%This is a plain tex paper.
\magnification=\magstep1
\vbadness=10000
\hbadness=10000
\tolerance=10000
\newcount\sectionnumber \sectionnumber
=0 \newcount\subsectionnumber
\def\section#1{\advance\sectionnumber by 1 \bigbreak
{\bf\number\sectionnumber \ #1}
\medskip\par\nobreak \subsectionnumber=0}
\def\s{\number\sectionnumber}
\def\citex#1{[#1]}
\def\mult{{\rm mult}}
\def\Ad{{\rm ad}}
\def\Tr{{\rm Tr}}
\def\Aut{{\rm Aut}}
\def\half{{1/2}}
\def\te{\theta_{E_8}}
\def\tll{\theta_{\Lambda_2}}
\def\tl{\theta_{\Lambda}}
\def\ddp{\eta_{2+}}
\def\ddm{\eta_{2-}}
\def\im{{\rm Im}}
\def\C{{\bf C}}
\def\Z{{\bf Z}}
\def\R{{\bf R}}
\def\vv{V_{II_{1,1}}}
\proclaim Introduction to the monster Lie algebra.
Groups, Combinatorics, and geometry, p. 99--107, edited by M. W. Liebeck
and J. Saxl, L.M.S. lecture note series 165, C.U.P. 1992.
Richard E. Borcherds,
\bigskip
I would like to thank J. M. E. Hyland for suggesting many improvements
to this paper.
The monster sporadic simple group, of order
$$2^{46}3^{20}5^97^611^213^317.19.23.29.31.41.47.59.71,$$ acts
on a
graded vector space
$V=\oplus_{n\in \Z}V_n$ constructed by Frenkel, Lepowsky and Meurman
\citex{Fre}
The dimension of $V_n$ is equal to the coefficient $c(n)$
of the elliptic modular function
$$j(\tau)-744=\sum_nc(n)q^n= q^{-1} +
196884q+21493760q^2+\cdots$$ (where we write $q$ for $e^{2\pi i\tau}$,
and $\im(\tau)>0$).
The main problem is to describe
$V$ as a graded representation of the monster, or in other words
to calculate the trace ${\rm Tr}(g|V_n)$ of each element $g$ of the monster on
each space $V_n$. The best way to describe this information is
to define the Thompson series
$$T_g(q)= \sum_{n\in \Z} {\rm Tr}(g|V_n)q^n$$ for each element $g$ of
the monster, so that our problem is to work out what these Thompson
series are. For example, if 1 is the identity element of the monster
then ${\rm Tr}(1|V_n)=\dim(V_n)=c(n)$, so that the Thompson series
$T_1(q) = j(\tau)-744$ is the elliptic modular function. McKay,
Thompson, Conway and Norton conjectured \citex{Con} that the
Thompson series $T_g(q)$ are always Hauptmoduls for certain modular
groups of genus 0 (I will explain what this means in a moment.) In
this paper we describe the proof of this in \citex{Bor} using
an infinite dimensional Lie algebra acted on by the monster called the
monster Lie algebra. These Hauptmoduls are known explicitly, so this
gives a complete description of $V$ as a representation of the
monster.
We now recall the definition of a Hauptmodul. The group $SL_2(\Z)$
acts on the upper half plane $H=\{\tau\in \C|{\rm Im}(\tau)>0\}$ by
${a\ b\choose c\ d}(\tau) = {a\tau+b\over c\tau+d}$.
The elliptic modular function $j(\tau)$ is more or less
the simplest function defined on $H$ that is invariant under $SL_2(\Z)$
in much the same way that the function $e^{2\pi i \tau}$ is the simplest function invariant under $\tau\mapsto \tau+1$.
The element ${1\ 1\choose 0\ 1}$ of $SL_2(\Z)$
takes $\tau$ to $\tau+1$, so in particular $j(\tau)$ is periodic and
can be written as a Laurent series in $q=e^{2\pi i\tau}$. The exact
expression for $j$ is
$$j(\tau) = {(1+240\sum_{n>0}\sigma_3(n)q^n)^3\over q\prod_{n>0}(1-q^n)^{24}}
$$
where $\sigma_3(n) =\sum_{d|n}d^3$ is the sum of the cubes of the divisors of
$n$; see any book on modular forms or elliptic functions, for example \citex{Ser}.
Another way of thinking about $j$ is that it is
an isomorphism from the quotient space $H/SL_2(\Z)$
to the complex plane, which can be thought of as the Riemann sphere
minus the point at infinity.
There is nothing particularly special about the group $SL_2(\Z)$
acting on $H$.
We can consider functions invariant
under any group $G$ acting on $H$, for example some group $G$ of finite index
in $SL_2(\Z)$. If $G$ satisfies some mild conditions
then the quotient $H/G$ will again be a compact Riemann surface
with a finite number
of points removed. If this compact Riemann surface is a
sphere, rather than something of higher genus, then we say that
$G$ is a genus 0 group.
When this happens there is a more or less unique map from $H/G$ to the complex plane $\C$, which induces a function from $H$ to $\C$ which is invariant under the group $G$ acting on $H$. When this function is correctly normalized it
is called a Hauptmodul for the genus 0 group $G$.
For example, $j(\tau)-744 $
is the
Hauptmodul for the genus 0 group $SL_2(\Z)$.
For example, we could take $G$ to be the group $\Gamma_0(2)$, where
$\Gamma_0(N) = \{{a\ b\choose c\ d}\in SL_2(\Z)|c\equiv 0 \bmod N\}$.
The quotient $H/G$ is then a sphere with 2 points removed, so that $G$
is a genus 0 group. Its Hauptmodul starts off $T_{2-}(q) =q^{-1} +
276q-2048q^2+\cdots$, and is equal to the Thompson series for a
certain element of the monster of order 2 (of type 2B in atlas
notation). Similarly $\Gamma_0(N)$ is a genus 0 subgroup for several
other values of $N$ which correspond to elements of the
monster. (However the genus of $\Gamma_0(N)$ tends to infinity as $N$
increases, so there are only a finite number of integers $N$ for which
it has genus 0; more generally Thompson has shown that there are only
a finite number of conjugacy classes of genus 0 subgroups of
$SL_2(\R)$ which are commensurable with $SL_2(\Z)$.) The first person
to notice any connection between the monster and genus 0 subgroups of
$SL_2(\R)$ was probably Ogg, who observed that the primes $p$ for
which the normalizer of $\Gamma_0(p)$ in $SL_2(\R)$ has genus 0 are
exactly those which divide the order of the monster.
So the problem we want to solve is to calculate the Thompson
series $T_g(\tau)$ and show that they are Hauptmoduls of
genus 0 subgroups of $SL_2(\R)$. The difficulty with doing this is as follows.
Frenkel, Lepowsky, and Meurman constructed $V$ as the sum of two subspaces
$V^+$ and $V^-$, which are the $+1$ and $-1$ eigenspaces of
an element of order 2 in the monster. If an element $g$ of the monster
commutes with this element of order 2, then it is not difficult
to to work out its Thompson series $T_g(q) = \sum_n{\rm Tr}(g|V_n)q^n$ as the
sum of two series given by its traces on $V^+$ and $V^-$
and it would be tedious but straightforward to check
that these all gave
Hauptmoduls. However if an element of the monster is not conjugate to
something that commutes
with this element of order 2 then there is no obvious direct
way of working out its Thompson series, because it muddles up $V^+$ and $V^-$
in a very complicated way.
We now give a bird's eye view of the rest of this paper. We first
construct a Lie algebra from $V$, called the monster Lie algebra. This
is a generalized Kac-Moody algebra, so we recall some facts about such
algebras, and in particular show how each such algebra gives an
identity called its denominator formula. A well known example of this
is that the denominator formulas of affine Kac-Moody algebras are the
Macdonald identities. The denominator formula for the monster Lie
algebra turns out to be the product formula for the elliptic modular
function $j$, and we use this to work out the structure of the monster
Lie algebra, and in particular to find its ``simple roots''. Once we
know the simple roots, we can write down a sort of twisted denominator
formula for each element of the monster group. These twisted
denominator formulas imply some relations between the coefficients of
the Thompson series of the monster, which are strong enough to
characterize them and verify that they are indeed Hauptmoduls.
We now introduce the
the monster Lie algebra.
This is a $\Z^2$-graded Lie algebra, whose piece of degree
$(m,n)\in \Z^2$ is isomorphic as a module over the monster to $V_{mn}$
if $(m,n)\ne (0,0)$ and to $\R^2$ if $(m,n) = (0,0)$, so for
small degrees it looks something like
$$\matrix{&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\cr
\cdots&0&0&0&0&V_3&V_6&V_9&\cdots\cr
\cdots&0&0&0&0&V_2&V_4&V_6&\cdots\cr
\cdots&0&0&V_{-1}&0&V_1&V_2&V_3&\cdots\cr
\cdots&0&0&0&\R^2&0&0&0&\cdots\cr
\cdots&V_3&V_2&V_1&0&V_{-1}&0&0&\cdots\cr
\cdots&V_6&V_4&V_2&0&0&0&0&\cdots\cr
\cdots&V_9&V_6&V_3&0&0&0&0&\cdots\cr
&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\cr
}.$$
This is a very big Lie algebra because the dimension of $V_n$ increases exponentially fast; most infinite dimensional Lie algebras that occur in mathematics can be graded so that the dimensions of the graded pieces have only polynomial growth.
Very briefly, the construction of this Lie algebra goes as follows
(this construction is not used later, so it does not matter if the
reader finds this meaningless). The graded vector space $V$ is a
vertex algebra; roughly speaking this means that $V$ has an infinite
number of products on it which satisfy some rather complicated
identities. (The Griess product on the 196884 dimensional space $V_1$
is a special case of one of the vertex algebra products.) We can also
construct a vertex algebra from any even lattice, and the tensor
product of two vertex algebras is a vertex algebra. We take the tensor
product of the monster vertex algebra $V$ with that of the two
dimensional even Lorentzian lattice (defined by the matrix ${0,1\choose
1,0}$) to get a new vertex algebra $W$. The monster Lie algebra is
then the space of physical states of the vertex algebra $W$; this is a
subquotient of $W$, which is roughly a space of highest weight vectors
for an action of the Virasoro algebra on $W$. The subquotient can be
identified exactly using the no-ghost theorem from string theory, and
turns out to be as described above.
We need to know what the structure of the monster Lie algebra is.
It turns out to be something called a generalized Kac-Moody algebra,
so we will now have a digression to explain what these are.
To motivate Kac-Moody algebras we first look at the structure of
finite dimensional simple Lie algebras. A typical finite dimensional
Lie algebra is $sl_4(\R)$, the algebra of $4\times 4$ matrices of
trace 0 with bracket given by $[a,b]=ab-ba$. If we look at this Lie
algebra we see that it has three subalgebras $sl_2(\R)$ as indicated
below (as well as many others):
$$
\pmatrix{
*&*&&\cr
*&*&&\cr
\cr
\cr
},
\pmatrix{
\cr
&*&*&\cr
&*&*&\cr
\cr
},
\pmatrix{
\cr
\cr
&&*&*\cr
&&*&*\cr
}.
$$
The Dynkin diagram of $sl_4(\R)$ is constructed by drawing a point for
each of these $sl_2$'s, and connecting them by various numbers of
lines depending on how the $sl_2$'s are arranged; for example, if the two
$sl_2$'s commute then we draw no lines between the corresponding
points, and if the $sl_2$'s overlap in one corner we draw a single
line between the points. The Dynkin diagram of $sl_4$ is therefore
$\bullet-\bullet-\bullet$, where the two outer dots are not joined
because the corresponding $sl_2$'s of $sl_4$ commute, and so on.
Conversely we can reconstruct $sl_4$ from its Dynkin diagram by writing
down some generators and relations from the Dynkin diagram. Roughly
speaking, $sl_4$ is generated by an $sl_2$ for each point of the
Dynkin diagram together with some relations depending on the lines
between points; for example, if two points are not joined we add
relations saying that the corresponding $sl_2$'s commute. We can
construct all other finite dimensional split semisimple Lie algebras
from their Dynkin diagrams (usually denoted by $A_n$, $B_n$, $C_n$,
$D_n$, $E_6$, $E_7$, $E_8$, $F_4$, and $G_2$) in a similar way. If we
are given some graph which is not one of these Dynkin diagrams of
finite dimensional Lie algebras we can still write down generators and
relations; the main difference is that these Lie algebras will be
infinite dimensional rather than finite dimensional. We can think of
a Kac-Moody algebra as a Lie algebra generated by a copy of $sl_2$ for
each point of its Dynkin diagram.
The simplest example of an infinite dimensional Kac-Moody algebra is
the Lie algebra $sl_2(\R[z,z^{-1}])$ of $2\times 2$ matrices whose
entries are Laurent polynomials. (This is a slight simplification; the
Kac-Moody algebra is really a central extension of this Lie algebra,
but we will ignore this.) The Dynkin diagram is $\bullet=\bullet$, so
this algebra is generated by 2 copies of $sl_2(\R)$, which are the
subalgebras of elements of the form ${a,b\choose c,-a}$ and ${a,bz\choose
cz^{-1},-a}$. The algebra $sl_2(\R[z,z^{-1}])$ can be $\Z^2$-graded by
letting ${0,z^n\choose 0,0}$ have degree $(2n+1,1)$, letting
${z^n,0\choose 0,-z^n}$ have degree (2n,0), and letting ${0,0\choose
z^n,0}$ have degree $(2n-1,-1)$. The picture we get looks like
\halign{$#$&&$#$\cr
\cdots&\pmatrix{0&0\cr z^{-2}&0\cr}&&\pmatrix{0&0\cr z^{-1}&0\cr}&&
\pmatrix{0&0\cr 1&0\cr}&&\pmatrix{0&0\cr z&0\cr}&\cdots\cr
\cdots&&\pmatrix{z^{-1}&0\cr 0&-z^{-1}\cr}&&\pmatrix{1&0\cr 0&-1\cr}&&
\pmatrix{z&0\cr 0&-z\cr}&&\cdots\cr
\cdots&\pmatrix{0&z^{-1}\cr 0&0\cr}&&\pmatrix{0&1\cr 0&0\cr}&&
\pmatrix{0&z\cr 0&0\cr}&&\pmatrix{0&z^2\cr 0&0\cr}&\cdots\cr
}
where on each point of $\Z^2$ we have written something that spans
the corresponding subspace of the Lie algebra.
We now describe the denominator formula of Kac-Moody algebras, and as
an example show that for the Lie algebra $sl_2(\R[z,z^{-1}])$ it
becomes the Jacobi triple product identity. The characters $Ch(V)$ of
finite dimensional irreducible representations $V$ of finite
dimensional simple Lie algebras are described by the Weyl character
formula
$$
Ch(V)=
{\sum_{w\in W} \det(w)e^{w(\lambda-\rho)}
\over
e^{-\rho} \prod_{\alpha>0} (1-e^\alpha)^{\mult(\alpha)}
}
$$
and the same formula
turns out to be true for the characters of lowest weight representations of Kac-Moody algebras. The only case we will use this is when $V$ is the trivial one dimensional representation with character 1, when the character formula becomes the denominator formula
$$
e^{-\rho} \prod_{\alpha>0} (1-e^\alpha)^{\mult(\alpha)}
=
\sum_{w\in W} \det(w)e^{-w(\rho)}
.
$$
Rather than explain precisely what all the terms in this formula mean,
we will just describe them in the case of $sl_2(\R[z,z^{-1}])$. The
product is over the ``positive roots'' $\alpha$; for
$sl_2(\R[z,z^{-1}])$ these are the vectors $\alpha=(m,n)$ with $m>0$
for which there is something in $sl_2(\R[z,z^{-1}])$ of that degree,
in other words the vectors $(2m,0)$, $(2m-1,1)$, and $(2m-1,-1)$ for
$m>0$. The multiplicity $\mult(\alpha)$ is the dimension of the
subspace of that degree, which for $sl_2(\R[z,z^{-1}])$ is always
1. If we put $e^{(m,n)}=x^my^n$, then the product becomes
$$\prod_{m>0} (1-x^{2m})(1-x^{2m-1}y)(1-x^{2m-1}y^{-1}).$$ The sum is
a sum
over all elements of the ``Weyl group'' $W$, which is a reflection
group generated by one reflection for each point of the Dynkin
diagram. In this case the Weyl group is the infinite dihedral group,
which has an infinite cyclic subgroup of index 2, so the sum over the
Weyl group is essentially a sum over the integers $\Z$. If we work out
explicitly what it is, the Weyl denominator formula becomes
$$\prod_{m>0} (1-x^{2m})(1-x^{2m-1}y)(1-x^{2m-1}y^{-1})
=\sum_{n\in \Z} (-1)^nx^{n^2}y^n.$$
which is the Jacobi triple product identity.
Not surprisingly, we can do the same with $sl_2$ replaced by any other
simple finite dimensional Lie algebra, and we then obtain some
identities called the Macdonald identities. (In fact there is more
than one Macdonald identity for some finite dimensional Lie algebras,
because there are sometimes ways of ``twisting'' this construction.)
Generalized Kac-Moody
algebras are rather like Kac-Moody algebras except that we are allowed
to glue together the $sl_2$'s in more complicated ways, and are also
allowed to use Heisenberg Lie algebras as well as $sl_2$'s to generate
the algebra. In technical terms, the roots of a Kac-Moody algebra may
be either real (norm $ >0$) or imaginary (norm $\le 0$) but all the
simple roots must be real. Generalized Kac-Moody algebras may also
have imaginary simple roots. They have a denominator formula which is
similar to the one above, except that it has some extra correction
terms corresponding to the imaginary simple roots. (The simple roots
of a generalized Kac-Moody algebra correspond to a minimal set of
generators for the subalgebra corresponding to the positive roots; for
example, the simple roots of $sl_2(\R[z,z^{-1}]) $ are $(1,1)$ and
$(1,-1)$. For Kac-Moody algebras the simple roots also correspond to
the points of the Dynkin diagram and to the generators of the Weyl
group.)
We now return to the monster Lie algebra.
This is a generalized Kac-Moody algebra, and we are going to write down its
denominator formula, which says that a product over the positive roots
is a sum over the Weyl group. The positive roots are the vectors
$(m,n)$ with $m>0$, $n>0$, and the vector $(1,-1)$, and the
root $(m,n)$ has multiplicity $c(mn)$. The Weyl group
has order 2 and its nontrivial element maps $(m,n) $ to $(n,m)$,
so it exchanges $p=e^{(1,0)}$ and $q=e^{(0,1)}$.
The denominator formula
is the product formula for the $j$ function
$$p^{-1}\prod_{m>0,n\in \Z} (1-p^mq^n)^{c(mn)}=j(p)-j(q).$$
(The left side is apparently not antisymmetric in $p$ and $q$
but in fact it is
because of the factor of $p^{-1}(1-p^1q^{-1})$
in the product.) The reason why we get $j(p)$ and $j(q)$ rather than
just a monomial
in $p$ and $q$ on the right hand side (as we would for ordinary Kac-Moody
algebras) is because of the correction due to the imaginary simple roots
of $M$. The simple roots of $M$ correspond to a set of generators
of the subalgebra $E$ of the elements of $M$ whose degree is to the
right of the $y$ axis (so the roots of $E$ are the positive roots of $M$),
and turn out to be the vectors $(1,n)$ each with multiplicity $c(n)$.
In the picture of the monster Lie algebra given earlier,
the simple roots are given by the column just to the right of
the one containing $\R^2$. The sum of the simple root spaces
is isomorphic to the space $V$. The simple root $(1,-1)$ is real of norm 2,
and the simple roots $(1,n)$ for
$n>0$ are imaginary of norm $-2n$ and have multiplicity $c(n)=\dim(V_n)$.
As these multiplicities are exactly the coefficients of the $j$
function, it is not surprising that $j$ appears in the correction
caused by the imaginary simple roots.
This discussion is slightly misleading because we have implied that
we obtain the product formula of the $j$ function
as the denominator formula of the monster Lie algebra by using
our knowledge of the simple roots; in fact we really have to use this
argument in reverse, using the product formula for the $j$ function
in order to work out what the simple roots of the monster Lie algebra are.
We will now extract information about the coefficients of
the Thompson series $T_g(\tau)$ from a twisted denominator formula
for the monster Lie algebra. For an arbitrary generalized Kac-Moody
algebra there is a more high powered version of the Weyl denominator
formula which states that
$$\Lambda(E)=H_*(E),$$
where $E$ is the subalgebra corresponding to the positive roots.
Here $\Lambda(E)= \Lambda^0(E)\ominus\Lambda^1(E) \oplus\Lambda^2(E)\ldots$
is the alternating sum of the exterior
powers of $E$,
and similarly $H_*(E)$ is the alternating sum of the homology
groups $H_i(E)$ of the Lie algebra $E$ (see \citex{Car}).
This identity is true for any Lie algebra
because the $H_i$'s are the homology groups of
a complex whose terms are the $\Lambda^i$'s. The left
hand side corresponds to a product over the positive roots
because $\Lambda(A\oplus B)=\Lambda(A)\otimes\Lambda(B)$,
$\Lambda(A) = 1-A$ if $A$ is one dimensional,
and $E$ is the sum of ${\rm mult}(\alpha)$ one dimensional
spaces for each positive root $\alpha$.
It is more difficult
to identify $H_*$ with a sum over the Weyl group, and we do this
roughly as follows. For Kac-Moody
algebras $H_i$ turns out to have dimension equal to the number
of elements in the Weyl group of length $i$; for finite dimensional
Lie algebras this was first observed by Bott, and was used by
Kostant to give a homological proof of the Weyl character formula.
The sum over the homology groups can therefore be identified with a sum
over the Weyl group.
For generalized Kac-Moody algebras things are
a bit more complicated. The sum over the homology groups
becomes a sum over the Weyl group, which is generated by an involution for
each simple root,
and the things we sum contain terms
corresponding to the imaginary
simple roots.
Anyway, we can work out the homology groups of $E$ explicitly
provided we know the simple roots of our Lie algebra; for example,
the first homology group $H_1$ is the sum of the simple root spaces.
For the monster
Lie algebra we have worked out the simple roots
using its denominator formula, which is the product formula for the $j$
function, and the homology groups of $E$
turn out to be $H_0=\R$, $H_1=\sum_{n\in \Z}V_npq^n$,
$H_2 = \sum_{m>0}V_mp^{m+1}$, and all the higher homology groups are 0.
Each homology group is a $\Z^2$-graded representation of the monster,
and we use the $p$'s and $q$'s to keep track of the grading.
If we substitute these values into the formula $\Lambda(E) = H_*$
we find that
$$\Lambda(\sum_{n\in \Z,m>0}V_{mn }p^mq^n)
= \sum_mV_mp^{m+1}-\sum_nV_npq^n.$$
Both sides of this are virtual graded representations of the monster.
If we replace everything by its dimension we recover the product
formula for the $j$ function. More generally, we can take the trace of
some element of the monster on both sides, which after some calculation
gives the identity
$$p^{-1}\exp\Bigl(-\sum_{i>0}\sum_{m>0,n\in \Z}{\rm Tr}(g^i|V_{mn})p^{mi}q^{ni}/i\Bigr)
= \sum_{m\in \Z}{\rm Tr}(g|V_m)p^m -\sum_{n\in \Z}{\rm Tr}(g|V_n)q^n$$
where ${\rm Tr}(g|V_n)$ is the trace of $g$ on the vector space $V_n$.
These relations between the coefficients of the Thompson
series turn out to be strong enough to
characterize them and check that they are Hauptmoduls for genus 0 subgroups.
Unfortunately this final step of the proof is rather messy.
The relations above determine the Thompson series from their first few
coefficients. Norton and Koike checked that certain modular functions of genus 0 also satisfy the same recursion relations. We can therefore prove that
the Thompson series $T_g(q)$ are modular functions of genus 0 by checking that the first few coefficients of both functions are the same.
Norton has conjectured
that Hauptmoduls with integer coefficients are essentially the same
as functions satisfying relations similar to the ones above, and a conceptual
proof or explanation of this would be a big improvement to the final step of the
proof. (It should be possible to prove this conjecture
by a very long and tedious
case by case check, because all functions which are either Hauptmoduls
or which satisfy
the relations above can be listed explicitly.)
\bigskip
\parindent = 0pt \everypar={\hangindent = .3in}
\medbreak {\bf Bibliography.}\medskip\par\nobreak
\item{[Bor]}{R. E. Borcherds, Monstrous moonshine and monstrous Lie
superalgebras, Invent. Math.
109, 405-444 (1992). }
\item{[Car]}{H. Cartan, S. Eilenberg, Homological algebra,
Princeton University Press 1956.}
\item{[Con]}{J. H. Conway, S. Norton, Monstrous moonshine,
Bull. London. Math. Soc. 11 (1979) 308-339.}
\item{[Fre]}{I. B. Frenkel, J. Lepowsky, A. Meurman,
Vertex operator algebras and the monster, Academic press 1988. }
\item{[Ser]}{J. P. Serre, A course in arithmetic. Graduate texts in
mathematics 7, Springer-Verlag, 1973.}
\bye