Domenico Fiorenza Vertex algebras avant Borcherds

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:

or equivalently, in the D’Andrea-Kac notation

or even, directly in terms of Laurent coefficients,

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

$n$-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:

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

with

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

which we can look at as a map

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

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

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

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 $\colon\,\,\colon$,

such that

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

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

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

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

and so

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

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

Moreover, multilinearity of regularization gives

and so

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

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

for any $B_1,\cdots B_m$, where

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

Advanced tricks

Lie derivatives

Assume a linear map

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

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

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 $B_1,\dots,B_m$,

hence

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

where

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

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

there exist

such that

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

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 $\Sigma$:

Then

Ward identities

Now add a field $A$. Then

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

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

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

The holomorphic case

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

And we obtain

That is

OPEs

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

We write

to mean

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

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

In other words

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.,

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

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

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

In other words,

Similarly,

The TT OPE

Set

Then

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

Therefore we have found the $T T$ OPE

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

Set

Then

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

Therefore we find