aainstitutetext: New High Energy Theory Center, Department of Physics and
Astronomy, Rutgers University, 126 Frelinghuysen Rd., Piscataway NJ 08855,
USAbbinstitutetext: Department of Physics and Astronomy, Rutgers University,
126 Frelinghuysen Rd., Piscataway NJ 08855, USA

# Non-Chiral Vertex Operator Algebra Associated To Lorentzian Lattices And
Narain CFTs

Ranveer Kumar Singh b Madhav Sinha<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract

Frenkel, Lepowsky, and Meurman constructed a vertex operator algebra (VOA)
associated to any even, integral, Euclidean lattice. In the language of
physics, these are examples of chiral conformal field theories (CFT). In this
paper, we define non-chiral vertex operator algebra and some associated
notions. We then give a construction of a non-chiral VOA associated to an
even, integral, Lorentzian lattice and construct their irreducible modules. We
obtain the moduli space of such modular invariant non-chiral VOAs based on
self-dual Lorentzian lattices of signature $(m,n)$ assuming the validity of a
technical result about automorphisms of the lattice. We finally show that
Narain conformal field theories in physics are examples of non-chiral VOA. Our
formalism helps us to identify the chiral algebra of Narain CFTs in terms of a
particular sublattice and break its partition function into sum of characters.

##  1 Introduction

The study of vertex (operator) algebras started with the work of Borcherds
Borcherds:1983sq  and Frenkel, Lepowsky, and Meurman Frenkel:1988xz  in
relation to Monstrous Moonshine and two dimensional conformal field theory.
The first non-trivial examples of the theory were constructed starting from an
even, integral Euclidean lattice, see Dolan:1994st ; Moore:2023zmv  for a more
physical construction. These are examples of what are called chiral conformal
field theory in physics, see Subsection 1.1.1 for an introduction. There are
ample examples of non-chiral conformal field theories which cannot be
described mathematically in the language of vertex operator algebra. One large
class of such theories is the Narain CFTs based on Lorentzian lattices.

In this paper, we define the notion of non-chiral vertex operator algebra and
study various related notions. Our definition is based on the notion111See
also Kapustin:2000aa ; rosellen2004opealgebras  for some earlier related but
different discussion on non-chiral VOAs. of full field algebras introduced by
Huang and Kong in Huang:2005gz  and full vertex algebras introduced by
Moriwaki in Moriwaki:2020cxf . We use formal calculus as well as complex
analysis to formulate our axioms. We replace the Jacobi identity axiom of
vertex (operator) algebras by a locality axiom which is general enough to
imply the duality and hence operator product expansion of vertex operators.
Various well-known examples of non-chiral conformal field theories in physics
are examples of our definition. More concretely, we construct a class of
examples of non-chiral VOAs based on Lorentzian lattices which cover the
moduli space of Narain CFTs as examples of our definition. We then define
modules and intertwiners of non-chiral VOAs on the lines of Frenkel:1993xz .
In the rest of this section, we describe a dictionary between non-chiral VOAs
of this paper and the notion of (non-chiral) conformal field theory in
physics.

###  1.1 The dictionary from non-chiral VOA to non-chiral CFTs

In this section, we give a dictionary between conformal field theories as
generally understood in physics and the non-chiral vertex operator algebra
definition in this paper. This dictionary, for the case of chiral conformal
field theory, is well-known to experts. We extend it to the case of non-chiral
VOA. We start by describing the defining data of a conformal field theory in
physics.

####  1.1.1 Physical definition of a CFT

We begin with the Belavin-Polyakov-Zamalodchikov (BPZ) definition of a
conformal field theory Belavin:1984vu . We follow DiFrancesco:1997nk ;
Moore:1988qv ; Moore:1989vd ; Blumenhagen:2009zz ; Gaberdiel:1999mc ;
Mukhi:2022bte  for this exposition.

A bosonic CFT is an inner product space $\mathscr{H}$ which222Physically, one
always has a Hilbert space. is decomposable as a direct sum of tensor product

$\mathscr{H}=\bigoplus_{\begin{subarray}{c}h,\bar{h}\\\
h-\bar{h}\in\mathbb{Z}\end{subarray}}V(h,c)\otimes\overline{V}(\bar{h},\bar{c})\leavevmode\nobreak\
,$ (1)

of irreducible highest weight modules of
$\text{Vir}_{c}\times\overline{\text{Vir}}_{\bar{c}}$, where where
$\text{Vir}_{c}\text{ and }\overline{\text{Vir}}_{\bar{c}}$ are two copies of
the Virasoro algebra with central charge $c,\bar{c}$ :

$\begin{split}&\left[L_{m},L_{n}\right]=(m-n)L_{m+n}+\frac{c}{12}m\left(m^{2}-1\right)\delta_{m+n,0}\,,\\\
&\left[\bar{L}_{m},\bar{L}_{n}\right]=(m-n)\bar{L}_{m+n}+\frac{\bar{c}}{12}m\left(m^{2}-1\right)\delta_{m+n,0}\,,\\\
&\left[L_{m},\bar{L}_{n}\right]=0\,,\end{split}$ (2)

such that the following are satisfied :

  1. 1.

Identity Property : There is a unique vector $|0\rangle\in
V(0,c)\otimes\overline{V}(0,\bar{c})$ which is invariant under the
$\mathfrak{sl}(2)\times\overline{\mathfrak{sl}(2)}$-subalgebra of
$\text{Vir}_{c}\times\overline{\text{Vir}}_{\bar{c}}$ generated by
$L_{0},\bar{L}_{0},L_{\pm 1}$, and $\bar{L}_{\pm 1}$.

  2. 2.

For each vector $\alpha\in\mathscr{H}$ there is an operator
$\phi_{\alpha}(z,\bar{z})$ acting on $\mathscr{H}$, parameterized by
$z\in\mathbb{C}$. Also, for every operator $\phi_{\alpha}$ there exists a
conjugate operator $\phi_{\alpha^{\vee}}$, partially characterized by the
requirement that the operator product expansion (OPE) of $\phi_{\alpha}$ and
$\phi_{\alpha^{\vee}}$ contains a descendant of the identity operator.

  3. 3.

$L_{n}$ Property : For $\alpha\in V(h,c)\otimes\overline{V}(\bar{h},\bar{c})$
a highest weight state of the
$(\text{Vir}_{c}\times\overline{\text{Vir}}_{\bar{c}})$-action, we have

$\begin{split}\left[L_{n},\phi_{\alpha}(z,\bar{z})\right]=\left(z^{n+1}\frac{d}{dz}+h(n+1)z^{n}\right)\phi_{\alpha}(z,\bar{z})\,,\\\
\left[\bar{L}_{n},\phi_{\alpha}(z,\bar{z})\right]=\left(\bar{z}^{n+1}\frac{d}{d\bar{z}}+\bar{h}(n+1)\bar{z}^{n}\right)\phi_{\alpha}(z,\bar{z})\,,\end{split}$
(3)

for real numbers $h$ and $\bar{h}$.

  4. 4.

Duality Property : The inner products $\left\langle
0\left|\phi_{\alpha_{1}}\left(z_{1},\bar{z}_{1}\right)\ldots\phi_{\alpha_{n}}\left(z_{n},\bar{z}_{n}\right)\right|0\right\rangle$
exist for $\left|z_{1}\right|>\ldots>\left|z_{n}\right|>0$ and admit an
unambiguous real-analytic continuation, independent of ordering333For
Fermionic fields, one has to keep track of signs while commuting them past
each other., to
$\mathbb{C}^{n}\setminus\\{z_{1},\dots,z_{n}=0,\infty;z_{i}=z_{j}\\}$.

  5. 5.

Modular invariance property : The torus partition function and correlation
functions, given in terms of traces exist and are modular invariant.

These axioms do not characterise a CFT but are necessary for a well defined
CFT. A particular subset of operators which only depend only on $z$ or
$\bar{z}$ are of interest. Such operators are called chiral ($z$ dependent)
and anti-chiral ($\bar{z}$ dependent) operators. The set of chiral and anti-
chiral operators form an algebra which we denote by $\mathscr{A}$ and
$\overline{\mathscr{A}}$ respectively. For any CFT,
$\mathds{1}\in\mathscr{A}\otimes\overline{\mathscr{A}}$, $T(z)\in\mathscr{A}$
and $\overline{T}(\bar{z})\in\overline{\mathscr{A}}$ where $T$ and
$\overline{T}$ are the holomorphic and anti-holomorphic stress tensor with
modes $L_{n}$ and $\bar{L}_{n}$ respectively. Let $\\{\mathcal{O}^{i}(z)\\}$
be a basis of $\mathscr{A}$. The OPE of chiral operators takes the form

$\mathcal{O}^{i}(z)\mathcal{O}^{j}(w)=\sum_{k}\frac{c_{ijk}}{(z-w)^{h_{ijk}}}\mathcal{O}^{k}(w)\,,$
(4)

for some coefficients $c_{ijk}$ where $h_{ijk}=h_{i}+h_{j}-h_{k}$. By the
usual contour integral manipulations we can write the above OPE as the algebra
of modes:

$\left[\mathcal{O}_{n}^{i},\mathcal{O}_{m}^{j}\right]=\sum_{k}c_{ijk}(n,m)\mathcal{O}_{n+m}^{k}\,\,,$
(5)

where

$\mathcal{O}^{j}(z)=\sum_{m}\mathcal{O}_{m}^{j}z^{-m-h_{j}}\,.$ (6)

Similar OPE holds for anti-chiral operators. This is called the chiral algebra
(anti-chiral algebra) of the CFT. In the following, we will only speak of the
chiral algebra but all statements hold for anti-chiral algebra equally well.
The chiral algebra of any CFT contains the (universal enveloping algebra of)
Virasoro algebra since the stress tensor is always a chiral operator. Other
examples of chiral algebra include the affine-Kac Moody algebra Goddard:1986bp
and the $W_{3}$ algebra Zamolodchikov:1985wn . Only the zero mode the of
chiral operator $\mathcal{O}(z)$ commute with the Hamiltonian
$(L_{0}+\bar{L}_{0})$ and are hence called the symmetry-generating algebra. In
the inner product space (1), one can talk about subspaces $\mathscr{H}_{i}$
which form irreducible representations of the chiral algebra $\mathscr{A}$.
For this reason the full chiral algebra is sometimes also called the spectrum-
generating algebra. We can thus decompose the physical Hilbert space
$\mathscr{H}$ as:

$\mathscr{H}=\bigoplus_{i,\bar{i}}N_{i,\bar{i}}\mathscr{H}_{i}\otimes\overline{\mathscr{H}}_{\bar{i}}\,,$
(7)

where $\mathscr{H}_{i},\overline{\mathscr{H}}_{\bar{i}}$ are irreducible
representations of $\mathscr{A},\overline{\mathscr{A}}$ respectively and
$N_{i,\bar{i}}\in\mathbb{N}_{0}=\mathbb{N}\cup\\{0\\}$, the set of non-
negative integers, is the number of times
$\mathscr{H}_{i}\otimes\overline{\mathscr{H}}_{\bar{i}}$ appears in
$\mathscr{H}$. For the index value $i,\bar{i}=0$, we take $\mathscr{H}_{0}$
and $\overline{\mathscr{H}}_{0}$ to be the subspace of $\mathscr{H}$ which
contains states corresponding to $\mathscr{A}$ and $\overline{\mathscr{A}}$
respectively modulo the null states. Hence,
$N_{i,0}=\delta_{i,0},N_{0,\bar{i}}$ $=\delta_{0,\bar{i}}$.

Let us now describe the relation between the two decompositions (1) and (7). A
general state $|h,\bar{h}\rangle\in\mathscr{H}$ is called a Virasoro primary
of conformal dimension $(h,\bar{h})$ if

$\begin{split}&L_{0}|h,\bar{h}\rangle=h|h,\bar{h}\rangle\quad\bar{L}_{0}|h,\bar{h}\rangle=\bar{h}|h,\bar{h}\rangle\\\
&L_{n}|h,\bar{h}\rangle=\bar{L}_{n}|h,\bar{h}\rangle=0,\quad n>0.\end{split}$
(8)

This follows from the OPE

$\begin{split}&T(z)\phi(w,\bar{w})=\frac{h}{(z-w)^{2}}\phi(w,\bar{w})+\frac{\partial_{w}\phi(w,\bar{w})}{z-w}+\text{
hol. },\\\
&\overline{T}(\bar{z})\phi(w,\bar{w})=\frac{\bar{h}}{(\bar{z}-\bar{w})^{2}}\phi(w,\bar{w})+\frac{\partial_{\bar{w}}\phi(w,\bar{w})}{\bar{z}-\bar{w}}+\text{
anti-hol. },\end{split}$ (9)

where $\phi(w,\bar{w})$ is the operator corresponding to the
$|h,\bar{h}\rangle$. One can identify the state $|h,\bar{h}\rangle$ with

$|h,\bar{h}\rangle\equiv\lim_{z,\bar{z}\to
0}\phi(z,\bar{z})|0\rangle\leavevmode\nobreak\ .$ (10)

We will consider $|h,\bar{h}\rangle$ as the tensor product of states
$|h\rangle,|\bar{h}\rangle$:
$|h,\bar{h}\rangle\equiv|h\rangle\otimes|\bar{h}\rangle$. The Verma modules
$V(h,c)$ and $\overline{V}(\bar{h},\bar{c})$ is given by

$\begin{split}&V(h,c):=\text{Span}_{\mathbb{C}}\\{L_{-n_{1}}\dots
L_{-n_{k}}|h\rangle:n_{1},\dots,n_{k}>0,k\in\mathbb{N}\\}\,,\\\
&\overline{V}(\bar{h},\bar{c}):=\text{Span}_{\mathbb{C}}\\{\bar{L}_{-n_{1}}\dots\bar{L}_{-n_{k}}|\bar{h}\rangle:n_{1},\dots,n_{k}>0,k\in\mathbb{N}\\}\,.\end{split}$
(11)

The commutators (2) and the Virasoro primary condition makes the Verma module
$V(h,c)\otimes\overline{V}(\bar{h},\bar{c})$ into a
$(\text{Vir}_{c}\times\overline{\text{Vir}}_{\bar{c}})$-representation. In
general, this representation is reducible and one has to quotient out singular
or null states 444A singular or null state is a vector which is not a highest
weight vector but is annihilated by $L_{n},\bar{L}_{n},\leavevmode\nobreak\
n>0$, see (di1996conformal, , Section 7.1.3) for a detailed discussion. from
these Verma modules to make them irreducible. We will assume that this has
already been done and that $V(h,c)\otimes\overline{V}(\bar{h},\bar{c})$ is an
irreducible $(\text{Vir}_{c}\times\overline{\text{Vir}}_{\bar{c}})$-module.
Note that from the OPE (9) we must have $\bar{h}=0$ ($h=0$) for a chiral
(anti-chiral field). Now we can
identify$\mathscr{H}_{0}\otimes\overline{\mathscr{H}}_{0}$ as

$\mathscr{H}_{0}\otimes\overline{\mathscr{H}}_{0}\cong\left(\bigoplus_{h\in\cal{S}}V(h,c)\right)\otimes\left(\bigoplus_{\bar{h}\in\overline{\cal{S}}}\overline{V}(\bar{h},\bar{c})\right)\subset\mathscr{H},$
(12)

where

$\begin{split}\mathcal{S}:=\\{h\in\mathbb{Z}:|h\rangle\otimes|0\rangle\equiv|h,0\rangle\in\mathscr{H}\\}\leavevmode\nobreak\
,\\\
\overline{\mathcal{S}}:=\\{\bar{h}\in\mathbb{Z}:|0\rangle\otimes|\bar{h}\rangle\equiv|0,\bar{h}\rangle\in\mathscr{H}\\}\leavevmode\nobreak\
.\end{split}$ (13)

A chiral primary is a Virasoro primary $|h,\bar{h}\rangle$ satisfying

$\mathcal{O}_{m}|h,\bar{h}\rangle=\overline{\mathcal{O}}_{m}|h,\bar{h}\rangle=0,\quad
m>0\,,$ (14)

and is an eigenstate of $\mathcal{O}_{0}$ and $\overline{\mathcal{O}}_{0}$ for
every $\mathcal{O}\in\mathscr{A}$ and
$\overline{\mathcal{O}}\in\overline{\mathcal{A}}$, the operator corresponding
to it will be called the chiral primary field. Then each irreducible factor in
(7) is a subspace of a Verma module constructed over a chiral primary by the
modes of every chiral and anti-chiral operator.

One can then talk about characters of the CFT defined for each irreducible
factor $\mathscr{H}_{i}\otimes\overline{\mathscr{H}}_{\bar{i}}$:

$\chi_{i,\bar{i}}(\tau,\bar{\tau})=\text{Tr}_{\mathscr{H}_{i}\otimes\overline{\mathscr{H}}_{\bar{i}}}q^{L_{0}-\frac{c}{24}}\bar{q}^{\bar{L}_{0}-\frac{\bar{c}}{24}},\quad
q=e^{2\pi i\tau},\leavevmode\nobreak\ \leavevmode\nobreak\ \bar{q}=e^{-2\pi
i\bar{\tau}}\leavevmode\nobreak\ ,$ (15)

where $\tau\in\mathbb{H}:=\\{x+iy\in\mathbb{C}:y>0\\}$ is in the upper half
plane. From the fact that
$\mathscr{H}_{i}\otimes\overline{\mathscr{H}}_{\bar{i}}$ is built over some
highest vector $|h_{i},\bar{h}_{i}\rangle$, we see that the character has the
form

$\chi_{i,\bar{i}}(\tau,\bar{\tau})=q^{h_{i}-\frac{c}{24}}\bar{q}^{\bar{h}_{i}-\frac{\bar{c}}{24}}\sum_{n,m=0}^{\infty}a(n)\bar{a}(m)q^{n}\bar{q}^{m}\leavevmode\nobreak\
,$ (16)

for some integers $a(n),\bar{a}(m)$. Thus we can separate the character into
chiral and anti-chiral characters:

$\chi_{i,\bar{i}}(\tau,\bar{\tau})=\chi_{i}(\tau)\overline{\chi}_{\bar{i}}(\bar{\tau})$
(17)

where

$\chi_{i}(\tau)=q^{h_{i}-\frac{c}{24}}\sum_{n=0}^{\infty}a(n)q^{n},\quad\overline{\chi}_{\bar{i}}(\bar{\tau})=\bar{q}^{\bar{h}_{i}-\frac{\bar{c}}{24}}\sum_{m=0}^{\infty}\bar{a}(m)\bar{q}^{m}\leavevmode\nobreak\
.$ (18)

The partition function of the CFT is then given by sum over chiral characters

$Z(\tau,\bar{\tau}):=\text{Tr}_{\mathscr{H}}q^{L_{0}-\frac{c}{24}}\bar{q}^{\bar{L}_{0}-\frac{\bar{c}}{24}}=\sum_{i,\bar{i}}N_{i,\bar{i}}\chi_{i}(\tau)\overline{\chi}_{\bar{i}}(\bar{\tau})\leavevmode\nobreak\
.$ (19)

Modular invariance of the partition function implies that the chiral
characters form a weight-zero weakly holomorphic vector valued modular form
555See for example Choie2019  for the definition of vector valued modular
forms.. Using this property of chiral characters, one can classify CFTs with
finitely many primary fields and given central charge Mukhi:2022bte ;
Gaberdiel:2016zke ; Das:2022uoe ; Das:2023qns ; Hampapura:2015cea ;
Mathur:1988gt ; Mathur:1988na ; Mathur:1988rx . A CFT is called a chiral CFT
if the partition function and its correlation functions decompose into a
product of an analytic function of $\tau$ and an analytic function of
$\bar{\tau}$. In such a case, we say that the CFT admits a holomorphic
factorisation. A non-chiral conformal field theory does not admit a
holomorphic factorisation.

####  1.1.2 Non-chiral vertex operator algebras

We now briefly describe the notion of non-chiral vertex operator algebras
studied in this paper. We refer to Section 2 for precise and detailed
discussions. Just as vertex operator algebras, we start with a vector space
$V$ which is $(\mathbb{R}\times\mathbb{R})$-graded. We have the vertex
operator map $Y_{V}:V\longrightarrow\text{End}(V)\\{x,\bar{x}\\}$ where
$x,\bar{x}$ are two formal variables. We require the existence of a vacuum
vector 1 and conformal vectors $\omega,\bar{\omega}$ such that
$Y_{V}(\textbf{1},x,\bar{x})=\mathds{1}$ and the coefficients of the formal
series for $Y_{V}(\omega,x,\bar{x}),Y_{V}(\bar{\omega},x,\bar{x})$ satisfies
two copies of the Virasoro algebra. The vertex operators are required to
satisfy certain translation and $L(0)$ axioms similar to VOAs. It turns out
that that the Jacobi identity for VOAs is difficult to formulate for non-
chiral VOAs Moriwaki:2020cxf . The appropriate locality axiom is motivated
from full field algebras of Huang:2005gz . Roughly speaking, locality of
vertex operators says that the matrix elements of the product
$Y_{V}(v,z_{1},\bar{z}_{1})Y_{V}(w,z_{2},\bar{z}_{2})$ and
$Y_{V}(v,z_{1},\bar{z}_{1})Y_{V}(w,z_{2},\bar{z}_{2})$, defined when
$|z_{1}|>|z_{2}|$ and $|z_{2}|>|z_{1}|$ respectively, are equal to the same
function, which is multi-valued and analytic in
$z_{1},\bar{z}_{1},z_{2},\bar{z}_{2}$ and single valued when
$\bar{z}_{1},\bar{z}_{2}$ are complex-conjugates of $z_{1},z_{2}$, defined on
$\mathbb{C}^{4}$ minus a diagonal subset. This version of locality turns into
a statement of analytic continuation of matrix elements for chiral vertex
operators and allows us to use contour integration for manipulating the modes
of the vertex operators. Moreover, locality allows us to prove duality of
vertex operators which in turn gives us the operator product expansion of the
product of two vertex operators. Modules and intertwiners of a non-chiral VOA
are then defined analogous to modules of a VOA.

####  1.1.3 The Dictionary

Let us now describe the dictionary between non-chiral VOA and its modules and
a non-chiral CFT. Given a bosonic CFT, its chiral and anti-chiral algebra is a
non-chiral VOA according to our definition. Note that our notion of non-chiral
VOA allows for more general structure in the sense that the chiral algebra of
a CFT always has the structure of a tensor product
$\mathscr{H}_{0}\otimes\overline{\mathscr{H}}_{0}$ as described above but non-
chiral VOAs are allowed to have more general vector spaces. The chiral and
anti-chiral operators are identified with the vertex operators of the VOA. For
general non-chiral VOA, chiral and anti-chiral vertex operators form only a
subsector of the set of vertex operators, see Definition 2.3 and Theorem 2.3
below. Next, the irreducibles
$\mathscr{H}_{i}\otimes\overline{\mathscr{H}}_{\bar{i}}$ must be identified
with modules of the VOA. Again in our generic construction we allow for the
modules to have more general structure rather than a tensor product. The
chiral primary field corresponding to an irreducible
$\mathscr{H}_{i}\otimes\overline{\mathscr{H}}_{\bar{i}}$ is identified with
the intertwiner $\mathcal{Y}^{\leavevmode\nobreak\
(i,\bar{i})}_{(i,\bar{i})(0,0)}(w,x,\bar{x})$ of type
${(i,\bar{i})\choose(i,\bar{i})(0,0)}$ where $(0,0)$ indicates the VOA as a
module for itself. The state-operator correspondence for the space (7)
corresponds to the vertex operator map for the VOA and its modules along with
the intertwining operators of type ${(i,\bar{i})\choose(i,\bar{i})(0,0)}$. The
full dictionary is summarised in Table 1. We hope to expand the dictionary to
include fusion rules and rationality on the CFT side to the notion of tensor
product of modules and (strong) regularity of non-chiral VOAs in a future
publication.

Non-chiral CFT | Non-chiral VOA
---|---
Chiral space of states $\mathscr{H}_{0}\otimes\overline{\mathscr{H}}_{0}$ (12) |  $(\mathbb{R}\times\mathbb{R})$-graded vector space $V$ (32)
State-operator map $\alpha\mapsto\phi_{\alpha}$ (2) on Page 3 | Vertex operator map $v\mapsto Y_{V}(v,x,\bar{x})$ (35)
Stress tensor $T(z),\overline{T}(\bar{z})$ | Conformal vertex operators $Y_{V}(\omega,x),Y_{V}(\bar{\omega},\bar{x})$ (46)
Duality of operators (4) on Page 3 | Locality of vertex operators Def. 2.2 (9)
Chiral and anti-chiral algebra (5) | Chiral and anti-chiral algebra (93)
Irreducible representation $\mathscr{H}_{i}\otimes\overline{\mathscr{H}}_{\bar{i}}$ of $\mathscr{A}\otimes\overline{\mathscr{A}}$ | Irreducible module $(W,Y_{W})$ of non-chiral VOA $(V,Y_{V})$ Def. 4.1
Characters $\chi_{i,\bar{i}}(\tau,\bar{\tau})$ (15) | Graded dimension $\chi_{W}(\tau,\bar{\tau})$ (301)
Chiral primary operator $\phi_{\alpha}$ for $\alpha\in\mathscr{H}_{i}\otimes\overline{\mathscr{H}}_{i}$ | Intertwiner $\mathcal{Y}^{\leavevmode\nobreak\ (i,\bar{i})}_{(i,\bar{i})(0,0)}(w,x,\bar{x})$ of type ${(i,\bar{i})\choose(i,\bar{i})(0,0)}$ Def. 4.2
Table 1: Dictionary between a CFT and non-chiral VOA

###  1.2 Lorentzian Lattice Vertex Operator Algebra (LLVOA)

One of the main constructions of this paper is a concrete example of a non-
chiral VOA based on an even integral Lorentzian lattice. The construction is
similar in spirit to Euclidean lattice VOAs but differs in that it is
inherently non-chiral and does not admit holomorphic factorisation thus
providing the first non-trivial example of a non-chiral VOA. We call it the
Lorentzian lattice vertex operator algebra (LLVOA). The modules for a non-
chiral VOA introduced in this paper can be defined analogous to modules for
usual VOAs. Infact, we are able to construct modules for the LLVOA in one to
one correspondence with certain cosets of the lattice. Collecting the VOA and
its modules, we define the partition function of the non-chiral CFT (see
Definition 4.3) thus obtained. We find that the modules of the LLVOA
constructed using the cosets of the lattice give rise to a modular invariant
partition function. We further attempt to classify all possible modular
invariant non-chiral CFT based on even, integral self-dual Lorentzian lattices
of a given signature. This leads us to a conjecture about automorphisms of
Lorentzian lattices which we prove for signature $(m,m)$ but are unable to
prove for general signature $(m,n)$ with $m\neq n$. Following the physics
terminology, we call the equivalence classes of non-chiral CFTs based on
Lorentzian lattices as the moduli space of LLVOAs. As expected from physical
arguments, the moduli space in signature $(m,n)$ is given by (see Theorem 5.2)

$\mathcal{M}_{m,n}\cong\frac{\mathrm{O}(m,n,\mathbb{R})}{\mathrm{O}(m,\mathbb{R})\times\mathrm{O}(n,\mathbb{R})\times\mathrm{O}(m,n,\mathbb{Z})}.$
(20)

The paper is organised as follows: in Section 2, we introduce the notion of
non-chiral VOA and prove some elementary consequences of the definition. Then
in Section 3, we construct the LLVOA and prove that it is an example of a non-
chiral VOA. In Section 4 we define the notion of modules and intertwining
operators and prove some important consequences and results required later. In
Section 5 we construct the modules of LLVOAs. We formulate a precise
conjecture about automorphism of Lorentzian lattices and under that
assumption, prove that the moduli space of LLVOAs in signature $(m,n)$ is
given by (20). Finally in Section 6 we review the physical construction of
Narain CFTs and comment on their relation to LLVOAs. Appendix A deals with the
independence of central extensions of lattices on the chosen basis of the
lattice. Appendix C contains some technical results about modules of
Heisenberg algebras and Appendix D contains the proof of Conjecture 3.3.1 for
the special case $m=n$.

##  2 Non-Chiral Vertex Operator Algebra

###  2.1 Formal calculus

We begin by collecting some notatations about formal calculus. The reader is
referred to (Frenkel:1988xz, , Chapter 3,8) and (lepowsky2012introduction, ,
Chapter 2) for more details.

Let $x$ be a formal variable. For a vector space $V$, we define the following:

$V[x]=\left\\{\,\sum_{n\in\mathbb{N}_{0}}v_{n}x^{n}:v_{n}\in V,\text{ where
only finitely many $v_{n}\neq 0$}\right\\}\,,$ (21)
$V[[x]]=\left\\{\,\sum_{n\in\mathbb{N}_{0}}v_{n}x^{n}:v_{n}\in V\right\\}\,,$
(22)

$V\\{x\\}=\left\\{\sum_{n\in\mathbb{F}}v_{n}x^{n}\,:\,v_{n}\in V\right\\}\,,$
(23)

where $\mathbb{N}_{0}=\mathbb{N}\cup\\{0\\}$ and $\mathbb{F}$ is an arbitrary
field of characteristic not 2. We will mostly be interested in the case
$\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$. For a complex number $s\in\mathbb{C}$
and formal variables $x_{1},x_{2}$, we will define

$(x_{1}+x_{2})^{s}:=\sum_{n=0}^{\infty}{s\choose
n}x_{1}^{s-n}x_{2}^{n}\leavevmode\nobreak\ ,$ (24)

where the binomial coefficient is defined as

${s\choose n}=\frac{s(s-1)\cdots(s-n+1)}{n!}.$ (25)

Note, in a series like this, we will always have non-negative integral powers
of the second variable. With this formula, one can check that

$\begin{split}\left(1-\frac{x_{1}}{x_{2}}\right)^{s}x_{2}^{s}&=(x_{2}-x_{1})^{s}.\end{split}$
(26)

If we replace $x_{1},x_{2}$ by complex variables $z_{1},z_{2}$ then by
definition stein2010complex

$\begin{split}&(z_{1}-z_{2})^{s}:=\exp(s\log(z_{1}-z_{2})),\\\
&(z_{2}-z_{1})^{s}:=\exp(s\log(z_{2}-z_{1})).\end{split}$ (27)

Using the fact that

$\log(1-z)=-\sum_{n=0}^{\infty}\frac{z^{n}}{n},\quad|z|<1,$ (28)

and the identity 666This identity can be proven by using the relation
$(1-x)^{s}=\exp(s\log(1-x))$ for any real $x$ with $|x|<1$ and
$s\in\mathbb{C}$.

$(-1)^{k}{s\choose
k}=\sum_{\ell=1}^{k}\frac{(-s)^{\ell}}{\ell!}\sum_{\begin{subarray}{c}n_{1}+\dots+n_{\ell}=k\\\
n_{1},\dots,n_{\ell}\geq 1\end{subarray}}\frac{1}{n_{1}\cdots n_{\ell}},$ (29)

it is easy to see that 777Another way of proving this identity is to use
Taylor’s theorem.

$\begin{split}&(z_{1}-z_{2})^{s}=\exp(s\log(z_{1}-z_{2}))=\sum_{n\geq
0}{s\choose
n}(-1)^{n}z_{1}^{s-n}z_{2}^{n},\quad|z_{1}|>|z_{2}|\leavevmode\nobreak\ ,\\\
&(z_{2}-z_{1})^{s}=\exp(s\log(z_{2}-z_{1}))=\sum_{n\geq 0}{s\choose
n}(-1)^{n}z_{2}^{s-n}z_{1}^{n},\quad|z_{2}|>|z_{1}|\leavevmode\nobreak\
,\end{split}$ (30)

which is consistent with the definition (24). Let $f(x)=\sum v_{n}x^{n}\,\in
V[[x,x^{-1}]]$, then we have the formal version of Taylor’s theorem:

$\mathrm{e}^{x_{0}\,\frac{d}{dx}}\,f(x)=f(x+x_{0})\,,$ (31)

One can prove this by expanding both sides and comparing terms of equal power
of $x,x_{0}$. As before, we need to expand the RHS in non-negative integral
powers of $x_{0}$.

###  2.2 Definition of non-chiral VOA

Let

$V=\coprod\limits_{(h,\bar{h})\in\mathbb{R}\times\mathbb{R}}V_{(h,\bar{h})}\,,$
(32)

be an $\mathbb{R}\times\mathbb{R}$-graded complex vector space vector space.
Let

$\overline{V}=\prod\limits_{(h,\bar{h})\in\mathbb{R}\times\mathbb{R}}V_{(h,\bar{h})}\,,$
(33)

denote the algebraic completion of $V$. Let

$V^{\prime}=\coprod\limits_{(h,\bar{h})\in\mathbb{R}\times\mathbb{R}}V^{\prime}_{(h,\bar{h})}\,,$
(34)

be the contragradient of $V$ where $V^{\prime}_{(h,\bar{h})}$ is the dual of
$V_{(h,\bar{h})}$. A series $\sum f_{n}$ in $\overline{V}$ is said to be
absolutely convergent if for every $f^{\prime}\in V^{\prime}$ the series
$\sum|\langle f^{\prime},f_{n}\rangle|$ is convergent. Here, $\langle
f^{\prime},f_{n}\rangle=f^{\prime}(f_{n})\in\mathbb{C}$ is just the action of
the linear functional on $f^{\prime}$ on $f_{n}$. A non-chiral vertex operator
algebra of central charge $(c,\bar{c})$ is a quintuple
$(V,Y_{V},\omega,\bar{\omega},\mathbf{1})$ where $V$ is an
$\mathbb{R}\times\mathbb{R}$-graded complex vector space and $Y_{V}$ is a
linear map, called the vertex operator map,

$\begin{split}Y_{V}:\,&V\otimes V\longrightarrow V\\{x^{\pm 1},\bar{x}^{\pm
1}\\}\,,\\\ &u\otimes v\longmapsto Y_{V}(u,x,\bar{x})v\,,\end{split}$ (35)

or equivalently a map

$\begin{split}Y_{V}:\,&\mathbb{C}^{\times}\times\mathbb{C}^{\times}\longrightarrow\text{Hom}(V\otimes
V,\overline{V})\\\ &(z,\bar{z})\longmapsto Y_{V}(\cdot,z,\bar{z}):u\otimes
v\longmapsto Y_{V}(u,z,\bar{z})v,\end{split}$ (36)

which is multi-valued and analytic if $z,\bar{z}$ are independent complex
variables and single valued when $\bar{z}$ is the complex conjugate of $z$.
The vertex operator $Y_{V}(u,x,\bar{x})$ is expanded as a formal power series

$Y_{V}(u,x,\bar{x})=\sum_{m,n\in\mathbb{R}}u_{m,n}x^{-m-1}\bar{x}^{-n-1}\in\text{End}(V)\\{x^{\pm
1},\bar{x}^{\pm 1}\\}\,,$ (37)

and when $u\in V_{(h,\bar{h})}$, it can also be expanded as

$Y_{V}(u,x,\bar{x})=\sum_{m,n\in\mathbb{R}}x_{m,n}(u)x^{-m-h}\bar{x}^{-n-\bar{h}}\in\text{End}(V)\\{x^{\pm
1},\bar{x}^{\pm 1}\\}\,,$ (38)

so that

$x_{m,n}(u)=u_{m+h-1,n+\bar{h}-1},\quad m,n\in\mathbb{Z}\,.$ (39)

We call $x_{m,n}(u)$ the modes of the vertex operators $Y_{V}(u,x,\bar{x})$.
The degree $(h,\bar{h})$ is called the conformal weight of $u\in
V_{(h,\bar{h})}$ and we write

$\text{wt}(u)=h,\quad\overline{\text{wt}}(u)=\bar{h}\,.$ (40)

The vector $\mathbf{1}\in V_{(0,0)}$ is called the vacuum vector and
$\omega\in V_{(2,0)},\bar{\omega}\in V_{(0,2)}$ are chiral and anti-chiral
conformal vectors respectively. This data is required to satisfy the following
properties:

  1. 1.

Identity property: The vertex operator corresponding to the vacuum vector acts
as identity, i.e.

$Y_{V}(\mathbf{1},x,\bar{x})u=u,\quad\forall\leavevmode\nobreak\ u\in V.$ (41)

  2. 2.

Grading-restriction property: For every
$(h,\bar{h})\in\mathbb{R}\times\mathbb{R}$,

$\text{dim}(V_{(h,\bar{h})})<\infty\,,$ (42)

and there exists $M\in\mathbb{R}$, such that

$V_{(h,\bar{h})}=0,\quad\text{ for }h<M\text{ or }\bar{h}<M.$ (43)

  3. 3.

Single-valuedness property: For every homogenous subspace $V_{(h,\bar{h})}$

$h-\bar{h}\in\mathbb{Z}.$ (44)

  4. 4.

Creation property: For any $v\in V$

$\lim_{x,\bar{x}\to 0}Y_{V}(v,x,\bar{x})\mathbf{1}=v\leavevmode\nobreak\ ,$
(45)

that is $Y_{V}(v,x,\bar{x})\mathbf{1}$ involves only non-negative powers of
$x,\bar{x}$ and the constant term is $v$.

  5. 5.

Virasoro property: The vertex operators $Y_{V}(\omega,x,\bar{x})$ and
$Y_{V}(\bar{\omega},x,\bar{x})$, called conformal vertex operators, have
Laurent series in $x,\bar{x}$ given by

$\begin{split}&T(x):=Y_{V}(\omega,x,\bar{x})=\sum_{n\in\mathbb{Z}}L(n)x^{-n-2},\\\
&\bar{T}(\bar{x}):=Y_{V}(\bar{\omega},x,\bar{x})=\sum_{n\in\mathbb{Z}}\bar{L}(n)\bar{x}^{-n-2},\end{split}$
(46)

where $L(n),\bar{L}(n)$ are operators which satisfy the Virasoro algebra with
central charge $c,\bar{c}$ respectively:

$\begin{split}&\left[L(m),L(n)\right]=(m-n)L(m+n)+\frac{c}{12}m\left(m^{2}-1\right)\delta_{m+n,0},\\\
&\left[\bar{L}(m),\bar{L}(n)\right]=(m-n)\bar{L}(m+n)+\frac{\bar{c}}{12}m\left(m^{2}-1\right)\delta_{m+n,0},\end{split}$
(47)

and

$\left[L(m),\bar{L}(n)\right]=0.$ (48)

  6. 6.

Grading property: The operator $(L(0),\bar{L}(0))$ is the grading operator on
$V$, that is for $v\in V_{(h,\bar{h})}$

$L(0)v=hv,\quad\bar{L}(0)v=\bar{h}v.$ (49)

  7. 7.

$L(0)$-property :

$\begin{split}&[L(0),Y_{V}(u,x,\bar{x})]=x\frac{\partial}{\partial
x}Y_{V}(u,x,\bar{x})+Y_{V}(L(0)u,x,\bar{x}),\\\
&[\bar{L}(0),Y_{V}(u,x,\bar{x})]=\bar{x}\frac{\partial}{\partial\bar{x}}Y_{V}(u,x,\bar{x})+Y_{V}(\bar{L}(0)u,x,\bar{x}).\end{split}$
(50)

  8. 8.

Translation property: For any $u\in V$

$\begin{split}&{\left[L(-1),Y_{V}(u,x,\bar{x})\right]=Y_{V}\left(L(-1)u,x,\bar{x}\right)=\frac{\partial}{\partial
x}Y_{V}(u,x,\bar{x}),}\\\
&{\left[\bar{L}(-1),Y_{V}(u,x,\bar{x})\right]=Y_{V}\left(\bar{L}(-1)u,x,\bar{x}\right)=\frac{\partial}{\partial\bar{x}}Y_{V}(u,x,\bar{x})}.\\\
&\end{split}$ (51)

  9. 9.

Locality property: For $u_{1},\dots,u_{n}\in V$, there is an operator-valued
function 888We thank Yi-Zhi Huang for discussions on this point.

$m_{n}(u_{1},...,u_{n};z_{1},\bar{z}_{1},...,z_{n},\bar{z}_{n})\leavevmode\nobreak\
,$

defined on 999The matrix elements of this operator are called correlation
functions in Physics.

$\\{(z_{1},\dots,z_{n},\bar{z}_{1},\dots,\bar{z}_{n})\in\mathbb{C}^{2n}\,\lvert\,z_{i},\bar{z}_{i}\neq
0,z_{i}\neq z_{j},\bar{z}_{i}\neq\bar{z}_{j}\\}\,,$ (52)

which is multi-valued and analytic when $\bar{z}_{1},...,\bar{z}_{n}$ are
viewed as independent variables and is single-valued when
$\bar{z}_{1},...,\bar{z}_{n}$ are equal to the complex conjugates of
$z_{1},...,z_{n}$. Moreover, for any permutation $\sigma\in S_{n}$, the
product of vertex operators

$Y_{V}\left(u_{\sigma(1)},z_{\sigma(1)},\bar{z}_{\sigma(1)}\right)\cdots
Y_{V}\left(u_{\sigma(n)},z_{\sigma(n)},\bar{z}_{\sigma(n)}\right)\leavevmode\nobreak\
,$ (53)

is the expansion of
$m_{n}\left(u_{1},\dots,u_{n};z_{1},\bar{z}_{1},\dots,z_{n},\bar{z}_{n}\right)$
in the domain
$\left|z_{\sigma(1)}\right|>\left|z_{\sigma(2)}\right|>\dots>|z_{\sigma(n)}|>0$.
Here, $\bar{z}_{\sigma(1)},...,\bar{z}_{\sigma(n)}$ are complex conjugates of
$z_{\sigma(1)},...,z_{\sigma(n)}$ respectively. If a function $m_{n}$
satisfying above properties exists, we say that the vertex operators
$Y_{V}\left(u_{1},z_{1},\bar{z}_{1}\right),\dots,Y_{V}\left(u_{n},z_{n},\bar{z}_{n}\right)$
are mutually local with respect to each other.

We will often denote the non-chiral VOA by $(V,Y_{V})$ or simply by $V$. For a
homogeneous vector $u\in V_{(h,\bar{h})}$, the sum in (37) and (38) runs only
over the set $\\{(m,n)\in\mathbb{R}^{2}\lvert\leavevmode\nobreak\
m-n\in\mathbb{Z}\\}$. To see this, first note that the $L(0)$-property 7
implies the commutator

$\begin{split}&\left[L(0),x_{m,n}(u)\right]=-mx_{m,n}(u),\\\
&\left[\bar{L}(0),x_{m,n}(u)\right]=-nx_{m,n}(u).\end{split}$ (54)

Equivalently,

$\begin{split}&\left[L(0),u_{m,n}\right]=(h-m-1)u_{m,n},\\\
&\left[\bar{L}(0),u_{m,n}\right]=(\bar{h}-n-1)u_{m,n}.\end{split}$ (55)

This implies that

$\begin{split}&\text{wt}\leavevmode\nobreak\
x_{m,n}(u)=-m,\quad\overline{\text{wt}}\leavevmode\nobreak\ x_{m,n}(u)=-n,\\\
&\text{wt}\leavevmode\nobreak\
u_{m,n}=h-m-1,\quad\overline{\text{wt}}\leavevmode\nobreak\
u_{m,n}=\bar{h}-n-1.\end{split}$ (56)

The single-valuedness property 3 implies that $m-n\in\mathbb{Z}$ in both the
sums. We will thus write the expansions of the vertex operators as

$\begin{split}Y_{V}(u,x,\bar{x})&=\sum_{\begin{subarray}{c}m,n\in\mathbb{R}\\\
(m-n)\in\mathbb{Z}\end{subarray}}u_{m,n}x^{-m-1}\bar{x}^{-n-1}\in\text{End}(V)\\{x^{\pm
1},\bar{x}^{\pm 1}\\}\\\ &=\sum_{\begin{subarray}{c}m,n\in\mathbb{R}\\\
(m-n)\in\mathbb{Z}\end{subarray}}x_{m,n}(u)x^{-m-h}\bar{x}^{-n-\bar{h}}.\end{split}$
(57)

The single-valuedness property 3 implies that the vertex operators (35) is
single-valued. To prove this, we must show that

$Y_{V}(u,z,\bar{z})=Y_{V}(u,e^{2\pi i}z,e^{-2\pi i}\bar{z}).$ (58)

From Remark 2.2, we have

$\begin{split}Y_{V}(u,e^{2\pi i}z,e^{-2\pi
i}\bar{z})&=\sum_{\begin{subarray}{c}m,n\in\mathbb{R}\\\
(m-n)\in\mathbb{Z}\end{subarray}}u_{m,n}z^{-m-1}\bar{z}^{-n-1}e^{2\pi
i(-m+n)}\\\ &=Y_{V}(u,z,\bar{z}).\end{split}$ (59)

For $v\in V_{(h,\bar{h})}$, if in the expansion (38) the index runs over
$m\in\mathbb{Z}-h,\leavevmode\nobreak\ n\in\mathbb{Z}-\bar{h}$, then assuming
that $z,\bar{z}$ are independent complex variables, we can use Cauchy’s
residue theorem to write

$x_{r,s}(v)=\frac{1}{(2\pi i)^{2}}\oint\oint dzd\bar{z}\leavevmode\nobreak\
Y_{V}(v,z,\bar{z})z^{r+h-1}\bar{z}^{s+\bar{h}-1}\leavevmode\nobreak\ ,$ (60)

where the contour of the integration is a circle around $z=0$ and $\bar{z}=0$
respectively. The creation property implies also the injectivity condition,
i.e.

$Y_{V}(v,x,\bar{x})=0\text{ implies }v=0,\text{ for }v\in V\,.$ (61)

###  2.3 Some consequences of the definition

We now prove some consequences of the definition. For any $v\in V$, we have

$Y_{V}(v,x,\bar{x})\mathbf{1}=\mathrm{e}^{\bar{x}\,\bar{L}(-1)}\mathrm{e}^{x\,L(-1)}v\,.$
(62)

We use the translation property 8 and Taylor’s theorem (31). For another
formal variable $x_{0},\bar{x}_{0}$, Taylor’s theorem gives

$Y_{V}\left(\mathrm{e}^{xL(-1)}\mathrm{e}^{\bar{x}\bar{L}(-1)}v,x_{0},\bar{x}_{0}\right)=\mathrm{e}^{x\frac{d}{dx}}\mathrm{e}^{\bar{x}\frac{d}{d\bar{x}}}Y_{V}(v,x_{0},\bar{x}_{0})=Y_{V}(v,x+x_{0},\bar{x}+\bar{x}_{0})\,.$
(63)

Now applying this operator on $\mathbf{1}$, taking limit $x_{0},\bar{x}_{0}\to
0$ we get

$\lim_{x_{0},\bar{x}_{0}\to
0}Y_{V}\left(\mathrm{e}^{xL(-1)}\mathrm{e}^{\bar{x}\bar{L}(-1)}v,x_{0},\bar{x}_{0}\right)\mathbf{1}=\lim_{x_{0},\bar{x}_{0}\to
0}Y_{V}(v,x+x_{0},\bar{x}+\bar{x}_{0})\mathbf{1}\,,$ (64)

and then using (45) we obtain (62). For any $v\in V$ we have

${\rm e}^{x_{2}L(-1)}{\rm
e}^{\bar{x}_{2}\bar{L}(-1)}Y_{V}(v,x_{1},\bar{x}_{1}){\rm e}^{-x_{2}L(-1)}{\rm
e}^{-\bar{x}_{2}\bar{L}(-1)}=Y_{V}(v,x_{1}+x_{2},\bar{x}_{1}+\bar{x_{2}})\,.$
(65)

Using the BCH formula (224) and translation property 8, we have

$\begin{split}{\rm e}^{x_{2}L(-1)}Y_{V}(v,x_{1},\bar{x}_{1}){\rm
e}^{-x_{2}L(-1)}&=\sum_{n=0}^{\infty}\frac{\left[(x_{2}\,L(-1))^{n},Y_{V}(v,x_{1},\bar{x}_{1})\right]}{n!}\,,\\\
&=\sum_{n=0}^{\infty}\frac{1}{n!}x_{2}^{n}\frac{\partial^{n}}{\partial
x_{1}^{n}}Y_{V}(v,x_{1},\bar{x}_{1})\,,\\\ &={\rm
e}^{x_{2}\frac{\partial}{\partial x_{1}}}Y_{V}(v,x_{1},\bar{x}_{1})\,,\\\
&=Y_{V}(v,x_{1}+x_{2},\bar{x}_{1})\,,\end{split}$ (66)

where in the last step we used Taylor’s theorem (31). Similarly we have

${\rm e}^{\bar{x}_{2}\bar{L}(-1)}Y_{V}(v,x_{1},\bar{x}_{1}){\rm
e}^{-\bar{x}_{2}\bar{L}(-1)}=Y_{V}(v,x_{1},\bar{x}_{1}+\bar{x}_{2}).$ (67)

Since $L(-1)$ and $\bar{L}(-1)$ commute, the result follows. We now prove
skew-symmetry which will be useful in proving the duality of vertex operators.
For any $u,v\in V$, we have

$Y_{V}(u,z,\bar{z})v=e^{zL(-1)}e^{\bar{z}\bar{L}(-1)}Y_{V}(v,-z,-\bar{z})u.$
(68)

Using Lemma locality property 9, 2.3 and Lemma 2.3, we have

$\begin{split}Y_{V}(u,z,\bar{z})Y_{V}(v,z^{\prime},\bar{z}^{\prime})\textbf{1}&\sim
Y_{V}(v,z^{\prime},\bar{z}^{\prime})Y_{V}(u,z,\bar{z})\textbf{1}\\\
&=Y_{V}(v,z^{\prime},\bar{z}^{\prime})\mathrm{e}^{\bar{z}\,\bar{L}(-1)}\mathrm{e}^{z\,L(-1)}u\\\
&=\mathrm{e}^{\bar{z}\,\bar{L}(-1)}\mathrm{e}^{z\,L(-1)}Y_{V}(v,z^{\prime}-z,\bar{z}^{\prime}-\bar{z})u\leavevmode\nobreak\
.\end{split}$ (69)

Now taking $z^{\prime},\bar{z}^{\prime}\to 0$ and using the creation property,
we obtain the required result. The following proposition shows the uniqueness
of vertex operators. The proof is on the lines of Dolan:1994st . Let
$U:V\longrightarrow V\\{x,\bar{x}\\}$ be a linear operator which is local with
respect to every other vertex operator, in the sense of Property 9, and
satisfies

$U(x,\bar{x})\mathbf{1}=\mathrm{e}^{\bar{x}\,\bar{L}(-1)}\mathrm{e}^{x\,L(-1)}v,$
(70)

for some $v\in V$, then

$U(z,\bar{z})=Y_{V}(v,z,\bar{z})\,,$ (71)

for a non-zero complex number $z$. For any $w\in V$, from Lemma 2.3 and
locality property 9 we have

$\begin{split}U(z_{1},\bar{z}_{1})\mathrm{e}^{\bar{z}_{2}\,\bar{L}(-1)}\mathrm{e}^{z_{2}\,L(-1)}w&=U(z_{1},\bar{z}_{1})Y_{V}(w,z_{2},\bar{z}_{2})\mathbf{1}\,,\\\
&\sim Y_{V}(w,z_{2},\bar{z}_{2})U(z_{1},\bar{z}_{1})\mathbf{1}\,,\\\
&=Y_{V}(w,z_{2},\bar{z}_{2})\mathrm{e}^{\bar{z}_{1}\,\bar{L}(-1)}\mathrm{e}^{z_{1}\,L(-1)}v\,,\\\
&=Y_{V}(w,z_{2},\bar{z}_{2})Y_{V}(v,z_{1},\bar{z}_{1})\mathbf{1}\,,\\\ &\sim
Y_{V}(v,z_{1},\bar{z}_{1})Y_{V}(w,z_{2},\bar{z}_{2})\mathbf{1}\leavevmode\nobreak\
,\end{split}$ (72)

where $\sim$ indicates equality up to analytic extension in the sense of
Property 9. Now taking $z_{2},\bar{z}_{2}\to 0$ we obtain,

$U(z_{1},\bar{z}_{1})w=Y_{V}(v,z_{1},\bar{z}_{1})w.$ (73)

As the two operators in (73) are equal for all $w\in V$, they are equal as
operators. We now prove the duality of vertex operators. For any $v,w\in V$ we
have

$Y_{V}(v,z_{1},\bar{z}_{1})Y_{V}(w,z_{2},\bar{z}_{2})=Y_{V}(Y_{V}(v,z_{1}-z_{2},\bar{z}_{1}-\bar{z}_{2})w,z_{2},\bar{z}_{2})\,,$
(74)

in the domain $|z_{1}|>|z_{2}|>|z_{1}-z_{2}|>0$, where the RHS is defined by

$Y_{V}(Y_{V}(v,z_{1}-z_{2},\bar{z}_{1}-\bar{z}_{2})w,z_{2},\bar{z}_{2})=\sum_{\begin{subarray}{c}m,n\in\mathbb{R}\\\
(m-n)\in\mathbb{Z}\end{subarray}}Y_{V}(v_{m,n}\cdot
w,z_{2},\bar{z}_{2})(z_{1}-z_{2})^{-m-1}(\bar{z}_{1}-\bar{z}_{2})^{-n-1}\,.$
(75)

The proof is on the lines of (Huang:2015eda, , Page 23). For any $u\in V,$ we
have

$\begin{split}Y_{V}(v,z_{1},\bar{z}_{1})&Y_{V}(w,z_{2},\bar{z}_{2})\mathrm{e}^{\bar{z}_{3}\,\bar{L}(-1)}\mathrm{e}^{z_{3}\,L(-1)}u\\\
&=Y_{V}(v,z_{1},\bar{z}_{1})Y_{V}(w,z_{2},\bar{z}_{2})Y_{V}(u,z_{3},\bar{z}_{3})\mathbf{1}\,,\\\
&\sim
Y_{V}(u,z_{3},\bar{z}_{3})Y_{V}(v,z_{1},\bar{z}_{1})Y_{V}(w,z_{2},\bar{z}_{2})\mathbf{1}\,,\\\
&=Y_{V}(u,z_{3},\bar{z}_{3})Y_{V}(v,z_{1},\bar{z}_{1})\mathrm{e}^{\bar{z}_{2}\,\bar{L}(-1)}\mathrm{e}^{z_{2}\,L(-1)}w\,,\\\
&=Y_{V}(u,z_{3},\bar{z}_{3})\mathrm{e}^{\bar{z}_{2}\,\bar{L}(-1)}\mathrm{e}^{z_{2}\,L(-1)}Y_{V}(v,z_{1}-z_{2},\bar{z}_{1}-\bar{z}_{2})w\,,\\\
&=Y_{V}(u,z_{3},\bar{z}_{3})Y_{V}\left(Y_{V}(v,z_{1}-z_{2},\bar{z}_{1}-\bar{z}_{2})w,z_{2},\bar{z}_{2}\right)\mathbf{1}\,,\\\
&\sim
Y_{V}\left(Y_{V}(v,z_{1}-z_{2},\bar{z}_{1}-\bar{z}_{2})w,z_{2},\bar{z}_{2}\right)Y_{V}(u,z_{3},\bar{z}_{3})\mathbf{1},\end{split}$
(76)

where we used Lemma 2.3, Lemma 2.3, and Locality property 9. Now taking
$z_{3},\bar{z}_{3}\to 0$ and using Proposition 2.3, we obtain the duality
relation. Note that the sum on the RHS of (75) converges. Indeed for any $u\in
V$, using skew-symmetry 101010We thank Yi-Zhi Huang for clarification on this
point.
(Lemma 2.3) we have

$\begin{split}Y_{V}(Y_{V}(v,&z_{1}-z_{2},\bar{z}_{1}-\bar{z}_{2})w,z_{2},\bar{z}_{2})u\\\
&=\sum_{\begin{subarray}{c}m,n\in\mathbb{R}\\\
(m-n)\in\mathbb{Z}\end{subarray}}Y_{V}(v_{m,n}\cdot
w,z_{2},\bar{z}_{2})(z_{1}-z_{2})^{-m-1}(\bar{z}_{1}-\bar{z}_{2})^{-n-1}u\\\
&=\sum_{\begin{subarray}{c}m,n\in\mathbb{R}\\\
(m-n)\in\mathbb{Z}\end{subarray}}\mathrm{e}^{\bar{z}_{2}\,\bar{L}(-1)}\mathrm{e}^{z_{2}\,L(-1)}Y_{V}(u,-z_{2},-\bar{z}_{2})v_{m,n}\cdot
w(z_{1}-z_{2})^{-m-1}(\bar{z}_{1}-\bar{z}_{2})^{-n-1}\\\
&=\mathrm{e}^{\bar{z}_{2}\,\bar{L}(-1)}\mathrm{e}^{z_{2}\,L(-1)}Y_{V}(u,-z_{2},-\bar{z}_{2})\sum_{\begin{subarray}{c}m,n\in\mathbb{R}\\\
(m-n)\in\mathbb{Z}\end{subarray}}v_{m,n}\cdot
w(z_{1}-z_{2})^{-m-1}(\bar{z}_{1}-\bar{z}_{2})^{-n-1}\\\
&=\mathrm{e}^{\bar{z}_{2}\,\bar{L}(-1)}\mathrm{e}^{z_{2}\,L(-1)}Y_{V}(v,-z_{2},-\bar{z}_{2})Y_{V}(u,z_{1}-z_{2},\bar{z}_{1}-\bar{z}_{2})w\leavevmode\nobreak\
.\end{split}$ (77)

Since the RHS of the last line is well defined in $|z_{2}|>|z_{1}-z_{2}|$, the
operator
$Y_{V}(Y_{V}(v,z_{1}-z_{2},\bar{z}_{1}-\bar{z}_{2})w,z_{2},\bar{z}_{2})$ is
well defined in $|z_{2}|>|z_{1}-z_{2}|$. Proposition 2.3 shows that a product
of two vertex operators can be written as a sum of single vertex operator:

$Y_{V}(v,z_{1},\bar{z}_{1})Y_{V}(w,z_{2},\bar{z}_{2})=\sum_{\begin{subarray}{c}m,n\in\mathbb{R}\\\
(m-n)\in\mathbb{Z}\end{subarray}}Y_{V}(v_{m,n}\cdot
w,z_{2},\bar{z}_{2})(z_{1}-z_{2})^{-m-1}(\bar{z}_{1}-\bar{z}_{2})^{-n-1}.$
(78)

In physics, we usually ignore the non-singular terms in the expansion above
and call it the operator product expansion. The sum in the operator product
expansion has finitely many terms with negative powers of $(z_{1}-z_{2})$ and
$(\bar{z}_{1}-\bar{z}_{2})$. To see this, we first expand the vertex operator
$Y_{V}(v,x,\bar{x})$ for $v\in V_{(h,\bar{h})}$ as

$Y_{V}(v,x,\bar{x})=\sum_{\begin{subarray}{c}m,n\in\mathbb{R}\\\
(m-n)\in\mathbb{Z}\end{subarray}}x_{m,n}(v)x^{-m-h}\bar{x}^{-n-\bar{h}},$ (79)

Since

$\text{wt}\leavevmode\nobreak\
x_{m,n}(v)=-m,\quad\overline{\text{wt}}\leavevmode\nobreak\ x_{m,n}(v)=-n,$
(80)

for $w\in V_{(h^{\prime},\bar{h}^{\prime})}$ we have

$x_{m,n}(v)\cdot w\in V_{(h^{\prime}-m,\bar{h}^{\prime}-n)}.$ (81)

Due to the grading-restriction property 2, there exists $M\in\mathbb{Z}$ such
that

$x_{m,n}(v)\cdot w=0,\quad m,n>M.$ (82)

Thus the operator product expansion is upper truncated. The operator product
expansion of the conformal vertex operator $T(x)$ with itself is given by

$\begin{split}&T(x_{1})T(x_{2})=\frac{c}{2}\frac{1}{(x_{1}-x_{2})^{4}}+\frac{2\,T(x_{2})}{(x_{1}-x_{2})^{2}}+\frac{1}{(x_{1}-x_{2})}\frac{\partial}{\partial
x_{2}}T(x_{2})+G_{1}(x_{1},x_{2})\\\
&\bar{T}(\bar{x}_{1})\bar{T}(\bar{x}_{2})=\frac{\bar{c}}{2}\frac{1}{(\bar{x}_{1}-\bar{x}_{2})^{4}}+\frac{2\,\bar{T}(\bar{x}_{2})}{(\bar{x}_{1}-\bar{x}_{2})^{2}}+\frac{1}{(\bar{x}_{1}-\bar{x}_{2})}\frac{\partial}{\partial\bar{x}_{2}}\bar{T}(\bar{x}_{2})+G_{2}(\bar{x}_{1},\bar{x}_{2})\\\
&T(x_{1})\bar{T}(\bar{x}_{2})=G_{3}(x_{1},\bar{x}_{2}),\end{split}$ (83)

where $G_{1}(x_{1},x_{2}),G_{3}(x_{1},x_{2})\in\mathrm{End}(V)[[x_{2}^{\pm
1},(x_{1}-x_{2})]],G_{2}(\bar{x}_{1},\bar{x}_{2})\in\mathrm{End}(V)[[\bar{x}_{2}^{\pm
1},(\bar{x}_{1}-\bar{x}_{2})]].$ The proof is straightforward using the
Virasoro algebra (47), see (Frenkel:2004jn, , Chapter 3) for more details. Of
particular interest are the chiral and anti-chiral vertex operators. A vector
$u\in V$ is called a chiral (anti-chiral) vector if the corresponding vertex
operator $Y_{V}(u,x,\bar{x})$ belongs in End$(V)\\{x^{\pm 1}\\}$
(End$(V)\\{\bar{x}^{\pm 1}\\}$) or equivalently only depends on $z$
($\bar{z}$). Such vertex operators will be called chiral (anti-chiral) vertex
operators. From the translation property 8 we see that the vertex operator
corresponding to $v$ is chiral if and only if $\bar{L}(-1)v=0$ and anti-chiral
if and only if $L(-1)v=0$. The algebra of the modes of chiral (anti-chiral)
vertex operators is called the chiral (anti-chiral) algebra in physics, see
Corollary 2.3. In the locality property 9 involving a chiral (resp. anti-
chiral) vertex operator $Y_{V}(u_{1},z_{1})(\text{resp.
}Y_{V}(u_{1},\bar{z}_{1}))$ and another vertex operator
$Y_{V}(u_{2},z_{2},\bar{z}_{2})$, we will often denote the function $m$ by

$R(Y_{V}(u_{1},z_{1})Y_{V}(u_{2},z_{2},\bar{z}_{2}))\quad(\text{resp.
}R(Y_{V}(u_{1},\bar{z}_{1})Y_{V}(u_{2},z_{2},\bar{z}_{2})))$ (84)

so that

$R(Y_{V}(u_{1},z_{1})Y_{V}(u_{2},z_{2},\bar{z}_{2}))=\begin{cases}Y_{V}\left(u_{1},z_{1}\right)Y_{V}\left(u_{2},z_{2},\bar{z}_{2}\right)&\text{for
}|z_{1}|>|z_{2}|,\\\
Y_{V}\left(u_{2},z_{2},\bar{z}_{2}\right)Y_{V}\left(u_{1},z_{1}\right)&\text{for
}|z_{2}|>|z_{1}|\end{cases}$ (85)

and

$R(Y_{V}(u_{1},\bar{z}_{1})Y_{V}(u_{2},z_{2},\bar{z}_{2}))=\begin{cases}Y_{V}\left(u_{1},\bar{z}_{1}\right)Y_{V}\left(u_{2},z_{2},\bar{z}_{2}\right)&\text{for
}|z_{1}|>|z_{2}|,\\\
Y_{V}\left(u_{2},z_{2},\bar{z}_{2}\right)Y_{V}\left(u_{1},\bar{z}_{1}\right)&\text{for
}|z_{2}|>|z_{1}|\end{cases}$ (86)

respectively. In physics, this is called radial ordering. Here
$z_{2},\bar{z}_{2}$ are complex conjugates of each other. Let $u,v$ be
homogeneous chiral and anti-chiral vector. Then the associated chiral and
anti-chiral vertex operator has an expansion of the form

$\begin{split}&Y_{V}(u,x)=\sum_{n\in\mathbb{Z}}x_{n}(u)x^{-n-(\mathrm{wt}\,u-\overline{\mathrm{wt}}\,u)}\in\mathrm{End}(V)[[x^{\pm
1}]]\,,\\\
&Y_{V}(v,\bar{x})=\sum_{n\in\mathbb{Z}}\bar{x}_{n}(v)\bar{x}^{-n-(\overline{\mathrm{wt}}\,v-\mathrm{wt}\,v)}\in\mathrm{End}(V)[[\bar{x}^{\pm
1}]]\,,\end{split}$ (87)

where

$x_{n}(u):=x_{n-\overline{\mathrm{wt}}\,u,-\overline{\mathrm{wt}}\,u}(u),\quad\bar{x}_{n}(v):=x_{-\mathrm{wt}\,v,n-\mathrm{wt}\,v}(v).$
(88)

From the expansion (57), we see that $Y_{V}(u,x,\bar{x})$ will be independent
of $\bar{x}$ if and only if

$x_{m,n}(u)=0\text{ unless }n=-\overline{\mathrm{wt}}\,u\,.$ (89)

But as $m-n\in\mathbb{Z}$ we then have

$x_{m,n}(u)=0\text{ unless }n=-\overline{\mathrm{wt}}\,u,\leavevmode\nobreak\
m\in\mathbb{Z}-\overline{\mathrm{wt}}\,u.$ (90)

This gives us the required expansion. The proof for anti-chiral vector $v$ is
similar. The fact that $Y_{V}(u,x)\in\mathrm{End}(V)[[x^{\pm
1}]],Y_{V}(v,\bar{x})\in\mathrm{End}(V)[[\bar{x}^{\pm 1}]]$ follows from the
single-valuedness property 3. By the above lemma, for chiral and anti-chiral
vertex operators, the requirements in Remark 2.2 is satisfied and hence we can
write

$\begin{split}&x_{n}(u)=\frac{1}{2\pi i}\oint dz\leavevmode\nobreak\
Y_{V}(u,z)z^{n+(\mathrm{wt}\,u-\overline{\mathrm{wt}}\,u)-1}\,,\\\
&\bar{x}_{n}(v)=\frac{1}{2\pi i}\oint d\bar{z}\leavevmode\nobreak\
Y_{V}(v,\bar{z})\bar{z}^{n+(\overline{\mathrm{wt}}\,v-\mathrm{wt}\,v)-1}\,,\end{split}$
(91)

where $u,v$ are chiral and anti-chiral vectors respectively and the contour of
integration is a circle around $z=0,\bar{z}=0$ respectively. We now derive the
commutator of the modes of two vertex operators and the Borcherd’s identity
Borcherds:1983sq . Let $u_{i}\in V_{(h_{i},\bar{h}_{i})}$ and $u_{j}\in
V_{(h_{j},h_{j}^{\prime})}$ $(v_{i}\in
V_{(h_{i}^{\prime},\bar{h}^{\prime}_{i})}$ and $v_{j}\in
V_{(h_{j}^{\prime},\bar{h}^{\prime}_{j})})$ be homogeneous chiral (resp. anti-
chiral) vectors with corresponding vertex operators

$\begin{split}&Y_{V}(u_{i},x)=\sum_{n\in\mathbb{Z}}x_{n}(u_{i})x^{-n-(h_{i}-\bar{h}_{i})},\quad\quad
Y_{V}(u_{j},x)=\sum_{n\in\mathbb{Z}}x_{n}(u_{j})x^{-n-(h_{j}-\bar{h}_{j})},\\\
&Y_{V}(v_{i},\bar{x})=\sum_{n\in\mathbb{Z}}\bar{x}_{n}(v_{i})\bar{x}^{-n-(\bar{h}^{\prime}_{i}-h^{\prime}_{i})},\quad\quad
Y_{V}(v_{j},\bar{x})=\sum_{n\in\mathbb{Z}}\bar{x}_{n}(v_{j})\bar{x}^{-n-(\bar{h}^{\prime}_{j}-h^{\prime}_{j})}.\end{split}$
(92)

Then the vectors $x_{p}(u_{i})\cdot u_{j}$ and $\bar{x}_{p}(v_{k})\cdot
v_{\ell}$ are chiral and anti-chiral vectors respectively. Further, we have

$\begin{split}&[x_{n}(u_{i}),x_{k}(u_{j})]=\sum_{\begin{subarray}{c}p\geq-(h_{i}-\bar{h}_{i})+1\end{subarray}}{n+(h_{i}-\bar{h}_{i})-1\choose
p+(h_{i}-\bar{h}_{i})-1}x_{k+n}(x_{p}(u_{i})\cdot u_{j})\,,\\\
&[\bar{x}_{n}(v_{i}),\bar{x}_{k}(v_{j})]=\sum_{\begin{subarray}{c}p\geq-(\bar{h}^{\prime}_{i}-h^{\prime}_{i})+1\end{subarray}}{n+(\bar{h}^{\prime}_{i}-h^{\prime}_{i})-1\choose
p+(\bar{h}^{\prime}_{i}-h^{\prime}_{i})-1}\bar{x}_{k+n}(\bar{x}_{p}(v_{i})\cdot
v_{j})\,,\\\ &[x_{n}(u_{i}),\bar{x}_{k}(v_{j})]=0\,.\end{split}$ (93)

In particular,

$\begin{split}&[L(n),x_{k}(u_{i})]=\sum_{\begin{subarray}{c}p\geq-1\end{subarray}}{n+1\choose
p+1}x_{k+n}(L(p)\cdot u_{i})\,,\\\ &[L(n),\bar{x}_{k}(v_{i})]=0\,,\\\
&[\bar{L}(n),\bar{x}_{k}(v_{i})]=\sum_{\begin{subarray}{c}p\geq-1\end{subarray}}{n+1\choose
p+1}\bar{x}_{k+n}(\bar{L}(p)\cdot v_{i})\,,\\\
&[\bar{L}(n),x_{k}(u_{i})]=0\,.\end{split}$ (94)

More generally, for $m\in\mathbb{Z}\,\,\text{ and }m_{+}\in\mathbb{Z}_{\geq
0}$ we have the Borcherd’s identity:

$\begin{split}\sum_{r\geq 0}{m\choose
r}\Big{(}(-1)^{r}&x_{n+m-r}(u_{i})x_{k+r}(u_{j})-(-1)^{m+r}x_{k+m-r}(u_{j})x_{n+r}(u_{i})\Big{)}\\\
&=\sum_{p\geq 1-(h_{i}-\bar{h}_{i})}{n+(h_{i}-\bar{h}_{i})-1\choose
p+(h_{i}-\bar{h}_{i})-1}x_{k+n+m+\bar{h}_{i}-\bar{h}_{j}}(x_{p+m}(u_{i})\cdot
u_{j}),\end{split}$ (95) $\begin{split}\sum_{r\geq 0}{m\choose
r}\Big{(}(-1)^{r}&\bar{x}_{n+m-r}(v_{i})\bar{x}_{k+r}(v_{j})-(-1)^{m+r}\bar{x}_{k+m-r}(v_{j})\bar{x}_{n+r}(v_{i})\Big{)}\\\
&=\sum_{p\geq
1-(\bar{h}^{\prime}_{i}-h^{\prime}_{i})}{n+(\bar{h}^{\prime}_{i}-h^{\prime}_{i})-1\choose
p+(\bar{h}^{\prime}_{i}-h^{\prime}_{i})-1}\bar{x}_{k+n+h^{\prime}_{i}-h^{\prime}_{j}}(\bar{x}_{p+m}(v_{i})\cdot
v_{j}),\end{split}$ (96) $\sum_{r\geq 0}{m_{+}\choose
r}\Big{(}(-1)^{r}x_{n+m_{+}-r}(u_{i})\bar{x}_{k+r}(v_{j})-(-1)^{m_{+}+r}\bar{x}_{k+m_{+}-r}(v_{j})x_{n+r}(u_{i})\Big{)}=0.$
(97)

We first show that $x_{p}(u_{i})\cdot u_{j}$ and $\bar{x}_{p}(v_{k})\cdot
v_{\ell}$ are chiral and anti-chiral vectors respectively. Indeed by the
translation property 8

$[\bar{L}(-1),x_{p}(u_{i})]=0\,,$ (98)

which implies that

$\bar{L}(-1)\cdot(x_{p}(u_{i})\cdot
u_{j})=x_{p}(u_{i})\cdot\bar{L}(-1)u_{j}=0\,.$ (99)

Similarly

$L(-1)\cdot(\bar{x}_{p}(v_{k})\cdot v_{\ell})=0\,.$ (100)

Now, we will follow the usual contour integration procedure, see for example
(Frenkel:2004jn, , Section 3.3.10). First note that

$Y_{V}(u_{i},z_{1})Y_{V}(u_{j},z_{2}),Y_{V}(u_{j},z_{2})Y_{V}(u_{i},z_{1}),R(Y_{V}(u_{i},z_{1})Y_{V}(u_{j},z_{2}))\,,$
(101)

are single-valued and analytic in $z_{1},z_{2}$ since their partial derivative
with respect to $\bar{z}_{1},\bar{z}_{2}$ is zero. So we can use Cauchy’s
residue theorem to integrate over $z_{1},z_{2}$ on any contour. Now let
$r_{1}>r_{2}>r_{3}>0$ be real numbers. Let $C^{a}_{i}(z)$ denote a contour in
the variable $z_{i}$, in counterclockwise direction, of radius $a$ and
centered around $z$. Further, $C_{i}^{r}:=C_{i}^{r}(0)$. Let $f(z_{1},z_{2})$
be a rational function analytic in $z_{1},z_{2}$ with poles only at
$z_{1}=0,z_{2}=0,z_{1}=z_{2}$. The integrals

$\begin{split}\oint_{C^{r_{2}}_{2}}dz_{2}\oint_{C^{r_{1}}_{1}}dz_{1}\leavevmode\nobreak\
Y_{V}(u_{i},z_{1})Y_{V}(u_{j},z_{2})f(z_{1},z_{2})\quad\text{and}\\\
\oint_{C^{r_{2}}_{2}}dz_{2}\oint_{C^{r_{3}}_{1}}dz_{1}\leavevmode\nobreak\
Y_{V}(u_{j},z_{2})Y_{V}(u_{i},z_{1})f(z_{1},z_{2})\,,\end{split}$ (102)

are well-defined. By the locality property 9 and the OPE (75), we see that

$\begin{split}&\oint_{C^{r_{2}}_{2}}dz_{2}\oint_{C^{r_{1}}_{1}}dz_{1}\leavevmode\nobreak\
Y_{V}(u_{i},z_{1})Y_{V}(u_{j},z_{2})f(z_{1},z_{2})\\\
&-\oint_{C^{r_{2}}_{2}}dz_{2}\oint_{C^{r_{3}}_{1}}dz_{1}\leavevmode\nobreak\
Y_{V}(u_{j},z_{2})Y_{V}(u_{i},z_{1})f(z_{1},z_{2})\\\
&=\oint_{C^{r_{2}}_{2}}dz_{2}\oint_{C^{r_{1}}_{1}-C^{r_{3}}_{1}}dz_{1}\leavevmode\nobreak\
R(Y_{V}(u_{i},z_{1})Y_{V}(u_{j},z_{2}))f(z_{1},z_{2})\\\
&=\oint_{C^{r_{2}}_{2}}dz_{2}\oint_{C^{\delta}_{1}(z_{2})}dz_{1}\leavevmode\nobreak\
Y_{V}(Y_{V}(u_{i},z_{1}-z_{2})u_{j},z_{2})f(z_{1},z_{2})\\\
&=\oint_{C^{r_{2}}_{2}}dz_{2}\oint_{C^{\delta}_{1}(z_{2})}dz_{1}\leavevmode\nobreak\
\sum_{p\in\mathbb{Z}}Y_{V}(x_{p}(u_{i})\cdot
u_{j},z_{2})(z_{1}-z_{2})^{-p-(h_{i}-\bar{h}_{i})}f(z_{1},z_{2})\,,\end{split}$
(103)

where $\delta$ is some small real number, see (Frenkel:2004jn, , Section
3.3.10) for details of the change in contour. If we now choose

$f=z_{1}^{n+(h_{i}-\bar{h}_{i})-1}z_{2}^{k+(h_{j}-\bar{h}_{j})-1}\,,$ (104)

then using (91) the LHS is111111There is a factor of $(2\pi i)^{2}$ which
cancels on both sides, so we ignore it. $[x_{n}(u_{i}),x_{k}(u_{j})]$ while
Cauchy’s residue theorem gives the RHS to be

$\begin{split}\oint_{C^{r_{2}}_{2}}dz_{2}\leavevmode\nobreak\
\sum_{p\geq-(h_{i}-\bar{h}_{i})+1}{n+(h_{i}-\bar{h}_{i})-1\choose
p+(h_{i}-\bar{h}_{i})-1}&Y_{V}(x_{p}(u_{i})\cdot
u_{j},z_{2})z_{2}^{k+(h_{j}-\bar{h}_{j})+n-p-1}\,,\end{split}$ (105)

where we used the identity

$\oint_{C^{\delta}_{1}(z_{2})}dz_{1}\leavevmode\nobreak\
\frac{z_{1}^{n+(h_{i}-\bar{h}_{i})-1}}{(z_{1}-z_{2})^{p+(h_{i}-\bar{h}_{i})}}={n+(h_{i}-\bar{h}_{i})-1\choose
p+(h_{i}-\bar{h}_{i})-1}z_{2}^{n-p}.$ (106)

Note, that it is necessary that $(h_{i}-\bar{h}_{i})\in\mathbb{Z}$, which is
true by the single valuedness property 3, for (106) to hold. Finally, using
(91) and the fact that

$x_{p}(u_{i}).u_{j}\in V_{h_{j}-p+\bar{h}_{i},\bar{h}_{i}+\bar{h}_{j}},$ (107)

the RHS becomes

$\sum_{\begin{subarray}{c}p\geq-(h_{i}-\bar{h}_{i})+1\end{subarray}}{n+(h_{i}-\bar{h}_{i})-1\choose
p+(h_{i}-\bar{h}_{i})-1}x_{k+n}(x_{p}(u_{i})\cdot u_{j}).$ (108)

The second commutator is similar. To prove the third commutator, note that
since

$\partial_{\bar{z}_{1}}R(Y_{V}(u_{i},z_{1})Y_{V}(v_{j},\bar{z}_{2}))=\partial_{z_{2}}R(Y_{V}(u_{i},z_{1})Y_{V}(v_{j},\bar{z}_{2}))=0,$
(109)

$R(Y_{V}(u_{i},z_{1})Y_{V}(v_{j},\bar{z}_{2}))$ cannot have any dependence on
$(z_{1}-z_{2})$ or $(\bar{z}_{1}-\bar{z}_{2})$. Moreover, from the proof of
the OPE in Proposition 2.3, we see that it cannot also have
$(z_{1}-\bar{z}_{2})$ dependence as well. This implies that the contour
integral on the RHS of (103) vanishes and we get

$[x_{m}(u_{i}),\bar{x}_{n}(v_{j})]=0\,.$ (110)

The three Borcherd’s identity follow by using

$\begin{split}&f_{1}=z_{1}^{n+(h_{i}-\bar{h}_{i})-1}z_{2}^{k+(h_{j}-\bar{h}_{j})-1}(z_{1}-z_{2})^{m}\,,\\\
&f_{2}=\bar{z}_{1}^{n+(\bar{h}^{\prime}_{i}-h^{\prime}_{i})-1}\bar{z}_{2}^{k+(\bar{h}^{\prime}_{i}-h^{\prime}_{i})-1}(\bar{z}_{1}-\bar{z}_{2})^{m}\,,\\\
&f_{3}=z_{1}^{n+(h_{i}-\bar{h}_{i})-1}\bar{z}_{2}^{k+(\bar{h}_{j}-h^{\prime}_{j})-1}(z_{1}-\bar{z}_{2})^{m}\,,\end{split}$
(111)

where for the second Borcherd’s identity, we need to integrate against
$d\bar{z}_{1},d\bar{z}_{2}$ on the curves
$C_{\bar{z}_{1}}^{r_{1}},C_{\bar{z}_{2}}^{r_{2}}$ respectively and for the
third Borcherd’s identity, we need to integrate against $dz_{1},d\bar{z}_{2}$
on the curves $C_{z_{1}}^{r_{1}},C_{\bar{z}_{2}}^{r_{2}}$ respectively. For
$n=0,-1$ in (94) we obtain the $L(0)$-property 7 and the translation property
8 of chiral vertex operators. Note that we already used these properties in
proving the OPE. The commutator of the modes of chiral and anti-chiral vertex
operators is closed. The algebra in (93) thus obtained is called the chiral
and anti-chiral algebra respectively of the non-chiral VOA $(V,Y_{V})$. Let
$(V,Y_{V})$ be a non-chiral VOA with central charge $(c,\bar{c})$. The graded
dimension or character of $V$ is defined by

$\chi_{V}(\tau,\bar{\tau}):=\text{Tr}_{V}\,q^{L(0)-\frac{c}{24}}\bar{q}^{\bar{L}(0)-\frac{\bar{c}}{24}}=\sum_{(h,\bar{h})\in\mathbb{R}\times\mathbb{R}}\left(\text{dim}\,V_{(h,\bar{h})}\right)q^{h-\frac{c}{24}}\bar{q}^{\bar{h}-\frac{\bar{c}}{24}}\,,$
(112)

where $q=e^{2\pi i\tau},\leavevmode\nobreak\ \bar{q}=e^{-2\pi i\bar{\tau}}$
and $\tau\in\mathbb{H}:=\\{\tau=x+iy:y>0\\}.$ Note that the single-valuedness
property implies that
$\chi_{V}(\tau+1,\bar{\tau}+1)=\chi_{V}(\tau,\bar{\tau})$ if

$c-\bar{c}=24k\,,$ (113)

for some integer $k$.

Let
$(V_{1},Y_{V_{1}},\omega_{V_{1}},\bar{\omega}_{V_{1}},\mathbf{1}_{V_{1}}),(V_{2},Y_{V_{2}},\omega_{V_{2}},\bar{\omega}_{V_{2}},\mathbf{1}_{V_{2}})$
be two non-chiral VOAs with the same central charge. Then a map
$f:V_{1}\rightarrow V_{2}$ is called a non-chiral VOA homomorphism if it is a
grading-preserving linear map such that

$f(Y_{V_{1}}(u,x,\bar{x})v)=Y_{V_{2}}(f(u),x,\bar{x})f(v)\text{ for }\,u,v\in
V_{1}\,,$ (114)

or equivalently,

$f\left(u_{n,m}\cdot v\right)=f(u)_{n,m}f(v)\text{ for }u,v\in V_{1},\quad
n,m\in\mathbb{R}\,,$ (115)

and such that

$f(\mathbf{1}_{V_{1}})=\mathbf{1}_{V_{2}},\quad
f(\omega_{V_{1}})=\omega_{V_{2}},\quad
f(\bar{\omega}_{V_{1}})=\bar{\omega}_{V_{2}}\,.$ (116)

An isomorphism of non-chiral VOAs is a bijective homomorphism. An endomorphism
of a non-chiral VOA $V$ is a homomorphism from $V$ to itself, and an
automorphism of $V$ is a bijective endomorphism. In particular, an
automorphism can be defined as a linear isomorphism $f:V\to V$ such that

$\begin{gathered}f\circ Y_{V}(v,x,\bar{x})\circ
f^{-1}=Y_{V}(f(v),x,\bar{x})\text{ for }v\in V\,,\\\ f(\omega)=\omega,\quad
f(\bar{\omega})=\bar{\omega}\,.\end{gathered}$ (117)

It follows that $f$ is grading-preserving and
$f(\mathbf{1}_{V})=\mathbf{1}_{V}$.

It is easy to see that the graded dimension of isomorphic non-chiral VOAs are
identical.

##  3 Lorentzian Lattice Vertex Operator Algebra (LLVOA)

In this section, we will construct a non-chiral vertex operator algebra
corresponding to an even, integral Lorentzian lattice
$\Lambda\subset\mathbb{R}^{m,n}$. In the first subsection, we recall some
basic facts about Lorentzian lattices and set up the notations for the rest of
the paper. We also record some results we will need later. In the next
subsection, we gather the ingredients needed to construct a non-chiral vertex
operator algebra, i.e. we will construct a vector space $V_{\Lambda}$
associated to the lattice, a vertex operator map $Y_{V_{\Lambda}}$ for this
vector space, a vacuum $\mathbf{1}$, and conformal vectors $\omega_{L}$,
$\omega_{R}$. In the last subsection, we will prove that
$(V_{\Lambda},Y_{V_{\Lambda}},\omega_{L},\omega_{R},\mathbf{1})$ is a non-
chiral VOA, which we will call the Lorentzian lattice vertex operator algebra
(LLVOA).

###  3.1 Lorentzian lattices

We begin with some basic definitions. Let $\mathbb{R}^{m}$ be the Euclidean
space equipped with a symmetric bilinear form $\langle\cdot,\cdot\rangle_{m}$.
Let $\mathbb{R}^{m,n}$ denote the $(m+n)$-dimensional vector space
$\mathbb{R}^{m+n}$ equipped with the symmetric bilinear form

$\bm{x}\circ\bm{x}^{\prime}:=\langle\vec{x},\vec{x}^{\prime}\rangle_{m}-\langle\vec{y},\vec{y}^{\prime}\rangle_{n}\,,$
(118)

where

$\bm{x}=(x^{1},\dots,x^{m},y^{1},\dots,y^{n})\equiv(\vec{x},\vec{y})\,,$ (119)

and similarly $\bm{x}^{\prime}$. We will omit the subscript on
$\langle\cdot,\cdot\rangle_{m}$ to make the notation lighter.

  1. 1.

A $d=(m+n)$-dimensional Lorentzian lattice of signature $(m,n)$ is a subset
$\Lambda\subset\mathbb{R}^{m,n}$ which is also a free $\mathbb{Z}$-module
spanned by $m+n$ vectors $\lambda_{j}\in\mathbb{R}^{m,n},1\leq j\leq m+n$,
linearly independent in $\mathbb{R}^{m,n}$. More explicitly

$\Lambda=\left\\{\sum_{j=1}^{m+n}n_{j}\lambda_{j}:n_{j}\in\mathbb{Z}\right\\}.$
(120)

$\\{\lambda_{j}\\}_{j=1}^{m+n}$ is called an integral basis of $\Lambda$. When
$n=0$ we call $\Lambda$ a Euclidean lattice. We will simply refer them as
lattices when we do not need to specify their signature.

  2. 2.

The dual lattice of a lattice $\Lambda$, denoted by $\Lambda^{\star}$, is
defined as

$\Lambda^{\star}=\\{\bm{x}^{\prime}\in\mathbb{R}^{m,n}:\bm{x}\circ\bm{x}^{\prime}\in\mathbb{Z}\leavevmode\nobreak\
\forall\leavevmode\nobreak\ \bm{x}\in\Lambda\\}\,.$ (121)

The lattice $\Lambda$ is said to be integral if
$\Lambda\subseteq\Lambda^{\star}$, i.e. $\bm{x}\circ\bm{y}\in\mathbb{Z}$ for
all $\bm{x},\bm{x}^{\prime}\in\Lambda$ and self-dual if
$\Lambda=\Lambda^{\star}$. The lattice $\Lambda$ is said to be even if

$\bm{x}\circ\bm{x}=||\vec{x}||^{2}-||\vec{y}||^{2}\in 2\mathbb{Z}\,,$ (122)

for all $\bm{x}=(\vec{x},\vec{y})\in\Lambda$, where
$||\vec{x}||^{2}:=\langle\vec{x},\vec{x}\rangle$.

  3. 3.

A generator matrix for $\Lambda$ is an $(m+n)\times(m+n)$ matrix such that the
$\mathbb{Z}$-span of its rows is $\Lambda$.

  4. 4.

A lattice homomorphism of two lattices
$f:\Lambda\longrightarrow\tilde{\Lambda}$ of the same signature is simply a
$\mathbb{Z}$-module morphism which also preserves the bilinear form:

$f(\bm{x})\circ
f(\bm{x}^{\prime})=\bm{x}\circ\bm{x}^{\prime},\quad\forall\leavevmode\nobreak\
\leavevmode\nobreak\ \bm{x},\bm{x}^{\prime}\in\Lambda\,.$ (123)

A bijective lattice homomorphism is called a lattice isomorphism. Two lattices
are said to be isomorphic if there exists a lattice isomorphism between them.

  5. 5.

An automorphism of the lattice $\Lambda$ is a lattice isomorphism from the
$\Lambda$ to itself. The group of all automorphisms (the group operation being
composition) is called the automorphism group of $\Lambda$ and denoted by
Aut$(\Lambda)$.

A generator matrix for the lattice $\Lambda$ in (120) is given by

$\mathcal{G}_{\Lambda}=\begin{pmatrix}\lambda_{1}^{1}&\lambda_{1}^{2}&\cdots&\lambda_{1}^{m+n}\\\
\vdots&\vdots&\cdots&\vdots\\\
\lambda_{m+n}^{1}&\lambda_{m+n}^{1}&\cdots&\lambda_{m+n}^{m+n}\end{pmatrix}\,,$
(124)

where $\lambda_{i}=(\lambda_{i}^{1},\dots,\lambda_{i}^{m+n})$ is a basis
vector of $\Lambda$. It is not hard to show that two generator matrices
$\mathcal{G}_{\Lambda},\mathcal{G}^{\prime}_{\Lambda}$ generate the same
lattice if and only if they are related by an $(m+n)\times(m+n)$ unimodular
matrix121212A matrix $U$ is called unimodular if det$(U)=\pm 1$.
$U\in\mathrm{GL}(m+n,\mathbb{Z})$:

$\mathcal{G}_{\Lambda}=U\mathcal{G}^{\prime}_{\Lambda}\,.$ (125)

Indeed $U$ is the change of basis matrix between the primed and unprimed
generator matrices since it is invertible and since it is also integral, it
preserves the lattice. If we take the symmetric bilinear form
$\langle\cdot,\cdot\rangle$ on $\mathbb{R}^{m},\mathbb{R}^{n}$ to be the
standard inner product, that is,

$\langle\vec{x},\vec{x}^{\prime}\rangle=\sum_{i=1}^{m}x^{i}x^{\prime i}\,,$
(126)

where $\vec{x}=(x^{1},\dots,x^{m})\in\mathbb{R}^{m}$ and similarly
$\vec{x}^{\prime}$ and analogous inner product on $\mathbb{R}^{n}$, then a
lattice isomorphism between lattices of signature $(m,n)$ can be identified
with an element of $\mathrm{O}(m,n,\mathbb{R})$ where
$\mathrm{O}(m,n,\mathbb{R})$ is the group of matrices, $A$, satisfying

$A^{T}g_{m,n}A=g_{m,n},\quad g_{m,n}=\begin{pmatrix}\mathds{1}_{m}&0\\\
0&-\mathds{1}_{n}\end{pmatrix}\,.$ (127)

We have the following theorem: _(serre1973course, , Chapter V)_ An even, self-
dual lattice of signature $(m,n)$ exists if and only if $(m-n)\equiv 0\bmod
8$. Moreover, there is a unique such lattice when $n\geq 1$ up to an
$\mathrm{O}(m,n,\mathbb{R})$ transformation.

The canonical choice of an even, self-dual lattice of signature
$\mathbb{R}^{m,n}$, denoted by $\mathrm{II}_{m,n}$, is

$\mathrm{II}_{m,n}=\left\\{(a_{1},\dots,a_{m+n})\in\mathbb{R}^{m,n}:a_{i}\in\mathbb{Z}\text{
or }a_{i}\in\mathbb{Z}+\frac{1}{2},\leavevmode\nobreak\
\sum_{i=1}^{m+n}a_{i}\in 2\mathbb{Z}\right\\}\,.$ (128)

A generator matrix for this lattice is

$\mathcal{G}_{\mathrm{II}_{m,n}}=\begin{pmatrix}[cccc|cccccc]1&0&\cdots&0&0&0&\cdots&0&0&-1\\\
0&1&\cdots&0&0&0&\cdots&0&0&-1\\\
\vdots&\vdots&\cdots&\vdots&\vdots&\vdots&\cdots&\vdots&\vdots&\vdots\\\
0&0&\cdots&1&0&0&\cdots&0&0&-1\\\ \hline\cr 0&0&\cdots&0&1&0&\cdots&0&0&-1\\\
0&0&\cdots&0&0&1&\cdots&0&0&-1\\\
\vdots&\vdots&\cdots&\vdots&\vdots&\vdots&\cdots&\vdots&\vdots&\vdots\\\
0&0&\cdots&0&0&0&\cdots&1&0&-1\\\ 0&0&\cdots&0&0&0&\cdots&0&0&2\\\
\frac{1}{2}&\frac{1}{2}&\cdots&\frac{1}{2}&\frac{1}{2}&\frac{1}{2}&\cdots&\frac{1}{2}&\frac{1}{2}&\frac{1}{2}\end{pmatrix}\,.$
(129)

We will use this lattice to elucidate many of the notations which we now
introduce. Consider a $d$-dimensional even, integral, Lorentzian lattice
$\Lambda\subset\mathbb{R}^{m,n}$ with Lorentzian inner product, denoted as
before by $\circ$, where $m+n=d$. We will often write a vector
$\lambda\in\Lambda$ as $\lambda=(\alpha^{\lambda},\beta^{\lambda})$, where
$\alpha^{\lambda}\in\mathbb{R}^{m}$ and $\beta^{\lambda}\in\mathbb{R}^{n}$.
Then we can write

$\lambda_{1}\circ\lambda_{2}=\langle\alpha^{\lambda_{1}},\alpha^{\lambda_{2}}\rangle-\langle\beta^{\lambda_{1}},\beta^{\lambda_{2}}\rangle\in\mathbb{Z}.$
(130)

Note that in general
$\langle\alpha^{\lambda_{1}},\alpha^{\lambda_{2}}\rangle,\langle\beta^{\lambda_{1}},\beta^{\lambda_{2}}\rangle\not\in\mathbb{Z}$.
We define the $\mathbb{Z}$-modules

$\begin{split}&\Lambda_{1}=\\{\alpha^{\lambda}\,\lvert\,\lambda=(\alpha^{\lambda},\beta^{\lambda})\in\Lambda\text{
for some }\beta^{\lambda}\in\mathbb{R}^{n}\\}\subset\mathbb{R}^{m},\\\
&\Lambda_{2}=\\{\beta^{\lambda}\,\lvert\,\lambda=(\alpha^{\lambda},\beta^{\lambda})\in\Lambda\text{
for some
}\alpha^{\lambda}\in\mathbb{R}^{m}\\}\subset\mathbb{R}^{n}.\end{split}$ (131)

Let
$\\{\lambda_{i}\equiv(\alpha^{\lambda_{i}},\beta^{\lambda_{i}})\\}_{i=1}^{d}$
be a basis of $\Lambda$. Then it is easy to see that

$\Lambda_{1}=\text{Span}_{\mathbb{Z}}\\{\alpha^{\lambda_{i}}\\}_{i=1}^{d},\quad\Lambda_{2}=\text{Span}_{\mathbb{Z}}\\{\beta^{\lambda_{i}}\\}_{i=1}^{d}.$
(132)

Note that in general $\Lambda_{1}$ and $\Lambda_{2}$ are not lattices, they
are just finitely generated $\mathbb{Z}$ modules possibly with non-trivial
torsion. For the lattice $\mathrm{II}_{m,n}$ in (129), it is easy to see that

$\begin{split}(\mathrm{II}_{m,n})_{1}=\mathbb{Z}^{m}\bigcup\left(\mathbb{Z}+\frac{1}{2}\right)^{m},\\\
(\mathrm{II}_{m,n})_{2}=\mathbb{Z}^{n}\bigcup\left(\mathbb{Z}+\frac{1}{2}\right)^{n}.\end{split}$
(133)

We further identify even, integral, Euclidean sublattices of $\Lambda$ as
follows:

$\begin{split}&\Lambda_{1}^{0}:=\\{(\alpha,0)\in\Lambda\,\lvert\,\alpha\in\mathbb{R}^{m}\\},\\\
&\Lambda_{2}^{0}:=\\{(0,\beta)\in\Lambda\,\lvert\,\beta\in\mathbb{R}^{n}\\}\,.\end{split}$
(134)

These can be identified naturally with submodules of $\Lambda_{1}$ and
$\Lambda_{2}$ respectively. Clearly, $\Lambda_{1}^{0},\Lambda_{2}^{0}$ are
sublattices of $\Lambda$ since any submodule of a finitely generated free
module is free. We also introduce the notation

$\Lambda_{0}:=\Lambda_{1}^{0}\oplus\Lambda_{2}^{0}\,.$ (135)

Note that the direct sum of $\Lambda_{1}^{0}$ and $\Lambda_{2}^{0}$ is
meaningful as they are two $\mathbb{Z}$-modules. For the lattice
$\mathrm{II}_{m,n}$ in (129),
$(\mathrm{II}_{m,n})_{0}=(\mathrm{II}_{m,n})_{1}^{0}\oplus(\mathrm{II}_{m,n})_{2}^{0}$
is easily seen to be generated by

$\mathcal{G}_{(\mathrm{II}_{m,n})_{0}}:=\left(\begin{array}[]{cc}\mathcal{G}_{m}&0_{m\times
n}\\\ 0_{n\times m}&\mathcal{G}_{n}\,,\end{array}\right)$ (136)

where

$\mathcal{G}_{m}:=\begin{pmatrix}1&0&0&\cdots&0&-1\\\ 0&1&0&\cdots&0&-1\\\
\vdots&\vdots&\vdots&\cdots&\vdots&\vdots\\\ 0&0&0&\cdots&1&-1\\\
0&0&0&\cdots&0&2\end{pmatrix}_{m\times m}\,,$ (137)

and $\mathcal{G}_{n}$ is defined similarly. It is useful to characterize the
automorphisms of $\Lambda$. We will take the symmetric bilinear form on
$\mathbb{R}^{m},\mathbb{R}^{n}$ to be the standard inner product for brevity.
We have the following important result. Let $\Lambda\in\mathbb{R}^{m,n}$ be an
integral Lorentzian lattice. Then
$\mathrm{Aut}(\Lambda)\cong\mathrm{O}_{\Lambda}(m,n,\mathbb{Z})$ where

$\mathrm{O}_{\Lambda}(m,n,\mathbb{Z}):=\\{A\in\mathrm{GL}(m+n,\mathbb{Z}):\mathcal{G}_{\Lambda}^{-1}A\mathcal{G}_{\Lambda}\in\mathrm{O}(m,n,\mathbb{R})\\}\,.$
(138)

Choose an integral basis $\\{\lambda_{i}\\}$ of $\Lambda$. Then the group of
$\mathbb{Z}$-module automorphisms of $\Lambda$ can be identified with
$\mathrm{GL}(m+n,\mathbb{Z})$. Now given any
$\lambda,\lambda^{\prime}\in\Lambda$ there exists coulumn vectors
$\vec{n},\vec{n}^{\prime}\in\mathbb{Z}^{m+n}$ such that
$\lambda=\vec{n}^{T}\mathcal{G}_{\Lambda}$ and
$\lambda^{\prime}=\vec{n}^{{}^{\prime}T}\mathcal{G}_{\Lambda}$. Any module
automorphism $A\in\mathrm{GL}(m+n,\mathbb{Z})$ acts by

$A(\lambda)=\vec{n}^{T}A\mathcal{G}_{\Lambda}\,.$ (139)

For $A$ to preserve inner product, we must have

$\begin{split}A(\lambda)\circ
A(\lambda^{\prime})&=\vec{n}^{T}A\mathcal{G}_{\Lambda}g_{m,n}\mathcal{G}_{\Lambda}^{T}A^{T}\vec{n}^{\prime}\,,\\\
&=\vec{n}^{T}\mathcal{G}_{\Lambda}g_{m,n}\mathcal{G}_{\Lambda}^{T}\vec{n}^{\prime}\,,\end{split}$
(140)

which implies

$A\mathcal{G}_{\Lambda}g_{m,n}\mathcal{G}_{\Lambda}^{T}A^{T}=\mathcal{G}_{\Lambda}g_{m,n}\mathcal{G}_{\Lambda}^{T}\,.$
(141)

Since $\\{\lambda_{i}\\}$ is a basis for $\mathbb{R}^{m,n}$ and $A$ must
preserve the inner products of $\lambda_{i}$’s, we must have that
$A=\mathcal{G}_{\Lambda}O\mathcal{G}_{\Lambda}^{-1}$ for some
$O\in\mathrm{O}(m,n,\mathbb{R})$.

###  3.2 Construction of the LLVOA

Let $\Lambda\subset\mathbb{R}^{m,n}$ be a $d=(m+n)$-dimensional Lorentzian
lattice. We denote by $\mathbb{C}[\Lambda]$ the group algebra of the lattice
$\Lambda$ and denote the element $\lambda\in\Lambda$ embedded in
$\mathbb{C}[\Lambda]$ by ${\rm e}^{\lambda}$. The multiplication in
$\mathbb{C}[\Lambda]$ is defined131313Technically speaking,
$\mathbb{C}[\Lambda]$ is the group algebra of the formal group ${\rm
e}^{\Lambda}:=\\{{\rm e}^{\lambda}:\lambda\in\Lambda\\}$ with group
multiplication given by ${\rm e}^{\lambda_{1}}\cdot{\rm e}^{\lambda_{2}}={\rm
e}^{\lambda_{1}+\lambda_{2}}$, i.e. $\mathbb{C}[\Lambda]=\mathbb{C}[{\rm
e}^{\Lambda}]$ by

${\rm e}^{\lambda_{1}}\cdot{\rm e}^{\lambda_{2}}={\rm
e}^{\lambda_{1}+\lambda_{2}}\,.$ (142)

Define the vector space

$\mathfrak{h}_{i}:=\Lambda_{i}\otimes_{\mathbb{Z}}\mathbb{C},\quad i=1,2\,,$
(143)

and extend the bilinear form on $\Lambda_{i}$ to $\mathfrak{h}_{i}$
$\mathbb{C}$-linearly. Here $\Lambda_{i}$ is defined as in (131). We define
the Lie algebra

$\begin{split}\hat{\mathfrak{h}}&:=\left(\bigoplus_{\begin{subarray}{c}r,s\in\mathbb{Z}\end{subarray}}(\mathfrak{h}_{1}\otimes
t^{r})\oplus(\mathfrak{h}_{2}\otimes\bar{t}^{\,s})\right)\oplus(\mathbb{C}\textbf{k}\oplus\mathbb{C}\bar{\textbf{k}}).\end{split}$
(144)

Introduce the notation

$\alpha(r):=\alpha\otimes
t^{r},\quad\beta(s):=\beta\otimes\bar{t}^{\,s},\quad\alpha\in\mathfrak{h}_{1},\beta\in\mathfrak{h}_{2}.$
(145)

The non-zero Lie bracket on $\hat{\mathfrak{h}}$ is

$\begin{split}[\,\alpha(r_{1}),\alpha^{\prime}(r_{2})\,]&=r_{1}\,\left\langle\alpha,\alpha^{\prime}\right\rangle\,\delta_{r_{1}+r_{2},0}\,\textbf{k},\\\
[\,\beta(s_{1}),\beta^{\prime}(s_{2})\,]&=s_{1}\,\left\langle\beta,\beta^{\prime}\right\rangle\,\delta_{s_{1}+s_{2},0}\,\bar{\textbf{k}}\
.\end{split}$ (146)

Note that

$\hat{\mathfrak{h}}=\hat{\mathfrak{h}}_{1}^{\star}\oplus\hat{\mathfrak{h}}_{2}^{\star}\oplus\hat{\mathfrak{h}}_{1}^{0}\oplus\hat{\mathfrak{h}}_{2}^{0}\,,$
(147)

where $\hat{\mathfrak{h}}_{1}^{\star},\hat{\mathfrak{h}}_{2}^{\star}$ are the
standard Heisenberg algebras associated to the abelian Lie algebras
$\mathfrak{h}_{1},\mathfrak{h}_{2}$ respectively (Frenkel:1988xz, , Chapter 1)
and

$\hat{\mathfrak{h}}_{1}^{0}:=\mathfrak{h}_{1}\otimes
t^{0}\cong\mathfrak{h}_{1},\quad\hat{\mathfrak{h}}_{2}^{0}:=\mathfrak{h}_{2}\otimes\bar{t}^{0}\cong\mathfrak{h}_{2}.$
(148)

Define

$\begin{split}\hat{\mathfrak{h}}^{-}:=\left(\bigoplus_{r,s<0}(\mathfrak{h}_{1}\otimes
t^{r})\oplus(\mathfrak{h}_{2}\otimes\bar{t}^{\,s})\right),&\quad\hat{\mathfrak{h}}^{0}:=(\mathfrak{h}_{1}\otimes
t^{0})\oplus(\mathfrak{h}_{2}\otimes\bar{t}^{0})\oplus\mathbb{C}\textbf{k}\oplus\mathbb{C}\bar{\textbf{k}}\,,\\\
\hat{\mathfrak{h}}^{+}:=\Bigg{(}\bigoplus_{r,s>0}(\mathfrak{h}_{1}&\otimes
t^{r})\oplus(\mathfrak{h}_{2}\otimes\bar{t}^{\,s})\Bigg{)}.\end{split}$ (149)

Note that

$\hat{\mathfrak{h}}=\hat{\mathfrak{h}}^{-}\oplus\hat{\mathfrak{h}}^{0}\oplus\hat{\mathfrak{h}}^{+}.$
(150)

We now define the space

$\begin{split}V_{\Lambda}&:=S(\hat{\mathfrak{h}}^{-})\otimes\mathbb{C}[\Lambda_{1}^{0}\oplus\Lambda_{2}^{0}]\,,\end{split}$
(151)

where $\Lambda_{1}^{0}$ and $\Lambda_{2}^{0}$ is defined as in (134) and for
any Lie algebra $\mathfrak{g}$, $S(\mathfrak{g})$ is the symmetric algebra for
$\mathfrak{g}$, and
$\mathbb{C}[\Lambda_{1}^{0}\oplus\Lambda_{2}^{0}]\equiv\mathbb{C}[\Lambda_{0}]$
is considered as a subspace of $\mathbb{C}[\Lambda]$. The space $V_{\Lambda}$
is generated by elements of the form

$\begin{split}&\left(\alpha_{1}(-m_{1})\cdot\alpha_{2}(-m_{2})\cdots\alpha_{k}(-m_{k})\cdot\beta_{1}(-\bar{m}_{1})\cdot\beta_{2}(-\bar{m}_{2})\cdots\beta_{\bar{k}}(-\bar{m}_{\bar{k}})\right)\otimes{\rm
e}^{(\alpha,\beta)}\end{split}$ (152)

for $m_{i},\,\bar{m}_{i}>0,\,k,\bar{k}\geq 0,(\alpha,\beta)\in\Lambda_{0}$,
$\alpha_{i}\in\mathfrak{h}_{1}$, and $\beta_{i}\in\mathfrak{h}_{2}$. The space
$V_{\Lambda}$ is a natural module of $\hat{\mathfrak{h}}^{-}$. We define the
action of $\hat{\mathfrak{h}}^{0}$ on $\mathbb{C}[\Lambda]$, and hence on
$\mathbb{C}[\Lambda_{0}]$, by

$\begin{split}\alpha^{\prime}(0)\,e^{\lambda}&=\langle\alpha^{\prime},\alpha^{\lambda}\rangle\,e^{\lambda}\,,\\\
\beta^{\prime}(0)\,e^{\lambda}&=\langle\beta^{\prime},\beta^{\lambda}\rangle\,e^{\lambda}\,,\end{split}$
(153)

where $\alpha^{\prime}(0)\in\hat{\mathfrak{h}}_{1}^{0},\leavevmode\nobreak\
\beta^{\prime}(0)\in\hat{\mathfrak{h}}_{2}^{0}$. The central elements k and
$\bar{\textbf{k}}$ act on $\mathbb{C}[\Lambda]$ as identity. Let
$\hat{\mathfrak{h}}^{+}$ act on $\mathbb{C}[\Lambda]$ by 0. We can extend the
action of these subspaces of $\hat{\mathfrak{h}}$ to $V_{\Lambda}$ by using
the Lie bracket given in (146). This makes $V_{\Lambda}$ into an
$\hat{\mathfrak{h}}$-module.

We define a $\mathbb{Z}$-bilinear map
$\epsilon:\Lambda\times\Lambda\rightarrow\mathbb{Z}$, which acts on the basis
as huang

$\epsilon\left(\lambda_{i},\lambda_{j}\right)=\begin{cases}\lambda_{i}\circ\lambda_{j}&i>j\,,\\\
0&i\leq j\,,\end{cases}\ \ \\\ $ (154)

where $\\{\lambda_{i}\\}_{i=1}^{m+n}$ is an integral basis of $\Lambda$. The
action of $\epsilon$ on general vectors is defined by the
$\mathbb{Z}$-bilinearity of $\epsilon$. Consider
$\hat{\Lambda}=\mathbb{Z}_{2}\times\Lambda$, with the multiplication on it
given by

$\begin{split}(\theta,\lambda)\cdot(\tau,\lambda^{\prime})=&\left(\theta\tau\,(-1)^{\epsilon(\lambda,\lambda^{\prime})},\lambda+\lambda^{\prime}\right).\end{split}$
(155)

We now consider the $\mathbb{Z}_{2}$ central extension of the lattice
$\Lambda$:

$0\longrightarrow\mathbb{Z}_{2}\longrightarrow\hat{\Lambda}\longrightarrow\Lambda\longrightarrow
0\ .$ (156)

We denote elements $(1,\lambda),(\theta,0)\in\hat{\Lambda}$ by ${\rm
e}_{\lambda}=(1,\lambda)$ and $\theta=(\theta,0)$ respectively. Then it is
easy to check that

$(\theta,\lambda)=\theta\,{\rm e}_{\lambda}={\rm e}_{\lambda}\theta\,,$ (157)

and

${\rm e}_{\lambda}{\rm e}_{\mu}=(-1)^{\epsilon(\lambda,\mu)}\,{\rm
e}_{\lambda+\mu}\ .$ (158)

Using the above relation, it can be shown that (see Lemma A for proof)

${\rm e}_{\lambda}{\rm e}_{\mu}=(-1)^{\lambda\circ\mu}\,{\rm e}_{\mu}\,{\rm
e}_{\lambda}\,.$ (159)

This property requires that the lattice be even and integral. Note that we
chose an integral basis of $\Lambda$ to define the central extension. In
Appendix A we show that a cocycle $\tilde{\epsilon}$ defined analogous to
(154) for a different choice of basis is cohomologous to $\epsilon$, and hence
gives rise to an isomorphic central extension. $\hat{\Lambda}$ acts on
$\mathbb{C}[\Lambda]$ as follows

$(\theta,\lambda^{\prime})\,e^{\lambda}=\theta(-1)^{\epsilon(\lambda^{\prime},\lambda)}\,e^{\lambda+\lambda^{\prime}}\,.$
(160)

In particular for $(\theta,\lambda^{\prime})=(1,\lambda^{\prime})={\rm
e}_{\lambda^{\prime}}$, we have

${\rm e}_{\lambda^{\prime}}\,{\rm
e}^{\lambda}=(-1)^{\epsilon(\lambda^{\prime},\lambda)}\,{\rm
e}^{\lambda+\lambda^{\prime}}\,.$ (161)

Note that the same cocycle $\epsilon$ restricted to $\Lambda_{0}$

$\epsilon:\Lambda_{0}\times\Lambda_{0}\longrightarrow\mathbb{Z}\,,$ (162)

defines a central extension
$\hat{\Lambda}_{0}:=\mathbb{Z}_{2}\times\Lambda_{0}\subset\hat{\Lambda}$:

$0\longrightarrow\mathbb{Z}_{2}\longrightarrow\hat{\Lambda}_{0}\longrightarrow\Lambda_{0}\longrightarrow
0\ .$ (163)

Moreover, the action (160) restricted to $\hat{\Lambda}_{0}$ makes
$\mathbb{C}[\Lambda_{0}]$ into a $\hat{\Lambda}_{0}$-module. This makes
$V_{\Lambda}$ into a $\hat{\Lambda}_{0}$-module where $\hat{\Lambda}_{0}$ acts
only on $\mathbb{C}[\Lambda_{0}]$. Let $x,\bar{x}$ be formal variables. For
any vector $\lambda=(\alpha^{\lambda},\beta^{\lambda})$, define the operators
$x^{\alpha^{\lambda}},\bar{x}^{\beta^{\lambda}}$ by the following actions

$\begin{split}x^{\alpha^{\lambda}}(u\otimes{\rm
e}^{\lambda^{\prime}})=x^{\langle\alpha^{\lambda},\,\alpha^{\lambda^{\prime}}\rangle}(u\otimes{\rm
e}^{\lambda^{\prime}})\,,\\\ \bar{x}^{\beta^{\lambda}}(u\otimes{\rm
e}^{\lambda^{\prime}})=\bar{x}^{\langle\beta^{\lambda},\,\beta^{\lambda^{\prime}}\rangle}(u\otimes{\rm
e}^{\lambda^{\prime}})\,,\end{split}$ (164)

where $u\in S(\hat{\mathfrak{h}}^{-}),\lambda^{\prime}\in\Lambda_{0}$. Note
that $x^{\alpha^{\lambda}},\bar{x}^{\beta^{\lambda}}$ acts as
$x^{\alpha^{\lambda}(0)},\bar{x}^{\beta^{\lambda}(0)}$. For
$\lambda=(\alpha^{\lambda},\beta^{\lambda})\in\Lambda_{0}$, define the vertex
operators

$\begin{split}Y_{V_{\Lambda}}({\rm
e}^{\lambda},x,\bar{x}):=&\left[\exp\left(-\sum_{r<0}\frac{\alpha^{\lambda}(r)}{r}x^{-r}\right)\exp\left(-\sum_{r>0}\frac{\alpha^{\lambda}(r)}{r}x^{-r}\right)\right.\\\
&\left.\exp\left(-\sum_{r<0}\frac{\beta^{\lambda}(r)}{r}\bar{x}^{-r}\right)\exp\left(-\sum_{r>0}\frac{\beta^{\lambda}(r)}{r}\bar{x}^{-r}\right)\right]{\rm
e}_{\lambda}x^{\alpha^{\lambda}}\bar{x}^{\beta^{\lambda}}\ .\end{split}$ (165)

From the Lie bracket in (146), it is easy to show

$[\alpha^{\lambda}(r),\beta^{\lambda}(s)]=0$ (166)

for all $r,s\in\mathbb{Z}$, so that the order of exponentials with
$\alpha^{\lambda}(r)$ and $\beta^{\lambda}(r)$ does not matter. For a formal
variable $x$, we introduce the notation

$\alpha^{\lambda}(x)=\underbrace{\sum_{r>0}\alpha^{\lambda}(r)x^{-r-1}}_{\alpha^{\lambda}(x)^{+}}+\underbrace{\sum_{r<0}\alpha^{\lambda}(r)x^{-r-1}}_{\alpha^{\lambda}(x)^{-}}+\,\alpha^{\lambda}(0)x^{-1}\
,$ (167)

Similarly, we can also define $\beta^{\lambda}(\bar{x})$. We define the formal
integration as the map by

$\int dx\leavevmode\nobreak\ x^{r}=\frac{x^{r+1}}{r+1},\quad n\neq-1\,.$ (168)

We can then write the vertex operator as

$\begin{split}Y_{V_{\Lambda}}({\rm e}^{\lambda},x,\bar{x})=\exp\left(\int
dx\leavevmode\nobreak\ \alpha^{\lambda}(x)^{-}\right)&\exp\left(\int
dx\leavevmode\nobreak\ \alpha^{\lambda}(x)^{+}\right)\\\ \times&\exp\left(\int
d\bar{x}\leavevmode\nobreak\ \beta^{\lambda}(\bar{x})^{-}\right)\exp\left(\int
d\bar{x}\leavevmode\nobreak\ \beta^{\lambda}(\bar{x})^{+}\right){\rm
e}_{\lambda}\,x^{\alpha^{\lambda}}\bar{x}^{\beta^{\lambda}}.\end{split}$ (169)

For a general vector $v$ of the form (152), the vertex operator is defined as

$Y_{V_{\Lambda}}(v,x,\bar{x})=\typecolon\prod_{r=1}^{k}\prod_{s=1}^{\bar{k}}\left(\frac{1}{(m_{r}-1)!}\frac{d^{m_{r}-1}\alpha_{r}(x)}{dx^{m_{r}-1}}\right)\left(\frac{1}{(\bar{m}_{s}-1)!}\frac{d^{\bar{m}_{s}-1}\beta_{s}(\bar{x})}{d\bar{x}^{\bar{m}_{s}-1}}\right)Y_{V_{\Lambda}}({\rm
e}^{\lambda},x,\bar{x})\typecolon\,,$ (170)

where the normal ordering is defined as

$\begin{split}&\typecolon\alpha^{\lambda}(p)\,\alpha^{\lambda^{\prime}}(q)\typecolon=\typecolon\alpha^{\lambda^{\prime}}(q)\,\alpha^{\lambda}(p)\typecolon=\begin{cases}\alpha^{\lambda}(p)\,\alpha^{\lambda^{\prime}}(q)&p\leq
q\,,\\\ \alpha^{\lambda^{\prime}}(q)\,\alpha^{\lambda}(p)&p\geq
q\,,\end{cases}\\\
&\typecolon\alpha^{\lambda}(p)\mathrm{e}_{\lambda^{\prime}}\typecolon=\typecolon\mathrm{e}_{\lambda^{\prime}}\,\alpha^{\lambda}(p)\typecolon=\mathrm{e}_{\lambda^{\prime}}\,\alpha^{\lambda}(p)\,,\\\
&\typecolon x^{\alpha^{\lambda}}\,{\rm
e}_{\lambda^{\prime}}\typecolon=\typecolon{\rm
e}_{\lambda^{\prime}}\,x^{\alpha^{\lambda}}\typecolon={\rm
e}_{\lambda^{\prime}}\,x^{\alpha^{\lambda}}\,,\end{split}$ (171)

and similarly for $\beta^{\lambda}$ and $\bar{x}^{\beta^{\lambda}}$. The
vertex operator for general vectors in $V_{\Lambda}$ is defined by linear
extension to all of $V_{\Lambda}$. Using the central extension (156), (170)
can be used to define vertex operators even if ${\rm
e}^{\lambda}\in\mathbb{C}[\Lambda]$. These vertex operators will act on
vectors of the form (152) with ${\rm e}^{\lambda}\in\mathbb{C}[\Lambda]$
rather than $\mathbb{C}[\Lambda_{0}]$. This will be crucial when we construct
module vertex operators and intertwining operators on the modules of
$V_{\Lambda}$. The vacuum vector is given by $\mathbf{1}=\mathrm{e}^{0}$. The
conformal vector is constructed below, see (187).

###  3.3 Proof of axioms

We now prove that
$(V_{\Lambda},Y_{V_{\Lambda}},\omega_{L},\omega_{R},\mathbf{1})$ is a non-
chiral VOA.

Proof of identity property 1: From the definition (165), it is clear that
$Y_{V_{\Lambda}}(\mathbf{1},x,\bar{x})={\rm e}_{0}=\mathds{1}.$

Proof of grading-restriction property 2: The grading on $V_{\Lambda}$ is given
by defining the conformal weight of vector $v$ of the form (152) by

$h_{v}=\frac{\langle\alpha,\alpha\rangle}{2}+\sum_{i=1}^{k}m_{i},\quad\bar{h}_{v}=\frac{\langle\beta,\beta\rangle}{2}+\sum_{j=1}^{\bar{k}}\bar{m}_{j}\,,$
(172)

where $m_{i}$ and $\bar{m}_{j}$ are positive integers appearing in (152). Note
that for ${\rm e}^{\lambda}$ with $\lambda=(\alpha,\beta)\in\Lambda_{0}$, we
have $(\alpha,0)\in\Lambda$,
$(\alpha,0)\circ(\alpha,0)=\langle\alpha,\alpha\rangle\in 2\mathbb{Z}_{+}$.
Then we have that $h_{v},\bar{h}_{v}\geq 0$ so that $V_{(h,\bar{h})}=0$ for
$h$ or $\bar{h}<0$, i.e. $M=0$ in (43).
Similarly, $\langle\beta,\beta\rangle\in 2\mathbb{Z}_{+}$, which implies that
both $h_{v}$ and $\bar{h}_{v}$ are positive integers 141414Note that this
argument also works for a general ${\rm e}^{\lambda}\in\mathbb{C}[\Lambda]$
with $\lambda\in\Lambda$ since
$\langle\alpha,\alpha\rangle,\langle\beta,\beta\rangle\geq 0$ even in this
case..

We will now show that dim$(V_{(h,\bar{h})})<\infty$. Note that
$\Lambda_{1}^{0}$ and $\Lambda_{2}^{0}$ are lattices. It suffices to show that
there exist only finitely many vectors of the form (152), satisfying the
conditions in (172). We first show that for any $h,\bar{h}\in\mathbb{R}$ the
number of distinct $\lambda=(\alpha,\beta)\in\Lambda_{0}$ satisfying

$\langle\alpha,\alpha\rangle\leq 2h\text{ and }\langle\beta,\beta\rangle\leq
2\bar{h}\,,\text{ where }\alpha\in\Lambda_{1}^{0},\beta\in\Lambda_{2}^{0}\,,$
(173)

can be only finitely many. Consider the sets

$X_{1}=\\{\alpha\in\Lambda_{1}^{0}\leavevmode\nobreak\
\lvert\leavevmode\nobreak\ \langle\alpha,\alpha\rangle\leq 2h\\},$ (174)
$X_{2}=\\{\beta\in\Lambda_{2}^{0}\leavevmode\nobreak\
\lvert\leavevmode\nobreak\ \langle\beta,\beta\rangle\leq 2\bar{h}\\},$ (175)

which have finite cardinality, say $N_{1}$ and $N_{2}$, due to the fact that
$\Lambda_{1}^{0}$ and $\Lambda_{2}^{0}$ are discrete. Then the set

$X=\\{\lambda=(\alpha,\beta)\in\Lambda_{0}\leavevmode\nobreak\
\lvert\leavevmode\nobreak\ \langle\alpha,\alpha\rangle\leq
2h,\leavevmode\nobreak\ \langle\beta,\beta\rangle\leq 2\bar{h}\\}$ (176)

is finite because the map

$\begin{split}&X\longrightarrow X_{1}\times X_{2}\\\
&\lambda=(\alpha,\beta)\longmapsto(\alpha,\beta)\end{split}$ (177)

is injective. More precisely $\\#X\leq N_{1}N_{2}$. Now, as there are only
finitely many combinations of positive integers $\\{m_{i}\\}_{i=1}^{k}$ and
$\\{\bar{m}_{i}\\}_{i=1}^{\bar{k}}$ such that

$h_{v}-\frac{\langle\alpha,\alpha\rangle}{2}=\sum_{i=1}^{k}m_{i},\quad\bar{h}_{v}-\frac{\langle\beta,\beta\rangle}{2}=\sum_{j=1}^{\bar{k}}\bar{m}_{j}\,,$
(178)

hence there are only finitely many generating vectors possible, which implies
that dim$(V_{h,\bar{h}})<\infty$.

Proof of single-valuedness property 3: For the general vector $v$ of the form
(152) we have

$\begin{split}h_{v}-\bar{h}_{v}&=\frac{\langle\alpha,\alpha\rangle-\langle\beta,\beta\rangle}{2}+\sum_{i=1}^{k}m_{i}-\sum_{j=1}^{\bar{k}}\bar{m}_{j}\\\
&=\frac{\lambda\circ\lambda}{2}+\sum_{i=1}^{k}m_{i}-\sum_{j=1}^{\bar{k}}\bar{m}_{j}\in\mathbb{Z},\end{split}$
(179)

where we used the fact that $\Lambda$ is an even Lorentzian lattice.

Proof of creation property 4: We want to show that for any state $v\in
V_{\Lambda}$,

$\lim_{x,\,\bar{x}\to 0}Y_{V_{\Lambda}}(v,x,\bar{x})\,\mathbf{1}=v.$ (180)

Let us first consider the case when $v={\rm e}^{\lambda}$, then the
$Y_{V_{\Lambda}}$ operator is given in (165). One then has to expand the
exponentials, we ignore the terms when $\alpha^{\lambda}(n)$ and
$\beta^{\lambda}(n)$ have $n>0$, as they annihilate $\mathbb{C}[\Lambda_{0}]$.
The two exponentials that remain will only have positive powers of $x$ and
$\bar{x}$, which vanish when we take the limit. Hence

$\lim_{x,\,\bar{x}\to 0}Y_{V_{\Lambda}}({\rm
e}^{\lambda},x,\bar{x})\,\mathbf{1}={\rm e}_{\lambda}\cdot\mathbf{1}.$ (181)

Here, we used the fact that $\mathbf{1}={\rm e}^{0}$ so that the action of
$x^{\alpha^{\lambda}}(\bar{x}^{\beta^{\lambda}})$ on this is by identity,
since
$\langle\alpha^{\lambda},0\rangle=0\left(\langle\beta^{\lambda},0\rangle=0\right)$.
Hence

$\lim_{x,\,\bar{x}\to 0}Y_{V_{\Lambda}}({\rm
e}^{\lambda},x,\bar{x})\,\mathbf{1}={\rm e}_{\lambda}\,{\rm
e}^{0}=(-1)^{\epsilon(0,\lambda)}{\rm e}^{\lambda}={\rm e}^{\lambda}\,,$ (182)

where we have used (154), (160) and that ${\rm e}_{\lambda}=(1,\lambda)$. We
now prove (180) for a general vector $v$ of the form (152). The normal
ordering in the definition (170) and the fact that $\hat{\mathfrak{h}}^{+}$
annihilates $\mathbb{C}[\Lambda_{0}]$ forces the product to take the form

$\frac{d^{m_{r}-1}\alpha_{r}(x)}{dx^{m_{r}-1}}\to\sum_{p_{r}\leq-
m_{r}}(m_{r}-1)!\alpha_{r}(p_{r})x^{-p_{r}-m_{r}},$ (183)

and

$\frac{d^{\bar{m}_{s}-1}\beta_{s}(\bar{x})}{d\bar{x}^{\bar{m}_{s}-1}}\to\sum_{q_{s}\leq-\bar{m}_{s}}(\bar{m}_{s}-1)!\beta_{s}(q_{s})\bar{x}^{-q_{s}-\bar{m}_{s}}.$
(184)

Thus we have

$\begin{split}&\typecolon\prod_{r=1}^{k}\prod_{s=1}^{\bar{k}}\left(\frac{1}{(m_{r}-1)!}\frac{d^{m_{r}-1}\alpha_{r}(x)}{dx^{m_{r}-1}}\right)\left(\frac{1}{(\bar{m}_{s}-1)!}\frac{d^{\bar{m}_{s}-1}\beta_{s}(\bar{x})}{d\bar{x}^{\bar{m}_{s}-1}}\right)Y_{V_{\Lambda}}({\rm
e}^{\lambda},x,\bar{x})\typecolon\mathbf{1}\\\
&\to\sum_{\begin{subarray}{c}p_{1}\leq-m_{1}\\\ ...\\\ p_{k}\leq-
m_{k}\end{subarray}}\typecolon\alpha_{1}(p_{1})\alpha_{2}(p_{2})\dots\alpha_{k}(p_{k})x^{-(p_{1}+\dots
p_{k})-(m_{1}+\dots+m_{k})}\\\
&\times\sum_{\begin{subarray}{c}q_{1}\leq-\bar{m}_{1}\\\ ...\\\
q_{\bar{k}}\leq-\bar{m}_{\bar{k}}\end{subarray}}\beta_{1}(q_{1})\beta_{2}(q_{2})\dots\beta_{\bar{k}}(q_{\bar{k}})\bar{x}^{-(q_{1}+\dots
q_{\bar{k}})-(\bar{m}_{1}+\dots+\bar{m}_{\bar{k}})}Y_{V_{\Lambda}}({\rm
e}^{\lambda},x,\bar{x})\typecolon\mathbf{1}.\end{split}$ (185)

When we take $x,\bar{x}\to 0$ only the $p_{r}=-m_{r},q_{s}=-\bar{m}_{s}$ terms
in the sum survives. Combining this fact with the proof of (180) for $v={\rm
e}^{\lambda}$, we get

$\begin{split}\lim_{x,\bar{x}\to 0}&Y_{V_{\Lambda}}(v,x,\bar{x})\mathbf{1}\\\
&=\left(\alpha_{1}(-m_{1})\cdot\alpha_{2}(-m_{2})\cdots\alpha_{k}(-m_{k})\,\beta_{1}(-\bar{m}_{1})\cdot\beta_{2}(-\bar{m}_{2})\cdots\beta_{\ell}(-\bar{m}_{\bar{k}})\right)\otimes{\rm
e}^{\lambda}=v,\end{split}$ (186)

where we also use the fact that $Y_{V_{\Lambda}}({\rm e}^{\lambda},x,\bar{x})$
can only contribute terms with $x^{n}$ and $\bar{x}^{m}$, where $n$ and $m$
are greater than 0.

Proof of Virasoro property 5: The conformal vector is given by

$\omega_{\Lambda}:=\frac{1}{2}\sum_{i=1}^{\text{dim}(\mathfrak{h}_{1})}\left(u_{i}(-1)^{2}\right)\otimes\mathbf{1}\,+\frac{1}{2}\sum_{i=1}^{\text{dim}(\mathfrak{h}_{2})}\left(v_{i}(-1)^{2}\right)\otimes\mathbf{1}\equiv\omega_{L}+\omega_{R},$
(187)

where
$u_{i}\in\Lambda_{1}\otimes_{\mathbb{Z}}\mathbb{C},v_{i}\in\Lambda_{2}\otimes_{\mathbb{Z}}\mathbb{C}$
are orthonormal basis of $\mathfrak{h}_{1}$ and $\mathfrak{h}_{2}$
respectively151515Note that a different choice of orthonormal basis will give
isomorphic LLVOAs. Indeed if $\\{u_{i}^{\prime}\\}$ and $\\{v_{j}^{\prime}\\}$
are orthonormal bases of $\mathfrak{h}_{1}$ and $\mathfrak{h}_{2}$
respectively, different from $\\{u_{i}\\}$ of and $\\{v_{j}\\}$. Then the map
$f:V_{\Lambda}\longrightarrow V_{\Lambda}$ which acts trivially on
$\mathbb{C}[\Lambda_{0}]$ and maps $u_{i}\mapsto
u_{i}^{\prime},\leavevmode\nobreak\ v_{i}\mapsto v_{i}^{\prime}$ is a non-
chiral VOA isomorphism.:

$\langle u_{i},u_{j}\rangle=\delta_{i,j},\quad\langle
v_{i},v_{j}\rangle=\delta_{i,j}.$ (188)

Since an integral basis of $\Lambda$ is also a basis of $\mathbb{R}^{m,n}$, it
is clear that $\text{dim}(\mathfrak{h}_{1})=m$ and
$\text{dim}(\mathfrak{h}_{2})=n$. One can check that the conformal vertex
operator is given by

$\begin{split}Y_{V_{\Lambda}}(\omega,x,\bar{x})&=Y_{V_{\Lambda}}(\omega_{L},x,\bar{x})+Y_{V_{\Lambda}}(\omega_{R},x,\bar{x})\\\
&=\sum_{p\in\mathbb{Z}}L_{\Lambda}(p)x^{-p-2}+\sum_{p\in\mathbb{Z}}\bar{L}_{\Lambda}(p)\bar{x}^{-p-2},\end{split}$
(189)

where the Virasoro generators are given by (Frenkel:1988xz, , Section 8.7)

$\begin{split}L_{\Lambda}(p)=\frac{1}{2}\sum_{i=1}^{m}\sum_{k\in\mathbb{Z}}\typecolon
u_{i}(k)u_{i}(p-k)\typecolon\\\
\bar{L}_{\Lambda}(p)=\frac{1}{2}\sum_{i=1}^{n}\sum_{k\in\mathbb{Z}}\typecolon
v_{i}(k)v_{i}(p-k)\typecolon.\end{split}$ (190)

Using the Lie brackets

$\begin{split}\left[\,u_{i}(p),u_{j}(q)\,\right]&=p\,\delta_{i,j}\,\delta_{p+q,0}\,\textbf{k},\\\
\left[\,v_{i}(p),v_{j}(q)\,\right]&=p\,\delta_{i,j}\,\delta_{p+q,0}\,\bar{\textbf{k}},\end{split}$
(191)

one can show that the Virasoro generators indeed satisfy the Virasoro algebra
(47) with central charge
$m=\text{dim}(\mathfrak{h}_{1}),n=\text{dim}(\mathfrak{h}_{2})$ respectively,
see (Frenkel:1988xz, , Chapter 2).

Proof of grading property 6: From (190) and normal ordering (171), we have

$\begin{split}&L_{\Lambda}(0)=\frac{1}{2}\sum_{i=1}^{m}\sum_{r\in\mathbb{Z}}\typecolon
u_{i}(r)u_{i}(-r)\typecolon=\sum_{i=1}^{m}\sum_{r>0}u_{i}(-r)u_{i}(r)+\frac{1}{2}u_{i}(0)^{2},\\\
&\bar{L}_{\Lambda}(0)=\frac{1}{2}\sum_{i=1}^{n}\sum_{r\in\mathbb{Z}}\typecolon
v_{i}(r)v_{i}(-r)\typecolon=\sum_{i=1}^{n}\sum_{r>0}v_{i}(-r)v_{i}(r)+\frac{1}{2}v_{i}(0)^{2}.\end{split}$
(192)

Then for $v\in V_{\Lambda}$ of the form (152), using the Lie bracket (191) and
the action (153), we have

$\begin{split}L_{\Lambda}(0)v&=\sum_{j=1}^{k}\left[\cdots\left(m_{j}\sum_{i=1}^{m}\langle
u_{i},\alpha_{j}\rangle u_{i}(-m_{j})\right)\cdots\right]\otimes{\rm
e}^{\lambda}+\frac{1}{2}\sum_{i=1}^{m}\langle u_{i},\alpha\rangle^{2}v\,,\\\
&=\sum_{j=1}^{k}m_{j}\left[\cdots\alpha_{j}(-m_{j})\cdots\right]\otimes{\rm
e}^{\lambda}+\frac{1}{2}\sum_{i=1}^{m}\langle\langle u_{i},\alpha\rangle
u_{i},\alpha\rangle v\,,\\\
&=\left[\sum_{j=1}^{k}m_{j}+\frac{\langle\alpha,\alpha\rangle}{2}\right]v\,,\end{split}$
(193)

where we used the fact that $\\{u_{i}\\}$ is an orthonormal basis of
$\mathfrak{h}_{1}$. Similarly

$\bar{L}_{\Lambda}(0)v=\left[\sum_{\bar{j}=1}^{\bar{k}}\bar{m}_{\bar{j}}+\frac{\langle\beta,\beta\rangle}{2}\right]v\,.$
(194)

Proof of $L(0)$-property 7: Let $(\alpha,\beta)\in\Lambda_{0}$. Then

$\displaystyle{\left[L_{\Lambda}(0),\alpha(x)^{\pm}\right]}$
$\displaystyle=\frac{1}{2}\sum_{r\in\mathbb{Z}_{\pm}}\sum_{s\in\mathbb{Z}}\sum_{i=1}^{m}\left[\typecolon
u_{i}(s)u_{i}(-s)\typecolon,\alpha(r)\right]x^{-r-1}$ (195)
$\displaystyle=\frac{1}{2}\sum_{r\in\mathbb{Z}_{\pm}}\sum_{i=1}^{m}\left(\sum_{s\geq
1}\left[u_{i}(-s)u_{i}(s),\alpha(r)\right]+\left[u_{i}(0)^{2},\alpha(r)\right]\right.$
$\displaystyle\left.\hskip
142.26378pt+\sum_{s\leq-1}\left[u_{i}(s)u_{i}(-s),\alpha(r)\right]\right)x^{-r-1}$
$\displaystyle=\frac{1}{2}\sum_{r\in\mathbb{Z}_{\pm}}\sum_{s\neq
0}\left(s\alpha(-s)\delta_{s+r,0}-n\alpha(s)\right)\delta_{r-s,0}x^{-r-1}$
$\displaystyle=\sum_{r\in\mathbb{Z}_{\pm}}\sum_{s\neq
0}n\alpha(-s)\delta_{s+r,0}x^{-r-1}$
$\displaystyle=-\sum_{r\in\mathbb{Z}_{\pm}}r\alpha(r)x^{-r-1},$

where we used

$\displaystyle\sum_{i=1}^{m}\left[u_{i}(-s)u_{i}(s),\alpha(r)\right]$
$\displaystyle=\sum_{i=1}^{m}u_{i}(-s)\left[u_{i}(s),\alpha(r)\right]+\left[u_{i}(-s),\alpha(r)\right]u_{i}(s)$
(196) $\displaystyle=\sum_{i=1}^{m}\left(s\left\langle
u_{i},\alpha\right\rangle\delta_{s+r,0}u_{i}(-s)-s\left\langle
u_{i},\alpha\right\rangle\delta_{r-s,0}u_{i}(s)\right)$
$\displaystyle=s\alpha(-s)\delta_{s+r,0}-s\alpha(s)\delta_{r-s,0}.$

Rearranging terms, we get

$\left[L_{\Lambda}(0),\alpha(x)^{\pm}\right]=x\frac{d\alpha(x)^{\pm}}{dx}+\alpha(x)^{\pm}.$
(197)

Similarly

$\left[\bar{L}_{\Lambda}(0),\beta(\bar{x})^{\pm}\right]=\bar{x}\frac{d}{d\bar{x}}\beta(\bar{x})^{\pm}+\beta(\bar{x})^{\pm}.$
(198)

Note that the same proof also shows that

$\begin{split}&\left[L_{\Lambda}(0),\alpha(x)\right]=x\frac{d}{dx}\alpha(x)+\alpha(x)\\\
&\left[\bar{L}_{\Lambda}(0),\beta(\bar{x})\right]=\bar{x}\frac{d}{d\bar{x}}\beta(\bar{x})+\beta(\bar{x}).\end{split}$
(199)

Next using (197) we have

$\displaystyle{\left[L_{\Lambda}(0),\int dx\leavevmode\nobreak\
\alpha(x)^{\pm}\right]}$ $\displaystyle=\int
dx\left[L_{\Lambda}(0),\alpha(x)^{\pm}\right]$ (200) $\displaystyle=\int
dx\left(x\frac{d}{dx}\alpha(x)^{\pm}+\alpha(x)^{\pm}\right)$
$\displaystyle=x\alpha(x)^{\pm},$

where we used integration by parts for the formal integration. Similarly

$\left[\bar{L}_{\Lambda}(0),\int d\bar{x}\leavevmode\nobreak\
\beta(\bar{x})^{\pm}\right]=\bar{x}\beta(\bar{x})^{\pm}.$ (201)

By $\mathrm{BCH}$ formula (224) we have

$\displaystyle L_{\Lambda}(0)\exp\left(\int dx\leavevmode\nobreak\
\alpha(x)^{\pm}\right)$ $\displaystyle=\exp\left(-\int dx\leavevmode\nobreak\
\alpha(x)^{\pm}\right)\sum_{n=0}^{\infty}\frac{1}{n!}\left[\left(\int
dx\leavevmode\nobreak\ \alpha(x)^{\pm}\right)^{n},L_{\Lambda}(0)\right]$ (202)
$\displaystyle=\exp\left(\int dx\leavevmode\nobreak\
\alpha(x)^{\pm}\right)L_{\Lambda}(0)+\sum_{n=1}^{\infty}\frac{1}{n!}\left[\left(-\int
dx\leavevmode\nobreak\ \alpha(x)^{\pm}\right)^{n},L_{\Lambda}(0)\right].$

By (200) and the fact that $[\alpha(r),\alpha(s)]=0$ for $r,s\leq 0$ or
$r,s\geq 0$, we get

$\left[L_{\Lambda}(0),\exp\left(\int dx\leavevmode\nobreak\
\alpha(x)^{\pm}\right)\right]=x\exp\left(\int dx\leavevmode\nobreak\
\alpha(x)^{\pm}\right)\alpha(x)^{\pm}.$ (203)

Similarly

$\left[\bar{L}_{\Lambda}(0),\exp\left(\int d\bar{x}\leavevmode\nobreak\
\beta(x)^{\pm}\right)\right]=\bar{x}\exp\left(\int
d\bar{x}\leavevmode\nobreak\ \beta(\bar{x})^{\pm}\right)\beta(\bar{x})^{\pm}.$
(204)

Finally, it is clear that for
$\lambda^{\prime}=\left(\alpha^{\prime},\beta^{\prime}\right)$ and $u\in
S\left(\mathfrak{h}^{-}\right)$ we have

$\displaystyle{\left[L_{\Lambda}(0),{\rm
e}_{\lambda}x^{\alpha}\right]\left(u\otimes{\rm e}^{\lambda^{\prime}}\right)}$
$\displaystyle=(-1)^{\epsilon\left(\lambda,\lambda^{\prime}\right)}\left(\frac{\left\langle\alpha+\alpha^{\prime},\alpha+\alpha^{\prime}\right\rangle}{2}-\frac{\left\langle\alpha^{\prime},\alpha^{\prime}\right\rangle}{2}\right)x^{\left\langle\alpha,\alpha^{\prime}\right\rangle}\left(u\otimes{\rm
e}^{\lambda+\lambda^{\prime}}\right)$ (205)
$\displaystyle=\left(\frac{\langle\alpha,\alpha\rangle}{2}+\left\langle\alpha,\alpha^{\prime}\right\rangle\right)(-1)^{\epsilon(\lambda,\lambda^{\prime})}x^{\left\langle\alpha,\alpha^{\prime}\right\rangle}\left(u\otimes{\rm
e}^{\lambda+\lambda^{\prime}}\right)$
$\displaystyle=\left(\frac{\langle\alpha,\alpha\rangle}{2}{\rm
e}_{\lambda}x^{\alpha}+{\rm
e}_{\lambda}x\frac{d}{dx}x^{\alpha}\right)\left(u\otimes{\rm
e}^{\lambda^{\prime}}\right)$

Putting all this together, we obtain

$\displaystyle{\left[L_{\Lambda}(0),Y_{V_{\Lambda}}\left({\rm
e}^{\lambda},x,\bar{x}\right)\right]=}$ $\displaystyle
x\frac{d}{dx}\left[\exp\left(\int dx\leavevmode\nobreak\
\alpha(x)^{-}\right)\exp\left(\int dx\leavevmode\nobreak\
\alpha(x)^{+}\right)\right]$ (206) $\displaystyle\times\exp\left(\int
d\bar{x}\leavevmode\nobreak\ \beta(\bar{x})^{-}\right)\exp\left(\int
d\bar{x}\leavevmode\nobreak\ \beta(\bar{x})^{+}\right){\rm
e}_{\lambda}x^{\alpha}\bar{x}^{\beta}$ $\displaystyle+\exp\left(\int
dx\leavevmode\nobreak\ \alpha(x)^{-}\right)\exp\left(\int
dx\leavevmode\nobreak\ \alpha(x)^{+}\right)$
$\displaystyle\times\exp\left(\int d\bar{x}\leavevmode\nobreak\
\beta(\bar{x})^{-}\right)\exp\left(\int d\bar{x}\leavevmode\nobreak\
\beta(\bar{x})^{+}\right)$
$\displaystyle\times\left(\frac{\langle\alpha,\alpha\rangle}{2}{\rm
e}_{\lambda}x^{\alpha}+{\rm e}_{\lambda}x\frac{d}{dx}x^{\alpha}\right)$
$\displaystyle=x\frac{d}{dx}Y_{V_{\Lambda}}\left({\rm
e}^{\lambda},x,\bar{x}\right)+\frac{\left\langle\alpha,\alpha\right\rangle}{2}Y_{V_{\Lambda}}\left({\rm
e}^{\lambda},x,\bar{x}\right).$

Similarly

$\left[\bar{L}_{\Lambda}(0),Y_{V_{\Lambda}}\left({\rm
e}^{\lambda},x,\bar{x}\right)\right]=\bar{x}\frac{d}{d\bar{x}}Y_{V_{\Lambda}}\left({\rm
e}^{\lambda},x,\bar{x}\right)+\frac{\left\langle\beta,\beta\right\rangle}{2}Y_{V_{\Lambda}}\left({\rm
e}^{\lambda},x,\bar{x}\right).$ (207)

For general vertex operators, we observe that

$\displaystyle{\left[L_{\Lambda}(0),\frac{d^{r}}{dx^{r}}\alpha(x)\right]}$
$\displaystyle=\frac{d^{r}}{dx^{r}}\left(x\frac{d}{dx}\alpha(x)+\alpha(x)\right)$
(208)
$\displaystyle=\frac{d^{r-1}}{dx^{r-1}}\left(\frac{d}{dx}\alpha(x)+x\frac{d^{2}}{dx^{2}}\alpha(x)\right)+\frac{d^{r}}{dx^{r}}\alpha(x)$
$\displaystyle=x\frac{d^{r}}{dx^{r}}\alpha(x)+(r+1)\frac{d^{r}}{dx^{r}}\alpha(x).$

This implies that for a general vector of the form (152) we have

$\displaystyle{\left[L_{\Lambda}(0),Y_{V_{\Lambda}}(v,x,\bar{x})\right]}$
$\displaystyle=x\frac{d}{dx}Y_{V_{\Lambda}}(v,x,\bar{x})+\left(\sum_{i=1}^{k}m_{i}+\frac{\langle\alpha,\alpha\rangle}{2}\right)Y_{V_{\Lambda}}(v,x,\bar{x})$
(209)
$\displaystyle=x\frac{d}{dx}Y_{V_{\Lambda}}(v,x,\bar{x})+Y_{V_{\Lambda}}\left(L_{\Lambda}(0)v,x,\bar{x}\right).$

Similarly

$\left[\bar{L}_{\Lambda}(0),Y_{V_{\Lambda}}(v,x,\bar{x})\right]=\bar{x}\frac{d}{d\bar{x}}Y_{V_{\Lambda}}(v,x,\bar{x})+Y_{V_{\Lambda}}\left(\bar{L}_{\Lambda}(0)v,x,\bar{x}\right).$
(210)

Proof of translation property 8: Observe that

$\begin{split}\left[L_{\Lambda}(-1),\alpha(-r)\right]=&\frac{1}{2}\sum_{i=1}^{m}\sum_{s\in\mathbb{Z}}[\typecolon
u_{i}(s)u_{i}(-1-s)\typecolon,\alpha(-r)]\\\
=&\frac{1}{2}\sum_{i=1}^{m}\sum_{s\in\mathbb{Z}}[u_{i}(-1-s)\,u_{i}(s),\alpha(-r)]\\\
=&\frac{1}{2}\sum_{i=1}^{m}\sum_{s\in\mathbb{Z}}\Bigl{(}u_{i}(-1-s)[u_{i}(s),\alpha(-r)]+[u_{i}(-1-s),\alpha(-r)]\,u_{i}(s)\Bigr{)}\\\
=&\frac{1}{2}\sum_{i=1}^{m}\sum_{s\in\mathbb{Z}}\Bigl{(}s\delta_{s-r,0}\langle
u_{i},\alpha\rangle\,u_{i}(-1-s)-(1+s)\delta_{r+s+1,0}\langle
u_{i},\alpha\rangle\,u_{i}(s)\Bigr{)}\\\
=&\frac{1}{2}\Bigl{(}r\,\alpha(-1-r)-(-r)\,\alpha(-1-r)\Bigr{)}=r\,\alpha(-\,r-1).\end{split}$
(211)

Using the above commutator, it is easy to see that

$\begin{split}&L_{\Lambda}(-1)\left(\alpha_{1}(-m_{1})\cdots\alpha_{k}(-m_{k})\,\beta_{1}(-\bar{m}_{1})\cdot\beta_{2}(-\bar{m}_{2})\cdots\beta_{\bar{k}}(-\bar{m}_{\bar{k}})\right)\otimes{\rm
e}^{(\alpha,\beta)}\\\
&=\sum_{i=1}^{m}\left(\alpha_{1}(-m_{1})\cdots\alpha_{k}(-m_{k})\,\beta_{1}(-\bar{m}_{1})\cdots\beta_{\bar{k}}(-\bar{m}_{\bar{k}})\right)\otimes
L_{\Lambda}(-1){\rm e}^{(\alpha,\beta)}\\\
&+\sum_{i=1}^{k}m_{i}\left(\alpha_{1}(-m_{1})\cdots\alpha_{i}(-1-m_{i})\cdots\alpha_{k}(-m_{k})\,\beta_{1}(-\bar{m}_{1})\cdots\beta_{\bar{k}}(-\bar{m}_{\bar{k}})\right)\otimes{\rm
e}^{(\alpha,\beta)}.\end{split}$ (212)

Now since

$\begin{split}L_{\Lambda}(-1){\rm
e}^{(\alpha,\beta)}&=\frac{1}{2}\sum_{i=1}^{m}\sum_{s\in\mathbb{Z}}\typecolon
u_{i}(s)u_{i}(-1-s)\typecolon{\rm e}^{(\alpha,\beta)}\\\
&=\sum_{i=1}^{m}u_{i}(0)u_{i}(-1){\rm e}^{(\alpha,\beta)}\\\
&=\sum_{i=1}^{m}\langle u_{i},\alpha\rangle u_{i}(-1){\rm
e}^{(\alpha,\beta)}\\\ &=\alpha(-1){\rm e}^{(\alpha,\beta)},\end{split}$ (213)

hence we get the action of $L_{\Lambda}(-1)$ on generating vectors $v$ of the
form (152) to be:

$\begin{split}L_{\Lambda}(-1)v&=\sum_{i=1}^{k}m_{i}\left(\alpha_{1}(-m_{1})\cdots\alpha_{i}(-1-m_{i})\cdots\alpha_{k}(-m_{k})\,\beta_{1}(-\bar{m}_{1})\cdots\beta_{\bar{k}}(-\bar{m}_{\bar{k}})\right)\otimes{\rm
e}^{(\alpha,\beta)}\\\
&+\left(\alpha(-1)\alpha_{1}(-m_{1})\cdots\alpha_{i}(-1-m_{i})\cdots\alpha_{k}(-m_{k})\,\beta_{1}(-\bar{m}_{1})\cdots\beta_{\bar{k}}(-\bar{m}_{\bar{k}})\right)\otimes{\rm
e}^{(\alpha,\beta)}.\end{split}$ (214)

The proof of the translation property now follows from exact same calculation
as in (huang, , Proposition 2.2).

####  3.3.1 Proof of locality of vertex operators

In this section, we will prove the locality of two vertex operators and defer
the proof of product of multiple vertex operators to Appendix B. The vertex
operators $Y_{V_{\Lambda}}({\rm e}^{\lambda},x,\bar{x})$ for
$\lambda\in\Lambda_{0}$ satisfy the locality property 9. More precisely there
exists, multi-valued, operator-valued functions $f(z_{1},z_{2})$ and
$g(\bar{z}_{1},\bar{z}_{2})$ analytic in $z_{1},z_{2}$ and
$\bar{z}_{1},\bar{z}_{2}$ respectively with possible singularities at
$\\{(z_{1},z_{2})\in\mathbb{C}^{2}\,\lvert\,z_{1},z_{2}\neq 0,z_{1}\neq
z_{2}\\}$, such that $f(z_{1},z_{2})g(\bar{z}_{1},\bar{z}_{2})$ is single-
valued when $\bar{z}_{1},\bar{z}_{2}$ are the complex conjugates of
$z_{1},z_{2}$ respectively and equals

$\begin{split}Y_{V_{\Lambda}}({\rm
e}^{\lambda},z_{1},\bar{z}_{1})Y_{V_{\Lambda}}({\rm
e}^{\lambda^{\prime}},z_{2},\bar{z}_{2})\ {\textrm{w}hen}\ \lvert
z_{1}\rvert>\lvert z_{2}\rvert\leavevmode\nobreak\ ,\\\ Y_{V_{\Lambda}}({\rm
e}^{\lambda^{\prime}},z_{2},\bar{z}_{2})Y_{V_{\Lambda}}({\rm
e}^{\lambda},z_{1},\bar{z}_{1})\ {\textrm{w}hen}\ \lvert z_{2}\rvert>\lvert
z_{1}\rvert\leavevmode\nobreak\ .\end{split}$ (215)

We will closely follow the proofs of results in (huang, , Section 2). We begin
by proving that

$\begin{split}\left[\alpha(x_{1}),\alpha^{\prime}(x_{2})\right]=\langle\alpha,\alpha^{\prime}\rangle\left[(x_{2}-x_{1})^{-2}-(-x_{2}+x_{1})^{-2}\right]\\\
\left[\beta(\bar{x}_{1}),\beta^{\prime}(\bar{x}_{2})\right]=\langle\beta,\beta^{\prime}\rangle\left[(\bar{x}_{2}-\bar{x}_{1})^{-2}-(-\bar{x}_{2}+\bar{x}_{1})^{-2}\right]\end{split}$
(216)

where
$\lambda=(\alpha,\beta),\lambda^{\prime}=(\alpha^{\prime},\beta^{\prime})$. We
have

$\displaystyle{\left[\alpha(x_{1}),\alpha^{\prime}(x_{2})\right]}$
$\displaystyle=\sum_{r,s\in\mathbb{Z}}[\alpha(r),\alpha^{\prime}(s)]x_{1}^{-r-1}x_{2}^{-s-1}$
(217)
$\displaystyle=\sum_{r,s\in\mathbb{Z}}\langle\alpha,\alpha^{\prime}\rangle\,r\,\delta_{r+s,0}\,x_{1}^{-r-1}x_{2}^{-s-1}$
$\displaystyle=-\langle\alpha,\alpha^{\prime}\rangle\sum_{s\in\mathbb{Z}}s\,x_{1}^{s-1}x_{2}^{-s-1}$
$\displaystyle=-\langle\alpha,\alpha^{\prime}\rangle\frac{\partial}{\partial
x_{1}}\sum_{s\in\mathbb{Z}}x_{1}^{s}x_{2}^{-s-1}$
$\displaystyle=-\langle\alpha,\alpha^{\prime}\rangle\frac{\partial}{\partial
x_{1}}\left(\left(x_{1}-x_{2}\right)^{-1}-\left(-x_{2}+x_{1}\right)^{-1}\right)$
$\displaystyle=\langle\alpha,\alpha^{\prime}\rangle\left(\left(x_{1}-x_{2}\right)^{-2}-\left(-x_{2}+x_{1}\right)^{-2}\right),$

Note that this commutator is also true for complex variables
$x_{1}=z_{1},x_{2}=z_{2}$. Indeed from (30)

$\begin{split}\alpha(z_{1})\alpha^{\prime}(z_{2})&=\typecolon\alpha(z_{1})\alpha(z_{2})\typecolon+\langle\alpha,\alpha^{\prime}\rangle\sum_{s\in\mathbb{Z}}sz_{1}^{-s-1}z_{2}^{s-1}\\\
&=\typecolon\alpha(z_{1})\alpha(z_{2})\typecolon+\frac{\langle\alpha,\alpha^{\prime}\rangle}{(z_{1}-z_{2})^{2}},\quad|z_{1}|>|z_{2}|,\\\
&\end{split}$ (218)

and

$\begin{split}\alpha^{\prime}(z_{2})\alpha(z_{1})&=\typecolon\alpha^{\prime}(z_{2})\alpha(z_{1})\typecolon-\langle\alpha^{\prime},\alpha\rangle\sum_{s\in\mathbb{Z}}sz_{2}^{-s-1}z_{1}^{s-1}\\\
&=\typecolon\alpha^{\prime}(z_{2})\alpha(z_{1})\typecolon-\frac{\langle\alpha^{\prime},\alpha\rangle}{(z_{2}-z_{1})^{2}},\quad|z_{2}|>|z_{1}|.\end{split}$
(219)

It is easy to see that

$\typecolon\alpha(z_{1})\alpha^{\prime}(z_{2})\typecolon=\typecolon\alpha^{\prime}(z_{2})\alpha(z_{1})\typecolon\leavevmode\nobreak\
,$ (220)

which gives us the commutator. In particular

$\left[\alpha(z_{1}),\alpha^{\prime}(z_{2})\right]=0.$ (221)

The other Lie bracket in (216) can be proved similarly. Using (216), we can
show that161616we will use $\exp$ and ${\rm e}$ interchangeably.

$\left[\alpha^{\prime}\left(x_{2}\right)^{-},{\rm
e}^{\int\alpha\left(x_{1}\right)^{+}dx_{1}}\right]=\frac{\langle\alpha,\alpha^{\prime}\rangle}{x_{1}-x_{2}}{\rm
e}^{\int\alpha\left(x_{1}\right)^{+}dx_{1}}\leavevmode\nobreak\ .$ (222)

Integrating both sides of (222) gives us

$\displaystyle\left[-\int\alpha^{\prime}\right.$
$\displaystyle\left.\left(x_{2}\right)^{-}dx_{2},{\rm
e}^{\int\alpha\left(x_{1}\right)^{+}dx_{1}}\right]$ (223)
$\displaystyle=\left(\langle\alpha,\alpha^{\prime}\rangle\log\left(x_{1}-x_{2}\right)-\langle\alpha,\alpha^{\prime}\rangle\log
x_{1}\right){\rm e}^{\int\alpha\left(x_{1}\right)^{+}dx_{1}}$
$\displaystyle=\langle\alpha,\alpha^{\prime}\rangle\left(\,\log(x_{1}-x_{2})-\log(x_{1})\,\right){\rm
e}^{\int\alpha\left(x_{1}\right)^{+}dx_{1}}=\langle\alpha,\alpha^{\prime}\rangle\log\left(1-\frac{x_{2}}{x_{1}}\right){\rm
e}^{\int\alpha\left(x_{1}\right)^{+}dx_{1}}.$

One can write analogous formulas for
$[\beta^{\prime}(\bar{x}_{1})^{\pm},\exp(\int\beta(x_{2})^{\mp})]$. Using the
BCH identity

$\exp(X)Y\exp(-X)=\sum_{s=0}^{\infty}\frac{\left[(X)^{s},Y\right]}{s!},$ (224)

where

$[X^{s},Y]=[\underbrace{X\ldots,[X,[X}_{s\text{
times}},Y]]\ldots],\quad[X^{0},Y]\equiv Y,$ (225)

we get

$\begin{split}\exp\left(-\int dx_{2}\leavevmode\nobreak\
\alpha^{\prime}(x_{2})^{-}\right)\exp\left(\int dx_{1}\leavevmode\nobreak\
\alpha(x_{1})^{+}\right)&\exp\left(\int dx_{2}\leavevmode\nobreak\
\alpha^{\prime}(x_{2})^{-}\right)\\\
&=\left(1-\frac{x_{2}}{x_{1}}\right)^{\langle\alpha,\alpha^{\prime}\rangle}\exp\left(\int
dx_{1}\leavevmode\nobreak\ \alpha(x_{1})^{+}\right).\end{split}$ (226)

To show the locality of vertex operator, we will also require the identities

$\begin{split}x_{1}^{\alpha}{\rm
e}_{\lambda^{\prime}}=x_{1}^{\langle\alpha,\alpha^{\prime}\rangle}{\rm
e}_{\lambda^{\prime}}x_{1}^{\alpha},\\\ \bar{x}_{2}^{\beta}{\rm
e}_{\lambda^{\prime}}=\bar{x}_{2}^{\langle\beta,\beta^{\prime}\rangle}{\rm
e}_{\lambda^{\prime}}\bar{x}_{2}^{\beta},\end{split}$ (227)

the first of which is shown below

$\begin{split}x_{1}^{\alpha}\,{\rm e}_{\lambda^{\prime}}\,(u\otimes{\rm
e}^{\lambda^{\prime\prime}})=(-1)^{\epsilon(\lambda^{\prime},\,\lambda^{\prime\prime})}\,x_{1}^{\alpha}(u\otimes\,{\rm
e}^{\lambda^{\prime}+\lambda^{\prime\prime}})=&(-1)^{\epsilon(\lambda^{\prime},\,\lambda^{\prime\prime})}\,x_{1}^{\langle\alpha,\alpha^{\prime}\rangle+\langle\alpha,\alpha^{\prime\prime}\rangle}(u\otimes\,{\rm
e}^{\lambda^{\prime}+\lambda^{\prime\prime}})\\\ {\rm
e}_{\lambda^{\prime}}\,x_{1}^{\alpha}\,(u\otimes{\rm
e}^{\lambda^{\prime\prime}})=x_{1}^{\langle\alpha,\alpha^{\prime\prime}\rangle}{\rm
e}_{\lambda^{\prime}}\,(u\otimes{\rm
e}^{\lambda^{\prime\prime}})=&(-1)^{\epsilon(\lambda^{\prime},\lambda^{\prime\prime})}x_{1}^{\langle\alpha,\alpha^{\prime\prime}\rangle}\,(u\otimes{\rm
e}^{\lambda^{\prime}+\lambda^{\prime\prime}}).\end{split}$ (228)

We now have171717We will often write $\int dx\leavevmode\nobreak\
\alpha(x)=\int\alpha(x)$ to simplify the expressions.

$\begin{split}&Y_{V_{\Lambda}}({\rm
e}^{\lambda},x_{1},\bar{x}_{1})Y_{V_{\Lambda}}({\rm
e}^{\lambda^{\prime}},x_{2},\bar{x}_{2})\\\
&=\exp\left(\int\alpha(x_{1})^{-}\right)\exp\left(\int\alpha(x_{1})^{+}\right)\exp\left(\int\beta(\bar{x}_{1})^{-}\right)\exp\left(\int\beta(\bar{x}_{1})^{+}\right){\rm
e}_{\lambda}x_{1}^{\alpha}\bar{x}_{1}^{\beta}\\\
&\quad\exp\left(\int\alpha^{\prime}(x_{2})^{-}\right)\exp\left(\int\alpha^{\prime}(x_{2})^{+}\right)\exp\left(\int\beta^{\prime}(\bar{x}_{2})^{-}\right)\exp\left(\int\beta^{\prime}(\bar{x}_{2})^{+}\right){\rm
e}_{\lambda^{\prime}}x_{2}^{\alpha^{\prime}}\bar{x}_{2}^{\beta^{\prime}}\,.\end{split}$
(229)

Now, utilizing (166) and the fact that ${\rm e_{\lambda}}$, $x_{1}^{\alpha},$
and $\bar{x}_{1}^{\beta}$ commute with exponential of integrals

$\begin{split}&=\exp\left(\int\alpha(x_{1})^{-}\right)\left[\exp\left(\int\alpha(x_{1})^{+}\right)\exp\left(\int\alpha^{\prime}(x_{2})^{-}\right)\right]\exp\left(\int\alpha^{\prime}(x_{2})^{+}\right)\\\
&\quad\exp\left(\int\beta(\bar{x}_{1})^{-}\right)\left[\exp\left(\int\beta(\bar{x}_{1})^{+}\right)\exp\left(\int\beta^{\prime}(\bar{x}_{2})^{-}\right)\right]\exp\left(\int\beta^{\prime}(\bar{x}_{2})^{+}\right){\rm
e}_{\lambda}x_{1}^{\alpha}\bar{x}_{1}^{\beta}{\rm
e}_{\lambda^{\prime}}x_{2}^{\alpha^{\prime}}\bar{x}_{2}^{\beta^{\prime}}\
.\end{split}$ (230)

After which we use (226) and (227) to write

$\begin{split}&=\left(1-\frac{x_{2}}{x_{1}}\right)^{\langle\alpha,\alpha^{\prime}\rangle}\left(1-\frac{\bar{x}_{2}}{\bar{x}_{1}}\right)^{\langle\beta,\beta^{\prime}\rangle}x_{1}^{\langle\alpha,\alpha^{\prime}\rangle}\bar{x}_{1}^{\langle\beta,\beta^{\prime}\rangle}\exp\left(\int\alpha(x_{1})^{-}\right)\exp\left(\int\alpha^{\prime}(x_{2})^{-}\right)\\\
&\quad\exp\left(\int\alpha(x_{1})^{+}\right)\exp\left(\int\alpha^{\prime}(x_{2})^{+}\right)\exp\left(\int\beta(\bar{x}_{1})^{-}\right)\exp\left(\int\beta^{\prime}(\bar{x}_{2})^{-}\right)\\\
&\quad\exp\left(\int\beta(\bar{x}_{1})^{+}\right)\exp\left(\int\beta^{\prime}(\bar{x}_{2})^{+}\right){\rm
e}_{\lambda}{\rm
e}_{\lambda^{\prime}}x_{1}^{\alpha}\bar{x}_{1}^{\beta}x_{2}^{\alpha^{\prime}}\bar{x}_{2}^{\beta^{\prime}}\
.\end{split}$ (231)

Finally we use (26) to collect terms to get

$\begin{split}&=\left(x_{1}-x_{2}\right)^{\langle\alpha,\alpha^{\prime}\rangle}\left(\bar{x}_{1}-\bar{x}_{2}\right)^{\langle\beta,\beta^{\prime}\rangle}\exp\left(\int\alpha(x_{1})^{-}\right)\exp\left(\int\alpha^{\prime}(x_{2})^{-}\right)\exp\left(\int\alpha(x_{1})^{+}\right)\\\
&\quad\exp\left(\int\alpha^{\prime}(x_{2})^{+}\right)\exp\left(\int\beta(\bar{x}_{1})^{-}\right)\exp\left(\int\beta^{\prime}(\bar{x}_{2})^{-}\right)\exp\left(\int\beta(\bar{x}_{1})^{+}\right)\exp\left(\int\beta^{\prime}(\bar{x}_{2})^{+}\right)\\\
&\quad{\rm e}_{\lambda}{\rm
e}_{\lambda^{\prime}}x_{1}^{\alpha}\bar{x}_{1}^{\beta}x_{2}^{\alpha^{\prime}}\bar{x}_{2}^{\beta^{\prime}}\\\
&\equiv\left(x_{1}-x_{2}\right)^{\langle\alpha,\alpha^{\prime}\rangle}\left(\bar{x}_{1}-\bar{x}_{2}\right)^{\langle\beta,\beta^{\prime}\rangle}F(x_{1},x_{2})\bar{F}(\bar{x}_{1},\bar{x}_{2}),\end{split}$
(232)

where we used (166), (226) and (227) and
$F(x_{1},x_{2})\bar{F}(\bar{x}_{1},\bar{x}_{2})$ contains the operator part of
$Y_{V_{\Lambda}}({\rm e}^{\lambda},x_{1},\bar{x}_{1})Y_{V_{\Lambda}}({\rm
e}^{\lambda^{\prime}},x_{2},\bar{x}_{2})$. Similarly we have

$\begin{split}&Y_{V_{\Lambda}}({\rm
e}^{\lambda^{\prime}},x_{2},\bar{x}_{2})Y_{V_{\Lambda}}({\rm
e}^{\lambda},x_{1},\bar{x}_{1})\\\
&=\left(x_{2}-x_{2}\right)^{\langle\alpha,\alpha^{\prime}\rangle}\left(\bar{x}_{2}-\bar{x}_{1}\right)^{\langle\beta,\beta^{\prime}\rangle}\exp\left(\int\alpha^{\prime}(x_{2})^{-}\right)\exp\left(\int\alpha(x_{1})^{-}\right)\exp\left(\int\alpha^{\prime}(x_{2})^{+}\right)\\\
&\quad\exp\left(\int\alpha(x_{1})^{+}\right)\exp\left(\int\beta^{\prime}(\bar{x}_{2})^{-}\right)\exp\left(\int\beta(\bar{x}_{1})^{-}\right)\exp\left(\int\beta^{\prime}(\bar{x}_{2})^{+}\right)\exp\left(\int\beta(\bar{x}_{1})^{+}\right)\\\
&\quad(-1)^{\lambda\circ\lambda^{\prime}}{\rm e}_{\lambda}{\rm
e}_{\lambda^{\prime}}x_{1}^{\alpha}\bar{x}_{1}^{\beta}x_{2}^{\alpha^{\prime}}\bar{x}_{2}^{\beta^{\prime}}\\\
&=(-1)^{\lambda\circ\lambda^{\prime}}\left(x_{2}-x_{1}\right)^{\langle\alpha,\alpha^{\prime}\rangle}\left(\bar{x}_{2}-\bar{x}_{1}\right)^{\langle\beta,\beta^{\prime}\rangle}F(x_{1},x_{2})\bar{F}(\bar{x}_{1},\bar{x}_{2}),\end{split}$
(233)

where we used (159). Note that
$(-x_{1}+x_{2})^{\langle\alpha,\alpha^{\prime}\rangle}=(x_{2}-x_{1})^{\langle\alpha,\alpha^{\prime}\rangle}$
when181818Recall that when $s\in\mathbb{C}$, $(-x_{1}+x_{2})^{s}$ is to be
expanded in positive integral powers of $x_{2}$ as in (24).
$\langle\alpha,\alpha^{\prime}\rangle\geq 0$. To prove locality, we take
complex variables $x_{1}=z_{1},x_{2}=z_{2}$ and
$\bar{x}_{1}=\bar{z}_{1},\bar{x}_{2}=\bar{z}_{2}$. Note that when we plug
complex variable in place of formal variable, we must consider
$(x_{1}-x_{2})^{s}$ as a formal series so that

$\begin{split}Y_{V_{\Lambda}}({\rm
e}^{\lambda},z_{1},\bar{z}_{1})Y_{V_{\Lambda}}({\rm
e}^{\lambda^{\prime}},z_{2},\bar{z}_{2})=\left(\sum_{p\geq
0}(-1)^{p}z_{1}^{\langle\alpha,\alpha^{\prime}\rangle-p}z_{2}^{p}\right)&\left(\sum_{q\geq
0}(-1)^{q}\bar{z}_{1}^{\langle\beta,\beta^{\prime}\rangle-q}\bar{z}_{2}^{q}\right)\\\
&\quad\times F(z_{1},z_{2})\bar{F}(\bar{z}_{1},\bar{z}_{2}),\end{split}$ (234)

and similarly $Y_{V_{\Lambda}}({\rm
e}^{\lambda^{\prime}},z_{2},\bar{z}_{2})Y_{V_{\Lambda}}({\rm
e}^{\lambda},z_{1},\bar{z}_{1})$. To complete the proof of locality, consider
the operator valued functions

$\begin{split}f(z_{1},z_{2})=\exp\left(\langle\alpha,\alpha^{\prime}\rangle\log(z_{1}-z_{2})\right)F(z_{1},z_{2}),\\\
g(\bar{z}_{1},\bar{z}_{2})=\exp\left(\langle\beta,\beta^{\prime}\rangle\log(\bar{z}_{1}-\bar{z}_{2})\right)\bar{F}(\bar{z}_{1},\bar{z}_{2})\,.\end{split}$
(235)

Then by (30) for $|z_{1}|>|z_{2}|$ we see that

$f(z_{1},z_{2})g(\bar{z}_{1},\bar{z}_{2})=\left(\sum_{p\geq
0}(-1)^{p}z_{1}^{\langle\alpha,\alpha^{\prime}\rangle-p}z_{2}^{p}\right)\left(\sum_{q\geq
0}(-1)^{q}\bar{z}_{1}^{\langle\beta,\beta^{\prime}\rangle-q}\bar{z}_{2}^{q}\right)F(z_{1},z_{2})\bar{F}(\bar{z}_{1},\bar{z}_{2}).$
(236)

For $|z_{2}|>|z_{1}|$ we have

$\begin{split}f(z_{1}&,z_{2})g(\bar{z}_{1},\bar{z}_{2})=\exp\left(\langle\alpha,\alpha^{\prime}\rangle\log(-(z_{2}-z_{1}))\right)\exp\left(\langle\beta,\beta^{\prime}\rangle\log(-(\bar{z}_{2}-\bar{z}_{1}))\right)\\\
=&\,{\rm
e}^{i\pi(\langle\alpha,\alpha^{\prime}\rangle-\langle\beta,\beta^{\prime}\rangle)}\exp\left(\langle\alpha,\alpha^{\prime}\rangle\log(z_{2}-z_{1})\right)\exp\left(\langle\beta,\beta^{\prime}\rangle\log(\bar{z}_{2}-\bar{z}_{1})\right)F(z_{1},z_{2})\bar{F}(\bar{z}_{1},\bar{z}_{2})\\\
=&(-1)^{\lambda\circ\lambda^{\prime}}\left(\sum_{p\geq
0}(-1)^{p}z_{2}^{\langle\alpha,\alpha^{\prime}\rangle-p}z_{1}^{p}\right)\left(\sum_{q\geq
0}(-1)^{q}\bar{z}_{2}^{\langle\beta,\beta^{\prime}\rangle-q}\bar{z}_{1}^{q}\right)F(z_{1},z_{2})\bar{F}(\bar{z}_{1},\bar{z}_{2}),\end{split}$
(237)

where we used the fact that in the principal branch of logarithm to write

$\log(-z)=\log|z|+i(\pi+\text{Arg}(z)),\quad\log(-\bar{z})=\log|z|-i(\pi+\text{Arg}(z))$
(238)

with

$-\pi<\pi+\text{Arg}(z)<\pi.$ (239)

From the calculations above, it is easy to see that the following formal
commutativity axiom holds for the vertex operators: there exists
$K,\bar{K}\in\mathbb{N}$ such that

$(x_{1}-x_{2})^{K}(\bar{x}_{1}-\bar{x}_{2})^{\bar{K}}\left[Y_{V_{\Lambda}}({\rm
e}^{\lambda},x_{1},\bar{x}_{1}),Y_{V_{\Lambda}}({\rm
e}^{\lambda^{\prime}},x_{2},\bar{x}_{2})\right]=0.$ (240)

Indeed we have

$\begin{split}\left[Y_{V_{\Lambda}}({\rm
e}^{\lambda},x_{1},\bar{x}_{1}),\right.&\left.Y_{V_{\Lambda}}({\rm
e}^{\lambda^{\prime}},x_{2},\bar{x}_{2})\right]=\left(\left(x_{1}-x_{2}\right)^{\langle\alpha,\alpha^{\prime}\rangle}\left(\bar{x}_{1}-\bar{x}_{2}\right)^{\langle\beta,\beta^{\prime}\rangle}\right.\\\
&\left.-(-1)^{\lambda\circ\lambda^{\prime}}\left(x_{2}-x_{1}\right)^{\langle\alpha,\alpha^{\prime}\rangle}\left(\bar{x}_{2}-\bar{x}_{1}\right)^{\langle\beta,\beta^{\prime}\rangle}\right)F(x_{1},x_{2})\bar{F}(\bar{x}_{1},\bar{x}_{2})\\\
=&\left(\left(x_{1}-x_{2}\right)^{\langle\alpha,\alpha^{\prime}\rangle}\left(\bar{x}_{1}-\bar{x}_{2}\right)^{\langle\beta,\beta^{\prime}\rangle}\right.\\\
&\left.-\left(-x_{2}+x_{1}\right)^{\langle\alpha,\alpha^{\prime}\rangle}\left(-\bar{x}_{2}+\bar{x}_{1}\right)^{\langle\beta,\beta^{\prime}\rangle}\right)F(x_{1},x_{2})\bar{F}(\bar{x}_{1},\bar{x}_{2})\end{split}$
(241)

Since $(\alpha,\beta),(\alpha^{\prime},\beta^{\prime})\in\Lambda_{0}$, we have
$\langle\alpha,\alpha^{\prime}\rangle,\langle\beta,\beta^{\prime}\rangle\in\mathbb{Z}$
and we can choose $K,\bar{K}\in\mathbb{N}$ large enough such that

$K+\langle\alpha,\alpha^{\prime}\rangle\in\mathbb{N},\quad\bar{K}+\langle\beta,\beta^{\prime}\rangle\in\mathbb{N}.$
(242)

We then get

$\begin{split}(x_{1}-x_{2})^{K}(\bar{x}_{1}-\bar{x}_{2})^{\bar{K}}&\left[Y_{V_{\Lambda}}({\rm
e}^{\lambda},x_{1},\bar{x}_{1}),Y_{V_{\Lambda}}({\rm
e}^{\lambda^{\prime}},x_{2},\bar{x}_{2})\right]\\\
=\left((x_{1}-x_{2})^{K+\langle\alpha,\alpha^{\prime}\rangle}\right.&\left.(\bar{x}_{1}-\bar{x}_{2})^{\bar{K}+\langle\beta,\beta^{\prime}\rangle}\right.\\\
-&\left.\left(-x_{2}+x_{1}\right)^{K+\langle\alpha,\alpha^{\prime}\rangle}\left(-\bar{x}_{2}+\bar{x}_{1}\right)^{\bar{K}+\langle\beta,\beta^{\prime}\rangle}\right)F(x_{1},x_{2})\bar{F}(\bar{x}_{1},\bar{x}_{2})\\\
=\left((x_{1}-x_{2})^{K+\langle\alpha,\alpha^{\prime}\rangle}\right.&\left.(\bar{x}_{1}-\bar{x}_{2})^{\bar{K}+\langle\beta,\beta^{\prime}\rangle}\right.\\\
-&\left.\left(x_{1}-x_{2}\right)^{K+\langle\alpha,\alpha^{\prime}\rangle}\left(\bar{x}_{1}-\bar{x}_{2}\right)^{\bar{K}+\langle\beta,\beta^{\prime}\rangle}\right)F(x_{1},x_{2})\bar{F}(\bar{x}_{1},\bar{x}_{2})\\\
=0.\end{split}$ (243)

We now prove the locality for general vertex operators. The vertex operators
$Y_{V_{\Lambda}}(v,x,\bar{x})$, where $v$ is the general vector of
$V_{\Lambda}$, satisfy the locality property 9. More precisely there exists,
multi-valued, operator-valued functions $f(z_{1},z_{2})$ and
$g(\bar{z}_{1},\bar{z}_{2})$ analytic in $z_{1},z_{2}$ and
$\bar{z}_{1},\bar{z}_{2}$ respectively with possible singularities at
$\\{(z_{1},z_{2})\in\mathbb{C}^{2}\,\lvert\,z_{1},z_{2}\neq 0,z_{1}\neq
z_{2}\\}$, such that $f(z_{1},z_{2})g(\bar{z}_{1},\bar{z}_{2})$ is single-
valued when $\bar{z}_{1},\bar{z}_{2}$ are the complex conjugates of
$z_{1},z_{2}$ respectively and equals

$\begin{split}Y_{V_{\Lambda}}(v,z_{1},\bar{z}_{1})Y_{V_{\Lambda}}(w,z_{2},\bar{z}_{2})\
{\textrm{w}hen}\ \lvert z_{1}\rvert>\lvert z_{2}\rvert,\\\
Y_{V_{\Lambda}}(w,z_{2},\bar{z}_{2})Y_{V_{\Lambda}}(v,z_{1},\bar{z}_{1})\
{\textrm{w}hen}\ \lvert z_{2}\rvert>\lvert z_{1}\rvert.\end{split}$ (244)

We will prove the locality for the spanning set of vectors of the form (152).
Explicitly, we will prove the locality for vertex operators of the form

$\begin{split}&Y_{V_{\Lambda}}(v,x,\bar{x})=\typecolon\prod_{r=1}^{k}\prod_{s=1}^{\bar{k}}\left(\frac{1}{(m_{r}-1)!}\frac{d^{m_{r}-1}\alpha_{r}(x)}{dx^{m_{r}-1}}\right)\left(\frac{1}{(\bar{m}_{s}-1)!}\frac{d^{\bar{m}_{s}-1}\beta_{s}(\bar{x})}{d\bar{x}^{\bar{m}_{s}-1}}\right)Y_{V_{\Lambda}}({\rm
e}^{\lambda},x,\bar{x})\typecolon\\\
&Y_{V_{\Lambda}}(w,x,\bar{x})=\typecolon\prod_{p=1}^{\ell}\prod_{q=1}^{\bar{\ell}}\left(\frac{1}{(n_{p}-1)!}\frac{d^{n_{p}-1}\alpha_{p}^{\prime}(x)}{dx^{n_{p}-1}}\right)\left(\frac{1}{(\bar{n}_{q}-1)!}\frac{d^{\bar{n}_{q}-1}\beta_{q}^{\prime}(\bar{x})}{d\bar{x}^{\bar{n}_{q}-1}}\right)Y_{V_{\Lambda}}({\rm
e}^{\lambda},x,\bar{x})\typecolon,\end{split}$ (245)

see Dolan:1989vr  for a similar calculation. Following the exact same steps as
in the proof of (huang, , Eq. (2.14)) with appropriate modifications, we can
show that

$\left[\alpha^{\prime}\left(x_{1}\right)^{+},{\rm
e}^{\int\alpha\left(x_{2}\right)^{-}dx_{2}}\right]=\left(\frac{\langle\alpha,\alpha^{\prime}\rangle}{x_{1}-x_{2}}-\frac{\langle\alpha,\alpha^{\prime}\rangle}{x_{1}}\right){\rm
e}^{\int\alpha\left(x_{1}\right)^{-}dx_{1}}.$ (246)

Differentiating on both the sides of (246) with respect to $x_{1}$ we obtain

$\left[\frac{1}{s!}\frac{d^{s}\alpha^{\prime}\left(x_{1}\right)^{+}}{dx_{1}^{s}},{\rm
e}^{\int\alpha\left(x_{2}\right)^{-}dx_{2}}\right]=(-1)^{s}\left(\frac{\langle\alpha,\alpha^{\prime}\rangle}{(x_{1}-x_{2})^{s+1}}-\frac{\langle\alpha,\alpha^{\prime}\rangle}{x_{1}^{s+1}}\right){\rm
e}^{\int\alpha\left(x_{1}\right)^{-}dx_{1}}.$ (247)

Differentiating both sides of (222) with respect to $x_{2}$ we obtain

$\left[\frac{1}{s!}\frac{d^{s}\alpha^{\prime}\left(x_{2}\right)^{-}}{dx_{2}^{s}},{\rm
e}^{\int\alpha\left(x_{1}\right)^{+}dx_{1}}\right]=\frac{\langle\alpha,\alpha^{\prime}\rangle}{(x_{1}-x_{2})^{s+1}}{\rm
e}^{\int\alpha\left(x_{1}\right)^{+}dx_{1}}.$ (248)

Analogous formula holds for
$[\beta^{\prime}(\bar{x}_{1})^{\pm},\exp(\int\beta(x_{2})^{\mp})]$. In
addition, we need

$\alpha(0){\rm
e}_{\lambda^{\prime}}x^{\alpha^{\prime}}=\langle\alpha,\alpha^{\prime}\rangle{\rm
e}_{\lambda^{\prime}}x^{\alpha^{\prime}}+{\rm
e}_{\lambda^{\prime}}x^{\alpha^{\prime}}\alpha(0),\quad\lambda^{\prime}=(\alpha^{\prime},\beta^{\prime}).$
(249)

This follows from the following calculation: for $u\in
S(\hat{\mathfrak{h}}^{-})$,
$\lambda^{\prime}=(\alpha^{\prime},\beta^{\prime}),\lambda^{\prime\prime}=(\alpha^{\prime\prime},\beta^{\prime\prime})$
we have

$\displaystyle\alpha(0){\rm
e}_{\lambda^{\prime}}x^{\alpha^{\prime}}\left(u\otimes{\rm
e}^{\lambda^{\prime\prime}}\right)$ (250)
$\displaystyle\quad=(-1)^{\epsilon(\lambda^{\prime},\lambda^{\prime\prime})}x^{\langle\alpha^{\prime},\alpha^{\prime\prime}\rangle}\langle\alpha,\alpha^{\prime}+\alpha^{\prime\prime}\rangle\left(u\otimes{\rm
e}^{\lambda^{\prime}+\lambda^{\prime\prime}}\right)$
$\displaystyle\quad=(-1)^{\epsilon(\lambda^{\prime},\lambda^{\prime\prime})}x^{\langle\alpha^{\prime},\alpha^{\prime\prime}\rangle}\langle\alpha,\alpha^{\prime}\rangle\left(u\otimes{\rm
e}^{\lambda^{\prime}+\lambda^{\prime\prime}}\right)+(-1)^{\epsilon(\lambda^{\prime},\lambda^{\prime\prime})}x^{\langle\alpha^{\prime},\alpha^{\prime\prime}\rangle}\langle\alpha,\alpha^{\prime\prime}\rangle\left(u\otimes{\rm
e}^{\lambda^{\prime}+\lambda^{\prime\prime}}\right)$
$\displaystyle\quad=\langle\alpha,\alpha^{\prime}\rangle\,{\rm
e}_{\lambda^{\prime}}x^{\alpha^{\prime}}\left(u\otimes{\rm
e}^{\lambda^{\prime\prime}}\right)+{\rm
e}_{\lambda^{\prime}}x^{\alpha^{\prime}}\alpha(0)\left(u\otimes{\rm
e}^{\lambda^{\prime\prime}}\right).$

Analogous formulas for $\beta(0){\rm
e}_{\lambda^{\prime}}\bar{x}^{\beta^{\prime}}$ is

$\beta(0){\rm
e}_{\lambda^{\prime}}\bar{x}^{\beta^{\prime}}=\langle\beta,\beta^{\prime}\rangle\,{\rm
e}_{\lambda^{\prime}}\bar{x}^{\beta^{\prime}}+{\rm
e}_{\lambda^{\prime}}\bar{x}^{\beta^{\prime}}\beta(0),$ (251)

which can be proved as follows:

$\begin{split}\beta(0)&{\rm
e}_{\lambda^{\prime}}\bar{x}^{\beta^{\prime}}\left(u\otimes{\rm
e}^{\lambda^{\prime\prime}}\right)\\\
=&\,(-1)^{\epsilon(\lambda^{\prime},\lambda^{\prime\prime})}\bar{x}^{\langle\beta^{\prime},\beta^{\prime\prime}\rangle}\,\langle\beta,\beta^{\prime}+\beta^{\prime\prime}\rangle\left(u\otimes{\rm
e}^{\lambda^{\prime}+\lambda^{\prime\prime}}\right)\\\
=&\left(\,(-1)^{\epsilon(\lambda^{\prime},\lambda^{\prime\prime})}\bar{x}^{\langle\beta^{\prime},\beta^{\prime\prime}\rangle}\,\langle\beta,\beta^{\prime}\rangle\left(u\otimes{\rm
e}^{\lambda^{\prime}+\lambda^{\prime\prime}}\right)+(-1)^{\epsilon(\lambda^{\prime},\lambda^{\prime\prime})}\bar{x}^{\langle\beta^{\prime},\beta^{\prime\prime}\rangle}\,\langle\beta,\beta^{\prime\prime}\rangle\left(u\otimes{\rm
e}^{\lambda^{\prime}+\lambda^{\prime\prime}}\right)\right)\\\
=\,&\langle\beta,\beta^{\prime}\rangle\,{\rm
e}_{\lambda^{\prime}}\bar{x}^{\beta^{\prime}}\left(u\otimes{\rm
e}^{\lambda^{\prime\prime}}\right)+{\rm
e}_{\lambda^{\prime}}\bar{x}^{\beta^{\prime}}\beta(0)\left(u\otimes{\rm
e}^{\lambda^{\prime\prime}}\right)\ .\end{split}$ (252)

Let us now consider the product of two vertex operators, as in (244). Using
the normal ordering from (171) we have

$\begin{split}&Y_{V_{\Lambda}}(v,x_{1},\bar{x}_{1})Y_{V_{\Lambda}}(w,x_{2},\bar{x}_{2})=\exp\left(\int\alpha(x_{1})^{-}\right)\exp\left(\int\beta(\bar{x}_{1})^{-}\right)\\\
&\times\typecolon\prod_{r=1}^{k}\prod_{s=1}^{\bar{k}}\left(\frac{1}{(m_{r}-1)!}\frac{d^{m_{r}-1}\alpha_{r}(x_{1})}{dx_{1}^{m_{r}-1}}\right)\left(\frac{1}{(\bar{m}_{s}-1)!}\frac{d^{\bar{m}_{s}-1}\beta_{s}(\bar{x}_{1})}{d\bar{x}_{1}^{\bar{m}_{s}-1}}\right)\typecolon\\\
&\times\exp\left(\int\alpha(x_{1})^{+}\right)\exp\left(\int\alpha^{\prime}(x_{2})^{-}\right)\exp\left(\int\beta(\bar{x}_{1})^{+}\right)\exp\left(\int\beta^{\prime}(\bar{x}_{2})^{-}\right){\rm
e}_{\lambda}x_{1}^{\alpha}\bar{x}_{1}^{\beta}\\\
&\times\typecolon\prod_{p=1}^{\ell}\prod_{q=1}^{\bar{\ell}}\left(\frac{1}{(n_{p}-1)!}\frac{d^{n_{p}-1}\alpha_{p}^{\prime}(x_{2})}{dx_{2}^{n_{p}-1}}\right)\left(\frac{1}{(\bar{n}_{q}-1)!}\frac{d^{\bar{n}_{q}-1}\beta_{q}^{\prime}(\bar{x}_{2})}{d\bar{x}_{2}^{\bar{n}_{q}-1}}\right)\typecolon\\\
&\times\exp\left(\int\alpha^{\prime}(x_{2})^{+}\right)\exp\left(\int\beta^{\prime}(\bar{x}_{2})^{+}\right){\rm
e}_{\lambda^{\prime}}x_{2}^{\alpha^{\prime}}\bar{x}_{2}^{\beta^{\prime}}\,,\end{split}$
(253)

where we have used that ${\rm e}_{\lambda}x_{1}^{\alpha}\bar{x}_{1}^{\beta}$
commutes with the exponential of integrals and the exponential of $\alpha$ and
$\beta$ commute with each other. Now, using (226) we get

$\begin{split}&Y_{V_{\Lambda}}(v,x_{1},\bar{x}_{1})Y_{V_{\Lambda}}(w,x_{2},\bar{x}_{2})=\left(1-\frac{x_{2}}{x_{1}}\right)^{\langle\alpha,\alpha^{\prime}\rangle}\left(1-\frac{\bar{x}_{2}}{\bar{x}_{1}}\right)^{\langle\beta,\beta^{\prime}\rangle}\exp\left(\int\alpha(x_{1})^{-}\right)\exp\left(\int\beta(\bar{x}_{1})^{-}\right)\\\
&\times\typecolon\prod_{r=1}^{k}\prod_{s=1}^{\bar{k}}\left(\frac{1}{(m_{r}-1)!}\frac{d^{m_{r}-1}\alpha_{r}(x_{1})}{dx_{1}^{m_{r}-1}}\right)\left(\frac{1}{(\bar{m}_{s}-1)!}\frac{d^{\bar{m}_{s}-1}\beta_{s}(\bar{x}_{1})}{d\bar{x}_{1}^{\bar{m}_{s}-1}}\right)\typecolon\\\
&\times\exp\left(\int\alpha^{\prime}(x_{2})^{-}\right)\exp\left(\int\beta^{\prime}(\bar{x}_{2})^{-}\right)\exp\left(\int\alpha(x_{1})^{+}\right)\exp\left(\int\beta(\bar{x}_{1})^{+}\right){\rm
e}_{\lambda}x_{1}^{\alpha}\bar{x}_{1}^{\beta}\\\
&\times\typecolon\prod_{p=1}^{\ell}\prod_{q=1}^{\bar{\ell}}\left(\frac{1}{(n_{p}-1)!}\frac{d^{n_{p}-1}\alpha_{p}^{\prime}(x_{2})}{dx_{2}^{n_{p}-1}}\right)\left(\frac{1}{(\bar{n}_{q}-1)!}\frac{d^{\bar{n}_{q}-1}\beta_{q}^{\prime}(\bar{x}_{2})}{d\bar{x}_{2}^{\bar{n}_{q}-1}}\right)\typecolon\\\
&\times\exp\left(\int\alpha^{\prime}(x_{2})^{+}\right)\exp\left(\int\beta^{\prime}(\bar{x}_{2})^{+}\right){\rm
e}_{\lambda^{\prime}}x_{2}^{\alpha^{\prime}}\bar{x}_{2}^{\beta^{\prime}}\,.\end{split}$
(254)

Further, using (247), (248), (249), and (251) successively on the product in
normal order, (227), and the formal variable identity (26) we get

$\begin{split}&Y_{V_{\Lambda}}(v,x_{1},\bar{x}_{1})Y_{V_{\Lambda}}(w,x_{2},\bar{x}_{2})=\left(x_{1}-x_{2}\right)^{\langle\alpha,\alpha^{\prime}\rangle}\left(\bar{x}_{1}-\bar{x}_{2}\right)^{\langle\beta,\beta^{\prime}\rangle}\\\