = k1M .

What are the restricted g-modules with K = k?

Example 6.4. Let U be an arbitrary g-module, then Indgg[t]+CK U can be extended to g[t] by letting

(gtn)|U = 0, for n >> 0 and K|U = k. This is a highest weight module if and only if U is a highestweight g-module.

Let I be the unique maximal ideal of V k(g), then Vk(g) = V k(g)/I is a simple vertex algebra. What

are its representations? They are the V k(g)-modules M such that IMM = 0. So we need to study I: ifa ∈ I, then aM (z) = 0.

Example 6.5. Consider the case k ∈ Z+, then Vk(g) is an irreducible g-module with K = k1 andan|0〉 = 0 for a ∈ V and n ∈ Z+. Let ei, fi, hi be the Chevalley generators of g and let (θ|θ) = 2, whereθ is the highest root of g. Then g|0〉 = 0, so fi|0〉 = 0, hi|0〉 = 0 and ei|0〉 = 0, for i = 1, . . . , r = rank(g).Also g ⊂ g and one set e0 = e−θt, f0 = eθt

−1 and h0 = k − θ. Then e0|0〉 = 0 and h0|0〉 = k|0〉. HenceVk(g) is an irreducible highest weight module over g with highest weight vector |0〉 of weight Λ0 suchthat Λ0(hi) = kδ0,i. We denote this module by L(Λ0).

Proposition 6.6. I is a left g-module generated by fk+10 |0〉.

Hence, for k ∈ Z+, a Vk(g)-moduleM is actually a Vk(g)-module if and only if (eMθ (z))k+1 = 0. Thus

we need to nd all restricted g-modules M with K = k such that (eθ(z))k+1 = 0 in M .

Recall that Vk(g) = V k(g)/I, where I is the maximal submodule. But G, the group with Lie algebra g,

acts by . . . of g such that K is xed and g[t] is an invariant subalgebra. Hence G acts by automorphismsof V k(g), hence of Vk(g), so eσθ (z)k+1 = 0 on M for each σ ∈ G.

What does the relation a(z)k+1 = 0, for a ∈ g, mean? If a(z) =∑

(atn)z−n−1, then a(z)k+1 = 0 ifand only if ∑

n1+...+nk+1=N

(atn1) . . . (atnk+1) = 0 on M, ∀N ∈ Z. (6.2)

We can ask which of the irreducible g-modules L(Λ), with Λ(K) = k, are Vkg-modules. We know thatL(Λ0) is. Apply both sides of (6.2) to vΛ and take N = −k − 1, then we have a term

(at−1)kvΛ + (. . .)vΛ,

where in the parenthesis one of the factors is atn with n ∈ Z+. Take a = eθ, then (eθtn)vΛ = 0 for

n ∈ Z+, that means (eθt−1)k+1vΛ = 0. So fk+1

0 vΛ = 0. If we now take N = 0 and a = ealpha, α > 0, in(6.2) we get

ek+1−α vΛ + . . . (e−αt)vΛ . . . = ek+1

−α vΛ = 0.

Then fk+1i vΛ = 0. This two conditions are satised if and only if ∆(hi) ∈ Z+, for all i.

Then only the integrable g-modules L(Λ), with Λ(K) = k, can be Vk(g)-modules.

Theorem 6.7. Let k ∈ Z+, then all irreducible Vk(g)-modules are L(∆) ∆(K)=k∆(hi)∈Z+

. Moreover, every

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.

In the case of Virasoro vertex algebras V c and Vc we have that any restricted moduleM over the Virasoroalgebra with C = c is a V c-module.

Theorem 6.8. Vc = V c, that is V c is an irreducible module over the Virasoro algebra, if and only if cis not of the form

cp,q = 1− 6(p− q)2

pq, (6.3)

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 highestweight 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 ∈ EndV be an energy operator, that we remember is a diagonalizable operator, such that any

H-eigenvector a has eigenvalue ∆a. Setting

a(z) =∑

n∈−∆a+Zanz−n−∆a ,

we have [H, an] = −nan. Assume that specH ⊂ Z+.

Denition 6.10. A positive energy V-module M is a graded representation M = ⊕j∈Z+Mj such that

aMn Mj ⊂Mj−n. (6.4)

The condition means that aMn M0 = 0 for n > 0 and aM0 M0 ⊂M0. For example, if L ∈ V is an energymomentum state, then H = L0 is an energy operator and if we have a highest weight representation Mof V k(g) or V c, then the eigenspace decomposition with respect to LM0

M =⊕

j∈h+Z+

Mj

is such that (6.4) holds (respect to the usual decomposition we have Mj = Mj−h). Write aM (z) =∑aMn z

−n−∆a , then Borcherds identity reads∑j∈Z+

(−1)j(n

j

)(aMm+n−jb

Mk+j−n − (−1)nbMk−ja

Mm+j

)=∑j∈Z+

(m+ ∆a − 1

j

)(a(n+j)b)

Mm+k.

Pick v ∈M0 and apply both sides of Borcherds identity to v. For m = 1, n = −1 and k = −1 we get

aM0 bM0 (v) =∑j∈Z+

(∆a

j

)(a(j−1)b)(v). (6.5)

Exercise 27. Let m = 1, n = −2 and k = −1, then for v ∈M0, Borcherds identity is∑j∈Z+

(∆a

j

)(a(j−2)b)(v) = 0. (6.6)

Solution. The left hand side of Borcherds identity applied to v is zero since both k + j − n and m + jare greater than zero. The right hand side is∑

j∈Z+

(∆a

j

)(a(j−2)b)(v),

thus proving identity (6.6).

Dene the product ∗ on V by RHS of (6.5):

a ∗ b =∑j∈Z+

(∆a

j

)a(j−1)b.

Exercise 28. Prove that ∑j∈Z+

(∆a

j

)a(j−2)b = ((T +H)a) ∗ b.

Solution. Recall that (Ta)(n) = −na(n−1) and (Ha)(n) = (∆a−n− 1)a(n). Then, using the denition ofthe product ∗ on V we get:

((T +H)a) ∗ b =∑j∈Z+

(∆a

j

)a(j−2)b+

∑j∈Z+

((∆a

j

)(∆a − j)−

(∆a

j + 1

)(j + 1)

)a(j−1)b .

The claim follows since(

∆a

j

)∆a−jj+1 (j + 1) =

(∆a

j+1

)(j + 1).

Thus the map (V, ∗) −→ EndM0, dened by a −→ aM0 |M0is a homomorphism such that the kernel

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 isa two-sided ideal of the algebra (V, ∗) such that Zhu(V ) = (V

/J, ∗) is a unital associative algebra with

1 = im(|0〉).

Theorem 6.12. Given a positive energy V -module M we have a Zhu(V )-module M0 given by a −→aM0 |M0

. Thus we get a functor from the category of positive energy V -modules to the category of Zhu(V )-

modules. Conversely, given a Zhu(V )-moduleM0, one can construct the corresponding almost irreduciblepositive 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 submodulehas a nonzero intersection with M0

Proof. Letting n = 0 in Borcherds identity we get the commutator formula:

[aMm , bMk ] =

∑j∈Z

(m+ ∆a − 1

j

)(a(j)b)

Mm+k,

with m, k ∈ Z. So we get a Z-graded, with deg an = n, Lie algebra

g =⊕j∈Z

gj .

Let g∈Z+= ⊕j∈Z+

gj , given a representation of the Zhu algebra in M0, we get in particular a represen-tation of the Lie algebra g0 in M , hence of g∈Z+

.

Exercise 29. Prove that a ∗ b− b ∗ a|M0= [a0, b0]|M0

.

Consider Indgg∈Z+

= M = ⊕j∈Z+Mj .

Theorem 6.13. LetM = M/(maximal g-submodule intersecting M0 trivially), then the whole Borcherds

identity holds, hence M is a V -module.

Example 6.14. V = V k(g) is spanned by elements (a1t−k1) . . . (ast−ks)|0〉 = a1(−k1) . . . a

s(−ks)|0〉, where

ai ∈ g and ki ≥ 1. But vectors of the form a(−2)b+ . . . ∈ J , hence mod J we can remove factors with

ki > 1, then V = a1(−1) . . . a

s(−1)|0〉+ J . We have a surjective homomorphism

U(g) −→ Zhu(V k(g))

a1 . . . as −→ a1(−1) . . . a

s(−1)|0〉 ∈ V

/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 Zhu(Vk(g)) = U(g)/(maximal ideal). If k ∈ Z+, then the element

(eθt−1)k+1|0〉 ∈ V k(g) is zero in Vk(g) and it goes to ek+1

θ ∈ U(g) under the previous isomorphism.So we have the following theorem:

Theorem 6.17. If k ∈ Z+, then Zhu(Vk(g)) ∼= U(g)/(ek+1θ ).

Let J = (ek+1θ ) and let G be the Lie group corresponding to g, so J contains all the elements

(eσθ )k+1, σ ∈ G. Since g is simple, the span of eσθ is g, hence there is a basis of elements a1 . . . ad ofg such that alla ai are nilpotent elements of g and (ai)k+1 ∈ J . By Poincaré-Birkho-Witt theorem,

elments (a1)k1 . . . (ad)kd 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 V is sum of irreducibles. The case k 6∈ Z+ is still open.

Example 6.18. Since Vc = V c/I, then Zhu(Vc) ∼= C[x]

/(f(x)). But we know that Vc = V c unless

c 6= cp,q = 1− 6(p−q)2pq , p, q ≥ 2, (p, q) = 1. If c = cp,q, then V

c is not simple and f(x) is a polynomial of

degree (p−1)(q−1)2 . If p = 3 and q = 2, then deg f(x) = 1, so f(x) = x and cp,q = 0. If p = 4 and 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 anilpotent element of g, then we associate to them a so called W -algebra, Wk(g, f). Let us see how thisis 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 toadx, g decomposes as

g =⊕i∈ 1

2Z

gi, g = g− ⊕ g0 ⊕ g+.

We have a non-degenerate pairing between g− and g+, so g− = g∗+. In particular f ∈ g∗+. Let G+ bethe Lie group corresponding to g+: it acts on g+ and g∗+ and we have G+ · f = O ⊂ g∗+. G+ also actson g and µ : g∗ −→ g∗+ is the moment map for this action. We consider

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 nitevertex algebras.

Some famous examples are

Virasoro V c ∼=Wk(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).

30