Chapter 3 of 11

§Vertex operator algebras

Although vertex algebra contains many familiar properties a chiral CFT constains, it does not contain the most important structure: the energy-momentum tensor and Virasoro algebra. A vertex operator algebra (VOA) is a vertex algebra with so-called conformal structure, which is essentially the energy-momentum tensor and Virasoro algebra. We now define VOA.

§3.1 Definition of VOA

Definition 3.1 (Vertex operator algebra)

Let $(V,Y,\mathbf1,T)$ be a vertex algebra. A vertex operator algebra (VOA) structure on $V$ is specified by a conformal vector $\omega\in V$ whose field

Y(\omega,z)=\sum_{n\in\mathbb Z}L_nz^{-n-2}, \qquad L_n=\omega_{(n+1)},

satisfies the following conditions.

Virasoro relations. For some $c\in\mathbb C$ $C$ , called the central charge,

$[L_m,L_n]=(m-n)L_{m+n} +\frac{c}{12}(m^3-m)\delta_{m+n,0}\operatorname{id}_V.$
Translation compatibility. $L_{-1}=T$ $T$ .
Conformal grading. $L_0$ acts diagonally with integral eigenvalues,

$V=\bigoplus_{h\in\mathbb Z}V_h, \qquad V_h=\{a\in V:L_0a=ha\},$

where $\dim V_h<\infty$ and $V_h=0$ for sufficiently negative $h$ $h$ . A state in $V_h$ has conformal weight $h$ $h$ .

We denote the VOA by $(V,Y,\mathbf1,\omega)$ $(V, Y, 1, ω)$ . The VOA is called CFT type if they satisfy additional conditions $V_h=0$ for $h<0$ and $V_0=\mathbb C\mathbf1$ $C 1$ .

§3.2 Vacuum and conformal vector

Proposition 3.2 (Vacuum and conformal weights)

The vacuum and conformal vector satisfy

L_n\mathbf1=0\quad(n\geq-1), \qquad \omega=L_{-2}\mathbf1, \qquad \mathbf1\in V_0,\quad\omega\in V_2.

Proof.

Apply Proposition 2.3 to $\omega$ and compare coefficients in $Y(\omega,z)\mathbf1=e^{zT}\omega$ $e^{z T} ω$ . Then $L_0\mathbf1=0$ and

L_0\omega=[L_0,L_{-2}]\mathbf1=2\omega.

§3.3 Virasoro action on fields

Proposition 3.3 (Virasoro covariance)

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

[L_m,Y(a,z)] =\sum_{j\geq0}\binom{m+1}{j}z^{m+1-j}Y(L_{j-1}a,z).

The sum is finite by lower truncation. In particular,

\begin{aligned} [L_{-1},Y(a,z)]&=\partial_zY(a,z),\\ [L_0,Y(a,z)]&=z\partial_zY(a,z)+Y(L_0a,z),\\ [L_1,Y(a,z)]&=z^2\partial_zY(a,z)+2zY(L_0a,z)+Y(L_1a,z). \end{aligned}

Proof.

Use Proposition 2.10 with first state $\omega$ and field coordinates $u,z$ $u, z$ . Taking $\operatorname{Res}_x$ turns each of the first two delta kernels into $1$ $1$ . Multiplying by $u^{m+1}$ and taking $\operatorname{Res}_u$ therefore gives

\operatorname{Res}_u u^{m+1}[Y(\omega,u),Y(a,z)] =[L_m,Y(a,z)].

On the right-hand side, the $u$ $u$ -residue of the delta kernel is

\operatorname{Res}_u u^{m+1}u^{-1} \delta\!\left(\frac{z+x}{u}\right) =\iota_{z,x}(z+x)^{m+1} =\sum_{j\geq0}\binom{m+1}{j}z^{m+1-j}x^j.

Thus the remaining $x$ $x$ -residue is

\begin{aligned} [L_m,Y(a,z)] &=\operatorname{Res}_x\left( \sum_{j\geq0}\binom{m+1}{j}z^{m+1-j}x^j\right) \left(\sum_{n\in\mathbb Z}Y(\omega_{(n)}a,z)x^{-n-1}\right)\\ &=\sum_{j\geq0}\binom{m+1}{j}z^{m+1-j}Y(\omega_{(j)}a,z). \end{aligned}

The residue selects $n=j$ $j$ . Since $\omega_{(j)}=L_{j-1}$ $L_{j - 1}$ , this is the claimed formula; lower truncation makes the sum finite even for $m<-1$ $- 1$ . The three special cases use $Y(L_{-1}a,z)=\partial_zY(a,z)$ $\partial_{z} Y (a, z)$ .

Remark 3.4 (Global conformal transformations)

The modes $L_{-1},L_0,L_1$ span an $\mathfrak{sl}_2$ subalgebra with no central term. Their actions on fields describe infinitesimal translations, dilations, and special conformal transformations, respectively. The corresponding coordinate flows are

z\longmapsto z+t,\qquad z\longmapsto e^t z, \qquad z\longmapsto\frac{z}{1-tz}.

Together they generate the Möbius transformations, the global holomorphic conformal transformations of the Riemann sphere. For $a\in V_h$ with $L_1a=0$ (a quasiprimary state), the three commutators act on $Y(a,z)$ as

\partial_z,\qquad z\partial_z+h,\qquad z^2\partial_z+2hz.

The terms involving $h$ account for the conformal weight of the field. For a general homogeneous state, the additional term $Y(L_1a,z)$ describes mixing with another field. See Keller, §7.3.

§3.4 Weights and dilations

Proposition 3.5 (Weight compatibility)

For $a\in V_h$ $V_{h}$ , $b\in V_k$ $V_{k}$ , and $n\in\mathbb Z$ $Z$ ,

[L_0,a_{(n)}]=(h-n-1)a_{(n)}, \qquad a_{(n)}b\in V_{h+k-n-1}.

In particular, $T V_h\subseteq V_{h+1}$ and $L_nV_h\subseteq V_{h-n}$ $V_{h - n}$ . For $q\in\mathbb C^\times$ $C^{\times}$ ,

q^{L_0}Y(a,z)q^{-L_0}=q^hY(a,qz).

Proof.

Compare coefficients in $[L_0,Y(a,z)]=(z\partial_z+h)Y(a,z)$ $h) Y (a, z)$ . Applying the resulting commutator to $b$ gives the weight of $a_{(n)}b$ $a_{(n)} b$ . The statements for $T=L_{-1}$ and $L_n$ follow from the Virasoro relations. Conjugation by $q^{L_0}$ multiplies $a_{(n)}$ by $q^{h-n-1}$ $q^{h - n - 1}$ , proving the last identity.

Thus $L_0$ generates dilations, and conformal weights determine the powers allowed in each OPE. If $V_\ell=0$ for $\ell<B$ $B$ , then $a_{(n)}b=0$ whenever $n>h+k-B-1$ $1$ .

§3.5 Products involving the conformal vector

Proposition 3.6 (Products of the conformal vector)

The nonnegative products of $\omega$ with itself are

\omega_{(0)}\omega=T\omega,\qquad \omega_{(1)}\omega=2\omega,\qquad \omega_{(2)}\omega=0,\qquad \omega_{(3)}\omega=\frac c2\mathbf1,

and $\omega_{(n)}\omega=0$ for $n\geq4$ $4$ . Equivalently, the exact identity in $V((x))$ is

Y(\omega,x)\omega =\frac{c\mathbf1}{2x^4}+\frac{2\omega}{x^2} +\frac{T\omega}{x} +\sum_{k\geq0}L_{-k-2}\omega\,x^k.

Proof.

For $n\geq0$ $0$ , use $\omega=L_{-2}\mathbf1$ and $L_n\mathbf1=0$ to obtain

\begin{aligned} L_n\omega &=[L_n,L_{-2}]\mathbf1\\ &=(n+2)L_{n-2}\mathbf1 +\frac{c}{12}(n^3-n)\delta_{n,2}\mathbf1. \end{aligned}

For $n=0,1,2$ $0, 1, 2$ , this gives $2\omega,0,c\mathbf1/2$ $2 ω, 0, c 1 /2$ , respectively; for $n\geq3$ it vanishes. Together with $L_{-1}\omega=T\omega$ and $\omega_{(j)}=L_{j-1}$ $L_{j - 1}$ , these are the claimed products. Expanding $Y(\omega,x)\omega=\sum_nL_n\omega\,x^{-n-2}$ gives the full identity, including its regular terms.

Applying the state-field map coefficientwise gives

\begin{aligned} Y(Y(\omega,x)\omega,w) &=\frac{c\,\operatorname{id}_V}{2x^4} +\frac{2Y(\omega,w)}{x^2} +\frac{\partial_wY(\omega,w)}{x}\\ &\quad+\sum_{k\geq0}Y(L_{-k-2}\omega,w)x^k. \end{aligned}

After acting on any $b\in V$ $V$ , Theorem 2.8 identifies this with the expansion at $x=0$ of the common expression for the two products of $Y(\omega,u)$ and $Y(\omega,w)$ $Y (ω, w)$ , with $u=w+x$ $x$ . This is the precise expansion underlying the stress-tensor OPE.

Definition 3.7 (Primary and quasiprimary states)

A homogeneous state $a\in V_h$ is primary if $L_na=0$ for all $n>0$ $0$ , and quasiprimary if $L_1a=0$ $0$ .

Proposition 3.8 (Primary-field covariance)

If $a\in V_h$ is primary, then

Y(\omega,x)a =\frac{ha}{x^2}+\frac{Ta}{x} +\sum_{k\geq0}L_{-k-2}a\,x^k,

and, for every $m\in\mathbb Z$ $Z$ ,

[L_m,Y(a,z)] =\bigl(z^{m+1}\partial_z+(m+1)hz^m\bigr)Y(a,z).

For quasiprimary $a$ $a$ , the latter identity still holds for $m=-1,0,1$ $- 1, 0, 1$ .

Proof.

In $Y(\omega,x)a=\sum_nL_na\,x^{-n-2}$ $\sum_{n} L_{n} a x^{- n - 2}$ , the primary-state conditions leave only $L_0a=ha$ and $L_{-1}a=Ta$ among the singular terms. The same conditions in Proposition 3.3 give the commutator. For $m=-1,0,1$ $- 1, 0, 1$ , only $L_{-1}a,L_0a,L_1a$ can contribute.

The three modes $L_{-1},L_0,L_1$ describe infinitesimal translations, dilations, and special conformal transformations. The conformal vector itself is quasiprimary since $L_1\omega=0$ $0$ ; it is not primary when $c\ne0$ $0$ , since $L_2\omega=c\mathbf1/2$ $c 1 /2$ .

§3.6 The weight-one Lie algebra

Proposition 3.9 (The weight-one Lie algebra)

If $V$ is of CFT type, then $V_1$ is a Lie algebra with bracket

[a,b]:=a_{(0)}b,\qquad a,b\in V_1.

Proof.

The bracket is bilinear, and weight compatibility gives $a_{(0)}b\in V_1$ $V_{1}$ . Taking the $z^{-1}$ coefficient of skew-symmetry gives

a_{(0)}b=-b_{(0)}a+T(b_{(1)}a).

Higher terms vanish by nonnegative weights. Since $b_{(1)}a\in V_0=\mathbb C\mathbf1$ and $T\mathbf1=0$ $0$ , the bracket is antisymmetric.

Now we prove the Jacobi identity holds for the Lie algebra. Taking $\operatorname{Res}_x\operatorname{Res}_z\operatorname{Res}_w$ in Proposition 2.10 gives

[a_{(0)},b_{(0)}]=(a_{(0)}b)_{(0)},

where the bracket on the left is the commutator of endomorphisms. Applying this identity to $d\in V_1$ yields

a_{(0)}(b_{(0)}d)-b_{(0)}(a_{(0)}d) =(a_{(0)}b)_{(0)}d,

or, in terms of the bracket on states,

[a,[b,d]]-[b,[a,d]]=[[a,b],d].

Together with antisymmetry, this is the Jacobi identity.

Proposition 3.10 (The invariant form on weight one)

Let $V$ be of CFT type. The bilinear form on $V_1$ defined by

a_{(1)}b=\langle a,b\rangle\mathbf1

is symmetric and invariant under the Lie bracket:

\langle a,b\rangle=\langle b,a\rangle,\qquad \langle[a,b],d\rangle=\langle a,[b,d]\rangle \quad(a,b,d\in V_1).

Proof.

Weight compatibility gives $a_{(1)}b\in V_0=\mathbb C\mathbf1$ $C 1$ , and the $x^{-2}$ coefficient of skew-symmetry gives $a_{(1)}b=b_{(1)}a$ $b_{(1)} a$ . The zero-mode commutator identity, obtained from Proposition 2.10, is

[a_{(0)},Y(b,x)]=Y(a_{(0)}b,x).

Applying it to $d$ and comparing coefficients of $x^{-2}$ gives

0=a_{(0)}(b_{(1)}d) =\bigl(\langle[a,b],d\rangle+\langle b,[a,d]\rangle\bigr)\mathbf1.

Using this identity with $a,b$ interchanged proves invariance.

The form may be degenerate; the term quadratic Lie algebra is usually reserved for a Lie algebra with a nondegenerate invariant symmetric bilinear form. Its normalization here is fixed by $Y$ and the vacuum. See Lam, Vertex operator algebras and finite groups.

§3.7 The weight-one current algebra

Proposition 3.11 (Current algebra)

Let $V$ be of CFT type. For $a,b\in V_1$ $V_{1}$ ,

Y(a,x)b-\frac{\langle a,b\rangle\mathbf1}{x^2} -\frac{[a,b]}x\in V[[x]].

The corresponding operators on $V$ satisfy

[a_{(m)},b_{(n)}] =[a,b]_{(m+n)} +m\langle a,b\rangle\delta_{m+n,0}\operatorname{id}_V \qquad(m,n\in\mathbb Z).

Proof.

The state $a_{(j)}b$ has weight $1-j$ $j$ , so it vanishes for $j\ge2$ $2$ . This proves the first assertion. Taking coefficients in the Jacobi identity gives

[a_{(m)},b_{(n)}] =\sum_{j\ge0}\binom mj(a_{(j)}b)_{(m+n-j)}.

Only $j=0,1$ remain. Substitute $a_{(0)}b=[a,b]$ $[a, b]$ , $a_{(1)}b=\langle a,b\rangle\mathbf1$ $⟨ a, b ⟩ 1$ , and $Y(\mathbf1,x)=\operatorname{id}_V$ $id_{V}$ .

Thus the fields $Y(a,x)$ $Y (a, x)$ , $a\in\mathfrak g:=V_1$ $V_{1}$ , give $V$ a representation of the affine current algebra

\widehat{\mathfrak g} =(\mathfrak g\otimes\mathbb C[t,t^{-1}])\oplus\mathbb C K,

with $K$ central and bracket

[a\otimes t^m,b\otimes t^n] =[a,b]\otimes t^{m+n} +m\langle a,b\rangle\delta_{m+n,0}K.

The action is $a\otimes t^m\mapsto a_{(m)}$ and $K\mapsto\operatorname{id}_V$ $id_{V}$ . In particular, the zero modes give an action of $\mathfrak g$ on each graded piece of $V$ $V$ . Their exponentials will give VOA automorphisms in Proposition 4.6.

§3.8 References

Christoph A. Keller, Introduction to Vertex Operator Algebras, §5.
Christophe Nozaradan, Introduction to Vertex Algebras, §3.7.

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