# Contents

## Idea

If you open your favourite book in conformal field theory, within the first 10 pages you will almost surely find an expression like this:

$T(z)T(w)\sim \frac{D/2}{(z-w)^4}+\frac{2}{(z-w)^2}T(w)+\frac{1}{(z-w)}\partial T(w)$

or equivalently, in the D’Andrea-Kac notation

$[T_\lambda T]= \frac{D}{12}{\lambda^3}+2\lambda T+\partial T,$

or even, directly in terms of Laurent coefficients,

$[L_m,L_n]= (m-n)L_{m+n}+D\frac{m^3-m}{12}\delta_{m,-n}.$

Following Borcherds, expressions like the above are at the basis of the modern axiomatization of vertex algebras. It’s worth recalling how they historically arose from path-integral heuristics.

## The basic rules

### The players

• $\Sigma$, a Riemann surface
• $\mathcal{M}$, a sheaf on $\Sigma$;
• $M= \mathcal{M}(\Sigma)$ (we pretend $M$ to be a differentiable manifold, endowed with a volume form $d vol_M$)
• $\hat\mathcal{M}_\Sigma\to {\Sigma}$, the etale space of $\mathcal{M}$
• $ev\colon M\times \Sigma\to \hat{\mathcal{M}}_\Sigma$, the map sending a global section $X$ and a point $p$ to the germ $X_p$ of $X$ at $p$
• $\mathcal{F}\subseteq C^\infty(M\times \Sigma;\mathbb{C})\cap ev^*\{\hat{\mathcal{M}}_\Sigma\to \mathbb{C}\}$, a set of local fields, i.e., smooth functions on $M\times \Sigma$ whose value at $(X,p)$ only depends on the germ $X_p$.
• $S\colon M\to \mathbb{R}$, the action.

Notation: Let $V$ be a (topological) vector space; if $f\colon M\to V$ is integrable w.r.t. the measure $e^{-S}d vol_M$, we write

$\langle f\rangle =\int_M f\, e^{-S}d vol_M$

### $n$-point functions

To a pair $(A,X)$ consisting of a field and a global section, we can associate a distribution $A_X\in \mathcal{D}'(\Sigma)$ as follows:

$A_X\colon f\mapsto \int_\Sigma f(p)A(X,p)d{\rm vol}_\Sigma$

Distributions are an algebra (w.r.t. tensor product)

$\mathcal{D}'(\Sigma^m)\otimes \mathcal{D}'(\Sigma^n)\to \mathcal{ D}'(\Sigma^{m+n})$

with

$\varphi\otimes\psi\colon f\mapsto \int_{\Sigma^{m+n}}\varphi(\vec{p}_1)\psi(\vec{p}_2)f(\vec{p}_1,\vec{p}_2)d{\rm vol}_{\Sigma^{m+n}}$

so we get a map $\mathcal{F}^n\times M \to \mathcal{D}'(\Sigma^n)$

$(A_1,\dots,A_n;X)\mapsto A_{1,X}\otimes\cdots \otimes A_{n,X}$

which we can look at as a map

$M\to{\rm Hom}(\mathcal{F}^{\otimes n},\mathcal{D}'(\Sigma^n)).$

Integrating over $M$ we get the $n$-point function

$\langle\,\rangle_n\colon\mathcal{F}^{\otimes n}\to\mathcal {D}'(\Sigma^n)$

Explicitly, for $A_1,\dots,A_n$ in $\mathcal{F}$, the $n$-point function $\langle A_1\cdots A_n\rangle_n$ is the distribution on $\Sigma^n$ whose density at $(p_1,\dots, p_n)$ is

$\langle A_1(X,p_1)\cdots A_n(X,p_n)\rangle_n= \int_M A_1(X,p_1)\cdots A_n(X,p_n) e^{-S(X)}d vol_M(X)$

We can also look at the map $\mathcal{F}^n\times M \to \mathcal{ D}'(\Sigma^n)$ as to a map $\mathcal{F}^n\to C^\infty(M, \mathcal {D}'(\Sigma^n)$; we denote denote by $A_1\cdots A_n$ the image of $(A_1,\dots,A_n)$ by this map.

### Off-diagonal regularity

Let $\Delta_n\hookrightarrow \Sigma^n$ be the big diagonal

$\Delta_n=\bigl\{\text{non-injective maps} \{1,\dots,n\}\to \Sigma\}\bigr\}$

Consider the restrictions $\mathcal{D}'(\Sigma^n)\to \mathcal {D}'(\Sigma^n\setminus \Delta_n)$. We ask the restrictions of the $n$-point functions to be regular distributions on $\Sigma^n\setminus \Delta_n$, i.e., for fixed $A_1,\dots,A_n$, the density $\langle A_1(p_1)\cdots A_n(p_n)\rangle_n$ is a smooth function on on $\Sigma^n\setminus \Delta_n$. Singularities may appear when $p_i,p_j\to p$ for $i\neq j$. In particular, $1$-point functions are smooth.

### Regularization

We further assume to have an operator $\colon\,\,\colon$,

$\mathcal{F}^{\otimes n}\to C^\infty(M\times \Sigma^n;\mathbb{C})$
$A_1\otimes \cdots\otimes A_n\mapsto :A_1\cdots A_n:$

such that

$\langle A_1\cdots A_n\rangle_n=\langle:A_1\cdots A_n:\rangle+\text{singular distribution},$

where the $\langle\,\,\rangle$ on the right-hand side is integration over $M$ with respect to the measure $e^{-S}d{\rm vol}_M$:

$\langle\,\,\rangle: C^\infty(M\times \Sigma^n;\mathbb{C})\to C^\infty( \Sigma^n;\mathbb{C}).$

Via the embedding $C^\infty(\Sigma^n;\mathbb{C})\hookrightarrow \mathcal{D}'(\Sigma^n;\mathbb{C})$ induced by a choice of a volume form on $\Sigma$, we can look at both $A_1\cdots A_n$ and $:A_1\cdots A_n:$ as $\mathcal{D}'(\Sigma^n;\mathbb{C})$-valed functions on $M$. With this identification, the difference $A_1\cdots A_n - :A_1\cdots A_n:$ is an $\mathcal{D}'(\Sigma^n;\mathbb{C})$-valed functions on $M$ whose integral over $M$ is the singular part of $\langle A_1\cdots A_n\rangle_n$. Moreover, we also require $\langle:A:\rangle=\langle A\rangle_1$ for $1$-point functions. This rules out the trivial regularization given by $:A_1\cdots A_n: = 0$ for any $A_1,\dots, A_n$. For the 2-point functions regularization gives

$A B = : A B: +\,\varphi_{A B}$

for a suitable function $\varphi_{A B}:M \to \mathcal{D}'(\Sigma^2;\mathbb{C})$. One the 2-point regularizations have been chosen, we can define higher order regularizations iteratively, as follows:

$:A:=A$

$:A B:=A B-\varphi_{A B}$

$:A B C:= A B C -\varphi_{A B} C-\varphi_{B C} A-\varphi_{A C} B$

$:A B C D:=A B C D-\varphi_{A B} C D-\varphi_{A C} B D-\cdots- \varphi_{C D} A B +\varphi_{A B}\varphi_{C D}+\varphi_{A C}\varphi_{B D}+\varphi_{A D}\varphi_{B C}$

and so on. Therefore, in general, if $\mathcal{A}=A_1\cdots A_n$, then

${:}\mathcal{A}:=\mathcal{A}-\text{contractions}$

Remark: If $\mathcal{A}=A_1\cdots A_n$ and $\mathcal{B}=B_1\cdots B_m$, then

${:}\mathcal{A}\mathcal{B}:=\mathcal{A}\mathcal{B}-all contractions$
${:}\mathcal{A}:=\mathcal{A}-\mathcal{A}contractions$
${:}\mathcal{B}:=\mathcal{B}-\mathcal{B}contractions$

and so

${:}\mathcal{A}::\mathcal{B}:=:\mathcal{A}\mathcal {B}:+\mathcal{ A}\mathcal{B}contractions.$

Therefore singularities of $\langle:\mathcal{A}::\mathcal{B}:\rangle$ come entirely from the $\mathcal{A}\mathcal{B}$-contractions of $\mathcal{A}\mathcal{B}$.

Example: $:A_1 A_2: :B_1 B_2:=\varphi_{A_1 B_1}\varphi_{A_2 B_2}+\varphi_{A_1 B_2}\varphi_{A_2 B_1}+\varphi_{A_1 B_1}:A_2 B_2:+\varphi_{A_1 B_2}:A_2 B_1:+\varphi_{A_2 B_1}:A_1 B_2:+\varphi_{A_2 B_2}:A_1 B_1:+:A_1 A_2 B_1 B_2:$

Notation: One defines

$R_{A B}(z)=:A B:(z,z);$

it is a smooth function on $\Sigma$. In physicists’ notation, one writes $:A(z)B(w):$ for $:AB:(z,w)$, and so

$R_{AB}(z)=\lim_{w\to z}:A(z)B(w):$

Moreover, multilinearity of regularization gives

$\partial_z:A(z)B(w):=:\partial_zA(z)B(w):$

and so

$\partial_z R_{AB}(z)=R_{\partial_z A\, B}(z)+R_{ A\partial_z B}(z)$

## The game

### The rules of the game

Promote each element $A$ of $\mathcal{F}$ to an operator $\hat{A}$, its quantization. By this we mean that $\hat{A}$ is an element of some associative algebra. More precisely, consider the free associative algebra

$T(\mathcal{F})=\bigoplus_{n\geq 0}\mathcal{F}^{\otimes n}$

generated by $\mathcal{F}$, modulo the following relations:

$\hat{A}_1\cdots \hat{A}_n=0 \qquad iff\qquad \langle A_1\cdots A_n\, B_1\cdots B_m\rangle_{m+n}\biggr\vert_{\Sigma^{m+n}\setminus \Delta^{\ge n}_{m+n}}=0$

for any $B_1,\cdots B_m$, where

$\Delta_n^{\geq k}=\{(p_1,\dots,p_n)\,|\, p_i=p_j for some i\neq j with j\geq k\}$

We will adopt the following shorthand notation for the above relations:

$\langle\mathcal{A}\cdots\rangle=0.$

### Lie derivatives

Assume a linear map

$i\colon C^\infty_0(\Sigma)\to H^0(M;T M)$

is given, where $T M$ denotes the tangent bundle of $M$. Then each $1$-form $\omega$ on $M$ gives a distribution $(\omega|i)$ on $\Sigma$ by

$f\mapsto\langle (\omega|i_f)\rangle=\int_M(\omega|i_f) e^{-S}d vol_M.$

For any vector field $v$ on $M$, the Lie derivative $\mathcal{L}_v$ satisfies

$\langle (d S|v)\rangle+\langle div(v)\rangle=\int_M \mathcal{L}_v(e^{-S}d vol_M)=0.$

Hence, if the vector fields $i_f$ are divergence-free, the $1$-point function $(dS|i)$ is zero. If $f$ is supported away from $q_1,\dots,q_m$, then, for any $B_1,\dots,B_m$,

$\mathcal{L}_{i_f}(B_1(X,q_1)\cdots B_m(X,q_m))=0$

hence

$\langle (d S|i)\cdots\rangle=0$

so we recover within this formalism a version of Ehrenfest’s theorem.

### Noether’s theorem

Let $v$ be a symmetry of the action, i.e., a vector field such $(d S|v)=0$, and assume furthermore that $div(v)=0$, so that

$div_S(v)=0,$

where

$div_S(v)=\mathcal {L}_{v}(e^{-S}d vol_M).$

Then, for any $X$ in $M$, the map $C^\infty(M;\mathbb{C})\to \mathbb{C}$ given by

$\rho\mapsto div_S(\rho v)_X$

is a distribution on $\Sigma$ which is zero on constant functions. From the exact sequence

$0\to \mathbb{C}\to C^\infty(\Sigma)\stackrel{d}{\to}\{Exact\, 1-forms on \Sigma\}\to 0,$

there exist

$j_{v,X}:\{Exact\, 1-forms on \Sigma\}\to \mathbb{C}$

such that

$div_S(\rho v)_X=(j_{v,X}|d\rho).$

Assume $j_{v,X}$ extends to a 1-current $j_{v,X}: \{1-forms on \Sigma\}\to \mathbb{C}$. Then

$0=\int_M\mathcal{L}_{\rho v}(e^{-S}d vol_M)= \langle(j_{v,X}|d\rho)\rangle= \langle(\partial j_{v}|\rho)\rangle$

Hence $\langle \partial j_{v}\rangle$ is the zero distribution and, more in general, $\langle \partial j_{v}\cdots\rangle=0, i.e. \partial\hat{j_v}=0$. Identify $j_{v,X}$ with a 1-form via the canonical pairing of 1-forms on $\Sigma$:

$(j_{v,X}|\omega)=\int_\Sigma j_{v,X} \wedge \omega$

Then

$\langle d j_{v}\cdots\rangle=0.$

### Ward identities

Now add a field $A$. Then

$0=\int_M\mathcal{L}_{\rho v}A(p)(e^{-S}d vol_M)= \langle(\partial j_X|\rho)A(p)\rangle+\langle\mathcal{L}_{\rho v}A(p)\rangle.$

If $\rho$ is a bump function at $p$, then $\mathcal{L}_{\rho v}A(p)=\mathcal {L}_{v}A(p)$ and so

$\langle\mathcal{L}_v A(p)\rangle=\langle A(p)\int_\Sigma \rho d j_{v}\rangle=\langle A(p)\int_{B_p} d j_{v}\rangle =\int_{\partial B_p}\langle A(p) j_{v} \rangle,$

where $B_p$ denotes a little disk centered at $p$ (the support of the bump function $\rho$). Let

$Q_{v,p}=\int_{\partial B_p}\langle j_v\cdots \rangle$

be the charge of $v$ at $p$. Then

$\langle\mathcal{L}_v\cdots\rangle\bigr\vert_p=Q_{v,p}$

### The holomorphic case

If $\langle A(p) j_v \rangle$ is holomorphic in $B_p\setminus{p}$, then

$\int_{\partial B_p}\langle A(p) j_v \rangle=Res_{z\to p}\langle A(p) j_v(z) \rangle dz$

And we obtain

$\langle\mathcal{L}_v A(p)\rangle=Res_{z\to p}\langle A(p) j_v \rangle,$

That is

$Q_{v,p}= Res_{z\to p}\langle j_v(z)\cdots\rangle dz.$

### OPEs

Assume $\langle A(z)B(w)\cdots \rangle$ is a holomorphic function of $z$ for $z\neq w$.

We write

(1)$A(z)B(w)\sim \sum_{k\geq 1} C_k(w)\frac{1}{(z-w)^k}$

to mean

$\langle A(z)B(w)\cdots \rangle= \sum_{k\geq 1} \langle C_k(w)\cdots \rangle \frac{1}{(z-w)^k}+\langle:A(z)B(w):\cdots \rangle$

The expression in equation (1) is called operator product exapansion of $A$ and $B$ ($A B$ OPE for short). Note that $:A(z)B(w):$ is a holomorphic function of $z$ also at $z=w$.

Since $Res_{w\to z}:A(z)j_v(w):dw=0$, to compute $\langle\mathcal {L}_v A(z)\rangle$ one only needs the $A j_v$ OPE

$A(z)j_v(w)\sim\sum_{k\geq 1} C_k(w)\frac{1}{(z-w)^k}.$

### The algebra of currents

Assume two conserved currents $j_1$ and $j_2$ are given, and let $Q_{1,p}, Q_{2,p}$ be the associated charges at $p$. Then the commutator $[Q_{1,p}, Q_{2,p}]$ acts as

$Res_{z\to p}Res_{w\to p}\langle j_1(z)j_2(w)\cdots\rangle-Res_{w\to p}Res_{z\to p}\langle j_1(z)j_2(w)\cdots\rangle= Res_{w\to p}Res_{z\to w}\langle j_1(z)j_2(w))\cdots\rangle =Res_{w\to p}\langle (Res_{z\to w} j_1(z)j_2(w)))\cdots\rangle.$

In other words

$[j_1,j_2](w)=Res_{z\to w} j_1(z)j_2(w)$

and the Lie bracket is completely determined by the OPE of $j_1(z)j_2(w)$.

## Let’s play

Now we specialize the above general setup to conformal field theory on the complex plane. So our $\Sigma$ will be the complex plane $\mathbb{C}$, the sheaf $\mathcal{M}$ will be the sheaf of smooth fuctions on $\mathbb{C}$ with values in $\mathbb{R}^D$, and the action will be the Polyakov action for the standard Euclidean metric both on the source $\mathbb{C}$ and on the target $\mathbb{R}^D$, i.e.,

$S[X]=\frac{1}{2\pi}\int_\Sigma \partial X^\mu\overline{\partial}X_\mu$

The tangent space of $M=C^\infty(\mathbb{C};\mathbb{R}^D)$ at each point $X$ is identified with the subspace $C^\infty_0(\mathbb{C};\mathbb{R}^D)$ of compactly supported functions. For any $\mu=1,\dots, D$, a linear map $i_\mu: C^\infty_0(\mathbb{C},\mathbb{R})\to H^0(M, T M)$ is given by $i_{\mu,f}\colon X^\nu\mapsto X^\nu+\delta^{\nu}_\mu\epsilon f$. Postulating the volume form on $M$ is such that the vector fields $i_{\mu,f}$ are divergence-free, we have $\langle(d S|i_\mu)\cdots\rangle=0$ for any $\mu$. One then computes $(d S|i_\mu)=\overline{\partial}\partial X^\mu$, hence

$\langle \overline{\partial}\partial X^\mu(z)\cdots\rangle=0$

by the general argument above. This in particular means that the distribution $\langle \overline{\partial}_z\partial_z X^\mu(z)X^\nu(w)$ is supported at $z=w$, and indeed one computes $\langle \overline{\partial}_z\partial_z X^\mu(z)X^\nu(w)\cdots\rangle=-\pi\langle \delta^{\mu\nu}\delta(z-w)\cdots \rangle$. Since $\delta(z-w)=\overline{\partial}_z\partial_z\log|z-w|^2$, this is conveniently rewritten as

$\overline{\partial}_z\partial_z\langle (X^\mu(z) X_\mu(w)+\frac{1}{2}\delta^{\mu\nu}\log|z-w|^2)\cdots\rangle=0$

In particular, $\langle X^\mu(z)X_\mu(w)+\frac{1}{2}\delta^{\mu\nu}\log|z-w|^2\rangle$ is an harmonic function and so we have the regularization

$:X^\mu(z)X^\nu(w):=X^\mu(z) X^\nu(w)+\frac{1}{2}\delta^{\mu\nu}\log|z-w|^2$

In other words,

$\varphi_{X^\mu X^\nu}=-\frac{1}{2}\delta^{\mu\nu}\log|z-w|^2.$

Similarly,

$\varphi_{\partial_z X^\mu\partial_z X^\nu}=-\frac{1}{2}\delta^{\mu\nu}\frac{1}{(z-w)^2}.$

### The TT OPE

Set

$T(z)=R_{\partial_z X^\mu\partial_z X_\mu}(z)=\lim_{z'\to z}:\partial_z X^\mu(z)\partial_{z'}X_\mu(z')$

Then

$\overline{\partial}_zT(z)=(R_{\overline\partial_z \partial_z X^\mu\partial_z X_\mu}+R_{\partial X^\mu\overline{\partial}_z\partial_z X_\mu})(z)=0,$

that is, $T(z)$ is holomorphic! We have

$T(z)T(w)=\lim_{(z',w')\to (z,w)}:\partial_zX^\mu(z)\partial_{z'}X_\mu(z'): :\partial_wX^\nu(w)\partial_{w'}X_\nu(w'): = \frac{D/2}{(z-w)^4}+ 2\frac{1}{(z-w)^2}:\partial_zX^\mu(z)\partial_w X_\mu(w):+\cdots$
$\phantom{T(z)T(w)} =\frac{D/2}{(z-w)^4}+\frac{2}{(z-w)^2}T(w)+\frac{1}{(z-w)}\partial_w T(w)+\cdots$

Therefore we have found the $T T$ OPE

$T(z)T(w)\sim \frac{D/2}{(z-w)^4}+\frac{2}{(z-w)2}T(w)+\frac{1}{(z-w)}\partial_w T(w)\phantom{+\cdots}$

### The Virasoro algebra

Fix $w=0$. Holomorphic vector fields on $\mathbb{C}\setminus0$ act as symmetries of the action. The charge associated with the vector field $z^{n+1}\frac{\partial}{\partial z}$ is

$j_n(z)dz=z^{n+1}T(z)dz.$

Set

$L_n=Res_{z\to 0}\langle j_n(z)\cdots\rangle dz$

Then

$[L_m,L_n]=Res_{z\to 0}\left(Res_{w\to z}\langle j_m(w)j_n(z)\cdots\rangle dw\right)dz=Res_{z\to 0}\left( Res_{w\to z} z^{n+1}w^{m+1}\langle T(w)T(z)\cdots \rangle dw\right)dz$

Look at the expression $Res_{w\to z} z^{n+1}w^{m+1}\langle T(w)T(z)\cdots \rangle dw=z^{n+1}Res_{w\to z}\bigl(z+(w-z)\bigr)^{m+1}\langle T(w)T(z)\cdots\rangle dw$ and use the $T T$ OPE to find

$Res_{w\to z}\bigl(z+(w-z)\bigr)^{m+1}\langle T(w)T(z)\cdots\rangle dw= Res_{w\to z}\bigl(z+(w-z)\bigr)^{m+1}( \frac{2}{(w-z)^2}\langle T(z)\cdots\rangle+ \frac{1}{(w-z)} \langle \partial_z T(z)\cdots\rangle+\frac{D/2}{(w-z)^4}\langle\cdots\rangle)$
$\phantom{Res_{w\to z}\bigl(z+(w-z)\bigr)^{m+1}\langle T(w)T(z)\cdots\rangle dw}= \frac{D}{2}\binom{m+1}{3}z^{m-2} \langle\cdots\rangle+2\binom{m+1}{1}z^m\langle T(z)\cdots\rangle + z^{m+1}\langle \partial_z T(z)\cdots\rangle$

Therefore we find

$[L_m,L_n]=Res_{z\to 0} \biggl(D \frac{m^3-m}{12}z^{n+m-1}+2(m+1)z^{m+n+1}\langle T(z)\cdots\rangle+ z^{n+m+2}\partial_z \langle T(z)\cdots\rangle\biggr)dz$
$\phantom{[L_m,L_n]}= D\frac{m^3-m}{12}\delta_{m,-n}-2(m+1)Res_{z\to 0}j_{n+m}(z)dz+Res_{z\to 0}(\partial_z z^{n+m+2})\langle{T}(z)\cdots\rangle dz$
$\phantom{[L_m,L_n]}= D\frac{m^3-m}{12}\delta_{m,-n}+(m-n)Res_{z\to 0}j_{m+n}(z)dz$
$\phantom{[L_m,L_n]}=(m-n)L_{m+n}+D\frac{m^3-m}{12}\delta_{m,-n}.$
Revised on May 19, 2010 at 13:43:39 by Domenico Fiorenza