# A lattice theoretical interpretation of generalized deep holes of the Leech
lattice vertex operator algebra

Ching Hung Lam(1) 111Partially supported by grant AS-IA-107-M02 of Academia
Sinica and MoST grant 110-2115-M-001-011-MY3 of Taiwan and Masahiko
Miyamoto(2) 222Partially supported by the Grants-in-Aids for Scientific
Research, No.18K18708, The Ministry of Education, Science and Culture, Japan.
$(1)$ Institute of Mathematics, Academia Sinica, Taipei 10617, Taiwan $(2)$
Institute of Mathematics, University of Tsukuba, Tsukuba, Ibaraki 305-8571,
Japan

###### Abstract

We give a lattice theoretical interpretation of generalized deep holes of the
Leech lattice VOA $V_{\Lambda}$. We show that a generalized deep hole defines
a “true” automorphism invariant deep hole of the Leech lattice. We also show
that there is a correspondence between the set of isomorphism classes of
holomorphic VOA $V$ of central charge $24$ having non-abelian $V_{1}$ and the
set of equivalence classes of pairs $(\tau,\tilde{\beta})$ satisfying certain
conditions, where $\tau\in Co_{0}$ and $\tilde{\beta}$ is a $\tau$-invariant
deep hole of squared length $2$. It provides a new combinatorial approach
towards the classification of holomorphic VOAs of central charge $24$. In
particular, we give an explanation for an observation of G. Höhn, which
relates the weight one Lie algebras of holomorphic VOAs of central charge $24$
to certain codewords associated with the glue codes of Niemeier lattices.

##  1 Introduction

The classification of strongly regular holomorphic vertex operator algebras
(abbreviated as VOA) of central charge $24$ with non-trivial weight one space
is basically completed. It has been shown that there are exactly $70$ strongly
regular holomorphic VOAs of CFT-type with central charge $24$ and non-zero
weight one space and their VOA structures are uniquely determined by the Lie
algebra structures of their weight one space. Moreover, the possible Lie
algebra structures for their weight one subspaces have been given in
Schellekens’ list [44]. The main tool is the so-called orbifold construction
[6, 10, 17, 19, 35, 36, 37]. Nevertheless, their constructions and the
uniqueness proofs were done by case by case analysis [9, 10, 13, 17, 18, 19,
22, 23, 25, 26, 27, 28, 29, 30, 31, 32, 36, 41]. A simplified and uniform
construction and proof of uniqueness is somehow expected. Recently, it is
proved in [16] (see also [7]) that for any holomorphic VOA of central charge
$24$ with a semisimple weight one Lie algebra, the VOA obtained by an orbifold
construction by an inner automorphism $g$ defined by a $W$-element is always
isomorphic to the Leech lattice VOA $V_{\Lambda}$. By taking its reverse
automorphism $\tilde{g}$ of $V_{\Lambda}$, there is a direct orbifold
construction from the Leech lattice VOA $V_{\Lambda}$ to $V$. As a
consequence, a relatively simpler proof for the Schellekens’ list is obtained.
Möller and Scheithauer [38] called an automorphism $\tilde{g}\in{\rm
Aut}(V_{\Lambda})$ _a generalized deep hole_ with a slightly different
definition. They obtained a classification of generalized deep holes and gave
a new proof for the classification of holomorphic VOA of central charge 24
with non-trivial weight one spaces; we do not need these classification
results in this article.

As it is well-known, a deep hole of a lattice $L$ is an element $v$ of
${\mathbb{R}}L$ such that a distance $\min\\{|\\!|\alpha-v|\\!|\mid\alpha\in
L\\}$ from $L$ is the largest (covering radius) among elements $v$ in
${\mathbb{R}}L$ and so it has a geometrical meaning. Furthermore, deep holes
of the Leech lattice $\Lambda$ have many interesting geometrical and algebraic
meanings. Therefore, it is natural to expect some geometrical properties for
generalized deep holes.

Recall that any automorphism $\tilde{g}\in{\rm Aut}(V_{\Lambda})$ can be
written as

$\tilde{g}=\widehat{\tau}\exp(2\pi i\beta(0)),$

where $\tau\in Co.0=O(\Lambda)$, $\beta\in{\mathbb{R}}\Lambda^{\tau}$ and
$\widehat{\tau}$ denotes a standard lift of $\tau$ in $O(\widehat{\Lambda})$
[2, 14]. In this paper, we will show that a holomorphic VOA of central charge
$24$ with non-abelian weight one Lie algebra naturally defines a pair
$(\tau,\tilde{\beta})$, where $\tau\in Co_{0}$ is the same isometry as defined
by $\tilde{g}$ and $\tilde{\beta}$ is a ($\tau$-invariant) deep hole of the
Leech lattice related to $\beta$. In particular, a generalized deep hole for
the Leech lattice VOA defines a “true” deep hole of the Leech lattice. It
provides a new combinatorial approach towards the classification of
holomorphic VOAs of central charge $24$. The key observation is that there is
a “duality” associated with the lattice $\Lambda^{\tau}$ which changes the
level; we call it $\ell$-_duality_.

######  Main Theorem 1 (see Theorem 4.2).

If $\tilde{g}=\widehat{\tau}\exp(2\pi i\beta(0))$ is a generalized deep hole,
then there is an isometry:

$\varphi_{\tau}:\sqrt{\ell}(\Lambda^{\tau})^{\ast}\to\Lambda^{\tau},$

where $\ell=|\widehat{\tau}|$.

Using this $\ell$-duality $\varphi$ (extended to
${\mathbb{C}}\Lambda^{\tau}$), it is easy to show that
$\langle\varphi_{\tau}(\sqrt{\ell}\beta),\varphi_{\tau}(\sqrt{\ell}\beta)\rangle\in
2{\mathbb{Z}}$ and its neighbor lattice
$N=\Lambda_{\varphi_{\tau}(\sqrt{\ell}\beta)}+{\mathbb{Z}}\varphi_{\tau}(\sqrt{\ell}\beta)$
defines a Niemeier lattice with $N\ncong\Lambda$.

Let $V=(V_{\Lambda})^{[\tilde{g}]}$ be the VOA obtained by an orbifold
construction from $V_{\Lambda}$ by $\tilde{g}$. Let ${\cal H}$ be a Cartan
subalgebra of $V_{1}$. Then there is an even lattice $L$ such that

${\rm Comm}({\rm Comm}(M({\cal H}),V),V)\cong V_{L}.$

The lattice $L$ (or $L^{*}$) encoded the information of the root system of
$V_{1}$.

######  Main Theorem 2 (see Theorem 4.8).

Via $\ell$-duality, we have
$N^{\tau}\cong\sqrt{\ell}\varphi_{\tau}(L^{\ast})$, where $L^{\ast}$ denotes
the dual lattice of $L$.

In addition, we prove the following theorem.

######  Main Theorem 3 (see Theorem 5.1 and Proposition 6.1).

Let $\varphi$ be the isometry defined in the Main Theorem 1. Then we can
choose $\beta\in{\mathbb{C}}\Lambda^{\tau}$ such that
$\tilde{g}=\widehat{\tau}\exp(2\pi i\beta(0))$ and
$\tilde{\beta}=\sqrt{\ell}\varphi(\beta)$ is a deep hole of $\Lambda$ of
squared length $2$.

The classification of generalized deep holes is thus equivalent to the
classification of the pairs $(\tau,\tilde{\beta})$ with $\tau\in O(\Lambda)$
and $\tilde{\beta}$ a $\tau$-invariant deep hole of squared length $2$
satisfying certain conditions and up to some equivalence.

Let ${\cal T}$ be the set of pairs $(\tau,\tilde{\beta})$ satisfying certain
conditions (see Section 7, (C1)–(C3)). We define a relation $\sim$ on ${\cal
T}$ as follows:
$(\tau,\tilde{\beta})\sim(\tau^{\prime},\tilde{\beta}^{\prime})$ if and only
if
(1) $\tilde{\beta}$ and $\tilde{\beta}^{\prime}$ are equivalent deep holes of
the Leech lattice $\Lambda$, i.e., there are $\sigma\in O(\Lambda)$ and
$\lambda\in\Lambda$ such that
$\tilde{\beta}^{\prime}=\sigma(\tilde{\beta}-\lambda)$;
(2) $\tau$ is conjugate to $\sigma^{-1}\tau^{\prime}\sigma$ in $O(N)$.

The following is another main theorem of this article.

######  Main Theorem 4 (see Theorem 7.1).

There is a one-to-one correspondence between the set of isomorphism classes of
holomorphic VOA $V$ of central charge $24$ having non-abelian $V_{1}$ and the
set ${\cal T}/\sim$ of equivalence classes of pairs $(\tau,\tilde{\beta})$ by
$\sim$.

Since a deep hole determines a unique Niemeier lattice, up to isometry and
there are only $23$ Niemeier lattices with non-trivial root system, it is
straightforward to list all possible choices for $(\tau,\tilde{\beta})$ and to
determine the corresponding Lie algebra structures for $V_{1}$. Therefore, one
can complete the classification of holomorphic VOAs of central charge 24 with
non-abelian $V_{1}$ by a purely combinatorial method. Indeed, we will provide
an explanation for an observation of Höhn, which relates the weight one Lie
algebras of holomorphic VOAs of central charge $24$ to certain codewords
associated with the glue codes of Niemeier lattices [21, Theorem 3.1 and Table
3] (see Section 8 for details).

##  2 Some previous results

We first recall several results from [7]. Let $V$ be a holomorphic VOA of
central charge 24 with $V_{1}\not=0$. Suppose $V_{1}=\oplus_{j=1}^{t}{\cal
G}_{j,k_{j}}$ is a semisimple Lie algebra. We use $k_{j}$ (resp.
$h^{\vee}_{j}$ and $r_{j}$) to denote the level (resp. the dual Coxeter number
and the lace number) of ${\cal G}_{j}$, $j=1,\dots,t$. Let ${\cal H}$ be a
Cartan subalgebra of $V_{1}$. It is easy to see that ${\rm Comm}({\rm
Comm}(M({\cal H}),V),V)$ is isomorphic to a lattice VOA $V_{L}$ for an even
lattice $L\subseteq{\cal H}$. Let $\rho_{j}$ be a Weyl vector of ${\cal
G}_{j,k_{j}}$ and set $\alpha=\sum_{j=1}^{t}\rho_{j}/h_{j}^{\vee}$, which we
call a W-element. From the property of lattice VOA, we have a decomposition

$V=\oplus_{\delta\in L^{\ast}}U(\delta)\otimes M({\cal H})e^{\delta},$

where $U(\delta)$ are ${\rm Comm}(M({\cal H}),V)$-modules and $M({\cal
H})e^{\delta}$ are $M({\cal H})$-module. Since $V$ is holomorphic, we have
$U(\delta)\not=0$ for every $\delta\in L^{\ast}$.

Set $g=\exp(2\pi i\alpha(0))$. In [7] (see also [16]), it is proved the VOA
$V^{[\alpha]}$ obtained by an orbifold construction from $V$ by the inner
automorphism $g$ is isomorphic to the Leech lattice VOA, i.e.,
$V^{[\alpha]}\cong V_{\Lambda}$. Namely, the integer weights submodule
$T^{1}_{{\mathbb{Z}}}$ of the simple $g$-twisted $V$-module $T^{1}$ is nonzero
and $V^{g}\oplus
T^{1}_{{\mathbb{Z}}}\oplus\cdots\oplus(T^{1}_{{\mathbb{Z}}})^{\boxtimes(|g|-1)}\cong
V_{\Lambda}$.

In this case, there is an automorphism $\tilde{g}$ of $V_{\Lambda}$ acting on
$(T^{1}_{{\mathbb{Z}}})^{\boxtimes m}$ as $e^{2\pi im/n}$, $n=|g|$. Moreover,
the orbifold construction from $V_{\Lambda}$ by an automorphism $\tilde{g}$
gives $V$, i.e., $V_{\Lambda}^{[\tilde{g}]}\cong V$. That $\tilde{g}\in{\rm
Aut}(V^{[g]})$ is called the reverse automorphism of $g$.

Since $\tilde{g}\in{\rm Aut}(V_{\Lambda})$, we have
$\tilde{g}=\hat{\tau}\exp(2\pi i\beta(0))$ for some $\tau\in O(\Lambda)=Co.0$
and $\beta\in{\mathbb{C}}\Lambda$ [2, 14]. By choosing a suitable (standard)
lift $\hat{\tau}$ of $\tau$, we may choose
$\beta\in{\mathbb{C}}\Lambda^{\tau}$ [17, 26].

######  Remark 2.1.

By the definition of W-element, it is clear that the order of $g$ on
$V_{1}=\oplus_{j=1}^{t}{\cal G}_{j,k_{j}}$ is ${\rm
LCM}(\\{r_{i}h_{i}^{\vee}\mid i\\}$; indeed, one can show that $|g|={\rm
LCM}(\\{r_{i}h_{i}^{\vee}\mid i\\}$ on $V$ [16, Proposition 5.1].

Set $n=|g|=|\tilde{g}|$ and let $K_{0},N_{0}\in{\mathbb{Z}}$ with ${\rm
GCD}(K_{0},N_{0})=1$ such that
$\langle\alpha,\alpha\rangle=\frac{2K_{0}}{N_{0}}$. As we have shown in [7],
$n=|\tau|N_{0}$ and $N_{0}|h^{\vee}_{j}$; thus, $|\tau|={\rm
LCM}(\\{r_{i}h_{j}^{\vee}/N_{0}\mid j\\})$.

It is also proved in [7, Proposition 4.2] that
$k_{i}/h_{i}^{\vee}=\frac{K_{0}-N_{0}}{N_{0}}$; therefore we have

$\frac{2}{r_{i}k_{i}}=\frac{2}{r_{i}h^{\vee}_{i}}\frac{N_{0}}{K_{0}-N_{0}}=\frac{2}{(K_{0}-N_{0})r_{i}h^{\vee}/N_{0}}$

and

${{\rm LCM}(\\{r_{i}k_{i}\mid i\\})}={(K_{0}-N_{0}){\rm
LCM}(\\{r_{i}h^{\vee}/N_{0}\mid i\\})}={(K_{0}-N_{0})|\tau|}.$

As a consequence, we have the following result.

######  Lemma 2.2.

${\rm LCM}(\\{r_{i}k_{i}\mid i\\})=(K_{0}-N_{0})|\tau|$.

Next we recall an important result from [7].

######  Proposition 2.3 ([7, Propositions 3.25 and 4.3]).

We have $N_{0}\alpha\in L^{\ast}$. Moreover,

${\rm Comm}(M({\cal H}),V^{[g]})=\oplus_{j=0}^{|\tau|-1}U(jN_{0}\alpha)$

and ${\cal H}+{\rm Comm}(M({\cal H}),V)_{1}$ is a Cartan subalgebra of
$V^{[g]}\cong V_{\Lambda}$.

Since $U(jN_{0}\alpha)\otimes e^{jN_{0}\alpha}\in V$, $U(jN_{0}\alpha)$
appears in $g^{-jN_{0}}$-twisted $V$-module and so we may choose $\beta$ and
$\hat{\tau}$ so that $\langle\beta,\alpha\rangle\equiv
1/N_{0}|\tau|\pmod{{\mathbb{Z}}}$ and $\widehat{\tau}$ acts on
$U(jN_{0}\alpha)$ as a multiple by $e^{-2\pi ij/|\tau|}$.

As we discussed, a $W$-element of a holomorphic VOA of central charge $24$
with a semisimple weight one Lie algebra defines an automorphism
$\tilde{g}=\hat{\tau}\exp(2\pi i\beta(0))$ for some $\tau\in{\rm
Aut}(\Lambda)=Co.0$ and $\beta\in{\mathbb{C}}\Lambda^{\tau}$ such that
$V=V_{\Lambda}^{[\tilde{g}]}$. One main question is to determine the isometry
$\tau\in Co.0$ arising in this manner.

Denote

${\cal P}=\left\\{\tau\in
Co.0\,\left|\begin{array}[]{l}{}^{\exists}\beta\in{\mathbb{Q}}\Lambda^{\tau}\mbox{
s.t. }\widehat{\tau}\exp(2\pi i\beta(0))\mbox{ can be realized as the reverse
}\\\ \mbox{ automorphism of some orbifold construction given by a
$W$-element}\end{array}\right.\right\\}.$

In Section 6, we will show that ${\cal P}\subset{\cal
P}_{0}=\\{1A,2A,2C,3B,4C,5B,6E,6G,7B,8E,10F\\}$. The main idea is to analyze
the conformal weights of the irreducible $\hat{\tau}$-twisted modules and
$\tilde{g}$-twisted modules.

##  3 Irreducible twisted modules for lattice VOAs

First we review some basic properties about the irreducible twisted modules
for lattice VOAs. Let $P$ be an even unimodular lattice. Let $\tau\in O(P)$ be
of order $n$ and $\hat{\tau}\in O(\hat{P})$ be a standard lift of $\tau$. We
also use $\pi=\pi_{\tau}:P\to(P^{\langle\tau\rangle})^{*}$ to denote the
natural projection. More preciously,

$\pi(x)=\pi_{\tau}(x)=\frac{1}{n}\sum_{i=0}^{n-1}\tau^{i}(x).$ (3.1)

By [12], $V_{P}$ has a unique irreducible $\hat{\tau}$-twisted $V_{P}$-module,
up to isomorphism. Such a module $V_{P}[\hat{\tau}]$ was constructed in [11]
explicitly; as a vector space,

$V_{P}[\hat{\tau}]\cong M(1)[\tau]\otimes{\mathbb{C}}[\pi(P)]\otimes T,$

where $M(1)[\tau]$ is the “$\tau$-twisted” free bosonic space,
${\mathbb{C}}[\pi(P)]$ is the group algebra of $\pi(P)$ and $T$ is an
irreducible module for a certain “$\tau$-twisted” central extension of $P$.
(see [34, Propositions 6.1 and 6.2] and [11, Remark 4.2] for detail). Recall
that

$\dim T=|P_{\tau}/(1-\tau)P|^{1/2}$

and that the conformal weight $\phi(\tau)$ of $T$ is given by

$\phi(\tau):=\frac{1}{4n^{2}}\sum_{j=1}^{n-1}j(n-j)\dim\mathfrak{h}_{(j)},$
(3.2)

where
$\mathfrak{h}_{(j)}=\\{x\in\mathfrak{h}\mid\tau(x)=\exp((j/n)2\pi\sqrt{-1})x\\}$.
Note that $M(1)[\tau]$ is spanned by vectors of the form

$x_{1}(-m_{1})\dots x_{s}(-m_{s})1,$

where $m_{i}\in(1/n){\mathbb{Z}}_{>0}$ and $x_{i}\in\mathfrak{h}_{(nm_{i})}$
for $1\leq i\leq s$.

In addition, the conformal weight of $x_{1}(-m_{1})\dots x_{s}(-m_{s})\otimes
e^{\alpha}\otimes t\in V_{P}[\hat{\tau}]$ is given by

$\sum_{i=1}^{s}m_{i}+\frac{(\alpha|\alpha)}{2}+\phi(\tau),$

where $x_{1}(-m_{1})\dots x_{s}(-m_{s})\in M(1)[\tau]$,
$e^{\alpha}\in{\mathbb{C}}[\pi(P)]$ and $t\in T$. Note that
$m_{i}\in(1/n){\mathbb{Z}}_{>0}$ and that the conformal weight of
$V_{P}[\hat{\tau}]$ is $\phi(\tau)$.

Since $\sum_{j=1}^{n-1}j(n-j)=n(n^{2}-1)/6$, we have

$\phi(\tau)=\frac{1}{24}\sum_{i=1}^{d}a_{i}\frac{(n_{i}^{2}-1)}{n_{i}}=\frac{1}{24}\\{\sum
a_{i}n_{i}-\sum_{i=1}^{d}\frac{a_{i}}{n_{i}}\\}=1-\frac{1}{24}\sum_{i=1}^{d}\frac{a_{i}}{n_{i}}$

if $\tau\in O(P)$ has the the frame shape $\prod_{j=1}^{d}n_{j}^{a_{j}}$ by
(3.2).

######  Remark 3.1.

Let $v\in{\mathbb{Q}}\otimes_{\mathbb{Z}}P^{\tau}\subset\mathfrak{h}_{(0)}$.
Then $\exp(2\pi iv(0))$ has finite order on $V_{P}$ and commutes with
$\hat{\tau}$. Set $g=\hat{\tau}\exp(2\pi iv(0))$. Then the unique irreducible
$g$-twisted for $V_{P}$ is given by

$V_{P}[g]\cong M(1)[\tau]\otimes{\mathbb{C}}[-v+\pi(P)]\otimes T$

as a vector space [4]. In this case, the conformal weight of $V_{P}[g]$ is
given by

$\frac{1}{2}\min\\{(\beta|\beta)\mid\beta\in-v+\pi(P)\\}+\phi(\tau).$

##  4 Leech lattice and $\ell$-duality

###  4.1 $\ell$-duality

In this subsection, we study the properties of the lattice $\Lambda^{\tau}$
for

$\tau\in{\cal P}_{0}=\\{1A,2A,2C,3B,4C,5B,6E,6G,7B,8E,10F\\}.$

######  Lemma 4.1.

For $\tau\in{\cal P}_{0}$, we have $\det(\Lambda^{\tau})=\ell^{{\rm
rank}(\Lambda^{\tau})/2}$.

[Proof] For $\tau\in{\cal P}_{0}$, it is easy to check that
$(1-\tau)(\Lambda^{*}_{\tau})=\Lambda_{\tau}$. It follows that
$\det(\Lambda^{\tau})=\det(\Lambda_{\tau})=\prod m^{a_{m}}$, where $\prod
m^{a_{m}}$ is the frame shape of $\tau$. The result then follows by a direct
calculation (cf. Table 1 in Section 6). ∎

Below is one of our key observations.

######  Theorem 4.2.

If $\tau\in{\cal P}_{0}$, then there is an isometry

$\varphi=\varphi_{\tau}:\sqrt{\ell}(\Lambda^{\tau})^{\ast}\to\Lambda^{\tau},$

where $\ell=|\hat{\tau}|$.

######  Remark 4.3.

It is known that $\ell=|\tau|$ or $2|\tau|$ and $\ell=2|\tau|$ if and only if
$<\tau>$ contains a $2C$-element, see [2].

In [20], Harada and Lang have determined the structure of $\Lambda^{\tau}$ for
$\tau\in Co.0$. In particular, the Gram matrix for the lattice
$\Lambda^{\tau}$ has been given explicitly. It is straightforward to check
that Theorem 4.2 holds for

$\tau\in{\cal P}_{0}=\\{1A,2A,2C,3B,4C,5B,6E,6G,7B,8E,10F\\}.$

Note that the Gram matrix of the dual lattice $L^{*}$ is equal to the inverse
of the Gram matrix of $L$. Therefore, it suffices to check that the Gram
matrix of $\Lambda^{\tau}$ is equal to $\ell$ times the Gram matrix of
$(\Lambda^{\tau})^{*}$ for any $\tau\in{\cal P}_{0}$.

######  Remark 4.4.

Let $\hat{\tau}$ be a standard lift of $\tau$ in ${\rm Aut}(V_{\Lambda})$ and
let $\phi(\tau)$ be the conformal weight of the irreducible
$\widehat{\tau}$-twisted module of $V_{\Lambda}$. By direct calculations, it
is straightforward to verify that for any $\tau\in{\cal P}_{0}$, we have

$\phi(\tau^{m})=1-\frac{1}{|\widehat{\tau^{m}}|}\qquad\mbox{ for all
}m||\tau|.$

In other words, $\widehat{\tau^{m}}$ is of type $0$ for any $\tau\in{\cal
P}_{0}$ and $m\in{\mathbb{Z}}$ as defined in [17].

###  4.2 Reverse automorphisms and associated Niemeier lattices

From now on, we assume that $\tilde{g}=\widehat{\tau}\exp(2\pi
i\beta(0))\in{\rm Aut}(V_{\Lambda})$ is the reverse automorphism associated
with an orbifold construction defined by a W-element and assume that
$\tau\in{\cal P}_{0}$ and $\tau(\beta)=\beta$.

First we discuss some relations between the lattice $L$ (see Section 2)
associated with the roots of $V_{1}$ and the fixed point sublattice
$\Lambda^{\tau}$ of the Leech lattice.

######  Notation 4.1.

Let $X$ be an even lattice and let
$\beta\in{\mathbb{Q}}\otimes_{\mathbb{Z}}X$. Denote $X_{\beta}=\\{x\in
X\mid\langle x,\beta\rangle\in{\mathbb{Z}}\\}$. Now suppose that
$\langle\beta,\beta\rangle=2k/n$ for some positive integers $k$ and $n$ with
$(k,n)=1$. Let $\bar{X}=\\{x\in X\mid\langle
x,n\beta\rangle\in{\mathbb{Z}}\\}$. For $x\in\bar{L}$, we define
$x^{[\beta]}=x-m\beta$ if $\langle x,n\beta\rangle\equiv mk\mod n$ and $0\leq
m\leq n-1$ and $\bar{X}^{[\beta]}=\\{x^{[\beta]}\mid x\in\bar{X}\\}$. Then
$\bar{X}^{[\beta]}$ is also an even lattice and
$\det(\bar{X})=\det(\bar{X}^{[\beta]})$.

Recall that $\langle\alpha,\alpha\rangle=2K_{0}/N_{0}$ and $N_{0}\alpha\in
L^{*}$ for a W-element $\alpha$ (cf. Proposition 2.3). Thus we have
$\bar{L}=L$. Since $V^{[g]}\cong V_{\Lambda}$, we have
$V^{g}\cong(V_{\Lambda})^{\tilde{g}}$ and

$V_{L_{\alpha}}={\rm Com}({\rm Com}(M(\widehat{{\cal
H}}),V^{g}),V^{g})\cong{\rm Com}({\rm Com}(M(\widehat{{\cal
H}}),(V_{\Lambda})^{\tilde{g}}),(V_{\Lambda})^{\tilde{g}})=V_{\Lambda_{\beta}^{\tau}}.$

It implies $L_{\alpha}\cong\Lambda^{\tau}_{\beta}$. Moreover,
$\Lambda^{\tau}=L^{[\alpha]}+{\mathbb{Z}}N_{0}\alpha$.

Now consider the irreducible $\tilde{g}$-twisted module

$V_{\Lambda}[\tilde{g}]\cong{\mathbb{C}}[-\beta+(\Lambda^{\tau})^{*}]\otimes
M(1)[\tau]\otimes T.$

Since $(V_{\Lambda}[\tilde{g}])_{\mathbb{Z}}\neq 0$ and
$\phi(\tau)=1-1/|\widehat{\tau}|$ for $\tau\in{\cal P}_{0}$, there exists
$\beta^{\prime}\in-\beta+(\Lambda^{\tau})^{*}$ such that
$\langle\beta^{\prime},\beta^{\prime}\rangle/2\equiv
1/|\widehat{\tau}|\mod{\mathbb{Z}}$. Without loss, we may assume
$\beta=-\beta^{\prime}$. In this case, $\beta\in L^{*}$. By a general result
(cf. [21, 24]), we also have

(1)
$(\mathcal{D}(L),q)\cong(\mathrm{Irr}(V_{\Lambda_{\tau}}^{\hat{\tau}}),-q^{\prime})$
as quadratic spaces;

(2) $\det(L)=\det(\Lambda^{\tau})\times|\tau|^{2}$.

The quadratic space structure of
$(\mathrm{Irr}(V_{\Lambda_{\tau}}^{\hat{\tau}}),q^{\prime})$ has also been
determined in [24]. In particular, it has proved that the exponent of
$L^{*}/L$ is $\ell=|\widehat{\tau}|$ and
$q(L^{*})\subset\frac{1}{\ell}{\mathbb{Z}}$ . Thus we have the following
lemma.

######  Lemma 4.5.

We have $\ell\beta\in L$ and $L=\Lambda_{\beta}^{\tau}+{\mathbb{Z}}\ell\beta$.
Moreover, $\sqrt{\ell}L^{*}$ is an even lattice.

By Theorem 4.2, there is an isometry
$\varphi=\varphi_{\tau}:\sqrt{\ell}(\Lambda^{\tau})^{*}\to\Lambda^{\tau}$ with
$\ell=|\widehat{\tau}|$ and it induces an isometry from
${\mathbb{C}}\Lambda^{\tau}\to{\mathbb{C}}\Lambda^{\tau}$.

######  Definition 4.6.

Set $\tilde{\beta}=\sqrt{\ell}\varphi(\beta)$ and
$N:=\Lambda^{[\tilde{\beta}]}=\Lambda_{\tilde{\beta}}+{\mathbb{Z}}\tilde{\beta}$.

By our assumption, $\langle\tilde{\beta},\tilde{\beta}\rangle\in
2{\mathbb{Z}}$ and $N$ is an even unimodular lattice.

######  Theorem 4.7.

Let $n=|g|=|\tilde{g}|$. Suppose $m\varphi(\sqrt{\ell}\beta)\in\Lambda$. Then
we have $n|m$. Moreover,
$[\Lambda^{[\tilde{\beta}]}:\Lambda_{\tilde{\beta}}]=n$.

[Proof] If $m\varphi(\sqrt{\ell}\beta)\in\Lambda$, then
$m\varphi(\sqrt{\ell}\beta)\in\Lambda^{\tau}=\varphi(\sqrt{\ell}(\Lambda^{\tau})^{*})$,
which is equivalent to
$m\sqrt{\ell}\beta\in\sqrt{\ell}(\Lambda^{\tau})^{\ast}$ (or equivalently,
$m\beta\in(\Lambda^{\tau})^{\ast}$).

Since
$[\Lambda^{\tau}:\Lambda^{\tau}_{\beta}]=n(=[(\Lambda^{\tau}_{\beta})^{*}:(\Lambda^{\tau})^{*}])$
and $(\Lambda^{\tau}_{\beta})^{*}=(\Lambda^{\tau})^{*}+{\mathbb{Z}}\beta$,
$m\beta\in(\Lambda^{\tau})^{\ast}$ implies $n$ divides $m$.

For the second statement, it suffices to show
$n\sqrt{\ell}\varphi(\beta)\in\Lambda^{\tau}$. Let $k=|\tau|$. Then
$\widehat{\tau}^{k}=\exp(2\pi i\delta(0))$ for some
$\delta\in{\mathbb{C}}\Lambda^{\tau}$. Since $n=|\tau|N_{0}=kN_{0}$ and
$\tilde{g}^{n}=\widehat{\tau}^{n}\exp(2\pi i\beta(0))^{n}=1$, we have $n\beta-
N_{0}\delta\in\Lambda^{\tau}$.

As $\widehat{\tau}$ is a standard lift,
$\widehat{\tau}(e^{\gamma})=e^{\gamma}$ for any $\gamma\in\Lambda^{\tau}$.
Thus, $\langle\delta,\Lambda^{\tau}\rangle\subset{\mathbb{Z}}$ and
$\delta\in(\Lambda^{\tau})^{*}$. By using the isometry $\varphi$, we have
$n\sqrt{\ell}\varphi(\beta)-N_{0}\sqrt{\ell}\varphi(\delta)\in\sqrt{\ell}\varphi(\Lambda^{\tau})<\sqrt{\ell}\varphi((\Lambda^{\tau})^{*})=\Lambda^{\tau}$.
Since $\delta\in(\Lambda^{\tau})^{*}$, we have
$\sqrt{\ell}\varphi(\delta)\in\Lambda^{\tau}$ and
$n\sqrt{\ell}\varphi(\beta)\in\Lambda^{\tau}$ as desired. ∎

######  Theorem 4.8.

Let
$N=\Lambda^{[\tilde{\beta}]}=\Lambda_{\tilde{\beta}}+{\mathbb{Z}}\tilde{\beta}$.
Then $\varphi$ induces an isometry from $\sqrt{\ell}L^{*}$ to $N^{\tau}$. In
particular, we have $N^{\tau}\cong\sqrt{\ell}L^{*}$.

[Proof] Since $\tau$ fixes $\varphi(\sqrt{\ell}\beta)$, we may assume that
$\tau$ acts on $N=\Lambda^{[\varphi(\sqrt{\ell}\beta)]}$ and
$N^{\tau}=\Lambda^{\tau}_{\tilde{\beta}}+{\mathbb{Z}}\tilde{\beta}$. Note also
that $\langle\beta,\beta\rangle\in(2/\ell){\mathbb{Z}}$.

By Lemma 4.5, we have $L=\Lambda_{\beta}^{\tau}+{\mathbb{Z}}\ell\beta$. Thus,

$\begin{split}L^{*}&=\\{x\in(\Lambda_{\beta}^{\tau})^{*}\mid\langle
x,\ell\beta\rangle\in{\mathbb{Z}}\\}\\\
&=\\{x\in(\Lambda^{\tau})^{*}+{\mathbb{Z}}\beta\mid\langle
x,\ell\beta\rangle\in{\mathbb{Z}}\\}\\\
&=<\\{x\in(\Lambda^{\tau})^{*}\mid\langle
x,\ell\beta\rangle\in{\mathbb{Z}}\\},\beta>.\end{split}$

Then

$\begin{split}\sqrt{\ell}\varphi(L^{*})&=<\\{\sqrt{\ell}\varphi(x)\in\sqrt{\ell}\varphi(\Lambda^{\tau})^{*}\mid\langle\sqrt{\ell}\varphi(x),\sqrt{\ell}\varphi(\beta)\rangle\in{\mathbb{Z}}\\},\sqrt{\ell}\varphi(\beta)>\\\
&=<\\{y\in\Lambda^{\tau}\mid\langle
y,\tilde{\beta}\rangle\in{\mathbb{Z}}\\},\tilde{\beta}>\\\
&=\Lambda^{\tau}_{\tilde{\beta}}+{\mathbb{Z}}\tilde{\beta}=N^{\tau}\end{split}$

as desired. ∎

######  Lemma 4.9.

Under the above conditions, we have
$\Lambda^{[\varphi(\sqrt{\ell}\beta)]}\not\cong\Lambda$.

[Proof] Since $\tau$ fixes $\varphi(\sqrt{\ell}\beta)$, we may assume that
$\tau$ acts on $N=\Lambda^{[\varphi(\sqrt{\ell}\beta)]}$. Then $\tau$ induces
an isometry $\tau^{\prime}$ of $N=\Lambda^{[\widetilde{\beta}]}$.

Suppose $N\cong\Lambda$. Since all elements in $O(\Lambda)$ are determined by
its frame shape up to conjugate, we may assume $\tau^{\prime}$ is conjugate to
$\tau$. In particular, $N^{\tau^{\prime}}$ is isometric to $\Lambda^{\tau}$
and so $\sqrt{\ell}(N^{\tau^{\prime}})^{\ast}\cong N^{\tau^{\prime}}$. By
Theorem 4.8, $N^{\tau^{\prime}}\cong\sqrt{\ell}L^{\ast}$ and so
$(N^{\tau})^{\ast}=\frac{1}{\sqrt{\ell}}L$. Then
$\Lambda^{\tau}\cong\sqrt{\ell}(\Lambda^{\tau})^{\ast}\cong\sqrt{\ell}(N^{\tau^{\prime}})^{\ast}\cong
L$. It is not possible since $\det(L)=\det(\Lambda^{\tau}).|\tau|^{2}$. ∎

As a consequence of the above lemma and Theorem 4.7, we have the following
lemma.

######  Lemma 4.10.

Let $h$ be the Coxeter number of $N=\Lambda^{[\varphi(\sqrt{\ell}\beta)]}$ and
$n=|g|=|\tilde{g}|$. Then we have $n\geq h$.

###  4.3 Roots of $V_{1}$

Next we study the structure of the root lattice of $V_{1}$. Let
$V_{1}=\oplus_{j=1}^{t}{\cal G}_{j,k_{j}}$. If $u\in L^{*}$ is a root of
${\cal G}_{j}$, then

$\langle u,u\rangle=\begin{cases}\frac{2}{k_{j}}&\text{ if }u\text{ is a long
root,}\\\ \frac{2}{r_{j}k_{j}}&\text{ if }u\text{ is a short
root.}\end{cases}$

As a corollary of $\ell$-duality, we have:

######  Lemma 4.11.

$|\tau|(K_{0}-N_{0})$ divides $\ell$.

[Proof] Let $e^{u}\otimes t\in{\cal G}_{j,k_{j}}$ be a short root vector. Then
$\langle u,u\rangle=\frac{2}{r_{j}k_{j}}$ and we have

$\langle
u,u\rangle=\frac{2}{r_{j}k_{j}}=\frac{2}{r_{j}h_{j}^{\vee}}\times\frac{h_{j}^{\vee}}{k_{j}}=\frac{2}{r_{j}h_{j}^{\vee}}\times\frac{N_{0}}{K_{0}-N_{0}}.$

Since $\langle u,u\rangle\in\frac{{\mathbb{Z}}}{\ell}$ and ${\rm
LCM}(\\{r_{j}h_{j}^{\vee}/N_{0}:j\\})=|\tau|$, we have that $K_{0}-N_{0}$
divides $\ell/|\tau|$. ∎

######  Remark 4.12.

Let $e^{u}\otimes t\in{\cal G}_{j,k_{j}}$ be a root vector associated with a
simple short root. Then $\langle\alpha,u\rangle=\frac{1}{r_{j}h_{j}^{\vee}}$
and $e^{u}\otimes t$ belongs to $\tilde{g}^{s_{j}}$-twisted modules, where
$s_{j}r_{j}h^{\vee}_{j}=n$. More precisely, it is belong to $s_{j}$-power of
$\tilde{g}$-twisted module $T^{1}$ by fusion products.

######  Lemma 4.13.

Suppose there is $j$ such that $r_{j}h^{\vee}_{j}=n=|\tau|N_{0}$. Then there
is a root $u\in L^{*}$ of ${\cal G}_{j}$ such that $\langle
u,u\rangle=2/\ell$.

[Proof] Let $u$ be a simple short root of ${\cal G}_{j}$. Since
$r_{j}h^{\vee}_{j}=n=|\tau|N_{0}$, we have $\langle
u,u\rangle=\frac{2}{|\tau|(K_{0}-N_{0})}$ and $e^{u}\otimes t\in T^{1}_{1}$.
It implies $\langle u,u\rangle/2\in 1/\ell+{\mathbb{Z}}$. Note also that
$\ell/|\tau|=1$ or $2$ and $|\tau|(K_{0}-N_{0})$ divides $\ell$. Therefore,
$\ell=|\tau|(K_{0}-N_{0})$ and $\langle u,u\rangle=2/\ell$. ∎

######  Definition 4.14.

We call a root $\delta$ of $V_{1}$ satisfying
$\langle\delta,\delta\rangle=\frac{2}{\ell}$ a shortest root and call a simple
component ${\cal G}_{j}$ of $V_{1}$ containing a shortest root a full
component. Note that a shortest root exists if $r_{j}h_{j}^{\vee}=n$ for some
$j$.

##  5 Deep hole

Let $V$ be a holomorphic VOA of central charge $24$ and $\alpha$ a W-element
of $V_{1}$. Let $g=\exp(2\pi i\alpha(0))\in{\rm Aut}(V)$ and let
$\tilde{g}=\hat{\tau}\exp(2\pi i\beta(0))\in{\rm Aut}(V_{\Lambda})$ be the
reverse automorphism of $g$, where $\beta\in{\mathbb{C}}\Lambda^{\tau}$.

In this section, we assume that $\tau\in{\cal P}_{0}$ and try to relate the
automorphism $\tilde{g}=\hat{\tau}\exp(2\pi i\beta(0)\in{\rm
Aut}(V_{\Lambda})$ to a deep hole of the Leech lattice. In Section 6, we will
prove that $\tau\in{\cal P}_{0}$.

Since $\tau\in{\cal P}_{0}$, there is an isometry
$\varphi:\sqrt{\ell}(\Lambda^{\tau})^{*}\to\Lambda^{\tau}$. Set
$\tilde{\beta}=\sqrt{\ell}\varphi(\beta)$ and
$N=\Lambda^{[\tilde{\beta}]}=\Lambda_{\tilde{\beta}}+{\mathbb{Z}}\tilde{\beta}$.
Note that $\langle\beta,\beta\rangle\in 2{\mathbb{Z}}/\ell$ and
$\langle\tilde{\beta},\tilde{\beta}\rangle\in 2{\mathbb{Z}}$. Moreover, the
root sublattice $R$ of $N$ is non-zero by Lemma 4.9.

One of the main aims in this paper is to prove the following theorem.

######  Theorem 5.1.

The vector $\tilde{\beta}=\varphi(\sqrt{\ell}\beta)$ is a deep hole of
$\Lambda$.

###  5.1 Coinvariant sublattices $\Lambda_{\tau}$ and $N_{\tau}$

In this subsection, we will discuss another main observation of this article,
which is related to the structure of the coinvariant sublattice $N_{\tau}$.

Since $\tau$ fixes $\tilde{\beta}=\varphi(\sqrt{\ell}\beta)$, we have
$(\Lambda_{\tilde{\beta}})_{\tau}=\Lambda_{\tau}$ and
$N_{\tau}>\Lambda_{\tau}$. Moreover, $[N_{\tau}:\Lambda_{\tau}]=|\tau|$
because
$\det(N_{\tau})=\det(N^{\tau})=\det(\Lambda^{\tau})/|\tau|^{2}=\det(\Lambda_{\tau})/|\tau|^{2}$.
For $\tau\in{\cal P}_{0}$, we note that $\Lambda_{\tau}\cong L_{B}(c)$ as
defined in Appendix A.

######  Lemma 5.2.

For $\tau\in{\cal P}_{0}\setminus\\{2C\\}$, we have $N_{\tau}\cong L_{A}(C)$,
where $C=\langle c\rangle$ is defined as in Table 2 in Appendix A.

[Proof] It follows from the fact that $\Lambda_{\tau}\cong L_{B}(C)$ and
$\mathcal{D}(\Lambda_{\tau})$ has only one non-zero singular element of order
$|\tau|$, up to isometry if $\tau\in{\cal P}_{0}\setminus\\{2C\\}$ (see
Appendix A). ∎

Next we consider the case when $\tau=2C$. In this case,
$\Lambda_{\tau}\cong\sqrt{2}D_{12}^{+}$. We use the standard model for root
lattices of type $D$, i.e,

$D_{12}=\left\\{(x_{1},\cdots,x_{12})\in{\mathbb{Z}}^{12}\mid\sum_{i=1}^{12}x_{i}=0\mod
2\right\\}$

and
$D_{12}^{+}=\mathrm{Span}_{\mathbb{Z}}\\{D_{12},\frac{1}{2}(1,\cdots,1)\\}$.
Note that $D_{12}^{+}$ is an odd lattice.

######  Lemma 5.3.

For $\tau\in 2C$, there are two classes of non-zero singular elements, up to
isometry; they correspond to vectors of the form $\sqrt{2}(1,0,\cdots,0)$ and
$\frac{\sqrt{2}}{2}(1^{4}0^{8})$, respectively. There are two index $2$
overlattices of $\Lambda_{\tau}$. They are isometric to
$L_{A}(1^{12})\cong\sqrt{2}D_{12}^{*}$ or the overlattice
$\mathrm{Span}_{\mathbb{Z}}\\{\sqrt{2}D_{12}^{+},\frac{\sqrt{2}}{2}(1^{4}0^{8})\\}$,
which has the root sublattice $A_{1}^{4}$.

Let
$X=\mathrm{Span}_{\mathbb{Z}}\\{\sqrt{2}D_{12}^{+},\frac{\sqrt{2}}{2}(1^{4}0^{8})\\}$.
Then

$\begin{split}X^{*}&=\\{\alpha\in\frac{\sqrt{2}}{2}D_{12}^{+}\mid\langle\alpha,\frac{\sqrt{2}}{2}(1^{4}0^{8})\rangle\in{\mathbb{Z}}\\}\\\
&=\frac{\sqrt{2}}{2}\mathrm{Span}_{\mathbb{Z}}\\{D_{4}+D_{8},\frac{1}{2}(1^{4}0^{8})\\}\cong\frac{\sqrt{2}}{2}({\mathbb{Z}}^{4}+D_{8}).\end{split}$

Therefore, $2X^{*}\cong A_{1}^{4}+\sqrt{2}D_{8}$. Then the quadratic form of
$2X^{*}$ is not isometric to $L$, which is not possible and hence we have the
following lemma.

######  Lemma 5.4.

For $\tau\in 2C$, $N_{\tau}\cong L_{A}(1^{12})\cong\sqrt{2}D_{12}^{*}$.

As a consequence, for any $\tau\in{\cal P}_{0}$, $N_{\tau}\cong L_{A}(c)$ as
described in Appendix A. We also note that $\tau$ acts on $L_{A}(c)$ as
$g_{\Delta,c}$ defined in (A.4). In particular, $\tau$ is contained in the
Weyl group of $R$.

######  Remark 5.5.

If the frame shape of $\tau$ is $\prod k^{m_{k}}$, then $N_{\tau}$ contains
$\oplus A_{k-1}^{m_{k}}$ and $\tau$ acts on $\oplus A_{k-1}^{m_{k}}$ as a
product of the Coxeter elements. In particular, $\tau$ preserves any
irreducible component of the root system of $N$.

######  Corollary 5.6.

If $N_{0}=1$, then the Coxeter number of $N$ is greater than or equal to
$|\tau|$.

[Proof] For $\tau\in{\cal P}_{0}$, it is easy to check that its frame shape of
$\tau$ contains a positive power of $|\tau|$. Therefore, $N_{\tau}$ contains a
root system $A_{|\tau|-1}$ whose Coxeter number is $|\tau|$. ∎

###  5.2 Affine root system and shortest roots

One important fact that we will prove is the following:

######  Lemma 5.7.

Let $\beta\in{\mathbb{C}}\Lambda$. Set
$S=\\{\lambda\in\Lambda\mid\langle\lambda-\beta,\lambda-\beta\rangle=2\\}$. If
$\tilde{S}=\\{\lambda-\beta\mid\lambda\in S\\}$ contains an affine fundamental
root system, then $\beta$ is a deep hole.

[Proof] For $\lambda\in\Lambda$, let
$\lambda^{*}=(\lambda,1,\frac{\langle\lambda,\lambda\rangle}{2}-1)\in\Lambda+\Pi_{1,1}$,
which is called a Leech root. Set
$\hat{\beta}=(\beta,1,\frac{\langle\beta,\beta\rangle}{2})\in{\mathbb{Q}}(\Lambda+\Pi_{1,1})$.
Note that $\hat{\beta}$ is an isotropic vector. Then
$\lambda^{*}-\hat{\beta}=(\lambda-\beta,0,\frac{\langle\lambda,\lambda\rangle}{2}-1-\frac{\langle\beta,\beta\rangle}{2})$
and we have

$\begin{split}\langle\lambda-\beta,\lambda-\beta\rangle=\langle\lambda^{*}-\hat{\beta},\lambda^{*}-\hat{\beta}\rangle=\langle\lambda^{*},\lambda^{*}\rangle+\langle\hat{\beta},\hat{\beta}\rangle-2\langle\lambda^{*},\hat{\beta}\rangle=2-2\langle\lambda^{*},\hat{\beta}\rangle.\end{split}$

Thus, $\langle\lambda-\beta,\lambda-\beta\rangle=2$ if and only if
$\langle\lambda^{*},\hat{\beta}\rangle=0$.

Suppose that $\beta$ is not a deep hole. Then there is a $\delta\in\Lambda$
such that $\langle\delta-\beta,\delta-\beta\rangle<2$. In this case,
$\langle\delta^{*},\hat{\beta}\rangle>0$. Since
$\delta^{\ast}\not\in\\{\lambda^{\ast}\mid\lambda\in S\\}$ and the minimal
squared norm of $\Lambda$ is $4$,
$\langle\delta^{\ast},\lambda^{\ast}\rangle\leq 0$ for all $\lambda\in S$. On
the other hand, since $\tilde{S}$ contains an affine fundamental root system,
there exists a subset $S^{\prime}\subset S$ and positive real numbers
$n_{\lambda}\in{\mathbb{R}}_{\geq 0}$ such that $\sum_{\lambda\in
S^{\prime}}n_{\lambda}\lambda=m\beta$, where $m=\sum_{\lambda\in
S^{\prime}}n_{\lambda}$. By direct calculations, we have

$\begin{split}\sum_{\lambda\in
S^{\prime}}n_{\lambda}\left(\frac{\langle\lambda,\lambda\rangle}{2}-1\right)&=\sum_{\lambda\in
S^{\prime}}n_{\lambda}\frac{\langle\lambda,\lambda\rangle-\langle\lambda-\beta,\lambda-\beta\rangle}{2}\\\
&=\sum_{\lambda\in
S^{\prime}}n_{\lambda}\left(\langle\lambda,\beta\rangle-\frac{\langle\beta,\beta\rangle}{2}\right)=\frac{m\langle\beta,\beta\rangle}{2}.\end{split}$

Thus, we also have

$\sum_{\lambda\in S^{\prime}}n_{\lambda}\lambda^{*}=\left(\sum_{\lambda\in
S^{\prime}}n_{\lambda}\lambda,\sum_{\lambda\in
S^{\prime}}n_{\lambda},\sum_{\lambda\in
S^{\prime}}n_{\lambda}\left(\frac{\langle\lambda,\lambda\rangle}{2}-1\right)\right)=m\hat{\beta}$

but it implies $\langle\delta^{\ast},\hat{\beta}\rangle\leq 0$, which is a
contradiction. ∎

Therefore, in order to prove Theorem 5.1, it is enough to show that the
Coxeter number $h$ of $N$ is greater than or equal to $n=|g|$, since
$R_{1}=\\{\lambda-\tilde{\beta}\mid\langle\lambda-\tilde{\beta},\lambda-\tilde{\beta}\rangle=2,\lambda\in\Lambda_{\tilde{\beta}}\\}$
will contain an affine fundamental root system in this case.

###  5.3 Existence of shortest roots

The purpose of this subsection is to show the following proposition.

######  Proposition 5.8.

There is a root $\delta$ of $V_{1}$ such that
$\langle\delta,\delta\rangle=\frac{2}{\ell}$.

[Proof] By Lemma 4.13, it suffices to show that there is $j$ such that
$r_{j}h_{j}^{\vee}=n$.

Recall that ${\rm LCM}(r_{j}h^{\vee}_{j}\mid j)=n$. If $|\tau|$ is a prime
power or $V_{1}$ is simple, then there is $j$ such that $r_{j}h_{j}^{\vee}=n$.
Therefore, we may assume that $|\tau|$ is not a prime power and ${\rm
rank}(V_{1})>4$, that is, $\tau$ is $6E=1^{2}2^{2}3^{2}6^{2}$ or
$6G=2^{3}6^{3}$ and $\mathrm{rank}(V_{1})=8$ or $6$. We have already shown
$N_{0}|h_{j}^{\vee}$ for all $j$. By Corollary 5.6, we may also assume
$N_{0}\not=1$. Suppose there is no $j$ such that $r_{j}h_{j}^{\vee}=n=6N_{0}$.
Then there exist $k$ and $l$ such that $r_{k}h_{k}^{\vee}=3N_{0}$ and
$r_{l}h_{l}^{\vee}=2N_{0}$. If one of ${\cal G}_{j}$ is of type $G_{2}$, then
$N_{0}=4$ and there is also a component ${\cal G}_{i}$ such that
$r_{i}h_{i}^{\vee}=8$, that is, ${\cal G}_{i}=A_{7}$, or $C_{3}$. Another
possible component is $A_{3}$. Therefore, possible choices for $\tau$ and
$V_{1}$ are $\tau=6E$ and $V_{1}=G_{2}C_{3}^{2}$, or $G_{2}C_{3}A_{3}$.
However, since $\dim V_{1}=120\not=14+2\times 21$ nor $14+21+15$, we have a
contradiction. Therefore, there is no component of type $G_{2}$. In order to
get $r_{k}h_{k}^{\vee}=3N_{0}$, $r_{k}=1$, $h_{k}^{\vee}=3N_{0}$ and $V_{1}$
contains a component of type $A_{3N_{0}-1}$. Since $N_{0}\not=1$,
$\mathrm{rank}(V_{1})\leq 8$ and $V_{1}$ has a component with
$r_{i}h_{i}^{\vee}=2N_{0}$, we have $N_{0}=2$ and $V_{1}=A_{3}+A_{5}$ (i.e.,
$\mathrm{rank}(V_{1})=8$ and $\tau=6E$). However, $\dim V_{1}=72\not=15+35$,
which is a contradiction. This completes the proof of Proposition 5.8. ∎

As a corollary, $(\Lambda_{\tilde{\beta}}-\tilde{\beta})_{2}\not=\emptyset$
and so we may choose $\beta$ so that
$\langle\beta,\beta\rangle=\frac{2}{\ell}$. In particular, $\tilde{\beta}$ has
squared length $2$.

###  5.4 Proof of Theorem 5.1 (part 2)

######  Notation 5.9.

For an even lattice $K$, we use $R(K)$ to denote the sublattice generated by
$K_{2}=\\{a\in K\mid\langle a,a\rangle=2\\}$. A component lattice of $R(K)$
means a sublattice generated by an irreducible component of the roots in
$K_{2}$.

######  Lemma 5.10.

Let $X\subseteq R(N^{\tau})$ be a component lattice and $\tilde{X}\subseteq
R(N)$ be a component lattice containing $X$. If $X\not=\tilde{X}$, then ${\rm
rank}(\tilde{X})\geq{\rm rank}(X)+2$ or $\tau$ acts on $\tilde{X}$ trivially.

[Proof] If $\tilde{X}\not=X$ and $\tau$ does not act on $\tilde{X}$ trivially,
then there are $\mu_{1}\in X$ and
$\mu_{2}\in\tilde{X}-(\tilde{X}^{\tau})^{\perp}$ such that
$\langle\mu_{1},\mu_{2}\rangle=-1$. Since $\mu_{2}\not\in N^{\tau}$ and ${\rm
rank}(R(N_{\tau}))={\rm rank}(N_{\tau})$ (see Remark 5.5), there is a
$\mu_{3}\in R(N_{\tau})$ such that $\langle\mu_{2},\mu_{3}\rangle=-1$. Since
$\mu_{3}\in\tilde{X}$, ${\rm rank}(\tilde{X})\geq{\rm rank}(X)+2$, as we
desired. ∎

Now let’s start the proof of Theorem 5.1.

Suppose Theorem 5.1 is false and let $V$ be a counterexample. As we explained,
it is enough to show that the Coxeter number of $N$ is greater than or equal
to $n$. By Proposition 5.8, there is a simple component ${\cal G}_{j}$ of
$V_{1}$ such that $r_{j}h^{\vee}_{j}=n$ and we may choose
$\langle\tilde{\beta},\tilde{\beta}\rangle=2$, or equivalently
$\langle\beta,\beta\rangle=\frac{2}{\ell}$. Suppose there is a full component
${\cal G}_{j}$ which is simply laced and let $L_{j}$ be its root lattice. Then
$\sqrt{\ell}\phi(L_{j})\subseteq R^{\tau}$ and so $h\geq n$, which is a
contradiction.

Therefore, there are no simply laced full components in $V_{1}$ and ${\rm
GCD}(6,|\tau|)\not=1$ since the lacing number divides $|\tau|$. We also have
$N_{0}\not=1$ by Corollary 5.6. Furthermore, if ${\rm rank}(V_{1})=4$, then we
have shown $N_{0}=1$ in [7] and we have a contradiction. Therefore, we may
assume $|\tau|=2,3,4,6,8$. We will need a few lemmas.

######  Lemma 5.11.

If $|\tau|=2$, then $N_{0}$ is odd.

[Proof] Suppose $N_{0}$ is even. Then $K_{0}$ is odd since ${\rm
GCD}(K_{0},N_{0})=1$. Since $K_{0}-N_{0}=1$ or $2$, $K_{0}=N_{0}+1$. Since
$|\tau|=2$, there are no components of type $G_{2}$. Moreover, $N_{0}$ is even
and $n/N_{0}=2$; thus $4|n$ and there are no full components of type $B$ and
$F$. Therefore, the only full components of $V_{1}$ are of type
$C_{N_{0}-1,1}$. Since $N_{0}$ divides the dual Coxeter numbers, the other
components are of the type $A_{N_{0}-1}$ or $D_{N_{0}/2+1}$. Recall that ${\rm
rank}(\Lambda^{\tau})=12$ or $16$ and $N_{0}$ is even. By a direct
calculation, it is easy to verify that the possible cases are only
$V_{1}=C_{3,1}^{a}A_{3,1}^{b}$ ($N_{0}=4$) with $a+b=4$ and $C_{7}D_{5}$
($N_{0}=8$). Since $\dim V_{1}=24(N_{0}+1)$, none of them is possible. Note
that $\dim C_{3}=21,\dim A_{3}=15,\dim C_{7}=105,$ and $\dim D_{5}=45$. ∎

######  Lemma 5.12.

Let $V$ be a counterexample of Theorem 5.1. Then $\tau$ and $(V_{1},N_{0})$
will be one of the following:

  1. (1)

$\tau=2A=1^{8}2^{8}$ and
$(V_{1},N_{0})=(C_{4}^{4},5),(C_{6}^{2}B_{4},7),(C_{8}F_{4}^{2},9),(C_{10}B_{6},11)$;

  2. (2)

$\tau=2C=2^{12}$ and
$(V_{1},N_{0})=(B_{2}^{6},5),(C_{4}A_{4}^{2},5),(B_{3}^{4},5),(B_{4}^{3},7),(F_{4}A_{8},9)$,
$(B_{6}^{2},11)$, $(B_{12},23)$;

  3. (3)

$\tau=4F=1^{4}2^{2}4^{2}$ and $(V_{1},N_{0})=(C_{7,1}A_{3,1},4)$;

  4. (4)

$\tau=6E=1^{2}2^{2}3^{2}6^{2}$ and $(V_{1},N_{0})=(C_{5}G_{2}A_{1},2)$;

  5. (5)

$\tau=6G=2^{3}6^{3}$ and $(V_{1},N_{0})=(F_{4,6}A_{2,2},3)$.

[Proof] We may assume that ${\cal G}_{1}$ is a full component.
Case 1: If ${\cal G}_{1}=F_{4}$, then $\dim F_{4}=52$ and $n=18$ and so
$(\tau,{\rm rank}\Lambda^{\tau},\dim V_{1})$ are $(2A,16,240),(2C,12,132)$,
$(6G,6,60),(6E,8,72)$ and the possible choices for $({\cal G}_{j},\dim{\cal
G}_{j})$ are $(F_{4},52)$, $(C_{8},136)$, $(A_{8},80)$ and $(A_{2},8)$. Hence
the possible cases are $\tau=2A$ and $V_{1}=F_{4}^{2}C_{8}$; $\tau=2C$ and
$V_{1}=F_{4}A_{8}$; and $\tau=6G$ and $V_{1}=F_{4}A_{2}$.

From now on, we may assume that there are no full components of type $F_{4}$.
Case 2: If ${\cal G}_{1}=G_{2}$, then $n=12$ and $|\tau|=3$ $(N_{0}=4,\dim
V_{1}=120)$ or $|\tau|=6$ ($N_{0}=2,\dim V_{1}=72$ or $60$). If $|\tau|=3$,
then $\tau=3B$, $4|h^{\vee}_{i}$, $h_{i}^{\vee}<12$ and ${\rm
rank}(\Lambda^{\tau})=12$. Then $V_{1}=G_{2}^{a}+A_{3}^{b}$ and $2a+3b=12,\dim
V_{1}=14a+15b=120$, which has no solution. If $|\tau|=6$, then
$V_{1}=G_{2}^{a}+C_{5}^{b}+A_{1}^{c}+A_{3}^{d}+A_{5}^{e}$. Using dimensions
and rank$=2a+5b+c+3d+5e=6$ or $8$, the solution is only
$V_{1}=C_{5}G_{2}A_{1}$.

From now on, we may assume that all full components are of type $C_{m}$ or of
type $B_{m}$.
Case 3: If ${\cal G}_{1}=C_{8}$, then $n=18$ and $|\tau|=2$ or $6$.
(3.1) If $|\tau|=2$, then $N_{0}=9$ and so possible components are
$C_{8},B_{5},A_{8}$ and so $V_{1}=C_{8}^{2},C_{8}A_{8}$ but
$\dim(V_{1})=24K_{0}=240$. Neither one is possible.
(3.2) If $|\tau|=6$, then $N_{0}=3$ and $V_{1}=C_{8}$, but $\dim
V_{1}=96\not=\dim C_{8}$, a contradiction.
Case 4: If ${\cal G}_{1}=C_{2m}$ and $0<m\not=4$, then $n=2(2m+1)$. First we
show that $m<7$. Suppose $m\geq 7$. Then ${\rm rank}(V_{1})\geq{\rm
rank}(C_{2m})\geq 14$ and so $\tau=2A$ and $N_{0}=2m+1$ and $m=7$ or $8$. In
this case, $\dim V_{1}=24(2m+2)$ and $\dim C_{2m}=2m(4m+1)$. If $m=7$, then
there is no components ${\cal G}_{j}$ of rank 2 (or less than 2) with
$h_{j}^{\vee}$ divisible by $N_{0}=15$. So, $m=8$ and $V_{1}=C_{16}$; however,
$\dim V_{1}=24(N_{0}+1)=24\times 18\not=\dim C_{16}=16\times 33$.

For $m<7$ and $m\neq 4$, we have $(2m+1,3)=1$, $N_{0}=2m+1$ and $|\tau|=2$.
So, the possible components of $V_{1}$ are $C_{2m},A_{2m},B_{m+1}$. Since
$K_{0}-N_{0}=1$ for $\tau=2A$ and $K_{0}-N_{0}=2$ for $\tau=2C$, $\dim
V_{1}=24(2m+2)$ for $\tau=2A$ and $\dim V_{1}=12(2m+3)$ for $\tau=2C$. If
$\tau=2A$, then by solving the relation $2ma+2mb+(m+1)c=16$ and
$2m(4m+1)a+2m(2m+2)b+(m+1)(2m+3)c=24(2m+2)$ and $a>0$ and $b,c\geq 0$, we have
$V_{1}=C_{10}B_{6},C_{6}^{2}B_{4},C_{4}^{4}$.
If $\tau=2C$, then by solving the relation $2ma+2mb+(m+1)c=12$ and
$2m(4m+1)a+2m(2m+2)b+(m+1)2m+3)=12(2m+3)$, we have $V_{1}=C_{4,2}A_{4,2}^{2}$
or $C_{2,2}^{6}$.
We next assume ${\cal G}_{1}=C_{2m+1}$. Then since $2m+2$ is even,
$|\tau|\not=2$.
Case 5: If ${\cal G}_{1}=C_{3}$, then $n=8$ and $|\tau|=4$ and $N_{0}=2$ and
${\cal G}$ is a direct sum of $C_{3}$ or $A_{1}$, or $A_{3}$. Since $\dim
C_{3}=21$, $\dim A_{3}=15$, $\dim A_{1}=3$ and ${\rm rank}(V_{1})=10$, it is
impossible to have $\dim V_{1}=72$ and we have a contradiction.
Case 6: If ${\cal G}_{1}=C_{5}$, then $n=12$ and $(\tau,N_{0})=(4F,3),(6,2)$.
If $\tau=4F$, then the other possible components are $A_{2},D_{4},A_{5},C_{2}$
since there is no component of type $G_{2}$. Since ${\rm rank}(V_{1})=10$ is
even, it has another component of type $C_{5}$ or $A_{5}$, but since $\dim
V_{1}=96$, $\dim C_{5}=55$, and $\dim A_{5}=35$, we have a contradiction. If
$|\tau|=6$, then ${\rm rank}(V_{1})=8$ or $6$ and possible components are
$A_{1},A_{3}$. Since $\dim V_{1}=72$ or $60$ and $\dim A_{1}=3$, $\dim
A_{3}=15$, we have a contradiction.
Case 7: If ${\cal G}_{1}=C_{7}$, then $n=16$ and $(\tau,N_{0})=(4F,4),(8,2)$.
If $N_{0}=2$, then $\dim V_{1}=72$ and $\dim C_{7}=105$, a contradiction. If
$\tau=4F$, then $N_{0}=4$ and the only solution is $V_{1}=C_{7}A_{3}$.
Case 8: If ${\cal G}_{1}=C_{2m+1}$ and $m\geq 4$, then ${\rm rank}(V_{1})\geq
2m+1\geq 9$ and so ${\cal G}_{1}=C_{9}$. However, since ${\rm rank}(V_{1})=10$
and $N_{0}=5$, we have a contradiction.
Case 9: We may assume that all full components are of type $B$. Say, ${\cal
G}_{1}=B_{m}$, then $n=2(2m-1)$. Since $2m-1$ is odd, $|\tau|=2,6$.

If $|\tau|=6$, then $N_{0}=1/3(2m-1)$ and $m\leq 8$. Since $N_{0}\neq 1$ is
odd, the only possible values for $(N_{0},m)$ are $(3,5)$ and $(5,8)$. If
$\tau=6G$, then ${\rm rank}V_{1}=6$ and $(N_{0},m)=(3,5)$. The only possible
choice for $V_{1}$ is $B_{5}+A_{1}$, but since the coxeter number of $A_{1}$
does not divide $3$, a contradiction.
If $\tau=6E$, then ${\rm rank}V_{1}=8$, the possible choice of $V_{1}$ is
$B_{8}$. Since $2C\not\in<6E>$, $K_{0}=N_{0}+1=4$ and $\dim
V_{1}=24K_{0}=96\not=\dim B_{8}=8\times 17$, which is a contradiction.
Therefore, we have $|\tau|=2$ and $N_{0}=2m-1$. Since $2m-1$ is odd, the non-
full components are all of type $A_{2m-2}$ and so $V_{1}=B_{m}^{\otimes
a}\oplus A_{2m-2}^{\otimes b}$. We note $\dim B_{m}=m(2m+1)$ and $\dim
A_{2m-2}=4m(m-1)$. If $K_{0}=2m$, then $am+b(2m-2)=16$ and $\dim
V_{1}=48m=am(2m+1)+4bm(m-1)$ It is easy to see that there is no solution for
$a\in{\mathbb{Z}}_{>0}$ and $b\in{\mathbb{Z}}_{\geq 0}$. Hence $K_{0}-N_{0}=2$
and $\tau=2C$. In this case, $am+b(2m-2)=12$ and $\dim
V_{1}=12(2m+1)=am(2m+1)+4bm(m-1)$. The solutions are $b=0$ and $am=12$;
therefore, $V_{1}=(B_{2,2})^{\oplus 6}$, $(B_{3,2})^{\oplus 4}$,
$(B_{4,2})^{\oplus 3}$, $(B_{6,2})^{\oplus 2}$, or $B_{12,2}$, as we desired.
This completes the proof of Lemma 5.12. ∎

We will show that none of the Lie algebras in the list in Lemma 5.12 satisfy
the desired condition and we will get a contradiction.

First we note that $V_{1}$ contains a full component of type $C_{m}$ or
$F_{4}$ unless $\tau=2C$.

####  5.4.1 The case $\tau\not=2C$

We note the short root lattices of $C_{m}$ are $D_{m}$. Namely, if $C_{m}$ is
a full component of $V_{1}$, then $D_{m}$ is a sublattice of $N^{\tau}$. Let
$X$ be a component of the root lattice $R(N)$ of $N$ containing the above
$D_{m}$. For $m\geq 4$, a lattice containing $D_{m}$ is of type $D$ or $E$,
that is, $X$ is of type $D$ or $E$. Recall that $n=r_{j}h_{j}^{\vee}=2(m+1)$
for $C_{m}$ and the Coxeter numbers of $D_{m}$ and $E_{k}$ are $2(m-1)$ and
$12$, $18$ and $30$. Since $V$ is a counterexample, the Coxeter number of $N$
is strictly less than $n$ and the possible choice of $X$ is $D_{m}$ or
$D_{m+1}$.

######  Lemma 5.13.

If $X=D_{m+1}$, then $\tau$ acts on $X$ trivially. In particular, if $V_{1}$
is a direct sum of full components, then $X=D_{m}$.

If $\tau=2A$, then the list of lemma 5.15 says that $V_{1}$ is a direct sum of
full components. Therefore, if $C_{m}$ is a full component, then $D_{m}$ is a
connected component of $R(N)$. Therefore, the possible cases are as follows:

$V_{1}$ | $C_{4,1}^{4}$ | $C_{6,1}^{2}B_{4,1}$ | $C_{8,1}F_{4,1}^{2}$ | $C_{10,1}B_{6,1}$
---|---|---|---|---
$R(N)$ | $D_{4}^{6}$ | $D_{6}^{4}$ | $D_{8}^{3}$ | $D_{10}E_{7}^{2}$
$R(N^{\tau})$ contains | $D_{4}^{4}$ | $D_{6}^{2}A_{1}^{4}$ | $D_{8}D_{4}^{2}$ | $D_{10}A_{1}^{6}$

It contradicts the fact that $N_{\tau}=L_{A}(1^{8})$.

If $\tau=4F$, then $R(N)$ contains $D_{7}\oplus A_{3}^{4}A_{1}^{2}$. The
possible choice of $R(N)$ is $D_{7}E_{6}A_{11}$ or $D_{8}^{3}$, but neither of
them contains $D_{7}A_{3}^{4}A_{1}^{2}$, which is a contradiction.

If $\tau=6E$, then $R(N)$ contains $D_{5}A_{2}A_{5}^{2}A_{2}^{2}A_{1}^{2}$.
The possible choice of $R(N)$ is $D_{5}^{2}A_{7}^{2}$ or $D_{6}^{4}$ or
$D_{6}A_{9}^{2}$, but none of them contains
$D_{5}A_{5}^{2}A_{2}^{3}A_{1}^{2}$, which is a contradiction.

If $\tau=6G=2^{3}6^{3}$, then $n=18$ and $V_{1}=F_{4,6}A_{2,2}$. Hence $X$
contains $D_{4}A_{5}^{3}A_{1}^{3}$. Since $h<18$, the possible choice of
$R(N)$ is
$D_{4}^{6},D_{5}^{2}A_{7}^{2},D_{6}A_{9}^{2},D_{6}^{4},E_{6}^{4},A_{11}D_{7}E_{6},D_{8}^{3}$,
$D_{9}A_{15}$. However, none of them contains $D_{4}A_{5}^{3}A_{1}^{3}$, which
is a contradiction.

####  5.4.2 The case $\tau=2C$

By the previous lemma,
$V_{1}=B_{2}^{6},C_{4}A_{4}^{2},B_{3}^{4},B_{4}^{3},F_{4}A_{8},B_{6}^{2},B_{12}$.
Since $\tau=2C$, we have $\ell=4$. Since $v+v^{\tau}\in\Lambda^{\tau}$ and
$\\{\frac{v+v^{\tau}}{2}\mid
v\in\Lambda\\}=\pi(\Lambda)=(\Lambda^{\tau})^{\ast}$ and
$\sqrt{\ell}\phi((\Lambda^{\tau})^{\ast})=\Lambda^{\tau}$, we have
$2(\Lambda^{\tau})^{\ast}=\Lambda^{\tau}$, that is, we can take $\phi=1$ and
$\tilde{\beta}=2\beta$.

By the reverse construction, there is a Niemeier lattice $P$ such that

$V^{[N_{0}\alpha]}=(V_{\Lambda})^{[(\widehat{\tau}\exp(2\pi
i\beta(0)))^{2}]}=V_{P}.$

Since $N_{0}\not=1$, $P\not=\Lambda$. Set $\widehat{\tau}^{2}=\exp(2\pi
i\delta(0))$, which has order $2$. Therefore,

$V_{P}=(V_{\Lambda})^{[\tilde{g}^{2}]}=(V_{\Lambda})^{[\exp(2\pi
i(\delta(0)+2\beta(0))]}.$

Since $N_{0}$ is odd, $(\widehat{\tau}\exp(2\pi
i\beta(0)))^{2N_{0}}=\exp(\delta+2N_{0}\beta)=1$, there is a
$\mu\in\Lambda^{\tau}$ such that $\delta=2N_{0}\beta+\mu$ and so
$V_{P}=(V_{\Lambda})^{[\exp(2\pi i(N_{0}+1)\tilde{\beta}(0)]}$. Note also that
$V=V_{P}^{[\widetilde{g}]}$ and $\widetilde{g}$ acts on $V_{P}$ with order
$2$.

Case $V_{1}=B_{m}^{12/m}$:

We first note that all components have level $2$ and $V^{\exp(2\pi
iN_{0}\alpha)}=(V_{P})^{\widetilde{g}}$; thus $V_{P}$ contains the Lie
subalgebras generated by the long roots. Since the long roots of $B_{m}$ is of
the type $D_{m}$, $(V_{P})_{1}$ contains a Lie subalgebra of type
$D_{m,2}^{\oplus(12/m)}$ and the fixed point sublattice
$P^{\tau}\supseteq\sqrt{2}D_{m}^{\oplus 12/m}$. Then $P$ must contain
$D_{m,1}^{\oplus 24/m}$ as a sublattice. Since $N_{0}=2m-1$ and
$|\widetilde{g}^{2}|=N_{0}$, the Coxeter number of $P$ is less than or equal
to $2m-1$ and thus the root system of $P$ is $D_{m}^{\oplus 24/m}$. Since
$2m-1$ is odd and $2\delta\in\Lambda$, there is $\mu\in\Lambda$ such that
$\delta=2(2m-1)\beta+\mu$ and $4(2m-1)\beta\in\Lambda$. Thus, we have
$V_{P}=(V_{\Lambda})^{[4\beta]}$, that is, every element $\mu\in P$ is written
$\mu=\mu_{1}+4s\beta$ with $\langle\mu_{1},4\beta\rangle\in{\mathbb{Z}}$, that
is, $\mu_{1}\in\Lambda_{4\beta}$. Since $\langle
4\beta,2\beta\rangle\in{\mathbb{Z}}$, $\mu_{1}\in\Lambda_{2\beta}$ if
$\langle\mu,2\beta\rangle\in{\mathbb{Z}}$ (that is, $\mu\in\Lambda_{2\beta}$).

Since $V_{P}=V^{[\widetilde{g}]}$ can be constructed by a
${\mathbb{Z}}_{2}$-orbifold construction, we have

$L^{*}=\\{x\in(P^{\tau})^{*}\mid\langle
x,2\beta\rangle\in{\mathbb{Z}}\\}+{\mathbb{Z}}\beta.$

by the same argument as in Theorem 4.8.

For each $B_{m}\subseteq V_{1}$, we have a fundamental root system
$d_{1},...,d_{m-1},d_{m}$ with long roots $d_{j}$ $(j=1,\dots,m-1)$, i.e,
$\langle d_{j},\alpha\rangle=\frac{1}{2m-1}$ for $j=1,\dots,m-1$ and $\langle
d_{m},\alpha\rangle=\frac{1}{2(2m-1)}$. Without loss, we may assume
$d_{m}=\beta$, which is a short root. Note also that
$\tilde{d}_{m}=d_{m-1}+2d_{m}$ is also a long root and
$\\{d_{1},...,d_{m-1},\tilde{d}_{m}\\}$ forms a fundamental root system for
the long roots. Therefore, in $P$, we have a root sublattice of type
$D_{m}^{2}$ with the fundamental system $\\{a_{1},\dots,a_{m}\\}$ and
$\\{b_{1},\dots,b_{m}\\}$,

$\begin{array}[]{rclrcl}a_{1}-\cdots-&a_{m-2}&-a_{m-1},&b_{1}-\cdots-&b_{m-2}&-b_{m-1}\cr&|&&&|&\cr&a_{m}&&&b_{m}&\cr\end{array}$

such that $\tau$ permutes $a_{i}$ and $b_{i}$ and
$\frac{a_{i}+b_{i}}{2}=d_{i}$ for $i=1,...,m-1$ and
$\frac{a_{m}+b_{m}}{2}=2\beta+d_{m-1}=2\beta+\frac{a_{m-1}+b_{m-1}}{2}$.
Therefore $2\beta=\frac{a_{m}-a_{m-1}+b_{m}-b_{m-1}}{2}$ and $\langle
2\beta,a_{i}\rangle\in{\mathbb{Z}},\langle 2\beta,b_{i}\rangle\in{\mathbb{Z}}$
for all $i=1,...,m$, that is, $R(P)\subseteq N=\Lambda^{[2\beta]}$. Since
$2\beta\in N$ and has norm $2$, $N$ contains $D_{2m}$ and so the Coxeter
number of $N$ is greater than or equal to $2(2m-1)$, which contradicts the
choice of $V$.

Case $V_{1}=C_{4,2}A_{4,2}^{2}$:

Since $V_{1}=C_{4,2}A_{4,2}^{2}$ and $\det(L)=2^{10}4^{2}$,
$L=\sqrt{2}(D_{4}+E_{8})$ and $N^{\tau}\cong D_{4}+\sqrt{2}E_{8}$ in this
case. In particular, $N^{\tau}$ contains a root sublattice of type $D_{4}$.
Therefore, the possible types for $N$ are $D_{4}^{6}$, $D_{4}A_{5}^{4}$, and
$D_{5}^{2}A_{7}^{2}$. However, the lattices orthogonal to a sublattice of the
type $D_{4}+\sqrt{2}E_{8}$ do not contain $L_{A}(1^{12})$.

Case $V_{1}=F_{4,2}A_{8,2}$:

In this case, $n=18$ and $N_{0}=9$.

Let $a_{1},a_{2},b_{3},b_{4}$ be the fundamental roots for $F_{4}$ such that
$a_{1},a_{2}$ are short roots. We may assume $\beta=a_{1}$. Since $|\tau|=2$,
there is a Niemeier lattice $P$ such that
$V_{P}=(V_{\Lambda})^{[\tilde{g}^{2}]}$, which is also equal to
$V^{[9\alpha]}$. Set $s=g^{9}=\exp(2\pi i9\alpha(0))$. Then
$(F_{4,2}A_{8,2})^{<s>}$ contains $B_{4,2}A_{8,2}$ and the Coxeter number of
$P$ is less than or equal to $N_{0}=9$. Therefore, the root system of $P$ is
$A_{8}^{3}$. Since $V=V_{P}^{[\tilde{g}]}$, $\tau$ acts on one $A_{8}$ as a
diagram automorphism and permutes the other two $A_{8}$’s.

Since $N_{0}=9$ is odd, we have

$P=\Lambda_{4\beta}+{\mathbb{Z}}4\beta\quad\text{ and
}\quad\tilde{\beta}=2\beta.$

by the similar argument as in the case $V_{1}=B_{m}^{12/m}$. By the choice of
$\beta$, the two $A_{8}$’s permuted by $\tau$ is orthogonal to $\beta$; hence,
$N=\Lambda^{[\tilde{\beta}]}=\Lambda_{2\beta}+{\mathbb{Z}}2\beta$ contains
$A_{8}^{2}$.

Since $F_{4}$ is a full component, $N$ also contains $D_{4}$ as a root
sublattice, which is orthogonal to the above $A_{8}^{2}$. Therefore, $N\supset
D_{4}+A_{8}+A_{8}$. Moreover, the Coxeter number of $N$ is strictly less than
$n=18$ by our assumption. Then $N$ has the type $A_{9}^{2}D_{6}$. In this
case, $N_{\tau}$, the sublattice orthogonal to $D_{4}+\sqrt{2}A_{8}$, does not
contain a root sublattice of the type $A_{1}^{12}$ and we have a
contradiction.

This completes the proof of Theorem 5.1. ∎

##  6 Classification of $\tau$

In this section, we will show that $\tau\in Co.0$ defined by $W$-elements of
holomorphic VOAs of central charge 24 are contained in

${\cal P}_{0}=1A\cup 2A\cup 2C\cup 3B\cup 4C\cup 5B\cup 6E\cup 6G\cup 7B\cup
8E\cup 10F.$

Combining the results in Section 3 - 5, one can associated a pair of
$\tau\in{\cal P}_{0}$ and a $\tau$-invariant deep hole $\tilde{\beta}$ with
any holomorphic VOA $V$ of central charge $24$ with non-abelian weight one Lie
algebra $V_{1}$.

Set

${\cal P}=\left\\{\tau\in
Co.0\,\left|\begin{array}[]{l}{}^{\exists}\beta\in{\mathbb{Q}}\Lambda^{\tau}\mbox{
s.t. }\widehat{\tau}\exp(2\pi i\beta(0))\mbox{ can be realized as the reverse
}\\\ \mbox{automorphism of some orbifold construction given by a
$W$-element}\end{array}\right.\right\\}.$

The main result of this section is the following.

######  Proposition 6.1.

${\cal P}\subseteq{\cal P}_{0}$.

Since $\dim((V_{\Lambda})^{[\tilde{g}]})_{1}>24$, it is easy to see that ${\rm
rank}(\Lambda^{\tau})\geq 4$ and $|\tau|\leq 15$. Therefore, $\tau$ is in one
of the conjugacy classes in the following list (cf. [7]).

Table 1:  Conjugacy classes of $\tau\in Co.0$ with $\dim\Lambda^{\tau}\geq 4$ Class | frame shape | $\dim{\cal H}_{0}$ | $\phi(\tau)$ | Power | $\Lambda^{\tau}$ | Check for ${\cal P}_{0}$
---|---|---|---|---|---|---
$2A$ | $1^{8}2^{8}$ | $16$ | $1-1/2$ |  |  | $\in{\cal P}_{0}$
$-2A$ | $2^{16}/1^{8}$ | $8$ | $1-0$ |  |  | Lemma 6.8
$2C$ | $2^{12}$ | $12$ | $1-1/4$ |  |  | $\in{\cal P}_{0}$
$3B$ | $1^{6}3^{6}$ | $12$ | $1-1/3$ |  |  | $\in{\cal P}_{0}$
$3C$ | $3^{9}/1^{3}$ | $6$ | $1-0$ |  | $3E_{6}^{-1}$ | Lemma 6.8
$3D$ | $3^{8}$ | $8$ | $1-1/9$ |  | $3E_{8}$ | Lemma 6.3
$-4A$ | $1^{8}4^{8}/2^{8}$ | $8$ | $1-1/4$ | $-2A$ |  | Lemma 6.8
$4C$ | $1^{4}2^{2}4^{4}$ | $10$ | $1-1/4$ |  |  | $\in{\cal P}_{0}$
$-4C$ | $2^{6}4^{4}/1^{4}$ | $6$ | $1-0$ | $2A$ |  | Lemma 6.8
$4D$ | $2^{4}4^{4}$ | $8$ | $1-1/8$ | $2A$ | $2D_{4}+2D_{4}$ | Lemma 6.13
$4F$ | $4^{6}$ | $6$ | $1-1/16$ | $2C$ | $4I_{6}$ | Lemma 6.13
$5B$ | $1^{4}5^{4}$ | $8$ | $1-1/5$ |  |  | $\in{\cal P}_{0}$
$5C$ | $5^{5}/1^{1}$ | $4$ | $1-0$ |  |  | Lemma 6.8
$6C$ | $1^{4}2^{1}6^{5}/3^{4}$ | $6$ | $1-1/6$ | $-2A,3B$ |  | Lemma 6.8
$-6C$ | $2^{5}3^{4}6^{1}/1^{4}$ | $6$ | $1-0$ | $2A,3B$ | $6E_{6}^{-1}$ | Lemma 6.11
$-6D$ | $1^{5}3^{1}6^{4}/2^{4}$ | $6$ | $1-1/6$ | $2A,3C$ | $3E_{6}^{-1}$ | Lemma 6.8
$6E$ | $1^{2}2^{2}3^{2}6^{2}$ | $8$ | $1-1/6$ |  |  | $\in{\cal P}_{0}$
$-6E$ | $2^{4}6^{4}/1^{2}3^{2}$ | $4$ | $1-0$ | $-2A$ |  | Lemma 6.8
$6F$ | $3^{3}6^{3}/1^{1}2^{1}$ | $4$ | $1-0$ |  | $3D_{4}$ | Lemma 6.2
$6G$ | $2^{3}6^{3}$ | $6$ | $1-1/12$ |  |  | $\in{\cal P}_{0}$
$6I$ | $6^{4}$ | $4$ | $1-1/36$ | $2C,3D$ | $6I_{4}$ | Lemma 6.2
$7B$ | $1^{3}7^{3}$ | $6$ | $1-1/7$ |  |  | $\in{\cal P}_{0}$
$8E$ | $1^{2}2^{1}4^{1}8^{2}$ | $6$ | $1-1/8$ |  |  | $\in{\cal P}_{0}$
$10D$ | $1^{2}2^{1}10^{3}/5^{2}$ | $4$ | $1-1/10$ | $-2A$ |  | Lemma 6.8
$-10D$ | $2^{3}5^{2}10^{1}/1^{2}$ | $4$ | $1-0$ |  |  | Lemma 6.2
$-10E$ | $1^{3}5^{1}10^{2}/2^{2}$ | $4$ | $1-1/10$ | $5C$ |  | Lemma 6.8
$10F$ | $2^{2}10^{2}$ | $4$ | $1-1/20$ |  |  | $\in{\cal P}_{0}$
$-12E$ | $1^{2}3^{2}4^{2}12^{2}/2^{2}6^{2}$ | $4$ | $1-1/12$ | $-2A,-6E$ |  | Lemma 6.8
$-12H$ | $1^{1}2^{2}3^{1}12^{2}/4^{2}$ | $4$ | $1-1/12$ | $3C$ |  | Lemma 6.2
$12I$ | $1^{2}4^{1}6^{2}12^{1}/3^{2}$ | $4$ | $1-1/12$ | $-4C$ |  | Lemma 6.2
$-12I$ | $2^{2}3^{2}4^{1}12^{1}/1^{2}$ | $4$ | $1-0$ |  |  | Lemma 6.2
$12J$ | $2^{1}4^{1}6^{1}12^{1}$ | $4$ | $1-1/24$ | $2A,3B,6E$ |  | Lemma 6.2
$14B$ | $1^{1}2^{1}7^{1}14^{1}$ | $4$ | $1-1/14$ | $2A,7B$ |  | Lemma 6.2
$15D$ | $1^{1}3^{1}5^{1}15^{1}$ | $4$ | $1-1/15$ |  |  | Lemma 6.2

First we treat the cases when ${\rm rank}(V_{1})=4$, which will eliminate many
cases with $\dim\Lambda^{\tau}=4$.

######  Lemma 6.2.

If ${\rm rank}V_{1}=4$, then $\tau\in{\cal P}_{0}$.

[Proof] Suppose ${\rm rank}(V_{1})=4$. By [13], we know that $\dim V_{1}>24$
and $h_{j}^{\vee}/k_{j}=(\dim V_{1}-24)/24$ for any component ${\cal G}_{j}$;
thus $V_{1}$ must be of the type $B_{4,14},C_{4,10},D_{4,36}$ or
$G_{2,24}^{\oplus 2}$ and $|\tau|=14,10,6,12$, respectively. In these cases,
$N_{0}=1$. If $|\tau|=14$, we have $\tau=14B$. Then $\ell=14$ and
$\det(\Lambda^{\tau})={14}^{2}$. In this case, $\det(L)=14^{4}$ and
$\det(\sqrt{14}L^{*})=1$, which is not possible since $\sqrt{14}L^{*}$ is even
(see Section 4.2). If $|\tau|=10$, then $V_{1}=C_{4,10}$ and $N_{0}=1$ and so
there are $\mu\in L^{\ast}$ such that $\langle\alpha,\mu\rangle\equiv 1/10$
and $1/5$ $\pmod{{\mathbb{Z}}}$. Therefore, since we may choose $\beta$ from
$L^{\ast}$, both of $\tau$ and $\tau^{2}$ are of type zero and $\phi(\tau)<1$
and $\phi(\tau^{2})<1$, which implies $\tau=10F$ and $\tau\in{\cal P}_{0}$. If
$|\tau|=6$, $V_{1}=D_{4,36}$ and so there is $\mu\in L^{\ast}$ such that
$\langle\tau,\mu\rangle\equiv 1/6\pmod{{\mathbb{Z}}}$ and $\phi(\tau)<1$.
Therefore, $\tau=6I$ and $N_{0}=1$. Since $6\beta\in\Lambda^{\tau}$ which has
the Gram matrix $6I_{4}$, $\langle\beta,\beta\rangle\in 2{\mathbb{Z}}/12$.
However, $\phi(6I)=1/36$; thus $\tilde{g}=\widehat{\tau}\exp(2\pi i\beta(0))$
is not type zero and we have a contradiction. ∎

From now on, we may assume $\dim V^{\langle\tau\rangle}>4$.

######  Lemma 6.3.

$\tau\not\in 3D$.

[Proof] Recall that $\Lambda^{3D}\cong\sqrt{3}E_{8}$ (cf. [20]). If ${\rm
GCD}(N_{0},3)=1$, then $3N_{0}\beta\in\Lambda^{3D}$ and so
$\langle\beta,\beta\rangle\in 6{\mathbb{Z}}/9N_{0}^{2}=2{\mathbb{Z}}/3N_{0}$,
which contradicts $\phi(3D)+\frac{\langle\beta,\beta\rangle}{2}\in
1-1/9+\frac{2{\mathbb{Z}}}{3N_{0}}\subset{\mathbb{Z}}$. Therefore, $3|N_{0}$.
Since $\dim{\mathbb{C}}\Lambda^{3D}=8$, $N_{0}|h^{\vee}_{j}$ for all $j$ and
$|g|=N_{0}|\tau|={\rm LCM}(\\{r_{j}h_{j}^{\vee}\\})$, the only possible case
is $N_{0}=3$ and $V_{1}=A_{8}$. Since
$K_{0}/N_{0}=\langle\alpha,\alpha\rangle/2=\dim V_{1}/(\dim V_{1}-24)$, we
have $K_{0}/3=80/56=10/7$, which is not possible. ∎

Suppose Proposition 6.1 is false and let $V$ be a counterexample such that
$k={\rm rank}(V_{1})$ is maximal among all counterexamples. We also choose $V$
so that ${\rm Comm}(M({\cal H}),V)$ is largest among all counterexamples with
${\rm rank}(V_{1})=k$.

Let $\alpha$ be a $W$-element of $V_{1}$ and set $g=\exp(2\pi i\alpha(0))$. As
we discussed, an orbifold construction from $V$ by $g$ gives $V_{\Lambda}$.
Let $\tilde{g}=\widehat{\tau}\exp(2\pi i\beta(0))$ be the reverse automorphism
of $g$ corresponding to the above orbifold construction. Since $V$ is a
counterexample, $\tau\not\in{\cal P}_{0}$.

If $m$ divides $|\tau|$ and $m\not=1$, then we have

${\rm rank}(V^{[g^{m}]}_{1})>{\rm rank}V_{1}\quad\text{ or }\quad{\rm
Comm}(M({\cal H}),V^{[g^{m}]})>{\rm Comm}(M({\cal H}),V)$

by Proposition 2.3. Therefore, $V^{[g^{m}]}$ is not a counterexample for any
$1\not=m\mid|\tau|$. Let $L$ be an even lattice such that ${\rm Comm}({\rm
Comm}(M({\cal H}),V),V)\cong V_{L}$, where ${\cal H}$ is a Cartan subalgebra
of $V_{1}=\mathfrak{g}_{1,k_{1}}\oplus\cdots\oplus\mathfrak{g}_{r,k_{r}}$. Let
$\alpha=\sum_{i=1}^{r}\rho_{i}/h_{i}^{\vee}$ be a W-element of $V_{1}$.
Suppose $\langle\alpha,\alpha\rangle=\frac{2K_{0}}{N_{0}}$ with
$K_{0},N_{0}\in{\mathbb{Z}}$ and ${\rm GCD}(K_{0},N_{0})=1$.

Now consider the VOA $V^{[g^{N_{0}}]}$ obtained by an orbifold construction
from $V$ by $g^{N_{0}}=\exp(2\pi iN_{0}\alpha(0))$. Then $V^{[g^{N_{0}}]}$
also contains an abelian Lie subalgebra ${\cal
H}+\oplus_{j=0}^{|\tau|-1}U(jN_{0}\alpha)_{1}$ of rank 24. By viewing it as a
Cartan subalgebra of $(V^{[g^{N_{0}}]})_{1}$, there is an even unimodular
lattice $P$ of rank $24$ such that $V^{[g^{N_{0}}]}\cong V_{P}$. Furthermore,
there is $\mu\in{\mathbb{C}}\Lambda^{\tau}$ such that
$(V_{\Lambda})^{[\exp(2\pi i\mu(0))]}=V_{P}$ since
$(V^{[g^{N_{0}}]})^{[g]}=V_{\Lambda}$. In particular, $\tau$ fixes $\mu$ and
we may view $\tau$ as an isometry of $P$.

######  Remark 6.4.

Let $\tilde{g}^{\prime}\in{\rm Aut}(V_{P})$ be the reverse automorphism of
$g^{N_{0}}\in{\rm Aut}(V)$. By our choice, ${\cal
H}+\oplus_{j=0}^{|\tau|-1}U(jN_{0}\alpha)_{1}$ is a Cartan subalgebra of
$V_{P}$ and thus, there is a $\delta\in{\mathbb{C}}P^{\tau}$ and a standard
lift $\widehat{\tau}^{\prime}$ of $\tau$ such that
$\tilde{g}^{\prime}=\widehat{\tau}^{\prime}\exp(2\pi i\delta(0))$ and $\langle
N_{0}\alpha,\delta\rangle\equiv 1/|\tau|\pmod{{\mathbb{Z}}}$. Then
$V=(V_{P})^{[\widehat{\tau}^{\prime}\exp(2\pi i\delta(0))]}$ and the actions
of $\widehat{\tau}$ and $\widehat{\tau}^{\prime}$ on $U(jN_{0}\alpha)$ are
same; namely, they both act on $U(jN_{0}\alpha)$ as a multiple by a scalar
$e^{-2\pi ij/|\tau|}$.

Since $N_{0}\alpha\in L^{\ast}$, $V^{<g^{N_{0}}>}$ contains $V_{L}$ and
$L\subseteq{\cal H}={\mathbb{C}}\Lambda^{\tau}$. We may also view $V_{L}$ as a
subspace of $V_{P}^{<\widehat{\tau}^{\prime}\exp(2\pi i\delta(0))>}$. As a
conclusion, we have the following.

######  Theorem 6.5.

The VOA $V^{[e^{2\pi iN_{0}\alpha(0)}]}$ is isomorphic to a lattice VOA
$V_{P}$ for an even unimodular lattice $P$ of rank $24$. Furthermore,
$P^{\tau}=L+{\mathbb{Z}}N_{0}\alpha$ and $|g^{N_{0}}|=|\tau|$.

As we have shown in Theorem 6.5, $V^{[\exp(2\pi iN_{0}\alpha(0)]}$ is a
Niemeier lattice VOA $V_{P}$ and there is $\delta\in{\mathbb{C}}P^{\tau}$ such
that $\widehat{\tau}\exp(2\pi i\delta(0))$ is the reverse automorphism of
$\exp(2\pi iN_{0}\alpha(0))$.

######  Lemma 6.6.

By taking a suitable standard lift $\widehat{\tau}$ of $\tau$, we may assume
$\delta\in L^{\ast}$.

[Proof] By the definition of W-element, there is $\mu\in L^{\ast}$ such that
$\langle\alpha,\mu\rangle\equiv 1/N_{0}|\tau|\pmod{{\mathbb{Z}}}$. Therefore,
$\delta-\mu\in(P^{\tau})^{\ast}$ and there is $\xi\in P$ such that
$\pi(\xi)=\delta-\mu$. Set $\xi=\xi^{\prime}+\pi(\xi)$ with
$\xi^{\prime}\in{\cal H}^{\perp}$ Since $\exp(2\pi i\xi(0))=1$ and $\tau$ acts
on $({\cal H})^{\perp}$ fixed point freely, there is $\tilde{h}\in{\cal
H}^{\perp}$ such that

$\begin{array}[]{l}\exp(2\pi i\tilde{h}(0))^{-1}(\widehat{\tau}\exp(2\pi
i\delta(0))\exp(2\pi i\tilde{h}(0)=\exp(2\pi
i\tilde{h}(0))^{-1}\widehat{\tau}\exp(2\pi i\tilde{h}(0))\exp(2\pi
i\delta(0))\cr=\widehat{\tau}\exp(2\pi i\xi^{\prime}(0))\exp(2\pi
i\delta(0))=\widehat{\tau}\exp(2\pi i\xi(0))\exp(-2\pi i\pi(\xi)(0))\exp(2\pi
i\delta(0))\cr=\widehat{\tau}\exp(2\pi i\mu(0)).\end{array}$

Therefore, $\widehat{\tau}\exp(2\pi
i\delta(0))=\widehat{\tau}\exp(\mu(0))\exp(\pi\xi(0))$ is conjugate to
$\widehat{\tau}\exp(2\pi i\mu(0)$. Replacing $\widehat{\tau}$ by $\exp(2\pi
i\tilde{h}(0))^{-1}(\widehat{\tau}\exp(2\pi i\delta(0))\exp(2\pi
i\tilde{h}(0))$, we may assume $\delta\in L^{\ast}$. ∎

By Theorem 6.5, we have $P^{\tau}=L+{\mathbb{Z}}N_{0}\alpha$ and
$rN_{0}\alpha\in L$ if and only if $|\tau|$ divides $r$ and
$\langle\delta,N_{0}\alpha\rangle\equiv 1/|\tau|\pmod{{\mathbb{Z}}}$.
Therefore, $m\delta\not\in\pi(P)$ if $|\tau|\not|m$. Hence if $|\tau|\not|m$,
then the lowest weight of $\widehat{\tau}\exp(2\pi i\delta(0))$-twisted module
is greater than $\phi(\tau)$. Therefore, we have:

######  Lemma 6.7.

If $\phi(\tau^{m})\geq 1$, then $\widehat{\tau}^{m}\exp(2\pi
im\delta(0))$-twisted module does not contain a weight one element.

######  Lemma 6.8.

$<\tau>$ does not contain $-2A,3C,5C$.

[Proof] Let $m$ be a divisor of $|\tau|$. Suppose $|\tau^{m}|=p$ is a prime
and $\phi(\tau^{m})\geq 1$.

Let $\tilde{V}:=V^{[g^{pN_{0}}]}$. From the property of orbifold construction
and the choice of $\widehat{\tau}^{\prime}$ and $\delta$, we have
$\tilde{V}=(V_{P})^{[(\widehat{\tau}^{\prime})^{m}\exp(2\pi im\delta(0))]}$.
From Corollary 4.9 in [7], $\tau\not=-2A,3C,5C$ and so $m\not=1$ and
$\tilde{V}$ is not a counterexample.

Let $\tilde{\alpha}$ be a W-element of $\tilde{V}_{1}$ and
$\widehat{\sigma}\exp(2\pi i\mu(0))$ the reverse automorphism of $\exp(2\pi
i\tilde{\alpha}(0))$. Then $\sigma\in{\cal P}_{0}$ and we can use the results
in §3 $\sim$ 5.

Next we will show that ${\mathbb{C}}P^{\sigma}={\mathbb{C}}P^{\tau^{m}}$. As
we explained in Section 2, ${\cal H}+\oplus_{j=0}^{m-1}U(jpN_{0}\alpha)$ is a
Cartan subalgebra of $\tilde{V}_{1}$. Since $\langle\delta,\alpha\rangle\equiv
1/|\tau|\pmod{{\mathbb{Z}}}$, $\widehat{\tau^{\prime}}$ acts on
$U(jN_{0}\alpha)$ as a multiple by a scalar $e^{-2\pi ij/|\tau|}$ for all
$j\in{\mathbb{Z}}$. In particular, ${\mathbb{C}}P^{\tau^{m}}={\cal
H}+\oplus_{j=0}^{|\tau|-1}U(jpN_{0}\alpha)_{1}$ and so
${\mathbb{C}}P^{\sigma}={\mathbb{C}}P^{\tau^{m}}$.

We next show that if the order $p$ of $\tau^{m}$ is an odd prime, we have
$|\widehat{\sigma}|=p$ and if the order of $\tau^{m}$ is $2$ we have
$|\widehat{\sigma}|=2$ or $4$.

Since $\phi(\tau^{m})\geq 1$, $\widehat{\tau^{\prime}}^{m}\exp(2\pi
im\delta(0))$-twisted module contains no weight one element and so we have

$\tilde{V}_{1}=(V_{P})^{<\widehat{\tau^{\prime}}^{m}\exp(2\pi
im\delta(0))>}_{1}.$

Since $|\tau^{m}|=p$ is a prime, the root vectors of
$(V_{P})^{<\widehat{\tau^{\prime}}^{m}\exp(2\pi im\delta(0))>}_{1}$ are given
by

(1) $e^{\mu}$ with $\mu\in P^{\tau}$ or

(2) $e^{\mu}+\cdots+(\widehat{\tau^{\prime}}\exp(2\pi
im\delta(0)))^{p-1}(e^{\mu})$ with $\mu\in P$. In this case, it corresponds to
the root associated with $\pi(\mu)\in(P^{\tau})^{*}$ and
$\langle\pi(\mu),\pi(\mu)\rangle=2/\tilde{r}_{j}\tilde{k}_{j}$. Note also that

$\langle\pi(\mu),\pi(\mu)\rangle=\langle\pi(\mu),\frac{1}{p}\sum_{j=0}^{p-1}\tau^{mj}\mu\rangle=\langle\mu,\frac{1}{p}\sum_{j=0}^{p-1}\tau^{mj}\mu\rangle\in{\mathbb{Z}}/p$

Therefore, if $p$ is an odd prime, then $\tilde{r}_{j}\tilde{k}_{j}=1$ or $p$
and if $p=2$, then $\tilde{r}_{j}\tilde{k}_{j}=1,2,4$.

On the other hand, since $\sigma\in{\cal P}_{0}$, we have shown
$|\widehat{\sigma}|={\rm LCM}(\\{\tilde{r}_{j}\tilde{k}_{j}\mid j\\})$. If
$\tilde{r}_{j}\tilde{k}_{j}=1$ for all $j$, then we can easily get a
contradiction since ${\rm rank}(\widetilde{V}_{1})=8,6,4$, respectively.

Hence if the order $p$ of $\tau^{m}$ is an odd prime, we have
$|\widehat{\sigma}|=p$ and if the order of $\tau^{m}$ is $2$ we have
$|\widehat{\sigma}|=2$ or $4$. From the choice of $\sigma$, we have ${\rm
rank}({\mathbb{C}}\Lambda^{\sigma})={\rm
rank}({\mathbb{C}}\Lambda^{\tau^{m}})$. Therefore, if
$\tau^{m}=-2A,then\dim\Lambda^{-2A}=8$, but there is no $\sigma\in{\cal
P}_{0}$ of order $2$ or $4$ such that $\Lambda^{\sigma}=8$. If $\tau^{m}=3C$,
then ${\rm rank}\Lambda^{3C}=6$, but there is no $\sigma\in{\cal P}_{0}$ of
order $3$ with ${\rm rank}\Lambda^{\sigma}=6$. If $\tau=5C$, then ${\rm
rank}\Lambda^{5C}=4$, but there is no $\sigma\in{\cal P}_{0}$ of order $5$
with ${\rm rank}\Lambda^{\sigma}=4$. ∎

######  Lemma 6.9.

$\tau\not\in-4C$.

[Proof] Suppose $\tau\in-4C$. We note $V=(V_{P})^{[\widehat{\tau}\exp(2\pi
i\delta(0))]}$. Since $\phi(\tau)\geq 1$, $\widehat{\tau}\exp(2\pi
i\delta(0))$-twisted module does not contain elements of weight one. Recall
that $\widehat{\tau}\exp(2\pi i\delta(0))$ is the reverse automorphism of
$\exp(2\pi iN_{0}\alpha(0))$; the above statement means that there is no root
$\mu\in V_{1}$ such that $|\tau|\langle\mu,N_{0}\alpha\rangle=1$. Since
$|\tau|=4$, ${\rm LCM}(\\{r_{j}h^{\vee}_{j}/N_{0}\mid j\\})$ divides $2$,
which contradicts ${\rm LCM}(\\{r_{j}h^{\vee}_{j}/N_{0}\mid j\\})=|\tau|$. ∎

Therefore the remaining possibility of $\tau\not\in{\cal P}_{0}$ are
$4D,4F,-6C$. Then since $4D^{2}=2A,4F^{2}=2C,(-6C)^{2}=3B,(-6C)^{3}=2A$, non-
trivial powers of $\tau$ are in ${\cal P}_{0}$.

Furthermore, $\phi(4D)=1-1/8$, $\phi(4F)=1/16$, $\phi(-6C)=1$. Namely,
$\widehat{\tau}$ is not of type zero or $\phi(\tau)=1$.

We note that since $V=V_{\Lambda}^{[\widehat{\tau}\exp(2\pi i\beta(0))]}$,
$\tilde{V}=V^{[\widetilde{g}^{m}]}$ is a holomorphic VOA for any $m||\tau|$.
Assume $m\not=1$. Unfortunately, $\widetilde{g}^{m}$ is not necessary to be a
reverse automorphism for an automorphism defined by W-element in $\tilde{V}$.
Let $\tilde{\alpha}$ be a W-element of $\tilde{V}_{1}$ and
$\widehat{\sigma}\exp(2\pi i\mu(0))$ be a reverse automorphism of $\exp(2\pi
i\tilde{\alpha}(0))$.

######  Lemma 6.10.

Let $1\not=m||\tau|$. If $\tau^{m}=2A,2C,3B$, then $\sigma$ is conjugate to
$\tau^{m}$.

[Proof] If $\tau^{m}=2A,2C,3B$, then
$\tilde{V}:=V_{P}^{[\widehat{\tau}^{m}\exp(2\pi im\delta(0))]}$ is not a
counterexample. Let $\tilde{\alpha}$ be a W-element of $\tilde{V}_{1}$ and
$\widehat{\sigma}\exp(2\pi i\mu(0))$ a reverse automorphism for $\exp(2\pi
i\tilde{\alpha}(0))$. In particular, $\sigma\in{\cal P}_{0}$.

As we have shown, ${\rm rank}({\mathbb{C}}\Lambda)^{\tau^{m}}={\rm
rank}\tilde{V}_{1}={\rm rank}({\mathbb{C}}\Lambda)^{\sigma}$. Let
$\tilde{{\cal H}}$ be a Cartan subalgebra of $\tilde{V}$ and set ${\rm
Comm}({\rm Comm}(\tilde{{\cal H}},\tilde{V}),\tilde{V})=V_{\tilde{L}}$ with an
even lattice $\tilde{L}$.

Since $\sigma\in{\cal P}_{0}$, $\sqrt{|\widehat{\sigma}|}(\tilde{L})^{\ast}$
is an even lattice by Lemma 5.1. On the other hand, since $\tau^{m}\in{\cal
P}_{0}$ and $\tilde{V}=(V_{P})^{[\widehat{\tau}^{m}\exp(2m\pi i\delta(0))]}$,
we have $\phi(\tau^{m})=1-\frac{1}{|\widehat{\tau}^{m}|}$ and
$e^{m\delta}\otimes t\in\tilde{V}$ with ${\rm
wt}(e^{m\delta})\equiv\frac{1}{|\widehat{\tau}^{m}|}\pmod{{\mathbb{Z}}}$ from
an orbifold construction by an automorphism with positive frame shape.
Therefore $|\widehat{\tau}^{m}|$ have to divide $|\widehat{\sigma}|$.

If $\tau^{m}=2A$, then ${\rm rank}({\mathbb{C}}\Lambda)^{\tau^{m}}=16$ and
$\sigma=2A$. If $\tau^{m}=2C$, then ${\rm
rank}({\mathbb{C}}\Lambda)^{\sigma}=12$ and so $\sigma=2C$ or $3B$. If
$\sigma=3B$, then $|\widehat{\sigma}|=3$ is not a multiple of
$|\widehat{\tau^{m}}|=2$. If $\tau^{m}=3B$, then ${\rm
rank}({\mathbb{C}}\Lambda^{\sigma})=12$ and so $\sigma=2C$ or $3B$. Since
$|\widehat{\tau}^{m}|=3$ does not divide $|\widehat{\sigma}|=4$ if
$\sigma=2C$. Hence $\sigma=3B$. ∎

As a corollary, for $\mu\in L^{\ast}$, if
$\langle\mu,N_{0}\alpha\rangle\in\frac{{\mathbb{Z}}}{m}$, then $S(\mu)\otimes
e^{\mu}\in V^{[|\tau|g/m]}$ and so we have
$|\widehat{\tau}^{m}|\langle\mu,\mu\rangle\in 2{\mathbb{Z}}$.

######  Lemma 6.11.

$\tau\not=-6C$.

[Proof] Since $\phi(-6C)=1$, $\widehat{\tau}\exp(2\pi i\delta(0))$-twisted
module does not contain weight one elements. Therefore,
$r_{j}h^{\vee}_{j}/N_{0}=2$ or $3$ and the both have to appear. Recall that
$(-6G)^{3}=2A$ and $(-6G)^{2}=3B$. If $r_{j}h^{\vee}_{j}/N_{0}=2$, then
$\exp(2\pi iN_{0}\alpha(0))^{3}$ fixes all elements in ${\cal G}_{j}$ and so
$\sqrt{3}L_{j}^{\ast}$ is an even lattice since $2A\in{\cal P}_{0}$, where
$L_{j}^{\ast}$ is the co-root lattice of ${\cal G}_{j}$. If
$r_{j}h^{\vee}_{j}/N_{0}=3$, then $\exp(2\pi iN_{0}\alpha(0))^{2}$ fixes all
elements in ${\cal G}_{j}$ and so $\sqrt{2}L_{j}^{\ast}$ is an even lattice.
Therefore $\sqrt{6}L^{\ast}$ is an even lattice and $K_{0}-N_{0}=1$ and
$h_{j}^{\vee}=k_{j}N_{0}$. If $N_{0}=1$, then $r_{j}h^{\vee}_{j}=2$ and $3$
and so ${\cal G}_{j}=A_{1}$ or $A_{2}$, which contradicts to $\dim V_{1}>24$
since ${\rm rank}V_{1}=6$. If $N_{0}=2$, $r_{j}h^{\vee}_{j}=4$ and $6$ and so
${\cal G}_{1}=A_{3}$ and ${\cal G}_{2}=A_{5}$, which contradicts ${\rm
rank}V_{1}=6$. If $N_{0}=3$, then one of $r_{2}h_{2}^{\vee}=9$ and so ${\cal
G}_{2}=A_{8}$, which contradict ${\rm rank}V_{1}=6$. If $N_{0}=4$, then one of
$r_{1}h_{1}^{\vee}=8$ and ${\cal G}_{2}=G_{2}$. So, ${\cal
G}_{1}=A_{7},C_{3},D_{5}$. Since ${\rm rank}V_{1}=6$ and ${\cal
G}_{j}\not=A_{1}$, we have a contradiction. If $N_{0}>4$,
$r_{1}h_{2}^{\vee}\geq 10$ and $r_{2}h_{2}^{\vee}\geq 15$, which contradicts
${\rm rank}V_{1}=6$. ∎

So the remaining cases are $\tau=4D$ or $4F$. Recall that a short root in
${\cal G}_{j}$ is called a shortest root if $r_{j}h_{j}^{\vee}={\rm
LCM}(r_{i}h_{i}^{\vee}|i)=|\tau|N_{0}$ (or equivalently $r_{j}k_{j}={\rm
LCM}(r_{i}k_{i}|i)$). Since $|\tau|=4$ and ${\rm
LCM}(r_{i}h_{i}^{\vee}/N_{0}\mid i)=|\tau|=4$, there is a shortest root. Say,
${\cal G}_{1}$ contains a shortest root.

######  Lemma 6.12.

If $\tau=4D$ or $4F$, then $\exp(2N_{0}\pi i\alpha(0))^{2}$ fixes a root
vector associated with a shortest root.

[Proof] Since ${\rm LCM}(\\{r_{j}h_{j}^{\vee}/N_{0}\mid i\\}=|\tau|$ and
$h_{j}^{\vee}$ is divisible by $N_{0}$, $r_{j}$ divides $|\tau|$. Therefore,
${\cal G}_{j}\not=G_{2}$.

If ${\cal G}_{1}\not\cong B_{m}$, then there is a shortest root which is a sum
of two shortest roots in the fundamental root system and thus $\exp(2N_{0}\pi
i\alpha(0))^{2}$ fixes a root vector associated with a shortest root in ${\cal
G}_{1}$. So we may assume ${\cal G}_{1}=B_{m}$; however,
$r_{1}h_{1}^{\vee}=2(2m-1)$ in this case, which contradicts that $|\tau|=4$
divides $r_{1}h_{1}^{\vee}$. ∎

######  Lemma 6.13.

$\tau\not=4D$ nor $4F$.

[Proof] Suppose false. Set $\tilde{V}=V^{[\exp(4N_{0}\pi i\alpha(0))]}$. We
may view that $\tau$ is an automorphism of $\tilde{V}$ of order $2$. Let
$\tilde{{\cal H}}$ be a Cartan subalgebra of $\tilde{V}_{1}$. Then there is an
even lattice $\tilde{L}$ such that ${\rm Comm}({\rm Comm}(M(\tilde{{\cal
H}}),\tilde{V}),\tilde{V})\cong V_{\tilde{L}}$. Let $\tilde{\alpha}$ be a
W-element of $\tilde{V}_{1}$ and $\widehat{\sigma}\exp(2\pi i\xi(0))$ the
reverse automorphism of $\exp(2\pi i\tilde{\alpha}(0))$. Since $4D^{2}=2A$ and
$4F^{2}=2C$, we have $\sigma=2A$ and $2C$, respectively by the same argument
as in the proof of Lemma 6.10.

Since $\tilde{V}^{[\exp(2\pi i\alpha(0)]}=V_{\Lambda}$, there is no root
orthogonal to $\alpha$ in $\tilde{V}$. Therefore, we can define positive roots
by the condition $\langle\mu,\alpha\rangle>0$. Note that $\alpha$ is fixed by
$\tau$. Therefore, if $\mu$ is a positive root, then so is $\tau^{i}(\mu)$ for
any $i$. By Lemma 6.12, there is a shortest root $\mu\in L^{\ast}$ such that
$s\otimes e^{\mu}\in V^{<\exp(2\pi
iN_{0}\alpha(0))^{2}>}\subseteq\tilde{V}=V^{[\exp(4\pi iN_{0}\alpha(0))]}$.

Since $\tau$ has positive frame shape, we may choose $\delta$ in $L^{\ast}$.
Note that $\phi(\tau)=1-\frac{1}{2\ell}$ for $\tau\in 4C$ or $4D$. If
$\langle\delta,\delta\rangle/2\in\frac{1}{\ell}{\mathbb{Z}}$, then the
irreducible $\widehat{\tau}\exp(2\pi i\delta(0))$-twisted module has no
subspaces with integral weights, which is not possible. Therefore, it suffices
to show that $\sqrt{\ell}L^{\ast}$ is an even lattice or equivalently,
$\ell\langle\mu,\mu\rangle\in 2{\mathbb{Z}}$ since $\mu$ is a shortest root.
Without loss, we may assume $\langle\mu,\alpha\rangle>0$. Then we have one of
the following two cases:
(1) $\mu$ is still a root in $\tilde{V}=V^{[\exp(4N_{0}\pi i\alpha(0))]}$ or
(2) $s\otimes e^{\mu}=s^{\prime}\otimes
e^{\mu^{\prime}}+\tau(s^{\prime})\otimes e^{\tau(\mu^{\prime})}$ with a
positive root $\mu^{\prime}\in{\mathbb{C}}\Lambda^{<\tau^{2}>}$ and
$\mu=\frac{\mu^{\prime}+\tau(\mu^{\prime})}{2}$. In particular,
$\mu^{\prime}+\tau(\mu^{\prime})$ is not a root. Note that $\mu^{\prime}$ and
$\tau(\mu^{\prime})$ are both positive roots of the same length and hence
$\frac{2\langle\mu^{\prime},\tau(\mu^{\prime})\rangle}{\langle\mu^{\prime},\mu^{\prime}\rangle}=0,-1$
or $2$ but $-1$ is not possible since $\mu^{\prime}+\tau(\mu^{\prime})$ is not
a root. We have
$t=\frac{\langle\mu^{\prime},\tau(\mu^{\prime})\rangle}{\langle\mu^{\prime},\mu^{\prime}\rangle}\in{\mathbb{Z}}$.

By Lemma 6.10, $\tau^{2}$ is conjugate to $\sigma$ and thus
$|\widehat{\tau}|=2|\widehat{\sigma}|$. Since $\tilde{V}$ is not a
counterexample, $\sigma\in{\cal P}_{0}$ and so
$|\widehat{\sigma}|\langle\mu,\mu\rangle\in 2{\mathbb{Z}}$ for the case (1)
and $|\widehat{\sigma}|\langle\mu^{\prime},\mu^{\prime}\rangle\in
2{\mathbb{Z}}$ for the case (2). For the case (1),
$\ell\langle\mu,\mu\rangle\in 4{\mathbb{Z}}$. For the case (2),
$|\widehat{\sigma}|\langle\tau(\mu^{\prime}),\mu^{\prime}\rangle=t|\widehat{\sigma}|\langle\mu^{\prime},\mu^{\prime}\rangle\in
2{\mathbb{Z}}$. Then we have
$\ell\langle\mu,\mu\rangle=2|\widehat{\sigma}|\langle\frac{\mu^{\prime}+\tau(\mu^{\prime})}{2},\frac{\mu^{\prime}+\tau(\mu^{\prime})}{2}\rangle=|\widehat{\sigma}|\langle\mu^{\prime}+\tau(\mu^{\prime}),\mu^{\prime}\rangle\in
2{\mathbb{Z}}$ as desired. ∎

This completes the proof of ${\cal P}\subseteq{\cal P}_{0}$.

##  7 Reverse construction

As we have already shown, one can define a pair of $\tau\in{\cal P}_{0}$ and a
$\tau$-invariant deep hole $\tilde{\beta}$ for any holomorphic VOA $V$ of
central charge $24$ with non-abelian $V_{1}$ by choosing a $W$-element
$\alpha$ of $V_{1}$ and considering the reverse automorphism $\tilde{g}\in{\rm
Aut}(\Lambda)=Co.0$ of $g=\exp(2\pi i\alpha(0))\in{\rm Aut}(V)$. The pair
$(\tau,\tilde{\beta})$ satisfies the following conditions:

  * (C1)

$\tau\in{\cal P}_{0}$ and $\tilde{\beta}$ is a $\tau$-invariant deep hole of
Leech lattice $\Lambda$ with $\langle\tilde{\beta},\tilde{\beta}\rangle=2$;

  * (C2)

the Coxeter number $h$ of
$N=\Lambda_{\tilde{\beta}}+{\mathbb{Z}}\tilde{\beta}$ is divisible by
$|\tau|$;

  * (C3)

$N_{\tau}=L_{A}(c_{\tau})$ with $c_{\tau}$ as defined in Appendix,

where
$\Lambda_{\tilde{\beta}}=\\{\lambda\in\Lambda\mid\langle\lambda,\tilde{\beta}\rangle\in{\mathbb{Z}}\\}$
and $N_{\tau}$ denotes an orthogonal complement of $N^{\tau}$ in $N$. We also
use the same notation $\tau$ to denote the isometry of $N$ induced from $\tau$
on $\Lambda$.

Let ${\cal T}$ be the set of pairs satisfying the conditions (C1) to (C3). In
this section, we will study the reverse construction. Take a pair
$(\tau,\tilde{\beta})\in{\cal T}$. Assume that $\tau$ has the frame shape
$\prod m^{a_{m}}$. Then $N_{\tau}>\oplus_{m\neq 1}A_{m-1}^{a_{m}}$ and the
conformal weight of the $\widehat{\tau}$-twisted module of $V_{\Lambda}$ is
given by $\phi(\tau)=1-\frac{1}{\ell}$, where $\ell=|\widehat{\tau}|$.

Let $\beta=\frac{1}{\sqrt{\ell}}\varphi^{-1}(\tilde{\beta})$ and define
$\tilde{g}=\widehat{\tau}\exp(2\pi i\beta(0))\in{\rm Aut}(V_{\Lambda})$. Note
that $\langle\beta,\beta\rangle=\frac{2}{\ell}$. Set
$N=\Lambda_{\tilde{\beta}}+{\mathbb{Z}}\tilde{\beta}$ and let $h$ be the
Coxeter number of $N$. Since $-\tilde{\beta}$ is a deep hole of $\Lambda$,
$N^{(1)}$ contains an affine fundamental root system of rank $24$ and
$N=\oplus_{k=0}^{h-1}N^{(k)}$, where
$N^{(k)}=\Lambda_{\tilde{\beta}}-k\tilde{\beta}$.

We may assume that $\hat{\tau}^{|\tau|}=\exp(2\pi i\delta(0))$ for some
$\delta$ fixed by $\tau$ and take $\delta=0$ if $\hat{\tau}^{|\tau|}=1$. For
$i\equiv k\mod 2|\tau|$, the irreducible $\tilde{g}^{k}$-twisted modules are
as follows:

$T^{k}=V_{\Lambda}[\tilde{g}^{k}]=\begin{cases}{\mathbb{C}}[-k\beta+(\Lambda^{\tau^{i}})^{*}]\otimes
M(1)[\tau^{i}]\otimes T_{\tau^{i}}&\text{ if }0\leq i<|\tau|,\\\
{\mathbb{C}}[-k\beta+\delta+(\Lambda^{\tau^{i}})^{*}]\otimes
M(1)[\tau^{i}]\otimes T_{\tau^{i}}&\text{ if }|\tau|\leq i<2|\tau|\\\
\end{cases}$

(see [34, Propositions 6.1 and 6.2] and [11, Remark 4.2] for detail). In
particular, $T^{1}$ contains a weight one element $e^{\beta}\otimes t$ with
$t\in T_{\tau}$, which is the lowest weight element of $T^{1}$, and so
$T^{1}_{{\mathbb{Z}}}\not=0$ and the orbifold construction gives a holomorphic
VOA $V:=V^{[\tilde{g}]}$ of central charge $24$ and $V_{1}$ is non-abelian
since $V_{1}$ contains a root vector $e^{\beta}\otimes t$.

One of the main purposes of this section is to determine an equivalent
relation $\sim$ on ${\cal T}$ so that two pairs $(\tau,\tilde{\beta})$ and
$(\tau^{\prime},\tilde{\beta}^{\prime})$ define isomorphic VOAs if and only if
$(\tau,\tilde{\beta})\sim(\tau^{\prime},\tilde{\beta}^{\prime})$. It is clear
that $\tau$ and $\tau^{\prime}$ are conjugates in $O(\Lambda)$ and
$\tilde{\beta}$ and $\tilde{\beta}^{\prime}$ are equivalent deep holes of the
Leech lattice $\Lambda$ if $(\tau,\tilde{\beta})$ and
$(\tau^{\prime},\tilde{\beta}^{\prime})$ define isomorphic VOAs.

For two equivalent deep holes $\tilde{\beta}$ and $\tilde{\beta}^{\prime}$,
there are $\sigma\in O(\Lambda)$ and $\lambda\in\Lambda$ such that
$\tilde{\beta}^{\prime}=\sigma(\tilde{\beta}-\lambda)$. Since
$\tilde{\beta}^{\prime}$ is $\tau^{\prime}$-invariant, $\tilde{\beta}-\lambda$
is $\sigma^{-1}\tau^{\prime}\sigma$-invariant. Moreover,
$\langle\tilde{\beta},\lambda\rangle\in{\mathbb{Z}}$ since
$\langle\tilde{\beta},\tilde{\beta}\rangle=\langle\tilde{\beta}^{\prime},\tilde{\beta}^{\prime}\rangle=2$.
In this case,
$\Lambda_{\tilde{\beta}}+\tilde{\beta}=\Lambda_{\tilde{\beta}-\lambda}+(\tilde{\beta}-\lambda)$.
Therefore, up to the action of $O(\Lambda)$, we may identify
$N=\Lambda_{\tilde{\beta}}+{\mathbb{Z}}\tilde{\beta}$ with
$N^{\prime}=\Lambda_{\tilde{\beta}^{\prime}}+{\mathbb{Z}}\tilde{\beta}^{\prime}$.
We define an equivalent relation on ${\cal T}$ as follows:

$(\tau,\tilde{\beta})\sim(\tau^{\prime},\tilde{\beta}^{\prime})$ if and only
if
(1) $\tilde{\beta}$ and $\tilde{\beta}^{\prime}$ are equivalent deep holes of
the Leech lattice $\Lambda$, i.e., there are $\sigma\in O(\Lambda)$ and
$\lambda\in\Lambda$ such that
$\tilde{\beta}^{\prime}=\sigma(\tilde{\beta}-\lambda)$;
(2) $\tau$ is conjugate to $\sigma^{-1}\tau^{\prime}\sigma$ in $O(N)$.
Note that $(\sigma^{-1}\tau^{\prime}\sigma,\tilde{\beta}-\lambda)\in{\cal T}$
and
$N=\Lambda_{\tilde{\beta}}+{\mathbb{Z}}\tilde{\beta}=\Lambda_{\tilde{\beta}-\lambda}+{\mathbb{Z}}(\tilde{\beta}-\lambda)$.
Moreover, $\tau$ and $\tau^{\prime}$ are conjugate in $O(\Lambda)$ since they
have the same frame shape by (2).

We will prove the following main theorem.

######  Theorem 7.1.

There is a one-to-one correspondence between the set of isomorphism classes of
holomorphic VOA $V$ of central charge $24$ having non-abelian $V_{1}$ and the
set ${\cal T}/\sim$ of equivalence classes of pairs $(\tau,\tilde{\beta})$ by
$\sim$.

###  7.1 W-element and the automorphism $\tilde{g}$

Let $(\tau,\tilde{\beta})\in{\cal T}$. Set
$\beta=\frac{1}{\sqrt{\ell}}\varphi^{-1}(\tilde{\beta})$ and define
$\tilde{g}=\widehat{\tau}\exp(2\pi i\beta(0))\in{\rm Aut}(V_{\Lambda})$. Let
$V=V^{[\tilde{g}]}$ be the holomorphic VOA obtained by an orbifold
construction from $V_{\Lambda}$ and $\tilde{g}$. We will show that there is a
$W$-element $\alpha$ of $V_{1}$ such that $\tilde{g}$ can be viewed as a
reverse automorphism of $\exp(2\pi i\alpha(0))\in{\rm Aut}(V)$.

First we will prove the following theorem.

######  Theorem 7.2.

Let $h$ be the Coxeter number of
$N=\Lambda_{\tilde{\beta}}+{\mathbb{Z}}\tilde{\beta}$. Then $|\tilde{g}|=h$.

We need several lemmas.

######  Lemma 7.3.

$s\beta\in\pi(\Lambda)$ if and only if $h|s$.

[Proof] Since $\sqrt{\ell}\varphi((\Lambda^{\tau})^{*})=\Lambda^{\tau}$, we
have $s\beta\in\pi(\Lambda)=(\Lambda^{\tau})^{\ast}$ if and only if
$s\tilde{\beta}\in\Lambda^{\tau}$ if and only if $h|s$. ∎

######  Lemma 7.4.

We have $\tilde{g}^{2h}=1$. Furthermore, $\tilde{g}^{h}=1$ if
$2C\not\in<\tau>$.

[Proof] By (C3) and $\tau\in{\cal P}_{0}$, $N_{\tau}$ contains a root
sublattice $S\cong A_{|\tau|-1}$. For a root $\mu$ of $S$, there is
$\lambda\in\Lambda$ and $m\in{\mathbb{Z}}$ such that
$\mu=\lambda+m\tilde{\beta}\in N_{\tau}\subseteq{\mathbb{C}}\Lambda_{\tau}$.
Then

$0=\sum_{i=0}^{|\tau|-1}\tau^{i}\mu=\sum_{i=0}^{|\tau|-1}\tau^{i}\lambda+|\tau|m\tilde{\beta}\in\Lambda^{\tau}.$

Since $h$ is the smallest positive integer satisfying
$h\tilde{\beta}\in\Lambda^{\tau}$ and $\tau$ is fixed point free on $S$ with
order $|\tau|$, we have $m=h/|\tau|$. Moreover, we also have
$\frac{h}{|\tau|}\tilde{\beta}\in\pi(\Lambda)\in(\Lambda^{\tau})^{\ast}$ and
so
$\langle\frac{h\tilde{\beta}}{|\tau|},\Lambda^{\tau}\rangle\subseteq{\mathbb{Z}}$.
Hence
${\mathbb{Z}}\supseteq\frac{h}{|\tau|}\langle\sqrt{\ell}\varphi(\beta),\sqrt{\ell}\varphi((\Lambda^{\tau})^{\ast})\rangle=\langle\frac{h\ell}{|\tau|}\beta,(\Lambda^{\tau})^{\ast}\rangle$
and so $\frac{\ell}{|\tau|}h\beta\in\Lambda$, which means
$(\exp(2\pi\beta(0))^{\frac{\ell h}{|\tau|}}=1$. In particular, if
$2C\not\in<\tau>$, then $\ell=|\tau|$ and $\exp(2\pi i\beta(0))^{h}=1$. ∎

From now on, we may assume $\tau=2C,6G,10F$. In particular, $|\tau|=2s$ for
some $s=1,3$ or $5$. Note that $\sigma=\tau^{s}$ is a $2C$-element.

######  Lemma 7.5.

Let $\tau\in 2C,6G$, or $10F$. Let $\tilde{\beta}$ be a $\tau$-invariant deep
hole satisfying (C1)-(C3). Let $h$ be the Coxeter number of
$N=\Lambda_{\beta}+{\mathbb{Z}}\beta$. Then $h/|\tau|$ is odd.

[Proof] By our assumptions, $N_{\tau}\cong L_{A}(c_{\tau})$ as defined in
Appendix; namely,

$N_{\tau}\supset\bigoplus_{m_{i}||\tau|,m_{i}\neq 1}A_{m_{i}-1}^{a_{i}}$

as a full rank sublattice if the frame shape of $\tau$ is $\prod
m_{i}^{a_{i}}$. Moreover,

$[N_{\tau}:\bigoplus_{m_{i}||\tau|,m_{i}\neq
1}A_{m_{i}-1}^{a_{i}}]=|\tau|\quad\text{ and }\quad
R(N_{\tau})=\bigoplus_{m_{i}||\tau|,m_{i}\neq 1}A_{m_{i}-1}^{a_{i}}.$

Since there are only $16$ Niemeier lattices whose Coxeter number is even, it
is straightforward to determine all Niemeier lattices $N$ such that
$L_{A}(c_{\tau})\subset N$ as a direct summand. Indeed,
$R(N)=A_{1}^{24},D_{4}^{6},A_{5}^{4}D_{4}$, $D_{6}^{4},A_{9}^{2}D_{6}$,
$D_{8}^{3}$, $A_{17}E_{7}$, $D_{12}^{2}$, or $D_{24}$ if $\tau\in 2C$;
$R(N)=A_{5}^{4}D_{4}$ or $A_{17}E_{7}$ if $\tau\in 6G$; $R(N)=A_{9}^{2}D_{6}$
if $\tau\in 10F$.

It turns out that $h/|\tau|$ is odd for all possible cases. ∎

The following lemma can be verified easily using the definition of $A_{m-1}$.

######  Lemma 7.6.

Let $m=2s$ be an even integer and let $\tau$ be a Coxeter element in
$\mathrm{Weyl}(A_{m-1})$. Then the $(-1)$-eigenlattice of $\tau^{s}$ on
$A_{m-1}$ is isometric to
$A_{1}^{s}={\mathbb{Z}}x_{1}\oplus\cdots\oplus{\mathbb{Z}}x_{s}$. Moreover,
the vector $\frac{1}{2}(x_{1}+\cdots+x_{s})\in s\gamma+A_{m-1}$, where
$\gamma+A_{m-1}$ is a generator of $A_{m-1}^{*}/A_{m-1}$.

######  Lemma 7.7.

For $\tau=2C,6G,10F$ and $v,u\in\Lambda^{\tau}$, $\langle v,u\rangle\in
2{\mathbb{Z}}$.

[Proof] We note $\Lambda^{\tau}\subseteq\Lambda^{\sigma}$. Since
$\ell(\sigma)=4$, $\Lambda^{\sigma}=2(\Lambda^{\sigma})^{\ast}$. Hence
$\frac{v}{2}\in(\Lambda^{\sigma})^{\ast}$ and $\langle
v,u\rangle=2\langle\frac{v}{2},u\rangle\in 2{\mathbb{Z}}$. ∎

Now now on, we set $p=h/|\tau|$, which is an odd integer.

######  Proposition 7.8.

For $\tau=2C,6G,10F$ and
$v\in(\Lambda_{\tilde{\beta}}+sp\tilde{\beta})^{\tau}$, we have $\langle
v,v\rangle\equiv 2\pmod{4}$.

[Proof] We first note that $(\Lambda_{\tilde{\beta}}+\tilde{\beta})_{2}$ forms
an affine fundamental system of roots of $N$. Choose a fundamental root system
$X$ from $\Lambda_{\tilde{\beta}}+{\mathbb{Z}}\tilde{\beta}$ such that
$\tilde{\beta}\in X$. Let $\rho$ be the corresponding Weyl vector and consider
$\xi=\exp(2\pi i\rho(0)/h)\in{\rm Aut}(V_{N})$.

Although $\tau$ does not fix $X$, $\tau$ preserves
$(\Lambda_{\tilde{\beta}}+\tilde{\beta})_{2}$ and $\xi$ acts on
$\\{e^{\alpha}\mid\alpha\in(\Lambda_{\tilde{\beta}}+\tilde{\beta})_{2}\\}$
with the same eigenvalue; hence the commutator $[\tau,\xi]=1$. Therefore we
can induce $\tau$ to an automorphism of $(V_{N})^{[\xi^{j}]}$ for any $j$. As
it is well-known, $(V_{N})^{[\xi]}\cong V_{\Lambda}$. Now consider the VOA
obtained by an orbifold construction using $V_{N}$ and $\xi^{2}$. Then
$(V_{N})^{[\xi^{2}]}\cong V_{M}$ for some Niemeier lattice $M$ and the Coxeter
number of $M$ is $2$, i.e, $R(M)=A_{1}^{24}$. It is clear that

$\Lambda_{\tilde{\beta}}+{\mathbb{Z}}sp\tilde{\beta}=\Lambda_{\tilde{\beta}}\cup(\Lambda_{\tilde{\beta}}+sp\tilde{\beta})\subset
M.$

By Lemma 7.6, it is easy to check that

$L_{A}(c_{2C})\cong
M_{\sigma}<\Lambda_{\tilde{\beta}}+{\mathbb{Z}}sp\tilde{\beta}=\Lambda_{\tilde{\beta}}\cup(\Lambda_{\tilde{\beta}}+sp\tilde{\beta})\subset
M.$

Then
$(\Lambda_{\tilde{\beta}}+sp\tilde{\beta})^{\tau}<M^{\tau}<M^{\tau^{s}}=(L_{A}(c_{2C}))^{\perp}$.

Therefore, it suffices to check for the case $\tau\in 2C$ and $N\cong
N(A_{1}^{24})$. In this case, we may assume $N$ contains
$A_{1}^{24}=\oplus_{i=1}^{24}{\mathbb{Z}}x_{i}$ and $N/(A_{1}^{24})$ is the
binary Golay code $G_{24}$ of length $24$ and $\tau(x_{i})=x_{i}$ for
$i=1,...,12$ and $\tau(x_{i})=-x_{i}$ for $i=13,...,24$. We may also assume
$\tilde{\beta}=x_{1}$. By (C3),
$N_{\tau}=<x_{13},...,x_{24},\frac{x_{13}+...+x_{24}}{2}>$. Then
$\\{13,...,24\\}$ is a dodecad of $G_{24}$ and so
$N^{\tau}=(N_{\tau})^{\perp}=<x_{1},...,x_{12},\frac{x_{1}+...+x_{12}}{2}>$.

For a Weyl vector $\rho$ of $N$, we may choose $x_{i}$ such that $\exp(2\pi
i\rho(0)/2)(x_{i})=\sqrt{-1}x_{i}$ and so $\exp(2\pi
i\rho(0)/2)(\frac{x_{i}+...+x_{12}}{2})=\sqrt{-1}^{12}(\frac{x_{i}+...+x_{12}}{2})=(\frac{x_{i}+...+x_{12}}{2})$.
Namely,

$(\Lambda_{x_{1}})^{\tau}=<\frac{x_{1}+...+x_{12}}{2},x_{i}+x_{1}\mid
i=1,...,12>.$

We note
$\langle\frac{x_{1}+...+x_{12}}{2},\frac{x_{1}+...+x_{12}}{2}\rangle+2\langle\beta,\frac{x_{1}+...+x_{12}}{2}\rangle=6+2\equiv
0\pmod{4}$ and $\langle
x_{1}+x_{i},x_{1}+x_{i}\rangle+2\langle\beta,x_{1}+x_{i}\rangle=4+4\equiv
0\pmod{4}$. Since $(\Lambda_{x_{1}})^{\tau}$ is spanned by such elements and
$\Lambda_{x_{1}}$ is self orthogonal modulo $2$, $\langle v,v\rangle\equiv
2\pmod{4}$ for all $v\in\Lambda_{\tilde{\beta}}+\tilde{\beta}$, as we desired.
∎

We come back to the proof of Theorem 7.2. We still assume $\tau=2C,6G$, or
$10F$ and $s,p$ are odd. The main idea is to study the possibilities of
weights in $T^{1}$ modulo ${\mathbb{Z}}$. Recall that

$T^{1}=V_{\Lambda}[\tilde{g}]={\mathbb{C}}[-\beta+(\Lambda^{\tau})^{*}]\otimes
M(1)[\tau]\otimes T_{\tau}$

and the weights of $M(1)[\tau]\otimes T_{\tau}$ are in
$1-\frac{1}{\ell}+\frac{1}{|\tau|}{\mathbb{Z}}$. Therefore, we will focus on
the set $-\beta+(\Lambda^{\tau})^{*}$.

Since $\Lambda_{\tilde{\beta}}+p\tilde{\beta}$ contains a root $\mu$ in
$N_{\tau}$, $p\tilde{\beta}\in\pi(\Lambda)$ as we explained and we have
$|\Lambda^{\tau}/\Lambda_{\tilde{\beta}}^{\tau}|=|(\Lambda_{\tilde{\beta}}^{\tau})^{*}/(\Lambda^{\tau})^{*}|=p$.
For $\lambda\in\Lambda$, set
$\mu=\sqrt{\ell}\varphi(\pi(\lambda))\in\Lambda^{\tau}$. We assume
$\langle\mu,\tilde{\beta}\rangle=\frac{t}{p}$, that is,
$\langle\pi(\lambda),\beta\rangle=\frac{t}{4sp}$. Since
$p\mu\in\Lambda^{\tau}_{\tilde{\beta}}$,

$\langle
p\mu+\tilde{\beta},p\mu+\tilde{\beta}\rangle=p^{2}\langle\mu,\mu\rangle+2p\langle\mu,\tilde{\beta}\rangle+2=\langle
p\mu,p\mu\rangle+2t+2$

and $\langle p\mu,p\mu\rangle+2t\equiv 0\mod 4$ by Proposition 7.8. Since $p$
is odd, we have $\langle\mu,\mu\rangle+2t=4m$ for some $m\in{\mathbb{Z}}$;
namely, $\langle\pi(\lambda),\pi(\lambda)\rangle=\frac{4m-2t}{4s}$. Hence

$\langle\pi(\lambda)+\beta,\pi(\lambda)+\beta\rangle=\frac{4m-2t}{4s}+2\frac{t}{4sp}+\frac{2}{4s}=\frac{2mp+t(1-p)+p}{2sp}$

and so ${\rm wt}(e^{\pi(\lambda)+\beta})=\frac{2mp+t(1-p)+p}{4sp}$. Since
$2mp+t(1-p)+p$ is always odd, the possible weights modulo ${\mathbb{Z}}$ are
$\frac{\mbox{odd}}{4ps}$, that is, there are $h=2sp$ distinct weights modulo
${\mathbb{Z}}$ at most. That means $|\tilde{g}|\leq h$ and this completes the
proof of Proposition 7.2. ∎

We next show that if we start the orbifold construction from $V$ by using a
$W$-element, we will come back to the original $(\tau,\tilde{\beta})$ and $N$.

Let $V=V_{\Lambda}^{[\tilde{g}]}$. Then we have:

$V=\bigoplus_{j=0}^{|\tilde{g}|-1}(T^{j}_{{\mathbb{Z}}})^{<\tilde{g}>}.$

Define $L$ by ${\rm Comm}({\rm Comm}({\cal H}),V),V)\cong V_{L}$. Since $V$ is
holomorphic, as a $V_{L}$-module, $V$ contains a submodule isomorphic to
$V_{L+\mu}$ for all $\mu\in L^{\ast}$.

Note that $\tilde{g}^{|\tau|}=\exp(2\pi i(\delta+|\tau|\beta)(0))$ is an inner
automorphism. Then

$L=\bigcup_{j=0}^{|\tilde{g}|/|\tau|-1}j(\delta+|\tau|\beta)+\Lambda_{\beta}^{\tau}$
(7.1)

Recall that $|\tilde{g}|=h$ and then
$(h/|\tau|)(\delta+|\tau|\beta)\in\Lambda_{\beta}^{\tau}$. Since $p=h/|\tau|$
is odd and $2\delta\in\Lambda^{\tau}$, we have $\delta\equiv
h\tilde{\beta}\mod\Lambda_{\beta}^{\tau}$.

######  Proposition 7.9.

$\sqrt{\ell}L^{\ast}\cong\Lambda_{\tilde{\beta}}^{\tau}+{\mathbb{Z}}\tilde{\beta}=N^{\tau}$.

[Proof] Set $p=h/|\tau|=2k+1$. Then
$(\delta+|\tau|\beta)+\Lambda^{\tau}_{\beta}=(k+1)\ell\beta+\Lambda^{\tau}_{\beta}$.
Since $2h\beta=(2k+1)\ell\beta\in\Lambda^{\tau}$,
$L=\bigcup_{j=0}^{2k}j(k+1)\ell\beta+\Lambda_{\beta}^{\tau}=\Lambda^{\tau}_{\beta}+{\mathbb{Z}}\ell\beta$.
By the same argument as in Theorem 4.8, we have
$\sqrt{\ell}L^{*}=\Lambda_{\tilde{\beta}}^{\tau}+{\mathbb{Z}}\tilde{\beta}=N^{\tau}$.
∎

######  Lemma 7.10.

${\rm Comm}_{V}({\cal H})=V_{\Lambda_{\tau}}^{\hat{\tau}}$.

[Proof] Suppose false. Then there exits $0<j<|\tilde{g}|=h$ such that
$V_{\Lambda_{\beta}^{\tau}}$ appears as a submodule of $T^{j}_{\mathbb{Z}}$.
It implies $j\beta\in(\Lambda^{\tau})^{*}$. It contradicts that $h$ is the
smallest integer such that $h\tilde{\beta}\in\Lambda^{\tau}$ (or equivalently,
$h\beta\in\pi(\Lambda)=(\Lambda^{\tau})^{\ast}$). ∎

As a corollary, we have

######  Lemma 7.11.

${\rm rank}(V_{1})={\rm rank}(\Lambda^{\tau})$.

[Proof] Since ${\cal H}\subseteq V_{\Lambda}^{<\tilde{g}>}$, we may assume
${\cal H}\subseteq V_{1}$. By the lemma above, we have ${\rm Comm}_{V}({\cal
H})\cap V_{1}=0$ . ∎

As we have shown, $(\Lambda_{\tilde{\beta}}+\tilde{\beta})_{2}$ is a
fundamental affine root system $\tilde{X}$ of $R(N)$. Note that
$\tilde{\beta}\in\tilde{X}$ by our convention. Let $\rho$ be a Weyl vector of
$X$. Then $\exp(2\pi i\rho(0)/h)$ is an automorphism of $N$ of order $h$ and
$N_{\rho/h}=\Lambda_{\tilde{\beta}}$. If $v$ is an extended root in
$\Lambda_{\tilde{\beta}}+\tilde{\beta}$, then $\langle v,\rho\rangle=-(h-1)$.
Therefore, $\rho^{\perp}\subseteq\Lambda_{\tilde{\beta}}=\\{x\in
N\mid\langle\rho,x\rangle\in h{\mathbb{Z}}\\}$. Furthermore, for any $w\in
R(N)\cap N^{(k)}$, $\langle\rho,w\rangle\cong k\pmod{h}$.

Set $\alpha=\sqrt{\ell}\varphi^{-1}(\pi(\rho)/h)$. Then

$\langle\alpha,\beta\rangle=\langle\sqrt{\ell}\varphi^{-1}(\pi(\rho)/h),\frac{1}{\sqrt{\ell}}\varphi^{-1}(\tilde{\beta})\rangle=\frac{1}{h}\langle\pi(\rho),\tilde{\beta}\rangle=\frac{1}{h}\langle\rho,\tilde{\beta}\rangle=\frac{1}{h}.$

More generally,
$\langle\alpha,L^{\ast}\rangle=\langle\varphi^{-1}(\pi(\rho)/h),\varphi^{-1}(N^{\tau})\rangle=\frac{1}{h}\langle\rho,N^{\tau}\rangle<\frac{{\mathbb{Z}}}{h}$.

Since ${\cal H}$ is a Cartan subalgebra of $V_{1}$ and $V_{1}$ contains a root
vector associated with $\beta$, i.e., a weight one element $e^{\beta}\otimes
T_{\tau}\in T^{1}$ with $\langle\beta,\beta\rangle=\frac{2}{\ell}$. As we have
shown in the previous sections, the isometry in $Co.0$ determined by the
reverse automorphism defined by a $W$-element of $V_{1}$ is $\tau$ itself and
$\beta$ is one of shortest roots of $V_{1}$.

######  Lemma 7.12.

Let $\gamma$ be a root of $V_{1}$. Then $\langle\alpha,\gamma\rangle\neq 0$.

[Proof] Since $\rho^{\perp}\cap N^{\tau}\subseteq\Lambda^{\tau}$, we have
$\alpha^{\perp}\cap L^{\ast}\subseteq\pi(\Lambda)$. and we have the desired
result. ∎

In other words, $\alpha$ is a regular element in ${\cal H}$ and we can define
a positive root $\mu$ by $\langle\mu,\alpha\rangle>0$. Note that there is a
correspondence between $L^{*}$ and $N^{\tau}$ through the map
$\sqrt{\ell}\varphi_{\tau}$. Since $\langle\rho,v\rangle\in{\mathbb{Z}}$ for
all $v\in N$, $\langle\alpha,\mu\rangle\in\frac{1}{h}{\mathbb{Z}}$ for all
$\mu\in L^{\ast}$. Since $\langle\alpha,\beta\rangle=\frac{1}{h}$, we may
assume $\beta$ is a simple short root of a full component. Therefore, there is
a $W$-element $\widetilde{\alpha}$ such that
$\langle\widetilde{\alpha},\beta\rangle=\frac{1}{h}$. In this case,
$\widehat{\tau}\exp(2\pi i\beta(0))$ is a reverse automorphism of
$V_{\Lambda}$ for $\exp(2\pi i\widetilde{\alpha}(0))$ and we can recover the
pair $(\tau,\tilde{\beta})$ and a Niemeier lattice
$N=\Lambda_{\tilde{\beta}}+{\mathbb{Z}}\tilde{\beta}$.

###  7.2 The relation $\sim$

Next we will study the relation $\sim$ on ${\cal T}$. Let
$(\tau,\tilde{\beta})$ and $(\tau^{\prime},\tilde{\beta}^{\prime})$ be two
pairs in ${\cal T}$. Suppose
$(\tau,\tilde{\beta})\sim(\tau^{\prime},\tilde{\beta}^{\prime})$. As we
mentioned, up to the action of $O(\Lambda)$, one can assume $N=N^{\prime}$ and
$\tilde{\beta}^{\prime}=\tilde{\beta}-\lambda$ for some
$\lambda\in\Lambda_{\tilde{\beta}}$.

######  Lemma 7.13.

Suppose $\tilde{\beta}^{\prime}=\tilde{\beta}-\lambda$ is also fixed by
$\tau$. Then the VOAs defined by $(\tau,\tilde{\beta})$ and
$(\tau,\tilde{\beta}^{\prime})$ are isomorphic.

[Proof] By our assumption, $\lambda\in\Lambda^{\tau}$ and thus
$\bar{\lambda}=\frac{1}{\sqrt{\ell}}\varphi^{-1}(\lambda)\in(\Lambda^{\tau})^{*}$.
Then $\widehat{\tau}\exp(2\pi i(\beta-\bar{\lambda})(0))=\tilde{g}\exp(-2\pi
i\bar{\lambda}(0))$ is conjugate to $\tilde{g}$ by [26, Lemma 4.5 (2)].
Therefore, they define isomorphic VOAs by orbifold constructions. ∎

In other words, the choice of $\tilde{\beta}$ is somewhat ambiguous and we can
choose any $\tau$-invariant root in the affine fundamental root system as the
starting point.

######  Proposition 7.14.

If $(\tau,\tilde{\beta})\sim(\tau^{\prime},\tilde{\beta}^{\prime})$, then the
VOAs obtained by an orbifold constructions from $V_{\Lambda}$ associated with
$\tilde{g}=\widehat{\tau}\exp(2\pi i\beta(0))$ and
$\tilde{g}^{\prime}=\widehat{\tau^{\prime}}\exp(2\pi i\beta^{\prime}(0))$ are
isomorphic, where $\beta=\frac{1}{\sqrt{\ell}}\varphi^{-1}(\tilde{\beta})$ and
$\beta^{\prime}=\frac{1}{\sqrt{\ell}}\varphi^{-1}(\tilde{\beta}^{\prime})$.

[Proof] Up to the action of $O(\Lambda)$, one may assume $N=N^{\prime}$ and
$\tilde{\beta}^{\prime}=\tilde{\beta}-\lambda$ for some
$\lambda\in\Lambda_{\tilde{\beta}}$. Since $\tau$ is conjugate to
$\tau^{\prime}$ in $O(N)$, there is a $\mu\in O(N)$ such that
$\tau=\mu\tau^{\prime}\mu^{-1}$. Then $\mu\tilde{\beta}^{\prime}$ is fixed by
$\tau$. Note that $h$ is still the smallest integer such that
$h\mu\tilde{\beta}^{\prime}\in\Lambda^{\tau}$. Since
$N=\cup_{j=1}^{h-1}(\Lambda_{\tilde{\beta}}+j\tilde{\beta})$,
$\mu\tilde{\beta}^{\prime}\in(\Lambda_{\tilde{\beta}}+k\tilde{\beta})$ for
some $k$ with $(k,h)=1$. Therefore, $<\tilde{g}^{k}>=<\tilde{g}>$ and
$V^{[\tilde{g}]}\cong V^{[\tilde{g}^{k}]}$.

Recall that $N_{\tau}\cong L_{A}(c_{\tau})>R=\oplus A_{m_{i}-1}^{a_{i}}$ and
$\tau$ acts on $N_{\tau}$ as a Coxeter element of the root system $R$.
Therefore, for $(k,|\tau|)=1$, $\tau$ is conjugate to $\tau^{k}$ by an element
$s\in Weyl(R)$. In this case,
$s\mu\tilde{\beta}^{\prime}=\mu\tilde{\beta}^{\prime}$ and $s\tau
s^{-1}=\tau^{k}$. Therefore, $(\tau,\mu\tilde{\beta}^{\prime})$ and
$(\tau^{k},\mu\tilde{\beta}^{\prime})$ define isomorphic VOAs. Moreover, by
Lemma 7.13, the VOAs obtained by an orbifold constructions from $V_{\Lambda}$
associated with $\tilde{g}^{k}=\widehat{\tau}^{k}\exp(2\pi ik\beta(0))$ and
$\widehat{\tau}^{k}\exp(2\pi i\mu\beta^{\prime}(0))$ are isomorphic and we
have the desired result. ∎

Therefore, there is a one-to-one correspondence between the set of isomorphic
classes of holomorphic VOA $V$ of central charge $24$ with non-abelian $V_{1}$
and the set of equivalent classes $(\sim)$ of pairs $(\tau,\tilde{\beta})$
with $\tau\in{\cal P}_{0}$ and $\tau$-invariant deep hole $\tilde{\beta}$ of
$\Lambda$ satisfying the conditions in Theorem 7.1.

######  Remark 7.15.

Since there is a correspondence between deep holes, up to equivalence, and
Niemeier lattices, each pair in ${\cal T}$ will define a unique pair
$(N,\tau)$, where $N$ is a Niemeier lattice with $R(N)\neq\emptyset$ and
$\tau\in O(N)$ with a positive frame shape. Therefore, the classification of
holomorphic VOAs of central charge $24$ with non-abelian weight one Lie
algebras can also be reduced to a classification of the possible pairs of
$(N,\tau)$, up to some equivalence.

##  8 Höhn’s observation and Lie algebra structure of $V_{1}$

In [21], Höhn observed that there is an interesting bijection between certain
equivalence classes of cyclic subgroups of the glue codes of the Niemeier
lattices with roots and the Lie algebra structures of the weight one subspace
of a holomorphic vertex operator algebras of central charge $24$ corresponding
to the $69$ semisimple cases in Schellekens’ list. In this section, we will
provide an explanation for Höhn’s observation using our main theorem. In
particular, we will give a pure combinatorial classification of holomorphic
vertex operator algebras of central charge $24$ with non-trivial weight one
spaces. Indeed, the orbit lattice $N(Z)$ described by Höhn is essentially the
lattice $L=\sqrt{\ell}(N^{\tau})^{*}$ and the glue vector $v$ is the vector
$\lambda_{c}$ as we defined in the appendix. The rescaling of the levels of
the Lie algebras are handled uniformly using the $\ell$-duality map
$\sqrt{\ell}\varphi_{\tau}$.

###  8.1 Roots of $V_{1}$ and $N^{\tau}$

First we study the relationship between the root system of $V_{1}$ and the
lattice $N^{\tau}\cong\sqrt{\ell}L^{*}$. Recall from [15] that

$L_{\mathfrak{g}_{i}}(k_{i},0)=\bigoplus_{j\in
Q_{i}/k_{i}Q^{i}_{l}}V_{\sqrt{k_{i}}Q^{i}_{l}+\frac{1}{\sqrt{k_{i}}}\beta_{j}}\otimes
M^{0,\beta_{j}},$ (8.1)

where $Q^{i}$ and $Q^{i}_{l}$ denote the root lattice and long root lattice of
$\mathfrak{g}_{i}$, respectively and $\beta_{j}$ is a representative of $j\in
Q^{i}/k_{i}Q^{i}_{l}$.

By (8.1), the roots of the Lie algebra
$V_{1}=\bigoplus_{i=1}^{t}\mathfrak{g}_{i}$ can be represented by elements in
$L^{*}$. Indeed, we can view them as roots of the lattice $\sqrt{\ell}L^{*}$.

First we recall the definition for the root system of an even lattice $K$ (cf.
[42, 43]).

######  Definition 8.1.

A vector $v\in K$ is primitive if the sublattice spanned by $v$ is a direct
summand of $L$. A primitive vector $v$ is called a root if $2\langle
v,K\rangle/\langle v,v\rangle<{\mathbb{Z}}$. The set of roots

$R(K)=\\{v\in K\mid v\text{ is primitive},\ 2\langle v,K\rangle/\langle
v,v\rangle<{\mathbb{Z}}\\}$

is called the root system of $K$.

######  Definition 8.2.

Let $K$ be an even lattice. The level $\ell$ of $K$ is the smallest positive
integer $\ell$ such that $\sqrt{\ell}K^{*}$ is still even.

As in [42], we denote the scaled root system as follows.

${{}^{\alpha}}\\!A_{n},{{}^{\alpha}}\\!D_{n},{{}^{\alpha}}\\!E_{n}$ |  | roots of length $2\alpha$
---|---|---
${{}^{\alpha}}\\!B_{n}$ |  | short roots of length $\alpha$
${{}^{\alpha}}\\!C_{n}$ |  | short roots of length $2\alpha$
${{}^{\alpha}}\\!G_{2}$ |  | short roots of length $2\alpha$
${{}^{\alpha}}\\!F_{4}$ |  | short roots of length $2\alpha$

We also consider the reduced discriminant group of an arbitrary scaled root
system $R$, which is a subgroup of the discriminant group of the sublattice
generated by $R$ (see [42, Definition 2.1]). For an irreducible root system
${}^{\alpha}\\!X$ of type $A$, $D$ or $E$, the reduced discriminant group of
${}^{\alpha}\\!X$ is simply the discriminant group $\mathcal{D}(X)$.

For non-simply laced root system, the corresponding root lattices and reduced
discriminant groups are as follows.

Root system |  | root lattice | Reduced discriminant
---|---|---|---
${{}^{\alpha}}\\!B_{n}$ |  | $\sqrt{\alpha}{\mathbb{Z}}^{n}$ | ${\mathbb{Z}}_{2}$
${{}^{\alpha}}\\!C_{n}$ |  | $\sqrt{\alpha}D_{n}$ | ${\mathbb{Z}}_{2}$
${{}^{\alpha}}\\!G_{2}$ |  | $\sqrt{\alpha}A_{2}\cong\sqrt{3\alpha}A_{2}^{*}$ | $1$
${{}^{\alpha}}\\!F_{4}$ |  | $\sqrt{\alpha}D_{4}\cong\sqrt{2\alpha}D_{4}^{*}$ | $1$

For explicit description of the reduced discriminant groups for the root
systems of ${{}^{\alpha}}\\!B_{n}$ and ${{}^{\alpha}}\\!C_{n}$, we use the
following standard model for their root lattices, namely,

${{}^{\alpha}}\\!B_{n}=\bigoplus_{i=1}^{n}{\mathbb{Z}}e_{i},\quad\text{ where
}(e_{i},e_{j})=\alpha\delta_{i,j},$

roots: $e_{i}$ of norm $\alpha$ and $\pm e_{i}\pm e_{j}$ of norm $2\alpha$;
and

${{}^{\alpha}}\\!C_{n}=\\{\sum_{i=1}^{n}x_{i}e_{i}\mid\sum_{i=1}^{n}x_{i}\equiv
0\mod 2\\}\quad\text{ where }(e_{i},e_{j})=\alpha\delta_{i,j},$

roots: $\pm e_{i}\pm e_{j}$ of norm $2\alpha$ and $2e_{i}$ of norm $2\alpha$.

The reduced discriminant group of ${}^{\alpha}\\!B_{n}$ is $0$ if $\alpha$ is
odd and the reduced discriminant group of $R$ is given by
$\langle\bar{g}\rangle\cong{\mathbb{Z}}_{2}$ where

$\begin{cases}g=\frac{1}{2}(e_{1}+\cdots+e_{n})&\text{ if
}R=^{2\alpha}\\!\\!B_{n},\\\ g=e_{1}&\text{ if
}R=^{\alpha}\\!\\!C_{n}.\end{cases}$

######  Remark 8.3 (see [42]).

An element $v\in K$ is a root if and only if $v\in\langle v,v\rangle/2\cdot
K^{*}$ and $v$ is primitive in $K$. In particular, $\langle v,v\rangle/2$
divides the exponent of $\mathcal{D}(K)=K^{*}/K$.

######  Lemma 8.4.

Let $v\in K$ such that $v\in\langle v,v\rangle/2\cdot K^{*}$. Assume that
$v/n\in K$ for some $n\in{\mathbb{Z}}$ with $n>1$. Then $n=2$. Moreover,
$\frac{v}{2}$ is a root and $K={\mathbb{Z}}(\frac{v}{2})\perp(K\cap\langle
v\rangle^{\perp})$.

[Proof] Since $v/n\in K$ and $v\in\langle v,v\rangle/2\cdot K^{*}$, we have

$\langle\frac{v}{n},v\rangle\in\frac{\langle v,v\rangle}{2}{\mathbb{Z}},\text{
or equivalently, }2/n\in{\mathbb{Z}};$

hence, $n=2$. Then $\frac{v}{2}$ is primitive in $K$ and $\frac{2\langle
v/2,x\rangle}{\langle v/2,v/2\rangle}=\frac{4\langle v,x\rangle}{\langle
v,v\rangle}\in{\mathbb{Z}}$ for any $x\in K$; hence, it is a root.

Let $p_{v}:K\to({\mathbb{Z}}\frac{v}{2})^{*}$ be the natural projection. For
any $x\in K$, $p_{v}(x)=r_{x}v$ for some $r_{x}\in{\mathbb{Q}}$. Then
$\frac{2\langle v,x\rangle}{\langle v,v\rangle}=\frac{2\langle
v,r_{x}v\rangle}{\langle v,v\rangle}=2r_{x}\in{\mathbb{Z}}$. That means
$r_{x}\in\frac{1}{2}{\mathbb{Z}}$ and $p_{v}(K)={\mathbb{Z}}(\frac{v}{2})$ as
desired. ∎

The following results can be found in [42].

######  Theorem 8.5.

Let $K$ be an integral lattice and $R$ a root system contained in $K$. Let
$\mathcal{K}=K/(\langle R\rangle\perp(R^{\perp}\cap K))$ be the glue code of
$K$ over $R$. Then the elements of $R$ are actually roots of $K$ if and only
if $\mathcal{K}$ is contained in the reduced discriminant group of $R$, and
furthermore $K$ contains no elements of the form $e/2$, where $e$ is a basis
vector of an $A_{1}$-component.

Recall that $\ell$ is divisible by $|\tau|(K_{0}-N_{0})={\rm
LCM}(\\{r_{i}k_{i}\\}_{i=1}^{t})$ by Lemmas 2.2 and 4.11.

######  Lemma 8.6.

Let $\beta$ be a root of $\mathfrak{g}_{i,k_{i}}$ and $M=\sqrt{\ell}L^{*}$.
Then

$v=\frac{\sqrt{\ell}}{\sqrt{k_{i}}}\beta\in\frac{\langle v,v\rangle}{2}M^{*}.$

In particular, $v=\frac{\sqrt{\ell}}{\sqrt{k_{i}}}\beta$ is a root of $M$ if
it is primitive in $M$ or ${\mathbb{Z}}(v/2)$ is an orthogonal summand of $M$.

###### Proof.

Let $M=\sqrt{\ell}L^{*}$ and $v=\frac{\sqrt{\ell}}{\sqrt{k_{i}}}\beta$. Then
$M^{*}=\frac{1}{\sqrt{\ell}}L$ and

$\frac{\langle v,v\rangle}{2}=\begin{cases}\frac{\ell}{r_{i}k_{i}}&\text{ if
$\beta$ is a short root,}\\\ \frac{\ell}{k_{i}}&\text{ if $\beta$ is a long
root.}\end{cases}$

Therefore,

$v=\frac{\sqrt{\ell}}{\sqrt{k_{i}}}\beta=\begin{cases}\frac{\langle
v,v\rangle}{2}\frac{1}{\sqrt{\ell}}\sqrt{k_{i}}(r_{i}\beta)&\text{ if $\beta$
is a short root,}\\\ \frac{\langle
v,v\rangle}{2}\frac{1}{\sqrt{\ell}}\sqrt{k_{i}}\beta&\text{ if $\beta$ is a
long root.}\end{cases}$

Since $L>\sqrt{k_{i}}Q^{i}_{l}$ and $r_{i}\beta\in Q^{i}_{l}$ (resp. $\beta\in
Q^{i}_{l}$) if $\beta$ is a short root (resp., a long root),
$v\in\frac{\langle v,v\rangle}{2}M^{*}$ as desired. ∎

######  Lemma 8.7.

Let $\mathfrak{g}_{i,k_{i}}$ be a simple Lie subalgebra of $V_{1}$. Suppose
there is a root $\beta$ of $\mathfrak{g}_{i,k_{i}}$ such that
$v=\frac{\sqrt{\ell}}{\sqrt{k_{i}}}\beta$ is not a root of $N^{\tau}$. Then
$\beta$ is a long root and the long root lattice of $\mathfrak{g}_{i}$ is of
the type $A_{1}^{r}$, where $r={\rm rank}(\mathfrak{g}_{i})$. In particular,
$\mathfrak{g}_{i}$ is of type $A_{1}$ or $C_{r}$.

[Proof] Without loss, we may assume ${\rm rank}(V_{1})>1$. By Lemma 8.4,
${\mathbb{Z}}(v/2)$ is an orthogonal summand of $N^{\tau}$. Suppose $\beta$ is
a short root. Then for any short root $\beta^{\prime}$,
$v^{\prime}=\frac{\sqrt{\ell}}{\sqrt{k_{i}}}\beta^{\prime}$ is not a root of
$N^{\tau}$. Therefore,
$\frac{\sqrt{\ell}}{2\sqrt{k_{i}}}Q_{i}\cong\sqrt{s}A_{1}^{r}$; however,
${\mathbb{Z}}\frac{\sqrt{\ell}}{2\sqrt{k_{i}}}\alpha$ is not an orthogonal
summand of $\sqrt{s}A_{1}^{r}$ for any long root $\alpha$. Therefore, $\beta$
is a long root and the long root lattice is of type $A_{1}^{r}$. ∎

By the lemmas above, the lattice $N^{\tau}=\sqrt{\ell}L^{*}$ contains the
information of the (scaled) root system of $\mathfrak{g}$.

###  8.2 Orbit diagrams and Lie algebra structures of $V_{1}$

Let $V$ be a holomorphic VOA of central charge $24$ and $\alpha$ a W-element
of $V_{1}$. Let $g=\exp(2\pi i\alpha(0))\in{\rm Aut}(V)$ and let
$\tilde{g}=\hat{\tau}\exp(2\pi i\beta(0))\in{\rm Aut}(V_{\Lambda})$ be the
reverse automorphism of $g$, where $\beta\in{\mathbb{C}}\Lambda^{\tau}$. Let
$\varphi_{\tau}:\sqrt{\ell}(\Lambda^{\tau})^{*}\to\Lambda^{\tau}$ be the
isometry described in Theorem 4.2. Recall that the vector
$\tilde{\beta}=\varphi(\sqrt{\ell}\beta)$ is a deep hole of $\Lambda$ and
$N=\Lambda^{[\tilde{\beta}]}\ncong\Lambda$. Moreover, the Coxeter number $h$
of $N$ is equal to $n={\rm LCM}(r_{i}h_{i}^{\vee})$ and
$N^{\tau}\cong\sqrt{\ell}L^{*}$.

Let $R$ be the set of roots in $N$ and set
$R_{k}=\\{\mu\in-k\tilde{\beta}+\Lambda_{\tilde{\beta}}\mid\langle\mu,\mu\rangle=2\\}.$
Then $R=\cup_{k=1}^{h-1}R_{k}$. Since the Leech lattice does not contain a
root, $\langle\mu,\nu\rangle\leq 0$ for $\mu,\nu\in R_{k}$ with $\mu\neq\nu$
for each $k$. Indeed, $\langle\mu,\nu\rangle=0$ or $-1$ for any $\mu,\nu\in
R_{k}$ with $\mu\neq\nu$.

One can associate a (simply laced) Dynkin diagram with $R_{k}$ for each $k$.
Namely, the nodes are labeled by elements of $R_{k}$ and two nodes $x$ and $y$
are connected if and only if $\langle x,y\rangle=-1$. By abuse of notations,
we often use $R_{k}$ to denote both the Dynkin diagram and the subset of
roots. Note that $\tau$ acts on $R_{k}$ for each $k$ and acts as a diagram
automorphism associated with the diagram defined by $R_{k}$.

Since $\tilde{\beta}$ is a deep hole of the Leech lattice $\Lambda$, $R_{1}$
is a disjoint union of the affine diagrams associated with the root system of
$N$. Moreover, $\tau$ acts on $R_{1}$. Since $N_{\tau}\cong L_{A}(c)$ and
$\tau$ acts as $g_{\Delta,c}$ on a root sublattice of $N_{\tau}$,
$\tau\in\mathrm{Weyl}(R)$ and preserves all irreducible components of $R(N)$.
Note also that $g_{\Delta,c}$ induces an isometry in $O(\Lambda)$ if and only
if $\lambda_{c}\in N$ [5]. Therefore, the vector $\lambda_{c}$ corresponds to
a codeword of the glue code $N/R$. In particular,
$P_{Q_{i}}(\lambda_{c})\in(Q_{i})^{*}$ for every irreducible root sublattice
$Q_{i}$ of $N$. Note also that $\tau$ has a positive frame when viewing as an
isometry of $N$.

Therefore, for each irreducible component, we can consider the quotient
diagram as follows: we identify an orbit of nodes as one node and two nodes
are connected if the nodes in the corresponding orbits are connected. By
removing the node associated with the extended node, one obtain a usual Dynkin
diagram (see Table LABEL:affinediagram).

Note that the fixed sublattice is the (scaled) root lattice of the quotient
diagram. A fixed node (or fixed simple root) corresponds to a simple short
root of a full component.

Table 2: Diagram automorphisms of affine diagrams Type | $A_{n}$ | $D_{2k}$ | $D_{2k}$ | $D_{2k+1}$ | $D_{2k+1}$ | $E_{6}$ | $E_{7}$
---|---|---|---|---|---|---|---
Root subsystem | $(A_{\frac{n+1}{k}-1})^{k}$ | $A_{1}^{k}$ | $A_{1}^{2}$ | $A_{3}A_{1}^{k-1}$ | $A_{1}^{2}$ | $A_{2}^{2}$ | $A_{1}^{3}$
Frame Shape | $1^{-1}(\frac{n+1}{k})^{k}$ | $2^{k}$ | $1^{2k-4}2^{2}$ | $1^{-1}2^{k-1}4$ | $1^{2k-3}2^{2}$ | $3^{2}$ | $1^{1}2^{3}$
Quotient diagram | $A_{k-1}$ | $B_{k}$ | $C_{2k-2}$ | $C_{k-1}$ | $C_{2k-1}$ | $G_{2}$ | $F_{4}$
Fixed sublattice | $\sqrt{\frac{n+1}{k}}A_{k-1}$ | $A_{1}^{k}$ | $D_{2k-2}$ | $A_{1}^{k-1}$ | $D_{2k-1}$ | $A_{2}$ | $D_{4}$
Fixed simple roots | $\emptyset$ | $A_{1}$ | $A_{2k-3}$ | $\emptyset$ | $A_{2k-2}$ | $A_{1}$ | $A_{2}$

######  Remark 8.8.

Let ${\cal G}_{i}$ be a simple Lie subalgebra of $V_{1}$ with
$r_{i}h_{i}^{\vee}=n=h$. Then $\ell=r_{i}k_{i}$ and
$k_{i}/h_{i}^{\vee}=\ell/h$. Therefore, the level $k_{j}$ of the simple Lie
subalgebra ${\cal G}_{j}$ is given by $k_{j}=\ell h^{\vee}_{j}/h$ for any $j$.
Note also that the short roots of ${\cal G}_{i}$ will correspond to an
irreducible (connected) component $S_{i}$ of $N^{\tau}_{2}$. Moreover,
$S_{i}\cap R_{1}$ corresponds to the simple short roots of ${\cal G}_{i}$.
Therefore, $S_{i}$ and $S_{i}\cap R_{1}$ determines the type of ${\cal G}_{i}$
uniquely.

######  Remark 8.9.

In [39], a notion of generalized hole diagrams is introduced. It was also
shown that a generalized hole diagram determines a generalized deep hole up to
conjugacy and that there are exactly $70$ such diagrams. This notion of
generalized hole diagrams essentially corresponds to the diagram associated
with simple short roots of the full components (i.e., elements in $R_{1}\cap
N^{\tau}_{2}$).

###  8.3 Possible pairs for $(N,\tau)$

Next we will discuss the possible choices for the pair $(N,\tau)$ for each
$\tau\in{\cal P}_{0}$.

####  8.3.1 $\tau\in 2A$

For $\tau\in 2A$ of $O(\Lambda)$, we have $\ell=|\hat{\tau}|=|\tau|=2$. In
this case, $\Lambda_{\tau}\cong\sqrt{2}E_{8}$. Let
$N=\Lambda^{[\tilde{\beta}]}$. Then the Coxeter number of $N$ is divisible by
$2$ and

$N_{\tau}=\mathrm{Span}_{\mathbb{Z}}\\{A_{1}^{8},\frac{1}{2}(\alpha_{1}+\cdots+\alpha_{8})\\},$

where ${\mathbb{Z}}\alpha_{1}+\cdots+{\mathbb{Z}}\alpha_{8}\cong A_{1}^{8}$.
The vector $v=\frac{1}{2}(\alpha_{1}+\cdots+\alpha_{8})$ corresponds to a
codeword $c\in N/R$. The possible choices for $(N,\tau)$, the codeword $c$ and
the corresponding root systems and Lie algebra structures for $V_{1}$ are
listed in Table LABEL:T:2to6.

Table 3: $(N,\tau)$ for the case $2A$ Type | Codeword $c$ | Embedding | $R(N^{\tau})$ | $V_{1}$
---|---|---|---|---
$A_{1}^{24}$ | $(1^{8},0^{8})$ | $A_{1}^{8}\hookrightarrow A_{1}^{8}$ | $A_{1}^{16}$ | $A_{1,2}^{16}$
$A_{3}^{8}$ |  $(22022000$) | $(A_{1}^{2})^{4}\hookrightarrow A_{3}^{4}$ | $A_{3}^{4}(\sqrt{2}A_{1})^{4}$ | $A_{3,2}^{4}A_{1,1}^{4}$
$D_{4}^{6}$ | $(233200)$ | $(A_{1}^{2})^{4}\hookrightarrow D_{4}^{4}$ | $D_{4}^{2}C_{2}^{4}$ | $D_{4,2}^{2}C_{2,1}^{4}$
$A_{5}^{4}D_{4}$ | $(3300|1)$ | $(A_{1}^{3})^{2}+A_{1}^{2}\hookrightarrow A_{5}^{2}+D_{4}$ | $A_{5}^{2}C_{2}(\sqrt{2}A_{2})^{2}$ | $A_{5,2}^{2}C_{2,1}A_{2,1}^{2}$
$A_{7}^{2}D_{5}^{2}$ | $(44|00)$ | $(A_{1}^{4})^{2}\hookrightarrow A_{7}^{2}$ | $D_{5}^{2}(\sqrt{2}A_{3})^{2}$ | $D_{5,2}^{2}A_{3,1}^{2}$
$A_{7}^{2}D_{5}^{2}$ | $(20|33)$ | $(A_{1}^{4})+(A_{1}^{2})^{2}\hookrightarrow A_{7}+D_{5}^{2}$ | $A_{7}C_{3}^{2}(\sqrt{2}A_{3})$ | $A_{7,2}C_{3,1}^{2}A_{3,1}$
$D_{6}^{4}$ | $(2222)$ | $(A_{1}^{2})^{4}\hookrightarrow D_{6}^{4}$ | $C_{4}^{4}$ | $C_{4,1}^{4}$
$D_{6}^{4}$ | $(1230)$ | $(A_{1}^{2})+(A_{1}^{3})^{2}\hookrightarrow D_{6}+D_{6}^{2}$ | $D_{6}C_{4}B_{3}^{2}$ | $D_{6,2}C_{4,1}B_{3,1}^{2}$
$A_{9}^{2}D_{6}$ | $(05|3)$ | $(A_{1}^{5})+(A_{1}^{3})\hookrightarrow A_{9}+D_{6}$ | $A_{9}(\sqrt{2}A_{4})B_{3}$ | $A_{9,2}A_{4,1}B_{3,1}$
$A_{11}D_{7}E_{6}$ | $(620)$ | $A_{1}^{6}+A_{1}^{2}\hookrightarrow A_{11}+D_{7}$ | $E_{6}C_{5}(\sqrt{2}A_{5})$ | $E_{6,2}C_{5,1}A_{5,1}$
$D_{8}^{3}$ | $(033)$ | $(A_{1}^{4})^{2}\hookrightarrow D_{8}^{2}$ | $D_{8}B_{4}^{2}$ | $D_{8,2}B_{4,1}^{2}$
$D_{8}^{3}$ |  $(221$) | $(A_{1}^{2})^{2}+A_{1}^{4}\hookrightarrow D_{8}^{2}+D_{8}$ | $C_{6}^{2}B_{4}$ | $C_{6,1}^{2}B_{4,1}$
$A_{15}D_{9}$ | $(80)$ | $A_{1}^{8}\hookrightarrow A_{15}$ | $D_{9}(\sqrt{2}A_{7})$ | $D_{9,2}A_{7,1}$
$E_{7}^{2}D_{10}$ | $(11|2)$ | $(A_{1}^{3})^{2}+A_{1}^{2}\hookrightarrow E_{7}^{2}+D_{10}$ | $C_{8}F_{4}^{2}$ | $C_{8,1}F_{4,1}^{2}$
$E_{7}^{2}D_{10}$ | $(01|1)$ | $A_{1}^{3}+A_{1}^{5}\hookrightarrow E_{7}+D_{10}$ | $E_{7}B_{5}F_{4}$ | $E_{7,2}B_{5,1}F_{4,1}$
$D_{12}^{2}$ | $(21)$ | $A_{1}^{2}+A_{1}^{6}\hookrightarrow D_{12}+D_{12}$ | $C_{10}B_{6}$ | $C_{10,1}B_{6,1}$
$E_{8}D_{16}$ | $(01)$ | $A_{1}^{8}\hookrightarrow D_{16}$ | $B_{8}E_{8}$ | $B_{8,1}E_{8,2}$

####  8.3.2 $\tau\in 3B$

For $\tau\in 3B$ of $O(\Lambda)$, we have $\ell=|\hat{\tau}|=|\tau|=3$. In
this case, $\Lambda_{\tau}\cong K_{12}$ is the Coxeter-Todd lattice of rank
$12$. Let $N=\Lambda^{[\tilde{\beta}]}$. Then the Coxeter number of $N$ is
divisible by $3$ and
$N_{\tau}=\mathrm{Span}_{\mathbb{Z}}\\{A_{2}^{6},(\gamma_{1},\dots,\gamma_{6})\\}$,
where $\gamma_{i}+A_{2}$ is a generator of $A_{2}^{*}/A_{2}$ for each $i$. The
vector $(\gamma_{1},\dots,\gamma_{6})$ corresponds to a codeword $c\in N/R$.
The possible choices for $(N,\tau)$, $c$ and the corresponding root systems
and Lie algebra structures for $V_{1}$ are listed in Table LABEL:table3to8.

Table 4: $(N,\tau)$ for the case $3B$ Type | Codeword $c$ | Embedding | $R(N^{\tau})$ | $V_{1}$
---|---|---|---|---
$A_{2}^{12}$ | $(1^{6}0^{6})$ | $A_{2}^{6}\hookrightarrow A_{2}^{6}$ | $A_{2}^{6}$ | $A_{2,3}^{6}$
$A_{5}^{4}D_{4}$ | $(2220|0)$ | $(A_{2}^{2})^{3}\hookrightarrow A_{5}^{3}$ | $A_{5}D_{4}(\sqrt{3}A_{1})^{3}$ | ${A_{5,3}}{D_{4,3}}A_{1,1}^{3}$
$A_{8}^{3}$ | $(630)$ | $(A_{2}^{3})^{2}\hookrightarrow A_{8}^{2}$ | $A_{8}(\sqrt{3}A_{2})^{2}$ | $A_{8,3}A_{2,1}^{2}$
$E_{6}^{4}$ | $(0111)$ | $(A_{2}^{3})^{3}\hookrightarrow E_{6}^{3}$ | $E_{6}G_{2}^{3}$ | $E_{6,3}{G_{2,1}}^{3}$
$A_{11}D_{7}E_{6}$ | $(401)$ | $A_{2}^{4}+A_{2}^{2}\hookrightarrow A_{11}E_{6}$ | $D_{7}(\sqrt{3}A_{3})G_{2}$ | ${D_{7,3}}{A_{3,1}}{G_{2,1}}$
$A_{17}E_{7}$ | $(60)$ | $A_{2}^{6}\hookrightarrow A_{17}$ | $E_{7}{(\sqrt{3}A_{5})}$ | $E_{7,3}A_{5,1}$

####  8.3.3 $\tau\in 5B$

For $\tau\in 5B$ of $O(\Lambda)$, we have $\ell=|\hat{\tau}|=|\tau|=5$ and the
Coxeter number of $N=\Lambda^{[\tilde{\beta}]}$ is divisible by $5$. In this
case,
$N_{\tau}\cong\mathrm{Span}_{\mathbb{Z}}\\{A_{4}^{4},(\gamma_{1},\gamma_{2},\gamma_{3},\gamma_{4})\\}$,
where $\gamma_{i}+A_{4}$ is a generator of $A_{4}^{*}/A_{4}$ for each $i$.
Again, the vector $(\gamma_{1},\gamma_{2},\gamma_{3},\gamma_{4})$ corresponds
to a codeword $c\in N/R$. That means $5=|\tau|$ divides $|N/R|$, also. The
possible choices for $(N,\tau)$, $c$ and the corresponding root systems and
Lie algebra structures are listed in Table LABEL:table5to6.

Table 5: $(N,\tau)$ for the case $5B$ Type | Codeword $c$ | Embedding | $R(N^{\tau})$ | $V_{1}$
---|---|---|---|---
$A_{4}^{6}$ | $(123400)$ | $A_{4}^{4}\hookrightarrow A_{4}^{4}$ | $A_{4}^{2}$ | $A_{4,5}^{2}$
$A_{9}^{2}D_{6}$ | $(24|0)$ | $(A_{4}^{2})^{2}\hookrightarrow A_{9}^{2}$ | $D_{6}(\sqrt{5}A_{1}^{2})$ | $D_{6,5}A_{1,1}^{2}$

####  8.3.4 $\tau\in 7B$

For $\tau\in 7B$ of $O(\Lambda)$, we have $\ell=|\hat{\tau}|=7$. Let
$N=\Lambda^{[\tilde{\beta}]}$. Then the Coxeter number of $N$ is divisible by
$7$ and

$N_{\tau}=\mathrm{Span}_{\mathbb{Z}}\\{A_{6}^{3},(\gamma_{1},\gamma_{2},\gamma_{3})\\},$

where $\gamma_{i}+A_{6}$ is a generator of $A_{6}^{*}/A_{6}$ for each $i$. The
vector $(\gamma_{1},\gamma_{2},\gamma_{3})$ corresponds to a codeword $c\in
N/R$. The possible choices for $(N,\tau)$, $c$ and the corresponding root
systems and Lie algebra structures are listed in Table LABEL:table7to5

Table 6: $(N,\tau)$ for the case $7B$ Type | Codeword $c$ | Embedding | $R(N^{\tau})$ | $V_{1}$
---|---|---|---|---
$A_{6}^{4}$ | $(0124)$ | $A_{6}^{3}\hookrightarrow A_{6}^{3}$ | $A_{6}$ | $A_{6,7}$

####  8.3.5 $\tau\in 2C$

For $\tau\in 2C$ of $O(\Lambda)$, we have $\ell=|\hat{\tau}|=2|\tau|=4$.

Let $N=\Lambda^{[\tilde{\beta}]}$. Then the Coxeter number of $N$ is divisible
by $2$ and

$N_{\tau}=\mathrm{Span}_{\mathbb{Z}}\\{A_{1}^{12},\frac{1}{2}(\alpha,\dots,\alpha)\\},$

where ${\mathbb{Z}}\alpha\cong A_{1}$. The vector
$\frac{1}{2}(\alpha,\dots,\alpha)$ again corresponds to a codeword $c\in N/R$.
The possible choices for $(N,\tau)$, $c$ and the corresponding root systems
and Lie algebra structures are listed in Table LABEL:2c1.

Table 7: $(N,\tau)$ for the case $2C$ Type | Codeword $c$ | Embedding | $R(N^{\tau})$ | $V_{1}$
---|---|---|---|---
$A_{1}^{24}$ | $(1^{12}0^{12})$ | $A_{1}^{12}\hookrightarrow A_{1}^{12}$ | $A_{1}^{12}$ | $A_{1,4}^{12}$
$D_{4}^{6}$ | $(111111)$ | $(A_{1}^{2})^{6}\hookrightarrow D_{4}^{6}$ | $B_{2}^{6}$ | $B_{2,2}^{6}$
$D_{6}^{4}$ | $(2222)$ | $(A_{1}^{3})^{4}\hookrightarrow D_{6}^{4}$ | $B_{3}^{4}$ | $B_{3,2}^{4}$
$D_{8}^{3}$ | $(111)$ | $(A_{1}^{4})^{3}\hookrightarrow D_{8}^{3}$ | $B_{4}^{3}$ | $B_{4,2}^{3}$
$D_{12}^{2}$ | $(11)$ | $(A_{1}^{6})^{2}\hookrightarrow D_{12}^{2}$ | $B_{2}^{6}$ | $B^{2}_{6,2}$
$D_{24}$ | $(1)$ | $A_{1}^{12}\hookrightarrow D_{24}$ | $B_{12}$ | $B_{12,2}$
$A_{5}^{4}D_{4}$ | $(3333|0)$ | $(A_{1}^{3})^{4}\hookrightarrow A_{5}^{4}$ | $D_{4}\sqrt{2}A_{2}^{4}$ | $D_{4,4}A^{4}_{2,2}$
$A_{9}^{2}D_{6}$ | $(55|2)$ | $(A_{1}^{5})^{2}+A_{1}^{2}\hookrightarrow A_{9}^{2}+D_{6}$ | $C_{4}\sqrt{2}A_{4}^{2}$ | $C_{4,2}A^{2}_{4,2}$
$A_{17}E_{7}$ | $(9|1)$ | $A_{1}^{9}+A_{1}^{3}\hookrightarrow A_{17}+E_{7}$ | $F_{4}\sqrt{2}A_{8}$ | $A_{8,2}F_{4,2}$

####  8.3.6 $\tau\in 4C$

For $\tau\in 4C$ of $O(\Lambda)$, we have $\ell=|\hat{\tau}|=|\tau|=4$. Let
$N=\Lambda^{[\tilde{\beta}]}$. Then the Coxeter number of $N$ is divisible by
$4$ and the coinvariant lattice

$N_{\tau}=\mathrm{Span}_{\mathbb{Z}}\\{A_{3}^{4}A_{1}^{2},(\lambda,\lambda,\lambda,\lambda,\frac{1}{2}\alpha,\frac{1}{2}\alpha)\\},$

where $\lambda=1/2(1,1,1,-3)\in A_{3}^{*}$ and ${\mathbb{Z}}\alpha=A_{1}$.
Note that $N^{\tau}$ contains $A_{3}^{4}A_{1}^{2}$ as an index $4$ sublattice.
The possible choices for $(N,\tau)$, $c$ and the corresponding root systems
and Lie algebra structures are listed in Table LABEL:4Ctable.

Table 8: $(N,\tau)$ for the case $4C$ Type | Codeword $c$ | Embedding | $R(N^{\tau})$ | $V_{1}$
---|---|---|---|---
$A_{3}^{8}$ | $(32001011)$ | $A_{3}^{4}+A_{1}^{2}\hookrightarrow A_{3}^{4}+A_{3}$ | $A_{3}^{3}\sqrt{2}A_{1}$ | $A_{3,4}^{3}A_{1,2}$
$A_{7}^{2}D_{5}^{2}$ | $(02|13)$ | $A_{3}^{2}+(A_{3}A_{1})^{2}\hookrightarrow A_{7}+D_{5}^{2}$ | $A_{7}2A_{1}A_{1}^{2}$ | $A_{7,4}A_{1,1}^{3}$
$A_{7}^{2}D_{5}^{2}$ | $(22|20)$ | $A_{3}^{2}+A_{3}^{2}+A_{1}^{2}\hookrightarrow A_{7}+A_{7}+D_{5}$ | $D_{5}C_{3}2A_{1}^{2}$ | $D_{5,4}C_{3,2}A_{1,1}^{2}$
$A_{11}D_{7}E_{6}$ | $(310)$ | $A_{3}^{3}+A_{3}A_{1}^{2}\hookrightarrow A_{11}+D_{7}$ | $E_{6}B_{2}2A_{2}$ | $E_{6,4}B_{2,1}A_{2,1}$
$A_{15}D_{9}$ | $(4|2)$ | $A_{3}^{4}+A_{1}^{2}\hookrightarrow A_{15}+D_{9}$ | $C_{7}2A_{3}$ | $C_{7,2}A_{3,1}$

####  8.3.7 $\tau\in 6E$

For $\tau\in 6E$ of $O(\Lambda)$, we have $\ell=|\hat{\tau}|=6$. Let
$N=\Lambda^{[\tilde{\beta}]}$. Then the Coxeter number of $N$ is divisible by
$6$ and the coinvariant sublattice

$N_{\tau}\cong\mathrm{Span}_{\mathbb{Z}}\\{A_{5}^{2}A_{2}^{2}A_{1}^{2},(\beta,\beta,\gamma,\gamma,\frac{1}{2}\alpha,\frac{1}{2}\alpha)\\},$

where $\beta=\frac{1}{6}(1^{5},-5)\in A_{5}^{*}$,
$\gamma=\frac{1}{3}(1,1,-2)\in A_{3}^{*}$ and $\langle\alpha,\alpha\rangle=2$.
Note that $N^{\tau}$ is an index $6$ sublattice of $E_{8}\perp E_{8}$. The
possible choices for $(N,\tau)$, $c$ and the corresponding root systems and
Lie algebra structures are listed in Table LABEL:6Etable.

Table 9: $(N,\tau)$ for the case $6E$ Type | Codeword $c$ | Embedding | $R(N^{\tau})$ | $V_{1}$
---|---|---|---|---
$A_{5}^{4}D_{4}$ | $(0255|1)$ | $A_{5}^{2}+A_{2}^{2}+A_{1}^{2}\hookrightarrow A_{5}^{2}+A_{5}+D_{4}$ | $A_{5}\sqrt{3}A_{1}B_{2}$ | $A_{5,6}B_{2,3}A_{1,1}$
$A_{11}D_{7}E_{6}$ | $(222)$ | $A_{5}^{2}+A_{1}^{2}+A_{2}^{2}\hookrightarrow A_{11}+D_{7}+E_{6}$ | $\sqrt{6}A_{1}C_{5}G_{2}$ | $C_{5,3}G_{2,2}A_{1,1}$

###  8.4 $\tau\in 8E$

For $\tau\in 8E$, $\ell=|\hat{\tau}|=8$. Let $N=\Lambda^{[\tilde{\beta}]}$.
Then the Coxeter number of $N$ is divisible by $8$ and the coinvariant lattice
$N_{\tau}$ is an index $8$ over-lattice of the lattice $A_{7}^{2}A_{3}A_{1}$.
More precisely,

$N_{\tau}=\mathrm{Span}_{\mathbb{Z}}\\{A_{7}^{2}A_{3}A_{1},(\gamma_{3},\gamma_{1},\beta,\alpha)\\},$

where $\gamma_{3}=\frac{1}{8}(3^{5},-5^{3})$,
$\gamma_{1}=\frac{1}{8}(1^{7},-7)$ are in $A_{7}^{*}$,
$\beta\in\frac{1}{2}(1,1,1)\in A_{3}^{*}$ and $\alpha=\frac{1}{2}(1,1)\in
A_{1}^{*}$.

The possible choices for $(N,\tau)$, $c$ and the corresponding root systems
and Lie algebra structures are listed in Table LABEL:6Etable.

Type | Codeword $c$ | Embedding | $R(N^{\tau})$ | $V_{1}$
---|---|---|---|---
$A_{7}^{2}D_{5}^{2}$ | $(37|10)$ | $A_{7}^{2}+A_{3}A_{1}\hookrightarrow A_{7}^{2}+D_{5}$ | $D_{5}A_{1}$ | $D_{5,8}A_{1,2}$

###  8.5 $\tau\in 6G$

For $\tau\in 6G$, $\ell=|\hat{\tau}|=12$. Let $N=\Lambda^{[\tilde{\beta}]}$.
Then the Coxeter number of $N$ is divisible by $6$ and $N_{\tau}$ contains
$A_{5}^{3}A_{1}^{3}$ as an index $6$ sublattice. The possible choices for
$(N,\tau)$, $c$ and the corresponding root systems and Lie algebra structures
are listed in the following table.

Type | Codeword $c$ | Embedding | $R(N^{\tau})$ | $V_{1}$
---|---|---|---|---
$A_{5}^{4}D_{4}$ | $(31110)$ | $A_{5}^{3}+A_{1}^{3}\hookrightarrow A_{5}^{3}+A_{5}$ | $D_{4}\sqrt{2}A_{2}$ | $D_{4,12}A_{2,6}$