Chapter 2 of 10

§Vertex algebras

§2.1 Definition and properties of a vertex algebra

We first give the definition in terms of locality, then derive its consequences and compare equivalent formulations. Throughout, we work over $\mathbb C$ with ordinary, purely even vertex algebras. We assume familiarity with formal calculus; all variables are formal.

For endomorphisms $A,B\in\operatorname{End}(V)$ $End (V)$ , the bracket $[A,B]=AB-BA$ denotes their commutator; for fields it is interpreted coefficientwise.

Definition 2.1 (Vertex algebra)

A vertex algebra is a tuple $(V,Y,\mathbf1,T)$ consisting of a complex vector space $V$ $V$ , a nonzero vacuum vector $\mathbf1\in V$ $V$ , a linear translation operator $T\colon V\to V$ $V$ , and a linear state-field correspondence (also called the state-operator correspondence)

Y\colon V\longrightarrow \operatorname{End}(V)[[z,z^{-1}]], \qquad a\longmapsto Y(a,z)=\sum_{n\in\mathbb Z}a_{(n)}z^{-n-1},

satisfying the following conditions for all $a,b\in V$ $V$ .

Lower truncation. $Y(a,z)b\in V((z))$ $V ((z))$ ; equivalently, $a_{(n)}b=0$ for sufficiently large $n$ $n$ , with a bound that may depend on $a$ and $b$ $b$ .
Vacuum and creation.

$Y(\mathbf1,z)=\operatorname{id}_V, \qquad Y(a,z)\mathbf1\in a+zV[[z]].$
Translation covariance and invariance of the vacuum.

$T\mathbf1=0, \qquad [T,Y(a,z)]=\partial_zY(a,z).$
Locality. There exists an integer $N=N(a,b)\geq0$ such that

$(z-w)^N[Y(a,z),Y(b,w)]=0.$

Remark 2.2 (State-field correspondence in physics)

In physics, the map $Y$ is often left implicit by writing $a(z)=Y(a,z)$ $Y (a, z)$ , $|0\rangle=\mathbf1$ $1$ , and $|a\rangle=a$ $a$ . The creation condition in Definition 2.1 then becomes

\left.a(z)|0\rangle\right|_{z=0}=|a\rangle,

often written as $\lim_{z\to0}a(z)|0\rangle=|a\rangle$ $∣ a ⟩$ . Here evaluation at zero means taking the constant term of a formal power series; an analytic limit requires a separate analytic interpretation.

The vacuum and translation axioms express the identity field, creation of states from the vacuum, and infinitesimal translations. Locality encodes compatibility between fields, whose commutator is annihilated by a sufficiently high power of $z-w$ $w$ . But note that this does not require the commutator itself to vanish. We will explain its connection with OPEs below.

§2.2 Vacuum and translation

§Creation and translation identities

Proposition 2.3 (Creation from the vacuum)

For every $a\in V$ $V$ ,

Y(a,z)\mathbf1=e^{zT}a.

Consequently, for $k\geq0$ $0$ ,

a_{(-k-1)}\mathbf1=\frac{T^ka}{k!}, \qquad a_{(n)}\mathbf1=0\quad(n\geq0).

In particular, $a_{(-1)}\mathbf1=a$ and $a_{(-2)}\mathbf1=Ta$ $T a$ .

Proof.

The creation axiom gives $G(z):=Y(a,z)\mathbf1\in V[[z]]$ with $G(0)=a$ $a$ . Using $T\mathbf1=0$ $0$ ,

\partial_zG(z)=[T,Y(a,z)]\mathbf1=TG(z).

Coefficient recursion gives $G(z)=\sum_{k\geq0}z^kT^ka/k!$ $\sum_{k \geq 0} z^{k} T^{k} a / k!$ . Comparing modes proves the remaining identities.

Thus inserting a field on the vacuum produces the corresponding translated state.

The operator $T$ acts on states, while its commutator acts on fields. Exponentiating this commutator translates the insertion coordinate.

Proposition 2.4 (Formal translation of fields)

For a formal parameter $x$ $x$ ,

e^{xT}Y(a,z)e^{-xT}=Y(a,z+x).

The right-hand side means its Taylor expansion in nonnegative powers of $x$ $x$ :

Y(a,z+x):=\sum_{k\geq0}\frac{x^k}{k!}\partial_z^kY(a,z).

Proof.

Set

F(x):=e^{xT}Y(a,z)e^{-xT} \in\operatorname{End}(V)[[z,z^{-1}]][[x]].

Translation covariance gives

\partial_xF(x) =e^{xT}[T,Y(a,z)]e^{-xT} =\partial_zF(x).

Writing $F(x)=\sum_{k\geq0}F_kx^k$ yields $(k+1)F_{k+1}=\partial_zF_k$ with $F_0=Y(a,z)$ $Y (a, z)$ , hence the Taylor expansion above.

§Skew-symmetry

Proposition 2.5 (Skew-symmetry)

For all $a,b\in V$ $V$ ,

Y(a,z)b=e^{zT}Y(b,-z)a.

Proof.

Choose $N\geq0$ so that locality holds for $Y(a,z)$ and $Y(b,w)$ and, by lower truncation,

u^NY(b,u)a\in V[[u]].

Applying locality to the vacuum gives

\begin{aligned} (z-w)^NY(a,z)Y(b,w)\mathbf1 &=(z-w)^NY(b,w)Y(a,z)\mathbf1. \end{aligned}

Using Proposition 2.3 on both sides, we obtain

(z-w)^NY(a,z)e^{wT}b =(z-w)^NY(b,w)e^{zT}a.

By Proposition 2.4, moving $e^{zT}$ to the left replaces $Y(b,w)$ by $Y(b,w-z)$ $z)$ , initially expanded in nonnegative powers of $z$ $z$ . Our choice of $N$ removes its poles:

(z-w)^NY(a,z)e^{wT}b =(z-w)^Ne^{zT}Y(b,w-z)a.

Both sides have only nonnegative powers of $w$ $w$ . Taking the constant term in $w$ is therefore legitimate and yields

\begin{aligned} z^NY(a,z)b &=z^Ne^{zT}Y(b,-z)a. \end{aligned}

Canceling $z^N$ in $V((z))$ proves the identity.

§Translation of states and differentiation of fields

We now use skew-symmetry to identify the field corresponding to $Ta$ $T a$ .

Proposition 2.6 (Translation and derivatives)

For $a\in V$ and $n\in\mathbb Z$ $Z$ ,

Y(Ta,z)=\partial_zY(a,z), \qquad (Ta)_{(n)}=[T,a_{(n)}]=-n\,a_{(n-1)}.

Proof.

Fix $b\in V$ $V$ . By Proposition 2.5 and translation covariance,

\begin{aligned} \partial_zY(a,z)b &=e^{zT}\bigl(TY(b,-z)-[T,Y(b,-z)]\bigr)a\\ &=e^{zT}Y(b,-z)Ta\\ &=Y(Ta,z)b. \end{aligned}

The minus sign comes from $\partial_zY(b,-z)=-[T,Y(b,-z)]$ $- [T, Y (b, - z)]$ ; the last equality uses skew-symmetry again. Comparing modes gives $(Ta)_{(n)}=-n a_{(n-1)}$ $- n a_{(n - 1)}$ , and translation covariance gives the same formula for $[T,a_{(n)}]$ $[T, a_{(n)}]$ .

§2.3 Products and iterates

§Locality and a common expansion

Proposition 2.7 (Common expansion of two products)

For $a,b,c\in V$ $V$ , there is a unique

F(z,w)\in V[[z,w]][z^{-1},w^{-1},(z-w)^{-1}]

such that

\begin{aligned} Y(a,z)Y(b,w)c&=\iota_{z,w}F(z,w),\\ Y(b,w)Y(a,z)c&=\iota_{w,z}F(z,w). \end{aligned}

Proof.

The two products lie in $V((z))((w))$ and $V((w))((z))$ $V ((w)) ((z))$ , respectively. Locality makes them equal after multiplication by $(z-w)^N$ $w)^{N}$ . Their common value belongs to

V((z))((w))\cap V((w))((z)) =V[[z,w]][z^{-1},w^{-1}].

Write this common value as $z^{-r}w^{-s}P(z,w)$ $z^{- r} w^{- s} P (z, w)$ , with $r,s\geq0$ and $P\in V[[z,w]]$ $V [[z, w]]$ , and set

F(z,w)=\frac{P(z,w)}{z^rw^s(z-w)^N}.

Expanding $(z-w)^{-N}$ in the two orders recovers the products. Injectivity of either expansion map gives uniqueness.

Theorem 2.8 (The iterate is the third expansion)

For the same $F$ as in Proposition 2.7,

Y(Y(a,x)b,w)c=\iota_{w,x}F(w+x,w).

The iterate is understood coefficientwise and belongs to $V((w))((x))$ $V ((w)) ((x))$ .

Proof.

We consider

\begin{aligned} \iota_{x,w}\!\left(e^{-wT}F(x+w,w)\right) &=e^{-wT}Y(a,x+w)Y(b,w)c\\ &=Y(a,x)e^{-wT}Y(b,w)c\\ &=Y(a,x)Y(c,-w)b. \end{aligned}

The last two equalities use Proposition 2.4 and Proposition 2.5. Since $e^{-wT}F(x+w,w)$ belongs to $V[[x,w]][x^{-1},w^{-1},(x+w)^{-1}]$ $w)^{- 1}]$ , uniqueness in Proposition 2.7, applied to $a,c,b$ with variables $x,-w$ $x, - w$ , identifies it as the common expression for the two product orders. Hence its other expansion is

\iota_{w,x}\!\left(e^{-wT}F(x+w,w)\right) =Y(c,-w)Y(a,x)b.

Multiplying by $e^{wT}$ and applying skew-symmetry coefficientwise yields

\begin{aligned} \iota_{w,x}F(x+w,w) &=e^{wT}Y(c,-w)Y(a,x)b\\ &=Y(Y(a,x)b,w)c. \end{aligned}

The operators $e^{\pm wT}$ act on vector coefficients and commute with the indicated expansions.

Thus the two products and the iterate are expansions of one expression. Schematically,

Y(a,z)Y(b,w)c\sim Y(b,w)Y(a,z)c\sim Y(Y(a,x)b,w)c, \qquad z=x+w.

Here $\sim$ means equality of the singular parts at $x=z-w=0$ $0$ , discarding nonnegative powers of $x$ $x$ . For the two products, this is understood through their common expression $F(w+x,w)$ expanded in $x$ $x$ . The common singular part is

\sum_{n\geq0}Y(a_{(n)}b,w)c\,x^{-n-1},

a finite sum by lower truncation. This is the singular operator product expansion (OPE).

§Weak associativity

Proposition 2.9 (Weak associativity)

For $a,b,c\in V$ $V$ , some integer $L\geq0$ satisfies

\begin{aligned} &(x+w)^L Y(a,x+w)Y(b,w)c\\ &\qquad=(x+w)^L Y(Y(a,x)b,w)c. \end{aligned}

On the left, $Y(a,x+w)$ is expanded in nonnegative powers of $w$ $w$ . After multiplication by the chosen factor, both sides lie in $V[[x,w]][x^{-1},w^{-1}]$ $V [[x, w]] [x^{- 1}, w^{- 1}]$ .

Proof.

Write the common expression from Theorem 2.8 as

F(z,w)=\frac{P(z,w)}{z^rw^s(z-w)^N}.

Take $L=r$ $r$ . Both sides then expand

\frac{P(x+w,w)}{w^sx^N}.

Its denominator is a monomial, so the two expansion orders agree.

§Jacobi identity

The two products and the iterate are now three expansions of one expression. The formal delta identity combines them into the VA Jacobi identity.

Proposition 2.10 (Jacobi identity)

For $a,b,c\in V$ $V$ ,

\begin{aligned} &x^{-1}\delta\!\left(\frac{z-w}{x}\right)Y(a,z)Y(b,w)c\\ &\quad-x^{-1}\delta\!\left(\frac{w-z}{-x}\right)Y(b,w)Y(a,z)c\\ &\qquad=z^{-1}\delta\!\left(\frac{w+x}{z}\right)Y(Y(a,x)b,w)c. \end{aligned}

The binomials in the three delta kernels are expanded in nonnegative powers of $w$ $w$ , $z$ $z$ , and $x$ $x$ , respectively. Since the identity holds for every $c$ $c$ , we may omit $c$ and read the operator sums pointwise on $V$ $V$ .

Proof.

By Theorem 2.8, the three expressions multiplying the delta kernels are the three expansions of

F(z,w)=\frac{P(z,w)}{z^rw^s(z-w)^N}, \qquad P\in V[[z,w]].

Multiply the standard identity

x^{-1}\delta\!\left(\frac{z-w}{x}\right) -x^{-1}\delta\!\left(\frac{w-z}{-x}\right) =z^{-1}\delta\!\left(\frac{w+x}{z}\right)

by $P(z,w)/(x^Nz^rw^s)$ $P (z, w) / (x^{N} z^{r} w^{s})$ . This series has a uniform lower bound in each variable, so the products are well-defined. Delta substitution turns the first two terms into the product expansions of $F$ and the third into its iterate expansion, with exactly the conventions stated above. This proves the identity.

§2.4 Fields of products

Proposition 2.11 (Iterate formula)

For $a,b,c\in V$ and $m\in\mathbb Z$ $Z$ ,

\begin{aligned} Y(a_{(m)}b,z)c =\operatorname{Res}_u\Bigl(&\iota_{u,z}(u-z)^m Y(a,u)Y(b,z)c\\ &-\iota_{z,u}(u-z)^m Y(b,z)Y(a,u)c\Bigr). \end{aligned}

The first kernel is expanded in nonnegative powers of $z$ $z$ , and the second in nonnegative powers of $u$ $u$ .

Proof.

Apply Proposition 2.10 with field coordinates $u,z$ and relative coordinate $x$ $x$ . Multiply by $x^m$ and take $\operatorname{Res}_x\operatorname{Res}_u$ $Res_{x} Res_{u}$ . For the iterate term, the $u$ $u$ -residue of $u^{-1}\delta((z+x)/u)$ is $1$ $1$ , leaving

\begin{aligned} \operatorname{Res}_x x^mY(Y(a,x)b,z)c &=\operatorname{Res}_x x^m \sum_{j\in\mathbb Z}Y(a_{(j)}b,z)c\,x^{-j-1}\\ &=Y(a_{(m)}b,z)c. \end{aligned}

For the two product terms,

\begin{aligned} \operatorname{Res}_x x^{m-1}\delta\!\left(\frac{u-z}{x}\right) &=\iota_{u,z}(u-z)^m,\\ \operatorname{Res}_x x^{m-1}\delta\!\left(\frac{z-u}{-x}\right) &=\iota_{z,u}(u-z)^m. \end{aligned}

Taking the remaining $u$ $u$ -residue proves the formula.

For $m\ge0$ $0$ , the kernel is a polynomial, so its two expansions agree and the formula simplifies to

Y(a_{(m)}b,z) =\operatorname{Res}_u (u-z)^m[Y(a,u),Y(b,z)].

These are the singular OPE coefficients. Negative products are expressed by normal ordering as follows.

Definition 2.12 (Normal ordering)

For a field $A(z)=\sum_j A_{(j)}z^{-j-1}$ $\sum_{j} A_{(j)} z^{- j - 1}$ , write

A_+(z)=\sum_{j<0}A_{(j)}z^{-j-1},\qquad A_-(z)=\sum_{j\ge0}A_{(j)}z^{-j-1}.

Here $A_+$ is the creation part and $A_-$ the annihilation part. The normal-ordered product with a field $B(z)$ is

:A(z)B(z):=A_+(z)B(z)+B(z)A_-(z).

For several fields, multiple normal ordering means right nesting:

:A_1(z)\cdots A_k(z): :=:A_1(z)\bigl(:A_2(z)\cdots A_k(z):\bigr):, \qquad :A_1(z):=A_1(z).

Thus $:ABC:=:A(:BC:):$ $) :$ . This is a choice of parentheses, not an associativity identity; in general $:A(:BC:):\ne:(:AB:)C:$ $) C :$ .

Corollary 2.13 (Fields of negative products)

For $a,b\in V$ and $k\ge0$ $0$ ,

Y(a_{(-k-1)}b,z) =\frac1{k!}:\bigl(\partial_z^kY(a,z)\bigr)Y(b,z):.

In particular, $Y(a_{(-1)}b,z)=:Y(a,z)Y(b,z):$ $Y (a, z) Y (b, z) :$ .

Proof.

For $m=-1$ $- 1$ , the kernels in Proposition 2.11 are

\iota_{u,z}(u-z)^{-1}=\sum_{j\ge0}z^ju^{-j-1},\qquad \iota_{z,u}(u-z)^{-1}=-\sum_{j\ge0}u^jz^{-j-1}.

Their residues give, respectively, the creation part of $Y(a,z)$ on the left and the annihilation part on the right:

Y(a_{(-1)}b,z) =Y(a,z)_+Y(b,z)+Y(b,z)Y(a,z)_-.

Translation gives $Y(T^ka,z)=\partial_z^kY(a,z)$ and, on comparing the coefficient of $z^0$ $z^{0}$ , $(T^ka)_{(-1)}=k!a_{(-k-1)}$ $k! a_{(- k - 1)}$ . Apply the case $k=0$ to $T^ka,b$ and divide by $k!$ $k!$ .

With the right-nesting convention, repeated application gives, for $k_1,\ldots,k_s\ge0$ $0$ ,

Y\bigl(a^1_{(-k_1-1)}\cdots a^s_{(-k_s-1)}b,z\bigr) =\frac{:\bigl(\partial_z^{k_1}Y(a^1,z)\bigr)\cdots \bigl(\partial_z^{k_s}Y(a^s,z)\bigr)Y(b,z):} {k_1!\cdots k_s!}.

Thus fields of states built from negative products are nested normal-ordered derivatives of the original fields. The general iterate formula also handles nonnegative products through their singular OPE coefficients. These identities hold in any vertex algebra, without a conformal vector; see Nozaradan, §4.3.3.

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