Domenico Fiorenza
Vertex algebras avant Borcherds

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:

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

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

or even, directly in terms of Laurent coefficients,

$$[L_m,L_n]=(m-n)L_{m+n}+D\frac{m^3-m}{12}{\delta }_{m,-n}.$$[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
• $ℳ$, a sheaf on $\Sigma$;
• $M=ℳ(\Sigma )$ (we pretend $M$ to be a differentiable manifold, endowed with a volume form $d{\mathrm{vol}}_M$)
• ${\widehat{ℳ}}_{\Sigma }\to \Sigma$, the etale space of $ℳ$
• $\mathrm{ev}:M\times \Sigma \to {\widehat{ℳ}}_{\Sigma }$, the map sending a global section $X$ and a point $p$ to the germ $X_p$ of $X$ at $p$
• $ℱ\subseteq C^{\mathrm{\infty }}(M\times \Sigma ;ℂ)\cap {\mathrm{ev}}^{*}\{{\widehat{ℳ}}_{\Sigma }\to ℂ\}$, a set of local fields, i.e., smooth functions on $M\times \Sigma$ whose value at $(X,p)$ only depends on the germ $X_p$.
• $S:M\to ℝ$, the action.

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

$$⟨f⟩={\int }_{M}f{e}^{-S}d{\mathrm{vol}}_M$$\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 $A_X\in 𝒟\prime (\Sigma )$ as follows:

$$A_X:f\mapsto {\int }_{\Sigma }f(p)A(X,p)d{rm\mathrm{vol}}_{\Sigma }$$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)

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

with

$$\phi \otimes \psi :f\mapsto {\int }_{{\Sigma }^{m+n}}\phi ({\stackrel{\rightharpoonup }{p}}_1)\psi ({\stackrel{\rightharpoonup }{p}}_2)f({\stackrel{\rightharpoonup }{p}}_1,{\stackrel{\rightharpoonup }{p}}_2)d{rm\mathrm{vol}}_{{\Sigma }^{m+n}}$$\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 ${ℱ}^n\times M\to 𝒟\prime ({\Sigma }^n)$

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

which we can look at as a map

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

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

$$⟨{⟩}_n:{ℱ}^{\otimes n}\to 𝒟\prime ({\Sigma }^n)$$\langle\,\rangle_n\colon\mathcal{F}^{\otimes n}\to\mathcal {D}'(\Sigma^n)

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

$$⟨A_1(X,p_1)\cdots A_n(X,p_n){⟩}_n={\int }_M{A}_1(X,p_1)\cdots A_n(X,p_n)e^{-S(X)}d{\mathrm{vol}}_M(X)$$\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 ${ℱ}^n\times M\to 𝒟\prime ({\Sigma }^n)$ as to a map ${ℱ}^n\to C^{\mathrm{\infty }}(M,𝒟\prime ({\Sigma }^n)$; 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

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

Consider the restrictions $𝒟\prime ({\Sigma }^n)\to 𝒟\prime ({\Sigma }^n\setminus {\Delta }_n)$. 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 $⟨A_1(p_1)\cdots A_n(p_n){⟩}_n$ is a smooth function on on ${\Sigma }^n\setminus {\Delta }_n$. Singularities may appear when $p_i,p_j\to p$ for $i\ne j$. In particular, $1$-point functions are smooth.

Regularization

We further assume to have an operator $::$,

$${ℱ}^{\otimes n}\to C^{\mathrm{\infty }}(M\times {\Sigma }^n;ℂ)$$\mathcal{F}^{\otimes n}\to C^\infty(M\times \Sigma^n;\mathbb{C})

$$A_1\otimes \cdots \otimes A_n\mapsto :A_1\cdots A_n:$$A_1\otimes \cdots\otimes A_n\mapsto :A_1\cdots A_n:

such that

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

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

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

Via the embedding $C^{\mathrm{\infty }}({\Sigma }^n;ℂ)\hookrightarrow 𝒟\prime ({\Sigma }^n;ℂ)$ 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 $𝒟\prime ({\Sigma }^n;ℂ)$-valed functions on $M$. With this identification, the difference $A_1\cdots A_n-:A_1\cdots A_n:$ is an $𝒟\prime ({\Sigma }^n;ℂ)$-valed functions on $M$ whose integral over $M$ is the singular part of $⟨A_1\cdots A_n{⟩}_n$. Moreover, we also require $⟨:A:⟩=⟨A{⟩}_1$ 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

$$AB=:AB:+{\phi }_{AB}$$A B = : A B: +\,\varphi_{A B}

for a suitable function ${\phi }_{AB}:M\to 𝒟\prime ({\Sigma }^2;ℂ)$. One the 2-point regularizations have been chosen, we can define higher order regularizations iteratively, as follows:

$:A:=A$

$:AB:=AB-{\phi }_{AB}$

$:ABC:=ABC-{\phi }_{AB}C-{\phi }_{BC}A-{\phi }_{AC}B$

$:ABCD:=ABCD-{\phi }_{AB}CD-{\phi }_{AC}BD-\cdots -{\phi }_{CD}AB+{\phi }_{AB}{\phi }_{CD}+{\phi }_{AC}{\phi }_{BD}+{\phi }_{AD}{\phi }_{BC}$

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

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

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

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

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

$$:ℬ:=ℬ-ℬ\mathrm{contractions}$${:}\mathcal{B}:=\mathcal{B}-\mathcal{B}contractions

and so

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

Therefore singularities of $⟨:𝒜::ℬ:⟩$ come entirely from the $𝒜ℬ$-contractions of $𝒜ℬ$.

Example: $:A_1{A}_2::B_1{B}_2:={\phi }_{A_1{B}_1}{\phi }_{A_2{B}_2}+{\phi }_{A_1{B}_2}{\phi }_{A_2{B}_1}+{\phi }_{A_1{B}_1}:A_2{B}_2:+{\phi }_{A_1{B}_2}:A_2{B}_1:+{\phi }_{A_2{B}_1}:A_1{B}_2:+{\phi }_{A_2{B}_2}:A_1{B}_1:+:A_1{A}_2{B}_1{B}_2:$

Notation: One defines

$$R_{AB}(z)=:AB:(z,z);$$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 $:\mathrm{AB}:(z,w)$, and so

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

Moreover, multilinearity of regularization gives

$${\partial }_z:A(z)B(w):=:{\partial }_{\mathrm{zA}}(z)B(w):$$\partial_z:A(z)B(w):=:\partial_zA(z)B(w):

and so

$${\partial }_z{R}_{\mathrm{AB}}(z)=R_{{\partial }_{z}AB}(z)+R_{A{\partial }_{z}B}(z)$$\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 $ℱ$ 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

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

generated by $ℱ$, modulo the following relations:

$${\hat{A}}_1\cdots {\hat{A}}_n=0\mathrm{iff}⟨A_1\cdots A_n{B}_1\cdots B_m{⟩}_{m+n}{\mid }_{{\Sigma }^{m+n}\setminus {\Delta }_{m+n}^{\ge n}}=0$$\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

$${\Delta }_n^{\ge k}=\{(p_1,\dots ,p_n)\mid p_i=p_j\mathrm{for}\mathrm{some}i\ne j\mathrm{with}j\ge k\}$$\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:

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

Advanced tricks

Lie derivatives

Assume a linear map

$$i:C_0^{\mathrm{\infty }}(\Sigma )\to H^0(M;TM)$$i\colon C^\infty_0(\Sigma)\to H^0(M;T M)

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

$$f\mapsto ⟨(\omega \mid i_f)⟩={\int }_M(\omega \mid i_f)e^{-S}d{\mathrm{vol}}_M.$$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 ${ℒ}_v$ satisfies

$$⟨(dS\mid v)⟩+⟨\mathrm{div}(v)⟩={\int }_M{ℒ}_v(e^{-S}d{\mathrm{vol}}_M)=0.$$\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 $(\mathrm{dS}\mid i)$ is zero. If $f$ is supported away from $q_1,\dots ,q_m$, then, for any $B_1,\dots ,B_m$,

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

hence

$$⟨(dS\mid i)\cdots ⟩=0$$\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 $(dS\mid v)=0$, and assume furthermore that $\mathrm{div}(v)=0$, so that

$${\mathrm{div}}_S(v)=0,$$div_S(v)=0,

where

$${\mathrm{div}}_S(v)={ℒ}_v(e^{-S}d{\mathrm{vol}}_M).$$div_S(v)=\mathcal {L}_{v}(e^{-S}d vol_M).

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

$$\rho \mapsto {\mathrm{div}}_S(\rho v{)}_X$$\rho\mapsto div_S(\rho v)_X

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

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

there exist

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

such that

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

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

$$0={\int }_M{ℒ}_{\rho v}(e^{-S}d{\mathrm{vol}}_M)=⟨(j_{v,X}\mid d\rho )⟩=⟨(\partial j_v\mid \rho )⟩$$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 $⟨\partial j_v⟩$ is the zero distribution and, more in general, $⟨\partial j_v\cdots ⟩=0,i.e.\partial \widehat{j_v}=0$. Identify $j_{v,X}$ with a 1-form via the canonical pairing of 1-forms on $\Sigma$:

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

Then

$$⟨d{j}_v\cdots ⟩=0.$$\langle d j_{v}\cdots\rangle=0.

Ward identities

Now add a field $A$. Then

$$0={\int }_M{ℒ}_{\rho v}A(p)(e^{-S}d{\mathrm{vol}}_M)=⟨(\partial j_X\mid \rho )A(p)⟩+⟨{ℒ}_{\rho v}A(p)⟩.$$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 ${ℒ}_{\rho v}A(p)={ℒ}_{v}A(p)$ and so

$$⟨{ℒ}_{v}A(p)⟩=⟨A(p){\int }_{\Sigma }\rho d{j}_v⟩=⟨A(p){\int }_{B_p}d{j}_v⟩={\int }_{\partial B_p}⟨A(p)j_v⟩,$$\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

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

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

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

The holomorphic case

If $⟨A(p)j_v⟩$ is holomorphic in $B_p\setminus p$, then

$${\int }_{\partial B_p}⟨A(p)j_v⟩={\mathrm{Res}}_{z\to p}⟨A(p)j_v(z)⟩\mathrm{dz}$$\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

$$⟨{ℒ}_{v}A(p)⟩={\mathrm{Res}}_{z\to p}⟨A(p)j_v⟩,$$\langle\mathcal{L}_v A(p)\rangle=Res_{z\to p}\langle A(p) j_v \rangle,

That is

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

OPEs

Assume $⟨A(z)B(w)\cdots ⟩$ is a holomorphic function of $z$ for $z\ne w$.

We write

(1)$$A(z)B(w)\sim \sum _{k\ge 1}{C}_k(w)\frac{1}{(z-w{)}^k}$$A(z)B(w)\sim \sum_{k\geq 1} C_k(w)\frac{1}{(z-w)^k}

to mean

$$⟨A(z)B(w)\cdots ⟩=\sum _{k\ge 1}⟨C_k(w)\cdots ⟩\frac{1}{(z-w{)}^k}+⟨:A(z)B(w):\cdots ⟩$$\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 $B$ ($AB$ OPE for short). Note that $:A(z)B(w):$ is a holomorphic function of $z$ also at $z=w$.

Since ${\mathrm{Res}}_{w\to z}:A(z)j_v(w):\mathrm{dw}=0$, to compute $⟨{ℒ}_{v}A(z)⟩$ one only needs the $A{j}_v$ OPE

$$A(z)j_v(w)\sim \sum _{k\ge 1}{C}_k(w)\frac{1}{(z-w{)}^k}.$$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

$${\mathrm{Res}}_{z\to p}{\mathrm{Res}}_{w\to p}⟨j_1(z)j_2(w)\cdots ⟩-{\mathrm{Res}}_{w\to p}{\mathrm{Res}}_{z\to p}⟨j_1(z)j_2(w)\cdots ⟩={\mathrm{Res}}_{w\to p}{\mathrm{Res}}_{z\to w}⟨j_1(z)j_2(w))\cdots ⟩={\mathrm{Res}}_{w\to p}⟨({\mathrm{Res}}_{z\to w}{j}_1(z)j_2(w)))\cdots ⟩.$$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

$$[j_1,j_2](w)={\mathrm{Res}}_{z\to w}{j}_1(z)j_2(w)$$[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 $ℂ$, the sheaf $ℳ$ will be the sheaf of smooth fuctions on $ℂ$ with values in ${ℝ}^D$, and the action will be the Polyakov action for the standard Euclidean metric both on the source $ℂ$ and on the target ${ℝ}^D$, i.e.,

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

The tangent space of $M=C^{\mathrm{\infty }}(ℂ;{ℝ}^D)$ at each point $X$ is identified with the subspace $C_0^{\mathrm{\infty }}(ℂ;{ℝ}^D)$ of compactly supported functions. For any $\mu =1,\dots ,D$, a linear map $i_{\mu }:C_0^{\mathrm{\infty }}(ℂ,ℝ)\to H^0(M,TM)$ is given by $i_{\mu ,f}:X^{\nu }\mapsto X^{\nu }+{\delta }_{\mu }^{\nu }ϵf$. Postulating the volume form on $M$ is such that the vector fields $i_{\mu ,f}$ are divergence-free, we have $⟨(dS\mid i_{\mu })\cdots ⟩=0$ for any $\mu$. One then computes $(dS\mid i_{\mu })=\overline{\partial }\partial X^{\mu }$, hence

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

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

$${\overline{\partial }}_z{\partial }_z⟨(X^{\mu }(z)X_{\mu }(w)+\frac{1}{2}{\delta }^{\mu \nu }\log \mid z-w{\mid }^2)\cdots ⟩=0$$\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, $⟨X^{\mu }(z)X_{\mu }(w)+\frac{1}{2}{\delta }^{\mu \nu }\log \mid z-w{\mid }^2⟩$ is an harmonic function and so we have the regularization

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

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

Similarly,

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

The TT OPE

Set

$$T(z)=R_{{\partial }_z{X}^{\mu }{\partial }_z{X}_{\mu }}(z)=\underset{z\prime \to z}{\lim }:{\partial }_z{X}^{\mu }(z){\partial }_{z\prime }{X}_{\mu }(z\prime )$$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

$${\overline{\partial }}_{\mathrm{zT}}(z)=(R_{{\overline{\partial }}_z{\partial }_z{X}^{\mu }{\partial }_z{X}_{\mu }}+R_{\partial X^{\mu }{\overline{\partial }}_z{\partial }_z{X}_{\mu }})(z)=0,$$\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

$$T(z)T(w)=\underset{(z\prime ,w\prime )\to (z,w)}{\lim }:{\partial }_{\mathrm{zX}}^{\mu }(z){\partial }_{z\prime }{X}_{\mu }(z\prime )::{\partial }_{\mathrm{wX}}^{\nu }(w){\partial }_{w\prime }{X}_{\nu }(w\prime ):=\frac{D/2}{(z-w{)}^4}+2\frac{1}{(z-w{)}^2}:{\partial }_{\mathrm{zX}}^{\mu }(z){\partial }_w{X}_{\mu }(w):+\cdots$$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

$$\phantom{T(z)T(w)}=\frac{D/2}{(z-w{)}^4}+\frac{2}{(z-w{)}^2}T(w)+\frac{1}{(z-w)}{\partial }_{w}T(w)+\cdots$$\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 $TT$ OPE

$$T(z)T(w)\sim \frac{D/2}{(z-w{)}^4}+\frac{2}{(z-w)2}T(w)+\frac{1}{(z-w)}{\partial }_{w}T(w)\phantom{+\cdots }$$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 $ℂ\setminus 0$ act as symmetries of the action. The charge associated with the vector field $z^{n+1}\frac{\partial }{\partial z}$ is

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

Set

$$L_n={\mathrm{Res}}_{z\to 0}⟨j_n(z)\cdots ⟩\mathrm{dz}$$L_n=Res_{z\to 0}\langle j_n(z)\cdots\rangle dz

Then

$$[L_m,L_n]={\mathrm{Res}}_{z\to 0}({\mathrm{Res}}_{w\to z}⟨j_m(w)j_n(z)\cdots ⟩\mathrm{dw})\mathrm{dz}={\mathrm{Res}}_{z\to 0}({\mathrm{Res}}_{w\to z}{z}^{n+1}{w}^{m+1}⟨T(w)T(z)\cdots ⟩\mathrm{dw})\mathrm{dz}$$[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 ${\mathrm{Res}}_{w\to z}{z}^{n+1}{w}^{m+1}⟨T(w)T(z)\cdots ⟩\mathrm{dw}=z^{n+1}{\mathrm{Res}}_{w\to z}(z+(w-z){)}^{m+1}⟨T(w)T(z)\cdots ⟩\mathrm{dw}$ and use the $TT$ OPE to find

$${\mathrm{Res}}_{w\to z}(z+(w-z){)}^{m+1}⟨T(w)T(z)\cdots ⟩\mathrm{dw}={\mathrm{Res}}_{w\to z}(z+(w-z){)}^{m+1}(\frac{2}{(w-z{)}^2}⟨T(z)\cdots ⟩+\frac{1}{(w-z)}⟨{\partial }_{z}T(z)\cdots ⟩+\frac{D/2}{(w-z{)}^4}⟨\cdots ⟩)$$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)

$$\phantom{{\mathrm{Res}}_{w\to z}(z+(w-z){)}^{m+1}⟨T(w)T(z)\cdots ⟩\mathrm{dw}}=\frac{D}{2}\left(\genfrac{}{}{0ex}{}{m+1}{3}\right)z^{m-2}⟨\cdots ⟩+2\left(\genfrac{}{}{0ex}{}{m+1}{1}\right)z^m⟨T(z)\cdots ⟩+z^{m+1}⟨{\partial }_{z}T(z)\cdots ⟩$$\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

$$[L_m,L_n]={\mathrm{Res}}_{z\to 0}(D\frac{m^3-m}{12}{z}^{n+m-1}+2(m+1)z^{m+n+1}⟨T(z)\cdots ⟩+z^{n+m+2}{\partial }_z⟨T(z)\cdots ⟩)\mathrm{dz}$$[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

$$\phantom{[L_m,L_n]}=D\frac{m^3-m}{12}{\delta }_{m,-n}-2(m+1){\mathrm{Res}}_{z\to 0}{j}_{n+m}(z)\mathrm{dz}+{\mathrm{Res}}_{z\to 0}({\partial }_z{z}^{n+m+2})⟨T(z)\cdots ⟩\mathrm{dz}$$\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

$$\phantom{[L_m,L_n]}=D\frac{m^3-m}{12}{\delta }_{m,-n}+(m-n){\mathrm{Res}}_{z\to 0}{j}_{m+n}(z)\mathrm{dz}$$\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

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